Question

A uniform steel rod of cross-sectional area A is attached to rigid supports and is unstressed at a temperature of 458F. The steel is assumed to be elastoplastic with sY 5 36 ksi and E 5 29 3 106 psi. Knowing that a 5 6.5 3 1026 /8F, determine the stress in the bar (a) when the temperature is raised to 3208F, (b) after the temperature has returned to 458F.

Answer 1

Explanation:

As the given rod is attached to rigid supports as a result, the deformation occurring due to the change in temperature will cause stress in the rod.

Let us assume that P is the compressive force in the rod due to change in temperature.

So,        $\Delta T = \frac{\sigma_{y}}{E_{a}}$

                          = $\frac{36 \times 10^{3}}{(29 \times 10^{6} \times 6.5 \times 10^{-6})}$

                          = $190.98^{o}F$

Now, we will calculate the actual change in temperature as follows.

             $\Delta T = 320 - 45 = 275^{o}F$

This means that the actual change in temperature is more than required for yielding.

(a)   Formula to calculate yielding stress is as follows.

             $\sigma' = \frac{P'}{A}$

           $\sigma' = -\frac{AE \alpha \Delta T}{A}$

                        = $-E \alpha \Delta T$

                        = $-29 \times 10^{6} \times 6.5 \times 10^{-6} \times 275$

                        = $-51.8375 \times 10^{3} psi$

Hence, stress in the bar when temperature is raised to $320^{o}F$ is $-51.8375 \times 10^{3} psi$.

(b)  Now, we will calculate the residual stress as follows.

            $\sigma_{r} = -\sigma_{y} - \sigma'$

           $\sigma_{r} = -36 + 51.837 ksi$

                          = 15.837 ksi

Therefore, stress in the bar when the temperature has returned to $45^{o}F$ is 15.837 ksi.

Answer 2

Answer:

(a). The stress is $-51.83\times10^{3}\ psi$

(b). The residual stress is 15.83 ksi

Explanation:

Given that,

Initial temperature = 45°F

Raised temperature = 320°F

$\sigma_{y}=36\ ksi$

$E=29\times10^{6}\ psi$

$\alpha=6.5\times10^{-6}\ /^{\circ}F$

We need to calculate the actual change in temperature

$\Delta T_{a}=320-45=275^{\circ}F$

We need to calculate the stress

Using formula of stress

$\sigma'=\dfrac{P}{A}$

$\sigma'=-\dfrac{AE\alpha\Delta T}{A}$

$\sigma'=-E\alpha\Delta T$

Put the value into the formula

$\sigma'=-29\times10^{6}\times6.5\times10^{-6}\times275$

$\sigma'=-51.83\times10^{3}\ psi$

(b). We need to calculate the residual stress

Using formula of stress

$\sigma_{r}=-\sigma_{y}-\sigma'$

Put the value into the formula

$\sigma_{r}=-36+51.83\times10^{3}$

$\sigma_{r}=15.83\ ksi$

Hence, (a). The stress is $-51.83\times10^{3}\ psi$

(b). The residual stress is 15.83 ksi