Solution to Problem 406 | Shear and Moment Diagrams

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

$R_C = 1710 \, \text{lb}$
 

$\Sigma M_C = 0$

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

$R_A = 670 \, \text{lb}$
 

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:

1. V_AB = 670 - 60x for segment AB is linear; at x = 0, V_AB= 670 lb; at x = 4 ft, V_AB = 430 lb.
2. 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.
3. 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:

1. 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.
2. 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.
3. 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.