Solution to Problem 413 | Shear and Moment Diagrams

$6R_E = 1200 + 1 \, [ \, 6(100) \, ]$

$R_E = 300 \, \text{lb}$
 

$\Sigma M_E = 0$

$6R_B + 1200 = 5 \, [ \, 6(100) \, ]$

$R_B = 300 \, \text{lb}$
 

Segment AB:
$V_{AB} = -100x \, \text{lb}$

$M_{AB} = -100x(x/2)$

$M_{AB} = -50x^2 \, \text{lb}\cdot\text{ft}$
 

Segment BC:
$V_{BC} = -100x + 300 \, \text{lb}$

$M_{BC} = -100x(x/2) + 300(x - 2)$

$M_{BC} = -50x^2 + 300x - 600 \, \text{lb}\cdot\text{ft}$
 

Segment CD:
$V_{CD} = -100(6) + 300$

$V_{CD} = -300 \, \text{lb}$

$M_{CD} = -100(6)(x - 3) + 300(x - 2)$

$M_{CD} = -600x + 1800 + 300x - 600$

$M_{CD} = -300x + 1200 \, \text{lb}\cdot\text{ft}$
 

Segment DE:
$V_{DE} = -100(6) + 300$

$V_{DE} = -300 \, \text{lb}$

$M_{DE} = -100(6)(x - 3) + 1200 + 300(x - 2)$

$M_{DE} = -600x + 1800 + 1200 + 300x - 600$

$M_{DE} = -300x + 2400 \, \text{lb}\cdot\text{ft}$
 

To draw the Shear Diagram:

1. V_AB = -100x is linear; at x = 0, V_AB = 0; at x = 2 ft, V_AB = -200 lb.
2. V_BC = 300 - 100x is also linear; at x = 2 ft, V_BC = 100 lb; at x = 4 ft, V_BC = -300 lb. When V_BC = 0, x = 3 ft, or V_BC =0 at 1 ft from B.
3. The shear is uniformly distributed at -300 lb along segments CD and DE.

To draw the Moment Diagram:

1. M_AB = -50x² is a second degree curve; at x= 0, M_AB = 0; at x = ft, M_AB = -200 lb·ft.
2. M_BC = -50x² + 300x - 600 is also second degree; at x = 2 ft; M_BC = -200 lb·ft; at x = 6 ft, M_BC = -600 lb·ft; at x = 3 ft, M_BC = -150 lb·ft.
3. M_CD = -300x + 1200 is linear; at x = 6 ft, M_CD = -600 lb·ft; at x = 7 ft, M_CD = -900 lb·ft.
4. M_DE = -300x + 2400 is again linear; at x = 7 ft, M_DE = 300 lb·ft; at x = 8 ft, M_DE = 0.