Question

1. A shaft 80 mm diameter transmits power at maximum shear stress of 63 MPa Find the length of a 20
mm wide key required to mount a pulley on the shaft so that the stress in the key does not exceed 42
MPa.
[Ans. 152 mm)​

Answer 1

Answer:

The length= 188.49 mm

Explanation:

For the shaft,

d=80mm

τshaft=63MPa

Torque transmitted by the shaft= $\frac{\pi }{16}$×80³×63---1

                                                    = 6333450.79

Considering shearing of key, the tangential shearing force acting at the circumference of shaft

F= Area resisting shearing × shear stress

 = l×w×τ

Torque transmitted by shaft = F ×$\frac{d}{2}$ = l×w×τ×$\frac{d}{2}$ ---2

From 1 and 2

6333450.79= l×20×42×$\frac{80}{2}$

→l=188.495 mm

The length of the key = 188.495 mm.

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