A key of 14 mm width, 9 mm height and 100 mm length is mounted on
| A key of 14 mm width, 9 mm height and 100 mm length is mounted on a shaft of 50 mm diameter. If allowable shear stress for the key material is 50 MPa, what is the maximum torque that can be transmitted?
A. 3500 Nm
B. 4500 Nm
C. 2250 Nm
D. 1750 Nm
Please scroll down to see the correct answer and solution guide.
Right Answer is: D
SOLUTION
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.
