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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.

Respuesta :

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,        [tex]\Delta T = \frac{\sigma_{y}}{E_{a}}[/tex]

                          = [tex]\frac{36 \times 10^{3}}{(29 \times 10^{6} \times 6.5 \times 10^{-6})}[/tex]

                          = [tex]190.98^{o}F[/tex]

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

             [tex]\Delta T = 320 - 45 = 275^{o}F[/tex]

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

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

             [tex]\sigma' = \frac{P'}{A}[/tex]

           [tex]\sigma' = -\frac{AE \alpha \Delta T}{A}[/tex]

                        = [tex]-E \alpha \Delta T[/tex]

                        = [tex]-29 \times 10^{6} \times 6.5 \times 10^{-6} \times 275[/tex]

                        = [tex]-51.8375 \times 10^{3} psi[/tex]

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

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

            [tex]\sigma_{r} = -\sigma_{y} - \sigma'[/tex]

           [tex]\sigma_{r} = -36 + 51.837 ksi[/tex]

                          = 15.837 ksi

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

Answer:

(a). The stress is [tex]-51.83\times10^{3}\ psi[/tex]

(b). The residual stress is 15.83 ksi

Explanation:

Given that,

Initial temperature = 45°F

Raised temperature = 320°F

[tex]\sigma_{y}=36\ ksi[/tex]

[tex]E=29\times10^{6}\ psi[/tex]

[tex]\alpha=6.5\times10^{-6}\ /^{\circ}F[/tex]

We need to calculate the actual change in temperature

[tex]\Delta T_{a}=320-45=275^{\circ}F[/tex]

We need to calculate the stress

Using formula of stress

[tex]\sigma'=\dfrac{P}{A}[/tex]

[tex]\sigma'=-\dfrac{AE\alpha\Delta T}{A}[/tex]

[tex]\sigma'=-E\alpha\Delta T[/tex]

Put the value into the formula

[tex]\sigma'=-29\times10^{6}\times6.5\times10^{-6}\times275[/tex]

[tex]\sigma'=-51.83\times10^{3}\ psi[/tex]

(b). We need to calculate the residual stress

Using formula of stress

[tex]\sigma_{r}=-\sigma_{y}-\sigma'[/tex]

Put the value into the formula

[tex]\sigma_{r}=-36+51.83\times10^{3}[/tex]

[tex]\sigma_{r}=15.83\ ksi[/tex]

Hence, (a). The stress is [tex]-51.83\times10^{3}\ psi[/tex]

(b). The residual stress is 15.83 ksi

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