(a)

Without temperature change:

$\sigma = \dfrac{P}{A} = \dfrac{1200}{0.25} = 4800 \, \text{ psi }$
$\sigma = 4.8 \, \text{ksi} \lt 10 \, \text{ksi}$

A drop of temperature is needed to increase the stress to 10 ksi. See figure below.

$\delta = \delta_T + \delta_{st}$

$\dfrac{\sigma \, L}{E} = \alpha \, L \, (\Delta T) + \dfrac{PL}{AE}$

$\sigma = \alpha \, E \, (\Delta T) + \dfrac{P}{A}$

$10\,000 = (6.5 \times 10^{-6})(29 \times 10^6)(\Delta T) + \dfrac{1200}{0.25}$

$\Delta T = 27.59^\circ \text{F}$

Required temperature: *(temperature must drop from 40°F)*

$T = 40 - 27.59 = 12.41^\circ \text{F}$ *answer*

(b) Temperature at which the stress will be zero:

From the figure below:

$\delta = \delta_T$

$\dfrac{PL}{AE} = \alpha \, L \, (\Delta T)$

$P = \alpha AE(T_f \, - \, T_i)$

$1200 = (6.5 \times 10^{-6})(0.25) (29 \times 10^6)(T_f - 40)$

$T_f = 65.46^\circ\text{F}$ *answer*

## Comments

## shouldn't you have to

shouldn't you have to increase the temperature to increase the tensile stress and decrease the temperature to decrease the tensile stress?

## No. Temperature change itself

No. Temperature change itself won't induce stress. Stress will occur only if you restrain the expansion or contraction due to temperature change.