# Solution to Problem 443 | Relationship Between Load, Shear, and Moment

**Problem 443**

Beam carrying the triangular loads shown in Fig. P-443.

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 443**

By symmetry:

$R_1 = R_2 = \frac{1}{2}(\frac{1}{2}Lw_o)$

$R_1 = R_2 = \frac{1}{2}(\frac{1}{2}Lw_o)$

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

**To draw the Shear Diagram**

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

V_{B}= ¼ Lwo - 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} - Load in AB is linear, thus, V
_{AB}is second degree or parabolic curve. The load is from 0 at A to w_{o}(w_{o}is downward or -w_{o}) at B, thus the slope of V_{AB}is decreasing. - V
_{BC}is also parabolic since the load in BC is linear. The magnitude of load in BC is from -w_{o}to 0 or increasing, thus the slope of V_{BC}is increasing.

**To draw the Moment Diagram**

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

M_{B}= 0 + 2/3 (L/2)(1/4 Lw_{o}) = 1/12 Lw_{o} - M
_{C}= M_{B}+ Area in shear diagram

M_{C}= 1/12 Lw_{o}- 2/3 (L/2)(1/4 Lw_{o}) = 0 - M
_{AC}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.

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