Question

A steel tube (G 11.5 106 psi) has an outer diameter d2 2.0 in. and an inner diameter d1 1.5 in. When twisted by a torque T, the tube develops a maximum normal strain of 170 106 . What is the magnitude of the applied torque T?

Answer 1

Explanation:

Given data:

G = 11.5×10⁶psi

d₂ = 2.0 inch

d₁ = 1.5 inch

ε$_{max}$ = 170 × 10⁻⁶

Y$_{max}$ = 2ε

T/J = τ$_{max}$ /R

$\frac{Td_{2} }{2J}$ = τ$_{max}$         (1)

τ$_{max}$ = G Y

from 1 and 2

$\frac{T d_{2} }{2J} = G Y_{max}$

T = $\frac{2 G Y_{max}J }{d_{2} }$

$\frac{2* 11.5*10^{6}*0.006895*10^{6}*340*10*^{-6} *\frac{\pi }{32}[2^{4}-1.5^{4}]*(0.0254)^{4} }{2* 0.0254}$

  = 474.14 Nm