Solution to Problem 406 | Shear and Moment Diagrams

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Problem 406
Beam loaded as shown in Fig. P-406.

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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 406

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$\Sigma M_A = 0$

$12R_C = 4(900) + 18(400) + 9[(60)(18)]$

$R_C = 1710 \, \text{lb}$

$\Sigma M_C = 0$

$12R_A + 6(400) = 8(900) + 3[60(18)]$

$R_A = 670 \, \text{lb}$

Segment AB:
$V_{AB} = 670 - 60x \, \text{lb}$

$M_{AB} = 670x - 60x(x/2)$

$M_{AB} = 670x - 30x^2 \, \text{lb}\cdot\text{ft}$

Segment BC:
$V_{BC} = 670 - 900 - 60x$

$V_{BC} = -230 - 60x \, \text{lb}$

$M_{BC} = 670x - 900(x - 4) - 60x(x/2)$

$M_{BC} = 3600 - 230x - 30x^2 \, \text{lb}\cdot\text{ft}$

Segment CD:
$V_{CD} = 670 + 1710 - 900 - 60x$

$V_{CD} = 1480 - 60x \, \text{lb}$

$M_{CD} = 670x + 1710(x - 12) - 900(x - 4) - 60x(x/2)$

$M_{CD} = -16920 + 1480x - 30x^2 \, \text{lb}\cdot\text{ft}$

To draw the Shear Diagram:

V_AB = 670 - 60x for segment AB is linear; at x = 0, V_AB= 670 lb; at x = 4 ft, V_AB = 430 lb.
For segment BC, V_BC = -230 - 60x is also linear; at x= 4 ft, V_BC = -470 lb, at x = 12 ft, V_BC = -950 lb.
V_CD = 1480 - 60x for segment CD is again linear; at x = 12, V_CD = 760 lb; at x = 18 ft, V_CD = 400 lb.

To draw the Moment Diagram:

M_AB = 670x - 30x² for segment AB is a second degree curve; at x = 0, M_AB = 0; at x = 4 ft, M_AB = 2200 lb·ft.
For BC, M_BC = 3600 - 230x - 30x², is a second degree curve; at x = 4 ft, M_BC = 2200 lb·ft, at x = 12 ft, M_BC = -3480 lb·ft; When M_BC = 0, 3600 - 230x - 30x² = 0, x = -15.439 ft and 7.772 ft. Take x = 7.772 ft, thus, the moment is zero at 3.772 ft from B.
For segment CD, M_CD = -16920 + 1480x - 30x² is a second degree curve; at x = 12 ft, M_CD = -3480 lb·ft; at x = 18 ft, M_CD = 0.

Submitted by Romel Verterra on January 22, 2009 - 6:41am

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