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      <text>11.7.3 Beam Curvature</text>
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      <text><![CDATA[<h4>Engineering Example</h4>
Here we shall find a formula for the curvature of a bent beam.
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<h4 class="HELM_example">Example</h4>
<p>When a horizontal beam is acted on by forces which bend it, each small segment of the beam will be slightly curved and can be regarded as an arc of a circle. The radius \(R\) of that circle is called the <b>radius of curvature</b> of the beam at the point concerned. If the shape of the beam is described by an equation of the form \(y = f(x)\) then there is a formula for the radius of curvature \(R\) which involves only the first and second derivatives \( \dfrac{dy}{dx}\) and \( \dfrac{d^2y}{dx^2}\). </p>
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The equation is found as follows.
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Start with the equation of a circle in the simple implicit form
\[x^2 + y^2 = R^2\]
and perform implicit differentiation twice. Now use the result of the first implicit
differentiation to find a simple expression for the quantity \(1+ \left( \dfrac{dy}{dx}\right)^2\) in terms of
\(R\) and \(y\); this can then be used to simplify the result of the second differentiation,
and will lead to a formula for \( \dfrac{1}{R}\) (called the <b>curvature</b>) in terms of \( \dfrac{dy}{dx}\) and \( \dfrac{d^2y}{dx^2}\).
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Have a go before looking at the solution below.
<h4 class="HELM_solution">Solution</h4>
Differentiating \( x^2 + y^2 = R^2\) gives
\[ 2x + 2y\dfrac{dy}{dx} = 0. \hspace{5cm} (1)\]
Differentiating again gives 
\[ 2 + 2 \left( \dfrac{dy}{dx} \right)^2 + 2y \dfrac{d^2y}{dx^2} = 0. \hspace{3cm} (2)\]
From \( (1)\),
\[ \dfrac{dy}{dx} = -\dfrac{x}{y} \hspace{0.5cm} \therefore \hspace{0.5cm} 1 + \left( \dfrac{dy}{dx} \right) ^2 = 1 + \dfrac{x^2}{y^2} = \dfrac{y^2 + x^2}{y^2} = \left( \dfrac{R}{y} \right)^2. \hspace{2cm} (3) \]
So 
\[1 + \left( \dfrac{dy}{dx} \right)^2 = \left( \dfrac{R}{y} \right)^2.\]
Thus \( (2) \) becomes
\[ 2 \left( \dfrac{R}{y} \right)^2 + 2y \left( \dfrac{d^2y}{dx^2} \right) = 0 \hspace{0.5cm} \therefore \hspace{0.5cm}  \dfrac{d^2y}{dx^2} = \dfrac{R^2}{y^3} = \left( \dfrac{1}{R} \right) \left( \dfrac{R}{y} \right) ^3 \]
so
\[  \dfrac{d^2y}{dx^2} = -\dfrac{1}{R} \left( \dfrac{R}{y} \right)^3. \hspace{4cm} (4) \]
Rearranging \( (4) \) to make \( \dfrac{1}{R} \) the subject and substituting for \( \left( \dfrac{R}{y} \right) \) from \( (3) \) gives the result:
\[ \dfrac{1}{R} = \dfrac{\dfrac{d^2y}{dx^2}}{ \left[ 1 + \left( \dfrac{dy}{dx} \right)^2 \right] ^{3/2}} \]
The equation usually found in textbooks omits the minus sign but the sign indicates whether the circle is above or below the curve, as you will see by sketching a few examples. When the gradient is small (as for a slightly deflected horizontal beam), i.e. \( \dfrac{dy}{dx}\) is small, the denominator in the equation for \( (1/R)\) is close to \(1\), and so the second derivative alone is often used to estimate the radius of curvature in the theory of bending beams.]]></text>
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