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
