\u03b8 TL GJ 410 10 3 N mm250 mm 80 10 3 N mm 2 982 mm 4 0013 rad Using \u03c0 rad 180 \u03b8

# Θ tl gj 410 10 3 n mm250 mm 80 10 3 n mm 2 982 mm 4

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θ = TL GJ = (4.10 * 10 3 N # mm)(250 mm) (80 * 10 3 N > mm 2 )(982 mm 4 ) = 0.013 rad Using π rad = 180 , θ = (0.013 rad)(180 ° > π rad) = 0.75 ° 3–10 TORSION IN MEMBERS HAVING NONCIRCULAR CROSS SECTIONS The 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 Stress τ max = T > Q (3–12) Deflection for Noncircular Sections θ = 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 Q for Z p and K for J . Refer to Figure 3–10 for the methods of deter- mining the values for K and Q for 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 Problem 3–8 A 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 * 10 6 psi. Solution Objective Compute the maximum shear stress and the angle of twist in the shaft.
96 PART ONE Principles of Design and Stress Analysis Colored dot ( ) denotes location of H9270 max FIGURE 3–10 Methods for determining values for K and Q for several types of cross sections Given Torque = T = 15000 lb # in; length = L = 8.00 in. The shaft is square; thus, a = 1.75 in. G = 11.5 * 10 6 psi. Analysis Results Figure 3–10 shows the methods for calculating the values for Q and K for use in Equations (3–12) and (3–13). Q = 0.208 a 3 = (0.208)(1.75 in) 3 = 1.115 in 3 K = 0.141 a 4 = (0.141)(1.75 in) 4 = 1.322 in 4 Now the stress and the deflection can be computed. τ max = T Q = 15 000 lb # in (1.115 in 3 ) = 13 460 psi θ = TL GK = (15 000 lb # in)(8.00 in) (11.5 * 10 6 lb > in 2 )(1.322 in 4 ) = 0.0079 rad Convert the angle of twist to degrees: θ = (0.0079 rad)(180 ° > π rad) = 0.452 °
CHAPTER THREE Stress and Deformation Analysis 97 The shear stress computed by this approach is the aver- age stress in the tube wall. However, if the wall thickness t is 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.