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      <text>13.2.2.6 Engineering example 2</text>
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      <text><![CDATA[<h3>Engineering example 2</h3>
<h4 class="HELM_example">Example</h4>
<p><strong>Torsion of a mild-steel bar</strong></p>

<p>For materials such as mild-steel, the relationship between applied shear stress and shear strain (deformation) can be described as follows.</p>
<ul>
    <li> For small values of the shear strain, the shear stress (\(\tau\)) and shear strain (\(\omega\)) are proportional to one another, i.e. \(\omega = \dfrac{1}{G} \times \tau\ (1)\) (where \(G\) is the <em>shear modulus</em>).<br>This is known as <strong>elastic behaviour</strong>.</li><br>
    <li> There is a maximum shear stress that the material is capable of supporting. If the shear strain is increased further, the shear stress remains roughly constant.<br>This is known as <strong>plastic behaviour</strong>.</li>
</ul>

<p>The figure below summarises the relationship between shear stress and shear strain; the point \((\omega_Y,\tau_Y)\) is known as the <strong>yield point</strong>.</p>
<figure class="HELM_figure">
    <img src="@@PLUGINFILE@@/13.2%20Engineering%20example.%20Shear%20and%20stress..svg" alt="A graph showing how shear stress increases linearly with shear strain up to the yield point, where it becomes constant." width="469" height="264" class="img-responsive atto_image_button_text-bottom">
</figure>
<p>Now suppose that one end of a bar of circular cross section is twisted through an angle \(\theta\), then the shear strain on the surface is given by \(\omega_S = \dfrac{R\,\theta}{L}\ (2)\) (where \(R\) and \(L\) are the radius and length of the bar respectively),
    while the shear strain, at a distance \(r\) from the central core, is given by \(\omega = \dfrac{r\,\theta}{L}\ (3)\).</p>

<p>The torque transmitted by a bar is given by the integral \(T = \displaystyle \int^R_0 2 \pi r^2 \tau(r) \, \mathrm{d}r\ (4)\).</p>

<p>As the shear strain is a function of distance from the central axis of the bar, it may be that the shear strain on the surface is greater than the critical shear strain \(\omega_Y\). In this scenario the shear stress is given by:</p>
\[\tau = \begin{cases}\displaystyle \frac{\tau_Y}{\omega_Y} \omega, &amp; \omega \leq \omega_Y \\ \tau_Y, &amp;\omega &gt; \omega_Y \end{cases} \quad (5) \]
<p> i.e. the regions near the central axis exhibit elasticity, but in those regions near the surface the elastic limit has been exceeded and the metal exhibits plasticity (see the figure below).</p>

<figure class="HELM_figure">
<img src="@@PLUGINFILE@@/13.2%20Engineering%20example.%20Twist%20on%20cross%20section%20of%20bar..svg" alt="The outer section of the bar where the elastic limit is exceeded exhibit plasticity, which corresponds to the yield point." width="284" height="269" class="img-responsive atto_image_button_text-bottom">
</figure>
  
<p><strong>Problem in words</strong> Find an expression for the torque transmitted by a bar as a function of the angle \(\theta\) through which one end is turned.
</p>
<p><strong>Mathematical statement of problem</strong>
</p>
<p>Using Equations \((3)\) to \((5)\), find a formula for \(T\) in terms of the variable \(\theta\).</p>
<p><strong>Mathematical analysis</strong></p>
<p> Substituting \((3)\) into \((5)\): </p>
\[\begin{aligned} \tau &amp;= \begin{cases}\displaystyle \frac{\tau_Y}{\omega_Y}\, \frac{r\theta}{L}, &amp; \displaystyle\frac{r\theta}{L} \leq \omega_Y \\ \tau_Y, &amp; \displaystyle\frac{r\theta}{L} &gt; \omega_Y \end{cases} \\ &amp;= \begin{cases}\displaystyle
\frac{\tau_Y}{\omega_Y}\, \frac{r\theta}{L}, &amp; \displaystyle r \leq \frac{L\omega_Y}{\theta} = r_e \\ \tau_Y, &amp; r &gt; \displaystyle\frac{L\omega_Y}{\theta} = r_e \end{cases} \end{aligned}\]

<p>For small values of \(\theta\), \(r_e \geq R\) so that the whole of the bar will be in the elastic region, i.e. \(\tau = \dfrac{\tau_Y}{\omega_Y} \, \dfrac{r\theta}{L}\).</p>

<p>Now \((4)\) becomes:</p>
\[\begin{aligned} T &amp;= \int_0^R 2\pi r^2 \frac{\tau_Y}{\omega_Y}\,\frac{r\theta}{L}\,\mathrm{d}r \\ &amp;= 2\pi \frac{\tau_Y}{\omega_Y}\,\frac{\theta}{L} \int_0^R r^3\,\mathrm{d}r \\ &amp;= 2\pi \frac{\tau_Y}{\omega_Y}\,\frac{\theta}{L} \left[\frac{r^4}{4}\right]_0^R
\\ &amp;= \frac{\pi}{2}\,\frac{\tau_Y}{\omega_Y}\,\frac{\theta}{L} R^4, \quad (6) \end{aligned}\]
<p>i.e. the torque is directly proportional to the twist, \(\theta\). For larger \(\theta\), \(r_e &lt; R\), so that \((4)\) becomes:</p>

\[\begin{aligned} T &amp;= \int_0^{r_e} 2\pi r^2 \frac{\tau_Y}{\omega_Y}\,\frac{r\theta}{L} \,\mathrm{d}r + \int_{r_e}^R 2\pi r^2 \tau_Y \,\mathrm{d}r \\ &amp;= 2\pi \frac{\tau_Y}{\omega_Y}\,\frac{\theta}{L} \int_0^{r_e} r^3 \,\mathrm{d}r + 2\pi\tau_Y
\int_{r^e}^R r^2 \,\mathrm{d}r \\ &amp;= 2\pi \frac{\tau_Y}{\omega_Y}\,\frac{\theta}{L} \left[\frac{r^4}{4}\right]_0^{r_e} + 2\pi \tau_Y \left[\frac{r^3}{3}\right]_{r_e}^R \\ &amp;= \frac{\pi}{2}\,\frac{\tau_Y}{\omega_Y}\,\frac{\theta}{L} r_e^4 + \frac{2\pi}{3}
\tau_Y (R^3 - r_e^3) \end{aligned}\]

<p>But \(r_e = L\omega_Y / \theta\), so:</p>

\[\begin{aligned} T &amp;= \frac{\pi}{2}\,\frac{\tau_Y}{\omega_Y}\,\frac{\theta}{L}\,\frac{L^4\omega_Y^4}{\theta^4} + \frac{2\pi}{3}\tau_Y R^3 - \frac{2\pi}{3} \tau_Y \frac{L^3\omega_Y^3}{\theta^3} \\ &amp;= \frac{2\pi}{3}\tau_Y R^3 + \pi \left(\frac{1}{2}\tau_Y
- \frac{2}{3}\tau_Y\right) \frac{L^3\omega_Y^3}{\theta^3} \\ &amp;= \frac{2\pi}{3}\tau_Y R^3 - \frac{\pi}{6} \tau_Y \frac{L^3\omega_Y^3}{\theta^3} \quad (7) \end{aligned}\]

<p>Equation \((6)\) will apply when \(r_e \geq R\), i.e. \(L\omega_Y/\theta \geq R\) or \(\theta \leq L\omega_Y/R\), so that combining \((6)\) and \((7)\) gives overall:</p>

\[ T = \begin{cases}\displaystyle \frac{\pi}{2}\,\frac{\tau_Y}{\omega_Y}\,\frac{\theta}{L} R^4, &amp; \displaystyle \theta \leq \frac{L \omega_Y}{R} \\ \displaystyle \frac{2\pi}{3} \tau_Y R^3 - \frac{\pi}{6}\tau_Y \frac{L^3\omega_Y^3}{\theta^3}, &amp;
\displaystyle \theta &gt; \frac{L\omega_Y}{R} \end{cases} \quad (8) \]

<p>At the critical value of \(\theta\), i.e. when the outer edge begins to exhibit plasticity, both formulae in \((8)\) give \(T_{\mbox{crit}} = \dfrac{\pi}{2} \tau_Y R^3\). Furthermore, the first derivatives are both \(\dfrac{dT}{d\theta} = \dfrac{\pi}{2}
    \dfrac{\tau_Y}{\omega_Y} \dfrac{R^4}{L}\), i.e. the curves join smoothly. The second derivatives, though, are not equal (zero in one case). In the theoretical limit as \(\theta \rightarrow \infty\), \(T = \dfrac{2\pi}{3} \tau_Y R^3\) so this is the
    total torsional torque which can be carried by the bar. (The critical torque above is three-quarters of this value.) However, clearly \(\theta \rightarrow \infty\) is merely a theoretical limit since the bar would, in fact, shear at a finite value
    of \(\theta\).</p>]]></text>
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