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and strains are important in many situations like engines and power plants +- quantified by coefficient of thermal expansion, $\alpha$ (units are K$^{-1}$) + + $$\delta l_\text{thermal} = l\alpha\Delta T$$ + + $$\epsilon\text{thermal} = \frac{\delta l_\text{thermal}}{l} = \alpha\Delta T$$ + +- for isotropic values $\alpha$ is the same in all directions +- using principle of superposition (total effects of combined body = sum of effects of individual + loads) we can state for a uniaxial bar: + + $$\delta l_\text{total} = \delta l_\text{elastic} + \delta l_\text{thermal} = \frac{FL}{AE} + l\alpha\Delta T$$ + +## Typical Values of $\alpha$: + +Material | $\alpha \times10^{-6}$ +--------- | -------- +Concrete | 10 +Steel | 11 +Aluminium | 23 +Nylon | 144 +Rubber | 162 + +# Resistive Heating of a Bar + +# Single Bar Assembly + +The bar shown in figure \ref{img:tss1-bar} is subjected to a temperature rise of $\Delta T$ and +restricted from expanding by constraints at each end: + +![\label{img:tss1-bar}](./images/tss1-004.png) + +Since the bar cannot extend we know that + +$$\delta l_\text{total} = \delta l_\text{elastic} + \delta l_\text{thermal} = \frac{FL}{AE} + l\alpha \Delta T = 0$$ + +Cancelling through $l$ and rearranging for $F$ gives + +$$F = -AE\alpha\Delta T$$ + +So we can determine the stress, $\sigma$, by + +$$\sigma = \frac FA = -E\alpha\Delta T$$ + +## Compound Bar Assembly +![\label{img:tss1-compound}](./images/tss1-003.jpg) +![\label{img:tss1-fbd}](./images/tss1-004.jpg) + +- aluminium bar will 'want to' expand more than the steel +- because of the rigid end blocks, it is therefore in compression +- steel bar 'wants to' expand less than the aluminium bar +- but the rigid block, which is attached to aluminium bar forces it to expand more + so the steel is in tension + +- considering the bar analytically, the change in length are given by + + $$\frac{F_sl}{A_sE_s} + l\alpha_s\Delta T = \frac{F_al}{A_aE_a} + l\alpha_a\Delta T$$ + +- considering equilibrium using the free body diagram (figure \ref{img:tss1-fbd}): + + $$F_s = -F_a$$ + +- substituting in for $F_s$ and rearranging gives + + $$\sigma_a = \frac{F_a}{A_a} = \frac{\Delta T(\alpha_s-\alpha_a)}{\frac{1}{E_a} + \frac{A_a}{A_sE_s}}$$ + +- $\alpha_s < \alpha_a \rightarrow \sigma_a < 0$ i.e. the bar is in compression +- from force equilibrium we know $A_a\sigma_a = -A_s\sigma_s$ and can use that to find steel is in + tension + +# Generalised Hooke's Law Including Thermal Strains + +$$\epsilon_\text{total} = \epsilon_\text{mechanical} + \epsilon_\text{thermal}$$ + +\begin{align*} +\epsilon_{\text{total},x} &= \frac{1}{E} (\sigma_x - \nu(\sigma_y+\sigma_z)) + \alpha\Delta T \\ +\epsilon_{\text{total},y} &= \frac{1}{E} (\sigma_y - \nu(\sigma_z+\sigma_z)) + \alpha\Delta T \\ +\epsilon_{\text{total},z} &= \frac{1}{E} (\sigma_z - \nu(\sigma_x+\sigma_y)) + \alpha\Delta T +\end{align*} + +# An Initially Straight Uniform Beam + +![\label{img:tss1-beam}](./images/tss3-beam.svg) + +- assume $\Delta T = \Delta T(y)$ (it is purely a function of $y$) +- $\alpha$, axial force $P$, and pure bending about $z-z$ axis $M$ are also applied +- $\sigma_y = \sigma_z = \tau_{xz} = \tau_{yz} = 0$ because the cross sectional dimensions are small + compared with length +- $\tau_{xy} = 0$ because $M$ does not vary with $x$ --- $S = \frac{\mathrm{d}M}{\mathrm{d}x} = 0$ + +## Compatibility + +\begin{equation} +\epsilon_x = \bar\epsilon = \frac{y}{R} \label{eqn:tss-compat} +\end{equation} + +where $\bar\epsilon$ is mean strain (at $y = 0$) and $R$ is radius of curvature + +## Stress-Strain + +From generalised Hooke's Law equation: + +\begin{equation} +\epsilon_x = \frac{\sigma_x}{E} + \alpha\Delta T \label{eqn:tss-ss} +\end{equation} + +($\sigma_y = \sigma_z = 0$) + +Substitute (\ref{eqn:tss-compat}) into (\ref{eqn:tss-ss}) to get: + +\begin{equation} +\sigma_x = E\left(\bar\epsilon + \frac{y}{R} - \alpha\Delta T\right) \label{eqn:tss-ss2} +\end{equation} + +## Axial Force Equilibrium + +\begin{equation} +P = \int_A \sigma_x \mathrm{d} A \label{eqn:tss-afe} +\end{equation} + +sub (\ref{eqn:tss-ss2}) into (\ref{eqn:tss-afe}) to get: + +$$P = E\int_a \bar\epsilon + \frac{y}{R} - \alpha\Delta T \mathrm{d}A$$ + +knowing that $\int_A y\mathrm{d}A = 0$ (as axis passes through centroid) rearrange to get + +\begin{equation} +P = E\bar\epsilon A - E\alpha \int_A \Delta T\mathrm{d} A \label{eqn:tss-afe2} +\end{equation} + +## Moment Equilibrium + +\begin{equation} +M = \int_Ay\sigma_x\mathrm{d}A \label{eqn:tss-me} +\end{equation} + +sub (\ref{eqn:tss-ss2}) into (\ref{eqn:tss-me}) to get + +$$M = E\int_A \left(\bar\epsilon + \frac{y}{R} - \alpha\Delta T\right)y \mathrm{d}A$$ + +which, knowing $\int_Ay^2\mathrm{d}A = I$ and $\int_A y\mathrm{d}A = 0$ still, rearranges to: + +\begin{equation} +M = \frac{EI}{R} - E\alpha\int_A\Delta Ty\mathrm{d}a \label{eqn:tss-me2} +\end{equation} + +# Thin Walled Cylinders + +- often temperature variations can be approximated as linear through their thickness: + + $$\Delta T(y) = \Delta T_\text{wall}\frac{y}{t}$$ + + and for a thin cylinder + + $$\sigma_r \approx 0$$ + +- it is convenient to consider effect of temperature change and temperature gradient separately +- if the cylinder is not restrained, the uniform temperature change causes dimensional changes + but no stress +- stresses due to axial restraint are easily calculated + +using a cylindrical coordinate system: + +$$\epsilon_\theta = \frac1E(\sigma_\theta-\nu\sigma_z) + \alpha\Delta T$$ +$$\epsilon_z = \frac1E(\sigma_z-\nu\sigma_\theta) + \alpha\Delta T$$ + +as $\sigma_r \approx 0$ + +- away from the end of the cylinder, sections remain plane and circular +- from compatibility considerations (with no mean temperature change), the hoop and axial strains + must both be zero, giving: + + $$\epsilon_\theta = 0 = \frac1E(\sigma_\theta-\nu\sigma_z) + \alpha\Delta T_\text{wall}\frac{y}{t}$$ + $$\epsilon_z = 0 = \frac1E(\sigma_z-\nu\sigma_\theta) + \alpha\Delta T_\text{wall}\frac{y}{t}$$ + +- solving gives + + $$\sigma_\theta = \sigma_z = \frac{-E\alpha\Delta T_\text{wall}}{1-\nu}\frac{y}{t}$$ + + so at $y=\frac{t}{2}$ you get + + $$\sigma = \frac{-E\alpha\Delta T_\text{wall}}{2(1-\nu)}$$ + + showing that it is under compression, and at $y=-\frac{t}{2}$ it will be under tension +