$\text{Total load} = 2\,[\,\frac{1}{2}(L/2)(w_o)\,]$

$\text{Total load} = \frac{1}{2}Lw_o$

**By symmetry**

$R_1 = R_2 = \frac{1}{2} \times \text{Total load}$

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

**To draw the Shear Diagram**

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

V_{B} = ¼ Lw_{o} - 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}
- The shear diagram in AB is second degree curve. The shear in AB is from -w
_{o} (downward w_{o}) to zero or increasing, thus, the slope of shear at AB is increasing (upward parabola).
- The shear diagram in BC is second degree curve. The shear in BC is from zero to -w
_{o} (downward w_{o}) or decreasing, thus, the slope of shear at BC is decreasing (downward parabola)

**To draw the Moment Diagram**

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

M_{B} = 0 + 1/3 (L/2)(¼ Lw_{o}) = 1/24 L^{2}w_{o}
- M
_{C} = M_{B} + Area in shear diagram

M_{C} = 1/24 L^{2}w_{o} - 1/3 (L/2)(¼ Lw_{o}) = 0
- The shear diagram from A to C is decreasing, thus, the moment diagram is a concave downward third degree curve.