Answer:
d= 21.6 mm
Explanation:
Given that
P= 28 KW
τ = 53 MPa
N= 2516 RPM
We know that
[tex]P =\dfrac{2\pi NT}{60}[/tex]
[tex]T =\dfrac{60\times P}{2\pi N}[/tex]
By putting the values
[tex]T =\dfrac{60\times 28\times 1000}{2\pi \times 2516}[/tex]
T=106.27 N.m
We know that for solid shaft
[tex]\tau =\dfrac{16T}{\pi d^3}[/tex]
[tex]53\times 10^6 =\dfrac{16\times 106.27}{\pi d^3}[/tex]
[tex]d^3=\dfrac{16\times 106.27}{\pi \times 53\times 10^6 }[/tex]
[tex]d=\left(\dfrac{16\times 106.27}{\pi \times 53\times 10^6 }\right)^{\frac{1}{3}}[/tex]
d= 0.0216 m
d= 21.6 mm
