Question

Find the minimum diameter of a solid circular shaft if it has a maximum shear stress of 53 MPa and it is to transmit 28 kW of power at a rotational speed of 2,516 rpm. Enter your numerical value below for minimum diameter, d, in units of millimeters. Enter your answer to one decimal place and the result must be within +/- 1% of the correct value to be counted as correct.

Answer 1

Answer:

d= 21.6 mm

Explanation:

Given that

P= 28 KW

τ = 53 MPa

N= 2516 RPM

We know that

$P =\dfrac{2\pi NT}{60}$

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

By  putting the values

$T =\dfrac{60\times 28\times 1000}{2\pi \times 2516}$

T=106.27 N.m

We know that for solid shaft

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

$53\times 10^6 =\dfrac{16\times 106.27}{\pi d^3}$

$d^3=\dfrac{16\times 106.27}{\pi \times 53\times 10^6 }$

$d=\left(\dfrac{16\times 106.27}{\pi \times 53\times 10^6 }\right)^{\frac{1}{3}}$

d= 0.0216 m

d= 21.6 mm