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Question

A solid steel shaft of diameter 50 mm is to be designed using an allowable shear stress τallow=50MPa and allowable angle of twist per unit length θ=1o per meter. The maximum permissible torque (in Nm) T that may be applied to the shaft is ? (assuming G = 100GPa).

A
1227
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B
1070
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C
2454
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D
2140
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Solution

The correct option is B 1070
Torque based on allowable shear stress.

TJ=τr

T=τ(d/2)×π32d4

=τ×π16d3

T=(50×106)×π16×(0.05)3=1227Nm

Torque based on allowable angle of twist.

TJ=Gθl

T=(100×109)×(1×π180)×π32×(0.05)4=1070Nm

So max. permissible torque, T = 1070 Nm

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