A t2 1 B Li L2 t = 5 mm t2 4 mm t3 = 3 mm Li = 100 mm L2 150 mm h = 120 mm G = 30 GPa A 5-m long tube with double-cell cross section shown above is subjected to torque 7=200 Nm. The tube is fixed at one end and free at the other end. Determine the shear stress at Points A, B, C above and the angle of twist at the free end. Example Double Cell 91 92 h - There are now two closed-loop cells, so there are two shear flows q and q2. But, we have only on equilibrium equation q = This is a statically indeterminate structure and, hence, compatibility equation is needed. T 2Am B t3 A t Li L2 a) Torque in cell 1 Torque in cell 2 T = 2 Am1 91 = 2 Lh q T2 = 2 Am2 92 = 2 Lh 92 Compatibility of rate of twist: 01-02 Equilibrium: T = T+ T (f)-(f), 1 Lm q 1 ds = + + 91419h 9141 (91-92)h] + ...(1) = 2GAm1 t G 2Lh t3 t t1 t 0 1 1 Lm q ds = + 1 9242 (92-91)h 92L2 92h + + 2GAM20 t G 2Lh t3 t2 ti t