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mmme2053 thermal stress and strain add notes, lecture notes, lecture slides, worked examples

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Akbar Rahman 2022-11-10 15:54:53 +00:00
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---
author: Akbar Rahman
date: \today
title: MMME2053 // Thermal Stress and Strain
tags: [ uni, mmme2053, thermal, stress, strain ]
uuid: 0e38e4c3-c8b6-4890-86d8-c4411d4da97b
---
# Introduction
- changes of a temperature in a body cause expansion and contraction
- thermal stresses 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