# Solution to Problem 413 | Shear and Moment Diagrams

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

$R_E = 300 \, \text{lb}$

$\Sigma M_E = 0$

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

$R_B = 300 \, \text{lb}$

**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.

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