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ISRO IPRC Technical Assistant Mechanical held on 21/04/2018

Option 4 : 1750 Nm

ISRO VSSC Technical Assistant Mechanical held on 09/06/2019

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80 Questions
320 Marks
120 Mins

__Concept:__

For a rectangular key shear stress **\(\tau = \frac{F}{A} = \frac{F}{{b\; × \;l}}\)**

where F = force acting on key, b = width, l = length

The torque transmitted by the shaft is given by, \(T = F × \frac{d}{2}\)

where d = diameter of the shaft

__Calculation:__

**Given:**

b = 14 mm, l = 100 mm, d = 50 mm = 0.05 m, allowable stress of key = 50 MPa

For rectangular key shear stress is

\(\tau = \frac{F}{A} = \frac{F}{{b\; × \;l}}\)

\( \frac{F}{{b\; × \;l}}≤ 50~ MPa\)

F ≤ 50 × b × l

F = 50 × 14 × 100 = 70000 N = **70 kN**

The torque transmitted by the shaft

\(T = F × \frac{d}{2}\)

T = 70000 × 0.025 = **1750 Nm**

**Hence max torque that can be transmitted is 1750 Nm.**