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

Step-by-step 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

`(Td_(2) )/(2J)` = τ
`_(max)` (1)

τ
`_(max)` = G Y

from 1 and 2

`(T d_(2) )/(2J) = G Y_(max)`

T =
`(2 G Y_(max)J )/(d_(2) )`

`(2* 11.5*10^(6)*0.006895*10^(6)*340*10*^(-6) *(\pi )/(32)[2^(4)-1.5^(4)]*(0.0254)^(4) )/(2* 0.0254)`

= 474.14 Nm