**Problem 417**

Beam carrying the triangular loading shown in Fig. P-417.

**Solution 417**

**By symmetry:**

$R_1 = R_2 = \frac{1}{2}(\frac{1}{2}Lw_o) = \frac{1}{4}Lw_o$

$\dfrac{y}{x} = \dfrac{w_o}{L/2}$

$y = \dfrac{2w_o}{L}x$

$F = \frac{1}{2}xy = \frac{1}{2}x \left( \dfrac{2w_o}{L} \right)$

$F = \dfrac{w_o}{L}x^2$

$V = R_1 - F$

$V = \frac{1}{4}Lw_o - \dfrac{w_o}{L}x^2$

$M = R_1x - F(\frac{1}{3}x)$

$M = \frac{1}{4}Lw_ox - \left( \dfrac{w_o}{L}x^2 \right)(\frac{1}{3}x)$

**To draw the Shear Diagram:**

V = Lw_{o}/4 - w_{o}x^{2}/L is a second degree curve; at x = 0, V = Lw_{o}/4; at x = L/2, V = 0. The other half of the diagram can be drawn by the concept of symmetry.

**To draw the Moment Diagram**

M = Lw_{o}x/4 - w_{o}x_{3}/3L is a third degree curve; at x = 0, M = 0; at x = L/2, M = L_{2}w_{o}/12. The other half of the diagram can be drawn by the concept of symmetry.

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