Solution to Problem 404 | Shear and Moment Diagrams | Strength of Materials Review at MATHalino

Problem 404
Beam loaded as shown in Fig. P-404.

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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 404

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$\Sigma M_A = 0$
$12R_D + 4800 = 3(2000)$

$R_D = 100 \, \text{lb}$

$\Sigma M_D = 0$

$12R_A = 9(2000) + 4800$

$R_A = 1900 \, \text{lb}$

Segment AB:
$V_{AB} = 1900 \, \text{lb}$

$M_{AB} = 1900x \, \text{lb}\cdot\text{ft}$

Segment BC:
$V_{BC} = 1900 - 2000$

$V_{BC} = -100 \, \text{lb}$

$M_{BC} = 1900x - 2000(x - 3)$

$M_{BC} = 1900x - 2000x + 6000$

$M_{BC} = -100x + 6000 \, \text{lb}\cdot\text{ft}$

Segment CD:
$V_{CD} = 1900 - 2000$

$V_{CD} = -100 \, \text{lb}$

$M_{CD} = 1900x - 2000(x - 3) - 4800$

$M_{CD} = 1900x - 2000x + 6000 - 4800$

$M_{CD} = -100x + 1200 \, \text{lb}\cdot\text{ft}$

To draw the shear diagram:

At segment AB, the shear is uniformly distributed at 1900 lb.
A shear of -100 lb is uniformly distributed over segments BC and CD.

To draw the Moment Diagram:

M_AB = 1900x is linear; at x = 0, M_AB = 0; at x = 3 ft, M_AB = 5700 lb·ft.
For segment BC, M_BC = -100x + 6000 is linear; at x = 3 ft, M_BC = 5700 lb·ft; at x = 9 ft, M_BC = 5100 lb·ft.
M_CD = -100x + 1200 is again linear; at x = 9 ft, M_CD = 300 lb·ft; at x = 12 ft, M_CD = 0.

Solution to Problem 404 | Shear and Moment Diagrams

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