Maximum elastic torque:
A rectangular section torsion member had dimension of 100mm by 200 mm and is built of a steel for which the shear yield point is τ_{y} = 100 MPa. Find out T_{p} for the cross-section and the ratio of T_{p} to T_{y}, where T_{y} is the maximum elastic torque.
Solution
Here, 2a = 100 mm and 2b = 200 mm
τ_{Y} = 100 MPa, = 100 N/mm^{2}
For
b / a = 100 /50 = 2, T _{Y} = 3.936 τ a^{3}
Now, T_{Y} = 3.936 × 100 × (50)^{3} = 49200 Nm
Also,
Tp = (20/3) τ_{ Y} a^{3 } = (20 /3)× 100 × (50)^{3 } = 83333.33 Nm
Therefore, we have,
T_{p} / T_{Y} = 83333.33/49200 = 1.6938