Strength Of Materials Miscellaneous
- 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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τ1 = 240 MPa
When shaft diameter is doubled, maximum shear stress = τ2
Let d be the original diameterτ1 = 
Tr1 
IP1 τ2 Tr2 IP2
... (since torque in both the cases is constant)τ1 = r1 × IP2 = d × π(2d)4 τ2 IP1 r2 { 2 × (πd2 / 32) } { 32 × (2d / 2) } = πd × 16d4 = 8 πd4 . d . 2 ⇒ τ2 = τ1 = 240 = 30 MPa 8 8 Correct Option: C
τ1 = 240 MPa
When shaft diameter is doubled, maximum shear stress = τ2
Let d be the original diameterτ1 = Tr1 IP1 τ2 Tr2 IP2
... (since torque in both the cases is constant)τ1 = r1 × IP2 = d × π(2d)4 τ2 IP1 r2 { 2 × (πd2 / 32) } { 32 × (2d / 2) } = πd × 16d4 = 8 πd4 . d . 2 ⇒ τ2 = τ1 = 240 = 30 MPa 8 8
