# Fourth Edition MECHANICS OF MATERIALS OF Fourth Edition MECHANICS OF MATERIALS CHAPTER Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf Torsion Lecture Notes: J. Walt Oler

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MECHANICS OFFourth EditionMECHANICS OF MATERIALSCHAPTERMATERIALSFerdinand P. BeerE. Russell Johnston, Jr.John T. DeWolfTorsionLecture Notes:J. Walt OlerTexas Tech Universityy 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfContentsIntroductionTorsional Loads on Circular ShaftsStatically Indeterminate ShaftsSample Problem 3.4Net Torque Due to Internal StressesAxial Shear ComponentsDesign of Transmission ShaftsStress ConcentrationsShaft DeformationsShearing StrainStresses in Elastic RangePlastic DeformationsElastoplastic MaterialsResidual StressesStresses in Elastic RangeNormal StressesTorsional Failure ModesResidual StressesExamples 3.08, 3.09Torsion of Noncircular MembersSample Problem 3.1Angle of Twist in Elastic RangeThin-Walled Hollow ShaftsExample 3.10 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 2MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfTorsional Loads on Circular Shafts Interested in stresses and strains of circular shafts subjected to twisting lcouples or torques. Turbine exerts torque T on the shaft. Generator creates an equal and Shaft transmits the torque to the generator. Generator creates an equal and opposite torque T. 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 3MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfNet Torque Due to Internal Stresses Net of the internal shearing stresses is an internal torque, equal and opposite to the dAdFT applied torque, Although the net torque due to the shearing stresses is known, the distribution of the stresses is not. Distribution of shearing stresses is statically indeterminate must consider shaft deformations. Unlike the normal stress due to axial loads, the distribution of shearing stresses due to torsional l d t b d if 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 4loads cannot be assumed uniform.MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfAxial Shear Components Torque applied to shaft produces shearing stresses on the faces perpendicular to the iaxis. Conditions of equilibrium require the existence of equal stresses on the faces of the The existence of the axial shear componentsexistence of equal stresses on the faces of the two planes containing the axis of the shaft.The existence of the axial shear components is demonstrated by considering a shaft made up of axial slats. The slats slide with respect to each other when equal and opposite torques are applied to the ends of the shaft. 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 5MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfShaft Deformations From observation, the angle of twist, of the shaft is proportional to the applied torque and to the shaft lengthto the shaft length.LT When subjected to torsion, every cross section of a circular shaft remains plane and undistorted. Cross-sections for hollow and solid circular shafts remain plain and undistorted because a circular shaft is axisymmetric. Cross-sections of noncircular (non-axisymmetric) shafts are distorted when subjected to torsion. 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 6MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfShearing Strain Consider an interior section of the shaft. As a torsional load is applied, an element on the i t i li d d f i t h btan in radianxy xy interior cylinder deforms into a rhombus. Since the ends of the element remain planar, the shear strain is equal to angle of twist For small value of , It follows thatthe shear strain is equal to angle of twist.AA L AA or AA LL (and are in radians)AA AA Shear strain is proportional to twist and radiusmaxmax and LcL(Shown next page) 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 7cLtan AAL MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfShearing StraincAt shaft surface, Max shearing strainmaxcL Shearing strain at a distance from the shaft axisShearing strain at a distance from the shaft axisShear strain inside the shaft, LL maxmaxShear strain at the shaft's surface, LcL cLL maxmax LLc c 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 8maxc MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfStresses in Elastic Range Multiplying previous eq. by the shear modulus,max GcG cFrom Hookes Law, G , soThe shearing stress varies linearly421 cJ maxcThe shearing stress varies linearly with the radial position in the section.F h ll li dmax max2From or c c For hollow cylinder 414221 ccJ 211 min max2Substitute ccc 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 9J = Polar moment of inertiamin 1max 2 ccMECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfStresses in Elastic Range Recall that the sum of the moments from the internal stress distribution is equal to the torque on the shaft at the sectionJcdAcdAT max2max the torque on the shaft at the section,Th lt k th l ti t idAJ 2max c The results are known as the elastic torsion formulas,andmaxTTc dAdFT and max JJ 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 10MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfNormal Stresses Elements with faces parallel and perpendicular to the shaft axis are subjected to shear stresses only (Element a). Normal stresses, shearing stresses or a combination of both may be found for other orientations (Element b). Consider an element at 45o to the shaft axis, max0max450max0max22245cos2o AAAFAAF0 2AA Element a is in pure shear. Element c is subjected to a tensile stress on Note that all stresses for elements a and c have jtwo faces and compressive stress on the other two. 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 11the same magnitude.MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfTorsional Failure Modes Ductile materials generally fail in shear. Brittle materials are weaker inshear. 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 planesspecimen breaks along planes perpendicular to the direction in which tension is a maximum, i.e., along surfaces at 45o to the shaft 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 12gaxis.MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolf 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 13MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolf 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 14MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfSample Problem 3.1SOLUTION: Cut sections through shafts ABd BC d f iand BC and perform static equilibrium analyses to find torque loadings.Shaft BC is hollow with inner and outer Apply elastic torsion formulas to find minimum and maximum stress on shaft BC.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, Given allowable shearing stress and applied torque, invert the elastic torsion formula to find the g ,determine (a) the minimum and maximum shearing stress in shaft BC, (b) the required diameter d of shafts AB and CDrequired diameter. 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 15if the allowable shearing stress in these shafts is 65 MPa.MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfSOLUTIONSample Problem 3.1SOLUTION: Cut sections through shafts AB and BCand perform static equilibrium analysis to find torque loadingsto find torque loadings. CDABABxTTTMmkN6mkN60 mkN20mkN14mkN60BCBCxTTM 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 16MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfA l l ti t i f l t Gi ll bl h i t dSample Problem 3.1 Apply elastic torsion formulas to find minimum and maximum stress on shaft BC. Given allowable shearing stress and applied torque, invert the elastic torsion formula to find the required diameter. 444142 045.0060.022 ccJ mkN665 34max MPaTcJTc46 m1092.1322 109213m060.0mkN204622max JcTBCm109.38 33242cccJ mm8772 cdMPa2.86m1092.13 46 Jmm45min1min cmm8.772 cddAB = dBC 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 17MPa7.64mm60MPa2.86min2max cMPa7.64MPa2.86minmaxMECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolf290 N-mFixed end 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 18MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolf2.8 kN-mFree end(T = 0)(T = 0) 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 19MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolf 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 20MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolf 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 21MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolf 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 22MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolf 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 23MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolf 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 24Why ? Show on next page !!!CCD ABBrT Tr MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfGear CGear C Gear BGear Brc=240 mmFFTCDrB=80 mmBCT=TAB= 900 N-mFCDTCD = (F)(rc) TAB = 900 = (F)(rB)CD ABT TF C BCCD ABr rrT Tr 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 25BrMECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolf 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 26MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfAngle of Twist in Elastic Range Recall that the angle of twist and maximum shearing strain are related,Lc max and max JTJTc Lmax In the elastic range, the shearing strain and shear are related by Hookes Law,T max JJmaxmaxTc cG JG L Equating the expressions for shearing strain and solving for the angle of twistsolving for the angle of twist,JGTLThe relation shows that, within the elastic range, the angle of twist is proportional to the torque T applied to the shaft. This is accordance with the experimental evidence. The 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 27previous eqs also provides us with a convenient method for determining the modulus of rigidity of a given materials.MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfAngle of Twist in Elastic Range If the torsional loading or shaft cross-section changes along the length, the angle of rotation is found as the sum of segment rotationsfound as the sum of segment rotationsi iiiiGJLT 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 28MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolf 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 29MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolf 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 30MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolf(A to B)(A to C) 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 31See from A to CMECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfA/BB/CA to B B to CB/CA/BA to C 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 32A to CMECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolf 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 33MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolf 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 34MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfSample Problem 3.4SOLUTION: Apply a static equilibrium analysis on th t h ft t fi d l ti hithe two shafts to find a relationship between TCD and T0 . Apply a kinematic analysis to relate Two solid steel shafts are connected Find the maximum allowable torque on each shaft choose the smallest.the angular rotations of the gears.Two solid steel shafts are connected by gears. Knowing that for each shaft G = 77 GPa and that the allowable shearing stress is 55 MPa, determine Find the corresponding angle of twist for each shaft and the net angular rotation of end Ao eac s a t c oose t e s a est.g(a) the largest torque T0 that may be applied to the end of shaft AB, (b) the corresponding angle through which rotation of end A. 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 35end A of shaft AB rotates.MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfSample Problem 3.4SOLUTION: Apply a static equilibrium analysis on the two shafts to find a relationship Apply a kinematic analysis to relate the angular rotations of the gears.the two shafts to find a relationship between TCD and T0 .the angular rotations of the gears.60CCD ABBrT TrT T 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 36222.73CD OCD OT TT TMECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfFi d th T f th i Fi d th di l f t i t f hSample Problem 3.4 Find the T0 for the maximum allowable torque on each shaft choose the smallest. Find the corresponding angle of twist for each shaft and the net angular rotation of end A.D = 0 (Fix)8.0510.261.8 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 37inlb5610 TMECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfD = 0 (Fix) 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 38MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolf 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 39MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfDesign of Transmission Shafts Determine torque applied to shaft at specified power and speed,fTTP 2fPPTfTTP22 Designer must select shaft Find shaft cross-section which will not exceed the maximum allowable shearing stress, Designer must select shaft material and cross-section to meet performance specifications without exceeding allowable g h ftlid3maxTJJTcwithout exceeding allowable shearing stress. Angular velocity 2 f 1 Hz(rpm) Hz60 rpmf shafts hollow2shaftssolid2max414222max3TccccJccAngular velocity, 2 f 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 40p 2 max22 ccMECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfWatts = N-m/s 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 41MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfStatically Indeterminate Shafts Given the shaft dimensions and the applied torque, we would like to find the torque reactions at A and B.at A and B. From a free-body analysis of the shaft,which is not sufficient to find the end torques. The problem is statically indeterminate. Divide the shaft into two components which Divide the shaft into two components which must have compatible deformations, Substitute into the original equilibrium equation, 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 42MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfStress Concentrations The derivation of the torsion formula,JTcmaxassumed a circular shaft with uniform cross-section loaded through rigid end plates. The use of flange couplings, gears, and pulleys attached to shafts by keys in keyways and cross-section discontinuities Experimental or numerically determined t ti f t li dkeyways, and cross section discontinuities can cause stress concentrations.JTcKmaxconcentration factors are applied as 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 43MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolf 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 44MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolf 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 45MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfPlastic Deformations With the assumption of a linearly elastic material,JTcmax If the yield strength is exceeded or the material has a nonlinear shearing-stress-strain curve, this expression does not hold. Shearing strain varies linearly regardless of material properties. Application of shearing-stress-strain curve allows determination of stress distribution The integral of the moments from the internal stress distribution is equal to the torque on the shaft at the ticurve allows determination of stress distribution. ccddT02022 section, 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 46MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfElastoplastic Materials At the maximum elastic torque,YYY ccJT 321cL YY As the torque is increased, a plastic region ( ) develops around an elastic core ( )Y YY YL 3341343341332 11cTccT YYYY YLY cc 334134 1YYTT As , the torque approaches a limiting value,0Ytorque plastic TT YP 34 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 47MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfResidual Stresses Plastic region develops in a shaft when subjected to a large enough torque.Wh th t i d th d ti f t When the torque is removed, the reduction of stress and strain at each point takes place along a straight line to a generally non-zero residual stress. On a T- curve, the shaft unloads along a straight line to an angle greater than zero. Residual stress found from principle of superposition.p p p p 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 48 0 dAJTcm MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfExample 3.08/3.09SOLUTION: Solve Eq. (3.32) for Y/c and evaluate the elastic core radius. Solve Eq. (3.36) for the angle of twist.A solid circular shaft is subjected to a torque at each end. Assuming that the shaft is made of an l l i i l i hmkN6.4 T Evaluate Eq. (3.16) for the angle which the shaft untwists when the torque is removed. The permanent t i t i th diff b t thelastoplastic material with and determine (a) the radius of the elastic core, (b) the angle of twist of the shaft When theMPa150YGPa77G Find the residual stress distribution by twist is the difference between the angles of twist and untwist.angle of twist of the shaft. When the torque is removed, determine (c) the permanent twist, (d) the distribution of residual stressesa superposition of the stress due to twisting and untwisting the shaft. 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 49of residual stresses.MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfSOLUTIONExample 3.08/3.09SOLUTION: Solve Eq. (3.32) for Y/c and evaluate the elastic core radius13 Solve Eq. (3.36) for the angle of twist YY31341 334134YYYY TTccTT m1025 321421 cJ 49-3Pa1077m10614m2.1mN1068.3YYYYJGLTcc m10614m10254922 YYYYJTcTcJ o3338.50rad103.148rad104.93rad104.93Pa1077m10614Y m1025m10614Pa101503496YYYTcTJ 630.0o50.8mkN68.3 630.06836.43431 Y 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 5068.3 cmm8.15YMECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfExample 3.08/3.09 Evaluate Eq. (3.16) for the angle which the shaft untwists when the torque is removed The Find the residual stress distribution by a superposition of the stress due to twisting and untwisting the shaftthe torque is removed. The permanent twist is the difference between the angles of twist and untwist.twisting and untwisting the shaft m10614m1025mN106.449-33maxJTc 3 21N1064JGTLMPa3.187 394936.69rad108.116Pa1077m1014.6m2.1mN106.4o1 8169.650.8 p 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 511.81o81.1pMECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfTorsion of Noncircular Members Previous torsion formulas are valid for axisymmetric or circular shafts. Pl ti f i l Planar cross-sections of noncircular shafts do not remain planar and stress and strain distribution do not vary linearlyTLT For uniform rectangular cross-sections,linearly. At large values of a/b, the maximum Gabcabc 3221max shear stress and angle of twist for other open sections are the same as a rectangular bar. 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 52MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfThin-Walled Hollow Shafts Summing forces in the x-direction on AB, flowshear 0qtttxtxtFBBAABBAAxshear stress varies inversely with thicknessqBBAA dAqpdsqdstpdFpdM 20 Compute the shaft torque from the integral of the moments due to shear stresstATqAdAqdMT2220 tA2 tdsGATL24 Angle of twist (from Chapter 11) 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 53tGA4MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfExample 3.10Extruded aluminum tubing with a rectangular cross-section has a torque loading of 2.7 kN .m. Determine the shearing stress in each of the four walls with (a) uniform wall thickness of 4 mm and wall thicknesses of (b) 3 mm on AB and CD and 5 mm on CD and BD.SOLUTION:D t i th h fl th h th Determine the shear flow through the tubing walls. Find the corresponding shearing stress p g gwith each wall thickness . 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 54MECHANICS OF MATERIALSFourthEditionBeer Johnston DeWolfExample 3.10SOLUTION: Determine the shear flow through the tubing walls Find the corresponding shearing stress with each wall thickness.tubing walls.With a uniform wall thickness, With a variable wall thicknessWith a variable wall thickness 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 55

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