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      <text>19.4.4.2 Engineering Example 4</text>
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      <text><![CDATA[<h5>Engineering Example 4</h5>
<p><br></p>
<h5>Deflection of a uniformly loaded beam<br></h5>

<h6>Introduction<br></h6>
<p>A uniformly loaded beam of length \(L\) is supported at both ends as shown in Figure 9. The deflection \(y(x)\) is a function of horizontal position \(x\) and obeys the ordinary differential equation (ODE)</p>
<p>\(\qquad \dfrac{d^4y}{dx^4}(x) = \dfrac{1}{EI} q(x) \qquad \qquad \qquad\) (1)</p>
<p>where \(E\) is Young's modulus, \(I\) is the moment of inertia and \(q(x)\) is the load per unit length at point \(x.\) We assume in this problem that \(q(x) = q\) a constant. The boundary conditions are (i) no deflection at \(x=0\) and \(x=L\) (ii) no
    curvature of the beam at \(x=0\) and \(x=L.\)</p>

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    <img src="@@PLUGINFILE@@/beam.svg" alt="Image of the beam described on the text." width="479" height="152" class="img-responsive atto_image_button_text-bottom">
</figure>

<p><br></p>
<h6>Problem in words<br></h6>
<p>Find the deflection of a beam, supported so that that there is no deflection and no curvature of the beam at its ends, subject to a uniformly distributed load, as a function of position along the beam.</p>

<p><br></p>
<h6>Mathematical statement of problem<br></h6>
<p>Find the equation of the curve \(y(x)\) assumed by the bending beam that satisfies the ODE (1). Use the coordinate system shown in the previous figure where the origin is at the left extremity of the beam. In this coordinate system, the boundary conditions,
    which require that there is no deflection at \(x=0\) and \(x=L,\) and that there is no curvature of the beam at \(x=0\) and \(x=L,\) are</p>
<p>\( \qquad\) (a) \(\, y(0)=0\)</p>
<p>\( \qquad\) (b) \(\, y(L)=0\)</p>
<p>\( \qquad\) (c) \(\, \dfrac{d^2y}{dx^2}\big|_{x=0} =0\)</p>
<p>\( \qquad\) (d) \(\, \dfrac{d^2y}{dx^2}\big|_{x=L} =0\)</p>
<p>\( \qquad\) (e) \(\, \dfrac{d^2y}{dx^2}\big|_{x=L} = 0\)</p>
<p>Note that \(\dfrac{dy(x)}{dx}\) and \(\dfrac{d^2y(x)}{dx^2}\) are respectively the slope and the radius of curvature of the curve at point \((x,y).\)</p>

<p><br></p>
<h6>Mathematical analysis<br></h6>
<p>Integrating Equation (1) leads to:</p>
<p>\(\qquad EI \dfrac{d^3y}{dx^3}(x) = qx + A \qquad \qquad \qquad \qquad \qquad \qquad\) (2)</p>
<p>Integrating a second time:</p>
<p>\(\qquad EI \dfrac{d^2y}{dx^2}(x) = qx^2/2 + Ax+ B \qquad \qquad \qquad \qquad \qquad\)(3)</p>
<p>Integrating a third time:</p>
<p>\(\qquad EI \dfrac{dy}{dx}(x) = qx^3/6 + Ax^2/2 + Bx+ C \qquad \qquad \qquad \qquad \) (4)</p>
<p>Integrating a fourth time:</p>
<p>\(\qquad EI y(x) = qx^4/24 + Ax^3/6 + Bx^2/2 + Cx+D. \qquad \qquad \qquad\) (5)</p>
<p>The boundary conditions (a) and (b) enable determination of the constants of integration \(A, B, C, D.\) Indeed, the boundary condition (a), \(y(0)=0,\) and Equation (5) gives</p>
<p>\(\qquad EI y(0) = q \times (0)^4/24 + A \times (0)^3/6 + B \times (0)^2/2 + C \times (0)+ D = 0\)</p>
<p>which yields \(D=0.\)</p>

<p>The boundary condition (b), \(y(L)=0,\) and Equation (5) give</p>
<p>\(\qquad EI y(L) = q L^4/24 + A L^3/6 + B L^2/2 + CL + D.\)</p>
<p>Using the newly found value for \(D\) one writes</p>
<p>\(\qquad q L^4/24 + A L^3/6 + B L^2/2 + CL = 0 \qquad \qquad \qquad\) (6)</p>
<p>The boundary condition (c) obtained from the definition of the radius of curvature, \(\dfrac{d^2y}{dx^2}(0) = 0,\) and Equation (3) give</p>
<p>\(\qquad I \dfrac{d^2y}{dx^2} (0) = q \times (0)^2/2 + A \times (0) + B\)</p>
<p>in which yields \(B=0.\)</p>

<p>The boundary condition (d), \(\dfrac{d^2y}{dx^2}(L) = 0,\) and Equation (3) give</p>
<p>\(\qquad EI\dfrac{d^2y}{dx^2} (L) = q L^2/2 + A L = 0\)</p>
<p>which yields \(A =-q L /2.\)</p>
<p>The expressions for \(A, B, D\) are introduced in Equation (6) to find the last unknown constant \(C.\) This leads to \(q L^4/24 - q L^4/12 + CL = 0\) or \(C = q L^3/24.\) Finally, Equation (5) and the values of constants lead to the solution</p>
<p>\(\qquad y(x) = [qx^4/24 - q L x^3/12 + q L^3 x/24] / EI . \qquad \qquad \qquad\) (7)</p>

<p><br></p>
<h6>Interpretation</h6>
<p>The predicted deflection is zero at both ends as required, and you may check that it is symmetrical about the centre of the beam by switching to the coordinate system \((X,Y)\) with \(L/2 - x = X\) and \(y = Y\) and verifying that the deflection \(Y(X)\) is symmetrical about the vertical axis, i.e.  \(Y(X) = Y(-X).\)</p>]]></text>
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