A torsional bar is subjected to two cxtcrnal torques (10Nm and 5Nm) and a distributed torque (15NM/m) at the locations indicated in the picture. The angle of twist at A (z = 0) is 0. The shcar modulus is 2GP.. Compute the following quantities: (a) the reaction moment at A, (b) the angle of twist at B, and (c) the angle of twist at C. OD=5cm OD=4cm 15Nm/m X 10Nm 5 Nm B A y 15 cm A 10cm 8cm 8 cm 15 cm В

How do we find the moment and torque equation?

For the circular shaft (G=12×106 psi) shown below: a. Compute the reaction torque and draw the torque (T) diagram. b. Find the max shear stress (Tmax) along the shaft. c. Compute the angle of twist AE at E. 1000 lb in 500 lb in 800 lb. in A B 2 in 2 in + 3 in D 5 in to E AC CD 2 DE

2. The tapered cylindrical bar in Fig. 2 is subjected to opposing torques T on its free ends. For the tapered section only, the angle of twist 0 can be expressed as: 32 TI(d? + d,d2 + dž) Gd d? 37n (a) Derive an expression for the effective spring constant keff for the entire bar, treating each segment as a torsional spring in series. Fig. 2 (b) For the bar shown, let; I, = 12 = 12 in., 1= 24 in., d1= 0.25 in., d2= 1 in., T = 1000 lbf-in., and G = 13.5 x 103 kpsi. Calculate the angle twist 0 for the entire bar using the expression for kef determined in (a). Verify that this value is the sum the angles of twist for each individual %3D %3D segment.