Slide 1Design of Transmission ShaftsStress ConcentrationsPlastic DeformationsElastoplastic MaterialsResidual StressesExample 3.08/3.09Slide 8Slide 9Torsion of Noncircular MembersThin-Walled Hollow ShaftsExample 3.10Slide 13© 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourthEditionBeer • Johnston • DeWolf3 - 1•Find the T0 for the maximum allowable torque on each shaft – choose the smallest.    in.lb561in.5.0in.5.08.28000in.lb663in.375.0in.375.080000420max0420maxTTpsiJcTTTpsiJcTCDCDABABinlb5610T•Find the corresponding angle of twist for each shaft and the net angular rotation of end A.         oo/ooo642/o642/2.2226.826.895.28.28.295.2rad514.0psi102.11in.5.0.in24in.lb5618.22.22rad387.0psi102.11in.375.0.in24in.lb561BABACBCDCDDCABABBAGJLTGJLTo48.10ASample Problem 3.4© 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourthEditionBeer • Johnston • DeWolf3 - 2Design of Transmission Shafts•Principal transmission shaft performance specifications are: power speed•Determine torque applied to shaft at specified power and speed,fPPTfTTP22•Find shaft cross-section which will not exceed the maximum allowable shearing stress,   shafts hollow2shafts solid2max414222max3maxTccccJTccJJTc•Designer must select shaft material and cross-section to meet performance specifications without exceeding allowable shearing stress.© 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourthEditionBeer • Johnston • DeWolf3 - 3Stress Concentrations•The derivation of the torsion formula,assumed a circular shaft with uniform cross-section loaded through rigid end plates.JTcmaxJTcKmax•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 cross-section discontinuities can cause stress concentrations© 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourthEditionBeer • Johnston • DeWolf3 - 4Plastic Deformations•With the assumption of a linearly elastic material,JTcmax ccddT02022•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 shearing-stress-strain curve allows determination of stress distribution.•If the yield strength is exceeded or the material has a nonlinear shearing-stress-strain curve, this expression does not hold.© 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourthEditionBeer • Johnston • DeWolf3 - 5•As , the torque approaches a limiting value,0Ytorque plastic TTYP34Elastoplastic Materials•As the torque is increased, a plastic region ( ) develops around an elastic core ( )YYY334134334133211cTccTYYYY3341341YYTTYLY•At the maximum elastic torque,YYYccJT321cLYY© 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourthEditionBeer • Johnston • DeWolf3 - 6Residual Stresses•Plastic region develops in a shaft when subjected to a large enough torque.•On a T- 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 non-zero residual stress.•Residual stresses found from principle of superposition 0dAJTcm© 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourthEditionBeer • Johnston • DeWolf3 - 7Example 3.08/3.09A 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YGPa77GmkN6.4 TSOLUTION:•Solve Eq. (3.32) for Y/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© 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourthEditionBeer • Johnston • DeWolf3 - 8SOLUTION:•Solve Eq. (3.32) for Y/c and evaluate the elastic core radius31341334134YYYYTTccTT   mkN68.3m1025m10614Pa10150m10614m1025349649321421YYYYYTcJTJcTcJ630.068.36.43431cYmm8.15Y•Solve Eq. (3.36) for the angle of twist    o33349-38.50rad103.148630.0rad104.93rad104.93Pa1077m10614m2.1mN1068.3YYYYYYYJGLTcco50.8Example 3.08/3.09© 2006 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALSFourthEditionBeer • Johnston • DeWolf3 - 9•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    o394931.8169.650.86.69rad108.116Pa1077m1014.6m2.1mN106.4pφJGTLo81.1p•Find the residual stress distribution by a superposition of the stress due to twisting and untwisting the shaft  MPa3.187m10614m1025mN106.449-33maxJTcExample 3.08/3.09© 2006 The