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Torsional DeformationsProblem 3.2-1A copper rod of length L18.0 in. is to be twisted by torques T(see figure) until the angle of rotation between the ends of the rod is 3.0°. If the allowable shear strain in the copper is 0.0006 rad, what is the maximum permissible diameter of the rod?3TorsionProblem 3.2-2A plastic bar of diameter d50 mm is to be twisted bytorques T(see figure) until the angle of rotation between the ends of thebar is 5.0°. If the allowable shear strain in the plastic is 0.012 rad, what is theminimum permissible length of the bar?L
Problem 3.2-3A circular aluminum tube subjected to pure torsion by torques T(see figure) has an outer radius r2equal to twice the innerradius r1(a)If the maximum shear strain in the tube is measured as 400 106rad, what is the shear strain 1at the innersurface? (b)If the maximum allowable rate of twist is 0.15 degrees per foot and the maximum shear strain is to be kept at 400 106rad by adjusting the torque T, what is theminimum required outer radius (r2)min?Solution 3.2-3Circular aluminum tubeLTr1r2Tr22r1max400106rad(a) SHEAR STRAIN AT INNER SURFACEFrom Eq. (3-5b):g1200106rad—g112g212(400106rad)218.2106rad in.(0.15ft)¢180raddegree≤¢112ftin.≤uallow0.15ft(b) MINIMUM OUTER RADIUSFrom Eq. (3-5a):(r2)min1.83in.—(r2)mingmaxuallow400106rad218.2106rad in.gmaxr2fLr2ur1r2Problem 3.2-4A circular steel tube of length L0.90 m is loaded intorsion by torques T(see figure). (a)If the inner radius of the tube is r140 mm and the measuredangle of twist between the ends is 0.5°, what is the shear strain 1(in radians) at the inner surface? (b)If the maximum allowable shear strain is 0.0005 rad and the angleof twist is to be kept at 0.5° by adjusting the torque T, what is themaximum permissible outer radius (r2)max?Problems 3.2-3, 3.2-4, and 3.2-5. 182CHAPTER 3Torsion
SECTION 3.2Torsional Deformations183Solution 3.2-4Circular steel tubeL0.90 mr140 mm(a) SHEAR STRAIN AT INNER SURFACEFrom Eq. (3-5b):g1388106rad—gming1r1fL(40mm)(0.008727rad)900mmgmax0.0005rad0.008727radf0.5(0.5 )¢180raddegree≤(b) MAXIMUM OUTER RADIUSFrom Eq. (3-5a):(r2)max51.6mm—(r2)max(0.0005rad)(900mm)0.008727radr2gmaxLfgmaxg2r2fL;r1r2Problem 3.2-5Solve the preceding problem if the length L50 in., the inner radius r11.5 in., the angle of twist is 0.6°, and the allowableshear strain is 0.0004 rad.Solution 3.2-5Circular steel tubeL50 in.r11.5 in.(a) SHEAR STRAIN AT INNER SURFACEFrom Eq. (3-5b):g1314106rad—gming1r1fL(1.5in.)(0.010472rad)50in.