A torque of 1200 Nm is applied to the cross section shown be

A torque of 1200 N-m is applied to the cross section shown below. Determine the shear flows q_1, q_2, q_3, q_12, and q_23. The shear modulus for the material is 26 GPa.

Solution

using closed thin wall theory, we can get the values of q1, q2, q3, q12 and q23.

Acording to closed thin wall theory shear flow, q = T / (2A_E)

Where, T = Torque applied

A_E = Area Enclosed defined from the wall midline

now, q1 = 1200000 / (2x60x75) = 133.33 N /mm

Similarly q3 = 1200000 / (2((35x60)+(pi/8 x 60²) = 170 N / mm