$\Sigma F_V = 0$

$4w_o + 2 \,[ \, \frac{1}{2}w_o(1) \, ] = 20(4) + 2(50)$

$5w_o = 180$

$w_o = 36 \, \text{kN/m}$

**To draw the Shear Diagram**

- V
_{A} = 0
- V
_{B} = V_{A} + Area in load diagram

V_{B} = 0 + ½ (36)(1) = 18 kN

V_{B2} = V_{B} - 50 = 18 - 50

V_{B2} = -32 kN
- The net uniformly distributed load in segment BC is 36 - 20 = 16 kN/m upward.

V_{C} = V_{B2} + Area in load diagram

V_{C} = -32 + 16(4) = 32 kN

V_{C2} = V_{C} - 50 = 32 - 50

V_{C2} = -18 kN
- V
_{D} = V_{C2} + Area in load diagram

V_{D} = -18 + ½ (36)(1) = 0
- The shape of shear at AB and CD are parabolic spandrel with vertex at A and D, respectively.
- The location of zero shear is obviously at the midspan or 2 m from B.

**To draw the Moment Diagram**

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

M_{B} = 0 + 1/3 (1)(18)

M_{B} = 6 kN·m
- M
_{midspan} = M_{B} + Area in shear diagram

M_{midspan} = 6 - ½ (32)(2)

M_{midspan} = -26 kN·m
- M
_{C} = Mmidspan + Area in shear diagram

M_{C} = -26 + ½ (32)(2)

M_{C} = 6 kN·m
- M
_{D} = M_{C} + Area in shear diagram

M_{D} = 6 - 1/3 (1)(18) = 0
- The moment diagram at AB and CD are 3rd degree curve while at BC is 2nd degree curve.