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    <name>
      <text>17.1.1.8 Engineering Example 2</text>
    </name>
    <questiontext format="html">
      <text><![CDATA[<h4>
    Engineering Example 2 </h4>
<h5><br>The web-flange junction </h5>

<h5>Introduction</h5>
<p>In problems of torsion, the <strong>torsion constant</strong>, \(J\), which is a function of the shape and structure of the element under consideration, is an important quantity. A common beam section is the <strong>thick \( I\) -section</strong> shown
    here,
</p>
<figure class="HELM_figure">
    <img src="@@PLUGINFILE@@/17.1%20Engineering%20example%202%20fig1.svg" alt="A common beam section called the thick I-section." width="199" height="179" class="img-responsive atto_image_button_text-bottom">

</figure>
<p>for which the torsion constant is given by</p>
\[J = 2J_1 + J_2 + 2 \alpha D^4\]
<p>where the \(J_1\) and \(J_2\) terms refer to the flanges and web respectively, and the \(D^4\) term refers to the web-flange junction. In fact</p>
\[\alpha = \min \left[ \dfrac{t_f}{t_w}, \dfrac{t_w}{t_f} \right] \left( 0.15 + 0.1 \, \dfrac{r}{t_f} \right)\]
<p>where \(t_f\) and \(t_w\) are the thicknesses of the flange and web respectively, and \(r\) is the radius of the concave circle element between the flange and the web. \(D\) is the diameter of the circle of the web-flange junction.</p>
<figure class="HELM_figure">
    <img src="@@PLUGINFILE@@/17.1%20Engineering%20example%202%20fig2.svg" alt="Zoom in of the thick I-section" width="364" height="127" class="img-responsive atto_image_button_text-bottom">
</figure>
<p>As \(D\) occurs in the form \(D^4\), the torsion constant is very sensitive to it. Calculation of \(D\) is therefore a crucial part of the calculation of \(J\).</p>

<h5> Problem in words</h5>
Find \(D\), the diameter of the circle within the web--flange junction as a function of the other dimensions of the structural element.

<h5>Mathematical statement of problem</h5>
<ol type="a">
    <li> Find \(D\), the diameter of the circle, in terms of \(t_f\) and \(t_w\) (the thicknesses of the flange and the web respectively) in the case where \(r=0\). When \(t_f = 3 \mbox{ cm}\) and \(t_w = 2 \mbox{ cm}\), find \(D\).</li>
    <li> For \(r \neq 0\), find \(D\) in terms of \(t_f\), \(t_w\) and \(r\). In the special case where \(t_f = 3 \mbox{ cm}\), \(t_w = 2 \mbox{ cm}\) and \(r = 0.4 \mbox{ cm}\), find \(D\).</li>
</ol>

<h5>Mathematical analysis</h5>
<ol type="a">
    <li>Consider a co-ordinate system based on the midpoint of the outer surface of the flange.
        <figure class="HELM_figure">
            <img src="@@PLUGINFILE@@/17.1%20Engineering%20example%202%20fig3.svg" alt="A coordinate system based on the midpoint of the outer surface of the flange" width="401" height="177" class="img-responsive atto_image_button_text-bottom">
        </figure>
        <p>The centre of the circle will lie at \(( 0, -R)\) where \(R\) is the radius of the circle, i.e. \(R = D/2\). The equation of the circle is </p>
        \[\begin{equation}\tag{1}x^2 + (y + R)^2 = R^2.\end{equation}\]
        <p>In addition, the circle passes through the `corner' at point \(A = (t_w/2, -t_f)\), so </p>
        \[\begin{equation}\tag{2} \left( \dfrac{t_w}{2} \right)^2 + (-t_f + R)^2 = R^2.\end{equation}\] On expanding \[\dfrac{t_w^2}{4} + t_f^2 - 2 R t_f + R^2 = R^2 \] giving \[2 R t_f = \dfrac{t_w^2}{4} + t_f^2\qquad\implies\quad R = \dfrac{(t_w^2/4) + t_f^2}{2t_f}=
        \frac{t_w^2}{8t_f} + \dfrac{t_f}{2} \] so that \[\begin{equation}\tag{3}D = 2R = \dfrac{t_w^2}{4t_f} + t_f\end{equation}.\] Setting \(t_f = 3 \mbox{ cm}\), \(t_w = 2 \mbox{ cm}\) gives \[s D = \frac{2^2}{4 \times 3} + 3 = 3.33 \mbox{ cm}\]</li>
    <li>
        <p>Again using a co-ordinate system based on the mid-point of the outer surface of the flange, consider now the case \(r \neq 0\). </p>
        <figure class="HELM_figure">
            <img src="@@PLUGINFILE@@/17.1%20Engineering%20example%202%20fig4.svg" alt="A picture of a co-ordinate system based on the mid-point of the outer surface of the flange" width="357" height="175" class="img-responsive atto_image_button_text-bottom">
        </figure>
        <p>Point \(B = (t_w/2+r,-t_f-r)\) lies, not on the circle described by (1), but on the slightly larger circle with the same centre, and radius \(R+r\). The equation of this circle is </p>
        \[\begin{equation}\tag{4} x^2 + (y + R)^2 = (R+r)^2 .\end{equation}\]
        <p>Putting the co-ordinates of point \(B\) into equation (4) gives</p>
        \[\begin{equation}\tag{5}\left( \frac{t_w}{2}+r\right)^2 + \left(-t_f-r+R\right)^2 = \left(R+r\right)^2 \end{equation}\] which, on expanding gives \[ \dfrac{t_w^2}{4} + t_wr + r^2 + t_f^2 + r^2 + R^2 + 2t_fr - 2t_fR - 2rR = R^2 + 2rR + r^2 \] Cancelling
        and gathering terms gives \[\dfrac{t_w^2}{4} + t_wr + r^2 + t_f^2 + 2t_fr = 4rR + 2t_fR = 2R \left( 2r + t_f\right)\] so that \[2R = D = \dfrac{(t_w^2/4) + t_wr + r^2 + t_f^2 + 2t_fr}{(2r + t_f)}\] so \[\begin{equation}\tag{6} D=\dfrac{t_w^2 +
        4t_wr + 4r^2 + 4t_f^2 + 8t_fr}{(8r + 4t_f)}\end{equation}\]
        <p>Now putting \(t_f =3 \mbox{ cm}, t_w = 2 \mbox{ cm}\) and \(r = 0.4 \mbox{ cm}\) makes</p>
        \[D = \frac{2^2 + (4\times2\times0.4) + (4\times0.4^2) + (4\times3^2) + (8\times3\times0.4)}{(8\times0.4) + (4\times3)} = \frac{53.44}{15.2} = 3.52 \mbox{ cm}\]</li>

</ol>

<h5>Interpretation</h5>
<p>Note that setting \(r = 0\) in Equation (6) recovers the special case of \(r = 0\) given by equation (3). The value of \(D\) is now available to be used in calculations of the torsion constant, \(J\).</p>]]></text>
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