Solution to Problem 436 | Relationship Between Load, Shear, and Moment | Strength of Materials Review at MATHalino

Problem 436
A distributed load is supported by two distributed reactions as shown in Fig. P-436.

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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 436

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$\Sigma M_{midpoint\,\,of\,\,CD} = 0$

$4w_1 (11) = 440(8)(5)$

$w_1 = 400 \, \text{lb/ft}$

$\Sigma M_{midpoint\,\,of\,\,AB} = 0$

$2w_2 (11) = 440(8)(6)$

$w_2 = 960 \, \text{lb/ft}$

To draw the Shear Diagram

V_A = 0
V_B = V_A + Area in load diagram
V_B = 0 + 400(4) = 1600 lb
V_C = V_B + Area in load diagram
V_C = 1600 - 440(8) = -1920 lb
V_D = V_C + Area in load diagram
V_D = -1920 + 960(2) = 0
Location of zero shear:
x / 1600 = (8 - x) / 1920
x = 40/11 ft = 3.636 ft from B

To draw the Moment Diagram

M_A = 0
M_B = M_A + Area in shear diagram
M_B = 0 + ½ (1600)(4) = 3200 lb·ft
M_x = M_B + Area in shear diagram
M_x = 3200 + ½ (1600)(40/11)
M_x = 6109.1 lb·ft
M_C = M_x + Area in shear diagram
M_C = 6109.1 - ½ (8 - 40/11)(1920)
M_C = 1920 lb·ft
M_D = M_C + Area in shear diagram
M_D = 1920 - ½ (1920)(2) = 0

Solution to Problem 436 | Relationship Between Load, Shear, and Moment

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