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- 1. 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.
- 2. 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 ϭAB ϭ 2040 psi A Solve for P2: P2 ϭ 5600 lb P2 B C 1
- 3. 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
- 4. 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
- 5. 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 ␣
- 6. 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
- 7. 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
- 8. 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)
- 9. 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
- 10. 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
- 11. 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)(j2 ) ϭ g A2jdj Fx ϭ Ύ B dF ϭ D g g ϭ mass density Ύ x L g gA2 2 A2jdj ϭ (L Ϫ x 2) g 2g (a) TENSILE STRESS IN BAR AT DISTANCE x Fx g2 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 g2L2 — 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
- 12. 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
- 13. 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.
- 14. 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
- 15. 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
- 16. 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
- 17. 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
- 18. 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
- 19. 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?
- 20. 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
- 21. 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
- 22. 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
- 23. 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
- 24. 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
- 25. 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
- 26. 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 )
- 27. 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
- 28. 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
- 29. 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
- 30. 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
- 31. 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
- 32. 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
- 33. 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)
- 34. 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
- 35. 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
- 36. 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
- 37. 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
- 38. 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 2rh tϭ P P ϭ As 2rh (b) DOWNWARD DISPLACEMENT ␦ ␥ ϭ shear strain at distance r Rubber cylinder gϭ t P ϭ G 2rhG 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 2hG ␦ϭ As ϭ shear area at distance r Ύ Pdr 2rhG d␦ ϭ Ύ b2 d2 Ύ b2 d2 Pdr 2rhG dr P bր2 ϭ [ln r] dր2 r 2hG P b ln 2hG 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
- 39. 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
- 40. 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
- 41. 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
- 42. 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
- 43. 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
- 44. 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
- 45. 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
- 46. 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
- 47. 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
- 48. 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
- 49. 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 7d 2 B d Ϫ ¢d Ϫ ≤ R ϭ 4 4 64 P ϭ sallow A ϭ 7d2 s 64 allow P B 7sallow SOLVE FOR d: d2 ϭ 64 P 7sallow dϭ8 1200 kN ϭ 153 mm B 7 (150 MPa) SUBSTITUTE NUMERICAL VALUES: dmin ϭ 8
- 50. 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 3tallow 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
- 51. 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
- 52. 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
- 53. 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
- 54. 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) ¢ 2allow 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) ¢ 2allow 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
- 55. 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
- 56. 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
- 57. 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.)
- 58. 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 sallowt(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
- 59. 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
- 60. 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
- 61. 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)
- 62. 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
- 63. 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. 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
- 65. 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. 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
- 67. 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. 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

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