Question

A. What is the minimum diameter of a solid steel shaft that will not twist through more than 3° in a 6-m length when subjected to a torque of 12 kN·m?  Use G = 83 GPa.

B. What maximum shearing stress is developed?
C. Determine the bearing stress of the punch out plate of the precious problem, Answer in MPa

Answer 1

Ans

Given data:

Twist, $\theta = 3 ^o$

Length of the shaft, L = 6 m

Torque, T = 12 kN-m

G = 83 GPa

(A)

From torsion formula, we have

$\frac{T}{J} = \frac{G\theta}{L}$

$\Rightarrow \frac{12 * 10^3}{\frac{\pi}{32}d^4} = \frac{83*10^9*3*\frac{\pi}{180}}{6}$

= 1.68755 * 107

$\Rightarrow d = 0.11398 m$

  (ans)

(B)

For maximum shearing stress, we have

$\tau_{max} = \frac{16T}{\pi d^3}$

$= \frac{16*12*10^3}{\pi *0.11398^3}$

(ans)