Thermal Effects

Abstract

Thermal effects refers to changes in geometry that may occur due to exposure to changes in temperature.

Appendix: Solutions

• 11.3.1 Determine the change in volume of a cube, of length, h, on a side that is subjected to a uniform temperature change $\Delta T$ if the cube is free to expand in all directions.

Solution

From Eq. (11.2)

$$\varepsilon_{xx} = \varepsilon_{xx}^{\sigma } + \varepsilon_{xx}^{T}$$

Likewise:

$$\varepsilon_{yy} = \varepsilon_{yy}^{\sigma } + \varepsilon_{yy}^{T}$$

$$\varepsilon_{zz} = \varepsilon_{zz}^{\sigma } + \varepsilon_{zz}^{T}$$

With all stresses being zero, the strains are the free thermal strains.

$$\varepsilon_{xx} = \varepsilon_{yy} = \varepsilon_{zz} = \alpha\Delta T$$

The expanded lengths are:

$$\begin{aligned}\Delta V & = \left[ {h\left( {1 + \alpha\Delta T} \right)} \right]^{3} - h^{3} \\\Delta V & = h^{3} \left( {1 + \alpha\Delta T} \right)^{3} - h^{3} \\\Delta V & = h^{3} \left( {1 + 3\alpha\Delta T + 3\alpha^{2}\Delta T^{2} + \alpha^{3}\Delta T^{3} } \right) - h^{3} \\\Delta V & = h^{3} \left( {3\alpha\Delta T + 3\alpha^{2}\Delta T^{2} + \alpha^{3}\Delta T^{3} } \right) \\ \end{aligned}$$

• 11.3.2 Determine the total change in length, $\delta$, for an axial rod of length, L, that is subjected to a temperature change, $\Delta T$, and an axial load, P, if the rod has modulus, E, coefficient of thermal expansion, $\alpha$, Poisson’s ratio, $\nu$, and a cross-sectional area A.

Solution

$$\delta = \frac{PL}{AE} + L\alpha\Delta T$$

• 11.3.3 What is the lateral strain $\varepsilon_{yy}$ for the rod in Exercise 11.3.2?

Solution

$$\varepsilon_{yy} = \frac{1}{E}\left( {\sigma_{yy} - \nu \sigma_{xx} - \nu \sigma_{zz} } \right) + \alpha\Delta T$$

$$\begin{aligned} \varepsilon_{yy} & = \frac{1}{E}\left( {0 - \nu \frac{P}{A} - 0} \right) + \alpha\Delta T \\ \varepsilon_{yy} & = - \frac{\nu P}{AE} + \alpha\Delta T \\ \end{aligned}$$

• 11.3.4 What is the axial stress in a rod fixed at its ends between frictionless walls if it is subjected to a temperature change $\Delta T$?

Solution

$$\begin{aligned} \varepsilon_{xx} & = \frac{1}{E}\left( {\sigma_{xx} - \nu \sigma_{yy} - \nu \sigma_{zz} } \right) + \alpha\Delta T \\ \varepsilon_{xx} & = \frac{1}{E}\left( {\sigma_{xx} - 0 - 0} \right) + \alpha\Delta T = 0 \\ \sigma_{xx} & = - E\alpha\Delta T \\ \end{aligned}$$