Question

A steel shaft transmits 20 kW of power at 10 Hz. Determine the smallest safe diameter of the shaft if the shear stress τ_max is not to exceed 50 MPa and the angle of twist θ is limited to 5°in a length of 4 m. Use G = 70 GPa.

Answer 1

● Given data :-

T=20 KW =20000 N-m

_max = 50 MPa=50×10^6 N/m²

=5° =0.08726 radians

L=4 m

G=70 GPa=70 ×10^9 N/m²

Solution:-

◆ So now for

(T/J)=(G/L)

20000/J=((70×10^9×0.08726)/4)

J=1.3097×10^(-5)

πd⁴/32=1.30×10^-5

d=0.1074 m =107.4 mm.

◆ Now for

(50 ×10^6)/r =(70×10^9)×(0.08726)/4

r=0.03274 m

So the diameter is d=2×0.03274 =0.06548 mm.

● So we From above calculation minimum diameter for the shaft is 65.48 mm.

Thank you.