By symmetry:

$R_1 = R_2 = \frac{1}{2}(\frac{1}{2}Lw_o)$

$R_1 = R_2 = \frac{1}{4}Lw_o$

**To draw the Shear Diagram**

- V
_{A} = R1 = ¼ Lw_{o}
- V
_{B} = VA + Area in load diagram

V_{B} = ¼ Lwo - ½ (L/2)(w_{o}) = 0
- V
_{C} = V_{B} + Area in load diagram

V_{C} = 0 - ½ (L/2)(w_{o}) = -¼ Lw_{o}
- Load in AB is linear, thus, V
_{AB} is second degree or parabolic curve. The load is from 0 at A to w_{o} (w_{o} is downward or -w_{o}) at B, thus the slope of V_{AB} is decreasing.
- V
_{BC} is also parabolic since the load in BC is linear. The magnitude of load in BC is from -w_{o} to 0 or increasing, thus the slope of V_{BC} is increasing.

**To draw the Moment Diagram**

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

M_{B} = 0 + 2/3 (L/2)(1/4 Lw_{o}) = 1/12 Lw_{o}
- M
_{C} = M_{B} + Area in shear diagram

M_{C} = 1/12 Lw_{o} - 2/3 (L/2)(1/4 Lw_{o}) = 0
- M
_{AC} is third degree because the shear diagram in AC is second degree.
- The shear from A to C is decreasing, thus the slope of moment diagram from A to C is decreasing.