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Alex73 [517]
3 years ago
13

A uniform steel rod of cross-sectional area A is attached to rigid supports and is unstressed at a temperature of 458F. The stee

l 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.
Physics
2 answers:
Vedmedyk [2.9K]3 years ago
6 0

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.

artcher [175]3 years ago
4 0

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

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