Solution to Problem 408 | Shear and Moment Diagrams

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shear and moment diagrams

uniformly distributed load

Problem 408
Beam loaded as shown in Fig. P-408.

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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 408

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$\Sigma M_A = 0$

$6R_D = 1[2(50)] + 5[2(20)]$

$R_D = 50 \, \text{kN}$

$\Sigma M_D = 0$

$6R_A = 5[2(50)] + 1[2(20)]$

$R_A = 90 \, \text{kN}$

Segment AB:
$V_{AB} = 90 - 50x \, \text{kN}$

$M_{AB} = 90x - 50x(x/2)$

$M_{AB} = 90x - 25x^2 \, \text{kN}\cdot\text{m}$

Segment BC:
$V_{BC} = 90 - 50(2)$

$V_{BC} = -10 \, \text{kN}$

$M_{BC} = 90x - 2(50)(x - 1)$

$M_{BC} = -10x + 100 \, \text{kN}\cdot\text{m}$

Segment CD:
$V_{CD} = 90 - 2(50) - 20(x - 4)$

$V_{CD} = -20x + 70 \, \text{kN}$

$M_{CD} = 90x - 2(50)(x - 1) - 20(x - 4)(x - 4)/2$

$M_{CD} = 90x - 100(x - 1) - 10(x - 4)^2$

$M_{CD} = -10x^2 + 70x - 60 \, \text{kN}\cdot\text{m}$

To draw the Shear Diagram:

V_AB = 90 - 50x is linear; at x = 0, V_BC = 90 kN; at x = 2 m, V_BC = -10 kN. When V_AB = 0, x = 1.8 m.
V_BC = -10 kN along segment BC.
V_CD = -20x + 70 is linear; at x = 4 m, V_CD = -10 kN; at x = 6 m, V_CD = -50 kN.

To draw the Moment Diagram:

M_AB = 90x - 25x² is second degree; at x = 0, M_AB = 0; at x = 1.8 m, M_AB = 81 kN·m; at x = 2 m, MAB = 80 kN·m.
M_BC = -10x + 100 is linear; at x = 2 m, M_BC = 80 kN·m; at x = 4 m, M_BC = 60 kN·m.
M_CD = -10x² + 70x - 60; at x = 4 m, M_CD = 60 kN·m; at x = 6 m, M_CD = 0.

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