Question

9.3-11 A circular tube of aluminum is subjected to torsion by torques T applied at the ends (see figure). The bar is 24 in. long, and the inside and outside diameters are 1.25 in. and 1.75 in., respectively. It is determined by measurement that the angle of twist is 4° when the torque is 6200 lb-in. (a) Calculate the maximum shear stress Tmax in the tube, the shear modulus of elasticity G, and the maximum shear strain y max (in radians). (b) If the maximum shear strain in the tube is limited to 2.5 x 10-3 and the inside diameter is increased to 1.375 in., what is the maximum permissible torque? T T 24 in. ho 1.25 in. 1.75 in.

Answer 1

T outer dia 1 24 in do = 1.75in inner di= 1.25in Polar moment of inertia I (464-4,4)= (1-754- 1.234) = 0.6811 int J = 32 T- 6200 lb-in Augleog Twist o = 4° 0.0698 elad, Torsional & τ 11 T J GO e outer dig Imax. 0ca- Shear max stress T J do 7965.056 Tmax= 175 x 6200 lb-in in 2 0.6811 6200 x 24 Go GE 7 Il OJ 3.13 x 106 eb inz 0.0698 0-680 Tmax= rmax G -3 2-544 X107 Imax. Nmax a 7965.056 3.13 x 106 di = 1•375 in do= lors in 2-5x10 3 i if Vmax.= e 6 = A (doh dil) 0.5698 in 4 3.138106 eb in 2 JE 2.5x10-3 3.13 x10r = 7825 25 in 2 X Tmax= V max Tmax Tmax. do 2 J Tmax- 5095.64 ebrin 7825 x 0.5698 0.875

Answer 2