Question

1. (20 pts.) For the figure shown: a. What is the largest torque that can be applied to the shaft if the shearing stress is not to exceed 120 MPa b. What is the corresponding minimum value of the shearing stress in the shaft? c. What torque should be applied to the end of the shaft to produce a twist of 2 degrees? Use value G-71-GPa for the modulus of rigidity of steel 60 mm 40 mm 1.5 m

Answer 1

T OI 60mm 1.5m H Llomm 1 Section properties Y nou) ( 60% + (06-07) TT 32 1.021x106 mm y- 32 6 mm У J: 11021X106 polar moment of inertia shear stress T ITE Formula. used, 7% = 7 -

Tmax Vmax=80mm rmin=20mm Emin I may tmax I. rmay ។ T x 30 1.021x106 Tmux - 2,938 x105xT . < ?permisside 2,958 x165x T < 120 N mm2 Ampa = 1/mm 2) (pa= "/me T < 4.084 xlo' N-mm T < 4.084 K Mam max. Torque a ppies, Emin - T rmin 4.084 x106 x 20 1:02.1 X166 80,01 Nome Emin = 80.01 Mpa

GO T ។ е TL 4 ។ 2 180° = (radians) 1 I radians 180 2xTI x 71x103 x 1.021x106 T = 180) 71apa 71x10² mpa G=71x1034 (1,5x103) mm2 T = 1.687 X10 N-mm 1.687 Karm Torque applica 20 twist to produce