Solution to Problem 444 | Relationship Between Load, Shear, and Moment

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

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

1. V_A = R1 = ¼ Lw_o
2. V_B = V_A + Area in load diagram
   V_B = ¼ Lw_o - ½ (L/2)(w_o) = 0
3. V_C = V_B + Area in load diagram
   V_C = 0 - ½ (L/2)(w_o) = -¼ Lw_o
4. 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).
5. 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

1. MA = 0
2. M_B = M_A + Area in shear diagram
   M_B = 0 + 1/3 (L/2)(¼ Lw_o) = 1/24 L²w_o
3. M_C = M_B + Area in shear diagram
   M_C = 1/24 L²w_o - 1/3 (L/2)(¼ Lw_o) = 0
4. The shear diagram from A to C is decreasing, thus, the moment diagram is a concave downward third degree curve.