$\text{Total load} = 2\,[\,\frac{1}{2}(L/2)(w_o)\,]$
$\text{Total load} = \frac{1}{2}Lw_o$
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
- VA = R1 = ¼ Lwo
- VB = VA + Area in load diagram
VB = ¼ Lwo - ½ (L/2)(wo) = 0
- VC = VB + Area in load diagram
VC = 0 - ½ (L/2)(wo) = -¼ Lwo
- The shear diagram in AB is second degree curve. The shear in AB is from -wo (downward wo) 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 -wo (downward wo) or decreasing, thus, the slope of shear at BC is decreasing (downward parabola)
To draw the Moment Diagram
- MA = 0
- MB = MA + Area in shear diagram
MB = 0 + 1/3 (L/2)(¼ Lwo) = 1/24 L2wo
- MC = MB + Area in shear diagram
MC = 1/24 L2wo - 1/3 (L/2)(¼ Lwo) = 0
- The shear diagram from A to C is decreasing, thus, the moment diagram is a concave downward third degree curve.