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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 τy = 100 MPa. Find out Tp for the cross-section and the ratio of Tp to Ty, where Ty is the maximum elastic torque.
Solution
Here, 2a = 100 mm and 2b = 200 mm
τY = 100 MPa, = 100 N/mm2
For
b / a = 100 /50 = 2, T Y = 3.936 τ a3
Now, TY = 3.936 × 100 × (50)3 = 49200 Nm
Also,
Tp = (20/3) τ Y a3 = (20 /3)× 100 × (50)3 = 83333.33 Nm
Therefore, we have,
Tp / TY = 83333.33/49200 = 1.6938
Can I have a quote for answers relating to this question (6 and 7) on my paper. Thanks
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