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

**Problem 444**

Beam loaded as shown in Fig. P-444.

Without writing shear and moment equations, draw the shear and moment diagrams for the beams specified in the following problems. Give numerical values at all change of loading positions and at all points of zero shear.

**Solution 444**

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

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

**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.

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