Solution to Problem 414 | Shear and Moment Diagrams

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Problem 414
Cantilever beam carrying the load shown in Fig. P-414.

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Write shear and moment equations for the beams in the following problems. In each problem, let x be the distance measured from left end of the beam. Also, draw shear and moment diagrams, specifying values at all change of loading positions and at points of zero shear. Neglect the mass of the beam in each problem.

Solution 414

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Segment AB:
$V_{AB} = -2x \, \text{kN}$

$M_{AB} = -2x(x/2)$

$M_{AB} = -x^2 \, \text{kN}\cdot\text{m}$

Segment BC:
$\dfrac{y}{x - 2} = \dfrac{2}{3}$

$y = \frac{2}{3}(x - 2)$

$F_1 = 2x$

$F_2 = \frac{1}{2}(x - 2)y$

$F_2 = \frac{1}{2}(x - 2) \, [ \, \frac{2}{3}(x - 2) \, ]$

$F_2 = \frac{1}{3} (x - 2)^2$

$V_{BC} = -F_1 - F_2$

$V_{BC} = -2x - \frac{1}{3} (x - 2)^2$

$M_{BC} = -(x/2)F_1 - \frac{1}{3}(x - 2)F_2$

$M_{BC} = -(x/2)(2x) - \frac{1}{3}(x - 2) \, [ \, \frac{1}{3} (x - 2)^2 \, ]$

$M_{BC} = -x^2 - \frac{1}{9}(x - 2)^3$

To draw the Shear Diagram:

V_AB = -2x is linear; at x = 0, V_AB = 0; at x = 2 m, V_AB = -4 kN.
V_BC = -2x - 1/3 (x - 2)² is a second degree curve; at x = 2 m, V_BC = -4 kN; at x = 5 m; V_BC = -13 kN.

To draw the Moment Diagram:

M_AB = -x² is a second degree curve; at x = 0, M_AB = 0; at x = 2 m, M_AB = -4 kN·m.
M_BC = -x² -1/9 (x - 2)³ is a third degree curve; at x = 2 m, M_BC = -4 kN·m; at x = 5 m, M_BC = -28 kN·m.

Submitted by Romel Verterra on January 26, 2009 - 10:20pm

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