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3 torsion- Mechanics of Materials - 4th - Beer
1.
MECHANICS OF MATERIALS Fourth Edition Ferdinand
P. Beer E. Russell Johnston, Jr. John T. DeWolf Lecture Notes: J. Walt Oler Texas Tech University CHAPTER © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 Torsion
2.
© 2006 The
McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf 3 - 2 Contents Introduction Torsional Loads on Circular Shafts Net Torque Due to Internal Stresses Axial Shear Components Shaft Deformations Shearing Strain Stresses in Elastic Range Normal Stresses Torsional Failure Modes Sample Problem 3.1 Angle of Twist in Elastic Range Statically Indeterminate Shafts Sample Problem 3.4 Design of Transmission Shafts Stress Concentrations Plastic Deformations Elastoplastic Materials Residual Stresses Example 3.08/3.09 Torsion of Noncircular Members Thin-Walled Hollow Shafts Example 3.10
3.
© 2006 The
McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf 3 - 3 Torsional Loads on Circular Shafts • Interested in stresses and strains of circular shafts subjected to twisting couples or torques • Generator creates an equal and opposite torque T’ • Shaft transmits the torque to the generator • Turbine exerts torque T on the shaft
4.
© 2006 The
McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf 3 - 4 Net Torque Due to Internal Stresses ( )∫ ∫== dAdFT τρρ • Net of the internal shearing stresses is an internal torque, equal and opposite to the applied torque, • Although the net torque due to the shearing stresses is known, the distribution of the stresses is not. • Unlike the normal stress due to axial loads, the distribution of shearing stresses due to torsional loads can not be assumed uniform. • Distribution of shearing stresses is statically indeterminate – must consider shaft deformations.
5.
© 2006 The
McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf 3 - 5 Axial Shear Components • Torque applied to shaft produces shearing stresses on the faces perpendicular to the axis. • The existence of the axial shear components is demonstrated by considering a shaft made up of axial slats. • Conditions of equilibrium require the existence of equal stresses on the faces of the two planes containing the axis of the shaft. • The slats slide with respect to each other when equal and opposite torques are applied to the ends of the shaft.
6.
© 2006 The
McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf 3 - 6 • From observation, the angle of twist of the shaft is proportional to the applied torque and to the shaft length. L T ∝ ∝ φ φ Shaft Deformations • When subjected to torsion, every cross-section of a circular shaft remains plane and undistorted. • Cross-sections of noncircular (non- axisymmetric) shafts are distorted when subjected to torsion. • Cross-sections for hollow and solid circular shafts remain plain and undistorted because a circular shaft is axisymmetric.
7.
© 2006 The
McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf 3 - 7 Shearing Strain • Consider an interior section of the shaft. As a torsional load is applied, an element on the interior cylinder deforms into a rhombus. • Shear strain is proportional to twist and radius maxmax and γ ρ γ φ γ cL c == L L ρφ γρφγ == or • It follows that • Since the ends of the element remain planar, the shear strain is equal to angle of twist.
8.
© 2006 The
McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf 3 - 8 Stresses in Elastic Range J c dA c dAT max2max τ ρ τ ρτ ∫ =∫ == • Recall that the sum of the moments from the internal stress distribution is equal to the torque on the shaft at the section, 4 2 1 cJ π= ( )4 1 4 22 1 ccJ −= π andmax J T J Tc ρ ττ == • The results are known as the elastic torsion formulas, • Multiplying the previous equation by the shear modulus, maxγ ρ γ G c G = maxτ ρ τ c = From Hooke’s Law, γτ G= , so The shearing stress varies linearly with the radial position in the section.
9.
© 2006 The
McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf 3 - 9 Normal Stresses • Elements with faces parallel and perpendicular to the shaft axis are subjected to shear stresses only. Normal stresses, shearing stresses or a combination of both may be found for other orientations. ( ) max 0 0max 45 0max0max 2 2 245cos2 o τ τ σ ττ === =°= A A A F AAF • Consider an element at 45o to the shaft axis, • Element a is in pure shear. • Note that all stresses for elements a and c have the same magnitude • Element c is subjected to a tensile stress on two faces and compressive stress on the other two.
10.
© 2006 The
McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf 3 - 10 Torsional Failure Modes • Ductile materials generally fail in shear. Brittle materials are weaker in tension than shear. • When subjected to torsion, a ductile specimen breaks along a plane of maximum shear, i.e., a plane perpendicular to the shaft axis. • When subjected to torsion, a brittle specimen breaks along planes perpendicular to the direction in which tension is a maximum, i.e., along surfaces at 45o to the shaft axis.
11.
© 2006 The
McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf 3 - 11 Shaft BC is hollow with inner and outer diameters of 90 mm and 120 mm, respectively. Shafts AB and CD are solid of diameter d. For the loading shown, determine (a) the minimum and maximum shearing stress in shaft BC, (b) the required diameter d of shafts AB and CD if the allowable shearing stress in these shafts is 65 MPa. Sample Problem 3.1 SOLUTION: • Cut sections through shafts AB and BC and perform static equilibrium analyses to find torque loadings. • Given allowable shearing stress and applied torque, invert the elastic torsion formula to find the required diameter. • Apply elastic torsion formulas to find minimum and maximum stress on shaft BC.
12.
© 2006 The
McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf 3 - 12 SOLUTION: • Cut sections through shafts AB and BC and perform static equilibrium analysis to find torque loadings. ( ) CDAB ABx TT TM =⋅= −⋅==∑ mkN6 mkN60 ( ) ( ) mkN20 mkN14mkN60 ⋅= −⋅+⋅==∑ BC BCx T TM Sample Problem 3.1
13.
© 2006 The
McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf 3 - 13 • Apply elastic torsion formulas to find minimum and maximum stress on shaft BC. ( ) ( ) ( )[ ] 46 444 1 4 2 m1092.13 045.0060.0 22 − ×= −=−= ππ ccJ ( )( ) MPa2.86 m1092.13 m060.0mkN20 46 2 2max = × ⋅ === −J cTBCττ MPa7.64 mm60 mm45 MPa2.86 min min 2 1 max min = == τ τ τ τ c c MPa7.64 MPa2.86 min max = = τ τ • Given allowable shearing stress and applied torque, invert the elastic torsion formula to find the required diameter. m109.38 mkN6 65 3 3 2 4 2 max − ×= ⋅ === c c MPa c Tc J Tc ππ τ mm8.772 == cd Sample Problem 3.1
14.
© 2006 The
McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf 3 - 14 Angle of Twist in Elastic Range • Recall that the angle of twist and maximum shearing strain are related, L cφ γ =max • In the elastic range, the shearing strain and shear are related by Hooke’s Law, JG Tc G == max max τ γ • Equating the expressions for shearing strain and solving for the angle of twist, JG TL =φ • If the torsional loading or shaft cross-section changes along the length, the angle of rotation is found as the sum of segment rotations ∑= i ii ii GJ LT φ
15.
© 2006 The
McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf 3 - 15 • Given the shaft dimensions and the applied torque, we would like to find the torque reactions at A and B. Statically Indeterminate Shafts • From a free-body analysis of the shaft, which is not sufficient to find the end torques. The problem is statically indeterminate. ftlb90 ⋅=+ BA TT ftlb90 12 21 ⋅=+ AA T JL JL T • Substitute into the original equilibrium equation, AB BA T JL JL T GJ LT GJ LT 12 21 2 2 1 1 21 0 ==−=+= φφφ • Divide the shaft into two components which must have compatible deformations,
16.
© 2006 The
McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf 3 - 16 Sample Problem 3.4 Two solid steel shafts are connected by gears. Knowing that for each shaft G = 11.2 x 106 psi and that the allowable shearing stress is 8 ksi, determine (a) the largest torque T0 that may be applied to the end of shaft AB, (b) the corresponding angle through which end A of shaft AB rotates. SOLUTION: • Apply a static equilibrium analysis on the two shafts to find a relationship between TCD and T0 . • Find the corresponding angle of twist for each shaft and the net angular rotation of end A. • Find the maximum allowable torque on each shaft – choose the smallest. • Apply a kinematic analysis to relate the angular rotations of the gears.
17.
© 2006 The
McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf 3 - 17 SOLUTION: • Apply a static equilibrium analysis on the two shafts to find a relationship between TCD and T0 . ( ) ( ) 0 0 8.2 in.45.20 in.875.00 TT TFM TFM CD CDC B = −== −== ∑ ∑ • Apply a kinematic analysis to relate the angular rotations of the gears. CB CC B C B CCBB r r rr φφ φφφ φφ 8.2 in.875.0 in.45.2 = == = Sample Problem 3.4
18.
© 2006 The
McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf 3 - 18 • Find the T0 for the maximum allowable torque on each shaft – choose the smallest. ( ) ( ) ( ) ( ) in.lb561 in.5.0 in.5.08.2 8000 in.lb663 in.375.0 in.375.0 8000 0 4 2 0 max 0 4 2 0 max ⋅= == ⋅= == T T psi J cT T T psi J cT CD CD AB AB π π τ τ inlb5610 ⋅=T • Find the corresponding angle of twist for each shaft and the net angular rotation of end A. ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( ) oo / oo o 64 2 / o 64 2 / 2.2226.8 26.895.28.28.2 95.2rad514.0 psi102.11in.5.0 .in24in.lb5618.2 2.22rad387.0 psi102.11in.375.0 .in24in.lb561 +=+= === == × ⋅ == == × ⋅ == BABA CB CD CD DC AB AB BA GJ LT GJ LT φφφ φφ φ φ π π o 48.10=Aφ Sample Problem 3.4
19.
© 2006 The
McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf 3 - 19 Design of Transmission Shafts • Principal transmission shaft performance specifications are: power speed • Determine torque applied to shaft at specified power and speed, f PP T fTTP πω πω 2 2 == == • Find shaft crosssection which will not exceed the maximum allowable shearing stress, ( ) ( ) ( )shaftshollow 2 shaftssolid 2 max 4 1 4 2 22 max 3 max τ π τ π τ T cc cc J T c c J J Tc =−= == = • Designer must select shaft material and crosssection to meet performance specifications without exceeding allowable shearing stress.
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McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf 3 - 20 Stress Concentrations • The derivation of the torsion formula, assumed a circular shaft with uniform crosssection loaded through rigid end plates. J Tc =maxτ J Tc K=maxτ • Experimental or numerically determined concentration factors are applied as • The use of flange couplings, gears and pulleys attached to shafts by keys in keyways, and crosssection discontinuities can cause stress concentrations
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McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf 3 - 21 Plastic Deformations • With the assumption of a linearly elastic material, J Tc =maxτ ( ) ∫=∫= cc ddT 0 2 0 22 ρτρπρπρρτ • The integral of the moments from the internal stress distribution is equal to the torque on the shaft at the section, • Shearing strain varies linearly regardless of material properties. Application of shearingstress strain curve allows determination of stress distribution. • If the yield strength is exceeded or the material has a nonlinear shearingstressstrain curve, this expression does not hold.
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McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf 3 - 22 • As , the torque approaches a limiting value,0→Yρ torqueplasticTT YP == 3 4 Elastoplastic Materials • As the torque is increased, a plastic region ( ) develops around an elastic core ( )Yττ = Y Y τ ρ ρ τ = −= −= 3 3 4 1 3 4 3 3 4 13 3 2 11 c T c cT Y Y Y Y ρρ τπ −= 3 3 4 1 3 4 1 φ φY YTT φ γ ρ YL Y = • At the maximum elastic torque, YYY c c J T τπτ 3 2 1== c L Y Y γ φ =
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McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf 3 - 23 Residual Stresses • Plastic region develops in a shaft when subjected to a large enough torque. • On a T-φ curve, the shaft unloads along a straight line to an angle greater than zero. • When the torque is removed, the reduction of stress and strain at each point takes place along a straight line to a generally nonzero residual stress. • Residual stresses found from principle of superposition ( ) 0=∫ dAτρ J Tc m =′τ
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McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf 3 - 24 Example 3.08/3.09 A solid circular shaft is subjected to a torque at each end. Assuming that the shaft is made of an elastoplastic material with and determine (a) the radius of the elastic core, (b) the angle of twist of the shaft. When the torque is removed, determine (c) the permanent twist, (d) the distribution of residual stresses. MPa150=Yτ GPa77=G mkN6.4 ⋅=T SOLUTION: • Solve Eq. (3.32) for ρY/c and evaluate the elastic core radius • Find the residual stress distribution by a superposition of the stress due to twisting and untwisting the shaft • Evaluate Eq. (3.16) for the angle which the shaft untwists when the torque is removed. The permanent twist is the difference between the angles of twist and untwist • Solve Eq. (3.36) for the angle of twist
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McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf 3 - 25 SOLUTION: • Solve Eq. (3.32) for ρY/c and evaluate the elastic core radius 3 1 341 3 3 4 1 3 4 −=⇒ −= Y YY Y T T cc TT ρρ ( ) ( )( ) mkN68.3 m1025 m10614Pa10150 m10614 m1025 3 496 49 3 2 14 2 1 ⋅= × ×× = =⇒= ×= ×== − − − − Y Y Y Y Y T c J T J cT cJ τ τ ππ 630.0 68.3 6.4 34 3 1 = −= c Yρ mm8.15=Yρ • Solve Eq. (3.36) for the angle of twist ( )( ) ( )( ) o3 3 3 49- 3 8.50rad103.148 630.0 rad104.93 rad104.93 Pa1077m10614 m2.1mN1068.3 =×= × = ×= ×× ⋅× == =⇒= − − − φ φ φ ρ φ φ ρ φ φ Y Y Y Y YY Y JG LT cc o 50.8=φ Example 3.08/3.09
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McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf 3 - 26 • Evaluate Eq. (3.16) for the angle which the shaft untwists when the torque is removed. The permanent twist is the difference between the angles of twist and untwist ( )( ) ( )( ) o 3 949 3 1.81 69.650.8 6.69rad108.116 Pa1077m1014.6 m2.1mN106.4 = °−°= ′−= °=×= ×× ⋅× = =′ − − φφ φ pφ JG TL o 81.1=pφ • Find the residual stress distribution by a superposition of the stress due to twisting and untwisting the shaft ( )( ) MPa3.187 m10614 m1025mN106.4 49- 33 max = × ×⋅× ==′ − J Tc τ Example 3.08/3.09
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McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf 3 - 27 Torsion of Noncircular Members • At large values of a/b, the maximum shear stress and angle of twist for other open sections are the same as a rectangular bar. Gabc TL abc T 3 2 2 1 max == φτ • For uniform rectangular cross-sections, • Previous torsion formulas are valid for axisymmetric or circular shafts • Planar cross-sections of noncircular shafts do not remain planar and stress and strain distribution do not vary linearly
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McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf 3 - 28 Thin-Walled Hollow Shafts • Summing forces in the x-direction on AB, shear stress varies inversely with thickness ( ) ( ) flowshear 0 ==== ∆−∆==∑ qttt xtxtF BBAA BBAAx τττ ττ ( ) ( ) tA T qAdAqdMT dAqpdsqdstpdFpdM 2 22 2 0 0 = === ==== ∫∫ τ τ • Compute the shaft torque from the integral of the moments due to shear stress ∫= t ds GA TL 2 4 φ • Angle of twist (from Chapter 11)
29.
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McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf 3 - 29 Example 3.10 Extruded aluminum tubing with a rectangular cross-section has a torque loading of 24 kip- in. Determine the shearing stress in each of the four walls with (a) uniform wall thickness of 0.160 in. and wall thicknesses of (b) 0.120 in. on AB and CD and 0.200 in. on CD and BD. SOLUTION: • Determine the shear flow through the tubing walls. • Find the corresponding shearing stress with each wall thickness .
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McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourth Edition Beer • Johnston • DeWolf 3 - 30 SOLUTION: • Determine the shear flow through the tubing walls. ( )( ) ( ) in. kip 335.1 in.986.82 in.-kip24 2 in.986.8in.34.2in.84.3 2 2 === == A T q A • Find the corresponding shearing stress with each wall thickness. With a uniform wall thickness, in.160.0 in.kip335.1 == t q τ ksi34.8=τ With a variable wall thickness in.120.0 in.kip335.1 == ACAB ττ in.200.0 in.kip335.1 == CDBD ττ ksi13.11== BCAB ττ ksi68.6== CDBC ττ Example 3.10
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