Solution to Problem 443 | Relationship Between Load, Shear, and Moment | Strength of Materials Review at MATHalino

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

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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

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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

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.

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

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