A key of 14 mm width, 9 mm height and 100 mm length is mounted on a shaft of 50 mm diameter. If the allowable shear stress for the key material is 50 MPa, what is the maximum torque that can be transmitted?

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.