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Mechanics.of.materials.gere.6th.ch01 05.07.09.11-12
 

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    Mechanics.of.materials.gere.6th.ch01 05.07.09.11-12 Mechanics.of.materials.gere.6th.ch01 05.07.09.11-12 Document Transcript

    • http://www.elsolucionario.blogspot.com LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS LOS SOLUCIONARIOS CONTIENEN TODOS LOS EJERCICIOS DEL LIBRO RESUELTOS Y EXPLICADOS DE FORMA CLARA VISITANOS PARA DESARGALOS GRATIS.
    • 1 Tension, Compression, and Shear P1 Normal Stress and Strain A Problem 1.2-1 A solid circular post ABC (see figure) supports a load P1 ϭ 2500 lb acting at the top. A second load P2 is uniformly distributed around the shelf at B. The diameters of the upper and lower parts of the post are dAB ϭ 1.25 in. and dBC ϭ 2.25 in., respectively. dAB P2 B (a) Calculate the normal stress ␴AB in the upper part of the post. (b) If it is desired that the lower part of the post have the same compressive stress as the upper part, what should be the magnitude of the load P2? dBC C Solution 1.2-1 P1 ϭ 2500 lb Circular post in compression ALTERNATE SOLUTION FOR PART (b) P1 ϩ P2 P1 ϩ P2 ϭ ␲ 2 ABC 4 dBC P1 P1 sAB ϭ ϭ 2 sBC ϭ sAB AAB ␲ dAB 4 dAB ϭ 1.25 in. sBC ϭ dBC ϭ 2.25 in. (a) NORMAL STRESS IN PART AB P1 2500 lb sAB ϭ ϭ ϭ 2040 psi AAB ␲ (1.25 in.) 2 4 dBC 2 P1 ϩ P2 P1 ϭ 2 or P2 ϭ P1 B ¢ ≤ Ϫ 1R 2 dAB dBC dAB dBC ϭ 1.8 dAB ∴ P2 ϭ 2.24 P1 ϭ 5600 lb (b) LOAD P FOR EQUAL STRESSES 2 sBC ϭ P1 ϩ P2 2500 lb ϩ P2 ϭ ␲ 2 ABC 4 (2.25 in.) P1 ϭ␴AB ϭ 2040 psi A Solve for P2: P2 ϭ 5600 lb P2 B C 1
    • 2 CHAPTER 1 Tension, Compression, and Shear Problem 1.2-2 Calculate the compressive stress ␴c in the circular piston rod (see figure) when a force P ϭ 40 N is applied to the brake pedal. Assume that the line of action of the force P is parallel to the piston rod, which has diameter 5 mm. Also, the other dimensions shown in the figure (50 mm and 225 mm) are measured perpendicular to the line of action of the force P. 50 mm 5 mm 225 mm P = 40 N Piston rod Solution 1.2-2 Free-body diagram of brake pedal 50 mm ©MA ϭ 0 ‫۔ ە‬ A F EQUILIBRIUM OF BRAKE PEDAL 225 mm P = 40 N F ϭ compressive force in piston rod d ϭ diameter of piston rod F ϭ P¢ F(50 mm) Ϫ P(275 mm) ϭ 0 275 mm 275 ≤ ϭ (40 N) ¢ ≤ ϭ 220 N 50 mm 50 COMPRESSIVE STRESS IN PISTON ROD (d ϭ 5 mm) sc ϭ F 220 N ϭ ϭ 11.2 MPa A ␲ (5 mm) 2 4 ϭ 5 mm Problem 1.2-3 A steel rod 110 ft long hangs inside a tall tower and holds a 200-pound weight at its lower end (see figure). If the diameter of the circular rod is 1⁄4 inch, calculate the maximum normal stress ␴max in the rod, taking into account the weight of the rod itself. (Obtain the weight density of steel from Table H-1, Appendix H.) 110 ft 1 — in. 4 200 lb
    • SECTION 1.2 Solution 1.2-3 3 Normal Stress and Strain Long steel rod in tension P ϭ 200 lb smax ϭ L ϭ 110 ft d gL ϭ (490 lbրft3 )(110 ft) ¢ d ϭ 1⁄4 in. L WϩP P ϭ gL ϩ A A 1 ft2 ≤ 144 in.2 ϭ 374.3 psi P 200 lb ϭ␲ ϭ 4074 psi A 4 (0.25 in.) 2 Weight density: ␥ ϭ 490 lb/ft3 W ϭ Weight of rod ϭ ␥(Volume) ␴maxϭ 374 psi ϩ 4074 psi ϭ 4448 psi ϭ ␥AL Rounding, we get ␴max ϭ 4450 psi P = 200 lb Problem 1.2-4 A circular aluminum tube of length L ϭ 400 mm is loaded in compression by forces P (see figure). The outside and inside diameters are 60 mm and 50 mm, respectively. A strain gage is placed on the outside of the bar to measure normal strains in the longitudinal direction. Strain gage P P L = 400 mm (a) If the measured strain is ⑀ ϭ 550 ϫ 10Ϫ6, what is the shortening ␦ of the bar? (b) If the compressive stress in the bar is intended to be 40 MPa, what should be the load P? Solution 1.2-4 Aluminum tube in compression Strain gage P e ϭ 550 ϫ 10Ϫ6 L ϭ 400 mm P (b) COMPRESSIVE LOAD P d1 ϭ 50 mm ␴ ϭ 40 MPa ␲ 2 ␲ 2 A ϭ [d2 Ϫ d1 ] ϭ [ (60 mm) 2 Ϫ (50 mm) 2 ] 4 4 ϭ 863.9 mm2 (a) SHORTENING ␦ OF THE BAR P ϭ ␴A ϭ (40 MPa)(863.9 mm2) d2 ϭ 60 mm ␦ ϭ eL ϭ (550 ϫ 10Ϫ6)(400 mm) ϭ 0.220 mm ϭ 34.6 kN
    • 4 CHAPTER 1 Tension, Compression, and Shear y Problem 1.2-5 The cross section of a concrete pier that is loaded uniformly in compression is shown in the figure. 20 in. (a) Determine the average compressive stress ␴c in the concrete if the load is equal to 2500 k. (b) Determine the coordinates x and ෆ of the point where the y ෆ resultant load must act in order to produce uniform normal stress. 16 in. 16 in. 48 in. 16 in. O Solution 1.2-5 (a) AVERAGE COMPRESSIVE STRESS ␴c P ϭ 2500 k 16 in. x C 16 in. 2 1 y 16 in. O x 16 in. Concrete pier in compression y 48 in. 20 in. 20 in. 16 in. 3 4 x (b) COORDINATES OF CENTROID C 1 From symmetry, y ϭ (48 in.) ϭ 24 in. 2 © xi Ai (see Chapter 12, Eq. 12-7a) A 1 x ϭ (x1 A1 ϩ 2x2 A2 ϩ x3 A3 ) A xϭ USE THE FOLLOWING AREAS: A1 ϭ (48 in.)(20 in.) ϭ 960 in.2 1 A2 ϭ A4 ϭ (16 in.)(16 in.) ϭ 128 in.2 2 A3 ϭ (16 in.)(16 in.) ϭ 256 in.2 ϭ 1 [ (10 in.)(960 in.2 ) 1472 in.2 ϩ 2(25.333 in.)(128 in.2) A ϭ A1 ϩ A2 ϩ A3 ϩ A4 ϭ (960 ϩ 128 ϩ 256 ϩ 128) P 2500 k ϭ ϭ 1.70 ksi A 1472 in.2 sc ϭ ϩ(28 in.)(256 in.2)] in.2 ϭ 1472 in.2 ϭ15.8 in. Problem 1.2-6 A car weighing 130 kN when fully loaded is pulled slowly up a steep inclined track by a steel cable (see figure). The cable has an effective cross-sectional area of 490 mm2, and the angle ␣ of the incline is 30°. Calculate the tensile stress ␴t in the cable. Cable ␣
    • SECTION 1.2 Solution 1.2-6 5 Normal Stress and Strain Car on inclined track FREE-BODY DIAGRAM OF CAR W TENSILE STRESS IN THE CABLE W ϭ Weight of car st ϭ T W sin ␣ ϭ A A T ϭ Tensile force in cable R1 W ϭ 130 kN ␣ ϭ 30Њ A ϭ Effective area of cable R2 SUBSTITUTE NUMERICAL VALUES: ␣ ϭ Angle of incline ␣ A ϭ 490 mm2 st ϭ R1, R2 ϭ Wheel reactions (no friction force between wheels and rails) (130 kN)(sin 30Њ) 490 mm2 ϭ 133 MPa EQUILIBRIUM IN THE INCLINED DIRECTION ©FT ϭ 0 Q bϪ T Ϫ W sin ␣ ϭ 0 ϩ T ϭ W sin ␣ Problem 1.2-7 Two steel wires, AB and BC, support a lamp weighing 18 lb (see figure). Wire AB is at an angle ␣ ϭ 34° to the horizontal and wire BC is at an angle ␤ ϭ 48°. Both wires have diameter 30 mils. (Wire diameters are often expressed in mils; one mil equals 0.001 in.) Determine the tensile stresses ␴AB and ␴BC in the two wires. C A ␣ B ␤ Solution 1.2-7 Two steel wires supporting a lamp FREE-BODY DIAGRAM OF POINT B SUBSTITUTE NUMERICAL VALUES: ϪTAB(0.82904) ϩ TBC(0.66913) ϭ 0 TBC TAB ␤ ␣ ␣ ϭ 34Њ d ϭ 30 mils ϭ 0.030 in. y W = 18 lb 0 ␤ ϭ 48Њ Aϭ x ␲d 2 ϭ 706.9 ϫ 10 Ϫ6 in.2 4 TAB(0.55919) ϩ TBC(0.74314) Ϫ 18 ϭ 0 SOLVE THE EQUATIONS: TAB ϭ 12.163 lb TBC ϭ 15.069 lb TENSILE STRESSES IN THE WIRES TAB ϭ 17,200 psi A TBC sBC ϭ ϭ 21,300 psi A sAB ϭ EQUATIONS OF EQUILIBRIUM ⌺Fx ϭ 0 ⌺Fy ϭ 0 Ϫ TAB cos ␣ ϩ TBC cos ␤ ϭ 0 TAB sin ␣ ϩ TBC sin ␤ Ϫ W ϭ 0
    • 6 CHAPTER 1 Tension, Compression, and Shear Problem 1.2-8 A long retaining wall is braced by wood shores set at an angle of 30° and supported by concrete thrust blocks, as shown in the first part of the figure. The shores are evenly spaced, 3 m apart. For analysis purposes, the wall and shores are idealized as shown in the second part of the figure. Note that the base of the wall and both ends of the shores are assumed to be pinned. The pressure of the soil against the wall is assumed to be triangularly distributed, and the resultant force acting on a 3-meter length of the wall is F ϭ 190 kN. If each shore has a 150 mm ϫ 150 mm square cross section, what is the compressive stress ␴c in the shores? Solution 1.2-8 F 30° 1.5 m A C 0.5 m 4.0 m A ϭ area of one shore Shore F 30° 1.5 m A A ϭ (150 mm)(150 mm) C ϭ 22,500 mm2 0.5 m ϭ 0.0225 m2 4.0 m FREE-BODY DIAGRAM OF WALL AND SHORE SUMMATION OF MOMENTS ABOUT POINT A ©MA ϭ 0 ‫۔ ە‬ B ϪF(1.5 m)ϩCV (4.0 m)ϩCH (0.5 m) ϭ 0 30° A AH CH CV 30° C AV C ϭ compressive force in wood shore CH ϭ horizontal component of C CV ϭ vertical component of C CV ϭ C sin 30Њ 30° B F ϭ 190 kN Wall CH ϭ C cos 30Њ Retaining wall Concrete Shore thrust block Retaining wall braced by wood shores B F 1.5 m Soil or Ϫ (190 kN)(1.5 m) ϩ C(sin 30Њ)(4.0 m) ϩ C(cos 30Њ)(0.5 m) ϭ 0 ∴ C ϭ 117.14 kN COMPRESSIVE STRESS IN THE SHORES sc ϭ C 117.14 kN ϭ A 0.0225 m2 ϭ5.21 MPa
    • SECTION 1.2 Problem 1.2-9 A loading crane consisting of a steel girder ABC supported by a cable BD is subjected to a load P (see figure). The cable has an effective cross-sectional area A ϭ 0.471 in2. The dimensions of the crane are H ϭ 9 ft, L1 ϭ 12 ft, and L2 ϭ 4 ft. 7 Normal Stress and Strain D Cable H (a) If the load P ϭ 9000 lb, what is the average tensile stress in the cable? (b) If the cable stretches by 0.382 in., what is the average strain? Girder B A L1 C L2 P Solution 1.2-9 Loading crane with girder and cable EQUILIBRIUM D ©MA ϭ 0 ‫۔ ە‬ TV (12 ft) Ϫ (9000 lb)(16 ft) ϭ 0 TV ϭ 12,000 lb TH L1 12 ft ϭ ϭ TV H 9 ft 12 ∴ TH ϭ TV ¢ ≤ 9 H B A L1 H ϭ 9 ft C L2 P = 9000 lb L1 ϭ 12 ft L2 ϭ 4 ft A ϭ effective area of cable A ϭ 0.471 TH ϭ (12,000 lb) ¢ ϭ 16,000 lb 12 ≤ 9 TENSILE FORCE IN CABLE in.2 2 2 T ϭ ͙TH ϩ TV ϭ ͙(16,000 lb) 2 ϩ (12,000 lb) 2 P ϭ 9000 lb ϭ 20,000 lb FREE-BODY DIAGRAM OF GIRDER T (a) AVERAGE TENSILE STRESS IN CABLE TV sϭ TH A 12 ft B C P ϭ 9000 lb (b) AVERAGE STRAIN IN CABLE L ϭ length of cable 4 ft P = 9000 lb T ϭ tensile force in cable T 20,000 lb ϭ ϭ 42,500 psi A 0.471 in.2 L ϭ ͙H 2ϩ L2 ϭ 15 ft 1 ␦ ϭ stretch of cable ␦ ϭ 0.382 in. eϭ ␦ 0.382 in. ϭ ϭ 2120 ϫ 10 Ϫ6 L (15 ft)(12 in.րft)
    • 8 CHAPTER 1 Tension, Compression, and Shear Problem 1.2-10 Solve the preceding problem if the load P ϭ 32 kN; the cable has effective cross-sectional area A ϭ 481 mm2; the dimensions of the crane are H ϭ 1.6 m, L1 ϭ 3.0 m, and L2 ϭ 1.5 m; and the cable stretches by 5.1 mm. Figure is with Prob. 1.2-9. Solution 1.2-10 Loading crane with girder and cable D H ϭ 1.6 m B L1 A ϭ effective area of cable A ϭ 481 mm2 A L1ϭ 3.0 m L2 ϭ 1.5 m H P ϭ 32 kN C L2 P = 32 kN TENSILE FORCE IN CABLE FREE-BODY DIAGRAM OF GIRDER T TH A 3.0 m TV T = tensile force in cable B C 2 T ϭ ͙TH ϩ TV2 ϭ ͙(90 kN) 2 ϩ (48 kN) 2 ϭ 102 kN (a) AVERAGE TENSILE STRESS IN CABLE 1.5 m P = 32 kN EQUILIBRIUM ©MA ϭ 0 ‫۔ە‬ TV (3.0 m) Ϫ (32 kN)(4.5 m) ϭ 0 TV ϭ 48 kN TH L1 3.0 m ϭ ϭ TV H 1.6 m 3.0 ∴ TH ϭ TV ¢ ≤ 1.6 3.0 TH ϭ (48 kN) ¢ ≤ 1.6 sϭ T 102 kN ϭ ϭ 212 MPa A 481 mm2 (b) AVERAGE STRAIN IN CABLE L ϭ length of cable L ϭ ͙H 2 ϩ L2 ϭ 3.4 m 1 ␦ ϭ stretch of cable ␦ ϭ 5.1 mm eϭ ␦ 5.1 mm ϭ ϭ 1500 ϫ 10 Ϫ6 L 3.4 m ϭ 90 kN Problem 1.2-11 A reinforced concrete slab 8.0 ft square and 9.0 in. thick is lifted by four cables attached to the corners, as shown in the figure. The cables are attached to a hook at a point 5.0 ft above the top of the slab. Each cable has an effective cross-sectional area A ϭ 0.12 in2. Determine the tensile stress ␴t in the cables due to the weight of the concrete slab. (See Table H-1, Appendix H, for the weight density of reinforced concrete.) Cables Reinforced concrete slab
    • SECTION 1.2 Solution 1.2-11 9 Normal Stress and Strain Reinforced concrete slab supported by four cables W T ϭ tensile force in a cable Cable AB: A TV H ϭ T LAB H Cable TV ϭ T ¢ t H ͙H ϩ L2ր2 2 EQUILIBRIUM B (Eq. 1) ⌺Fvert ϭ 0 ↑ϩ ↓Ϫ L L ≤ W Ϫ 4TV ϭ 0 H ϭ height of hook above slab TV ϭ L ϭ length of side of square slab t ϭ thickness of slab W 4 (Eq. 2) COMBINE EQS. (1) & (2): T¢ ␥ ϭ weight density of reinforced concrete W ϭ weight of slab ϭ ␥L2t D ϭ length of diagonal of slab ϭ L͙2 H ͙H ϩ L ր2 Tϭ 2 2 ≤ϭ W 4 W ͙H 2 ϩ L2ր2 W ϭ ͙1 ϩ L2ր2H 2 4 H 4 DIMENSIONS OF CABLE AB TENSILE STRESS IN A CABLE A LAB H B L2 ϭ H2ϩ B 2 LAB ϭ length of cable D= L 2 2 A ϭ effective cross-sectional area of a cable st ϭ T W ϭ ͙1 ϩ L2ր2H2 A 4A SUBSTITUTE NUMERICAL VALUES AND OBTAIN ␴t : H ϭ 5.0 ft ␥ ϭ 150 FREE-BODY DIAGRAM OF HOOK AT POINT A L ϭ 8.0 ft A ϭ 0.12 lb/ft3 t ϭ 9.0 in. ϭ 0.75 ft in.2 W ϭ ␥L2t ϭ 7200 lb W TH st ϭ A T T W ͙1 ϩ L2ր2H2 ϭ 22,600 psi 4A TV T T T Problem 1.2-12 A round bar ACB of length 2L (see figure) rotates about an axis through the midpoint C with constant angular speed ␻ (radians per second). The material of the bar has weight density ␥. (a) Derive a formula for the tensile stress ␴x in the bar as a function of the distance x from the midpoint C. (b) What is the maximum tensile stress ␴max? ␻ A C L B x L
    • 10 CHAPTER 1 Tension, Compression, and Shear Solution 1.2-12 ␻ D Rotating Bar dM B C x ␰ d␰ L ␻ ϭ angular speed (rad/s) A ϭ cross-sectional area ␥ ϭ weight density Consider an element of mass dM at distance ␰ from the midpoint C. The variable ␰ ranges from x to L. g dM ϭ g A dj dF ϭ Inertia force (centrifugal force) of element of mass dM g dF ϭ (dM)(j␻2 ) ϭ g A␻2jdj Fx ϭ Ύ B dF ϭ D g g ϭ mass density Ύ x L g gA␻2 2 A␻2jdj ϭ (L Ϫ x 2) g 2g (a) TENSILE STRESS IN BAR AT DISTANCE x Fx g␻2 2 ϭ (L Ϫ x 2) — A 2g sx ϭ We wish to find the axial force Fx in the bar at Section D, distance x from the midpoint C. The force Fx equals the inertia force of the part of the rotating bar from D to B. (b) MAXIMUM TENSILE STRESS xϭ0 smax ϭ g␻2L2 — 2g Mechanical Properties of Materials Problem 1.3-1 Imagine that a long steel wire hangs vertically from a high-altitude balloon. (a) What is the greatest length (feet) it can have without yielding if the steel yields at 40 ksi? (b) If the same wire hangs from a ship at sea, what is the greatest length? (Obtain the weight densities of steel and sea water from Table H-1, Appendix H.) Solution 1.3-1 Hanging wire of length L W ϭ total weight of steel wire ␥S ϭ weight density of steel L Lmax ϭ ϭ 11,800 ft ϭ 490 lb/ft3 ␥W ϭ weight density of sea water ϭ 63.8 lb/ft3 A ϭ cross-sectional area of wire ␴max ϭ 40 ksi (yield strength) (b) WIRE HANGING IN SEA WATER F ϭ tensile force at top of wire F ϭ (gS Ϫ gW ) AL Lmax ϭ (a) WIRE HANGING IN AIR W ϭ ␥S AL W smax ϭ ϭ gSL A smax 40,000 psi ϭ (144 in.2րft2 ) gS 490 lbրft3 ϭ smax ϭ F ϭ (gS Ϫ gW )L A smax gS Ϫ gW 40,000 psi (144 in.2րft2 ) (490 Ϫ 63.8)lbրft3 ϭ 13,500 ft
    • SECTION 1.3 Mechanical Properties of Materials 11 Problem 1.3-2 Imagine that a long wire of tungsten hangs vertically from a high-altitude balloon. (a) What is the greatest length (meters) it can have without breaking if the ultimate strength (or breaking strength) is 1500 MPa? (b) If the same wire hangs from a ship at sea, what is the greatest length? (Obtain the weight densities of tungsten and sea water from Table H-1, Appendix H.) Solution 1.3-2 Hanging wire of length L W ϭ total weight of tungsten wire ␥T ϭ weight density of tungsten ϭ 190 L kN/m3 ␥W ϭ weight density of sea water ϭ 10.0 kN/m3 A ϭ cross-sectional area of wire ␴max ϭ 1500 MPa (breaking strength) (b) WIRE HANGING IN SEA WATER F ϭ tensile force at top of wire F ϭ (␥TϪ␥W)AL F ϭ (gT Ϫ gW )L A smax Lmax ϭ gT Ϫ gW smax ϭ ϭ (a) WIRE HANGING IN AIR ϭ 8300 m W ϭ ␥T AL smax ϭ W ϭ gTL A Lmax ϭ 1500 MPa (190 Ϫ 10.0) kNրm3 smax 1500 MPa ϭ gT 190 kNրm3 ϭ 7900 m Problem 1.3-3 Three different materials, designated A, B, and C, are tested in tension using test specimens having diameters of 0.505 in. and gage lengths of 2.0 in. (see figure). At failure, the distances between the gage marks are found to be 2.13, 2.48, and 2.78 in., respectively. Also, at the failure cross sections the diameters are found to be 0.484, 0.398, and 0.253 in., respectively. Determine the percent elongation and percent reduction in area of each specimen, and then, using your own judgment, classify each material as brittle or ductile. P Gage length P
    • 12 CHAPTER 1 Solution 1.3-3 Tension, Compression, and Shear Tensile tests of three materials 0.505 in P P Percent reduction in area ϭ ϭ ¢1 Ϫ 2.0 in Percent elongation ϭ L0 ϭ 2.0 in. L1 Ϫ L0 L1 (100) ϭ ¢ Ϫ 1 ≤100 L0 L0 Percent elongation ϭ ¢ where L1 is in inches. L1 Ϫ 1 ≤ (100) 2.0 (Eq. 1) A0 Ϫ A1 (100) A0 A1 ≤ (100) A0 d0 ϭ initial diameter d1 ϭ final diameter A1 d1 2 ϭ¢ ≤ A0 d0 d0 ϭ 0.505 in. Percent reduction in area ϭ B1 Ϫ ¢ d1 2 ≤ R (100) 0.505 (Eq. 2) where d1 is in inches. Material L1 (in.) d1 (in.) % Elongation (Eq. 1) A 2.13 0.484 6.5% 8.1% Brittle B 2.48 0.398 24.0% 37.9% Ductile C 2.78 0.253 39.0% 74.9% Ductile Problem 1.3-4 The strength-to-weight ratio of a structural material is defined as its load-carrying capacity divided by its weight. For materials in tension, we may use a characteristic tensile stress (as obtained from a stress-strain curve) as a measure of strength. For instance, either the yield stress or the ultimate stress could be used, depending upon the particular application. Thus, the strength-to-weight ratio RS/W for a material in tension is defined as ␴ RS/W ϭ ᎏᎏ ␥ Solution 1.3-4 % Reduction (Eq. 2) Brittle or Ductile? in which ␴ is the characteristic stress and ␥ is the weight density. Note that the ratio has units of length. Using the ultimate stress ␴U as the strength parameter, calculate the strength-to-weight ratio (in units of meters) for each of the following materials: aluminum alloy 6061-T6, Douglas fir (in bending), nylon, structural steel ASTM-A572, and a titanium alloy. (Obtain the material properties from Tables H-1 and H-3 of Appendix H. When a range of values is given in a table, use the average value.) Strength-to-weight ratio The ultimate stress ␴U for each material is obtained from Table H-3, Appendix H, and the weight density ␥ is obtained from Table H-1. The strength-to-weight ratio (meters) is sU (MPa) RSրW ϭ (103 ) g(kNրm3 ) Values of ␴U, ␥, and RS/W are listed in the table. ␴U (MPa) Aluminum alloy 6061-T6 Douglas fir Nylon Structural steel ASTM-A572 Titanium alloy ␥ (kN/m3) 310 26.0 11.9 ϫ 103 65 60 500 5.1 9.8 77.0 12.7 ϫ 103 6.1 ϫ 103 6.5 ϫ 103 1050 44.0 23.9 ϫ 103 RS/W (m) Titanium has a high strength-to-weight ratio, which is why it is used in space vehicles and high-performance airplanes. Aluminum is higher than steel, which makes it desirable for commercial aircraft. Some woods are also higher than steel, and nylon is about the same as steel.
    • SECTION 1.3 Problem 1.3-5 A symmetrical framework consisting of three pinconnected bars is loaded by a force P (see figure). The angle between the inclined bars and the horizontal is ␣ ϭ 48°. The axial strain in the middle bar is measured as 0.0713. Determine the tensile stress in the outer bars if they are constructed of aluminum alloy having the stress-strain diagram shown in Fig. 1-13. (Express the stress in USCS units.) Mechanical Properties of Materials A B C ␣ D P Solution 1.3-5 Symmetrical framework L ϭ length of bar BD A B C ␣ L1 ϭ distance BC ϭ L cot ␣ ϭ L(cot 48Њ) ϭ 0.9004 L L2 ϭ length of bar CD ϭ L csc ␣ ϭ L(csc 48Њ) ϭ 1.3456 L D Elongation of bar BD ϭ distance DE ϭ eBDL P eBDL ϭ 0.0713 L Aluminum alloy L3 ϭ distance CE ␣ ϭ 48Њ L3 ϭ ͙L2 ϩ (L ϩ eBD L) 2 1 eBD ϭ 0.0713 Use stress-strain diagram of Figure 1-13 C B ␣ ϭ ͙(0.9004L) 2 ϩ L2 (1 ϩ 0.0713) 2 ϭ 1.3994 L ␦ ϭ elongation of bar CD ␦ ϭ L3 Ϫ L2 ϭ 0.0538L L Strain in bar CD L2 L3 D ϭ ␦ 0.0538L ϭ 0.0400 ϭ L2 1.3456L From the stress-strain diagram of Figure 1-13: ⑀BDL E s Ϸ 31 ksi 13
    • 14 CHAPTER 1 Tension, Compression, and Shear Problem 1.3-6 A specimen of a methacrylate plastic is tested in tension at room temperature (see figure), producing the stress-strain data listed in the accompanying table. Plot the stress-strain curve and determine the proportional limit, modulus of elasticity (i.e., the slope of the initial part of the stress-strain curve), and yield stress at 0.2% offset. Is the material ductile or brittle? STRESS-STRAIN DATA FOR PROBLEM 1.3-6 P P Solution 1.3-6 Stress (MPa) 8.0 17.5 25.6 31.1 39.8 44.0 48.2 53.9 58.1 62.0 62.1 Strain 0.0032 0.0073 0.0111 0.0129 0.0163 0.0184 0.0209 0.0260 0.0331 0.0429 Fracture Tensile test of a plastic Using the stress-strain data given in the problem statement, plot the stress-strain curve: ␴PL ϭ proportional limit ␴PL Ϸ 47 MPa Modulus of elasticity (slope) Ϸ 2.4 GPa ␴Y ϭ yield stress at 0.2% offset 60 Stress (MPa) ␴Y Ϸ 53 MPa ␴Y ␴PL 40 slope ≈ 40 MPa = 2.4 GPa 0.017 Material is brittle, because the strain after the proportional limit is exceeded is relatively small. — 20 0.2% offset 0 0.01 0.02 0.03 Strain 0.04 Problem 1.3-7 The data shown in the accompanying table were obtained from a tensile test of high-strength steel. The test specimen had a diameter of 0.505 in. and a gage length of 2.00 in. (see figure for Prob. 1.3-3). At fracture, the elongation between the gage marks was 0.12 in. and the minimum diameter was 0.42 in. Plot the conventional stress-strain curve for the steel and determine the proportional limit, modulus of elasticity (i.e., the slope of the initial part of the stress-strain curve), yield stress at 0.1% offset, ultimate stress, percent elongation in 2.00 in., and percent reduction in area. TENSILE-TEST DATA FOR PROBLEM 1.3-7 Load (lb) 1,000 2,000 6,000 10,000 12,000 12,900 13,400 13,600 13,800 14,000 14,400 15,200 16,800 18,400 20,000 22,400 22,600 Elongation (in.) 0.0002 0.0006 0.0019 0.0033 0.0039 0.0043 0.0047 0.0054 0.0063 0.0090 0.0102 0.0130 0.0230 0.0336 0.0507 0.1108 Fracture
    • SECTION 1.3 Solution 1.3-7 L0 ϭ 2.00 in. ENLARGEMENT OF PART OF THE STRESS-STRAIN CURVE 2 ␲d0 ϭ 0.200 in.2 4 Stress (psi) CONVENTIONAL STRESS AND STRAIN sϭ P A0 eϭ Load P (lb) 1,000 2,000 6,000 10,000 12,000 12,900 13,400 13,600 13,800 14,000 14,400 15,200 16,800 18,400 20,000 22,400 22,600 ␴YP 70,000 ␦ L0 ␴YP ≈ 69,000 psi (0.1% offset) ␴PL Elongation ␦ (in.) 0.0002 0.0006 0.0019 0.0033 0.0039 0.0043 0.0047 0.0054 0.0063 0.0090 0.0102 0.0130 0.0230 0.0336 0.0507 0.1108 Fracture Stress ␴ (psi) 5,000 10,000 30,000 50,000 60,000 64,500 67,000 68,000 69,000 70,000 72,000 76,000 84,000 92,000 100,000 112,000 113,000 ␴PL ≈ 65,000 psi 60,000 0.1% pffset 50,000 psi Slope ≈ 0.00165 ≈ 30 × 106 psi Strain e 0.00010 0.00030 0.00100 0.00165 0.00195 0.00215 0.00235 0.00270 0.00315 0.00450 0.00510 0.00650 0.01150 0.01680 0.02535 0.05540 50,000 0 0.0020 0.0040 Strain RESULTS Proportional limit Ϸ 65,000 psi Modulus of elasticity (slope) Ϸ 30 ϫ 106 psi Yield stress at 0.1% offset Ϸ 69,000 psi Ultimate stress (maximum stress) Ϸ 113,000 psi Percent elongation in 2.00 in. ϭ L1 Ϫ L0 (100) L0 ϭ STRESS-STRAIN DIAGRAM 0.12 in. (100) ϭ 6% 2.00 in. Percent reduction in area 150,000 Stress (psi) ϭ A0 Ϫ A1 (100) A0 100,000 ϭ 0.200 in.2 Ϫ ␲ (0.42 in.) 2 4 (100) 0.200 in.2 ϭ 31% 50,000 0 15 Tensile test of high-strength steel d0 ϭ 0.505 in. A0 ϭ Mechanical Properties of Materials 0.0200 0.0400 Strain 0.0600
    • 16 CHAPTER 1 Tension, Compression, and Shear Elasticity, Plasticity, and Creep Problem 1.4-1 A bar made of structural steel having the stressstrain diagram shown in the figure has a length of 48 in. The yield stress of the steel is 42 ksi and the slope of the initial linear part of the stress-strain curve (modulus of elasticity) is 30 ϫ 103 ksi. The bar is loaded axially until it elongates 0.20 in., and then the load is removed. How does the final length of the bar compare with its original length? (Hint: Use the concepts illustrated in Fig. 1-18b.) ␴ (ksi) 60 40 20 0 0 0.002 0.004 0.006 ⑀ Solution 1.4-1 Steel bar in tension ␴ ELASTIC RECOVERY eE A ␴Y B eE ϭ sB 42 ksi ϭ 0.00140 ϭ Slope 30 ϫ 103 ksi RESIDUAL STRAIN eR ⑀E eR ϭ eB Ϫ eE ϭ 0.00417Ϫ0.00140 0 ⑀R ⑀B ⑀ ϭ 0.00277 L ϭ 48 in. PERMANENT SET Yield stress ␴Y ϭ 42 ksi eRL ϭ (0.00277)(48 in.) Slope ϭ 30 ϫ 103 ksi ␦ ϭ 0.20 in. ϭ 0.13 in. Final length of bar is 0.13 in. greater than its original length. STRESS AND STRAIN AT POINT B ␴B ϭ ␴Y ϭ 42 ksi eB ϭ ␦ 0.20 in. ϭ ϭ 0.00417 L 48 in. Problem 1.4-2 A bar of length 2.0 m is made of a structural steel having the stress-strain diagram shown in the figure. The yield stress of the steel is 250 MPa and the slope of the initial linear part of the stress-strain curve (modulus of elasticity) is 200 GPa. The bar is loaded axially until it elongates 6.5 mm, and then the load is removed. How does the final length of the bar compare with its original length? (Hint: Use the concepts illustrated in Fig. 1-18b.) ␴ (MPa) 300 200 100 0 0 0.002 0.004 ⑀ 0.006
    • SECTION 1.4 Solution 1.4-2 17 Steel bar in tension ␴ ELASTIC RECOVERY eE L ϭ 2.0 m ϭ 2000 mm A ␴Y B eE ϭ Yield stress ␴Y ϭ 250 MPa Slope ϭ 200 GPa ⑀B ⑀R 0 eR ϭ eB Ϫ eE ϭ 0.00325Ϫ0.00125 ϭ0.00200 ⑀ Permanent set ϭ eRL ϭ (0.00200)(2000 mm) STRESS AND STRAIN AT POINT B ϭ 4.0 mm ␴B ϭ ␴Y ϭ 250 MPa Final length of bar is 4.0 mm greater than its original length. ␦ 6.5 mm ϭ ϭ 0.00325 L 2000 mm Problem 1.4-3 An aluminum bar has length L ϭ 4 ft and diameter d ϭ 1.0 in. The stress-strain curve for the aluminum is shown in Fig. 1-13 of Section 1.3. The initial straight-line part of the curve has a slope (modulus of elasticity) of 10 ϫ 106 psi. The bar is loaded by tensile forces P ϭ 24 k and then unloaded. Solution 1.4-3 sB 250 MPa ϭ ϭ 0.00125 Slope 200 GPa RESIDUAL STRAIN eR ␦ ϭ 6.5 mm ⑀E eB ϭ Elasticity, Plasticity, and Creep (a) What is the permanent set of the bar? (b) If the bar is reloaded, what is the proportional limit? (Hint: Use the concepts illustrated in Figs. 1-18b and 1-19.) Aluminum bar in tension ␴ ␴B STRESS AND STRAIN AT POINT B B sB ϭ A P 24 k ϭ ϭ 31 ksi A ␲ (1.0 in.) 2 4 From Fig. 1-13: eB Ϸ 0.04 ELASTIC RECOVERY eE ⑀E 0 ⑀R ⑀B L ϭ 4 ft ϭ 48 in. d ϭ 1.0 in. P ϭ 24 k ⑀ eE ϭ sB 31 ksi ϭ ϭ 0.0031 Slope 10 ϫ 106 psi RESIDUAL STRAIN eR eR ϭ eB Ϫ eE ϭ 0.04 Ϫ 0.0031 ϭ 0.037 (Note: The accuracy in this result is very poor because eB is approximate.) See Fig. 1-13 for stress-strain diagram Slope from O to A is 10 ϫ 106 psi. (a) PERMANENT SET eRL ϭ (0.037)(48 in.) Ϸ 1.8 in. (b) PROPORTIONAL LIMIT WHEN RELOADED ␴B ϭ 31 ksi
    • 18 CHAPTER 1 Tension, Compression, and Shear Problem 1.4-4 A circular bar of magnesium alloy is 800 mm long. The stress-strain diagram for the material is shown in the figure. The bar is loaded in tension to an elongation of 5.6 mm, and then the load is removed. 200 ␴ (MPa) (a) What is the permanent set of the bar? (b) If the bar is reloaded, what is the proportional limit? (Hint: Use the concepts illustrated in Figs. 1-18b and 1-19.) 100 0 Solution 1.4-4 Slope ϭ B A 0.010 (sPL ) 1 88 MPa ϭ ϭ 44 GPa eA 0.002 STRESS AND STRAIN AT POINT B eB ϭ ⑀R ␦ 5.6 mm ϭ ϭ 0.007 L 800 mm From ␴-e diagram: ␴B ϭ (␴PL)2 ϭ 170 MPa ⑀E 0 0.005 ⑀ Magnesium bar in tension ␴ (␴PL )2 (␴PL )1 0 ⑀B ⑀ L ϭ 800 mm ELASTIC RECOVERY eE eE ϭ ␦ ϭ 5.6 mm (␴PL )1 ϭ initial proportional limit ϭ 88 MPa (from stress-strain diagram) (␴PL )2 ϭ proportional limit when the bar is reloaded INITIAL SLOPE OF STRESS-STRAIN CURVE From ␴-e diagram: At point A: (␴PL )1 ϭ 88 MPa eA ϭ 0.002 Problem 1.4-5 A wire of length L ϭ 4 ft and diameter d ϭ 0.125 in. is stretched by tensile forces P ϭ 600 lb. The wire is made of a copper alloy having a stress-strain relationship that may be described mathematically by the following equation: 18,000⑀ ␴ ϭ ᎏᎏ 0 Յ ⑀ Յ 0.03 (␴ ϭ ksi) 1 ϩ 300⑀ in which ⑀ is nondimensional and ␴ has units of kips per square inch (ksi). sB (sPL ) 2 170 MPa ϭ ϭ ϭ 0.00386 Slope Slope 44 GPa RESIDUAL STRAIN eR eR ϭ eB Ϫ eE ϭ 0.007 Ϫ 0.00386 ϭ 0.00314 (a) PERMANENT SET eRL ϭ (0.00314)(800 mm) ϭ 2.51 mm (b) PROPORTIONAL LIMIT WHEN RELOADED (␴PL)2 ϭ ␴B ϭ 170 MPa (a) Construct a stress-strain diagram for the material. (b) Determine the elongation of the wire due to the forces P. (c) If the forces are removed, what is the permanent set of the bar? (d) If the forces are applied again, what is the proportional limit?
    • SECTION 1.5 Solution 1.4-5 Wire stretched by forces P ALTERNATIVE FORM OF THE STRESS-STRAIN RELATIONSHIP L ϭ 4 ft ϭ 48 in. d ϭ 0.125 in. Solve Eq. (1) for e in terms of ␴: s eϭ 0 Յ s Յ 54 ksi (s ϭ ksi) (Eq. 2) 18,000 Ϫ 300s This equation may also be used when plotting the stress-strain diagram. P ϭ 600 lb COPPER ALLOY sϭ 18,000e 1 ϩ 300e 0 Յ e Յ 0.03 (s ϭ ksi) (Eq. 1) (b) ELONGATION ␦ OF THE WIRE (a) STRESS-STRAIN DIAGRAM (From Eq. 1) sϭ 60 ␴ = 54 ksi ␴B 19 Linear Elasticity, Hooke’s Law, and Poisson’s Ratio P 600 lb ϭ ϭ 48,900 psi ϭ 48.9 ksi A ␲ (0.125 in.) 2 4 From Eq. (2) or from the stress-strain diagram: B e ϭ 0.0147 40 ␦ ϭ eL ϭ (0.0147)(48 in.) ϭ 0.71 in. ␴ (ksi) STRESS AND STRAIN AT POINT B (see diagram) 20 ␴B ϭ 48.9 ksi ⑀E = ⑀B − ⑀R ⑀R 0 0.01 ELASTIC RECOVERY eE ⑀B ⑀ 0.02 eB ϭ 0.0147 0.03 eE ϭ sB 48.9 ksi ϭ ϭ 0.00272 Slope 18,000 ksi INITIAL SLOPE OF STRESS-STRAIN CURVE Take the derivative of ␴ with respect to e: ds (1 ϩ 300e)(18,000) Ϫ (18,000e)(300) ϭ de (1 ϩ 300e) 2 18,000 ϭ (1 ϩ 300e) 2 At e ϭ 0, ds ϭ 18,000 ksi de ∴ Initial slopeϭ18,000 ksi RESIDUAL STRAIN eR eR ϭ eB Ϫ eE ϭ 0.0147 Ϫ 0.0027 ϭ 0.0120 (c) Permanent set ϭ eR L ϭ (0.0120)(48 in.) ϭ 0.58 in. (d) Proportional limit when reloaded ϭ ␴B ␴Bϭ49 ksi Linear Elasticity, Hooke’s Law, and Poisson’s Ratio When solving the problems for Section 1.5, assume that the material behaves linearly elastically. Problem 1.5-1 A high-strength steel bar used in a large crane has diameter d ϭ 2.00 in. (see figure). The steel has modulus of elasticity E ϭ 29 ϫ 106 psi and Poisson’s ratio ␯ ϭ 0.29. Because of clearance requirements, the diameter of the bar is limited to 2.001 in. when it is compressed by axial forces. What is the largest compressive load Pmax that is permitted? d P P
    • 20 CHAPTER 1 Tension, Compression, and Shear Solution 1.5-1 STEEL BAR Steel bar in compression d ϭ 2.00 in. Max. ⌬d ϭ 0.001 in. E ϭ 29 ϫ 106 psi ␯ ϭ 0.29 LATERAL STRAIN e¿ ϭ AXIAL STRESS ␴ ϭ Ee ϭ (29 ϫ 106 psi)(Ϫ0.001724) ϭϪ50.00 ksi (compression) Assume that the yield stress for the high-strength steel is greater than 50 ksi. Therefore, Hooke’s law is valid. ¢d 0.001 in. ϭ ϭ 0.0005 d 2.00 in. AXIAL STRAIN e¿ 0.0005 eϭ Ϫ ϭ Ϫ ϭ Ϫ 0.001724 n 0.29 (shortening) MAXIMUM COMPRESSIVE LOAD Pmax ϭ sA ϭ (50.00 ksi) ¢ Problem 1.5-2 A round bar of 10 mm diameter is made of aluminum alloy 7075-T6 (see figure). When the bar is stretched by axial forces P, its diameter decreases by 0.016 mm. Find the magnitude of the load P. (Obtain the material properties from Appendix H.) Solution 1.5-2 d ϭ 10 mm d = 10 mm P P 7075-T6 Aluminum bar in tension ⌬d ϭ 0.016 mm AXIAL STRESS ␴ ϭ Ee ϭ (72 GPa)(0.004848) (Decrease in diameter) ϭ 349.1 MPa (Tension) 7075-T6 From Table H-2: E ϭ 72 GPa ␯ ϭ 0.33 From Table H-3: Yield stress ␴Y ϭ 480 MPa LATERAL STRAIN e¿ ϭ ϭ157 k ␲ ≤ (2.00 in.) 2 4 ¢d Ϫ0.016 mm ϭ ϭ Ϫ0.0016 d 10 mm Because ␴ < ␴Y , Hooke’s law is valid. LOAD P (TENSILE FORCE) P ϭ sA ϭ (349.1 MPa) ¢ ϭ 27.4 kN ␲ ≤ (10 mm) 2 4 AXIAL STRAIN eϭ Ϫ Ϫe¿ 0.0016 ϭ n 0.33 ϭ 0.004848 (Elongation) Problem 1.5-3 A nylon bar having diameter d1 ϭ 3.50 in. is placed inside a steel tube having inner diameter d2 ϭ 3.51 in. (see figure). The nylon bar is then compressed by an axial force P. At what value of the force P will the space between the nylon bar and the steel tube be closed? (For nylon, assume E ϭ 400 ksi and ␯ ϭ 0.4.) Steel tube d1 d2 Nylon bar
    • SECTION 1.5 Solution 1.5-3 Linear Elasticity, Hooke’s Law, and Poisson’s Ratio Nylon bar inside steel tube AXIAL STRAIN e¿ 0.002857 ϭ Ϫ0.007143 ϭϪ n 0.4 (Shortening) eϭ Ϫ d1 d2 AXIAL STRESS ␴ ϭ Ee ϭ (400 ksi)(Ϫ0.007143) COMPRESSION d1ϭ3.50 in. ⌬d1 ϭ 0.01 in. d2ϭ3.51 in. ϭϪ2.857 ksi (Compressive stress) Nylon: E ϭ 400 ksi ␯ ϭ 0.4 Assume that the yield stress is greater than ␴ and Hooke’s law is valid. LATERAL STRAIN e¿ ϭ ¢d1 (Increase in diameter) d1 e¿ ϭ 0.01 in. ϭ 0.002857 3.50 in. FORCE P (COMPRESSION) P ϭ sA ϭ (2.857 ksi) ¢ ϭ 27.5 k Problem 1.5-4 A prismatic bar of circular cross section is loaded by tensile forces P (see figure). The bar has length L ϭ 1.5 m and diameter d ϭ 30 mm. It is made of aluminum alloy with modulus of elasticity E ϭ 75 GPa and Poisson’s ratio ␯ ϭ 1⁄3. If the bar elongates by 3.6 mm, what is the decrease in diameter ⌬d? What is the magnitude of the load P? Solution 1.5-4 ␲ ≤ (3.50 in.) 2 4 d P P L Aluminum bar in tension L ϭ 1.5 m d ϭ 30 mm DECREASE IN DIAMETER E ϭ 75 GPa ␯ ϭ 1⁄3 ⌬d ϭ eЈd ϭ (0.0008)(30 mm) ϭ 0.024 mm ␦ ϭ 3.6 mm (elongation) AXIAL STRESS AXIAL STRAIN ␴ ϭ Ee ϭ (75 GPa)(0.0024) eϭ ␦ 3.6 mm ϭ ϭ 0.0024 L 1.5 m ϭ180 MPa (This stress is less than the yield stress, so Hooke’s law is valid.) LATERAL STRAIN e¿ ϭ Ϫ ne ϭ Ϫ( 1 )(0.0024) 3 ϭϪ0.0008 (Minus means decrease in diameter) LOAD P (TENSION) P ϭ sA ϭ (180 MPa) ¢ ϭ127 kN ␲ ≤ (30 mm) 2 4 21
    • 22 CHAPTER 1 Tension, Compression, and Shear Problem 1.5-5 A bar of monel metal (length L ϭ 8 in., diameter d ϭ 0.25 in.) is loaded axially by a tensile force P ϭ 1500 lb (see figure from Prob. 1.5-4). Using the data in Solution 1.5-5 L ϭ 8 in. Table H-2, Appendix H, determine the increase in length of the bar and the percent decrease in its cross-sectional area. Bar of monel metal in tension d ϭ 0.25 in. From Table H-2: E ϭ 25,000 ksi P ϭ 1500 lb ␯ ϭ 0.32 AXIAL STRESS P 1500 lb sϭ ϭ␲ ϭ 30,560 psi A 4 (0.25 in.) 2 DECREASE IN CROSS-SECTIONAL AREA Original area: A0 ϭ ␲d 2 4 Final area: ␲ (d Ϫ ¢d) 2 4 ␲ A1 ϭ [d2 Ϫ 2d¢d ϩ (¢d) 2 ] 4 Assume ␴ is less than the proportional limit, so that Hooke’s law is valid. A1 ϭ AXIAL STRAIN Decrease in area: eϭ s 30,560 psi ϭ ϭ 0.001222 E 25,000 ksi ⌬A ϭ A0 Ϫ A1 ¢A ϭ INCREASE IN LENGTH ␲ (¢d)(2d Ϫ ¢d) 4 ␦ ϭ e L ϭ (0.001222)(8 in.) ϭ 0.00978 in. PERCENT DECREASE IN AREA LATERAL STRAIN Percent ϭ e¿ ϭ Ϫ ne ϭ Ϫ(0.32)(0.001222) ϭ Ϫ0.0003910 ϭ DECREASE IN DIAMETER (¢d)(2d Ϫ ¢d) ¢A (100) ϭ (100) A0 d2 (0.0000978)(0.4999) (100) (0.25) 2 ϭ 0.078% ¢d ϭ Ϳe¿d Ϳ ϭ (0.0003910)(0.25 in.) ϭ 0.0000978 in. Problem 1.5-6 A tensile test is peformed on a brass specimen 10 mm in diameter using a gage length of 50 mm (see figure). When the tensile load P reaches a value of 20 kN, the distance between the gage marks has increased by 0.122 mm. (a) What is the modulus of elasticity E of the brass? (b) If the diameter decreases by 0.00830 mm, what is Poisson’s ratio? 10 mm 50 mm P P
    • SECTION 1.5 Solution 1.5-6 Linear Elasticity, Hooke’s Law, and Poisson’s Ratio Brass specimen in tension d ϭ 10 mm Gage length L ϭ 50 mm P ϭ 20 kN ␦ ϭ 0.122 mm (a) MODULUS OF ELASTICITY ⌬d ϭ 0.00830 mm Eϭ AXIAL STRESS P 20 kN ϭ ϭ 254.6 MPa A ␲ (10 mm) 2 4 Assume ␴ is below the proportional limit so that Hooke’s law is valid. sϭ eϭ (b) POISSON’S RATIO eЈ ϭ ␯e ⌬d ϭ eЈd ϭ ␯ed nϭ AXIAL STRAIN s 254.6 MPa ϭ 104 GPa ϭ e 0.002440 ¢d 0.00830 mm ϭ ϭ 0.34 ed (0.002440)(10 mm) ␦ 0.122 mm ϭ ϭ 0.002440 L 50 mm P Problem 1.5-7 A hollow steel cylinder is compressed by a force P (see figure). The cylinder has inner diameter d1 ϭ 3.9 in., outer diameter d2 ϭ 4.5 in., and modulus of elasticity E ϭ 30,000 ksi. When the force P increases from zero to 40 k, the outer diameter of the cylinder increases by 455 ϫ 10Ϫ6 in. (a) Determine the increase in the inner diameter. (b) Determine the increase in the wall thickness. (c) Determine Poisson’s ratio for the steel. d1 d2 Solution 1.5-7 Hollow steel cylinder d1 ϭ 3.9 in. (c) POISSON’S RATIO d2 ϭ 4.5 in. d1 d2 t ϭ 0.3 in. E ϭ 30,000 ksi t P ϭ 40 k (compression) ⌬d2 ϭ 455 ϫ 10Ϫ6 in. (increase) Axial stress: s ϭ ␲ 2 ␲ [d Ϫ d2 ] ϭ [ (4.5 in.) 2 Ϫ (3.9 in.) 2 ] 1 4 2 4 2 ϭ 3.9584 in. Aϭ sϭ LATERAL STRAIN ¢d2 455 ϫ 10 Ϫ6 in. e¿ ϭ ϭ ϭ 0.0001011 d2 4.5 in. P A P 40 k ϭ A 3.9584 in.2 ϭ 10.105 ksi (compression) (␴ Ͻ ␴Y ; Hooke’s law is valid) Axial strain: (a) INCREASE IN INNER DIAMETER ¢d1 ϭ e¿d1 ϭ (0.0001011)(3.9 in.) ϭ 394 ϫ 10 Ϫ6 in. (b) INCREASE IN WALL THICKNESS ¢t ϭ e¿t ϭ (0.0001011)(0.3 in.) ϭ 30 ϫ 10 Ϫ6 in. eϭ s 10.105 ksi ϭ E 30,000 ksi ϭ 0.000337 nϭ e¿ 0.0001011 ϭ e 0.000337 ϭ 0.30 23
    • 24 CHAPTER 1 Tension, Compression, and Shear Problem 1.5-8 A steel bar of length 2.5 m with a square cross section 100 mm on each side is subjected to an axial tensile force of 1300 kN (see figure). Assume that E ϭ 200 GPa and v ϭ 0.3. Determine the increase in volume of the bar. Solution 1.5-8 Length: L ϭ 2.5 m ϭ 2500 mm Side: b ϭ 100 mm ␯ ϭ 0.3 AXIAL STRESS sϭ P P ϭ A b2 sϭ 2.5 m DECREASE IN SIDE DIMENSION e¿ ϭ ne ϭ 195 ϫ 10 Ϫ6 ¢b ϭ e¿b ϭ (195 ϫ 10 Ϫ6 )(100 mm) Force: P ϭ 1300 kN 1300 kN ϭ 130 MPa (100 mm) 2 Stress ␴ is less than the yield stress, so Hooke’s law is valid. AXIAL STRAIN s 130 MPa eϭ ϭ E 200 GPa ϭ 650 ϫ 10 Ϫ6 INCREASE IN LENGTH ¢L ϭ eL ϭ (650 ϫ 10 Ϫ6 )(2500 mm) ϭ 1.625 mm 100 mm 1300 kN 1300 kN Square bar in tension Find increase in volume. E ϭ 200 GPa 100 mm ϭ 0.0195 mm FINAL DIMENSIONS L1 ϭ L ϩ ¢L ϭ 2501.625 mm b1 ϭ b Ϫ ¢b ϭ 99.9805 mm FINAL VOLUME V1 ϭ L1b2 ϭ 25,006,490 mm3 1 INITIAL VOLUME V ϭ Lb2 ϭ 25,000,000 mm3 INCREASE IN VOLUME ⌬V ϭ V1ϪV ϭ 6490 mm3
    • SECTION 1.6 25 Shear Stress and Strain Shear Stress and Strain Problem 1.6-1 An angle bracket having thickness t ϭ 0.5 in. is attached to the flange of a column by two 5⁄8-inch diameter bolts (see figure). A uniformly distributed load acts on the top face of the bracket with a pressure p ϭ 300 psi. The top face of the bracket has length L ϭ 6 in. and width b ϭ 2.5 in. Determine the average bearing pressure ␴b between the angle bracket and the bolts and the average shear stress ␶aver in the bolts. (Disregard friction between the bracket and the column.) p b L t Solution 1.6-1 Angle bracket bolted to a column p ϭ pressure acting on top of the bracket ϭ 300 psi F b F ϭ resultant force acting on the bracket ϭ pbL ϭ (300 psi) (2.5 in.) (6.0 in.) ϭ 4.50 k L BEARING PRESSURE BETWEEN BRACKET AND BOLTS Ab ϭ bearing area of one bolt ϭ dt ϭ (0.625 in.) (0.5 in.) ϭ 0.3125 in.2 t Two bolts d ϭ 0.625 in. t ϭ thickness of angle ϭ 0.5 in. b ϭ 2.5 in. L ϭ 6.0 in. sb ϭ F 4.50 k ϭ ϭ 7.20 ksi 2Ab 2(0.3125 in.2 ) AVERAGE SHEAR STRESS IN THE BOLTS As ϭ Shear area of one bolt ␲ ␲ ϭ d2 ϭ (0.625 in.) 2 ϭ 0.3068 in.2 4 4 taver ϭ F 4.50 k ϭ ϭ 7.33 ksi 2As 2(0.3068 in.2 )
    • 26 CHAPTER 1 Tension, Compression, and Shear Problem 1.6-2 Three steel plates, each 16 mm thick, are joined by two 20-mm diameter rivets as shown in the figure. P/2 P/2 P (a) If the load P ϭ 50 kN, what is the largest bearing stress acting on the rivets? (b) If the ultimate shear stress for the rivets is 180 MPa, what force Pult is required to cause the rivets to fail in shear? (Disregard friction between the plates.) Solution 1.6-2 P/2 P/2 P P Three plates joined by two rivets P sb ϭ P P 50 kN ϭ ϭ 2Ab 2dt 2(20 mm)(16 mm) t ϭ 78.1 MPa (b) ULTIMATE LOAD IN SHEAR P P Shear force on two rivets ϭ t ϭ thickness of plates ϭ 16 mm d ϭ diameter of rivets ϭ 20 mm P ϭ 50 kN ␶ULT ϭ 180 MPa (for shear in the rivets) (a) MAXIMUM BEARING STRESS ON THE RIVETS Maximum stress occurs at the middle plate. Ab ϭ bearing area for one rivet Shear force on one rivet ϭ P 2 P 4 Let A ϭ cross-sectional area of one rivet Pր4 P P Shear stress t ϭ ϭ ␲d 2 ϭ A ␲d 2 4( 4 ) or, P ϭ ␲d2␶ At the ultimate load: PULT ϭ ␲d 2tULT ϭ ␲(20 mm) 2 (180 MPa) ϭ 226 kN ϭ dt Problem 1.6-3 A bolted connection between a vertical column and a diagonal brace is shown in the figure. The connection consists of three 5⁄8-in. bolts that join two 1⁄4-in. end plates welded to the brace and a 5⁄8-in. gusset plate welded to the column. The compressive load P carried by the brace equals 8.0 k. Determine the following quantities: (a) The average shear stress ␶aver in the bolts, and (b) The average bearing stress ␴b between the gusset plate and the bolts. (Disregard friction between the plates.) P Column Brace End plates for brace Gusset plate
    • SECTION 1.6 Solution 1.6-3 Shear Stress and Strain 27 Diagonal brace P End plates (a) AVERAGE SHEAR STRESS IN THE BOLTS A ϭ cross-sectional area of one bolt ϭ ␲d2 ϭ 0.3068 in.2 4 V ϭ shear force acting on one bolt P Gusset plate 1 P P ¢ ≤ϭ 3 2 6 V P 8.0 k taver ϭ ϭ ϭ A 6A 6(0.3068 in.2 ) ϭ ϭ 4350 psi 3 bolts in double shear P ϭ compressive force in brace ϭ 8.0 k d ϭ diameter of bolts ϭ 5⁄8 in. ϭ 0.625 in. t1 ϭ thickness of gusset plate ϭ 5⁄8 in. ϭ 0.625 in. t2 ϭ thickness of end plates ϭ 1⁄4 in. ϭ 0.25 in. Problem 1.6-4 A hollow box beam ABC of length L is supported at end A by a 20-mm diameter pin that passes through the beam and its supporting pedestals (see figure). The roller support at B is located at distance L/3 from end A. (a) Determine the average shear stress in the pin due to a load P equal to 10 kN. (b) Determine the average bearing stress between the pin and the box beam if the wall thickness of the beam is equal to 12 mm. (b) AVERAGE BEARING STRESS AGAINST GUSSET PLATE Ab ϭ bearing area of one bolt ϭ t1d ϭ (0.625 in.)(0.625 in.) ϭ 0.3906 in.2 F ϭ bearing force acting on gusset plate from one bolt P ϭ 3 P 8.0 k sb ϭ ϭ ϭ 6830 psi 3Ab 3(0.3906 in.2 ) P Box beam A B L — 3 C 2L — 3 Box beam Pin at support A
    • 28 CHAPTER 1 Tension, Compression, and Shear Solution 1.6-4 Hollow box beam P A C B P ϭ 10 kN d ϭ diameter of pin ϭ 20 mm t ϭ wall thickness of box beam ϭ 12 mm (a) AVERAGE SHEAR STRESS IN PIN L — 3 2L — 3 R = 2P Double shear taver ϭ 2P 2P 4P ϭ 2 ϭ 31.8 MPa ␲ 2 ␲d 2¢ d ≤ 4 (b) AVERAGE BEARING STRESS ON PIN R =P 2 R =P 2 sb ϭ Problem 1.6-5 The connection shown in the figure consists of five steel plates, each 3⁄16 in. thick, joined by a single 1⁄4-in. diameter bolt. The total load transferred between the plates is 1200 lb, distributed among the plates as shown. (a) Calculate the largest shear stress in the bolt, disregarding friction between the plates. (b) Calculate the largest bearing stress acting against the bolt. Solution 1.6-5 2P P ϭ ϭ 41.7 MPa 2(dt) dt 360 lb 600 lb 480 lb 600 lb 360 lb Plates joined by a bolt d ϭ diameter of bolt ϭ 1⁄4 in. (a) MAXIMUM SHEAR STRESS IN BOLT Vmax 4Vmax tmax ϭ d2 ϭ ϭ 7330 psi ␲d2 ␲4 t ϭ thickness of plates ϭ ⁄16 in. 3 FREE-BODY DIAGRAM OF BOLT (b) MAXIMUM BEARING STRESS 360 lb 480 lb A B B A A B B A 360 lb 600 lb 600 lb Fmax ϭ maximum force applied by a plate against the bolt Fmax ϭ 600 lb sb ϭ Section A Ϫ A: V ϭ 360 lb Section B Ϫ B: V ϭ 240 lb Vmax ϭ max. shear force in bolt ϭ 360 lb Fmax ϭ 12,800 psi dt
    • SECTION 1.6 Problem 1.6-6 A steel plate of dimensions 2.5 ϫ 1.2 ϫ 0.1 m is hoisted by a cable sling that has a clevis at each end (see figure). The pins through the clevises are 18 mm in diameter and are located 2.0 m apart. Each half of the cable is at an angle of 32° to the vertical. For these conditions, determine the average shear stress ␶aver in the pins and the average bearing stress ␴b between the steel plate and the pins. Shear Stress and Strain P Cable sling 32° 32° Clevis 2.0 m Steel plate (2.5 × 1.2 × 0.1 m) Solution 1.6-6 Steel plate hoisted by a sling Dimensions of plate: 2.5 ϫ 1.2 ϫ 0.1 m TENSILE FORCE T IN CABLE Volume of plate: V ϭ (2.5) (1.2) (0.1) m ϭ 0.300 m3 ⌺Fvertical ϭ 0 Weight density of steel: ␥ ϭ 77.0 kN/m3 T cos 32Њ Ϫ Weight of plate: W ϭ ␥V ϭ 23.10 kN d ϭ diameter of pin through clevis ϭ18 mm Tϭ t ϭ thickness of plate ϭ 0.1 m ϭ 100 mm W ϭ0 2 W 23.10 kN ϭ ϭ 13.62 kN 2 cos 32Њ 2 cos 32Њ SHEAR STRESS IN THE PINS (DOUBLE SHEAR) FREE-BODY DIAGRAMS OF SLING AND PIN P=W ↑ϩ ↓ Ϫ T 32° taver ϭ T 13.62 kN ϭ ␲ 2Apin 2( 4 )(18 mm) 2 ϭ 26.8 MPa Pin H Cable W 2 32° 32° BEARING STRESS BETWEEN PLATE AND PINS Ab ϭ bearing area ϭ td T 13.62 kN sb ϭ ϭ td (100 mm)(18 mm) ϭ 7.57 MPa W 2 H H 2.0 m W 2 29
    • 30 CHAPTER 1 Tension, Compression, and Shear Problem 1.6-7 A special-purpose bolt of shank diameter d ϭ 0.50 in. passes through a hole in a steel plate (see figure). The hexagonal head of the bolt bears directly against the steel plate. The radius of the circumscribed circle for the hexagon is r ϭ 0.40 in. (which means that each side of the hexagon has length 0.40 in.). Also, the thickness t of the bolt head is 0.25 in. and the tensile force P in the bolt is 1000 lb. Steel plate d (a) Determine the average bearing stress ␴b between the hexagonal head of the bolt and the plate. (b) Determine the average shear stress ␶aver in the head of the bolt. Solution 1.6-7 P 2r t Bolt in tension d ϭ 0.50 in. (a) BEARING STRESS BETWEEN BOLT HEAD AND PLATE r ϭ 0.40 in. P 2r Abϭ bearing area t ϭ 0.25 in. Abϭ area of hexagon minus area of bolt P ϭ 1000 lb d 3r2 ͙3 ␲d2 Ϫ 2 4 3 ␲ 2 2 Ab ϭ (0.40 in.) ( ͙3) Ϫ ¢ ≤ (0.50 in.) 2 4 ϭ 0.4157 in.2Ϫ0.1963 in.2 ϭ t Area of one equilateral triangle r sb ϭ r2 ͙3 ϭ 4 Area of hexagon 2r ϭ 0.2194 in.2 ϭ P 1000 lb ϭ ϭ 4560 psi Ab 0.2194 in.2 (b) SHEAR STRESS IN HEAD OF BOLT 3r2 ͙3 2 As ϭ shear area As ϭ ␲dt taver ϭ P P 1000 lb ϭ ϭ As ␲dt ␲(0.50 in.)(0.25 in.) ϭ 2550 psi Problem 1.6-8 An elastomeric bearing pad consisting of two steel plates bonded to a chloroprene elastomer (an artificial rubber) is subjected to a shear force V during a static loading test (see figure). The pad has dimensions a ϭ 150 mm and b ϭ 250 mm, and the elastomer has thickness t ϭ 50 mm. When the force V equals 12 kN, the top plate is found to have displaced laterally 8.0 mm with respect to the bottom plate. What is the shear modulus of elasticity G of the chloroprene? b a V t
    • SECTION 1.6 Solution 1.6-8 Shear Stress and Strain 31 Bearing pad subjected to shear d = 8.0 mm taver ϭ V ␣ t = 50 mm b = 250 mm d 8.0 mm gaver ϭ ϭ ϭ 0.16 t 50 mm Gϭ V ϭ 12 kN V 12 kN ϭ ϭ 0.32 MPa ab (150 mm)(250 mm) t 0.32 MPa ϭ 2.0 MPa ϭ g 0.16 Width of pad: a ϭ 150 mm Length of pad: b ϭ 250 mm d ϭ 8.0 mm Problem 1.6-9 A joint between two concrete slabs A and B is filled with a flexible epoxy that bonds securely to the concrete (see figure). The height of the joint is h ϭ 4.0 in., its length is L ϭ 40 in., and its thickness is t ϭ 0.5 in. Under the action of shear forces V, the slabs displace vertically through the distance d ϭ 0.002 in. relative to each other. (a) What is the average shear strain ␥aver in the epoxy? (b) What is the magnitude of the forces V if the shear modulus of elasticity G for the epoxy is 140 ksi? A B L h t d A h B V V t Solution 1.6-9 Epoxy joint between concrete slabs d A (a) AVERAGE SHEAR STRAIN B V V t h ϭ 4.0 in. t ϭ 0.5 in. L ϭ 40 in. d ϭ 0.002 in. G ϭ 140 ksi h d gaver ϭ ϭ 0.004 t (b) SHEAR FORCES V Average shear stress : ␶aver ϭ G␥aver V ϭ ␶aver(hL) ϭ G␥aver(hL) ϭ (140 ksi)(0.004)(4.0 in.)(40 in.) ϭ 89.6 k
    • 32 CHAPTER 1 Tension, Compression, and Shear Problem 1.6-10 A flexible connection consisting of rubber pads (thickness t ϭ 9 mm) bonded to steel plates is shown in the figure. The pads are 160 mm long and 80 mm wide. (a) Find the average shear strain ␥aver in the rubber if the force P ϭ 16 kN and the shear modulus for the rubber is G ϭ 1250 kPa. (b) Find the relative horizontal displacement ␦ between the interior plate and the outer plates. 160 mm P — 2 Rubber pad X P P — 2 Rubber pad X 80 mm t = 9 mm t = 9 mm Section X-X Solution 1.6-10 Rubber pads bonded to steel plates P — 2 Thickness t P P — 2 Rubber pad Rubber pads: t ϭ 9 mm Length L ϭ 160 mm Width b ϭ 80 mm (a) SHEAR STRESS AND STRAIN IN THE RUBBER PADS Pր2 8 kN ϭ ϭ 625 kPa bL (80 mm)(160 mm) taver 625 kPa gaver ϭ ϭ ϭ 0.50 G 1250 kPa taver ϭ (b) HORIZONTAL DISPLACEMENT ␦ ϭ ␥avert ϭ (0.50)(9 mm) ϭ 4.50 mm G ϭ 1250 kPa P ϭ 16 kN Problem 1.6-11 A spherical fiberglass buoy used in an underwater experiment is anchored in shallow water by a chain [see part (a) of the figure]. Because the buoy is positioned just below the surface of the water, it is not expected to collapse from the water pressure. The chain is attached to the buoy by a shackle and pin [see part (b) of the figure]. The diameter of the pin is 0.5 in. and the thickness of the shackle is 0.25 in. The buoy has a diameter of 60 in. and weighs 1800 lb on land (not including the weight of the chain). d Pin Shackle (b) (a) Determine the average shear stress ␶aver in the pin. (b) Determine the average bearing stress ␴b between the pin and the shackle. (a)
    • SECTION 1.6 Solution 1.6-11 33 Shear Stress and Strain Submerged buoy d ϭ diameter of buoy ϭ 60 in. dp t T ϭ tensile force in chain dp ϭ diameter of pin ϭ 0.5 in. t ϭ thickness of shackle ϭ 0.25 in. T W ϭ weight of buoy EQUILIBRIUM T ϭ FBϪW ϭ 2376 lb (a) AVERAGE SHEAR STRESS IN PIN Ap ϭ area of pin ␲ Ap ϭ d2 ϭ 0.1963 in.2 4 p taver ϭ T ϭ 6050 psi 2Ap ϭ 1800 lb ␥W ϭ weight density of sea water Ab ϭ 2dpt ϭ 0.2500 in.2 ϭ 63.8 lb/ft3 FREE-BODY DIAGRAM OF BUOY FB W sb ϭ T ϭ 9500 psi Ab FB ϭ buoyant force of water pressure (equals the weight of the displaced sea water) V ϭ volume of buoy ϭ T (b) BEARING STRESS BETWEEN PIN AND SHACKLE ␲d 3 ϭ 65.45 ft3 6 FB ϭ ␥W V ϭ 4176 lb Problem 1.6-12 The clamp shown in the figure is used to support a load hanging from the lower flange of a steel beam. The clamp consists of two arms (A and B) joined by a pin at C. The pin has diameter d ϭ 12 mm. Because arm B straddles arm A, the pin is in double shear. Line 1 in the figure defines the line of action of the resultant horizontal force H acting between the lower flange of the beam and arm B. The vertical distance from this line to the pin is h ϭ 250 mm. Line 2 defines the line of action of the resultant vertical force V acting between the flange and arm B. The horizontal distance from this line to the centerline of the beam is c ϭ 100 mm. The force conditions between arm A and the lower flange are symmetrical with those given for arm B. Determine the average shear stress in the pin at C when the load P ϭ 18 kN. c Line 2 Arm A Arm B Line 1 P h Arm A C P
    • 34 CHAPTER 1 Tension, Compression, and Shear Solution 1.6-12 Clamp supporting a load P ©MC ϭ 0 ‫۔ ە‬ FREE-BODY DIAGRAM OF CLAMP VcϪHh ϭ 0 c Hϭ H H V V h Arm A Vc Pc ϭ ϭ 3.6 kN h 2h FREE-BODY DIAGRAM OF PIN Arm B P 4 C P 2 P 4 P H (from arm A) (from other half of arm B) H 2 h ϭ 250 mm c ϭ 100 mm (from half of arm B) H 2 SHEAR FORCE F IN PIN P ϭ 18 kN F P 4 From vertical equilibrium: P ϭ 9 kN 2 d ϭ diameter of pin at C ϭ12 mm Vϭ FREE-BODY DIAGRAMS OF ARMS A AND B c V= P 2 V= Fϭ P 2 h Arm B P 2 Arm A C H P H C P 2 ¢ ϭ 4.847 kN H H P 2 H 2 ≤ ϩ¢ ≤ B 4 2 H 2 AVERAGE SHEAR STRESS IN THE PIN taver ϭ F F ϭ 2 ϭ 42.9 MPa Apin ␲d 4
    • SECTION 1.6 Problem 1.6-13 A specially designed wrench is used to twist a circular shaft by means of a square key that fits into slots (or keyways) in the shaft and wrench, as shown in the figure. The shaft has diameter d, the key has a square cross section of dimensions b ϫ b, and the length of the key is c. The key fits half into the wrench and half into the shaft (i.e., the keyways have a depth equal to b/2). Derive a formula for the average shear stress ␶aver in the key when a load P is applied at distance L from the center of the shaft. Hints: Disregard the effects of friction, assume that the bearing pressure between the key and the wrench is uniformly distributed, and be sure to draw free-body diagrams of the wrench and key. Shear Stress and Strain c Shaft Key Lever L b P d Solution 1.6-13 Wrench with keyway FREE-BODY DIAGRAM OF KEY FREE-BODY DIAGRAM OF WRENCH b 2 P Plane of shear F C d b F L d b + 2 4 b 2 With friction disregarded, the bearing pressures between the wrench and the shaft are radial. Because the bearing pressure between the wrench and the key is uniformly distributed, the force F acts at the midpoint of the keyway. (Width of keyway ϭ b/2) ©MC ϭ 0 ‫۔ ە‬ d b PL Ϫ F ¢ ϩ ≤ ϭ 0 2 4 Fϭ 4PL 2d ϩ b taver ϭ ϭ F bc 4PL bc(2d ϩ b) F 35
    • 36 CHAPTER 1 Tension, Compression, and Shear Problem 1.6-14 A bicycle chain consists of a series of small links, each 12 mm long between the centers of the pins (see figure). You might wish to examine a bicycle chain and observe its construction. Note particularly the pins, which we will assume to have a diameter of 2.5 mm. In order to solve this problem, you must now make two measurements on a bicycle (see figure): (1) the length L of the crank arm from main axle to pedal axle, and (2) the radius R of the sprocket (the toothed wheel, sometimes called the chainring). Links Pin 12 mm 2.5 mm T F Sprocket (a) Using your measured dimensions, calculate the tensile force T in the chain due to a force F ϭ 800 N applied to one of the pedals. (b) Calculate the average shear stress ␶aver in the pins. R Chain L Solution 1.6-14 Bicycle chain Pin T 2 T 2 T F T 2 Sprocket T 2 12 mm R d = 2.5 mm F ϭ force applied to pedalϭ800 N Tϭ L ϭ length of crank arm MEASUREMENTS (FOR AUTHOR’S BICYCLE) (2) R ϭ 90 mm FL ϭ TR T ϭ taver ϭ ϭ (a) TENSILE FORCE T IN CHAIN ⌺Maxle ϭ 0 (800 N)(162 mm) ϭ 1440 N 90 mm (b) SHEAR STRESS IN PINS R ϭ radius of sprocket (1) L ϭ 162 mm Chain L FL R Substitute numerical values: Problem 1.6-15 A shock mount constructed as shown in the figure is used to support a delicate instrument. The mount consists of an outer steel tube with inside diameter b, a central steel bar of diameter d that supports the load P, and a hollow rubber cylinder (height h) bonded to the tube and bar. (a) Obtain a formula for the shear stress ␶ in the rubber at a radial distance r from the center of the shock mount. (b) Obtain a formula for the downward displacement ␦ of the central bar due to the load P, assuming that G is the shear modulus of elasticity of the rubber and that the steel tube and bar are rigid. Tր2 T 2T ϭ 2 ϭ Apin 2( ␲d ) ␲d 2 4 2FL ␲d2R Substitute numerical values: taver ϭ 2(800 N)(162 mm) ϭ 147 MPa ␲(2.5 mm) 2 (90 mm) Steel tube r P Steel bar d Rubber h b
    • SECTION 1.7 Solution 1.6-15 Allowable Stresses and Allowable Loads 37 Shock mount dr d r b r ϭ radial distance from center of shock mount to element of thickness dr Steel tube ϭ 2␲rh tϭ P P ϭ As 2␲rh (b) DOWNWARD DISPLACEMENT ␦ ␥ ϭ shear strain at distance r Rubber cylinder gϭ t P ϭ G 2␲rhG d␦ ϭ downward displacement for element dr P d d␦ ϭ gdr ϭ h ␦ϭ ␦ ␦ϭ b (a) SHEAR STRESS ␶ AT RADIAL DISTANCE r P 2␲hG ␦ϭ As ϭ shear area at distance r Ύ Pdr 2␲rhG d␦ ϭ Ύ b΋2 d΋2 Ύ b΋2 d΋2 Pdr 2␲rhG dr P bր2 ϭ [ln r] dր2 r 2␲hG P b ln 2␲hG d Allowable Stresses and Allowable Loads Problem 1.7-1 A bar of solid circular cross section is loaded in tension by forces P (see figure). The bar has length L ϭ 16.0 in. and diameter d ϭ 0.50 in. The material is a magnesium alloy having modulus of elasticity E ϭ 6.4 ϫ 106 psi. The allowable stress in tension is ␴allow ϭ 17,000 psi, and the elongation of the bar must not exceed 0.04 in. What is the allowable value of the forces P? Solution 1.7-1 d P P L Magnesium bar in tension L ␴max ϭ E⑀max ϭ (6.4 ϫ 106 psi)(0.00250) d P L ϭ 16.0 in. d ϭ 0.50 in. E ϭ 6.4 ϫ 106 psi ␴allow ϭ 17,000 psi ␦max ϭ 0.04 in. MAXIMUM LOAD BASED UPON ELONGATION ␦max 0.04 in. emax ϭ ϭ ϭ 0.00250 L 16 in. P ϭ 16,000 psi Pmax ϭ smax A ϭ (16,000 psi) ¢ ϭ 3140 lb ␲ ≤ (0.50 in.) 2 4 MAXIMUM LOAD BASED UPON TENSILE STRESS ␲ Pmax ϭ sallow A ϭ (17,000 psi) ¢ ≤ (0.50 in.) 2 4 ϭ 3340 lb ALLOWABLE LOAD Elongation governs. Pallow ϭ 3140 lb
    • 38 CHAPTER 1 Tension, Compression, and Shear Problem 1.7-2 A torque T0 is transmitted between two flanged shafts by means of four 20-mm bolts (see figure). The diameter of the bolt circle is d ϭ 150 mm. If the allowable shear stress in the bolts is 90 MPa, what is the maximum permissible torque? (Disregard friction between the flanges.) Solution 1.7-2 T0 d T0 Shafts with flanges F F T0 ϭ torque transmitted by bolts F F dB ϭ bolt diameter ϭ 20 mm d ϭ diameter of bolt circle d ALLOWABLE SHEAR FORCE IN ONE BOLT F ϭ tallowAbolt ϭ (90 MPa) ¢ ϭ 28.27 kN ϭ 150 mm ␲ ≤ (20 mm) 2 4 ␶allow ϭ 90 MPa MAXIMUM TORQUE F ϭ shear force in one bolt T0 ϭ 2Fd ϭ 2(28.27 kN)(150 mm) d T0 ϭ 4F ¢ ≤ ϭ 2Fd 2 ϭ 8.48 kNиm P Problem 1.7-3 A tie-down on the deck of a sailboat consists of a bent bar bolted at both ends, as shown in the figure. The diameter dB of the bar is 1⁄4 in., the diameter dW of the washers is 7⁄8 in., and the thickness t of the fiberglass deck is 3⁄8 in. If the allowable shear stress in the fiberglass is 300 psi, and the allowable bearing pressure between the washer and the fiberglass is 550 psi, what is the allowable load Pallow on the tie-down? dB dB t dW Solution 1.7-3 dW Bolts through fiberglass P 2 dB ϭ dB Fiberglass 1 in. 4 7 in. 8 3 t ϭ in. 8 dW ϭ t dW P1 ϭ 309.3 lb 2 P1 ϭ 619 lb ALLOWABLE LOAD BASED UPON BEARING PRESSURE ␴b ϭ 550 psi ␲ 2 (d Ϫ d2 ) B 4 W 2 2 P2 ␲ 7 1 ϭ sb Ab ϭ (550 psi) ¢ ≤ B ¢ in. ≤ Ϫ ¢ in. ≤ R 2 4 8 4 Bearing area Ab ϭ ALLOWABLE LOAD BASED UPON SHEAR STRESS IN FIBERGLASS ␶allow ϭ 300 psi Shear area As ϭ ␲dW t P1 ϭ tallow As ϭ tallow (␲d W t) 2 7 3 ϭ (300 psi)(␲) ¢ in. ≤¢ in. ≤ 8 8 ϭ 303.7 lb P2 ϭ 607 lb ALLOWABLE LOAD Bearing pressure governs. Pallow ϭ 607 lb
    • SECTION 1.7 Problem 1.7-4 An aluminum tube serving as a compression brace in the fuselage of a small airplane has the cross section shown in the figure. The outer diameter of the tube is d ϭ 25 mm and the wall thickness is t ϭ 2.5 mm. The yield stress for the aluminum is ␴Y ϭ 270 MPa and the ultimate stress is ␴U ϭ 310 MPa. Calculate the allowable compressive force Pallow if the factors of safety with respect to the yield stress and the ultimate stress are 4 and 5, respectively. Solution 1.7-4 t ϭ 2.5 mm d0 ϭ inner diameter ϭ 20 mm d t d Aluminum tube in compression d ϭ 25 mm t 39 Allowable Loads Atube ϭ ␲ 2 2 (d Ϫ d0 ) ϭ 176.7 mm2 4 YIELD STRESS ULTIMATE STRESS ␴Y ϭ 270 MPa ␴U ϭ 310 MPa F.S. ϭ 4 F.S. ϭ 5 270 MPa 4 ϭ 67.5 MPa sallow ϭ sallow ϭ 310 MPa 5 ϭ 62 MPa The ultimate stress governs. ALLOWABLE COMPRESSIVE FORCE Pallow ϭ ␴allow Atube ϭ (62 MPa )(176.7 mm2) ϭ11.0 kN Problem 1.7-5 A steel pad supporting heavy machinery rests on four short, hollow, cast iron piers (see figure). The ultimate strength of the cast iron in compression is 50 ksi. The outer diameter of the piers is d ϭ 4.5 in. and the wall thickness is t ϭ 0.40 in. Using a factor of safety of 3.5 with respect to the ultimate strength, determine the total load P that may be supported by the pad. Solution 1.7-5 t d Cast iron piers in compression Four piers d0 ϭ d Ϫ 2t ϭ 3.7 in. ␴U ϭ 50 ksi t Aϭ n ϭ 3.5 d sallow ϭ d ϭ 4.5 in. t ϭ 0.4 in. sU 50 ksi ϭ ϭ 14.29 ksi n 3.5 ␲ 2 ␲ (d Ϫ d2 ) ϭ [ (4.5 in.) 2 Ϫ (3.7 in.) 2 ] o 4 4 2 ϭ 5.152 in. P1 ϭ allowable load on one pier ϭ ␴allow A ϭ (14.29 ksi)(5.152 in.2) ϭ 73.62 k Total load P ϭ 4P1 ϭ 294 k
    • 40 CHAPTER 1 Tension, Compression, and Shear Problem 1.7-6 A long steel wire hanging from a balloon carries a weight W at its lower end (see figure). The 4-mm diameter wire is 25 m long. What is the maximum weight Wmax that can safely be carried if the tensile yield stress for the wire is ␴Y ϭ 350 MPa and a margin of safety against yielding of 1.5 is desired? (Include the weight of the wire in the calculations.) d L W Solution 1.7-6 Wire hanging from a balloon d ϭ 4.0 mm L ϭ 25 m d L ␴Y ϭ 350 MPa Margin of safety ϭ 1.5 W Factor of safety ϭ n ϭ 2.5 sY sallow ϭ ϭ 140 MPa n Weight density of steel: ␥ ϭ 77.0 kN/m3 Weight of wire: ␲d 2 W0 ϭ gAL ϭ g ¢ ≤ (L) 4 W0 ϭ (77.0 kNրm3 ) ¢ ϭ 24.19 N ␲ ≤ (4.0 mm) 2 (25 m) 4 Total load P ϭ Wmax ϩ W0 ϭ ␴allow A Wmax ϭ sallow A Ϫ W0 ϭ (140 MPa) ¢ ϭ (140 MPa) ¢ ␲d2 ≤ Ϫ 24.19 N 4 ␲ ≤ (4.0 mm) 2 Ϫ 24.19 N 4 ϭ 1759.3 N Ϫ 24.2 N ϭ 1735.1 N Wmax ϭ 1740 N
    • SECTION 1.7 Problem 1.7-7 A lifeboat hangs from two ship’s davits, as shown in the figure. A pin of diameter d ϭ 0.80 in. passes through each davit and supports two pulleys, one on each side of the davit. Cables attached to the lifeboat pass over the pulleys and wind around winches that raise and lower the lifeboat. The lower parts of the cables are vertical and the upper parts make an angle ␣ ϭ 15° with the horizontal. The allowable tensile force in each cable is 1800 lb, and the allowable shear stress in the pins is 4000 psi. If the lifeboat weighs 1500 lb, what is the maximum weight that should be carried in the lifeboat? Solution 1.7-7 T T Pulley Pin Cable ALLOWABLE TENSILE FORCE IN ONE CABLE BASED UPON SHEAR IN THE PINS 15° Pulley RH Pin Vallow ϭ tallow Apin ϭ (4000 psi) ¢ ϭ 2011 lb V ϭ 1.2175T RV T1 ϭ ALLOWABLE FORCE IN ONE CABLE BASED UPON T2 ϭ Tallow ϭ 1800 lb T Pin diameter d ϭ 0.80 in. T ϭ tensile force in one cable MAXIMUM WEIGHT Shear in the pins governs. Tallow ϭ 1800 lb Tmax ϭ T1 ϭ 1652 lb ␶allow ϭ 4000 psi Total tensile force in four cables W ϭ weight of lifeboat ϭ 4Tmax ϭ 6608 lb Wmax ϭ 4TmaxϪW ϭ1500 lb ⌺Fvert ϭ 0 ␲ ≤ (0.80 in.) 2 4 Vallow ϭ 1652 lb 1.2175 TENSION IN THE CABLE ⌺Fhoriz ϭ 0 ␣ = 15° Davit Lifeboat supported by four cables FREE-BODY DIAGRAM OF ONE PULLEY RH ϭ T cos 15Њ ϭ 0.9659T RV ϭ T Ϫ T sin 15Њ ϭ 0.7412T V ϭ shear force in pin V ϭ ͙(RH ) 2 ϩ (RV ) 2 ϭ 1.2175T 41 Allowable Stresses and Allowable Loads ϭ 6608 lbϪ1500 lb ϭ 5110 lb
    • 42 CHAPTER 1 Tension, Compression, and Shear Problem 1.7-8 A ship’s spar is attached at the base of a mast by a pin connection (see figure). The spar is a steel tube of outer diameter d2 ϭ 80 mm and inner diameter d1 ϭ 70 mm. The steel pin has diameter d ϭ 25 mm, and the two plates connecting the spar to the pin have thickness t ϭ 12 mm. The allowable stresses are as follows: compressive stress in the spar, 70 MPa; shear stress in the pin, 45 MPa; and bearing stress between the pin and the connecting plates, 110 MPa. Determine the allowable compressive force Pallow in the spar. Solution 1.7-8 P Pin Spar Connecting plate Pin connection for a ship’s spar P Spar: d2 ϭ 80 mm d1 ϭ 70 mm Spar Pin: d ϭ 25 mm Plates: t ϭ 12 mm Pin Plate ALLOWABLE LOAD P BASED UPON SHEAR IN THE PIN (DOUBLE SHEAR) ␶allow ϭ 45 MPa As ϭ 2 ¢ ␲d 2 ␲ ≤ ϭ (25 mm) 2 ϭ 981.7 mm2 4 2 P2 ϭ ␶allow As ϭ (45 MPa )(981.7 mm2) ϭ 44.2 kN ALLOWABLE LOAD P BASED UPON COMPRESSION IN THE SPAR ALLOWABLE LOAD P BASED UPON BEARING ␴b ϭ 110 MPa ␴c ϭ 70 MPa ␲ ␲ Ac ϭ (d2 Ϫ d2 ) ϭ [ (80 mm) 2 Ϫ (70 mm) 2 ] 2 1 4 4 ϭ1178.1 mm2 P1 ϭ ␴cAc ϭ (70 Mast Ab ϭ 2dt ϭ 2(25 mm)(12 mm) ϭ 600 mm2 P3 ϭ ␴bAb ϭ (110 MPa )(600 mm2) ϭ 66.0 kN ALLOWABLE COMPRESSIVE LOAD IN THE SPAR MPa )(1178.1 mm2) ϭ 82.5 kN Shear in the pin governs. Pallow ϭ 44.2 kN Problem 1.7-9 What is the maximum possible value of the clamping force C in the jaws of the pliers shown in the figure if a ϭ 3.75 in., b ϭ 1.60 in., and the ultimate shear stress in the 0.20-in. diameter pin is 50 ksi? What is the maximum permissible value of the applied load P if a factor of safety of 3.0 with respect to failure of the pin is to be maintained? P Pin P a b
    • SECTION 1.7 Allowable Loads Solution 1.7-9 Forces in pliers V ϭ shear force in pin (single shear) V V VϭR ∴ Cϭ and P ϭ a b 1ϩ 1ϩ a b FREE-BODY DIAGRAM OF ONE ARM Pin C R MAXIMUM CLAMPING FORCE Cult ␶ult ϭ 50 ksi Vult ϭ ␶ult Apin P a ϭ (50 ksi) ¢ b C ϭ clamping force ϭ1571 lb Vult 1571 lb Cult ϭ ϭ b 1.60 in. 1ϩ 1ϩ a 3.75 in. R ϭ reaction at pin a ϭ 3.75 in. b ϭ 1.60 in. ϭ1100 lb d ϭ diameter of pin ϭ 0.20 in. ©Mpin ϭ 0 Cϭ Pa b ⌺Fvert ϭ 0 ␲ ≤ (0.20 in.) 2 4 MAXIMUM LOAD Pult Cb Ϫ Pa ϭ 0 ‫۔ە‬ Pϭ Cb a C a ϭ P b ↑ϩ ↓Ϫ P ϩ C Ϫ R ϭ 0 R ϭ P ϩ C ϭ P ¢1 ϩ Pult ϭ a b ≤ ϭ C ¢1 ϩ ≤ a b Vult 1571 lb ϭ ϭ 469.8 lb a 3.75 in. 1ϩ 1ϩ b 1.60 in. ALLOWABLE LOAD Pallow Pallow ϭ Problem 1.7-10 A metal bar AB of weight W is suspended by a system of steel wires arranged as shown in the figure. The diameter of the wires is 2 mm, and the yield stress of the steel is 450 MPa. Determine the maximum permissible weight Wmax for a factor of safety of 1.9 with respect to yielding. Pult 469.8 lb ϭ n 3.0 ϭ157 lb 0.75 m 0.75 m 2.5 m 1.75 m 1.75 m W A B 43
    • 44 CHAPTER 1 Tension, Compression, and Shear Solution 1.7-10 Bar AB suspended by steel wires 3b FREE-BODY DIAGRAM OF WIRE ACE 3b 10 b W 2 E F CAB E ⌺Fhoriz ϭ 0 7b C TCD ϭ 2CAB D C TCD ϭ 7b A B W b = 0.25 m LAC ϭ LEC ϭ ͙(3b) 2 ϩ (7b) 2 ϭ b͙58 FREE-BODY DIAGRAM OF POINT A TAC A CAB W 2 ©Fvert ϭ 0 TAC ¢ 7b b͙58 W͙58 TAC ϭ 14 ©Fhoriz ϭ 0 CAB ϭ TAC ¢ 3W 14 ≤ 3b b͙58 ϭ W 2 ≤ ϭ CAB Problem 1.7-11 Two flat bars loaded in tension by forces P are spliced using two rectangular splice plates and two 5⁄8-in. diameter rivets (see figure). The bars have width b ϭ 1.0 in. (except at the splice, where the bars are wider) and thickness t ϭ 0.4 in. The bars are made of steel having an ultimate stress in tension equal to 60 ksi. The ultimate stresses in shear and bearing for the rivet steel are 25 ksi and 80 ksi, respectively. Determine the allowable load Pallow if a safety factor of 2.5 is desired with respect to the ultimate load that can be carried. (Consider tension in the bars, shear in the rivets, and bearing between the rivets and the bars. Disregard friction between the plates.) 3W 7 CAB A W 2 ALLOWABLE TENSILE FORCE IN A WIRE d ϭ 2 mm ␴Y ϭ 450 MPa sY ¢ F.S. ϭ 1.9 ␲d ≤ sY A 4 Tallow ϭ ϭ n n 450 MPa ␲ ϭ¢ ≤¢ ≤ (2 mm) 2 ϭ 744.1 N 1.9 4 2 MAXIMUM TENSILE FORCES IN WIRES TCD ϭ 3W 7 TAC ϭ W͙58 14 Force in wire AC is larger. MAXIMUM ALLOWABLE WEIGHT W Wmax ϭ 14 TAC ϭ ͙58 ϭ 1370 N 14 Tallow ͙58 ϭ 14 ͙58 (744.1 N) b = 1.0 in. P P Bar Splice plate t = 0.4 in. P P
    • SECTION 1.7 Solution 1.7-11 Allowable Stresses and Allowable Loads 45 Splice between two flat bars t P P P2 ϭ tult (2AR ) ϭ 2(25 ksi)(0.3068 in.2 ) ϭ 15.34 k ULTIMATE LOAD BASED UPON TENSION IN THE BARS ULTIMATE LOAD BASED UPON BEARING Cross-sectional area of bars: Ab ϭ bearing area ϭ dt A ϭ bt b ϭ 1.0 in. 5 P3 ϭ sbAb ϭ (80 ksi) ¢ in. ≤ (0.4 in.) ϭ 20.0 k 8 t ϭ 0.4 in. A ϭ 0.40 in.2 ULTIMATE LOAD P1 ϭ ␴ultA ϭ (60 ksi)(0.40 in.2) ϭ 24.0 k Shear governs. Pult ϭ 15.34 k ULTIMATE LOAD BASED UPON SHEAR IN THE RIVETS Double shear d ϭ diameter of rivets ALLOWABLE LOAD d ϭ 5⁄8 in. AR ϭ area of rivets Pallow ϭ AR ϭ 2 ␲d 2 ␲ 5 ϭ ¢ in. ≤ ϭ 0.3068 in.2 4 4 8 (a) Obtain a formula for the allowable load Pallow that the bar can carry in tension. (b) Calculate the value of Pallow if the bar is made of brass with diameter d ϭ 40 mm and ␴allow ϭ 80 MPa. Problem 1.7-12 A solid bar of circular cross section (diameter d) has a hole of diameter d/4 drilled laterally through the center of the bar (see figure). The allowable average tensile stress on the net cross section of the bar is ␴allow. d P Pult 15.34 k ϭ ϭ 6.14 k n 2.5 (Hint: Use the formulas of Case 15, Appendix D.) d — 4 d — 4 P d Solution 1.7-12 Bar with a hole CROSS SECTION OF BAR From Case 15, Appendix D: d — 4 A ϭ 2r2 ¢ ␣ Ϫ rϭ d bϭ dր8 r 1 ϭ arc cos ¢ ≤ 4 ␣ ϭ arc cos B d 2 aϭ r2 Ϫ ¢ ab ≤ r2 d 15 d ≤ ϭd ϭ ͙15 8 B 64 8 d 8 2 d d ¢ ≤ ¢ ͙15 ≤ d 2 1 8 8 A ϭ 2 ¢ ≤ B arc cos Ϫ R 2 4 (dր2) 2 ϭ d2 1 ͙15 ¢ arc cos Ϫ ≤ ϭ 0.5380 d 2 2 4 16 (a) ALLOWABLE LOAD IN TENSION Pallow ϭ ␴allow A ϭ 0.5380d2 ␴allow (b) SUBSTITUTE NUMERICAL VALUES ␴allow ϭ 80 MPa Pallow ϭ 68.9 kN d ϭ 40 mm
    • 46 CHAPTER 1 Tension, Compression, and Shear Problem 1.7-13 A solid steel bar of diameter d1 ϭ 2.25 in. has a hole of diameter d2 ϭ 1.125 in. drilled through it (see figure). A steel pin of diameter d2 passes through the hole and is attached to supports. Determine the maximum permissible tensile load Pallow in the bar if the yield stress for shear in the pin is ␶Y ϭ 17,000 psi, the yield stress for tension in the bar is ␴Y ϭ 36,000 psi, and a factor of safety of 2.0 with respect to yielding is required. (Hint: Use the formulas of Case 15, Appendix D.) Solution 1.7-13 d2 d1 d1 P Bar with a hole d1 ϭ 2.25 in. r ␣ C a d2 d1 ALLOWABLE LOAD BASED ON TENSION IN THE BAR d2 ϭ 1.125 in. P1 ϭ From Case 15, Appendix D: A ϭ 2r2 ¢ ␣ Ϫ ab ≤ r2 d1 r ϭ ϭ 1.125 in. 2 d2ր2 d2 ␣ ϭ arc cos ϭ arc cos d1ր2 d1 d2 1.125 in. 1 1 ϭ ϭ ␣ ϭ arc cos ϭ 1.0472 rad d1 2.25 in. 2 2 d2 a ϭ ϭ 0.5625 in. 2 r 36,000 psi sY (1.5546 in.2 ) Aϭ n 2.0 ϭ 28.0 k b b ϭ ͙r2 Ϫ a2 ϭ 0.9743 in. ␣ ab A ϭ 2r2¢ ␣ Ϫ 2 ≤ C a r (0.5625 in.)(0.9743 in.) A ϭ 2(1.125 in.) 2 B 1.0472 Ϫ R (1.125 in.) 2 2 ϭ 1.5546 in. ALLOWABLE LOAD BASED ON SHEAR IN THE PIN Double shear As ϭ 2Apin ϭ 2 ¢ ϭ 1.9880 in.2 P2 ϭ 17,000 psi tY As ϭ (1.9880 in.) 2 n 2.0 ϭ 16.9 k ALLOWABLE LOAD Shear in the pin governs. Pallow ϭ 16.9 k Problem 1.7-14 The piston in an engine is attached to a connecting rod AB, which in turn is connected to a crank arm BC (see figure). The piston slides without friction in a cylinder and is subjected to a force P (assumed to be constant) while moving to the right in the figure. The connecting rod, which has diameter d and length L, is attached at both ends by pins. The crank arm rotates about the axle at C with the pin at B moving in a circle of radius R. The axle at C, which is supported by bearings, exerts a resisting moment M against the crank arm. (a) Obtain a formula for the maximum permissible force Pallow based upon an allowable compressive stress ␴c in the connecting rod. (b) Calculate the force Pallow for the following data: ␴c ϭ 160 MPa, d ϭ 9.00 mm, and R ϭ 0.28L. 2 ␲d2 ␲ ≤ϭ (1.125 in.) 2 4 2 Cylinder P Piston Connecting rod A M d C B L R
    • SECTION 1.7 Solution 1.7-14 P Allowable Stresses and Allowable Loads Piston and connecting rod A M ␣ C R L B The maximum allowable force P occurs when cos ␣ has its smallest value, which means that ␣ has its largest value. LARGEST VALUE OF ␣ d ϭ diameter of rod AB L2 − R2 A C ␣ FREE-BODY DIAGRAM OF PISTON R L RP B P ␣ C The largest value of ␣ occurs when point B is the farthest distance from line AC. The farthest distance is the radius R of the crank arm. Therefore, BC ϭ R ͙L2 Ϫ R2 R 2 ϭ 1Ϫ¢ ≤ L B L P ϭ applied force (constant) Also, AC ϭ ͙L2 Ϫ R2 C ϭ compressive force in connecting rod cos ␣ ϭ RP ϭ resultant of reaction forces between cylinder and piston (no friction) ⌺Fhoriz ϭ 0 S d ϩ Ϫ P Ϫ C cos ␣ ϭ 0 P ϭ C cos ␣ MAXIMUM COMPRESSIVE FORCE C IN CONNECTING ROD Cmax ϭ ␴c Ac in which Ac ϭ area of connecting rod ␲d2 Ac ϭ 4 MAXIMUM ALLOWABLE FORCE P P ϭ Cmax cos ␣ ϭ sc Ac cos ␣ (a) MAXIMUM ALLOWABLE FORCE P ␲d 2 R 2 ≤ 1Ϫ¢ ≤ 4 B L Pallow ϭ sc Ac cos ␣ ϭ sc ¢ (b) SUBSTITUTE NUMERICAL VALUES ␴c ϭ 160 MPa R ϭ 0.28L Pallow ϭ 9.77 kN d ϭ 9.00 mm R/L ϭ 0.28 47
    • 48 CHAPTER 1 Tension, Compression, and Shear Design for Axial Loads and Direct Shear Problem 1.8-1 An aluminum tube is required to transmit an axial tensile force P ϭ 34 k (see figure). The thickness of the wall of the tube is to be 0.375 in. What is the minimum required outer diameter dmin if the allowable tensile stress is 9000 psi? Solution 1.8-1 Aluminum tube in tension P P SOLVE FOR d: d P d P dϭ P ϩt ␲tsallow SUBSTITUTE NUMERICAL VALUES: P ϭ 34 k dmin ϭ t ϭ 0.375 in. ␴allow ϭ 9000 psi 34 k ϩ 0.375 in. ␲(0.375 in.)(9000 psi) ϭ 3.207 in. ϩ 0.375 in. ␲ ␲ A ϭ [d2 Ϫ (d Ϫ 2t) 2 ] ϭ (4t)(d Ϫ t) 4 4 dmin ϭ 3.58 in. ϭ ␲t(d Ϫ t) P ϭ sallow A ϭ ␲t(d Ϫ t)sallow Problem 1.8-2 A steel pipe having yield stress ␴Y ϭ 270 MPa is to carry an axial compressive load P ϭ 1200 kN (see figure). A factor of safety of 1.8 against yielding is to be used. If the thickness t of the pipe is to be one-eighth of its outer diameter, what is the minimum required outer diameter dmin? Solution 1.8-2 d Steel pipe in compression d t =— 8 P ϭ 1200 kN ␴Y ϭ 270 MPa n ϭ 1.8 d Aϭ d t =— 8 P ␴allow ϭ 150 MPa ␲ 2 d 2 7␲d 2 B d Ϫ ¢d Ϫ ≤ R ϭ 4 4 64 P ϭ sallow A ϭ 7␲d2 s 64 allow P B 7␲sallow SOLVE FOR d: d2 ϭ 64 P 7␲sallow dϭ8 1200 kN ϭ 153 mm B 7␲ (150 MPa) SUBSTITUTE NUMERICAL VALUES: dmin ϭ 8
    • SECTION 1.8 Problem 1.8-3 A horizontal beam AB supported by an inclined strut CD carries a load P ϭ 2500 lb at the position shown in the figure. The strut, which consists of two bars, is connected to the beam by a bolt passing through the three bars meeting at joint C. If the allowable shear stress in the bolt is 14,000 psi, what is the minimum required diameter dmin of the bolt? 4 ft 4 ft B C A 3 ft P D Beam AB Bolt      Strut CD Solution 1.8-3 Beam ACB supported by a strut CD FREE-BODY DIAGRAM A 4 ft ␣ B 3 ft P (RD)V ©MA ϭ 0 ‫۔ە‬ (RD ) H ϭ 3 ft C 5 ft D FCD ϭ compressive force in strut ϭ RD D (RD)H 4 ft 4 ft C A Ϫ P(8 ft) ϩ (RD ) H (3 ft) ϭ 0 8 P 3 FCD ϭ (RD ) H ¢ (RD)H ␣ 5 5 8P 10P ≤ϭ¢ ≤¢ ≤ϭ 4 4 3 3 SHEAR FORCE ACTING ON BOLT Vϭ REACTION AT JOINT D FCD 5P ϭ 2 3 REQUIRED AREA AND DIAMETER OF BOLT D Aϭ V tallow ϭ 5P 3tallow Aϭ ␲d2 4 d2 ϭ 20P 3␲tallow SUBSTITUTE NUMERICAL VALUES: P ϭ 2500 lb (RD)V RD 49 Design for Axial Loads and Direct Shear d2 ϭ 0.3789 in.2 dmin ϭ 0.616 in. ␶allow ϭ 14,000 psi
    • 50 CHAPTER 1 Tension, Compression, and Shear Problem 1.8-4 Two bars of rectangular cross section (thickness t ϭ 15 mm) are connected by a bolt in the manner shown in the figure. The allowable shear stress in the bolt is 90 MPa and the allowable bearing stress between the bolt and the bars is 150 MPa. If the tensile load P ϭ 31 kN, what is the minimum required diameter dmin of the bolt? t t P P P Solution 1.8-4 P Bolted connection BASED UPON SHEAR IN THE BOLT t P P t One bolt in double shear. P ϭ 31 kN ␶allow ϭ 90 MPa Abolt ϭ d2 ϭ 2P ␲tallow d1 ϭ 2(31 kN) 2P ϭ B ␲tallow B ␲(90 MPa) ϭ 14.8 mm ␴b ϭ 150 MPa t ϭ 15 mm ␲d2 P ϭ 4 2tallow P 2tallow BASED UPON BEARING BETWEEN PLATE AND BOLT Find minimum diameter of bolt. Abearing ϭ dϭ P tsb P sb dt ϭ d2 ϭ P sb 31 kN ϭ 13.8 mm (15 mm) (150 MPa) MINIMUM DIAMETER OF BOLT Shear governs. dmin ϭ 14.8 mm
    • SECTION 1.8 Design for Axial Loads and Direct Shear Problem 1.8-5 Solve the preceding problem if the bars have thickness t ϭ 5⁄16 in., the allowable shear stress is 12,000 psi, the allowable bearing stress is 20,000 psi, and the load P ϭ 1800 lb. Solution 1.8-5 Bolted connection BASED UPON SHEAR IN THE BOLT t P P t One bolt in double shear. P ϭ 1800 lb ␶allow ϭ 12,000 psi ␴b ϭ 20,000 psi t ϭ 5⁄16 in. Find minimum diameter of bolt. Abolt ϭ d2 ϭ d1 ϭ ␲d2 P ϭ 4 2tallow P 2tallow 2(1800 lb) 2P ϭ ϭ 0.309 in. ␲tallow B ␲(12,000 psi) B 2P ␲tallow BASED UPON BEARING BETWEEN PLATE AND BOLT P P Abearing ϭ dt ϭ sb sb P 1800 lb dϭ d2 ϭ 5 ϭ 0.288 in. tsb ( 16 in.)(20,000 psi) MINIMUM DIAMETER OF BOLT Shear governs. dmin ϭ 0.309 in. Problem 1.8-6 A suspender on a suspension bridge consists of a cable that passes over the main cable (see figure) and supports the bridge deck, which is far below. The suspender is held in position by a metal tie that is prevented from sliding downward by clamps around the suspender cable. Let P represent the load in each part of the suspender cable, and let ␪ represent the angle of the suspender cable just above the tie. Finally, let ␴allow represent the allowable tensile stress in the metal tie. (a) Obtain a formula for the minimum required cross-sectional area of the tie. (b) Calculate the minimum area if P ϭ 130 kN, ␪ ϭ 75°, and ␴allow ϭ 80 MPa. Main cable Suspender Collar ␪ ␪ Tie Clamp P P 51
    • 52 CHAPTER 1 Solution 1.8-6 Tension, Compression, and Shear Suspender tie on a suspension bridge F F ␪ F ϭ tensile force in cable above tie P ϭ tensile force in cable below tie ␪ FORCE TRIANGLE cot u ϭ ␴allow ϭ allowable tensile stress in the tie Tie P F T ϭ P cot u P ␪ (a) MINIMUM REQUIRED AREA OF TIE Amin ϭ P T P T T P cot u ϭ sallow sallow (b) SUBSTITUTE NUMERICAL VALUES: P ϭ 130 kN FREE-BODY DIAGRAM OF HALF THE TIE Note: Include a small amount of the cable in the free-body diagram ␪ ϭ 75Њ ␴allow ϭ 80 MPa Amin ϭ 435 mm 2 T ϭ tensile force in the tie F T P Problem 1.8-7 A square steel tube of length L ϭ 20 ft and width b2 ϭ 10.0 in. is hoisted by a crane (see figure). The tube hangs from a pin of diameter d that is held by the cables at points A and B. The cross section is a hollow square with inner dimension b1 ϭ 8.5 in. and outer dimension b2 ϭ 10.0 in. The allowable shear stress in the pin is 8,700 psi, and the allowable bearing stress between the pin and the tube is 13,000 psi. Determine the minimum diameter of the pin in order to support the weight of the tube. (Note: Disregard the rounded corners of the tube when calculating its weight.) d A B Square tube Square tube Pin d A B L b2 b1 b2
    • SECTION 1.8 Solution 1.8-7 T Tube hoisted by a crane T ϭ tensile force in cable T W ϭ weight of steel tube d ϭ diameter of pin b1 ϭ inner dimension of tube ϭ 8.5 in. b2 Design for Axial Loads and Direct Shear b2 ϭ outer dimension of tube ϭ 10.0 in. b1 d L ϭ length of tube ϭ 20 ft ␶allow ϭ 8,700 psi ␴b ϭ 13,000 psi W ϭ gs AL 1 ft2 ϭ (490 lbրft3 )(27.75 in.2 ) ¢ ≤ (20 ft) 144 in. ϭ 1,889 lb DIAMETER OF PIN BASED UPON SHEAR Double shear. 2(8,700 psi) ¢ 2␶allow Apin ϭ W ␲d 2 ≤ ϭ 1889 lb 4 d2 ϭ 0.1382 in.2 d1 ϭ 0.372 in. DIAMETER OF PIN BASED UPON BEARING WEIGHT OF TUBE ␴b(b2 Ϫ b1) d ϭ W ␥s ϭ weight density of steel (13,000 psi)(10.0 in. Ϫ 8.5 in.) d ϭ 1,889 lb d2 ϭ 0.097 in. ϭ 490 lb/ft3 A ϭ area of tube ϭ b2 2 Ϫ b2 1 MINIMUM DIAMETER OF PIN ϭ (10.0 in.) Ϫ (8.5 in.) ϭ 27.75 in. 2 2 Shear governs. dmin ϭ 0.372 in. Problem 1.8-8 Solve the preceding problem if the length L of the tube is 6.0 m, the outer width is b2 ϭ 250 mm, the inner dimension is b1 ϭ 210 mm, the allowable shear stress in the pin is 60 MPa, and the allowable bearing stress is 90 MPa. Solution 1.8-8 T Tube hoisted by a crane T ϭ tensile force in cable T W ϭ weight of steel tube d ϭ diameter of pin b1 ϭ inner dimension of tube ϭ 210 mm b2 ϭ outer dimension of tube b2 ϭ 250 mm b1 d L ϭ length of tube ϭ 6.0 m ␶allow ϭ 60 MPa ␴b ϭ 90 MPa WEIGHT OF TUBE ␥s ϭ weight density of steel ϭ 77.0 kN/m3 A ϭ area of tube A ϭ b2 Ϫ b2 ϭ 18,400 mm2 2 1 W ϭ ␥sAL ϭ (77.0 kN/m3)(18,400 mm2)(6.0 m) ϭ 8.501 kN DIAMETER OF PIN BASED UPON SHEAR Double shear. 2(60 MPa) ¢ 2␶allow Apin ϭ W ␲ 2 ≤d ϭ 8.501 kN 4 d 2 ϭ 90.20 mm2 d1 ϭ 9.497 mm DIAMETER OF PIN BASED UPON BEARING ␴b(b2 Ϫ b1)d ϭ W (90 MPa )(40 mm)d ϭ 8.501 kN d2 ϭ 2.361 mm MINIMUM DIAMETER OF PIN Shear governs. dmin ϭ 9.50 mm 53
    • 54 CHAPTER 1 Tension, Compression, and Shear Problem 1.8-9 A pressurized circular cylinder has a sealed cover plate fastened with steel bolts (see figure). The pressure p of the gas in the cylinder is 290 psi, the inside diameter D of the cylinder is 10.0 in., and the diameter dB of the bolts is 0.50 in. If the allowable tensile stress in the bolts is 10,000 psi, find the number n of bolts needed to fasten the cover. Cover plate Steel bolt p Cylinder D Solution 1.8-9 Pressurized cylinder NUMBER OF BOLTS Bolt p D p ϭ 290 psi D ϭ 10.0 in. db ϭ 0.50 in. ␴allow ϭ 10,000 psi n ϭ number of bolts F ϭ total force acting on the cover plate from the internal pressure ␲D2 Fϭp¢ ≤ 4 P ϭ tensile force in one bolt Pϭ F ␲pD2 ϭ n 4n ␲ Ab ϭ area of one bolt ϭ d2 4 b P ϭ ␴allow Ab sallow ϭ nϭ ␲pD2 pD2 P ϭ ϭ 2 Ab (4n)( ␲ )d2 ndb b 4 pD2 d2 sallow b SUBSTITUTE NUMERICAL VALUES: nϭ (290 psi)(10 in.) 2 ϭ 11.6 (0.5 in.) 2 (10,000 psi) Use 12 bolts
    • SECTION 1.8 Design for Axial Loads and Direct Shear Problem 1.8-10 A tubular post of outer diameter d2 is guyed by two cables fitted with turnbuckles (see figure). The cables are tightened by rotating the turnbuckles, thus producing tension in the cables and compression in the post. Both cables are tightened to a tensile force of 110 kN. Also, the angle between the cables and the ground is 60°, and the allowable compressive stress in the post is ␴c ϭ 35 MPa. If the wall thickness of the post is 15 mm, what is the minimum permissible value of the outer diameter d2? Cable Turnbuckle d2 Post 60° Solution 1.8-10 30° Tubular post with guy cables 30° d2 ϭ outer diameter T d1 ϭ inner diameter T t ϭ wall thickness P ϭ 15 mm T ϭ tensile force in a cable ϭ 110 kN d2 ␴allow ϭ 35 MPa P ϭ compressive force in post ϭ 2T cos 30Њ AREA OF POST Aϭ ␲ 2 ␲ (d2 Ϫ d2 ) ϭ [d2 Ϫ (d2 Ϫ 2t) 2 ] 1 4 4 2 ϭ ␲t(d2 Ϫ t) EQUATE AREAS AND SOLVE FOR d2: 2T cos 30Њ ϭ ␲t(d2 Ϫ t) sallow d2 ϭ 2T cos 30Њ ϩt ␲tsallow SUBSTITUTE NUMERICAL VALUES: REQUIRED AREA OF POST (d2 ) min ϭ 131 mm P 2T cos 30Њ Aϭ ϭ sallow sallow Problem 1.8-11 A cage for transporting workers and supplies on a construction site is hoisted by a crane (see figure). The floor of the cage is rectangular with dimensions 6 ft by 8 ft. Each of the four lifting cables is attached to a corner of the cage and is 13 ft long. The weight of the cage and its contents is limited by regulations to 9600 lb. Determine the required cross-sectional area AC of a cable if the breaking stress of a cable is 91 ksi and a factor of safety of 3.5 with respect to failure is desired. 60° 55
    • 56 CHAPTER 1 Tension, Compression, and Shear Solution 1.8-11 Cage hoisted by a crane W From geometry: L2 ϭ ¢ b 2 c 2 ≤ ϩ ¢ ≤ ϩ h2 2 2 (13 ft)2 ϭ (3 ft)2 ϩ (4 ft)2 ϩ h2 Solving, h ϭ 12 ft B FORCE IN A CABLE T TV A A T ϭ force in one cable (cable AB) TV ϭ vertical component of T b c (Each cable carries the same load.) W ϭ 9600 lb W 9600 lb ϭ ϭ 2400 lb 4 4 T L 13 ft ϭ ϭ TV h 12 ft 13 T ϭ TV ϭ 2600 lb 12 Breaking stress of a cable: REQUIRED AREA OF CABLE Dimensions of cage: ∴ TV ϭ b ϭ 6 ft c ϭ 8 ft Length of a cable: L ϭ 13 ft Weight of cage and contents: ␴ult ϭ 91 ksi AC ϭ Factor of safety: n ϭ 3.5 sallow ϭ sult 91 ksi ϭ ϭ 26,000 psi n 3.5 GEOMETRY OF ONE CABLE (CABLE AB) Point B is above the midpoint of the cage B L = 13 ft h A b 2 c 2 b ϭ 3 ft 2 c ϭ 4 ft 2 h ϭ height from A to B T sallow ϭ 2,600 lb ϭ 0.100 in.2 26,000 psi (Note: The diameter of the cable cannot be calculated from the area AC, because a cable does not have a solid circular cross section. A cable consists of several strands wound together. For details, see Section 2.2.)
    • SECTION 1.8 Problem 1.8-12 A steel column of hollow circular cross section is supported on a circular steel base plate and a concrete pedestal (see figure). The column has outside diameter d ϭ 250 mm and supports a load P ϭ 750 kN. d P Column (a) If the allowable stress in the column is 55 MPa, what is the minimum required thickness t? Based upon your result, select a thickness for the column. (Select a thickness that is an even integer, such as 10, 12, 14, . . ., in units of millimeters.) (b) If the allowable bearing stress on the concrete pedestal is 11.5 MPa, what is the minimum required diameter D of the base plate if it is designed for the allowable load Pallow that the column with the selected thickness can support? Solution 1.8-12 P Base plate t D Hollow circular column P SUBSTITUTE NUMERICAL VALUES IN EQ. (1): t2 Ϫ 250 t ϩ d t (750 ϫ 103 N) ϭ0 ␲(55 Nրmm2 ) (Note: In this eq., t has units of mm.) t2 Ϫ 250t ϩ 4,340.6 ϭ 0 Solve the quadratic eq. for t: t ϭ 18.77 mm D tmin ϭ 18.8 mm Use t ϭ 20 mm d ϭ 250 mm (b) DIAMETER D OF THE BASE PLATE P ϭ 750 kN For the column, ␴allow ϭ 55 MPa (compression in column) A ϭ ␲t(d Ϫ t) Pallow ϭ ␴allow ␲t(d Ϫ t) D ϭ diameter of base plate ␴b ϭ 11.5 MPa (allowable pressure on concrete) P Aϭ sallow ϭ ␲t(d Ϫ t) ϭ ␲t2 Ϫ ␲td ϩ t2 Ϫ dt ϩ P sallow ␲d2 ␲ Ϫ (d Ϫ 2t) 2 4 4 ␲ (4t)(d Ϫ t) ϭ ␲t(d Ϫ t) 4 P ϭ ␲D2 Pallow ϭ sb 4 4(55 MPa)(20 mm)(230 mm) 11.5 MPa D2 ϭ 88,000 mm2 sallow Dmin ϭ 297 mm ϭ0 P ϭ0 ␲sallow Area of base plate ϭ ␲D2 sallow␲t(d Ϫ t) ϭ sb 4 4sallowt(d Ϫ t) D2 ϭ sb (a) THICKNESS t OF THE COLUMN Aϭ Pallow ϭ ␴allow A where A is the area of the column with t ϭ 20 mm. t ϭ thickness of column (Eq. 1) 57 Design for Axial Loads and Direct Shear D ϭ 296.6 mm
    • 58 CHAPTER 1 Tension, Compression, and Shear Problem 1.8-13 A bar of rectangular cross section is subjected to an axial load P (see figure). The bar has width b ϭ 2.0 in. and thickness t ϭ 0.25 in. A hole of diameter d is drilled through the bar to provide for a pin support. The allowable tensile stress on the net cross section of the bar is 20 ksi, and the allowable shear stress in the pin is 11.5 ksi. (a) Determine the pin diameter dm for which the load P will be a maximum. (b) Determine the corresponding value Pmax of the load. Solution 1.8-13 P b d t P Bar with pin connection t GRAPH OF EQS. (1) AND (2) P Load P (lb) d P2 20,000 Eq.(2) Width of bar b ϭ 2 in. Eq.(1) Thickness t ϭ 0.25 in. 10,000 Pmax ␴allow ϭ 20 ksi P1 ␶allow ϭ 11.5 ksi d ϭ diameter of pin (inches) 0.5 dm Diameter d (in.) 0 P ϭ axial load (pounds) (a) MAXIMUM LOAD OCCURS WHEN P1 ϭ P2 ALLOWABLE LOAD BASED UPON TENSION IN BAR 10,000 Ϫ 5,000d ϭ 18,064d 2 P1 ϭ ␴allow Anet ϭ ␴allow(b Ϫ d)t or 18,064d 2 ϩ 5,000d Ϫ 10,000 ϭ 0 ϭ (20,000 psi)(2 in. Ϫ d)(0.25 in.) ϭ 5,000(2 Ϫ d) ϭ 10,000 Ϫ 5,000d Eq. (1) P2 ϭ 2tallow ¢ ␲d 4 2 ≤ ϭ tallow ¢ ␲d 2 2 ≤ ␲d 2 ϭ (11,500 psi) ¢ ≤ ϭ 18,064d 2 2 Solve quadratic equation: d ϭ 0.6184 in. ALLOWABLE LOAD BASED UPON SHEAR IN PIN Double shear 1.0 dm ϭ 0.618 in. (b) MAXIMUM LOAD Substitute d ϭ 0.6184 in. into Eq. (1) or Eq. (2): Eq. (2) Pmax ϭ 6910 lb Problem 1.8-14 A flat bar of width b ϭ 60 mm and thickness t ϭ 10 mm is loaded in tension by a force P (see figure). The bar is attached to a support by a pin of diameter d that passes through a hole of the same size in the bar. The allowable tensile stress on the net cross section of the bar is ␴T ϭ 140 MPa, the allowable shear stress in the pin is ␶S ϭ 80 MPa, and the allowable bearing stress between the pin and the bar is ␴B ϭ 200 MPa. (a) Determine the pin diameter dm for which the load P will be a maximum. (b) Determine the corresponding value Pmax of the load. d P b t P
    • SECTION 1.8 Solution 1.8-14 59 Design for Axial Loads and Direct Shear Bar with a pin connection SHEAR IN THE PIN d PS ϭ 2tS Apin ϭ 2tS ¢ P b ϭ 2(80 MPa) ¢ ␲d 2 ≤ 4 ␲ 2 1 ≤ (d ) ¢ ≤ 4 1000 ϭ 0.040 ␲d2 ϭ 0.12566d2 t P BEARING BETWEEN PIN AND BAR PB ϭ ␴B td ϭ (200 MPa)(10 mm)(d) ¢ d b ϭ 60 mm ϭ 2.0 d t ϭ 10 mm 1 ≤ 1000 (Eq. 3) GRAPH OF EQS. (1), (2), AND (3) d ϭ diameter of hole and pin ␴T ϭ 140 MPa P (kN) ␶S ϭ 80 MPa 100 ␴B ϭ 200 MPa ␴ and ␶ are in N/mm2 (same as MPa) Sh Eq.(3) 25 b, t, and d are in mm PB ring Bea P max 50 P is in kN PS ea r P Tens T ion 75 UNITS USED IN THE FOLLOWING CALCULATIONS: Eq.(1) dm Eq.(2) 0 0 10 30 20 d (mm) TENSION IN THE BAR PT ϭ ␴T (Net area) ϭ ␴t(t)(b Ϫ d) 1 ϭ (140 MPa)(10 mm)(60 mm Ϫ d) ¢ ≤ 1000 ϭ 1.40 (60 Ϫ d) (Eq. 2) (Eq. 1) (a) PIN DIAMETER dm PT ϭ PB or 1.40(60 Ϫ d) ϭ 2.0 d Solving, dm ϭ 84.0 mm ϭ 24.7 mm 3.4 (b) LOAD Pmax Substitute dm into Eq. (1) or Eq. (3): Pmax ϭ 49.4 kN 40
    • 60 CHAPTER 1 Tension, Compression, and Shear Problem 1.8-15 Two bars AC and BC of the same material support a vertical load P (see figure). The length L of the horizontal bar is fixed, but the angle ␪ can be varied by moving support A vertically and changing the length of bar AC to correspond with the new position of support A. The allowable stresses in the bars are the same in tension and compression. We observe that when the angle ␪ is reduced, bar AC becomes shorter but the cross-sectional areas of both bars increase (because the axial forces are larger). The opposite effects occur if the angle ␪ is increased. Thus, we see that the weight of the structure (which is proportional to the volume) depends upon the angle ␪. Determine the angle ␪ so that the structure has minimum weight without exceeding the allowable stresses in the bars. (Note: The weights of the bars are very small compared to the force P and may be disregarded.) Solution 1.8-15 A θ B C L P Two bars supporting a load P Joint C A WEIGHT OF TRUSS T ␥ ϭ weight density of material W ϭ ␥(AACLAC ϩ ABCLBC) θ ϭ C C θ B C L P T ϭ tensile force in bar AC C ϭ compressive force in bar BC ©Fvert ϭ 0 ©Fhoriz ϭ 0 Tϭ P sin u Cϭ P tan u AREAS OF BARS P ϭ gPL 1 1 ¢ ϩ ≤ sallow sin u cos u tan u gPL 1 ϩ cos2u ¢ ≤ sallow sin u cos u ␥, P, L, and ␴allow are constants W varies only with ␪ Let k ϭ gPL sallow (k has units of force) W 1 ϩ cos2u ϭ k sin u cos u (Nondimensional) GRAPH OF EQ. (2): 12 AAC ϭ T P ϭ sallow sallow sin u W 9 k ABC ϭ C C ϭ sallow sallow tan u 6 LENGTHS OF BARS LAC ϭ L cos u LBC ϭ L Eq. (1) 3 0 30° ␪ 60° 90° Eq. (2)
    • SECTION 1.8 ANGLE ␪ THAT MAKES W A MINIMUM Use Eq. (2) Let f ϭ 1 ϩ cos2u sin u cos u df ϭ0 du df (sin u cos u) (2) (cos u) (Ϫsin u) Ϫ (1 ϩ cos2u) (Ϫsin2u ϩ cos2u) ϭ du sin2u cos2u ϭ Ϫsin2u cos2u ϩ sin2u Ϫ cos2u Ϫ cos4u sin2u cos2u SET THE NUMERATOR ϭ 0 AND SOLVE FOR ␪: Ϫsin2␪ cos2␪ ϩ sin2␪ Ϫ cos2␪ Ϫ cos4␪ ϭ 0 Replace sin2␪ by 1 Ϫ cos2␪: Ϫ(1 Ϫ cos2␪)(cos2␪) ϩ 1 Ϫ cos2␪ Ϫ cos2␪ Ϫ cos4␪ ϭ 0 Combine terms to simplify the equation: 1 Ϫ 3 cos2u ϭ 0 u ϭ 54.7Њ cos u ϭ 1 ͙3 Design for Axial Loads and Direct Shear 61
    • 2 Axially Loaded Members Changes in Lengths of Axially Loaded Members Problem 2.2-1 The T-shaped arm ABC shown in the figure lies in a vertical plane and pivots about a horizontal pin at A. The arm has constant cross-sectional area and total weight W. A vertical spring of stiffness k supports the arm at point B. Obtain a formula for the elongation ␦ of the spring due to the weight of the arm. Solution 2.2-1 k A B C b b b T-shaped arm F ϭ tensile force in the spring FREE-BODY DIAGRAM OF ARM ©MA ϭ 0 ‫۔ ە‬ F A C B W 3 W 3 F(b) Ϫ W 3 Fϭ 4W 3 W b W 3b W ¢ ≤Ϫ ¢ ≤ Ϫ (2b) ϭ 0 3 2 3 2 3 ␦ ϭ elongation of the spring b b ␦ϭ F 4W ϭ k 3k Problem 2.2-2 A steel cable with nominal diameter 25 mm (see Table 2-1) is used in a construction yard to lift a bridge section weighing 38 kN, as shown in the figure. The cable has an effective modulus of elasticity E ϭ 140 GPa. (a) If the cable is 14 m long, how much will it stretch when the load is picked up? (b) If the cable is rated for a maximum load of 70 kN, what is the factor of safety with respect to failure of the cable? 63
    • 64 CHAPTER 2 Axially Loaded Members Solution 2.2-2 Bridge section lifted by a cable (b) FACTOR OF SAFETY A ϭ 304 mm2 (from Table 2-1) PULT ϭ 406 kN (from Table 2-1) W ϭ 38 kN Pmax ϭ 70 kN E ϭ 140 GPa nϭ L ϭ 14 m PULT 406 kN ϭ ϭ 5.8 Pmax 70 kN (a) STRETCH OF CABLE ␦ϭ (38 kN)(14 m) WL ϭ EA (140 GPa)(304 mm2 ) ϭ 12.5 mm Problem 2.2-3 A steel wire and a copper wire have equal lengths and support equal loads P (see figure). The moduli of elasticity for the steel and copper are Es ϭ 30,000 ksi and Ec ϭ 18,000 ksi, respectively. Copper wire (a) If the wires have the same diameters, what is the ratio of the elongation of the copper wire to the elongation of the steel wire? (b) If the wires stretch the same amount, what is the ratio of the diameter of the copper wire to the diameter of the steel wire? Steel wire P P Solution 2.2-3 Steel wire and copper wire Copper wire Equal lengths and equal loads Steel: Es ϭ 30,000 ksi Copper: Ec ϭ 18,000 ksi (a) RATIO OF ELONGATIONS (EQUAL DIAMETERS) Steel wire P P ␦c ϭ PL Ec A ␦s ϭ PL Es A ␦c Es 30 ϭ ϭ ϭ 1.67 ␦s Ec 18 (b) RATIO OF DIAMETERS (EQUAL ELONGATIONS) ␦c ϭ ␦s Ec ¢ PL PL ϭ or Ec Ac ϭ Es As Ec Ac Es As ␲ 2 ␲ 2 ≤ d ϭ Es ¢ ≤ ds 4 c 4 d2 Es c ϭ d2 Ec s Es dc 30 ϭ ϭ ϭ 1.29 ds B Ec B 18
    • SECTION 2.2 Changes in Lengths of Axially Loaded Members Problem 2.2-4 By what distance h does the cage shown in the figure move downward when the weight W is placed inside it? Consider only the effects of the stretching of the cable, which has axial rigidity EA ϭ 10,700 kN. The pulley at A has diameter dA ϭ 300 mm and the pulley at B has diameter dB ϭ 150 mm. Also, the distance L1 ϭ 4.6 m, the distance L2 ϭ 10.5 m, and the weight W ϭ 22 kN. (Note: When calculating the length of the cable, include the parts of the cable that go around the pulleys at A and B.) L1 A L2 B Cage W Solution 2.2-4 Cage supported by a cable dA ϭ 300 mm A LENGTH OF CABLE dB ϭ 150 mm L1 1 1 L ϭ L1 ϩ 2L2 ϩ (␲dA ) ϩ (␲dB ) 4 2 L1 ϭ 4.6 m L2 ϭ 10.5 m L2 EA ϭ 10,700 kN W ϭ 22 kN ϭ 4,600 mm ϩ 21,000 mm ϩ 236 mm ϩ 236 mm ϭ 26,072 mm ELONGATION OF CABLE ␦ϭ B TL (11 kN)(26,072 mm) ϭ ϭ 26.8 mm EA (10,700 kN) LOWERING OF THE CAGE W TENSILE FORCE IN CABLE Tϭ W ϭ 11 kN 2 h ϭ distance the cage moves downward hϭ 1 ␦ ϭ 13.4 mm 2 Problem 2.2-5 A safety valve on the top of a tank containing steam under pressure p has a discharge hole of diameter d (see figure). The valve is designed to release the steam when the pressure reaches the value pmax. If the natural length of the spring is L and its stiffness is k, what should be the dimension h of the valve? (Express your result as a formula for h.) h d p 65
    • 66 CHAPTER 2 Axially Loaded Members Solution 2.2-5 Safety valve pmax ϭ pressure when valve opens L ϭ natural length of spring (L > h) k ϭ stiffness of spring h FORCE IN COMPRESSED SPRING F ϭ k(L Ϫ h) (From Eq. 2-1a) PRESSURE FORCE ON SPRING d h ϭ height of valve (compressed length of the spring) P ϭ pmax ¢ EQUATE FORCES AND SOLVE FOR h: FϭP k(L Ϫ h) ϭ hϭLϪ p ϭ pressure in tank ␲pmax d2 4 ␲pmax d2 4k d ϭ diameter of discharge hole Problem 2.2-6 The device shown in the figure consists of a pointer ABC supported by a spring of stiffness k ϭ 800 N/m. The spring is positioned at distance b ϭ 150 mm from the pinned end A of the pointer. The device is adjusted so that when there is no load P, the pointer reads zero on the angular scale. If the load P ϭ 8 N, at what distance x should the load be placed so that the pointer will read 3° on the scale? Solution 2.2-6 ␲d2 ≤ 4 P x A B C 0 k b Pointer supported by a spring ©MA ϭ 0 ‫۔ە‬ FREE-BODY DIAGRAM OF POINTER P x ϪPx ϩ (k␦)b ϭ 0 B A or ␦ϭ C Px kb Let ␣ ϭ angle of rotation of pointer F = k␦ b Pϭ8N k ϭ 800 N/m b ϭ 150 mm ␦ ϭ displacement of spring F ϭ force in spring ϭ k␦ ␦ Px tan ␣ ϭ ϭ 2 b kb xϭ kb2 tan ␣ P SUBSTITUTE NUMERICAL VALUES: ␣ ϭ 3Њ xϭ (800 Nրm)(150 mm) 2 tan 3Њ 8N ϭ 118 mm
    • SECTION 2.2 Changes in Lengths of Axially Loaded Members Problem 2.2-7 Two rigid bars, AB and CD, rest on a smooth horizontal surface (see figure). Bar AB is pivoted end A and bar CD is pivoted at end D. The bars are connected to each other by two linearly elastic springs of stiffness k. Before the load P is applied, the lengths of the springs are such that the bars are parallel and the springs are without stress. Derive a formula for the displacement ␦C at point C when the load P is acting. (Assume that the bars rotate through very small angles under the action of the load P.) Solution 2.2-7 b b b A B C P D Two bars connected by springs b b B DISPLACEMENT DIAGRAMS ␦B 2 A A b b B ␦B D C D C ␦C 2 ␦C P k ϭ stiffness of springs ␦B ϭ displacement of point B ␦C ϭ displacement at point C due to load P ␦C ϭ displacement of point C FREE-BODY DIAGRAMS ⌬1 ϭ elongation of first spring A b b F1 F1 b ␦B 2 ϭ ␦C Ϫ B ⌬2 ϭ shortening of second spring F2 F2 ϭ ␦B Ϫ ␦C 2 Also, ¢1 ϭ b C D F1 4P ϭ ; k 3k ¢2 ϭ P SOLVE THE EQUATIONS: F1 ϭ tensile force in first spring F2 ϭ compressive force in second spring ¢1 ϭ ¢1 ␦C Ϫ ␦B 4P ϭ 2 3k EQUILIBRIUM ‫۔ە‬ ¢2 ϭ ¢2 ␦B Ϫ ␦C 2P ϭ 2 3k ©MA ϭ 0 ϪbF1 ϩ 2bF2 ϭ 0 ©MD ϭ 0 2bP Ϫ 2bF1 ϩ bF2 ϭ 0 4P Solving, F1 ϭ 3 2P F2 ϭ 3 F1 ϭ 2F2 F2 ϭ 2F1 Ϫ 2P Eliminate ␦B and obtain ␦C : ␦C ϭ 20P 9k F2 2P ϭ k 3k 67
    • 68 CHAPTER 2 Axially Loaded Members Problem 2.2-8 The three-bar truss ABC shown in the figure has a span L ϭ 3 m and is constructed of steel pipes having cross-sectional area A ϭ 3900 mm2 and modulus of elasticity E ϭ 200 GPa. A load P acts horizontally to the right at joint C. C (a) If P ϭ 650 kN, what is the horizontal displacement of joint B? (b) What is the maximum permissible load Pmax if the displacement of joint B is limited to 1.5 mm? 45° A 45° L Solution 2.2-8 Truss with horizontal load C P From force triangle, L — 2 45° A FAB ϭ P (tension) 2 (a) HORIZONTAL DISPLACEMENT ␦B 45° B P ϭ 650 kN ␦B ϭ L RB ϭ Lϭ3m (650 kN)(3 m) 2(200 GPa)(3900 mm2 ) ϭ 1.25 mm A ϭ 3900 mm2 E ϭ 200 GPa ©MA ϭ 0 FAB LAB PL ϭ EA 2EA (b) MAXIMUM LOAD Pmax gives RB ϭ ␦max ϭ 1.5 mm P 2 Pmax P ϭ ␦max ␦ FREE-BODY DIAGRAM OF JOINT B Pmax ϭ P ¢ Pmax ϭ (650 kN) ¢ Force triangle: ϭ 780 kN FBC B FAB RB = P 2 FBC FAB ␦max ≤ ␦ 1.5 mm ≤ 1.25 mm P B
    • SECTION 2.2 Problem 2.2-9 An aluminum wire having a diameter d ϭ 2 mm and length L ϭ 3.8 m is subjected to a tensile load P (see figure). The aluminum has modulus of elasticity E ϭ 75 GPa. If the maximum permissible elongation of the wire is 3.0 mm and the allowable stress in tension is 60 MPa, what is the allowable load Pmax? Solution 2.2-9 P P L Pmax ϭ d P L ϭ EA ␦ L max (75 GPa)(3.142 mm2 ) (3.0 mm) 3.8 m ϭ 186 N d ϭ 2 mm L ϭ 3.8 m MAXIMUM LOAD BASED UPON STRESS E ϭ 75 GPa ␲d ϭ 3.142 mm2 4 2 sallow ϭ 60 MPa ␦ϭ PL EA sϭ P A Pmax ϭ Asallow ϭ (3.142 mm2 )(60 MPa) ϭ 189 N MAXIMUM LOAD BASED UPON ELONGATION ␦max ϭ 3.0 mm d Aluminum wire in tension P Aϭ 69 Changes in Lengths of Axially Loaded Members ALLOWABLE LOAD Elongation governs. Problem 2.2-10 A uniform bar AB of weight W ϭ 25 N is supported by two springs, as shown in the figure. The spring on the left has stiffness k1 ϭ 300 N/m and natural length L1 ϭ 250 mm. The corresponding quantities for the spring on the right are k2 ϭ 400 N/m and L2 ϭ 200 mm. The distance between the springs is L ϭ 350 mm, and the spring on the right is suspended from a support that is distance h ϭ 80 mm below the point of support for the spring on the left. At what distance x from the left-hand spring should a load P ϭ 18 N be placed in order to bring the bar to a horizontal position? Pmax ϭ 186 N h k1 L1 k2 L2 W A B P x L
    • 70 CHAPTER 2 Axially Loaded Members Solution 2.2-10 Bar supported by two springs Reference line ©MA ϭ 0 ‫۔ ە‬ F2L Ϫ PX Ϫ ©Fvert ϭ 0 cϩ h L1 WL ϭ0 2 (Eq. 1) Ϫ T F1 ϩ F2 Ϫ P Ϫ W ϭ 0 k1 L2 k2 ␦1 A ␦2 B P W SOLVE EQS. (1) AND (2): F1 ϭ P¢ 1 Ϫ L — 2 F1 ϭ (18) ¢ 1 Ϫ F2 ϭ (18) ¢ k1 ϭ 300 N/m k2 ϭ 400 N/m L ϭ 350 mm Px W ϩ L 2 x ≤ ϩ 12.5 ϭ 30.5 Ϫ 51.429x 0.350 x ≤ ϩ 12.5 ϭ 51.429x ϩ 12.5 0.350 ELONGATIONS OF THE SPRINGS h ϭ 80 mm ␦1 ϭ NATURAL LENGTHS OF SPRINGS L2 ϭ 200 mm F1 F1 ϭ ϭ 0.10167 Ϫ 0.17143x k1 300 ␦2 ϭ P ϭ 18 N F2 F2 ϭ ϭ 0.12857x ϩ 0.031250 k2 400 BAR AB REMAINS HORIZONTAL OBJECTIVE Find distance x for bar AB to be horizontal. Points A and B are the same distance below the reference line (see figure above). ∴ L1 ϩ ␦1 ϭ h ϩ L2 ϩ ␦2 FREE-BODY DIAGRAM OF BAR AB or F1 F2 A B 0.250 ϩ 0.10167 Ϫ 0.17143 x ϭ 0.080 ϩ 0.200 ϩ 0.12857 x ϩ 0.031250 SOLVE FOR x: 0.300 x ϭ 0.040420 W x x ϭ 135 mm L — 2 F2 ϭ UNITS: Newtons and meters L — 2 W ϭ 25 N P x W ≤ϩ L 2 SUBSTITUTE NUMERICAL VALUES: x L1 ϭ 250 mm (Eq. 2) L — 2 x ϭ 0.1347 m
    • SECTION 2.2 Problem 2.2-11 A hollow, circular, steel column (E ϭ 30,000 ksi) is subjected to a compressive load P, as shown in the figure. The column has length L ϭ 8.0 ft and outside diameter d ϭ 7.5 in. The load P ϭ 85 k. If the allowable compressive stress is 7000 psi and the allowable shortening of the column is 0.02 in., what is the minimum required wall thickness tmin? P t L d Solution 2.2-11 Column in compression P REQUIRED AREA BASED UPON ALLOWABLE SHORTENING ␦ϭ PL EA Aϭ (85 k)(96 in.) PL ϭ E␦allow (30,000 ksi)(0.02 in.) ϭ 13.60 in.2 t SHORTENING GOVERNS Amin ϭ 13.60 in.2 L d MINIMUM THICKNESS tmin Aϭ ␲ 2 [d Ϫ (d Ϫ 2t) 2 ] 4 or 4A Ϫ d 2 ϭ Ϫ (d Ϫ 2t) 2 ␲ P ϭ 85 k E ϭ 30,000 ksi L ϭ 8.0 ft d ϭ 7.5 in. ␴allow ϭ 7,000 psi ␦allow ϭ 0.02 in. d d 2 A tϭ Ϫ ¢ ≤ Ϫ ␲ 2 B 2 (d Ϫ 2t) 2 ϭ d 2 Ϫ P A Aϭ P 85 k ϭ ϭ 12.14 in.2 sallow 7,000 psi 4A ␲ d Ϫ 2t ϭ or d d 2 Amin tmin ϭ Ϫ ¢ ≤ Ϫ ␲ 2 B 2 tmin ϭ B d2Ϫ or 7.5 in. 7.5 in. 2 13.60 in.2 Ϫ ¢ ≤ Ϫ ␲ 2 B 2 SUBSTITUTE NUMERICAL VALUES REQUIRED AREA BASED UPON ALLOWABLE STRESS sϭ 71 Changes in Lengths of Axially Loaded Members tmin ϭ 0.63 in. 4A ␲
    • 72 CHAPTER 2 Axially Loaded Members Problem 2.2-12 The horizontal rigid beam ABCD is supported by vertical bars BE and CF and is loaded by vertical forces P1 ϭ 400 kN and P2 ϭ 360 kN acting at points A and D, respectively (see figure). Bars BE and CF are made of steel (E ϭ 200 GPa) and have cross-sectional areas ABE ϭ 11,100 mm2 and ACF ϭ 9,280 mm2. The distances between various points on the bars are shown in the figure. Determine the vertical displacements ␦A and ␦D of points A and D, respectively. 1.5 m 1.5 m B A 2.1 m C D 2.4 m P1 = 400 kN P2 = 360 kN F 0.6 m E Solution 2.2-12 Rigid beam supported by vertical bars 1.5 m 1.5 m B A 2.1 m C D 2.4 m P1 = 400 kN SHORTENING OF BAR BE FBE LBE (296 kN)(3.0 m) ␦BE ϭ ϭ EABE (200 GPa)(11,100 mm2 ) ϭ 0.400 mm P2 = 360 kN F SHORTENING OF BAR CF 0.6 m ␦CF ϭ E FCF LCF (464 kN)(2.4 m) ϭ EACF (200 GPa)(9,280 mm2 ) ϭ 0.600 mm ABE ϭ 11,100 mm2 ACF ϭ 9,280 mm2 DISPLACEMENT DIAGRAM E ϭ 200 GPa A 1.5 m B 1.5 m 2.1 m C D LBE ϭ 3.0 m LCF ϭ 2.4 m ␦A ␦BE P1 ϭ 400 kN; P2 ϭ 360 kN ␦CF ␦D FREE-BODY DIAGRAM OF BAR ABCD 1.5 m A 1.5 m B ␦BE Ϫ ␦A ϭ ␦CF Ϫ ␦BE or 2.1 m C D ␦A ϭ 2␦BE Ϫ ␦CF ␦A ϭ 2(0.400 mm) Ϫ 0.600 m ϭ 0.200 mm (Downward) P1 = 400 kN FBE FCF P2 = 360 kN ©MB ϭ 0 ‫۔ە‬ (400 kN)(1.5 m) ϩ FCF (1.5 m) Ϫ (360 kN)(3.6 m) ϭ 0 FCF ϭ 464 kN ©MC ϭ 0 ‫۔ ە‬ (400 kN)(3.0 m) Ϫ FBE (1.5 m) Ϫ (360 kN)(2.1 m) ϭ 0 FBE ϭ 296 kN 2.1 (␦ Ϫ ␦BE ) 1.5 CF 12 7 ␦D ϭ ␦ Ϫ ␦ 5 CF 5 BE 12 7 ϭ (0.600 mm) Ϫ (0.400 mm) 5 5 ␦D Ϫ ␦CF ϭ or ϭ 0.880 mm (Downward)
    • SECTION 2.2 Problem 2.2-13 A framework ABC consists of two rigid bars AB and BC, each having length b (see the first part of the figure). The bars have pin connections at A, B, and C and are joined by a spring of stiffness k. The spring is attached at the midpoints of the bars. The framework has a pin support at A and a roller support at C, and the bars are at an angle ␣ to the hoizontal. B b — 2 P B b — 2 k ␣ ␣ A C Solution 2.2-13 When a vertical load P is applied at joint B (see the second part of the figure) the roller support C moves to the right, the spring is stretched, and the angle of the bars decreases from ␣ to the angle ␪. Determine the angle ␪ and the increase ␦ in the distance between points A and C. (Use the following data; b ϭ 8.0 in., k ϭ 16 lb/in., ␣ ϭ 45°, and P ϭ 10 lb.) b — 2 b — 2 ␪ A ␪ C Framework with rigid bars and a spring WITH LOAD P B b — 2 b — 2 L2 ϭ span from A to C ϭ 2b cos ␪ k b — 2 73 Changes in Lengths of Axially Loaded Members ␣ A S2 ϭ length of spring b — 2 C ␣ L1 ϭ L2 ϭ b cos u 2 FREE-BODY DIAGRAM OF BC P WITH NO LOAD B F L1 ϭ span from A to C ϭ 2b cos ␣ h S1 ϭ length of spring h — 2 P — 2 F h — 2 L1 ϭ ϭ b cos ␣ 2 ␪ L2 — 2 P B C P — 2 h ϭ height from C to B ϭ b sin ␪ L2 ϭ b cos u 2 A ␪ ␪ L2 C F ϭ force in spring due to load P ©MB ϭ 0 ‫۔ ە‬ P L2 h ¢ ≤ Ϫ F ¢ ≤ ϭ 0 or P cos ␪ ϭ F sin ␪ 2 2 2 (Eq. 1) (Continued)
    • 74 CHAPTER 2 Axially Loaded Members DETERMINE THE ANGLE ␪ From Eq. (2): cos ␣ ϭ cos u Ϫ ⌬S ϭ elongation of spring Therefore, ␦ ϭ 2b ¢ cos u Ϫ cos u ϩ ϭ S2 Ϫ S1 ϭ b(cos ␪ Ϫ cos ␣) For the spring: F ϭ k(⌬S) F ϭ bk(cos ␪ Ϫ cos ␣) ϭ Substitute F into Eq. (1): P cos ␪ ϭ bk(cos ␪ Ϫ cos ␣)(sin ␪) or P cot u Ϫ cos u ϩ cos ␣ ϭ 0 bk 2P cot u b P cot u bk P cot u ≤ bk (Eq. 3) NUMERICAL RESULTS (Eq. 2) This equation must be solved numerically for the angle ␪. b ϭ 8.0 in. k ϭ 16 lb/in. ␣ ϭ 45Њ P ϭ 10 lb 0.078125 cot ␪ Ϫ cos ␪ ϩ 0.707107 ϭ 0 (Eq. 4) Substitute into Eq. (2): Solve Eq. (4) numerically: DETERMINE THE DISTANCE ␦ ␦ ϭ L2 Ϫ L1 ϭ 2b cos ␪ Ϫ 2b cos ␣ ϭ 2b(cos ␪ Ϫ cos ␣) u ϭ 35.1Њ Substitute into Eq. (3): ␦ ϭ 1.78 in. Problem 2.2-14 Solve the preceding problem for the following data: b ϭ 200 mm, k ϭ 3.2 kN/m, ␣ ϭ 45°, and P ϭ 50 N. Solution 2.2-14 Framework with rigid bars and a spring See the solution to the preceding problem. Eq. (2): P cot u Ϫ cos u ϩ cos ␣ ϭ 0 bk Eq. (3): ␦ϭ 2P cot u k NUMERICAL RESULTS b ϭ 200 mm k ϭ 3.2 kN/m ␣ ϭ 45Њ P ϭ 50 N Substitute into Eq. (2): 0.078125 cot ␪ Ϫ cos ␪ ϩ 0.707107 ϭ 0 Solve Eq. (4) numerically: u ϭ 35.1Њ Substitute into Eq. (3): ␦ ϭ 44.5 mm (Eq. 4)
    • SECTION 2.3 75 Changes in Lengths under Nonuniform Conditions Changes in Lengths under Nonuniform Conditions Problem 2.3-1 Calculate the elongation of a copper bar of solid circular cross section with tapered ends when it is stretched by axial loads of magnitude 3.0 k (see figure). The length of the end segments is 20 in. and the length of the prismatic middle segment is 50 in. Also, the diameters at cross sections A, B, C, and D are 0.5, 1.0, 1.0, and 0.5 in., respectively, and the modulus of elasticity is 18,000 ksi. (Hint: Use the result of Example 2-4.) A B C 3.0 k D 20 in. 20 in. 50 in. Solution 2.3-1 Bar with tapered ends A MIDDLE SEGMENT (L ϭ 50 in.) B C 3.0 k D 20 in. 50 in. 20 in. 3.0 k ␦2 ϭ (3.0 k)(50 in.) PL ϭ EA (18,000 ksi)( ␲ )(1.0 in.) 2 4 ϭ 0.01061in. dA ϭ dD ϭ 0.5 in. P ϭ 3.0 k dB ϭ dC ϭ 1.0 in. E ϭ 18,000 ksi END SEGMENT (L ϭ 20 in.) From Example 2-4: 4PL ␦ϭ ␲E dA dB ␦1 ϭ ELONGATION OF BAR NL ␦ϭ a ϭ 2␦1 ϩ ␦2 EA ϭ 2(0.008488 in.) ϩ (0.01061 in.) ϭ 0.0276 in. 4(3.0 k)(20 in.) ϭ 0.008488 in. ␲(18,000 ksi)(0.5 in.)(1.0 in.) Problem 2.3-2 A long, rectangular copper bar under a tensile load P hangs from a pin that is supported by two steel posts (see figure). The copper bar has a length of 2.0 m, a cross-sectional area of 4800 mm2, and a modulus of elasticity Ec ϭ 120 GPa. Each steel post has a height of 0.5 m, a cross-sectional area of 4500 mm2, and a modulus of elasticity Es ϭ 200 GPa. Steel post (a) Determine the downward displacement ␦ of the lower end of the copper bar due to a load P ϭ 180 kN. (b) What is the maximum permissible load Pmax if the displacement ␦ is limited to 1.0 mm? Copper bar P 3.0 k
    • 76 CHAPTER 2 Axially Loaded Members Solution 2.3-2 Copper bar with a tensile load (a) DOWNWARD DISPLACEMENT ␦ (P ϭ 180 kN) Steel post Ls ␦c ϭ Lc ϭ 0.625 mm ␦s ϭ Copper bar PLc (180 kN)(2.0 m) ϭ Ec Ac (120 GPa)(4800 mm2 ) (Pր2)Ls (90 kN)(0.5 m) ϭ Es As (200 GPa)(4500 mm2 ) ϭ 0.050 mm P ␦ ϭ ␦c ϩ ␦s ϭ 0.625 mm ϩ 0.050 mm Lc ϭ 2.0 m ϭ 0.675 mm Ac ϭ 4800 mm2 (b) MAXIMUM LOAD Pmax (␦max ϭ 1.0 mm) Ec ϭ 120 GPa Pmax ␦max ϭ P ␦ Ls ϭ 0.5 m As ϭ 4500 mm2 Pmax ϭ P ¢ Pmax ϭ (180 kN) ¢ Es ϭ 200 GPa ␦max ≤ ␦ 1.0 mm ≤ ϭ 267 kN 0.675 mm Problem 2.3-3 A steel bar AD (see figure) has a cross-sectional area of 0.40 in.2 and is loaded by forces P1 ϭ 2700 lb, P2 ϭ 1800 lb, and P3 ϭ 1300 lb. The lengths of the segments of the bar are a ϭ 60 in., b ϭ 24 in., and c ϭ 36 in. P1 (a) Assuming that the modulus of elasticity E ϭ 30 ϫ 106 psi, calculate the change in length ␦ of the bar. Does the bar elongate or shorten? (b) By what amount P should the load P3 be increased so that the bar does not change in length when the three loads are applied? Solution 2.3-3 A P3 ϭ 1300 lb P1 ϭ 2700 lb E ϭ 30 ϫ 106 psi AXIAL FORCES NAB ϭ P1 ϩ P2 Ϫ P3 ϭ 3200 lb NBC ϭ P2 Ϫ P3 ϭ 500 lb NCD ϭ ϪP3 ϭ Ϫ1300 lb P2 B 60 in. A ϭ 0.40 a C b D c Steel bar loaded by three forces P1 in.2 B A P2 P2 ϭ 1800 lb P3 C 24 in. D 36 in. (a) CHANGE IN LENGTH Ni Li ␦ ϭ a Ei Ai 1 ϭ (N L ϩ NBC LBC ϩ NCD LCD ) EA AB AB 1 [(3200 lb)(60 in.) ϭ 6 (30 ϫ 10 psi)(0.40 in.2 ) ϩ (500 lb)(24 in.) Ϫ (1300 lb)(36 in.)] ϭ 0.0131 in. (elongation) P3
    • SECTION 2.3 77 Changes in Lengths under Nonuniform Conditions The force P must produce a shortening equal to 0.0131 in. in order to have no change in length. (b) INCREASE IN P3 FOR NO CHANGE IN LENGTH P ∴ 0.0131 in. ϭ ␦ ϭ 120 in. P ϭ increase in force P3 ϭ PL EA P(120 in.) (30 ϫ 106 psi)(0.40 in.2 ) P ϭ 1310 lb Problem 2.3-4 A rectangular bar of length L has a slot in the middle half of its length (see figure). The bar has width b, thickness t, and modulus of elasticity E. The slot has width b/4. b — 4 P (a) Obtain a formula for the elongation ␦ of the bar due to the axial loads P. (b) Calculate the elongation of the bar if the material is high-strength steel, the axial stress in the middle region is 160 MPa, the length is 750 mm, and the modulus of elasticity is 210 GPa. Solution 2.3-4 L — 4 b L — 4 t ϭ thickness P L — 2 L — 4 L ϭ length of bar (a) ELONGATION OF BAR Ni Li P(Lր4) P(Lր2) P(Lր4) ␦ϭ a ϭ ϩ 3 ϩ EAi E(bt) E(bt) E( 4bt) ϭ P L — 2 L — 4 Bar with a slot b — 4 P t b PL 1 4 1 7PL ¢ ϩ ϩ ≤ϭ Ebt 4 6 4 6Ebt STRESS IN MIDDLE REGION P P 4P sϭ ϭ 3 ϭ or A ( 4bt) 3bt P 3s ϭ bt 4 Substitute into the equation for ␦: ␦ϭ ϭ 7PL 7L P 7L 3s ϭ ¢ ≤ϭ ¢ ≤ 6Ebt 6E bt 6E 4 7sL 8E (b) SUBSTITUTE NUMERICAL VALUES: s ϭ 160 MPa ␦ϭ L ϭ 750 mm E ϭ 210 GPa 7(160 MPa)(750 mm) ϭ 0.500 mm 8 (210 GPa)
    • 78 CHAPTER 2 Axially Loaded Members Problem 2.3-5 Solve the preceding problem if the axial stress in the middle region is 24,000 psi, the length is 30 in., and the modulus of elasticity is 30 ϫ 106 psi. Solution 2.3-5 Bar with a slot STRESS IN MIDDLE REGION b — 4 sϭ b P L — 4 P L — 2 L — 4 L ϭ length of bar (a) ELONGATION OF BAR Ni Li P(Lր4) P(Lր2) P(Lր4) ␦ϭ a ϭ ϩ ϩ 3 EAi E(bt) E(bt) E ( 4 bt) ϭ PL 1 4 1 7PL ¢ ϩ ϩ ≤ϭ Ebt 4 6 4 6Ebt 7sL 8E ϭ (b) SUBSTITUTE NUMERICAL VALUES: s ϭ 24,000 psi C P1 = 400 kN ␦ϭ 7(24,000 psi)(30 in.) ϭ 0.0210 in. 8(30 ϫ 106 psi) P1 = 400 kN P2 = 720 kN L ϭ length of each column ϭ 3.75 m E ϭ 206 GPa L A C L = 3.75 m P2 = 720 kN B L = 3.75 m A (a) SHORTENING ␦AC OF THE TWO COLUMNS Ni Li NAB L NBC L ␦AC ϭ a ϭ ϩ Ei Ai EAAB EABC L B L ϭ 30 in. E ϭ 30 ϫ 106 psi (a) Assuming that E ϭ 206 GPa, determine the total shortening ␦AC of the two columns due to the combined action of the loads P1 and P2. (b) How much additional load P0 can be placed at the top of the column (point C) if the total shortening ␦AC is not to exceed 4.0 mm? Steel columns in a building P 3s ϭ bt 4 7PL 7L P 7L 3s ϭ ¢ ≤ϭ ¢ ≤ 6Ebt 6E bt 6E 4 Problem 2.3-6 A two-story building has steel columns AB in the first floor and BC in the second floor, as shown in the figure. The roof load P1 equals 400 kN and the second-floor load P2 equals 720 kN. Each column has length L ϭ 3.75 m. The cross-sectional areas of the first- and secondfloor columns are 11,000 mm2 and 3,900 mm2, respectively. Solution 2.3-6 or SUBSTITUTE INTO THE EQUATION FOR ␦: ␦ϭ t ϭ thickness P P 4P ϭ 3 ϭ A ( 4 bt) 3bt AAB ϭ 11,000 mm2 ABC ϭ 3,900 mm2 ϭ (1120 kN)(3.75 m) (206 GPa)(11,000 mm2 ) ϩ (400 kN)(3.75 m) (206 GPa)(3,900 mm2 ) ϭ 1.8535 mm ϩ 1.8671 mm ϭ 3.7206 mm ␦AC ϭ 3.72 mm
    • SECTION 2.3 (b) ADDITIONAL LOAD P0 AT POINT C Changes in Lengths under Nonuniform Conditions 79 Solve for P0: (␦AC)max ϭ 4.0 mm P0 ϭ E␦0 AAB ABC ¢ ≤ L AAB ϩ ABC ␦0 ϭ additional shortening of the two columns due to the load P0 SUBSTITUTE NUMERICAL VALUES: ␦0 ϭ (␦AC)max Ϫ ␦AC ϭ 4.0 mm Ϫ 3.7206 mm E ϭ 206 ϫ 109 Nրm2 ϭ 0.2794 mm Also, ␦0 ϭ AAB ϭ 11,000 ϫ 10 Ϫ6 m2 L ϭ 3.75 m P0 L P0 L P0 L 1 1 ϩ ϭ ¢ ϩ ≤ EAAB EABC E AAB ABC ABC ϭ 3,900 ϫ 10 Ϫ6 m2 P0 ϭ 44,200 N ϭ 44.2 kN Problem 2.3-7 A steel bar 8.0 ft long has a circular cross section of diameter d1 ϭ 0.75 in. over one-half of its length and diameter d2 ϭ 0.5 in. over the other half (see figure). The modulus of elasticity E ϭ 30 ϫ 106 psi. d1 = 0.75 in. Bar in tension d1 = 0.75 in. P = 5000 lb 4.0 ft P ϭ 5000 lb E ϭ 30 ϫ 106 psi L ϭ 4 ft ϭ 48 in. (a) ELONGATION OF NONPRISMATIC BAR Ni Li PL 1 ␦ϭ a ϭ Ei Ai E a Ai ␦ϭ (5000 lb)(48 in.) 30 ϫ 106 psi 1 1 ϫ B␲ 2ϩ␲ 2R 4 (0.75 in) 4 (0.50 in.) ϭ 0.0589 in. 4.0 ft Original bar: Vo ϭ A1L ϩ A2L ϭ L(A1 ϩ A2) P = 5000 lb 4.0 ft 4.0 ft (b) ELONGATION OF PRISMATIC BAR OF SAME VOLUME d2 = 0.50 in. P d2 = 0.50 in. P (a) How much will the bar elongate under a tensile load P ϭ 5000 lb? (b) If the same volume of material is made into a bar of constant diameter d and length 8.0 ft, what will be the elongation under the same load P? Solution 2.3-7 ␦0 ϭ 0.2794 ϫ 10 Ϫ3 m Prismatic bar: Vp ϭ Ap(2L) Equate volumes and solve for Ap: Vo ϭ Vp Ap ϭ ϭ ␦ϭ L(A1 ϩ A2) ϭ Ap(2L) A1 ϩ A2 1 ␲ ϭ ¢ ≤ (d2 ϩ d2 ) 1 2 2 2 4 ␲ [ (0.75 in.) 2 ϩ (0.50 in.) 2 ] ϭ 0.3191 in.2 8 P(2L) (5000 lb)(2)(48 in.) ϭ EAp (30 ϫ 106 psi)(0.3191 in.2 ) ϭ 0.0501 in. NOTE: A prismatic bar of the same volume will always have a smaller change in length than will a nonprismatic bar, provided the constant axial load P, modulus E, and total length L are the same.
    • 80 CHAPTER 2 Axially Loaded Members Problem 2.3-8 A bar ABC of length L consists of two parts of equal lengths but different diameters (see figure). Segment AB has diameter d1 ϭ 100 mm and segment BC has diameter d2 ϭ 60 mm. Both segments have length L/2 ϭ 0.6 m. A longitudinal hole of diameter d is drilled through segment AB for one-half of its length (distance L/4 ϭ 0.3 m). The bar is made of plastic having modulus of elasticity E ϭ 4.0 GPa. Compressive loads P ϭ 110 kN act at the ends of the bar. If the shortening of the bar is limited to 8.0 mm, what is the maximum allowable diameter dmax of the hole? Solution 2.3-8 Bar with a hole A d2 B C d1 P L — 4 P ϭ 110 kN P L — 4 L — 2 L ϭ 1.2 m E ϭ 4.0 GPa d1 ϭ 100 mm P d d1 d2 P dmax ϭ maximum allowable diameter of the hole d2 ϭ 60 mm L — 4 L — 4 L — 2 SUBSTITUTE NUMERICAL VALUES INTO EQ. (1) FOR ␦ AND SOLVE FOR d ϭ dmax: d ϭ diameter of hole UNITS: Newtons and meters SHORTENING ␦ OF THE BAR 0.008 ϭ Ni Li P Li ␦ϭ a ϭ a Ei Ai E Ai Lր4 P Lր4 Lր2 ϩ ϭ C ϩ ␲ 2 ␲ 2S E ␲ 2 (d1 Ϫ d 2) d d 4 4 1 4 2 ϭ PL 1 1 2 ¢ ϩ ϩ ≤ ␲E d2 Ϫ d2 d2 d2 1 1 2 (110,000)(1.2) ␲(4.0 ϫ 109 ) ϫ B 761.598 ϭ (Eq. 1) NUMERICAL VALUES (DATA): ␦ ϭ maximum allowable shortening of the bar ϭ 8.0 mm 1 1 2 R 2 2ϩ 2ϩ (0.1) Ϫ d (0.1) (0.06) 2 1 1 2 ϩ 2ϩ 0.01 0.0036 0.01 Ϫ d 1 ϭ 761.598 Ϫ 100 Ϫ 555.556 ϭ 106.042 0.01 Ϫ d2 d2 ϭ 569.81 ϫ 10 Ϫ6 m2 d ϭ 0.02387 m dmax ϭ 23.9 mm P Problem 2.3-9 A wood pile, driven into the earth, supports a load P entirely by friction along its sides (see figure). The friction force f per unit length of pile is assumed to be uniformly distributed over the surface of the pile. The pile has length L, cross-sectional area A, and modulus of elasticity E. (a) Derive a formula for the shortening ␦ of the pile in terms of P, L, E, and A. (b) Draw a diagram showing how the compressive stress ␴c varies throughout the length of the pile. f L
    • SECTION 2.3 Solution 2.3-9 Changes in Lengths under Nonuniform Conditions Wood pile with friction P N(y) ϭ axial force P A d␦ ϭ f ␦ϭ Py ␴c = AL P f= L L dy y N(y) ϭ fy (Eq. 2) N(y) dy fy dy ϭ EA EA Ύ L d␦ ϭ 0 Compressive stress in pile f EA Ύ L ydy ϭ 0 fL2 PL ϭ 2EA 2EA PL ␦ϭ 2EA 0 Friction force per unit length of pile (b) COMPRESSIVE STRESS ␴c IN PILE sc ϭ FROM FREE-BODY DIAGRAM OF PILE: ©Fvert ϭ 0 cϩ T Ϫ fL Ϫ P ϭ 0 P fϭ L N(y) fy Py ϭ ϭ A A AL At the base (y ϭ 0): ␴c ϭ 0 (Eq. 1) (a) SHORTENING ␦ OF PILE: At the top(y ϭ L): sc ϭ P A See the diagram above. At distance y from the base: Problem 2.3-10 A prismatic bar AB of length L, cross-sectional area A, modulus of elasticity E, and weight W hangs vertically under its own weight (see figure). A (a) Derive a formula for the downward displacement ␦C of point C, located at distance h from the lower end of the bar. (b) What is the elongation ␦B of the entire bar? (c) What is the ratio ␤ of the elongation of the upper half of the bar to the elongation of the lower half of the bar? C L h B Solution 2.3-10 Prismatic bar hanging vertically W ϭ Weight of bar A dy (a) DOWNWARD DISPLACEMENT C y 81 L Consider an element at distance y from the lower end. h B ␦C ϭ Wy L d␦ ϭ L N(y) ϭ L Ύ d␦ ϭ Ύ h ␦C ϭ ␦C h N(y)dy Wydy ϭ EA EAL Wydy W ϭ (L2 Ϫ h2 ) EAL 2EAL W (L2 Ϫ h2 ) 2EAL (b) ELONGATION OF BAR (h ϭ 0) ␦B ϭ WL 2EA (c) RATIO OF ELONGATIONS Elongation of upper half of bar ¢ h ϭ ␦upper ϭ 3WL 8EA L ≤: 2 Elongation of lower half of bar: ␦lower ϭ ␦B Ϫ ␦upper ϭ bϭ ␦upper ␦lower ϭ WL 3WL WL Ϫ ϭ 2EA 8EA 8EA 3ր8 ϭ3 1ր8
    • 82 CHAPTER 2 Axially Loaded Members Problem 2.3-11 A flat bar of rectangular cross section, length L, and constant thickness t is subjected to tension by forces P (see figure). The width of the bar varies linearly from b1 at the smaller end to b2 at the larger end. Assume that the angle of taper is small. b2 t (a) Derive the following formula for the elongation of the bar: P b1 b2 PL ␦ ϭ ᎏᎏ ln ᎏᎏ Et(b2 Ϫ b1) b 1 L P (b) Calculate the elongation, assuming L ϭ 5 ft, t ϭ 1.0 in., P ϭ 25 k, b1 ϭ 4.0 in., b2 ϭ 6.0 in., and E ϭ 30 ϫ 106 psi. Solution 2.3-11 Tapered bar (rectangular cross section) dx x P 0 b b1 L0 x ≤ L0 b2 ϭ b1¢ A(x) ϭ bt ϭ b1t ¢ x ≤ L0 L0 ϩ L ≤ L0 From Eq. (1): (Eq. 1) ␦ϭ Ύ L0 ϭ PL0 d␦ ϭ Eb1t PL0 ln x Eb1t ͉ Ύ ϭ L0 L0 ϭ L ¢ (Eq. 3) b1 ≤ b2 Ϫ b1 b2 PL ln Et(b2 Ϫ b1 ) b1 L ϭ 5 ft ϭ 60 in. L0 ϩL dx x PL0 L0 ϩ L ln Eb1t L0 (Eq. 4) (Eq. 5) (b) SUBSTITUTE NUMERICAL VALUES: L0 L0 ϩL Solve Eq. (3) for L0: ␦ϭ PL0 dx Pdx ϭ EA(x) Eb1tx L0 ϩL L0 ϩ L b2 ϭ L0 b1 Substitute Eqs. (3) and (4) into Eq. (2): (a) ELONGATION OF THE BAR d␦ ϭ P L t ϭ thickness (constant) b ϭ b1¢ b2 t ϭ 10 in. P ϭ 25 k b1 ϭ 4.0 in. b2 ϭ 6.0 in. E ϭ 30 ϫ 106 psi From Eq. (5): ␦ ϭ 0.010 in. (Eq. 2)
    • SECTION 2.3 83 Changes in Lengths under Nonuniform Conditions Problem 2.3-12 A post AB supporting equipment in a laboratory is tapered uniformly throughout its height H (see figure). The cross sections of the post are square, with dimensions b ϫ b at the top and 1.5b ϫ 1.5b at the base. Derive a formula for the shortening ␦ of the post due to the compressive load P acting at the top. (Assume that the angle of taper is small and disregard the weight of the post itself.) P A A b b H B B 1.5b Solution 2.3-12 Tapered post P SHORTENING OF ELEMENT dy d␦ ϭ A b Pdy ϭ EAy y E¢ Pdy b (H ϩ 0.5y) 2 ≤ H2 2 SHORTENING OF ENTIRE POST H ␦ϭ by dy Ύ PH2 d␦ ϭ 2 Eb Ύ H 0 From Appendix C: 1.5 b B ␦ϭ Square cross sections ϭ b ϭ width at A 1.5b ϭ width at B by ϭ width at distance y ϭ b ϩ (1.5b Ϫ b) ϭ y H b (H ϩ 0.5y) H Ay ϭ cross-sectional area at distance y ϭ (by ) 2 ϭ b2 (H ϩ 0.5y) 2 H2 ϭ dy (H ϩ 0.5y) 2 Ύ (a ϩ bx) dx 2 ϭϪ H PH2 1 BϪ R (0.5)(H ϩ 0.5y) 0 Eb2 PH2 1 1 ϩ R 2 BϪ (0.5)(1.5H) 0.5H Eb 2PH 3Eb2 1 b(a ϩ bx) 1.5b
    • 84 CHAPTER 2 Axially Loaded Members Problem 2.3-13 A long, slender bar in the shape of a right circular cone with length L and base diameter d hangs vertically under the action of its own weight (see figure). The weight of the cone is W and the modulus of elasticity of the material is E. Derive a formula for the increase ␦ in the length of the bar due to its own weight. (Assume that the angle of taper of the cone is small.) Solution 2.3-13 d L Conical bar hanging vertically ELEMENT OF BAR d Ny dy L y TERMINOLOGY Ay ϭ cross-sectional area at element dy AB ϭ cross-sectional area at base of cone ␲ d2 ϭ 4 1 ϭ AB L 3 Vy ϭ volume of cone below element dy 1 A y 3 y Wy ϭ weight of cone below element dy Vy V (W) ϭ Ny ϭ Wy d␦ ϭ Ny dy E Ay ϭ AyyW ABL Wy dy 4W ϭ y dy E ABL ␲ d 2 EL ELONGATION OF CONICAL BAR ␦ϭ V ϭ volume of cone ϭ Wϭweight of cone ELONGATION OF ELEMENT dy Ny ϭ axial force acting on element dy ϭ dy Ny Ύ d␦ ϭ 4W ␲d 2 EL L Ύ y dy ϭ ␲ d E 2WL 2 0
    • SECTION 2.3 85 Changes in Lengths under Nonuniform Conditions Problem 2.3-14 A bar ABC revolves in a horizontal plane about a vertical axis at the midpoint C (see figure). The bar, which has length 2L and cross-sectional area A, revolves at constant angular speed ␻. Each half of the bar (AC and BC) has weight W1 and supports a weight W2 at its end. Derive the following formula for the elongation of one-half of the bar (that is, the elongation of either AC or BC): A ␻ C B W1 W2 W1 L L2␻2 ␦ ϭ ᎏᎏ (W1 ϩ 3W2) 3g EA W2 L in which E is the modulus of elasticity of the material of the bar and g is the acceleration of gravity. Solution 2.3-14 ␻ Rotating bar W1 B C W2 Centrifugal force produced by weight W2 ϭ¢ F(x) x dx ␰ d␰ W2 ≤ (L␻2 ) g AXIAL FORCE F(x) L Ύ F(x) ϭ ␻ ϭ angular speed jϭL jϭx W1␻2 W2L␻2 jdj ϩ g gL W1␻2 2 W2L␻2 ϭ (L Ϫ x2 ) ϩ g 2gL A ϭ cross-sectional area E ϭ modulus of elasticity ELONGATION OF BAR BC g ϭ acceleration of gravity F(x) ϭ axial force in bar at distance x from point C Consider an element of length dx at distance x from point C. To find the force F(x) acting on this element, we must find the inertia force of the part of the bar from distance x to distance L, plus the inertia force of the weight W2. Since the inertia force varies with distance from point C, we now must consider an element of length d␰ at distance ␰, where ␰ varies from x to L. Mass of element dj ϭ dj W1 ¢ ≤ L g Acceleration of element ϭ ␰␻2 Centrifugal force produced by element ϭ (mass)(acceleration) ϭ W1␻2 jdj gL Ύ EA W␻ W L␻ dx ϭΎ (L Ϫ x )dx ϩ Ύ 2gLEA gEA L ␦ϭ F(x) dx 0 L L 2 1 W1␻ B 2gLEA 2 0 2 ϭ x2 dx R ϩ 0 Ύ L L2 dx Ϫ 0 W1L2␻2 W2L2␻2 ϩ 3gEA gEA 2 2 L␻ ϭ (W ϩ 3W2 ) 3gEA 1 ϭ 2 2 2 Ύ 0 L W2L␻2 gEA Ύ 0 L dx
    • 86 CHAPTER 2 Axially Loaded Members Problem 2.3-15 The main cables of a suspension bridge [see part (a) of the figure] follow a curve that is nearly parabolic because the primary load on the cables is the weight of the bridge deck, which is uniform in intensity along the horizontal. Therefore, let us represent the central region AOB of one of the main cables [see part (b) of the figure] as a parabolic cable supported at points A and B and carrying a uniform load of intensity q along the horizontal. The span of the cable is L, the sag is h, the axial rigidity is EA, and the origin of coordinates is at midspan. (a) y (a) Derive the following formula for the elongation of cable AOB shown in part (b) of the figure: A 16h 2 qL3 ␦ ϭ ᎏᎏ (1 ϩ ᎏᎏ) 3L2 8hE A L — 2 L — 2 B h (b) Calculate the elongation ␦ of the central span of one of the main cables of the Golden Gate Bridge, for which the dimensions and properties are L ϭ 4200 ft, h ϭ 470 ft, q ϭ 12,700 lb/ft, and E ϭ 28,800,000 psi. The cable consists of 27,572 parallel wires of diameter 0.196 in. O q x (b) Hint: Determine the tensile force T at any point in the cable from a free-body diagram of part of the cable; then determine the elongation of an element of the cable of length ds; finally, integrate along the curve of the cable to obtain an equation for the elongation ␦. Solution 2.3-15 Cable of a suspension bridge y L — 2 A Equation of parabolic curve: L — 2 B yϭ D h O x 4hx2 L2 dy 8hx ϭ 2 dx L q FREE-BODY DIAGRAM OF HALF OF CABLE ©MB ϭ 0 ‫۔ە‬ y VB B HB D H Ϫ Hh ϩ Hϭ h O x qL2 8h ©Fhorizontal ϭ 0 q L — 2 qL L ¢ ≤ϭ0 2 4 HB ϭ H ϭ qL2 8h (Eq. 1) ©Fvertical ϭ 0 VB ϭ qL 2 (Eq. 2)
    • SECTION 2.3 (a) ELONGATION ␦ OF CABLE AOB FREE-BODY DIAGRAM OF SEGMENT DB OF CABLE y ␦ϭ VB HB B T 0 h− ␪ TH D L2 ␦ϭ 4hx2 L2 TV L — −x 2 D ␪ TV T ©Fvert ϭ 0 TH ϭ HB qL2 ϭ 8h qL qL L TV ϭ VB Ϫ q ¢ Ϫ x ≤ ϭ Ϫ ϩ qx 2 2 2 qL ≤ ϩ (qx) 2 B 8h (Eq. 3) ϭ ¢ qL2 64h2x2 1ϩ 8h B L4 (Eq. 4) (Eq. 5) T ds dy dx Tds d␦ ϭ EA 8hx ϭ dx 1 ϩ ¢ 2 ≤ B L ds ϭ ͙(dx) 2 ϩ (dy) 2 ϭ dx ϭ dx 1 ϩ B 2 64h2x2 L4 ␦ϭ 2 EA Ύ 0 qL2 64h2x2 ¢1 ϩ ≤ dx 8h L4 Lր2 qL3 16h2 ␦ϭ ¢1 ϩ ≤ 8hEA 3L2 B 1ϩ¢ L ϭ 4200 ft q ϭ 12,700 lb/ft A ϭ (27,572) ¢ (Eq. 7) h ϭ 470 ft E ϭ 28,800,000 psi ␲ ≤ (0.196 in.) 2 ϭ 831.90 in.2 4 Substitute into Eq. (7): ␦ ϭ 133.7 in ϭ 11.14 ft 2 2 ELONGATION d␦ OF AN ELEMENT OF LENGTH ds T Ύ 27,572 wires of diameter d ϭ 0.196 in. TENSILE FORCE T IN CABLE 2 2 T ϭ ͙TH ϩ TV ϭ 1 qL2 64h2x2 ¢1 ϩ ≤ dx EA 8h L4 (b) GOLDEN GATE BRIDGE CABLE L VB Ϫ TV Ϫ q ¢ Ϫ x ≤ ϭ 0 2 ϭ qx T ds For both halves of cable: x TH ©Fhoriz ϭ 0 Ύ d␦ ϭ Ύ EA Substitute for T from Eq. (5) and for ds from Eq. (6): 4hx2 q x Changes in Lengths under Nonuniform Conditions dy 2 ≤ dx (Eq. 6) 87
    • 88 CHAPTER 2 Axially Loaded Members Statically Indeterminate Structures P Problem 2.4-1 The assembly shown in the figure consists of a brass core (diameter d1 ϭ 0.25 in.) surrounded by a steel shell (inner diameter d2 ϭ 0.28 in., outer diameter d3 ϭ 0.35 in.). A load P compresses the core and shell, which have length L ϭ 4.0 in. The moduli of elasticity of the brass and steel are Eb ϭ 15 ϫ 106 psi and Es ϭ 30 ϫ 106 psi, respectively. Steel shell Brass core L (a) What load P will compress the assembly by 0.003 in.? (b) If the allowable stress in the steel is 22 ksi and the allowable stress in the brass is 16 ksi, what is the allowable compressive load Pallow? (Suggestion: Use the equations derived in Example 2-5.) Solution 2.4-1 d1 d2 d3 Cylindrical assembly in compression P Substitute numerical values: Es As ϩ Eb Ab ϭ (30 ϫ 106 psi)(0.03464 in.2 ) ϩ (15 ϫ 106 psi)(0.04909 in.2 ) Steel shell Brass core L d1 d2 d3 ϭ 1.776 ϫ 106 lb P ϭ (1.776 ϫ 106 lb) ¢ ϭ 1330 lb 0.003 in. ≤ 4.0 in. (b) ALLOWABLE LOAD ␴sϭ22 ksi ␴bϭ16 ksi Use Eqs. (2-12a and b) of Example 2-5. d1 ϭ 0.25 in. Ebϭ15 ϫ 106 psi For steel: d2 ϭ 0.28 in. Esϭ30 ϫ 106 psi ss ϭ d3 ϭ 0.35 in. L ϭ 4.0 in. As ϭ Ab ϭ ␲ 2 (d Ϫ d2 ) ϭ 0.03464 in.2 2 4 3 ␲ 2 d ϭ 0.04909 in.2 4 1 (a) DECREASE IN LENGTH (␦ ϭ 0.003 in.) Use Eq. (2-13) of Example 2-5. ␦ϭ PL Es As ϩ Eb Ab P ϭ (Es As ϩ Eb Ab ) ¢ or ␦ ≤ L PEs Es As ϩ Eb Ab Ps ϭ (Es As ϩ Eb Ab ) Ps ϭ (1.776 ϫ 106 lb) ¢ For brass: sb ϭ PEb Es As ϩ Eb Ab Steel governs. 22 ksi ≤ ϭ 1300 lb 30 ϫ 106 psi Ps ϭ (Es As ϩ Eb Ab ) Ps ϭ (1.776 ϫ 106 lb) ¢ ss Es sb Eb 16 ksi ≤ ϭ 1890 lb 15 ϫ 106 psi Pallow ϭ 1300 lb
    • SECTION 2.4 Statically Indeterminate Structures Problem 2.4-2 A cylindrical assembly consisting of a brass core and an aluminum collar is compressed by a load P (see figure). The length of the aluminum collar and brass core is 350 mm, the diameter of the core is 25 mm, and the outside diameter of the collar is 40 mm. Also, the moduli of elasticity of the aluminum and brass are 72 GPa and 100 GPa, respectively. P Aluminum collar Brass core 350 mm (a) If the length of the assembly decreases by 0.1% when the load P is applied, what is the magnitude of the load? (b) What is the maximum permissible load Pmax if the allowable stresses in the aluminum and brass are 80 MPa and 120 MPa, respectively? (Suggestion: Use the equations derived in Example 2-5.) Solution 2.4-2 25 mm 40 mm Cylindrical assembly in compression P ␦ϭ PL Ea Aa ϩ Eb Ab or P ϭ (Ea Aa ϩ Eb Ab ) ¢ ␦ ≤ L Substitute numerical values: 350 mm A B Ea Aa ϩ Eb Ab ϭ (72 GPa)(765.8 mm2) ϩ(100 GPa)(490.9 mm2) ϭ 55.135 MN ϩ 49.090 MN db da ϭ 104.23 MN 0.350 mm P ϭ (104.23 MN) ¢ ≤ 350 mm ϭ 104.2 kN A ϭ aluminum (b) ALLOWABLE LOAD B ϭ brass ␴a ϭ 80 MPa L ϭ 350 mm For aluminum: db ϭ 25 mm sa ϭ ␲ Aa ϭ (d2 Ϫ d2 ) b 4 a ϭ765.8 mm2 Ea ϭ 72 GPa ␴b ϭ 120 MPa Use Eqs. (2-12a and b) of Example 2-5. da ϭ40 mm Eb ϭ 100 GPa ϭ 490.9 mm2 (a) DECREASE IN LENGTH (␦ ϭ 0.1% of L ϭ 0.350 mm) Use Eq. (2-13) of Example 2-5. ␲ Ab ϭ d2 4 b PEa Ea Aa ϩ Eb Ab Pa ϭ (104.23 MN) ¢ For brass: sb ϭ 89 PEb Ea Aa ϩ Eb Ab Pb ϭ (104.23 MN) ¢ Aluminum governs. Pa ϭ (Ea Aa ϩ Eb Ab) ¢ 80 MPa ≤ ϭ 115.8 kN 72 GPa sa ≤ Ea Pb ϭ (Ea Aa ϩ Eb Ab ) ¢ 120 MPa ≤ ϭ 125.1 kN 100 GPa Pmax ϭ 116 kN sb ≤ Eb
    • 90 CHAPTER 2 Axially Loaded Members Problem 2.4-3 Three prismatic bars, two of material A and one of material B, transmit a tensile load P (see figure). The two outer bars (material A) are identical. The cross-sectional area of the middle bar (material B) is 50% larger than the cross-sectional area of one of the outer bars. Also, the modulus of elasticity of material A is twice that of material B. (a) What fraction of the load P is transmitted by the middle bar? (b) What is the ratio of the stress in the middle bar to the stress in the outer bars? (c) What is the ratio of the strain in the middle bar to the strain in the outer bars? Solution 2.4-3 A B P A Prismatic bars in tension A P B A FREE-BODY DIAGRAM OF END PLATE STRESSES: sA ϭ PA 2 PB PA 2 P PB EBP sB ϭ ϭ AB EA AA ϩ EB AB (a) LOAD IN MIDDLE BAR EQUATION OF EQUILIBRIUM ©Fhorizϭ 0 PA ϩ PB Ϫ P ϭ 0 PB EB AB 1 ϭ ϭ EA AA P EA AA ϩ EB AB ϩ1 EB AB EA AA 1 ϩ 1 4 Given: ϭ2 ϭ ϭ EB AB 1.5 3 (1) EQUATION OF COMPATIBILITY ␦A ϭ ␦B (2) FORCE-DISPLACEMENT RELATIONS ∴ AA ϭ total area of both outer bars ␦A ϭ PA L EA AA PA EAP ϭ AA EA AA ϩ EB AB ␦B ϭ PB L EB AB (3) (b) RATIO OF STRESSES Substitute into Eq. (2): PA L PB L ϭ EA AA EB AB PB 1 1 3 ϭ ϭ ϭ EA AA P 8 11 ¢ ≤¢ ≤ϩ1 ϩ1 EB AB 3 (4) sB EB 1 ϭ ϭ sA EA 2 (c) RATIO OF STRAINS SOLUTION OF THE EQUATIONS All bars have the same strain Solve simultaneously Eqs. (1) and (4): EA AAP PA ϭ EA AA ϩ EB AB EB ABP PB ϭ EA AA ϩ EB AB Ratio ϭ 1 (5) Substitute into Eq. (3): ␦ ϭ ␦A ϭ ␦B ϭ PL EA AA ϩ EB AB (6) (7)
    • SECTION 2.4 Statically Indeterminate Structures Problem 2.4-4 A bar ACB having two different cross-sectional areas A1 and A2 is held between rigid supports at A and B (see figure). A load P acts at point C, which is distance b1 from end A and distance b2 from end B. A (a) Obtain formulas for the reactions RA and RB at supports A and B, respectively, due to the load P. (b) Obtain a formula for the displacement ␦C of point C. (c) What is the ratio of the stress ␴1 in region AC to the stress ␴2 in region CB? Solution 2.4-4 A1 P C A2 B b2 b1 Bar with intermediate load C A B P b1 A1 FREE-BODY DIAGRAM RA Solve Eq. (1) and Eq. (5) simultaneously: C A b2 A2 B RB P RA ϭ b2 A1 P b1 A2 ϩ b2 A1 RB ϭ b1 A2 P b1 A2 ϩ b2 A1 (b) DISPLACEMENT OF POINT C EQUATION OF EQUILIBRIUM ©Fhoriz ϭ 0 RA ϩ RB ϭ P (Eq. 1) EQUATION OF COMPATIBILITY s1 ϭ ␦CB ϭ shortening of CB (Eq. 2) FORCE DISPLACEMENT RELATIONS RA b1 ␦AC ϭ EA1 RB b2 ␦CB ϭ EA2 (Eqs. 3&4) (a) SOLUTION OF EQUATIONS Substitute Eq. (3) and Eq. (4) into Eq. (2): RA b1 RB b2 ϭ EA1 EA2 RA b1 b1 b2 P ϭ EA1 E(b1 A2 ϩ b2 A1 ) (c) RATIO OF STRESSES ␦AC ϭ elongation of AC ␦AC ϭ ␦CB ␦C ϭ ␦AC ϭ (Eq. 5) RA (tension) A1 s2 ϭ RB (compression) A2 s1 b2 ϭ s2 b1 (Note that if b1 ϭ b2, the stresses are numerically equal regardless of the areas A1 and A2.) 91
    • 92 CHAPTER 2 Axially Loaded Members Problem 2.4-5 Three steel cables jointly support a load of 12 k (see figure). The diameter of the middle cable is 3⁄4 in. and the diameter of each outer cable is 1⁄2 in. The tensions in the cables are adjusted so that each cable carries one-third of the load (i.e., 4 k). Later, the load is increased by 9 k to a total load of 21 k. (a) What percent of the total load is now carried by the middle cable? (b) What are the stresses ␴M and ␴O in the middle and outer cables, respectively? (Note: See Table 2-1 in Section 2.2 for properties of cables.) Solution 2.4-5 Three cables in tension FORCE-DISPLACEMENT RELATIONS 1 in. 2 1 in. 2 ␦M ϭ 3 in. 4 ␦O ϭ Po L EAo (3, 4) SUBSTITUTE INTO COMPATIBILITY EQUATION: PM L PO L ϭ EAM EAO P AREAS OF CABLES (from Table 2-1) Middle cable: AM ϭ 0.268 PM L EAM PM PO ϭ AM AO (5) SOLVE SIMULTANEOUSLY EQS. (1) AND (5): PM ϭ P2 ¢ in.2 Outer cables: AO ϭ 0.119 in.2 AM 0.268 in.2 ≤ ϭ (9 k) ¢ ≤ AM ϩ 2AO 0.506 in.2 ϭ 4.767 k (for each cable) FIRST LOADING P1 ϭ 12 k ¢ Each cable carries SECOND LOADING Po ϭ P2 ¢ P1 or 4 k. ≤ 3 ϭ 2.117 k FORCES IN CABLES Middle cable: Force ϭ 4 k ϩ 4.767 k ϭ 8.767 k P2 ϭ 9 k (additional load) PO Ao 0.119 in.2 ≤ ϭ (9 k) ¢ ≤ AM ϩ 2AO 0.506 in.2 Outer cables: Force ϭ 4 k ϩ 2.117 k ϭ 6.117 k PM PO (for each cable) (a) PERCENT OF TOTAL LOAD CARRIED BY MIDDLE CABLE P2 = 9 k Percent ϭ EQUATION OF EQUILIBRIUM ©Fvert ϭ 0 2PO ϩ PM Ϫ P2 ϭ 0 (1) (b) STRESSES IN CABLES (␴ ϭ P/A) 8.767 k ϭ 32.7 ksi 0.268 in.2 6.117 k Outer cables: sO ϭ ϭ 51.4 ksi 0.119 in.2 Middle cable: sM ϭ EQUATION OF COMPATIBILITY ␦M ϭ ␦O 8.767 k (100%) ϭ 41.7% 21 k (2)
    • SECTION 2.4 Problem 2.4-6 A plastic rod AB of length L ϭ 0.5 m has a diameter d1 ϭ 30 mm (see figure). A plastic sleeve CD of length c ϭ 0.3 m and outer diameter d2 ϭ 45 mm is securely bonded to the rod so that no slippage can occur between the rod and the sleeve. The rod is made of an acrylic with modulus of elasticity E1 ϭ 3.1 GPa and the sleeve is made of a polyamide with E2 ϭ 2.5 GPa. d1 d2 C A D P b c L Plastic rod with sleeve A d1 C d2 D d1 B P P b c b L P ϭ 12 kN d1 ϭ 30 mm b ϭ 100 mm L ϭ 500 mm d2 ϭ 45 mm c ϭ 300 mm Rod: E1 ϭ 3.1 GPa Sleeve: E2 ϭ 2.5 GPa Rod: A1 ϭ ␲d2 1 ϭ 706.86 mm2 4 Sleeve: A2 ϭ ␲ 2 2 (d Ϫ d1 ) ϭ 883.57 mm2 4 2 E1A1 ϩ E2A2 ϭ 4.400 MN (a) ELONGATION OF ROD Part AC: ␦AC ϭ Pb ϭ 0.5476 mm E1A1 Part CD: ␦CD ϭ Pc E1A1E2A2 ϭ 0.81815 mm (From Eq. 2-13 of Example 2-5) ␦ ϭ 2␦AC ϩ ␦CD ϭ 1.91 mm B P (a) Calculate the elongation ␦ of the rod when it is pulled by axial forces P ϭ 12 kN. (b) If the sleeve is extended for the full length of the rod, what is the elongation? (c) If the sleeve is removed, what is the elongation? Solution 2.4-6 (b) SLEEVE AT FULL LENGTH ␦ ϭ ␦CD ¢ L 500 mm ≤ ϭ (0.81815 mm) ¢ ≤ c 300 mm ϭ 1.36 mm (c) SLEEVE REMOVED ␦ϭ 93 Statically Indeterminate Structures PL ϭ 2.74 mm E1A1 b
    • 94 CHAPTER 2 Axially Loaded Members Problem 2.4-7 The axially loaded bar ABCD shown in the figure is held between rigid supports. The bar has cross-sectional area A1 from A to C and 2A1 from C to D. A1 (a) Derive formulas for the reactions RA and RD at the ends of the bar. (b) Determine the displacements ␦B and ␦C at points B and C, respectively. (c) Draw a diagram in which the abscissa is the distance from the left-hand support to any point in the bar and the ordinate is the horizontal displacement ␦ at that point. Solution 2.4-7 B L — 4 C L — 4 D L — 2 (a) REACTIONS Solve simultaneously Eqs. (1) and (6): 2A1 A1 P A A Bar with fixed ends FREE-BODY DIAGRAM OF BAR RA 1.5A1 P RD B L — 4 C L — 4 RA ϭ D L — 2 RD ϭ P 3 (b) DISPLACEMENTS AT POINTS B AND C ␦B ϭ ␦AB ϭ EQUATION OF EQUILIBRIUM ©Fhoriz ϭ 0 2P 3 RA ϩ RD ϭ P RAL PL ϭ (To the right) 4EA1 6EA1 ␦C ϭ Ϳ␦CDͿ ϭ (Eq. 1) EQUATION OF COMPATIBILITY ␦AB ϩ ␦BC ϩ ␦CD ϭ 0 ϭ (Eq. 2) RDL 4EA1 PL (To the right) 12EA1 Positive means elongation. (c) DISPLACEMENT DIAGRAM FORCE-DISPLACEMENT EQUATIONS ␦AB ϭ RA (Lր4) EA1 ␦CD ϭ Ϫ ␦BC ϭ (RA Ϫ P)(Lր4) EA1 Displacement PL —— 6EA1 (Eqs. 3, 4) RD (Lր2) E(2A1 ) PL —— 12EA1 (Eq. 5) SOLUTION OF EQUATIONS Substitute Eqs. (3), (4), and (5) into Eq. (2): RAL (RA Ϫ P)(L) RDL ϩ Ϫ ϭ 0 (Eq. 6) 4EA1 4EA1 4EA1 A 0 B L — 4 C L — 2 D Distance from end A L
    • SECTION 2.4 Problem 2.4-8 The fixed-end bar ABCD consists of three prismatic segments, as shown in the figure. The end segments have crosssectional area A1 ϭ 840 mm2 and length L1 ϭ 200 mm. The middle segment has cross-sectional area A2 ϭ 1260 mm2 and length L2 ϭ 250 mm. Loads PB and PC are equal to 25.5 kN and 17.0 kN, respectively. A1 Solution 2.4-8 A1 A PC B D C L1 L2 L1 Bar with three segments A2 A1 PB A A2 PB (a) Determine the reactions RA and RD at the fixed supports. (b) Determine the compressive axial force FBC in the middle segment of the bar. 95 Statically Indeterminate Structures A1 B PC C L1 L2 PB ϭ 25.5 kN L1 ϭ 200 mm L2 ϭ 250 mm A1 ϭ 840 D PC ϭ 17.0 kN A2 ϭ 1260 mm2 mm2 m ϭ meter L1 FREE-BODY DIAGRAM PC PB RA A B C RD D EQUATION OF EQUILIBRIUM SOLUTION OF EQUATIONS ©Fhoriz ϭ 0 S d Ϫ ϩ Substitute Eqs. (3), (4), and (5) into Eq. (2): PB ϩ RD Ϫ PC Ϫ RA ϭ 0 or RA Ϫ RD ϭ PB Ϫ PC ϭ 8.5 kN (Eq. 1) Ϫ EQUATION OF COMPATIBILITY ␦AD ϭ elongation of entire bar ␦AD ϭ ␦AB ϩ ␦BC ϩ ␦CD ϭ 0 ␦BC ϭ ϭ ␦CD ϭ RAL1 RA 1 ϭ ¢ 238.095 ≤ m EA1 E (Eq. 3) (RA Ϫ PB )L2 EA2 RA PB 1 1 ¢ 198.413 ≤Ϫ ¢ 198.413 ≤ m m E E RDL1 RD 1 ϭ ¢ 238.095 ≤ m EA1 E PB RD 1 1 ¢ 198.413 ≤ϩ ¢ 238.095 ≤ϭ0 m m E E Simplify and substitute PB ϭ 25.5 kN: (Eq. 2) FORCE-DISPLACEMENT RELATIONS ␦AB ϭ RA RA 1 1 ¢ 238.095 ≤ ϩ ¢ 198.413 ≤ m m E E RA ¢ 436.508 1 1 ≤ ϩ RD ¢ 238.095 ≤ m m ϭ 5,059.53 kN m (a) REACTIONS RA AND RD Solve simultaneously Eqs. (1) and (6). (Eq. 4) (Eq. 5) From (1): RD ϭ RA Ϫ 8.5 kN Substitute into (6) and solve for RA: RA ¢ 674.603 1 kN ≤ ϭ 7083.34 m m RA ϭ 10.5 kN RD ϭ RA Ϫ 8.5 kN ϭ 2.0 kN (b) COMPRESSIVE AXIAL FORCE FBC FBC ϭ PB Ϫ RA ϭ PC Ϫ RD ϭ 15.0 kN (Eq. 6)
    • 96 CHAPTER 2 Axially Loaded Members Problem 2.4-9 The aluminum and steel pipes shown in the figure are fastened to rigid supports at ends A and B and to a rigid plate C at their junction. The aluminum pipe is twice as long as the steel pipe. Two equal and symmetrically placed loads P act on the plate at C. A Steel pipe L P (a) Obtain formulas for the axial stresses ␴a and ␴s in the aluminum and steel pipes, respectively. (b) Calculate the stresses for the following data: P ϭ 12 k, cross-sectional area of aluminum pipe Aa ϭ 8.92 in.2, cross-sectional area of steel pipe As ϭ 1.03 in.2, modulus of elasticity of aluminum Ea ϭ 10 ϫ 106 psi, and modulus of elasticity of steel Es ϭ 29 ϫ 106 psi. P C Aluminum pipe 2L B Solution 2.4-9 Pipes with intermediate loads RA A SOLUTION OF EQUATIONS A Es As L P 1 P P C Substitute Eqs. (3) and (4) into Eq. (2): RAL RB (2L) Ϫ ϭ0 Es As Ea Aa P C (Eq. 5) Solve simultaneously Eqs. (1) and (5): 2 Ea Aa 2L RA ϭ 4Es As P Ea Aa ϩ 2Es As RB ϭ 2Ea Aa P Ea Aa ϩ 2Es As (Eqs. 6, 7) (a) AXIAL STRESSES B Aluminum: sa ϭ B RB Steel: ss ϭ RA ϩ RB ϭ 2P (Eq. 1) (Eq. 9) (b) NUMERICAL RESULTS P ϭ 12 k EQUATION OF COMPATIBILITY ␦AB ϭ ␦AC ϩ ␦CB ϭ 0 (Eq. 2) (A positive value of ␦ means elongation.) Aa ϭ 8.92 in.2 Ea ϭ 10 ϫ 106 psi Substitute into Eqs. (8) and (9): ss ϭ 9,350 psi (tension) (Eqs. 3, 4)) As ϭ 1.03 in.2 Es ϭ 29 ϫ 106 psi sa ϭ 1,610 psi (compression) FORCE-DISPLACEMENT RELATIONS RB (2L) ␦BC ϭ Ϫ Ea Aa 4Es P RA ϭ As Ea Aa ϩ 2Es As (tension) EQUATION OF EQUILIBRIUM RAL ␦AC ϭ Es As (Eq. 8) (compression) Pipe 1 is steel. Pipe 2 is aluminum. ©Fvert ϭ 0 2EaP RB ϭ Aa Ea Aa ϩ 2Es As
    • SECTION 2.4 Statically Indeterminate Structures Problem 2.4-10 A rigid bar of weight W ϭ 800 N hangs from three equally spaced vertical wires, two of steel and one of aluminum (see figure). The wires also support a load P acting at the midpoint of the bar. The diameter of the steel wires is 2 mm, and the diameter of the aluminum wire is 4 mm. What load Pallow can be supported if the allowable stress in the steel wires is 220 MPa and in the aluminum wire is 80 MPa? (Assume Es ϭ 210 GPa and Ea ϭ 70 GPa.) S A S Rigid bar of weight W P Solution 2.4-10 Rigid bar hanging from three wires SOLUTION OF EQUATIONS Substitute (3) and (4) into Eq. (2): S A Fs L FAL ϭ Es As EAAA S (Eq. 5) Solve simultaneously Eqs. (1) and (5): FA ϭ (P ϩ W) ¢ W = 800 N P STEEL WIRES ds ϭ 2 mm ␴s ϭ 220 MPa Es ϭ 210 GPa ALUMINUM WIRES Fs ϭ (P ϩ W) ¢ EAAA ≤ EAAA ϩ 2Es As Es As ≤ EAAA ϩ 2Es As (Eq. 6) (Eq. 7) STRESSES IN THE WIRES sA ϭ ␴A ϭ 80 MPa EA ϭ 70 GPa FA (P ϩ W)EA ϭ AA EAAA ϩ 2Es As (Eq. 8) ss ϭ dA ϭ 4 mm Fs (P ϩ W)Es ϭ As EAAA ϩ 2Es As (Eq. 9) FREE-BODY DIAGRAM OF RIGID BAR ALLOWABLE LOADS (FROM EQS. (8) AND (9)) FS FA FS PA ϭ AA ϭ (Eq. 2) ␦A ϭ FAL EAAA ␲ (4 mm) 2 ϭ 12.5664 mm2 4 PA ϭ 1713 N Ps ϭ 1504 N FORCE DISPLACEMENT RELATIONS Fs L Es As ␲ (2 mm) 2 ϭ 3.1416 mm2 4 (Eq. 1) EQUATION OF COMPATIBILITY ␦s ϭ (Eq. 11) As ϭ ©Fvert ϭ 0 ␦s ϭ ␦A ss (EAAA ϩ 2Es As ) Ϫ W Es SUBSTITUTE NUMERICAL VALUES INTO EQS. (10) AND (11): EQUATION OF EQUILIBRIUM 2Fs ϩ FA Ϫ P Ϫ W ϭ 0 (Eq. 10) Ps ϭ P+W sA (EAAA ϩ 2Es As ) Ϫ W EA Steel governs. (Eqs. 3, 4) Pallow ϭ 1500 N 97
    • 98 CHAPTER 2 Axially Loaded Members Problem 2.4-11 A bimetallic bar (or composite bar) of square cross section with dimensions 2b ϫ 2b is constructed of two different metals having moduli of elasticity E1 and E2 (see figure). The two parts of the bar have the same cross-sectional dimensions. The bar is compressed by forces P acting through rigid end plates. The line of action of the loads has an eccentricity e of such magnitude that each part of the bar is stressed uniformly in compression. E2 P b b e E1 b b (a) Determine the axial forces P1 and P2 in the two parts of the bar. (b) Determine the eccentricity e of the loads. (c) Determine the ratio ␴1/␴2 of the stresses in the two parts of the bar. Solution 2.4-11 2b Bimetallic bar in compression E2 P2 P1 E1 P2 P1 b2 E2 b b E1 2b FREE-BODY DIAGRAM (Plate at right-hand end) b 2 b 2 (a) AXIAL FORCES Solve simultaneously Eqs. (1) and (3): P2 P P1 ϭ e P1 P1 ϩ P2 ϭ P ©M ϭ 0 ‫۔ ە‬ Pe ϩ P1¢ b b ≤ Ϫ P2¢ ≤ ϭ 0 2 2 (Eq. 1) (Eq. 2) P2 P1 ϭ E2 E1 eϭ (Eq. 3) b(E2 Ϫ E1 ) 2(E2 ϩ E1 ) (c) RATIO OF STRESSES s1 ϭ ␦2 ϭ ␦1 or PE2 E1 ϩ E2 Substitute P1 and P2 into Eq. (2) and solve for e: EQUATION OF COMPATIBILITY P2L P1L ϭ E2A E1A P2 ϭ (b ECCENTRICITY OF LOAD P EQUATIONS OF EQUILIBRIUM ©F ϭ 0 PE1 E1 ϩ E2 P1 A s2 ϭ P2 A s1 P1 E1 ϭ ϭ s2 P2 E2 P e
    • SECTION 2.4 Statically Indeterminate Structures Problem 2.4-12 A circular steel bar ABC (E = 200 GPa) has crosssectional area A1 from A to B and cross-sectional area A2 from B to C (see figure). The bar is supported rigidly at end A and is subjected to a load P equal to 40 kN at end C. A circular steel collar BD having cross-sectional area A3 supports the bar at B. The collar fits snugly at B and D when there is no load. Determine the elongation ␦AC of the bar due to the load P. (Assume L1 ϭ 2L3 ϭ 250 mm, L2 ϭ 225 mm, A1 ϭ 2A3 ϭ 960 mm2, and A2 ϭ 300 mm2.) A A1 L1 B L3 A3 D A2 L2 C P Solution 2.4-12 Bar supported by a collar FREE-BODY DIAGRAM OF BAR ABC AND COLLAR BD RA SOLVE SIMULTANEOUSLY EQS. (1) AND (3): RA ϭ A A1 RD ϭ PL1A3 L1A3 ϩ L3A1 CHANGES IN LENGTHS (Elongation is positive) L1 RD ␦AB ϭ B B RD PL3A1 L1A3 ϩ L3A1 L3 A3 L2 A2 D ELONGATION OF BAR ABC ␦AC ϭ ␦AB ϩ ␦AC RD C PL1L3 RAL1 ϭ EA1 E(L1A3 ϩ L3A1 ) SUBSTITUTE NUMERICAL VALUES: P P ϭ 40 kN EQUILIBRIUM OF BAR ABC L1 ϭ 250 mm ©Fvert ϭ 0 RA ϩ RD Ϫ P ϭ 0 (Eq. 1) (Eq. 2) A3 ϭ 480 mm2 FORCE-DISPLACEMENT RELATIONS RAL1 EA1 ␦BD ϭ Ϫ RESULTS: RDL3 EA3 RA ϭ RD ϭ 20 kN ␦AB ϭ 0.02604 mm Substitute into Eq. (2): RAL1 RDL3 Ϫ ϭ0 EA1 EA3 A1 ϭ 960 mm2 A2 ϭ 300 mm2 (Elongation is positive.) ␦AB ϭ L2 ϭ 225 mm L3 ϭ 125 mm COMPATIBILITY (distance AD does not change) ␦AB(bar) ϩ ␦BD(collar) ϭ 0 E ϭ 200 GPa (Eq. 3) ␦BC ϭ 0.15000 mm ␦AC ϭ ␦AB ϩ ␦AC ϭ 0.176 mm ␦BC ϭ PL2 EA2 99
    • 100 CHAPTER 2 Axially Loaded Members Problem 2.4-13 A horizontal rigid bar of weight W ϭ 7200 lb is supported by three slender circular rods that are equally spaced (see figure). The two outer rods are made of aluminum (E1 ϭ 10 ϫ 106 psi) with diameter d1 ϭ 0.4 in. and length L1 ϭ 40 in. The inner rod is magnesium (E2 ϭ 6.5 ϫ 106 psi) with diameter d2 and length L2. The allowable stresses in the aluminum and magnesium are 24,000 psi and 13,000 psi, respectively. If it is desired to have all three rods loaded to their maximum allowable values, what should be the diameter d2 and length L2 of the middle rod? d2 L2 d1 d1 L1 W = weight of rigid bar Solution 2.4-13 Bar supported by three rods BAR 1 ALUMINUM E1 ϭ 10 ϫ 106 psi d1 ϭ 0.4 in. d2 ϭ 2 2(24,000 psi) (0.4 in.) 2 4(7200 lb) Ϫ ␲(13,000 psi) 13,000 psi ϭ 0.70518 in.2 Ϫ 0.59077 in.2 ϭ 0.11441 in.2 L1 ϭ 40 in. 2 SUBSTITUTE NUMERICAL VALUES: ␴1 ϭ 24,000 psi W = 7200 lb EQUATION OF COMPATIBILITY ␦1 ϭ ␦2 d2 ϭ ? 1 BAR 2 MAGNESIUM E2 ϭ 6.5 ϫ 106 psi 1 d2 ϭ 0.338 in. FORCE-DISPLACEMENT RELATIONS L2 ϭ ? ␴2 ϭ 13,000 psi ␦1 ϭ FREE-BODY DIAGRAM OF RIGID BAR EQUATION OF EQUILIBRIUM F1 ␦2 ϭ F1 ©Fvert ϭ 0 F2 2F1 ϩ F2 Ϫ W ϭ 0 (Eq. 1) ␲d2 1 4 F2 ϭ ␴2A2 A2 ϭ (Eq. 5) L1 L2 ≤ ϭ s2 ¢ ≤ E1 E2 L2 ϭ (40 in.) ¢ 24,000 psi 6.5 ϫ 106 psi ≤¢ ≤ 13,000 psi 10 ϫ 106 psi ϭ 48.0 in. Diameter d1 is known; solve for d2: 4W 2s1d2 2 Ϫ ␲s2 s2 s1E2 ≤ s2E1 SUBSTITUTE NUMERICAL VALUES: ␲d2 2 4 ␲d2 ␲d2 1 2 ≤ ϩ s2¢ ≤ϭW 4 4 d2 ϭ 2 s1 ¢ L2 ϭ L1 ¢ Substitute into Eq. (1): 2s1¢ F2L2 L2 ϭ s2 ¢ ≤ E2A2 E2 (Eq. 4) Length L1 is known; solve for L2: FULLY STRESSED RODS A1 ϭ F1L1 L1 ϭ s1 ¢ ≤ E1A1 E1 Substitute (4) and (5) into Eq. (3): W F1 ϭ ␴1A1 (Eq. 3) (Eq. 2) (Eq. 6)
    • SECTION 2.4 Problem 2.4-14 A rigid bar ABCD is pinned at point B and supported by springs at A and D (see figure). The springs at A and D have stiffnesses k1 ϭ 10 kN/m and k2 ϭ 25 kN/m, respectively, and the dimensions a, b, and c are 250 mm, 500 mm, and 200 mm, respectively. A load P acts at point C. If the angle of rotation of the bar due to the action of the load P is limited to 3°, what is the maximum permissible load Pmax? Solution 2.4-14 a = 250 mm A b = 500 mm B C D P c = 200 mm k 2 = 25 kN/m k1 = 10 kN/m Rigid bar supported by springs a EQUATION OF COMPATIBILITY b A 101 Statically Indeterminate Structures B C k1 ␦A ␦D ϭ a b D k2 P (Eq. 2) FORCE-DISPLACEMENT RELATIONS c ␦A ϭ NUMERICAL DATA FA k1 ␦D ϭ FD k2 (Eqs. 3, 4) a ϭ 250 mm SOLUTION OF EQUATIONS b ϭ 500 mm Substitute (3) and (4) into Eq. (2): c ϭ 200 mm k1 ϭ 10 kN/m FA FD ϭ ak1 bk2 k2 ϭ 25 kN/m SOLVE SIMULTANEOUSLY EQS. (1) AND (5): ␲ umax ϭ 3Њ ϭ rad 60 FA ϭ FREE-BODY DIAGRAM AND DISPLACEMENT DIAGRAM a ␦D ϭ RB P FD bcP ϭ k2 a2k1 ϩ b2k2 Pϭ Pmax ϭ ␦A B C D ␪ FD ϭ uϭ ␦D cP ϭ 2 b a k1 ϩ b2k2 u 2 (a k1 ϩ b2k2 ) c umax 2 (a k1 ϩ b2k2 ) c SUBSTITUTE NUMERICAL VALUES: ␦D Pmax ϭ ␲ր60 rad [ (250 mm) 2 (10 kNրm) 200 mm ϩ (500 mm) 2 (25 kNրm) ] ␦C ϭ 1800 N EQUATION OF EQUILIBRIUM ©MB ϭ 0 ‫ ۔ە‬FA (a) Ϫ P(c) ϩ FD (b) ϭ 0 bck2P a k1 ϩ b2k2 2 MAXIMUM LOAD FD c A ack1P a k1 ϩ b2k2 2 ANGLE OF ROTATION b FA (Eq. 5) (Eq. 1)
    • 102 CHAPTER 2 Axially Loaded Members Problem 2.4-15 A rigid bar AB of length L ϭ 66 in. is hinged to a support at A and supported by two vertical wires attached at points C and D (see figure). Both wires have the same cross-sectional area (A ϭ 0.0272 in.2) and are made of the same material (modulus E ϭ 30 ϫ 106 psi). The wire at C has length h ϭ 18 in. and the wire at D has length twice that amount. The horizontal distances are c ϭ 20 in. and d ϭ 50 in. 2h h A (a) Determine the tensile stresses ␴C and ␴D in the wires due to the load P ϭ 340 lb acting at end B of the bar. (b) Find the downward displacement ␦B at end B of the bar. C D B c d P L Solution 2.4-15 Bar supported by two wires FREE-BODY DIAGRAM TD 2h TC h A C D B A C D B c d P RA P L h ϭ 18 in. 2h ϭ 36 in. DISPLACEMENT DIAGRAM A c ϭ 20 in. C D ␦C d ϭ 50 in. ␦D L ϭ 66 in. Eϭ 30 ϫ 106 psi A ϭ 0.0272 in.2 P ϭ 340 lb B ␦B EQUATION OF EQUILIBRIUM ©MA ϭ 0 ‫۔ ە‬ TC (c) ϩ TD (d) ϭ PL (Eq. 1) EQUATION OF COMPATIBILITY ␦C ␦D ϭ c d (Eq. 2)
    • SECTION 2.4 FORCE-DISPLACEMENT RELATIONS ␦C ϭ TC h EA ␦D ϭ TD (2h) EA Statically Indeterminate Structures 103 DISPLACEMENT AT END OF BAR (Eqs. 3, 4) ␦B ϭ ␦D ¢ 2hTD L L 2hPL2 ≤ϭ ¢ ≤ϭ d EA d EA(2c2 ϩ d2 ) SOLUTION OF EQUATIONS SUBSTITUTE NUMERICAL VALUES Substitute (3) and (4) into Eq. (2): (Eq. 10) 2c2 ϩ d2 ϭ 2(20 in.) 2 ϩ (50 in.) 2 ϭ 3300 in.2 TC h TD (2h) ϭ cEA dEA or TC 2TD ϭ c d (Eq. 5) (a) TENSILE FORCES IN THE WIRES 2cPL 2c2 ϩ d2 TD ϭ dPL 2c2 ϩ d2 sD ϭ (Eqs. 6, 7) TD dPL sD ϭ ϭ A A(2c2 ϩ d2 ) (Eq. 8) (50 in.)(340 lb)(66 in.) dPL 2 2 ϭ A(2c ϩ d ) (0.0272 in.2 )(3300 in.2 ) ϭ 12,500 psi TENSILE STRESSES IN THE WIRES TC 2cPL sC ϭ ϭ A A(2c2 ϩ d2 ) 2(20 in.)(340 lb)(66 in.) 2cPL 2 2 ϭ A(2c ϩ d ) (0.0272 in.2 )(3300 in.2 ) ϭ 10,000 psi Solve simultaneously Eqs. (1) and (5): TC ϭ sC ϭ (b) (Eq. 9) Problem 2.4-16 A trimetallic bar is uniformly compressed by an axial force P ϭ 40 kN applied through a rigid end plate (see figure). The bar consists of a circular steel core surrounded by brass and copper tubes. The steel core has diameter 30 mm, the brass tube has outer diameter 45 mm, and the copper tube has outer diameter 60 mm. The corresponding moduli of elasticity are Es ϭ 210 GPa, Eb ϭ 100 GPa, and Ec ϭ 120 GPa. Calculate the compressive stresses ␴s, ␴b, and ␴c in the steel, brass, and copper, respectively, due to the force P. ␦B ϭ ϭ 2hPL2 EA(2c2 ϩ d2 ) 2(18 in.)(340 lb)(66 in.) 2 (30 ϫ 106 psi)(0.0272 in.2 )(3300 in.2 ) ϭ 0.0198 in. P = 40 kN Copper tube Brass tube Steel core 30 mm 45 mm 60 mm
    • 104 CHAPTER 2 Axially Loaded Members Solution 2.4-16 Trimetallic bar in compression Copper SOLVE SIMULTANEOUSLY EQS. (1), (6), AND (7): Ps ϭ P Ps ϭ compressive force in steel core Pb ϭ compressive force in brass tube Eb Ab Es As ϩ Eb Ab ϩ Ec Ac Pc ϭ P Steel Es As Es As ϩ Eb Ab ϩ Ec Ac Pb ϭ P Brass Ec Ac Es As ϩ Eb Ab ϩ Ec ϩ Ac Pc ϭ compressive force in copper tube COMPRESSIVE STRESSES FREE-BODY DIAGRAM OF RIGID END PLATE Let ©EA ϭ EsAs ϩ EbAb ϩ EcAc ss ϭ Ps PEs ϭ As ©EA sc ϭ P Pc PEc ϭ Ac ©EA P s Pb Pc sb ϭ Pb PEb ϭ Ab ©EA SUBSTITUTE NUMERICAL VALUES: EQUATION OF EQUILIBRIUM ©Fvert ϭ 0 Ps ϩ Pb ϩ Pc ϭ P P ϭ 40 kN EQUATIONS OF COMPATIBILITY ␦s ϭ ␦b ␦c ϭ ␦s (Eqs. 2) Es ϭ 210 GPa Eb ϭ 100 GPa Ec ϭ 120 GPa d1 ϭ 30 mm (Eq. 1) d2 ϭ 45 mm ␲ As ϭ d2 ϭ 706.86 mm2 4 1 FORCE-DISPLACEMENT RELATIONS Pb L ␦b ϭ Eb Ab Pc L ␦c ϭ Ec Ac Ab ϭ ␲ 2 (d Ϫ d2 ) ϭ 883.57 mm2 1 4 2 Ac ϭ Ps L ␦s ϭ Es As ␲ 2 (d Ϫ d2 ) ϭ 1237.00 mm2 2 4 3 (Eqs. 3, 4, 5) SOLUTION OF EQUATIONS ©EAϭ385.238 ϫ 106 N Substitute (3), (4), and (5) into Eqs. (2): Pb ϭ Ps Eb Ab Es As Pc ϭ Ps Ec Ac Es As (Eqs. 6, 7) ss ϭ PEs ϭ 21.8 MPa ©EA sb ϭ PEb ϭ 10.4 MPa ©EA sc ϭ PEc ϭ 12.5 MPa ©EA d3 ϭ 60 mm
    • SECTION 2.5 Thermal Effects Thermal Effects Problem 2.5-1 The rails of a railroad track are welded together at their ends (to form continuous rails and thus eliminate the clacking sound of the wheels) when the temperature is 60°F. What compressive stress ␴ is produced in the rails when they are heated by the sun to 120°F if the coefficient of thermal expansion ␣ ϭ 6.5 ϫ 10Ϫ6/°F and the modulus of elasticity E ϭ 30 ϫ 106 psi? Solution 2.5-1 Expansion of railroad rails The rails are prevented from expanding because of their great length and lack of expansion joints. ¢T ϭ 120ЊF Ϫ 60ЊF ϭ 60ЊF s ϭ E␣(¢T ) Therefore, each rail is in the same condition as a bar with fixed ends (see Example 2-7). The compressive stress in the rails may be calculated from Eq. (2-18). ϭ (30 ϫ 106 psi)(6.5 ϫ 10 Ϫ6րЊF)(60ЊF) s ϭ 11,700 psi Problem 2.5-2 An aluminum pipe has a length of 60 m at a temperature of 10°C. An adjacent steel pipe at the same temperature is 5 mm longer than the aluminum pipe. At what temperature (degrees Celsius) will the aluminum pipe be 15 mm longer than the steel pipe? (Assume that the coefficients of thermal expansion of aluminum and steel are ␣a ϭ 23 ϫ 10Ϫ6/°C and ␣s ϭ 12 ϫ 10Ϫ6/°C, respectively.) Solution 2.5-2 Aluminum and steel pipes INITIAL CONDITIONS La ϭ 60 m T0 ϭ 10ЊC Ls ϭ 60.005 m T0 ϭ 10ЊC ␣a ϭ 23 ϫ 10Ϫ6/ЊC ␣s ϭ 12 ϫ 10Ϫ6/ЊC or, ␣a(⌬T)La ϩ La ϭ ⌬L ϩ ␣s(⌬T)Ls ϩ Ls Solve for ⌬T: ¢T ϭ ¢L ϩ (Ls Ϫ La ) ␣a La Ϫ ␣s Ls FINAL CONDITIONS Substitute numerical values: Aluminum pipe is longer than the steel pipe by the amount ⌬L ϭ 15 mm. ␣a La Ϫ ␣s Ls ϭ 659.9 ϫ 10Ϫ6 m/ЊC ⌬T ϭ increase in temperature ¢T ϭ ␦a ϭ ␣a(⌬T)La ␦s ϭ ␣s(⌬T)Ls ␦a La Steel pipe ␦s Ls From the figure above: ␦a ϩ La ϭ ⌬L ϩ ␦s ϩ Ls T ϭ T0 ϩ ¢T ϭ 10ЊC ϩ 30.31ЊC ϭ 40.3ЊC Aluminum pipe ⌬L 15 mm ϩ 5 mm ϭ 30.31ЊC 659.9 ϫ 10 Ϫ6 mրЊC 105
    • 106 CHAPTER 2 Axially Loaded Members Problem 2.5-3 A rigid bar of weight W ϭ 750 lb hangs from three equally spaced wires, two of steel and one of aluminum (see figure). The diameter of the wires is 1⁄8 in. Before they were loaded, all three wires had the same length. What temperature increase ⌬T in all three wires will result in the entire load being carried by the steel wires? (Assume Es ϭ 30 ϫ 106 psi, ␣s ϭ 6.5 ϫ 10Ϫ6/°F, and ␣a ϭ 12 ϫ 10Ϫ6/°F.) S A S W = 750 lb Solution 2.5-3 Bar supported by three wires ␦2 ϭ increase in length of a steel wire due to load W/2 S A ϭ S WL 2Es As ␦3 ϭ increase in length of aluminum wire due to temperature increase ⌬T S ϭ steel ϭ ␣a(⌬T)L W Rigid Bar For no load in the aluminum wire: ␦1 ϩ ␦2 ϭ ␦3 A ϭ aluminum W ϭ 750 lb dϭ As ϭ ␣s (¢T)L ϩ 1 in. 8 or ␲d2 ϭ 0.012272 in.2 4 ¢T ϭ Es ϭ 30 ϫ 106 psi W 2Es As (␣a Ϫ ␣s ) Substitute numerical values: EsAs ϭ 368,155 lb ¢T ϭ ␣s ϭ 6.5 ϫ 10Ϫ6/ЊF L ϭ Initial length of wires S A S ␦1 ␦3 W 2 750 lb (2)(368,155 lb)(5.5 ϫ 10 Ϫ6րЊF) ϭ 185ЊF ␣a ϭ 12 ϫ 10Ϫ6/ЊF ␦2 W 2 ␦1 ϭ increase in length of a steel wire due to temperature increase ⌬T ϭ ␣s (⌬T)L WL ϭ ␣a (¢T )L 2Es As NOTE: If the temperature increase is larger than ⌬T, the aluminum wire would be in compression, which is not possible. Therefore, the steel wires continue to carry all of the load. If the temperature increase is less than ⌬T, the aluminum wire will be in tension and carry part of the load.
    • SECTION 2.5 Problem 2.5-4 A steel rod of diameter 15 mm is held snugly (but without any initial stresses) between rigid walls by the arrangement shown in the figure. Calculate the temperature drop ⌬T (degrees Celsius) at which the average shear stress in the 12-mm diameter bolt becomes 45 MPa. (For the steel rod, use ␣ ϭ 12 ϫ 10Ϫ6/°C and E ϭ 200 GPa.) Solution 2.5-4 B 107 Thermal Effects 12 mm diameter bolt 15 mm Steel rod with bolted connection Solve for ¢T: ¢T ϭ R 15 mm ␲d2 B 4 where dB ϭ diameter of bolt AR ϭ ␲d2 R 4 where dR ϭ diameter of steel rod ¢T ϭ 2td2 B E␣d2 R AB ϭ 12 mm diameter bolt R ϭ rod B ϭ bolt P ϭ tensile force in steel rod due to temperature drop ⌬T AR ϭ cross-sectional area of steel rod From Eq. (2-17) of Example 2-7: P ϭ EAR␣(⌬T) Bolt is in double shear. V ϭ shear force acting over one cross section of the bolt 1 V ϭ Pր2 ϭ EAR␣(¢T) 2 2tAB EAR␣ SUBSTITUTE NUMERICAL VALUES: ␶ ϭ 45 MPa dB ϭ 12 mm ␣ ϭ 12 ϫ 10Ϫ6/ЊC ¢T ϭ dR ϭ 15 mm E ϭ 200 GPa 2(45 MPa)(12 mm) 2 (200 GPa)(12 ϫ 10 Ϫ6րЊC)(15 mm) 2 ¢T ϭ 24ЊC ␶ ϭ average shear stress on cross section of the bolt AB ϭ cross-sectional area of bolt tϭ EAR ␣(¢T) V ϭ AB 2AB Problem 2.5-5 A bar AB of length L is held between rigid supports and heated nonuniformly in such a manner that the temperature increase ⌬T at distance x from end A is given by the expression ⌬T ϭ ⌬TBx3/L3, where ⌬TB is the increase in temperature at end B of the bar (see figure). Derive a formula for the compressive stress ␴c in the bar. (Assume that the material has modulus of elasticity E and coefficient of thermal expansion ␣.) ∆TB ∆T 0 A B x L
    • 108 CHAPTER 2 Axially Loaded Members Solution 2.5-5 Bar with nonuniform temperature change d␦ ϭ Elongation of element dx ∆TB ∆T d␦ ϭ ␣(¢T )dx ϭ ␣(¢TB ) ¢ 0 A ␦ ϭ elongation of bar B x Ύ ␦ϭ L L L d␦ ϭ 0 At distance x: x3 ¢T ϭ ¢TB ¢ 3 ≤ L Ύ ␣(¢T ) B 0 x3 ≤ dx L3 ¢ 3≤ L x3 dx ϭ 1 ␣(¢TB )L 4 COMPRESSIVE FORCE P REQUIRED TO SHORTEN THE BAR ␦ BY THE AMOUNT Pϭ REMOVE THE SUPPORT AT END B OF THE BAR: A B COMPRESSIVE STRESS IN THE BAR dx x EA␦ 1 ϭ EA␣(¢TB ) L 4 sc ϭ L P E␣(¢TB ) ϭ A 4 Consider an element dx at a distance x from end A. Problem 2.5-6 A plastic bar ACB having two different solid circular cross sections is held between rigid supports as shown in the figure. The diameters in the left- and right-hand parts are 50 mm and 75 mm, respectively. The corresponding lengths are 225 mm and 300 mm. Also, the modulus of elasticity E is 6.0 GPa, and the coefficient of thermal expansion ␣ is 100 ϫ 10Ϫ6/°C. The bar is subjected to a uniform temperature increase of 30°C. Calculate the following quantities: (a) the compressive force P in the bar; (b) the maximum compressive stress ␴c; and (c) the displacement ␦C of point C. Solution 2.5-6 75 mm 50 mm C 225 mm 300 mm Bar with rigid supports 75 mm 50 mm 225 mm 300 mm ␣ ϭ 100 ϫ 10Ϫ6/ЊC E ϭ 6.0 GPa LEFT-HAND PART: L1 ϭ 225 mm L2 ϭ 300 mm A2 ϭ d2 ϭ 75 mm ␲ 2 ␲ d ϭ (75 mm) 2 ϭ 4417.9 mm2 4 2 4 (a) COMPRESSIVE FORCE P Remove the support at end B. d1 ϭ 50 mm ␲ ␲ A1 ϭ d2 ϭ (50 mm) 2 4 1 4 ϭ 1963.5 mm2 RIGHT-HAND PART: B C A ⌬T ϭ 30°C A C B C B A A L1 A1 L2 A2 P B
    • SECTION 2.5 ␦T ϭ elongation due to temperature sc ϭ ϭ 1.5750 mm ␦C ϭ Shortening of AC PL1 PL2 ϩ EA1 EA2 ϭ P(19.0986 ϫ 10Ϫ9 m/Nϩ11.3177 ϫ 10Ϫ9 m/N) ␦C ϭ PL1 Ϫ ␣(¢T )L1 EA1 ϭ 0.9890 mm Ϫ 0.6750 mm ϭ (30.4163 ϫ 10Ϫ9 m/N)P ␦C ϭ 0.314 mm (P ϭ newtons) Compatibility: ␦T ϭ ␦P (Positive means AC shortens and point C displaces to the left.) 1.5750 ϫ 10Ϫ3 m ϭ (30.4163 ϫ 10Ϫ9 m/N)P or P 51.78 kN ϭ ϭ 26.4 MPa A1 1963.5 mm2 (c) DISPLACEMENT OF POINT C ␦P ϭ shortening due to P P ϭ 51,781 N P ϭ 51.8 kN d1 Problem 2.5-7 A circular steel rod AB (diameter d1 ϭ 1.0 in., length A L1 ϭ 3.0 ft) has a bronze sleeve (outer diameter d2 ϭ 1.25 in., length L2 ϭ 1.0 ft) shrunk onto it so that the two parts are securely bonded (see figure). Calculate the total elongation ␦ of the steel bar due to a temperature rise ⌬T ϭ 500°F. (Material properties are as follows: for steel, Es ϭ 30 ϫ 106 psi and ␣s ϭ 6.5 ϫ 10Ϫ6/°F; for bronze, Eb ϭ 15 ϫ 106 psi and ␣b ϭ 11 ϫ 10Ϫ6/°F.) Solution 2.5-7 d2 B L2 L1 Steel rod with bronze sleeve d2 d1 A B SUBSTITUTE NUMERICAL VALUES: ␣s ϭ 6.5 ϫ 10Ϫ6/ЊF ␣b ϭ 11 ϫ 10Ϫ6/ЊF L2 Es ϭ 30 ϫ 106 psi Eb ϭ 15 ϫ 106 psi L1 L1 ϭ 36 in. d1 ϭ 1.0 in. L2 ϭ 12 in. ELONGATION OF THE TWO OUTER PARTS OF THE BAR ␦1 ϭ ␣s(⌬T)(L1 Ϫ L2) As ϭ ␲ 2 d ϭ 0.78540 in.2 4 1 d2 ϭ 1.25 in. ␲ 2 2 (d Ϫ d1 ) ϭ 0.44179 in.2 4 2 ϭ (6.5 ϫ 10Ϫ6/ЊF)(500ЊF)(36 in. Ϫ 12 in.) Ab ϭ ϭ 0.07800 in. ⌬T ϭ 500ЊF ELONGATION OF THE MIDDLE PART OF THE BAR The steel rod and bronze sleeve lengthen the same amount, so they are in the same condition as the bolt and sleeve of Example 2-8. Thus, we can calculate the elongation from Eq. (2-21): ␦2 ϭ 109 (b) MAXIMUM COMPRESSIVE STRESS P ϭ ␣(⌬T)(L1ϩL2) ϭ Thermal Effects (␣s Es As ϩ ␣b Eb Ab )(¢T)L2 Es As ϩ Eb Ab L2 ϭ 12.0 in. ␦2 ϭ 0.04493 in. TOTAL ELONGATION ␦ ϭ ␦1 ϩ ␦2 ϭ 0.123 in.
    • 110 CHAPTER 2 Axially Loaded Members Problem 2.5-8 A brass sleeve S is fitted over a steel bolt B (see figure), and the nut is tightened until it is just snug. The bolt has a diameter dB ϭ 25 mm, and the sleeve has inside and outside diameters d1 ϭ 26 mm and d2 ϭ 36 mm, respectively. Calculate the temperature rise ⌬T that is required to produce a compressive stress of 25 MPa in the sleeve. (Use material properties as follows: for the sleeve, ␣S ϭ 21 ϫ 10Ϫ6/°C and ES ϭ 100 GPa; for the bolt, ␣B ϭ 10 ϫ 10Ϫ6/°C and EB ϭ 200 GPa.) (Suggestion: Use the results of Example 2-8.) Solution 2.5-8 Bolt (B) d2 ϭ 36 mm Brass Sleeve Subscript S means “sleeve”. Subscript B means “bolt”. Use the results of Example 2-8. ␴S ϭ compressive force in sleeve EQUATION (2-20a): (␣S Ϫ ␣B )(¢T)ES EB AB (Compression) ES AS ϩ EB AB SOLVE FOR ⌬T: ¢T ϭ Sleeve (S) ␴S ϭ 25 MPa Steel Bolt or dB SUBSTITUTE NUMERICAL VALUES: B ¢T ϭ d1 Brass sleeve fitted over a Steel bolt S sS ϭ d2 sS (ES AS ϩ EB AB ) (␣S Ϫ ␣B )ES EB AB d1 ϭ 26 mm dB ϭ 25 mm ES ϭ 100 GPa EB ϭ 200 GPa ␣S ϭ 21 ϫ 10Ϫ6/ЊC ␣B ϭ 10 ϫ 10Ϫ6/ЊC AS ϭ ␲ 2 ␲ (d2 Ϫ d2 ) ϭ (620 mm2 ) 1 4 4 AB ϭ ␲ ␲ (dB ) 2 ϭ (625 mm2 ) 4 4 1ϩ ES AS ϭ 1.496 EB AB ¢T ϭ 25 MPa (1.496) (100 GPa)(11 ϫ 10 Ϫ6րЊC) ¢T ϭ 34ЊC (Increase in temperature) sS ES AS ¢1 ϩ ≤ ES (␣S Ϫ ␣B ) EB AB Problem 2.5-9 Rectangular bars of copper and aluminum are held by pins at their ends, as shown in the figure. Thin spacers provide a separation between the bars. The copper bars have cross-sectional dimensions 0.5 in. ϫ 2.0 in., and the aluminum bar has dimensions 1.0 in. ϫ 2.0 in. Determine the shear stress in the 7/16 in. diameter pins if the temperature is raised by 100°F. (For copper, Ec ϭ 18,000 ksi and ␣c ϭ 9.5 ϫ 10Ϫ6/°F; for aluminum, Ea ϭ 10,000 ksi and ␣a ϭ 13 ϫ 10Ϫ6/°F.) Suggestion: Use the results of Example 2-8. Copper bar Aluminum bar Copper bar
    • SECTION 2.5 Solution 2.5-9 111 Thermal Effects Rectangular bars held by pins C 0.5 in. × 2.0 in. 1.0 in. × 2.0 in. 0.5 in. × 2.0 in. A C Pin Diameter of pin: dP ϭ Area of pin: AP ϭ Copper Aluminum 7 in. ϭ 0.4375 in. 16 ␲ 2 d ϭ 0.15033 in.2 4 P Area of two copper bars: Ac ϭ 2.0 in.2 Area of aluminum bar: Aa ϭ 2.0 in.2 SUBSTITUTE NUMERICAL VALUES: (3.5 ϫ 10 Ϫ6րЊF)(100ЊF)(18,000 ksi)(2 in.2 ) Pa ϭ Pc ϭ 18 2.0 1ϩ¢ ≤ ¢ ≤ 10 2.0 ϭ 4,500 lb FREE-BODY DIAGRAM OF PIN AT THE LEFT END ⌬T ϭ 100ЊF Copper: Ec ϭ 18,000 ksi Aluminum: Ea ϭ 10,000 ksi Pc 2 Pa ␣c ϭ 9.5 ϫ 10Ϫ6/ЊF ␣a ϭ 13 ϫ 10Ϫ6/ЊF Pc 2 Use the results of Example 2-8. Find the forces Pa and Pc in the aluminum bar and copper bar, respectively, from Eq. (2-19). Replace the subscript “S” in that equation by “a” (for aluminum) and replace the subscript “B” by “c” (for copper), because ␣ for aluminum is larger than ␣ for copper. (␣a Ϫ ␣c )(¢T)Ea Aa Ec Ac Pa ϭ Pc ϭ Ea Aa ϩ Ec Ac Note that Pa is the compressive force in the aluminum bar and Pc is the combined tensile force in the two copper bars. Pa ϭ Pc ϭ V ϭ shear force in pin ϭ Pc /2 ϭ 2,250 lb ␶ ϭ average shear stress on cross section of pin tϭ V 2,250 lb ϭ AP 0.15033 in.2 t ϭ 15.0 ksi (␣a Ϫ ␣c )(¢T)Ec Ac Ec Ac 1ϩ Ea Aa Problem 2.5-10 A rigid bar ABCD is pinned at end A and supported by two cables at points B and C (see figure). The cable at B has nominal diameter dB ϭ 12 mm and the cable at C has nominal diameter dC ϭ 20 mm. A load P acts at end D of the bar. What is the allowable load P if the temperature rises by 60°C and each cable is required to have a factor of safety of at least 5 against its ultimate load? (Note: The cables have effective modulus of elasticity E ϭ 140 GPa and coefficient of thermal expansion ␣ ϭ 12 ϫ 10Ϫ6/°C. Other properties of the cables can be found in Table 2-1, Section 2.2.) dC dB A C B 2b 2b D b P
    • 112 CHAPTER 2 Solution 2.5-10 Axially Loaded Members Rigid bar supported by two cables FREE-BODY DIAGRAM OF BAR ABCD TB A TC C B RAH SUBSTITUTE EQS. (3) AND (4) INTO EQ. (2): 2b D 2b b RAV P TB ϭ force in cable B dB ϭ 12 mm TC ϭ force in cable C E ϭ 140 GPa ⌬T ϭ 60ЊC (Eq. 6) (Eq. 8) in which P has units of newtons. TB (2b) ϩ TC (4b) Ϫ P(5b) ϭ 0 (Eq. 1) 2b C b SOLVE EQS. (7) AND (8) FOR THE LOAD P: D PB ϭ 4.0096 TB ϩ 13,953 (Eq. 9) PC ϭ 0.8887 TC Ϫ 1,546 DISPLACEMENT DIAGRAM B (Eq. 7) TC ϭ 1.1253 P ϩ 1,740 2TB ϩ 4TC ϭ 5P 2b SUBSTITUTE NUMERICAL VALUES INTO EQ. (5): TB ϭ 0.2494 P Ϫ 3,480 EQUATION OF EQUILIBRIUM A (Eq. 5) SOLVE SIMULTANEOUSLY EQS. (1) AND (6): AC ϭ 173 mm2 ␣ ϭ 12 ϫ 10Ϫ6/ЊC or 2TB AC Ϫ TC AB ϭ ϪE␣(⌬T)AB AC in which TB and TC have units of newtons. mm2 ©MA ϭ 0 ‫۔ە‬ or TB(346) Ϫ TC(76.7) ϭ Ϫ1,338,000 dC ϭ 20 mm From Table 2-1: AB ϭ 76.7 TC L 2TB L ϩ ␣(¢T)L ϭ ϩ 2␣(¢T)L EAC EAB (Eq. 10) ALLOWABLE LOADS From Table 2-1: ␦B COMPATIBILITY: (TB)ULT ϭ 102,000 N Factor of safety ϭ 5 ␦C ␦C ϭ 2␦B (TC)ULT ϭ 231,000 N (Eq. 2) (TB)allow ϭ 20,400 N (TC)allow ϭ 46,200 N From Eq. (9): PB ϭ (4.0096)(20,400 N) ϩ 13,953 N FORCE-DISPLACEMENT AND TEMPERATURE- ϭ 95,700 N DISPLACEMENT RELATIONS From Eq. (10): PC ϭ (0.8887)(46,200 N) Ϫ 1546 N ␦B ϭ TB L ϩ ␣(¢T )L EAB (Eq. 3) ␦C ϭ TC L ϩ ␣(¢T )L EAC (Eq. 4) ϭ 39,500 N Cable C governs. Pallow ϭ 39.5 kN Problem 2.5-11 A rigid triangular frame is pivoted at C and held by two identical horizontal wires at points A and B (see figure). Each wire has axial rigidity EA ϭ 120 k and coefficient of thermal expansion ␣ ϭ 12.5 ϫ 10Ϫ6/°F. (a) If a vertical load P ϭ 500 lb acts at point D, what are the tensile forces TA and TB in the wires at A and B, respectively? (b) If, while the load P is acting, both wires have their temperatures raised by 180°F, what are the forces TA and TB? (c) What further increase in temperature will cause the wire at B to become slack? A b B b D C P 2b
    • SECTION 2.5 Thermal Effects Solution 2.5-11 Triangular frame held by two wires FREE-BODY DIAGRAM OF FRAME TA A (b) LOAD P AND TEMPERATURE INCREASE ⌬T Force-displacement and temperaturedisplacement relations: b TB B ␦A ϭ D C TAL ϩ ␣(¢T)L EA (Eq. 8) ␦B ϭ b TBL ϩ ␣(¢T)L EA (Eq. 9) 2b P Substitute (8) and (9) into Eq. (2): EQUATION OF EQUILIBRIUM TAL 2TBL ϩ ␣(¢T)L ϭ ϩ 2␣(¢T)L EA EA ©MC ϭ 0 ‫۔ە‬ P(2b) Ϫ TA(2b) Ϫ TB(b) ϭ 0 or 2TA ϩ TB ϭ 2P (Eq. 1) ␦A 1 TA ϭ [4P ϩ EA␣(¢T ) ] 5 B ␦B b EQUATION OF COMPATIBILITY ␦A ϭ 2␦B P ϭ 500 lb ⌬T ϭ 180ЊF C EA ϭ 120,000 lb ␣ ϭ 12.5 ϫ 10Ϫ6/ЊF 1 TA ϭ (2000 lb ϩ 270 lb) ϭ 454 lb 5 (a) LOAD P ONLY Force-displacement relations: TB L EA (Eq. 12) Substitute numerical values: (Eq. 2) ␦B ϭ (Eq. 11) 2 TB ϭ [P Ϫ EA␣(¢T ) ] 5 A b TA L EA (Eq. 10) Solve simultaneously Eqs. (1) and (10): DISPLACEMENT DIAGRAM ␦A ϭ TA Ϫ 2TB ϭ EA␣(⌬T) or (Eq. 3, 4) 2 TB ϭ (500 lb Ϫ 270 lb) ϭ 92 lb 5 (L ϭ length of wires at A and B.) (c) WIRE B BECOMES SLACK Substitute (3) and (4) into Eq. (2): Set TB ϭ 0 in Eq. (12): P ϭ EA␣(⌬T) TA L 2TB L ϭ EA EA or or TA ϭ 2TB (Eq. 5) Solve simultaneously Eqs. (1) and (5): TA ϭ 4P 5 TB ϭ 2P 5 Numerical values: P 500 lb ϭ EA␣ (120,000 lb)(12.5 ϫ 10 Ϫ6րЊF) ϭ 333.3ЊF Further increase in temperature: ¢T ϭ 333.3ЊF Ϫ 180ЊF P ϭ 500 lb ∴ TA ϭ 400 lb (Eqs. 6, 7) ¢T ϭ ϭ 153ЊF TB ϭ 200 lb 113
    • 114 CHAPTER 2 Axially Loaded Members Misfits and Prestrains Problem 2.5-12 A steel wire AB is stretched between rigid supports (see figure). The initial prestress in the wire is 42 MPa when the temperature is 20°C. A B (a) What is the stress ␴ in the wire when the temperature drops to 0°C? (b) At what temperature T will the stress in the wire become zero? (Assume ␣ ϭ 14 ϫ 10Ϫ6/°C and E ϭ 200 GPa.) Solution 2.5-12 Steel wire Steel wire with initial prestress A B From Eq. (2-18): ␴2 ϭ E␣(⌬T) s ϭ s1 ϩ s2 ϭ s1 ϩ E␣(¢T ) Initial prestress: ␴1 ϭ 42 MPa Initial temperature: T1 ϭ 20ЊC E ϭ 200 GPa ␣ ϭ 14 ϫ ϭ 42 MPa ϩ 56 MPa ϭ 98 MPa (b) TEMPERATURE WHEN STRESS EQUALS ZERO 10Ϫ6/ЊC (a) STRESS ␴ WHEN TEMPERATURE DROPS TO 0ЊC T2 ϭ 0ЊC ϭ 42 MPa ϩ (200 GPa)(14 ϫ 10 Ϫ6րЊC)(20ЊC) ⌬T ϭ 20ЊC Note: Positive ⌬T means a decrease in temperature and an increase in the stress in the wire. Negative ⌬T means an increase in temperature and a decrease in the stress. ␴ ϭ ␴1 ϩ ␴2 ϭ 0 ¢T ϭ Ϫ ␴1 ϩ E␣(⌬T) ϭ 0 s1 E␣ (Negative means increase in temp.) ¢T ϭ Ϫ 42 MPa ϭ Ϫ 15ЊC (200 GPa)(14 ϫ 10 Ϫ6րЊC) T ϭ 20ЊC ϩ 15ЊC ϭ 35ЊC Stress ␴ equals the initial stress ␴1 plus the additional stress ␴2 due to the temperature drop. Problem 2.5-13 A copper bar AB of length 25 in. is placed in position at room temperature with a gap of 0.008 in. between end A and a rigid restraint (see figure). Calculate the axial compressive stress ␴c in the bar if the temperature rises 50°F. (For copper, use ␣ ϭ 9.6 ϫ 10Ϫ6/°F and E ϭ 16 ϫ 106 psi.) 0.008 in. A 25 in. B
    • SECTION 2.5 Solution 2.5-13 S 115 Misfits and Prestrains Bar with a gap ␴c ϭ stress in the bar L ϭ 25 in. A ϭ EeC ϭ S ϭ 0.008 in. L ⌬T ϭ 50ЊF (increase) ␣ ϭ 9.6 ϫ B E␦C E ϭ [␣(¢T)L Ϫ S] L L Note: This result is valid only if ␣(⌬T)L Ն S. (Otherwise, the gap is not closed). 10Ϫ6/ЊF E ϭ 16 ϫ 106 psi Substitute numerical values: ␦ ϭ elongation of the bar if it is free to expand sc ϭ ϭ ␣(⌬T)L 16 ϫ 106 psi [ (9.6 ϫ 10 Ϫ6րЊF)(50ЊF)(25 in.) 25 in. Ϫ 0.008 in.] ϭ 2,560 psi ␦C ϭ elongation that is prevented by the support ϭ ␣(⌬T)L Ϫ S eC ϭ strain in the bar due to the restraint ϭ ␦C /L Problem 2.5-14 A bar AB having length L and axial rigidity EA is fixed at end A (see figure). At the other end a small gap of dimension s exists between the end of the bar and a rigid surface. A load P acts on the bar at point C, which is two-thirds of the length from the fixed end. If the support reactions produced by the load P are to be equal in magnitude, what should be the size s of the gap? Solution 2.5-14 2L — 3 s L — 3 A C B P Bar with a gap (load P) 2L — 3 S L — 3 COMPATIBILITY EQUATION ␦1 Ϫ ␦2 ϭ S or 2PL RBL Ϫ ϭS 3EA EA A L ϭ length of bar P B EQUILIBRIUM EQUATION S ϭ size of gap RA EA ϭ axial rigidity RB P RA ϭ reaction at end A (to the left) Reactions must be equal; find S. RB ϭ reaction at end B (to the left) FORCE-DISPLACEMENT RELATIONS 2L — 3 (Eq. 1) P ϭ RA ϩ RB ␦1 ϭ P ␦1 RB ␦2 P( 2L ) 3 EA Reactions must be equal. ␦2 ϭ RBL EA Substitute for RB in Eq. (1): 2PL PL PL Ϫ ϭ S or S ϭ 3EA 2EA 6EA NOTE: The gap closes when the load reaches the value P/4. When the load reaches the value P, equal to 6EAs/L, the reactions are equal (RA ϭ RB ϭ P/2). When the load is between P/4 and P, RA is greater than RB. If the load exceeds P, RB is greater than RA. ∴ RA ϭ RB P ϭ 2RB RB ϭ P 2
    • 116 CHAPTER 2 Axially Loaded Members Problem 2.5-15 Wires B and C are attached to a support at the left-hand end and to a pin-supported rigid bar at the right-hand end (see figure). Each wire has cross-sectional area A ϭ 0.03 in.2 and modulus of elasticity E ϭ 30 ϫ 106 psi. When the bar is in a vertical position, the length of each wire is L ϭ 80 in. However, before being attached to the bar, the length of wire B was 79.98 in. and of wire C was 79.95 in. Find the tensile forces TB and TC in the wires under the action of a force P ϭ 700 lb acting at the upper end of the bar. 700 lb B b C b b 80 in. Solution 2.5-15 Wires B and C attached to a bar P = 700 lb Elongation of wires: B b ␦B ϭ SB ϩ 2␦ (Eq. 2) C b ␦C ϭ SC ϩ ␦ (Eq. 3) b FORCE-DISPLACEMENT RELATIONS L = 80 in. ␦B ϭ P ϭ 700 lb TBL EA ␦C ϭ TC L EA A ϭ 0.03 in.2 SOLUTION OF EQUATIONS E ϭ 30ϫ106 psi (Eqs. 4, 5) Combine Eqs. (2) and (4): TBL ϭ SB ϩ 2␦ EA LB ϭ 79.98 in. LC ϭ 79.95 in. Combine Eqs. (3) and (5): EQUILIBRIUM EQUATION TCL ϭ SC ϩ ␦ EA P = 700 lb TB TC (Eq. 6) b ©Mpin ϭ 0 ‫۔ە‬ Eliminate ␦ between Eqs. (6) and (7): b TC(b) ϩ TB(2b) ϭ P(3b) TB Ϫ 2TC ϭ (Eq. 7) b Pin 2TB ϩ TC ϭ 3P (Eq. 1) EASB 2EASC Ϫ L L Solve simultaneously Eqs. (1) and (8): TB ϭ 6P EASB 2EASC ϩ Ϫ 5 5L 5L TC ϭ 3P 2EASB 4EASC Ϫ ϩ 5 5L 5L DISPLACEMENT DIAGRAM SB ϭ 80 in. Ϫ LB ϭ 0.02 in. SC ϭ 80 in. Ϫ LC ϭ 0.05 in. (Eq. 8) SUBSTITUTE NUMERICAL VALUES: B SB 2␦ C ␦ SC L = 80 in. EA ϭ 2250 lbրin. 5L TB ϭ 840 lb ϩ 45 lb Ϫ 225 lb ϭ 660 lb TC ϭ 420 lb Ϫ 90 lb ϩ 450 lb ϭ 780 lb (Both forces are positive, which means tension, as required for wires.)
    • SECTION 2.5 Problem 2.5-16 A rigid steel plate is supported by three posts of high-strength concrete each having an effective cross-sectional area A ϭ 40,000 mm2 and length L ϭ 2 m (see figure). Before the load P is applied, the middle post is shorter than the others by an amount s ϭ 1.0 mm. Determine the maximum allowable load Pallow if the allowable compressive stress in the concrete is ␴allow ϭ 20 MPa. (Use E ϭ 30 GPa for concrete.) Solution 2.5-16 Misfits and Prestrains P S s C C C L Plate supported by three posts P EQUILIBRIUM EQUATION Steel plate P 2P1 ϩ P2 ϭ P s C C C L P1 P2 (Eq. 1) P1 COMPATIBILITY EQUATION ␦1 ϭ shortening of outer posts ␦2 ϭ shortening of inner post s ϭ size of gap ϭ 1.0 mm L ϭ length of posts ϭ 2.0 m A ϭ 40,000 mm2 ␴allow ϭ 20 MPa E ϭ 30 GPa C ϭ concrete post DOES THE GAP CLOSE? Stress in the two outer posts when the gap is just closed: s 1.0 mm s ϭ Ee ϭ E ¢ ≤ ϭ (30 GPa) ¢ ≤ L 2.0 m ϭ 15 MPa Since this stress is less than the allowable stress, the allowable force P will close the gap. ␦1 ϭ ␦2 ϩ s (Eq. 2) FORCE-DISPLACEMENT RELATIONS ␦1 ϭ P1L EA ␦2 ϭ P2L EA (Eqs. 3, 4) SOLUTION OF EQUATIONS Substitute (3) and (4) into Eq. (2): P1L P2L ϭ ϩs EA EA or P1 Ϫ P2 ϭ EAs L (Eq. 5) Solve simultaneously Eqs. (1) and (5): P ϭ 3P1 Ϫ EAs L By inspection, we know that P1 is larger than P2. Therefore, P1 will control and will be equal to ␴allow A. Pallow ϭ 3sallow A Ϫ EAs L ϭ 2400 kN Ϫ 600 kN ϭ 1800 kN ϭ 1.8 MN 117
    • 118 CHAPTER 2 Axially Loaded Members Problem 2.5-17 A copper tube is fitted around a steel bolt and the nut is turned until it is just snug (see figure). What stresses ␴s and ␴c will be produced in the steel and copper, respectively, if the bolt is now tightened by a quarter turn of the nut? The copper tube has length L ϭ 16 in. and cross-sectional area Ac ϭ 0.6 in.2, and the steel bolt has cross-sectional area As ϭ 0.2 in.2 The pitch of the threads of the bolt is p ϭ 52 mils (a mil is one-thousandth of an inch). Also, the moduli of elasticity of the steel and copper are Es ϭ 30 ϫ 106 psi and Ec ϭ 16 ϫ 106 psi, respectively. Note: The pitch of the threads is the distance advanced by the nut in one complete turn (see Eq. 2-22). Solution 2.5-17 Copper tube Steel bolt Steel bolt and copper tube Copper tube FORCE-DISPLACEMENT RELATIONS ␦c ϭ Pc L Ec Ac ␦s ϭ Ps L Es As (Eq. 3, Eq. 4) Steel bolt SOLUTION OF EQUATIONS L ϭ 16 in. Substitute (3) and (4) into Eq. (2): p ϭ 52 mils ϭ 0.052 in. PsL PcL ϩ ϭ np EcAc EsAs 1 n ϭ (See Eq. 2-22) 4 Solve simultaneously Eqs. (1) and (5): Steel bolt: As ϭ 0.2 in.2 npEs As Ec Ac L(Es As ϩ Ec Ac ) Es ϭ 30 ϫ 106 psi Ps ϭ Pc ϭ Copper tube: Ac ϭ 0.6 in.2 Substitute numerical values: Ec ϭ 16 ϫ 106 psi Ps ϭ Pc ϭ 3,000 lb EQUILIBRIUM EQUATION Pc STRESSES Steel bolt: Ps ss ϭ Ps ϭ tensile force in steel bolt Pc ϭ Ps Ps 3,000 lb ϭ ϭ 15 ksi (tension) As 0.2 in.2 Copper tube: Pc ϭ compressive force in copper tube (Eq. 1) sc ϭ Pc 3,000 lb ϭ Ac 0.6 in.2 ϭ 5 ksi (compression) COMPATIBILITY EQUATION Ps Pc np ␦c ϭ shortening of copper tube ␦s ϭ elongation of steel bolt ␦c ϩ ␦s ϭ np (Eq. 5) (Eq. 2) (Eq. 6)
    • SECTION 2.5 Problem 2.5-18 A plastic cylinder is held snugly between a rigid plate and a foundation by two steel bolts (see figure). Determine the compressive stress ␴p in the plastic when the nuts on the steel bolts are tightened by one complete turn. Data for the assembly are as follows: length L ϭ 200 mm, pitch of the bolt threads p ϭ 1.0 mm, modulus of elasticity for steel Es ϭ 200 GPa, modulus of elasticity for the plastic Ep ϭ 7.5 GPa, cross-sectional area of one bolt As ϭ 36.0 mm2, and cross-sectional area of the plastic cylinder Ap ϭ 960 mm2. Solution 2.5-18 119 Misfits and Prestrains Steel bolt L Probs. 2.5-18 and 2.5-19 Plastic cylinder and two steel bolts FORCE-DISPLACEMENT RELATIONS ␦s ϭ S P S Ps L Es As ␦p ϭ Pp L Ep Ap L ϭ 200 mm SOLUTION OF EQUATIONS P ϭ 1.0 mm Substitute (3) and (4) into Eq. (2): Es ϭ 200 GPa As ϭ 36.0 mm2 Pp L Ps L ϩ ϭ np Es As Ep Ap (for one bolt) Ep ϭ 7.5 GPa Solve simultaneously Eqs. (1) and (5): Ap ϭ 960 mm2 Pp ϭ n ϭ 1 (See Eq. 2-22) EQUILIBRIUM EQUATION 2npEs As Ep Ap L(Ep Ap ϩ 2Es As ) STRESS IN THE PLASTIC CYLINDER Ps Ps sp ϭ Pp Ap ϭ 2 np Es As Ep L(Ep Ap ϩ 2Es As ) SUBSTITUTE NUMERICAL VALUES: N ϭ Es As Ep ϭ 54.0 ϫ 1015 N2/m2 Pp D ϭ Ep Ap ϩ 2Es As ϭ 21.6 ϫ 106 N Ps ϭ tensile force in one steel bolt Pp ϭ compressive force in plastic cylinder Pp ϭ 2Ps sp ϭ (Eq. 1) COMPATIBILITY EQUATION Ps S Pp Ps np P S ␦s ϭ elongation of steel bolt ␦p ϭ shortening of plastic cylinder ␦s ϩ ␦p ϭ np (Eq. 3, Eq. 4) (Eq. 2) 2np N 2(1)(1.0 mm) N ¢ ≤ϭ ¢ ≤ L D 200 mm D ϭ 25.0 MPa (Eq. 5)
    • 120 CHAPTER 2 Axially Loaded Members Problem 2.5-19 Solve the preceding problem if the data for the assembly are as follows: length L ϭ 10 in., pitch of the bolt threads p ϭ 0.058 in., modulus of elasticity for steel Es ϭ 30 ϫ 106 psi, modulus of elasticity for the plastic Ep ϭ 500 ksi, cross-sectional area of one bolt As ϭ 0.06 in.2, and cross-sectional area of the plastic cylinder Ap ϭ 1.5 in.2 Solution 2.5-19 Plastic cylinder and two steel bolts FORCE-DISPLACEMENT RELATIONS ␦s ϭ S P S Ps L Es As ␦p ϭ Pp L Ep Ap L ϭ 10 in. SOLUTION OF EQUATIONS p ϭ 0.058 in. Substitute (3) and (4) into Eq. (2): Es ϭ 30 ϫ 106 psi As ϭ 0.06 in.2 (for one bolt) Ep ϭ 500 ksi Ap ϭ 1.5 Pp L Ps L ϩ ϭ np Es As Ep Ap Solve simultaneously Eqs. (1) and (5): in.2 Pp ϭ n ϭ 1 (see Eq. 2-22) 2 np Es As Ep Ap L(Ep Ap ϩ 2Es As ) EQUILIBRIUM EQUATION STRESS IN THE PLASTIC CYLINDER Ps ϭ tensile force in one steel bolt sp ϭ Pp ϭ compressive force in plastic cylinder Pp ϭ 2Ps (Eq. 1) Ps Pp Ap 2np N 2(1)(0.058 in.) N ¢ ≤ϭ ¢ ≤ L D 10 in. D ϭ 2400 psi COMPATIBILITY EQUATION ␦s ϭ elongation of steel bolt ␦p ϭ shortening of plastic cylinder ␦s ϩ ␦p ϭ np (Eq. 2) Ps np P L(Ep Ap ϩ 2Es As ) D ϭ Ep Ap ϩ 2Es As ϭ 4350 ϫ 103 lb Pp S 2 np Es As Ep N ϭ Es As Ep ϭ 900 ϫ 109 lb2/in.2 Ps Pp ϭ SUBSTITUTE NUMERICAL VALUES: sP ϭ Ps (Eq. 3, Eq. 4) S (Eq. 5)
    • SECTION 2.5 Problem 2.5-20 Prestressed concrete beams are sometimes manufactured in the following manner. High-strength steel wires are stretched by a jacking mechanism that applies a force Q, as represented schematically in part (a) of the figure. Concrete is then poured around the wires to form a beam, as shown in part (b). After the concrete sets properly, the jacks are released and the force Q is removed [see part (c) of the figure]. Thus, the beam is left in a prestressed condition, with the wires in tension and the concrete in compression. Let us assume that the prestressing force Q produces in the steel wires an initial stress ␴0 ϭ 620 MPa. If the moduli of elasticity of the steel and concrete are in the ratio 12:1 and the cross-sectional areas are in the ratio 1:50, what are the final stresses ␴s and ␴c in the two materials? Misfits and Prestrains Steel wires Q Q (a) Concrete Q Q (b) (c) Solution 2.5-20 Prestressed concrete beam L ϭ length Steel wires ␴0 ϭ initial stress in wires Q ϭ Q ϭ 620 MPa As As ϭ total area of steel wires Concrete Ac ϭ area of concrete ϭ 50 As Ps Es ϭ 12 Ec Pc Psϭ final tensile force in steel wires Pc ϭ final compressive force in concrete EQUILIBRIUM EQUATION Ps ϭ Pc (Eq. 1) COMPATIBILITY EQUATION AND FORCE-DISPLACEMENT RELATIONS ␦1 ϭ initial elongation of steel wires s0L QL ϭ ϭ EsAs Es ␦2 ϭ final elongation of steel wires Ps L ϭ Es As ␦3 ϭ shortening of concrete Pc L ϭ Ec Ac ␦1 Ϫ ␦2 ϭ ␦3 STRESSES ss ϭ sc ϭ Ps ϭ As s0 Es As 1ϩ Ec Ac s0 Pc ϭ Ac Ac Es ϩ As Ec SUBSTITUTE NUMERICAL VALUES: s0 ϭ 620 MPa or s0 L Ps L Pc L Ϫ ϭ (Eq. 2, Eq. 3) Es Es As Ec Ac Solve simultaneously Eqs. (1) and (3): s0 As Ps ϭ Pc ϭ Es As 1ϩ Ec Ac 121 Es ϭ 12 Ec As 1 ϭ Ac 50 ss ϭ 620 MPa ϭ 500 MPa (Tension) 12 1ϩ 50 sc ϭ 620 MPa ϭ 10 MPa (Compression) 50 ϩ 12
    • 122 CHAPTER 2 Axially Loaded Members Stresses on Inclined Sections Problem 2.6-1 A steel bar of rectangular cross section (1.5 in. ϫ 2.0 in.) carries a tensile load P (see figure). The allowable stresses in tension and shear are 15,000 psi and 7,000 psi, respectively. Determine the maximum permissible load Pmax. Solution 2.6-1 2.0 in. P P 1.5 in. Rectangular bar in tension 2.0 in. P Maximum shear stress: tmax ϭ sx P ϭ 2 2A ␴allow ϭ 15,000 psi ␶allow ϭ 7,000 psi P Because ␶allow is less than one-half of ␴allow, the shear stress governs. 1.5 in. A ϭ 1.5 in. ϫ 2.0 in. Pmax ϭ 2␶allow A ϭ 2(7,000 psi) (3.0 in.2) ϭ 3.0 in.2 ϭ 42,000 lb Maximum Normal Stress: sx ϭ P A Problem 2.6-2 A circular steel rod of diameter d is subjected to a tensile force P ϭ 3.0 kN (see figure). The allowable stresses in tension and shear are 120 MPa and 50 MPa, respectively. What is the minimum permissible diameter dmin of the rod? Solution 2.6-2 Steel rod in tension P P ϭ 3.0 kN d P Aϭ ␲d 2 4 d P Because ␶allow is less than one-half of ␴allow, the shear stress governs. Maximum normal stress: sx ϭ P A tmax ϭ Maximum shear stress: tmax ϭ sx P ϭ 2 2A Solve for d: dmin ϭ 6.18 mm ␴allow ϭ 120 MPa ␶allow ϭ 50 MPa P 2A or 50 MPa ϭ 3.0 kN ␲d2 (2) ¢ ≤ 4 P = 3.0 kN
    • SECTION 2.6 Problem 2.6-3 A standard brick (dimensions 8 in. ϫ 4 in. ϫ 2.5 in.) is compressed lengthwise by a force P, as shown in the figure. If the ultimate shear stress for brick is 1200 psi and the ultimate compressive stress is 3600 psi, what force Pmax is required to break the brick? Solution 2.6-3 123 P 8 in. 4 in. 2.5 in. Standard brick in compression Maximum shear stress: P tmax ϭ 8 in. Stresses on Inclined Sections 4 in. 2.5 in. sx P ϭ 2 2A ␴ult ϭ 3600 psi ␶ult ϭ 1200 psi Because ␶ult is less than one-half of ␴ult, the shear stress governs. tmax ϭ A ϭ 2.5 in. ϫ 4.0 in. ϭ 10.0 in.2 or Pmax ϭ 2Atult Pmax ϭ 2(10.0 in.2 )(1200 psi) Maximum normal stress: sx ϭ P 2A P A ϭ 24,000 lb Problem 2.6-4 A brass wire of diameter d ϭ 2.42 mm is stretched tightly between rigid supports so that the tensile force is T ϭ 92 N (see figure). What is the maximum permissible temperature drop ⌬T if the allowable shear stress in the wire is 60 MPa? (The coefficient of thermal expansion for the wire is 20 ϫ 10Ϫ6/°C and the modulus of elasticity is 100 GPa.) Solution 2.6-4 d T Probs. 2.6-4 and 2.6-5 Brass wire in tension MAXIMUM SHEAR STRESS d T T tmax ϭ d ϭ 2.42 mm Aϭ Solve for temperature drop ⌬T: ␲d2 ϭ 4.60 mm2 4 ␣ ϭ 20 ϫ 10Ϫ6/ЊC E ϭ 100 GPa ␶allow ϭ 60 MPa Initial tensile force: T ϭ 92 N T A Stress due to temperature drop: ␴x ϭ E␣(⌬T) Stress due to initial tension: sx ϭ (see Eq. 2-18 of Section 2.5) T Total stress: sx ϭ ϩ E␣(¢T ) A sx 1 T ϭ B ϩ E␣(¢T )R 2 2 A ¢T ϭ 2tmax Ϫ TրA E␣ tmax ϭ tallow SUBSTITUTE NUMERICAL VALUES: ¢T ϭ ϭ 2(60 MPa) Ϫ (92 N)ր(4.60 mm2 ) (100 GPa)(20 ϫ 10 Ϫ6րЊC) 120 MPa Ϫ 20 MPa ϭ 50ЊC 2 MPaրЊC T
    • 124 CHAPTER 2 Axially Loaded Members Problem 2.6-5 A brass wire of diameter d ϭ 1/16 in. is stretched between rigid supports with an initial tension T of 32 lb (see figure). (a) If the temperature is lowered by 50°F, what is the maximum shear stress ␶max in the wire? (b) If the allowable shear stress is 10,000 psi, what is the maximum permissible temperature drop? (Assume that the coefficient of thermal expansion is 10.6 ϫ 10Ϫ6/°F and the modulus of elasticity is 15 ϫ 106 psi.) Solution 2.6-5 Brass wire in tension T d T 1 dϭ in. 16 Aϭ tmax ϭ ␲d2 4 (Eq. 1) tmax ϭ 9,190 psi (b) MAXIMUM PERMISSIBLE TEMPERATURE DROP IF ␶allow ϭ 10,000 psi ␣ ϭ 10.6 ϫ 10Ϫ6/ЊF E ϭ 15 ϫ 106 psi Solve Eq. (1) for ⌬T: Initial tensile force: T ϭ 32 lb T A Stress due to temperature drop: ␴x ϭ E␣(⌬T ) (see Eq. 2-18 of Section 2.5) T Total stress: sx ϭ ϩ E␣(¢T ) A sx 1 T ϭ B ϩ E␣(¢T )R 2 2 A Substitute numerical values: ϭ 0.003068 in.2 Stress due to initial tension: sx ϭ (a) MAXIMUM SHEAR STRESS WHEN TEMPERATURE DROPS 50ЊF 2tmax Ϫ TրA tmax ϭ tallow E␣ Substitute numerical values: ¢T ϭ ¢T ϭ 60.2ЊF Problem 2.6-6 A steel bar with diameter d ϭ 12 mm is subjected to a tensile load P ϭ 9.5 kN (see figure). (a) What is the maximum normal stress ␴max in the bar? (b) What is the maximum shear stress ␶max? (c) Draw a stress element oriented at 45° to the axis of the bar and show all stresses acting on the faces of this element. P d = 12 mm P = 9.5 kN
    • SECTION 2.6 Solution 2.6-6 Steel bar in tension d = 12 mm P 125 Stresses on Inclined Sections P = 9.5 kN (c) STRESS ELEMENT AT ␪ ϭ 45Њ 9,000 9,000 P ϭ 9.5 kN 9,000 0 (a) MAXIMUM NORMAL STRESS P 9.5 kN sx ϭ ϭ ␲ ϭ 84.0 MPa A 4 (12 mm) 2 ␪ = 45° y x 9,000 9,000 9,000 NOTE: All stresses have units of MPa. smax ϭ 84.0 MPa (b) MAXIMUM SHEAR STRESS The maximum shear stress is on a 45Њ plane and equals ␴x /2. tmax ϭ sx ϭ 42.0 MPa 2 Problem 2.6-7 During a tension test of a mild-steel specimen (see figure), the extensometer shows an elongation of 0.00120 in. with a gage length of 2 in. Assume that the steel is stressed below the proportional limit and that the modulus of elasticity E ϭ 30 ϫ 106 psi. (a) What is the maximum normal stress ␴max in the specimen? (b) What is the maximum shear stress ␶max? (c) Draw a stress element oriented at an angle of 45° to the axis of the bar and show all stresses acting on the faces of this element. 2 in. T T
    • 126 CHAPTER 2 Solution 2.6-7 Axially Loaded Members Tension test 2 in. T T Elongation: ␦ ϭ 0.00120 in. (b) MAXIMUM SHEAR STRESS (2 in. gage length) The maximum shear stress is on a 45Њ plane and equals ␴x /2. Strain: e ϭ ␦ 0.00120 in. ϭ ϭ 0.00060 L 2 in. tmax ϭ Hooke’s law : ␴x ϭ Ee ϭ (30 ϫ 106 psi)(0.00060) ϭ 18,000 psi sx ϭ 9,000 psi 2 (c) STRESS ELEMENT AT ␪ ϭ 45Њ NOTE: All stresses have units of psi. (a) MAXIMUM NORMAL STRESS ␴x is the maximum normal stress. 9,000 9,000 smax ϭ 18,000 psi ␪ = 45° y 9,000 0 x 9,000 9,000 9,000 Problem 2.6-8 A copper bar with a rectangular cross section is held without stress between rigid supports (see figure). Subsequently, the temperature of the bar is raised 50°C. Determine the stresses on all faces of the elements A and B, and show these stresses on sketches of the elements. (Assume ␣ ϭ 17.5 ϫ 10Ϫ6/°C and E ϭ 120 GPa.) Solution 2.6-8 45° A Copper bar with rigid supports 45° A B STRESSES ON ELEMENTS A AND B 105 A ⌬T ϭ 50ЊC (Increase) 105 52.5 ␴x ϭ E␣ (⌬T) (See Eq. 2-18 of Section 2.5) ϭ 105 MPa (Compression) MAXIMUM SHEAR STRESS sx 2 ϭ 52.5 MPa 0 ␪ = 45° B y E ϭ 120 GPa STRESS DUE TO TEMPERATURE INCREASE 52.5 52.5 ␣ ϭ 17.5 ϫ 10Ϫ6/ЊC tmax ϭ B x 52.5 52.5 NOTE: All stresses have units of MPa. 52.5
    • SECTION 2.6 Stresses on Inclined Sections Problem 2.6-9 A compression member in a bridge truss is fabricated from a wide-flange steel section (see figure). The cross-sectional area A ϭ 7.5 in.2 and the axial load P ϭ 90 k. P Determine the normal and shear stresses acting on all faces of stress elements located in the web of the beam and oriented at (a) an angle ␪ ϭ 0°, (b) an angle ␪ ϭ 30°, and (c) an angle ␪ ϭ 45°. In each case, show the stresses on a sketch of a properly oriented element. Solution 2.6-9 ␪ Truss member in compression P P ␪ ␶␪ ϭ Ϫ␴x sin ␪ cos ␪ ϭ Ϫ(Ϫ12.0 ksi)(sin 120Њ) (cos 120Њ) ϭ Ϫ5.2 ksi P ϭ 90 k A ϭ 7.5 in.2 sx ϭ Ϫ 3.0 P 90 k ϭϪ A 7.5 in.2 9.0 5.2 y ␪ = 30° x 0 3.0 ϭ Ϫ 12.0 ksi (Compression) 9.0 (a) ␪ ϭ 0Њ 5.2 NOTE: All stresses have units of ksi. y 12.0 ksi 12.0 ksi 0 x (b) ␪ ϭ 30Њ Use Eqs. (2-29a) and (2-29b): ␴␪ ϭ ␴x cos2␪ ϭ (Ϫ12.0 ksi)(cos 30Њ)2 (c) ␪ ϭ 45Њ ␴␪ ϭ ␴x cos2␪ ϭ (Ϫ12.0 ksi)(cos 45Њ)2 ϭ Ϫ6.0 ksi ␶␪ ϭ Ϫ␴x sin ␪ cos ␪ ϭ Ϫ(Ϫ12.0 ksi)(sin 45Њ) (cos 45Њ) ϭ 6.0 ksi 6.0 6.0 ϭ Ϫ9.0 ksi ␶␪ ϭ Ϫ␴x sin ␪ cos ␪ ϭ Ϫ(Ϫ12.0 ksi)(sin 30Њ)(cos 30Њ) 0 ϭ 5.2 ksi ␪ ϭ 30Њ ϩ 90Њ ϭ 120Њ ␴␪ ϭ ␴x cos2␪ ϭ (Ϫ12.0 ksi)(cos 120Њ)2 ϭ Ϫ3.0 ksi 6.0 ␪ = 45° 6.0 y 6.0 x 6.0 NOTE: All stresses have units of ksi. 127 P
    • 128 CHAPTER 2 Axially Loaded Members Problem 2.6-10 A plastic bar of diameter d ϭ 30 mm is compressed in a testing device by a force P ϭ 170 N applied as shown in the figure. Determine the normal and shear stresses acting on all faces of stress elements oriented at (a) an angle ␪ ϭ 0°, (b) an angle ␪ ϭ 22.5°, and (c) an angle ␪ ϭ 45°. In each case, show the stresses on a sketch of a properly oriented element. P = 170 N 100 mm 300 mm ␪ Plastic bar d = 30 mm Solution 2.6-10 Plastic bar in compression 100 mm 300 mm P = 170 N ␶␪ ϭ Ϫ␴x sin ␪ cos ␪ ϭ Ϫ(Ϫ962.0 kPa)(sin 22.5Њ)(cos 22.5Њ) ϭ 340 kPa Plastic bar d = 30 mm ␪ ␪ ϭ 22.5Њ ϩ 90Њ ϭ 112.5Њ ␴␪ ϭ ␴x cos2␪ ϭ (Ϫ962.0 kPa)(cos 112.5Њ)2 ϭ Ϫ141 kPa ␶␪ ϭ Ϫ␴x sin ␪ cos ␪ ϭ Ϫ(Ϫ962.0 kPa)(sin 112.5Њ)(cos 112.5Њ) FREE-BODY DIAGRAM ϭ Ϫ340 kPa 141 P = 170 N 100 mm 340 300 mm 0 F ϭ Compressive force in plastic bar 821 PLASTIC BAR (ROTATED TO THE HORIZONTAL) F x F NOTE: All stresses have units of kPa. ␴␪ ϭ ␴x cos2␪ ϭ (Ϫ962.0 kPa)(cos 45Њ)2 ϭ Ϫ481 kPa d = 30 mm sx ϭ Ϫ 141 (c) ␪ ϭ 45Њ ␪ 0 ␶␪ ϭ Ϫ␴x sin ␪ cos ␪ F 680 N ϭ Ϫ␲ 2 A 4 (30 mm) ϭ Ϫ(Ϫ962.0 kPa)(sin 45Њ)(cos 45Њ) ϭ 481 kPa ϭ Ϫ 962.0 kPa (Compression) 481 (a) ␪ ϭ 0Њ 962 kPa y 0 x 962 kPa (b) ␪ ϭ 22.5Њ Use Eqs. (2-29a) and (2-29b) ␴␪ ϭ ␴x cos2␪ ϭ (Ϫ962.0 kPa)(cos 22.5Њ)2 ϭ Ϫ821 kPa ␪ = 22.5° x 340 F ϭ 4P ϭ 4(170 N)ϭ680 N y 821 y F 0 481 481 481 y 481 ␪ = 45° x 481 NOTE: All stresses have units of kPa.
    • SECTION 2.6 Stresses on Inclined Sections Problem 2.6-11 A plastic bar fits snugly between rigid supports at room temperature (68°F) but with no initial stress (see figure). When the temperature of the bar is raised to 160°F, the compressive stress on an inclined plane pq becomes 1700 psi. p (a) What is the shear stress on plane pq? (Assume ␣ ϭ 60 ϫ 10Ϫ6/°F and E ϭ 450 ϫ 103 psi.) (b) Draw a stress element oriented to plane pq and show the stresses acting on all faces of this element. Solution 2.6-11 ␪ q Probs. 2.6-11 and 2.6-12 Plastic bar between rigid supports p ␪ q ␣ ϭ 60 ϫ 10Ϫ6/ЊF E ϭ 450 ϫ 103 psi (b) STRESS ELEMENT ORIENTED TO PLANE pq Temperature increase: ␪ ϭ 34.18Њ ␴␪ ϭ Ϫ1700 psi ␶␪ ϭ 1150 psi ⌬T ϭ 160ЊF Ϫ 68ЊF ϭ 92ЊF ␪ ϭ 34.18Њ ϩ 90Њ ϭ 124.18Њ ␴␪ ϭ ␴x cos2␪ ϭ (Ϫ2484 psi)(cos 124.18Њ)2 NORMAL STRESS ␴x IN THE BAR ␴xϭ ϪE␣(⌬T ) (See Eq. 2-18 in Section 2.5) ␴x ϭ Ϫ(450 ϫ 103 psi)(60 ϫ 10Ϫ6/ЊF)(92ЊF) ϭ Ϫ2484 psi (Compression) ␴␪ ϭ ␴x 784 1150 1700 y psi)(cos2␪) Ϫ 1700 psi ϭ 0.6844 Ϫ 2484 psi cos ␪ ϭ 0.8273 ϭ Ϫ(Ϫ2484 psi)(sin 124.18Њ)(cos 124.18Њ) For plane pq: ␴␪ ϭ Ϫ1700 psi Therefore, Ϫ1700 psi ϭ (Ϫ2484 cos2u ϭ ␶␪ ϭ Ϫ␴x sin ␪ cos ␪ ϭ Ϫ1150 psi ANGLE ␪ TO PLANE pq cos2␪ ϭ Ϫ784 psi ␪ ϭ 34.18Њ (a) SHEAR STRESS ON PLANE pq ␶␪ ϭ Ϫ␴x sin ␪ cos ␪ ϭ Ϫ(Ϫ2484 psi)(sin 34.18Њ)(cos 34.18Њ) ϭ 1150 psi (Counter clockwise) 0 1700 ␪ = 34.18° x 1150 784 NOTE: All stresses have units of psi. 129
    • 130 CHAPTER 2 Axially Loaded Members Problem 2.6-12 A copper bar is held snugly (but without any initial stress) between rigid supports (see figure). The allowable stresses on the inclined plane pq, for which ␪ ϭ 55°, are specified as 60 MPa in compression and 30 MPa in shear. (a) What is the maximum permissible temperature rise ⌬T if the allowable stresses on plane pq are not to be exceeded? (Assume ␣ ϭ 17 ϫ 10Ϫ6/°C and E ϭ 120 GPa.) (b) If the temperature increases by the maximum permissible amount, what are the stresses on plane pq? Solution 2.6-12 Copper bar between rigid supports p ␪ Shear stress governs. ␴x ϭ Ϫ63.85 MPa Due to temperature increase ⌬T: q ␴x ϭ ϪE␣(⌬T) (See Eq. 2-18 in Section 2.5) Ϫ63.85 MPa ϭ Ϫ(120 GPa)(17 ϫ 10Ϫ6/ЊC)(⌬T) ␣ ϭ 17 ϫ 10Ϫ6/ЊC ¢T ϭ 31.3ЊC E ϭ 120 GPa Plane pq: ␪ ϭ 55Њ (b) STRESSES ON PLANE pq Allowable stresses on plane pq: ␴allow ϭ 60 MPa (Compression) ␶allow ϭ 30 MPa (Shear) (a) MAXIMUM PERMISSIBLE TEMPERATURE RISE ⌬T ␴␪ ϭ ␴x cos2␪ Ϫ60 MPa ϭ ␴x (cos 55Њ)2 ␴x ϭ Ϫ182.4 MPa ␶␪ ϭ Ϫ␴x sin ␪ cos ␪ 30 MPa ϭ Ϫ␴x (sin 55Њ)(cos 55Њ) ␴x ϭ Ϫ63.85 MPa ␴␪ ϭ ␴x cos2␪ ϭ (Ϫ63.85 MPa)(cos 55Њ)2 ϭ Ϫ 21.0 MPa (Compression) ␶␪ ϭ Ϫ␴x sin ␪ cos ␪ ϭ Ϫ(Ϫ63.85 MPa)(sin 55Њ)(cos 55Њ) ϭ 30.0 MPa (Counter clockwise) ␴x ϭ Ϫ63.85 MPa Problem 2.6-13 A circular brass bar of diameter d is composed of two segments brazed together on a plane pq making an angle ␣ ϭ 36° with the axis of the bar (see figure). The allowable stresses in the brass are 13,500 psi in tension and 6500 psi in shear. On the brazed joint, the allowable stresses are 6000 psi in tension and 3000 psi in shear. If the bar must resist a tensile force P ϭ 6000 lb, what is the minimum required diameter dmin of the bar? P ␣ d p q P
    • SECTION 2.6 Solution 2.6-13 P d 131 Stresses on Inclined Sections Brass bar in tension p ␣ n ␪ = 54° Tensile stress: ␴␪ ϭ ␴x cos2␪ P sx ϭ q 6000 psi sallow 2 ϭ cos u (cos 54Њ) 2 ϭ 17,370 psi ␣ ϭ 36Њ (3) ␪ ϭ 90Њ Ϫ ␣ ϭ 54Њ Shear stress: ␶␪ ϭ Ϫ␴x sin ␪ cos ␪ P ϭ 6000 lb sx ϭ ` Aϭ ␲d2 4 3,000 psi tallow ` ϭ sin ucos u (sin 54Њ)(cos 54Њ) ϭ 6,310 psi (4) STRESS ␴x BASED UPON ALLOWABLE STRESSES ALLOWABLE STRESS IN THE BRASS Compare (1), (2), (3), and (4). Tensile stress (␪ ϭ 0Њ): ␴allow ϭ 13,500 psi ␴x ϭ 13,500 psi Shear stress on the brazed joint governs. (1) Shear stress (␪ ϭ 45Њ): ␶allow ϭ 6500 psi tmax ϭ ␴x ϭ 6310 psi DIAMETER OF BAR sx 2 P 6000 lb ϭ ϭ 0.951 in.2 sx 6310 psi Aϭ ␲d2 4 ␴x ϭ 2 ␶allow ϭ 13,000 psi 4A B␲ Aϭ (2) STRESS ␴x BASED UPON ALLOWABLE STRESSES ON THE BRAZED JOINT (␪ ϭ 54Њ) d2 ϭ 4A ␲ dmin ϭ dmin ϭ 1.10 in. ␴allow ϭ 6000 psi (tension) ␶allow ϭ 3000 psi (shear) Problem 2.6-14 Two boards are joined by gluing along a scarf joint, as shown in the figure. For purposes of cutting and gluing, the angle ␣ between the plane of the joint and the faces of the boards must be between 10° and 40°. Under a tensile load P, the normal stress in the boards is 4.9 MPa. (a) What are the normal and shear stresses acting on the glued joint if ␣ ϭ 20°? (b) If the allowable shear stress on the joint is 2.25 MPa, what is the largest permissible value of the angle ␣? (c) For what angle ␣ will the shear stress on the glued joint be numerically equal to twice the normal stress on the joint? P P ␣
    • 132 CHAPTER 2 Solution 2.6-14 Axially Loaded Members Two boards joined by a scarf joint y P P x ␣ 10Њ Յ ␣ Յ 40Њ ␪ ϭ 33.34Њ or Due to load P: ␴x ϭ 4.9 MPa ␣ ϭ 90Њ Ϫ ␪ (a) STRESSES ON JOINT WHEN ␣ ϭ 20Њ n ‹ ␣ ϭ 56.66Њ or 33.34Њ Since ␣ must be between 10Њ and 40Њ, we select ␣ ϭ 33.3Њ Note: If ␣ is between 10Њ and 33.3Њ, ␪= 90°Ϫ a |␶␪| Ͻ 2.25 MPa. ␣ If ␣ is between 33.3Њ and 40Њ, |␶␪| Ͼ 2.25 MPa. ␪ ϭ 90Њ Ϫ ␣ ϭ 70Њ ␴␪ ϭ ␴x cos2␪ ϭ (4.9 MPa)(cos 70Њ)2 ϭ 0.57 MPa (c) WHAT IS ␣ if ␶␪ ϭ 2␴␪? Numerical values only: ␶␪ ϭ Ϫ␴x sin ␪ cos ␪ |␶␪| ϭ ␴x sin ␪ cos ␪ ϭ (Ϫ4.9 MPa)(sin 70Њ)(cos 70Њ) ϭ Ϫ 1.58 MPa (b) LARGEST ANGLE ␣ IF ␶allow ϭ 2.25 MPa ␶allow ϭ Ϫ␴x sin ␪ cos ␪ The shear stress on the joint has a negative sign. Its numerical value cannot exceed ␶allow ϭ 2.25 MPa. Therefore, Ϫ2.25 MPa ϭ Ϫ(4.9 MPa)(sin ␪)(cos ␪) or sin ␪ cos ␪ ϭ 0.4592 From trigonometry: sin u cos u ϭ 1 sin 2u 2 Therefore: sin 2␪ ϭ 2(0.4592) ϭ 0.9184 Solving : 2␪ ϭ 66.69Њ or 56.66Њ 113.31Њ ` |␴␪| ϭ ␴x cos2␪ tu ` ϭ2 su ␴x sin ␪ cos ␪ ϭ 2␴xcos2␪ sin ␪ ϭ 2 cos ␪ or tan ␪ ϭ 2 ␪ ϭ 63.43Њ ␣ ϭ 90Њ Ϫ ␪ a ϭ 26.6Њ NOTE: For ␣ ϭ 26.6Њ and ␪ ϭ 63.4Њ, we find ␴␪ ϭ 0.98 MPa and ␶␪ ϭ Ϫ1.96 MPa. Thus, ` tu ` ϭ 2 as required. su
    • SECTION 2.6 Problem 2.6-15 Acting on the sides of a stress element cut from a bar in uniaxial stress are tensile stresses of 10,000 psi and 5,000 psi, as shown in the figure. Stresses on Inclined Sections 5,000 psi ␶␪ ␶␪ ␪ (a) Determine the angle ␪ and the shear stress ␶␪ and show all stresses on a sketch of the element. (b) Determine the maximum normal stress ␴max and the maximum shear stress ␶max in the material. ␶␪ ␶␪ 10,000 psi Solution 2.6-15 ␴␪ = 10,000 psi 5,000 psi Bar in uniaxial stress From Eq. (1) or (2): 5,000 psi sx ϭ 15,000 psi ␶␪ tu ϭ Ϫsx sin u cos u ϭ (Ϫ15,000 psi)(sin 35.26Њ)(cos 35.26Њ) ϭ Ϫ7,070 psi 10,000 psi ␪ Minus sign means that ␶␪ acts clockwise on the plane for which ␪ ϭ 35.26Њ. 10,000 psi ␶␪ 5,000 5,000 psi 0 ␴␪ ϭ 10,000 psi 10,000 psi su ϭ cos2u cos2u (1) PLANE AT ANGLE ␪ ϩ 90Њ suϩ90Њ 5,000 psi ϭ sin2u sin2u Equate (1) and (2): 10,000 psi 5,000 psi ϭ cos2u sin2u tan u ϭ 10,000 5,000 smax ϭ sx ϭ 15,000 psi suϩ90Њ ϭ 5,000 psi 1 2 7,070 (b) MAXIMUM NORMAL AND SHEAR STRESSES ϭ sx sin2u tan2u ϭ x NOTE: All stresses have units of psi. suϩ90Њ ϭ sx [cos(u ϩ 90Њ) ] 2 ϭ sx [Ϫsin u] 2 sx ϭ ␪ = 35.26° 7,070 ␴␪ ϭ ␴x cos2␪ sx ϭ 10,000 y (a) ANGLE ␪ AND SHEAR STRESS ␶␪ 1 ͙2 u ϭ 35.26Њ (2) tmax ϭ sx ϭ 7,500 psi 2 133
    • 134 CHAPTER 2 Axially Loaded Members Problem 2.6-16 A prismatic bar is subjected to an axial force that produces a tensile stress ␴␪ ϭ 63 MPa and a shear stress ␶␪ ϭ Ϫ21 MPa on a certain inclined plane (see figure). Determine the stresses acting on all faces of a stress element oriented at ␪ ϭ 30° and show the stresses on a sketch of the element. Solution 2.6-16 63 MPa ␪ 21 MPa Bar in uniaxial stress STRESS ELEMENT AT ␪ ϭ 30Њ su ϭ sx cos2u ϭ (70 MPa)(cos 30Њ) 2 ␪ ϭ 52.5 MPa tu ϭ Ϫsx sin u cos u ϭ (Ϫ70 MPa)(sin 30Њ)(cos 30Њ) ϭ Ϫ30.31 MPa ␴␪ ϭ 63 MPa Plane at ␪ ϭ 30Њ ϩ 90Њ ϭ 120Њ ␶␪ ϭ Ϫ 21 MPa su ϭ (70 MPa)(cos 120Њ) 2 ϭ 17.5 MPa INCLINED PLANE AT ANGLE ␪ tu ϭ ( Ϫ 70 MPa)(sin 120Њ)(cos 120Њ) ␴␪ ϭ ␴xcos2␪ 63 MPa ϭ ϭ 30.31 MPa ␴xcos2␪ 17.5 63 MPa sx ϭ cos2u (1) 30° y tu ϭ Ϫsx sin u cos u 30.31 Ϫ21 MPa ϭ Ϫsx sin u cos u 21 MPa sx ϭ sin u cos u 52.5 0 x 30.31 (2) 52.5 Equate (1) and (2): 63 MPa 21 MPa ϭ sin u cos u cos2u or tan u ϭ 21 1 ϭ 63 3 u ϭ 18.43Њ From (1) or (2): ␴xϭ70.0 MPa (tension) NOTE: All stresses have units of MPa.
    • SECTION 2.6 Problem 2.6-17 The normal stress on plane pq of a prismatic bar in tension (see figure) is found to be 7500 psi. On plane rs, which makes an angle ␤ ϭ 30° with plane pq, the stress is found to be 2500 psi. Determine the maximum normal stress ␴max and maximum shear stress ␶max in the bar. Stresses on Inclined Sections p r ␤ P P s q Solution 2.6-17 Bar in tension p r ␤ P P s q 7500 psi cos u1 ϭ ϭ ͙3 ϭ 1.7321 cos(u1 ϩ 30Њ) B 2500 psi Eq. (2-29a): SUBSTITUTE NUMERICAL VALUES INTO EQ. (2): ␴␪ ϭ ␴xcos2␪ ␤ ϭ 30Њ PLANE pq: ␴1 ϭ ␴xcos2␪1 PLANE rs: ␴2 ϭ ␴xcos2(␪1 ␴1 ϭ 7500 psi ϩ ␤) ␴2 ϭ 2500 psi Equate ␴x from ␴1 and ␴2: s1 s2 sx ϭ ϭ 2 2 cos u1 cos (u1 ϩ b) or cos2 u1 s1 ϭ cos2 (u1 ϩ b) s2 cos u1 s1 ϭ cos(u1 ϩ b) B s2 (Eq. 1) 135 Solve by iteration or a computer program: ␪1 ϭ 30Њ MAXIMUM NORMAL STRESS (FROM EQ. 1) smax ϭ sx ϭ 7500 psi s1 ϭ cos2u1 cos230Њ ϭ 10,000 psi MAXIMUM SHEAR STRESS (Eq. 2) tmax ϭ sx ϭ 5,000 psi 2
    • 136 CHAPTER 2 Axially Loaded Members Problem 2.6-18 A tension member is to be constructed of two pieces of plastic glued along plane pq (see figure). For purposes of cutting and gluing, the angle ␪ must be between 25° and 45°. The allowable stresses on the glued joint in tension and shear are 5.0 MPa and 3.0 MPa, P respectively. q (a) Determine the angle ␪ so that the bar will carry the largest load P. (Assume that the strength of the glued joint controls the design.) (b) Determine the maximum allowable load Pmax if the cross-sectional area of the bar is 225 mm2. Solution 2.6-18 ␪ p Bar in tension with glued joint ␪ p P P q 25Њ Ͻ ␪ Ͻ 45Њ (a) DETERMINE ANGLE ⍜ FOR LARGEST LOAD A ϭ 225 Point A gives the largest value of ␴x and hence the largest load. To determine the angle ␪ corresponding to point A, we equate Eqs. (1) and (2). mm2 On glued joint: ␴allow ϭ 5.0 MPa ␶allow ϭ 3.0 MPa 5.0 MPa 3.0 MPa ϭ sin u cos u cos2u ALLOWABLE STRESS ␴x IN TENSION su ϭ sx cos2u sx ϭ su 5.0 MPa ϭ cos2u cos2u (1) tan u ϭ 3.0 5.0 u ϭ 30.96Њ ␶␪ ϭ Ϫ ␴xsin ␪ cos ␪ (b) DETERMINE THE MAXIMUM LOAD Since the direction of ␶␪ is immaterial, we can write: |␶␪| ϭ ␴xsin ␪ cos ␪ From Eq. (1) or Eq. (2): or sx ϭ ͿtuͿ sin u cos u ϭ 3.0 MPa sin u cos u (2) (MPa) 15 Eq.(1) 10 A Eq.(2) 5 25° 0 15° 45° 30° 45° u 5.0 MPa 3.0 MPa ϭ ϭ 6.80 MPa 2 sin u cos u cos u Pmax ϭ sx A ϭ (6.80 MPa)(225 mm2 ) ϭ 1.53 kN GRAPH OF EQS. (1) AND (2) sx sx ϭ 60° 75° 90° P
    • SECTION 2.7 137 Strain Energy Strain Energy When solving the problems for Section 2.7, assume that the material behaves linearly elastically. Problem 2.7-1 A prismatic bar AD of length L, cross-sectional area A, and modulus of elasticity E is subjected to loads 5P, 3P, and P acting at points B, C, and D, respectively (see figure). Segments AB, BC, and CD have lengths L/6, L/2, and L/3, respectively. 5P A 3P B L — 6 P C L — 2 D L — 3 (a) Obtain a formula for the strain energy U of the bar. (b) Calculate the strain energy if P ϭ 6 k, L ϭ 52 in., A ϭ 2.76 in.2, and the material is aluminum with E ϭ 10.4 ϫ 106 psi. Solution 2.7-1 Bar with three loads 5P A 3P B L — 6 P C D L — 2 L — 3 Pϭ6k (a) STRAIN ENERGY OF THE BAR (EQ. 2-40) L ϭ 52 in. N2Li i Uϭ a 2Ei Ai E ϭ 10.4 ϫ 106 psi A ϭ 2.76 in.2 ϭ INTERNAL AXIAL FORCES NAB ϭ 3P NBC ϭ Ϫ2P NCD ϭ P LENGTHS LAB ϭ L 6 ϭ 1 L L L B(3P) 2 ¢ ≤ ϩ (Ϫ2P) 2 ¢ ≤ ϩ (P) 2 ¢ ≤ R 2EA 6 2 3 P2L 23 23P2L ¢ ≤ϭ 2EA 6 12EA (b) SUBSTITUTE NUMERICAL VALUES: LBC ϭ L 2 LCD ϭ L 3 Uϭ 23(6 k) 2 (52 in.) 12(10.4 ϫ 106 psi)(2.76 in.2 ) ϭ 125 in.-lb
    • 138 CHAPTER 2 Axially Loaded Members Problem 2.7-2 A bar of circular cross section having two different diameters d and 2d is shown in the figure. The length of each segment of the bar is L/2 and the modulus of elasticity of the material is E. 2d (a) Obtain a formula for the strain energy U of the bar due to the load P. (b) Calculate the strain energy if the load P ϭ 27 kN, the length L ϭ 600 mm, the diameter d ϭ 40 mm, and the material is brass with E ϭ 105 GPa. Solution 2.7-2 L — 2 2d L — 2 d L — 2 P ϭ 27 kN Add the strain energies of the two segments of the bar (see Eq. 2-40). d ϭ 40 mm Uϭ a ϭ iϭ1 2 Ei Ai ϭ P (Lր2) 1 1 B␲ ϩ R 2E (2d) 2 ␲ (d 2) 4 4 2 PL 1 1 5P L ϩ ≤ϭ ¢ ␲E 4d2 d2 4␲Ed2 2 L — 2 P (a) STRAIN ENERGY OF THE BAR N2 Li i P Bar with two segments P 2 d P L ϭ 600 mm E ϭ 105 GPa 2 Uϭ 5(27 kN) (600 mm) 4␲(105 GPa)(40 mm) 2 ϭ 1.036 N ؒ m ϭ 1.036 J 2 (b) SUBSTITUTE NUMERICAL VALUES: Problem 2.7-3 A three-story steel column in a building supports roof and floor loads as shown in the figure. The story height H is 10.5 ft, the cross-sectional area A of the column is 15.5 in.2, and the modulus of elasticity E of the steel is 30 ϫ 106 psi. Calculate the strain energy U of the column assuming P1 ϭ 40 k and P2 ϭ P3 ϭ 60 k. P1 P2 P3 H H H
    • SECTION 2.7 Solution 2.7-3 139 Strain Energy Three-story column Upper segment: N1 ϭ ϪP1 P1 Middle segment: N2 ϭ Ϫ(P1 ϩ P2) Lower segment: N3 ϭ Ϫ(P1 ϩ P2 ϩ P3) H P2 STRAIN ENERGY N2Li i Uϭ a 2Ei Ai H P3 ϭ ϭ H H [P2 ϩ (P1 ϩ P2 ) 2 ϩ (P1 ϩ P2 ϩ P3 ) 2 ] 2EA 1 H [Q] 2EA [Q] ϭ (40 k)2 ϩ (100 k)2 ϩ (160 k)2 ϭ 37,200 k2 H ϭ 10.5 ft E ϭ 30 ϫ 106 psi A ϭ 15.5 in.2 P1 ϭ 40 k 2EA ϭ 2(30 ϫ 106 psi)(15.5 in.2) ϭ 930 ϫ 106 lb Uϭ P2 ϭ P3 ϭ 60 k To find the strain energy of the column, add the strain energies of the three segments (see Eq. 2-40). (10.5 ft)(12 in.րft) [37,200 k2 ] 930 ϫ 106 lb ϭ 5040 in.-lb Problem 2.7-4 The bar ABC shown in the figure is loaded by a force P acting at end C and by a force Q acting at the midpoint B. The bar has constant axial rigidity EA. (a) Determine the strain energy U1 of the bar when the force P acts alone (Q ϭ 0). (b) Determine the strain energy U2 when the force Q acts alone (P ϭ 0). (c) Determine the strain energy U3 when the forces P and Q act simultaneously upon the bar. Solution 2.7-4 A P B L/2 (a) FORCE P ACTS ALONE (Q ϭ 0) P2L 2EA (b) FORCE Q ACTS ALONE (P ϭ 0) U2 ϭ A P B L/2 C L/2 Bar with two loads Q U1 ϭ Q Q2 (Lր2) Q2L ϭ 2EA 4EA C (c) FORCES P AND Q ACT SIMULTANEOUSLY Segment BC: UBC ϭ P2 (Lր2) P2L ϭ 2EA 4EA Segment AB: UAB ϭ (P ϩ Q) 2 (Lր2) 2EA L/2 ϭ U3 ϭ UBC ϩ UAB ϭ P2L PQL Q2L ϩ ϩ 4EA 2EA 4EA P2L PQL Q2L ϩ ϩ 2EA 2EA 4EA (Note that U3 is not equal to U1 ϩ U2. In this case, U3 > U1 ϩ U2. However, if Q is reversed in direction, U3 Ͻ U1 ϩ U2. Thus, U3 may be larger or smaller than U1 ϩ U2.)
    • 140 CHAPTER 2 Axially Loaded Members Problem 2.7-5 Determine the strain energy per unit volume (units of psi) and the strain energy per unit weight (units of in.) that can be stored in each of the materials listed in the accompanying table, assuming that the material is stressed to the proportional limit. DATA FOR PROBLEM 2.7-5 Weight density (lb/in.3) Mild steel Tool steel Aluminum Rubber (soft) Proportional limit (psi) 0.284 0.284 0.0984 0.0405 Material Modulus of elasticity (ksi) 30,000 30,000 10,500 0.300 36,000 75,000 60,000 300 Solution 2.7-5 Strain-energy density DATA: STRAIN ENERGY PER UNIT WEIGHT Material Weight density (lb/in.3) Modulus of elasticity (ksi) Proportional limit (psi) Mild steel 0.284 30,000 36,000 Uϭ P2L 2EA Weight W ϭ ␥AL ␥ ϭ weight density U s2 ϭ W 2gE Tool steel 0.284 30,000 75,000 uW ϭ Aluminum 0.0984 10,500 60,000 At the proportional limit: Rubber (soft) 0.0405 0.300 300 STRAIN ENERGY PER UNIT VOLUME uW ϭ s2 PL 2gE RESULTS 2 Uϭ PL 2EA uR (psi) Volume V ϭ AL Stress s ϭ P A 94 330 171 1740 Rubber (soft) (Eq. 1) 76 Aluminum u ϭ uR ϭ modulus of resistance 22 Tool steel At the proportional limit: s2 PL 2E uW (in.) Mild steel U s2 uϭ ϭ V 2E uR ϭ (Eq. 2) 150 3700
    • SECTION 2.7 Problem 2.7-6 The truss ABC shown in the figure is subjected to a horizontal load P at joint B. The two bars are identical with crosssectional area A and modulus of elasticity E. P B (a) Determine the strain energy U of the truss if the angle ␤ ϭ 60°. (b) Determine the horizontal displacement ␦B of joint B by equating the strain energy of the truss to the work done by the load. Strain Energy ␤ ␤ A C L Solution 2.7-6 Truss subjected to a load P P B ␤ ␤ A C L ␤ ϭ 60Њ Axial forces: NAB ϭ P (tension) LAB ϭ LBC ϭ L NBC ϭ ϪP (compression) sin b ϭ ͙3ր2 (a) STRAIN ENERGY OF TRUSS (EQ. 2-40) cos ␤ ϭ 1/2 N2Li (NAB ) 2L (NBC ) 2L i Uϭ a ϭ ϩ 2EiAi 2EA 2EA FREE-BODY DIAGRAM OF JOINT B B ␤ P ␤ FAB ©Fvert ϭ 0 ϭ (b) HORIZONTAL DISPLACEMENT OF JOINT B (EQ. 2-42) ␦B ϭ FBC ↑ϩ ↓Ϫ ϪFAB sin ␤ ϩ FBC sin ␤ ϭ 0 FAB ϭ FBC (Eq. 1) ϩ Ϫ ©Fhoriz ϭ 0 → ← ϪFAB cos ␤ Ϫ FBC cos ␤ ϩ P ϭ 0 FAB ϭ FBC ϭ P2L EA P P ϭ ϭP 2 cos b 2(1ր2) (Eq. 2) 2U 2 P2L 2PL ϭ ¢ ≤ϭ P P EA EA 141
    • 142 CHAPTER 2 Axially Loaded Members A Problem 2.7-7 The truss ABC shown in the figure supports a horizontal load P1 ϭ 300 lb and a vertical load P2 ϭ 900 lb. Both bars have cross-sectional area A ϭ 2.4 in.2 and are made of steel with E ϭ 30 ϫ 106 psi. (a) Determine the strain energy U1 of the truss when the load P1 acts alone (P2 ϭ0). (b) Determine the strain energy U2 when the load P2 acts alone (P1ϭ0). (c) Determine the strain energy U3 when both loads act simultaneously. 30° C B P1 = 300 lb P2 = 900 lb 60 in. Solution 2.7-7 Truss with two loads Force P1 alone A FAB 30° C B P1 LBC 0 FBC LAB 300 lb P2 alone P1 and P2 1800 lb 1800 lb Ϫ1558.8 lb Ϫ1258.8 lb (a) LOAD P1 ACTS ALONE U1 ϭ P2 (FBC ) 2LBC (300 lb) 2 (60 in.) ϭ 2EA 144 ϫ 106 lb ϭ 0.0375 in.-lb P1 ϭ 300 lb (b) LOAD P2 ACTS ALONE P2 ϭ 900 lb U2 ϭ A ϭ 2.4 in.2 E ϭ 30 ϫ 106 psi LBC ϭ 60 in. ϭ ␤ ϭ 30Њ 1 sin ␤ ϭ sin 30Њ ϭ 2 cos b ϭ cos 30Њ ϭ ͙3 2 LBC 120 LAB ϭ ϭ in. ϭ 69.282 in. cos 30Њ ͙3 2EA ϭ 2(30 ϫ 106 psi)(2.4 in.2) ϭ 144 ϫ 106 lb FORCES FAB AND FBC IN THE BARS From equilibrium of joint B: FAB ϭ 2P2 ϭ 1800 lb FBC ϭ P1 Ϫ P2 ͙3 ϭ 300 lb Ϫ 1558.8 lb ϭ 1 B (FAB ) 2LAB ϩ (FBC ) 2LBC R 2EA 1 B (1800 lb) 2 (69.282 in.) 2EA ϩ ( Ϫ 1558.8 lb) 2 (60 in.) R 370.265 ϫ 106 lb2-in. ϭ 2.57 in.-lb 144 ϫ 106 lb (c) LOADS P1 AND P2 ACT SIMULTANEOUSLY U3 ϭ ϭ 1 B (FAB ) 2LAB ϩ (FBC ) 2LBC R 2EA 1 B (1800 lb) 2 (69.282 in.) 2EA ϩ ( Ϫ 1258.8 lb) 2 (60 in.) R 319.548 ϫ 106 lb2-in. 144 ϫ 106 lb ϭ 2.22 in.-lb ϭ NOTE: The strain energy U3 is not equal to U1 ϩ U2.
    • SECTION 2.7 Problem 2.7-8 The statically indeterminate structure shown in the figure consists of a horizontal rigid bar AB supported by five equally spaced springs. Springs 1, 2, and 3 have stiffnesses 3k, 1.5k, and k, respectively. When unstressed, the lower ends of all five springs lie along a horizontal line. Bar AB, which has weight W, causes the springs to elongate by an amount ␦. 1 (a) Obtain a formula for the total strain energy U of the springs in terms of the downward displacement ␦ of the bar. (b) Obtain a formula for the displacement ␦ by equating the strain energy of the springs to the work done by the weight W. (c) Determine the forces F1, F2, and F3 in the springs. (d) Evaluate the strain energy U, the displacement ␦, and the forces in the springs if W ϭ 600 N and k ϭ 7.5 N/mm. k 1.5k 3k 1.5k 3 2 2 A 2 3 W 2 1 W k1 ϭ 3k (c) FORCES IN THE SPRINGS k2 ϭ 1.5k F1 ϭ 3k␦ ϭ k3 ϭ k ␦ ϭ downward displacement of rigid bar For a spring: U ϭ k␦2 2 Uϭ2¢ 3k␦ 2 ϭ 5k␦2 ≤ϩ2¢ 1.5k␦ 2 2 ≤ϩ 2 k␦ 2 3W 20 W 10 W ϭ 600 N k ϭ 7.5 N/mm ϭ 7500 N/mm U ϭ 5k␦2 ϭ 5k ¢ W␦ 2 W ϭ 8.0 mm 10k F1 ϭ 3W ϭ 180 N 10 F2 ϭ 3W ϭ 90 N 20 F3 ϭ W ϭ 60 N 10 Strain energy of the springs equals 5k␦ 2 W ␦ϭ 10k W 2 W2 ≤ ϭ 10k 20k ϭ 2.4 N ؒ m ϭ 2.4 J ␦ϭ Work done by the weight W equals and F2 ϭ 1.5k␦ ϭ (d) NUMERICAL VALUES (b) DISPLACEMENT ␦ W␦ ∴ ϭ 5k␦2 2 3W 10 Eq. (2-38b) (a) STRAIN ENERGY U OF ALL SPRINGS 2 F3 ϭ k␦ ϭ 1 3k B Solution 2.7-8 Rigid bar supported by springs 1 143 Strain Energy NOTE: W ϭ 2F1 ϩ 2F2 ϩ F3 ϭ 600 N (Check)
    • 144 CHAPTER 2 Axially Loaded Members Problem 2.7-9 A slightly tapered bar AB of rectangular cross section and length L is acted upon by a force P (see figure). The width of the bar varies uniformly from b2 at end A to b1 at end B. The thickness t is constant. A B b1 b2 (a) Determine the strain energy U of the bar. (b) Determine the elongation ␦ of the bar by equating the strain energy to the work done by the force P. L Solution 2.7-9 Tapered bar of rectangular cross section A B b2 b(x) x b1 P dx L (b2 Ϫ b1 )x b(x) ϭ b2 Ϫ L Apply this integration formula to Eq. (1): A(x) ϭ tb(x) Uϭ ϭ t B b2 Ϫ (b2 Ϫ b1 )x R L ϭ (a) STRAIN ENERGY OF THE BAR Uϭ ϭ Ύ [N(x) ] 2dx 2E A(x) Ύ L 0 (Eq. 2-41) P2dx P2 ϭ 2Et b(x) 2Et From Appendix C: Uϭ L Ύ b Ϫ (b Ϫ b ) 0 dx 2 2 x 1 L Ύ a ϩ bx ϭ b ln (a ϩ bx) dx 1 (1) L (b2 Ϫ b1 )x P2 1 B ln B b2 Ϫ RR 1 2Et Ϫ(b2 Ϫ b1 )( L ) L 0 P2 ϪL ϪL B ln b1 Ϫ ln b2 R 2Et (b2 Ϫ b1 ) (b2 Ϫ b1 ) b2 P2L ln 2Et(b2 Ϫ b1 ) b1 (b) ELONGATION OF THE BAR (EQ. 2-42) ␦ϭ b2 2U PL ϭ ln P Et(b2 Ϫ b1 ) b1 NOTE: This result agrees with the formula derived in Prob. 2.3-11. Problem 2.7-10 A compressive load P is transmitted through a rigid plate to three magnesium-alloy bars that are identical except that initially the middle bar is slightly shorter than the other bars (see figure). The dimensions and properties of the assembly are as follows: length L ϭ 1.0 m, cross-sectional area of each bar A ϭ 3000 mm2, modulus of elasticity E ϭ 45 GPa, and the gap s ϭ 1.0 mm. (a) Calculate the load P1 required to close the gap. (b) Calculate the downward displacement ␦ of the rigid plate when P ϭ 400 kN. (c) Calculate the total strain energy U of the three bars when P ϭ 400 kN. (d) Explain why the strain energy U is not equal to P␦/2. (Hint: Draw a load-displacement diagram.) P s L P
    • SECTION 2.7 Solution 2.7-10 Strain Energy Three bars in compression P (c) STRAIN ENERGY U FOR P ϭ 400 kN s = 1.0 mm EA␦2 Uϭ a 2L L Outer bars: ␦ ϭ 1.321 mm Middle bar: ␦ ϭ 1.321 mm Ϫ s ϭ 0.321 mm Uϭ s ϭ 1.0 mm L ϭ 1.0 m EA [2(1.321 mm) 2 ϩ (0.321 mm) 2 ] 2L For each bar: 1 ϭ (135 ϫ 106 Nրm)(3.593 mm2 ) 2 A ϭ 3000 mm2 ϭ 243 N ؒ m ϭ 243 J E ϭ 45 GPa EA ϭ 135 ϫ 106 Nրm L (a) LOAD P1 REQUIRED TO CLOSE THE GAP In general, ␦ ϭ PL EA␦ and P ϭ EA L For two bars, we obtain: P1 ϭ 2 ¢ EAs ≤ ϭ 2(135 ϫ 106 Nրm)(1.0 mm) L P1 ϭ 270 kN (b) DISPLACEMENT ␦ FOR P ϭ 400 kN Since P Ͼ P1, all three bars are compressed. The force P equals P1 plus the additional force required to compress all three bars by the amount ␦ Ϫ s. EA P ϭ P1 ϩ 3¢ ≤ (␦ Ϫ s) L or 400 kN ϭ 270 kN ϩ 3(135 ϫ 106 N/m)(␦ Ϫ 0.001 m) Solving, we get ␦ ϭ 1.321 mm (d) LOAD-DISPLACEMENT DIAGRAM U ϭ 243 J ϭ 243 N . m P␦ 1 ϭ (400 kN)(1.321 mm) ϭ 264 N ؒ m 2 2 P␦ because the 2 load-displacement relation is not linear. The strain energy U is not equal to B 400 kN 400 270 kN 300 A Load P (kN) 200 ␦ = 1.0 mm ␦ = 1.321 mm 100 0 0.5 1.0 1.5 Displacement ␦ (mm) 2.0 U ϭ area under line OAB. P␦ ϭ area under a straight line from O to B, 2 which is larger than U. 145
    • 146 CHAPTER 2 Axially Loaded Members Problem 2.7-11 A block B is pushed against three springs by a force P (see figure). The middle spring has stiffness k1 and the outer springs each have stiffness k2. Initially, the springs are unstressed and the middle spring is longer than the outer springs (the difference in length is denoted s). s k2 P k1 B k2 (a) Draw a force-displacement diagram with the force P as ordinate and the displacement x of the block as abscissa. (b) From the diagram, determine the strain energy U1 of the springs when x ϭ 2s. (c) Explain why the strain energy U1 is not equal to P␦/2, where ␦ ϭ 2s. Solution 2.7-11 x Block pushed against three springs s k2 P k1 B k2 x (b) STRAIN ENERGY U1 WHEN x ϭ 2s Force P0 required to close the gap: P0 ϭ k1s (1) U1 ϭ Area below force-displacement curve + + FORCE-DISPLACEMENT RELATION BEFORE GAP IS CLOSED = P ϭ k1x 1 1 1 ϭ P0 s ϩ P0 s ϩ (P1 Ϫ P0 )s ϭ P0 s ϩ P1 s 2 2 2 (0 Յ x Յ s)(0 Յ P Յ P0) (2) FORCE-DISPLACEMENT RELATION AFTER GAP IS CLOSED All three springs are compressed. Total stiffness equals k1 ϩ 2k2. Additional displacement equals x Ϫ s. Force P equals P0 plus the force required to compress all three springs by the amount x Ϫ s. U1 ϭ (2k1 ϩ k2 )s2 (c) STRAIN ENERGY U1 IS NOT EQUAL TO P ϭ P0 ϩ (k1 ϩ 2k2 )(x Ϫ s) For ␦ ϭ 2s: ϭ k1s ϩ (k1 ϩ 2k2 )x Ϫ k1s Ϫ 2k2s (x Ն s); (P Ն P0) P ϭ (k1 ϩ 2k2 )x Ϫ 2k2s (3) P1 ϭ force P when x ϭ 2s Substitute x ϭ 2s into Eq. (3): P1 ϭ 2(k1 ϩ k2)s (4) B P␦ is not equal to the strain energy because 2 the force-displacement relation is not linear. A Displacement x 0 s Slope = k1 2s P␦ 1 ϭ P1 (2 s) ϭ P1s ϭ 2(k1 ϩ k2 )s2 2 2 P␦ ϭ area under a straight line from O to B, which 2 is larger than U1. Slope = k1 + 2k2 P0 P␦ 2 U1 ϭ area under line OAB. Eq (3) Eq (2) (5) (This quantity is greater than U1.) Thus, (a) FORCE-DISPLACEMENT DIAGRAM Force P P1 ϭ k1s2 ϩ (k1 ϩ k2 )s2
    • SECTION 2.7 Problem 2.7-12 A bungee cord that behaves linearly elastically has an unstressed length L0 ϭ 760 mm and a stiffness k ϭ 140 N/m. The cord is attached to two pegs, distance b ϭ 380 mm apart, and pulled at its midpoint by a force P ϭ 80 N (see figure). A b (a) How much strain energy U is stored in the cord? (b) What is the displacement ␦C of the point where the load is applied? (c) Compare the strain energy U with the quantity P␦C/2. (Note: The elongation of the cord is not small compared to its original length.) 147 Strain Energy B C P Solution 2.7-12 Bungee cord subjected to a load P. DIMENSIONS BEFORE THE LOAD P IS APPLIED L1 b 2 ϭ ¢ ≤ ϩ x2 2 B 2 A From triangle ACD: L0 2 d b D C L1 ϭ ͙b2 ϩ 4x2 L0 2 (3) EQUILIBRIUM AT POINT C Let F ϭ tensile force in bungee cord B L0 ϭ 760 mm (2) L0 ϭ 380 mm 2 b ϭ 380 mm F F C Bungee cord: P = 80 N k ϭ 140 N/m L0 = 760 mm L1ր2 F ϭ x Pր2 (1) Fϭ¢ ϭ DIMENSIONS AFTER THE LOAD P IS APPLIED C P/2 F From triangle ACD: 1 d ϭ ͙L2 Ϫ b2 ϭ 329.09 mm 0 2 P P L1 1 ≤¢ ≤¢ ≤ 2 2 x P b 2 1ϩ¢ ≤ 2B 2x (4) ELONGATION OF BUNGEE CORD Let ␦ ϭ elongation of the entire bungee cord A L1 2 C x b D P ϭ 80 N P L1 2 ␦ϭ F P b2 ϭ 1ϩ 2 k 2kB 4x (5) Final length of bungee cord ϭ original length ϩ ␦ L1 ϭ L0 ϩ ␦ ϭ L0 ϩ P b2 1ϩ 2 2kB 4x (6) B Let x ϭ distance CD Let L1 ϭ stretched length of bungee cord (Continued)
    • 148 CHAPTER 2 Axially Loaded Members SOLUTION OF EQUATIONS From Eq. (5): Combine Eqs. (6) and (3): ␦ϭ L1 ϭ L0 ϩ or P b2 1 ϩ 2 ϭ ͙b2 ϩ 4x2 2kB 4x L1 ϭ L0 ϩ L0 ϭ ¢ 1 Ϫ 1 U ϭ (140 Nրm)(305.81 mm) 2 ϭ 6.55 N ؒ m 2 U ϭ 6.55 J P ͙b2 ϩ 4x2 ϭ ͙b2 ϩ 4x2 4kx P ≤ ͙b2 ϩ 4x2 4kx (7) ϭ 168.8 mm SUBSTITUTE NUMERICAL VALUES INTO EQ. (7): (80 N)(1000 mmրm) 760 mm ϭ B 1 Ϫ R 4(140 Nրm)x U ϭ 6.55 J ϫ ͙(380 mm) 2 ϩ 4x2 142.857 ≤ ͙144,400 ϩ 4x2 x (8) (9) (9) P␦C 1 ϭ (80 N)(168.8 mm) ϭ 6.75 J 2 2 The two quantities are not the same. The work done by the load P is not equal to P␦C /2 because the loaddisplacement relation (see below) is non-linear when the displacements are large. (The work done by the load P is equal to the strain energy because the bungee cord behaves elastically and there are no energy losses.) Solve for x (Use trial & error or a computer program): x ϭ 497.88 mm (a) STRAIN ENERGY U OF THE BUNGEE CORD k ϭ 140 Nրm (c) COMPARISON OF STRAIN ENERGY U WITH THE P␦C /2 QUANTITY Units: x is in millimeters k␦2 Uϭ 2 (b) DISPLACEMENT ␦C OF POINT C ␦C ϭ x Ϫ d ϭ 497.88 mm Ϫ 329.09 mm This equation can be solved for x. 760 ϭ ¢ 1 Ϫ P b2 1 ϩ 2 ϭ 305.81 mm 2kB 4x U ϭ area OAB under the curve OA. P␦C ϭ area of triangle OAB, which is greater 2 than U. P ϭ 80 N Load P Large displacements 80 N 0 A B ␦C Small displacements Displacement
    • SECTION 2.8 Impact Loading Impact Loading The problems for Section 2.8 are to be solved on the basis of the assumptions and idealizations described in the text. In particular, assume that the material behaves linearly elastically and no energy is lost during the impact. Collar Problem 2.8-1 A sliding collar of weight W ϭ 150 lb falls from a height h ϭ 2.0 in. onto a flange at the bottom of a slender vertical rod (see figure). The rod has length L ϭ 4.0 ft, cross-sectional area A ϭ 0.75 in.2, and modulus of elasticity E ϭ 30 ϫ 106 psi. L Rod Calculate the following quantities: (a) the maximum downward displacement of the flange, (b) the maximum tensile stress in the rod, and (c) the impact factor. h Flange Probs. 2.8-1, 2.8-2, and 2.8-3 Solution 2.8-1 Collar falling onto a flange (a) DOWNWARD DISPLACEMENT OF FLANGE ␦st ϭ WL ϭ 0.00032 in. EA Eq. of (2-53): W L h Flange ␦max ϭ ␦st B 1 ϩ ¢ 1 ϩ ϭ 0.0361 in. 2h 1ր2 ≤ R ␦st (b) MAXIMUM TENSILE STRESS (EQ. 2-55) smax ϭ W ϭ 150 lb E␦max ϭ 22,600 psi L (c) IMPACT FACTOR (EQ. 2-61) h ϭ 2.0 in. L ϭ 4.0 ft ϭ 48 in. E ϭ 30 ϫ 106 psi A ϭ 0.75 in.2 Impact factor ϭ ␦max 0.0361 in. ϭ ␦st 0.00032 in. ϭ 113 149
    • 150 CHAPTER 2 Axially Loaded Members Problem 2.8-2 Solve the preceding problem if the collar has mass M ϭ 80 kg, the height h ϭ 0.5 m, the length L ϭ 3.0 m, the cross-sectional area A ϭ 350 mm2, and the modulus of elasticity E ϭ 170 GPa. Solution 2.8-2 Collar falling onto a flange (a) DOWNWARD DISPLACEMENT OF FLANGE ␦st ϭ W WL ϭ 0.03957 mm EA Eq. (2-53): L ␦max ϭ ␦st B 1 ϩ ¢ 1 ϩ ϭ 6.33 mm h Flange 2h 1ր2 ≤ R ␦st (b) MAXIMUM TENSILE STRESS (EQ. 2-55) smax ϭ M ϭ 80 kg W ϭ Mg ϭ E␦max ϭ 359 MPa L (c) IMPACT FACTOR (EQ. 2-61) (80 kg)(9.81 m/s2) Impact factor ϭ ϭ 784.8 N ␦max 6.33 mm ϭ ␦st 0.03957 mm ϭ 160 h ϭ 0.5 m L ϭ 3.0 m E ϭ 170 GPa A ϭ 350 mm2 Problem 2.8-3 Solve Problem 2.8-1 if the collar has weight W ϭ 50 lb, the height h ϭ 2.0 in., the length L ϭ 3.0 ft, the cross-sectional area A ϭ 0.25 in.2, and the modulus of elasticity E ϭ 30,000 ksi. Solution 2.8-3 Collar falling onto a flange (a) DOWNWARD DISPLACEMENT OF FLANGE ␦st ϭ WL ϭ 0.00024 in. EA Eq. (2-53): W ϭ 0.0312 in. L h Flange ␦max ϭ ␦st B 1 ϩ ¢ 1 ϩ (b) MAXIMUM TENSILE STRESS (EQ. 2-55) smax ϭ E␦max ϭ 26,000 psi L (c) IMPACT FACTOR (EQ. 2-61) W ϭ 50 lb h ϭ 2.0 in. L ϭ 3.0 ft ϭ 36 in. E ϭ 30,000 psi A ϭ 0.25 in.2 2h 1ր2 ≤ R ␦st Impact factor ϭ ␦max 0.0312 in. ϭ ␦st 0.00024 in. ϭ 130
    • SECTION 2.8 Problem 2.8-4 A block weighing W ϭ 5.0 N drops inside a cylinder from a height h ϭ 200 mm onto a spring having stiffness k ϭ 90 N/m (see figure). (a) Determine the maximum shortening of the spring due to the impact, and (b) determine the impact factor. Block Cylinder k Prob. 2.8-4 and 2.8-5 Solution 2.8-4 Block dropping onto a spring W h k W ϭ 5.0 N h ϭ 200 mm k ϭ 90 N/m (a) MAXIMUM SHORTENING OF THE SPRING ␦st ϭ W 5.0 N ϭ ϭ 55.56 mm k 90 Nրm Eq. (2-53): ␦max ϭ ␦st B 1 ϩ ¢ 1 ϩ ϭ 215 mm (b) IMPACT FACTOR (EQ. 2-61) Impact factor ϭ ␦max 215 mm ϭ ␦st 55.56 mm ϭ 3.9 2h 1ր2 ≤ R ␦st Impact Loading h 151
    • 152 CHAPTER 2 Axially Loaded Members Problem 2.8-5 Solve the preceding problem if the block weighs W ϭ 1.0 lb, h ϭ 12 in., and k ϭ 0.5 lb/in. Solution 2.8-5 Block dropping onto a spring (a) MAXIMUM SHORTENING OF THE SPRING ␦st ϭ W h W 1.0 lb ϭ ϭ 2.0 in. k 0.5 lbրin. Eq. (2-53): ␦max ϭ ␦st B 1 ϩ ¢ 1 ϩ ϭ 9.21 in. k (b) IMPACT FACTOR (EQ. 2-61) Impact factor ϭ W ϭ 1.0 lb h ϭ 12 in. 2h 1ր2 ≤ R ␦st ␦max 9.21 in. ϭ ␦st 2.0 in. ϭ 4.6 k ϭ 0.5 lb/in. Problem 2.8-6 A small rubber ball (weight W ϭ 450 mN) is attached by a rubber cord to a wood paddle (see figure). The natural length of the cord is L0 ϭ 200 mm, its cross-sectional area is A ϭ 1.6 mm2, and its modulus of elasticity is E ϭ 2.0 MPa. After being struck by the paddle, the ball stretches the cord to a total length L1 ϭ 900 mm. What was the velocity v of the ball when it left the paddle? (Assume linearly elastic behavior of the rubber cord, and disregard the potential energy due to any change in elevation of the ball.) Solution 2.8-6 Rubber ball attached to a paddle WHEN THE RUBBER CORD IS FULLY STRETCHED: Uϭ EA␦2 EA ϭ (L Ϫ L0 ) 2 2L0 2L0 1 CONSERVATION OF ENERGY KE ϭ U g ϭ 9.81 m/s2 Aϭ E ϭ 2.0 MPa 1.6 mm2 L0 ϭ 200 mm L1 ϭ 900 mm v2 ϭ Wv2 EA ϭ (L Ϫ L0 ) 2 2g 2L0 1 gEA B WL0 gEA (L Ϫ L0 ) 2 WL0 1 W ϭ 450 mN v ϭ (L1 Ϫ L0 ) (9.81 mրs2 )(2.0 MPa)(1.6 mm2 ) B (450 mN)(200 mm) SUBSTITUTE NUMERICAL VALUES: WHEN THE BALL LEAVES THE PADDLE Wv2 KE ϭ 2g v ϭ (700 mm) ϭ 13.1 mրs
    • SECTION 2.8 Impact Loading Problem 2.8-7 A weight W ϭ 4500 lb falls from a height h onto a vertical wood pole having length L ϭ 15 ft, diameter d ϭ 12 in., and modulus of elasticity E ϭ 1.6 ϫ 106 psi (see figure). If the allowable stress in the wood under an impact load is 2500 psi, what is the maximum permissible height h? W = 4,500 lb h d = 12 in. L = 15 ft Solution 2.8-7 Weight falling on a wood pole STATIC STRESS W h d sst ϭ W 4500 lb ϭ ϭ 39.79 psi A 113.10 in.2 MAXIMUM HEIGHT hmax Eq. (2-59): L or smax ϭ sst B 1 ϩ ¢ 1 ϩ smax 2hE 1ր2 Ϫ 1 ϭ ¢1 ϩ ≤ sst Lsst 2hE 1ր2 ≤ R Lsst Square both sides and solve for h: h ϭ hmax ϭ W ϭ 4500 lb d ϭ 12 in. L ϭ 15 ft ϭ 180 in. Aϭ ␲d2 ϭ 113.10 in.2 4 E ϭ 1.6 ϫ 106 psi ␴allow ϭ 2500 psi (ϭ ␴max) Find hmax Lsmax smax ¢ Ϫ 2≤ sst 2E SUBSTITUTE NUMERICAL VALUES: hmax ϭ (180 in.)(2500 psi) 2500 psi ¢ Ϫ 2≤ 39.79 psi 2(1.6 ϫ 106 psi) ϭ 8.55 in. 153
    • 154 CHAPTER 2 Axially Loaded Members Problem 2.8-8 A cable with a restrainer at the bottom hangs vertically from its upper end (see figure). The cable has an effective cross-sectional area A ϭ 40 mm2 and an effective modulus of elasticity E ϭ 130 GPa. A slider of mass M ϭ 35 kg drops from a height h ϭ 1.0 m onto the restrainer. If the allowable stress in the cable under an impact load is 500 MPa, what is the minimum permissible length L of the cable? Cable Slider L h Restrainer Probs. 2.8-8 and 2.8-9 Solution 2.8-8 Slider on a cable STATIC STRESS sst ϭ W 343.4 N ϭ ϭ 8.585 MPa A 40 mm2 MINIMUM LENGTH Lmin Eq. (2-59): L W h or smax 2hE 1ր2 Ϫ 1 ϭ ¢1 ϩ ≤ sst Lsst Aϭ 40 mm2 h ϭ 1.0 m E ϭ 130 GPa ␴allow ϭ ␴max ϭ 500 MPa 2hE 1ր2 ≤ R Lsst Square both sides and solve for L: L ϭ Lmin ϭ W ϭ Mg ϭ (35 kg)(9.81 m/s2) ϭ 343.4 N smax ϭ sst B 1 ϩ ¢ 1 ϩ 2Ehsst smax (smax Ϫ 2sst ) SUBSTITUTE NUMERICAL VALUES: Lmin ϭ Find minimum length Lmin Problem 2.8-9 Solve the preceding problem if the slider has weight W ϭ 100 lb, h ϭ 45 in., A ϭ 0.080 in.2, E ϭ 21 ϫ 106 psi, and the allowable stress is 70 ksi. 2(130 GPa)(1.0 m)(8.585 MPa) (500 MPa) [500 MPa Ϫ 2(8.585 MPa) ] ϭ 9.25 mm
    • SECTION 2.8 Solution 2.8-9 Slider on a cable MINIMUM LENGTH Lmin smax ϭ sst B 1 ϩ ¢ 1 ϩ Eq. (2-59): or smax 2hE 1ր2 Ϫ 1 ϭ ¢1 ϩ ≤ sst Lsst L ϭ Lmin ϭ W ϭ 100 lb A ϭ 0.080 in.2 E ϭ 21 ϫ 106 psi h h ϭ 45 in ␴allow ϭ ␴max ϭ 70 ksi Find minimum length Lmin 2Ehsst smax (smax Ϫ 2sst ) SUBSTITUTE NUMERICAL VALUES: Lmin ϭ 2(21 ϫ 106 psi)(45 in.)(1250 psi) (70,000 psi) [70,000 psi Ϫ 2(1250 psi) ] ϭ 500 in. STATIC STRESS sst ϭ 2hE 1ր2 ≤ R Lsst Square both sides and solve for L: L W W 100 lb ϭ ϭ 1250 psi A 0.080 in.2 Problem 2.8-10 A bumping post at the end of a track in a railway yard has a spring constant k ϭ 8.0 MN/m (see figure). The maximum possible displacement d of the end of the striking plate is 450 mm. What is the maximum velocity vmax that a railway car of weight W ϭ 545 kN can have without damaging the bumping post when it strikes it? Solution 2.8-10 v k d Bumping post for a railway car STRAIN ENERGY WHEN SPRING IS COMPRESSED TO THE v MAXIMUM ALLOWABLE AMOUNT k d k ϭ 8.0 MN/m W ϭ 545 kN d ϭ maximum displacement of spring d ϭ ␦max ϭ 450 mm Uϭ k␦2 kd2 max ϭ 2 2 CONSERVATION OF ENERGY KE ϭ U k B Wրg Wv2 kd2 ϭ 2g 2 v2 ϭ kd2 Wրg v ϭ vmax ϭ d 8.0 MNրm B (545 kN)ր(9.81 mրs2 ) Find vmax SUBSTITUTE NUMERICAL VALUES: KINETIC ENERGY BEFORE IMPACT vmax ϭ (450 mm) KE ϭ 155 Impact Loading Mv2 Wv2 ϭ 2 2g ϭ 5400 mmրs ϭ 5.4 mրs
    • 156 CHAPTER 2 Axially Loaded Members Problem 2.8-11 A bumper for a mine car is constructed with a spring of stiffness k ϭ 1120 lb/in. (see figure). If a car weighing 3450 lb is traveling at velocity v ϭ 7 mph when it strikes the spring, what is the maximum shortening of the spring? Solution 2.8-11 v k Bumper for a mine car v k k ϭ 1120 lb/in. W ϭ 3450 lb v ϭ 7 mph ϭ 123.2 in./sec g ϭ 32.2 ft/sec2 ϭ 386.4 in./sec2 Find the shortening ␦max of the spring. Conservation of energy KE ϭ U Solve for ␦max: KINETIC ENERGY JUST BEFORE IMPACT Mv2 Wv2 KE ϭ ϭ 2 2g STRAIN ENERGY WHEN SPRING IS FULLY COMPRESSED Uϭ k␦2 max 2 Wv2 k␦2 max ϭ 2g 2 ␦max ϭ Wv2 B gk (3450 lb)(123.2 in.րsec) 2 B (386.4 in.րsec2 )(1120 lbրin.) SUBSTITUTE NUMERICAL VALUES: ␦max ϭ ϭ 11.0 in.
    • SECTION 2.8 Impact Loading Problem 2.8-12 A bungee jumper having a mass of 55 kg leaps from a bridge, braking her fall with a long elastic shock cord having axial rigidity EA ϭ 2.3 kN (see figure). If the jumpoff point is 60 m above the water, and if it is desired to maintain a clearance of 10 m between the jumper and the water, what length L of cord should be used? Solution 2.8-12 Bungee jumper SOLVE QUADRATIC EQUATION FOR ␦max: ␦max ϭ h ϭ WL WL 2 WL 1ր2 ϩ B¢ ≤ ϩ 2L ¢ ≤R EA EA EA WL 2EA 1ր2 B 1 ϩ ¢1 ϩ ≤ R EA W VERTICAL HEIGHT C h ϭ C ϩ L ϩ ␦max hϪCϭLϩ W ϭ Mg ϭ (55 kg)(9.81 m/s2) ϭ 539.55 N SOLVE FOR L: Lϭ EA ϭ 2.3 kN Height: h ϭ 60 m WL 2EA 1ր2 B 1 ϩ ¢1 ϩ ≤ R EA W hϪC W 2EA 1ր2 1ϩ B 1 ϩ ¢1 ϩ ≤ R EA W Clearance: C ϭ 10 m SUBSTITUTE NUMERICAL VALUES: Find length L of the bungee cord. 539.55 N W ϭ ϭ 0.234587 EA 2.3 kN P.E. ϭ Potential energy of the jumper at the top of bridge (with respect to lowest position) ϭ W(L ϩ ␦max) Numerator ϭ h Ϫ C ϭ 60 m Ϫ 10 m ϭ 50 m Denominator ϭ 1 ϩ (0.234587) U ϭ strain energy of cord at lowest position ϭ EA␦2 max 2L ϭ1.9586 Lϭ CONSERVATION OF ENERGY P.E. ϭ U or ␦2 Ϫ max W(L ϩ ␦max ) ϭ EA␦2 max 2L 2WL 2WL2 ␦max Ϫ ϭ0 EA EA 50 m ϭ 25.5 m 1.9586 ϫ B 1 ϩ ¢1 ϩ 1ր2 2 ≤ R 0.234587 157
    • 158 CHAPTER 2 Axially Loaded Members Problem 2.8-13 A weight W rests on top of a wall and is attached to one end of a very flexible cord having cross-sectional area A and modulus of elasticity E (see figure). The other end of the cord is attached securely to the wall. The weight is then pushed off the wall and falls freely the full length of the cord. W W (a) Derive a formula for the impact factor. (b) Evaluate the impact factor if the weight, when hanging statically, elongates the band by 2.5% of its original length. Solution 2.8-13 Weight falling off a wall CONSERVATION OF ENERGY P.E. ϭ U or ␦2 Ϫ max W ϭ Weight W(L ϩ ␦max ) ϭ EA␦2 max 2L 2WL 2WL2 ␦max Ϫ ϭ0 EA EA Properties of elastic cord: SOLVE QUADRATIC EQUATION FOR ␦max: E ϭ modulus of elasticity ␦max ϭ A ϭ cross-sectional area WL WL 2 WL 1ր2 ϩ B¢ ≤ ϩ 2L ¢ ≤R EA EA EA L ϭ original length STATIC ELONGATION ␦max ϭ elongation of elastic cord ␦st ϭ P.E. ϭ potential energy of weight before fall (with respect to lowest position) P.E. ϭ W(L ϩ ␦max) Let U ϭ strain energy of cord at lowest position Uϭ EA␦2 max 2L WL EA IMPACT FACTOR ␦max 2EA 1ր2 ϭ 1 ϩ B1 ϩ R ␦st W NUMERICAL VALUES ␦st ϭ (2.5%)(L) ϭ 0.025L ␦st ϭ WL EA W ϭ 0.025 EA EA ϭ 40 W Impact factor ϭ 1 ϩ [1 ϩ 2(40) ] 1ր2 ϭ 10 Problem 2.8-14 A rigid bar AB having mass M ϭ 1.0 kg and length L ϭ 0.5 m is hinged at end A and supported at end B by a nylon cord BC (see figure). The cord has cross-sectional area A ϭ 30 mm2, length b ϭ 0.25 m, and modulus of elasticity E ϭ 2.1 GPa. If the bar is raised to its maximum height and then released, what is the maximum stress in the cord? C b A B W L
    • SECTION 2.8 Solution 2.8-14 Impact Lading Falling bar AB From line AD : sin 2u ϭ 2h 2h ϭ AD L From Appendix C: sin 2u ϭ 2 sin u cos u C ∴ b A and B W 2h b L 2bL ϭ2¢ ≤¢ ≤ϭ 2 2 2 2 2 L b ϩ L2 ͙b ϩ L ͙b ϩ L hϭ bL2 b2 ϩ L2 (Eq. 1) CONSERVATION OF ENERGY L P.E. ϭ potential energy of raised bar AD ϭ W ¢h ϩ RIGID BAR: W ϭ Mg ϭ (1.0 kg)(9.81 m/s2) ϭ 9.81 N ␦max ≤ 2 U ϭ strain energy of stretched cord ϭ L ϭ 0.5 m P.E. ϭ U NYLON CORD: A ϭ 30 mm2 W ¢h ϩ For the cord: ␦max ϭ b ϭ 0.25 m E ϭ 2.1 GPa ␦max EA␦2 max ≤ϭ 2 2b EA␦2 max 2b (Eq. 2) smaxb E Substitute into Eq. (2) and rearrange: Find maximum stress ␴max in cord BC. GEOMETRY OF BAR AB AND CORD BC s2 Ϫ max s2 Ϫ max C CG ␪ h A ␪ ␦max 2 b B ␦max CG L W 2WL2E smax Ϫ ϭ0 A A(b2 ϩ L2 ) W 8L2EA B1 ϩ 1 ϩ R 2A B W(b2 ϩ L2 ) SOLVE FOR ␴max: smax ϭ SUBSTITUTE NUMERICAL VALUES: smax ϭ 33.3 MPa CD ϭ CB ϭ b AD ϭ AB ϭ L h ϭ height of center of gravity of raised bar AD ␦max ϭ elongation of cord From triangle ABC: sin u ϭ cos u ϭ (Eq. 3) Substitute from Eq. (1) into Eq. (3): D h W 2WhE smax Ϫ ϭ0 A bA b ͙b ϩ L2 L 2 ͙b2 ϩ L2 (Eq. 4) 159
    • 160 CHAPTER 2 Axially Loaded Members Stress Concentrations The problems for Section 2.10 are to be solved by considering the stress-concentration factors and assuming linearly elastic behavior. P P d b Problem 2.10-1 The flat bars shown in parts (a) and (b) of the figure are subjected to tensile forces P ϭ 3.0 k. Each bar has thickness t ϭ 0.25 in. (a) For the bar with a circular hole, determine the maximum stresses for hole diameters d ϭ 1 in. and d ϭ 2 in. if the width b ϭ 6.0 in. (b) For the stepped bar with shoulder fillets, determine the maximum stresses for fillet radii R ϭ 0.25 in. and R ϭ 0.5 in. if the bar widths are b ϭ 4.0 in. and c ϭ 2.5 in. (a) R P c b (b) Probs. 2.10-1 and 2.10-2 Solution 2.10-1 Flat bars in tension R = radius P d b P P P c b (b) (a) P ϭ 3.0 k t ϭ 0.25 in. (a) BAR WITH CIRCULAR HOLE (b ϭ 6 in.) Obtain K from Fig. 2-63 FOR d ϭ 1 in.: c ϭ b Ϫ d ϭ 5 in. P 3.0 k snom ϭ ϭ ϭ 2.40 ksi ct (5 in.)(0.25 in.) 1 d/b ϭ K Ϸ 2.60 6 ␴max ϭ k␴nom Ϸ 6.2 ksi FOR d ϭ 2 in.: c ϭ b Ϫ d ϭ 4 in. P 3.0 k ϭ ϭ 3.00 ksi ct (4 in.)(0.25 in.) 1 d/b ϭ K Ϸ 2.31 3 ␴max ϭ K␴nom Ϸ 6.9 ksi snom ϭ (b) STEPPED BAR WITH SHOULDER FILLETS b ϭ 4.0 in. snom ϭ c ϭ 2.5 in.; Obtain k from Fig. 2-64 P 3.0 k ϭ ϭ 4.80 ksi ct (2.5 in.)(0.25 in.) FOR R ϭ 0.25 in.: R/c ϭ 0.1 b/c ϭ 1.60 k Ϸ 2.30 ␴max ϭ K␴nom Ϸ 11.0 ksi FOR R ϭ 0.5 in.: R/c ϭ 0.2 K Ϸ 1.87 b/c ϭ 1.60 ␴max ϭ K␴nom Ϸ 9.0 ksi P
    • SECTION 2.10 Stress Concentrations Problem 2.10-2 The flat bars shown in parts (a) and (b) of the figure are subjected to tensile forces P ϭ 2.5 kN. Each bar has thickness t ϭ 5.0 mm. (a) For the bar with a circular hole, determine the maximum stresses for hole diameters d ϭ 12 mm and d ϭ 20 mm if the width b ϭ 60 mm. (b) For the stepped bar with shoulder fillets, determine the maximum stresses for fillet radii R ϭ 6 mm and R ϭ 10 mm if the bar widths are b ϭ 60 mm and c ϭ 40 mm. Solution 2.10-2 Flat bars in tension R = radius P d b P P c b P (b) (a) P ϭ 2.5 kN t ϭ 5.0 mm (a) BAR WITH CIRCULAR HOLE (b ϭ 60 mm) Obtain K from Fig. 2-63 FOR d ϭ 12 mm: c ϭ b Ϫ d ϭ 48 mm P 2.5 kN ϭ ϭ 10.42 MPa ct (48 mm)(5 mm) 1 d/b ϭ K Ϸ 2.51 5 snom ϭ ␴max ϭ K␴nom Ϸ 26 MPa FOR d ϭ 20 mm: c ϭ b Ϫ d ϭ 40 mm P 2.5 kN ϭ ϭ 12.50 MPa ct (40 mm)(5 mm) 1 d/b ϭ K Ϸ 2.31 3 ␴max ϭ K␴nom Ϸ 29 MPa snom ϭ (b) STEPPED BAR WITH SHOULDER FILLETS b ϭ 60 mm c ϭ 40 mm; Obtain K from Fig. 2-64 snom ϭ P 2.5 kN ϭ ϭ 12.50 MPa ct (40 mm)(5 mm) FOR R ϭ 6 mm: R/c ϭ 0.15 K Ϸ 2.00 ␴max ϭ K␴nom Ϸ 25 MPa FOR R ϭ 10 mm: R/c ϭ 0.25 K Ϸ 1.75 b/c ϭ 1.5 b/c ϭ 1.5 ␴max ϭ K␴nom Ϸ 22 MPa 161
    • 162 CHAPTER 2 Axially Loaded Members Problem 2.10-3 A flat bar of width b and thickness t has a hole of diameter d drilled through it (see figure). The hole may have any diameter that will fit within the bar. What is the maximum permissible tensile load Pmax if the allowable tensile stress in the material is ␴t? Solution 2.10-3 P b P d Flat bar in tension d b b d t ϭ thickness P* 0 P K 3.00 0.333 0.1 P 2.73 0.330 0.2 2.50 0.320 ␴t ϭ allowable tensile stress 0.3 2.35 0.298 Find Pmax 0.4 2.24 0.268 Find K from Fig. 2-64 Pmax ϭ snomct ϭ ϭ st smax ct ϭ (b Ϫ d)t K K st d bt ¢ 1 Ϫ ≤ K b Because ␴t, b, and t are constants, we write: P* ϭ Pmax 1 d ϭ ¢1 Ϫ ≤ st bt K b Problem 2.10-4 A round brass bar of diameter d1 ϭ 20 mm has upset ends of diameter d2 ϭ 26 mm (see figure). The lengths of the segments of the bar are L1 ϭ 0.3 m and L2 ϭ 0.1 m. Quarter-circular fillets are used at the shoulders of the bar, and the modulus of elasticity of the brass is E ϭ 100 GPa. If the bar lengthens by 0.12 mm under a tensile load P, what is the maximum stress ␴max in the bar? We observe that Pmax decreases as d/b increases. Therefore, the maximum load occurs when the hole becomes very small. ( d S0 b Pmax ϭ P and K S 3) st bt 3 d2 L2 d2 d1 L1 Probs. 2.10-4 and 2.10-5 L2 P
    • SECTION 2.10 Solution 2.10-4 Round brass bar with upset ends d2 = 26 mm P d1 = 20 mm L1 L2 P ϭ ␦ ϭ 0.12 mm ␦EA2 P ␦E ϭ ϭ A1 A1 2L2 A1 ϩ L1 A2 2L2 ( A2 ) ϩ L1 ␦E d 2L2 ( d1 ) 2 2 ϩ L1 SUBSTITUTE NUMERICAL VALUES: L2 ϭ 0.1 m snom ϭ L1 ϭ 0.3 m R ϭ radius of fillets ϭ PL2 PL1 ≤ϩ EA2 EA1 Solve for P: Use Fig. 2-65 for the stress-concentration factor: snom ϭ L2 E ϭ 100 GPa ␦ϭ2¢ Stress Concentrations 26 mm Ϫ 20 mm ϭ 3 mm 2 (0.12 mm)(100 GPa) 2(0.1 m)( 20 ) 2 ϩ 0.3 m 26 ϭ 28.68 MPa R 3 mm ϭ ϭ 0.15 D1 20 mm Use the dashed curve in Fig. 2-65. K Ϸ 1.6 smax ϭ Ksnom Ϸ (1.6)(28.68 MPa) ␦EA1 A2 Pϭ 2L2 A1 ϩ L1 A2 Ϸ 46 MPa Problem 2.10-5 Solve the preceding problem for a bar of monel metal having the following properties: d1 ϭ 1.0 in., d2 ϭ 1.4 in., L1 ϭ 20.0 in., L2 ϭ 5.0 in., and E ϭ 25 ϫ 106 psi. Also, the bar lengthens by 0.0040 in. when the tensile load is applied. Solution 2.10-5 Round bar with upset ends d2 = 1.4 in. P d1 = 1.0 in L1 L2 P L2 Use Fig. 2-65 for the stress-concentration factor. snom ϭ ϭ E ϭ 25 ϫ 106 psi ␦ ϭ 0.0040 in. snom ϭ L2 ϭ 5 in. R ϭ radius of fillets Rϭ 1.4 in. Ϫ 1.0 in. 2 ϩ L1 (0.0040 in.)(25 ϫ 106 psi) 2(5 in.)( 1.0 ) 2 ϩ 20 in. 1.4 ϭ 3,984 psi R 0.2 in. ϭ 0.2 ϭ D1 1.0 in. Use the dashed curve in Fig. 2-65. K Ϸ 1.53 ϭ 0.2 in. PL2 PL1 ≤ϩ EA2 EA1 Solve for P: P ϭ ␦E d 2L2 ( d1 ) 2 2 SUBSTITUTE NUMERICAL VALUES: L1 ϭ 20 in. ␦ ϭ 2¢ ␦EA2 P ␦E ϭ ϭ A1 2L2A1 ϩ L1A2 2L2 ( A1 ) ϩ L1 A2 ␦EA1A2 2L2A1 ϩ L1A2 smax ϭ Ksnom Ϸ (1.53)(3984 psi) Ϸ 6100 psi 163
    • 164 CHAPTER 2 Axially Loaded Members Problem 2.10-6 A prismatic bar of diameter d0 ϭ 20 mm is being compared with a stepped bar of the same diameter (d1 ϭ 20 mm) that is enlarged in the middle region to a diameter d2 ϭ 25 mm (see figure). The radius of the fillets in the stepped bar is 2.0 mm. P1 (a) Does enlarging the bar in the middle region make it stronger than the prismatic bar? Demonstrate your answer by determining the maximum permissible load P1 for the prismatic bar and the maximum permissible load P2 for the enlarged bar, assuming that the allowable stress for the material is 80 MPa. (b) What should be the diameter d0 of the prismatic bar if it is to have the same maximum permissible load as does the stepped bar? P2 d0 d1 P1 d2 d1 P2 Soluton 2.10-6 Prismatic bar and stepped bar Stepped bar: See Fig. 2-65 for the stressconcentration factor. P1 R ϭ 2.0 mm D1 ϭ 20 mm D2 ϭ 25 mm RրD ϭ 0.10 D2 րD ϭ 1.25 K Ϸ 1.75 1 1 P2 P2 P2 snom ϭ ␲ 2 ϭ A1 4 d1 d0 d1 P1 P2 ϭ snom A1 ϭ d2 d1 d0 ϭ 20 mm smax K st smax A ϭ A K 1 K 1 80 MPa ␲ ≤¢ ≤ (20 mm) 2 1.75 4 Ϸ 14.4 kN Enlarging the bar makes it weaker, not stronger. The ratio of loads is P1րP2 ϭ K ϭ 1.75 d1 ϭ 20 mm d2 ϭ 25 mm (b) DIAMETER OF PRISMATIC BAR FOR THE SAME ALLOWABLE LOAD Fillet radius: R ϭ 2 mm Allowable stress: ␴t ϭ 80 MPa (a) COMPARISON OF BARS Prismatic bar: P1 ϭ st A0 ϭ st ¢ ϭ (80 MPa) ¢ P2 ϭ¢ snom ϭ P1 ϭ P2 2 ␲d0 ≤ 4 ␲ ≤ (20 mm) 2 ϭ 25.1 kN 4 st ¢ d0 ϭ 2 2 ␲d0 st ␲d1 ≤ϭ ¢ ≤ 4 K 4 d1 ͙K Ϸ 20 mm ͙1.75 2 d0 ϭ 2 d1 K Ϸ 15.1 mm
    • SECTION 2.10 Problem 2.10-7 A stepped bar with a hole (see figure) has widths b ϭ 2.4 in. and c ϭ 1.6 in. The fillets have radii equal to 0.2 in. What is the diameter dmax of the largest hole that can be drilled through the bar without reducing the load-carrying capacity? Solution 10-7 P 165 Stress Concentrations b d c P Stepped bar with a hole P d P c b b ϭ 2.4 in. BASED UPON HOLE (Use Fig. 2-63) c ϭ 1.6 in. b ϭ 2.4 in. Fillet radius: R ϭ 0.2 in. d ϭ diameter of the hole (in.) c1 ϭ b Ϫ d Pmax ϭ snom c1t ϭ Find dmax ϭ BASED UPON FILLETS (Use Fig. 2-64) b ϭ 2.4 in. b/c ϭ 1.5 c ϭ 1.6 in. R ϭ 0.2 in. R/c ϭ 0.125 K Ϸ 2.10 Pmax ϭ snomct ϭ smax (b Ϫ d)t K d 1 ¢ 1 Ϫ ≤ btsmax K b d/b K Pmax րbtsmax 0.3 0.125 2.66 0.329 0.4 0.167 2.57 0.324 0.5 0.208 2.49 0.318 0.6 0.250 2.41 0.311 0.7 0.292 2.37 0.299 d (in.) smax smax c ct ϭ ¢ ≤ (bt) K K b Ϸ 0.317 bt smax Based upon hole 0.33 Based upon fillets 0.32 Pmax bt␴max 0.317 0.31 dmax ≈ 0.51 in. d (in.) 0.30 0.3 0.4 0.5 0.6 0.7 0.8
    • 166 CHAPTER 2 Axially Loaded Members Nonlinear Behavior (Changes in Lengths of Bars) Problem 2.11-1 A bar AB of length L and weight density ␥ hangs vertically under its own weight (see figure). The stress-strain relation for the material is given by the Ramberg-Osgood equation (Eq. 2-71): ␴ ␴␣ ␴ ⑀ ϭ ᎏᎏ ϩ ᎏ0 ᎏᎏ ᎏ E E ␴0 A ΂ ΃ m Derive the following formula L ␥L2 ␥L ␴0␣L ␦ ϭ ᎏᎏ ϩ ᎏᎏ ᎏᎏ 2E (m ϩ1)E ␴0 ΂ ΃ m for the elongation of the bar. Solution 2.11-1 B Bar hanging under its own weight STRAIN AT DISTANCE x Let A ϭ cross-sectional area Let N ϭ axial force at distance x dx s s0␣ s m gx s0␣ gx m ϩ ¢ ≤ ϭ ϩ ¢ ≤ E E s0 E E s0 ELONGATION OF BAR L N ϭ ␥Ax N s ϭ ϭ gx A x eϭ ␦ϭ Ύ L e dx ϭ 0 ␴ 45,000 1 618 0 s0␣ gx dx ϩ E E s0␣L gL2 gL m ϭ ϩ ¢ ≤ 2E (m ϩ 1)E s0 Problem 2.11-2 A prismatic bar of length L ϭ 1.8 m and cross-sectional area A ϭ 480 mm2 is loaded by forces P1 ϭ 30 kN and P2 ϭ 60 kN (see figure). The bar is constructed of magnesium alloy having a stress-strain curve described by the following Ramberg-Osgood equation: ΂ 1␴ ΃ 70 ⑀ ϭ ᎏᎏ ϩ ᎏᎏ ᎏᎏ Ύ L 10 Ύ 0 A (␴ ϭ MPa) ¢ L gx m ≤ dx s0 Q.E.D. B 2L — 3 P1 C L — 3 in which ␴ has units of megapascals. (a) Calculate the displacement ␦C of the end of the bar when the load P1 acts alone. (b) Calculate the displacement when the load P2 acts alone. (c) Calculate the displacement when both loads act simultaneously. Solution 2.11-2 A B 2L — 3 L ϭ 1.8 m Axially loaded bar P1 C L — 3 P2 A ϭ 480 mm2 P1 ϭ 30 kN P2 ϭ 60 kN RambergϪOsgood Equation: eϭ 1 s 10 s ϩ ¢ ≤ (s ϭ MPa) 45,000 618 170 Find displacement at end of bar. P2
    • SECTION 2.11 (a) P1 ACTS ALONE (c) BOTH P1 AND P2 ARE ACTING P1 30 kN AB: s ϭ ϭ ϭ 62.5 MPa A 480 mm2 AB: s ϭ e ϭ 0.001389 P1 ϩ P2 90 kN ϭ ϭ 187.5 MPa A 480 mm2 e ϭ 0.008477 2L ␦c ϭ e ¢ ≤ ϭ 1.67 mm 3 ␦AB ϭ e ¢ BC: s ϭ (b) P2 ACTS ALONE ABC: s ϭ 167 Nonlinear Behavior P2 60 kN ϭ ϭ 125 MPa A 480 mm2 2L ≤ ϭ 10.17 mm 3 P2 60 kN ϭ ϭ 125 MPa A 480 mm2 e ϭ 0.002853 ␦BC ϭ e ¢ e ϭ 0.002853 ␦c ϭ eL ϭ 5.13 mm L ≤ ϭ 1.71 mm 3 ␦C ϭ ␦AB ϩ ␦BC ϭ 11.88 mm (Note that the displacement when both loads act simultaneously is not equal to the sum of the displacements when the loads act separately.) Problem 2.11-3 A circular bar of length L ϭ 32 in. and diameter d ϭ 0.75 in. is subjected to tension by forces P (see figure). The wire is made of a copper alloy having the following hyperbolic stress-strain relationship: d P P 18,000⑀ 1 ϩ 300⑀ ␴ ϭ ᎏᎏ 0 Յ ⑀ Յ 0.03 (␴ ϭ ksi) L (a) Draw a stress-strain diagram for the material. (b) If the elongation of the wire is limited to 0.25 in. and the maximum stress is limited to 40 ksi, what is the allowable load P? Solution 2.11-3 Copper bar in tension d P P (b) ALLOWABLE LOAD P Max. elongation ␦max ϭ 0.25 in. Max. stress ␴max ϭ 40 ksi L L ϭ 32 in. Based upon elongation: d ϭ 0.75 in. ␲d2 ϭ 0.4418 in.2 4 emax ϭ ␦max 0.25 in. ϭ ϭ 0.007813 L 32 in. (a) STRESS-STRAIN DIAGRAM smax ϭ 18,000 emax ϭ 42.06 ksi 1 ϩ 300 emax Aϭ sϭ 18,000e 1 ϩ 300e 0 Յ e Յ 0.03 (s ϭ ksi) Slope = 18,000 ksi ␴ 60 (ksi) smax ϭ 40 ksi Asymptote equals 60 ksi 40 20 ⑀ 0 0.01 0.02 0.03 BASED UPON STRESS: Stress governs. P ϭ ␴max A ϭ (40 ksi)(0.4418 in.2) ϭ 17.7 k
    • 168 CHAPTER 2 Axially Loaded Members Problem 2.11-4 A prismatic bar in tension has length L ϭ 2.0 m and cross-sectional area A ϭ 249 mm2. The material of the bar has the stress-strain curve shown in the figure. Determine the elongation ␦ of the bar for each of the following axial loads: P ϭ 10 kN, 20 kN, 30 kN, 40 kN, and 45 kN. From these results, plot a diagram of load P versus elongation ␦ (load-displacement diagram). 200 ␴ (MPa) 100 0 Solution 2.11-4 0 0.005 ⑀ 0.010 Bar in tension P P 50 L 40 L ϭ 2.0 m 30 A ϭ 249 mm2 P (kN) 20 STRESS-STRAIN DIAGRAM 10 (See the problem statement for the diagram) 0 10 5 15 ␦ (mm) 20 LOAD-DISPLACEMENT DIAGRAM P (kN) ␴ ϭ P/A (MPa) e (from diagram) ␦ ϭ eL (mm) 10 40 0.0009 1.8 20 80 0.0018 3.6 30 120 0.0031 6.2 40 161 0.0060 12.0 45 181 0.0081 NOTE: The load-displacement curve has the same shape as the stress-strain curve. 16.2 Problem 2.11-5 An aluminum bar subjected to tensile forces P has length L ϭ 150 in. and cross-sectional area A ϭ 2.0 in.2 The stress-strain behavior of the aluminum may be represented approximately by the bilinear stress-strain diagram shown in the figure. Calculate the elongation ␦ of the bar for each of the following axial loads: P ϭ 8 k, 16 k, 24 k, 32 k, and 40 k. From these results, plot a diagram of load P versus elongation ␦ (load-displacement diagram). ␴ 12,000 psi E2 = 2.4 × 106 psi E1 = 10 × 106 psi 0 ⑀
    • SECTION 2.11 Solution 2.11-5 Nonlinear Behavior Aluminum bar in tension P P LOAD-DISPLACEMENT DIAGRAM P (k) ␴ ϭ P/A (psi) L ϭ 150 in. 8 4,000 0.00040 0.060 A ϭ 2.0 in.2 16 8,000 0.00080 0.120 STRESS-STRAIN DIAGRAM 24 12,000 0.00120 0.180 32 16,000 0.00287 0.430 40 20,000 0.00453 0.680 L ␴ e ␦ ϭ eL (from Eq. 1 or Eq. 2) (in.) E2 ␴1 40 k 40 E1 30 ⑀1 0 ⑀ 24 k P (k) 20 0.68 in. 0.18 in. 10 E1 ϭ 10 ϫ 106 psi E2 ϭ 2.4 ϫ 106 psi ␦ (in.) 0 ␴1 ϭ 12,000 psi e1 ϭ 12,000 psi s1 ϭ E1 10 ϫ 106 psi ϭ 0.0012 For 0 Յ s Յ s1: s s ϭ (s ϭ psi) E2 10 ϫ 106 psi For s Ն s1: eϭ e ϭ e1 ϩ Eq. (1) s Ϫ s1 s Ϫ 12,000 ϭ 0.0012 ϩ E2 2.4 ϫ 106 ϭ s Ϫ 0.0038 2.4 ϫ 106 (s ϭ psi) Eq. (2) 0.2 0.4 0.6 0.8 169
    • 170 CHAPTER 2 Axially Loaded Members Problem 2.11-6 A rigid bar AB, pinned at end A, is supported by a wire CD and loaded by a force P at end B (see figure). The wire is made of high-strength steel having modulus of elasticity E ϭ 210 GPa and yield stress ␴Y ϭ 820 MPa. The length of the wire is L ϭ 1.0 m and its diameter is d ϭ 3 mm. The stress-strain diagram for the steel is defined by the modified power law, as follows: ␴ ϭ E⑀ ΂ E⑀ ΃ ␴ ␴ ϭ ␴Y ᎏᎏ C L A 0 Յ ␴ Յ ␴Y n D B ␴ Ն ␴Y P Y (a) Assuming n ϭ 0.2, calculate the displacement ␦B at the end of the bar due to the load P. Take values of P from 2.4 kN to 5.6 kN in increments of 0.8 kN. (b) Plot a load-displacement diagram showing P versus ␦B. Solution 2.11-6 2b Rigid bar supported by a wire From Eq. (2): e ϭ C A B D Stress in wire: s ϭ P 2b (6) (0 Յ s Յ sY ) (1) (s Ն sY ) (n ϭ 0.2) ␦B ϭ 3 3 ␦ ϭ eL 2 2 0.002425 3.64 679.1 0.003234 4.85 848.8 0.004640 6.96 1018.6 0.01155 17.3 5.6 1188.4 0.02497 37.5 e ϭ 0.0039048 P ϭ 3.864 kN ␦B ϭ 5.86 mm (b) LOAD-DISPLACEMENT DIAGRAM (3) Obtain e from stress-strain equations: s From Eq. (1): e ϭ (0 Յ s Յ sY ) E ␦B (mm) Eq. (3) For ␴ ϭ ␴Y ϭ 820 MPa: (2) (a) DISPLACEMENT ␦B AT END OF BAR ␦ ϭ elongation of wire 509.3 4.8 STRESS-STRAIN DIAGRAM e Eq. (4) or (5) 4.0 ␲d2 ϭ 7.0686 mm2 4 ␴ (MPa) Eq. (6) 3.2 d ϭ 3 mm Ee n ≤ sY F 3P ϭ A 2A 2.4 L ϭ 1.0 m s ϭ sY ¢ 3P 2 P (kN) ␴Y ϭ 820 MPa s ϭ Ee (5) PROCEDURE: Assume a value of P Calculate ␴ from Eq. (6) Calculate e from Eq. (4) or (5) Calculate ␦B from Eq. (3) b Wire: E ϭ 210 GPa Aϭ sY s 1րn ¢ ≤ E sY Axial force in wire: F ϭ L b (4) 8 6 P (kN) 4 P = 3.86 kN ␴Y = 820 MPa 2 ␦B = 5.86 mm 0 20 40 60 ␦B (mm)
    • SECTION 2.12 171 Elastoplastic Analysis Elastoplastic Analysis The problems for Section 2.12 are to be solved assuming that the material is elastoplastic with yield stress ␴Y , yield strain ⑀Y , and modulus of elasticity E in the linearly elastic region (see Fig. 2-70). ␪ A Problem 2.12-1 Two identical bars AB and BC support a vertical load P (see figure). The bars are made of steel having a stress-strain curve that may be idealized as elastoplastic with yield stress ␴Y. Each bar has cross-sectional area A. Determine the yield load PY and the plastic load P . P ␪ C B P Solution 2.12-1 A Two bars supporting a load P ␪ ␪ C ␴YA ␴ YA ␪ B B P P Structure is statically determinate. The yield load PY and the plastic lead PP occur at the same time, namely, when both bars reach the yield stress. JOINT B ⌺Fvert ϭ 0 (2␴Y A) sin ␪ ϭ P PY ϭ PP ϭ 2sY A sin u Problem 2.12-2 A stepped bar ACB with circular cross sections is held between rigid supports and loaded by an axial force P at midlength (see figure). The diameters for the two parts of the bar are d1 ϭ 20 mm and d2 ϭ 25 mm, and the material is elastoplastic with yield stress ␴Y ϭ 250 MPa. Determine the plastic load PP. A d1 C L — 2 d2 P L — 2 B
    • 172 CHAPTER 2 Solution 2.12-2 Axially Loaded Members Bar between rigid supports d1 A d2 C B SUBSTITUTE NUMERICAL VALUES: PP ϭ (250 MPa) ¢ P L — 2 d1 ϭ 20 mm L — 2 d2 ϭ 25 mm ϭ (250 MPa) ¢ ␴Y ϭ 250 MPa ϭ 201 kN DETERMINE THE PLASTIC LOAD PP: ␲ 2 2 ≤ (d1 ϩ d2 ) 4 ␲ ≤ [ (20 mm) 2 ϩ (25 mm) 2 ] 4 At the plastic load, all parts of the bar are stressed to the yield stress. P Point C: FAC FCB FAC ϭ ␴Y A1 FCB ϭ ␴Y A2 P ϭ FAC ϩ FCB PP ϭ sYA1 ϩ sYA2 ϭ sY (A1 ϩ A2 ) Problem 2.12-3 A horizontal rigid bar AB supporting a load P is hung from five symmetrically placed wires, each of cross-sectional area A (see figure). The wires are fastened to a curved surface of radius R. R (a) Determine the plastic load PP if the material of the wires is elastoplastic with yield stress ␴Y. (b) How is PP changed if bar AB is flexible instead of rigid? (c) How is PP changed if the radius R is increased? A B P Solution 2.12-3 Rigid bar supported by five wires F F F F A F B P A B P (a) PLASTIC LOAD PP At the plastic load, each wire is stressed to the yield stress. ∴ PP ϭ 5sY A F ϭ ␴Y A (b) BAR AB IS FLEXIBLE At the plastic load, each wire is stressed to the yield stress, so the plastic load is not changed. (c) RADIUS R IS INCREASED Again, the forces in the wires are not changed, so the plastic load is not changed.
    • SECTION 2.12 173 Elastoplastic Analysis Problem 2.12-4 A load P acts on a horizontal beam that is supported by four rods arranged in the symmetrical pattern shown in the figure. Each rod has cross-sectional area A and the material is elastoplastic with yield stress ␴Y. Determine the plastic load PP. ␣ ␣ P Solution 2.12-4 Beam supported by four rods F ␣ F ␣ F F P F ϭ ␴Y A P At the plastic load, all four rods are stressed to the yield stress. Sum forces in the vertical direction and solve for the load: PP ϭ 2F ϩ 2F sin ␣ PP ϭ 2sY A (1 ϩ sin ␣) Problem 2.12-5 The symmetric truss ABCDE shown in the figure is constructed of four bars and supports a load P at joint E. Each of the two outer bars has a cross-sectional area of 0.307 in.2, and each of the two inner bars has an area of 0.601 in.2 The material is elastoplastic with yield stress ␴Y ϭ 36 ksi. Determine the plastic load PP. 21 in. A 54 in. 21 in. C B D 36 in. E P
    • 174 CHAPTER 2 Axially Loaded Members Solution 2.12-5 Truss with four bars 21 in. 27 in. B A 27 in. C 3 PLASTIC LOAD PP At the plastic load, all bars are stressed to the yield stress. D FAE ϭ ␴Y AAE 5 4 5 21 in. 36 in. 3 PP ϭ 4 E 6 8 s A ϩ s A 5 Y AE 5 Y BE SUBSTITUTE NUMERICAL VALUES: AAE ϭ 0.307 in.2 ABE ϭ 0.601 in.2 P LAE ϭ 60 in. FBE ϭ ␴Y ABE ␴Y ϭ 36 ksi LBE ϭ 45 in. 6 8 PP ϭ (36 ksi)(0.307 in.2 ) ϩ (36 ksi)(0.601 in.2 ) 5 5 JOINT E Equilibrium: FBE 2FAE ¢ FAE E or P Pϭ 3 4 ≤ ϩ 2FBE ¢ ≤ ϭ P 5 5 ϭ 13.26 k ϩ 34.62 k ϭ 47.9 k 6 8 FAE ϩ FBE 5 5 b Problem 2.12-6 Five bars, each having a diameter of 10 mm, support a load P as shown in the figure. Determine the plastic load PP if the material is elastoplastic with yield stress ␴Y ϭ 250 MPa. b b b 2b P Solution 2.12-6 b Truss consisting of five bars b b F b F F F F At the plastic load, all five bars are stressed to the yield stress F ϭ ␴Y A 2b d ϭ 10 mm Aϭ P ␲d2 ϭ 78.54 mm2 4 ␴Y ϭ 250 MPa PP ϭ 2F ¢ ϭ P 1 ͙2 ≤ ϩ 2F ¢ Sum forces in the vertical direction and solve for the load: 2 ͙5 ≤ϩF sY A (5͙2 ϩ 4͙5 ϩ 5) 5 ϭ 4.2031 sY A Substitute numerical values: PP ϭ (4.2031)(250 MPa)(78.54 mm2) ϭ 82.5 kN
    • SECTION 2.12 Problem 2.12-7 A circular steel rod AB of diameter d ϭ 0.60 in. is stretched tightly between two supports so that initially the tensile stress in the rod is 10 ksi (see figure). An axial force P is then applied to the rod at an intermediate location C. B A d (a) Determine the plastic load P if the material is elastoplastic P with yield stress ␴Y ϭ 36 ksi. (b) How is PP changed if the initial tensile stress is doubled to 20 ksi? Solution 2.12-7 175 Elastoplastic Analysis A P B C Bar held between rigid supports A P B POINT C: ␴⌼A C d P C d ϭ 0.6 in. ␴Y ϭ 36 ksi Initial tensile stress ϭ 10 ksi PP ϭ 2sY A ϭ (2)(36 ksi) ¢ ϭ 20.4 k ␴⌼A ␲ ≤ (0.60 in.) 2 4 (a) PLASTIC LOAD PP (B) INITIAL TENSILE STRESS IS DOUBLED The presence of the initial tensile stress does not affect the plastic load. Both parts of the bar must yield in order to reach the plastic load. PP is not changed. Problem 2.12-8 A rigid bar ACB is supported on a fulcrum at C and loaded by a force P at end B (see figure). Three identical wires made of an elastoplastic material (yield stress ␴Y and modulus of elasticity E) resist the load P. Each wire has cross-sectional area A and length L. (a) Determine the yield load P and the corresponding yield Y displacement ␦Y at point B. (b) Determine the plastic load PP and the corresponding displacement ␦P at point B when the load just reaches the value PP. (c) Draw a load-displacement diagram with the load P as ordinate and the displacement ␦B of point B as abscissa. L A L C B P a a a a
    • 176 CHAPTER 2 Axially Loaded Members Solution 2.12-8 Rigid bar supported by wires (b) PLASTIC LOAD PP L A C B ␴Y A ␴Y A P L a a a a A C B (a) YIELD LOAD PY ␴Y A Yielding occurs when the most highly stressed wire reaches the yield stress ␴Y. ␴Y A 2 At the plastic load, all wires reach the yield stress. ©MC ϭ 0 ␴Y A PP ϭ A C PP B 4sY A 3 At point A: ␦A ϭ (sY A) ¢ PY ␴Y A 2 At point B: ⌺MC ϭ 0 ␦B ϭ 3␦A ϭ ␦P ϭ PY ϭ sY A At point A: ␦A ϭ ¢ sY A sY L L ≤¢ ≤ϭ 2 EA 2E At point B: ␦B ϭ 3␦A ϭ ␦Y ϭ sYL L ≤ϭ EA E 3sY L E (c) LOAD-DISPLACEMENT DIAGRAM P PP ϭ PP 4 P 3 Y ␦P ϭ 2␦Y PY 3sYL 2E 0 Problem 2.12-9 The structure shown in the figure consists of a horizontal rigid bar ABCD supported by two steel wires, one of length L and the other of length 3L/4. Both wires have cross-sectional area A and are made of elastoplastic material with yield stress ␴Y and modulus of elasticity E. A vertical load P acts at end D of the bar. (a) Determine the yield load P and the corresponding yield Y displacement ␦Y at point D. (b) Determine the plastic load P and the corresponding displacement P ␦P at point D when the load just reaches the value P . P (c) Draw a load-displacement diagram with the load P as ordinate and the displacement ␦D of point D as abscissa. ␦Y ␦P ␦B L A 3L 4 B C D P 2b b b
    • SECTION 2.12 Solution 2.12-9 177 Elastoplastic Analysis Rigid bar supported by two wires STRESSES L A sB ϭ 3L 4 B C FB A sC ϭ FC A ∴ sC ϭ 2sB (7) Wire C has the larger stress. Therefore, it will yield first. D (a) YIELD LOAD P 2b b sC ϭ sY b A ϭ cross-sectional area FC ϭ sY A ␴Y ϭ yield stress sC sY ϭ 2 2 FB ϭ (From Eq. 7) 1 s A 2 Y From Eq. (3): E ϭ modulus of elasticity 1 2 ¢ sY A ≤ ϩ 3(sY A) ϭ 4P 2 DISPLACEMENT DIAGRAM A sB ϭ B C ␦C ␦B P ϭ PY ϭ sY A D From Eq. (4): ␦D ␦B ϭ FBL sY L ϭ EA 2E From Eq. (2): COMPATIBILITY: ␦D ϭ ␦Y ϭ 2␦B ϭ 3 ␦C ϭ ␦B 2 (1) ␦D ϭ 2␦B (2) sY L E (b) PLASTIC LOAD At the plastic load, both wires yield. ␴B ϭ ␴Y ϭ ␴C FREE-BODY DIAGRAM FB FC FB ϭ FC ϭ ␴Y A From Eq. (3): 2(␴Y A) ϩ 3(␴Y A) ϭ 4P A B C D 5 P ϭ PP ϭ sY A 4 From Eq. (4): P 2b b b ␦B ϭ EQUILIBRIUM: ©MA ϭ 0 ‫۔ە‬ From Eq. (2): FB (2b) ϩ FC (3b) ϭ P(4b) 2FB ϩ 3FC ϭ 4P (3) FORCE-DISPLACEMENT RELATIONS ␦B ϭ FBL EA 3 FC ¢ L ≤ 4 ␦C ϭ EA ␦D ϭ ␦P ϭ 2␦B ϭ 2sY L E (c) LOAD-DISPLACEMENT DIAGRAM P (4, 5) PP PP ϭ PY Substitute into Eq. (1): 3FCL 3FBL ϭ 4EA 2EA FC ϭ 2FB FBL sY L ϭ EA E 5 P 4 Y ␦P ϭ 2␦Y (6) 0 ␦Y ␦P ␦D
    • 178 CHAPTER 2 Axially Loaded Members Problem 2.12-10 Two cables, each having a length L of approximately 40 m, support a loaded container of weight W (see figure). The cables, which have effective cross-sectional area A ϭ 48.0 mm2 and effective modulus of elasticity E ϭ 160 GPa, are identical except that one cable is longer than the other when they are hanging separately and unloaded. The difference in lengths is d ϭ 100 mm. The cables are made of steel having an elastoplastic stress-strain diagram with ␴Y ϭ 500 MPa. Assume that the weight W is initially zero and is slowly increased by the addition of material to the container. L (a) Determine the weight WY that first produces yielding of the shorter cable. Also, determine the corresponding elongation ␦Y of the shorter cable. (b) Determine the weight WP that produces yielding of both cables. Also, determine the elongation ␦P of the shorter cable when the weight W just reaches the value WP. (c) Construct a load-displacement diagram showing the weight W as ordinate and the elongation ␦ of the shorter cable as abscissa. (Hint: The load displacement diagram is not a single straight line in the region 0 Յ W Յ WY.) Solution 2.12-10 Two cables supporting a load L ϭ 40 m A ϭ 48.0 mm2 F1 ϭ sY A d ϭ difference in length ϭ 100 mm 2 (b) PLASTIC LOAD WP E ϭ 160 GPa 1 WP ϭ 2sY A ϭ 48 kN ␴Y ϭ 500 MPa L INITIAL STRETCHING OF CABLE 1 Initially, cable 1 supports all of the load. Let W1 ϭ load required to stretch cable 1 to the same length as cable 2 W1 ϭ W s1 ϭ EA d ϭ 19.2 kN L ␦1 ϭ 100 mm (elongation of cable 1 ) W1 Ed ϭ ϭ 400 MPa (s1 6 sY ∴ OK) A L (a) YIELD LOAD WY Cable 1 yields first. F1 ϭ ␴Y A ϭ 24 kN ␦1Y ϭ total elongation of cable 1 F1L sYL ␦1Y ϭ ϭ ϭ 0.125 m ϭ 125 mm EA E ␦Y ϭ ␦1Y ϭ 125 mm ␦2Y ϭ elongation of cable 2 ϭ ␦1Y Ϫ d ϭ 25 mm F2 ϭ EA ␦ ϭ 4.8 kN L 2Y WY ϭ F1 ϩ F2 ϭ 24 kN ϩ 4.8 kN ϭ 28.8 kN W F2 ϭ sY A ␦2P ϭ elongation of cable 2 sYL L ϭ F2 ¢ ≤ ϭ ϭ 0.125 mm ϭ 125 mm EA E ␦1P ϭ ␦2P ϩ d ϭ 225 mm ␦P ϭ ␦1P ϭ 225 mm (c) LOAD-DISPLACEMENT DIAGRAM WP W 50 (kN) 40 30 20 WY W1 10 ␦1 ␦Y 0 WY ϭ 1.5 W1 WP ϭ 1.667 WY 100 ␦P 200 300 ␦Y ϭ 1.25 ␦1 ␦P ϭ 1.8 ␦Y 0 Ͻ W Ͻ W1: slope ϭ 192,000 N/m W1 Ͻ W Ͻ WY: slope ϭ 384,000 N/m WY Ͻ W Ͻ WP: slope ϭ 192,000 N/m ␦ (mm)
    • SECTION 2.12 Problem 2.12-11 A hollow circular tube T of length L ϭ 15 in. is uniformly compressed by a force P acting through a rigid plate (see figure). The outside and inside diameters of the tube are 3.0 and 2.75 in., repectively. A concentric solid circular bar B of 1.5 in. diameter is mounted inside the tube. When no load is present, there is a clearance c ϭ 0.010 in. between the bar B and the rigid plate. Both bar and tube are made of steel having an elastoplastic stress-strain diagram with E ϭ 29 ϫ 103 ksi and ␴Y ϭ 36 ksi. P c T (a) Determine the yield load PY and the corresponding shortening ␦Y of the tube. (b) Determine the plastic load PP and the corresponding shortening ␦P of the tube. (c) Construct a load-displacement diagram showing the load P as ordinate and the shortening ␦ of the tube as abscissa. (Hint: The load-displacement diagram is not a single straight line in the region 0 Յ P Յ PY.) Solution 2.12-11 179 Elastoplastic Analysis T B T L B Tube and bar supporting a load P Clearance = c T T B T L B BAR: d ϭ 1.5 in. L ϭ 15 in. c ϭ 0.010 in. E ϭ 29 ϫ 103 ksi ␴Y ϭ 36 ksi TUBE: d2 ϭ 3.0 in. d1 ϭ 2.75 in. AT ϭ ␲ 2 (d Ϫ d2 ) ϭ 1.1290 in.2 1 4 2 AB ϭ ␲d2 ϭ 1.7671 in.2 4 INITIAL SHORTENING OF TUBE T Initially, the tube supports all of the load. Let P1 ϭ load required to close the clearance P1 ϭ EAT c ϭ 21,827 lb L Let ␦1 ϭ shortening of tube s1 ϭ P1 ϭ 19,330 psi AT ␦1 ϭ c ϭ 0.010 in. (s1 6 sY ∴ OK) (Continued)
    • 180 CHAPTER 2 Axially Loaded Members (a) YIELD LOAD PY Because the tube and bar are made of the same material, and because the strain in the tube is larger than the strain in the bar, the tube will yield first. ␦TY ϭ shortening of tube at the yield stress FTL sYL ϭ ϭ 0.018621 in. EAT E 0 ϭ ␦TY Ϫ c ϭ 0.008621 in. EAB FB ϭ ␦ ϭ 29,453 lb L BY PY ϭ FT ϩ FB ϭ 40,644 lb ϩ 29,453 lb ϭ 70,097 lb PY ϭ 70,100 lb P1 0.01 PY ϭ 3.21 P1 0.02 ␦Y ϭ 1.86 ␦1 PP ϭ 1.49 PY ␦Y ␦P ϭ 1.54 ␦Y 0 Ͻ P ϽP1: slope ϭ 2180 k/in. P1 ϽP Ͻ PY: slope ϭ 5600 k/in. (b) PLASTIC LOAD PP FB ϭ ␴Y AB PP ϭ FT ϩ FB ϭ sY (AT ϩ AB ) ϭ 104,300 lb ␦BP ϭ shortening of bar sYL L ≤ϭ ϭ 0.018621 in. EAB E ␦TP ϭ ␦BP ϩ c ϭ 0.028621 in. ␦P ϭ ␦TP ϭ 0.02862 in. 40 ␦1 ␦BY ϭ shortening of bar ϭ FB ¢ PY 60 20 ␦Y ϭ ␦TY ϭ 0.01862 in. FT ϭ ␴Y AT PP P 100 (kips) 80 FT ϭ ␴Y AT ϭ 40,644 lb ␦TY ϭ (c) LOAD-DISPLACEMENT DIAGRAM PY Ͻ P Ͻ PP: slope ϭ 3420 k/in. ␦P 0.03 ␦ (in.)
    • 3 Torsion Torsional Deformations d Problem 3.2-1 A copper rod of length L ϭ 18.0 in. is to be twisted by torques T (see figure) until the angle of rotation between the ends of the rod is 3.0°. If the allowable shear strain in the copper is 0.0006 rad, what is the maximum permissible diameter of the rod? Solution 3.2-1 T T L Probs. 3.2-1 and 3.2-2 Copper rod in torsion d T T L L ϭ 18.0 in. f ϭ 3.0Њ ϭ (3.0) ¢ ϭ 0.05236 rad ␲ ≤ rad 180 ␥allow ϭ 0.0006 rad From Eq. (3-3): gmax ϭ rf df ϭ L 2L dmax ϭ 2Lgallow (2)(18.0 in.)(0.0006 rad) ϭ f 0.05236 rad dmax ϭ 0.413 in. Find dmax Problem 3.2-2 A plastic bar of diameter d ϭ 50 mm is to be twisted by torques T (see figure) until the angle of rotation between the ends of the bar is 5.0°. If the allowable shear strain in the plastic is 0.012 rad, what is the minimum permissible length of the bar? Solution 3.2-2 Plastic bar in torsion d ϭ 50 mm ␲ f ϭ 5.0Њ ϭ (5.0) ¢ ≤ rad ϭ 0.08727 rad 180 d T L ␥allow ϭ 0.012 rad Lmin ϭ Find Lmin From Eq. (3-3): gmax ϭ rf df ϭ L 2L T df (50 mm)(0.08727 rad) ϭ 2gallow (2)(0.012 rad) Lmin ϭ 182 mm 181
    • 182 CHAPTER 3 Torsion Problem 3.2-3 A circular aluminum tube subjected to pure torsion by torques T (see figure) has an outer radius r2 equal to twice the inner radius r1. T (a) If the maximum shear strain in the tube is measured as 400 ϫ 10Ϫ6 rad, what is the shear strain ␥1 at the inner surface? (b) If the maximum allowable rate of twist is 0.15 degrees per foot and the maximum shear strain is to be kept at 400 ϫ 10Ϫ6 rad by adjusting the torque T, what is the minimum required outer radius (r2)min? Solution 3.2-3 r2 ␥max ϭ 400 ϫ 10Ϫ6 rad ϭ (0.15Њրft) ¢ r1 ␲ rad 1 ft ≤¢ ≤ 180 degree 12 in. ϭ 218.2 ϫ 10 Ϫ6 radրin. (a) SHEAR STRAIN AT INNER SURFACE From Eq. (3-5b): g1 ϭ L r2 r1 Problems 3.2-3, 3.2-4, and 3.2-5 Circular aluminum tube r2 ϭ 2r1 uallow ϭ 0.15Њրft T 1 1 g ϭ (400 ϫ 10 Ϫ6 rad) 2 2 2 g1 ϭ 200 ϫ 10 Ϫ6 rad Problem 3.2-4 A circular steel tube of length L ϭ 0.90 m is loaded in torsion by torques T (see figure). (a) If the inner radius of the tube is r1 ϭ 40 mm and the measured angle of twist between the ends is 0.5°, what is the shear strain ␥1 (in radians) at the inner surface? (b) If the maximum allowable shear strain is 0.0005 rad and the angle of twist is to be kept at 0.5° by adjusting the torque T, what is the maximum permissible outer radius (r 2)max? (b) MINIMUM OUTER RADIUS From Eq. (3-5a): gmax ϭ r2 f ϭ r2u L (r2 ) min ϭ gmax 400 ϫ 10 Ϫ6 rad ϭ uallow 218.2 ϫ 10 Ϫ6 rad րin. (r2 ) min ϭ 1.83 in.
    • SECTION 3.2 Solution 3.2-4 Torsional Deformations Circular steel tube r2 L ϭ 0.90 m (b) MAXIMUM OUTER RADIUS r1 r1 ϭ 40 mm f ϭ 0.5Њ ϭ (0.5Њ) ¢ ϭ 0.008727 rad gmax ϭ 0.0005 rad ␲ rad ≤ 180 degree (a) SHEAR STRAIN AT INNER SURFACE From Eq. (3-5a): gmax ϭ g2 ϭ r2 (r2 ) max ϭ gmax L f ; r2 ϭ L f (0.0005 rad)(900 mm) 0.008727 rad (r2 ) max ϭ 51.6 mm From Eq. (3-5b): f (40 mm)(0.008727 rad) ϭ L 900 mm gmin ϭ g1 ϭ r1 g1 ϭ 388 ϫ 10 Ϫ6 rad Problem 3.2-5 Solve the preceding problem if the length L ϭ 50 in., the inner radius r1 ϭ 1.5 in., the angle of twist is 0.6°, and the allowable shear strain is 0.0004 rad. Solution 3.2-5 Circular steel tube r2 L ϭ 50 in. r1 r1 ϭ 1.5 in. f ϭ 0.6Њ ϭ (0.6Њ) ¢ ϭ 0.010472 rad gmax ϭ 0.0004 rad ␲ rad ≤ 180 degree (a) SHEAR STRAIN AT INNER SURFACE From Eq. (3-5b): gmin ϭ g1 ϭ r1 f (1.5 in.)(0.010472 rad) ϭ L 50 in. g1 ϭ 314 ϫ 10 Ϫ6 rad (b) MAXIMUM OUTER RADIUS From Eq. (3-5a): gmax ϭ g2 ϭ r2 (r2 ) max ϭ gmaxL f ; r2 ϭ L f (0.0004 rad)(50 in.) 0.010472 rad (r2 ) max ϭ 1.91 in. 183
    • 184 CHAPTER 3 Torsion Circular Bars and Tubes Problem 3.3-1 A prospector uses a hand-powered winch (see figure) to raise a bucket of ore in his mine shaft. The axle of the winch is a steel rod of diameter d ϭ 0.625 in. Also, the distance from the center of the axle to the center of the lifting rope is b ϭ 4.0 in. If the weight of the loaded bucket is W ϭ 100 lb, what is the maximum shear stress in the axle due to torsion? P W d b W Solution 3.3-1 Hand-powered winch Axle d ϭ 0.625 in. MAXIMUM SHEAR STRESS IN THE AXLE b ϭ 4.0 in. From Eq. (3-12): W ϭ 100 lb d tmax ϭ 16T ␲d 3 tmax ϭ (16)(400 lb-in) ␲(0.625in.) 3 Torque T applied to the axle: T ϭ Wb ϭ 400 lb-in. b W tmax ϭ 8,340 psi Problem 3.3-2 When drilling a hole in a table leg, a furniture maker uses a hand-operated drill (see figure) with a bit of diameter d ϭ 4.0 mm. (a) If the resisting torque supplied by the table leg is equal to 0.3 Nиm, what is the maximum shear stress in the drill bit? (b) If the shear modulus of elasticity of the steel is G ϭ 75 GPa, what is the rate of twist of the drill bit (degrees per meter)? d
    • SECTION 3.3 Solution 3.3-2 185 Circular Bars and Tubes Torsion of a drill bit d T T (b) RATE OF TWIST From Eq. (3-14): d ϭ 4.0 mm T ϭ 0.3 N и m G ϭ 75 GPa (a) MAXIMUM SHEAR STRESS uϭ uϭ From Eq. (3-12): tmax ϭ 16T ␲d 3 tmax ϭ T GIP 0.3 N ؒ m ␲ (75 GPa) ¢ ≤ (4.0 mm) 4 32 16(0.3 N ؒ m) ␲(4.0 mm) 3 u ϭ 0.1592 rad րmϭ 9.12Њրm tmax ϭ 23.8 MPa Problem 3.3-3 While removing a wheel to change a tire, a driver applies forces P ϭ 25 lb at the ends of two of the arms of a lug wrench (see figure). The wrench is made of steel with shear modulus of elasticity G ϭ 11.4 ϫ 106 psi. Each arm of the wrench is 9.0 in. long and has a solid circular cross section of diameter d ϭ 0.5 in. (a) Determine the maximum shear stress in the arm that is turning the lug nut (arm A). (b) Determine the angle of twist (in degrees) of this same arm. P 9.0 in. A 9.0 in. d = 0.5 in. P = 25 lb Solution 3.3-3 Lug wrench P ϭ 25 lb P L ϭ 9.0 in. d L d ϭ 0.5 in. L P T ϭ torque acting on arm A Arm A T T G ϭ 11.4 ϫ 106 psi T ϭ P(2L) ϭ 2(25 lb)(9.0 in.) ϭ 450 lb-in. (a) MAXIMUM SHEAR STRESS From Eq. (3-12): tmax ϭ 16T (16)(450 lb-in.) ϭ ␲d 3 ␲(0.5 in.) 3 tmax ϭ 18,300 psi (b) ANGLE OF TWIST From Eq. (3-15): (450 lb-in.)(9.0 in.) ␲ (11.4 ϫ 106 psi) ¢ ≤ (0.5 in.) 4 32 f ϭ 0.05790 rad ϭ 3.32Њ fϭ TL ϭ GIP
    • 186 CHAPTER 3 Torsion Problem 3.3-4 An aluminum bar of solid circular cross section is twisted by torques T acting at the ends (see figure). The dimensions and shear modulus of elasticity are as follows: L ϭ 1.2 m, d ϭ 30 mm, and G ϭ 28 GPa. d T T (a) Determine the torsional stiffness of the bar. (b) If the angle of twist of the bar is 4°, what is the maximum shear stress? What is the maximum shear strain (in radians)? Solution 3.3-4 L Aluminum bar in torsion d T From Eq. (3-11): T tmax ϭ L L ϭ 1.2 m d ϭ 30 mm G ϭ 28 GPa tmax ϭ ␾ ϭ 4Њ ϭ (a) TORSIONAL STIFFNESS GIPf d Tr Td ϭ ϭ¢ ≤¢ ≤ IP 2IP L 2IP Gdf 2L (28 GPa)(30 mm)(0.069813 rad) 2(1.2 m) ϭ 24.43 MPa GIP G␲d 4 (28 GPa)(␲)(30 mm) 4 kT ϭ ϭ ϭ L 32L 32(1.2 m) tmax ϭ 24.4 MPa k T ϭ 1860 N . m MAXIMUM SHEAR STRAIN Hooke’s Law: (b) MAXIMUM SHEAR STRESS tmax 24.43 MPa ϭ G 28 GPa f ϭ 4Њ ϭ (4Њ)(p ր180)rad ϭ 0.069813 rad gmax ϭ From Eq. (3-15): gmax ϭ 873 ϫ 10 Ϫ6 rad fϭ TL GIP Tϭ GIPf L Problem 3.3-5 A high-strength steel drill rod used for boring a hole in the earth has a diameter of 0.5 in. (see figure).The allowable shear stress in the steel is 40 ksi and the shear modulus of elasticity is 11,600 ksi. What is the minimum required length of the rod so that one end of the rod can be twisted 30° with respect to the other end without exceeding the allowable stress? T d = 0.5 in. T L
    • SECTION 3.3 Solution 3.3-5 Steel drill rod T T d G␲d 4f ; substitute T into Eq. (1): 32L G ϭ 11,600 psi Tϭ d ϭ 0.5 in. tmax ϭ ¢ f ϭ 30Њ ϭ (30Њ) ¢ ␶allow ϭ 40 ksi ␲ ≤ rad ϭ 0.52360 rad 180 Lmin ϭ ϭ MINIMUM LENGTH From Eq. (3-12): tmax ϭ TL 32TL ϭ GIP G␲d 4 From Eq. (3-15): f ϭ L 16T ␲d 3 (1) Gdf 16 G␲d 4f ≤ϭ 3≤ ¢ 32L 2L ␲d Gdf 2tallow (11,600 ksi)(0.5 in.)(0.52360 rad) 2(40 ksi) Lmin ϭ 38.0 in. Problem 3.3-6 The steel shaft of a socket wrench has a diameter of 8.0 mm. and a length of 200 mm (see figure). If the allowable stress in shear is 60 MPa, what is the maximum permissible torque Tmax that may be exerted with the wrench? Through what angle ␾ (in degrees) will the shaft twist under the action of the maximum torque? (Assume G ϭ 78 GPa and disregard any bending of the shaft.) Solution 3.3-6 Circular Bars and Tubes d = 8.0 mm T L = 200 mm Socket wrench ANGLE OF TWIST d T From Eq. (3-15): f ϭ L d ϭ 8.0 mm L ϭ 200 mm ␶allow ϭ 60 MPa G ϭ 78 GPa MAXIMUM PERMISSIBLE TORQUE From Eq. (3-12): tmax ϭ 16T ␲d 3 ␲d 3tmax Tmax ϭ 16 ␲(8.0 mm) 3 (60 MPa) Tmax ϭ 16 Tmax ϭ 6.03 N # m TmaxL GIP From Eq. (3-12): Tmax ϭ fϭ¢ ␲d 3tmax L ≤¢ ≤ 16 GIP ␲d 3tmax 16 IP ϭ ␲d 4 32 fϭ ␲d 3tmaxL(32) 2tmaxL ϭ Gd 16G(␲d 4 ) fϭ 2(60 MPa)(200 mm) ϭ 0.03846 rad (78 GPa)(8.0 mm) f ϭ (0.03846 rad) ¢ 180 degրrad ≤ ϭ 2.20Њ ␲ 187
    • 188 CHAPTER 3 Torsion Problem 3.3-7 A circular tube of aluminum is subjected to torsion by torques T applied at the ends (see figure). The bar is 20 in. long, and the inside and outside diameters are 1.2 in. and 1.6 in., respectively. It is determined by measurement that the angle of twist is 3.63° when the torque is 5800 lb-in. Calculate the maximum shear stress ␶max in the tube, the shear modulus of elasticity G, and the maximum shear strain ␥max (in radians). T T 20 in. 1.2 in. 1.6 in. Solution 3.3-7 Aluminum tube in torsion L ϭ 20 in. d1 ϭ 1.2 in. SHEAR MODULUS OF ELASTICITY d1 d2 ϭ 1.6 in. T ϭ 5800 lb-in. ␾ ϭ 3.63Њ ϭ 0.063355 rad ␲ 4 4 IP ϭ (d2 Ϫd1 ) ϭ 0.43982 in.4 32 MAXIMUM SHEAR STRESS tmax ϭ Tr (5800 lb-in.)(0.8 in.) ϭ IP 0.43982 in.4 tmax ϭ 10,550 psi d2 fϭ TL GIP Gϭ (5800 lb-in.)(20 in.) (0.063355 rad)(0.43982 in.4 ) Gϭ TL fIP G ϭ 4.16 ϫ 106 psi MAXIMUM SHEAR STRAIN gmax ϭ tmax G gmax ϭ ¢ gmax ϭ rf Tr fIP ≤¢ ≤ϭ IP TL L (0.8 in.)(0.063355 rad) 20 in. gmax ϭ 0.00253 rad Problem 3.3-8 A propeller shaft for a small yacht is made of a solid steel bar 100 mm in diameter. The allowable stress in shear is 50 MPa, and the allowable rate of twist is 2.0° in 3 meters. Assuming that the shear modulus of elasticity is G ϭ 80 GPa, determine the maximum torque Tmax that can be applied to the shaft.
    • SECTION 3.3 Solution 3.3-8 189 Circular Bars and Tubes Propeller shaft d T d ϭ 100 mm G ϭ 80 GPa MAX. TORQUE BASED UPON RATE OF TWIST ␶allow ϭ 50 MPa 1 ␲ u ϭ 2Њ in 3 m ϭ (2Њ) ¢ ≤ radրm 3 180 ϭ 0.011636 radրm MAX. TORQUE BASED UPON SHEAR STRESS tϭ 16T ␲d 3 T1 ϭ T ␲d 3tallow 16 uϭ T GIP T2 ϭ GIPu ϭ G ¢ ϭ (80 GPa) ¢ ␲d 4 ≤u 32 ␲ ≤ (100 mm) 4 (0.011636 rad րm) 32 T2 ϭ 9140 N ؒ m RATE OF TWIST GOVERNS Tmax ϭ 9140 N ؒ m ␲(100 mm) 3 (50 MPa) ϭ 16 T1 ϭ 9820 N ؒ m Problem 3.3-9 Three identical circular disks A, B, and C are welded to the ends of three identical solid circular bars (see figure). The bars lie in a common plane and the disks lie in planes perpendicular to the axes of the bars. The bars are welded at their intersection D to form a rigid connection. Each bar has diameter d1 ϭ 0.5 in. and each disk has diameter d2 ϭ 3.0 in. Forces P1, P2, and P3 act on disks A, B, and C, respectively, thus subjecting the bars to torsion. If P1 ϭ 28 lb, what is the maximum shear stress ␶max in any of the three bars? P3 C 135° P1 P3 d1 A D 135° P1 90° d2 P2 P2 B
    • 190 CHAPTER 3 Solution 3.3-9 Torsion Three circular bars T3 THE THREE TORQUES MUST BE IN EQUILIBRIUM T1 C 135° T1 T3 A 45° 135° T2 90° T3 is the largest torque T3 ϭ T1 ͙2 ϭ P1d2 ͙2 B MAXIMUM SHEAR STRESS (Eq. 3-12) T2 tmax ϭ 16T 16T3 16P1 d2 ͙2 ϭ 3 ϭ 3 ␲d 3 ␲d1 ␲d1 tmax ϭ d1 ϭ diameter of bars 16(28 lb)(3.0 in.) ͙2 ϭ 4840 psi ␲(0.5 in.) 3 ϭ 0.5 in. d2 ϭ diameter of disks ϭ 3.0 in. P1 ϭ 28 lb T1 ϭ P1d2 T2 ϭ P2d2 T3 ϭ P3d2 Problem 3.3-10 The steel axle of a large winch on an ocean liner is subjected to a torque of 1.5 kNиm (see figure). What is the minimum required diameter dmin if the allowable shear stress is 50 MPa and the allowable rate of twist is 0.8°/m? (Assume that the shear modulus of elasticity is 80 GPa.) Solution 3.3-10 d T Axle of a large winch T T ϭ 1.5 kN и m d T G ϭ 80 GPa ␶allow ϭ 50 MPa ␲ uallow ϭ 0.8Њրm ϭ (0.8Њ) ¢ ≤ radրm 180 ϭ 0.013963 rad րm MIN. DIAMETER BASED UPON SHEAR STRESS 16T 16T tϭ d3 ϭ 3 ␲tallow ␲d 16(1.5 kN ؒ m) d ϭ ϭ 152.789 ϫ 10 Ϫ6 m3 ␲(50 MPa) 3 d ϭ 0.05346 m T dmin ϭ 53.5 mm MIN. DIAMETER BASED UPON RATE OF TWIST uϭ d4 ϭ T 32T ϭ GIp G␲d 4 d4 ϭ 32T ␲Guallow 32(1.5 kN ؒ m) ␲(80 GPa)(0.013963 radրm) ϭ 0.00001368 m4 d ϭ 0.0608 m dmin ϭ 60.8 mm RATE OF TWIST GOVERNS dmin ϭ 60.8 mm
    • SECTION 3.3 191 Circular Bars and Tubes Problem 3.3-11 A hollow steel shaft used in a construction auger has outer diameter d2 ϭ 6.0 in. and inner diameter d1 ϭ 4.5 in. (see figure). The steel has shear modulus of elasticity G ϭ 11.0 ϫ 106 psi. For an applied torque of 150 k-in., determine the following quantities: (a) shear stress ␶2 at the outer surface of the shaft, (b) shear stress ␶1 at the inner surface, and (c) rate of twist ␪ (degrees per unit of length). d2 Also, draw a diagram showing how the shear stresses vary in magnitude along a radial line in the cross section. d1 d2 Solution 3.3-11 Construction auger d2 ϭ 6.0 in. C r2 ϭ 3.0 in. d1 ϭ 4.5 in. r1 ϭ 2.25 in. G ϭ 11 ϫ d1 d2 106 psi T ϭ 150 k-in. ␲ 4 4 IP ϭ (d2 Ϫ d1 ) ϭ 86.98 in.4 32 (a) SHEAR STRESS AT OUTER SURFACE Tr2 (150 k-in.)(3.0 in.) t2 ϭ ϭ IP 86.98 in.4 ϭ 5170 psi (b) SHEAR STRESS AT INNER SURFACE t1 ϭ Tr1 r1 ϭ t2 ϭ 3880 psi r2 IP (c) RATE OF TWIST (150 k-in.) T uϭ ϭ GIP (11 ϫ 106 psi)(86.98 in.4 ) u ϭ 157 ϫ 10 Ϫ6 rad րin. ϭ 0.00898Њրin. (d) SHEAR STRESS DIAGRAM 5170 psi 3880 psi C 0 0.75 1.50 2.25 3.00
    • 192 CHAPTER 3 Torsion Problem 3.3-12 Solve the preceding problem if the shaft has outer diameter d2 ϭ 150 mm and inner diameter d1 ϭ 100 mm. Also, the steel has shear modulus of elasticity G ϭ 75 GPa and the applied torque is 16 kNиm. Solution 3.3-12 Construction auger d2 ϭ 150 mm r2 ϭ 75 mm d1 ϭ 100 mm r1 ϭ 50 mm (b) SHEAR STRESS AT INNER SURFACE C t1 ϭ G ϭ 75 GPa T ϭ 16 kN и m ␲ 4 4 IP ϭ (d2 Ϫ d1 ) ϭ 39.88 ϫ 106 mm4 32 (a) SHEAR STRESS AT OUTER SURFACE Tr1 r1 ϭ t2 ϭ 20.1 MPa r2 IP d1 d2 (c) RATE OF TWIST uϭ T 16 kN ؒ m ϭ GIP (75 GPa)(39.88 ϫ 106 mm4 ) u ϭ 0.005349 rad րm ϭ 0.306Њրm (d) SHEAR STRESS DIAGRAM Tr2 (16 kN ؒ m)(75 mm) t2 ϭ ϭ IP 39.88 ϫ 106 mm4 30.1 MPa 20.1 MPa ϭ 30.1 MPa C 25 0 Problem 3.3-13 A vertical pole of solid circular cross section is twisted by horizontal forces P ϭ 1100 lb acting at the ends of a horizontal arm AB (see figure). The distance from the outside of the pole to the line of action of each force is c ϭ 5.0 in. If the allowable shear stress in the pole is 4500 psi, what is the minimum required diameter dmin of the pole? 50 75 r (mm) c P c B A P d
    • SECTION 3.3 Solution 3.3-13 Circular Bars and Tubes Vertical pole c P c tmax ϭ B A P(2c ϩ d)d 16P(2c ϩ d) ϭ ␲d 4ր16 ␲d 3 (␲␶max)d 3 Ϫ (16P)d Ϫ 32Pc ϭ 0 P d P ϭ 1100 lb SUBSTITUTE NUMERICAL VALUES: c ϭ 5.0 in. UNITS: Pounds, Inches ␶allow ϭ 4500 psi (␲)(4500)d 3 Ϫ (16)(1100)d Ϫ 32(1100)(5.0) ϭ 0 Find dmin or d3 Ϫ 1.24495d Ϫ 12.4495 ϭ 0 TORSION FORMULA tmax ϭ d ϭ 2.496 in. Solve numerically: Tr Td ϭ IP 2IP T ϭ P(2c ϩ d) dmin ϭ 2.50 in. IP ϭ ␲d 4 32 Problem 3.3-14 Solve the preceding problem if the horizontal forces have magnitude P ϭ 5.0 kN, the distance c ϭ 125 mm, and the allowable shear stress is 30 MPa. Solution 3.3-14 Vertical pole TORSION FORMULA c P c B A P tmax ϭ Tr Td ϭ IP 2IP T ϭ P(2c ϩ d) IP ϭ d tmax ϭ ␲d 4 32 P(2c ϩ d)d 16P(2c ϩ d) ϭ ␲d 4ր16 ␲d 3 (␲␶max)d 3 Ϫ (16P)d Ϫ 32Pc ϭ 0 SUBSTITUTE NUMERICAL VALUES: P ϭ 5.0 kN UNITS: Newtons, Meters c ϭ 125 mm (␲)(30 ϫ 106)d 3 Ϫ (16)(5000)d Ϫ 32(5000)(0.125) ϭ 0 ␶allow ϭ 30 MPa or Find dmin d 3 Ϫ 848.826 ϫ 10Ϫ6d Ϫ 212.207 ϫ 10Ϫ6 ϭ 0 Solve numerically: d ϭ 0.06438 m dmin ϭ 64.4 mm 193
    • 194 CHAPTER 3 Torsion Problem 3.3-15 A solid brass bar of diameter d ϭ 1.2 in. is subjected to torques T1, as shown in part (a) of the figure. The allowable shear stress in the brass is 12 ksi. d T1 (a) What is the maximum permissible value of the torques T1? (b) If a hole of diameter 0.6 in. is drilled longitudinally through the bar, as shown in part (b) of the figure, what is the maximum permissible value of the torques T2? (c) What is the percent decrease in torque and the percent decrease in weight due to the hole? T1 (a) d T2 T2 (b) Solution 3.3-15 Brass bar in torsion (c) PERCENT DECREASE IN TORQUE (a) SOLID BAR d ϭ 1.2 in. d ␶allow ϭ 12 ksi d1 1 ϭ d2 2 Find max. torque T1 tmax ϭ 16T ␲d 3 T1 ϭ 4 4 d1 4 T2 ␲(d2 Ϫ d1 )tallow 16 ϭ и ϭ1Ϫ¢ ≤ 3 T1 16d2 d2 ␲d2tallow ␲d 3tallow 16 ␲(1.2 in.) 3 (12 ksi) T1 ϭ 16 ϭ 4072 lb-in. T2 ϭ 0.9375 T1 % decrease ϭ 6.25% PERCENT DECREASE IN WEIGHT 2 2 W2 A2 d2 Ϫ d1 d1 2 ϭ ϭ ϭ1Ϫ¢ ≤ 2 W1 A1 d2 d2 (b) BAR WITH A HOLE d1 1 ϭ d2 2 W2 3 ϭ W1 4 tmax ϭ d2 ϭ d ϭ 1.2 in. % decrease ϭ 25% d1 ϭ 0.6 in. d1 d2 NOTE: The hollow bar weighs 25% less than the solid bar with only a 6.25% decrease in strength. 16Td2 Tr Tdր2 ϭ␲ 4 4 ϭ 4 4 IP 32 (d2 Ϫ d1 ) ␲(d2 Ϫ d1 ) T2 ϭ 4 4 ␲(d2 Ϫ d1 )tallow 16d2 T2 ϭ ␲[ (1.2 in.) 4 Ϫ (0.6 in.) 4 ] (12 ksi) 16(1.2 in.) T2 ϭ 3817 lb-in.
    • SECTION 3.3 Circular Bars and Tubes Problem 3.3-16 A hollow aluminum tube used in a roof structure has an outside diameter d2 ϭ 100 mm and an inside diameter d1 ϭ 80 mm (see figure). The tube is 2.5 m long, and the aluminum has shear modulus G ϭ 28 GPa. (a) If the tube is twisted in pure torsion by torques acting at the ends, what is the angle of twist ␾ (in degrees) when the maximum shear stress is 50 MPa? (b) What diameter d is required for a solid shaft (see figure) to resist the same torque with the same maximum stress? (c) What is the ratio of the weight of the hollow tube to the weight of the solid shaft? Solution 3.3-16 ϭ 100 mm d1 ϭ 80 mm L ϭ 2.5 m G ϭ 28 GPa FOR THE SOLID SHAFT: tmax ϭ Tr Td2 tmax ϭ ϭ , Ip 2Ip Tϭ 2Iptmax fϭ 2tmaxL Gd2 fϭ 2(50 MPa)(2.5 m) ϭ 0.08929 rad (28 GPa)(100 mm) f ϭ 5.12Њ (b) DIAMETER OF A SOLID SHAFT ␶max is the same as for tube. Torque is the same. Tϭ d3 ϭ 4 4 d2 Ϫ d1 d2 (100 mm) 4 Ϫ (80 mm) 4 ϭ 590,400 mm3 100 mm d ϭ 83.9 mm d2 2Iptmax L TL fϭ ϭ¢ ≤¢ ≤ GIp d2 GIp For the tube: T ϭ 16T 16 2tmax ␲ 4 4 ¢ ≤¢ ≤ (d Ϫ d1 ) 3ϭ 32 2 ␲d ␲d 3 d2 Solve for d 3: d 3 ϭ ␶max ϭ 50 MPa (a) ANGLE OF TWIST FOR THE TUBE d d Hollow aluminum tube d2 d1 d2 d1 d2 2IPtmax d2 2tmax ␲ 4 4 ¢ ≤ (d Ϫ d1 ) d2 32 2 (c) RATIO OF WEIGHTS 2 2 Wtube Atube d2 Ϫ d1 ϭ ϭ Wsolid Asolid d2 Wtube (100 mm) 2 Ϫ (80 mm) 2 ϭ ϭ 0.51 Wsolid (83.9 mm) 2 The weight of the tube is 51% of the weight of the solid shaft, but they resist the same torque. 195
    • 196 CHAPTER 3 Torsion Problem 3.3-17 A circular tube of inner radius r1 and outer radius r2 is subjected to a torque produced by forces P ϭ 900 lb (see figure). The forces have their lines of action at a distance b ϭ 5.5 in. from the outside of the tube. If the allowable shear stress in the tube is 6300 psi and the inner radius r1 ϭ 1.2 in., what is the minimum permissible outer radius r2? P P P r2 r1 P b Solution 3.3-17 b Circular tube in torsion P SOLUTION OF EQUATION r2 r1 UNITS: Pounds, Inches Substitute numerical values: P b 2r2 2r2 b 6300 psi ϭ 4(900 lb)(5.5 in. ϩ r2 )(r2 ) ␲[ (r4 ) Ϫ (1.2 in.) 4 ] 2 P ϭ 900 lb or b ϭ 5.5 in. r4 Ϫ 2.07360 2 Ϫ 0.181891 ϭ 0 r2 (r2 ϩ 5.5) or ␶allow ϭ 6300 psi r1 ϭ 1.2 in. Find minimum permissible radius r2 2 r4 Ϫ 0.181891 r2 Ϫ1.000402 r2 Ϫ2.07360 ϭ 0 2 Solve numerically: TORSION FORMULA T ϭ 2P(bϩr2) IP ϭ ␲ 4 (r Ϫ r4 ) 1 2 2 tmax ϭ Tr2 2P(b ϩ r2 )r2 4P(b ϩ r2 )r2 ϭ ␲ 4 ϭ 4 IP ␲(r4 Ϫ r4 ) 2 1 2 (r2 Ϫ r1 ) All terms in this equation are known except r2. r2 ϭ 1.3988 in. MINIMUM PERMISSIBLE RADIUS r2 ϭ 1.40 in.
    • SECTION 3.4 Nonuniform Torsion 197 Nonuniform Torsion Problem 3.4-1 A stepped shaft ABC consisting of two solid circular segments is subjected to torques T1 and T2 acting in opposite directions, as shown in the figure. The larger segment of the shaft has diameter d1 ϭ 2.25 in. and length L1 ϭ 30 in.; the smaller segment has diameter d2 ϭ 1.75 in. and length L2 ϭ 20 in. The material is steel with shear modulus G ϭ 11 ϫ 106 psi, and the torques are T1 ϭ 20,000 lb-in. and T2 ϭ 8,000 lb-in. Calculate the following quantities: (a) the maximum shear stress ␶max in the shaft, and (b) the angle of twist ␾C (in degrees) at end C. Solution 3.4-1 T1 d1 d2 B A L1 C L2 Stepped shaft T1 d1 L1 A d1 ϭ 2.25 in. B SEGMENT BC T2 L2 L1 ϭ 30 in. d2 ϭ 1.75 in. d2 L2 ϭ 20 in. TBC ϭ ϩT2 ϭ 8,000 lb-in. C tBC ϭ 16 TBC 16(8,000 lb-in.) ϭ 7602 psi 3 ϭ ␲d2 ␲(1.75 in.) 3 fBC ϭ TBC L2 ϭ G(Ip ) BC (8,000 lb-in.)(20 in.) ␲ (11 ϫ 106 psi) ¢ ≤ (1.75 in.) 4 32 G ϭ 11 ϫ 106 psi ϭ ϩ0.015797 rad T1 ϭ 20,000 lb-in. (a) MAXIMUM SHEAR STRESS T2 ϭ 8,000 lb-in. Segment BC has the maximum stress SEGMENT AB tmax ϭ 7600 psi TAB ϭ T2ϪT1 ϭ Ϫ12,000 lb-in. 16 TAB 16(12,000 lb-in.) tAB ϭ ` ` ϭ ϭ 5365 psi ␲d3 ␲(2.25 in.) 3 1 TABL1 fAB ϭ ϭ G(Ip ) AB (Ϫ12,000 lb-in.)(30 in.) ␲ (11 ϫ 106 psi) ¢ ≤ (2.25 in.) 4 32 ϭ Ϫ0.013007 rad (b) ANGLE OF TWIST AT END C ␾C ϭ ␾AB ϩ ␾BC ϭ (Ϫ0.013007 ϩ 0.015797) rad fC ϭ 0.002790 rad ϭ 0.16Њ T2
    • 198 CHAPTER 3 Torsion Problem 3.4-2 A circular tube of outer diameter d3 ϭ 70 mm and inner diameter d2 ϭ 60 mm is welded at the right-hand end to a fixed plate and at the left-hand end to a rigid end plate (see figure). A solid circular bar of diameter d1 ϭ 40 mm is inside of, and concentric with, the tube. The bar passes through a hole in the fixed plate and is welded to the rigid end plate. The bar is 1.0 m long and the tube is half as long as the bar. A torque T ϭ 1000 N и m acts at end A of the bar. Also, both the bar and tube are made of an aluminum alloy with shear modulus of elasticity G ϭ 27 GPa. Tube Fixed plate End plate Bar T A Tube (a) Determine the maximum shear stresses in both the bar and tube. (b) Determine the angle of twist (in degrees) at end A of the bar. Bar d1 d2 d3 Solution 3.4-2 Bar and tube TORQUE Tube T ϭ 1000 N и m (a) MAXIMUM SHEAR STRESSES Bar Bar: tbar ϭ T A 16T ϭ 79.6 MPa ␲d3 1 Tube: ttube ϭ T(d3ր2) ϭ 32.3 MPa (Ip ) tube (b) ANGLE OF TWIST AT END A TUBE d3 ϭ 70 mm d2 ϭ 60 mm Ltube ϭ 0.5 m G ϭ 27 GPa Bar: fbar ϭ Tube: ftube ϭ ␲ 4 4 (Ip ) tube ϭ (d3 Ϫ d2 ) 32 fA ϭ 9.43Њ BAR (Ip ) bar ϭ 4 ␲d1 32 Lbar ϭ 1.0 m ϭ 251.3 ϫ 103 mm4 TL tube ϭ 0.0171 rad G(Ip ) tube ␾A ϭ ␾bar ϩ ␾tube ϭ 0.1474 ϩ 0.0171 ϭ 0.1645 rad ϭ 1.0848 ϫ 106 mm4 d1 ϭ 40 mm TL bar ϭ 0.1474 rad G(Ip ) bar G ϭ 27 GPa
    • SECTION 3.4 Problem 3.4-3 A stepped shaft ABCD consisting of solid circular segments is subjected to three torques, as shown in the figure. The torques have magnitudes 12.0 k-in., 9.0 k-in., and 9.0 k-in. The length of each segment is 24 in. and the diameters of the segments are 3.0 in., 2.5 in., and 2.0 in. The material is steel with shear modulus of elasticity G ϭ 11.6 ϫ 103 ksi. 12.0 k-in. 3.0 in. (a) Calculate the maximum shear stress ␶max in the shaft. (b) Calculate the angle of twist ␾D (in degrees) at end D. Solution 3.4-3 24 in. 9.0 k-in. 2.5 in. 9.0 k-in. 2.0 in. C D tAB ϭ TAB rAB ϭ 5660 psi (Ip ) AB TBC rBC ϭ 5870 psi (Ip ) BC tCD ϭ B A TCD rCD ϭ 5730 psi (Ip ) CD G ϭ 11.6 ϫ 103 ksi rAB ϭ 1.5 in. rBC ϭ 1.25 in. rCD ϭ 1.0 in. LAB ϭ LBC ϭ LCD ϭ 24 in. tmax ϭ 5870 psi (b) ANGLE OF TWIST AT END D TORQUES TAB ϭ 12.0 ϩ 9.0 ϩ 9.0 ϭ 30 k-in. TBC ϭ 9.0 ϩ 9.0 ϭ 18 k-in. TCD ϭ 9.0 k-in. POLAR MOMENTS OF INERTIA ␲ (3.0 in.) 4 ϭ 7.952 in.4 32 ␲ (Ip ) BC ϭ (2.5 in.) 4 ϭ 3.835 in.4 32 (Ip ) CD ϭ 24 in. (a) SHEAR STRESSES tBC ϭ 3.0 in. ␲ (2.0 in.) 4 ϭ 1.571 in.4 32 fAB ϭ TAB LAB ϭ 0.007805 rad G(Ip ) AB fBC ϭ TBC LBC ϭ 0.009711 rad G(Ip ) BC fCD ϭ TCD LCD ϭ 0.011853 rad G(Ip ) CD ␾D ϭ ␾AB ϩ ␾BC ϩ ␾CD ϭ 0.02937 rad fD ϭ 1.68Њ 9.0 k-in. 2.0 in. D C Stepped shaft 12.0 k-in. (Ip ) AB ϭ 9.0 k-in. 2.5 in. B A 199 Nonuniform Torsion 24 in.
    • 200 CHAPTER 3 Torsion Problem 3.4-4 A solid circular bar ABC consists of two segments, as shown in the figure. One segment has diameter d1 ϭ 50 mm and length L1 ϭ 1.25 m; the other segment has diameter d2 ϭ 40 mm and length L2 ϭ 1.0 m. What is the allowable torque Tallow if the shear stress is not to exceed 30 MPa and the angle of twist between the ends of the bar is not to exceed 1.5°? (Assume G ϭ 80 GPa.) Solution 3.4-4 A T C B L1 d1 = 50 mm A L 1 = 1.25 m ␶allow ϭ 30 MPa L2 G ϭ 80 GPa ALLOWABLE TORQUE BASED UPON SHEAR STRESS Segment BC has the smaller diameter and hence the larger stress. Tallow ϭ d2 = 40 mm L 2 = 1.0 m B T C ALLOWABLE TORQUE BASED UPON ANGLE OF TWIST ␾allow ϭ 1.5Њ ϭ 0.02618 rad 16T ␲d 3 d2 Bar consisting of two segments T tmax ϭ d1 T TiLi TL1 TL2 L2 T L1 fϭ a ϭ ϩ ϭ ¢ ϩ ≤ GIPi GIP1 GIP2 G IP1 IP2 fϭ 32T L1 L2 ¢ 4 ϩ 4≤ ␲G d1 d2 Tallow ϭ 3 ␲d2tallow ϭ 3.77 N # m 16 ␲fallowG ϭ 348 N ؒ m L1 L 2 32 ¢ 4 ϩ 4≤ d1 d2 ANGLE OF TWIST GOVERNS Tallow ϭ 348 N ؒ m Problem 3.4-5 A hollow tube ABCDE constructed of monel metal is subjected to five torques acting in the directions shown in the figure. The magnitudes of the torques are T1 ϭ 1000 lb-in., T2 ϭ T4 ϭ 500 lb-in., and T3 ϭ T5 ϭ 800 lb-in. The tube has an outside diameter d2 ϭ 1.0 in. The allowable shear stress is 12,000 psi and the allowable rate of twist is 2.0°/ft. Determine the maximum permissible inside diameter d1 of the tube. T2 = T1 = 1000 lb-in. 500 lb-in. A B T3 = T4 = 800 lb-in. 500 lb-in. C D d2 = 1.0 in. T5 = 800 lb-in. E
    • SECTION 3.4 Solution 3.4-5 Nonuniform Torsion Hollow tube of monel metal REQUIRED POLAR MOMENT OF INERTIA BASED UPON d1 ALLOWABLE SHEAR STRESS d2 d2 ϭ 1.0 in. tmax ϭ ␶allow ϭ 12,000 psi ␪allow ϭ 2Њ/ft ϭ 0.16667Њ/in. IP ϭ Tmax (d2ր2) ϭ 0.05417 in.4 tallow REQUIRED POLAR MOMENT OF INERTIA BASED UPON ϭ 0.002909 rad/in. ALLOWABLE ANGLE OF TWIST From Table H-2, Appendix H: G ϭ 9500 ksi uϭ TORQUES Tmax GIP IP ϭ Tmax ϭ 0.04704 in.4 Guallow SHEAR STRESS GOVERNS T2 T1 A Tmaxr IP T3 B C T4 T5 D E Required IP ϭ 0.05417 in.4 IP ϭ T1 ϭ 1000 lb-in. T2 ϭ 500 lb-in. T4 ϭ 500 lb-in. T3 ϭ 800 lb-in. T5 ϭ 800 lb-in. INTERNAL TORQUES ␲ 4 4 (d Ϫ d1 ) 32 2 4 4 d1 ϭ d2 Ϫ 32IP 32(0.05417 in.4 ) ϭ (1.0 in.) 4 Ϫ ␲ ␲ ϭ 0.4482 in.4 TAB ϭ Ϫ T1 ϭ Ϫ 1000 lb-in. d1 ϭ 0.818 in. TBC ϭ Ϫ T1 ϩ T2 ϭ Ϫ 500 lb-in. (Maximum permissible inside diameter) TCD ϭ Ϫ T1 ϩ T2 Ϫ T3 ϭ Ϫ 1300 lb-in. TDE ϭ Ϫ T1 ϩ T2 Ϫ T3 ϩ T4 ϭ Ϫ 800 lb-in. Largest torque (absolute value only): Tmax ϭ 1300 lb-in. Problem 3.4-6 A shaft of solid circular cross section consisting of two segments is shown in the first part of the figure. The left-hand segment has diameter 80 mm and length 1.2 m; the right-hand segment has diameter 60 mm and length 0.9 m. Shown in the second part of the figure is a hollow shaft made of the same material and having the same length. The thickness t of the hollow shaft is d/10, where d is the outer diameter. Both shafts are subjected to the same torque. If the hollow shaft is to have the same torsional stiffness as the solid shaft, what should be its outer diameter d? 80 mm 1.2 m 60 mm 0.9 m d 2.1 m d t=— 10 201
    • 202 CHAPTER 3 Torsion Solution 3.4-6 Solid and hollow shafts SOLID SHAFT CONSISTING OF TWO SEGMENTS 80 mm TORSIONAL STIFFNESS kT ϭ 60 mm T f Torque T is the same for both shafts. ‹ For equal stiffnesses, ␾1 ϭ ␾2 1.2 m f1 ϭ © TLi ϭ GIPi 0.9 m 98,741 m Ϫ3 ϭ T(1.2 m) T(0.9 m) ϩ ␲ ␲ G ¢ ≤ (80 mm) 4 G ¢ ≤ (60 mm) 4 32 32 d4 ϭ 3.5569 m d4 3.5569 ϭ 36.023 ϫ 10 Ϫ6 m4 98,741 d ϭ 0.0775 m ϭ 77.5 mm 32T ϭ (29,297 mϪ3 ϩ 69,444 mϪ3 ) ␲G ϭ 32T (98,741 m Ϫ3 ) ␲G HOLLOW SHAFT d = outer diameter d t=— 10 2.1 m d0 ϭ inner diameter ϭ 0.8d TL ϭ GIp G¢ T(2.1 m) ␲ ≤ [d 4 Ϫ (0.8d) 4 ] 32 32T 2.1 m 32T 3.5569 m ϭ ¢ ≤ϭ ¢ ≤ ␲G 0.5904 d 4 ␲G d4 f2 ϭ UNITS: d ϭ meters Problem 3.4-7 Four gears are attached to a circular shaft and transmit the torques shown in the figure. The allowable shear stress in the shaft is 10,000 psi. (a) What is the required diameter d of the shaft if it has a solid cross section? (b) What is the required outside diameter d if the shaft is hollow with an inside diameter of 1.0 in.? 8,000 lb-in. 19,000 lb-in. 4,000 lb-in. A 7,000 lb-in. B C D
    • SECTION 3.4 Solution 3.4-7 203 Nonuniform Torsion Shaft with four gears (b) HOLLOW SHAFT 4,000 lb-in. 19,000 lb-in. 8,000 lb-in. A B ␶allow ϭ 10,000 psi TAB ϭ Ϫ8000 lb-in. C 7,000 lb-in. D TBC ϭ ϩ11,000 lb-in. TCD ϭ ϩ7000 lb-in. Inside diameter d0 ϭ 1.0 in. d Tmax¢ ≤ Tr 2 tmax ϭ tallow ϭ Ip Ip 10,000 psi ϭ (a) SOLID SHAFT tmax ϭ d3 ϭ 16T ␲d 3 ¢ (11,000 lb-in.) ¢ d ≤ 2 ␲ ≤ [d 4 Ϫ (1.0 in.) 4 ] 32 UNITS: d ϭ inches 10,000 ϭ 16Tmax 16(11,000 lb-in.) ϭ ϭ 5.602 in.3 ␲tallow ␲(10,000 psi) 56,023 d d4 Ϫ 1 or Required d ϭ 1.78 in. d4 Ϫ 5.6023 d Ϫ 1 ϭ 0 Solving, d ϭ 1.832 Required d ϭ 1.83 in. Problem 3.4-8 A tapered bar AB of solid circular cross section is twisted by torques T (see figure). The diameter of the bar varies linearly from dA at the left-hand end to dB at the right-hand end. For what ratio dB /dA will the angle of twist of the tapered bar be one-half the angle of twist of a prismatic bar of diameter dA? (The prismatic bar is made of the same material, has the same length, and is subjected to the same torque as the tapered bar.) Hint: Use the results of Example 3-5. Solution 3.4-8 dA dB Problems 3.4-8, 3.4-9 and 3.4-10 T TL b ϩ b ϩ 1 ¢ ≤ G(IP ) A 3b3 dB L TAPERED BAR (From Eq. 3-27) 2 bϭ ANGLE OF TWIST dB dA f1 ϭ 1 f 2 2 or PRISMATIC BAR TL G(IP ) A L B A dA f2 ϭ T Tapered bar AB T f1 ϭ B A T b2 ϩ b ϩ 1 1 ϭ 2 3b3 3b3 Ϫ 2b2 Ϫ 2b Ϫ 2 ϭ 0 SOLVE NUMERICALLY: bϭ dB ϭ 1.45 dA
    • 204 CHAPTER 3 Torsion Problem 3.4-9 A tapered bar AB of solid circular cross section is twisted by torques T ϭ 36,000 lb-in. (see figure). The diameter of the bar varies linearly from dA at the left-hand end to dB at the right-hand end. The bar has length L ϭ 4.0 ft and is made of an aluminum alloy having shear modulus of elasticity G ϭ 3.9 ϫ 106 psi. The allowable shear stress in the bar is 15,000 psi and the allowable angle of twist is 3.0°. If the diameter at end B is 1.5 times the diameter at end A, what is the minimum required diameter dA at end A? (Hint: Use the results of Example 3-5). Solution 3.4-9 Tapered bar T B A dA dB ϭ 1.5 dA T ϭ 36,000 lb-in. T dB L MINIMUM DIAMETER BASED UPON ALLOWABLE ANGLE OF TWIST (From Eq. 3-27) L ϭ 4.0 ft ϭ 48 in. ␤ ϭ dB /dA ϭ 1.5 G ϭ 3.9 ϫ 106 psi fϭ ␶allow ϭ 15,000 psi ϭ 0.0523599 rad MINIMUM DIAMETER BASED UPON ALLOWABLE SHEAR STRESS 16T tmax ϭ 3 ␲dA (36,000 lb-in.)(48 in.) (0.469136) ␲ 4 6 (3.9 ϫ 10 psi) ¢ ≤ dA 32 2.11728 in.4 ϭ 4 dA ϭ ␾allow ϭ 3.0Њ 3 dA 16(36,000 lb-in.) 16T ϭ ϭ ␲tallow ␲(15,000 psi) ϭ 12.2231 in.3 dA ϭ 2.30 in. b2 ϩ b ϩ 1 TL TL ¢ ≤ϭ (0.469136) 3 G(IP ) A G(IP ) A 3b 4 dA ϭ 2.11728 in.4 2.11728 in.4 ϭ fallow 0.0523599 rad ϭ 40.4370 in.4 dA ϭ 2.52 in. ANGLE OF TWIST GOVERNS Min. dA ϭ 2.52 in.
    • SECTION 3.4 Nonuniform Torsion Problem 3.4-10 The bar shown in the figure is tapered linearly from end A to end B and has a solid circular cross section. The diameter at the smaller end of the bar is dA ϭ 25 mm and the length is L ϭ 300 mm. The bar is made of steel with shear modulus of elasticity G ϭ 82 GPa. If the torque T ϭ 180 N и m and the allowable angle of twist is 0.3°, what is the minimum allowable diameter dB at the larger end of the bar? (Hint: Use the results of Example 3-5.) Solution 3.4-10 Tapered bar T B A dA T dB L (0.3Њ) ¢ dA ϭ 25 mm L ϭ 300 mm G ϭ 82 GPa T ϭ 180 N и m ␾allow ϭ 0.3Њ DIAMETER BASED UPON ALLOWABLE ANGLE OF TWIST b2 ϩ b ϩ 1 (180 N # m)(0.3 m) ¢ ≤ ␲ 3b3 (82 GPa) ¢ ≤ (25 mm) 4 32 b2 ϩ b ϩ 1 3b3 0.914745␤3 Ϫ ␤2 Ϫ 1 ϭ 0 SOLVE NUMERICALLY: (From Eq. 3-27) ␤ ϭ 1.94452 dB dA b ϩbϩ1 TL ¢ ≤ G(IP ) A 3b3 Min. dB ϭ bdA ϭ 48.6 mm 2 fϭ ϭ 0.304915 ϭ Find dB bϭ ␲ rad ≤ 180 degrees (IP ) A ϭ ␲ 4 d 32 A Problem 3.4-11 A uniformly tapered tube AB of hollow circular cross section is shown in the figure. The tube has constant wall thickness t and length L. The average diameters at the ends are dA and dB ϭ 2dA. The polar moment of inertia may be represented by the approximate formula IP Ϸ ␲d 3t/4 (see Eq. 3-18). Derive a formula for the angle of twist ␾ of the tube when it is subjected to torques T acting at the ends. B A T T L t t dA dB = 2dA 205
    • 206 CHAPTER 3 Torsion Solution 3.4-11 Tapered tube t ϭ thickness (constant) T T dA, dB ϭ average diameters at the ends A B dB ϭ 2dA L IP ϭ ␲d 3t (approximate formula) 4 ANGLE OF TWIST L L dA O B dB = 2dA d(x) dx x Take the origin of coordinates at point O. For element of length dx: x x d(x) ϭ (d ) ϭ dA 2L B L df ϭ IP (x) ϭ ␲[d(x) ] 3t ␲td3 3 A ϭ x 4 4L3 Tdx ϭ GIP (x) For entire bar: Ύ fϭ 2L df ϭ L Problem 3.4-12 A prismatic bar AB of length L and solid circular cross section (diameter d) is loaded by a distributed torque of constant intensity t per unit distance (see figure). Tdx 4TL3 # dx ϭ 3 3 ␲tdA ␲GtdA x3 G ¢ 3 ≤ x3 4L 4TL3 3 ␲GtdA Ύ 2L L dx 3TL 3 3 ϭ x 2␲GtdA t A (a) Determine the maximum shear stress ␶max in the bar. (b) Determine the angle of twist ␾ between the ends of the bar. B L Solution 3.4-12 Bar with distributed torque (a) MAXIMUM SHEAR STRESS t Tmax ϭ tL A B dx L t ϭ intensity of distributed torque d ϭ diameter G ϭ shear modulus of elasticity x tmax ϭ 16Tmax 16tL ϭ ␲d3 ␲d3 (b) ANGLE OF TWIST ␲d 4 32 T(x)dx 32 tx dx df ϭ ϭ GIp ␲Gd 4 T(x) ϭ tx fϭ Ύ 0 L df ϭ IP ϭ 32t ␲Gd 4 L Ύ x dx ϭ ␲Gd 0 16tL2 4
    • SECTION 3.4 Problem 3.4-13 A prismatic bar AB of solid circular cross section (diameter d) is loaded by a distributed torque (see figure). The intensity of the torque, that is, the torque per unit distance, is denoted t(x) and varies linearly from a maximum value tA at end A to zero at end B. Also, the length of the bar is L and the shear modulus of elasticity of the material is G. Nonuniform Torsion t(x) A (a) Determine the maximum shear stress ␶max in the bar. (b) Determine the angle of twist ␾ between the ends of the bar. Solution 3.4-13 L Bar with linearly varying torque t(x) A B dx L x (a) MAXIMUM SHEAR STRESS 16Tmax 16TA 8tAL ϭ ϭ ␲d3 ␲d3 ␲d3 tmax ϭ t(x) = TA x L tA (b) ANGLE OF TWIST T(x) ϭ torque at distance x from end B T(x) ϭ t(x) ϭ intensity of distributed torque tA ϭ maximum intensity of torque d ϭ diameter G ϭ shear modulus TA ϭ maximum torque ϭ 1 tAL 2 df ϭ fϭ 207 t(x)x tAx2 ϭ 2 2L IP ϭ ␲d 4 32 T(x) dx 16tAx2 dx ϭ GIP ␲GLd 4 Ύ 0 L df ϭ 16tA ␲GLd 4 Ύ 0 L x2dx ϭ 16tAL2 3␲Gd 4 B
    • 208 CHAPTER 3 Torsion Problem 3.4-14 A magnesium-alloy wire of diameter d ϭ 4 mm and length L rotates inside a flexible tube in order to open or close a switch from a remote location (see figure). A torque T is applied manually (either clockwise or counterclockwise) at end B, thus twisting the wire inside the tube. At the other end A, the rotation of the wire operates a handle that opens or closes the switch. A torque T0 ϭ 0.2 N и m is required to operate the switch. The torsional stiffness of the tube, combined with friction between the tube and the wire, induces a distributed torque of constant intensity t ϭ 0.04 Nиm/m (torque per unit distance) acting along the entire length of the wire. T0 = torque Flexible tube B d A t (a) If the allowable shear stress in the wire is ␶allow ϭ 30 MPa, what is the longest permissible length Lmax of the wire? (b) If the wire has length L ϭ 4.0 m and the shear modulus of elasticity for the wire is G ϭ 15 GPa, what is the angle of twist ␾ (in degrees) between the ends of the wire? Solution 3.4-14 Wire inside a flexible tube t T0 d T L d ϭ 4 mm (b) ANGLE OF TWIST ␾ T0 ϭ 0.2 N и m Lϭ4m t ϭ 0.04 N и m/m ␾1 ϭ angle of twist due to distributed torque t ϭ (a) MAXIMUM LENGTH Lmax ␶allow ϭ 30 MPa 16T ␲d3 16tL2 (from problem 3.4-12) ␲Gd 4 ␾2 ϭ angle of twist due to torque T0 Equilibrium: T ϭ tL ϩ T0 From Eq. (3-12): tmax ϭ G ϭ 15 GPa ␲d tmax 16 3 Tϭ ␲d3tmax 16 1 (␲d 3tmax Ϫ 16T0 ) Lϭ 16t 1 (␲d3tallow Ϫ 16T0 ) Lmax ϭ 16t tL ϩ T0 ϭ Substitute numerical values: Lmax ϭ 4.42 m ϭ T0L 32 T0L ϭ (from Eq. 3-15) GIP ␲Gd4 ␾ ϭ total angle of twist ϭ ␾1 ϩ ␾2 fϭ 16L (tL ϩ 2T0 ) ␲Gd 4 Substitute numerical values: f ϭ 2.971 rad ϭ 170Њ T
    • SECTION 3.5 Pure Shear 209 Pure Shear Problem 3.5-1 A hollow aluminum shaft (see figure) has outside diameter d2 ϭ 4.0 in. and inside diameter d1 ϭ 2.0 in. When twisted by torques T, the shaft has an angle of twist per unit distance equal to 0.54°/ft. The shear modulus of elasticity of the aluminum is G ϭ 4.0 ϫ 106 psi. d2 T T L (a) Determine the maximum tensile stress ␴max in the shaft. (b) Determine the magnitude of the applied torques T. d1 d2 Problems 3.5-1, 3.5-2, and 3.5-3 Solution 3.5-1 Hollow aluminum shaft T T d1 d2 d2 ϭ 4.0 in. G ϭ 4.0 ϫ d1 ϭ 2.0 in. 106 ␪ ϭ 0.54Њ/ft psi (a) MAXIMUM TENSILE STRESS ␴max occurs on a 45Њ plane and is equal to ␶max. MAXIMUM SHEAR STRESS smax ϭ tmax ϭ 6280 psi ␶max ϭ Gr␪ (from Eq. 3-7a) (b) APPLIED TORQUE r ϭ d2 /2 ϭ 2.0 in. u ϭ (0.54Њրft) ¢ 1 ft ␲ rad ≤¢ ≤ 12 in. 180 degree ϭ 785.40 ϫ 10 Ϫ6 radրin. ␶max ϭ (4.0 ϫ 106 psi)(2.0 in.)(785.40 ϫ 10Ϫ6 rad/in.) ϭ 6283.2 psi Use the torsion formula tmax ϭ Tϭ tmaxIP r IP ϭ ␲ [ (4.0 in.) 4 Ϫ (2.0 in.) 4 ] 32 ϭ 23.562 in.4 Tϭ (6283.2 psi)(23.562 in.4 ) 2.0 in. ϭ 74,000 lb-in. Tr IP
    • 210 CHAPTER 3 Torsion Problem 3.5-2 A hollow steel bar (G ϭ 80 GPa) is twisted by torques T (see figure). The twisting of the bar produces a maximum shear strain ␥max ϭ 640 ϫ 10Ϫ6 rad. The bar has outside and inside diameters of 150 mm and 120 mm, respectively. (a) Determine the maximum tensile strain in the bar. (b) Determine the maximum tensile stress in the bar. (c) What is the magnitude of the applied torques T ? Solution 3.5-2 Hollow steel bar T T d1 d2 G ϭ 80 GPa ␥max ϭ 640 ϫ 10Ϫ6 rad d2 ϭ 150 mm d1 ϭ 120 mm ␲ 4 4 (d Ϫ d1 ) 32 2 ␲ ϭ [ (150 mm) 4 Ϫ (120 mm) 4 ] 32 IP ϭ ϭ 29.343 ϫ 106 mm4 (b) MAXIMUM TENSILE STRESS ␶max ϭ G␥max ϭ (80 GPa)(640 ϫ 10Ϫ6) ϭ 51.2 MPa smax ϭ tmax ϭ 51.2 MPa (c) APPLIED TORQUES Torsion formula: tmax ϭ (a) MAXIMUM TENSILE STRAIN gmax emax ϭ ϭ 320 ϫ 10 Ϫ6 2 Tϭ 2IPtmax 2(29.343 ϫ 106 mm4 )(51.2 MPa) ϭ d2 150 mm ϭ 20,030 N и m ϭ 20.0 kN # m Problem 3.5-3 A tubular bar with outside diameter d2 ϭ 4.0 in. is twisted by torques T ϭ 70.0 k-in. (see figure). Under the action of these torques, the maximum tensile stress in the bar is found to be 6400 psi. (a) Determine the inside diameter d1 of the bar. (b) If the bar has length L ϭ 48.0 in. and is made of aluminum with shear modulus G ϭ 4.0 ϫ 106 psi, what is the angle of twist ␾ (in degrees) between the ends of the bar? (c) Determine the maximum shear strain ␥max (in radians)? Tr Td2 ϭ IP 2IP
    • SECTION 3.5 Solution 3.5-3 Pure Shear 211 Tubular bar T T d1 d2 L d2 ϭ 4.0 in. T ϭ 70.0 k-in. ϭ 70,000 lb-in. ␴max ϭ 6400 psi L ϭ 48 in. ␶max ϭ ␴max ϭ 6400 psi fϭ (a) INSIDE DIAMETER d1 TL GIp From torsion formula, T ϭ Tr Td2 Torsion formula: tmax ϭ ϭ IP 2IP ∴ fϭ Td2 (70.0 k-in.)(4.0 in.) IP ϭ ϭ 2tmax 2(6400 psi) ϭ ϭ 21.875 in.4 Also, Ip ϭ ␲ 4 ␲ 4 4 (d2 Ϫ d1 ) ϭ [ (4.0 in.) 4 Ϫ d1 ] 32 32 Equate formulas: ␲ 4 [256 in.4 Ϫ d1 ] ϭ 21.875 in.4 32 G ϭ 4.0 ϫ 106 psi 2IPtmax d2 2IPtmax L 2Ltmax ¢ ≤ϭ d2 GIP Gd2 2(48 in.)(6400 psi) ϭ 0.03840 rad (4.0 ϫ 106 psi)(4.0 in.) f ϭ 2.20Њ (c) MAXIMUM SHEAR STRAIN gmax ϭ 6400 psi tmax ϭ G 4.0 ϫ 106 psi ϭ 1600 ϫ 10 Ϫ6 rad Solve for d1: d1 ϭ 2.40 in. (b) ANGLE OF TWIST ␾ Problem 3.5-4 A solid circular bar of diameter d ϭ 50 mm (see figure) is twisted in a testing machine until the applied torque reaches the value T ϭ 500 N и m. At this value of torque, a strain gage oriented at 45° to the axis of the bar gives a reading ⑀ ϭ 339 ϫ 10Ϫ6. What is the shear modulus G of the material? Solution 3.5-4 d = 50 mm Strain gage T 45° Bar in a testing machine T T 45° Strain gage at 45Њ: emax ϭ 339 ϫ SHEAR STRESS (FROM EQ. 3-12) 10Ϫ6 tmax ϭ d ϭ 50 mm T ϭ 500 N . m SHEAR MODULUS SHEAR STRAIN (FROM EQ. 3-33) ␥max ϭ 2emax ϭ 678 ϫ 16T 16(500 N . m) ϭ ϭ 20.372 MPa ␲d 3 ␲(0.050 m) 3 10Ϫ6 Gϭ tmax 20.372 MPa ϭ ϭ 30.0 GPa gmax 678 ϫ 10 Ϫ6 T = 500 N·m
    • 212 CHAPTER 3 Torsion Problem 3.5-5 A steel tube (G ϭ 11.5 ϫ 106 psi) has an outer diameter d2 ϭ 2.0 in. and an inner diameter d1 ϭ 1.5 in. When twisted by a torque T, the tube develops a maximum normal strain of 170 ϫ 10Ϫ6. What is the magnitude of the applied torque T? Solution 3.5-5 Steel tube T T d1 d2 G ϭ 11.5 ϫ 106 psi d2 ϭ 2.0 in. d1 ϭ 1.5 in. emax ϭ 170 ϫ 10Ϫ6 IP ϭ ␲ 2 ␲ 4 (d Ϫ d1 ) ϭ [ (2.0 in.) 4 Ϫ (1.5 in.) 4 ] 32 2 32 ϭ 1.07379 in.4 Equate expressions: Td2 ϭ Ggmax 2IP SOLVE FOR TORQUE Tϭ SHEAR STRAIN (FROM EQ. 3-33) 2GIPgmax d2 2(11.5 ϫ 106 psi)(1.07379 in.4 )(340 ϫ 10 Ϫ6 ) 2.0 in. ␥max ϭ 2emax ϭ 340 ϫ 10Ϫ6 ϭ SHEAR STRESS (FROM TORSION FORMULA) ϭ 4200 lb-in. tmax ϭ Tr Td2 ϭ IP 2IP Also, ␶max ϭ G␥max Problem 3.5-6 A solid circular bar of steel (G ϭ 78 GPa) transmits a torque T ϭ 360 Nиm. The allowable stresses in tension, compression, and shear are 90 MPa, 70 MPa, and 40 MPa, respectively. Also, the allowable tensile strain is 220 ϫ 10Ϫ6. Determine the minimum required diameter d of the bar.
    • SECTION 3.5 Solution 3.5-6 Pure Shear 213 Solid circular bar of steel T ϭ 360 N . m G ϭ 78 GPa DIAMETER BASED UPON ALLOWABLE TENSILE STRAIN gmax ϭ 2emax; tmax ϭ Ggmax ϭ 2Gemax ALLOWABLE STRESSES tmax ϭ Tension: 90 MPa Compression: 70 MPa Shear: 40 MPa Allowable tensile strain: emax ϭ 220 ϫ 10Ϫ6 d3 ϭ DIAMETER BASED UPON ALLOWABLE STRESS 16T ␲d 3 d3 ϭ 16T 16T ϭ ␲tmax 2␲Gemax 16(360 N . m) 2␲(78 GPa)(220 ϫ 10 Ϫ6 ) ϭ 53.423 ϫ 10 Ϫ6 m3 The maximum tensile, compressive, and shear stresses in a bar in pure torsion are numerically equal. Therefore, the lowest allowable stress (shear stress) governs. TENSILE STRAIN GOVERNS tallow ϭ 40 MPa dmin ϭ 37.7 mm tmax ϭ 16T ␲d 3 d3 ϭ d ϭ 0.0377 m ϭ 37.7 mm 16(360 N . m) 16T ϭ ␲tallow ␲(40 MPa) d3 ϭ 45.837 ϫ 10 Ϫ6 m3 d ϭ 0.0358 m ϭ 35.8 mm Problem 3.5-7 The normal strain in the 45° direction on the surface of a circular tube (see figure) is 880 ϫ 10Ϫ6 when the torque T ϭ 750 lb-in. The tube is made of copper alloy with G ϭ 6.2 ϫ 106 psi. If the outside diameter d2 of the tube is 0.8 in., what is the inside diameter d1? Solution 3.5-7 Strain gage T = 750 lb-in. d 2 = 0.8 in. T 45° Circular tube with strain gage T T d1 d2 45° d2 ϭ 0.80 in. T ϭ 750 lb-in. G ϭ 6.2 ϫ 106 psi Strain gage at 45Њ: emax ϭ 880 ϫ 10Ϫ6 MAXIMUM SHEAR STRAIN ␥max ϭ 2emax tmax ϭ Ggmax ϭ 2Gemax T(d2ր2) IP Td2 ␲ 4 4 (d2 Ϫ d1 ) ϭ 32 4Gemax 4 4 d2 Ϫ d1 ϭ 8Td2 ␲Gemax 4 4 d1 ϭ d2 Ϫ 8Td2 ␲Gemax INSIDE DIAMETER MAXIMUM SHEAR STRESS tmax ϭ IP ϭ IP ϭ Td2 Td2 ϭ 2tmax 4Gemax Substitute numerical values: 8(750 lb-in.)(0.80 in.) 4 d1 ϭ (0.8 in.) 4 Ϫ ␲(6.2 ϫ 106 psi)(880 ϫ 10 Ϫ6 ) ϭ 0.4096 in.4 Ϫ 0.2800 in.4 ϭ 0.12956 in.4 d1 ϭ 0.60 in.
    • 214 CHAPTER 3 Torsion Problem 3.5-8 An aluminum tube has inside diameter d1 ϭ 50 mm, shear modulus of elasticity G ϭ 27 GPa, and torque T ϭ 4.0 kN и m. The allowable shear stress in the aluminum is 50 MPa and the allowable normal strain is 900 ϫ 10Ϫ6. Determine the required outside diameter d2. Solution 3.5-8 Aluminum tube T T d1 d2 d1 ϭ 50 mm G ϭ 27 GPa NORMAL STRAIN GOVERNS T ϭ 4.0 kN и m ␶allow ϭ 50 MPa eallow ϭ 900 ϫ 10Ϫ6 ␶allow ϭ 48.60 MPa Determine the required diameter d2. REQUIRED DIAMETER ALLOWABLE SHEAR STRESS (␶allow)1 ϭ 50 MPa ALLOWABLE SHEAR STRESS BASED ON NORMAL STRAIN g t emax ϭ ϭ 2 2G t ϭ 2Gemax (␶allow)2 ϭ 2Geallow ϭ 2(27 GPa)(900 ϫ 10Ϫ6) ϭ 48.6 MPa tϭ Tr IP (4000 N # m)(d2ր2) ␲ 4 [d Ϫ (0.050 m) 4 ] 32 2 Rearrange and simplify: 4 d2 Ϫ (419.174 ϫ 10 Ϫ6 )d2 Ϫ 6.25 ϫ 10 Ϫ6 ϭ 0 Solve numerically: d2 ϭ 0.07927 m d2 ϭ 79.3 mm Problem 3.5-9 A solid steel bar (G ϭ 11.8 ϫ 106 psi) of diameter d ϭ 2.0 in. is subjected to torques T ϭ 8.0 k-in. acting in the directions shown in the figure. (a) Determine the maximum shear, tensile, and compressive stresses in the bar and show these stresses on sketches of properly oriented stress elements. (b) Determine the corresponding maximum strains (shear, tensile, and compressive) in the bar and show these strains on sketches of the deformed elements. 48.6 MPa ϭ T d = 2.0 in. T = 8.0 k-in.
    • SECTION 3.5 Solution 3.5-9 Solid steel bar d = 2.0 in. T 215 Pure Shear T = 8.0 k-in. G = 11.8x106 psi (b) MAXIMUM STRAINS gmax ϭ T ϭ 8.0 k-in. G ϭ 11.8 ϫ 106 psi ϭ 432 ϫ 10 Ϫ6 rad (a) MAXIMUM STRESSES tmax ϭ 5093 psi tmax ϭ G 11.8 ϫ 106 psi emax ϭ 16T 16(8000 lb-in.) ϭ ␲d 3 ␲(2.0 in.) 3 gmax ϭ 216 ϫ 10 Ϫ6 2 et ϭ 216 ϫ 10 Ϫ6 ec ϭ Ϫ 216 ϫ 10 Ϫ6 ϭ 5093 psi st ϭ 5090 psi sc ϭ Ϫ5090 psi ␥max = 4.32 × 10−6 rad ␴c = 5090 psi 45° y 0 x Tmax = 5090 psi 45° y x ⑀c = 216 × 10−6 0 ␴t = 5090 psi ⑀t = 216 × 10−6 1 Problem 3.5-10 A solid aluminum bar (G ϭ 27 GPa) of diameter d ϭ 40 mm is subjected to torques T ϭ 300 N и m acting in the directions shown in the figure. (a) Determine the maximum shear, tensile, and compressive stresses in the bar and show these stresses on sketches of properly oriented stress elements. (b) Determine the corresponding maximum strains (shear, tensile, and compressive) in the bar and show these strains on sketches of the deformed elements. d = 40 mm T 1 T = 300 N·m
    • 216 CHAPTER 3 Solution 3.5-10 Torsion Solid aluminum bar d = 40 mm T = 300 N · m T G = 27 GPa (b) MAXIMUM STRAINS (a) MAXIMUM STRESSES gmax ϭ 16T 16(300 N ؒ m) tmax ϭ 3 ϭ ␲d ␲(0.040 m) 3 ϭ 884 ϫ 10 Ϫ6 rad gmax emax ϭ ϭ 442 ϫ 10 Ϫ6 2 et ϭ 442 ϫ 10 Ϫ6 ec ϭ Ϫ442 ϫ 10 Ϫ6 ϭ 23.87 MPa st ϭ 23.9 MPa tmax 23.87 MPa ϭ G 27 GPa sc ϭ Ϫ 23.9 MPa ␴t = 23.9 MPa ␥max = 884 × 10−6 rad 45° y y = 0 x ␶max 23.9 MPa x 45° 0 ⑀c = 442 × 10−6 ␴c = 23.9 MPa 1 1 ⑀t = 442 × 10−6 Transmission of Power Problem 3.7-1 A generator shaft in a small hydroelectric plant turns at 120 rpm and delivers 50 hp (see figure). 120 rpm (a) If the diameter of the shaft is d ϭ 3.0 in., what is the maximum shear stress ␶max in the shaft? (b) If the shear stress is limited to 4000 psi, what is the minimum permissible diameter dmin of the shaft? Solution 3.7-1 n ϭ 120 rpm d 50 hp Generator shaft H ϭ 50 hp d ϭ diameter (a) MAXIMUM SHEAR STRESS ␶max d ϭ 3.0 in. TORQUE Hϭ 2␲nT 33,000 Tϭ 33,000 H (33,000)(50 hp) ϭ 2␲n 2␲(120 rpm) H ϭ hp n ϭ rpm ϭ 2188 lb-ft ϭ 26,260 lb-in. T ϭ lb-ft tmax ϭ 16T 16(26,260 lb-in.) ϭ ␲d 3 ␲(3.0 in.) 3 tmax ϭ 4950 psi (b) MINIMUM DIAMETER dmin tallow ϭ 4000 psi d3 ϭ 16(26,260 lb-in.) 16T ϭ ϭ 334.44 in.3 ␲tallow ␲(4000 psi) dmin ϭ 3.22 in.
    • SECTION 3.7 Problem 3.7-2 A motor drives a shaft at 12 Hz and delivers 20 kW of power (see figure). 12 Hz d (a) If the shaft has a diameter of 30 mm, what is the maximum shear stress ␶max in the shaft? 20 kW (b) If the maximum allowable shear stress is 40 MPa, what is the minimum permissible diameter dmin of the shaft? Solution 3.7-2 f ϭ 12 Hz Motor-driven shaft P ϭ 20 kW ϭ 20,000 N и m/s tmax ϭ TORQUE P ϭ 2␲f T P ϭ watts f ϭ Hz ϭ sϪ1 T ϭ Newton meters Tϭ P 20,000 W ϭ ϭ 265.3 N . m 2␲f 2␲(12 Hz) 16T 16(265.3 N . m) ϭ ␲d 3 ␲(0.030 m) 3 ϭ 50.0 MPa (b) MINIMUM DIAMETER dmin ␶allow ϭ 40 MPa d3 ϭ (a) MAXIMUM SHEAR STRESS ␶max 16(265.3 N . m) 16T ϭ ␲tallow ␲(40 MPa) ϭ 33.78 ϫ 10 Ϫ6 m3 d ϭ 30 mm dmin ϭ 0.0323 m ϭ 32.3 mm Problem 3.7-3 The propeller shaft of a large ship has outside diameter 18 in. and inside diameter 12 in., as shown in the figure. The shaft is rated for a maximum shear stress of 4500 psi. 18 in. 100 rpm (a) If the shaft is turning at 100 rpm, what is the maximum horsepower that can be transmitted without exceeding the allowable stress? (b) If the rotational speed of the shaft is doubled but the power requirements remain unchanged, what happens to the shear stress in the shaft? Solution 3.7-3 Tϭ 2tallow IP d2 2(4500 psi)(8270.2 in.4 ) 18 in. ϭ 4.1351 ϫ 106 lb-in. ϭ 344,590 lb-ft. Tϭ (a) HORSEPOWER n ϭ 100 rpm n ϭ rpm TORQUE T(d2 ր2) IP 12 in. 18 in. Hollow propeller shaft d2 ϭ 18 in. d1 ϭ 12 in. ␶allow ϭ 4500 psi ␲ 4 4 IP ϭ (d2 Ϫ d1 ) ϭ 8270.2 in.4 32 tmax ϭ 217 Transmission of Power 2␲ nT 33,000 T ϭ lb-ft H ϭ hp Hϭ 2␲(100 rpm)(344,590 lb-ft) 33,000 ϭ 6560 hp Hϭ (b) ROTATIONAL SPEED IS DOUBLED Hϭ 2␲nT 33,000 If n is doubled but H remains the same, then T is halved. If T is halved, so is the maximum shear stress. ∴ Shear stress is halved
    • 218 CHAPTER 3 Torsion Problem 3.7-4 The drive shaft for a truck (outer diameter 60 mm and inner diameter 40 mm) is running at 2500 rpm (see figure). 2500 rpm 60 mm (a) If the shaft transmits 150 kW, what is the maximum shear stress in the shaft? (b) If the allowable shear stress is 30 MPa, what is the maximum power that can be transmitted? 40 mm 60 mm Solution 3.7-4 Drive shaft for a truck d2 ϭ 60 mm IP ϭ d1 ϭ 40 mm n ϭ 2500 rpm tmax ϭ ␲ 4 (d Ϫ d4 ) ϭ 1.0210 ϫ 10 Ϫ6 m4 1 32 2 Td2 (572.96 N . m)(0.060 m) ϭ 2 IP 2(1.0210 ϫ 10 Ϫ6 m4 ) ϭ 16.835 MPa tmax ϭ 16.8 MPa (a) MAXIMUM SHEAR STRESS ␶max P ϭ power (watts) P ϭ 150 kW ϭ 150,000 W (b) MAXIMUM POWER Pmax T ϭ torque (newton meters) tallow ϭ 30 MPa n ϭ rpm 2␲nT 60P Pϭ Tϭ 60 2␲n 60(150,000 W) Tϭ ϭ 572.96 N . m 2␲(2500 rpm) . Pmax ϭ P tallow 30 MPa ϭ (150 kW) ¢ ≤ tmax 16.835 MPa ϭ 267 kW Problem 3.7-5 A hollow circular shaft for use in a pumping station is being designed with an inside diameter equal to 0.75 times the outside diameter. The shaft must transmit 400 hp at 400 rpm without exceeding the allowable shear stress of 6000 psi. Determine the minimum required outside diameter d. Solution 3.7-5 Hollow shaft d ϭ outside diameter d0 ϭ inside diameter ϭ 0.75 d H ϭ 400 hp ␲ 4 [d Ϫ (0.75 d) 4 ] ϭ 0.067112 d 4 32 TORQUE Hϭ 2␲nT 33,000 Tϭ T ϭ lb-ft 33,000 H (33,000)(400 hp) ϭ 2␲n 2␲(400 rpm) ϭ 5252.1 lb-ft ϭ 63,025 lb-in. n ϭ 400 rpm ␶allow ϭ 6000 psi IP ϭ H ϭ hp n ϭ rpm MINIMUM OUTSIDE DIAMETER tmax ϭ Td 2IP IP ϭ 0.067112 d 4 ϭ Td Td ϭ 2tmax 2tallow (63,025 lb-in.)(d) 2(6000 psi) d 3 ϭ 78.259 in.3 dmin ϭ 4.28 in.
    • SECTION 3.7 219 Transmission of Power Problem 3.7-6 A tubular shaft being designed for use on a construction site must transmit 120 kW at 1.75 Hz. The inside diameter of the shaft is to be one-half of the outside diameter. If the allowable shear stress in the shaft is 45 MPa, what is the minimum required outside diameter d? Solution 3.7-6 Tubular shaft d ϭ outside diameter T ϭ newton meters d0 ϭ inside diameter Tϭ ϭ 0.5 d P ϭ 120 kW ϭ 120,000 W f ϭ 1.75 Hz MINIMUM OUTSIDE DIAMETER ␶allow ϭ 45 MPa IP ϭ tmax ϭ ␲ 4 [d Ϫ (0.5 d) 4 ] ϭ 0.092039 d 4 32 P ϭ watts Td 2IP IP ϭ Td Td ϭ 2tmax 2tallow (10,913.5 N . m)(d) 2(45 MPa) 0.092039 d4 ϭ TORQUE P ϭ 2␲f T P 120,000 W ϭ ϭ 10,913.5 N . m 2␲f 2␲(1.75 Hz) d3 ϭ 0.0013175 m3 f ϭ Hz d ϭ 0.1096 m dmin ϭ 110 mm Problem 3.7-7 A propeller shaft of solid circular cross section and diameter d is spliced by a collar of the same material (see figure). The collar is securely bonded to both parts of the shaft. What should be the minimum outer diameter d1 of the collar in order that the splice can transmit the same power as the solid shaft? Solution 3.7-7 d1 SOLID SHAFT 16T1 ␲d 3 T2 ϭ d T EQUATE TORQUES T1 ϭ ␲d 3tmax 16 HOLLOW COLLAR IP ϭ d Splice in a propeller shaft T tmax ϭ d1 ␲ 4 (d1 Ϫ d 4) 32 tmax ϭ T2r T2 (d1ր2) ϭ IP IP 2tmaxIP 2tmax ␲ 4 ϭ ¢ ≤ (d1 Ϫ d 4) d1 d1 32 ␲tmax 4 ϭ (d1 Ϫ d 4) 16 d1 For the same power, the torques must be the same. For the same material, both parts can be stressed to the same maximum stress. ∴ T1 ϭ T2 or ¢ ␲d3tmax ␲tmax 4 ϭ (d1 Ϫ d 4) 16 16d1 d1 4 d1 ≤ Ϫ Ϫ1ϭ0 d d MINIMUM OUTER DIAMETER Solve Eq. (1) numerically: Min. d1 ϭ 1.221 d (Eq. 1)
    • 220 CHAPTER 3 Torsion Problem 3.7-8 What is the maximum power that can be delivered by a hollow propeller shaft (outside diameter 50 mm, inside diameter 40 mm, and shear modulus of elasticity 80 GPa) turning at 600 rpm if the allowable shear stress is 100 MPa and the allowable rate of twist is 3.0°/m? Solution 3.7-8 d2 ϭ 50 mm G ϭ 80 GPa Hollow propeller shaft d1 ϭ 40 mm n ϭ 600 rpm BASED UPON ALLOWABLE RATE OF TWIST uϭ tallow ϭ 100 MPa uallow ϭ 3.0Њրm ␲ 4 4 IP ϭ (d2 Ϫ d1 ) ϭ 362.3 ϫ 10 Ϫ9 m4 32 T1 (d2ր2) IP T1 ϭ T2 ϭ GIPuallow T2 ϭ (80 GPa)(362.3 ϫ 10 Ϫ9 m4 )(3.0Њրm) ␲ ϫ¢ rad րdegree ≤ 180 BASED UPON ALLOWABLE SHEAR STRESS tmax ϭ T2 GIP T2 ϭ 1517 N . m 2tallow IP d2 SHEAR STRESS GOVERNS 2(100 MPa)(362.3 ϫ 10 Ϫ9 m4 ) T1 ϭ 0.050 m ϭ 1449 N . m Tallow ϭ T1 ϭ 1449 N . m MAXIMUM POWER Pϭ 2␲nT 2␲(600 rpm)(1449 N . m) ϭ 60 60 P ϭ 91,047 W Pmax ϭ 91.0 kW Problem 3.7-9 A motor delivers 275 hp at 1000 rpm to the end of a shaft (see figure). The gears at B and C take out 125 and 150 hp, respectively. Determine the required diameter d of the shaft if the allowable shear stress is 7500 psi and the angle of twist between the motor and gear C is limited to 1.5°. (Assume G ϭ 11.5 ϫ 106 psi, L1 ϭ 6 ft, and L2 ϭ 4 ft.) Motor C A d B L1 L2
    • SECTION 3.7 Solution 3.7-9 Motor-driven shaft 150 hp 125 hp 275 hp A B L1 L2 C L1 ϭ 6 ft DIAMETER BASED UPON ALLOWABLE SHEAR STRESS The larger torque occurs in segment AB tmax ϭ L2 ϭ 4 ft ϭ d ϭ diameter n ϭ 1000 rpm G ϭ 11.5 ϫ 106 psi IP ϭ TORQUES ACTING ON THE SHAFT Tϭ H ϭ hp n ϭ rpm T ϭ lb-ft ϭ 33,000(275 hp) 2␲(1000 rpm) ϭ 1444 lb-ft ϭ 17,332 lb-in. 125 At point B: TB ϭ TA ϭ 7878 lb-in. 275 150 TA ϭ 9454 lb-in. At point C: TC ϭ 275 FREE-BODY DIAGRAM 6 ft TA ϭ 17,332 lb-in. TC ϭ 9454 lb-in. 16(17,332 lb-in.) ϭ 11.77 in.3 ␲(7500 psi) ␲d 4 32 fAB ϭ At point A: TA ϭ A 16TAB ␲tallow fϭ TL 32TL ϭ GIP ␲Gd 4 Segment AB: 33,000 H 2␲n TA = 17,332 lb-in. d3 ϭ DIAMETER BASED UPON ALLOWABLE ANGLE OF TWIST (fAC ) allow ϭ 1.5Њ ϭ 0.02618 rad 2␲nT 33,000 16TAB ␲d 3 d ϭ 2.27 in. ␶allow ϭ 7500 psi Hϭ Transmission of Power TC = 9454 lb-in. B 4 ft TB = 7878 lb-in. C fAB ϭ 32 TAB LAB ␲Gd 4 32(17,330 lb-in.)(6 ft)(12 in.րft) ␲(11.5 ϫ 106 psi)d 4 1.1052 d4 Segment BC: fBC ϭ ϭ 32 TBC LBC ␲Gd 4 32(9450 lb-in.)(4 ft)(12 in.րft) ␲(11.5 ϫ 106 psi)d 4 fBC ϭ 0.4018 d4 From A to C: fAC ϭ fAB ϩ fBC ϭ (␾AC) allow ϭ 0.02618 rad ∴ 0.02618 ϭ 1.5070 d4 d ϭ diameter Angle of twist governs TB ϭ 7878 lb-in. d ϭ 2.75 in. INTERNAL TORQUES TAB ϭ 17,332 lb-in. TBC ϭ 9454 lb-in. 1.5070 d4 and d ϭ 2.75 in. 221
    • 222 CHAPTER 3 Torsion Problem 3.7-10 The shaft ABC shown in the figure is driven by a motor that delivers 300 kW at a rotational speed of 32 Hz. The gears at B and C take out 120 and 180 kW, respectively. The lengths of the two parts of the shaft are L1 ϭ 1.5 m and L2 ϭ 0.9 m. Determine the required diameter d of the shaft if the allowable shear stress is 50 MPa, the allowable angle of twist between points A and C is 4.0°, and G ϭ 75 GPa. Solution 3.7-10 Motor-driven shaft 300 kW A 120 kW B L1 180 kW L2 C L1 ϭ 1.5 m INTERNAL TORQUES TAB ϭ 1492 N и m TBC ϭ 895.3 N и m L2 ϭ 0.9 m DIAMETER BASED UPON ALLOWABLE SHEAR STRESS d ϭ diameter The larger torque occurs in segment AB f ϭ 32 Hz tmax ϭ ␶allow ϭ 50 MPa 16 TAB ␲ d3 d3 ϭ d3 ϭ 0.0001520 m3 G ϭ 75 GPa 16 TAB 16(1492 N . m) ϭ ␲tallow ␲(50 MPa) d ϭ 0.0534 m ϭ 53.4 mm (fAC ) allow ϭ 4Њ ϭ 0.06981 rad DIAMETER BASED UPON ALLOWABLE ANGLE OF TWIST TORQUES ACTING ON THE SHAFT IP ϭ P ϭ 2␲f T P ϭ watts f ϭ Hz ␲d 4 32 fϭ TL 32TL ϭ GIP ␲Gd 4 Segment AB: T ϭ newton meters fAB ϭ 300,000 W At point A: TA ϭ ϭ 1492 N ؒ m 2␲(32 Hz) 32 TAB LAB 32(1492 N . m)(1.5 m) ϭ ␲Gd 4 ␲(75 GPa)d 4 fAB ϭ P Tϭ 2␲f 0.3039 ϫ 10 Ϫ6 d4 Segment BC: 120 At point B: TB ϭ T ϭ 596.8 N ؒ m 300 A fBC ϭ 180 T ϭ 895.3 N ؒ m 300 A 32 TBC LBC 32(895.3 N . m)(0.9 m) ϭ ␲Gd 4 ␲(75 GPa)d 4 fBC ϭ 0.1094 ϫ 10 Ϫ6 d4 At point C: TC ϭ FREE-BODY DIAGRAM TA = 1492 N . m A 1.5 m TA ϭ 1492 N и m TB ϭ 596.8 N и m TC ϭ 895.3 N и m d ϭ diameter TC = 895.3 N . m B 0.9 m TB = 596.8 N . m C From A to C: fAC ϭ fAB ϩ fBC ϭ (␾AC)allow ϭ 0.06981 rad 0.4133 ϫ 10 Ϫ6 d4 and d ϭ 0.04933 m ϭ 49.3 mm ∴ 0.06981 ϭ SHEAR STRESS GOVERNS d ϭ 53.4 mm 0.4133 ϫ 10 Ϫ6 d4
    • SECTION 3.8 223 Statically Indeterminate Torsional Members Statically Indeterminate Torsional Members Problem 3.8-1 A solid circular bar ABCD with fixed supports is acted upon by torques T0 and 2T0 at the locations shown in the figure. Obtain a formula for the maximum angle of twist ␾max of the bar. (Hint: Use Eqs. 3-46a and b of Example 3-9 to obtain the reactive torques.) T0 TA B A 2T0 C 3L — 10 D 3L — 10 TD 4L — 10 L Solution 3.8-1 TA Circular bar with fixed ends ANGLE OF TWIST AT SECTION B T0 A B TB T0 TA LA B A 2T0 C TD D LB 3L — 10 L 3L — 10 4L — 10 From Eqs. (3-46a and b): T0LB TA ϭ L TB ϭ T0LA L fB ϭ fAB ϭ 9T0L TA (3Lր10) ϭ GIP 20GIP ANGLE OF TWIST AT SECTION C fC ϭ fCD ϭ APPLY THE ABOVE FORMULAS TO THE GIVEN BAR: TA ϭ T0 ¢ TD ϭ T0 ¢ 15T0 7 4 ≤ ϩ 2T0 ¢ ≤ ϭ 10 10 10 15T0 3 6 ≤ ϩ 2T0 ¢ ≤ ϭ 10 10 10 TD (4Lր10) 3T0L ϭ GIP 5GIP MAXIMUM ANGLE OF TWIST fmax ϭ fC ϭ Problem 3.8-2 A solid circular bar ABCD with fixed supports at ends A and D is acted upon by two equal and oppositely directed torques T0, as shown in the figure. The torques are applied at points B and C, each of which is located at distance x from one end of the bar. (The distance x may vary from zero to L/2.) (a) For what distance x will the angle of twist at points B and C be a maximum? (b) What is the corresponding angle of twist ␾max? (Hint: Use Eqs. 3-46a and b of Example 3-9 to obtain the reactive torques.) 3T0 L 5GIP T0 TA B A T0 C D x x L TD
    • 224 CHAPTER 3 Torsion Solution 3.8-2 Circular bar with fixed ends T0 TA (a) ANGLE OF TWIST AT SECTIONS B AND C A TD B fB ϭ fAB ϭ LA LB T0 dfB ϭ (L Ϫ 4x) dx GIPL dfB ϭ 0; L Ϫ 4x ϭ 0 dx L or x ϭ 4 L From Eqs. (3-46a and b): TA ϭ T0 LB L TB ϭ T0 LA L (b) MAXIMUM ANGLE OF TWIST fmax ϭ (fB ) max ϭ (fB ) xϭ L ϭ 4 APPLY THE ABOVE FORMULAS TO THE GIVEN BAR: T0 TA T0 TAx ϭ (L Ϫ 2x)(x) GIP GIPL T0 B A C T0L 8GIP D TD x x L TA ϭ T0 (L Ϫ x) T0 x T0 Ϫ ϭ (L Ϫ 2x) L L L TD ϭ TA Disk Problem 3.8-3 A solid circular shaft AB of diameter d is fixed against rotation at both ends (see figure). A circular disk is attached to the shaft at the location shown. What is the largest permissible angle of rotation ␾max of the disk if the allowable shear stress in the shaft is ␶allow? (Assume that a Ͼ b. Also, use Eqs. 3-46a and b of Example 3-9 to obtain the reactive torques.) Solution 3.8-3 TA A A d B a Shaft fixed at both ends To d Disk B TB Since a Ͼ b, the larger torque (and hence the larger stress) is in the right hand segment. tmax ϭ a b T0 ϭ L ϭ aϩb TB (dր2) T0 ad ϭ IP 2LIP 2LIPtmax ad (T0 ) max ϭ 2LIPtallow ad aϾb Assume that a torque T0 acts at the disk. The reactive torques can be obtained from Eqs. (3-46a and b): TA ϭ T0b L b TB ϭ T0a L ANGLE OF ROTATION OF THE DISK (FROM Eq. 3-49) fϭ T0ab GLIP fmax ϭ (T0 ) maxab 2btallow ϭ GLIP Gd
    • SECTION 3.8 Statically Indeterminate Torsional Members Problem 3.8-4 A hollow steel shaft ACB of outside diameter 50 mm and inside diameter 40 mm is held against rotation at ends A and B (see figure). Horizontal forces P are applied at the ends of a vertical arm that is welded to the shaft at point C. Determine the allowable value of the forces P if the maximum permissible shear stress in the shaft is 45 MPa. (Hint: Use Eqs. 3-46a and b of Example 3-9 to obtain the reactive torques.) 225 200 mm A P 200 mm C B P 600 mm 400 mm Solution 3.8-4 Hollow shaft with fixed ends GENERAL FORMULAS: From Eqs. (3-46a and b): TA A TA ϭ TB ϭ TO B LA TB T0 LB L T0 LA L LB L APPLY THE ABOVE FORMULAS TO THE GIVEN SHAFT TA The larger torque, and hence the larger shear stress, occurs in part CB of the shaft. A TO ‹ Tmax ϭ TB ϭ 0.24 P C B T0 ϭ P(400 mm) 400 mm LB ϭ 400 mm LA ϭ 600 mm L ϭ LA ϩ LB ϭ 1000 mm d1 ϭ 40 mm ␶allow ϭ 45 MPa T0 LB P(0.4 m)(400 mm) TA ϭ ϭ ϭ 0.16 P L 1000 mm TB ϭ T0 LA P(0.4 m)(600 mm) ϭ ϭ 0.24 P L 1000 mm UNITS: P ϭ Newtons SHEAR STRESS IN PART CB tmax ϭ 600 mm d2 ϭ 50 mm TB T ϭ Newton meters Tmax (dր2) IP Tmax ϭ 2tmaxIP d (Eq. 1) UNITS: Newtons and meters ␲max ϭ 45 ϫ 106N/m2 ␲ 4 4 IP ϭ (d2 Ϫ d1 ) ϭ 362.26 ϫ 10 Ϫ9m4 32 d ϭ d2 ϭ 0.05 mm Substitute numerical values into (Eq. 1): 2(45 ϫ 106 Nրm2 )(362.26 ϫ 10 Ϫ9m4 ) 0.05 m .m ϭ 652.07 N 652.07 N . m Pϭ ϭ 2717 N 0.24 m 0.24P ϭ Pallow ϭ 2710 N
    • 226 CHAPTER 3 Torsion Problem 3.8-5 A stepped shaft ACB having solid circular cross sections with two different diameters is held against rotation at the ends (see figure). If the allowable shear stress in the shaft is 6000 psi, what is the maximum torque (T0)max that may be applied at section C? (Hint: Use Eqs. 3-45a and b of Example 3-9 to obtain the reactive torques.) Solution 3.8-5 1.50 in. 0.75 in. C A B T0 6.0 in. 15.0 in. Stepped shaft ACB dB dA TA C A B TB T0 LA LB dA ϭ 0.75 in. ALLOWABLE TORQUE BASED UPON SHEAR STRESS CB dB ϭ 1.50 in. IN SEGMENT LA ϭ 6.0 in. tCB ϭ LB ϭ 15.0 in. 16TB 3 ␲dB 1 1 3 ␲dBtCB ϭ ␲d3 t 16 16 B allow ␶allow ϭ 6000 psi TB ϭ Find (T0)max Combine Eqs. (2) and (5) and solve for T0: REACTIVE TORQUES (from Eqs. 3-45a and b) TA ϭ T0 ¢ LBIPA ≤ LBIPA ϩ LAIPB LAIPB TB ϭ T0 ¢ ≤ LBIPA ϩ LAIPB (T0 ) CB ϭ (1) ϭ (2) ϭ LAIPB 1 ␲d 3t ¢1 ϩ ≤ 16 A allow LBIPA 4 LAdB 1 3 ␲dAtallow ¢ 1 ϩ 4≤ 16 LBdA Substitute numerical values: (T0)AC ϭ 3678 lb-in. Substitute numerical values: (T0 ) max ϭ 3680 lb-in. NOTE: From Eqs. (4) and (6) we find that (3) Combine Eqs. (1) and (3) and solve for T0: (T0 ) AC ϭ (6) SEGMENT AC GOVERNS 16TA tAC ϭ ␲d 1 1 3 ␲dA tAC ϭ ␲d3 t 16 16 A allow 4 LBdA 1 ␲d3 tallow ¢ 1 ϩ 4≤ 16 B LAdB (T0)CB ϭ 4597 lb-in. ALLOWABLE TORQUE BASED UPON SHEAR STRESS IN SEGMENT AC TA ϭ LBIPA 1 3 ␲dB tallow ¢ 1 ϩ ≤ 16 LAIPB (5) (T0 ) AC LA dB ϭ¢ ≤¢ ≤ (T0 ) CB LB dA which can be used as a partial check on the results. (4)
    • SECTION 3.8 Problem 3.8-6 A stepped shaft ACB having solid circular cross sections with two different diameters is held against rotation at the ends (see figure). If the allowable shear stress in the shaft is 43 MPa, what is the maximum torque (T0)max that may be applied at section C? (Hint: Use Eqs. 3-45a and b of Example 3-9 to obtain the reactive torques.) Solution 3.8-6 227 Statically Indeterminate Torsional Members 20 mm 25 mm B C A T0 225 mm 450 mm Stepped shaft ACB dB dA TA C A B TB T0 LA LB dA ϭ 20 mm ALLOWABLE TORQUE BASED UPON SHEAR STRESS CB dB ϭ 25 mm IN SEGMENT LA ϭ 225 mm tCB ϭ LB ϭ 450 mm 16TB 3 ␲dB 1 1 ␲d 3t ϭ ␲d 3t 16 B CB 16 B allow ␶allow ϭ 43 MPa TB ϭ Find (T0)max Combine Eqs. (2) and (5) and solve for T0: REACTIVE TORQUES (from Eqs. 3-45a and b) (T0 ) CB ϭ TA ϭ T0 ¢ LBIPA ≤ LBIPA ϩ LAIPB (1) ϭ LAIPB TB ϭ T0 ¢ ≤ LBIPA ϩ LAIPB (2) ϭ LAIPB 1 ␲d3 tallow ¢ 1 ϩ ≤ A 16 LBIPA 4 LAdB 1 3 ␲dAtallow ¢ 1 ϩ 4≤ 16 LBdA Substitute numerical values: (T ) ϭ 150.0 N . m 0 AC (6) Substitute numerical values: (T ) ϭ 240.0 N . m (T0 ) max ϭ 150 N . m NOTE: From Eqs. (4) and (6) we find that (T0 ) AC LA dB ϭ¢ ≤¢ ≤ (T0 ) CB LB dA (3) Combine Eqs. (1) and (3) and solve for T0: (T0 ) AC ϭ 4 LBdA 1 ␲d3 tallow ¢ 1 ϩ 4≤ 16 B LAdB SEGMENT AC GOVERNS 16TA tAC ϭ 3 ␲dA 1 1 3 ␲dAtAC ϭ ␲d3 t 16 16 A allow LBIPA 1 3 ␲dBtallow ¢ 1 ϩ ≤ 16 LAIPB 0 CB ALLOWABLE TORQUE BASED UPON SHEAR STRESS IN SEGMENT AC TA ϭ (5) which can be used as a partial check on the results. (4)
    • 228 CHAPTER 3 Torsion Problem 3.8-7 A stepped shaft ACB is held against rotation at ends A and B and subjected to a torque T0 acting at section C (see figure). The two segments of the shaft (AC and CB) have diameters dA and dB, respectively, and polar moments of inertia IPA and IPB, respectively. The shaft has length L and segment AC has length a. dA A C IPB B T0 (a) For what ratio a/L will the maximum shear stresses be the same in both segments of the shaft? (b) For what ratio a/L will the internal torques be the same in both segments of the shaft? (Hint: Use Eqs. 3-45a and b of Example 3-9 to obtain the reactive torques.) Solution 3.8-7 dB IPA a L Stepped shaft dB dA TA B C A TB T0 a L SEGMENT AC: dA, IPA SEGMENT CB: dB, IPB LBIPA dA LAIPB dB ϭ IPA IPB LA ϭ a LB ϭ L Ϫ a or REACTIVE TORQUES (from Eqs. 3-45a and b) LBIPA LAIPB TA ϭ T0 ¢ ≤; TB ϭ T0 ¢ ≤ LBIPA ϩ LAIPB LBIPA ϩ LAIPB dA a ϭ L dA ϩ dB Solve for a րL: (b) EQUAL TORQUES TA ϭ TB or TA (dAր2) tAC ϭ IPA TB (dB ր2) tCB ϭ IPB or tAC ϭ tCB TAdA TB dB ϭ IPA IPB LB IPA ϭ LAIPB (L Ϫ a) IPA ϭ aIPB Solve for a րL: (Eq. 1) Substitute TA and TB into Eq. (1): Problem 3.8-8 A circular bar AB of length L is fixed against rotation at the ends and loaded by a distributed torque t(x) that varies linearly in intensity from zero at end A to t0 at end B (see figure). Obtain formulas for the fixed-end torques TA and TB. LB dA ϭ LA dB (LϪa)dA ϭ adB (a) EQUAL SHEAR STRESSES or or IPA a ϭ L IPA ϩ IPB 4 dA a ϭ 4 4 L dA ϩ dB or t0 t(x) TA TB A B x L
    • SECTION 3.8 Solution 3.8-8 229 Statically Indeterminate Torsional Members Fixed-end bar with triangular load t0 ELEMENT OF DISTRIBUTED LOAD t(x) TA TB A t(x)dx dTA dTB B B x x dx dx L t0 x L T0 ϭ Resultant of distributed torque t(x) ϭ T0 ϭ Ύ L t(x)dx ϭ 0 Ύ 0 EQUILIBRIUM TA ϩ TB ϭ T0 ϭ t0 L 2 L t0 x t0 L dx ϭ L 2 dTA ϭ Elemental reactive torque dTB ϭ Elemental reactive torque From Eqs. (3-46a and b): dTA ϭ t(x)dx ¢ LϪx ≤ L dTB ϭ t(x)dx ¢ x ≤ L REACTIVE TORQUES (FIXED-END TORQUES) TA ϭ TB ϭ Ύ dTA ϭ Ύ dTB ϭ ¢ t0 ≤ ¢ Ύ L Ύ L 0 0 t0 L LϪx ≤ dx ϭ L 6 ¢ t0 ≤ ¢ ≤ NOTE: TA ϩ TB ϭ Problem 3.8-9 A circular bar AB with ends fixed against rotation has a hole extending for half of its length (see figure). The outer diameter of the bar is d2 ϭ 3.0 in. and the diameter of the hole is d1 ϭ 2.4 in. The total length of the bar is L ϭ 50 in. At what distance x from the left-hand end of the bar should a torque T0 be applied so that the reactive torques at the supports will be equal? x L x L t0 L x dx ϭ L 3 t0 L 2 25 in. A 25 in. T0 3.0 in. B x 2.4 in. 3.0 in.
    • 230 CHAPTER 3 Torsion Solution 3.8-9 Bar with a hole L/2 d2 A TA L/2 T0 B TB x L ϭ 50 in. Substitute Eq. (1) into Eq. (2) and simplify: L/2 ϭ 25 in. fB ϭ d2 ϭ outer diameter ϭ 3.0 in. COMPATIBILITY ␾B ϭ 0 d1 ϭ diameter of hole ∴ ϭ 2.4 in. T0 ϭ Torque applied at distance x xϭ EQUILIBRIUM ∴ TA ϭ TB ϭ T0 2 (1) REMOVE THE SUPPORT AT END B xϭ TB T0 xϭ IPB ␾B ϭ Angle of twist at B IPA ϭ Polar moment of inertia at left-hand end IPB ϭ Polar moment of inertia at right-hand end TB (Lր2) TB (Lր2) T0 (x Ϫ Lր2) ϩ Ϫ GIPB GIPA GIPB Ϫ T0 (Lր2) GIPA d1 4 L B2 ϩ ¢ ≤ R 4 d2 SUBSTITUTE NUMERICAL VALUES: IPA fB ϭ IPB L ¢3 Ϫ ≤ 4 IPA 4 4 IPB d2 Ϫ d1 d1 4 ϭ ϭ1Ϫ¢ ≤ 4 IPA d2 d2 L/2 x x 3L L ϭ Ϫ IPB 4IPB 4IPA SOLVE FOR x: Find x so that TA ϭ TB TA ϩ TB ϭ T0 T0 L L x L L B ϩ Ϫ ϩ Ϫ R G 4IPB 4IPA IPB 2IPB 2IPA (2) 50 in. 2.4 in. 4 B2 ϩ ¢ ≤ R ϭ 30.12 in. 4 3.0 in.
    • SECTION 3.8 Problem 3.8-10 A solid steel bar of diameter d1 ϭ 25.0 mm is enclosed by a steel tube of outer diameter d3 ϭ 37.5 mm and inner diameter d2 ϭ 30.0 mm (see figure). Both bar and tube are held rigidly by a support at end A and joined securely to a rigid plate at end B. The composite bar, which has a length L ϭ 550 mm, is twisted by a torque T ϭ 400 N и m acting on the end plate. Tube A B T Bar (a) Determine the maximum shear stresses ␶1 and ␶2 in the bar and tube, respectively. (b) Determine the angle of rotation ␾ (in degrees) of the end plate, assuming that the shear modulus of the steel is G ϭ 80 GPa. (c) Determine the torsional stiffness kT of the composite bar. (Hint: Use Eqs. 3-44a and b to find the torques in the bar and tube.) End plate L d1 d2 d3 Solution 3.8-10 Bar enclosed in a tube Tube (2) A B T = 400 N . m Bar (1) End plate L = 550 mm d1 d2 ϭ 30.0 mm IP1 ≤ ϭ 100.2783 N # m IP1 ϩ IP2 Tube: T2 ϭ T ¢ IP2 ≤ ϭ 299.7217 N # m IP1 ϩ IP2 T1 (d1΋2) ϭ 32.7 MPa IP1 Tube: t2 ϭ d3 ϭ 37.5 mm G ϭ 80 GPa POLAR MOMENTS OF INERTIA Bar: IP1 ϭ Bar: T1 ϭ T ¢ Bar: t1 ϭ d3 d1 ϭ 25.0 mm TORQUES IN THE BAR (1) AND TUBE (2) FROM EQS. (3-44A AND B) (a) MAXIMUM SHEAR STRESSES d2 ␲ 4 d ϭ 38.3495 ϫ 10 Ϫ9 m4 32 1 ␲ 4 4 Tube: IP2 ϭ ¢ d Ϫ d2 ≤ ϭ 114.6229 ϫ 10 Ϫ9 m4 32 3 231 Statically Indeterminate Torsional Members T2 (d3 ր2) ϭ 49.0 MPa IP2 (b) ANGLE OF ROTATION OF END PLATE fϭ T1L T2L ϭ ϭ 0.017977 rad GIP1 GIP2 f ϭ 1.03Њ (c) TORSIONAL STIFFNESS kT ϭ T ϭ 22.3 kN # m f
    • 232 CHAPTER 3 Torsion Problem 3.8-11 A solid steel bar of diameter d1 ϭ 1.50 in. is enclosed by a steel tube of outer diameter d3 ϭ 2.25 in. and inner diameter d2 ϭ 1.75 in. (see figure). Both bar and tube are held rigidly by a support at end A and joined securely to a rigid plate at end B. The composite bar, which has length L ϭ 30.0 in., is twisted by a torque T ϭ 5000 lb-in. acting on the end plate. (a) Determine the maximum shear stresses ␶1 and ␶2 in the bar and tube, respectively. (b) Determine the angle of rotation ␾ (in degrees) of the end plate, assuming that the shear modulus of the steel is G ϭ 11.6 ϫ 106 psi. (c) Determine the torsional stiffness kT of the composite bar. (Hint: Use Eqs. 3-44a and b to find the torques in the bar and tube.) Solution 3.8-11 Bar enclosed in a tube Tube (2) A TORQUES IN THE BAR (1) AND TUBE (2) FROM EQS. (3-44A AND B) B T = 5000 lb-in. Bar (1) End plate L = 30.0 in. d1 d3 d2 ϭ 1.75 in. IP1 ≤ ϭ 1187.68 lb-in. IP1 ϩ IP2 Tube: T2 ϭ T ¢ IP2 ≤ ϭ 3812.32 lb-in. IP1 ϩ IP2 (a) MAXIMUM SHEAR STRESSES T1 (d1ր2) Bar: t1 ϭ ϭ 1790 psi IP1 T2 (d3 ր2) Tube: t2 ϭ ϭ 2690 psi IP2 d2 d1 ϭ 1.50 in. Bar: T1 ϭ T ¢ d3 ϭ 2.25 in. (b) ANGLE OF ROTATION OF END PLATE T1L T2L ϭ ϭ 0.00618015 rad GIP1 GIP2 G ϭ 11.6 ϫ 106 psi fϭ POLAR MOMENTS OF INERTIA f ϭ 0.354Њ Bar: IP1 ϭ ␲ 4 d ϭ 0.497010 in.4 32 1 ␲ 4 4 Tube: IP2 ϭ ¢ d Ϫ d2 ≤ ϭ 1.595340 in.4 32 3 (c) TORSIONAL STIFFNESS kT ϭ T ϭ 809 k-in. f
    • SECTION 3.8 Problem 3.8-12 The composite shaft shown in the figure is manufactured by shrink-fitting a steel sleeve over a brass core so that the two parts act as a single solid bar in torsion. The outer diameters of the two parts are d1 ϭ 40 mm for the brass core and d2 ϭ 50 mm for the steel sleeve. The shear moduli of elasticity are Gb ϭ 36 GPa for the brass and Gs ϭ 80 GPa for the steel. Assuming that the allowable shear stresses in the brass and steel are ␶b ϭ 48 MPa and ␶s ϭ 80 MPa, respectively, determine the maximum permissible torque Tmax that may be applied to the shaft. (Hint: Use Eqs. 3-44a and b to find the torques.) Solution 3.8-12 T d1 d2 Total torque: T ϭ TB ϩ TS Eq. (3-44a): TB ϭ T ¢ GBIPB ≤ GBIPB ϩ GS IPS ϭ 0.237918 T Brass core B d1 ϭ 40 mm Eq. (3-44b): TS ϭ T ¢ d2 ϭ 50 mm GS ϭ 80 GPa Allowable stresses: ␶B ϭ 48 MPa Steel sleeve Brass core TORQUES d1 d2 GB ϭ 36 GPa T Composite shaft shrink fit Steel sleeve S T ϭ TB ϩ TS ␶S ϭ 80 MPa GSIPS ≤ GS IPB ϩ GS IPS ϭ 0.762082 T (CHECK) ALLOWABLE TORQUE T BASED UPON BRASS CORE BRASS CORE (ONLY) tB ϭ TB TB (d1ր2) IPB TB ϭ 2tBIPB d1 Substitute numerical values: TB ϭ 0.237918 T TB ϭ 2(48 MPa)(251.327 ϫ 10 Ϫ9 m4 ) 40 mm ␲ 4 d ϭ 251.327 ϫ 10 Ϫ9 m4 32 1 GBIPB ϭ 9047.79 N # m2 ALLOWABLE TORQUE T BASED UPON STEEL SLEEVE STEEL SLEEVE (ONLY) tS ϭ IPB ϭ T ϭ 2535 N и m TS TS (d2ր2) IPS TS ϭ 2tS IPS d2 SUBSTITUTE NUMERICAL VALUES: TS ϭ 0.762082 T TS IPS ϭ ␲ 4 4 (d Ϫ d1 ) ϭ 362.265 ϫ 10 Ϫ9 m4 32 2 GSIPS ϭ 28,981.2 N # m 2 ϭ 233 Statically Indeterminate Torsional Members 2(80 MPa)(362.265 ϫ 10 Ϫ9 m4 ) 50 mm T ϭ 1521 N и m STEEL SLEEVE GOVERNS Tmax ϭ 1520 N # m
    • 234 CHAPTER 3 Torsion Problem 3.8-13 The composite shaft shown in the figure is manufactured by shrink-fitting a steel sleeve over a brass core so that the two parts act as a single solid bar in torsion. The outer diameters of the two parts are d1 ϭ 1.6 in. for the brass core and d2 ϭ 2.0 in. for the steel sleeve. The shear moduli of elasticity are Gb ϭ 5400 ksi for the brass and Gs ϭ 12,000 ksi for the steel. Assuming that the allowable shear stresses in the brass and steel are ␶b ϭ 4500 psi and ␶s ϭ 7500 psi, respectively, determine the maximum permissible torque Tmax that may be applied to the shaft. (Hint: Use Eqs. 3-44a and b to find the torques.) Solution 3.8-13 Composite shaft shrink fit TORQUES Steel sleeve S Total torque: T ϭ TB ϩ TS Eq. (3-44a): TB ϭ T ¢ d1 d2 Brass core B ϭ 0.237918 T Eq. (3-44b): TS ϭ T ¢ d1 ϭ 1.6 in. d2 ϭ 2.0 in. GB ϭ 5,400 psi GS ϭ 12,000 psi GS IPS ≤ GB IPB ϩ GS IPS ϭ 0.762082 T Allowable stresses: ␶B ϭ 4500 psi GB IPB ≤ GB IPB ϩ GS IPS T ϭ TB ϩ TS (CHECK) ␶S ϭ 7500 psi ALLOWABLE TORQUE T BASED UPON BRASS CORE BRASS CORE (ONLY) tB ϭ TB TB (d1ր2) IPB TB ϭ 2tB IPB d1 Substitute numerical values: TB ϭ 0.237918 T TB IPB ϭ ϭ ␲ 4 d ϭ 0.643398 in.4 32 1 2(4500 psi)(0.643398 in.4 ) 1.6 in. T ϭ 15.21 k-in. GBIPB ϭ 3.47435 ϫ 106 lb-in.2 ALLOWABLE TORQUE T BASED UPON STEEL SLEEVE STEEL SLEEVE (ONLY) tS ϭ TS TS (d2ր2) IPS TS ϭ 2tS IPS d2 Substitute numerical values: TS ␲ 4 4 IPS ϭ (d2 Ϫ d1 ) ϭ 0.927398 in.4 32 GSIPS ϭ 11.1288 ϫ 106 lb-in.2 TS ϭ 0.762082 T ϭ 2(7500 psi)(0.927398 in.4 ) 2.0 in. T ϭ 9.13 k-in. STEEL SLEEVE GOVERNS Tmax ϭ 9.13 k-in.
    • SECTION 3.8 Problem 3.8-14 A steel shaft (Gs ϭ 80 GPa) of total length L ϭ 4.0 m is encased for one-half of its length by a brass sleeve (Gb ϭ 40 GPa) that is securely bonded to the steel (see figure). The outer diameters of the shaft and sleeve are d1 ϭ 70 mm and d2 ϭ 90 mm, respectively. Brass sleeve d2 = 90 mm Steel shaft d1 = 70 mm T T A B C Brass sleeve Composite shaft d2 = 90 mm Steel shaft d1 = 70 mm T T A B C L = 2.0 m 2 L = 2.0 m 2 GSIPS ϭ 188.574 ϫ 103 N и m2 PROPERTIES OF THE BRASS SLEEVE (b) d2 ϭ 90 mm d1 ϭ 70 mm Gb ϭ 40 GPa Allowable shear stress: ␶b ϭ 70 MPa ␲ 4 4 (d Ϫ d1 ) ϭ 4.0841 ϫ 10 Ϫ6 m4 32 2 TORQUES IN THE COMPOSITE BAR AB TS ϭ Torque in the steel shaft AB Tb ϭ Torque in the brass sleeve AB GS IPS ≤ GSIPS ϩ Gb IPb TS ϭ T (0.53582) Tb ϭ T Ϫ TS ϭ T (0.46418) (Eq. 1) (Eq. 2) ANGLE OF TWIST OF THE COMPOSITE BAR AB fAB ϭ TS (Lր2) Tb (Lր2) ϭ GS IPS Gb IPb ϭ (5.6828 ϫ 10 Ϫ6 )T UNITS: T ϭ N и m ␾ ϭ rad (Eqs. 3 and 4) (b) ALLOWABLE TORQUE T2 BASED UPON SHEAR STRESS IN THE BRASS SLEEVE tb ϭ T(d2 ր2) tb ϭ 70 MPa Ipb Tb ϭ 0.46418 T (From Eq. 2) (0.46418T )(0.045 m) 70 MPa ϭ 4.0841 ϫ 10 Ϫ6 m4 Solve for T (Equal to T2 ): T2 ϭ 13.69 kN . m GbIPB ϭ 163.363 ϫ 103 N и m2 From Eq. (3-44a): TS ϭ T ¢ ANGLE OF TWIST OF THE ENTIRE SHAFT ABC (a) ALLOWABLE TORQUE T1 BASED UPON ANGLE OF TWIST fallow ϭ 8.0Њ ϭ 0.13963 rad f ϭ (16.2887 ϫ 10 Ϫ6 ) T ϭ 0.13963 rad T1 ϭ 8.57 kN . m ␲ 4 IPS ϭ d ϭ 2.3572 ϫ 10 Ϫ6 m4 32 1 IPB ϭ ANGLE OF TWIST OF PART BC OF THE STEEL SHAFT T(Lր2) fBC ϭ ϭ (10.6059 ϫ 10 Ϫ6 )T (Eq. 4) GS IPS ␾ ϭ ␾AB ϩ ␾BC ␾ ϭ (16.2887 ϫ 10Ϫ6) T UNITS: ␾ ϭ rad TϭNиm PROPERTIES OF THE STEEL SHAFT (s) d1 ϭ 70 mm GS ϭ 80 GPa Allowable shear stress: ␶S ϭ 110 MPa (Eq. 3) 235 (a) Determine the allowable torque T1 that may be applied to the ends of the shaft if the angle of twist ␾ between the ends is limited to 8.0°. (b) Determine the allowable torque T2 if the shear stress in the brass is limited to ␶b ϭ 70 MPa. (c) Determine the allowable torque T3 if the shear stress in the steel is limited to ␶s ϭ 110 MPa. (d) What is the maximum allowable torque Tmax if all three of the preceding conditions must be satisfied? L = 2.0 m 2 L = 2.0 m 2 Solution 3.8-14 Statically Indeterminate Torsional Members (c) ALLOWABLE TORQUE T3 BASED UPON SHEAR STRESS IN THE STEEL SHAFT BC T(d2ր2) tS ϭ tS ϭ 110 MPa IPS T(0.035 m) 2.3572 ϫ 10 Ϫ6 m4 Solve for T (Equal to T3 ): T3 ϭ 7.41 kN . m 110 MPa ϭ (d) MAXIMUM ALLOWABLE TORQUE Shear stress in steel governs Tmax ϭ 7.41 kN . m
    • 236 CHAPTER 3 Torsion Strain Energy in Torsion Problem 3.9-1 A solid circular bar of steel (G ϭ 11.4 ϫ 106 psi) with length L ϭ 30 in. and diameter d ϭ 1.75 in. is subjected to pure torsion by torques T acting at the ends (see figure). d T T (a) Calculate the amount of strain energy U stored in the bar when the maximum shear stress is 4500 psi. (b) From the strain energy, calculate the angle of twist ␾ (in degrees). Solution 3.9-1 Steel bar (a) STRAIN ENERGY d T L T Uϭ L ϭ ␲d 3tmax 2 L T 2L 32 ϭ¢ ≤ ¢ ≤¢ ≤ 2GIP 16 2G ␲d 4 ␲d 2Lt2 max 16G G ϭ 11.4 ϫ 106 psi Substitute numerical values: L ϭ 30 in. (Eq. 2) U ϭ 32.0 in.-lb d ϭ 1.75 in. (b) ANGLE OF TWIST ␶max ϭ 4500 psi 16 T tmax ϭ ␲d 3 ␲d 4 IP ϭ 32 ␲d 3tmax Tϭ 16 (Eq. 1) Uϭ Tf 2 fϭ 2U T Substitute for T and U from Eqs. (1) and (2): fϭ 2Ltmax Gd Substitute numerical values: f ϭ 0.013534 rad ϭ 0.775Њ Problem 3.9-2 A solid circular bar of copper (G ϭ 45 GPa) with length L ϭ 0.75 m and diameter d ϭ 40 mm is subjected to pure torsion by torques T acting at the ends (see figure). (a) Calculate the amount of strain energy U stored in the bar when the maximum shear stress is 32 MPa. (b) From the strain energy, calculate the angle of twist ␾ (in degrees) (Eq. 3)
    • SECTION 3.9 Solution 3.9-2 T (a) STRAIN ENERGY Uϭ L G ϭ 45 GPa ϭ L ϭ 0.75 m IP ϭ ␲d2Lt2 max 16G (Eq. 2) U ϭ 5.36 J ␶max ϭ 32 MPa 16T ␲d 3 ␲d 3tmax 2 L T 2L 32 ϭ¢ ≤ ¢ ≤¢ ≤ 2GIP 16 2G ␲d 4 Substitute numerical values: d ϭ 40 mm tmax ϭ 237 Copper bar d T Strain Energy in Torsion (b) ANGLE OF TWIST ␲d 3tmax 16 ␲d 4 32 Tϭ (Eq. 1) Uϭ Tf 2 fϭ 2U T Substitute for T and U from Eqs. (1) and (2): fϭ 2Ltmax Gd (Eq. 3) Substitute numerical values: f ϭ 0.026667 rad ϭ 1.53Њ d2 Problem 3.9-3 A stepped shaft of solid circular cross sections T (see figure) has length L ϭ 45 in., diameter d2 ϭ 1.2 in., and 6 psi. diameter d1 ϭ 1.0 in. The material is brass with G ϭ 5.6 ϫ 10 Determine the strain energy U of the shaft if the angle of twist is 3.0°. Solution 3.9-3 d1 T L — 2 L — 2 Stepped shaft d2 d1 T STRAIN ENERGY T L — 2 d1 ϭ 1.0 in. d2 ϭ 1.2 in. L — 2 16 T 2(Lր2) 16 T 2(Lր2) T 2L Uϭ a ϭ ϩ 4 4 2GIP ␲Gd2 ␲Gd1 ϭ 8T 2L 1 1 ¢ 4 ϩ 4≤ ␲G d2 d1 Also, U ϭ (Eq. 1) Tf 2 (Eq. 2) L ϭ 45 in. Equate U from Eqs. (1) and (2) and solve for T: G ϭ 5.6 ϫ 106 psi (brass) Tϭ ␾ ϭ 3.0Њ ϭ 0.0523599 rad Uϭ 4 4 ␲Gd1 d2 f 4 4 16L(d1 ϩ d2 ) 4 4 Tf ␲Gf2 d1 d2 ϭ ¢ 4 4≤ 2 32L d1 ϩ d2 f ϭ radians SUBSTITUTE NUMERICAL VALUES: U ϭ 22.6 in.-lb
    • 238 CHAPTER 3 Torsion Problem 3.9-4 A stepped shaft of solid circular cross sections (see figure) has length L ϭ 0.80 m, diameter d2 ϭ 40 mm, and diameter d1 ϭ 30 mm. The material is steel with G ϭ 80 GPa. Determine the strain energy U of the shaft if the angle of twist is 1.0°. Soluton 3.9-4 Stepped shaft d2 Equate U from Eqs. (1) and (2) and solve for T: d1 T T Tϭ L — 2 L — 2 d1 ϭ 30 mm L ϭ 0.80 m Uϭ d2 ϭ 40 mm 4 4 ␲G d1 d2 f 4 4 16L(d1 ϩ d2 ) 4 4 Tf ␲Gf2 d1 d2 ϭ ¢ 4 4≤ 2 32L d1 ϩ d2 f ϭ radians SUBSTITUTE NUMERICAL VALUES: U ϭ 1.84 J G ϭ 80 GPa (steel) ␾ ϭ 1.0Њ ϭ 0.0174533 rad STRAIN ENERGY 16T 2(Lր2) 16T 2(Lր2) T 2L Uϭ a ϭ ϩ 4 4 2GIP ␲Gd2 ␲Gd1 ϭ 8T 2L 1 1 ¢ 4 ϩ 4≤ ␲G d2 d1 Also, U ϭ (Eq. 1) Tf 2 (Eq. 2) Problem 3.9-5 A cantilever bar of circular cross section and length L is fixed at one end and free at the other (see figure). The bar is loaded by a torque T at the free end and by a distributed torque of constant intensity t per unit distance along the length of the bar. (a) What is the strain energy U1 of the bar when the load T acts alone? (b) What is the strain energy U2 when the load t acts alone? (c) What is the strain energy U3 when both loads act simultaneously? Solution 3.9-5 Cantilever bar with distributed torque G ϭ shear modulus t IP ϭ polar moment of inertia T ϭ torque acting at free end L T t ϭ torque per unit distance t L T
    • SECTION 3.9 239 Strain Energy in Torsion (c) BOTH LOADS ACT SIMULTANEOUSLY (a) LOAD T ACTS ALONE (Eq. 3-51a) T 2L 2GIP t (b) LOAD t ACTS ALONE dx U1 ϭ T At distance x from the free end: From Eq. (3-56) of Example 3-11: U2 ϭ x T(x) ϭ T ϩ tx t 2L3 6GIP U3 ϭ Ύ 0 ϭ L [T(x) ] 2 1 dx ϭ 2GIP 2GIP L Ύ (T ϩ tx) dx 2 0 T 2L TtL2 t 2L3 ϩ ϩ 2GIP 2GIP 6GIP NOTE: U3 is not the sum of U1 and U2. 2T0 Problem 3.9-6 Obtain a formula for the strain energy U of the statically indeterminate circular bar shown in the figure. The bar has fixed supports at ends A and B and is loaded by torques 2T0 and T0 at points C and D, respectively. Hint: Use Eqs. 3-46a and b of Example 3-9, Section 3.8, to obtain the reactive torques. Solution 3.9-6 L — 4 From Eq. (3-46a): ϩ TB ϭ 3T0 Ϫ TA ϭ 7T0 4 L — 2 L — 4 TB D L — 2 L — 4 n Ti2Li Uϭ a iϭ1 2Gi IPi T0 ¢ 5T0 4 L ≤ 7T0 4 ϭ L 4 ϭ ϭ INTERNAL TORQUES TAC ϭ Ϫ L — 4 D STRAIN ENERGY (from Eq. 3-53) REACTIVE TORQUES L C B C TA ϭ B T0 A TA 3L ≤ 4 A Statically indeterminate bar 2T0 (2T0 ) ¢ T0 TCD ϭ T0 4 TDB ϭ 5T0 4 Uϭ 1 L 2 L 2 L B T 2 ¢ ≤ ϩ TCD¢ ≤ ϩ TDB¢ ≤ R 2GIp AC 4 2 4 7T0 2 L T0 2 L 5T0 2 L 1 B ¢Ϫ ≤ ¢ ≤ϩ¢ ≤ ¢ ≤ϩ¢ ≤ ¢ ≤R 2GIP 4 4 4 2 4 4 2 19T0L 32GIP
    • 240 CHAPTER 3 Torsion Problem 3.9-7 A statically indeterminate stepped shaft ACB is fixed at ends A and B and loaded by a torque T0 at point C (see figure). The two segments of the bar are made of the same material, have lengths LA and LB, and have polar moments of inertia IPA and IPB. Determine the angle of rotation ␾ of the cross section at C by using strain energy. Hint: Use Eq. 3-51b to determine the strain energy U in terms of the angle ␾. Then equate the strain energy to the work done by the torque T0. Compare your result with Eq. 3-48 of Example 3-9, Section 3.8. Solution 3.9-7 A A IPA T0 C IPB LA LB Statically indeterminate bar IPA C LA T0 ␾ IPB WORK DONE BY THE TORQUE T0 B Wϭ T0f 2 LB EQUATE U AND W AND SOLVE FOR ␾ T0f Gf2 IPA IPB ¢ ϩ ≤ϭ 2 LA LB 2 STRAIN ENERGY (FROM EQ. 3-51B) n GIPif2 GIPAf2 GIPBf2 i Uϭ a ϭ ϩ 2Li 2LA 2LB iϭ1 ϭ fϭ Gf2 IPA IPB ¢ ϩ ≤ 2 LA LB T0LALB G(LBIPA ϩ LAIPB ) (This result agrees with Eq. (3-48) of Example 3-9, Section 3.8.) Problem 3.9-8 Derive a formula for the strain energy U of the cantilever bar shown in the figure. The bar has circular cross sections and length L. It is subjected to a distributed torque of intensity t per unit distance. The intensity varies linearly from t ϭ 0 at the free end to a maximum value t ϭ t 0 at the support. t0 t L B
    • SECTION 3.9 Solution 3.9-8 Strain Energy in Torsion Cantilever bar with distributed torque t0 t(x) = t0 dx x L x ␰ x ϭ distance from right-hand end of the bar d␰ L STRAIN ENERGY OF ELEMENT dx ELEMENT d␰ Consider a differential element d␰ at distance ␰ from the right-hand end. dU ϭ t0 2 4 [T(x) ] 2dx 1 ϭ ¢ ≤ x dx 2GIP 2GIP 2L ϭ dT STRAIN ENERGY OF ENTIRE BAR d␰ Uϭ Ύ L dU ϭ 0 dT ϭ external torque acting on this element dT ϭ t(␰)d␰ j ϭ t0 ¢ ≤ dj L Uϭ ELEMENT dx AT DISTANCE x T(x) T(x) dx T(x) ϭ internal torque acting on this element T(x) ϭ total torque from x ϭ 0 to x ϭ x T(x) ϭ Ύ x dT ϭ 0 t0x2 ϭ 2L x 0¢ Ύt 0 2 t0 x4 dx 8L GIP 2 j ≤ dj L t 2L3 0 40GIP 2 t0 8L2GIP L Ύ x dx 4 0 2 t0 L5 ϭ 2 ¢ ≤ 8L GIP 5 241
    • 242 CHAPTER 3 Torsion Problem 3.9-9 A thin-walled hollow tube AB of conical shape has constant thickness t and average diameters dA and dB at the ends (see figure). B A T T (a) Determine the strain energy U of the tube when it is subjected to pure torsion by torques T. (b) Determine the angle of twist ␾ of the tube. L t t Note: Use the approximate formula IP Ϸ ␲d 3t/4 for a thin circular ring; see Case 22 of Appendix D. dB dA Solution 3.9-9 Thin-walled, hollow tube B A T x Therefore, T d(x) Ύ dx L 0 t ϭ thickness dB ϭ average diameter at end B d(x) ϭ average diameter at distance x from end A d(x) ϭ dA ϩ ¢ dB Ϫ dA ≤x L ϭ POLAR MOMENT OF INERTIA IP (x) ϭ ␲d t 4 3 dB Ϫ dA ␲[d(x) ] 3t ␲t ϭ B dA ϩ ¢ ≤ xR 4 4 L Uϭ ϭ T dx P Ύ 0 WϭU L dx 3 dB Ϫ dA x ≤ R B dA ϩ ¢ L From Appendix C: Ύ (a ϩ bx) dx 3 ϭϪ 2T 2 L(dA ϩ dB ) T 2L dA ϩ dB ؒ ϭ ¢ 2 2 ≤ 2 2 ␲Gt ␲Gt dA dB 2dAdB Work of the torque T: W ϭ Ύ 2GI (x) 2T2 ␲Gt 2(dB Ϫ dA ) (b) ANGLE OF TWIST 2 0 1 L(dA ϩ dB ) 2 2 2dA dB Uϭ (a) STRAIN ENERGY (FROM EQ. 3-54) L 1 2b(a ϩ bx) 2 L Substitute this expression for the integral into the equation for U (Eq. 1): 3 IP ϭ dx 3 dB Ϫ dA B dA ϩ ¢ ≤ xR L 4 2 dB Ϫ dA B dA ϩ ¢ ≤x R L L 0 L L ϭϪ ϩ 2(dB Ϫ dA )(dB ) 2 2(dB Ϫ dA )(dA ) 2 ϭϪ dA ϭ average diameter at end A L (Eq. 1) Tf T 2L(dA ϩ dB ) ϭ 2 2 2 ␲Gt dAdB Solve for ␾: fϭ Tf 2 2TL(dA ϩ dB ) 2 2 ␲Gt dAdB
    • SECTION 3.9 Problem 3.9-10 A hollow circular tube A fits over the end of a solid circular bar B, as shown in the figure. The far ends of both bars are fixed. Initially, a hole through bar B makes an angle ␤ with a line through two holes in tube A. Then bar B is twisted until the holes are aligned, and a pin is placed through the holes. When bar B is released and the system returns to equilibrium, what is the total strain energy U of the two bars? (Let IPA and IPB represent the polar moments of inertia of bars A and B, respectively. The length L and shear modulus of elasticity G are the same for both bars.) Strain Energy in Torsion IPA IPB Tube A Bar B L L ␤ Tube A Bar B Solution 3.9-10 Circular tube and bar IPA ␤ IPB Tube A Tube A Bar B Bar B L L COMPATIBILITY TUBE A ␾A ϩ ␾B ϭ ␤ ␾A FORCE-DISPLACEMENT RELATIONS T fA ϭ T ϭ torque acting on the tube TL GIPA fB ϭ TL GIPB Substitute into the equation of compatibility and solve for T: bG IPAIPB ¢ ≤ L IPA ϩ IPB ␾A ϭ angle of twist Tϭ BAR B STRAIN ENERGY T ␾B T 2L T 2L T 2L Uϭ a ϭ ϩ 2GIP 2GIPA 2GIPB ϭ T 2L 1 1 ¢ ϩ ≤ 2G IPA IPB Substitute for T and simplify: Uϭ T ϭ torque acting on the bar ␾B ϭ angle of twist b2G IPA IPB ¢ ≤ 2L IPA ϩ IPB 243
    • 244 CHAPTER 3 Torsion Problem 3.9-11 A heavy flywheel rotating at n revolutions per minute is rigidly attached to the end of a shaft of diameter d (see figure). If the bearing at A suddenly freezes, what will be the maximum angle of twist ␾ of the shaft? What is the corresponding maximum shear stress in the shaft? (Let L ϭ length of the shaft, G ϭ shear modulus of elasticity, and Im ϭ mass moment of inertia of the flywheel about the axis of the shaft. Also, disregard friction in the bearings at B and C and disregard the mass of the shaft.) Hint: Equate the kinetic energy of the rotating flywheel to the strain energy of the shaft. Solution 3.9-11 A d n (rpm) B C Rotating flywheel d ϭ diameter of shaft Shaft Flywheel Uϭ ␲Gd 4f2 64L UNITS: G ϭ (force)/(length)2 IP ϭ (length)4 d ϭ diameter n ϭ rpm ␾ ϭ radians L ϭ length U ϭ (length)(force) KINETIC ENERGY OF FLYWHEEL K.E. ϭ vϭ 1 I v2 2 m 2␲n 60 K.E. ϭ U 1 2␲n 2 Im ¢ ≤ 2 60 ␲ n Im 1800 fϭ 2␲ImL 2n 2 15d B G MAXIMUM SHEAR STRESS 2 2 ϭ ␲2n2Im ␲Gd 4f2 ϭ 1800 64L Solve for ␾: n ϭ rpm K.E. ϭ EQUATE KINETIC ENERGY AND STRAIN ENERGY tϭ T(dր2) IP fϭ TL GIP UNITS: Eliminate T: Im ϭ Gdf 2L 2␲ImL Gd 2n tmax ϭ ؒ 2 2L 15d B G (force)(length)(second)2 ␻ ϭ radians per second K.E. ϭ (length)(force) STRAIN ENERGY OF SHAFT (FROM EQ. 3-51b) 2 GIPf 2L ␲ 4 IP ϭ d 32 Uϭ tϭ tmax ϭ 2␲GIm n 15d B L
    • SECTION 3.10 Thin-Walled Tubes 245 Thin-Walled Tubes Problem 3.10-1 A hollow circular tube having an inside diameter of 10.0 in. and a wall thickness of 1.0 in. (see figure) is subjected to a torque T ϭ 1200 k-in. Determine the maximum shear stress in the tube using (a) the approximate theory of thin-walled tubes, and (b) the exact torsion theory. Does the approximate theory give conservative or nonconservative results? Solution 3.10-1 10.0 in. 1.0 in. Hollow circular tube APPROXIMATE THEORY (EQ. 3-63) t1 ϭ 10.0 in. T 1200 k-in. ϭ 6314 psi 2 ϭ 2␲r t 2␲(5.5 in.) 2 (1.0 in.) tapprox ϭ 6310 psi 1.0 in. EXACT THEORY (EQ. 3-11) t2 ϭ T ϭ 1200 k-in. t ϭ 1.0 in. ϭ r ϭ radius to median line T(d2 ր2) ϭ IP Td2 ␲ 4 4 2 ¢ ≤ d2 Ϫ d1 32 16(1200 k-in.)(12.0 in.) ␲[ (12.0 in.) 4 Ϫ (10.0 in.) 4 ] ϭ 6831 psi r ϭ 5.5 in. d2 ϭ outside diameter ϭ 12.0 in. texact ϭ 6830 psi d1 ϭ inside diameter ϭ 10.0 in. Because the approximate theory gives stresses that are too low, it is nonconservative. Therefore, the approximate theory should only be used for very thin tubes. Problem 3.10-2 A solid circular bar having diameter d is to be replaced by a rectangular tube having cross-sectional dimensions d ϫ 2d to the median line of the cross section (see figure). Determine the required thickness tmin of the tube so that the maximum shear stress in the tube will not exceed the maximum shear stress in the solid bar. Solution 3.10-2 t d d 2d Bar and tube Am ϭ (d)(2d) ϭ 2d 2 SOLID BAR tmax ϭ d t 16T ␲d 3 (Eq. 3-12) tmax ϭ T T ϭ 2tAm 4td 2 (Eq. 3-64) (Eq. 3-61) EQUATE THE MAXIMUM SHEAR STRESSES AND SOLVE FOR t RECTANGULAR TUBE 16T T ϭ ␲d 3 4td 2 t tmin ϭ d 2d ␲d 64 If t Ͼ tmin, the shear stress in the tube is less than the shear stress in the bar.
    • 246 CHAPTER 3 Torsion Problem 3.10-3 A thin-walled aluminum tube of rectangular cross section (see figure) has a centerline dimensions b ϭ 6.0 in. and h ϭ 4.0 in. The wall thickness t is constant and equal to 0.25 in. t h (a) Determine the shear stress in the tube due to a torque T ϭ 15 k-in. (b) Determine the angle of twist (in degrees) if the length L of the tube is 50 in. and the shear modulus G is 4.0 ϫ 106 psi. b Use with Prob. 3.10-4 Solution 3.10-3 Thin-walled tube Eq. (3-64): Am ϭ bh ϭ 24.0 in.2 t Eq. (3-71) with t1 ϭ t2 ϭ t: h J ϭ 28.8 in.4 (a) SHEAR STRESS (EQ. 3-61) b tϭ b ϭ 6.0 in. h ϭ 4.0 in. T ϭ 1250 psi 2tAm (b) ANGLE OF TWIST (EQ. 3-72) t ϭ 0.25 in. TL ϭ 0.0065104 rad GJ ϭ 0.373Њ fϭ T ϭ 15 k-in. L ϭ 50 in. G ϭ 4.0 ϫ 106 psi Problem 3.10-4 A thin-walled steel tube of rectangular cross section (see figure) has centerline dimensions b ϭ150 mm and h ϭ 100 mm. The wall thickness t is constant and equal to 6.0 mm. (a) Determine the shear stress in the tube due to a torque T ϭ 1650 N и m. (b) Determine the angle of twist (in degrees) if the length L of the tube is 1.2 m and the shear modulus G is 75 GPa. Solution 3.10-4 Thin-walled tube b ϭ 150 mm h ϭ 100 mm t h t ϭ 6.0 mm T ϭ 1650 N и m L ϭ 1.2 m b Eq. (3-64): Am ϭ bh ϭ 0.015 Eq. (3-71) with t1 ϭ t2 ϭ t: J ϭ 10.8 ϫ 10Ϫ6 m4 G ϭ 75 GPa tϭ T ϭ 9.17 MPa 2tAm (b) ANGLE OF TWIST (Eq. 3-72) fϭ TL ϭ 0.002444 rad GJ ϭ 0.140Њ m2 Jϭ (a) SHEAR STRESS (Eq. 3-61) 2b2h2t bϩh Jϭ 2b2h2t bϩh
    • SECTION 3.10 Problem 3.10-5 A thin-walled circular tube and a solid circular bar of the same material (see figure) are subjected to torsion. The tube and bar have the same cross-sectional area and the same length. What is the ratio of the strain energy U1 in the tube to the strain energy U2 in the solid bar if the maximum shear stresses are the same in both cases? (For the tube, use the approximate theory for thin-walled bars.) Solution 3.10-5 Am ϭ ␲r2 r Bar (2) J ϭ 2␲r3t A ϭ 2␲rt SOLID BAR (2) r2 2 A ϭ ␲r2 T T ϭ tmax ϭ 2tAm 2␲r2t Tϭ ϭ Tube (1) THIN-WALLED TUBE (1) t U1 ϭ 247 Thin-Walled Tubes tmax ϭ 2␲r 2t␶max T 2L (2␲r 2ttmax ) 2L ϭ 2GJ 2G(2␲r 3t) IP ϭ ␲ 4 r 2 2 Tr2 2T ϭ 3 IP ␲r2 Tϭ 3 ␲r2 tmax 2 3 2 (␲r2 tmax ) 2L ␲r2 t2 L T 2L max ϭ ϭ 2GIP ␲ 4 4G 8G ¢ r2 ≤ 2 At2 L max But ␲r2 ϭ A ∴ U2 ϭ 2 4G U2 ϭ ␲rtt2 L max G A 2␲ At2 L І U1 ϭ max 2G But rt ϭ RATIO U1 ϭ2 U2 Problem 3.10-6 Calculate the shear stress ␶ and the angle of twist ␾ (in degrees) for a steel tube (G ϭ 76 GPa) having the cross section shown in the figure. The tube has length L ϭ 1.5 m and is subjected to a torque T ϭ 10 kN и m. t = 8 mm r = 50 mm r = 50 mm b = 100 mm Solution 3.10-6 Steel tube t = 8 mm r = 50 mm SHEAR STRESS r = 50 mm G ϭ 76 GPa L ϭ 1.5 m tϭ 10 kN . m T ϭ 2tAm 2(8 mm)(17,850 mm2 ) ϭ 35.0 MPa T ϭ 10 kN . m b = 100 mm Am ϭ ␲(50 mm)2ϩ2(100 mm)(50 mm) ϭ 17,850 mm2 Lm ϭ 2bϩ2␲r ϭ 2(100 mm)ϩ2␲(50 mm) ϭ 514.2 mm 4tA2 4(8 mm)(17,850 mm2 ) 2 m ϭ Lm 514.2 mm ϭ 19.83 ϫ 106 mm4 Jϭ Am ϭ ␲r2ϩ2br ANGLE OF TWIST fϭ (10 kN . m) (1.5 m) TL ϭ GJ (76 GPa)(19.83 ϫ 106 mm4 ) ϭ 0.00995 rad ϭ 0.570Њ
    • 248 CHAPTER 3 Torsion Problem 3.10-7 A thin-walled steel tube having an elliptical cross section with constant thickness t (see figure) is subjected to a torque T ϭ 18 k-in. Determine the shear stress ␶ and the rate of twist ␪ (in degrees per inch) if G ϭ 12 ϫ 106 psi, t ϭ 0.2 in., a ϭ 3 in., and b ϭ 2 in. (Note: See Appendix D, Case 16, for the properties of an ellipse.) t 2b 2a Solution 3.10-7 Elliptical tube t FROM APPENDIX D, CASE 16: Am ϭ ␲ab ϭ ␲(3.0 in.)(2.0 in.) ϭ 18.850 in.2 2b Lm Ϸ p[1.5(a ϩ b) Ϫ ͙ab] ϭ ␲[1.5(5.0 in.) Ϫ ͙6.0 in.2 ] ϭ 15.867 in. 4tA2 4(0.2 in.)(18.850 in.2 ) 2 m ϭ Lm 15.867 in. 4 ϭ 17.92 in. Jϭ 2a T ϭ 18 k-in. G ϭ 12 ϫ 106 psi SHEAR STRESS t ϭ constant t ϭ 0.2 in a ϭ 3.0 in. b ϭ 2.0 in. tϭ T 18 k-in. ϭ 2tAm 2(0.2 in.)(18.850 in.2 ) ϭ 2390 psi ANGLE OF TWIST PER UNIT LENGTH (RATE OF TWIST) uϭ f T 18 k-in. ϭ ϭ 6 L GJ (12 ϫ 10 psi)(17.92 in.4 ) u ϭ 83.73 ϫ 10 Ϫ6 radրin. ϭ 0.0048Њրin. Problem 3.10-8 A torque T is applied to a thin-walled tube having a cross section in the shape of a regular hexagon with constant wall thickness t and side length b (see figure). Obtain formulas for the shear stress ␶ and the rate of twist ␪. t b
    • SECTION 3.10 Solution 3.10-8 Thin-Walled Tubes Regular hexagon SHEAR STRESS t tϭ T T͙3 ϭ 2tAm 9b2t ANGLE OF TWIST PER UNIT LENGTH (RATE OF TWIST) 4A2 t 4A2 t 9b3t m m J ϭ Lm ϭ ϭ Lm 2 dS Ύ b t 0 T 2T 2T uϭ ϭ 3 ϭ GJ G(9b t) 9Gb3t b ϭ Length of side t ϭ Thickness (radians per unit length) Lm ϭ 6b FROM APPENDIX D, CASE 25: ␤ ϭ 60Њ n ϭ 6 b 6b2 nb2 cot ϭ cot 30Њ 4 2 4 Am ϭ ϭ 3͙3b2 2 Problem 3.10-9 Compare the angle of twist ␾1 for a thin-walled circular tube (see figure) calculated from the approximate theory for thin-walled bars with the angle of twist ␾2 calculated from the exact theory of torsion for circular bars. t r C (a) Express the ratio ␾1/␾2 in terms of the nondimensional ratio ␤ ϭ r/t. (b) Calculate the ratio of angles of twist for ␤ ϭ 5, 10, and 20. What conclusion about the accuracy of the approximate theory do you draw from these results? Solution 3.10-9 Thin-walled tube (a) RATIO t f1 4r2 ϩ t2 t2 ϭ ϭ1ϩ 2 f2 4r2 4r r C Let b ϭ APPROXIMATE THEORY f1 ϭ TL GJ J ϭ 2␲r3t f1 ϭ TL 2␲Gr3t TL GIP From Eq. (3-17): Ip ϭ TL 2TL f2 ϭ ϭ GIP ␲Grt(4r2 ϩ t2 ) f1 1 ϭ1ϩ 2 f2 4b ␤ ␾1/␾2 ␲rt (4r2 ϩ t2 ) 2 5 1.0100 10 EXACT THEORY f2 ϭ (b) r t 1.0025 20 1.0006 As the tube becomes thinner and ␤ becomes larger, the ratio ␾1/␾2 approaches unity. Thus, the thinner the tube, the more accurate the approximate theory becomes. 249
    • 250 CHAPTER 3 Torsion Problem 3.10-10 A thin-walled rectangular tube has uniform thickness t and dimensions a ϫ b to the median line of the cross section (see figure). How does the shear stress in the tube vary with the ratio ␤ ϭ a/b if the total length Lm of the median line of the cross section and the torque T remain constant? From your results, show that the shear stress is smallest when the tube is square (␤ ϭ 1). t b a Solution 3.10-10 Rectangular tube t T, t, and Lm are constants. Let k ϭ b 2T ϭ constant tL2 m tϭk (1 ϩ b) 2 b 8 a 6 ␶ k 4 t ϭ thickness (constant) 2 a, b ϭ dimensions of the tube 0 bϭ a b ¢ ≤ t k Lm ϭ 2(a ϩ b) ϭ constant SHEAR STRESS Lm 2(1 ϩ b) bL2 m Am ϭ 4(1 ϩ b) 2 tϭ min 3 a ␤ϭb 8T tL2 m ALTERNATE SOLUTION Am ϭ ab ϭ bb2 Lm ϭ 2b(1 ϩ b) ϭ constant bϭ tmin ϭ 2 From the graph, we see that ␶ is minimum when ␤ ϭ 1 and the tube is square. T ϭ constant T tϭ 2tAm ϭ4 1 Am ϭ b B Lm R 2(1 ϩ b) T(4)(1 ϩ b) 2 2T(1 ϩ b) 2 T ϭ ϭ 2tAm 2tbL2 tL2 b m m tϭ 2 2T (1 ϩ b) 2 B R b tL2 m dt 2T b(2)(1 ϩ b) Ϫ (1 ϩ b) 2 (1) ϭ 2B R ϭ0 db tLm b2 or 2␤ (1ϩ␤)Ϫ(1ϩ␤)2 ϭ 0 І␤ϭ1 Thus, the tube is square and ␶ is either a minimum or a maximum. From the graph, we see that ␶ is a minimum.
    • SECTION 3.10 Problem 3.10-11 A tubular aluminum bar (G ϭ 4 ϫ 106 psi) of square cross section (see figure) with outer dimensions 2 in. ϫ 2 in. must resist a torque T ϭ 3000 lb-in. Calculate the minimum required wall thickness tmin if the allowable shear stress is 4500 psi and the allowable rate of twist is 0.01 rad/ft. Thin-Walled Tubes t 2 in. 2 in. Solution 3.10-11 Square aluminum tube THICKNESS t BASED UPON SHEAR STRESS t tϭ 2 in. T 2tAm tAm ϭ T 2t t(b Ϫ t) 2 ϭ UNITS: t ϭ in. t(2.0 in. Ϫ t) 2 ϭ 2 in. b ϭ in. T ϭ lb-in. ␶ ϭ psi 3000 lb-in. 1 ϭ in.3 2(4500 psi) 3 3t(2 Ϫ t) 2 Ϫ 1 ϭ 0 Solve for t: t ϭ 0.0915 in. Outer dimensions: 2.0 in. ϫ 2.0 in. THICKNESS t BASED UPON RATE OF TWIST G ϭ 4 ϫ 106 psi uϭ T ϭ 3000 lb-in. T T ϭ GJ Gt(b Ϫ t) 3 t(b Ϫ t) 3 ϭ T Gu UNITS: t ϭ in. ␶allow ϭ 4500 psi uallow ϭ 0.01 radրft ϭ 0.01 radրin. 12 G ϭ psi ␪ ϭ rad/in. t(2.0 in. Ϫ t) 3 ϭ 3000 lb-in (4 ϫ 10 psi)(0.01ր12 rad րin.) Let b ϭ outer dimension ϭ 2.0 in. Centerline dimension ϭ b Ϫ t Am ϭ (b Ϫ t)2 T 2t Lm ϭ 4(b Ϫ t) 4tA2 4t(b Ϫ t) 4 m Jϭ ϭ ϭ t(b Ϫ t) 3 Lm 4(b Ϫ t) ϭ 6 9 10 10t(2 Ϫ t) 3 Ϫ 9 ϭ 0 Solve for t: t ϭ 0.140 in. ANGLE OF TWIST GOVERNS tmin ϭ 0.140 in. 251
    • 252 CHAPTER 3 Torsion Problem 3.10-12 A thin tubular shaft of circular cross section (see figure) with inside diameter 100 mm is subjected to a torque of 5000 N и m. If the allowable shear stress is 42 MPa, determine the required wall thickness t by using (a) the approximate theory for a thin-walled tube, and (b) the exact torsion theory for a circular bar. 100 mm t Solution 3.10-12 Thin tube (b) EXACT THEORY tϭ 100 mm Tr2 Ip Ip ϭ ␲ 4 ␲ (r2 Ϫ r4 ) ϭ [ (50 ϩ t) 4 Ϫ (50) 4 ] 1 2 2 (5,000 N . m)(50 ϩ t) ␲ [ (50 ϩ t) 4 Ϫ (50) 4 ] 2 (50 ϩ t) 4 Ϫ (50) 4 (5000 N . m)(2) ϭ 50 ϩ t (␲)(42 MPa) 42 MPa ϭ t T ϭ 5,000 N и m d1 ϭ inner diameter ϭ 100 mm ␶allow ϭ 42 MPa ϭ t is in millimeters. r ϭ Average radius Solve for t: t ϭ 7.02 mm t 2 r1 ϭ Inner radius ϭ 50 mm ϩ The approximate result is 5% less than the exact result. Thus, the approximate theory is nonconservative and should only be used for thin-walled tubes. ϭ 50 mm r2 ϭ Outer radius ϭ 50 mm ϩ t Am ϭ ␲r2 (a) APPROXIMATE THEORY tϭ T T T ϭ 2 ϭ 2tAm 2t(␲r ) 2␲r2t 42 MPa ϭ or t ¢ 50 ϩ 5,000 N . m t 2 2␲ ¢ 50 ϩ ≤ t 2 t 2 5,000 N . m 5 ϫ 106 ≤ ϭ ϭ mm3 2 2␲(42 MPa) 84␲ Solve for t: t ϭ 6.66 mm 5 ϫ 106 mm3 21␲
    • SECTION 3.10 Problem 3.10-13 A long, thin-walled tapered tube AB of circular cross section (see figure) is subjected to a torque T. The tube has length L and constant wall thickness t. The diameter to the median lines of the cross sections at the ends A and B are dA and dB, respectively. Derive the following formula for the angle of twist of the tube: ΂ T ΃ B d(x) dA x dB dx 4T fϭ ␲GT Ύ L 0 t ϭ thickness Ύ (a ϩ bx) dB ϭ average diameter at end B T ϭ torque fϭ d(x) ϭ average diameter at distance x from end A. dB Ϫ dA ≤x L ␲d 3t 4 3 dB Ϫ dA ␲t ␲t [d(x) ] 3 ϭ B dA ϩ ¢ ≤ xR 4 4 L For element of length dx: Tdx ϭ GJ(x) 4Tdx 3 dB Ϫ dA G␲t B dA ϩ ¢ ≤ xR L dx 3 dB Ϫ dA B dA ϩ ¢ ≤ xR L From table of integrals (see Appendix C): dx dA ϭ average diameter at end A df ϭ dB dA For entire tube: L J(x) ϭ t t Thin-walled tapered tube A J ϭ 2␲r3t ϭ T L Hint: If the angle of taper is small, we may obtain approximate results by applying the formulas for a thin-walled prismatic tube to a differential element of the tapered tube and then integrating along the axis of the tube. d(x) ϭ dA ϩ ¢ B A 2TL dA ϩ dB ␾ ϭ ᎏᎏ ᎏᎏᎏᎏ 2 2 dAdB ␲ Gt Solution 3.10-13 ϭ 3 ϭϪ 4T Ϫ ␲Gt C 2¢ 1 2b(a ϩ bx) 2 2 dB Ϫ dA dB Ϫ dA S ≤¢ dA ϩ ؒ x≤ L L 0 1 4T L L BϪ 2ϩ 2R ␲Gt 2(dB Ϫ dA )dB 2(dB Ϫ dA )dA fϭ 253 Thin-Walled Tubes 2TL dA ϩ dB ¢ 2 2 ≤ ␲Gt dAdB L
    • 254 CHAPTER 3 Torsion Stress Concentrations in Torsion The problems for Section 3.11 are to be solved by considering the stress-concentration factors. D2 Problem 3.11-1 A stepped shaft consisting of solid circular segments T having diameters D1 ϭ 2.0 in. and D2 ϭ 2.4 in. (see figure) is subjected to torques T. The radius of the fillet is R ϭ 0.1 in. If the allowable shear stress at the stress concentration is 6000 psi, what is the maximum permissible torque Tmax? Solution 3.11-1 R D1 T Probs. 3.11-1 through 3.11-5 Stepped shaft in torsion D2 R USE FIG. 3-48 FOR THE STRESS-CONCENTRATION D1 T FACTOR T D2 2.4 in. ϭ ϭ 1.2 D1 2.0 in. R 0.1 in. ϭ ϭ 0.05 D1 2.0 in. K Ϸ 1.52 D1 ϭ 2.0 in. D2 ϭ 2.4 in. Tmax ϭ R ϭ 0.1 in. ϭ ␶allow ϭ 6000 psi tmax ϭ K tnom ϭ K ¢ ␲D3tmax 1 16K 16 Tmax ≤ ␲ D3 1 ␲(2.0 in.) 3 (6000 psi) ϭ 6200 lb-in. 16(1.52) ∴ Tmax Ϸ 6200 lb-in. Problem 3.11-2 A stepped shaft with diameters D1 ϭ 40 mm and D2 ϭ 60 mm is loaded by torques T ϭ 1100 N и m (see figure). If the allowable shear stress at the stress concentration is 120 MPa, what is the smallest radius Rmin that may be used for the fillet? Solution 3.11-2 Stepped shaft in torsion D2 T USE FIG. 3-48 FOR THE STRESS-CONCENTRATION FACTOR R D1 T tmax ϭ Ktnom ϭ K ¢ Kϭ D1 ϭ 40 mm D2 ϭ 60 mm T ϭ 1100 N и m ␶allow ϭ 120 MPa 16T ≤ ␲D3 1 ␲D3tmax ␲(40 mm) 3 (120 MPa) 1 ϭ ϭ 1.37 16(1100 N # m) 16T D2 60 mm ϭ ϭ 1.5 D1 40 mm From Fig. (3-48) with we get D2 ϭ 1.5 and K ϭ 1.37, D1 R Ϸ 0.10 D1 ∴ Rmin Ϸ 0.10(40 mm) ϭ 4.0 mm
    • SECTION 3.11 Stress Concentrations in Torsion Problem 3.11-3 A full quarter-circular fillet is used at the shoulder of a stepped shaft having diameter D2 ϭ 1.0 in. (see figure). A torque T ϭ 500 lb-in. acts on the shaft. Determine the shear stress ␶max at the stress concentration for values as follows: D1 ϭ 0.7, 0.8, and 0.9 in. Plot a graph showing ␶max versus D1. Solution 3.11-3 Stepped shaft in torsion D2 R D1 (in.) D2/D1 D1 T T R (in.) R/D1 K ␶max (psi) 0.7 1.43 0.15 0.214 1.20 8900 0.8 1.25 0.10 0.125 1.29 6400 0.9 1.11 0.05 0.056 1.41 4900 D2 ϭ 1.0 in. T ϭ 500 lb-in. D1 ϭ 0.7, 0.8, and 0.9 in. 10,000 Full quarter-circular fillet (D2 ϭ D1 ϩ 2R) ␶max D2 Ϫ D1 D1 Rϭ ϭ 0.5 in. Ϫ 2 2 (psi) 5000 USE FIG. 3-48 FOR THE STRESS-CONCENTRATION FACTOR tmax ϭ Ktnom ϭ K ¢ ϭK 16T ≤ ␲D3 1 16(500 lb-in.) K ϭ 2546 3 ␲D3 D1 1 0 0.6 0.8 1 D1 (in.) Note that ␶max gets smaller as D1 gets larger, even though K is increasing. 255
    • 256 CHAPTER 3 Torsion Problem 3.11-4 The stepped shaft shown in the figure is required to transmit 600 kW of power at 400 rpm. The shaft has a full quarter-circular fillet, and the smaller diameter D1 ϭ 100 mm. If the allowable shear stress at the stress concentration is 100 MPa, at what diameter D2 will this stress be reached? Is this diameter an upper or a lower limit on the value of D2? Solution 3.11-4 Stepped shaft in torsion D2 R D1 T P ϭ 600 kW n ϭ 400 rpm D1 ϭ 100 mm ␶allow ϭ 100 MPa Full quarter-circular fillet POWER P ϭ 2␲nT (Eq. 3-42 of Section 3.7) 60 P ϭ watts n ϭ rpm T ϭ Newton meters 60P 60(600 ϫ 10 W) ϭ ϭ 14,320 N # m 2␲n 2␲(400 rpm) T Use the dashed line for a full quarter-circular fillet. R Ϸ 0.075 D1 R Ϸ 0.075 D1 ϭ 0.075 (100 mm) ϭ 7.5 mm D2 ϭ D1 ϩ 2R ϭ 100 mm ϩ 2(7.5 mm) ϭ 115 mm ∴ D2 Ϸ 115 mm 3 Tϭ USE FIG. 3-48 FOR THE STRESS-CONCENTRATION FACTOR tmax ϭ Ktnom ϭ K ¢ Kϭ ϭ tmax (␲D3 ) 1 16T 16T ≤ ␲D3 1 (100 MPa)(␲)(100 mm) 3 ϭ 1.37 16(14,320 N # m) This value of D2 is a lower limit (If D2 is less than 115 mm, R/D1 is smaller, K is larger, and ␶max is larger, which means that the allowable stress is exceeded.)
    • SECTION 3.11 Stress Concentrations in Torsion Problem 3.11-5 A stepped shaft (see figure) has diameter D2 ϭ 1.5 in. and a full quarter-circular fillet. The allowable shear stress is 15,000 psi and the load T ϭ 4800 lb-in. What is the smallest permissible diameter D1? Solution 3.11-5 Stepped shaft in torsion D2 R D1 T D2 ϭ 1.5 in. Use trial-and-error. Select trial values of D1 ␶allow ϭ 15,000 psi D1 (in.) T ϭ 4800 lb-in. Full quarter-circular fillet D2 ϭ D1 ϩ 2R Rϭ T R (in.) R/D1 K ␶max(psi) 1.30 1.35 1.40 0.100 0.075 0.050 0.077 0.056 0.036 1.38 1.41 1.46 15,400 14,000 13,000 D2 Ϫ D1 D1 ϭ 0.75 in. Ϫ 2 2 USE FIG. 3-48 FOR THE STRESS-CONCENTRATION FACTOR tmax ϭ Ktnom ϭ K ¢ 16T ≤ ␲D3 1 K 16(4800 lb-in.) ϭ 3B R ␲ D1 ϭ 24,450 K D3 1 ␶max (psi) 16,000 ␶allow 15,000 D1=1.31in. 14,000 13,000 1.30 1.40 From the graph, minimum D1 Ϸ 1.31 in. D1(in.) 257
    • 4 Shear Forces and Bending Moments Shear Forces and Bending Moments 800 lb Problem 4.3-1 Calculate the shear force V and bending moment M at a cross section just to the left of the 1600-lb load acting on the simple beam AB shown in the figure. A B 30 in. Solution 4.3-1 1600 lb 60 in. 120 in. 30 in. Simple beam 800 lb Free-body diagram of segment DB 1600 lb D A 1600 lb V B D B M 30 in. 30 in. 60 in. 30 in. RA ⌺MA ϭ 0: ⌺MB ϭ 0: RB RB RB ϭ 1400 lb RA ϭ 1000 lb ©FVERT ϭ 0: V ϭ 1600 lb Ϫ 1400 lb ϭ 200 lb M ϭ (1400 lb)(30 in.) ϭ 42,000 lb-in. ©MD ϭ 0: 6.0 kN Problem 4.3-2 Determine the shear force V and bending moment M at the midpoint C of the simple beam AB shown in the figure. A 1.0 m Solution 4.3-2 6.0 kN B C V A ⌺MA ϭ 0: ⌺MB ϭ 0: 2.0 m Free-body diagram of segment AC 2.0 kN/m C 1.0 m RA 1.0 m 4.0 m B Simple beam 6.0 kN A 2.0 kN/m C 1.0 m RB ϭ 4.5 kN RA ϭ 5.5 kN 2.0 m 1.0 m RB M 1.0 m RA ©FVERT ϭ 0: ©MC ϭ 0: V ϭ Ϫ0.5 kN M ϭ 5.0 kN ؒ m 259
    • 260 CHAPTER 4 Shear Forces and Bending Moments Problem 4.3-3 Determine the shear force V and bending moment M at the midpoint of the beam with overhangs (see figure). Note that one load acts downward and the other upward. P P b Solution 4.3-3 Beam with overhangs P P A B ©MA ϭ 0: P L RB ϭ P ¢ 1 ϩ A b L b RA b RB 2b ≤ L b (downward) M C L/2 RA V Free-body diagram (C is the midpoint) ©MB ϭ 0 ©FVERT ϭ 0: V ϭ RA Ϫ P ϭ P ¢ 1 ϩ 1 RA ϭ [P(L ϩ b ϩ b) ] L 2b ϭ P ¢1 ϩ ≤ L ©MC ϭ 0: M ϭ P ¢1 ϩ 2b 2bP ≤ϪP ϭ L L 2b L L ≤ ¢ ≤ Ϫ P ¢b ϩ ≤ L 2 2 PL PL Mϭ ϩ Pb Ϫ Pb Ϫ ϭ0 2 2 (upward) Problem 4.3-4 Calculate the shear force V and bending moment M at a cross section located 0.5 m from the fixed support of the cantilever beam AB shown in the figure. 4.0 kN 1.0 m Solution 4.3-4 1.5 kN/m A B 1.0 m 2.0 m Free-body diagram of segment DB Point D is 0.5 m from support A. 4.0 kN V 1.5 kN/m D B M 0.5 m B 1.0 m 2.0 m Cantilever beam 4.0 kN 1.0 m 1.5 kN/m A 1.0 m 2.0 m ©FVERT ϭ 0: V ϭ 4.0 kN ϩ (1.5 kNրm)(2.0 m) ϭ 4.0 kN ϩ 3.0 kN ϭ 7.0 kN ©MD ϭ 0: M ϭ Ϫ(4.0 kN)(0.5 m) Ϫ (1.5 kNրm)(2.0 m)(2.5 m) ϭ Ϫ2.0 kN ؒ m Ϫ 7.5 kN ؒ m ϭ Ϫ9.5 kN ؒ m
    • SECTION 4.3 Problem 4.3-5 Determine the shear force V and bending moment M at a cross section located 16 ft from the left-hand end A of the beam with an overhang shown in the figure. 400 lb/ft 200 lb/ft B A 10 ft Solution 4.3-5 10 ft C 6 ft 6 ft Beam with an overhang 400 lb/ft 200 lb/ft B A 10 ft 261 Shear Forces and Bending Moments 10 ft RA Free-body diagram of segment AD C 6 ft 400 lb/ft 10 ft V RA ϭ 2460 lb ⌺MA ϭ 0: RB ϭ 2740 lb M 6 ft RA RB ⌺MB ϭ 0: D A 6 ft Point D is 16 ft from support A. ©FVERT ϭ 0: V ϭ 2460 lb Ϫ (400 lbրft)(10 ft) ϭ Ϫ1540 lb ©MD ϭ 0: M ϭ (2460 lb)(16 ft) Ϫ (400 lbրft)(10 ft)(11 ft) ϭ Ϫ4640 lb-ft Problem 4.3-6 The beam ABC shown in the figure is simply P = 4.0 kN 1 supported at A and B and has an overhang from B to C. The loads consist of a horizontal force P1 ϭ 4.0 kN acting at the end of a vertical arm and a vertical force P2 ϭ 8.0 kN acting at 1.0 m A the end of the overhang. Determine the shear force V and bending moment M at a cross section located 3.0 m from the left-hand support. (Note: Disregard the widths of the beam and vertical arm and use centerline dimensions when making calculations.) Solution 4.3-6 P2 = 8.0 kN B 4.0 m C 1.0 m Beam with vertical arm P1 = 4.0 kN P2 = 8.0 kN 1.0 m A Free-body diagram of segment AD Point D is 3.0 m from support A. B 4.0 kN • m A 3.0 m 4.0 m RA ⌺MB ϭ 0: RA ϭ 1.0 kN (downward) ⌺MA ϭ 0: RB ϭ 9.0 kN (upward) 1.0 m RB RA ©FVERT ϭ 0: ©MD ϭ 0: M D V V ϭ ϪRA ϭ Ϫ 1.0 kN M ϭ ϪRA (3.0 m) Ϫ 4.0 kN ؒ m ϭ Ϫ7.0 kN ؒ m
    • 262 CHAPTER 4 Shear Forces and Bending Moments Problem 4.3-7 The beam ABCD shown in the figure has overhangs at each end and carries a uniform load of intensity q. For what ratio b/L will the bending moment at the midpoint of the beam be zero? q A D B b Solution 4.3-7 C L b Beam with overhangs q A D B b Free-body diagram of left-hand half of beam: Point E is at the midpoint of the beam. q C L b RC RB M = 0 (Given) A b L ≤ 2 V RB From symmetry and equilibrium of vertical forces: RB ϭ RC ϭ q ¢ b ϩ L/2 E ©ME ϭ 0 ‫۔ ە‬ L 1 L 2 ϪRB ¢ ≤ ϩ q ¢ ≤ ¢ b ϩ ≤ ϭ 0 2 2 2 L L 1 L 2 Ϫq ¢ b ϩ ≤ ¢ ≤ ϩ q ¢ ≤ ¢ b ϩ ≤ ϭ 0 2 2 2 2 Solve for b/L : b 1 ϭ L 2 Problem 4.3-8 At full draw, an archer applies a pull of 130 N to the bowstring of the bow shown in the figure. Determine the bending moment at the midpoint of the bow. 70° 1400 mm 350 mm
    • SECTION 4.3 Solution 4.3-8 Shear Forces and Bending Moments Archer’s bow B Free-body diagram of segment BC B ␤ ␤ P C H 2 T H A b C M ©MC ϭ 0 ‫۔ ە‬ b P ϭ 130 N ␤ ϭ 70° H ϭ 1400 mm ϭ 1.4 m H ≤ ϩ T(sin b) (b) Ϫ M ϭ 0 2 H M ϭ T ¢ cosb ϩ b sin b≤ 2 P H ϭ ¢ ϩ b tan b≤ 2 2 Substitute numerical values: b ϭ 350 mm 130 N 1.4 m B ϩ (0.35 m)(tan 70Њ)R 2 2 M ϭ 108 N ؒ m Mϭ ϭ 0.35 m Free-body diagram of point A T ␤ P T(cos b) ¢ A T T ϭ tensile force in the bowstring ⌺FHORIZ ϭ 0: 2T cos ␤Ϫ P ϭ 0 Tϭ P 2 cos b 263
    • 264 CHAPTER 4 Shear Forces and Bending Moments Problem 4.3-9 A curved bar ABC is subjected to loads in the form of two equal and opposite forces P, as shown in the figure. The axis of the bar forms a semicircle of radius r. Determine the axial force N, shear force V, and bending moment M acting at a cross section defined by the angle ␪. Solution 4.3-9 M P cos ␪ r ␪ A A V r ␪ P O N P ␪ P C A Curved bar B P M B P O ©FN ϭ 0 Q bϪ ϩ B A N Ϫ P sin uϭ 0 N ϭ P sin u V ␪ P C N O P sin ␪ ͚FV ϭ 0 ϩ Ϫ R a ©MO ϭ 0 ‫۔ ە‬ V Ϫ P cos u ϭ 0 V ϭ P cos u M Ϫ Nr ϭ 0 M ϭ Nr ϭ Pr sin u Problem 4.3-10 Under cruising conditions the distributed load acting on the wing of a small airplane has the idealized variation shown in the figure. Calculate the shear force V and bending moment M at the inboard end of the wing. 1600 N/m 2.6 m Solution 4.3-10 1.0 m 2.6 m Airplane wing 1600 N/m M 900 N/m Loading (in three parts) 900 N/m 700 N/m 1 V 2 900 N/m A B 2.6 m 2.6 m 1.0 m 3 B Bending Moment Shear Force ⌺FVERT ϭ 0 A Ϫ cϩ T 1 V ϩ (700 Nրm)(2.6 m) ϩ (900 Nրm)(5.2 m) 2 1 ϩ (900 Nրm)(1.0 m) ϭ 0 2 V ϭ Ϫ6040 N ϭ Ϫ6.04 kN (Minus means the shear force acts opposite to the direction shown in the figure.) ©MA ϭ 0 ‫۔ە‬ ϪM ϩ 1 2.6 m (700 Nրm)(2.6 m) ¢ ≤ 2 3 ϩ (900 Nրm)(5.2 m)(2.6 m) 1 1.0 m ϩ (900 Nրm)(1.0 m) ¢ 5.2 m ϩ ≤ϭ0 2 3 M ϭ 788.67 N • m ϩ 12,168 N • m ϩ 2490 N • m ϭ 15,450 N • m ϭ 15.45 kN ؒ m
    • SECTION 4.3 Problem 4.3-11 A beam ABCD with a vertical arm CE is supported as a simple beam at A and D (see figure). A cable passes over a small pulley that is attached to the arm at E. One end of the cable is attached to the beam at point B. What is the force P in the cable if the bending moment in the beam just to the left of point C is equal numerically to 640 lb-ft? (Note: Disregard the widths of the beam and vertical arm and use centerline dimensions when making calculations.) E P Cable A 8 ft B 6 ft Solution 4.3-11 265 Shear Forces and Bending Moments C D 6 ft 6 ft Beam with a cable E P Free-body diagram of section AC P Cable A P B 6 ft 4P __ 9 UNITS: P in lb M in lb-ft 8 ft C 6 ft P D 4P __ 9 3P __ 5 M C N 6 ft 4P __ 9 6 ft 4P __ 5 A B 6 ft V ©MC ϭ 0 ‫۔ە‬ 4P 4P (6 ft) ϩ (12 ft) ϭ 0 5 9 8P Mϭ Ϫ lb-ft 15 Numerical value of M equals 640 lb-ft. M Ϫ 8P lb-ft 15 and P ϭ 1200 lb ∴ 640 lb-ft ϭ Problem 4.3-12 A simply supported beam AB supports a trapezoidally distributed load (see figure). The intensity of the load varies linearly from 50 kN/m at support A to 30 kN/m at support B. Calculate the shear force V and bending moment M at the midpoint of the beam. 50 kN/m 30 kN/m A B 3m
    • 266 CHAPTER 4 Solution 4.3-12 Shear Forces and Bending Moments Beam with trapezoidal load Free-body diagram of section CB 50 kN/m 30 kN/m A Point C is at the midpoint of the beam. 40 kN/m B 30 kN/m V M 3m RA B C 1.5 m RB ⌺FVERT ϭ 0 Reactions ©MB ϭ 0 ‫ ە‬Ϫ RA (3 m) ϩ (30 kNրm)(3 m)(1.5 m) cϩ TϪ V Ϫ (30 kNրm)(1.5 m) Ϫ 1 (10 kNրm)(1.5 m) 2 ϩ (20 kNրm)(3 m)( 1΋2 )(2 m) ϭ 0 RA ϭ 65 kN 55 kN ϩ 55 kNϭ 0 V ϭ Ϫ2.5 kN ϩ ©FVERT ϭ 0 c RA ϩ RB Ϫ 1΋2 (50 kNրm ϩ 30 kNրm)(3 m) ϭ 0 RB ϭ 55 kN ©MC ϭ 0 ‫۔ە‬ Ϫ M Ϫ (30 kN/m)(1.5 m)(0.75 m) Ϫ 1΋2 (10 kNրm)(1.5 m)(0.5 m) ϩ (55 kN)(1.5 m) ϭ 0 M ϭ 45.0 kN ؒ m q1 = 3500 lb/ft Problem 4.3-13 Beam ABCD represents a reinforced-concrete foundation beam that supports a uniform load of intensity q1 ϭ 3500 lb/ft (see figure). Assume that the soil pressure on the underside of the beam is uniformly distributed with intensity q2. (a) Find the shear force VB and bending moment MB at point B. (b) Find the shear force Vm and bending moment Mm at the midpoint of the beam. Solution 4.3-13 C D 3.0 ft q2 8.0 ft 3.0 ft Foundation beam q1 = 3500 lb/ft A B A B (b) V and M at midpoint E C D 3500 lb/ft B A 3.0 ft q2 8.0 ft E 3.0 ft Vm 2000 lb/ft ⌺FVERT ϭ 0: q2(14 ft) ϭ q1(8 ft) 8 ∴ q2 ϭ q ϭ 2000 lbրft 14 1 (a) V and M at point B B A MB ⌺FVERT ϭ 0: 2000 lb/ft 3 ft ©MB ϭ 0: VB VB ϭ 6000 lb MB ϭ 9000 lb-ft 3 ft Mm 4 ft ⌺FVERT ϭ 0: Vm ϭ (2000 lb/ft)(7 ft) Ϫ (3500 lb/ft)(4 ft) Vm ϭ 0 ⌺ME ϭ 0: Mm ϭ (2000 lb/ft)(7 ft)(3.5 ft) Ϫ (3500 lb/ft)(4 ft)(2 ft) Mm ϭ 21,000 lb-ft
    • SECTION 4.3 E Problem 4.3-14 The simply-supported beam ABCD is loaded by a weight W ϭ 27 kN through the arrangement shown in the figure. The cable passes over a small frictionless pulley at B and is attached at E to the end of the vertical arm. Calculate the axial force N, shear force V, and bending moment M at section C, which is just to the left of the vertical arm. (Note: Disregard the widths of the beam and vertical arm and use centerline dimensions when making calculations.) Cable 1.5 m A B C 2.0 m 2.0 m W = 27 kN Solution 4.3-14 Beam with cable and weight E Cable A B 2.0 m Free-body diagram of pulley at B 1.5 m 27 kN C 2.0 m D 21.6 kN 2.0 m 27 kN RA ϭ 18 kN 10.8 kN 27 kN RD RA RD ϭ 9 kN Free-body diagram of segment ABC of beam 10.8 kN 21.6 kN A 2.0 m B M C N 2.0 m V 18 kN ©FHORIZ ϭ 0: N ϭ 21.6 kN (compression) ©FVERT ϭ 0: V ϭ 7.2 kN ©MC ϭ 0: M ϭ 50.4 kN ؒ m 267 Shear Forces and Bending Moments D 2.0 m
    • 268 CHAPTER 4 Shear Forces and Bending Moments Problem 4.3-15 The centrifuge shown in the figure rotates in a horizontal plane (the xy plane) on a smooth surface about the z axis (which is vertical) with an angular acceleration ␣. Each of the two arms has weight w per unit length and supports a weight W ϭ 2.0 wL at its end. Derive formulas for the maximum shear force and maximum bending moment in the arms, assuming b ϭ L/9 and c ϭ L/10. y c L b W x ␣ W Solution 4.3-15 Rotating centrifuge c L b W __ (L + b + c)␣ g x Tangential acceleration ϭ r␣ Substitute numerical data: W Inertial force Mr ␣ ϭ g r␣ Maximum V and M occur at x ϭ b. Ύ Ύ W ϭ 2.0 wL Mmax bϭ 91wL2␣ 30g 229wL3␣ ϭ 75g Vmax ϭ Lϩb W w␣ (L ϩ b ϩ c)␣ ϩ x dx g g b W␣ ϭ (L ϩ b ϩ c) g wL␣ ϩ (L ϩ 2b) 2g W␣ Mmax ϭ (L ϩ b ϩ c)(L ϩ c) g Lϩb w␣ ϩ x(x Ϫ b)dx g b W␣ ϭ (L ϩ b ϩ c)(L ϩ c) g w L2␣ ϩ (2L ϩ 3b) 6g Vmax ϭ w␣x __ g L 9 cϭ L 10
    • SECTION 4.5 269 Shear-Force and Bending-Moment Diagrams Shear-Force and Bending-Moment Diagrams When solving the problems for Section 4.5, draw the shear-force and bending-moment diagrams approximately to scale and label all critical ordinates, including the maximum and minimum values. Probs. 4.5-1 through 4.5-10 are symbolic problems and Probs. 4.5-11 through 4.5-24 are numerical problems. The remaining problems (4.5-25 through 4.5-30) involve specialized topics, such as optimization, beams with hinges, and moving loads. Problem 4.5-1 Draw the shear-force and bending-moment diagrams for a simple beam AB supporting two equal concentrated loads P (see figure). a P P A B L Solution 4.5-1 Simple beam a P P A L RB = P P V 0 ᎐P Pa M 0 a B RA = P a
    • 270 CHAPTER 4 Shear Forces and Bending Moments Problem 4.5-2 A simple beam AB is subjected to a counterclockwise couple of moment M0 acting at distance a from the left-hand support (see figure). Draw the shear-force and bending-moment diagrams for this beam. M0 A B a L Solution 4.5-2 Simple beam M0 A RA = B a M0 L RB = L M0 L V 0 M M0 L M0a L 0 ᎐ M0 (1᎐ a ) L q Problem 4.5-3 Draw the shear-force and bending-moment diagrams for a cantilever beam AB carrying a uniform load of intensity q over one-half of its length (see figure). A B L — 2 Solution 4.5-3 Cantilever beam MA = 3qL2 8 q A B RA = qL 2 L — 2 L — 2 qL — 2 V M 0 0 Ϫ 3qL2 8 qL2 Ϫ 8 L — 2
    • SECTION 4.5 Problem 4.5-4 The cantilever beam AB shown in the figure is subjected to a concentrated load P at the midpoint and a counterclockwise couple of moment M1 ϭ PL/4 at the free end. Draw the shear-force and bending-moment diagrams for this beam. Solution 4.5-4 271 Shear-Force and Bending-Moment Diagrams PL M1 = —– 4 P B A L — 2 L — 2 Cantilever beam P A B MA L/2 RA V M PL M1 ϭ 4 RA ϭ P L/2 MA ϭ P 0 PL 4 0 PL Ϫ 4 Problem 4.5-5 The simple beam AB shown in the figure is subjected to a concentrated load P and a clockwise couple M1 ϭ PL/4 acting at the third points. Draw the shear-force and bending-moment diagrams for this beam. A B Simple beam PL M1 = —– 4 P A 5P RA = —– 12 B L — 3 L — 3 L — 3 7P RB = —– 12 5P/12 V 0 Ϫ7P/12 5PL/36 M 7PL/36 0 ϪPL/18 PL M1 = —– 4 P L — 3 Solution 4.5-5 PL 4 L — 3 L — 3
    • 272 CHAPTER 4 Shear Forces and Bending Moments Problem 4.5-6 A simple beam AB subjected to clockwise couples M1 and 2M1 acting at the third points is shown in the figure. Draw the shear-force and bending-moment diagrams for this beam. M1 2M1 A B L — 3 Solution 4.5-6 L — 3 L — 3 Simple beam M1 2M1 A B 3M1 RA = —– L L — 3 L — 3 3M1 RB = —– L L — 3 0 3M1 Ϫ —– L V M1 M 0 ϪM1 ϪM1 Problem 4.5-7 A simply supported beam ABC is loaded by a vertical load P acting at the end of a bracket BDE (see figure). Draw the shear-force and bending-moment diagrams for beam ABC. B A C D E P L — 4 L — 4 L — 2 L Solution 4.5-7 Beam with bracket P A PL —– 4 C B P RA = —– 2 V P RC = —– 2 P —– 2 P Ϫ —– 2 0 PL —– 8 M 3L — 4 L — 4 0 3PL —– 8
    • SECTION 4.5 Problem 4.5-8 A beam ABC is simply supported at A and B and has an overhang BC (see figure). The beam is loaded by two forces P and a clockwise couple of moment Pa that act through the arrangement shown. Draw the shear-force and bending-moment diagrams for beam ABC. Solution 4.5-8 273 Shear-Force and Bending-Moment Diagrams P P A Pa C B a a a a Beam with overhang P P C upper beam: a Pa a a P P P P B lower beam: C a a a 2P P V 0 M 0 ϪP ϪPa Problem 4.5-9 Beam ABCD is simply supported at B and C and has overhangs at each end (see figure). The span length is L and each overhang has length L/3. A uniform load of intensity q acts along the entire length of the beam. Draw the shear-force and bending-moment diagrams for this beam. Solution 4.5-9 q A D B L 3 C L L 3 Beam with overhangs q A D L /3 B C L 5qL RB = __ 6 qL __ 2 V 0 L/3 5qL RC = __ 6 qL/3 qL – __ 3 5qL __2 72 qL – __ 2 M 0 –qL2/18 X1 –qL2/18 x1 ϭ L ͙5 ϭ 0.3727L 6
    • 274 CHAPTER 4 Shear Forces and Bending Moments Problem 4.5-10 Draw the shear-force and bending-moment diagrams for a cantilever beam AB supporting a linearly varying load of maximum intensity q0 (see figure). q0 A B L Solution 4.5-10 Cantilever beam q0 x q=q0 __ L q0 L 2 MB = __ 6 B A q0 L RB = __ 2 L x V 0 q0 x2 V = – __ 2L q0 L – __ 2 q0 x3 M = – __ 6L q0 L2 – __ 6 M 0 Problem 4.5-11 The simple beam AB supports a uniform load of intensity q ϭ 10 lb/in. acting over one-half of the span and a concentrated load P ϭ 80 lb acting at midspan (see figure). Draw the shear-force and bending-moment diagrams for this beam. P = 80 lb q = 10 lb/in. A B L — = 40 in. 2 Solution 4.5-11 Simple beam P = 80 lb 10 lb/in. A RA =140 lb B 40 in. 40 in. RB = 340 lb 140 V (lb) 60 0 6 in. –340 5600 M (lb/in.) 0 46 in. Mmax = 5780 L — = 40 in. 2
    • SECTION 4.5 Problem 4.5-12 The beam AB shown in the figure supports a uniform load of intensity 3000 N/m acting over half the length of the beam. The beam rests on a foundation that produces a uniformly distributed load over the entire length. Draw the shear-force and bending-moment diagrams for this beam. 3000 N/m A B 0.8 m Solution 4.5-12 275 Shear-Force and Bending-Moment Diagrams 1.6 m 0.8 m Beam with distributed loads 3000 N/m A B 1500 N/m 0.8 m 1.6 m 0.8 m 1200 V (N) 0 960 480 M –1200 480 (N . m) 0 Problem 4.5-13 A cantilever beam AB supports a couple and a concentrated load, as shown in the figure. Draw the shear-force and bending-moment diagrams for this beam. 200 lb 400 lb-ft A B 5 ft Solution 4.5-13 5 ft Cantilever beam 200 lb 400 lb-ft A B MA = 1600 lb-ft 5 ft RA = 200 lb 5 ft +200 V (lb) 0 0 M (lb-ft) –1600 –600 –1000
    • 276 CHAPTER 4 Shear Forces and Bending Moments Problem 4.5-14 The cantilever beam AB shown in the figure is subjected to a uniform load acting throughout one-half of its length and a concentrated load acting at the free end. Draw the shear-force and bending-moment diagrams for this beam. 2.0 kN/m 2.5 kN B A 2m Solution 4.5-14 2m Cantilever beam 2.0 kN/m M A = 14 kN . m 2.5 kN B A 2m R A = 6.5 kN 2m 6.5 V (kN) 2.5 0 0 M (kN . m) –5.0 –14.0 Problem 4.5-15 The uniformly loaded beam ABC has simple supports at A and B and an overhang BC (see figure). Draw the shear-force and bending-moment diagrams for this beam. 25 lb/in. A C B 72 in. Solution 4.5-15 Beam with an overhang 25 lb/in. A C B 72 in. RA = 500 lb 48 in. RB = 2500 lb 1200 500 V 0 (lb) 20 in. –1300 5000 M 0 (lb-in.) 20 in. 40 in. –28,800 48 in.
    • SECTION 4.5 Problem 4.5-16 A beam ABC with an overhang at one end supports a uniform load of intensity 12 kN/m and a concentrated load of magnitude 2.4 kN (see figure). Draw the shear-force and bending-moment diagrams for this beam. 12 kN/m 2.4 kN A C B 1.6 m Solution 4.5-16 277 Shear-Force and Bending-Moment Diagrams 1.6 m 1.6 m Beam with an overhang 2.4 kN 12 kN/m A C B 1.6 m RA = 13.2 kN 1.6 m 1.6 m RB = 8.4 kN 13.2 V (kN) 2.4 0 1.1m Mmax –6.0 5.76 M (kN . m) 0 Mmax = 7.26 0.64 m 1.1m Problem 4.5-17 The beam ABC shown in the figure is simply supported at A and B and has an overhang from B to C. The loads consist of a horizontal force P1 ϭ 400 lb acting at the end of the vertical arm and a vertical force P2 ϭ 900 lb acting at the end of the overhang. Draw the shear-force and bending-moment diagrams for this beam. (Note: Disregard the widths of the beam and vertical arm and use centerline dimensions when making calculations.) Solution 4.5-17 –3.84 P1 = 400 lb P2 = 900 lb 1.0 ft A B 4.0 ft C 1.0 ft Beam with vertical arm P1 = 400 lb P2 = 900 lb 1.0 ft A B 900 C V (lb) 0 4.0 ft 1.0 ft RA = 125 lb A 400 lb-ft RB = 1025 lb 900 lb B C 125 lb 1025 lb M (lb) Ϫ125 0 Ϫ400 Ϫ900
    • 278 CHAPTER 4 Shear Forces and Bending Moments Problem 4.5-18 A simple beam AB is loaded by two segments of uniform load and two horizontal forces acting at the ends of a vertical arm (see figure). Draw the shear-force and bending-moment diagrams for this beam. 8 kN 4 kN/m 4 kN/m 1m A B 1m 8 kN 2m Solution 4.5-18 2m 2m 2m Simple beam 4 kN/m 6.0 4 kN/m 16 kN . m A V (kN) B 2m 2m 2m 0 1.5 m Ϫ2.0 2m RA = 6 kN RB = 10 kN Ϫ10.0 16.0 12.0 4.5 M (kN . m) 0 4.0 1.5 m Problem 4.5-19 A beam ABCD with a vertical arm CE is supported as a simple beam at A and D (see figure). A cable passes over a small pulley that is attached to the arm at E. One end of the cable is attached to the beam at point B. The tensile force in the cable is 1800 lb. Draw the shear-force and bending-moment diagrams for beam ABCD. (Note: Disregard the widths of the beam and vertical arm and use centerline dimensions when making calculations.) E Cable A B 6 ft 1800 lb 6 ft Free-body diagram of beam ABCD 1440 Cable B 1800 B C 1440 5760 lb-ft 8 ft C D 1800 A 1080 D 720 800 6 ft RD = 800 lb 6 ft 6 ft RD = 800 lb Note: All forces have units of pounds. 640 V (lb) D Beam with a cable E 1800 lb A 8 ft C 6 ft Solution 4.5-19 1800 lb 4800 0 M 0 (lb-ft) Ϫ800 Ϫ800 Ϫ960 Ϫ4800 800
    • SECTION 4.5 Problem 4.5-20 The beam ABCD shown in the figure has overhangs that extend in both directions for a distance of 4.2 m from the supports at B and C, which are 1.2 m apart. Draw the shear-force and bending-moment diagrams for this overhanging beam. 279 Shear-Force and Bending-Moment Diagrams 10.6 kN/m 5.1 kN/m 5.1 kN/m A D B C 4.2 m 4.2 m 1.2 m Solution 4.5-20 Beam with overhangs 32.97 6.36 10.6 kN/m 5.1 kN/m V 0 (kN) 5.1 kN/m A Ϫ6.36 D B 4.2 m RB = 39.33 kN Ϫ32.97 C M 0 (kN . m) 4.2 m RC = 39.33 kN 1.2 m Ϫ59.24 Ϫ61.15 Ϫ61.15 4.0 k Problem 4.5-21 The simple beam AB shown in the figure supports a concentrated load and a segment of uniform load. Draw the shear-force and bending-moment diagrams for this beam. 2.0 k/ft C A 5 ft 10 ft 20 ft Solution 4.5-21 Simple beam 4.0 k A RA = 8 k V (k) 5 ft B 5 ft 10 ft RB = 16 k 8 0 2.0 k/ft C 4 12 ft 8 ft C Ϫ16 60 64 Mmax = 64 k-ft 40 M (k-ft) 0 12 ft C 8 ft B
    • 280 CHAPTER 4 Shear Forces and Bending Moments Problem 4.5-22 The cantilever beam shown in the figure supports a concentrated load and a segment of uniform load. Draw the shear-force and bending-moment diagrams for this cantilever beam. 3 kN B 0.8 m Solution 4.5-22 1.0 kN/m A 0.8 m 1.6 m Cantilever beam 4.6 3 kN MA = 6.24 kN . m V (kN) 1.0 kN/m A 1.6 0 B 0.8 m 0.8 m 1.6 m M (kN . m) RA = 4.6 kN 0 Ϫ1.28 Ϫ2.56 Ϫ6.24 180 lb/ft Problem 4.5-23 The simple beam ACB shown in the figure is subjected to a triangular load of maximum intensity 180 lb/ft. Draw the shear-force and bending-moment diagrams for this beam. A B C 6.0 ft 7.0 ft Solution 4.5-23 Simple beam 240 180 lb/ft V (lb) 0 x1 = 4.0 ft B Ϫ300 A Ϫ390 C Mmax = 640 6.0 ft RA = 240 lb 1.0 ft RB = 390 lb 360 M (lb-ft) 0 Problem 4.5-24 A beam with simple supports is subjected to a trapezoidally distributed load (see figure). The intensity of the load varies from 1.0 kN/m at support A to 3.0 kN/m at support B. Draw the shear-force and bending-moment diagrams for this beam. 3.0 kN/m 1.0 kN/m A B 2.4 m
    • SECTION 4.5 Solution 4.5-24 Simple beam 2.0 3.0 kN/m V (kN) 1.0 kN/m 0 A x1 = 1.2980 m x B Ϫ2.8 2.4 m RA = 2.0 kN RB = 2.8 kN x2 V ϭ 2.0 Ϫ x Ϫ 2.4 Set V ϭ 0: 281 Shear-Force and Bending-Moment Diagrams Mmax = 1.450 (x ϭ meters; V ϭ kN) M (kN . m) x1 ϭ 1.2980 m 0 Problem 4.5-25 A beam of length L is being designed to support a uniform load of intensity q (see figure). If the supports of the beam are placed at the ends, creating a simple beam, the maximum bending moment in the beam is qL2/8. However, if the supports of the beam are moved symmetrically toward the middle of the beam (as pictured), the maximum bending moment is reduced. Determine the distance a between the supports so that the maximum bending moment in the beam has the smallest possible numerical value. Draw the shear-force and bending-moment diagrams for this condition. Solution 4.5-25 q A B a L Beam with overhangs q A B (L Ϫ a)/ 2 (L Ϫ a)/ 2 a RB = qL/2 RA = qL/2 Solve for a: a ϭ (2 Ϫ ͙2)L ϭ 0.5858L q M1 ϭ M2 ϭ (L Ϫ a) 2 8 2 qL ϭ (3 Ϫ 2͙2) ϭ 0.02145qL2 8 0.2929 qL M2 V 0 M 0 M1 0.2071 qL 0.2071L Ϫ 0.2071 qL M1 The maximum bending moment is smallest when M1ϭ M2 (numerically). q(L Ϫ a) 2 M1 ϭ 8 qL2 qL a M2 ϭ RA ¢ ≤ Ϫ ϭ (2a Ϫ L) 2 8 8 (L Ϫ a) 2 ϭ L(2a Ϫ L) M1 ϭ M2 0.2929L Ϫ 0.2929 qL 0.02145 qL2 M 0 x1 Ϫ 0.02145 qL2 x1 = 0.3536 a = 0.2071 L x1 Ϫ 0.02145 qL2
    • 282 CHAPTER 4 Shear Forces and Bending Moments Problem 4.5-26 The compound beam ABCDE shown in the figure consists of two beams (AD and DE) joined by a hinged connection at D. The hinge can transmit a shear force but not a bending moment. The loads on the beam consist of a 4-kN force at the end of a bracket attached at point B and a 2-kN force at the midpoint of beam DE. Draw the shear-force and bending-moment diagrams for this compound beam. 4 kN 1m 2 kN B C D A E 2m Solution 4.5-26 1m 2m 2m 2m Compound beam 4 kN Hinge 4 kN . m B 2 kN C D A 2m 2m 2m RA = 2.5 kN 1m 1m RC = 2.5 kN 2.5 V (kN) E RE = 1 kN 1.0 0 D Ϫ1.5 Ϫ1.0 5.0 M 0 (kN . m) 1.0 D 1.0 2.67 m Ϫ2.0 Problem 4.5-27 The compound beam ABCDE shown in the figure consists of two beams (AD and DE) joined by a hinged connection at D. The hinge can transmit a shear force but not a bending moment. A force P acts upward at A and a uniform load of intensity q acts downward on beam DE. Draw the shear-force and bending-moment diagrams for this compound beam. Solution 4.5-27 Compound beam P q B C D A E L P L L q B C D A E Hinge L RB = 2P + qL P V L L RC = P + 2qL 2L RE = qL qL 0 D PL –qL −P−qL M qL 2 D 0 −qL2 L L 2L
    • SECTION 4.5 Problem 4.5-28 The shear-force diagram for a simple beam is shown in the figure. Determine the loading on the beam and draw the bendingmoment diagram, assuming that no couples act as loads on the beam. 12 kN V 0 –12 kN 2.0 m Solution 4.5-28 283 Shear-Force and Bending-Moment Diagrams Simple beam (V is given) 1.0 m 6.0 kN/m 1.0 m 12 kN B A 2m 1m 1m 12 RA = 12kN −12 RB = 12kN 12 V (kN) 0 M (kN . m) 0 Problem 4.5-29 The shear-force diagram for a beam is shown in the figure. Assuming that no couples act as loads on the beam, determine the forces acting on the beam and draw the bendingmoment diagram. 652 lb 580 lb 572 lb 500 lb V 0 –128 lb –448 lb 4 ft Solution 4.5-29 Forces on a beam (V is given) 16 ft 4 ft 652 580 572 Force diagram V (lb) 20 lb/ft 0 –128 2448 4 ft 652 lb 700 lb 16 ft 4 ft 1028 lb 500 lb –448 M (lb-ft) 0 14.50 ft –2160 500
    • 284 CHAPTER 4 Shear Forces and Bending Moments Problem 4.5-30 A simple beam AB supports two connected wheel loads P and 2P that are distance d apart (see figure). The wheels may be placed at any distance x from the left-hand support of the beam. P x (a) Determine the distance x that will produce the maximum shear force in the beam, and also determine the maximum shear force Vmax. (b) Determine the distance x that will produce the maximum bending moment in the beam, and also draw the corresponding bendingmoment diagram. (Assume P ϭ 10 kN, d ϭ 2.4 m, and L ϭ 12 m.) Solution 4.5-30 d A B L Moving loads on a beam P 2P x d P ϭ 10 kN d ϭ 2.4 m L ϭ 12 m A B L (a) Maximum shear force By inspection, the maximum shear force occurs at support B when the larger load is placed close to, but not directly over, that support. 2P P x=L−d d A B Reaction at support B: P 2P P xϩ (x ϩ d) ϭ (2d ϩ 3x) L L L Bending moment at D: MD ϭ RB (L Ϫ x Ϫ d) P ϭ (2d ϩ 3x)(L Ϫ x Ϫ d) L P ϭ [Ϫ3x2 ϩ (3L Ϫ 5d)x ϩ 2d(L Ϫ d) ] L RB ϭ dMD P ϭ (Ϫ6x ϩ 3L Ϫ 5d) ϭ 0 dx L L 5d Solve for x: x ϭ ¢ 3 Ϫ ≤ ϭ 4.0 m 6 L Substitute x into Eq (1): Mmax ϭ RA = Pd L d RB = P(3 − L) x ϭ L Ϫ d ϭ 9.6 m d Vmax ϭ RB ϭ P ¢ 3 Ϫ ≤ ϭ 28 kN L ϭ (b) Maximum bending moment By inspection, the maximum bending moment occurs at point D, under the larger load 2P. d A ϫ¢ L 5d ≤ ¢3 Ϫ ≤ ϩ 2d(L Ϫ d)R 6 L PL d 2 ¢ 3 Ϫ ≤ ϭ 78.4 kN ؒ m 12 L 0 4.0 m D Mmax = 78.4 M (kN . m) B Note: L P L 2 5d 2 B Ϫ 3¢ ≤ ¢ 3 Ϫ ≤ ϩ (3L Ϫ 5d) L 6 L 64 2P P x 2P RB 2.4 m 5.6 m P d ¢ 3 ϩ ≤ ϭ 16 kN 2 L P d RB ϭ ¢ 3 Ϫ ≤ ϭ 14 kN 2 L RA ϭ Eq.(1)
    • 5 Stresses in Beams (Basic Topics) Longitudinal Strains in Beams d Problem 5.4-1 Determine the maximum normal strain ⑀max produced in a steel wire of diameter d ϭ 1/16 in. when it is bent around a cylindrical drum of radius R ϭ 24 in. (see figure). Solution 5.4-1 R Steel wire R ϭ 24 in. R dϭ 1 in. 16 From Eq. (5-4): y emax ϭ r d dր2 d ϭ R ϩ dր2 2R ϩ d Substitute numerical values: ϭ Cylinder emax ϭ 1ր16 in. ϭ 1300 ϫ 10Ϫ6 2(24 in.) ϩ 1ր16 in. Problem 5.4-2 A copper wire having diameter d ϭ 3 mm is bent into a circle and held with the ends just touching (see figure). If the maximum permissible strain in the copper is ⑀max ϭ 0.0024, what is the shortest length L of wire that can be used? Solution 5.4-2 Copper wire L = length d ϭ 3 mm d ␳ d = diameter ␧max ϭ 0.0024 L L ϭ 2␲r r ϭ 2␲ From Eq. (5-4): emax ϭ y dր2 ␲d ϭ ϭ r Lր2␲ L L min ϭ ␲ (3 mm) ␲d ϭ ϭ 3.93 m emax 0.0024 285
    • 286 CHAPTER 5 Stresses in Beams (Basic Topics) Problem 5.4-3 A 4.5 in. outside diameter polyethylene pipe designed to carry chemical wastes is placed in a trench and bent around a quartercircular 90° bend (see figure). The bent section of the pipe is 46 ft long. Determine the maximum compressive strain ⑀max in the pipe. 90° Solution 5.4-3 Polyethylene pipe Angle equals 90º or ␲/2 radians, r ϭ ␳ ϭ radius of curvature d r ϭ radius L r L ϭ length of 90º bend L ϭ 46 ftϭ 552 in. d ϭ 4.5 in. 2␲r ␲r ϭ Lϭ 4 2 rϭ L 2L ϭ ␲ ␲ր2 emax ϭ ␦ A B C 0′ M0 ␳ ␪ ␦ A B L M0 Cantilever beam L ␳ y dր2 ϭ r 2Lր␲ ␲d ␲ 4.5 in. ϭ ¢ ≤ ϭ 6400 ϫ 10Ϫ6 4L 4 552 in. Problem 5.4-4 A cantilever beam AB is loaded by a couple M0 at its free end (see figure). The length of the beam is L ϭ 1.5 m and the longitudinal normal strain at the top surface is 0.001. The distance from the top surface of the beam to the neutral surface is 75 mm. Calculate the radius of curvature ␳, the curvature ␬, and the vertical deflection ␦ at the end of the beam. Solution 5.4-4 emax ϭ L ϭ length of beam L ϭ 1.5 m ␧max ϭ 0.001 y y ϭ 75 mm emax ϭ r y 75 mm ∴ rϭ ϭ ϭ 75 m emax 0.001 1 k ϭ ϭ 0.01333 mϪ1 r Assume that the deflection curve is nearly flat. Then the distance BC is the same as the length L of the beam. ∴ sin u ϭ L 1.5 m ϭ 0.02 ϭ r 75 m ␪ ϭ arcsin 0.02 ϭ 0.02 rad ␦ ϭ ␳ (1 Ϫ cos ␪) ϭ (75 m)(1 Ϫ cos (0.02 rad)) ϭ 15.0 mm L NOTE: ϭ 100, which confirms that the deflection ␦ curve is nearly flat.
    • SECTION 5.4 Problem 5.4-5 A thin strip of steel of length L ϭ 20 in. and thickness t ϭ 0.2 in. is bent by couples M0 (see figure). The deflection ␦ at the midpoint of the strip (measured from a line joining its end points) is found to be 0.25 in. Determine the longitudinal normal strain ⑀ at the top surface of the strip. Solution 5.4-5 287 Longitudinal Strains in Beams M0 M0 ␦ t L — 2 L — 2 Thin strip of steel The deflection curve is very flat (note that L/␦ ϭ 80) and therefore ␪ is a very small angle. 0′ sin u ϭ ␪ ␪ ␳ ␳ M0 For small angles, u ϭ sin u ϭ M0 ␦ L — 2 L ϭ 20 in. t ϭ 0.2 in. ␦ ϭ 0.25 in. L — 2 Lր2 r Lր2 (␪ is in radians) r ␦ ϭ ␳ Ϫ ␳ cos ␪ ϭ ␳(1 Ϫ cos ␪) ϭ r ¢ 1 Ϫ cos L ≤ 2r Substitute numerical values (␳ ϭ inches): 0.25 ϭ r ¢ 1 Ϫ cos 10 ≤ r Solve numerically: ␳ ϭ 200.0 in. NORMAL STRAIN y tր2 0.1 in. eϭ ϭ ϭ 500 ϫ 10Ϫ6 ϭ r r 200 in. (Shortening at the top surface) Problem 5.4-6 A bar of rectangular cross section is loaded and supported as shown in the figure. The distance between supports is L ϭ 1.2 m and the height of the bar is h ϭ 100 mm. The deflection ␦ at the midpoint is measured as 3.6 mm. What is the maximum normal strain ⑀ at the top and bottom of the bar? h P ␦ P a L — 2 L — 2 a
    • 288 CHAPTER 5 Solution 5.4-6 Stresses in Beams (Basic Topics) Bar of rectangular cross section h ␦ P P a L — 2 ␳ L — 2 ␳ ␪ ␪ a 0′ L ϭ 1.2 m h ϭ 100 mm ␦ ϭ 3.6 mm Note that the deflection curve is nearly flat (L/␦ ϭ 333) and ␪ is a very small angle. sin u ϭ uϭ Lր2 r Substitute numerical values (␳ ϭ meters): 0.0036 ϭ r ¢ 1 Ϫ cos 0.6 ≤ r Solve numerically: ␳ ϭ 50.00 m NORMAL STRAIN Lր2 (radians) r L ␦ ϭ r (1 Ϫ cos u) ϭ r ¢ 1 Ϫ cos ≤ 2r eϭ y hր2 50 mm ϭ ϭ ϭ 1000 ϫ 10Ϫ6 r r 50,000 mm (Elongation on top; shortening on bottom) Normal Stresses in Beams Problem 5.5-1 A thin strip of hard copper (E ϭ 16,400 ksi) having length L ϭ 80 in. and thickness t ϭ 3/32 in. is bent into a circle and held with the ends just touching (see figure). (a) Calculate the maximum bending stress ␴max in the strip. (b) Does the stress increase or decrease if the thickness of the strip is increased? Solution 5.5-1 E ϭ 16,400 ksi Copper strip bent into a circle L ϭ 80 in. (a) MAXIMUM BENDING STRESS L ϭ 2␲r ϭ 2␲r rϭ From Eq. (5-7): s ϭ smax ϭ 3 t = — in. 32 L 2␲ Ey 2␲Ey ϭ r L 2␲E(tր2) ␲Et ϭ L L t ϭ 3/32 in. Substitute numerical values: smax ϭ ␲ (16,400 ksi)(3ր32 in.) ϭ 60.4 ksi 80 in. (b) CHANGE IN STRESS If the thickness t is increased, the stress ␴max increases.
    • SECTION 5.5 Normal Stresses in Beams Problem 5.5-2 A steel wire (E ϭ 200 GPa) of diameter d ϭ 1.0 mm is bent around a pulley of radius R0 ϭ 400 mm (see figure). (a) What is the maximum stress ␴max in the wire? (b) Does the stress increase or decrease if the radius of the pulley is increased? R0 d Solution 5.5-2 Steel wire bent around a pulley E ϭ 200 GPa d ϭ 1.0 mm R0 ϭ 400 mm From Eq. (5-7): Ey (200 GPa) (0.5 mm) ϭ 250 MPa ϭ r 400.5 mm (a) MAXIMUM STRESS IN THE WIRE smax ϭ d r ϭ R0 ϩ ϭ 400 mm ϩ 0.5 mm ϭ 400.5 mm 2 (b) CHANGE IN STRESS If the radius is increased, the stress ␴max decreases. d y ϭ ϭ 0.5 mm 2 Problem 5.5-3 A thin, high-strength steel rule (E ϭ 30 ϫ 106 psi) having thickness t ϭ 0.15 in. and length L ϭ 40 in. is bent by couples M0 into a circular arc subtending a central angle ␣ ϭ 45° (see figure). (a) What is the maximum bending stress ␴max in the rule? (b) Does the stress increase or decrease if the central angle is increased? Solution 5.5-3 L = length t M0 M0 ␣ Thin steel rule bent into an arc Substitute numerical values: L (30 ϫ 106 psi) (0.15 in.) (0.78540 rad) 2 (40 in.) ϭ 44,200 psi ϭ 44.2 ksi smax ϭ ␳ Eϭ tϭ Lϭ ␣ϭ ␣ (a) MAXIMUM BENDING STRESS L ϭ r␣ smax ϭ rϭ L ␣ ␣ ϭ radians Ey E(tր2) Et␣ ϭ ϭ r Lր␣ 2L 30 ϫ 106 psi 0.15 in. 40 in. 45º ϭ 0.78540 rad (b) CHANGE IN STRESS If the angle ␣ is increased, the stress ␴max increases. 289
    • 290 CHAPTER 5 Stresses in Beams (Basic Topics) Problem 5.5-4 A simply supported wood beam AB with span length L ϭ 3.5 m carries a uniform load of intensity q ϭ 6.4 kN/m (see figure). Calculate the maximum bending stress ␴max due to the load q if the beam has a rectangular cross section with width b ϭ 140 mm and height h ϭ 240 mm. q A L Solution 5.5-4 Simple beam with uniform load L ϭ 3.5 m q ϭ 6.4 kN/m b ϭ 140 mm h ϭ 240 mm 2 Substitute numerical values: smax ϭ 2 Mmax ϭ qL 8 smax ϭ 3(6.4 kNրm)(3.5 m) 2 ϭ 7.29 MPa 4(140 mm)(240 mm) 2 Mmax 3qL2 ϭ S 4bh2 Sϭ bh 6 Problem 5.5-5 Each girder of the lift bridge (see figure) is 180 ft long and simply supported at the ends. The design load for each girder is a uniform load of intensity 1.6 k/ft. The girders are fabricated by welding three steel plates so as to form an I-shaped cross section (see figure) having section modulus S ϭ 3600 in3. What is the maximum bending stress ␴max in a girder due to the uniform load? Solution 5.5-5 Bridge girder q L ϭ 180 ft S ϭ 3600 q ϭ 1.6 k/ft in.3 Mmax ϭ qL2 8 smax ϭ Mmax qL2 ϭ S 8S smax ϭ (1.6 k րft)(180 ft) 2 (12 in.րft) ϭ 21.6 ksi 8(3600 in.3 ) L h B b
    • SECTION 5.5 Problem 5.5-6 A freight-car axle AB is loaded approximately as shown in the figure, with the forces P representing the car loads (transmitted to the axle through the axle boxes) and the forces R representing the rail loads (transmitted to the axle through the wheels). The diameter of the axle is d ϭ 80 mm, the distance between centers of the rails is L, and the distance between the forces P and R is b ϭ 200 mm. Calculate the maximum bending stress ␴max in the axle if P ϭ 47 kN. Solution 5.5-6 P B A d R b R L b Sϭ MAXIMUM BENDING STRESS smax ϭ ␲d 3 32 Mmax 32Pb ϭ S ␲d 3 Substitute numerical values: smax ϭ 32(47 kN)(200 mm) ϭ 187 MPa ␲(80 mm) 3 Problem 5.5-7 A seesaw weighing 3 lb/ft of length is occupied by two children, each weighing 90 lb (see figure). The center of gravity of each child is 8 ft from the fulcrum. The board is 19 ft long, 8 in. wide, and 1.5 in. thick. What is the maximum bending stress in the board? Solution 5.5-7 P Seesaw b ϭ 8 in. h ϭ 1.5 in. q ϭ 3 lb/ft P ϭ 90 lb P q h b d L d L Mmax ϭ Pd ϩ d ϭ 8.0 ft L ϭ 9.5 ft qL2 ϭ 720 lb-ft ϩ 135.4 lb-ft 2 ϭ 855.4 lb-ft ϭ 10,264 lb-in. bh2 ϭ 3.0 in3. 6 M 10,264 lb-in. smax ϭ ϭ ϭ 3420 psi S 3.0 in.3 Sϭ 291 P Freight-car axle Diameter d ϭ 80 mm Distance b ϭ 200 mm Load P ϭ 47 kN M max ϭ Pb Normal Stresses in Beams d
    • 292 CHAPTER 5 Stresses in Beams (Basic Topics) Problem 5.5-8 During construction of a highway bridge, the main girders are cantilevered outward from one pier toward the next (see figure). Each girder has a cantilever length of 46 m and an I-shaped cross section with dimensions as shown in the figure. The load on each girder (during construction) is assumed to be 11.0 kN/m, which includes the weight of the girder. Determine the maximum bending stress in a girder due to this load. 50 mm 2400 mm 25 mm 600 mm Solution 5.5-8 Bridge girder Mmax ϭ smax ϭ q qL2 1 ϭ (11.0 kNրm)(46 m) 2 ϭ 11,638 kN ؒ m 2 2 Mmax c I tf h1 h2 tw Iϭ L ˇ ˇ h c ϭ ϭ 1200 mm 2 bh3 b1h3 1 Ϫ 12 12 1 1 (0.6 m)(2.4 m) 3 Ϫ (0.575 m)(2.3 m) 3 12 12 ϭ 0.6912 m4 Ϫ 0.5830 m4 ϭ 0.1082 m4 ϭ b Lϭ qϭ bϭ tf ϭ h1 ϭ b1 ϭ 46 m 11.0 kN/m 600 mm h ϭ 2400 mm 50 mm tw ϭ 25 mm h Ϫ 2tf ϭ 2300 mm b Ϫ tw ϭ 575 mm Problem 5.5-9 The horizontal beam ABC of an oil-well pump has the cross section shown in the figure. If the vertical pumping force acting at end C is 8.8 k, and if the distance from the line of action of that force to point B is 14 ft, what is the maximum bending stress in the beam due to the pumping force? smax ϭ Mmax c (11,638 kN ؒ m)(1.2 m) ϭ I 0.1082 m4 ϭ 129 MPa ˇ A ˇ B C 0.875 in. 20.0 in. 0.625 in. 8.0 in.
    • SECTION 5.5 Solution 5.5-9 Beam in an oil-well pump Mmax ϭ PL ϭ (8.8 k)(14 ft) ϭ 123,200 lb-ft ϭ 1,478,400 lb-in. tf smax ϭ h1 P h2 tw Iϭ L ϭ b Lϭ Pϭ bϭ tf ϭ h1 ϭ b1 ϭ 1 1 (8.0 in.)(20.0 in.) 3 Ϫ (7.375 in.)(18.25 in.) 3 12 12 smax ϭ ϭ 9250 psi ϭ 9.25 ksi P ϭ 175 kN b ϭ 300 mm h ϭ 250 mm L ϭ 1500 mm a ϭ 500 mm bh2 Sϭ ϭ 3.125 ϫ 10Ϫ3 m3 6 M1 M1 0 M2 qa2 Pa2 ϭ 2 L ϩ 2a 2 q L PL ¢ ϩ a≤ Ϫ 2 2 2 2 P L PL ¢ ϩ a≤ Ϫ L ϩ 2a 2 2 P ϭ (2a Ϫ L) 4 ϭ P a P a L q Substitute numerical values: M1 ϭ 17,500 N ؒ m M2 ϭ Ϫ21,875 N ؒ m Mmax ϭ 21,875 N ؒ m smax ϭ b h MAXIMUM BENDING STRESS BENDING-MOMENT DIAGRAM M2 ϭ Mmax c (1.4784 ϫ 106 lb-in.)(10.0 in.) ϭ I 1,597.7 in.4 Railroad tie (or sleeper) 2P qϭ L ϩ 2a M1 ϭ h c ϭ ϭ 10.0 in. 2 ϭ 5,333.3 in.4 Ϫ 3,735.7 in.4 ϭ 1,597.7 in.4 14 ft 8.8 k 8.0 in. h ϭ 20.0 in. 0.875 in. tw ϭ 0.625 in. h Ϫ 2tf ϭ 18.25 in. b Ϫ tw ϭ 7.375 in. Solution 5.5-10 Mmax c I bh3 b1h3 1 Ϫ 12 12 Problem 5.5-10 A railroad tie (or sleeper) is subjected to two rail loads, each of magnitude P ϭ 175 kN, acting as shown in the figure. The reaction q of the ballast is assumed to be uniformly distributed over the length of the tie, which has cross-sectional dimensions b ϭ 300 mm and h ϭ 250 mm. Calculate the maximum bending stress ␴max in the tie due to the loads P, assuming the distance L ϭ 1500 mm and the overhang length a ϭ 500 mm. DATA 293 Normal Stresses in Beams Mmax 21,875 N ؒ m ϭ ϭ 7.0 MPa 5 3.125 ϫ 10Ϫ3 m3 ˇ ˇ (Tension on top; compression on bottom)
    • 294 CHAPTER 5 Stresses in Beams (Basic Topics) Problem 5.5-11 A fiberglass pipe is lifted by a sling, as shown in the figure. The outer diameter of the pipe is 6.0 in., its thickness is 0.25 in., and its weight density is 0.053 lb/in.3 The length of the pipe is L ϭ 36 ft and the distance between lifting points is s ϭ 11 ft. Determine the maximum bending stress in the pipe due to its own weight. s L Solution 5.5-11 Pipe lifted by a sling q a s t a d1 d2 L L ϭ 36 ft ϭ 432 in. s ϭ 11 ft ϭ 132 in. ␥ ϭ 0.053 lb/in.3 d2 ϭ 6.0 in. t ϭ 0.25 in. d1 ϭ d2 Ϫ 2t ϭ 5.5 in. ␲ A ϭ (d 2 Ϫ d 2 ) ϭ 4.5160 in.2 1 4 2 ␲ (d 4 Ϫ d 4 ) ϭ 18.699 in.4 1 64 2 q ϭ ␥A ϭ (0.053 lb/in.3)(4.5160 in.2) ϭ 0.23935 lb/in. Iϭ a ϭ (L Ϫ s)/2 ϭ 150 in. BENDING-MOMENT DIAGRAM MAXIMUM BENDING STRESS 0 M1 M2 M1 qa2 ϭ Ϫ2,692.7 lb-in. 2 qL L M2 ϭ Ϫ ¢ Ϫ s ≤ ϭ Ϫ2,171.4 lb-in. 4 2 M1 ϭ Ϫ Mmax ϭ 2,692.7 lb-in. smax ϭ Mmax c I smax ϭ (2,692.7 lb-in.)(3.0 in.) ϭ 432 psi 18.699 in.4 (Tension on top) cϭ d2 ϭ 3.0 in. 2
    • SECTION 5.5 Normal Stresses in Beams Problem 5.5-12 A small dam of height h ϭ 2.0 m is constructed of vertical wood beams AB of thickness t ϭ 120 mm, as shown in the figure. Consider the beams to be simply supported at the top and bottom. Determine the maximum bending stress ␴max in the beams, assuming that the weight density of water is ␥ ϭ 9.81 kN/m3. A h t B Solution 5.5-12 Vertical wood beam MAXIMUM BENDING MOMENT t q0 x q ϭ q0 (᎐᎐) L A A B h x RA L RA ϭ B q0 q0 L 6 ˇ ˇ q0 x3 6L q0 Lx q0 x3 ϭ Ϫ 6 6L dM q0L q0 x2 L ϭ Ϫ ϭ0 xϭ dx 6 2L ͙3 Substitute x ϭ Lր ͙3 into the equation for M: M ϭ RAx Ϫ ˇ h ϭ 2.0 m t ϭ 120 mm ␥ ϭ 9.81 kN/m3 (water) Let b ϭ width of beam perpendicular to the plane of the figure Let q0 ϭ maximum intensity of distributed load q0 ϭ gbh bt Sϭ 6 2 ˇ ˇ ˇ Mmax ϭ ˇ ˇ ˇ ˇ ˇ q0 L L q0 q0 L2 L3 ¢ ≤Ϫ ¢ ≤ϭ 6 ͙3 6L 3͙3 9͙3 ˇ ˇ ˇ ˇ ˇ For the vertical wood beam: L ϭ h; Mmax ϭ q0 h2 ˇ ˇ 9͙3 Maximum bending stress smax ϭ 2q0 h2 Mmax 2gh3 ϭ ϭ S 3͙3 bt 2 3͙3 t 2 ˇ ˇ SUBSTITUTE NUMERICAL VALUES: ␴max ϭ 2.10 MPa NOTE: For b ϭ 1.0 m, we obtain q0 ϭ 19,620 N/m, S ϭ 0.0024 m3, Mmax ϭ 5,034.5 N ؒ m, and ␴max ϭ Mmax/S ϭ 2.10 MPa 295
    • 296 CHAPTER 5 Stresses in Beams (Basic Topics) Problem 5.5-13 Determine the maximum tensile stress ␴t (due to pure bending by positive bending moments M) for beams having cross sections as follows (see figure): (a) a semicircle of diameter d, and (b) an isosceles trapezoid with bases b1 ϭ b and b2 ϭ 4b/3, and altitude h. b1 C d b2 (a) Solution 5.5-13 C (b) Maximum tensile stress (a) SEMICIRCLE (b) TRAPEZOID b1 C c C c d b2 From Appendix D, Case 10: (9␲ 2 Ϫ 64)r4 (9␲ 2 Ϫ 64)d 4 ϭ 72␲ 1152␲ 4r 2d cϭ ϭ 3␲ 3␲ IC ϭ st ϭ Mc 768M M ϭ 2 3 ϭ 30.93 3 IC (9␲ Ϫ 64)d d 4b 3 From Appendix D, Case 8: b1 ϭ b IC ϭ ϭ cϭ h3 (b2 ϩ 4b1b2 ϩ b2 ) 1 2 36(b1 ϩ b2 ) 73bh3 756 h(2b1 ϩ b2 ) 10h ϭ 3(b1 ϩ b2 ) 21 st ϭ Problem 5.5-14 Determine the maximum bending stress ␴max (due to pure bending by a moment M) for a beam having a cross section in the form of a circular core (see figure). The circle has diameter d and the angle ␤ ϭ 60°. (Hint: Use the formulas given in Appendix D, Cases 9 and 15.) b2 ϭ Mc 360M ϭ IC 73bh2 C ␤ ␤ d h h
    • SECTION 5.5 Solution 5.5-14 Circular core From Appendix D, Cases 9 and 15: ␤ C y ␤ y MAXIMUM BENDING STRESS ␲r4 r4 ab 2ab3 Ϫ ¢␣ Ϫ 2 ϩ 4 ≤ 4 2 r r d 2 ␣ ϭ radians ␣ϭ smax ϭ a ϭ r sin ␤ b ϭ r cos ␤ Mc Iy smax ϭ ␲ Ϫb 2 64M sin b d (4b Ϫ sin 4b) ϭ ϭ smax ϭ 576M (8␲͙3 ϩ 9)d 3 ϭ 10.96 M d3 ␲d 4 d 4 ␲ 1 Ϫ ¢ Ϫ b ϩ sin 4b ≤ 64 32 2 4 d4 (4b Ϫ sin 4b) 128 Solution 5.5-15 x P P A B L Substitute x into the equation for M: d A d Wheel loads on a beam P P B L Lϭ dϭ Pϭ Sϭ 24 ft ϭ 288 in. 5 ft ϭ 60 in. 3k 16.2 in.3 MAXIMUM BENDING MOMENT P P P (L Ϫ x) ϩ (L Ϫ x Ϫ d) ϭ (2L Ϫ d Ϫ 2x) L L L P M ϭ RAx ϭ (2Lx Ϫ dx Ϫ 2x2 ) L dM P L d ϭ (2L Ϫ d Ϫ 4x) ϭ 0 x ϭ Ϫ dx L 2 4 RA ϭ 3 ␲d 4 d 4 ␲ 1 Ϫ ¢ Ϫ b Ϫ ¢ sin 2b ≤ (Ϫcos 2b) ≤ 64 32 2 2 Problem 5.5-15 A simple beam AB of span length L ϭ 24 ft is subjected to two wheel loads acting at distance d ϭ 5 ft apart (see figure). Each wheel transmits a load P ϭ 3.0 k, and the carriage may occupy any position on the beam. Determine the maximum bending stress ␴max due to the wheel loads if the beam is an I-beam having section modulus S ϭ 16.2 in.3 RA d sin b 2 4 ␲d 4 d 4 ␲ ϭ Ϫ ¢ Ϫ b Ϫ (sin b cos b)(1 Ϫ 2 cos2b) ≤ 64 32 2 ϭ c ϭ r sin b ϭ For ␤ ϭ 60º ϭ ␲/3 rad: ␲d d ␲ Ϫ ¢ Ϫ b Ϫ sin b cos b ϩ 2 sin b cos3 b ≤ 64 32 2 4 Iy ϭ Iy ϭ rϭ d ␤ ϭ radians Normal Stresses in Beams Mmax ϭ P d 2 ¢L Ϫ ≤ 2L 2 MAXIMUM BENDING STRESS smax ϭ Mmax P d 2 ϭ ¢L Ϫ ≤ S 2LS 2 Substitute numerical values: smax ϭ 3k (288 in. Ϫ 30 in.) 2 2(288 in.)(16.2 in.3 ) ϭ 21.4 ksi C 297
    • 298 CHAPTER 5 Stresses in Beams (Basic Topics) Problem 5.5-16 Determine the maximum tensile stress ␴t and maximum compressive stress ␴c due to the load P acting on the simple beam AB (see figure). Data are as follows: P ϭ 5.4 kN, L ϭ 3.0 m, d ϭ 1.2 m, b ϭ 75 mm, t ϭ 25 mm, h ϭ 100 mm, and h1 ϭ 75 mm. t P d A L Solution 5.5-16 h1 h B b Simple beam of T-section t P A d c1 h1 B h C c2 L RA b RB P ϭ 5.4 kN L ϭ 3.0 m MAXIMUM BENDING MOMENT b ϭ 75 mm t ϭ 25 mm Mmax ϭ RA(L Ϫ d) ϭ RB(d) ϭ 3888 N ؒ m d ϭ 1.2 m h ϭ 100 mm h1 ϭ 75 mm PROPERTIES OF THE CROSS SECTION A ϭ 3750 mm2 c1 ϭ 62.5 mm c2 ϭ 37.5 mm IC ϭ 3.3203 ϫ 106 mm4 REACTIONS OF THE BEAM RA ϭ 2.16 kN RB ϭ 3.24 kN MAXIMUM TENSILE STRESS st ϭ Mmax c2 (3888 N ؒ m)(0.0375 m) ϭ IC 3.3203 ϫ 106 mm4 ˇ ˇ ϭ 43.9 MPa MAXIMUM COMPRESSIVE STRESS sc ϭ Mmax c1 (3888 N ؒ m)(0.0625 m) ϭ IC 3.3203 ϫ 106 mm4 ˇ ˇ ϭ 73.2 MPa 200 lb Problem 5.5-17 A cantilever beam AB, loaded by a uniform load and a concentrated load (see figure), is constructed of a channel section. Find the maximum tensile stress ␴t and maximum compressive stress ␴c if the cross section has the dimensions indicated and the moment of inertia about the z axis (the neutral axis) is I ϭ 2.81 in.4 (Note: The uniform load represents the weight of the beam.) 20 lb/ft B A 5.0 ft 3.0 ft y z C 0.606 in. 2.133 in.
    • SECTION 5.5 Solution 5.5-17 299 Normal Stresses in Beams Cantilever beam (channel section) I ϭ 2.81 in.4 200 lb ϭ 1000 lb-ft ϩ 640 lb-ft ϭ 1640 lb-ft ϭ 19,680 lb-in. B 5.0 ft c2 ϭ 2.133 in. Mmax ϭ (200 lb)(5.0 ft) ϩ (20 lbրft)(8.0 ft) ¢ 20 lb/ft A c1 ϭ 0.606 in. 8.0 ft ≤ 2 3.0 ft 0.606 in. MAXIMUM TENSILE STRESS Mc1 (19,680 lb-in.)(0.606 in.) st ϭ ϭ I 2.81 in.4 ϭ 4,240 psi 2.133 in. MAXIMUM COMPRESSIVE STRESS 8.0 ft y z C sc ϭ Mc2 (19,680 lb-in.)(2.133 in.) ϭ I 2.81 in.4 ϭ 14,940 psi Problem 5.5-18 A cantilever beam AB of triangular cross section has length L ϭ 0.8 m, width b ϭ 80 mm, and height h ϭ 120 mm (see figure). The beam is made of brass weighing 85 kN/m3. (a) Determine the maximum tensile stress ␴t and maximum compressive stress ␴c due to the beam’s own weight. (b) If the width b is doubled, what happens to the stresses? (c) If the height h is doubled, what happens to the stresses? Solution 5.5-18 L ϭ 0.8 m b ϭ 80 mm ␥ ϭ 85 kN/m3 h ϭ 120 mm bh3 36 Mmax ϭ c1 ϭ h 3 qL2 gbhL2 ϭ 2 4 c2 ϭ 2h 3 2 Tensile stress: st ϭ h L y C h/3 2h h 3 Compressive stress: ␴c ϭ 2␴t Substitute numerical values: ␴t ϭ 1.36 MPa ␴c ϭ 2.72 MPa (a) MAXIMUM STRESSES Iz ϭ IC ϭ B b z L bh ≤ 2 b Triangular beam q q ϭ gA ϭ g ¢ A Mc1 3gL ϭ Iz h (b) WIDTH b IS DOUBLED No change in stresses. (c) HEIGHT h IS DOUBLED Stresses are reduced by half.
    • 300 CHAPTER 5 Stresses in Beams (Basic Topics) Problem 5.5-19 A beam ABC with an overhang from B to C supports a uniform load of 160 lb/ft throughout its length (see figure). The beam is a channel section with dimensions as shown in the figure. The moment of inertia about the z axis (the neutral axis) equals 5.14 in.4 Calculate the maximum tensile stress ␴t and maximum compressive stress ␴c due to the uniform load. 160 lb/ft A C B 10 ft 5 ft y 0.674 in. z C Solution 5.5-19 2.496 in. Beam with an overhang q ϭ 160 lb/ft y A C B z C L ϭ 10 ft 0.674 in. 2.496 in. b ϭ 5 ft AT CROSS SECTION OF MAXIMUM POSITIVE BENDING MOMENT M1 st ϭ 0 3.75 ft Iz ϭ c1 ϭ RA ϭ M1 ϭ M2 ϭ M2 5.14 in.4 0.674 in. c2 ϭ 2.496 in. 600 lb RB ϭ 1800 lb 1125 lb-ft ϭ 13,500 lb-in. 2000 lb-ft ϭ 24,000 lb-in. M1c2 (13,500 lb-in.)(2.496 in.) ϭ ϭ 6,560 psi Iz 5.14 in.4 sc ϭ M1c1 (13,500 lb-in.)(0.674 in.) ϭ ϭ 1,770 psi Iz 5.14 in.4 AT CROSS SECTION OF MAXIMUM NEGATIVE BENDING MOMENT st ϭ M2c1 (24,000 lb-in.)(0.674 in.) ϭ ϭ 3,150 psi Iz 5.14 in.4 sc ϭ M2c2 (24,000 lb-in.)(2.496 in.) ϭ ϭ 11,650 psi Iz 5.14 in.4 MAXIMUM STRESSES ␴t ϭ 6,560 psi ␴c ϭ 11,650 psi Problem 5.5-20 A frame ABC travels horizontally with an acceleration a0 (see figure). Obtain a formula for the maximum stress ␴max in the vertical arm AB, which has length L, thickness t, and mass density ␳. A t a0 = acceleration L B C
    • SECTION 5.5 Solution 5.5-20 301 Normal Stresses in Beamss Accelerating frame L ϭ length of vertical arm t ϭ thickness of vertical arm ␳ ϭ mass density a0 ϭ acceleration Let b ϭ width of arm perpendicular to the plane of the figure Let q ϭ inertia force per unit distance along vertical arm VERTICAL ARM TYPICAL UNITS FOR USE IN THE PRECEDING EQUATION SI UNITS: ␳ ϭ kg/m3 ϭ N ؒ s2/m4 L ϭ meters (m) a0 ϭ m/s2 t ϭ meters (m) t ␴max ϭ N/m2 (pascals) q q ϭ rbta0 L Mmax ϭ qL2 rbta0 L2 ϭ 2 2 Mmax 3rL2a0 smax ϭ ϭ S t bt 2 Sϭ 6 Problem 5.5-21 A beam of T-section is supported and loaded as shown in the figure. The cross section has width b ϭ2 1/2 in., height h ϭ 3 in., and thickness t ϭ 1/2 in. Determine the maximum tensile and compressive stresses in the beam. USCS UNITS: ␳ ϭ slug/ft3 ϭ lb-s2/ft4 L ϭ ft a0 ϭ ft/s2 ␴max ϭ lb/ft2 (Divide by 144 to obtain psi) 1 t = — in. 2 P = 625 lb q = 80 lb/ft L1 = 4 ft 1 t = — in. 2 L2 = 8 ft Solution 5.5-21 P A L2 M1 ϭ RA L1 ϭ 9,000 lb – in. C RB RA t c2 b h PROPERTIES OF THE CROSS SECTION b ϭ 2.5 in. h ϭ 3.0 in. t ϭ 0.5 in. A ϭ bt ϩ (h Ϫ t)t ϭ 2.50 in.2 c1 ϭ 2.0 in. c2 ϭ 1.0 in. IC ϭ REACTIONS RA ϭ 187.5 lb (upward) RB ϭ 837.5 lb (upward) ᎐qL2 3 M2 ϭ ᎐᎐᎐᎐ ϭ ᎐12,000 lb – in. 2 L3 L1 ϭ 4 ft ϭ 48 in. L2 ϭ 8 ft ϭ 96 in. L3 ϭ 5 ft ϭ 60 in. P ϭ 625 lb q ϭ 80 lb/ft ϭ 6.6667 lb/in. C 1 b = 2 — in. 2 L3 = 5 ft BENDING-MOMENT DIAGRAM q B t h= 3 in. Beam of T-section L1 c1 t ϭ ft 25 4 in. ϭ 2.0833 in.4 12 AT CROSS SECTION OF MAXIMUM POSITIVE BENDING MOMENT st ϭ M1c2 ϭ 4,320 psi IC sc ϭ M1c1 ϭ 8,640 psi IC AT CROSS SECTION OF MAXIMUM NEGATIVE BENDING MOMENT st ϭ M2c1 ϭ 11,520 psi IC sc ϭ M2c2 ϭ 5,760 psi IC MAXIMUM STRESSES ␴t ϭ 11,520 psi ␴c ϭ 8,640 psi
    • 302 CHAPTER 5 Stresses in Beams (Basic Topics) Problem 5.5-22 A cantilever beam AB with a rectangular cross section has a longitudinal hole drilled throughout its length (see figure). The beam supports a load P ϭ 600 N. The cross section is 25 mm wide and 50 mm high, and the hole has a diameter of 10 mm. Find the bending stresses at the top of the beam, at the top of the hole, and at the bottom of the beam. 10 mm 50 mm A B 12.5 mm 37.5 mm P = 600 N L = 0.4 m 25 mm Solution 5.5-22 Rectangular beam with a hole y MOMENT OF INERTIA ABOUT THE NEUTRAL AXIS (THE z AXIS) c1 z C c2 – y B B MAXIMUM BENDING MOMENT M ϭ PL ϭ (600 N)(0.4 m) ϭ 240 N ؒ m PROPERTIES OF THE CROSS SECTION A1 ϭ area of rectangle ϭ (25 mm)(50 mm) ϭ 1250 mm2 A2 ϭ area of hole ␲ ϭ (10 mm) 2 ϭ 78.54 mm2 4 A ϭ area of cross section ϭ A1 Ϫ A2 ϭ 1171.5 mm2 Using line B-B as reference axis: ∑Ai yi ϭ A1(25 mm) Ϫ A2(37.5 mm) ϭ 28,305 mm3 3 a Ai yi 28,305 mm ϭ ϭ 24.162 mm A 1171.5 mm2 Distances to the centroid C: yϭ c2 ϭ y ϭ 24.162 mm c1 ϭ 50 mm Ϫ c2 ϭ 25.838 mm All dimensions in millimeters. Rectangle: Iz ϭ Ic ϩ Ad 2 1 ϭ (25)(50) 3 ϩ (25)(50)(25 Ϫ 24.162) 2 12 ϭ 260,420 ϩ 878 ϭ 261,300 mm4 Hole: ␲ Iz ϭ Ic ϩ Ad 2 ϭ (10) 4 ϩ (78.54)(37.5 Ϫ 24.162) 2 64 ϭ 490.87 ϩ 13,972 ϭ 14,460 mm4 Cross-section: I ϭ 261,300 Ϫ 14,460 ϭ 246,800 mm4 STRESS AT THE TOP OF THE BEAM Mc1 (240 N ؒ m)(25.838 mm) s1 ϭ ϭ I 246,800 mm4 ˇ ˇ ϭ 25.1 MPa (tension) STRESS AT THE TOP OF THE HOLE My s2 ϭ y ϭ c1 Ϫ 7.5 mm ϭ 18.338 mm I s2 ϭ (240 N ؒ m)(18.338 mm) ϭ 17.8 MPa 246,800 mm4 ˇ ˇ (tension) STRESS AT THE BOTTOM OF THE BEAM Mc2 (240 N ؒ m)(24.162 mm) s3 ϭ Ϫ ϭϪ I 246,800 mm4 ˇ ϭ Ϫ23.5 MPa (compression) ˇ
    • SECTION 5.5 Problem 5.5-23 A small dam of height h ϭ 6 ft is constructed of vertical wood beams AB, as shown in the figure. The wood beams, which have thickness t ϭ 2.5 in., are simply supported by horizontal steel beams at A and B. Construct a graph showing the maximum bending stress ␴max in the wood beams versus the depth d of the water above the lower support at B. Plot the stress ␴max (psi) as the ordinate and the depth d (ft) as the abscissa. (Note: The weight density ␥ of water equals 62.4 lb/ft3.) 303 Normal Stresses in Beams Steel beam A Wood beam t t Wood beam Steel beam h d B Side view Solution 5.5-23 Vertical wood beam in a dam A t h d B q0 MAXIMUM BENDING STRESS 1 Section modulus: S ϭ bt 2 6 h ϭ 6 ft t ϭ 2.5 in. ␥ ϭ 62.4 lb/ft3 Let b ϭ width of beam (perpendicular to the figure) Let q0 ϭ intensity of load at depth d q0 ϭ ␥ bd smax ϭ q0 C B RA RB d smax ϭ d B 3L V RA 0 C smax ϭ (62.4)d 3 d d d ¢1 Ϫ ϩ ≤ 6 9B 18 (2.5) 2 ϭ 0.1849d 3 (54 Ϫ 9d ϩ d͙2d) x0 ϭ d MC ϭ RA (L Ϫ d) ϭ gd 3 d 2d d ϩ ≤ 2 ¢1 Ϫ L 3LB 3L t SUBSTITUTE NUMERICAL VALUES: d ϭ depth of water (ft) (Max. d ϭ h ϭ 6 ft) L ϭ h ϭ 6 ft ␥ ϭ 62.4 lb/ft3 t ϭ 2.5 in. ␴max ϭ psi L ϭ h ϭ 6 ft q0 d 2 RA ϭ 6L q0 d d RB ϭ ¢3 Ϫ ≤ 6 L L Mmax 6 q0 d 2 d 2d d ϭ 2B ¢1 Ϫ ϩ ≤R S 6 L 3LB 3L bt q0 ϭ ␥ bd ANALYSIS OF BEAM A q0 d 6 2 ¢1 Ϫ 0 1 2 3 4 5 6 x0 ᎐RB 1000 ␴max (psi) Mmax ϭ q0 d 2 d 2d d ¢1 Ϫ ϩ ≤ 6 L 3LB 3L 0 9 59 171 347 573 830 830 psi MC 0 ␴max (psi) d (ft) d ≤ L Mmax M Top view 750 500 250 0 1 2 3 d (ft) 4 5 6
    • 304 CHAPTER 5 Stresses in Beams (Basic Topics) Design of Beams P Problem 5.6-1 The cross section of a narrow-gage railway bridge is shown in part (a) of the figure. The bridge is constructed with longitudinal steel girders that support the wood cross ties. The girders are restrained against lateral buckling by diagonal bracing, as indicated by the dashed lines. The spacing of the girders is s1 ϭ 50 in. and the spacing of the rails is s2 ϭ 30 in. The load transmitted by each rail to a single tie is P ϭ 1500 lb. The cross section of a tie, shown in part (b) of the figure, has width b ϭ 5.0 in. and depth d. Determine the minimum value of d based upon an allowable bending stress of 1125 psi in the wood tie. (Disregard the weight of the tie itself.) Solution 5.6-1 Railway cross tie P s2 P s2 Steel rail Wood tie d b Steel girder (b) s1 (a) P Steel rail Wood tie d b s1 s1 ϭ 50 in. b ϭ 5.0 in. s2 ϭ 30 in. d ϭ depth of tie P ϭ 1500 lb ␴allow ϭ 1125 psi P(s1 Ϫ s2 ) Mmax ϭ ϭ 15,000 lb-in. 2 bd 2 1 5d 2 Sϭ ϭ (50 in.)(d 2 ) ϭ d ϭ inches 6 6 6 Mmax ϭ sallow S 15,000 ϭ (1125) ¢ Solving, d 2 ϭ 16.0 in. dmin ϭ 4.0 in. Note: Symbolic solution: d 2 ϭ Problem 5.6-2 A fiberglass bracket ABCD of solid circular cross section has the shape and dimensions shown in the figure. A vertical load P ϭ 36 N acts at the free end D. Determine the minimum permissible diameter dmin of the bracket if the allowable bending stress in the material is 30 MPa and b ϭ 35 mm. (Disregard the weight of the bracket itself.) 3P(s1 Ϫ s2 ) bsallow 5b A B 2b D P Solution 5.6-2 Fiberglass bracket DATA P ϭ 36 N ␴allow ϭ 30 MPa CROSS SECTION d = diameter Iϭ b ϭ 35 mm ␲d 4 64 MAXIMUM BENDING MOMENT MINIMUM DIAMETER (96)(36 N)(35 mm) 96Pb d3 ϭ ϭ ␲sallow ␲(30 MPa) ϭ 1,283.4 mm3 Mmax ϭ P(3b) MAXIMUM BENDING STRESS Mmax c d 3Pbd 96 Pb smax ϭ cϭ sallow ϭ ϭ I 2 2I ␲d 3 5d 2 ≤ 6 dmin ϭ 10.9 mm C 2b
    • SECTION 5.6 Design of Beams P ϭ 2500 lb Problem 5.6-3 A cantilever beam of length L ϭ 6 ft supports a uniform load of intensity q ϭ 200 lb/ft and a concentrated load P ϭ 2500 lb (see figure). Calculate the required section modulus S if ␴allow ϭ 15,000 psi. Then select a suitable wide-flange beam (W shape) from Table E-1, Appendix E, and recalculate S taking into account the weight of the beam. Select a new beam size if necessary. q ϭ 200 lb/ft L = 6 ft Solution 5.6-3 Cantilever beam P ϭ 2500 lb q ϭ 200 lb/ft ␴allow ϭ 15,000 psi L ϭ 6 ft REQUIRED SECTION MODULUS qL2 Mmax ϭ PL ϩ ϭ 15,000 lb-ft ϩ 3,600 lb-ft 2 ϭ18,600 lb-ft ϭ 223,200 lb-in. Mmax 223,200 lb-in. Sϭ ϭ ϭ 14.88 in.3 sallow 15,000 psi TRIAL SECTION W 8 ϫ 21 S ϭ 18.2 in.3 M0 ϭ q0 L ϭ 378 lb-ft ϭ 4536 lb-in. 2 Mmax ϭ 223,200 ϩ 4,536 ϭ 227,700 lb-in. Required S ϭ Mmax 227,700 lb-in. ϭ ϭ 15.2 in.3 sallow 15,000 psi 15.2 in.3 Ͻ 18.2 in.3 Use І Beam is satisfactory. W 8 ϫ 21 P = 4000 lb Problem 5.6-4 A simple beam of length L ϭ 15 ft carries a uniform load of intensity q ϭ 400 lb/ft and a concentrated load P ϭ 4000 lb (see figure). Assuming ␴allow ϭ 16,000 psi, calculate the required section modulus S. Then select an 8-inch wide-flange beam (W shape) from Table E-1, Appendix E, and recalculate S taking into account the weight of the beam. Select a new 8-inch beam if necessary. Solution 5.6-4 q0 ϭ 21 lb/ft 2 7.5 ft q = 400 lb/ft L = 15 ft Simple beam P ϭ 4000 lb q ϭ 400 lb/ft L ϭ 15 ft ␴allow ϭ 16,000 psi use an 8-inch W shape TRIAL SECTION W 8 ϫ 28 REQUIRED SECTION MODULUS PL qL2 Mmax ϭ ϩ ϭ 15,000 lb-ft ϩ 11,250 lb-ft 4 8 ϭ 26,250 lb-ft ϭ 315,000 lb-in. Mmax 315,000 lb-in. Sϭ ϭ ϭ 19.69 in.3 sallow 16,000 psi q0 ϭ 28 lb/ft S ϭ 24.3 in.3 2 M0 ϭ q0 L ϭ 787.5 lb-ft ϭ 9450 lb-in. 8 Mmax ϭ 315,000 ϩ 9,450 ϭ 324,450 lb-in. Required S ϭ Mmax 324,450 lb-in. ϭ ϭ 20.3 in.3 sallow 16,000 psi 20.3 in.3 Ͻ 24.3 in.3 Use W 8 ϫ 28 І Beam is satisfactory. 305
    • 306 CHAPTER 5 Stresses in Beams (Basic Topics) Problem 5.6-5 A simple beam AB is loaded as shown in the figure on the next page. Calculate the required section modulus S if ␴allow ϭ 15,000 psi, L ϭ 24 ft, P ϭ 2000 lb, and q ϭ 400 lb/ft. Then select a suitable I-beam (S shape) from Table E-2, Appendix E, and recalculate S taking into account the weight of the beam. Select a new beam size if necessary. P q B A L — 4 L — 4 Solution 5.6-5 q L — 4 L — 4 Simple beam P ϭ 2000 lb q ϭ 400 lb/ft ␴allow ϭ 15,000 psi L ϭ 24 ft TRIAL SECTION S 10 ϫ 25.4 S ϭ 24.7 in.3 q0 ϭ 25.4 lb/ft 2 REQUIRED SECTION MODULUS M0 ϭ PL qL2 ϩ ϭ 12,000 lb-ft ϩ 7,200 lb-ft 4 32 ϭ 19,200 lb-ft ϭ 230,400 lb-in. Mmax ϭ Sϭ Mmax 230,400 lb-in. ϭ ϭ 15.36 in.3 sallow 15,000 psi q0 L ϭ 1829 lb-ft ϭ 21,950 lb-in. 8 Mmax ϭ 230,400 ϩ 21,950 ϭ 252,300 lb-in. Required S ϭ Mmax 252,300 lb-in. ϭ ϭ 16.8 in.3 sallow 15,000 psi . 16.8 in.3 Ͻ 24.7 in.3 І Beam is satisfactory. Use S 10 ϫ 25.4 Problem 5.6-6 A pontoon bridge (see figure) is constructed of two longitudinal wood beams, known as balks, that span between adjacent pontoons and support the transverse floor beams, which are called chesses. For purposes of design, assume that a uniform floor load of 8.0 kPa acts over the chesses. (This load includes an allowance for the weights of the chesses and balks.) Also, assume that the chesses are 2.0 m long and that the balks are simply supported with a span of 3.0 m. The allowable bending stress in the wood is 16 MPa. If the balks have a square cross section, what is their minimum required width bmin? Solution 5.6-6 Chess Pontoon Balk Pontoon bridge Chess Lc ϭ 2.0 m Pontoon FLOOR LOAD: w ϭ 8.0 kPa ALLOWABLE STRESS: ␴allow ϭ 16 MPa Lc ϭ length of chesses Balk Lb ϭ 3.0 m ϭ 2.0 m Lb ϭ length of balks ϭ 3.0 m
    • SECTION 5.6 LOADING DIAGRAM FOR ONE BALK Section modulus S ϭ q ϭ 8.0 kN/m b Mmax ϭ b Lb ϭ 3.0 m Sϭ ∴ W ϭ total load ϭ wLb Lc Design of Beams 307 b3 6 qL2 (8.0 kNրm)(3.0 m) 2 b ϭ ϭ 9,000 N ؒ m 8 8 ˇ ˇ Mmax 9,000 N ؒ m ϭ ϭ 562.5 ϫ 10Ϫ6 m3 sallow 16 MPa b3 ϭ 562.5 ϫ 10Ϫ6 m3 6 and b3 ϭ 3375 ϫ 10Ϫ6 m3 Solving, bmin ϭ 0.150 m ϭ 150 mm wLc W qϭ ϭ 2Lb 2 ϭ (8.0 kPa)(2.0 m) 2 ϭ 8.0 kN/m Problem 5.6-7 A floor system in a small building consists of wood planks supported by 2 in. (nominal width) joists spaced at distance s, measured from center to center (see figure). The span length L of each joist is 10.5 ft, the spacing s of the joists is 16 in., and the allowable bending stress in the wood is 1350 psi. The uniform floor load is 120 lb/ft2, which includes an allowance for the weight of the floor system itself. Calculate the required section modulus S for the joists, and then select a suitable joist size (surfaced lumber) from Appendix F, assuming that each joist may be represented as a simple beam carrying a uniform load. Solution 5.6-7 Planks s s L Joists s Floor joists q Mmax ϭ qL2 1 ϭ (13.333 lbրin.)(126 in.) 2 ϭ 26,460 lb-in. 8 8 Required S ϭ L ϭ 10.5 ft ␴allow ϭ 1350 psi L ϭ 10.5 ft ϭ 126 in. w ϭ floor load ϭ 120 lb/ft2 ϭ 0.8333 lb/in.2 s ϭ spacing of joists ϭ 16 in. q ϭ ws ϭ 13.333 lb/in. Mmax 26,460 lbրin. ϭ ϭ 19.6 in.3 sallow 1350 psi From Appendix F: Select 2 ϫ 10 in. joists
    • 308 CHAPTER 5 Stresses in Beams (Basic Topics) Problem 5.6-8 The wood joists supporting a plank floor (see figure) are 40 mm ϫ 180 mm in cross section (actual dimensions) and have a span length L ϭ 4.0 m. The floor load is 3.6 kPa, which includes the weight of the joists and the floor. Calculate the maximum permissible spacing s of the joists if the allowable bending stress is 15 MPa. (Assume that each joist may be represented as a simple beam carrying a uniform load.) Solution 5.6-8 Spacing of floor joists Planks h = 180 mm s s L Joists b = 40 mm s L ϭ 4.0 m w ϭ floor load ϭ 3.6 kPa s ϭ spacing of joists ␴allow ϭ 15 MPa q L ϭ 4.0 m q ϭ ws Sϭ bh2 6 Mmax ϭ Sϭ SPACING OF JOISTS qL2 wsL2 ϭ 8 8 Mmax wsL2 bh2 ϭ ϭ sallow 8sallow 6 smax ϭ 4 bh2sallow 3wL2 Substitute numerical values: 4(40 mm)(180 mm) 2 (15 MPa) smax ϭ 3(3.6 kPa) (4.0 m) 2 ϭ 0.450 m ϭ 450 mm
    • SECTION 5.6 Problem 5.6-9 A beam ABC with an overhang from B to C is constructed of a C 10 ϫ 30 channel section (see figure). The beam supports its own weight (30 lb/ft) plus a uniform load of intensity q acting on the overhang. The allowable stresses in tension and compression are 18 ksi and 12 ksi, respectively. Determine the allowable uniform load qallow if the distance L equals 3.0 ft. Design of Beams 309 q A C B L L 3.033 in. C 2.384 in. 0.649 in. 10.0 in. Solution 5.6-9 Beam with an overhang DATA C 10 ϫ 30 channel section c1 ϭ 2.384 in. c2 ϭ 0.649 in. ALLOWABLE BENDING MOMENT BASED UPON TENSION st I (18 ksi)(3.94 in.4 ) ϭ ϭ 29,750 lb-in. c1 2.384 in. Mt ϭ I ϭ 3.94 in.4 (from Table E-3) ALLOWABLE BENDING MOMENT q0 ϭ weight of beam ABC ϭ 30 lb/ft ϭ 2.5 lb/in. BASED UPON COMPRESSION q ϭ load on overhang Mc ϭ L ϭ length of overhang ϭ 3.0 ft = 36 in. ALLOWABLE BENDING MOMENT ALLOWABLE STRESSES ␴t ϭ 18 ksi ␴c ϭ 12 ksi MAXIMUM BENDING MOMENT (q ϩ q0 )L2 2 Tension on top; compression on bottom. Mmax occurs at support B. Mmax ϭ sc I (12 ksi)(3.94 in.4 ) ϭ ϭ 72,850 lb-in. c2 0.649 in. Tension governs. Mallow ϭ 29,750 lb-in. ALLOWABLE UNIFORM LOAD q (q ϩ q0 )L2 2Mallow qallow ϩ q0 ϭ 2 L2 2Mallow 2(29,750 lb-in.) qallow ϭ Ϫ q0 ϭ Ϫ 2.5 lbրin. L2 (36 in.) 2 Mmax ϭ ϭ 45.91 Ϫ 2.5 ϭ 43.41 lb/in. qallow ϭ (43.41)(12) ϭ 521 lb/ft Problem 5.6-10 A so-called “trapeze bar” in a hospital room provides a means for patients to exercise while in bed (see figure). The bar is 2.1 m long and has a cross section in the shape of a regular octagon. The design load is 1.2 kN applied at the midpoint of the bar, and the allowable bending stress is 200 MPa. Determine the minimum height h of the bar. (Assume that the ends of the bar are simply supported and that the weight of the bar is negligible.) C h
    • 310 CHAPTER 5 Stresses in Beams (Basic Topics) Solution 5.6-10 Trapeze bar (regular octagon) P L 2 C h b L P ϭ 1.2 kN L ϭ 2.1 m ␴allow ϭ 200 MPa Determine minimum height h. І IC ϭ 1.85948(0.41421h)4 ϭ 0.054738h4 SECTION MODULUS MAXIMUM BENDING MOMENT Mmax ϭ b ϭ 0.41421h PL (1.2 kN)(2.1 m) ϭ ϭ 630 N ؒ m 4 4 ˇ Sϭ IC 0.054738h4 ϭ ϭ 0.109476h3 hր2 hր2 ˇ PROPERTIES OF THE CROSS SECTION Use Appendix D, Case 25, with n ϭ 8 MINIMUM HEIGHT h M s 630N ؒ m 0.109476h3 ϭ ϭ 3.15 ϫ 10Ϫ6 m3 200 MPa h3 ϭ 28.7735 ϫ 10Ϫ6 m3 h ϭ 0.030643 m sϭ M S Sϭ ˇ b ␤ C ␤ 2 h 2 b ϭ length of one side 360Њ 360Њ ϭ ϭ 45Њ n 8 b b tan ϭ (from triangle) b 2 h 2 b h h cot ϭ 2 b 2 bϭ C For ␤ ϭ 45º: b 45Њ ϭ tan ϭ 0.41421 h 2 h 45Њ ϭ cot ϭ 2.41421 b 2 MOMENT OF INERTIA ˇ І hmin ϭ 30.6 mm ALTERNATIVE SOLUTION (n ϭ 8) Mϭ PL 4 b ϭ 45Њ b ϭ ( ͙2 Ϫ 1)h IC ϭ ¢ Sϭ¢ tan b ϭ ͙2 Ϫ 1 2 cot b ϭ ͙2 ϩ 1 2 h ϭ ( ͙2 ϩ 1)b 11 ϩ 8͙2 4 4͙2 Ϫ 5 4 ≤b ϭ¢ ≤h 12 12 4͙2 Ϫ 5 3 ≤h 6 h3 ϭ 3PL 2(4͙2 Ϫ 5)sallow Substitute numerical values: h3 ϭ 28.7735 ϫ 10Ϫ6 m3 hmin ϭ 30.643 mm b b nb4 ¢ cot ≤ ¢ 3 cot2 ϩ 1 ≤ 192 2 2 8b4 IC ϭ (2.41421) [3(2.41421) 2 ϩ 1] ϭ 1.85948b4 192 IC ϭ Problem 5.6-11 A two-axle carriage that is part of an overhead traveling crane in a testing laboratory moves slowly across a simple beam AB (see figure). The load transmitted to the beam from the front axle is 2000 lb and from the rear axle is 4000 lb. The weight of the beam itself may be disregarded. (a) Determine the minimum required section modulus S for the beam if the allowable bending stress is 15.0 ksi, the length of the beam is 16 ft, and the wheelbase of the carriage is 5 ft. (b) Select a suitable I-beam (S shape) from Table E-2, Appendix E. 4000 lb 5 ft A 2000 lb B 16 ft
    • SECTION 5.6 Solution 5.6-11 311 Moving carriage P2 P1 BENDING MOMENT UNDER LARGER LOAD P2 d x M ϭ RA x ϭ 125(43x Ϫ 3x2) A B L P1 ϭ load on front axle ϭ 2000 lb P2 ϭ load on rear axle ϭ 4000 lb L ϭ 16 ft d ϭ 5 ft ␴allow ϭ 15 ksi x ϭ distance from support A to the larger load P2 (feet) LϪx LϪxϪd RA ϭ P2 ¢ ≤ ϩ P1¢ ≤ L L x x 5 ϭ (4000 lb) ¢ 1 Ϫ ≤ ϩ (2000 lb) ¢ 1 Ϫ Ϫ ≤ 16 16 16 ϭ 125(43 Ϫ 3x) WEIGHT OF BEAM PER UNIT LENGTH ␲d 2 q ϭ g¢ ≤ 4 43 ϭ 7.1667 ft 6 43 43 2 ≤ Ϫ 3¢ ≤ R 6 6 ϭ19, 260 lb-ft ϭ 231,130 lb-in. (a) MINIMUM SECTION MODULUS Smin ϭ Mmax 231,130 lb-in. ϭ ϭ 15.41 in.3 sallow 15,000 psi (b) SELECT ON I-BEAM (S SHAPE) Table E-2. Select S 8 ϫ 23 (S ϭ 16.2 in.3) Problem 5.6-12 A cantilever beam AB of circular cross section and length L ϭ 450 mm supports a load P ϭ 400 N acting at the free end (see figure). The beam is made of steel with an allowable bending stress of 60 MPa. Determine the required diameter dmin of the beam, considering the effect of the beam’s own weight. ␥ ϭ weight density of steel ϭ 77.0 kN/m3 x ϭ xm ϭ Mmax ϭ (M) xϭxm ϭ 125 B (43) ¢ (x ϭ ft; RA ϭ lb) Solution 5.6-12 Cantilever beam DATA L ϭ 450 mm P ϭ 400 N ␴allow ϭ 60 MPa (x ϭ ft; M ϭ lb-ft) MAXIMUM BENDING MOMENT dM Set equal to zero and solve for x ϭ xm. dx dM ϭ 125(43 Ϫ 6x) ϭ 0 dx RA A B d P L MINIMUM DIAMETER Mmax ϭ ␴allow S PL ϩ ␲ gd 2L2 ␲d 3 ϭ sallow ¢ ≤ 8 32 Rearrange the equation: 32 PL ϭ0 ␲ (Cubic equation with diameter d as unknown.) sallow d 3 Ϫ 4g L2 d 2 Ϫ MAXIMUM BENDING MOMENT Mmax ϭ PL ϩ Design of Beams qL2 ␲ gd 3L2 ϭ PL ϩ 2 8 SECTION MODULUS S ϭ ␲d 32 3 Substitute numerical values (d ϭ meters): (60 ϫ 106 N/m2)d 3 Ϫ 4(77,000 N/m3)(0.45m)2d 2 Ϫ 32 (400 N)(0.45 m) ϭ 0 ␲ 60,000d 3 Ϫ 62.37d 2 Ϫ 1.833465 ϭ 0 Solve the equation numerically: d ϭ 0.031614 m dmin ϭ 31.61 mm
    • 312 CHAPTER 5 Stresses in Beams (Basic Topics) Problem 5.6-13 A compound beam ABCD (see figure) is supported at points A, B, and D and has a splice (represented by the pin connection) at point C. The distance a ϭ 6.0 ft and the beam is a W 16 ϫ 57 wide-flange shape with an allowable bending stress of 10,800 psi. Find the allowable uniform load qallow that may be placed on top of the beam, taking into account the weight of the beam itself. Solution 5.6-13 A B 4a C a 11qa RA ϭ 8 RBϭ D 4a C D a 4a 45qa 8 C qmax ϭ D 2a ᎐ qallow ϭ qmax Ϫ (weight of beam) W 16 ϫ 57 RDϭ 2qa 2qa2 11a 8 2sallow S 5a2 ␴allow ϭ 10,800 psi S ϭ 92.2 in.3 ALLOWABLE UNIFORM LOAD 121 qa2 128 B qmax ϭ 5q a2 ϭ sallow S 2 DATA: a ϭ 6 ft ϭ 72 in. 4a Pin connection at point C. 2(10,800 psi)(92.2 in.3 ) ϭ 76.833 lbրin. 5(72 in.) 2 ϭ 922 lb/ft qallow ϭ 922 lb/ft Ϫ 57 lb/ft ϭ 865 lb/ft 2a 5qa2 2 Problem 5.6-14 A small balcony constructed of wood is supported by three identical cantilever beams (see figure). Each beam has length L1 ϭ 2.1 m, width b, and height h ϭ 4b/3. The dimensions of the balcony floor are L1 ϫ L2, with L2 ϭ 2.5 m. The design load is 5.5 kPa acting over the entire floor area. (This load accounts for all loads except the weights of the cantilever beams, which have a weight density ␥ ϭ 5.5 kN/m3.) The allowable bending stress in the cantilevers is 15 MPa. Assuming that the middle cantilever supports 50% of the load and each outer cantilever supports 25% of the load, determine the required dimensions b and h. Solution 5.6-14q Cantilever beam for a balcony 4b h= — 3 L1 B Pin Mmax ϭ Pin A A Compound beam q M q b L1 ϭ 2.1 m L 2 ϭ 2.5 m Floor dimensions: L 1 ϫ L 2 Design load ϭ w ϭ 5.5 kPa ␥ ϭ 5.5 kN/m3 (weight density of wood beam) ␴allow ϭ 15 MPa 4b h= — 3 L2 b L1 MIDDLE BEAM SUPPORTS 50% OF THE LOAD. ∴ q ϭ w¢ L2 2.5 m ≤ ϭ (5.5 kPa) ¢ ≤ ϭ 6875 Nրm 2 2 WEIGHT OF BEAM q0 ϭ gbh ϭ 4gb2 4 ϭ (5.5 kNրm2 )b2 3 3 ϭ 7333b2 (N/m) (b ϭ meters)
    • SECTION 5.6 MAXIMUM BENDING MOMENT Rearrange the equation: (q ϩ 1 ϭ (6875 Nրm ϩ 7333b2 )(2.1 m) 2 2 2 ϭ15,159 ϩ 16,170b2 (N ؒ m) q0 )L2 1 Mmax ϭ Design of Beams bh2 8b3 ϭ 6 27 Mmax ϭ ␴allow S (120 ϫ 106)b3 Ϫ 436,590b2 Ϫ 409,300 ϭ 0 SOLVE NUMERICALLY FOR DIMENSION b Sϭ b ϭ 0.1517 m 8b3 15,159 ϩ 16,170b2 ϭ (15 ϫ 106 Nրm2 ) ¢ ≤ 27 4b ϭ 0.2023 m 3 REQUIRED DIMENSIONS b ϭ 152 mm Problem 5.6-15 A beam having a cross section in the form of an unsymmetric wide-flange shape (see figure) is subjected to a negative bending moment acting about the z axis. Determine the width b of the top flange in order that the stresses at the top and bottom of the beam will be in the ratio 4:3, respectively. hϭ h ϭ 202 mm y b 1.5 in. 1.25 in. z C 12 in. 1.5 in. 16 in. Solution 5.6-15 Unsymmetric wide-flange beam y AREAS OF THE CROSS SECTION (in.2) b c1 c2 A3 A1 1.25 in. z 1.5 in. 12 in. B 1.5 in. 16 in. Stresses at top and bottom are in the ratio 4:3. Find b (inches) h ϭ height of beam ϭ 15 in. LOCATE CENTROID stop c1 4 ϭ ϭ sbottom c2 3 4 60 c1 ϭ h ϭ ϭ 8.57143 in. 7 7 3 45 c2 ϭ h ϭ ϭ 6.42857 in. 7 7 A2 ϭ (12)(1.25) ϭ 15 in.2 A3 ϭ (16)(1.5) ϭ 24 in.2 C A2 B A1 ϭ 1.5b A ϭ A1 ϩ A2 ϩ A3 ϭ 39 ϩ 1.5b (in.2) FIRST MOMENT OF THE CROSS-SECTIONAL AREA ABOUT THE LOWER EDGE B-B QBB ϭ a yi Ai ϭ (14.25)(1.5b) ϩ (7.5)(15) ϩ (0.75)(24) ϭ 130.5 ϩ 21.375b (in.3) DISTANCE c2 c2 ϭ FROM LINE B-B TO THE CENTROID C QBB 130.5 ϩ 21.375b 45 ϭ ϭ in. A 39 ϩ 1.5b 7 SOLVE FOR b (39 ϩ 1.5b)(45) ϭ (130.5 ϩ 21.375b)(7) 82.125b ϭ 841.5 b ϭ 10.25 in. 313
    • 314 CHAPTER 5 Stresses in Beams (Basic Topics) Problem 5.6-16 A beam having a cross section in the form of a channel (see figure) is subjected to a bending moment acting about the z axis. Calculate the thickness t of the channel in order that the bending stresses at the top and bottom of the beam will be in the ratio 7:3, respectively. y t t C z t 50 mm 120 mm Solution 5.6-16 Channel beam AREAS OF THE CROSS SECTION (mm 2) A1 ϭ ht ϭ 50t A2 ϭ b1 t ϭ 120t Ϫ 2t 2 A ϭ 2A1 ϩ A2 ϭ 220t Ϫ 2t 2 ϭ 2t(110Ϫt) y z A1 A2 c1 c2 C A1 t b1 FIRST MOMENT OF THE CROSS-SECTIONAL AREA ABOUT B-B h t QBB ϭ ayi Ai ϭ (2) ¢ ≤ (50 t) ϩ ¢ ≤ (b1 )(t) 2 2 t ϭ 2(25)(50t) ϩ ¢ ≤ (120 Ϫ 2t)(t) 2 ϭ t (2500 ϩ 60t Ϫ t 2) (t ϭ mm; Q ϭ mm3) THE LOWER EDGE B B t h ϭ 50 mm t b ϭ 120 mm t ϭ thickness (constant) (t is in millimeters) b1 ϭ b Ϫ 2t ϭ 120 mm Ϫ 2t Stresses at the top and bottom are in the ratio 7:3. DISTANCE c2 FROM LINE B-B TO THE CENTROID C c2 ϭ Q BB t(2500 ϩ 60t Ϫ t 2 ) ϭ A 2t(110 Ϫ t) ϭ 2500 ϩ 60t Ϫ t 2 ϭ 15 mm 2(110 Ϫ t) Determine the thickness t. LOCATE CENTROID stop c1 7 ϭ ϭ sbottom c2 3 c1 ϭ 7 h ϭ 35 mm 10 c2 ϭ SOLVE FOR t 2(110 Ϫ t)(15) ϭ 2500 ϩ 60t Ϫ t 2 t 2 Ϫ 90t ϩ 800 ϭ 0 t ϭ 10 mm 3 h ϭ 15 mm 10 Problem 5.6-17 Determine the ratios of the weights of three beams that have the same length, are made of the same material, are subjected to the same maximum bending moment, and have the same maximum bending stress if their cross sections are (1) a rectangle with height equal to twice the width, (2) a square, and (3) a circle (see figures). h = 2b b a a d
    • SECTION 5.6 Solution 5.6-17 Ratio of weights of three beams Beam 1: Rectangle (h ϭ 2b) Beam 2: Square (a ϭ side dimension) Beam 3: Circle (d ϭ diameter) L, ␥, Mmax, and ␴max are the same in all three beams. M S ϭ section modulus S ϭ s Since M and ␴ are the same, the section moduli must be the same. bh2 2b3 3S 1ր3 (1) RECTANGLE: S ϭ ϭ bϭ¢ ≤ 6 3 2 3S 2ր3 A1 ϭ 2b2 ϭ 2 ¢ ≤ ϭ 2.6207S 2ր3 2 315 Design of Beams a3 a ϭ (6S) 1ր3 6 A2 ϭ a2 ϭ (6S)2/3 ϭ 3.3019S 2/3 (2) SQUARE: S ϭ ␲d 3 32S 1ր3 dϭ¢ ≤ ␲ 32 ␲d 2 ␲ 32S 2ր3 A3 ϭ ϭ ¢ ≤ ϭ 3.6905 S 2ր3 4 4 ␲ (3) CIRCLE: S ϭ Weights are proportional to the cross-sectional areas (since L and ␥ are the same in all 3 cases). W1 : W2 : W3 ϭ A1 : A2 : A3 A1 : A2 : A3 ϭ 2.6207 : 3.3019 : 3.6905 W1 : W2 : W3 ϭ 1 : 1.260 : 1.408 Problem 5.6-18 A horizontal shelf AD of length L ϭ 900 mm, width b ϭ 300 mm, and thickness t ϭ 20 mm is supported by brackets at B and C [see part (a) of the figure]. The brackets are adjustable and may be placed in any desired positions between the ends of the shelf. A uniform load of intensity q, which includes the weight of the shelf itself, acts on the shelf [see part (b) of the figure]. Determine the maximum permissible value of the load q if the allowable bending stress in the shelf is ␴allow ϭ 5.0 MPa and the position of the supports is adjusted for maximum load-carrying capacity. t A B D C b L (a) q A D B C L (b) Solution 5.6-18 Shelf with adjustable supports q t A D B b C Mmax ϭ M1 B A x M2 L ( 12 Ϫ 1) 2 Substitute x into the equation for either M1 or ƒ M2 ƒ : Solve for x: x ϭ L C M2 D x L ϭ 900 mm b ϭ 300 mm t ϭ 20 mm ␴allow ϭ 5.0 MPa For maximum load-carrying capacity, place the supports so that M1 ϭ ƒ M2 ƒ . Let x ϭ length of overhang qL qx2 M1 ϭ (L Ϫ 4x) ƒ M2 ƒ ϭ 8 2 2 qL qx ∴ (L Ϫ 4x) ϭ 8 2 qL2 (3 Ϫ 212) 8 Mmax ϭ sallow S ϭ sallow ¢ Eq. (1) bt 2 ≤ 6 Eq. (2) Equate Mmax from Eqs. (1) and (2) and solve for q: qmax ϭ 4bt 2sallow 3L2 (3 Ϫ 212) Substitute numerical values: qmax ϭ 5.76 kN/m
    • 316 CHAPTER 5 Stresses in Beams (Basic Topics) Problem 5.6-19 A steel plate (called a cover plate) having crosssectional dimensions 4.0 in. ϫ 0.5 in. is welded along the full length of the top flange of a W 12 ϫ 35 wide-flange beam (see figure, which shows the beam cross section). What is the percent increase in section modulus (as compared to the wide-flange beam alone)? Solution 5.6-19 4.0 ϫ 0.5 in. cover plate c1 1 W 12 ϫ 35 Beam with cover plate y z 4.0 ϫ 0.5 in. cover plate – y C c2 6.25 1 6.25 Moment of inertia about axis 1-1: 1 I1–1 ϭ I0 ϩ (4.0)(0.5) 3 ϩ (4.0)(0.5)(6.25 ϩ 0.25) 2 12 ϭ 369.5 in.4 Moment of inertia about z axis: I1–1 ϭ Iz ϩ A1 y 2 Iz ϭ I1–1 Ϫ A1 y 2 Iz ϭ 369.5 in.4 Ϫ (12.3 in.2)(1.057 in.)2 ϭ 355.8 in.4 All dimensions in inches. WIDE-FLANGE BEAM ALONE (AXIS 1-1 IS CENTROIDAL AXIS) W 12 ϫ 35 d ϭ 12.50 in. A0 ϭ 10.3 in.2 I0 ϭ 2.85 in.4 S0 ϭ 45.6 in.3 BEAM WITH COVER PLATE (z AXIS IS CENTROIDAL AXIS) A1 ϭ A0 ϩ (4.0 in.)(0.5 in.) ϭ 12.3 in.2 First moment with respect to axis 1-1: Q1 ϭ a yi Ai ϭ (6.25 in. ϩ 0.25 in.)(4.0 in.)(0.5 in.) ϭ 13.00 in.3 Q1 13.00 in.3 yϭ ϭ ϭ 1.057 in. A1 12.3 in.2 SECTION MODULUS (Use the smaller of the two section moduli) Iz 355.8 in.4 S1 ϭ ϭ ϭ 48.69 in.3 c2 7.307 in. INCREASE IN SECTION MODULUS S1 48.69 ϭ ϭ 1.068 S0 45.6 Percent increase ϭ 6.8% c1 ϭ 6.25 ϩ 0.5 Ϫ y ϭ 5.693 in. c2 ϭ 6.25 ϩ y ϭ 7.307 in. Problem 5.6-20 A steel beam ABC is simply supported at A and B and has an overhang BC of length L ϭ 150 mm (see figure on the next page). The beam supports a uniform load of intensity q ϭ 3.5 kN/m over its entire length of 450 mm. The cross section of the beam is rectangular with width b and height 2b. The allowable bending stress in the steel is ␴allow ϭ 60 MPa and its weight density is ␥ ϭ 77.0 kN/m3. (a) Disregarding the weight of the beam, calculate t he required width b of the rectangular cross section. (b) Taking into account the weight of the beam, calculate the required width b. q C A 2b B 2L L b
    • SECTION 5.6 Solution 5.6-20 Design of Beams 317 Beam with an overhang q C A 2b B Substitute numerical values: 3(3.5 kNրm)(150 mm) 2 b3 ϭ ϭ 0.98438 ϫ 10Ϫ6 m3 4(60 MPa) b 2L L 9qL RB ϭ 4 3qL RA ϭ 4 9qL2 32 M C B 0 A – qL2 2 L ϭ 150 mm q ϭ 3.5 kN/m ␴allow ϭ 60 MPa ␥ ϭ 77.0 kN/m3 b ϭ 0.00995 m ϭ 9.95 mm (b) INCLUDE THE WEIGHT OF THE BEAM q0 ϭ weight of beam per unit length q0 ϭ ␥(b)(2b) ϭ 2␥b 2 Mmax ϭ Sϭ qL2 Mmax ϭ 2 bh2 2b3 Sϭ ϭ 6 3 (a) DISREGARD THE WEIGHT OF THE BEAM Mmax ϭ ␴allow S b3 ϭ 3qL2 4sallow qL2 2b3 ϭ sallow ¢ ≤ 2 3 Problem 5.6-21 A retaining wall 5 ft high is constructed of horizontal wood planks 3 in. thick (actual dimension) that are supported by vertical wood piles of 12 in. diameter (actual dimension), as shown in the figure. The lateral earth pressure is p1 ϭ 100 lb/ft2 at the top of the wall and p2 ϭ 400 lb/ft2 at the bottom. Assuming that the allowable stress in the wood is 1200 psi, calculate the maximum permissible spacing s of the piles. (Hint: Observe that the spacing of the piles may be governed by the load-carrying capacity of either the planks or the piles. Consider the piles to act as cantilever beams subjected to a trapezoidal distribution of load, and consider the planks to act as simple beams between the piles. To be on the safe side, assume that the pressure on the bottom plank is uniform and equal to the maximum pressure.) 2b3 3 (q Ϫ q0 )L2 1 ϭ (q ϩ 2g b2 )L2 2 2 Mmax ϭ ␴allow S 1 2b3 (q ϩ 2g b2 ) L2 ϭ sallow ¢ ≤ 2 3 Rearrange the equation: 4␴allow b 3 Ϫ 6␥L2 b2 Ϫ 3qL2 ϭ 0 Substitute numerical values: (240 ϫ 106)b3 Ϫ 10,395b2 Ϫ 236.25 ϭ 0 (b ϭ meters) Solve the equation: b ϭ 0.00996 m ϭ 9.96 mm 3 in. p1 = 100 lb/ft2 12 in. diam. 12 in. diam. s 5 ft 3 in. Top view p2 = 400 lb/ft2 Side view
    • 318 CHAPTER 5 Stresses in Beams (Basic Topics) Solution 5.6-21 Retaining wall q1 t (1) PLANK AT THE BOTTOM OF THE DAM t ϭ thickness of plank ϭ 3 in. b ϭ width of plank (perpendicular to the plane of the figure) p2 ϭ maximum soil pressure ϭ 400 lb/ft 2 ϭ 2.778 lb/in.2 s ϭ spacing of piles q ϭ p2 b ␴allow ϭ 1200 psi S ϭ section modulus Mmax ϭ qs2 p2 bs2 ϭ 8 8 Mmax ϭ ␴allow S sϭ or Sϭ bt 2 6 p2 bs2 bt 2 ϭ sallow ¢ ≤ 8 6 4 sallow t 2 ϭ 72.0 in. B 3p2 Solve for s: h q s q2 Divide the trapezoidal load into two triangles (see dashed line). 1 2h 1 h sh2 Mmax ϭ (q1 )(h) ¢ ≤ ϩ (q2 )(h) ¢ ≤ ϭ (2p1 ϩ p2 ) 2 3 2 3 6 ␲d 3 Sϭ Mmax ϭ ␴allow S or 32 sh2 ␲d 3 (2p1 ϩ p2 ) ϭ sallow ¢ ≤ 6 32 Solve for s: 3␲ sallow d 3 sϭ ϭ 81.4 in. 16h2 (2p1 ϩ p2 ) PLANK GOVERNS smax ϭ 72.0 in. (2) VERTICAL PILE h ϭ 5 ft ϭ 60 in. p1 ϭ soil pressure at the top ϭ 100 lb/ft2 ϭ 0.6944 lb/in.2 q1 ϭ p1 s q2 ϭ p2 s d ϭ diameter of pile ϭ 12 in. Problem 5.6-22 A beam of square cross section (a ϭ length of each side) is bent in the plane of a diagonal (see figure). By removing a small amount of material at the top and bottom corners, as shown by the shaded triangles in the figure, we can increase the section modulus and obtain a stronger beam, even though the area of the cross section is reduced. (a) Determine the ratio ␤ defining the areas that should be removed in order to obtain the strongest cross section in bending. (b) By what percent is the section modulus increased when the areas are removed? y ␤a a z C a ␤a
    • SECTION 5.6 Solution 5.6-22 removed Beam of square cross section with corners y ␤a (1 ᎐ ␤) a am m 1 RATIO OF SECTION MODULI S ϭ (1 ϩ 3b)(1 Ϫ b) 2 S0 n C n1 1.10 (S) S0 max ϭ 1.0535 Eq. (1) a p p1 ␤a 1.00 a ϭ length of each side ␤a ϭ amount removed Beam is bent about the z axis. a 12 Eq. (1) GRAPH OF EQ. (1) q z 319 Design of Beams I0 a3 12 ϭ c0 12 0 .90 0.1 0.2 ␤ϭ1 9 0.3 ␤ ENTIRE CROSS SECTION (AREA 0) I0 ϭ a4 12 c0 ϭ S0 ϭ SQUARE mnpq (AREA 1) I1 ϭ (1 Ϫ b) 4a4 12 PARALLELOGRAM mm,n,n (AREA 2) 1 I2 ϭ (base)(height)3 3 (1 Ϫ b)a 3 ba4 1 I2 ϭ (ba12) B R ϭ (1 Ϫ b) 3 3 6 12 (a) VALUE OF ␤ FOR A MAXIMUM VALUE OF S/S0 d S ¢ ≤ϭ0 db S0 Take the derivative and solve this equation for ␤. 1 bϭ 9 (b) MAXIMUM VALUE OF S/S0 Substitute ␤ ϭ 1/9 into Eq. (1). (S/S0)max ϭ 1.0535 The section modulus is increased by 5.35% when the triangular areas are removed. I 12 a3 Sϭ ϭ (1 ϩ 3b)(1 Ϫ b) 2 c 12 REDUCED CROSS SECTION (AREA qmm,n,p,pq) a4 I ϭ I1 ϩ 2I2 ϭ (1 ϩ 3b)(1 Ϫ b) 3 12 cϭ 12 (1 Ϫ b)a Problem 5.6-23 The cross section of a rectangular beam having width b and height h is shown in part (a) of the figure. For reasons unknown to the beam designer, it is planned to add structural projections of width b/9 and height d to the top and bottom of the beam [see part (b) of the figure]. For what values of d is the bending-moment capacity of the beam increased? For what values is it decreased? b — 9 d h b (a) h d b — 9 (b)
    • 320 CHAPTER 5 Solution 5.6-23 Stresses in Beams (Basic Topics) Beam with projections d 1 2 h d h b — 9 S1 ϭ I1 bh2 ϭ c1 6 S2 S1 0 0.25 0.50 0.75 1.00 h d b (1) ORIGINAL BEAM bh3 h I1 ϭ c1 ϭ 12 2 S2 d versus S1 h Graph of 1.000 0.8426 0.8889 1.0500 1.2963 S2 S1 1.0 (2) BEAM WITH PROJECTIONS 1 8b 1 b I2 ϭ ¢ ≤ h3 ϩ ¢ ≤ (h ϩ 2d) 3 12 9 12 9 b ϭ [8h3 ϩ (h ϩ 2d) 3 ] 108 h 1 c2 ϭ ϩ d ϭ (h ϩ 2d) 2 2 I2 b [8h3 ϩ (h ϩ 2d) 3 ] S2 ϭ ϭ c2 54(h ϩ 2d) RATIO OF SECTION MODULI 2d 3 8 ϩ ¢1 ϩ ≤ S2 b[8h3 ϩ (h ϩ 2d) 3 ] h ϭ ϭ S1 2d 9(h ϩ 2d)(bh2 ) 9 ¢1 ϩ ≤ h EQUAL SECTION MODULI S2 d Set ϭ 1 and solve numerically for . S1 h d ϭ 0.6861 and h d ϭ0 h 0.5 0.2937 0 0.6861 0.5 1.0 d h Moment capacity is increased when d 7 0.6861 h Moment capacity is decreased when d 6 0.6861 h NOTES: S2 2d 3 2d ϭ 1 when ¢ 1 ϩ ≤ Ϫ 9 ¢ 1 ϩ ≤ ϩ 8 ϭ 0 S1 h h d ϭ 0.6861 and 0 or h 3 S2 d 14 Ϫ 1 ϭ 0.2937 is minimum when ϭ S1 h 2 ¢ S2 ≤ ϭ 0.8399 S1 min
    • SECTION 5.7 321 Nonprismatic Beams Nonprismatic Beams Problem 5.7-1 A tapered cantilever beam AB of length L has square cross sections and supports a concentrated load P at the free end (see figure on the next page). The width and height of the beam vary linearly from hA at the free end to hB at the fixed end. Determine the distance x from the free end A to the cross section of maximum bending stress if hB ϭ 3hA. What is the magnitude ␴max of the maximum bending stress? What is the ratio of the maximum stress to the largest stress ␴B at the support? Solution 5.7-1 B A hA hB x P L Tapered cantilever beam P A B x L SQUARE CROSS SECTIONS hA ϭ height and width at smaller end hB ϭ height and width at larger end hx ϭ height and width at distance x hB ϭ3 hA hx ϭ hA ϩ (hB Ϫ hA ) ¢ x 2x ≤ ϭ hA ¢ 1 ϩ ≤ L L h3 1 2x 3 A Sx ϭ (hx ) 3 ϭ ¢ 1 ϩ ≤ 6 6 L STRESS AT DISTANCE x Mx 6Px s1 ϭ ϭ Sx 2x 3 (hA ) 3 ¢ 1 ϩ ≤ L AT END A: x ϭ 0 ␴A ϭ 0 AT SUPPORT B: x ϭ L 2PL sB ϭ 9(hA ) 3 CROSS SECTION OF MAXIMUM STRESS ds1 ϭ0 Evaluate the derivative, set it equal dx to zero, and solve for x. Set xϭ L 4 MAXIMUM BENDING STRESS smax ϭ (s1 ) xϭLր4 ϭ Ratio of ␴max to ␴B smax ϭ2 sB 4PL 9(hA ) 3
    • 322 CHAPTER 5 Stresses in Beams (Basic Topics) Problem 5.7-2 A tall signboard is supported by two vertical beams consisting of thin-walled, tapered circular tubes (see figure). For purposes of this analysis, each beam may be represented as a cantilever AB of length L ϭ 8.0 m subjected to a lateral load P ϭ 2.4 kN at the free end. The tubes have constant thickness t ϭ 10.0 mm and average diameters dA ϭ 90 mm and dB ϭ 270 mm at ends A and B, respectively. Because the thickness is small compared to the diameters, the moment of inertia at any cross section may be obtained from the formula I ϭ ␲d 3t/8 (see Case 22, Appendix D), and therefore the section modulus may be obtained from the formula S ϭ ␲d 2t/4. At what distance x from the free end does the maximum bending stress occur? What is the magnitude ␴max of the maximum bending stress? What is the ratio of the maximum stress to the largest stress ␴B at the support? Solution 5.7-2 P = 2.4 kN Wind load B A x L = 8.0 m t = 10.0 mm dA = 90 mm dB = 270 mm Tapered circular tube P t B A x d L P ϭ 2.4 kN L ϭ 8.0 m t ϭ 10 mm d ϭ average diameter AT END A: x ϭ 0 ␴1 ϭ ␴A ϭ 0 AT SUPPORT B: x ϭ L ϭ 8.0 m ␴1 ϭ ␴B ϭ 33.53 MPa At end A: dA ϭ 90 mm At support B: dB ϭ 270 mm CROSS SECTION OF MAXIMUM STRESS ds1 ϭ0 Evaluate the derivative, set it equal to dx zero, and solve for x. Set AT DISTANCE x: dx ϭ dA ϩ (dB Ϫ dA ) ¢ Sx ϭ ␲ ␲ 2x 2 (dx ) 2 (t) ϭ (90) 2 ¢ 1 ϩ ≤ (10) 4 4 L ϭ 20,250␲ ¢ 1 ϩ Mx ϭ Px ϭ 2400x s1 ϭ x x 2x ≤ ϭ 90 ϩ 180 ϭ 90 ¢ 1 ϩ ≤ L L L Mx ϭ Sx 2x ≤ L xϭ L ϭ 4.0 m 2 MAXIMUM BENDING STRESS 2 Sx ϭ mm3 x ϭ meters, Mx ϭ N ؒ m 2400x 20.25␲ ¢ 1 ϩ 2x ≤ L 2 L ϭ meters, s1 ϭ MPa smax ϭ (s1 ) xϭLր2 ϭ 2400(4.0) (20.25 ␲)(1 ϩ 1) 2 ϭ 37.73 MPa RATIO OF ␴max to ␴B smax 9 ϭ ϭ 1.125 sB 8
    • SECTION 5.7 Problem 5.7-3 A tapered cantilever beam AB having rectangular cross sections is subjected to a concentrated load P ϭ 50 lb and a couple M0 ϭ 800 lb-in. acting at the free end (see figure). The width b of the beam is constant and equal to 1.0 in., but the height varies linearly from hA ϭ 2.0 in. at the loaded end to hB ϭ 3.0 in. at the support. At what distance x from the free end does the maximum bending stress ␴max occur? What is the magnitude ␴max of the maximum bending stress? What is the ratio of the maximum stress to the largest stress ␴B at the support? Solution 5.7-3 323 Nonprismatic Beams P = 50 lb A M0 = 800 lb-in. B hB = 3.0 in. hA = 2.0 in. x b = 1.0 in. b = 1.0 in. L = 20 in. Tapered cantilever beam P M0 A B x L P ϭ 50 lb M0 ϭ 800 lb-in. L ϭ 20 in. hA ϭ 2.0 in. hB ϭ 3.0 in. b ϭ 1.0 in. AT END A: x ϭ 0 ␴1 ϭ ␴A ϭ 1200 psi AT SUPPORT B: x ϭ L ϭ 20 in. ␴1 ϭ ␴B ϭ 1200 psi CROSS SECTION OF MAXIMUM STRESS UNITS: pounds and inches ds1 ϭ0 Evaluate the derivative, set it equal to dx zero, and solve for x. AT DISTANCE x: x ϭ 8.0 in. x x x hx ϭ hA ϩ (hB Ϫ hA ) ϭ 2 ϩ (1) ¢ ≤ ϭ 2 ϩ L L L Sx ϭ bh2 b x 2 1 x 2 x ϭ ¢2 ϩ ≤ ϭ ¢2 ϩ ≤ 6 6 L 6 L Mx ϭ Px ϩ M0 ϭ (50)(x) ϩ 800 ϭ 50(16 ϩ x) Mx 50(16 ϩ x)(6) 120,000(16 ϩ x) s1 ϭ ϭ ϭ Sx x 2 (40 ϩ x) 2 ¢2 ϩ ≤ L Set MAXIMUM BENDING STRESS (120,000)(24) smax ϭ (s1 ) xϭ8.0 ϭ ϭ 1250 psi (48) 2 RATIO OF ␴max TO ␴B smax 1250 25 ϭ ϭ ϭ 1.042 sB 1200 24 Problem 5.7-4 The spokes in a large flywheel are modeled as beams fixed at one end and loaded by a force P and a couple M0 at the other (see figure). The cross sections of the spokes are elliptical with major and minor axes (height and width, respectively) having the lengths shown in the figure. The cross-sectional dimensions vary linearly from end A to end B. Considering only the effects of bending due to the loads P and M0, determine the following quantities: (a) the largest bending stress ␴A at end A; (b) the largest bending stress ␴B at end B; (c) the distance x to the cross section of maximum bending stress; and (d) the magnitude ␴max of the maximum bending stress. P = 15 kN M0 = 12 kN· m B A x L = 1.10 m hA = 90 mm hB = 120 mm bA = 60 mm bB = 80 mm
    • 324 CHAPTER 5 Solution 5.7-4 Stresses in Beams (Basic Topics) Elliptical spokes in a flywheel P M0 A B x L = 1.10 m P ϭ 15 kN ϭ 15,000 N M0 ϭ 12 kN ؒ m ϭ 12,000 N ؒ m L ϭ 1.1 m (a) AT END A: x ϭ 0 sA ϭ (s1 ) xϭ0 ϭ UNITS: Newtons, meters ϭ 251.5 MPa AT END A: bA ϭ 0.06 m, hA ϭ 0.09 m AT SUPPORT B: bB ϭ 0.08 m, AT DISTANCE x: bx ϭ bA ϩ (bB Ϫ bA ) hx ϭ hA ϩ (hB Ϫ hA ) ␲ (b )(h ) 3 64 x x hB ϭ 0.12 m x x x ϭ 0.06 ϩ 0.02 ϭ 0.02 ¢ 3 ϩ ≤ L L L x x x ϭ 0.09 ϩ 0.03 ϭ 0.03 ¢ 3 ϩ ≤ L L L Case 16, Appendix D: I ϭ Ix ϭ Sx ϭ (80 ϫ 109 )(0.8) ϭ 251.5 ϫ 106 Nրm2 (3␲)(27) ␲ (bh3 ) 64 Ix ␲bxh2 x ϭ hx ր2 32 ␲ x x (0.02) ¢ 3 ϩ ≤ (0.03) 2 ¢ 3 ϩ ≤ 32 L L (b) AT END B: x ϭ L ϭ 1.1 m (80 ϫ 109 )(0.8 ϩ 1.1) sB ϭ (s1 ) xϭL ϭ (3␲)(3 ϩ 1) 3 ϭ 252.0 ϫ 106 N/m2 ϭ 252.0 MPa (c) CROSS SECTION OF MAXIMUM STRESS ds1 ϭ0 Evaluate the derivative, set it equal to dx zero, and solve for x. x ϭ 0.45 m Set (d) MAXIMUM BENDING STRESS 2 Sx ϭ ϭ 9␲ x 3 ¢3 ϩ ≤ 6 L 16 ϫ 10 smax ϭ (s1 ) xϭ0.45 ϭ ϭ 267.8 ϫ 106 N/m2 ϭ 267.8 MPa Mx ϭ M0 ϩ Px ϭ 12,000 N ؒ m ϩ (15,000 N)x ϭ 15,000(0.8 ϩ x) s1 ϭ ϭ Mx 15,000(0.8 ϩ x)(16 ϫ 106 ) ϭ Sx x 3 9␲ ¢ 3 ϩ ≤ L (80 ϫ 109 )(0.8 ϩ x) x 3 3␲ ¢ 3 ϩ ≤ L (80 ϫ 109 )(0.8 ϩ 0.45) 0.45 3 (3␲) ¢ 3 ϩ ≤ 1.1 Problem 5.7-5 Refer to the tapered cantilever beam of solid circular cross section shown in Fig. 5-24 of Example 5-9. (a) Considering only the bending stresses due to the load P, determine the range of values of the ratio dB /dA for which the maximum normal stress occurs at the support. (b) What is the maximum stress for this range of values?
    • SECTION 5.7 Solution 5.7-5 325 Nonprismatic Beams Tapered cantilever beam P B A dB dA x L FROM EQ. (5-32), EXAMPLE 5-9 s1 ϭ (a) GRAPH OF x/L VERSUS dB /dA (EQ. 2) x 3 ␲ B dA ϩ (dB Ϫ dA ) ¢ ≤ R L 32Px Eq. (1) x L 2 FIND THE VALUE OF x THAT MAKES ␴1 A MAXIMUM y¢ du dy ≤Ϫu¢ ≤ u ds1 dx dx N Let s1 ϭ ϭ ϭ y dx D y2 x 3 N ϭ ␲ B dA ϩ (dB Ϫ dA ) ¢ ≤ R [32P] L Ϫ[32Px] [␲] [3] B dA ϩ (dB Ϫ dA ) ¢ After simplification: N ϭ 32␲P B dA ϩ (dB Ϫ dA ) ¢ ds1 ϭ0 dx ∴ 32P B dA Ϫ 2(dB Ϫ dA ) ␲ B dA ϩ (dB Ϫ dA ) ¢ dA Ϫ 2(dB Ϫ dA ) ¢ dA x ϭ ϭ L 2(dB Ϫ dA ) Eq. (2) 0 x 2 1 ≤ R B (dB Ϫ dA ) R L L x 2 x ≤ R B dA Ϫ 2(dB Ϫ dA ) R L L x 6 D ϭ ␲ 2 B dA ϩ (dB Ϫ dA ) R L ds1 N ϭ ϭ dx D 1 x R L x ≤ϭ0 L 1 dB 2 ¢ Ϫ 1≤ dA Eq. (2) 1.5 2 2.5 Maximum bending stress occurs at the dB Յ 1.5 support when 1 Յ dA (b) MAXIMUM STRESS (AT SUPPORT B) Substitute x/L ϭ 1 into Eq. (1): smax ϭ x 4 ≤R L 1 32PL ␲d 3 B 3 dB dA
    • 326 CHAPTER 5 Stresses in Beams (Basic Topics) Fully Stressed Beams q Problems 5.7-6 to 5.7-8 pertain to fully stressed beams of rectangular cross section. Consider only the bending stresses obtained from the flexure formula and disregard the weights of the beams. B A hx Problem 5.7-6 A cantilever beam AB having rectangular cross sections with constant width b and varying height hx is subjected to a uniform load of intensity q (see figure). How should the height hx vary as a function of x (measured from the free end of the beam) in order to have a fully stressed beam? (Express hx in terms of the height hB at the fixed end of the beam.) hB x L hx hB b b Solution 5.7-6 Fully stressed beam with constant width and varying height hx ϭ height at distance x hB ϭ height at end B b ϭ width (constant) AT DISTANCE x: M ϭ qx2 2 3q B bsallow AT THE FIXED END (x ϭ L): hB ϭ L Sϭ bh2 x 6 Therefore, M 3qx2 ϭ 2 S bhx hx x ϭ hB L hx ϭ hB x L 3q hx ϭ x B bsallow sallow ϭ Problem 5.7-7 A simple beam ABC having rectangular cross sections with constant height h and varying width bx supports a concentrated load P acting at the midpoint (see figure). How should the width bx vary as a function of x in order to have a fully stressed beam? (Express bx in terms of the width bB at the midpoint of the beam.) P A h B C x L — 2 L — 2 h h bx bB
    • SECTION 5.7 Solution 5.7-7 327 Fully Stressed Beams Fully stressed beam with constant height and varying width h ϭ height of beam (constant) L bx ϭ width at distance x from end A ¢ 0 Յ x Յ ≤ 2 bB ϭ width at midpoint B (x ϭ L/2) Mϭ AT DISTANCE x M 3Px sallow ϭ ϭ S bxh2 Px 2 Sϭ 1 b h2 6 x 3Px bx ϭ sallow h2 AT MIDPOINT B (x ϭ L/2) bB ϭ 3PL 2sallowh2 Therefore, bx 2x 2bB x ϭ and bx ϭ bb L L L NOTE: The equation is valid for 0 Յ x Յ and the 2 beam is symmetrical about the midpoint. Problem 5.7-8 A cantilever beam AB having rectangular cross sections with varying width bx and varying height hx is subjected to a uniform load of intensity q (see figure). If the width varies linearly with x according to the equation bx ϭ bB x/L, how should the height hx vary as a function of x in order to have a fully stressed beam? (Express hx in terms of the height hB at the fixed end of the beam.) q B hB hx A x L hx hB bx bB Solution 5.7-8 hx ϭ hB ϭ bx ϭ bB ϭ Fully stressed beam with varying width and varying height height at distance x height at end B width at distance x width at end B bx ϭ bB ¢ x ≤ L qx 2 2 hx ϭ Sϭ 3qL x B bBsallow sallow ϭ hB ϭ Therefore, AT DISTANCE x Mϭ 3qL2 B bBsallow AT THE FIXED END (x ϭ L) bx h2 bB x x ϭ (h ) 2 6 6L x M 3qL x ϭ S bBh2 x hx x ϭ hB B L hx ϭ hB x BL
    • 328 CHAPTER 5 Stresses in Beams (Basic Topics) Shear Stresses in Rectangular Beams Problem 5.8-1 Eq. (5-39): The shear stresses ␶ in a rectangular beam are given by V h2 ¢ Ϫ y2 ≤ 1 2I 4 in which V is the shear force, I is the moment of inertia of the cross-sectional area, h is the height of the beam, and y1 is the distance from the neutral axis to the point where the shear stress is being determined (Fig. 5-30). By integrating over the cross-sectional area, show that the resultant of the shear stresses is equal to the shear force V. tϭ Solution 5.8-1 Resultant of the shear stresses h 2 dy1 y1 ␶ N.A. V ϭ shear force acting on the cross section R ϭ resultant of shear stresses ␶ Rϭ Ύ hր2 Ϫhր2 h 2 V ϭ tϭ 12V (b) bh3 Ύ hր2 0 ¢ 0 V h2 2 ¢ Ϫ y1 ≤ bdy1 2I 4 h Ϫ y2 ≤ dy1 1 4 2 12V 2h3 ϭ 3 ¢ ≤ϭV 24 h b Iϭ Ύ tbdy1 ϭ 2 hր2 bh3 12 ІRϭ V V h2 ¢ Ϫ y2 ≤ 1 2I 4 Q.E.D. Problem 5.8-2 Calculate the maximum shear stress ␶max and the maximum bending stress ␴max in a simply supported wood beam (see figure) carrying a uniform load of 18.0 kN/m (which includes the weight of the beam) if the length is 1.75 m and the cross section is rectangular with width 150 mm and height 250 mm. 18.0 kN/m 250 mm 1.75 m Solution 5.8-2 150 mm Wood beam with a uniform load q ϭ 18 kN/m h ϭ 250 mm L ϭ 1.75 m MAXIMUM SHEAR STRESS qL Vϭ A ϭ bh 2 tmax ϭ 3V 3qL 3(18 kNրm)(1.75 m) ϭ ϭ 2A 4bh 4(150 mm)(250 mm) ϭ 630 kPa b ϭ 150 mm MAXIMUM BENDING STRESS Mϭ qL2 8 smax ϭ Sϭ bh2 6 M 3qL2 3(18 kNրm)(1.75 m) 2 ϭ ϭ S 4bh2 4(150 mm)(250 mm) 2 ϭ 4.41 MPa
    • SECTION 5.8 Problem 5.8-3 Two wood beams, each of square cross section (3.5 in. ϫ 3.5 in., actual dimensions) are glued together to form a solid beam of dimensions 3.5 in. ϫ 7.0 in. (see figure). The beam is simply supported with a span of 6 ft. What is the maximum load Pmax that may act at the midpoint if the allowable shear stress in the glued joint is 200 psi? (Include the effects of the beam’s own weight, assuming that the wood weighs 35 lb/ft3.) 329 Shear Stresses in Rectangular Beams 3.5 in. P 7.0 in. 6 ft Solution 5.8-3 Simple beam with a glued joint P q h/2 h/2 b L/2 L ϭ 6 ft ϭ 72 in. b ϭ 3.5 in. h ϭ 7.0 in. tallow ϭ 200 psi g ϭ (35 lbրft3 ) ¢ 1 ft 35 lbրin.3 3≤ ϭ 1728 in. 1728 q ϭ weight of beam per unit distance ϭ ␥bh L/2 SUBSTITUTE NUMERICAL VALUES: Pmax ϭ (3.5 in.) (7.0 in.) 3 MAXIMUM LOAD Pmax P qL Vϭ ϩ A ϭ bh 2 2 4 35 ϫ B (200 psi) Ϫ ¢ lbրin.3 ≤ (72 in.) R 3 1728 ϭ 6500 lb (This result is based solely on the shear stress.) P qL 3¢ ϩ ≤ 3V 2 2 3 tmax ϭ ϭ ϭ (P ϩ qL) 2A 2bh 4bh Pmax ϭ 4 4 bht Ϫ qL ϭ bht Ϫ gbhL 3 3 4 ϭ bh ¢ t Ϫ gL ≤ 3 Problem 5.8-4 A cantilever beam of length L ϭ 2 m supports a load P ϭ 8.0 kN (see figure). The beam is made of wood with cross-sectional dimensions 120 mm ϫ 200 mm. Calculate the shear stresses due to the load P at points located 25 mm, 50 mm, 75 mm, and 100 mm from the top surface of the beam. From these results, plot a graph showing the distribution of shear stresses from top to bottom of the beam. P = 8.0 kN 200 mm L=2m 120 mm
    • 330 CHAPTER 5 Stresses in Beams (Basic Topics) Solution 5.8-4 Shear stresses in a cantilever beam P = 8.0 kN h = 200 mm L=2m Eq. (5-39): t ϭ V h2 2 ¢ Ϫ y1 ≤ 2I 4 V ϭ P ϭ 8.0 kN ϭ 8,000 N h ϭ 200 mm ( y1 ϭ mm) b = 120 mm Iϭ (200) 2 8,000 tϭ B Ϫ y2 R 1 4 2(80 ϫ 106 ) t ϭ 50 ϫ 10Ϫ6 (10,000 Ϫ y2 ) 1 bh3 ϭ 80 ϫ 106 mm4 12 Distance from the top surface (mm) 0 25 50 75 100 (N.A.) y1 (mm) 100 75 50 25 0 ␶ (MPa) 0 0.219 0.375 0.469 0.500 ␶ (kPa) 0 219 375 469 500 GRAPH OF SHEAR STRESS ␶ 0 219 375 469 Tmax = 500 kPa 469 375 219 (t ϭ Nրmm ϭ MPa) 2 N.A. (y1 ϭ mm; t ϭ MPa) 0 Problem 5.8-5 A steel beam of length L ϭ 16 in. and cross-sectional dimensions b ϭ 0.6 in. and h ϭ 2 in. (see figure) supports a uniform load of intensity q ϭ 240 lb/in., which includes the weight of the beam. Calculate the shear stresses in the beam (at the cross section of maximum shear force) at points located 1/4 in., 1/2 in., 3/4 in., and 1 in. from the top surface of the beam. From these calculations, plot a graph showing the distribution of shear stresses from top to bottom of the beam. Solution 5.8-5 q = 240 lb/in. h = 2 in. L = 16 in. Shear stresses in a simple beam q = 240 lb/in. h = 2.0 in. b = 0.6 in. L = 16 in. Eq. (5-39): t ϭ V h2 2 ¢ Ϫ y1 ≤ 2I 4 qL Vϭ ϭ 1920 lb 2 Distance from the top surface (in.) 0 0.25 0.50 0.75 1.00 (N.A.) y1 (in.) 1.00 0.75 0.50 0.25 0 GRAPH OF SHEAR STRESS ␶ 3 Iϭ bh ϭ 0.4 in.4 12 0 1050 1800 2250 Tmax = 2400 psi 2250 1800 1050 UNITS: pounds and inches tϭ b = 0.6 in. 1920 (2) 2 B Ϫ y2 R ϭ (2400)(1 Ϫ y2 ) 1 1 2(0.4) 4 N.A. (␶ ϭ psi; y1 ϭ in.) 0 ␶ (psi) 0 1050 1800 2250 2400
    • SECTION 5.8 331 Shear Stresses in Rectangular Beams Problem 5.8-6 A beam of rectangular cross section (width b and height h) supports a uniformly distributed load along its entire length L. The allowable stresses in bending and shear are ␴allow and ␶allow, respectively. (a) If the beam is simply supported, what is the span length L0 below which the shear stress governs the allowable load and above which the bending stress governs? (b) If the beam is supported as a cantilever, what is the length L0 below which the shear stress governs the allowable load and above which the bending stress governs? Solution 5.8-6 b ϭ width Beam of rectangular cross section h ϭ height UNIFORM LOAD L ϭ length q ϭ intensity of load ALLOWABLE STRESSES (b) CANTILEVER BEAM BENDING (a) SIMPLE BEAM Mmax ϭ qL2 2 smax ϭ ␴allow and ␶allow Mmax 3qL2 ϭ S bh2 BENDING qL2 Mmax ϭ 8 4sallow bh2 3L2 Vmax ϭ qL Mmax 3qL2 ϭ S 4bh2 qallow ϭ tmax ϭ (1) qallow ϭ sallow bh2 3L2 (3) A ϭ bh 3V 3qL ϭ 2A 2bh qallow ϭ 2tallow bh 3L (4) Equate (3) and (4) and solve for L0: SHEAR L0 ϭ qL Vmax ϭ A ϭ bh 2 3V 3qL tmax ϭ ϭ 2A 4bh qallow ϭ bh2 6 SHEAR bh2 Sϭ 6 smax ϭ Sϭ 4tallow bh 3L h sallow ¢ ≤ 2 tallow NOTE: If the actual length is less than L 0, the shear stress governs the design. If the length is greater than L 0, the bending stress governs. (2) Equate (1) and (2) and solve for L0: L0 ϭ h ¢ sallow ≤ tallow Problem 5.8-7 A laminated wood beam on simple supports is built up by gluing together three 2 in. ϫ 4 in. boards (actual dimensions) to form a solid beam 4 in. ϫ 6 in. in cross section, as shown in the figure. The allowable shear stress in the glued joints is 65 psi and the allowable bending stress in the wood is 1800 psi. If the beam is 6 ft long, what is the allowable load P acting at the midpoint of the beam? (Disregard the weight of the beam.) 3 ft P 2 in. 2 in. 2 in. L ϭ 6 ft 4 in.
    • 332 CHAPTER 5 Solution 5.8-7 Stresses in Beams (Basic Topics) Laminated wood beam on simple supports 2 in. 2 in. N.A. 2 in. 2 in. 4 in. L ϭ 6 ft ϭ 72 in. ␶allow ϭ 65 psi ␴allow ϭ 1800 psi ALLOWABLE LOAD BASED UPON BENDING STRESS ALLOWABLE LOAD BASED UPON SHEAR STRESS IN THE GLUED JOINTS tϭ VQ Ib Vϭ P 2 tϭ (Pր2)(16 in.3 ) P ϭ (72 in.4 )(4 in.) 36 Q ϭ (4 in.)(2 in.)(2 in.) ϭ 16 in.3 Iϭ bh3 1 ϭ (4 in.)(6 in.) 3 ϭ 72 in.4 12 12 (P ϭ lb; t ϭ psi) P1 ϭ 36tallow ϭ 36 (65 psi) ϭ 2340 lb M S Sϭ bh2 1 ϭ (4 in.)(6 in.) 2 ϭ 24 in.3 6 6 sϭ (18P lb-in.) 3P ϭ 4 24 in.3 P2 ϭ 4 4 s ϭ (1800 psi) ϭ 2400 lb 3 allow 3 Mϭ (P ϭ lb; s ϭ psi) ALLOWABLE LOAD Shear stress in the glued joints governs. Pallow ϭ 2340 lb Problem 5.8-8 A laminated plastic beam of square cross section is built up by gluing together three strips, each 10 mm ϫ 30 mm in cross section (see figure). The beam has a total weight of 3.2 N and is simply supported with span length L ϭ 320 mm. Considering the weight of the beam, calculate the maximum permissible load P that may be placed at the midpoint if (a) the allowable shear stress in the glued joints is 0.3 MPa, and (b) the allowable bending stress in the plastic is 8 MPa. Solution 5.8-8 PL 72 in. ϭ P¢ ≤ ϭ 18P (lb-in.) 4 4 sϭ P q 10 mm 10 mm 30 mm 10 mm L 30 mm Laminated plastic beam P q 10 mm 10 mm h = 30 mm 10 mm 10 mm N.A. L/2 L/2 b = 30 mm L ϭ 320 mm W ϭ 3.2 N W 3.2 N qϭ ϭ ϭ 10 Nրm L 320 mm Iϭ bh3 1 ϭ (30 mm)(30 mm) 3 ϭ 67,500 mm4 12 12 Sϭ bh2 1 ϭ (30 mm)(30 mm) 2 ϭ 4500 mm3 6 6
    • SECTION 5.8 (a) ALLOWABLE LOAD BASED UPON SHEAR ␶allow ϭ 0.3 MPa sϭ VQ P qL P tϭ Vϭ ϩ ϭ ϩ 1.6 N Ib 2 2 2 (V ϭ newtons; P ϭ newtons) Mmax S PL qL2 ϩ ϭ 0.08P ϩ 0.128 (N ؒ m) 4 8 (P ϭ newtons; M ϭ N ؒ m) Mmax ϭ Q ϭ (30 mm)(10 mm)(10 mm) ϭ 3000 mm3 Q 3000 mm3 1 ϭ ϭ 4 Ib (67,500 mm )(30 mm) 675 mm2 VQ Pր2 ϩ 1.6 N ϭ Ib 675 mm2 333 (b) ALLOWABLE LOAD BASED UPON BENDING STRESSES ␴allow ϭ 8 MPa IN GLUED JOINTS tϭ Shear Stresses in Rectangular Beams (␶ ϭ N/mm2 ϭ MPa) SOLVE FOR P: P ϭ 1350␶allow Ϫ 3.2 ϭ 405 N Ϫ 3.2 N ϭ 402 N sϭ ˇ ˇ (0.08P ϩ 0.128)(N ؒ m) 4.5 ϫ 10Ϫ6 m3 ˇ ˇ (␴ ϭ N/m2 ϭ Pa) SOLVE FOR P: P ϭ (56.25 ϫ 10Ϫ6) ␴allowϪ1.6 ϭ (56.25 ϫ 10Ϫ6)(8ϫ106 Pa) Ϫ 1.6 ϭ 450 Ϫ 1.6 ϭ 448 N Problem 5.8-9 A wood beam AB on simple supports with span length equal to 9 ft is subjected to a uniform load of intensity 120 lb/ft acting along the entire length of the beam and a concentrated load of magnitude 8800 lb acting at a point 3 ft from the right-hand support (see figure). The allowable stresses in bending and shear, respectively, are 2500 psi and 150 psi. (a) From the table in Appendix F, select the lightest beam that will support the loads (disregard the weight of the beam). (b) Taking into account the weight of the beam (weight density ϭ 35 lb/ft3), verify that the selected beam is satisfactory, or, if it is not, select a new beam. 8800 lb 3 ft 120 lb/ft A B 9 ft Solution 5.8-9 (a) DISREGARDING THE WEIGHT OF THE BEAM P d q 3473 V (lb) A B 0 L = 9 ft RA RA ϭ qL P ϩ 2 3 ᎐ 6047 RB q ϭ 120 lb/ft P ϭ 8800 lb d ϭ 3 ft ␴allow ϭ 2500 psi ␶allow ϭ 150 psi 2753 RA ϭ ᎐ 6407 (120 lbրft)(9 ft) 8800 lb ϩ ϭ 3473 lb 2 3 2 RB ϭ 540 lb ϩ (8800 lb) ϭ 6407 lb 3 RB ϭ qL 2P ϩ 2 3 Vmax ϭ RB ϭ 6407 lb (Continued)
    • 334 CHAPTER 5 Stresses in Beams (Basic Topics) Maximum bending moment occurs under the concentrated load. Mmax ϭ RB d Ϫ (b) CONSIDERING THE WEIGHT OF THE BEAM qBEAM ϭ 17.3 lb/ft (weight density ϭ 35 lb/ft 3) qd 2 2 RB ϭ 6407 lb ϩ 1 ϭ (6407 lb)(3 ft) Ϫ (120 lbրft)(3 ft) 2 2 Vmax ϭ 6485 lb ϭ 18,680 lb-ft ϭ 224,200 lb-in. 3Vmax 3(6407 lb) 3V tmax ϭ Areq ϭ ϭ ϭ 64.1 in.2 2A 2tallow 2(150 psi) (17.3 lbրft)(9 ft) ϭ 6407 ϩ 78 ϭ 6485 lb 2 Areq’d ϭ 3Vmax ϭ 64.9 in.2 2tallow 8 ϫ 10 beam is still satisfactory for shear. qTOTAL ϭ 120 lb/ft ϩ 17.3 lb/ft ϭ 137.3 lb/ft FROM APPENDIX F: Select 8 ϫ 10 in. beam (nominal dimensions) qd 2 1 lb ϭ (6485 lb)(3 ft) Ϫ ¢ 137.3 ≤ (3 ft) 2 2 2 ft ϭ 18,837 lb-ft ϭ 226,050 lb-in. Mmax 226,050 lb-in. Sreq’d ϭ ϭ ϭ 90.4 in.3 sallow 2500 psi A ϭ 71.25 in.2 8 ϫ 10 beam is still satisfactory for moment. M sϭ S Mmax ϭ RB d Ϫ Mmax 224,200 lb-in. Sreq ϭ ϭ ϭ 89.7 in.3 sallow 2500 psi S ϭ 112.8 in.3 Use 8 ϫ 10 in. beam Problem 5.8-10 A simply supported wood beam of rectangular cross section and span length 1.2 m carries a concentrated load P at midspan in addition to its own weight (see figure). The cross section has width 140 mm and height 240 mm. The weight density of the wood is 5.4 kN/m3. Calculate the maximum permissible value of the load P if (a) the allowable bending stress is 8.5 MPa, and (b) the allowable shear stress is 0.8 MPa. Solution 5.8-10 P 240 mm 0.6 m 0.6 m 140 mm Simply supported wood beam P q h b L/2 L/2 b ϭ 140 mm h ϭ 240 mm A ϭ bh ϭ 33,600 mm2 2 bh Sϭ ϭ 1344 ϫ 103 mm3 6 ␥ ϭ 5.4 kN/m3 L ϭ 1.2 m q ϭ ␥bh ϭ 181.44 N/m (a) ALLOWABLE LOAD P BASED UPON BENDING STRESS Mmax ␴allow ϭ 8.5 MPa s ϭ S PL qL2 P(1.2 m) (181.44 Nրm)(1.2 m) 2 Mmax ϭ ϩ ϭ ϩ 4 8 4 8 ϭ 0.3P ϩ 32.66 N ؒ m (P ϭ newtons; M ϭ N ؒ m) Mmax ϭ S␴allow ϭ (1344ϫ103 mm3 )(8.5 MPa) ϭ 11,424 N ؒ m Equate values of Mmax and solve for P: 0.3P ϩ 32.66 ϭ 11,424 P ϭ 37,970 N or P ϭ 38.0 kN (b) ALLOWABLE LOAD P BASED UPON SHEAR STRESS 3V 2A P qL P (181.44 Nրm)(1.2 m) ϭ ϩ Vϭ ϩ 2 2 2 2 P ϭ ϩ 108.86 (N) 2 2At 2 ϭ (33,600 mm2 )(0.8 MPa) ϭ 17,920 N Vϭ 3 3 Equate values of V and solve for P: P ϩ 108.86 ϭ 17,920 P ϭ 35,622 N 2 or P ϭ 35.6 kN ␶allow ϭ 0.8 MPa t ϭ NOTE: The shear stress governs and Pallow ϭ 35.6 kN
    • SECTION 5.8 Problem 5.8-11 A square wood platform, 8 ft ϫ 8 ft in area, rests on masonry walls (see figure). The deck of the platform is constructed of 2 in. nominal thickness tongue-and-groove planks (actual thickness 1.5 in.; see Appendix F) supported on two 8-ft long beams. The beams have 4 in. ϫ 6 in. nominal dimensions (actual dimensions 3.5 in. ϫ 5.5 in.). The planks are designed to support a uniformly distributed load w (lb/ft 2) acting over the entire top surface of the platform. The allowable bending stress for the planks is 2400 psi and the allowable shear stress is 100 psi. When analyzing the planks, disregard their weights and assume that their reactions are uniformly distributed over the top surfaces of the supporting beams. (a) Determine the allowable platform load w1 (lb/ft 2) based upon the bending stress in the planks. (b) Determine the allowable platform load w2 (lb/ft 2) based upon the shear stress in the planks. (c) Which of the preceding values becomes the allowable load wallow on the platform? (Hints: Use care in constructing the loading diagram for the planks, noting especially that the reactions are distributed loads instead of concentrated loads. Also, note that the maximum shear forces occur at the inside faces of the supporting beams.) Solution 5.8-11 Shear Stresses in Rectangular Beams 8 ft 335 8 ft Bea m ll Wa Wood platform with a plank deck Free-body diagram of one plank supported on the beams: q Plank Plank 3.5 in. 3.5 in. 89 in. 3.5 in. 8 ft. (96 in.) qϭ B w (lbրft2 ) wb R (b in.) ϭ (lbրin.) 144 144 in.2րft2 Load on one plank: 8 ft 3.5 in. 96 in. wb wb ≤ϭ¢ ≤ (48) ϭ 2 144 3 2; b ϭ in.) (R ϭ lb; w ϭ lb/ft Mmax occurs at midspan. Reaction Platform: 8 ft ϫ 8 ft t ϭ thickness of planks ϭ 1.5 in. w ϭ uniform load on the deck (lb/ft 2) ␴allow ϭ 2400 psi ␶allow ϭ 100 psi Find wallow (lb/ft 2) (a) ALLOWABLE LOAD BASED UPON BENDING STRESS IN THE PLANKS Let b ϭ width of one plank (in.) 1.5 in. b A ϭ 1.5b (in.2) b S ϭ (1.5 in.) 2 6 ϭ 0.375b (in.3) Mmax ϭ R ¢ R ϭ q¢ q(48 in.) 2 3.5 in. 89 in. ϩ ≤Ϫ 2 2 2 wb wb 89 ϭ (46.25) Ϫ (1152) ϭ wb 3 144 12 (M ϭ lb-in.; w ϭ lb/ft 2; b ϭ in.) Allowable bending moment: Mallow ϭ ␴allow S ϭ (2400 psi)(0.375 b) ϭ 900b (lb-in.) EQUATE Mmax AND Mallow AND SOLVE FOR w: 89 wb ϭ 900 b w1 ϭ 121 lb/ft2 12 (Continued)
    • 336 CHAPTER 5 Stresses in Beams (Basic Topics) (b) ALLOWABLE LOAD BASED UPON SHEAR STRESS EQUATE Vmax AND Vallow AND SOLVE FOR w: 89wb ϭ 100b w2 ϭ 324 lb/ft2 288 IN THE PLANKS See the free-body diagram in part (a). Vmax occurs at the inside face of the support. 89 in. wb 89 wb Vmax ϭ q ¢ ≤ ϭ 44.5q ϭ (44.5) ¢ ≤ϭ 2 144 288 2; b ϭ in.) (V ϭ lb; w ϭ lb/ft (c) ALLOWABLE LOAD Bending stress governs. wallow ϭ 121 lb/ft2 Allowable shear force: 2Atallow 2(1.5b)(100 psi) 3V tϭ Vallow ϭ ϭ ϭ 100b (lb) 2A 3 3 Problem 5.8-12 A wood beam ABC with simple supports at A and B and an overhang BC has height h ϭ 280 mm (see figure). The length of the main span of the beam is L ϭ 3.6 m and the length of the overhang is L/3 ϭ 1.2 m. The beam supports a concentrated load 3P ϭ 15 kN at the midpoint of the main span and a load P ϭ 5 kN at the free end of the overhang. The wood has weight density ␥ ϭ 5.5 kN/m3. (a) Determine the required width b of the beam based upon an allowable bending stress of 8.2 MPa. (b) Determine the required width based upon an allowable shear stress of 0.7 MPa. Solution 5.8-12 3P L — 2 A P L Rectangular beam with an overhang P 3P q A L — 2 L — 2 RA 7P qL ᎐ 18 6 RA V L — 3 b RB Pϩ qL 3 P 0 ᎐ 11P qL ᎐ 18 6 h= 280 mm C B ᎐ 11P 5qL ᎐ 9 6 L ϭ 3.6 m P ϭ 5 kN ␥ ϭ 5.5 kN/m3 (for the wood) q ϭ ␥bh h= 280 mm C B L — 3 b
    • SECTION 5.8 Shear Stresses in Rectangular Beams Mmax M 0 MB FIND b 7P 4qL ϩ RA ϭ 6 9 RB ϭ EQUATE MOMENTS AND SOLVE FOR b: 10,500 ϩ 1940.4b ϭ 107,150b b ϭ 0.0998 m ϭ 99.8 mm 17P 8qL ϩ 6 9 (b) REQUIRED WIDTH b BASED UPON SHEAR STRESS 11P 5qL ϩ 6 9 Mmax ϭ 7PL 7qL2 ϩ 12 72 MB ϭ Ϫ PL qL2 Ϫ 3 18 (a) REQUIRED WIDTH b BASED UPON BENDING STRESS 7PL 7qL2 7 Mmax ϭ ϩ ϭ (5000 N)(3.6 m) 12 72 12 7 ϩ (gbh)(3.6 m) 2 72 7 ϭ 10,500 N ؒ m ϩ (5500 Nրm3 )(b) 72 ϫ (0.280 m)(3.6 m)2 ϭ 10,500 ϩ 1940.4b (b ϭ meters) (M ϭ newton-meters) ˇ sϭ ˇ Mmax 6Mmax ϭ S bh2 Mmax ϭ ␴allow ϭ 8.2 MPa bh2sallow b ϭ (0.280 m) 2 (8.2 ϫ 106 Pa) 6 6 ϭ 107,150b 11P 5qL ϩ 6 9 11 5 ϭ (5000 N) ϩ (gbh)(3.6 m) 6 9 5 ϭ 9167 N ϩ (5500 Nրm3 )(b)(0.280 m)(3.6 m) 9 ϭ 9167 ϩ 3080b (b ϭ meters) 3Vmax 3Vmax tϭ ϭ (V ϭ newtons) 2A 2bh ␶allow ϭ 0.7 MPa 2bhtallow 2b Vmax ϭ ϭ (0.280 m)(0.7 ϫ 106 Nրm2 ) 3 3 ϭ 130,670b Vmax ϭ Vmax ϭ EQUATE SHEAR FORCES AND SOLVE FOR b: 9167 ϩ 3080b ϭ 130,670b b ϭ 0.0718 m ϭ 71.8 mm NOTE: Bending stress governs. b ϭ 99.8 mm 337
    • 338 CHAPTER 5 Stresses in Beams (Basic Topics) Shear Stresses in Circular Beams P Problem 5.9-1 A wood pole of solid circular cross section (d ϭ diameter) is subjected to a horizontal force P ϭ 450 lb (see figure). The length of the pole is L ϭ 6 ft, and the allowable stresses in the wood are 1900 psi in bending and 120 psi in shear. Determine the minimum required diameter of the pole based upon (a) the allowable bending stress, and (b) the allowable shear stress. Solution 5.9-1 d L d Wood pole of circular cross section (a) BASED UPON BENDING STRESS Mmax ϭ PL ϭ (450 lb)(72 in.) ϭ 32,400 lb-in. P d sϭ L d P ϭ 450 lb L ϭ 6 ft ϭ 72 in. ␴allow ϭ 1900 psi ␶allow ϭ 120 psi Find diameter d M 32M ϭ S ␲d 3 d3 ϭ 32Mmax ϭ 173.7 in.3 ␲sallow dmin ϭ 5.58 in. (b) BASED UPON SHEAR STRESS Vmax ϭ 450 lb 16Vmax 4V 16V ϭ d2 ϭ ϭ 6.366 in.2 2 3A 3␲d 3␲tallow dmin ϭ 2.52 in. (Bending stress governs.) tϭ Problem 5.9-2 A simple log bridge in a remote area consists of two parallel logs with planks across them (see figure). The logs are Douglas fir with average diameter 300 mm. A truck moves slowly across the bridge, which spans 2.5 m. Assume that the weight of the truck is equally distributed between the two logs. Because the wheelbase of the truck is greater than 2.5 m, only one set of wheels is on the bridge at a time. Thus, the wheel load on one log is equivalent to a concentrated load W acting at any position along the span. In addition, the weight of one log and the planks it supports is equivalent to a uniform load of 850 N/m acting on the log. Determine the maximum permissible wheel load W based upon (a) an allowable bending stress of 7.0 MPa, and (b) an allowable shear stress of 0.75 MPa. x W 850 N/m 300 m 2.5 m
    • SECTION 5.9 Solution 5.9-2 339 Shear Stresses in Circular Beams Log bridge W x q = 850 N/m L = 2.5 m Diameter d ϭ 300 mm ␴allow ϭ 7.0 MPa ␶allow ϭ 0.75 MPa Find allowable load W (a) BASED UPON BENDING STRESS Maximum moment occurs when wheel is at midspan (x ϭ L/2). WL qL2 W 1 ϩ ϭ (2.5 m) ϩ (850 Nրm)(2.5 m) 2 4 8 4 8 ϭ 0.625W ϩ 664.1 (N ؒ m) (W ϭ newtons) Mmax ϭ ␲d 3 ϭ 2.651 ϫ 10Ϫ3 m3 32 Mmax ϭ S␴allow ϭ (2.651 ϫ 10Ϫ3 m3)(7.0 MPa) ϭ 18,560 N ؒ m І 0.625W ϩ 664.1 ϭ 18,560 W ϭ 28,600 N ϭ 28.6 kN Sϭ (b) BASED UPON SHEAR STRESS Maximum shear force occurs when wheel is adjacent to support (x ϭ 0). qL 1 Vmax ϭ W ϩ ϭ W ϩ (850 Nրm)(2.5 m) 2 2 ϭ W ϩ 1062.5 N (W ϭ newtons) 2 ␲d Aϭ ϭ 0.070686 m2 4 4Vmax tmax ϭ 3A 3A tallow 3 Vmax ϭ ϭ (0.070686 m2 )(0.75 MPa) 4 4 ϭ 39,760 N І W ϩ1062.5 N ϭ 39,760 N W ϭ 38,700 N ϭ 38.7 kN Problem 5.9-3 A sign for an automobile service station is supported by two aluminum poles of hollow circular cross section, as shown in the figure. The poles are being designed to resist a wind pressure of 75 lb/ft2 against the full area of the sign. The dimensions of the poles and sign are h1 ϭ 20 ft, h2 ϭ 5 ft, and b ϭ 10 ft. To prevent buckling of the walls of the poles, the thickness t is specified as one-tenth the outside diameter d. (a) Determine the minimum required diameter of the poles based upon an allowable bending stress of 7500 psi in the aluminum. (b) Determine the minimum required diameter based upon an allowable shear stress of 2000 psi. b h2 d t=— 10 Wind load d h1
    • 340 CHAPTER 5 Stresses in Beams (Basic Topics) Solution 5.9-3 Wind load on a sign W h2 = 5ft d h1 = 20 ft b ϭ width of sign b ϭ 10 ft P ϭ 75 lb/ft 2 ␴allow ϭ 7500 psi ␶allow ϭ 2000 psi d ϭ diameter W ϭ wind force on one pole d b tϭ W ϭ ph2 ¢ ≤ ϭ 1875 lb 10 2 (a) REQUIRED DIAMETER BASED UPON BENDING STRESS Mmax ϭ W ¢ h1 ϩ Iϭ Iϭ h2 ≤ ϭ 506,250 lb-in. 2 ␲ 4 (d Ϫ d 4 ) 2 64 2 d2 ϭ d 4 d1 ϭ d Ϫ 2t ϭ d 5 ␲ 4d 4 ␲d 4 369 369␲d 4 B d4 Ϫ ¢ ≤ R ϭ ¢ ≤ϭ (in.4 ) 64 5 64 625 40,000 d (d ϭ inches) 2 M(dր2) 17.253M Mc ϭ sϭ ϭ I 369␲d 4ր40,000 d3 17.253Mmax (17.253)(506,250 lb-in.) d3 ϭ ϭ sallow 7500 psi cϭ ϭ 1164.6 in.3 (b) REQUIRED DIAMETER BASED UPON SHEAR STRESS Vmax ϭ W ϭ 1875 lb 4V r 2 ϩ r2r1 ϩ r 2 d 2 1 ¢ ≤ r2 ϭ 2 2 3A 2 r2 ϩ r1 d d d 2d r1 ϭ Ϫ t ϭ Ϫ ϭ 2 2 10 5 d 2 d 2d 2d 2 ¢ ≤ ϩ ¢ ≤¢ ≤ϩ¢ ≤ r 2 ϩ r2r1 ϩ r 2 2 2 5 5 61 2 1 ϭ ϭ 41 r2 ϩ r2 d 2 2d 2 2 1 ¢ ≤ ϩ¢ ≤ 2 5 tϭ ␲ 2 ␲ 4d 2 9␲d 2 (d 2 Ϫ d 2 ) ϭ B d 2 Ϫ ¢ ≤ R ϭ 1 4 4 5 100 4V 61 100 V tϭ ¢ ≤¢ ≤ ϭ 7.0160 2 3 41 9␲d 2 d 7.0160 Vmax (7.0160)(1875 lb) d2 ϭ ϭ ϭ 6.5775 in.2 tallow 2000 psi Aϭ d ϭ 2.56 in. (Bending stress governs.) d ϭ 10.52 in. Problem 5.9-4 Solve the preceding problem for a sign and poles having the following dimensions: h1 ϭ 6.0 m, h2 ϭ 1.5 m, b ϭ 3.0 m, and t ϭ d/10. The design wind pressure is 3.6 kPa, and the allowable stresses in the aluminum are 50 MPa in bending and 14 MPa in shear.
    • SECTION 5.9 Solution 5.9-4 Shear Stresses in Circular Beams Wind load on a sign W h2 = 1.5 m d h1 = 6.0 m b ϭ width of sign b ϭ 3.0 m p ϭ 3.6 kPa ␴allow ϭ 50 MPa ␶allow ϭ 16 MPa d ϭ diameter W ϭ wind force on one pole d b tϭ W ϭ ph2 ¢ ≤ ϭ 8.1 kN 10 2 (a) REQUIRED DIAMETER BASED UPON BENDING STRESS h2 Mmax ϭ W ¢ h1 ϩ ≤ ϭ 54.675 kN ؒ m 2 Mc ␲ 4 sϭ I ϭ (d 4 Ϫ d 4 ) d2 ϭ d d1 ϭ d Ϫ 2t ϭ d 1 I 64 2 5 ˇ Iϭ ␲ 4d 4 ␲d 4 369 369␲d 4 Bd4 Ϫ ¢ ≤ R ϭ ¢ ≤ϭ (m4 ) 64 5 64 625 40,000 d3 ϭ 17.253Mmax (17.253)(54.675 kN ؒ m) ϭ sallow 50 MPa ϭ 0.018866 m3 d ϭ 0.266 m ϭ 266 mm SHEAR STRESS Vmax ϭ W ϭ 8.1 kN tϭ ˇ ˇ 4V r 2 ϩ r1r2 ϩ r 2 2 1 ¢ ≤ 3A r2 ϩ r2 2 1 r2 ϭ d 2 d d d 2d r1 ϭ Ϫ t ϭ Ϫ ϭ 2 2 10 5 r 2 ϩ r1r2 ϩ r 2 2 1 ϭ r2 ϩ r2 2 1 ˇ d (d ϭ meters) 2 M(dր2) Mc 17.253M sϭ ϭ ϭ I 369␲d 4ր40,000 d3 cϭ (b) REQUIRED DIAMETER BASED UPON Aϭ tϭ ¢ ≤ d 2 2 ϩ¢ d 2d 2d 2 ≤¢ ≤ϩ¢ ≤ 2 5 5 61 ϭ 41 d 2 2d 2 ¢ ≤ ϩ¢ ≤ 2 5 ␲ 2 ␲ 4d 2 9␲d 2 (d 2 ϩ d 2 ) ϭ B d 2 Ϫ ¢ ≤ R ϭ 1 4 4 5 100 4V 61 100 V ¢ ≤¢ ≤ ϭ 7.0160 2 3 41 9␲d 2 d d2 ϭ 7.0160 Vmax (7.0160)(8.1 kN) ϭ tallow 14 MPa ϭ 0.004059 m2 d ϭ 0.06371 m ϭ 63.7 mm (Bending stress governs) 341
    • 342 CHAPTER 5 Stresses in Beams (Basic Topics) Shear Stresses in the Webs of Beams with Flanges Problem 5.10-1 through 5.10-6 A wide-flange beam (see figure) having the cross section described below is subjected to a shear force V. Using the dimensions of the cross section, calculate the moment of inertia and then determine the following quantities: (a) The maximum shear stress ␶max in the web. (b) The minimum shear stress ␶min in the web. (c) The average shear stress ␶aver (obtained by dividing the shear force by the area of the web) and the ratio ␶max/␶aver. (d) The shear force Vweb carried in the web and the ratio Vweb /V. y z O h1 h t Note: Disregard the fillets at the junctions of the web and flanges and determine all quantities, including the moment of inertia, by considering the cross section to consist of three rectangles. b Probs. 5.10-1 through 5.10-6 Problem 5.10-1 Dimensions of cross section: b ϭ 6 in., t ϭ 0.5 in., h ϭ 12 in., h1 ϭ 10.5 in., and V ϭ 30 k. Solution 5.10-1 Wide-flange beam (b) MINIMUM SHEAR STRESS IN THE WEB (Eq. 5-48b) t h1 b h b ϭ 6.0 in. t ϭ 0.5 in. h ϭ 12.0 in. h1 ϭ 10.5 in. V ϭ 30 k MOMENT OF INERTIA (Eq. 5-47) 1 I ϭ (bh3 Ϫ bh3 ϩ th3 ) ϭ 333.4 in.4 1 1 12 (a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 5-48a) V tmax ϭ (bh2 Ϫ bh2 ϩ th2 ) ϭ 5795 psi 1 1 8It tmin ϭ Vb 2 (h Ϫ h2 ) ϭ 4555 psi 1 8It (c) AVERAGE SHEAR STRESS IN THE WEB (Eq. 5-50) V taver ϭ ϭ 5714 psi th1 tmax ϭ 1.014 taver (d) SHEAR FORCE IN THE WEB (Eq. 5-49) th1 Vweb ϭ (2tmax ϩ tmin ) ϭ 28.25 k 3 Vweb ϭ 0.942 V
    • SECTION 5.10 Shear Stresses in the Webs of Beams with Flanges Problem 5.10-2 Dimensions of cross section: b ϭ 180 mm, t ϭ 12 mm, h ϭ 420 mm, h1 ϭ 380 mm, and V ϭ 125 kN. Solution 5.10-2 Wide-flange beam t h1 h b b ϭ 180 mm t ϭ 12 mm h ϭ 420 mm h1 ϭ 380 mm V ϭ 125 kN MOMENT OF INERTIA (Eq. 5-47) 1 I ϭ (bh3 Ϫ bh3 ϩ th3 ) ϭ 343.1 ϫ 106 mm4 1 1 12 (a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 5-48a) V tmax ϭ (bh2 Ϫ bh2 ϩ th2 ) ϭ 28.43 MPa 1 1 8It Problem 5.10-3 E); V ϭ 10 k. t b (c) AVERAGE SHEAR STRESS IN THE WEB (Eq. 5-50) V taver ϭ ϭ 27.41 MPa th1 tmax ϭ 1.037 taver (d) SHEAR FORCE IN THE WEB (Eq. 5-49) th1 Vweb ϭ (2tmax ϩ tmin ) ϭ 119.7 kN 3 Vweb ϭ 0.957 V Wide-flange shape, W 8 ϫ 28 (see Table E-1, Appendix Solution 5.10-3 h1 (b) MINIMUM SHEAR STRESS IN THE WEB (Eq. 5-48b) Vb 2 tmin ϭ (h Ϫ h2 ) ϭ 21.86 MPa 1 8It h Wide-flange beam W 8 ϫ 28 b ϭ 6.535 in. t ϭ 0.285 in. h ϭ 8.06 in. h1 ϭ 7.13 in. V ϭ 10 k MOMENT OF INERTIA (Eq. 5-47) 1 I ϭ (bh3 Ϫ bh3 ϩ th3 ) ϭ 96.36 in.4 1 1 12 (a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 5-48a) V tmax ϭ (bh2 Ϫ bh2 ϩ th2 ) ϭ 4861 psi 1 1 8It (b) MINIMUM SHEAR STRESS IN THE WEB (Eq. 5-48b) Vb 2 tmin ϭ (h Ϫ h2 ) ϭ 4202 psi 1 8It (c) AVERAGE SHEAR STRESS IN THE WEB (Eq. 5-50) V taver ϭ ϭ 4921 psi th1 tmax ϭ 0.988 taver (d) Shear force in the web (Eq. 5-49) th1 Vweb ϭ (2tmax ϩ tmin ) ϭ 9.432 k 3 Vweb ϭ 0.943 V 343
    • 344 CHAPTER 5 Stresses in Beams (Basic Topics) Problem 5.10-4 Dimensions of cross section: b ϭ 220 mm, t ϭ 12 mm, h ϭ 600 mm, h1 ϭ 570 mm, and V ϭ 200 kN. Solution 5.10-4 Wide-flange beam t h1 h b b ϭ 220 mm t ϭ 12 mm h ϭ 600 mm h1 ϭ 570 mm V ϭ 200 kN MOMENT OF INERTIA (Eq. 5-47) 1 I ϭ (bh3 Ϫ bh3 ϩ th3 ) ϭ 750.0 ϫ 106 mm4 1 1 12 (a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 5-48a) V tmax ϭ (bh2 Ϫ bh2 ϩ th2 ) ϭ 32.28 MPa 1 1 8It (b) MINIMUM SHEAR STRESS IN THE WEB (Eq. 5-48b) Vb 2 tmin ϭ (h Ϫ h2 ) ϭ 21.45 MPa 1 8It (c) AVERAGE SHEAR STRESS IN THE WEB (Eq. 5-50) V taver ϭ ϭ 29.24 MPa th1 tmax ϭ 1.104 taver (d) SHEAR FORCE IN THE WEB (Eq. 5-49) th1 Vweb ϭ (2tmax ϩ tmin ) ϭ 196.1 kN 3 Vweb ϭ 0.981 V Problem 5.10-5 Wide-flange shape, W 18 ϫ 71 (see Table E-1, Appendix E); V ϭ 21 k. Solution 5.10-5 t h1 b h Wide-flange beam W 18 ϫ 71 b ϭ 7.635 in. t ϭ 0.495 in. h ϭ 18.47 in. h1 ϭ 16.85 in. V ϭ 21 k MOMENT OF INERTIA (Eq. 5-47) 1 I ϭ (bh3 Ϫ bh3 ϩ th3 ) ϭ 1162 in.4 1 1 12 (a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 5-48a) V tmax ϭ (bh2 Ϫ bh2 ϩ th2 ) ϭ 2634 psi 1 1 8It (b) MINIMUM SHEAR STRESS IN THE WEB (Eq. 5-48b) Vb 2 tmin ϭ (h Ϫ h2 ) ϭ 1993 psi 1 8It (c) AVERAGE SHEAR STRESS IN THE WEB (Eq. 5-50) V taver ϭ ϭ 2518 psi th1 tmax ϭ 1.046 taver (d) SHEAR FORCE IN THE WEB (Eq. 5-49) th1 Vweb ϭ (2tmax ϩ tmin ) ϭ 20.19 k 3 Vweb ϭ 0.961 V
    • SECTION 5.10 Shear Stresses in the Webs of Beams with Flanges Problem 5.10-6 Dimensions of cross section: b ϭ 120 mm, t ϭ 7 mm, h ϭ 350 mm, h1ϭ 330 mm, and V ϭ 60 kN. Solution 5.10-6 Wide-flange beam t h1 h b b ϭ 120 mm t ϭ 7 mm h ϭ 350 mm h1 ϭ 330 mm V ϭ 60 kN MOMENT OF INERTIA (Eq. 5-47) 1 Iϭ (bh3 Ϫ bh3 ϩ th3 ) ϭ 90.34 ϫ 106 mm4 1 1 12 (a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 5-48a) tmax ϭ V (bh2 Ϫ bh2 ϩ th2 ) ϭ 28.40 MPa 1 1 8It (b) MINIMUM SHEAR STRESS IN THE WEB (Eq. 5-48b) Vb 2 tmin ϭ (h Ϫ h2 ) ϭ 19.35 MPa 1 8It (c) AVERAGE SHEAR STRESS IN THE WEB (Eq. 5-50) V taver ϭ ϭ 25.97 MPa th1 tmax ϭ 1.093 taver (d) SHEAR FORCE IN THE WEB (Eq. 5-49) th1 Vweb ϭ (2tmax ϩ tmin ) ϭ 58.63 kN 3 Vweb ϭ 0.977 V Problem 5.10-7 A cantilever beam AB of length L ϭ 6.5 ft supports a uniform load of intensity q that includes the weight of the beam (see figure). The beam is a steel W 10 ϫ 12 wide-flange shape (see Table E-1, Appendix E). Calculate the maximum permissible load q based upon (a) an allowable bending stress ␴allow ϭ 16 ksi, and (b) an allowable shear stress ␶allow ϭ 8.5 ksi. (Note: Obtain the moment of inertia and section modulus of the beam from Table E-1.) Solution 5.10-7 B A W 10 ϫ 12 L = 6.5 ft Cantilever beam qmax ϭ W 10 ϫ 12 From Table E-1: t b ϭ 3.960 in. t ϭ 0.190 in. b h ϭ 9.87 in. h1 ϭ 9.87 in. Ϫ2(0.210 in.) ϭ 9.45 in. I ϭ 53.8 in.4 S ϭ 10.9 in.3 L ϭ 6.5 ft ϭ 78 in. ␴allow ϭ 16,000 psi ␶allow ϭ 8,500 psi h1 q h (a) MAXIMUM LOAD BASED UPON BENDING STRESS qL2 Mmax 2Ss Mmax ϭ sϭ qϭ 2 2 S L 2Ssallow 2(10.9 in.3 )(16,000 psi) ϭ L2 (78 in.) 2 ϭ 57.33 lb/in. ϭ 688 lb/ft (b) MAXIMUM LOAD BASED UPON SHEAR STRESS Vmax ϭ qL qmax ϭ tmax ϭ Vmax (bh2 Ϫ bh2 ϩ th2 ) 1 1 8It (Eq. 5-48a) 8It(tallow ) Vmax ϭ L L(bh2 Ϫ bh2 ϩ th2 ) 1 1 Substitute numerical values: qmax ϭ 181.49 lb/in. ϭ 2180 lb/ft NOTE: Bending stress governs. qallow ϭ 688 lb/ft 345
    • 346 CHAPTER 5 Stresses in Beams (Basic Topics) Problem 5.10-8 A bridge girder AB on a simple span of length L ϭ 14 m supports a uniform load of intensity q that includes the weight of the girder (see figure). The girder is constructed of three plates welded to form the cross section shown. Determine the maximum permissible load q based upon (a) an allowable bending stress ␴allow ϭ 110 MPa, and (b) an allowable shear stress ␶allow ϭ 50 MPa. 450 mm q 30 mm A B L = 14 m 15 mm 1800 mm 30 mm 450 mm Solution 5.10-8 Bridge girder (simple beam) h1 h t b L ϭ 14 m b ϭ 450 mm t ϭ 15 mm h ϭ 1860 mm h1 ϭ 1800 mm ␴allow ϭ 110 MPa ␶allow ϭ 50 MPa c ϭ h/2 ϭ 930 mm Eq. (5-47): Iϭ 1 (bh3 Ϫ bh3 ϩ th3 ) 1 1 12 ϭ 29.897 ϫ 109 mm4 I 29.897 ϫ 109 mm4 Sϭ ϭ ϭ 32.147 ϫ 106 mm3 c 930 mm (a) MAXIMUM LOAD BASED UPON BENDING STRESS qL2 Mmax 8Ss Mmax ϭ sϭ qϭ 2 8 S L 8Ssallow 8(32.147 ϫ 106 mm3 )(110 MPa) qmax ϭ ϭ L2 (14 m) 2 3 N/m ϭ 144 kN/m ϭ 144.3 ϫ 10 (b) MAXIMUM LOAD BASED UPON SHEAR STRESS qL Vmax tmax ϭ (bh2 Ϫ bh2 ϩ th2 ) 1 1 2 8It 16It(tallow ) 2Vmax qmax ϭ ϭ L L(bh2 Ϫ bh2 ϩ th2 ) 1 1 Vmax ϭ (Eq. 5-48a) Substitute numerical values: qmax ϭ 173.8 ϫ 103 N/m ϭ 174 kN/m NOTE: Bending stress governs. Problem 5.10-9 A simple beam with an overhang supports a uniform load of intensity q ϭ 1200 lb/ft and a concentrated load P ϭ 3000 lb (see figure). The uniform load includes an allowance for the weight of the beam. The allowable stresses in bending and shear are 18 ksi and 11 ksi, respectively. Select from Table E-2, Appendix E, the lightest I-beam (S shape) that will support the given loads. Hint: Select a beam based upon the bending stress and then calculate the maximum shear stress. If the beam is overstressed in shear, select a heavier beam and repeat. qallow ϭ 144 kN/m P = 3000 lb 8 ft q = 1200 lb/ft A C B 12 ft 4 ft
    • SECTION 5.10 Solution 5.10-9 Shear Stresses in the Webs of Beams with Flanges Beam with an overhang Maximum bending moment: Mmax ϭ 22,820 lb-ft at x ϭ 6.167 ft P = 3000 lb 8 ft q = 1200 lb/ft REQUIRED SECTION MODULUS A C B 4 ft 12 ft RA x RB ␴allow ϭ 18 ksi ␶allow ϭ 11 ksi Select a beam of S shape RA ϭ 7400 lb RB ϭ 14,800 lb Maximum shear force: Vmax ϭ 10,000 lb at x ϭ 12 ft Sϭ Mmax (22,820 lb-ft)(12 in.րft) ϭ ϭ 15.2 in.3 sallow 18,000 psi From Table E-2: Lightest beam is S 8 ϫ 23 S ϭ 16.2 in.3 I ϭ 64.9 in.4 b ϭ 4.171 in. t ϭ 0.441 in. h ϭ 8.00 in. h1 ϭ 8.00 Ϫ 2(0.426) ϭ 7.148 in. MAXIMUM SHEAR STRESS (Eq. 5-48a) Vmax 2 tmax ϭ (bh Ϫ bh2 ϩ th2 ) 1 1 8It ϭ 3340 psi 6 11,000 psi І ok for shear Select S 8 ϫ 23 beam Problem 5.10-10 A hollow steel box beam has the rectangular cross section shown in the figure. Determine the maximum allowable shear force V that may act on the beam if the allowable shear stress is 36 MPa. 20 mm 450 10 mm mm 10 mm 20 mm 200 mm Solution 5.10-10 Rectangular box beam ␶allow ϭ 36 MPa Find Vallow tϭ VQ It Vallow ϭ tallowIt Q 1 1 (200) (450) 3 Ϫ (180)(410) 3 ϭ 484.9 ϫ 106 mm4 12 12 t ϭ 2(10 mm) ϭ 20 mm Iϭ Q ϭ (200) ¢ 450 450 410 410 ≤¢ ≤ Ϫ (180) ¢ ≤¢ ≤ 2 4 2 4 ϭ 1.280 ϫ 106 mm3 Vallow ϭ tallowIt Q (36 MPa) (484.9 ϫ 106 mm4 )(20 mm) 1.280 ϫ 106 mm3 ϭ 273 kN ϭ 347
    • 348 CHAPTER 5 Stresses in Beams (Basic Topics) Problem 5.10-11 A hollow aluminum box beam has the square cross section shown in the figure. Calculate the maximum and minimum shear stresses ␶max and ␶min in the webs of the beam due to a shear force V ϭ 28 k. 1.0 in. 1.0 in. 12 in. Solution 5.10-11 Square box beam t1 A A t1 V ϭ 28 k ϭ 28,000 lb t1 ϭ 1.0 in. b ϭ 12 in. b1 b1 ϭ 10 in. b tϭ VQ It t ϭ 2t1 ϭ 2.0 in. MOMENT OF INERTIA Iϭ 1 4 (b Ϫ b4 ) ϭ 894.67 in.4 1 12 MAXIMUM SHEAR STRESS IN THE WEB (AT NEUTRAL AXIS) Q ϭ A1 y1 Ϫ A2 y2 A2 ϭ b1 ¢ y1 ϭ b1 ≤ϭ 2 2 1 b b ¢ ≤ϭ 2 2 4 b2 1 A1 ϭ b ¢ y2 ϭ b b2 ≤ϭ 2 2 Qϭ¢ b2 b1 b2 b 1 1 3 ≤ ¢ ≤ Ϫ ¢ ≤ ¢ ≤ ϭ (b3 Ϫ b1 ) ϭ 91.0 in.3 2 4 2 4 8 tmax ϭ VQ (28,000 lb)(91.0 in.3 ) ϭ ϭ 1424 psi It (894.67 in.4 )(2.0 in.) ϭ 1.42 ksi MINIMUM SHEAR STRESS IN THE WEB (AT LEVEL A-A) bt1 b t1 Q ϭ Ay ϭ (bt1 ) ¢ Ϫ ≤ ϭ (b Ϫ t1 ) 2 2 2 t1 ϭ b Ϫ b1 2 Qϭ (12 in.) [ (12 in.) 2 Ϫ (10 in.) 2 ] ϭ 66.0 in.3 8 tmin ϭ b Q ϭ (b2 Ϫ b2 ) 1 8 VQ (28,000 lb)(66.0 in.3 ) ϭ 1033 psi ϭ It (894.67 in.4 )(2.0 in.) ϭ 1.03 ksi b1 1 b1 ¢ ≤ϭ 2 2 4 Problem 5.10-12 The T-beam shown in the figure has cross-sectional dimensions as follows: b ϭ 220 mm, t ϭ 15 mm, h ϭ 300 mm, and h1 ϭ 275 mm. The beam is subjected to a shear force V ϭ 60 kN. Determine the maximum shear stress ␶max in the web of the beam. y t h1 z C c Probs. 5.10-12 and 5.10-13 b h
    • SECTION 5.10 Solution 5.10-12 b ϭ 220 mm h1 ϭ 275 mm T-beam h ϭ 300 mm t ϭ 15 mm V ϭ 60 kN MOMENT OF INERTIA ABOUT THE z-AXIS 1 1 Iweb ϭ (15)(223.2) 3 ϩ (15)(76.79 Ϫ 25) 3 3 3 ϭ 56.29 ϫ 106 mm4 Find ␶max 1 25 2 (220)(25) 3 ϩ (220)(25) ¢ 76.79 Ϫ ≤ 12 2 ϭ 23.02 ϫ 106 mm4 LOCATE NEUTRAL AXIS (ALL DIMENSIONS IN MILLIMETERS) gAy cϭ ϭ gA ϭ b(h Ϫ h1 ) ¢ (220)(25) ¢ 223.2 h Ϫ h1 h1 ≤ ϩ th1¢ h Ϫ ≤ 2 2 b(h Ϫ h1 ) ϩ th1 Iflange ϭ I ϭ Iweb ϩ Iflange ϭ 79.31 ϫ 106 mm4 25 275 ≤ ϩ (15)(275) ¢ 300 Ϫ ≤ 2 2 ϭ 76.79 mm (220)(25) ϩ (15)(275) y 1.5 275 z C 76.79 Shear Stresses in the Webs of Beams with Flanges 25 FIRST MOMENT OF AREA ABOVE THE z AXIS 223.2 Q ϭ (15)(223.2) ¢ ≤ 2 ϭ 373.6 ϫ 103 mm3 MAXIMUM SHEAR STRESS VQ (60 kN)(373.6 ϫ 103 mm3 ) tmax ϭ ϭ It (79.31 ϫ 106 mm4 )(15 mm) ϭ 18.8 MPa 220 Problem 5.10-13 Calculate the maximum shear stress ␶max in the web of the T-beam shown in the figure if b ϭ 10 in., t ϭ 0.6 in., h ϭ 8 in., h1 ϭ 7 in., and the shear force V ϭ 5000 lb. Solution 5.10-13 T-beam y t h1 z C c b h 6.317 z 1.683 (10)(1)(0.5) ϩ (0.6)(7)(4.5) ϭ 1.683 in. 10(1) ϩ (0.6)(7) MOMENT OF INERTIA ABOUT THE z-AXIS 1 1 Iweb ϭ (0.6)(6.317) 3 ϩ (0.6)(1.683 Ϫ 1.0) 3 3 3 ϭ 50.48 in.4 0.6 7 C 1.0 Find ␶max LOCATE NEUTRAL AXIS (ALL DIMENSIONS IN INCHES) h Ϫ h1 h1 b(h Ϫ h1 ) ¢ ≤ ϩ th1¢ h Ϫ ≤ gAy 2 2 cϭ ϭ gA b(h Ϫ h1 ) ϩ th1 ϭ y b ϭ 10 in. t ϭ 0.6 in. h ϭ 8 in. h1 ϭ 7 in. V ϭ 5000 lb 10 1 (10)(1.0) 3 ϩ (10)(1.0)(1.683 Ϫ 0.5) 2 12 ϭ 14.83 in.4 Iflange ϭ I ϭ Iweb ϩ Iflange ϭ 65.31 in4. FIRST MOMENT OF AREA ABOVE THE z AXIS 6.317 Q ϭ (0.6)(6.317) ¢ ≤ ϭ 11.97 in.3 2 MAXIMUM SHEAR STRESS VQ (5000 lb)(11.97 in.3 ) tmax ϭ ϭ ϭ 1530 psi It (65.31 in.4 )(0.6 in.) 349
    • 350 CHAPTER 5 Stresses in Beams Built-Up Beams y Problem 5.11-1 A prefabricated wood I-beam serving as a floor joist has the cross section shown in the figure. The allowable load in shear for the glued joints between the web and the flanges is 65 lb/in. in the longitudinal direction. Determine the maximum allowable shear force Vmax for the beam. 0.75 in. z 0.625 in. 8 in. O 0.75 in. 5 in. Solution 5.11-1 Wood I-beam All dimensions in inches. Find Vmax based upon shear in the glued joints. Allowable load in shear for the glued joints is 65 lb/in. y 0.75 in. І fallow ϭ 65 lb/in. h1 = 8 h = 9.5 fallowI VQ Vmax ϭ I Q bh3 (b Ϫ t)h3 1 1 1 Iϭ Ϫ ϭ (5)(9.5) 3 Ϫ (4.375)(8) 3 12 12 12 12 ϭ 170.57 in.4 0.75 in. Q ϭ Qflange ϭ Af df ϭ (5)(0.75)(4.375) ϭ 16.406 in.3 fϭ z O t = 0.625 in. b=5 Vmax ϭ fallowI (65 lbրin.)(170.57 in.4 ) ϭ ϭ 676 lb Q 16.406 in.3 Problem 5.11-2 A welded steel girder having the cross section shown in the figure is fabricated of two 280 mm ϫ 25 mm flange plates and a 600 mm ϫ 15 mm web plate. The plates are joined by four fillet welds that run continuously for the length of the girder. Each weld has an allowable load in shear of 900 kN/m. Calculate the maximum allowable shear force Vmax for the girder. y 25 mm z 15 mm O 280 mm 600 mm 25 mm
    • SECTION 5.11 Solution 5.11-2 Built-Up Beams Welded steel girder All dimensions in millimeters. Allowable load in shear for one weld is 900 kN/m. y І fallow ϭ 2(900) ϭ 1800 kN/m 25 fallowI VQ Vmax ϭ I Q 3 3 (b Ϫ t)h1 bh 1 1 Iϭ Ϫ ϭ (280)(650) 3 Ϫ (265)(600) 3 12 12 12 12 ϭ 1638 ϫ 106 mm4 fϭ weld z h1 = 600 O h= 650 t = 15 Q ϭ Qflange ϭ Af df ϭ (280)(25)(312.5) ϭ 2.1875 ϫ 106 mm3 b = 280 Vmax ϭ 25 fallowI (1800 kNրm)(1638 ϫ 106 mm4 ) ϭ Q 2.1875 ϫ 106 mm3 ϭ 1.35 MN Problem 5.11-3 A welded steel girder having the cross section shown in the figure is fabricated of two 18 in. ϫ 1 in. flange plates and a 64 in. ϫ 3/8 in. web plate. The plates are joined by four longitudinal fillet welds that run continuously throughout the length of the girder. If the girder is subjected to a shear force of 300 kips, what force F (per inch of length of weld) must be resisted by each weld? y 1 in. z O 64 in. 3 — in. 8 18 in. Solution 5.11-3 Welded steel girder All dimensions in inches. y V ϭ 300 k F ϭ force per inch of length of one weld VQ VQ f ϭ shear flow f ϭ 2F ϭ Fϭ I 2I 1.0 weld z O h1 = 64 t = 0.375 b = 18 1 in. h= 66 bh3 (b Ϫ t)h3 1 1 1 Ϫ ϭ (18)(66) 3 Ϫ (17.625)(64) 3 12 12 12 12 ϭ 46,220 in.4 Iϭ Q ϭ Qflange ϭ Af df ϭ (18)(1.0)(32.5) ϭ 585 in.3 1.0 Fϭ VQ (300 k)(585 in.3 ) ϭ ϭ 1900 lbրin. 2I 2(46,220 in.4 ) 351
    • 352 CHAPTER 5 Stresses in Beams Problem 5.11-4 A box beam of wood is constructed of two 260 mm ϫ 50 mm boards and two 260 mm ϫ 25 mm boards (see figure). The boards are nailed at a longitudinal spacing s ϭ 100 mm. If each nail has an allowable shear force F ϭ 1200 N, what is the maximum allowable shear force Vmax? y 25 mm z 260 mm O 50 mm 50 mm 25 mm 260 mm Solution 5.11-4 Wood box beam All dimensions in millimeters. b ϭ 260 b1 ϭ 260 Ϫ 2(50) ϭ 160 h ϭ 310 h1 ϭ 260 s ϭ nail spacing ϭ 100 mm F ϭ allowable shear force for one nail ϭ 1200 N f ϭ shear flow between one flange and both webs 2F 2(1200 N) ϭ ϭ 24 kNրm s 100 mm fallowI VQ fϭ Vmax ϭ I Q fallow ϭ 1 (bh3 Ϫ b1h3 ) ϭ 411.125 ϫ 106 mm4 1 12 Q ϭ Qflange ϭ Af df ϭ (260)(25)(142.5) ϭ 926.25 ϫ 103 mm3 Iϭ Vmax ϭ fallowI (24 kNրm)(411.125 ϫ 106 mm4 ) ϭ Q 926.25 ϫ 103 mm3 ϭ 10.7 kN Problem 5.11-5 A box beam constructed of four wood boards of size 6 in. ϫ 1 in. (actual dimensions) is shown in the figure. The boards are joined by screws for which the allowable load in shear is F ϭ 250 lb per screw. Calculate the maximum permissible longitudinal spacing smax of the screws if the shear force V is 1200 lb. y 1 in. z 1 in. O 6 in. 1 in. 1 in. 6 in. Solution 5.11-5 Wood box beam All dimensions in inches. b ϭ 6.0 b1 ϭ 6.0 Ϫ 2(1.0) ϭ 4.0 h ϭ 8.0 h1 ϭ 6.0 F ϭ allowable shear force for one screw ϭ 250 lb V ϭ shear force ϭ 1200 lb s ϭ longitudinal spacing of the screws f ϭ shear flow between one flange and both webs VQ 2F 2FI ϭ ∴ smax ϭ s I VQ 1 I ϭ (bh3 Ϫ b1h3 ) ϭ 184 in.4 1 12 fϭ Q ϭ Qflange ϭ Af df ϭ (6.0)(1.0)(3.5) ϭ 21 in.3 smax ϭ 2FI 2(250 lb)(184 in.4 ) ϭ VQ (1200 lb)(21 in.3 ) ϭ 3.65 in.
    • SECTION 5.11 Problem 5.11-6 Two wood box beams (beams A and B) have the same outside dimensions (200 mm ϫ 360 mm) and the same thickness (t ϭ 20 mm) throughout, as shown in the figure on the next page. Both beams are formed by nailing, with each nail having an allowable shear load of 250 N. The beams are designed for a shear force V ϭ 3.2 kN. (a) What is the maximum longitudinal spacing sA for the nails in beam A? (b) What is the maximum longitudinal spacing sB for the nails in beam B? (c) Which beam is more efficient in resisting the shear force? Solution 5.11-6 Two wood box beams Cross-sectional dimensions are the same. All dimensions in millimeters. b ϭ 200 b1 ϭ 200 Ϫ 2(20) ϭ 160 h ϭ 360 h1 ϭ 360 Ϫ 2(20) ϭ 320 t ϭ 20 F ϭ allowable load per nail ϭ 250 N V ϭ shear force ϭ 3.2 kN 1 I ϭ (bh3 Ϫ b1h3 ) ϭ 340.69 ϫ 106 mm4 1 12 s ϭ longitudinal spacing of the nails f ϭ shear flow between one flange and both webs 2F VQ fϭ ϭ s I ∴ smax ϭ 2FI VQ Built-Up Beams y y A z B 360 mm O z t= 20 mm O t= 20 mm 200 mm 200 mm (a) BEAM A Q ϭ Ap dp ϭ (bt) ¢ hϪt 1 ≤ ϭ (200)(20) ¢ ≤ (340) 2 2 ϭ 680 ϫ 103 mm3 sA ϭ 353 2FI (2)(250 N)(340.7 ϫ 106 mm4 ) ϭ VQ (3.2 kN)(680 ϫ 103 mm3 ) ϭ 78.3 mm (b) BEAM B Q ϭ Af df ϭ (b Ϫ 2t)(t) ¢ ϭ 544 ϫ 103 mm3 hϪt 1 ≤ ϭ (160)(20) (340) 2 2 2FI (2)(250 N)(340.7 ϫ 106 mm4 ) ϭ VQ (3.2 kN)(544 ϫ 103 mm3 ) ϭ 97.9 mm sB ϭ (c) BEAM B IS MORE EFFICIENT because the shear flow on the contact surfaces is smaller and therefore fewer nails are needed. 3 — in. 16 Problem 5.11-7 A hollow wood beam with plywood webs has the cross-sectional dimensions shown in the figure. The plywood is attached to the flanges by means of small nails. Each nail has an allowable load in shear of 30 lb. Find the maximum allowable spacing s of the nails at cross sections where the shear force V is equal to (a) 200 lb and (b) 300 lb. 3 — in. 16 3 in. y z 3 in. 4 O 8 in. 3 in. 4 360 mm
    • 354 CHAPTER 5 Stresses in Beams Solution 5.11-7 Wood beam with plywood webs All dimensions in inches. b ϭ 3.375 b1 ϭ 3.0 h ϭ 8.0 h1 ϭ 6.5 F ϭ allowable shear force for one nail ϭ 30 lb s ϭ longitudinal spacing of the nails f ϭ shear flow between one flange and both webs VQ 2F 2FI fϭ ϭ ∴ smax ϭ s I VQ Iϭ 1 (bh3 Ϫ b1h3 ) ϭ 75.3438 in.4 1 12 Q ϭ Qflange ϭ Af df ϭ (3.0)(0.75)(3.625) ϭ 8.1563 in.3 (a) V ϭ 200 lb smax ϭ 2FI 2(30 lb)(75.344 in.4 ) ϭ VQ (200 lb)(8.1563 in.3 ) ϭ 2.77 in. (b) V ϭ 300 lb By proportion, smax ϭ (2.77 in.) ¢ 200 ≤ ϭ 1.85 in. 300 y Problem 5.11-8 A beam of T cross section is formed by nailing together two boards having the dimensions shown in the figure. If the total shear force V acting on the cross section is 1600 N and each nail may carry 750 N in shear, what is the maximum allowable nail spacing s? 200 mm 50 mm z C 200 mm 50 mm Solution 5.11-8 T-beam (nailed) A ϭ bt ϩ h1t ϭ t(b ϩ h1) ϭ (50)(400) ϭ 20 ϫ 103 mm2 y b c1 z C h1 c2 Q BB 3.25 ϫ 106 mm3 ϭ ϭ 162.5 mm A 20 ϫ 103 mm2 c1 ϭ h Ϫ c2 ϭ 250 Ϫ 162.5 ϭ 87.5 mm t c2 ϭ h t B B All dimensions in millimeters. V ϭ 1600 N F ϭ allowable load per nail F ϭ 750 N b ϭ 200 mm t ϭ 50 mm h ϭ 250 mm h1 ϭ 200 mm s ϭ nail spacing Find smax LOCATION OF NEUTRAL AXIS (z AXIS) Use the lower edge of the cross section (line B-B) as a reference axis. Q BB ϭ (h 1t) ¢ h1 t ≤ ϩ (bt) ¢ h Ϫ ≤ 2 2 ϭ (200)(50)(100) ϩ (200)(50)(225) ϭ 3.25 ϫ 106 mm3 MOMENT OF INERTIA ABOUT THE NEUTRAL AXIS 1 1 1 t 2 I ϭ tc3 ϩ t(h1 Ϫ c2 ) 3 ϩ bt3 ϩ bt ¢ c1 Ϫ ≤ 2 3 3 12 2 1 1 1 3 3 ϭ (50)(162.5) ϩ (50)(37.5) ϩ (200)(50) 3 3 3 12 ϩ (200)(50)(62.5) 2 ϭ 113.541 ϫ 106 mm4 FIRST MOMENT OF AREA OF FLANGE Q ϭ bt ¢ c1 Ϫ t ≤ ϭ (200)(50)(62.5) ϭ 625 ϫ 103 mm3 2 MAXIMUM ALLOWABLE SPACING OF NAILS VQ F ϭ s I Fallow I (750 N)(113.541 ϫ 106 mm4 ) smax ϭ ϭ VQ (1600 N)(625 ϫ 103 mm3 ) ϭ 85.2 mm fϭ
    • SECTION 5.11 Problem 5.11-9 The T-beam shown in the figure is fabricated by welding together two steel plates. If the allowable load for each weld is 2.0 k/in. in the longitudinal direction, what is the maximum allowable shear force V? Built-Up Beams y 0.5 in. 6 in. z C 0.5 in. 5 in. Solution 5.11-9 T-beam (welded) A ϭ bt ϩ h1t ϭ (5)(0.5) ϩ (6)(0.5) ϭ 5.5 in.2 y c1 t h1 h z C MOMENT OF INERTIA ABOUT THE NEUTRAL AXIS c2 B b QBB 11.125 in.3 ϭ ϭ 2.0227 in. A 5.5 in.2 c1 ϭ h Ϫ c2 ϭ 4.4773 in. c2 ϭ Bt All dimensions in inches. 1 1 1 t 2 I ϭ tc3 ϩ t(c2 Ϫ t) 3 ϩ bt3 ϩ (bt) ¢ c2 Ϫ ≤ 3 1 3 12 2 1 1 1 ϭ (0.5)(4.4773) 3 ϩ (0.5)(1.5227) 3 ϩ (5)(0.5) 3 3 3 12 ϩ (5)(0.5)(1.7727) 2 ϭ 23.455 in.4 F ϭ allowable load per inch of weld F ϭ 2.0 k/in. b ϭ 5.0 t ϭ 0.5 h ϭ 6.5 h1 ϭ 6.0 V ϭ shear force Find Vmax FIRST MOMENT OF AREA OF FLANGE LOCATION OF NEUTRAL AXIS (z AXIS) SHEAR FLOW AT WELDS Use the lower edge of the cross section (line B-B) as a reference axis. Q BB ϭ (bt) ¢ h1 t ≤ ϩ (h 1t) ¢ h Ϫ ≤ 2 2 ϭ (5)(0.5)(0.25) ϩ (6)(0.5)(3.5) ϭ 11.25 in.3 Q ϭ bt ¢ c2 Ϫ f ϭ 2F ϭ t ≤ ϭ (5)(0.5)(1.7727) ϭ 4.4318 in.3 2 VQ I MAXIMUM ALLOWABLE SHEAR FORCE Vmax ϭ 2FI 2(2.0 kրin.)(23.455 in.4 ) ϭ ϭ 21.2 k Q 4.4318 in.3 Problem 5.11-10 A steel beam is built up from a W 16 ϫ 77 wideflange beam and two 10 in. ϫ 1/2 in. cover plates (see figure on the next page). The allowable load in shear on each bolt is 2.1 kips. What is the required bolt spacing s in the longitudinal direction if the shear force V ϭ 30 kips? (Note: Obtain the dimensions and moment of inertia of the W shape from Table E-1.) y z 1 10 in. ϫ — in. 2 cover plates W 16 ϫ 77 O 355
    • 356 CHAPTER 5 Solution 5.11-10 Stresses in Beams Beam with cover plates MOMENT OF INERTIA ABOUT THE NEUTRAL AXIS t = 0.5 d W 16 ϫ 77 N.A. I ϭ Ibeam ϩ 2 B ϭ 1110 in.4 ϩ 2 B ϭ 1834 in.4 t = 0.5 1 3 d t 2 bt ϩ (bt) ¢ ϩ ≤ R 12 2 2 1 (10)(0.5) 3 ϩ (10)(0.5)(8.51) 2 R 12 FIRST MOMENT OF AREA OF A COVER PLATE b =10 All dimensions in inches. Wide-flange beam (W 16 ϫ 77): d ϭ 16.52 in. Ibeam ϭ 1110 in.4 Cover plates: b ϭ 10 in. t ϭ 0.5 in. F ϭ allowable load per bolt ϭ 2.1 k V ϭ shear force ϭ 30 k s ϭ spacing of bolts in the longitudinal direction Find smax Q ϭ bt ¢ dϩt ≤ ϭ (10)(0.5)(8.51) ϭ 42.55 in.3 2 MAXIMUM SPACING OF BOLTS fϭ VQ 2F ϭ s I smax ϭ sϭ 2FI VQ 2(2.1 k)(1834 in.4 ) ϭ 6.03 in. (30 k)(42.55 in.3 ) Problem 5.11-11 Two W 10 ϫ 45 steel wide-flange beams are bolted together to form a built-up beam as shown in the figure. What is the maximum permissible bolt spacing s if the shear force V ϭ 20 kips and the allowable load in shear on each bolt is F ϭ 3.1 kips? (Note: Obtain the dimensions and properties of the W shapes from Table E-1.) W 10 ϫ 45 W 10 ϫ 45 Solution 5.11-11 Built-up steel beam All dimensions in inches. W 10 ϫ 45: I1 ϭ 248 in.4 A ϭ 13.3 in.2 V ϭ 20 k F ϭ 3.1 k FIRST MOMENT OF AREA OF ONE BEAM d ϭ 10.10 in. Find maximum allowable bolt spacing smax. Q ϭ A¢ d ≤ ϭ (13.3)(5.05) ϭ 67.165 in.3 2 MAXIMUM SPACING OF BOLTS IN THE LONGITUDINAL DIRECTION MOMENT OF INERTIA OF BUILT-UP BEAM d 2 I ϭ 2 B I1 ϩ A¢ ≤ R ϭ 2[248 ϩ (13.3)(5.05) 2 ] 2 ϭ 1174.4 in.4 fϭ VQ 2F ϭ s I smax ϭ sϭ 2FI VQ 2(3.1 k)(1174.4 in.4 ) ϭ 5.42 in. (20 k)(67.165 in.3 )
    • SECTION 5.12 Beams with Axial Loads Beams with Axial Loads When solving the problems for Section 5.12, assume that the bending moments are not affected by the presence of lateral deflections. P = 25 lb Problem 5.12-1 While drilling a hole with a brace and bit, you exert a downward force P ϭ 25 lb on the handle of the brace (see figure). The diameter of the crank arm is d ϭ 7/16 in. and its lateral offset is b ϭ 4-7/8 in. Determine the maximum tensile and compressive stresses ␴t and ␴c, respectively, in the crank. Solution 5.12-1 P M d 7 d = — in. 16 7 b = 4 — in. 8 Brace and bit P ϭ 25 lb (compression) M ϭ Pb ϭ (25 lb)(4 7/8 in.) ϭ 121.9 lb-in. d ϭ diameter d ϭ 7/16 in. ␲d 2 Aϭ ϭ 0.1503 in.2 4 ␲d 3 Sϭ ϭ 0.008221 in.3 32 MAXIMUM STRESSES P M 25 lb 121.9 lb-in. st ϭ Ϫ ϩ ϭ Ϫ 2ϩ A S 0.1503 in. 0.008221 in.3 ϭ Ϫ166 psi ϩ 14,828 psi ϭ 14,660 psi P M sc ϭ Ϫ Ϫ ϭ Ϫ166 psi Ϫ 14,828 psi A S ϭ Ϫ14,990 psi Problem 5.12-2 An aluminum pole for a street light weighs 4600 N and supports an arm that weighs 660 N (see figure). The center of gravity of the arm is 1.2 m from the axis of the pole. The outside diameter of the pole (at its base) is 225 mm and its thickness is 18 mm. Determine the maximum tensile and compressive stresses ␴t and ␴c, respectively, in the pole (at its base) due to the weights. W2 = 660 N 1.2 m W1 = 4600 N 18 mm 225 mm 357
    • 358 CHAPTER 5 Solution 5.12-2 Stresses in Beams Aluminum pole for a street light W1 ϭ weight of pole ϭ 4600 N W2 ϭ weight of arm ϭ 660 N b ϭ distance between axis of pole and center of gravity of arm ϭ 1.2 m d2 ϭ outer diameter of poleϭ 225 mm d1 ϭ inner diameter of pole ϭ 225 mm Ϫ 2(18 mm) ϭ 189 mm ␲ 2 2 (d Ϫ d1 ) ϭ 11,706 mm2 4 2 ␲ 4 4 I ϭ (d2 Ϫ d1 ) ϭ 63.17 ϫ 106 mm4 64 d2 c ϭ ϭ 112.5 mm 2 AT BASE OF POLE (792 N ؒ m)(112.5 mm) P Mc 5260 N st ϭ Ϫ ϩ ϭϪ 2ϩ A I 11,706 mm 63.17 ϫ 106 mm4 PROPERTIES OF THE CROSS SECTION Aϭ MAXIMUM STRESSES ˇ P ˇ ϭ Ϫ0.4493 MPa ϩ 1.4105 MPa M ϭ 0.961 MPa ϭ 961 kPa P Mc sc ϭ Ϫ Ϫ ϭ Ϫ0.4493 MPa Ϫ 1.4105 MPa A I ϭ Ϫ1.860 MPa ϭ Ϫ1860 kPa d2 P ϭ W1 ϩ W2 ϭ 5260 N M ϭ W2 b ϭ 792 N ؒ m Problem 5.12-3 A curved bar ABC having a circular axis (radius r ϭ 12 in.) is loaded by forces P ϭ 400 lb (see figure). The cross section of the bar is rectangular with height h and thickness t. If the allowable tensile stress in the bar is 12,000 psi and the height h ϭ 1.25 in., what is the minimum required thickness tmin? h B C A P P 45° 45° r h t Solution 5.12-3 Curved bar B TENSILE STRESS M A st ϭ P e P ϭ r ϭ radius of curved bar e ϭ r Ϫ r cos 45º 1 ϭ r ¢1 Ϫ ≤ ͙2 Pr M ϭ Pe ϭ (2 Ϫ ͙2) 2 t ϭ thickness P r B 1 ϩ 3(2 Ϫ ͙2) R ht h MINIMUM THICKNESS tmin ϭ P r B 1 ϩ 3(2 Ϫ ͙2) R hsallow h SUBSTITUTE NUMERICAL VALUES: CROSS SECTION h ϭ height P M P 3Pr (2 Ϫ ͙2) ϩ ϭ ϩ A S ht th2 A ϭ ht 1 S ϭ th2 6 P ϭ 400 lb ␴allow ϭ 12,000 psi r ϭ 12 in. h ϭ 1.25 in. tmin ϭ 0.477 in.
    • SECTION 5.12 Problem 5.12-4 A rigid frame ABC is formed by welding two steel pipes at B (see figure). Each pipe has cross-sectional area A ϭ 11.31 ϫ 103 mm2, moment of inertia I ϭ 46.37 ϫ 106 mm4, and outside diameter d ϭ 200 mm. Find the maximum tensile and compressive stresses ␴t and ␴c, respectively, in the frame due to the load P ϭ 8.0 kN if L ϭ H ϭ 1.4 m. B d d P H A C d L Solution 5.12-4 359 Beams with Axial Loads L Rigid frame N M AXIAL FORCE: N ϭ RA sin ␣ ϭ P sin ␣ 2 BENDING MOMENT: M ϭ RAL ϭ PL 2 B V A d ␣ TENSILE STRESS N Mc P sin ␣ PLd st ϭ Ϫ ϩ ϭϪ ϩ A I 2A 4I RA REACTIONS: RA ϭ RC ϭ BAR AB: H tan ␣ ϭ L sin ␣ ϭ P ϭ 8.0 kN L ϭ H ϭ 1.4 m ␣ ϭ 45º sin ␣ ϭ 1ր 12 d ϭ 200 mm A ϭ 11.31 ϫ 103 mm2 I ϭ 46.37 ϫ 106 mm4 SUBSTITUTE NUMERICAL VALUES: Load P at midpoint B P 2 st ϭ Ϫ H ͙H 2 ϩ L2 d ϭ diameter c ϭ d/2 (8.0 kN)(1ր ͙2) (8.0 kN)(1.4 m)(200 mm) 3 2 ϩ 2(11.31 ϫ 10 mm ) 4(46.37 ϫ 106 mm4 ) ϭ Ϫ0.250 MPa ϩ 12.08 MPa ϭ 11.83 MPa (tension) N Mc sc ϭ Ϫ Ϫ ϭ Ϫ0.250 MPa Ϫ 12.08 MPa A I ϭ Ϫ12.33 MPa (compression) Problem 5.12-5 A palm tree weighing 1000 lb is inclined at an angle of 60° (see figure). The weight of the tree may be resolved into two resultant forces, a force P1 ϭ 900 lb acting at a point 12 ft from the base and a force P2 ϭ 100 lb acting at the top of the tree, which is 30 ft long. The diameter at the base of the tree is 14 in. Calculate the maximum tensile and compressive stresses ␴t and ␴c, respectively, at the base of the tree due to its weight. P2 = 100 lb 30 ft 12 ft P1 = 900 lb 60°
    • 360 CHAPTER 5 Stresses in Beams Solution 5.12-5 Palm tree M ϭ P1L1 cos 60º ϩ P2L2 cos 60º ϭ [(900 lb)(144 in.) ϩ (100 lb)(360 in.)] cos 60º ϭ 82,800 lb-in. N ϭ (P1 ϩ P2) sin 60º ϭ (1000 lb) sin 60º ϭ 866 lb P2 L2 L1 ␤ P1 FREE-BODY DIAGRAM P1 ϭ 900 lb P2 ϭ 100 lb L1 ϭ 12 ft ϭ 144 in. L2 ϭ 30 ft ϭ 360 in. d ϭ 14 in. V Aϭ ␲d 2 ϭ 153.94 in.2 4 Sϭ N M 866 lb 82,800 lb-in. st ϭ Ϫ ϩ ϭ Ϫ 2ϩ A S 153.94 in. 269.39 in.3 ϭ Ϫ5.6 psi ϩ 307.4 psi ϭ 302 psi MAXIMUM COMPRESSIVE STRESS ␲d ϭ 269.39 in.3 32 M N MAXIMUM TENSILE STRESS ␴c ϭ Ϫ5.6 psi Ϫ 307.4 psi ϭ Ϫ313 psi 3 Problem 5.12-6 A vertical pole of aluminum is fixed at the base and pulled at the top by a cable having a tensile force T (see figure). The cable is attached at the outer surface of the pole and makes an angle ␣ ϭ 25° at the point of attachment. The pole has length L ϭ 2.0 m and a hollow circular cross section with outer diameter d2 ϭ 260 mm and inner diameter d1 ϭ 200 mm. Determine the allowable tensile force Tallow in the cable if the allowable compressive stress in the aluminum pole is 90 MPa. Solution 5.12-6 Aluminum pole T sin ␣ T cos ␣ L ␣ T L d2 CROSS SECTION ␲ A ϭ (d 2 Ϫ d 2 ) ϭ 21,677 mm2 ϭ 21.677 ϫ 10Ϫ3 mm2 1 4 2 ␲ I ϭ (d 4 Ϫ d 4 ) ϭ 145,778 ϫ 103 mm4 1 64 2 ϭ 145.778 ϫ 10Ϫ6 m4 d2 V M N ␣ ϭ 25º L ϭ 2.0 m d2 ϭ 260 mm d1 ϭ 200 mm (␴c)allow ϭ 90 MPa d1 d2 cϭ d2 ϭ 130 mm ϭ 0.13 m 2 AT THE BASE OF THE POLE N ϭ T cos ␣ ϭ 0.90631T M ϭ (T cos ␣) ¢ (N, T ϭ newtons) d2 ≤ ϩ (T sin ␣)(L) 2 ϭ 0.11782 T ϩ 0.84524 T ϭ 0.96306 T (M ϭ newton meters)
    • SECTION 5.12 COMPRESSIVE STRESS sc ϭ (0.96306T)(0.13 m) N Mc 0.90631T ϩ ϭ Ϫ3 2ϩ A I 21.677 ϫ 10 m 145.778 ϫ 10Ϫ6 m4 ϭ 41.82 T ϩ 858.83 T ϭ 900.64 T (␴c ϭ pascals) 361 Beams with Axial Loads ALLOWABLE TENSILE FORCE (sc ) allow 90 ϫ 106 pascals Tallow ϭ ϭ 900.64 900.64 ϭ 99,900 N ϭ 99.9 kN Problem 5.12-7 Because of foundation settlement, a circular tower is leaning at an angle ␣ to the vertical (see figure). The structural core of the tower is a circular cylinder of height h, outer diameter d2, and inner diameter d1. For simplicity in the analysis, assume that the weight of the tower is uniformly distributed along the height. Obtain a formula for the maximum permissible angle ␣ if there is to be no tensile stress in the tower. h d1 d2 ␣ Solution 5.12-7 Leaning tower I d2 ϩ d2 2 1 ϭ A 16 h 2 cϭ W W ϭ weight of tower ␣ ϭ angle of tilt d2 2 AT THE BASE OF THE TOWER N ϭ W cos ␣ h 2 MϭW¢ h ≤ sin ␣ 2 TENSILE STRESS (EQUAL TO ZERO) V M N CROSS SECTION ␲ 2 (d Ϫ d 2 ) 1 4 2 ␲ I ϭ (d 4 Ϫ d 4 ) 1 64 2 ␲ ϭ (d 2 Ϫ d 2 )(d 2 ϩ d 2 ) 1 2 1 64 2 Aϭ d2 N Mc W cos ␣ W h st ϭ Ϫ ϩ ϭϪ ϩ ¢ sin ␣ ≤ ¢ ≤ ϭ 0 A I A I 2 2 ∴ cos ␣ hd2 sin ␣ ϭ A 4I MAXIMUM ANGLE ␣ ␣ ϭ arctan d2 ϩ d2 2 1 4hd2 tan ␣ ϭ d2 ϩ d2 4I 2 1 ϭ hd2A 4hd2
    • 362 CHAPTER 5 Stresses in Beams Problem 5.12-8 A steel bar of solid circular cross section is subjected to an axial tensile force T ϭ 26 kN and a bending moment M ϭ 3.2 kN и m (see figure). Based upon an allowable stress in tension of 120 MPa, determine the required diameter d of the bar. (Disregard the weight of the bar itself.) Solution 5.12-8 ␲d 2 4 Sϭ T Circular bar T ϭ 26 kN M ϭ 3.2 kN ؒ m ␴allow ϭ 120 MPa d ϭ diameter Aϭ M ␲d 3 32 (d ϭ meters) (120,000,000 Nրm2)(␲)d 3 Ϫ (104,000 N)d Ϫ 102,400 N ؒ m ϭ 0 SIMPLIFY THE EQUATION: TENSILE STRESS (15,000 ␲) d 3 Ϫ 13d Ϫ 12.8 ϭ 0 T M 4T 32M ϩ ϭ ϩ A S ␲d 2 ␲d 3 or ␲d 3 ␴allow Ϫ 4Td Ϫ 32M ϭ 0 (␲)(120 MPa)d 3 Ϫ 4(26 kN)d Ϫ 32(3.2 kN ؒ m) ϭ 0 st ϭ SOLVE NUMERICALLY FOR THE REQUIRED DIAMETER: d ϭ 0.0662 m ϭ 66.2 mm Problem 5.12-9 A cylindrical brick chimney of height H weighs w ϭ 825 lb/ft of height (see figure). The inner and outer diameters are d1 ϭ 3 ft and d2 ϭ 4 ft, respectively. The wind pressure against the side of the chimney is p ϭ 10 lb/ft2 of projected area. Determine the maximum height H if there is to be no tension in the brickwork. Solution 5.12-9 Brick chimney d2 p w H d1 d2 I 1 ϭ (d 2 ϩ d 2 ) 1 A 16 2 cϭ d2 2 AT BASE OF CHIMNEY H q W V M N M ϭ qH ¢ N ϭ W ϭ wH p ϭ wind pressure q ϭ intensity of load ϭ pd2 d2 ϭ outer diameter d1 ϭ inner diameter W ϭ total weight of chimney ϭ wH TENSILE STRESS (EQUAL TO ZERO) N Md2 st ϭ Ϫ ϩ ϭ0 A 2I or M 2I ϭ N Ad2 pd2H2 d 2 ϩ d 2 2 1 ϭ 2wH 8d2 CROSS SECTION ␲ A ϭ (d 2 Ϫ d 2 ) 1 4 2 ␲ ␲ Iϭ (d 4 Ϫ d 4 ) ϭ (d 2 Ϫ d 2 )(d 2 ϩ d 2 ) 1 1 2 1 64 2 64 2 H 1 ≤ ϭ pd2H 2 2 2 SOLVE FOR H Hϭ w(d 2 ϩ d 2 ) 2 1 4pd 2 2 SUBSTITUTE NUMERICAL VALUES w ϭ 825 lb/ft d2 ϭ 4 ft d1 ϭ 3 ft Hmax ϭ 32.2 ft p ϭ 10 lb/ft2
    • SECTION 5.12 Problem 5.12-10 A flying buttress transmits a load P ϭ 25 kN, acting at an angle of 60° to the horizontal, to the top of a vertical buttress AB (see figure). The vertical buttress has height h ϭ 5.0 m and rectangular cross section of thickness t ϭ 1.5 m and width b ϭ 1.0 m (perpendicular to the plane of the figure). The stone used in the construction weighs ␥ ϭ 26 kN/m3. What is the required weight W of the pedestal and statue above the vertical buttress (that is, above section A) to avoid any tensile stresses in the vertical buttress? Flying buttress P W 60° A A t — 2 h t h t B Solution 5.12-10 B Flying buttress FREE-BODY DIAGRAM OF VERTICAL BUTTRESS P W 60° CROSS SECTION A ϭ bt ϭ (1.0 m)(1.5 m) ϭ 1.5 m2 1 1 S ϭ bt 2 ϭ (1.0 m)(1.5 m) 2 ϭ 0.375 m3 6 6 AT THE BASE h WB t V M N P ϭ 25 kN h ϭ 5.0 m t ϭ 1.5 m b ϭ width of buttress perpendicular to the figure b ϭ 1.0 m ␥ ϭ 26 kN/m3 WB ϭ weight of vertical buttress ϭ bth␥ ϭ 195 kN N ϭ W ϩ WB ϩ P sin 60º ϭ W ϩ 195 kN ϩ (25 kN) sin 60º ϭ W ϩ 216.651 kN M ϭ (P cos 60º)h ϭ (25 kN)(cos 60º)(5.0 m) ϭ 62.5 kN ؒ m TENSILE STRESS (EQUAL TO ZERO) N M st ϭ Ϫ ϩ A S W ϩ 216.651 kN 62.5 kN ؒ m ϭϪ ϩ ϭ0 1.5 m2 0.375 m3 or ϪW Ϫ 216.651 kN ϩ 250 kN ϭ 0 W ϭ 33.3 kN Problem 5.12-11 A plain concrete wall (i.e., a wall with no steel reinforcement) rests on a secure foundation and serves as a small dam on a creek (see figure). The height of the wall is h ϭ 6.0 ft and the thickness of the wall is t ϭ 1.0 ft. (a) Determine the maximum tensile and compressive stresses ␴t and ␴c, respectively, at the base of the wall when the water level reaches the top (d ϭ h). Assume plain concrete has weight density ␥c ϭ 145 lb/ft3. (b) Determine the maximum permissible depth dmax of the water if there is to be no tension in the concrete. 363 Beams with Axial Loads ˇ ˇ t h d
    • 364 CHAPTER 5 Solution 5.12-11 Stresses in Beams Concrete wall t W h d F d/3 V W M h ϭ height of wall t ϭ thickness of wall b ϭ width of wall (perpendicular to the figure) ␥c ϭ weight density of concrete ␥w ϭ weight density of water d ϭ depth of water W ϭ weight of wall W ϭ bht␥c F ϭ resultant force for the water pressure MAXIMUM WATER PRESSURE ϭ ␥w d 1 1 F ϭ (d)(gw d)(b) ϭ bd 2gw 2 2 d 1 M ϭ F ¢ ≤ ϭ bd 3gw 3 6 1 A ϭ bt S ϭ bt 2 6 (a) STRESSES AT THE BASE WHEN d ϭ h h ϭ 6.0 ft ϭ 72 in. d ϭ 72 in. t ϭ 1.0 ft ϭ 12 in. 145 gc ϭ 145 lbրft3 ϭ lbրin.3 1728 62.4 gw ϭ 62.4 lbրft3 ϭ lbրin.3 1728 Substitute numerical values into Eqs. (1) and (2): ␴t ϭ Ϫ6.042 psi ϩ 93.600 psi ϭ 87.6 psi ␴c ϭ Ϫ6.042 psi Ϫ 93.600 psi ϭ Ϫ99.6 psi (b) MAXIMUM DEPTH FOR NO TENSION Set ␴t ϭ 0 in Eq. (1): Ϫhgc ϩ STRESSES AT THE BASE OF THE WALL (d ϭ DEPTH OF WATER) d 3gw ϭ0 t2 d 3 ϭ (72 in.)(12 in.) 2 ¢ dmax ϭ 28.9 in. 3 st ϭ Ϫ d gw W M ϩ ϭ Ϫhgc ϩ 2 A S t Eq. (1) sc ϭ Ϫ d 3gw W M Ϫ ϭ Ϫhgc Ϫ 2 A S t Eq. (2) d 3 ϭ ht 2 ¢ gc ≤ gw 145 ≤ ϭ 24,092 in.3 62.4
    • SECTION 5.12 Eccentric Axial Loads Eccentric Axial Loads P Problem 5.12-12 A circular post and a rectangular post are each compressed by loads that produce a resultant force P acting at the edge of the cross section (see figure). The diameter of the circular post and the depth of the rectangular post are the same. (a) For what width b of the rectangular post will the maximum tensile stresses be the same in both posts? (b) Under the conditions described in part (a), which post has the larger compressive stress? Solution 5.12-12 ␲d 2 4 Sϭ b d d EQUAL MAXIMUM TENSILE STRESSES ␲d 3 32 Mϭ Pd 2 P M 4P 16P 12P Tension: st ϭ Ϫ ϩ ϭ Ϫ 2 ϩ ϭ A S ␲d ␲d 2 ␲d 2 P M 4P 16P Compression: sc ϭ Ϫ Ϫ ϭ Ϫ 2 Ϫ A S ␲d ␲d 2 20P ϭϪ 2 ␲d RECTANGULAR POST bd 2 Pd Mϭ 6 2 P M P 3P 2P Tension: st ϭ Ϫ ϩ ϭ Ϫ ϩ ϭ A S bd bd bd A ϭ bd P Two posts in compression CIRCULAR POST Aϭ 365 Sϭ P M P 3P 4P Compression: sc ϭ Ϫ Ϫ ϭ Ϫ Ϫ ϭϪ A S bd bd bd 12P 2P ϭ ␲d 2 bd 6 1 ϭ ␲d b or (Eq. 1) (a) Determine the width b of the rectangular post ␲d From Eq. (1): b ϭ 6 (b) Compressive stresses 20P ␲d 2 4P 4P Rectangular post: sc ϭ Ϫ ϭ Ϫ bd (␲dր6)d Circular post: sc ϭ Ϫ ϭϪ 24P ␲d 2 Rectangular post has the larger compressive stress. Problem 5.12-13 Two cables, each carrying a tensile force P ϭ 1200 lb, are bolted to a block of steel (see figure). The block has thickness t ϭ 1 in. and width b ϭ 3 in. (a) If the diameter d of the cable is 0.25 in., what are the maximum tensile and compressive stresses ␴t and ␴c, respectively, in the block? (b) If the diameter of the cable is increased (without changing the force P), what happens to the maximum tensile and compressive stresses? b P t P
    • 366 CHAPTER 5 Stresses in Beams Solution 5.12-13 Steel block loaded by cables d P e t 2 P ϭ 1200 lb d ϭ 0.25 in. t d t ϭ 1.0 in. e ϭ ϩ ϭ 0.625 in. 2 2 b ϭ width of block ϭ 3.0 in. Steel block MAXIMUM COMPRESSIVE STRESS (AT BOTTOM OF BLOCK) t y ϭ Ϫ ϭ Ϫ0.5 in. 2 P Pey ϩ A I 1200 lb (1200 lb)(0.625 in.)(Ϫ0.5 in.) ϭ ϩ 3 in.2 0.25 in.4 ϭ 400 psi Ϫ 1500 psi ϭ Ϫ1100 psi sc ϭ CROSS SECTION OF BLOCK A ϭ bt ϭ 3.0 in.2 Iϭ 1 3 bt ϭ 0.25 in.4 12 (a) MAXIMUM TENSILE STRESS (AT TOP OF BLOCK) t y ϭ ϭ 0.5 in. 2 t (b) IF d IS INCREASED, the eccentricity e increases and both stresses increase in magnitude. P Pey ϩ A I 1200 lb (1200 lb)(0.625 in.)(0.5 in.) ϭ ϩ 3 in.2 0.25 in.4 ϭ 400 psi ϩ 1500 psi ϭ 1900 psi st ϭ Problem 5.12-14 A bar AB supports a load P acting at the centroid of the end cross section (see figure). In the middle region of the bar the cross-sectional area is reduced by removing one-half of the bar. (a) If the end cross sections of the bar are square with sides of length b, what are the maximum tensile and compressive stresses ␴t and ␴c, respectively, at cross section mn within the reduced region? (b) If the end cross sections are circular with diameter b, what are the maximum stresses ␴t and ␴c? b — 2 A b b b b — 2 m (a) b — 2 n B P b (b)
    • SECTION 5.12 Solution 5.12-14 Bar with reduced cross section (a) SQUARE BAR FOR TENSION: Cross section mn is a rectangle. b b2 A ϭ (b) ¢ ≤ ϭ 2 2 M ϭ P¢ b ≤ 4 cϭ Eccentric Axial Loads b 4 1 b 3 b4 I ϭ (b) ¢ ≤ ϭ 12 2 96 STRESSES ct ϭ 4r 2b ϭ ϭ 0.2122 b 3␲ 3␲ FOR COMPRESSION: b 2b cc ϭ r Ϫ ct ϭ Ϫ ϭ 0.2878 b 2 3␲ st ϭ P Mc 2P 6P 8P ϩ ϭ 2ϩ 2ϭ 2 A I b b b STRESSES sc ϭ P Mc 2P 6P 4P Ϫ ϭ 2 Ϫ 2 ϭϪ 2 A I b b b st ϭ (b) CIRCULAR BAR Cross section mn is a semicircle 1 ␲b2 ␲b2 ¢ ≤ϭ ϭ 0.3927 b2 2 4 8 From Appendix D, Case 10: Aϭ I ϭ 0.1098 ¢ M ϭ P¢ (0.2122 Pb)(0.2122 b) P Mct P ϩ ϭ 2ϩ A I 0.3927 b 0.006860 b 4 P P P ϭ 2.546 2 ϩ 6.564 2 ϭ 9.11 2 b b b (0.2122 Pb)(0.2878 b) P Mcc P sc ϭ Ϫ ϭ Ϫ A I 0.3927b 2 0.006860 b 4 P P P ϭ 2.546 2 Ϫ 8.903 2 ϭ Ϫ6.36 2 b b b b 4 ≤ ϭ 0.006860 b 4 2 2b ≤ ϭ 0.2122 Pb 3␲ Problem 5.12-15 A short column constructed of a W 10 ϫ 30 wide-flange shape is subjected to a resultant compressive load P ϭ 12 k having its line of action at the midpoint of one flange (see figure). (a) Determine the maximum tensile and compressive stresses ␴t and ␴c, respectively, in the column. (b) Locate the neutral axis under this loading condition. P = 12 k z C W 10 ϫ 30 y 367
    • 368 CHAPTER 5 Stresses in Beams Solution 5.12-15 Column of wide-flange shape y (a) MAXIMUM STRESSES P = 12 k P Pe(hր2) st ϭ Ϫ ϩ ϭ Ϫ1357 psi ϩ 1840 psi A I ϭ 480 psi e z h = 10.47 in. O N.A. W 10 ϫ 30 I ϭ 170 in.4 A ϭ 8.84 in.2 tf ϭ 0.510 in. P Pe(hր2) sc ϭ Ϫ Ϫ ϭ Ϫ1357 psi Ϫ 1840 psi A I ϭ Ϫ3200 psi (b) NEUTRAL AXIS (SEE FIGURE) y0 ϭ Ϫ h tf e ϭ Ϫ ϭ 4.98 in. 2 2 I ϭ Ϫ3.86 in. Ae P = 60 kN Problem 5.12-16 A short column of wide-flange shape is subjected to a compressive load that produces a resultant force P ϭ 60 kN acting at the midpoint of one flange (see figure). (a) Determine the maximum tensile and compressive stresses ␴t and ␴c , respectively, in the column. (b) Locate the neutral axis under this loading condition. y P 8 mm z C 12 mm 160 mm Solution 5.12-16 Column of wide-flange shape y P e (a) MAXIMUM STRESSES tf tw z O h N.A. b b ϭ 160 mm h ϭ 200 mm tw ϭ 8 mm tf ϭ 12 mm h tf P ϭ 60 kN e ϭ Ϫ ϭ 94 mm 2 2 A ϭ 2btf ϩ (h Ϫ 2tf) tw ϭ 5248 mm2 1 3 1 bh Ϫ (b Ϫ t w )(h Ϫ 2t f ) 3 12 12 ϭ 37.611 ϫ 106 mm4 Iϭ P Pe(hր2) st ϭ Ϫ ϩ A I (60 kN)(94 mm)(100 mm) 60 kN ϭϪ 2ϩ 5248 mm 37.611 ϫ 106 mm4 ϭ Ϫ11.43 MPa ϩ 15.00 MPa ϭ 3.57 MPa ␴c ϭ Ϫ11.43 MPa Ϫ 15.00 MPa ϭ Ϫ26.4 MPa (b) NEUTRAL AXIS (SEE FIGURE) I 37.611 ϫ 106 mm4 ϭϪ Ae (5248 mm2 )(94 mm) ϭ Ϫ76.2 mm y0 ϭ Ϫ 200 mm
    • SECTION 5.12 Problem 5.12-17 A tension member constructed of an L 4 ϫ 4 ϫ 3⁄4 inch angle section (see Appendix E) is subjected to a tensile load P ϭ 15 kips that acts through the point where the midlines of the legs intersect (see figure). Determine the maximum tensile stress ␴t in the angle section. Eccentric Axial Loads 2 3 3 L4ϫ4ϫ— 4 C 1 1 P ϫ 2 Solution 5.12-17 3 Angle section in tension c 2 3 C 1 1 c P B e c1 3 2 Bending occurs about axis 3-3. L4ϫ4ϫ Maximum tensile stress occurs at corner B. 3 4 st ϭ A ϭ 5.44 in.2 t ϭ thickness of legs c ϭ 1.27 in. ϭ 0.75 in. e ϭ eccentricity of load P t ϭ ¢ c Ϫ ≤ ͙2 2 ϭ (1.27 Ϫ 0.375) ͙2 ϭ 1.266 in. P ϭ 15 k (tensile load) c1 ϭ distance from centroid C to corner B of angle ϭ c͙2 ϭ (1.27 in.) ͙2 ϭ 1.796 in. I3 ϭ Ar 2 min MAXIMUM TENSILE STRESS (see Table E-4) rmin ϭ 0.778 in. I3 ϭ (5.44 in.2)(0.778 in.)2 ϭ 3.293 in.4 M ϭ Pe ϭ (15 k)(1.266 in.) ϭ 18.94 k-in. ϭ P Mc1 ϩ A I3 (18.99 k-in.)(1.796 in.) 15 k 2ϩ 5.44 in 3.293 in.4 ϭ 2.76 ksi ϩ 10.36 ksi ϭ 13.1 ksi 369
    • 370 CHAPTER 5 Stresses in Beams Problem 5.12-18 A short length of a C 8ϫ11.5 channel is subjected to an axial compressive force P that has its line of action through the midpoint of the web of the channel (see figure). (a) Determine the equation of the neutral axis under this loading condition. (b) If the allowable stresses in tension and compression are 10,000 psi and 8,000 psi, respectively, find the maximum permissible load Pmax. Solution 5.12-18 y P C 8 × 11.5 ϫ z C Channel in compression y 0.220 in. P z C 8 ϫ 11.5 A ϭ 3.38 in.2 Iz ϭ 1.32 in.4 c1 c2 h ϭ 2.260 in. c1 ϭ 0.571 in. tw ϭ 0.220 in. c2 ϭ 1.689 in. tw ϭ 0.571 Ϫ 0.110 ϭ 0.461 in. 2 (a) LOCATION OF THE NEUTRAL AXIS y0 ϭ Ϫ I 1.32 in.4 ϭϪ Ae (3.38 in.2 )(0.461 in.) ϭ Ϫ0.847 in. (b) MAXIMUM LOAD BASED UPON TENSILE STRESS ␴allow ϭ 10,000 psi st ϭ Ϫ ϭϪ (P ϭ pounds) P Pe c2 ϩ A I P(0.461 in.)(1.689 in.) P 2ϩ 3.38 in. 1.32 in.4 10,000 ϭ Ϫ 2.260 in. C MAXIMUM LOAD BASED UPON COMPRESSIVE STRESS ECCENTRICITY OF THE LOAD e ϭ c1 Ϫ ϫ P P ϩ ϭ 0.2941 P 3.38 1.695 P ϭ 34,000 lb ϭ 34 k ␴allow ϭ 8000 psi (P ϭ pounds) P Pe c1 Ϫ A I P(0.461 in.)(0.571 in.) P ϭϪ Ϫ 3.38 in.2 1.32 in.4 P P 8000 ϭ Ϫ ϭ 0.4953 P 3.38 5.015 P ϭ 16,200 lb ϭ 16.2 k sc ϭ Ϫ COMPRESSION GOVERNS. Pmax ϭ 16.2 k
    • 371 SECTION 5.13 Stress Concentrations Stress Concentrations The problems for Section 5.13 are to be solved considering the stress-concentration factors. M Problem 5.13-1 The beams shown in the figure are subjected to bending moments M ϭ 2100 lb-in. Each beam has a rectangular cross section with height h ϭ 1.5 in. and width b ϭ 0.375 in. (perpendicular to the plane of the figure). (a) For the beam with a hole at midheight, determine the maximum stresses for hole diameters d ϭ 0.25, 0.50, 0.75, and 1.00 in. (b) For the beam with two identical notches (inside height h1 ϭ 1.25 in.), determine the maximum stresses for notch radii R ϭ 0.05, 0.10, 0.15, and 0.20 in. M h d (a) 2R M M h h1 Probs. 5.13-1 through 5.13-4 (b) Solution 5.13-1 M ϭ 2100 lb-in. h ϭ 1.5 in. b ϭ 0.375 in. (b) BEAM WITH NOTCHES h1 ϭ 1.25 in. (a) BEAM WITH A HOLE d 1 Յ h 2 d 1 Ն h 2 Eq. (5-57): Eq. (5-56): d (in.) 0.25 0.50 0.75 1.00 6Mh b(h3 Ϫ d 3 ) 50,400 ϭ 3.375 Ϫ d 3 sC ϭ 12Md b(h3 Ϫ d 3 ) 67,200 d ϭ 3.375 Ϫ d 3 (1) Note: The larger the hole, the larger the stress. Eq. (5-58): 6M snom ϭ 2 ϭ 21,500 psi bh 1 R (in.) sB ϭ ␴C ␴B d Eq.(1) Eq.(2) ␴max h (psi) (psi) (psi) 0.1667 15,000 — 15,000 0.3333 15,500 — 15,500 0.5000 17,100 17,100 17,100 0.6667 — 28,300 28,300 h 1.5 in. ϭ ϭ 1.2 h1 1.25 in. (2) R h1 0.05 0.10 0.15 0.20 0.04 0.08 0.12 0.16 ␴max ϭ K␴nom K ␴max (Fig. 5-50) (psi) 3.0 2.3 2.1 1.9 65,000 49,000 45,000 41,000 Note: The larger the notch radius, the smaller the stress.
    • 372 CHAPTER 5 Stresses in Beams Problem 5.13-2 The beams shown in the figure are subjected to bending moments M ϭ 250 N и m. Each beam has a rectangular cross section with height h ϭ 44 mm and width b ϭ 10 mm (perpendicular to the plane of the figure). (a) For the beam with a hole at midheight, determine the maximum stresses for hole diameters d ϭ 10, 16, 22, and 28 mm. (b) For the beam with two identical notches (inside height h1 ϭ 40 mm), determine the maximum stresses for notch radii R ϭ 2, 4, 6, and 8 mm. Solution 5.13-2 M ϭ 250 N ؒ m h ϭ 44 mm b ϭ 10 mm (b) BEAM WITH NOTCHES d 1 Յ h 2 Eq. (5-57): d 1 Ն h 2 Eq. (5-56): 6Mh b(h3 Ϫ d 3 ) 6.6 ϫ 106 ϭ MPa (1) 85,180 Ϫ d 3 sC ϭ sB ϭ ϭ d (mm) 10 16 22 28 12Md b(h3 Ϫ d 3 ) 300 ϫ 10 3d MPa 85,180 Ϫ d 3 ␴C ␴B Eq.(1) Eq.(2) (MPa) (MPa) d h 0.227 0.364 0.500 0.636 h 44 mm ϭ ϭ 1.1 h1 40 mm 6M Eq. (5-58): snom ϭ 2 ϭ 93.8 MPa bh 1 ␴max ϭ K␴nom h1 ϭ 40 mm (a) BEAM WITH A HOLE 78 81 89 — — — 89 133 (2) ␴max (MPa) R (mm) 2 4 6 8 R h1 K (Fig. 5-50) ␴max (MPa) 0.05 0.10 0.15 0.20 2.6 2.1 1.8 1.7 240 200 170 160 Note: The larger the notch radius, the smaller the stress. 78 81 89 133 Note: The larger the hole, the larger the stress. Problem 5.13-3 A rectangular beam with semicircular notches, as shown in part (b) of the figure, has dimensions h ϭ 0.88 in. and h1 ϭ 0.80 in. The maximum allowable bending stress in the metal beam is ␴max ϭ 60 ksi, and the bending moment is M ϭ 600 lb-in. Determine the minimum permissible width bmin of the beam. Solution 5.13-3 h ϭ 0.88 in. ␴max ϭ 60 ksi h ϭ h1 ϩ 2R Beam with semicircular notches h1 ϭ 0.80 in. M ϭ 600 lb-in. 1 R ϭ (h Ϫ h1 ) ϭ 0.04 in. 2 R 0.04 in. ϭ ϭ 0.05 h1 0.80 in. From Fig. 5-50: K ഠ 2.57 smax ϭ Ksnom ϭ K ¢ 60 ksi ϭ 2.57 B Solve for b: 6M ≤ bh2 1 6(600 lb-in.) R b(0.80 in.) 2 bmin ഠ 0.24 in.
    • SECTION 5.13 Stress Concentrations 373 Problem 5.13-4 A rectangular beam with semicircular notches, as shown in part (b) of the figure, has dimensions h ϭ 120 mm and h1 ϭ 100 mm. The maximum allowable bending stress in the plastic beam is ␴max ϭ 6 MPa, and the bending moment is M ϭ 150 N и m. Determine the minimum permissible width bmin of the beam. Solution 5.13-4 h ϭ 120 mm ␴max ϭ 6 MPa h ϭ h1 ϩ 2R Beam with semicircular notches h1 ϭ 100 mm M ϭ 150 N ؒ m 1 R ϭ (h Ϫ h1 ) ϭ 10 mm 2 R 10 mm ϭ ϭ 0.10 h1 100 mm From Fig. 5-50: K ഠ 2.20 smax ϭ Ksnom ϭ K ¢ 6 MPa ϭ (2.20) B Solve for b: 6(150 N ؒ m) R b(100 mm) 2 ˇ ˇ bmin ഠ 33 mm Problem 5.13-5 A rectangular beam with notches and a hole (see figure) has dimensions h ϭ 5.5 in., h1 ϭ 5 in., and width b ϭ 1.6 in. The beam is subjected to a bending moment M ϭ 130 k-in., and the maximum allowable bending stress in the material (steel) is ␴max ϭ 42,000 psi. (a) What is the smallest radius Rmin that should be used in the notches? (b) What is the diameter dmax of the largest hole that should be drilled at the midheight of the beam? Solution 5.13-5 6M ≤ bh2 1 2R M M h1 h d Beam with notches and a hole h ϭ 5.5 in. h1 ϭ 5 in. b ϭ 1.6 in. M ϭ 130 k-in. ␴max ϭ 42,000 psi (b) LARGEST HOLE DIAMETER d 1 7 and use Eq. (5-56). h 2 12 Md sB ϭ b(h3 Ϫ d 3 ) Assume (a) MINIMUM NOTCH RADIUS h 5.5 in. ϭ 1.1 ϭ h1 5 in. 6M snom ϭ 2 ϭ 19,500 psi bh1 Kϭ smax 42,000 psi ϭ ϭ 2.15 snom 19,500 psi h From Fig. 5-50, with K ϭ 2.15 and ϭ 1.1, we get h1 R Ϸ 0.090 h1 І Rmin ഠ 0.090h1 ϭ 0.45 in. 42,000 psi ϭ 12(130 k-in.)d (1.6 in.) [ (5.5 in.) 3 Ϫ d3 ] d 3 ϩ 23.21d Ϫ 166.4 ϭ 0 Solve numerically: dmax ϭ 4.13 in. or
    • 7 Analysis of Stress and Strain Plane Stress Problem 7.2-1 An element in plane stress is subjected to stresses ␴x ϭ 6500 psi, ␴y ϭ 1700 psi, and ␶xy ϭ 2750 psi, as shown in the figure. Determine the stresses acting on an element oriented at an angle ␪ ϭ 60° from the x axis, where the angle ␪ is positive when counterclockwise. Show these stresses on a sketch of an element oriented at the angle ␪. Solution 7.2-1 y ␴y = 1700 psi O ␴x = 6500 psi x ␶xy = 2750 psi Plane stress (angle ␪) y 2920 psi 5280 psi ␪ ϭ 60Њ O x 3450 psi Problem 7.2-2 Solve the preceding problem for ␴x ϭ 80 MPa, ␴y ϭ 52 MPa, ␶xy ϭ 48 MPa, and ␪ ϭ 25° (see figure). ␴x ϭ 6500 psi ␴y ϭ 1700 psi ␶xy ϭ 2750 psi ␪ ϭ 60Њ sx ϩ sy sx Ϫ sy sx1 ϭ ϩ cos 2u ϩ txy sin 2u 2 2 ϭ 5280 psi sx Ϫ sy tx1y1 ϭ Ϫ sin 2u ϩ txy cos 2u 2 ϭ Ϫ3450 psi sy1 ϭ sx ϩ sy Ϫ sx1 ϭ 2920 psi 52 MPa 48 MPa 80 MPa 425
    • 426 CHAPTER 7 Solution 7.2-2 Analysis of Stress and Strain Plane stress (angle ␪) y 20.2 MPa 111.8 MPa ␪ ϭ 25Њ O x 20.1 MPa ␴x ϭ 80 MPa ␴y ϭ 52 MPa ␶xy ϭ 48 MPa ␪ ϭ 25Њ sx ϩ sy sx Ϫ sy sx1 ϭ ϩ cos 2u ϩ txy sin 2u 2 2 ϭ 111.8 MPa sx Ϫ sy tx1y1 ϭ Ϫ sin 2u ϩ txy cos 2u 2 ϭ 20.1 MPa sy1 ϭ sx ϩ sy Ϫ sx1 ϭ 20.2 MPa Problem 7.2-3 Solve Problem 7.2-1 for ␴x ϭ Ϫ9,900 psi, ␴y ϭ Ϫ3,400 psi, ␶xy ϭ 3,600 psi, and ␪ ϭ 50° (see figure). 3,400 psi 3,600 psi 9,900 psi Solution 7.2-3 Plane stress (angle ␪) y 10,760 psi 2,540 psi ␪ ϭ 50Њ O x 2,580 psi Problem 7.2-4 The stresses acting on element A in the web of a train rail are found to be 42 MPa tension in the horizontal direction and 140 MPa compression in the vertical direction (see figure). Also, shear stresses of magnitude 60 MPa act in the directions shown. Determine the stresses acting on an element oriented at a counterclockwise angle of 48° from the horizontal. Show these stresses on a sketch of an element oriented at this angle. ␴x ϭ Ϫ9900 psi ␴y ϭ Ϫ3400 psi ␶xy ϭ 3600 psi ␪ ϭ 50Њ sx ϩ sy sx Ϫ sy sx1 ϭ ϩ cos 2u ϩ txy sin 2u 2 2 ϭ Ϫ2540 psi sx Ϫ sy tx1y1 ϭ Ϫ sin 2u ϩ txy cos 2u 2 ϭ 2580 psi sy1 ϭ sx ϩ sy Ϫ sx1 ϭ Ϫ10,760 psi 140 MPa A A Side View 42 MPa 60 MPa Cross Section
    • SECTION 7.2 Plane Stress Plane stress (angle ␪) Solution 7.2-4 y 20.2 MPa 118.2 MPa ␪ ϭ 48Њ O x 84.2 MPa ␴x ϭ 42 MPa ␴y ϭ Ϫ140 MPa ␶xy ϭ Ϫ60 MPa ␪ ϭ 48Њ sx ϩ sy sx Ϫ sy sx1 ϭ ϩ cos 2u ϩ txy sin 2u 2 2 ϭ Ϫ118.2 MPa sx Ϫ sy tx1y1 ϭ Ϫ sin 2u ϩ txy cos 2u 2 ϭ Ϫ84.2 MPa sy1 ϭ sx ϩ sy Ϫ sx1 ϭ 20.2 MPa Problem 7.2-5 Solve the preceding problem if the normal and shear stresses acting on element A are 7,500 psi, 20,500 psi, and 4,800 psi (in the directions shown in the figure) and the angle is 30° (counterclockwise). 20,500 psi A 7,500 psi 4,800 psi Plane stress (angle ␪) Solution 7.2-5 y 9,340 psi 3,660 psi ␪ ϭ 30Њ O x 14,520 psi ␴x ϭ 7,500 psi ␴y ϭ Ϫ20,500 psi ␶xy ϭ Ϫ4,800 psi ␪ ϭ 30Њ sx ϩ sy sx Ϫ sy sx1 ϭ ϩ cos 2u ϩ txy sin 2u 2 2 ϭ Ϫ3,660 psi sx Ϫ sy tx1y1 ϭ Ϫ sin 2u ϩ txy cos 2u 2 ϭ Ϫ14, 520 psi sy1 ϭ sx ϩ sy Ϫ sx1 ϭ Ϫ9,340 psi Problem 7.2-6 An element in plane stress from the fuselage of an airplane is subjected to compressive stresses of magnitude 25.5 MPa in the horizontal direction and tensile stresses of magnitude 6.5 MPa in the vertical direction (see figure). Also, shear stresses of magnitude 12.0 MPa act in the directions shown. Determine the stresses acting on an element oriented at a clockwise angle of 40° from the horizontal. Show these stresses on a sketch of an element oriented at this angle. 6.5 MPa 25.5 MPa 12.0 MPa 427
    • 428 CHAPTER 7 Analysis of Stress and Strain Plane stress (angle ␪) Solution 7.2-6 y 17.8 MPa 18.5 MPa O x ␪ ϭ ᎐ 40Њ 0.5 MPa Problem 7.2-7 The stresses acting on element B in the web of a wide-flange beam are found to be 11,000 psi compression in the horizontal direction and 3,000 psi compression in the vertical direction (see figure). Also, shear stresses of magnitude 4,200 psi act in the directions shown. Determine the stresses acting on an element oriented at a counterclockwise angle of 41° from the horizontal. Show these stresses on a sketch of an element oriented at this angle. Solution 7.2-7 3,000 psi B B 2,280 psi 11,720 psi ␪ ϭ 41Њ x 3,380 psi 11,000 psi 4,200 psi Side View Cross Section Plane stress (angle ␪) y O ␴x ϭ Ϫ25.5 MPa ␴y ϭ 6.5 MPa ␶xy ϭ Ϫ12.0 MPa ␪ ϭ Ϫ40Њ sx ϩ sy sx Ϫ sy sx1 ϭ ϩ cos 2u ϩ txy sin 2u 2 2 ϭ Ϫ0.5 MPa sx Ϫ sy tx1y1 ϭ Ϫ sin 2u ϩ txy cos 2u 2 ϭ Ϫ17.8 MPa sy1 ϭ sx ϩ sy Ϫ sx1 ϭ Ϫ18.5 MPa ␴x ϭ Ϫ11,000 psi ␴y ϭ Ϫ3,000 psi ␶xy ϭ Ϫ4,200 psi ␪ ϭ 41Њ sx ϩ sy sx Ϫ sy sx1 ϭ ϩ cos 2u ϩ txy sin 2u 2 2 ϭ Ϫ11,720 psi sx Ϫ sy tx1y1 ϭ Ϫ sin 2u ϩ txy cos 2u 2 ϭ 3,380 psi sy1 ϭ sx ϩ sy Ϫ sx1 ϭ Ϫ2,280 psi Problem 7.2-8 Solve the preceding problem if the normal and shear stresses acting on element B are 54 MPa, 12 MPa, and 20 MPa (in the directions shown in the figure) and the angle is 42.5° (clockwise). 12 MPa 20 MPa B 54 MPa
    • SECTION 7.2 Solution 7.2-8 Plane stress (angle ␪) ␴x ϭ Ϫ54 MPa ␴y ϭ Ϫ12 MPa ␶xy ϭ 20 MPa ␪ ϭ Ϫ42.5Њ sx ϩ sy sx Ϫ sy sx1 ϭ ϩ cos 2u ϩ txy sin 2u 2 2 ϭ Ϫ54.8 MPa sx Ϫ sy tx1y1 ϭ Ϫ sin 2u ϩ txy cos 2u 2 ϭ Ϫ19.2 MPa sy1 ϭ sx ϩ sy Ϫ sx1 ϭ Ϫ11.2 MPa y 19.2 MPa 11.2 MPa O x ␪ ϭ – 42.5Њ 54.8 MPa y Problem 7.2-9 The polyethylene liner of a settling pond is subjected to stresses ␴x ϭ 350 psi, ␴y ϭ 112 psi, and ␶xy ϭ Ϫ120 psi, as shown by the plane-stress element in the first part of the figure. Determine the normal and shear stresses acting on a seam oriented at an angle of 30° to the element, as shown in the second part of the figure. Show these stresses on a sketch of an element having its sides parallel and perpendicular to the seam. Solution 7.2-9 Plane Stress 112 psi 30° 350 psi O x 120 psi Seam Plane stress (angle ␪) y 275 psi 187 psi ␪ ϭ 30 Њ O x 163 psi ␴x ϭ 350 psi ␴y ϭ 112 psi ␶xy ϭ Ϫ120 psi ␪ ϭ 30Њ sx ϩ sy sx Ϫ sy sx1 ϭ ϩ cos 2u ϩ txy sin 2u 2 2 ϭ 187 psi sx Ϫ sy tx1y1 ϭ Ϫ sin 2u ϩ txy cos 2u 2 ϭ Ϫ163 psi sy1 ϭ sx ϩ sy Ϫ sx1 ϭ 275 psi The normal stress on the seam equals 187 psi tension. The shear stress on the seam equals 163 psi, acting clockwise against the seam. 429
    • 430 CHAPTER 7 Analysis of Stress and Strain y Problem 7.2-10 Solve the preceding problem if the normal and shear stresses acting on the element are ␴x ϭ 2100 kPa, ␴y ϭ 300 kPa, and ␶xy ϭ Ϫ560 kPa, and the seam is oriented at an angle of 22.5° to the element (see figure). 300 kPa 22.5° 2100 kPa x O Seam 560 kPa Solution 7.2-10 Plane stress (angle ␪) y sx1 ϭ 960 kPa 1440 kPa ␪ ϭ 22.5Њ O x 1030 kPa ␴x ϭ 2100 kPa ␴y ϭ 300 kPa ␶xy ϭ Ϫ560 kPa ␪ ϭ 22.5Њ sx ϩ sy sx Ϫ sy cos 2u ϩ txy sin 2u 2 2 ϭ 1440 kPa sx Ϫ sy tx1y1 ϭ Ϫ sin 2u ϩ txy cos 2u 2 ϭ Ϫ1030 kPa sy1 ϭ sx ϩ sy Ϫ sx1 ϭ 960 kPa ϩ The normal stress on the seam equals 1440 kPa tension. The shear stress on the seam equals 1030 kPa, acting clockwise against the seam. 350 psi Problem 7.2-11 A rectangular plate of dimensions 3.0 in. ϫ 5.0 in. is formed by welding two triangular plates (see figure). The plate is subjected to a tensile stress of 500 psi in the long direction and a compressive stress of 350 psi in the short direction. Determine the normal stress ␴w acting perpendicular to the line of the weld and the shear stress ␶w acting parallel to the weld. (Assume that the normal stress ␴w is positive when it acts in tension against the weld and the shear stress ␶w is positive when it acts counterclockwise against the weld.) Solution 7.2-11 Biaxial stress (welded joint) sx1 ϭ y 125 psi 275 psi ␪ ϭ 30.96Њ O x 375 psi ␴x ϭ 500 psi ␴y ϭ Ϫ350 psi ␶xy ϭ 0 3 in. u ϭ arctan ϭ arctan 0.6 ϭ 30.96Њ 5 in. ld We 3 in. 500 psi 5 in. sx ϩ sy ϩ sx Ϫ sy cos 2u ϩ txy sin 2u 2 2 ϭ 275 psi sx Ϫ sy tx1y1 ϭ Ϫ sin 2u ϩ txy cos 2u ϭ Ϫ375 psi 2 sy1 ϭ sx ϩ sy Ϫ sx1 ϭ Ϫ125 psi STRESSES ACTING ON THE WELD 125 psi ␴w ␶w ␪ 375 psi ␴w ϭ Ϫ125 psi ␶w ϭ 375 psi ␪ ϭ30.96Њ
    • SECTION 7.2 Problem 7.2-12 Solve the preceding problem for a plate of dimensions 100 mm ϫ 250 mm subjected to a compressive stress of 2.5 MPa in the long direction and a tensile stress of 12.0 MPa in the short direction (see figure). 12.0 MPa ld We Solution 7.2-12 Biaxial stress (welded joint) y tx1y1 ϭ Ϫ 10.0 MPa 0.5 MPa ␪ ϭ 21.80 Њ O sin 2u ϩ txy cos 2u ϭ 5.0 MPa 2 sy1 ϭ sx ϩ sy Ϫ sx1 ϭ 10.0 MPa STRESSES ACTING ON THE WELD ␴x ϭ Ϫ2.5 MPa ␴y ϭ 12.0 MPa ␶xy ϭ 0 100 mm u ϭ arctan ϭ arctan 0.4 ϭ 21.80Њ 250 mm sx ϩ sy ϩ 2 ϭ Ϫ0.5 MPa sx Ϫ sy 2 10.0 MPa 5.0 MPa ␶w sx1 ϭ ␪ cos 2u ϩ txy sin 2u Biaxial stress y ␴ ␪ a Find angle ␪ for ␴ ϭ 0. ␴ ϭ normal stress on plane a-a sx ϩ sy ϩ 1600 psi a ␪ 3600 psi O x a STRESS ELEMENT a sx1 ϭ ␪ ϭ 21.80Њ ␴w ϭ 10.0 MPa ␶w ϭ Ϫ5.0 MPa Problem 7.2-13 At a point on the surface of a machine the material is in biaxial stress with ␴x ϭ 3600 psi and ␴y ϭ Ϫ1600 psi, as shown in the first part of the figure. The second part of the figure shows an inclined plane aa cut through the same point in the material but oriented at an angle ␪. Determine the value of the angle ␪ between zero and 90° such that no normal stress acts on plane aa. Sketch a stress element having plane aa as one of its sides and show all stresses acting on the element. Solution 7.2-13 2.5 MPa 100 mm 250 mm sx Ϫ sy ␴w x 5.0 MPa 431 Plane Stress ␴x ϭ 3600 psi ␴y ϭ Ϫ1600 psi ␶xy ϭ 0 sx1 ϭ 0 ␪ ϭ 56.31Њ sy1 ϭ sx ϩ sy Ϫ sx1 ϭ 2000 psi sx Ϫ sy tx1y1 ϭ Ϫ sin 2u ϩ txy cos 2u 2 ϭ Ϫ2400 psi y 2000 psi sx Ϫ sy cos 2u ϩ txy sin 2u 2 2 ϭ 1000 ϩ 2600 cos 2␪ (psi) 1000 For sx1 ϭ 0, we obtain cos 2u ϭ Ϫ 2600 І 2␪ ϭ 112.62Њ and ␪ ϭ 56.31Њ ␪ ϭ 56.31Њ O x 2400 psi
    • 432 CHAPTER 7 Analysis of Stress and Strain Problem 7.2-14 Solve the preceding problem for ␴x ϭ 32 MPa and ␴y ϭ Ϫ50 MPa (see figure). y 50 MPa a ␪ 32 MPa O x a Solution 7.2-14 Biaxial stress a ␴ ␪ ␴x ϭ 32 MPa ␴y ϭ Ϫ50 MPa a ␶xy ϭ 0 Find angles ␪ for ␴ ϭ 0. ␴ ϭ normal stress on plane a-a sx1 ϭ sx ϩ sy ϩ STRESS ELEMENT ␪ ϭ 38.66Њ sx1 ϭ 0 sy1 ϭ sx ϩ sy Ϫ sx1 ϭ Ϫ18 MPa sx Ϫ sy tx1y1 ϭ Ϫ sin 2u ϩ txy cos 2u 2 ϭ Ϫ40 MPa y 18 MPa sx Ϫ sy cos 2u ϩ txy sin 2u 2 2 ϭ Ϫ9 ϩ 41 cos 2␪ (MPa) 9 For sx1 ϭ 0, we obtain cos 2u ϭ 41 І 2␪ ϭ 77.32Њ and ␪ ϭ 38.66Њ ␪ ϭ 38.66Њ O Problem 7.2-15 An element in plane stress from the frame of a racing car is oriented at a known angle ␪ (see figure). On this inclined element, the normal and shear stresses have the magnitudes and directions shown in the figure. Determine the normal and shear stresses acting on an element whose sides are parallel to the xy axes; that is, determine ␴x, ␴y, and ␶xy. Show the results on a sketch of an element oriented at ␪ ϭ 0°. x 40 MPa y 4,180 psi 2,360 psi ␪ = 30° 15,220 psi O 4,900 psi Solution 7.2-15 Plane stress Transform from ␪ ϭ 30Њ to ␪ ϭ 0Њ. Let: ␴x ϭ Ϫ15,220 psi, ␴y ϭ Ϫ4,180 psi, ␶xy ϭ 2,360 psi, and ␪ ϭ Ϫ30Њ. sx1 ϭ sx ϩ sy y 14,500 psi O sx Ϫ sy ϩ cos 2u ϩ txy sin 2u 2 2 ϭ Ϫ14,500 psi sx Ϫ sy tx1y1 ϭ Ϫ sin 2u ϩ txy cos 2u ϭ Ϫ3,600 psi 2 sy1 ϭ sx ϩ sy Ϫ sx1 ϭ Ϫ4,900 psi x x 3,600 psi FOR ␪ ϭ 0: sx ϭ sx1 ϭ Ϫ14,500 psi sy ϭ sy1 ϭ Ϫ4,900 psi txy ϭ tx1y1 ϭ Ϫ3,600 psi
    • SECTION 7.2 Problem 7.2-16 the figure. Solve the preceding problem for the element shown in 433 Plane Stress y 26.7 MPa ␪ = 60° 66.7 MPa O x 25.0 MPa Solution 7.2-16 Plane stress Transform from ␪ ϭ 60Њ to ␪ ϭ 0Њ. Let: ␴x ϭ Ϫ26.7 MPa, ␴y ϭ 66.7 MPa, ␶xy ϭ Ϫ25.0 MPa, and ␪ ϭ Ϫ60Њ. sx1 ϭ sx ϩ sy ϩ sx Ϫ sy FOR ␪ ϭ 0: sx ϭ sx1 ϭ 65 MPa sy ϭ sy1 ϭ Ϫ25 MPa txy ϭ tx1y1 ϭ Ϫ28 MPa cos 2u ϩ txy sin 2u 2 2 ϭ 65 MPa sx Ϫ sy tx1y1 ϭ Ϫ sin 2u ϩ txy cos 2u ϭ Ϫ28 MPa 2 sy1 ϭ sx ϩ sy Ϫ sx1 ϭ Ϫ25 MPa 25 MPa y 65 MPa O x 28 MPa y Problem 7.2-17 A plate in plane stress is subjected to normal stresses ␴x and ␴y and shear stress ␶xy, as shown in the figure. At counterclockwise angles ␪ ϭ 40° and ␪ ϭ 80° from the x axis the normal stress is 5000 psi tension. If the stress ␴x equals 2000 psi tension, what are the stresses ␴y and ␶xy? Solution 7.2-17 Plane stress ␴x ϭ 2000 psi ␴y ϭ ? ␶xy ϭ ? At ␪ ϭ 40Њ and ␪ ϭ 80Њ; sx1 ϭ 5000 psi (tension) Find ␴y and ␶xy. sx1 ϭ sx ϩ sy 2 ϩ sx Ϫ sy 2 ϭ or ␶xy O sx1 ϭ 5000 ϭ or 2000 ϩ sy ϩ 2000 Ϫ sy cos 160Њ ϩ txy sin 160Њ 2 2 0.96985␴y ϩ 0.34202␶xy ϭ 4939.7 psi (2) SOLVE EQS. (1) AND (2): 2000 Ϫ sy ϩ cos 80Њ ϩ txy sin 80Њ 2 2 0.41318␴y ϩ 0.98481␶xy ϭ 3826.4 psi (1) ␴x = 2000 psi x FOR ␪ ϭ 80Њ: cos 2u ϩ txy sin 2u FOR ␪ ϭ 40Њ: sx1 ϭ 5000 2000 ϩ sy ␴y ␴y ϭ 4370 psi ␶xy ϭ 2050 psi
    • 434 CHAPTER 7 Analysis of Stress and Strain Problem 7.2-18 The surface of an airplane wing is subjected to plane stress with normal stresses ␴x and ␴y and shear stress ␶xy, as shown in the figure. At a counterclockwise angle ␪ ϭ 30° from the x axis the normal stress is 35 MPa tension, and at an angle ␪ ϭ 50° it is 10 MPa compression. If the stress ␴x equals 100 MPa tension, what are the stresses ␴y and ␶xy? Solution 7.2-18 Plane stress ␴x ϭ 100 MPa ␴y ϭ ? ␶xy ϭ ? At ␪ ϭ 30Њ, sx1 ϭ 35 MPa (tension) At ␪ ϭ 50Њ, sx1 ϭ Ϫ10 MPa (compression) Find ␴y and ␶xy sx1 ϭ sx ϩ sy 2 ϩ sx Ϫ sy 2 cos 2u ϩ txy sin 2u or ␶xy ␴x = 100 MPa x O FOR ␪ ϭ 50Њ: sx1 ϭ Ϫ10 ϭ or 100 ϩ sy ϩ 100 Ϫ sy cos 100Њ ϩ txy sin 100Њ 2 2 0.58682␴y ϩ 0.98481␶xy ϭ Ϫ51.318 MPa (2) ␴y ϭ Ϫ19.3 MPa sx1 ϭ 35 100 ϩ sy ␴y SOLVE EQS. (1) AND (2): FOR ␪ ϭ 30Њ: ϭ y ϩ ␶xy ϭ Ϫ40.6 MPa 100 Ϫ sy cos 60Њ ϩ txy sin 60Њ 2 2 0.25␴y ϩ 0.86603␶xy ϭ Ϫ40 MPa (1) y Problem 7.2-19 At a point in a structure subjected to plane stress, the stresses are ␴x ϭ Ϫ4000 psi, ␴y ϭ 2500 psi, and ␶xy ϭ 2800 psi (the sign convention for these stresses is shown in Fig. 7-1). A stress element located at the same point in the structure, but oriented at a counterclockwise angle ␪1 with respect to the x axis, is subjected to the stresses shown in the figure (␴b, ␶b, and 2000 psi). Assuming that the angle ␪1 is between zero and 90°, calculate the normal stress ␴b, the shear stress ␶b, and the angle ␪1. Solution 7.2-19 Plane stress ␴x ϭ Ϫ4000 psi ␴y ϭ 2500 psi ␶xy ϭ 2800 psi FOR ␪ ϭ ␪1: sx1 ϭ 2000 psi sy1 ϭ sb Find ␴b, ␶b, and ␪1 STRESS ␴b ␴b ϭ ␴x ϩ ␴y Ϫ 2000 psi ϭ Ϫ3500 psi ␶b 2000 psi ␪1 O ANGLE ␪1 sx1 ϭ ␶xy ϭ ␶b ␴b sx ϩ sy ϩ sx Ϫ sy cos 2u ϩ txy sin 2u 2 2 2000 psi ϭ Ϫ750 Ϫ 3250 cos 2␪1 ϩ 2800 sin 2␪1 or Ϫ65 cos 2␪1 ϩ 56 sin 2␪1 Ϫ 55 ϭ 0 Solve numerically: 2␪1ϭ 89.12Њ ␪1ϭ 44.56Њ SHEAR STRESS ␶b tb ϭ tx1y1 ϭ Ϫ ϭ 3290 psi sx Ϫ sy 2 sin 2u1 ϩ txy cos 2u1 x
    • SECTION 7.3 Principal Stresses and Maximum Shear Stresses Principal Stresses and Maximum Shear Stresses When solving the problems for Section 7.3, consider only the in-plane stresses (the stresses in the xy plane). Problem 7.3-1 An element in plane stress is subjected to stresses ␴x ϭ 6500 psi, ␴y ϭ 1700 psi, and ␶xy ϭ 2750 psi (see the figure for Problem 7.2-1). Determine the principal stresses and show them on a sketch of a properly oriented element. Solution 7.3-1 Principal stresses ␴x ϭ 6500 psi ␴y ϭ 1700 psi ␶xy ϭ 2750 psi PRINCIPAL STRESSES 2txy tan 2up ϭ Therefore, ␴1 ϭ 7750 psi and up1 ϭ 24.44Њ ␴2 ϭ 450 psi and up2 ϭ 114.44Њ ϭ 1.1458 sx Ϫ sy y 2␪pϭ 48.89Њ and ␪pϭ 24.44Њ 2␪pϭ 228.89Њ and ␪pϭ 114.44Њ sx1 ϭ sx ϩ sy 2 sx Ϫ sy ϩ 2 } 450 psi 7750 psi cos 2u ϩ txy sin 2u ␪p1 ϭ 24.44Њ O For 2␪pϭ 48.89Њ: sx1 ϭ 7750 psi For 2␪pϭ 228.89Њ: sx1 ϭ 450 psi x Problem 7.3-2 An element in plane stress is subjected to stresses ␴x ϭ 80 MPa, ␴y ϭ 52 MPa, and ␶xy ϭ 48 MPa (see the figure for Problem 7.2-2). Determine the principal stresses and show them on a sketch of a properly oriented element. Solution 7.3-2 Principal stresses ␴x ϭ 80 MPa ␴y ϭ 52 MPa ␶xy ϭ 48 MPa Therefore, ␴1 ϭ 116 MPa and up1 ϭ 36.87Њ ␴2 ϭ 16 MPa and up2 ϭ 126.87Њ } PRINCIPAL STRESSES y tan 2up ϭ 2txy sx Ϫ sy ϭ 3.429 16 MPa 116 MPa 2␪pϭ 73.74Њ and ␪pϭ 36.87Њ 2␪pϭ 253.74Њ and ␪pϭ 126.87Њ sx1 ϭ sx ϩ sy 2 ϩ sx Ϫ sy 2 cos 2u ϩ txy sin 2u For 2␪pϭ 73.74Њ: sx1 ϭ 116 MPa For 2␪pϭ 253.74Њ: sx1 ϭ 16 MPa ␪p1 ϭ 36.87Њ O x 435
    • 436 CHAPTER 7 Analysis of Stress and Strain Problem 7.3-3 An element in plane stress is subjected to stresses ␴x ϭ Ϫ9,900 psi, ␴y ϭ Ϫ3,400 psi, and ␶xy ϭ 3,600 psi (see the figure for Problem 7.2-3). Determine the principal stresses and show them on a sketch of a properly oriented element. Solution 7.3-3 Principal stresses ␴x ϭ Ϫ9900 psi ␴y ϭ Ϫ3400 psi ␶xy ϭ 3600 psi Therefore, ␴1 ϭ Ϫ1,800 psi and up1 ϭ 66.04Њ ␴2 ϭ Ϫ11,500 psi and up2 ϭ Ϫ23.96Њ PRINCIPAL STRESSES tan 2up ϭ 2txy sx Ϫ sy y ϭ Ϫ1.1077 2␪pϭ Ϫ47.92Њ and ␪pϭ Ϫ23.96Њ 2␪pϭ 132.08Њ and ␪pϭ 66.04Њ sx1 ϭ sx ϩ sy 2 ϩ sx Ϫ sy 2 11,500 psi 1,800 psi ␪p1 ϭ 66.04Њ cos 2u ϩ txy sin 2u O x For 2␪pϭ Ϫ47.92Њ: sx1 ϭ Ϫ11,500 psi For 2␪pϭ 132.08Њ: sx1 ϭ Ϫ1,800 psi Problem 7.3-4 An element in plane stress is subjected to stresses ␴x ϭ 42 MPa, ␴y ϭ Ϫ140 MPa, and ␶xy ϭ Ϫ60 MPa (see the figure for Problem 7.2-4). Determine the principal stresses and show them on a sketch of a properly oriented element. Solution 7.3-4 Principal stresses ␴x ϭ 42 MPa ␴y ϭ Ϫ140 MPa ␶xy ϭ Ϫ60 MPa Therefore, ␴1 ϭ 60 MPa and up1 ϭ Ϫ16.70Њ ␴2 ϭ Ϫ158 MPa and up2 ϭ 73.30Њ } PRINCIPAL STRESSES tan 2up ϭ 2txy sx Ϫ sy 2␪p ϭ Ϫ33.40Њ and ␪p ϭ Ϫ16.70Њ 2␪p ϭ 146.60Њ and ␪p ϭ 73.30Њ sx1 ϭ sx ϩ sy 2 ϩ y ϭ Ϫ0.6593 sx Ϫ sy 2 cos 2u ϩ txy sin 2u For 2␪p ϭ Ϫ33.40Њ: sx1 ϭ 60 MPa For 2␪p ϭ 146.60Њ: sx1 ϭ Ϫ158 MPa 60 MPa 158 MPa ␪p2 ϭ 73.30Њ O x }
    • SECTION 7.3 Principal Stresses and Maximum Shear Stresses Problem 7.3-5 An element in plane stress is subjected to stresses ␴x ϭ 7,500 psi, ␴y ϭ Ϫ20,500 psi, and ␶xy ϭ Ϫ4,800 psi (see the figure for Problem 7.2-5). Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented element. Solution 7.3-5 Maximum shear stresses ␴x ϭ 7,500 psi ␴y ϭ Ϫ20,500 psi ␶xy ϭ Ϫ4,800 psi tmax ϭ PRINCIPAL ANGLES 2txy tan 2up ϭ ϭ Ϫ0.3429 sx Ϫ sy 2␪p ϭ Ϫ18.92Њ and ␪p ϭ Ϫ9.46Њ 2␪p ϭ 161.08Њ and ␪p ϭ 80.54Њ sx1 ϭ sx ϩ sy 2 ϩ ϩ t2 ϭ 14,800 psi xy B 2 us1 ϭ up1 Ϫ 45Њ ϭ Ϫ54.46Њ and ␶ ϭ 14,800 psi MAXIMUM SHEAR STRESSES sx Ϫ sy 2 ¢ sx Ϫ sy ≤ 2 us2 ϭ up1 ϩ 45Њ ϭ 35.54Њ and ␶ ϭ Ϫ14,800 psi saver ϭ sx ϩ sy 2 } ϭ Ϫ6,500 psi y cos 2u ϩ txy sin 2u 6,500 psi 6,500 psi For 2␪p ϭ Ϫ18.92Њ: sx1 ϭ 8,300 psi For 2␪p ϭ 161.08Њ: sx1 ϭ Ϫ21,300 psi Therefore, up1 ϭ Ϫ9.46Њ ␪s2 ϭ 35.54Њ O x 14,800 psi Problem 7.3-6 An element in plane stress is subjected to stresses ␴x ϭ Ϫ25.5 MPa, ␴y ϭ 6.5 MPa, and ␶xy ϭ Ϫ12.0 MPa (see the figure for Problem 7.2-6). Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented element. Solution 7.3-6 Maximum shear stresses ␴x ϭ Ϫ25.5 MPa ␴y ϭ 6.5 MPa ␶xy ϭ Ϫ12.0 MPa PRINCIPAL ANGLES tan 2up ϭ 2txy sx Ϫ sy ϭ 0.7500 2␪p ϭ 36.87Њ and ␪pϭ 18.43Њ 2␪p ϭ 216.87Њ and ␪pϭ 108.43Њ sx1 ϭ sx ϩ sy 2 ϩ sx Ϫ sy 2 ϩ t2 ϭ 20.0 MPa xy B 2 us1 ϭ up1 Ϫ 45Њ ϭ 63.48Њ and ␶ ϭ 20.0 MPa us2 ϭ up1 ϩ 45Њ ϭ 153.43Њ and ␶ ϭ Ϫ20.0 MPa sx ϩ sy saver ϭ ϭ Ϫ9.5 MPa 2 MAXIMUM SHEAR STRESSES tmax ϭ ¢ sx Ϫ sy ≤ 2 } y 9.5 MPa cos 2u ϩ txy sin 2u For 2␪p ϭ 36.87Њ: sx1 ϭ Ϫ29.5 MPa For 2␪p ϭ 216.87Њ: sx1 ϭ 10.5 MPa Therefore, up1 ϭ 108.4Њ 9.5 MPa ␪s1 ϭ 63.43Њ O x 20.0 MPa 437
    • 438 CHAPTER 7 Analysis of Stress and Strain Problem 7.3-7 An element in plane stress is subjected to stresses ␴x ϭ Ϫ11,000 psi, ␴y ϭ Ϫ3,000 psi, and ␶xy ϭ Ϫ4200 psi (see the figure for Problem 7.2-7). Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented element. Solution 7.3-7 Maximum shear stresses ␴x ϭ Ϫ11,000 psi ␴y ϭ Ϫ3,000 psi ␶xy ϭ Ϫ4,200 psi tmax ϭ PRINCIPAL ANGLES 2txy tan 2up ϭ sx Ϫ sy ϭ 1.0500 2␪p ϭ 46.40Њ and ␪pϭ 23.20Њ 2␪p ϭ 226.40Њ and ␪pϭ 113.20Њ sx1 ϭ sx ϩ sy 2 ϩ ϩ t2 ϭ 5,800 psi xy B 2 us1 ϭ up1 Ϫ 45Њ ϭ 68.20Њ and ␶ ϭ 5,800 psi us2 ϭ up1 ϩ 45Њ ϭ 158.20Њ and ␶ ϭ Ϫ5,800 psi sx ϩ sy saver ϭ ϭ Ϫ7,000 psi 2 MAXIMUM SHEAR STRESSES sx Ϫ sy 2 ¢ sx Ϫ sy ≤ 2 } y 7,000 psi cos 2u ϩ txy sin 2u 7,000 psi For 2␪p ϭ 46.40Њ: sx1 ϭ Ϫ12,800 psi For 2␪p ϭ 226.40Њ: sx1 ϭ Ϫ1,200 psi Therefore, up1 ϭ 113.20Њ ␪s1 ϭ 68.20Њ O x 5,800 psi Problem 7.3-8 An element in plane stress is subjected to stresses ␴x ϭ Ϫ54 MPa, ␴y ϭ Ϫ12 MPa, and ␶xy ϭ 20 MPa (see the figure for Problem 7.2-8). Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented element. Solution 7.3-8 Maximum shear stresses ␴x ϭ Ϫ54 MPa ␴y ϭ Ϫ12 MPa ␶xy ϭ 20 MPa tmax ϭ PRINCIPAL ANGLES tan 2up ϭ 2txy sx Ϫ sy ϭ Ϫ0.9524 2␪p ϭ Ϫ43.60Њ and ␪p ϭ Ϫ21.80Њ 2␪p ϭ 136.40Њ and ␪p ϭ 68.20Њ sx1 ϭ sx ϩ sy 2 ϩ ϩ t2 ϭ 29.0 MPa xy B 2 us1 ϭ up1 Ϫ 45Њ ϭ 23.20Њ and ␶ ϭ 29.0 MPa us2 ϭ up1 ϩ 45Њ ϭ 113.20Њ and ␶ ϭ Ϫ29.0 MPa sx ϩ sy saver ϭ ϭ Ϫ33.0 MPa 2 MAXIMUM SHEAR STRESSES sx Ϫ sy 2 cos 2u ϩ txy sin 2u For 2␪p ϭ Ϫ43.60Њ: sx1 ϭ Ϫ62 MPa For 2␪p ϭ 136.40Њ: sx1 ϭ Ϫ4.0 MPa Therefore, up1 ϭ 68.20Њ ¢ sx Ϫ sy ≤ 2 y 33.0 MPa 33.0 MPa ␪s1 ϭ 23.20Њ O x 29.0 MPa }
    • SECTION 7.3 Problem 7.3-9 A shear wall in a reinforced concrete building is subjected to a vertical uniform load of intensity q and a horizontal force H, as shown in the first part of the figure. (The force H represents the effects of wind and earthquake loads.) As a consequence of these loads, the stresses at point A on the surface of the wall have the values shown in the second part of the figure (compressive stress equal to 1100 psi and shear stress equal to 480 psi). (a) Determine the principal stresses and show them on a sketch of a properly oriented element. (b) Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented element. Solution 7.3-9 Shear wall ␴x ϭ 0 ␴y ϭ Ϫ1100 psi ␶xy ϭ Ϫ480 psi 2txy sx Ϫ sy sx ϩ sy 2 ϩ ϭ Ϫ0.87273 sx Ϫ sy 2 1100 psi H 480 psi A A ϩ t2 ϭ 730 psi xy 2 B us1 ϭ up1 Ϫ 45Њ ϭ Ϫ65.56Њ and ␶ ϭ 730 psi us2 ϭ up1 ϩ 45Њ ϭ 24.44Њ and ␶ ϭ Ϫ730 psi sx ϩ sy saver ϭ ϭ Ϫ550 psi 2 tmax ϭ 2␪p ϭ Ϫ41.11Њ and ␪p ϭ Ϫ20.56Њ 2␪p ϭ 138.89Њ and ␪p ϭ 69.44Њ sx1 ϭ q (b) MAXIMUM SHEAR STRESSES (a) PRINCIPAL STRESSES tan 2up ϭ 439 Principal Stresses and Maximum Shear Stresses ¢ sx Ϫ sy ≤ 2 } cos 2u ϩ txy sin 2u y For 2␪p ϭ Ϫ41.11Њ: sx1 ϭ 180 psi For 2␪p ϭ 138.89Њ: sx1 ϭ Ϫ1280 psi Therefore, ␴1 ϭ 180 psi and up1 ϭ Ϫ20.56Њ ␴2 ϭ Ϫ1280 psi and up2 ϭ 69.44Њ 550 psi } 550 psi ␪s2 ϭ 24.44Њ y O 180 psi x 730 psi 1280 psi ␪p2 ϭ 69.44Њ O x Problem 7.3-10 A propeller shaft subjected to combined torsion and axial thrust is designed to resist a shear stress of 63 MPa and a compressive stress of 90 MPa (see figure). (a) Determine the principal stresses and show them on a sketch of a properly oriented element. (b) Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented element. 90 MPa 63 MPa
    • 440 CHAPTER 7 Analysis of Stress and Strain Solution 7.3-10 Propeller shaft ␴x ϭ Ϫ90 MPa ␴y ϭ 0 ␶xy ϭ Ϫ63 MPa (a) PRINCIPAL STRESSES 2txy tan 2up ϭ sx Ϫ sy sx ϩ sy 2 ϩ sx Ϫ sy 2 } ϩ t2 ϭ 77.4 MPa xy B 2 us1 ϭ up1 Ϫ 45Њ ϭ 72.23Њ and ␶ ϭ 77.4 MPa us2 ϭ up1 ϩ 45Њ ϭ 162.23Њ and ␶ ϭ Ϫ77.4 MPa sx ϩ sy saver ϭ ϭ Ϫ45 MPa 2 (b) MAXIMUM SHEAR STRESSES ϭ 1.4000 2␪p ϭ 54.46Њ and ␪p ϭ 27.23Њ 2␪p ϭ 234.46Њ and ␪p ϭ 117.23Њ sx1 ϭ Therefore, ␴1 ϭ 32.4 MPa and up1 ϭ 117.23Њ ␴2 ϭ Ϫ122.4 MPa and up2 ϭ 27.23Њ cos 2u ϩ txy sin 2u For 2␪p ϭ 54.46Њ: sx1 ϭ Ϫ122.4 MPa For 2␪p ϭ 234.46Њ: sx1 ϭ 32.4 MPa tmax ϭ ¢ sx Ϫ sy ≤ 2 } y y 45 MPa 45 MPa 32.4 MPa 122.4 MPa ␪s1 ϭ 72.23Њ ␪p2 ϭ 27.23Њ O O x x 77.4 MPa y Problems 7.3-11 through 7.3-16 An element in plane stress (see figure) is subjected to stresses ␴x, ␴y, and ␶xy. (a) Determine the principal stresses and show them on a sketch of a properly oriented element. (b) Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented element. ␴y ␶xy ␴x O ␴x ϭ 3500 psi, ␴y ϭ 1120 psi, ␶xy ϭ Ϫ1200 psi Data for 7.3-11 Solution 7.3-11 Plane stress ␴x ϭ 3500 psi ␴y ϭ 1120 psi ␶xy ϭ Ϫ1200 psi (a) PRINCIPAL STRESSES tan 2up ϭ 2txy sx Ϫ sy sx ϩ sy 2 ϩ Therefore, ␴1 ϭ 4000 psi and up1 ϭ Ϫ22.62Њ ␴2 ϭ 620 psi and up2 ϭ 67.38Њ sx Ϫ sy 2 cos 2u ϩ txy sin 2u For 2␪p ϭ Ϫ45.24Њ: sx1 ϭ 4000 psi For 2␪p ϭ 134.76Њ: sx1 ϭ 620 psi } y ϭ Ϫ1.0084 2␪p ϭ Ϫ45.24Њ and ␪p ϭ Ϫ22.62Њ 2␪p ϭ 134.76Њ and ␪p ϭ 67.38Њ sx1 ϭ x 4000 psi 620 psi ␪p2 ϭ 67.38Њ O x
    • SECTION 7.3 2 ¢ ≤ ϩ txy ϭ 1690 psi B 2 us1 ϭ up1 Ϫ 45Њ ϭ Ϫ67.62Њ and ␶ ϭ 1690 psi Principal Stresses and Maximum Shear Stresses (b) MAXIMUM SHEAR STRESSES sx Ϫ sy tmax ϭ us2 ϭ up1 ϩ 45Њ ϭ 22.38Њ and ␶ ϭ Ϫ1690 psi saver ϭ sx ϩ sy 2 y 2 2310 psi 2310 psi } ␪s2 ϭ 22.38Њ O ϭ 2310 psi x 1690 psi ␴x ϭ 2100 kPa, ␴y ϭ 300 kPa, ␶xy ϭ Ϫ560 kPa Data for 7.3-12 Solution 7.3-12 Plane stress ϩ t2 ϭ 1060 kPa xy B 2 us1 ϭ up1 Ϫ 45Њ ϭ Ϫ60.95Њ and ␶ ϭ 1060 kPa ␴x ϭ 2100 kPa ␴y ϭ 300 kPa ␶xy ϭ Ϫ560 kPa (b) MAXIMUM SHEAR STRESSES (a) PRINCIPAL STRESSES tmax ϭ tan 2up ϭ 2txy sx Ϫ sy ϭ Ϫ0.6222 sx ϩ sy 2 ϩ sx Ϫ sy 2 ≤ 2 cos 2u ϩ txy sin 2u } y For 2␪pϭ Ϫ31.89Њ: sx1 ϭ 2260 kPa 1200 kPa For 2␪pϭ 148.11Њ: sx1 ϭ 140 kPa 1200 kPa Therefore, ␴1 ϭ 2260 kPa and up1 ϭ Ϫ15.95Њ ␴2 ϭ 140 kPa and up2 ϭ 74.05Њ } y 2260 kPa 140 kPa ␪p2 ϭ 74.05Њ O sx Ϫ sy us2 ϭ up1 ϩ 45Њ ϭ 29.05Њ and ␶ ϭ Ϫ1060 kPa sx ϩ sy saver ϭ ϭ 1200 kPa 2 2␪p ϭ Ϫ31.89Њ and ␪p ϭ Ϫ15.95Њ 2␪p ϭ 148.11Њ and ␪p ϭ 74.05Њ sx1 ϭ ¢ x ␪s2 ϭ 29.05Њ O x 1060 kPa 441
    • 442 CHAPTER 7 Analysis of Stress and Strain ␴x ϭ 15,000 psi, ␴y ϭ 1,000 psi, ␶xy ϭ 2,400 psi Data for 7.3-13 Solution 7.3-13 Plane stress ␴x ϭ 15,000 psi ␴y ϭ 1,000 psi ␶xy ϭ 2,400 psi (a) PRINCIPAL STRESSES 2txy tan 2up ϭ sx Ϫ sy ϩ t2 ϭ 7,400 psi xy B 2 us1 ϭ up1 Ϫ 45Њ ϭ Ϫ35.54Њ and ␶ ϭ 7,400 psi us2 ϭ up1 ϩ 45Њ ϭ 54.46Њ and ␶ ϭ Ϫ7,400 psi sx ϩ sy saver ϭ ϭ 8,000 psi 2 (b) MAXIMUM SHEAR STRESSES tmax ϭ ϭ 0.34286 2␪p ϭ 18.92Њ and ␪p ϭ 9.46Њ 2␪p ϭ 198.92Њ and ␪p ϭ 99.46Њ sx ϩ sy sx Ϫ sy sx1 ϭ ϩ cos 2u ϩ txy sin 2u 2 2 For 2␪p ϭ 18.92Њ: sx1 ϭ 15,400 psi For 2␪p ϭ 198.92Њ: sx1 ϭ 600 psi ¢ sx Ϫ sy ≤ 2 } y 8000 psi 8000 psi Therefore, ␴1 ϭ 15,400 psi and up1 ϭ 9.46Њ ␴2 ϭ 600 psi and up2 ϭ 99.96Њ } ␪s2 ϭ 54.46Њ O y x 7400 psi 600 psi 15,400 psi ␪p1 ϭ 9.46Њ O x ␴x ϭ 16 MPa, ␴y ϭ Ϫ96 MPa, ␶xy ϭ Ϫ42 MPa Data for 7.3-14 Solution 7.3-14 Plane stress ␴x ϭ 16 MPa ␴y ϭ Ϫ96 MPa ␶xy ϭ Ϫ42 MPa (a) PRINCIPAL STRESSES tan 2up ϭ 2txy sx Ϫ sy y 30 MPa 110 MPa ϭ Ϫ0.7500 ␪p2 ϭ 71.57Њ 2␪p ϭ Ϫ36.87Њ and ␪p ϭ Ϫ18.43Њ 2␪p ϭ 143.13Њ and ␪p ϭ 71.57Њ sx1 ϭ sx ϩ sy 2 ϩ sx Ϫ sy 2 O cos 2u ϩ txy sin 2u For 2␪p ϭ Ϫ36.87Њ: sx1 ϭ 30 MPa For 2␪p ϭ 143.13Њ: sx1 ϭ Ϫ110 MPa Therefore, ␴1 ϭ 30 MPa and up1 ϭ Ϫ18.43Њ ␴2 ϭ Ϫ110 MPa and up2 ϭ 71.57Њ ¢ ≤ ϩ ϭ 70 MPa B 2 us1 ϭ up1 Ϫ 45Њ ϭ Ϫ63.43Њ and ␶ ϭ 70 MPa us2 ϭ up1 ϩ 45Њ ϭ 26.57Њ and ␶ ϭ Ϫ70 MPa sx ϩ sy saver ϭ ϭ Ϫ40 MPa 2 y } 40 MPa 40 MPa (b) MAXIMUM SHEAR STRESSES tmax ϭ sx Ϫ sy x ␪s2 ϭ 26.57Њ 2 t2 xy O } x 70 MPa
    • SECTION 7.3 ␴x ϭ Ϫ3000 psi, ␴y ϭ Ϫ12,000 psi, ␶xy ϭ 6000 psi Data for 7.3-15 Solution 7.3-15 Plane stress ␴x ϭ Ϫ3000 psi ␴y ϭ Ϫ12,000 psi ␶xy ϭ 6000 psi tmax ϭ sx Ϫ sy sx ϩ sy 2 ¢ sx Ϫ sy 2 ≤ 2 ϩ t2 ϭ 7500 psi xy us1 ϭ up1 Ϫ 45Њ ϭ Ϫ18.43Њ and ␶ ϭ 7500 psi 2txy ϭ 1.3333 us2 ϭ up1 ϩ 45Њ ϭ 71.57Њ and ␶ ϭ Ϫ7500 psi saver ϭ 2␪p ϭ 53.13Њ and ␪p ϭ 26.57Њ 2␪p ϭ 233.13Њ and ␪p ϭ 116.57Њ sx1 ϭ B (b) MAXIMUM SHEAR STRESSES (a) PRINCIPAL STRESSES tan 2up ϭ Principal Stresses and Maximum Shear Stresses ϩ sx Ϫ sy 2 sx ϩ sy 2 ϭ Ϫ7500 psi cos 2u ϩ txy sin 2u For 2␪p ϭ 53.13Њ: sx1 ϭ 0 For 2␪p ϭ 233.13Њ: sx1 ϭ Ϫ15,000 psi Therefore, ␴1 ϭ 0 and up1 ϭ 26.57Њ ␴2 ϭ Ϫ15,000 psi and up2 ϭ 116.57Њ } } y 7500 psi 7500 psi ␪s2 ϭ 71.57Њ x O y 7500 psi 15,000 psi ␪p1 ϭ 26.57Њ O Data for 7.3-16 x ␴x ϭ Ϫ100 MPa, ␴y ϭ 50 MPa, ␶xy ϭ Ϫ50 MPa Solution 7.3-16 Plane stress ␴x ϭ Ϫ100 MPa ␴y ϭ 50 MPa ␶xy ϭ Ϫ50 MPa (a) PRINCIPAL STRESSES tan 2up ϭ 2txy sx Ϫ sy ϭ 0.66667 y 2␪p ϭ 33.69Њ and ␪p ϭ 16.85Њ 2␪p ϭ 213.69Њ and ␪p ϭ 106.85Њ sx1 ϭ sx ϩ sy 2 ϩ sx Ϫ sy 2 65.1 MPa 115.1 MPa cos 2u ϩ txy sin 2u For 2␪p ϭ 33.69Њ: sx1 ϭ Ϫ115.1 MPa For 2␪p ϭ 213.69Њ: sx1 ϭ 65.1 MPa Therefore, ␴1 ϭ 65.1 MPa and up1 ϭ 106.85Њ ␴2 ϭ Ϫ115.1 MPa and up2 ϭ 16.85Њ } ␪p2 ϭ 16.85Њ O x 443
    • 444 CHAPTER 7 Analysis of Stress and Strain B 2 us1 ϭ up1 Ϫ 45Њ ϭ 61.85Њ and ␶ ϭ 90.1 MPa us2 ϭ up1 ϩ 45Њ ϭ 151.85Њ and ␶ ϭ Ϫ90.1 MPa sx ϩ sy saver ϭ ϭ Ϫ25.0 MPa 2 (b) MAXIMUM SHEAR STRESSES ¢ tmax ϭ sx Ϫ sy ≤ y 2 25.0 MPa ϩ t2 ϭ 90.1 MPa xy 25.0 MPa } ␪s1 ϭ 61.85Њ O x 90.1 MPa y Problem 7.3-17 At a point on the surface of a machine component the stresses acting on the x face of a stress element are ␴x ϭ 6500 psi and ␶xy ϭ 2100 psi (see figure). What is the allowable range of values for the stress ␴y if the maximum shear stress is limited to ␶0 ϭ 2900 psi? ␴y ␶xy = 2100 psi O Solution 7.3-17 Allowable range of values ␴x ϭ 6500 psi ␶xy ϭ 2100 psi ␴y ϭ ? Find the allowable range of values for ␴y if the maximum allowable shear stresses is ␶0 ϭ 2900 psi. tmax ϭ or B t2 ϭ ¢ max ¢ sx Ϫ sy 2 sx Ϫ sy 2 SOLVE FOR ␴y ≤ 2 ≤ Substitute numerical values: sy ϭ 6500 psi Ϯ 2͙(2900 psi) 2 Ϫ (2100 psi) 2 ϭ 6500 psi Ϯ 4000 psi Therefore, 2500 psi Յ ␴y Յ 10,500 psi 2 ϩ t2 xy Eq. (1) GRAPH OF ␶max ϩ t2 xy B From Eq. (1): Eq. (2) tmax ϭ sy ϭ sx Ϯ 2͙t2 Ϫ t2 max xy ␶max (ksi) ¢ 6500 Ϫ sy 2 ≤ 2 ϩ (2100) 2 6 Eq. (3) 4 2.9 ksi (ϭ ␶o) 2.1 ksi 2 2.5 ᎐5 ␴x = 6500 psi x 0 6.5 5 10.5 10 15 ␴y (ksi) Eq. (3)
    • SECTION 7.3 Principal Stresses and Maximum Shear Stresses Problem 7.3-18 At a point on the surface of a machine component the stresses acting on the x face of a stress element are ␴x ϭ 45 MPa and ␶xy ϭ 30 MPa (see figure). What is the allowable range of values for the stress ␴y if the maximum shear stress is limited to ␶0 ϭ 34 MPa? y ␴y ␶xy ␴x = 100 MPa x O Solution 7.3-18 Allowable range of values ␴x ϭ 45 MPa ␶xy ϭ 30 MPa ␴y ϭ ? Find the allowable range of values for ␴y if the maximum allowable shear stresses is ␶0 ϭ 34 MPa. tmax ϭ or B t2 ϭ ¢ max ¢ sx Ϫ sy 2 sx Ϫ sy 2 ≤ 2 ≤ SOLVE FOR ␴y sy ϭ sx Ϯ 2͙t2 Ϫ t2 max xy 2 ϩ t2 xy Substitute numerical values: Eq. (1) ϩ t2 xy sy ϭ 45 MPa Ϯ 2͙(34 MPa) 2 Ϫ (30 MPa) 2 ϭ 45 MPa Ϯ 32 MPa Therefore, 13 MPa Յ ␴y Յ 77 MPa Eq. (2) B GRAPH OF ␶max From Eq. (1): tmax ϭ ¢ 45 Ϫ sy 2 Eq. (3) 40 ≤ 2 ϩ (30) 2 Eq. (3) 34 MPa (ϭ ␶o) 30 MPa ␶max (MPa) 30 20 10 13 ᎐ 20 0 20 77 45 40 60 Problem 7.3-19 An element in plane stress is subjected to stresses ␴x ϭ 6500 psi and ␶xy ϭ Ϫ2800 psi (see figure). It is known that one of the principal stresses equals 7300 psi in tension. (a) Determine the stress ␴y. (b) Determine the other principal stress and the orientation of the principal planes; then show the principal stresses on a sketch of a properly oriented element. 80 100 ␴y (MPa) y ␴y 6500 psi O x 2800 psi 445
    • 446 CHAPTER 7 Analysis of Stress and Strain Solution 7.3-19 Plane stress ␴x ϭ 6500 psi ␶xy ϭ Ϫ2800 psi ␴y ϭ ? One principal stress ϭ 7300 psi (tension) (b) PRINCIPAL STRESSES tan 2up ϭ (a) STRESS ␴y Because ␴x is smaller than the given principal stress, we know that the given stress is the larger principal stress. ␴1 ϭ 7300 psi ϩ t2 xy 2 B 2 Substitute numerical values and solve for ␴y: ␴y ϭ Ϫ2500 psi s1 ϭ sx ϩ sy ϩ ¢ sx Ϫ sy ≤ 2 2txy sx Ϫ sy ϭ Ϫ0.62222 2␪p ϭ Ϫ31.891Њ and ␪p ϭ Ϫ15.945Њ 2␪p ϭ 148.109Њ and ␪p ϭ 74.053Њ sx1 ϭ sx ϩ sy 2 ϩ sx Ϫ sy 2 cos 2u ϩ txy sin 2u For 2␪p ϭ Ϫ31.891Њ: sx1 ϭ 7300 psi For 2␪p ϭ 148.109Њ: sx1 ϭ Ϫ3300 psi Therefore, ␴1 ϭ 7300 psi and up1 ϭ Ϫ15.95Њ ␴2 ϭ Ϫ3300 psi and up2 ϭ 74.05Њ } y 3300 psi 7300 psi ␪p2 ϭ 74.05Њ O x y Problem 7.3-20 An element in plane stress is subjected to stresses ␴x ϭ Ϫ68.5 MPa and ␶xy ϭ 39.2 MPa (see figure). It is known that one of the principal stresses equals 26.3 MPa in tension. (a) Determine the stress ␴y. (b) Determine the other principal stress and the orientation of the principal planes; then show the principal stresses on a sketch of a properly oriented element. Solution 7.3-20 Plane stress ␴x ϭ Ϫ68.5 MPa ␶xy ϭ 39.2 MPa ␴y ϭ ? One principal stress ϭ 26.3 MPa (tension) (a) STRESS ␴y Because ␴x is smaller than the given principal stress, we know that the given stress is the larger principal stress. ␴y 39.2 MPa O 68.5 MPa x ϩ t2 xy 2 B 2 Substitute numerical values and solve for ␴y: ␴y ϭ 10.1 MPa ␴1 ϭ 26.3 MPa s1 ϭ sx ϩ sy ϩ ¢ sx Ϫ sy ≤ 2
    • SECTION 7.4 Mohr’s Circle for Plane Stress (b) PRINCIPAL STRESSES tan 2up ϭ 2txy sx Ϫ sy ϭ Ϫ0.99746 y 2␪p ϭ Ϫ44.93Њ and ␪p ϭ Ϫ22.46Њ 2␪p ϭ 135.07Њ and ␪p ϭ 67.54Њ sx1 ϭ sx ϩ sy 2 ϩ sx Ϫ sy 2 26.3 MPa cos 2u ϩ txy sin 2u 84.7 MPa ␪p1 ϭ 67.54Њ For 2␪p ϭ Ϫ44.93Њ: sx1 ϭ Ϫ84.7 MPa For 2␪p ϭ 135.07Њ: sx1 ϭ 26.3 MPa O x Therefore, ␴1 ϭ 26.3 MPa and up1 ϭ 67.54Њ ␴2 ϭ Ϫ84.7 MPa and up2 ϭ Ϫ22.46Њ } Mohr’s Circle for Plane Stress The problems for Section 7.4 are to be solved using Mohr’s circle. Consider only the in-plane stresses (the stresses in the xy plane). y Problem 7.4-1 An element in uniaxial stress is subjected to tensile stresses ␴x ϭ 14,500 psi, as shown in the figure. Using Mohr’s circle, determine (a) the stresses acting on an element oriented at a counterclockwise angle ␪ ϭ 24° from the x axis and (b) the maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements. Solution 7.4-1 Uniaxial stress ␴x ϭ 14,500 psi ␴y ϭ 0 ␶xy ϭ 0 (a) ELEMENT AT ␪ ϭ 24Њ (All stresses in psi) 2␪ ϭ 48Њ ␪ ϭ 24Њ R ϭ 7250 psi Point C: sx1 ϭ 7250 psi 14,500 psi O Point D: sx1 ϭ R ϩ R cos 2u ϭ 12,100 psi tx1y1 ϭ R sin 2u ϭ Ϫ5390 psi Point DЈ: sx1 ϭ R Ϫ R cos 2u ϭ 2400 psi tx1y1 ϭ 5390 psi y S2 D (␪ ϭ 24Њ) 2␪s2 R O B (␪ ϭ 90Њ) ␴x A 2␪s1 = ᎐ 90Њ (␪ ϭ 0) 1 R D' ␶x1y1 2400 psi D D' 12,100 psi 2␪ C x O ␪ ϭ 24Њ x 5390 psi S1 14,500 447
    • 448 CHAPTER 7 Analysis of Stress and Strain y (b) MAXIMUM SHEAR STRESSES S2 7250 psi 7250 psi Point S1: 2us1 ϭ Ϫ90Њ us1 ϭ Ϫ45Њ ␶max ϭ R ϭ 7250 psi Point S2: 2us2 ϭ 90Њ us2 ϭ 45Њ ␶min ϭ ϪR ϭ Ϫ7250 psi ␴aver ϭ R ϭ 7250 psi ␪s2 ϭ 45Њ x 7250 psi O S1 y Problem 7.4-2 An element in uniaxial stress is subjected to tensile stresses ␴x ϭ 55 MPa, as shown in the figure. Using Mohr’s circle, determine (a) the stresses acting on an element oriented at an angle ␪ ϭ Ϫ30° from the x axis (minus means clockwise) and (b) the maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements. 55 MPa O Solution 7.4-2 Uniaxial stress ␴x ϭ 55 MPa ␴y ϭ 0 ␶xy ϭ 0 y (a) ELEMENT AT ␪ ϭ Ϫ30Њ (All stresses in MPa) 2␪ ϭ Ϫ60Њ ␪ ϭ Ϫ30Њ Point C: sx1 ϭ 27.5 MPa x D' 13.8 MPa R ϭ 27.5 MPa 23.8 MPa x ␪ ϭ –30Њ O S2 D' 41.2 MPa R O D B (␪ ϭ 90Њ) C 2␪ = ᎐60Њ A (␪ ϭ 0) (b) MAXIMUM SHEAR STRESSES ␴x1 R S1 ␶x1y1 55 MPa D (␪ ϭ ᎐30Њ) Point S1: 2us1 ϭ Ϫ90Њ us1 ϭ Ϫ45Њ ␶max ϭ R ϭ 27.5 MPa Point S2: 2us2 ϭ 90Њ us2 ϭ 45Њ ␶min ϭ ϪR ϭ Ϫ27.5 MPa ␴aver ϭ R ϭ 27.5 MPa y Point D: sx1 ϭ R ϩ R cos ƒ 2u ƒ ϭ R(1 ϩ cos 60Њ) ϭ 41.2 MPa tx1y1 ϭ R sin ƒ 2u ƒ ϭ R sin 60Њ ϭ 23.8 MPa Point DЈ: sx1 ϭ R Ϫ R cos ƒ 2u ƒ ϭ 13.8 MPa tx1y1 ϭ ϪR sin ƒ 2u ƒ ϭ Ϫ23.8 MPa S2 27.5 MPa ␪s2 ϭ 45Њ x 27.5 MPa O S1 27.5 MPa
    • SECTION 7.4 Problem 7.4-3 An element in uniaxial stress is subjected to compressive stresses of magnitude 5600 psi, as shown in the figure. Using Mohr’s circle, determine (a) the stresses acting on an element oriented at a slope of 1 on 2 (see figure) and (b) the maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements. y 1 2 O 5600 psi Solution 7.4-3 Uniaxial stress ␴x ϭ Ϫ5600 psi ␴y ϭ 0 ␶xy ϭ 0 1120 psi ␪ 1 ϭ 26.565Њ 2 2 D 4480 psi D' ␪ ϭ 26.57Њ 2␪ ϭ 53.130Њ ␪ ϭ 26.57Њ R ϭ 2800 psi 1 x y (a) ELEMENT AT A SLOPE OF 1 ON 2 (All stresses in psi) u ϭ arctan 449 Mohr’s Circle for Plane Stress x O 2240 psi Point C: sx1 ϭ Ϫ2800 psi (b) MAXIMUM SHEAR STRESSES S2 A (␪ ϭ 0) 2␪s2 = ᎐90Њ C 2␪ = 53.13Њ R 2␪s1 S1 D 5600 Point S1: 2us1 ϭ 90Њ us1 ϭ 45Њ ␶max ϭ R ϭ 2800 psi D' R R O B (␪ ϭ 90Њ) ␴x1 Point S2: 2us2 ϭ Ϫ90Њ us2 ϭ Ϫ45Њ ␶min ϭ ϪR ϭ Ϫ2800 psi ␴aver ϭ ϪR ϭ Ϫ2800 psi y S1 2800 psi ␶x1y1 2800 psi S2 Point D: sx1 ϭ ϪR Ϫ R cos 2u ϭ Ϫ4480 psi tx1y1 ϭ R sin 2u ϭ 2240 psi ␪s1 ϭ 45Њ O x 2800 psi Point DЈ: sx1 ϭ ϪR ϩ R cos 2u ϭ Ϫ1120 psi tx1y1 ϭ ϪR sin 2u ϭ Ϫ2240 psi Problem 7.4-4 An element in biaxial stress is subjected to stresses ␴x ϭ Ϫ60 MPa and ␴y ϭ 20 MPa, as shown in the figure. Using Mohr’s circle, determine (a) the stresses acting on an element oriented at a counterclockwise angle ␪ ϭ 22.5° from the x axis and (b) the maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements. y 20 MPa 60 MPa O x
    • 450 CHAPTER 7 Analysis of Stress and Strain Solution 7.4-4 Biaxial stress ␴x ϭ Ϫ60 MPa ␴y ϭ 20 MPa y ␶xy ϭ 0 8.28 MPa (a) ELEMENT AT ␪ ϭ 22.5Њ 28.28 MPa D' (b) MAXIMUM SHEAR STRESSES R (␪ ϭ 22.5Њ) B (␪ ϭ 90Њ) ␴x1 C O 2␪ R 2␪s1 Point S1: 2us1 ϭ 90Њ us1 ϭ 45Њ ␶max ϭ R ϭ 40 MPa Point S2: 2us2 ϭ Ϫ90Њ us2 ϭ Ϫ45Њ ␶min ϭ ϪR ϭ Ϫ40 MPa ␴aver ϭ Ϫ20 MPa S1 20 40 ␪ ϭ 22.5Њ x O R ϭ 40 MPa S2 D 48.28 MPa D' (All stresses in MPa) 2␪ ϭ 45Њ ␪ ϭ 22.5Њ 2R ϭ 60 ϩ 20 ϭ 80 MPa Point C: sx1 ϭ Ϫ20 MPa A (␪ ϭ 0) D 60 y 20 20 MPa ␶x1y1 20 MPa 40 MPa ␪s1 ϭ 45Њ Point D: sx1 ϭ Ϫ20 Ϫ R cos 2u ϭ Ϫ48.28 MPa tx1y1 ϭ R sin 2u ϭ 28.28 MPa Point DЈ: sx1 ϭ R cos 2u Ϫ 20 ϭ 8.28 MPa tx1y1 ϭ ϪR sin 2u ϭ Ϫ28.28 MPa S1 x S2 O 20 MPa y Problem 7.4-5 An element in biaxial stress is subjected to stresses ␴x ϭ 6000 psi and ␴y ϭ Ϫ1500 psi, as shown in the figure. Using Mohr’s circle, determine (a) the stresses acting on an element oriented at a counterclockwise angle ␪ ϭ 60° from the x axis and (b) the maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements. 1500 psi 6000 psi O Solution 7.4-5 (␪ ϭ 60Њ) S 2 D R Biaxial stress ␴x ϭ 6000 psi ␴y ϭ Ϫ1500 psi ␶xy ϭ 0 (a) ELEMENT AT ␪ ϭ 60Њ (All stresses in psi) 2␪ ϭ 120Њ ␪ ϭ 60Њ 2R ϭ 7500 psi R ϭ 3750 psi Point C: sx1 ϭ 2250 psi x 2␪ = 120Њ 60Њ B (␪ ϭ 90Њ) C O A (␪ ϭ 0) ␴x1 R 30Њ S1 D' 2250 1500 6000 ␶x1y1
    • SECTION 7.4 Point D: sx1 ϭ 2250 Ϫ R cos 60Њ ϭ 375 psi tx1y1 ϭ ϪR sin 60Њ ϭ Ϫ3248 psi Point DЈ: sx1 ϭ 2250 ϩ R cos 60Њ ϭ 4125 psi tx1y1 ϭ R sin 60Њ ϭ 3248 psi y 375 psi 451 Mohr’s Circle for Plane Stress (b) MAXIMUM SHEAR STRESSES Point S1: 2us1 ϭ Ϫ90Њ us1 ϭ Ϫ45Њ ␶max ϭ R ϭ 3750 psi Point S2: 2us2 ϭ 90Њ us2 ϭ 45Њ ␶min ϭ ϪR ϭ Ϫ3750 psi y ␴aver ϭ 2250 psi S2 2250 psi 4125 psi 2250 psi D ␪ ϭ 60Њ ␪s2 ϭ 45Њ x 3250 psi O x 3750 psi O S1 D' y Problem 7.4-6 An element in biaxial stress is subjected to stresses ␴x ϭ Ϫ24 MPa and ␴y ϭ 63 MPa, as shown in the figure. Using Mohr’s circle, determine (a) the stresses acting on an element oriented at a slope of 1 on 2.5 (see figure) and (b) the maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements. Solution 7.4-6 Biaxial stress ␴x ϭ Ϫ24 MPa ␴y ϭ 63 MPa ␶xy ϭ 0 (a) ELEMENT AT A SLOPE OF 1 ON 2.5 1 ϭ 21.801Њ (All stresses in MPa) u ϭ arctan 2.5 2␪ ϭ 43.603Њ 1 ␪ ϭ 21.801Њ ␪ 2R ϭ 87 MPa 2.5 R ϭ 43.5 MPa Point C: sx1 ϭ 19.5 MPa Point D: sx1 ϭ ϪR cos 2u ϩ 19.5 ϭ Ϫ12 MPa tx1y1 ϭ R sin 2u ϭ 30 MPa S2 D' R 43.603Њ C A (␪ ϭ 0) O B (␪ ϭ 90Њ) ␴x1 63 MPa 1 2.5 24 MPa x O Point DЈ: sx1 ϭ 19.5 ϩ R cos 2u ϭ 51 MPa tx1y1 ϭ ϪR sin 2u ϭ Ϫ30 MPa y 51 MPa D 12 MPa D' ␪ ϭ 21.80Њ x O 30 MPa (b) MAXIMUM SHEAR STRESSES Point S1: 2us1 ϭ 90Њ us1 ϭ 45Њ ␶max ϭ R ϭ 43.5 MPa Point S2: 2us2 ϭ Ϫ90Њ us2 ϭ Ϫ45Њ ␶min ϭ ϪR ϭ Ϫ43.5 MPa ␴aver ϭ 19.5 MPa y S1 19.5 MPa 2␪ ␪s1 ϭ 45Њ R D S2 S1 19.5 63 24 ␶x1y1 19.5 MPa O x 43.5 MPa
    • 452 CHAPTER 7 Analysis of Stress and Strain Problem 7.4-7 An element in pure shear is subjected to stresses ␶xy ϭ 3000 psi, as shown in the figure. Using Mohr’s circle, determine (a) the stresses acting on an element oriented at a counterclockwise angle ␪ ϭ 70° from the x axis and (b) the principal stresses. Show all results on sketches of properly oriented elements. y 3000 psi O Solution 7.4-7 Pure shear ␴x ϭ 0 ␴y ϭ 0 ␶xy ϭ 3000 psi x y (a) ELEMENT AT ␪ ϭ 70Њ (All stresses in psi) 2␪ ϭ 140Њ ␪ ϭ 70Њ R ϭ 3000 psi Origin O is at center of circle. D 1930 psi 1930 psi ␪ ϭ 70Њ B (␪ ϭ 90Њ) x O D' D 2300 psi R P2 O R 50Њ 2␪ = 140Њ P1 ␴x1 2␪p1 3000 psi D' A (␪ ϭ 0) ␶x1y1 Point D: sx1 ϭ R cos 50Њ ϭ 1928 psi tx1y1 ϭ ϪR sin 50Њ ϭ Ϫ2298 psi (b) PRINCIPAL STRESSES Point P1: 2up1 ϭ 90Њ up1 ϭ 45Њ ␴1 ϭ R ϭ 3000 psi Point P2: 2up2 ϭ Ϫ90Њ up2 ϭ Ϫ45Њ ␴2 ϭ ϪR ϭ Ϫ3000 psi y 3000 psi 3000 psi Point DЈ: sx1 ϭ ϪR cos 50Њ ϭ Ϫ1928 psi tx1y1 ϭ R sin 50Њ ϭ 2298 psi ␪p1 ϭ 45Њ P1 x O P2 Problem 7.4-8 An element in pure shear is subjected to stresses ␶xy ϭ Ϫ16 MPa, as shown in the figure. Using Mohr’s circle, determine (a) the stresses acting on an element oriented at a counterclockwise angle ␪ ϭ 20° from the x axis and (b) the principal stresses. Show all results on sketches of properly oriented elements. y O x 16 MPa
    • SECTION 7.4 Solution 7.4-8 ␴x ϭ 0 ␴y ϭ 0 Mohr’s Circle for Plane Stress Pure shear ␶xy ϭ Ϫ16 MPa y (a) ELEMENT AT ␪ ϭ 20Њ 10.3 MPa (All stresses in MPa) 2␪ ϭ 40Њ ␪ ϭ 20Њ R ϭ 16 MPa Origin O is at center of circle. D 10.3 MPa D' ␪ ϭ 20Њ x O D (␪ ϭ 20Њ) 12.3 MPa A (␪ ϭ 0) 2␪ R P2 (b) PRINCIPAL STRESSES 16 2␪p2 R P1 O 2␪p1 ␴x1 Point P1: 2up1 ϭ 270Њ up1 ϭ 135Њ ␴1 ϭ R ϭ 16 MPa Point P2: 2up2 ϭ 90Њ up2 ϭ 45Њ ␴2 ϭ ϪR ϭ Ϫ16 MPa y D' B (␪ ϭ 90Њ) ␶x1y1 Point D: sx1 ϭ ϪR sin 2u ϭ Ϫ10.28 MPa tx1y1 ϭ ϪR cos 2u ϭ Ϫ12.26 MPa P2 16 MPa 16 MPa ␪p2 ϭ 45Њ P1 O Point DЈ: sx1 ϭ R sin 2u ϭ 10.28 MPa tx1y1 ϭ R cos 2u ϭ 12.26 MPa x Problem 7.4-9 An element in pure shear is subjected to stresses ␶xy ϭ 4000 psi, as shown in the figure. Using Mohr’s circle, determine (a) the stresses acting on an element oriented at a slope of 3 on 4 (see figure) and (b) the principal stresses. Show all results on sketches of properly oriented elements. y 3 4 x O 4000 psi Solution 7.4-9 ␴x ϭ 0 ␴y ϭ 0 Pure shear ␶xy ϭ 4000 psi (a) ELEMENT AT A SLOPE OF 3 ON 4 3 (All stresses in psi) u ϭ arctan ϭ 36.870Њ 4 2␪ ϭ 73.740Њ ␪ ϭ 36.870Њ 3 R ϭ 4000 psi ␪ Origin O is at center of circle. B (␪ ϭ 90Њ) D' R 16.260Њ P1 P2 O 2␪ 4 R 2␪p1 R A (␪ ϭ 0) ␶x1y1 ␴x1 D 4000 453
    • 454 CHAPTER 7 Analysis of Stress and Strain Point D: sx1 ϭ R cos 16.260Њ ϭ 3840 psi tx1y1 ϭ R sin 16.260Њ ϭ 1120 psi (b) PRINCIPAL STRESSES Point P1: 2up1 ϭ 90Њ up1 ϭ 45Њ ␴1 ϭ R ϭ 4000 psi Point P2: 2up2 ϭ Ϫ90Њ up2 ϭ Ϫ45Њ ␴2 ϭ ϪR ϭ Ϫ4000 psi Point DЈ: sx1 ϭ ϪR cos 16.260Њ ϭ Ϫ3840 psi tx1y1 ϭ ϪR sin 16.260Њ ϭ Ϫ1120 psi y 3840 psi y D P1 3840 psi D' 4000 psi 4000 psi ␪ ϭ 36.87Њ ␪p1 ϭ 45Њ x O O 1120 psi x P2 y Problems 7.4-10 through 7.4-15 An element in plane stress is subjected to stresses ␴x, ␴y, and ␶xy (see figure). Using Mohr’s circle, determine the stresses acting on an element oriented at an angle ␪ from the x axis. Show these stresses on a sketch of an element oriented at the angle ␪. (Note: The angle ␪ is positive when counterclockwise and negative when clockwise.) ␴y ␶xy ␴x O Data for 7.4-10 ␴x ϭ 21 MPa, ␴y ϭ 11 MPa, ␶xy ϭ 8 MPa, ␪ ϭ 50° Solution 7.4-10 Plane stress (angle ␪) ␴x ϭ 21 MPa ␴y ϭ 11 MPa ␶xy ϭ 8 MPa ␪ ϭ 50Њ (All stresses in MPa) (␪ ϭ 90Њ) B 11 ␤ ϭ 2␪ Ϫ ␣ ϭ 100Њ Ϫ ␣ ϭ 42.01Њ Point D (␪ ϭ 50Њ): sx1 ϭ 16 ϩ R cos b ϭ 23.01 MPa tx1y1 ϭ ϪR sin b ϭ Ϫ6.31 MPa D (␪ ϭ 50Њ) R R 8 5 O ␣ 2␪ ϭ ␤ 100Њ ␴x1 C R 16 Point DЈ (␪ ϭ Ϫ40Њ): sx1 ϭ 16 Ϫ R cos b ϭ 8.99 MPa tx1y1 ϭ R sin b ϭ 6.31 MPa y 8 23.01 MPa 8.99 MPa 80Њ A (␪ ϭ 0) D' 21 ␶x1y1 R ϭ ͙(5) 2 ϩ (8) 2 ϭ 9.4340 MPa 8 ␣ ϭ arctan ϭ 57.99Њ 5 ␪ ϭ 50Њ D' O D x 6.31 MPa x
    • SECTION 7.4 Mohr’s Circle for Plane Stress ␴x ϭ 4500 psi, ␴y ϭ 14,100 psi, ␶xy ϭ Ϫ3100 psi, ␪ ϭ Ϫ55° Data for 7.4-11 Plane stress (angle ␪) Solution 7.4-11 ␴x ϭ 4500 psi ␴y ϭ 14,100 psi ␶xy ϭ Ϫ3100 psi ␪ ϭ Ϫ55Њ ␤ ϭ 180Њ Ϫ 110Њ Ϫ ␣ ϭ 37.14Њ Point D (␪ ϭ Ϫ55Њ): sx1 ϭ 9300 ϩ R cos b ϭ 13,850 psi tx1y1 ϭ ϪR sin b ϭ Ϫ3450 psi (All stresses in psi) (␪ ϭ 0) A 2␪ ϭ ᎐110Њ R R 3100 4800 ␣ ␤ C 3100 4500 O Point DЈ(␪ ϭ 35Њ): sx1 ϭ 9300 Ϫ R cos b ϭ 4750 psi tx1y1 ϭ R sin b ϭ 3450 psi D (␪ ϭ ᎐55Њ) y ␴x1 3450 psi R 9300 4750 psi B (␪ ϭ 90Њ) D' 14,100 D' x O ␶x1y1 ␪ ϭ ᎐55Њ R ϭ ͙(4800) ϩ (3100) ϭ 5714 psi 3100 ␣ ϭ arctan ϭ 32.86Њ 4800 2 Data for 7.4-12 2 13,850 psi D ␴x ϭ Ϫ44 MPa, ␴y ϭ Ϫ194 MPa, ␶xy ϭ Ϫ36 MPa, ␪ ϭ Ϫ35° Solution 7.4-12 Plane stress (angle ␪) ␴x ϭ Ϫ44 MPa ␴y ϭ Ϫ194 MPa ␶xy ϭ Ϫ36 MPa ␪ ϭ Ϫ35Њ (All stresses in MPa) Point D (␪ ϭ Ϫ35Њ): sx1 ϭ Ϫ119 ϩ R cos b ϭ Ϫ59.5 MPa tx1y1 ϭ R sin b ϭ 58.2 MPa Point DЈ(␪ ϭ 55Њ): sx1 ϭ Ϫ119 Ϫ R cos b ϭ Ϫ178.5 MPa tx1y1 ϭ ϪR sin b ϭ Ϫ58.2 MPa 44 D' A (␪ ϭ 0) R 75 ␣ 36 B R (␪ ϭ 90Њ) C 36 ␤ R 2␪ ϭ ᎐70Њ O ␴x1 y D' 178.5 MPa D (␪ ϭ ᎐35Њ) 119 194 R ϭ ͙(75) 2 ϩ (36) 2 ϭ 83.19 MPa 36 ϭ 25.64Њ ␣ ϭ arctan 75 ␤ ϭ 70Њ Ϫ ␣ ϭ 44.36Њ ␶x1y1 58.2 MPa x O ␪ ϭ ᎐35Њ 59.5 MPa D 455
    • 456 CHAPTER 7 Data for 7.4-13 Analysis of Stress and Strain ␴x ϭ Ϫ1520 psi, ␴y ϭ Ϫ480 psi, ␶xy ϭ 280 psi, ␪ ϭ 18° Plane stress (angle ␪) Solution 7.4-13 ␤ ϭ ␣ ϩ 36Њ ϭ 64.30Њ Point D (␪ ϭ 18Њ): sx1 ϭ Ϫ1000 Ϫ R cos b ϭ Ϫ1256 psi tx1y1 ϭ R sin b ϭ 532 psi ␴x ϭ Ϫ1520 psi ␴y ϭ Ϫ480 psi ␪ ϭ 18Њ ␶xy ϭ 280 psi (All stresses in psi) D' 480 B (␪ ϭ 90°) R ␣ ␤ 280 2␪ A ϭ 36Њ R (␪ ϭ 0) D (␪ ϭ 18Њ) Point DЈ(␪ ϭ 108Њ): sx1 ϭ Ϫ1000 ϩ R cos b ϭ Ϫ744 psi tx1y1 ϭ ϪR sin b ϭ Ϫ532 psi 280 520 C ␴x1 O y 280 744 psi 532 psi 1000 D' 1520 1256 psi ␪ ϭ 18Њ ␶x1y1 x D O R ϭ ͙(520) ϩ (280) ϭ 590.6 psi 280 ϭ 28.30Њ ␣ ϭ arctan 520 2 Data for 7.4-14 2 ␴x ϭ 31 MPa, ␴y ϭ Ϫ5 MPa, ␶xy ϭ 33 MPa, ␪ ϭ 45° Plane stress (angle ␪) Solution 7.4-14 ␴x ϭ 31 MPa ␴y ϭ Ϫ5 MPa ␶xy ϭ 33 MPa ␪ ϭ 45Њ (All stresses in MPa) Point D (␪ ϭ 45Њ): sx1 ϭ 13 ϩ R cos b ϭ 46.0 MPa tx1y1 ϭ ϪR sin b ϭ Ϫ18.0 MPa 5 (␪ ϭ 90°) B D (␪ ϭ 45°) 33 R C 18 O 13 D' R 31 ␤ ␴ x1 ␣ 33 A (␪ ϭ 0) R ϭ ͙(18) ϩ (33) ϭ 37.590 MPa 33 ϭ 61.390Њ ␣ ϭ arctan 18 ␤ ϭ 90Њ Ϫ ␣ ϭ 28.610Њ 2 y 33 D 20 MPa ␶x1y1 2 Point DЈ(␪ ϭ 135Њ): sx1 ϭ 13 Ϫ R cos b ϭ Ϫ20.0 MPa tx1y1 ϭ R sin b ϭ 18.0 MPa 46.0 MPa ␪ ϭ 45Њ D' O 18.0 MPa x
    • SECTION 7.4 Data for 7.4-15 Mohr’s Circle for Plane Stress ␴x ϭ Ϫ5750 psi, ␴y ϭ 750 psi, ␶xy ϭ Ϫ2100 psi, ␪ ϭ 75° Solution 7.4-15 Plane stress (angle ␪)) ␴x ϭ Ϫ5750 psi ␴y ϭ 750 psi ␶xy ϭ Ϫ2100 psi ␪ ϭ 75Њ (All stresses in psi) (␪ ϭ 0) A R ϭ ͙(3250) 2 ϩ (2100) 2 ϭ 3869 psi 2100 ␣ ϭ arctan ϭ 32.87Њ 3250 ␤ ϭ ␣ ϩ 30Њ ϭ 62.87Њ Point D (␪ ϭ 75Њ): sx1 ϭ Ϫ2500 ϩ R cos b ϭ Ϫ735 psi tx1y1 ϭ R sin b ϭ 3444 psi (␪ ϭ Ϫ15Њ) D' 30Њ Point DЈ(␪ ϭ Ϫ15Њ): sx1 ϭ Ϫ2500 Ϫ R cos b ϭ Ϫ4265 psi tx1y1 ϭ ϪR sin b ϭ Ϫ3444 psi R 2100 3250 C O ␣ 2␪ ϭ 150Њ ␴x1 2100 ␤ y B (␪ ϭ 90Њ) Point D: ␪ ϭ 75Њ R D 2500 5750 735 psi 3444 psi D ␪ ϭ 75° 750 x 4265 psi O ␶x1y1 D' y Problems 7.4-16 through 7.4-23 An element in plane stress is subjected to stresses ␴x, ␴y, and ␶xy (see figure). Using Mohr’s circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements. ␴y ␶xy ␴x O Data for 7.4-16 ␴x ϭ Ϫ31.5 MPa, ␴y ϭ 31.5 MPa, ␶xy ϭ 30 MPa Solution 7.4-16 Principal stresses S2 31.5 B (␪ ϭ 90Њ) R P2 30 A (␪ ϭ 0) O ␣ 31.5 2␪P1 2␪s1 31.5 S1 ␶x1y1 30 30 P1 ␴x1 ␴x ϭ Ϫ31.5 MPa ␴y ϭ 31.5 MPa ␶xy ϭ 30 MPa (All stresses in MPa) R ϭ ͙(31.5) 2 ϩ (30.0) 2 ϭ 43.5 MPa 30 ␣ ϭ arctan ϭ 43.60Њ 31.5 x 457
    • 458 CHAPTER 7 Analysis of Stress and Strain (a) PRINCIPAL STRESSES 2up1 ϭ 180Њ Ϫ ␣ ϭ 136.40Њ up1 ϭ 68.20Њ 2up2 ϭ Ϫ␣ ϭ Ϫ43.60Њ up2 ϭ Ϫ21.80Њ (b) MAXIMUM SHEAR STRESSES 2us1 ϭ 90Њ Ϫ ␣ ϭ 46.40Њ us1 ϭ 23.20Њ 2us2 ϭ 2us1 ϩ 180Њ ϭ 226.40Њ us2 ϭ 113.20Њ Point P1: ␴1 ϭ R ϭ 43.5 MPa Point P2: ␴2 ϭ ϪR ϭ Ϫ43.5 MPa Point S1: ␴aver ϭ 0 Point S2: ␴aver ϭ 0 ␶max ϭ R ϭ 43.5 MPa ␶min ϭ Ϫ43.5 MPa y y P1 43.5 MPa S1 ␪p1ϭ 68.20Њ 43.5 MPa P2 S2 ␪s1ϭ 23.20Њ O x O x 43.5 MPa Data for 7.4-17 ␴x ϭ 8400 psi, ␴y ϭ 0, ␶xy ϭ 1440 psi Solution 7.4-17 Principal stresses ␴x ϭ 8400 psi ␴y ϭ 0 (All stresses in psi) ␶xy ϭ 1440 psi 240 psi P2 P1 P2 S2 (␪ ϭ 90Њ) B y 8640 psi ␪p1ϭ 9.46Њ x R R O ␣ C R 4200 P1 ␴x1 1440 A (␪ ϭ 0) 2␪s1 4200 O S1 8400 ␶x1y1 R ϭ ͙(4200) 2 ϩ (1440) 2 ϭ 4440 psi 1440 ␣ ϭ arctan ϭ 18.92Њ 4200 (a) PRINCIPAL STRESSES 2up1 ϭ ␣ ϭ 18.92Њ up1 ϭ 9.46Њ 2up2 ϭ 180Њ ϩ ␣ ϭ 198.92Њ up2 ϭ 99.46Њ Point P1: ␴1 ϭ 4200 ϩ R ϭ 8640 psi Point P2: ␴2 ϭ 4200 Ϫ R ϭ Ϫ240 psi (b) MAXIMUM SHEAR STRESSES 2us1 ϭ Ϫ(90Њ Ϫ ␣) ϭ Ϫ71.08Њ us1 ϭ Ϫ35.54Њ us2 ϭ 54.46Њ 2us2 ϭ 90Њ ϩ ␣ ϭ 108.92Њ Point S1: ␴aver ϭ 4200 psi ␶max ϭ R ϭ 4440 psi Point S2: ␴aver ϭ 4200 psi ␶min ϭ Ϫ4440 psi y S1 4200 psi S2 4200 psi ␪s1ϭ 54.46Њ O 4440 psi x
    • SECTION 7.4 Data for 7.4-18 ␴x ϭ 0, ␴y ϭ Ϫ22.4 MPa, ␶xy ϭ Ϫ6.6 MPa Solution 7.4-18 Principal stresses ␴x ϭ 0 ␴y ϭ Ϫ22.4 MPa ␶xy ϭ Ϫ6.6 MPa (All stresses in MPa) y P1 S2 R 2␪p2 P2 24.2 MPa ␪p2ϭ 74.74Њ x A (␪ ϭ 0) O 1.8 MPa 6.6 11.2 6.6 R O C P1 ␴x1 (b) MAXIMUM SHEAR STRESSES B (␪ ϭ 90Њ) S1 2us1 ϭ Ϫ␣ Ϫ 90Њ ϭ Ϫ120.51Њ us1 ϭ Ϫ60.26Њ 2us2 ϭ 90Њ Ϫ ␣ ϭ 59.49Њ us2 ϭ 29.74Њ Point S1: ␴aver ϭ Ϫ11.2 MPa ␶max ϭ R ϭ 13.0 MPa Point S2: ␴aver ϭ Ϫ11.2 MPa ␶min ϭ Ϫ13.0 MPa 11.2 22.4 ␶x1y1 R ϭ ͙(11.2) 2 ϩ (6.6) 2 ϭ 13.0 MPa 6.6 ␣ ϭ arctan ϭ 30.51Њ 11.2 y 11.2 MPa S2 11.2 MPa (a) PRINCIPAL STRESSES S1 2up1 ϭ Ϫ␣ ϭ Ϫ30.51Њ up1 ϭ Ϫ15.26Њ 2up2 ϭ 180Њ Ϫ ␣ ϭ 149.49Њ up2 ϭ 74.74Њ Point P1: ␴1 ϭ R Ϫ 11.2 ϭ 1.8 MPa Point P2: ␴2 ϭ Ϫ11.2Ϫ R ϭ Ϫ24.2 MPa ␪s2ϭ 29.74Њ x O 13.0 MPa ␴x ϭ 1850 psi, ␴y ϭ 6350 psi, ␶xy ϭ 3000 psi Solution 7.4-19 Principal stresses S2 B (␪ ϭ 90Њ) R P2 ␣ 2250 3000 O A (␪ ϭ 0) 1850 ␶x1y1 P2 R ␣ Data for 7.4-19 Mohr’s Circle for Plane Stress R 2␪s1 3000 ␣ C 2␪p1 R S1 4100 6350 P1 ␴x1 ␴x ϭ 1850 psi ␴y ϭ 6350 psi ␶xy ϭ 3000 psi (All stresses in psi) R ϭ ͙(2250) 2 ϩ (3000) 2 ϭ 3750 psi 3000 ␣ ϭ arctan ϭ 53.13Њ 2260 459
    • 460 CHAPTER 7 Analysis of Stress and Strain (a) PRINCIPAL STRESSES (b) MAXIMUM SHEAR STRESSES 2up1 ϭ 180Њ Ϫ ␣ ϭ 126.87Њ up1 ϭ 63.43Њ 2up2 ϭ Ϫ␣ ϭ Ϫ53.13Њ up2 ϭ Ϫ26.57Њ Point P1: ␴1 ϭ 4100 ϩ R ϭ 7850 psi Point P2: ␴2 ϭ 4100 Ϫ R ϭ 350 psi 2us1 ϭ 90Њ Ϫ ␣ ϭ 36.87Њ us1 ϭ 18.43Њ us2 ϭ 108.43Њ 2us2 ϭ 270Њ Ϫ ␣ ϭ 216.87Њ Point S1: ␴aver ϭ 4100 psi ␶max ϭ R ϭ 3750 psi Point S2: ␴aver ϭ 4100 psi ␶min ϭ Ϫ3750 psi y y P1 4100 psi 7850 psi P2 3750 psi S2 ␪P1ϭ 63.43Њ 350 psi 4100 psi ␪s1ϭ 18.43Њ x S1 O x O ␴x ϭ 3100 kPa, ␴y ϭ 8700 kPa, ␶xy ϭ Ϫ4500 kPa Data for 7.4-20 Solution 7.4-20 Principal stresses y ␴x ϭ 3100 kPa ␴y ϭ 8700 kPa ␶xy ϭ Ϫ4500 kPa (All stresses in kPa) O 2␪p2 2800 C R ␣ R S1 3100 5900 ␶x1y1 P1 4500 P2 600 kPa ␪p2ϭ 29.05Њ x S2 R 4500 P2 P1 (␪ ϭ 0) A O 11,200 kPa ␴x1 4500 B (␪ ϭ 90Њ) 8700 (b) MAXIMUM SHEAR STRESSES us1 ϭ 74.05Њ 2us1 ϭ 90Њ ϩ ␣ ϭ 148.11Њ 2us2 ϭ 270Њ ϩ ␣ ϭ 328.11Њ us2 ϭ 164.05Њ Point S1: ␴aver ϭ 5900 kPa ␶max ϭ R ϭ 5300 kPa Point S2: ␴aver ϭ 5900 kPa ␶min ϭ Ϫ5300 kPa y R ϭ ͙(2800) 2 ϩ (4500) 2 ϭ 5300 kPa 4500 ␣ ϭ arctan ϭ 58.11Њ 2800 5300 kPa (a) PRINCIPAL STRESSES 2up1 ϭ ␣ ϩ 180Њ ϭ 238.11Њ up1 ϭ 119.05Њ 2up2 ϭ ␣ ϭ 58.11Њ up2 ϭ 29.05Њ Point P1: ␴1 ϭ 5900 ϩ R ϭ 11,200 kPa Point P2: ␴2 ϭ 5900 Ϫ R ϭ 600 kPa O S2 5900 kPa S1 ␪s1ϭ 74.05Њ x 5900 kPa
    • SECTION 7.4 Data for 7.4-21 Mohr’s Circle for Plane Stress ␴x ϭ Ϫ12,300 psi, ␴y ϭ Ϫ19,500 psi, ␶xy ϭ Ϫ7700 psi Solution 7.4-21 Principal stresses y ␴x ϭ Ϫ12,300 psi ␴y ϭ Ϫ19,500 psi ␶xy ϭ Ϫ7700 psi (All stresses in psi) 2␪s2 A (␪ ϭ 0) S2 R 2␪p2 P2 ␣ 24,400 psi ␪p2ϭ 57.53Њ P2 P1 3600 C x O 7700 7700 7400 psi O ␴x1 P1 (b) MAXIMUM SHEAR STRESSES B 12,300 15,900 19,500 S1 ␶x1y1 R ϭ ͙(3600) 2 ϩ (7700) 2 ϭ 8500 psi 7700 ␣ ϭ arctan ϭ 64.94Њ 3600 2us1 ϭ 270Њ Ϫ ␣ ϭ 205.06Њ us1 ϭ 102.53Њ us2 ϭ 12.53Њ 2us2 ϭ 90Њ Ϫ ␣ ϭ 25.06Њ Point S1: ␴aver ϭ Ϫ15,900 psi ␶max ϭ R ϭ 8500 psi Point S2: ␴aver ϭ Ϫ15,900 psi ␶min ϭ Ϫ8500 psi y 15,900 psi (a) PRINCIPAL STRESSES S1 2up1 ϭ Ϫ␣ ϭ Ϫ64.94Њ up1 ϭ Ϫ32.47Њ 2up2 ϭ 180Њ Ϫ ␣ ϭ 115.06Њ up2 ϭ 57.53Њ Point P1: ␴1 ϭ Ϫ15,900 ϩ R ϭ Ϫ7400 psi Point P2: ␴2 ϭ Ϫ15,900 Ϫ R ϭ Ϫ24,400 psi Data for 7.4-22 S2 O 15,900 psi ␪s2ϭ 12.53Њ x 8500 psi ␴x ϭ Ϫ3.1 MPa, ␴y ϭ 7.9 MPa, ␶xy ϭ Ϫ13.2 MPa Solution 7.4-22 Principal stresses 3.1 (␪ ϭ 0) A S2 ␴x ϭ Ϫ3.1 MPa ␴y ϭ 7.9 MPa ␶xy ϭ Ϫ13.2 MPa (All stresses in MPa) R 13.2 ␣ O P2 R S1 2.4 ␶x1y1 P1 C 5.5 7.9 ␴x1 13.2 B (␪ ϭ 90Њ) R ϭ ͙(5.5) 2 ϩ (13.2) 2 ϭ 14.3 MPa 13.2 ␣ ϭ arctan ϭ 67.38Њ 5.5 461
    • 462 CHAPTER 7 Analysis of Stress and Strain (a) PRINCIPAL STRESSES (b) MAXIMUM SHEAR STRESSES 2up1 ϭ 180Њ ϩ ␣ ϭ 247.38Њ up1 ϭ 123.69Њ 2up2 ϭ ␣ ϭ 67.38Њ up2 ϭ 33.69Њ Point P1: ␴1 ϭ 2.4 ϩ R ϭ 16.7 MPa Point P2: ␴2 ϭ ϪR ϩ 2.4 ϭ Ϫ11.9 MPa 2us1 ϭ ␣ ϩ 90Њ ϭ 157.38Њ us1 ϭ 78.69Њ 2us2 ϭ Ϫ90Њ ϩ ␣ ϭ Ϫ22.62Њ us2 ϭ Ϫ11.31Њ Point S1: ␴aver ϭ 2.4 MPa ␶max ϭ R ϭ 14.3 MPa Point S2: ␴aver ϭ 2.4 MPa ␶min ϭ Ϫ14.3 MPa y y S1 16.7 MPa P2 P1 2.4 MPa S2 11.9 MPa ␪s1ϭ 78.69Њ ␪p2ϭ 33.69Њ x O O x 2.4 MPa 14.3 MPa Data for 7.4-23 ␴x ϭ 700 psi, ␴y ϭ Ϫ2500 psi, ␶xy ϭ 3000 psi Solution 7.4-23 Principal stresses y ␴x ϭ 700 psi ␴y ϭ Ϫ2500 psi ␶xy ϭ 3000 psi (All stresses in psi) 4300 psi 2500 (␪ ϭ 90Њ) B S1 2500 psi s2 ␪p1ϭ 30.96Њ S2 x O R 3000 P2 1600 C R S1 900 P1 O ␣ ␴x1 3000 A (␪ ϭ 0) 700 ␶x1y1 R ϭ ͙(1600) 2 ϩ (3000) 2 ϭ 3400 psi 3000 ␣ ϭ arctan ϭ 61.93Њ 1600 (b) MAXIMUM SHEAR STRESSES 2us1 ϭ Ϫ90Њ ϩ ␣ ϭ Ϫ28.07Њ us1 ϭ Ϫ14.04Њ 2us2 ϭ 90Њ ϩ ␣ ϭ 151.93Њ us2 ϭ 75.96Њ Point S1: ␴aver ϭ Ϫ900 psi ␶max ϭ R ϭ 3400 psi Point S2: ␴aver ϭ Ϫ900 psi ␶min ϭ Ϫ3400 psi y 900 psi S2 ␪s2ϭ 75.96Њ (a) PRINCIPAL STRESSES 2up1 ϭ ␣ ϭ 61.93Њ up1 ϭ 30.96Њ 2up2 ϭ 180Њ ϩ ␣ ϭ 241.93Њ up2 ϭ 120.96Њ Point P1: ␴1 ϭ Ϫ900 ϩ R ϭ 2500 psi Point P2: ␴2 ϭ Ϫ900 Ϫ R ϭ Ϫ4300 psi O S1 3400 psi x 900 psi
    • SECTION 7.5 463 Hooke’s Law for Plane Stress Hooke’s Law for Plane Stress When solving the problems for Section 7.5, assume that the material is linearly elastic with modulus of elasticity E and Poisson’s ratio ␯. Problem 7.5-1 A rectangular steel plate with thickness t ϭ 0.25 in. is subjected to uniform normal stresses ␴x and ␴y, as shown in the figure. Strain gages A and B, oriented in the x and y directions, respectively, are attached to the plate. The gage readings give normal strains ⑀x ϭ 0.0010 (elongation) and ⑀y ϭ Ϫ0.0007 (shortening). Knowing that E ϭ 30 ϫ 106 psi and ␯ ϭ 0.3, determine the stresses ␴x and ␴y and the change ⌬t in the thickness of the plate. ␴y y B A O ␴x x Probs. 7.5-1 and 7.5-2 Solution 7.5-1 Rectangular plate in biaxial stress t ϭ 0.25 in. ␧x ϭ 0.0010 ␧y ϭ Ϫ0.0007 E ϭ 30 ϫ 106 psi ␯ ϭ 0.3 Substitute numerical values: Eq. (7-40a): E sx ϭ (ex ϩ ney ) ϭ 26,040 psi (1 Ϫ n2 ) Eq. (7-40b): E sy ϭ (ey ϩ nex ) ϭ Ϫ13,190 psi (1 Ϫ n2 ) Eq. (7-39c): n ez ϭ Ϫ (sx ϩ sy ) ϭ Ϫ128.5 ϫ 10Ϫ6 E ⌬t ϭ ␧zt ϭ Ϫ32.1 ϫ 10Ϫ6 in. (Decrease in thickness) Problem 7.5-2 Solve the preceding problem if the thickness of the steel plate is t ϭ 10 mm, the gage readings are ⑀x ϭ 480 ϫ 10Ϫ6 (elongation) and ⑀y ϭ 130 ϫ 10Ϫ6 (elongation), the modulus is E ϭ 200 GPa, and Poisson’s ratio is ␯ ϭ 0.30. Solution 7.5-2 Rectangular plate in biaxial stress t ϭ 10 mm ␧x ϭ 480 ϫ 10Ϫ6 ␧y ϭ 130 ϫ 10Ϫ6 E ϭ 200 GPa ␯ ϭ 0.3 Substitute numerical values: Eq. (7-40a): E sx ϭ (ex ϩ ney ) ϭ 114.1 MPa (1 Ϫ n2 ) Eq. (7-40b): E sy ϭ (ey ϩ nex ) ϭ 60.2 MPa (1 Ϫ n2 ) Eq. (7-39c): n ez ϭ Ϫ (sx ϩ sy ) ϭ Ϫ261.4 ϫ 10Ϫ6 E ⌬t ϭ ␧zt ϭ Ϫ2610 ϫ 10Ϫ6 mm (Decrease in thickness) Problem 7.5-3 Assume that the normal strains ⑀x and ⑀y for an element in plane stress (see figure) are measured with strain gages. (a) Obtain a formula for the normal strain ⑀z in the z direction in terms of ⑀x, ⑀y, and Poisson’s ratio ␯. (b) Obtain a formula for the dilatation e in terms of ⑀x, ⑀y, and Poisson’s ratio ␯. y ␴y ␶xy ␴x O x z
    • 464 CHAPTER 7 Solution 7.5-3 Analysis of Stress and Strain Plane stress Given: ␧x, ␧y, ␯ (b) DILATATION (a) NORMAL STRAIN ␧z n Eq. (7-34c): ez ϭ Ϫ (sx ϩ sy ) E E (ex ϩ ney ) Eq. (7-36a): sx ϭ (1 Ϫ n2 ) E (ey ϩ nex ) Eq. (7-36b): sy ϭ (1 Ϫ n2 ) Substitute ␴x and ␴y into the first equation and simplify: n ez ϭ Ϫ (e ϩ ey ) 1Ϫn x Eq. (7-47): e ϭ 1 Ϫ 2n (sx ϩ sy ) E Substitute ␴x and ␴y from above and simplify: eϭ 1 Ϫ 2n (e ϩ ey ) 1Ϫn x ␴y Problem 7.5-4 A magnesium plate in biaxial stress is subjected to tensile stresses ␴x ϭ 24 MPa and ␴y ϭ 12 MPa (see figure). The corresponding strains in the plate are ⑀x ϭ 440 ϫ 10Ϫ6 and ⑀y ϭ 80 ϫ 10Ϫ6. Determine Poisson’s ratio ␯ and the modulus of elasticity E for the material. y O x Probs. 7.5-4 through 7.5-7 Solution 7.5-4 Biaxial stress ␴y ϭ 12 MPa ␴x ϭ 24 MPa ␧x ϭ 440 ϫ 10Ϫ6 ␧y ϭ 80 ϫ 10Ϫ6 POISSON’S RATIO AND MODULUS OF ELASTICITY 1 Eq. (7-39a): ex ϭ (sx Ϫ nsy ) E 1 Eq. (7-39b): ey ϭ (sy Ϫ nsx ) E Substitute numerical values: E (440 ϫ 10Ϫ6) ϭ 24 MPa Ϫ ␯ (12 MPa) E (80 ϫ 10Ϫ6) ϭ 12 MPa Ϫ ␯ (24 MPa) Solve simultaneously: ␯ ϭ 0.35 E ϭ 45 GPa Problem 7.5-5 Solve the preceding problem for a steel plate with ␴x ϭ 10,800 psi (tension), ␴y ϭ Ϫ5400 psi (compression), ⑀x ϭ 420 ϫ 10Ϫ6 (elongation), and ⑀y ϭ Ϫ300 ϫ 10Ϫ6 (shortening). Solution 7.5-5 Biaxial stress ␴x ϭ 10,800 psi ␴y ϭ Ϫ5400 psi ␧x ϭ 420 ϫ 10Ϫ6 ␧y ϭ Ϫ300 ϫ 10Ϫ6 POISSON’S RATIO AND MODULUS OF ELASTICITY 1 Eq. (7-39a): ex ϭ (sx Ϫ nsy ) E 1 Eq. (7-39b): ey ϭ (sy Ϫ nsx ) E Substitute numerical values: E (420 ϫ 10Ϫ6) ϭ 10,800 psi Ϫ ␯ (Ϫ5400 psi) E (Ϫ300 ϫ 10Ϫ6) ϭ Ϫ5400 psi Ϫ ␯ (10,800 psi) Solve simultaneously: ␯ ϭ 1/3 E ϭ 30 ϫ 106 psi ␴x
    • SECTION 7.5 Hooke’s Law for Plane Stress Problem 7.5-6 A rectangular plate in biaxial stress (see figure) is subjected to normal stresses ␴x ϭ 90 MPa (tension) and ␴y ϭ Ϫ20 MPa (compression). The plate has dimensions 400 ϫ 800 ϫ 20 mm and is made of steel with E ϭ 200 GPa and ␯ ϭ 0.30. (a) Determine the maximum in-plane shear strain ␥max in the plate. (b) Determine the change ⌬t in the thickness of the plate. (c) Determine the change ⌬V in the volume of the plate. Solution 7.5-6 Biaxial stress ␴x ϭ 90 MPa ␴y ϭ Ϫ20 MPa E ϭ 200 GPa ␯ ϭ 0.30 Dimensions of Plate: 400 mm ϫ 800 mm ϫ 20 mm Shear Modulus (Eq. 7-38): Gϭ E ϭ 76.923 GPa 2(1 ϩ n) (a) MAXIMUM IN-PLANE SHEAR STRAIN Principal stresses: ␴1 ϭ 90 MPa ␴2 ϭ Ϫ20 MPa s1 Ϫ s2 Eq. (7-26): tmax ϭ ϭ 55.0 MPa 2 tmax Eq. (7-35): gmax ϭ ϭ 715 ϫ 10Ϫ6 G (b) CHANGE IN THICKNESS n Eq. (7-39c): ez ϭ Ϫ (sx ϩ sy ) ϭ Ϫ105 ϫ 10Ϫ6 E ⌬t ϭ ␧zt ϭ Ϫ2100 ϫ 10Ϫ6 mm (Decrease in thickness) (c) CHANGE IN VOLUME From Eq. (7-47): ¢V ϭ V0 ¢ 1 Ϫ 2n ≤ (sx ϩ sy ) E V0 ϭ (400)(800)(20) ϭ 6.4 ϫ 106 mm3 1 Ϫ 2n Also, ¢ ≤ (sx ϩ sy ) ϭ 140 ϫ 10Ϫ6 E І ⌬V ϭ (6.4 ϫ 106 mm3)(140 ϫ 10Ϫ6) ϭ 896 mm3 (Increase in volume) Problem 7.5-7 Solve the preceding problem for an aluminum plate with ␴x ϭ 12,000 psi (tension), ␴y ϭ Ϫ3,000 psi (compression), dimensions 20 ϫ 30 ϫ 0.5 in., E ϭ 10.5 ϫ 106 psi, and ␯ ϭ 0.33. Solution 7.5-7 Biaxial stress ␴x ϭ 12,000 psi ␴y ϭ Ϫ3,000 psi E ϭ 10.5 ϫ 106 psi ␯ ϭ 0.33 Dimensions of Plate: 20 in. ϫ 30 in. ϫ 0.5 in. Shear Modulus (Eq. 7-38): Gϭ E ϭ 3.9474 ϫ 106 psi 2(1 ϩ n) (a) MAXIMUM IN-PLANE SHEAR STRAIN Principal stresses: ␴1 ϭ 12,000 psi ␴2 ϭ Ϫ3,000 psi s1 Ϫ s2 Eq. (7-26): tmax ϭ ϭ 7,500 psi 2 tmax Eq. (7-35): gmax ϭ ϭ 1,900 ϫ 10Ϫ6 G (b) CHANGE IN THICKNESS n Eq. (7-39c): ez ϭ Ϫ (sx ϩ sy ) ϭ Ϫ282.9 ϫ 10Ϫ6 E ⌬t ϭ ␧zt ϭ Ϫ141 ϫ 10Ϫ6 in. (Decrease in thickness) (c) CHANGE IN VOLUME From Eq. (7-47): ¢V ϭ V0 ¢ 1 Ϫ 2n ≤ (sx ϩ sy ) E V0 ϭ (20)(30)(0.5) ϭ 300 in.3 1 Ϫ 2n Also, ¢ ≤ (sx ϩ sy ) ϭ 291.4 ϫ 10Ϫ6 E І ⌬V ϭ (300 in.3)(291.4 ϫ 10Ϫ6) ϭ 0.0874 in.3 (Increase in volume) 465
    • 466 CHAPTER 7 Analysis of Stress and Strain Problem 7.5-8 A brass cube 50 mm on each edge is compressed in two perpendicular directions by forces P ϭ 175 kN (see figure). Calculate the change ⌬V in the volume of the cube and the strain energy U stored in the cube, assuming E ϭ 100 GPa and ␯ ϭ 0.34. Solution 7.5-8 P = 175 kN P = 175 kN CHANGE IN VOLUME Biaxial stress-cube 1 Ϫ 2n (sx ϩ sy ) ϭ Ϫ448 ϫ 10Ϫ6 E V0 ϭ b3 ϭ (50 mm)3 ϭ 125 ϫ 103 mm3 ⌬V ϭ eV0 ϭ Ϫ56 mm3 (Decrease in volume) P Eq. (7-47): e ϭ P STRAIN ENERGY 1 2 (s ϩ s2 Ϫ 2nsxsy ) y 2E x ϭ 0.03234 MPa U ϭ uV0 ϭ (0.03234 MPa)(125 ϫ 103 mm3) ϭ 4.04 J Side b ϭ 50 mm P ϭ 175 kN E ϭ 100 GPa ␯ ϭ 0.34 (Brass) sx ϭ sy ϭ Ϫ Eq. (7-50): u ϭ (175 kN) P ϭ Ϫ70.0 MPa 2ϭϪ b (50 mm) 2 Problem 7.5-9 A 4.0-inch cube of concrete (E ϭ 3.0 ϫ 106 psi, ␯ ϭ 0.1) is compressed in biaxial stress by means of a framework that is loaded as shown in the figure. Assuming that each load F equals 20 k, determine the change ⌬V in the volume of the cube and the strain energy U stored in the cube. F F Solution 7.5-9 Biaxial stress – concrete cube CHANGE IN VOLUME A 1 Ϫ 2n (sx ϩ sy ) ϭ Ϫ0.0009429 E V0 ϭ b3 ϭ (4 in.)3 ϭ 64 in.3 ⌬V ϭ eV0 ϭ Ϫ0.0603 in.3 (Decrease in volume) Eq. (7-47): e ϭ F F Joint A: P ϭ F͙2 ϭ 28.28 kips sx ϭ sy ϭ Ϫ P ϭ Ϫ1768 psi b2 b ϭ 4 in. E ϭ 3.0 ϫ 106 psi ␯ ϭ 0.1 F ϭ 20 kips A F P F STRAIN ENERGY 1 2 (s ϩ s2 Ϫ 2nsxsy ) y 2E x ϭ 0.9377 psi U ϭ uV0 ϭ 60.0 in.-lb Eq. (7-50): u ϭ
    • SECTION 7.5 Problem 7.5-10 A square plate of width b and thickness t is loaded by normal forces Px and Py, and by shear forces V, as shown in the figure. These forces produce uniformly distributed stresses acting on the side faces of the plate. Calculate the change ⌬V in the volume of the plate and the strain energy U stored in the plate if the dimensions are b ϭ 600 mm and t ϭ 40 mm, the plate is made of magnesium with E ϭ 45 GPa and ␯ ϭ 0.35, and the forces are Px ϭ 480 kN, Py ϭ 180 kN, and V ϭ 120 kN. Py t b ϭ 600 mm E ϭ 45 GPa Px ϭ 480 kN Py ϭ 180 kN V ϭ 120 kN V y Px V V b O b x V Probs. 7.5-10 and 7.5-11 Solution 7.5-10 Py Square plate in plane stress t ϭ 40 mm ␯ ϭ 0.35 (magnesium) Px sx ϭ ϭ 20.0 MPa bt Py sy ϭ ϭ 7.5 MPa bt V txy ϭ ϭ 5.0 MPa bt STRAIN ENERGY t2 1 2 xy 2 Eq. (7-50): u ϭ (sx ϩ sy Ϫ 2nsxsy ) ϩ 2E 2G E Gϭ ϭ 16.667 GPa 2(1 ϩ n) Substitute numerical values: u ϭ 4653 Pa U ϭ uV0 ϭ 67.0 N . m ϭ 67.0 J CHANGE IN VOLUME 1 Ϫ 2n (sx ϩ sy ) ϭ 183.33 ϫ 10Ϫ6 E V0 ϭ b2t ϭ 14.4 ϫ 106 mm3 ⌬V ϭ eV0 ϭ 2640 mm3 (Increase in volume) Eq. (7-47): e ϭ Problem 7.5-11 Solve the preceding problem for an aluminum plate with b ϭ 12 in., t ϭ 1.0 in., E ϭ 10,600 ksi, ␯ ϭ 0.33, Px ϭ 90 k, Py ϭ 20 k, and V ϭ 15 k. Solution 7.5-11 Square plate in plane stress b ϭ 12.0 in. E ϭ 10,600 ksi STRAIN ENERGY Px ϭ 90 k Eq. (7-50): u ϭ Py ϭ 20 k V ϭ 15 k t ϭ 1.0 in. ␯ ϭ 0.33 (aluminum) Px sx ϭ ϭ 7500 psi bt Py sy ϭ ϭ 1667 psi bt V txy ϭ ϭ 1250 psi bt CHANGE IN VOLUME 1 Ϫ 2n (sx ϩ sy ) ϭ 294 ϫ 10Ϫ6 E V0 ϭ b2t ϭ 144 in.3 ⌬V ϭ eV0 ϭ 0.0423 in.3 (Increase in volume) Eq. (7-47): e ϭ 467 Hooke’s Law for Plane Stress t2 1 2 xy (sx ϩ s2 Ϫ 2nsxsy ) ϩ y 2E 2G E ϭ 3985 ksi 2(1 ϩ n) Substitute numerical values: u ϭ 2.591 psi U ϭ uV0 ϭ 373 in.-lb Gϭ Px
    • 468 CHAPTER 7 Analysis of Stress and Strain Problem 7.5-12 A circle of diameter d ϭ 200 mm is etched on a brass plate (see figure). The plate has dimensions 400 ϫ 400 ϫ 20 mm. Forces are applied to the plate, producing uniformly distributed normal stresses ␴x ϭ 42 MPa and ␴y ϭ 14 MPa. Calculate the following quantities: (a) the change in length ⌬ac of diameter ac; (b) the change in length ⌬bd of diameter bd; (c) the change ⌬t in the thickness of the plate; (d) the change ⌬V in the volume of the plate, and (e) the strain energy U stored in the plate. (Assume E ϭ 100 GPa and ␯ ϭ 0.34.) z y ␴y d ␴x a c ␴x b x ␴y Solution 7.5-12 Plate in biaxial stress ␴x ϭ 42 MPa ␴y ϭ 14 MPa Dimensions: 400 ϫ 400 ϫ 20 (mm) Diameter of circle: d ϭ 200 mm E ϭ 100 GPa ␯ ϭ 0.34 (Brass) (a) CHANGE IN LENGTH OF DIAMETER IN x DIRECTION 1 Eq. (7-39a): ex ϭ (sx Ϫ nsy ) ϭ 372.4 ϫ 10Ϫ6 E ⌬ac ϭ ␧x d ϭ 0.0745 mm (increase) (c) CHANGE IN THICKNESS n Eq. (7-39c): ez ϭ Ϫ (sx ϩ sy ) ϭ Ϫ190.4 ϫ 10Ϫ6 E ⌬t ϭ ␧zt ϭ Ϫ0.00381 mm (decrease) (d) CHANGE IN VOLUME 1 Ϫ 2n (sx ϩ sy ) ϭ 179.2 ϫ 10Ϫ6 E V0 ϭ (400)(400)(20) ϭ 3.2 ϫ 106 mm3 ⌬V ϭ eV0 ϭ 573 mm3 (increase) Eq. (7-47): e ϭ (b) CHANGE IN LENGTH OF DIAMETER IN y DIRECTION 1 Eq. (7-39b): ey ϭ (sy Ϫ nsx ) ϭ Ϫ2.80 ϫ 10Ϫ6 E ⌬bd ϭ ␧y d ϭ Ϫ560 ϫ 10Ϫ6 mm (decrease) (e) STRAIN ENERGY 1 2 (s ϩ s2 Ϫ 2nsxsy ) y 2E x ϭ 7.801 ϫ 10Ϫ3 MPa U ϭ uV0 ϭ 25.0 N . m ϭ 25.0 J Eq. (7-50): u ϭ Triaxial Stress When solving the problems for Section 7.6, assume that the material is linearly elastic with modulus of elasticity E and Poisson’s ratio ␯. Problem 7.6-1 An element of aluminum in the form of a rectangular parallelepiped (see figure) of dimensions a ϭ 6.0 in., b ϭ 4.0 in, and c ϭ 3.0 in. is subjected to triaxial stresses ␴x ϭ 12,000 psi, ␴y ϭ Ϫ4,000 psi, and ␴z ϭ Ϫ1,000 psi acting on the x, y, and z faces, respectively. Determine the following quantities: (a) the maximum shear stress ␶max in the material; (b) the changes ⌬a, ⌬b, and ⌬c in the dimensions of the element; (c) the change ⌬V in the volume; and (d) the strain energy U stored in the element. (Assume E ϭ 10,400 ksi and ␯ ϭ 0.33.) y a c b O z Probs. 7.6-1 and 7.6-2 x
    • SECTION 7.6 Solution 7.6-1 Triaxial Stress Triaxial stress ␴x ϭ 12,000 psi ␴y ϭ Ϫ4,000 psi ␴z ϭ Ϫ1,000 psi a ϭ 6.0 in. b ϭ 4.0 in. c ϭ 3.0 in. E ϭ 10,400 ksi ␯ ϭ 0.33 (aluminum) (a) MAXIMUM SHEAR STRESS ␴1 ϭ 12,000 psi ␴2 ϭ Ϫ1,000 psi ␴3 ϭ Ϫ4,000 psi s1 Ϫ s3 tmax ϭ ϭ 8,000 psi 2 (c) CHANGE IN VOLUME Eq. (7-56): 1 Ϫ 2n eϭ (sx ϩ sy ϩ sz ) ϭ 228.8 ϫ 10Ϫ6 E V ϭ abc ⌬V ϭ e (abc) ϭ 0.0165 in.3 (increase) (d) STRAIN ENERGY (b) CHANGES IN DIMENSIONS sx n Ϫ (sy ϩ sz ) ϭ 1312.5 ϫ 10Ϫ6 E E sy n Eq. (7-53b): ey ϭ Ϫ (sz ϩ sx ) ϭ Ϫ733.7 ϫ 10Ϫ6 E E sz n Eq. (7-53c): ez ϭ Ϫ (sx ϩ sy ) ϭ Ϫ350.0 ϫ 10Ϫ6 E E ⌬a ϭ a␧x ϭ 0.0079 in. (increase) ⌬b ϭ b␧y ϭ Ϫ0.0029 in. (decrease) ⌬c ϭ c␧z ϭ Ϫ0.0011 in. (decrease) 123 Eq. (7-53a): ex ϭ 1 Eq. (7-57a): u ϭ (sx ex ϩ sy ey ϩ sz ez ) 2 ϭ 9.517 psi U ϭ u (abc) ϭ 685 in.-lb Problem 7.6-2 Solve the preceding problem if the element is steel (E ϭ 200 GPa, ␯ ϭ 0.30) with dimensions a ϭ 300 mm, b ϭ 150 mm, and c ϭ 150 mm and the stresses are ␴x ϭ Ϫ60 MPa, ␴y ϭ Ϫ40 MPa, and ␴z ϭ Ϫ40 MPa. Solution 7.6-2 Triaxial stress ␴x ϭ Ϫ60 MPa ␴y ϭ Ϫ40 MPa ␴z ϭ Ϫ40 MPa a ϭ 300 mm b ϭ 150 mm c ϭ 150 mm E ϭ 200 GPa ␯ ϭ 0.30 (steel) ⌬a ϭ a␧x ϭ Ϫ0.0540 mm ⌬b ϭ b␧y ϭ Ϫ0.0075 mm ⌬c ϭ c␧z ϭ Ϫ0.0075 mm (decrease) (decrease) (decrease) (c) CHANGE IN VOLUME (a) MAXIMUM SHEAR STRESS ␴1 ϭ Ϫ40 MPa ␴2 ϭ Ϫ40 MPa ␴3 ϭ Ϫ60 MPa s1 Ϫ s3 tmax ϭ ϭ 10.0 MPa 2 Eq. (7-56): 1 Ϫ 2n eϭ (sx ϩ sy ϩ sz ) ϭ Ϫ280.0 ϫ 10Ϫ6 E V ϭ abc ⌬V ϭ e (abc) ϭ Ϫ1890 mm3 (decrease) (b) CHANGES IN DIMENSIONS (d) STRAIN ENERGY sx n Ϫ (sy ϩ sz ) ϭ Ϫ180.0 ϫ 10Ϫ6 E E sy n Eq. (7-53b): ey ϭ Ϫ (sz ϩ sx ) ϭ Ϫ50.0 ϫ 10Ϫ6 E E sz n Eq. (7-53c): ez ϭ Ϫ (sx ϩ sy ) ϭ Ϫ50.0 ϫ 10Ϫ6 E E Eq. (7-53a): ex ϭ 1 Eq. (7-57a): u ϭ (sx ex ϩ sy ey ϩ sz ez ) 2 ϭ 0.00740 MPa U ϭ u (abc) ϭ 50.0 N . m ϭ 50.0 J 469
    • 470 CHAPTER 7 Analysis of Stress and Strain Problem 7.6-3 A cube of cast iron with sides of length a ϭ 4.0 in. (see figure) is tested in a laboratory under triaxial stress. Gages mounted on the testing machine show that the compressive strains in the material are ⑀x ϭ Ϫ225 ϫ 10Ϫ6 and ⑀y ϭ ⑀z ϭ Ϫ37.5 ϫ 10Ϫ6. Determine the following quantities: (a) the normal stresses ␴x, ␴y, and ␴z acting on the x, y, and z faces of the cube; (b) the maximum shear stress ␶max in the material; (c) the change ⌬V in the volume of the cube; and (d) the strain energy U stored in the cube. (Assume E ϭ 14,000 ksi and ␯ ϭ 0.25.) Solution 7.6-3 Triaxial stress (cube) ␧x ϭ Ϫ225 ϫ 10Ϫ6 ␧y ϭ Ϫ37.5 ϫ 10Ϫ6 ␧z ϭ Ϫ37.5 ϫ 10Ϫ6 a ϭ 4.0 in. E ϭ 14,000 ksi ␯ ϭ 0.25 (cast iron) y a a a O z x Probs. 7.6-3 and 7.6-4 (c) CHANGE IN VOLUME (a) NORMAL STRESSES Eq. (7-55): e ϭ ␧x ϩ ␧y ϩ ␧z ϭ Ϫ0.000300 V ϭ a3 ⌬V ϭ ea3 ϭ Ϫ0.0192 in.3 (decrease) Eq. (7-54a): (d) STRAIN ENERGY E sx ϭ [ (1 Ϫ n)ex ϩ n(ey ϩ ez ) ] (1 ϩ n)(1 Ϫ 2n) ϭ Ϫ4200 psi In a similar manner, Eqs. (7-54 b and c) give ␴y ϭ Ϫ2100 psi ␴z ϭ Ϫ2100 psi 1 Eq. (7-57a): u ϭ (sx ex ϩ sy ey ϩ sz ez ) 2 ϭ 0.55125 psi U ϭ ua3 ϭ 35.3 in.-lb (b) MAXIMUM SHEAR STRESS ␴1 ϭ Ϫ2100 psi ␴2 ϭ Ϫ2100 psi ␴3 ϭ Ϫ4200 psi s1 Ϫ s3 tmax ϭ ϭ 1050 psi 2 Problem 7.6-4 Solve the preceding problem if the cube is granite (E ϭ 60 GPa, ␯ ϭ 0.25) with dimensions a ϭ 75 mm and compressive strains ⑀x ϭ Ϫ720 ϫ 10Ϫ6 and ⑀ y ϭ ⑀ z ϭ Ϫ270 ϫ 10Ϫ6. Solution 7.6-4 Triaxial stress (cube) ␧x ϭ Ϫ720 ϫ 10Ϫ6 ␧y ϭ Ϫ270 ϫ 10Ϫ6 ␧z ϭ Ϫ270 ϫ 10Ϫ6 a ϭ 75 mm E ϭ 60 GPa ␯ ϭ 0.25 (Granite) (c) CHANGE IN VOLUME (a) NORMAL STRESSES Eq. (7-55): e ϭ ␧x ϩ ␧y ϩ ␧z ϭ Ϫ1260 ϫ 10Ϫ6 V ϭ a3 ⌬V ϭ ea3 ϭ Ϫ532 mm3 (decrease) Eq. (7-54a): (d) STRAIN ENERGY E sx ϭ [ (1 Ϫ n)ex ϩ n(ex ϩ ez ) ] (1 ϩ n)(1 Ϫ 2n) ϭ Ϫ64.8 MPa In a similar manner, Eqs. (7-54 b and c) give ␴y ϭ Ϫ43.2 MPa ␴z ϭ Ϫ43.2 MPa 1 Eq. (7-57a): u ϭ (sx ex ϩ sy ey ϩ sz ez ) 2 ϭ 0.03499 MPa = 34.99 kPa U ϭ ua3 ϭ 14.8 N . m ϭ 14.8 J (b) MAXIMUM SHEAR STRESS ␴1 ϭ Ϫ43.2 MPa ␴2 ϭ Ϫ43.2 MPa ␴3 ϭ Ϫ64.8 MPa s1 Ϫ s3 tmax ϭ ϭ 10.8 MPa 2
    • SECTION 7.6 Problem 7.6-5 An element of aluminum in triaxial stress (see figure) is subjected to stresses ␴x ϭ 5200 psi (tension), ␴y ϭ Ϫ4750 psi (compression), and ␴z ϭ Ϫ3090 psi (compression). It is also known that the normal strains in the x and y directions are ⑀x ϭ 713.8 ϫ 10Ϫ6 (elongation) and ⑀y ϭ Ϫ502.3 ϫ 10Ϫ6 (shortening). What is the bulk modulus K for the aluminum? y ␴y ␴z ␴x ␴x O x ␴z ␴y Probs. 7.6-5 and 7.6-6 z Solution 7.6-5 Triaxial stress (bulk modulus) ␴x ϭ 5200 psi ␴y ϭ Ϫ4750 psi ␴z ϭ Ϫ3090 psi ␧x ϭ 713.8 ϫ 10Ϫ6 ␧y ϭ Ϫ502.3 ϫ 10Ϫ6 Find K. sx n Ϫ (sy ϩ sz ) E E sy n Eq. (7-53b): ey ϭ Ϫ (sz ϩ sx ) E E Eq. (7-53a): ex ϭ 471 Triaxial Stress Substitute numerical values and rearrange: (713.8 ϫ 10Ϫ6) E ϭ 5200 ϩ 7840 ␯ (Ϫ502.3 ϫ 10Ϫ6) E ϭ Ϫ4750 Ϫ 2110 ␯ Units: E ϭ psi (1) (2) Solve simultaneously Eqs. (1) and (2): E ϭ 10.801 ϫ 106 psi ␯ ϭ 0.3202 E Eq. (7-61): K ϭ ϭ 10.0 ϫ 106 psi 3(1 Ϫ 2n) Problem 7.6-6 Solve the preceding problem if the material is nylon subjected to compressive stresses ␴x ϭ Ϫ4.5 MPa, ␴y ϭ Ϫ3.6 MPa, and ␴z ϭ Ϫ2.1 MPa, and the normal strains are ⑀x ϭ Ϫ740 ϫ 10Ϫ6 and ⑀y ϭ Ϫ320 ϫ 10Ϫ6 (shortenings). Solution 7.6-6 Triaxial stress (bulk modulus) ␴x ϭ Ϫ4.5 MPa ␴y ϭ Ϫ3.6 MPa ␴z ϭ Ϫ2.1 MPa ␧x ϭ Ϫ740 ϫ 10Ϫ6 ␧y ϭ Ϫ320 ϫ 10Ϫ6 Find K. sx n Eq. (7-53a): ex ϭ Ϫ (sy ϩ sz ) E E sy n Eq. (7-53b): ey ϭ Ϫ (sz ϩ sx ) E E Problem 7.6-7 A rubber cylinder R of length L and cross-sectional area A is compressed inside a steel cylinder S by a force F that applies a uniformly distributed pressure to the rubber (see figure). (a) Derive a formula for the lateral pressure p between the rubber and the steel. (Disregard friction between the rubber and the steel, and assume that the steel cylinder is rigid when compared to the rubber.) (b) Derive a formula for the shortening ␦ of the rubber cylinder. Substitute numerical values and rearrange: (Ϫ740 ϫ 10Ϫ6) E ϭ Ϫ4.5 ϩ 5.7 ␯ (Ϫ320 ϫ 10Ϫ6) E ϭ Ϫ3.6 ϩ 6.6 ␯ Units: E ϭ MPa Solve simultaneously Eqs. (1) and (2): E ϭ 3,000 MPa ϭ 3.0 GPa ␯ ϭ 0.40 E Eq. (7-61): K ϭ ϭ 5.0 GPa 3(1 Ϫ 2n) (1) (2) F F S S R L
    • 472 CHAPTER 7 Analysis of Stress and Strain Solution 7.6-7 Rubber cylinder F (b) SHORTENING y R L x S z ␴y = – F A F sy ϭ Ϫ ␴x ϭ Ϫp A ␴z ϭ Ϫp ␧x ϭ ␧z ϭ 0 p p (a) LATERAL PRESSURE sx n Ϫ (sy ϩ sz ) E E F 0 ϭ Ϫp Ϫ n ¢ Ϫ Ϫ p ≤ OR A n F ¢ ≤ Solve for p: p ϭ 1Ϫn A Eq. (7-53a): ex ϭ Problem 7.6-8 A block R of rubber is confined between plane parallel walls of a steel block S (see figure). A uniformly distributed pressure p0 is applied to the top of the rubber block by a force F. (a) Derive a formula for the lateral pressure p between the rubber and the steel. (Disregard friction between the rubber and the steel, and assume that the steel block is rigid when compared to the rubber.) (b) Derive a formula for the dilatation e of the rubber. (c) Derive a formula for the strain-energy density u of the rubber. Solution 7.6-8 F F S R S Block of rubber p0 = pressure on top of the block F p x z ␴x ϭ Ϫp ␴y ϭ Ϫp0 ␴z ϭ 0 ␧x ϭ 0 ␧y 0 ␧z (a) LATERAL PRESSURE sx n Ϫ (sy ϩ sz ) E E 0 ϭ Ϫp Ϫ ␯ (Ϫp0) І p ϭ ␯p0 Eq. (7-53a): ex ϭ (b) DILATATION 1 Ϫ 2n (sx ϩ sy ϩ sz ) E 1 Ϫ 2n ϭ (Ϫp Ϫ p0 ) E Substitute for p: (1 ϩ n)(1 Ϫ 2n)p0 eϭϪ E Eq. (7-56): e ϭ p y OR sy n Ϫ (sz ϩ sx ) E E F n ϭϪ Ϫ (Ϫ2p) EA E Substitute for p and simplify: F (1 ϩ n)(Ϫ1 ϩ 2n) ey ϭ EA 1Ϫn (Positive ␧y represents an increase in strain, that is, elongation.) ␦ ϭ Ϫ␧yL (1 ϩ n)(1 Ϫ 2n) FL ␦ϭ ¢ ≤ (1 Ϫ n) EA (Positive ␦ represents a shortening of the rubber cylinder.) Eq. (7-53b): ey ϭ 0 (c) STRAIN ENERGY DENSITY Eq. (7-57b): 1 n u ϭ (s2 ϩ s2 ϩ s2 ) Ϫ (sxsy ϩ sxsz ϩ sysz ) y z 2E x E Substitute for ␴x , ␴y , ␴z, and p: (1 Ϫ n2 )p2 0 uϭ 2E
    • SECTION 7.6 Triaxial Stress Problem 7.6-9 A solid spherical ball of brass (E ϭ 15 ϫ 106 psi, ␯ ϭ 0.34) is lowered into the ocean to a depth of 10,000 ft. The diameter of the ball is 11.0 in. Determine the decrease ⌬ d in diameter, the decrease ⌬V in volume, and the strain energy U of the ball. Solution 7.6-9 Brass sphere E ϭ 15 ϫ 106 psi ␯ ϭ 0.34 Lowered in the ocean to depth h ϭ 10,000 ft Diameter d ϭ 11.0 in. Sea water: ␥ ϭ 63.8 lb/ft3 Pressure: ␴0 ϭ ␥h ϭ 638,000 lb/ft2 ϭ 4431 psi DECREASE IN DIAMETER s0 (1 Ϫ 2n) ϭ 94.53 ϫ 10Ϫ6 E ⌬d ϭ ␧0d ϭ 1.04 ϫ 10Ϫ3 in. (decrease) Eq. (7-59): e0 ϭ DECREASE IN VOLUME Eq. (7-60): e ϭ 3␧0 ϭ 283.6 ϫ 10Ϫ6 4 4 11.0 in. 3 V0 ϭ ␲r 3 ϭ (␲) ¢ ≤ ϭ 696.9 in.3 3 3 2 ⌬V ϭ eV0 ϭ 0.198 in.3 (decrease) STRAIN ENERGY Use Eq. (7-57b) with ␴x ϭ ␴y ϭ ␴z ϭ ␴0: 3(1 Ϫ 2n)s2 0 ϭ 0.6283 psi 2E U ϭ uV0 ϭ 438 in.-lb uϭ Problem 7.6-10 A solid steel sphere (E ϭ 210 GPa, ␯ ϭ 0.3) is subjected to hydrostatic pressure p such that its volume is reduced by 0.4%. (a) Calculate the pressure p. (b) Calculate the volume modulus of elasticity K for the steel. (c) Calculate the strain energy U stored in the sphere if its diameter is d ϭ 150 mm. Solution 7.6-10 Steel sphere E ϭ 210 GPa ␯ ϭ 0.3 Hydrostatic Pressure. V0 ϭ Initial volume ⌬V ϭ 0.004V0 ¢V Dilatation: e ϭ ϭ 0.004 V0 (a) PRESSURE 3s0 (1 Ϫ 2n) E Ee or s0 ϭ ϭ 700 MPa 3(1 Ϫ 2n) Pressure p ϭ ␴0 ϭ 700 MPa Eq. (7-60): e ϭ (b) VOLUME MODULUS OF ELASTICITY Eq. (7-63): K ϭ s0 700 MPa ϭ ϭ 175 GPa E 0.004 (c) STRAIN ENERGY (d ϭ diameter) d ϭ 150 mm r ϭ 75 mm From Eq. (7-57b) with ␴x ϭ ␴y ϭ ␴z ϭ ␴0: 3(1 Ϫ 2n)s2 0 uϭ ϭ 1.40 MPa 2E 3 4␲r V0 ϭ ϭ 1767 ϫ 10Ϫ6 m3 3 U ϭ uV0 ϭ 2470 N . m ϭ 2470 J 473
    • 474 CHAPTER 7 Analysis of Stress and Strain Problem 7.6-11 A solid bronze sphere (volume modulus of elasticity K ϭ 14.5 ϫ 106 psi) is suddenly heated around its outer surface. The tendency of the heated part of the sphere to expand produces uniform tension in all directions at the center of the sphere. If the stress at the center is 12,000 psi, what is the strain? Also, calculate the unit volume change e and the strain-energy density u at the center. Solution 7.6-11 Bronze sphere (heated) K ϭ 14.5 ϫ 106 psi ␴0 ϭ 12,000 psi (tension at the center) UNIT VOLUME CHANGE AT THE CENTER Eq. (7-62): e ϭ STRAIN AT THE CENTER OF THE SPHERE s0 (1 Ϫ 2n) E E Eq. (7-61): K ϭ 3(1 Ϫ 2n) Combine the two equations: s0 e0 ϭ ϭ 276 ϫ 10Ϫ6 3K s0 ϭ 828 ϫ 10Ϫ6 K STRAIN ENERGY DENSITY AT THE CENTER Eq. (7-59): e0 ϭ Eq. (7-57b) with ␴x ϭ ␴y ϭ ␴z ϭ ␴0: 3(1 Ϫ 2n)s2 s2 0 0 ϭ 2E 2K u ϭ 4.97 psi uϭ y Plane Strain When solving the problems for Section 7.7, consider only the in-plane strains (the strains in the xy plane) unless stated otherwise. Use the transformation equations of plane strain except when Mohr’s circle is specified (Problems 7.7-23 through 7.7-28). ␴y ␴x h b x Problem 7.7-1 A thin rectangular plate in biaxial stress is subjected to stresses