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θ=TLGJ=(4.10*103N#mm)(250mm)(80*103N>mm2)(982mm4)=0.013radUsing πrad=180∘,θ=(0.013 rad)(180°>πrad)=0.75°3–10 TORSION IN MEMBERS HAVING NONCIRCULAR CROSS SECTIONSThe behavior of members having noncircular cross sections when subjected to torsion is radically different from that for members having circular cross sections. However, the fac-tors of most use in machine design are the maximum stress and the total angle of twist for such members. The formulas for these factors can be expressed in similar forms to the for-mulas used for members of circular cross section (solid and hollow round shafts).The following two formulas can be used:➭Torsional Shear Stressτmax=T>Q(3–12)➭Deflection for Noncircular Sectionsθ=TL>GK(3–13)Note that Equations (3–12) and (3–13) are similar to Equations (3–10) and (3–11), with the substitution of Qfor Zpand Kfor J. Refer to Figure 3–10 for the methods of deter-mining the values for Kand Qfor several types of cross sec-tions useful in machine design. These values are appropriate only if the ends of the member are free to deform. If either end is fixed, as by welding to a solid structure, the resulting stress and angular twist are quite different. (See References 1–3, 6, and 7.)Example Problem3–8A 2.50-in-diameter shaft carrying a chain sprocket has one end milled in the form of a square to permit the use of a hand crank. The square is 1.75 in on a side. Compute the maximum shear stress on the square part of the shaft when a torque of 15 000 lb · in is applied.Also, if the length of the square part is 8.00 in, compute the angle of twist over this part. The shaft material is steel with G=11.5*106psi.Solution ObjectiveCompute the maximum shear stress and the angle of twist in the shaft.
96PART ONEPrinciples of Design and Stress Analysis Colored dot ( ) denoteslocation of H9270maxFIGURE 3–10 Methods for determining values for Kand Qfor several types of cross sectionsGivenTorque=T=15000lb#in; length=L=8.00in.The shaft is square; thus, a=1.75in.G=11.5*106psi.AnalysisResultsFigure 3–10 shows the methods for calculating the values for Qand Kfor use in Equations (3–12) and (3–13).Q=0.208a3=(0.208)(1.75in)3=1.115in3K=0.141a4=(0.141)(1.75in)4=1.322in4Now the stress and the deflection can be computed.τmax=TQ=15 000lb#in(1.115in3)=13 460 psiθ=TLGK=(15 000lb#in)(8.00in)(11.5*106lb>in2)(1.322in4)=0.0079radConvert the angle of twist to degrees:θ=(0.0079 rad)(180°>πrad)=0.452°
CHAPTER THREEStress and Deformation Analysis 97The shear stress computed by this approach is the aver-age stressin the tube wall. However, if the wall thickness tis small (a thin wall), the stress is nearly uniform through-out the wall, and this approach will yield a close approxi-mation of the maximum stress. For the analysis of tubular sections having nonuniform wall thickness, see References 1–3, 6, and 7.