$\Sigma M_D = 0$

$20R_1 = 1000(25) + 400(5)(22.5) + 2000(10) + 200(10)(5)$

$R_1 = 5000 \, \text{lb}$

$\Sigma M_B = 0$

$20R_2 + 1000(5) + 400(5)(2.5) = 2000(10) + 200(10)(15)$

$R_2 = 2000 \, \text{lb}$

**To draw the Shear Diagram**

- V
_{A} = -1000 lb
- V
_{B} = V_{A} + Area in load diagram; V_{B} = -1000 - 400(5) = -3000 lb; V_{B2} = -3000 + R1 = 2000 lb
- V
_{C} = V_{B2} + Area in load diagram; V_{C} = 2000 + 0 = 2000 lb; V_{C2} = 2000 - 2000 = 0
- V
_{D} = V_{C2} + Area in load diagram; V_{D} = 0 + 200(10) = 2000 lb

**To draw the Moment Diagram**

- M
_{A} = 0
- M
_{B} = M_{A} + Area in shear diagram

M_{B} = 0 - ½ (1000 + 3000)(5)

M_{B} = -10000 lb·ft
- M
_{C} = M_{B} + Area in shear diagram

M_{C} = -10000 + 2000(10) = 10000 lb·ft
- M
_{D} = M_{C} + Area in shear diagram

M_{D} = 10000 - ½ (10)(2000) = 0
- For segment BC, the location of zero moment can be accomplished by symmetry and that is 5 ft from B.
- The moment curve AB is a downward parabola with vertex at A'. A' is the location of zero shear for segment AB at point outside the beam.