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A solid circular shaft needs to be designed to transmit a torque of 50 N.m. If the allowable shear stress of the material is 140 MPa, assuming a factor of safety of 2, minimum allowable design diameter in mm is
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- 8
- 16
- 24
- 34
Correct Option: B
Here, T = 50N – m
τallowable = 140 MPa
FOS = 2;
| τsafe = | = 70 MPa | FOS |
| We know, | τsafe | Zp |
| Zp = | 16 |
| ⇒ d3 = | π τsafe |
= 15.38 mm ≈ 16 mm
