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Maximum shear stress developed on the surface of a solid circular shaft under pure torsion is 240 MPa. If the shaft diameter is doubled then the maximum shear stress developed corresponding to the same torque will be
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- 120 MPa
- 60 MPa
- 30 MPa
- 15 MPa
- 120 MPa
Correct Option: C
τ1 = 240 MPa
When shaft diameter is doubled, maximum shear stress = τ2
Let d be the original diameter
| = |  |  | ||||
| IP1 | ||||||
| τ2 | | | ||||
| IP2 | ||||||
... (since torque in both the cases is constant)
| = | × | = | × | |||||||
| τ2 | IP1 | r2 | { 2 × (πd2 / 32) } | { 32 × (2d / 2) } |
| = | = 8 | |
| πd4 . d . 2 |
| ⇒ τ2 = | = | = 30 MPa | ||
| 8 | 8 |
