$\Sigma M_B = 0$
$6R_E = 1200 + 1 \, [ \, 6(100) \, ]$
$R_E = 300 \, \text{lb}$
$\Sigma M_E = 0$
$6R_B + 1200 = 5 \, [ \, 6(100) \, ]$
$R_B = 300 \, \text{lb}$
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:
- V_{AB} = -100x is linear; at x = 0, V_{AB} = 0; at x = 2 ft, V_{AB} = -200 lb.
- 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.
- The shear is uniformly distributed at -300 lb along segments CD and DE.
To draw the Moment Diagram:
- M_{AB} = -50x^{2} is a second degree curve; at x= 0, M_{AB} = 0; at x = ft, M_{AB} = -200 lb·ft.
- M_{BC} = -50x^{2} + 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.
- M_{CD} = -300x + 1200 is linear; at x = 6 ft, M_{CD} = -600 lb·ft; at x = 7 ft, M_{CD} = -900 lb·ft.
- M_{DE} = -300x + 2400 is again linear; at x = 7 ft, M_{DE} = 300 lb·ft; at x = 8 ft, M_{DE} = 0.