Question

Determine the maximum shearing stress in a solid shaft of 1.5-in. diameter as it transmits 75 hp at a speed of 1800 rpm. (Round the final answer to two decimal places.) The maximum shearing stress in the solid shaft is ksi.

Answer 1

Answer:

$\tau=3.96\ ksi$

Explanation:

Given that

d= 1.5 in                      ( 1 in = 0.0254 m)

d= 0.0381 m

P= 75 hp                      ( 1 hp = 745.7 W)

P= 55927.5 W

N= 1800 rpm

We know that power P is given as

$P=\dfrac{2\pi N\ T}{60}$

T=Torque

N=Speed

$55927.5=\dfrac{2\times \pi \times 1800\ T}{60}$

T=296.85 N.m

The maximum shear stress is given as

$\tau=\dfrac{16 T}{\pi d^3}$

$\tau=\dfrac{16\times 296.85}{\pi \times 0.0381^3}$

$\tau=27.35\ MPa$

We know that 1 MPa =0.145 ksi

$\tau=3.96\ ksi$