Question

A 15-mm-diameter steel bar is to be used as a torsion spring. If the torsional stress in the bar is not to exceed 110 MPa when one end is twisted through an angle of 30°, what must be the length of the bar?

Answer 1

Answer:

The length of bar will be 2.82 m

Explanation:

Given that

d= 15 mm

r= 7.5 mm

Shear stress = 110 MPa

θ =  30°                                  (30°   = 30°  x π/180°  =0.523 rad)

θ = 0.523 rad

G for steel

G= 79.3 GPa

We know that

$\dfrac{\tau}{r}=\dfrac{G\theta }{L}$

$\dfrac{110}{7.5\times 10^{-3}}=\dfrac{79.3\times 10^3\times 0.523 }{L}$

L= 2. 82 m

The length of bar will be 2.82 m