Question 19.4: The temperature of the inside wall of a tube is 200 °C and t......
The temperature of the inside wall of a tube is 200 °C and the outside wall temperature is 40 °C. Calculate the stresses at the outside of the wall if the tube is made from a glass having a coefficient of thermal expansion of \alpha=8× 10^{-6} /°C, an elastic modulus of 10×10^{6} psi, and a Poisson’s ratio of 0.3.
Let x, y and z be the axial, hoop, and radial directions. The stress normal to the tube wall \sigma_{z} = 0 and symmetry requires that \sigma_{y}=\sigma_{x}. Let the reference position be the midwall where T = 120 °C. ΔT at the outside is 40 °C − 120 °C = −80 °C. The strains \varepsilon _{x} and \varepsilon _{y} must be zero relative to the midwall.
Substituting in Equation (19.9),
\varepsilon _{x}=\alpha \Delta T+(1/E)\left[\sigma_{x }-\upsilon (\sigma_{y}+\sigma_{z})\right]. (19.9)
0=\alpha \Delta T+(1/E)\left[\sigma_{x }-\upsilon (\sigma_{y}+\sigma_{z})\right], \alpha \Delta T +(1-\upsilon )\sigma_{x }/E=0, so \sigma_{x }=\alpha E \ \Delta T /(1-\upsilon ). Solving,
\sigma_{x } = (8 × 10^{-6} /°C)(80 \ °C)(10 × 10^{6} \ psi)/0.7 = 9140 \ psi.