Chapter 13, Problem 23P

To determine

To Calculate: The stress of the given aluminum bar.

Answer to Problem 23P

The stress that is required to keep the bar at this length is $\text{3}{\text{.5×10}}^{\text{7}}\raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.$ , if temperature is increased.

Explanation of Solution

Given info:

The initial temperature is, ${\text{T}}_{\text{1}}\text{=}{\text{15}}^{\text{°}}\text{C}$

The final temperature is, ${\text{T}}_{\text{2}}\text{=}{\text{35}}^{\text{°}}\text{C}$

The young modulus of the aluminum is, $\text{E}\text{=}{\text{70×10}}^{\text{9}}\raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.$

Formula used:

The thermal stress is given as,

  $\text{Stress =}\frac{\text{Force}}{\text{area}}\text{=}\text{αEΔT}$

Where, $\text{ΔT}$ is the temperaturechange, E is the young modulus of the aluminum and $\text{α}$ is the coefficient.

Calculation:

The thermal stress must be regained for the thermal expansion. So, the stress is given as,

  $\begin{array}{l}\text{Stress =}\text{αEΔT=}\left({\text{25×10}}^{\text{-6}}\raisebox{1ex}{$$}\!\left/ \!\raisebox{-1ex}{${}^{\text{°}}\text{C}$}\right.\right){\text{×70×10}}^{\text{9}}\raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.\text{×}\left({\text{35}}^{\text{°}}\text{C-}{\text{15}}^{\text{°}}\text{C}\right)\\ \text{=}\left({\text{25×10}}^{\text{-6}}\raisebox{1ex}{$$}\!\left/ \!\raisebox{-1ex}{${}^{\text{°}}\text{C}$}\right.\right){\text{×70×10}}^{\text{9}}\raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.\text{×}\left({\text{20}}^{\text{°}}\text{C}\right)\\ \text{=}\text{3}{\text{.5×10}}^{\text{7}}\raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.\end{array}$

Conclusion:

The stress is $\text{3}{\text{.5×10}}^{\text{7}}\raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.$ .