Solution to Problem 412 | Shear and Moment Diagrams | Strength of Materials Review at MATHalino

Problem 412
Beam loaded as shown in Fig. P-412.

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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 412

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$\Sigma M_A = 0$

$6R_C = 5 \, [ \, 6(800) \, ]$

$R_C = 4000 \, \text{lb}$

$\Sigma M_C = 0$

$6R_A = 1 \, [ \, 6(800) \, ]$

$R_A = 800 \, \text{lb}$

Segment AB:
$V_{AB} = 800 \, \text{lb}$

$M_{AB} = 800x \, \text{lb}\cdot\text{ft}$

Segment BC:
$V_{BC} = 800 - 800(x - 2)$

$V_{BC} = 2400 - 800x \, \text{lb}$

$M_{BC} = 800x - 800(x - 2)(x - 2)/2$

$M_{BC} = 800x - 400(x - 2)^2 \, \text{lb}\cdot\text{ft}$

Segment CD:
$V_{CD} = 800 + 4000 - 800(x - 2)$

$V_{CD} = 4800 - 800x + 1600$

$V_{CD} = 6400 - 800x \, \text{lb}$

$M_{CD} = 800x + 4000(x - 6) - 800(x - 2)(x - 2)/2$

$M_{CD} = 800x + 4000(x - 6) - 400(x - 2)^2 \, \text{lb}\cdot\text{ft}$

To draw the Shear Diagram:

800 lb of shear force is uniformly distributed along segment AB.
V_BC = 2400 - 800x is linear; at x = 2 ft, V_BC = 800 lb; at x = 6 ft, V_BC = -2400 lb. When V_BC = 0, 2400 - 800x = 0, thus x = 3 ft or V_BC = 0 at 1 ft from B.
V_CD = 6400 - 800x is also linear; at x = 6 ft, V_CD = 1600 lb; at x = 8 ft, V_BC = 0.

To draw the Moment Diagram:

M_AB = 800x is linear; at x = 0, M_AB = 0; at x = 2 ft, M_AB = 1600 lb·ft.
M_BC = 800x - 400(x - 2)² is second degree curve; at x = 2 ft, M_BC = 1600 lb·ft; at x = 6 ft, M_BC = -1600 lb·ft; at x = 3 ft, M_BC = 2000 lb·ft.
M_CD = 800x + 4000(x - 6) - 400(x - 2)² is also a second degree curve; at x = 6 ft, M_CD = -1600 lb·ft; at x = 8 ft, MCD = 0.

Solution to Problem 412 | Shear and Moment Diagrams

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