# Solution to Problem 660 | Deflections in Simply Supported Beams

**Problem 660**

A simply supported beam is loaded by a couple M at its right end, as shown in Fig. P-660. Show that the maximum deflection occurs at x = 0.577L.

**Solution 660**

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$EI \, t_{A/C} = \frac{1}{2}x \left( \dfrac{Mx}{L} \right)(\frac{2}{3}x)$

$EI \, t_{A/C} = \dfrac{Mx^3}{3L}$

$EI \, t_{B/C} = (Area_{BC}) \, \bar{X}_B$

$EI \, t_{B/C} = M(L - x) \frac{1}{2}(L - x) - \frac{1}{2}(L - x) \dfrac{M}{L}(L - x) \frac{2}{3}(L - x)$

$EI \, t_{B/C} = \frac{1}{2}M(L - x)^2 - \dfrac{M}{3L}(L - x)^3$

$EI \, t_{B/C} = \dfrac{M}{6L}(L - x)^2 [ \, 3L - 2(L - x) \, ]$

$EI \, t_{B/C} = \dfrac{M}{6L}(L - x)^2 (L + 2x)$

$EI \, t_{B/C} = \dfrac{M}{6}(L - x)^2 + \dfrac{Mx}{3L}(L - x)^2$

$EI \, t_{B/C} = \dfrac{M}{6}(L^2 - 2Lx + x^2) + \dfrac{Mx}{3L}(L^2 - 2Lx + x^2)$

$EI \, t_{B/C} = \dfrac{ML^2}{6} - \dfrac{MLx}{3} + \dfrac{Mx^2}{6} + \dfrac{MLx}{3} - \dfrac{2Mx^2}{3} + \dfrac{Mx^3}{3L}$

$EI \, t_{B/C} = \dfrac{ML^2}{6} - \dfrac{Mx^2}{2} + \dfrac{Mx^3}{3L}$

From the figure

$EI \, t_{A/C} = EI \, t_{B/C}$

$\dfrac{Mx^3}{3L} = \dfrac{ML^2}{6} - \dfrac{Mx^2}{2} + \dfrac{Mx^3}{3L}$

$\dfrac{Mx^2}{2} = \dfrac{ML^2}{6}$

$x^2 = \frac{1}{3}L^2$

$x = 0.577L$ (*okay!*)