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
- 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
- Load in AB is linear, thus, VAB is second degree or parabolic curve. The load is from 0 at A to wo (wo is downward or -wo) at B, thus the slope of VAB is decreasing.
- VBC is also parabolic since the load in BC is linear. The magnitude of load in BC is from -wo to 0 or increasing, thus the slope of VBC is increasing.
To draw the Moment Diagram
- MA = 0
- MB = MA + Area in shear diagram
MB = 0 + 2/3 (L/2)(1/4 Lwo) = 1/12 Lwo
- MC = MB + Area in shear diagram
MC = 1/12 Lwo - 2/3 (L/2)(1/4 Lwo) = 0
- MAC 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.
