Chapter 13, Problem 17P

An aluminum bar has the desired length when at 12°C. How much stress is required to keep it at this length if the temperature increases to 35°C? [See Table 9-1]

To determine

The stress required to keep the aluminum bar’s length steady even when the temperature has increased

Answer to Problem 17P

Solution:

The stress required to maintain the length of the aluminum bar is $3.97\times {10}^7\frac{N}{m^2}$

Explanation of Solution

Given:

Initial temperature, $Ti=12°C$

Final temperature, $Tf=35°C$

Formula used:

$\frac{F}{A}=\alpha E\Delta T$, where Δ is the coefficient of linear expansion of the material and E is the Young’s modulus of the material

Calculation:

The relation between thermal stress, the force per unit area required for maintaining the length of an object, and change in temperature (ΔT)is given by,

$\frac{F}{A}=\alpha E\Delta T$, where Δ is the coefficient of linear expansion of the material and E is the Young’s modulus of the material

In this case, $\Delta T=T_f–T_i=35°C-12°C=23°C$

For aluminum, $\alpha =25\times {10}^{-6}\frac{1}{°C},E=69\times {10}^9\frac{N}{m^2}$

Therefore, $\frac{F}{A}=(25\times {10}^{-6}\frac{1}{°C})\left(69\times {10}^9\frac{N}{m^2}\right)\left(23°C\right)$

or $=3.97\times {10}^7\frac{N}{m^2}$