Solution to Problem 270 Thermal Stress

 

 

Before temperature change:
$P_{br} = \sigma_{br} \, A_{br}$

$P_{br} = 2000(1.5)$

$P_{br} = 3000 \, \text{lb}$ compression
 

 

$\Sigma F_H = 0$

$P_{st} = P_{br} = 3000 \, \text{lb}$   tension

$\sigma_{st} = P_{st}/A_{st} = 3000/0.75$

$\sigma_{st} = 4000 \, \text{psi}$   tensile stress
 

$\delta = \dfrac{\sigma L}{E}$

$a = \delta_{br} = \dfrac{2000L}{12 \times 10^6} = 1.67 \times 10^{-4} L$   shortening

$b = \delta_{st} = \dfrac{4000L}{29 \times 10^6} = 1.38 \times 10^{-4} L$   lengthening
 

With temperature rise of 100°F: (Assuming complete freedom)
$\delta_T = \alpha \, L \, \Delta T$

$\delta_{Tbr} = (10.5 \times 10^{-6})L \, (100)$

$\delta_{Tbr} = 1.05 \times 10^{-3}L > a$
 

$\delta_{Tst} = (6.5 \times 10^{-6})L \, (100)$

$\delta_{Tst} = 6.5 \times 10^{-4}L$
 

$\delta_{Tbr} - a = 1.05 \times 10^{-3}L - 1.67 \times 10^{-4}L$

$\delta_{Tbr} - a = 8.83 \times 10^{-4}L$
 

$\delta_{Tst} + b = 6.5 \times 10^{-4}L + 1.38 \times 10^{-4}L$

$\delta_{Tst} + b = 7.88 \times 10^{-4}L$
 

$\delta_{Tbr} - a > \delta_{Tst} + b$   (see figure below)
 

 

$\delta_{Tbr} - a - d = b + \delta_{Tst} + c$

$(1.05 \times 10^{-3}L) - (1.67 \times 10^{-4}L) - \left( \dfrac{\sigma L}{E} \right)_{br} = (1.38 \times 10^{-4}L) + (6.5 \times 10^{-4}L) + \left( \dfrac{PL}{AE} \right)_{st}$

$(8.83 \times 10^{-4}L) - \dfrac{\sigma_{br} L}{12 \times 10^6} = (7.88 \times 10^{-4}L) + \dfrac{P_{st} L}{0.75(29 \times 10^6)}$

$9.5 \times 10^{-4} - \dfrac{P_{br}}{1.5(12 \times 10^6)} = \dfrac{P_{st}}{0.75(29 \times 10^6)}$

$P_{st} = 20\,662.5 - 1.2083P_{br}$     → Equation (1)
 

$\Sigma F_H = 0$

$P_{br} = P_{st}$     → Equation (2)
 

Equations (1) and (2)
$P_{st} = 20\,662.5 - 1.2083P_{st}$

$P_{st} = 9356.74 \, \text{lb}$

$P_{br} = 9356.74 \, \text{lb}$
 

$\sigma = P/A$

$\sigma_{br} = \dfrac{9356.74}{1.5} = 6237.83 \, \text{psi}$   compressive stress     answer

$\sigma_{st} = \dfrac{9356.74}{0.75} = 12 475.66 \, \text{psi}$   tensile stress     answer