Question

30 mm40 mm 30 mm T 800,000 N-mm 40 mm t 1 mm 60 mm G -78 Gpa, 0.3, E 203 GPa Problem 2 If a slit is cut lengthwise at one of the vertices making the cross section open, determine the maximum shear stress and the angle of twist. (a) Assume ends are unconstrained from warping (b Asse ends are constrained from warping with Cw 2 x 108.

Answer 1

T = 800,000 N - mm, L = 1200 mm = 800 N - m = 1.2 m G = 78 GPa = 78 times 10 Pa E = 203 GPa V = 0.3 sigma M_x = 0 Integral^L_0 t dx - T_A = 0 T_A = Integral^L_0 t dx = Integral^L_0 t_0 X/L dx = t_B L/2 = 1/2 (80) (1.2) = 480 N - m The max - stress in the shaft is T_max = 16 T_A/V = 16 (480)/0.3 = 7680/0.3 = 250 Mpa for the angle 'theta' of twist theta = Integral^L_0 T/GJ dx = t_B/2LGJ_0 Integral^L_0 (L^2 - x^2) dx = t_B L'/3 GJ = 800 (1.2)^2/3 times (78 times 10^9) times 203 = 1152/47502 times 10^9 = 0.00218 radius = 0.124 degree