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    <name>
      <text>12.3.3.1 Engineering Example</text>
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      <text><![CDATA[<h4 class="HELM_example">Example</h4>
<h4>Buckling of a strut</h4>
<p>The equation governing the buckling load \(P\) of a strut with one end fixed and the other end simply supported is
    given by</p>
\[\tan \mu L =\mu L,\] 
<p>where \(\mu =\sqrt{\dfrac{P}{EI}}\), \(L\) is the length of the strut and \(EI\)
    is the flexural rigidity of the strut. For safe design it is important that the load applied to the strut is less
    than the lowest buckling load. This equation has no exact solution and we must therefore use the method described in
    this Workbook to find the lowest buckling load \(P\).</p>
<figure class="HELM_figure"><img src="@@PLUGINFILE@@/12.3%20Buckling%20of%20a%20strut.svg" alt="A strut of length L whose shape is deflected by a load P." width="508" height="92" class="img-responsive atto_image_button_text-bottom"></figure>
<p>We let \(\mu L=x\) and so we need to solve the equation \(\tan x=x\). Before starting to apply the Newton&ndash;Raphson
    iteration we must first obtain an approximate solution by plotting graphs of \(y=\tan x\) and \(y=x\) using the same
    axes.</p>
<figure class="HELM_figure"><img src="@@PLUGINFILE@@/12.3%20Graph%20of%20tan%20x%20against%20x.svg" alt="A graph showing both tan x and x, which intersect at a point to the left of 3 pi over 2." width="400" height="201" class="img-responsive atto_image_button_text-bottom"></figure>
<p>From the graph it can be seen that the solution is near to but below \(x=3\pi /2\ (\sim 4.7)\). We therefore start
    the Newton&ndash;Raphson iteration with a value \(x_0=4.5\).</p>
<p>The equation is rewritten as \(\tan x-x=0\). Let \(f(x)=\tan
    x-x\). Then \(f'(x)=\sec^2 x-1=\tan^2x\).</p>
<p>The Newton&ndash;Raphson iteration is</p>
\[x_{n+1}=x_n-\dfrac{\tan x_n-x_n}{\tan^2 x_n}, \quad x_0=4.5,\] 
<p>so</p>
\[x_1=4.5-\dfrac{\tan(4.5)-4.5}{\tan^2 4.5} = 4.5-\dfrac{0.137332}{21.504847}=4.493614 \text{ to 7 sig.fig.}\]
<p>Rounding to \(\text{4 sig.fig.}\) and iterating:</p>
\[x_2=4.494-\dfrac{\tan(4.494)-4.494}{\tan^2 4.494}=4.494-\dfrac{0.004132}{20.229717}=4.493410 \text{ to 7 sig.fig.}\]
<p>So we conclude that the value of \(x\) is \(4.493\) to \(\text{4 sig.fig.}\) As \(x=\mu L=\left(\sqrt{P/EI}\right)L\) we find, after
    re-arrangement, that the smallest buckling load is given by \(P=20.19\dfrac{EI}{L^2}\).</p>]]></text>
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<file name="12.3 Graph of tan x against x.svg" path="/" encoding="base64">[encoded data, 139324 characters]</file>
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