Question 13.P.6: A uniform, simply-supported beam carries a distributed later......
A uniform, simply-supported beam carries a distributed lateral load varying in intensity from w_0 at one end to 2w_0 at the other. Calculate the greatest lateral deflection in the beam.
Step-by-Step
The vertical reactions at O and A are \frac{2}{3} w_0L and \frac{5}{6}w_0L. The bending moment at any section a distance æ from O is then
\begin{array}{l} M=\frac{2}{3} w_0 L z-\frac{1}{2} w_0 z^2-\frac{w_0 z^3}{6 L} \\ \text{Then} \\ E I \frac{d^2 v}{d z^2}=-\left[\frac{2}{3} w_0 L z-\frac{1}{2} w_0 z^2-\frac{w_0 z^3}{6 L}\right] \end{array}On integrating once.
\begin{array}{l} E I \frac{d^2 v}{d z^2}=-\left[\frac{w_0 L z^2}{3} -\frac{w_0 z^2}{6} -\frac{w_0 z^4}{24 L} + C_1\right] \end{array}where C_1 is a constant. On integrating further,
\begin{array}{l} E I v=-\left[\frac{w_0 L z^2}{9} -\frac{w_0 z^5}{24} -\frac{w_0 z^5}{120 L} + C_1z + C_2\right] \end{array}where C_2 is a further constant. If v = 0 at z = 0 and z = L, we have
C_1 = -\frac{11}{180}x_0L^3 and C_2 = 0Then
EIv = \frac{11}{180}w_0L^3z – \frac{w_0Lz^3}{9} – \frac{w_0z^4}{24} + \frac{w_0z^5}{120L}The greatest deflection occurs at dv/dz = 0, i.e. when
\frac{11}{180}w_0L^3 – \frac{w_0Lz^2}{3} + \frac{w_0z^3}{6} + \frac{w_0z^4}{24L}or when
15\left(\frac{z}{L}\right)^4 + 60\left(\frac{z}{L}\right)^3 – 120\left(\frac{z}{L}\right)^2 + 22 = 0The relevant root of this equation is z/L = 0.506 which gives the point of maximum deflection near to the mid-length. The maximum deflection is
v_{\max} = \frac{7.03}{360}\frac{w_0L^4}{EI} = 0.0195 \frac{w_0L^4}{EI}This is negligibly different from the deflection at mid-span, which is
(v)_{z = \frac{1}{2}L} = \frac{5w_0L^4}{256EI}Related Answered Questions
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