Solution to Problem 438 | Relationship Between Load, Shear, and Moment | Strength of Materials Review at MATHalino

Problem 438
The beam loaded as shown in Fig. P-438 consists of two segments joined by a frictionless hinge at which the bending moment is zero.

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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 438

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From the FBD of the section to the left of hinge

$\Sigma M_H = 0$

$4R_1 = 200(6)(3)$

$R_1 = 900 \, \text{lb}$

To draw the Shear Diagram

V_A = 0
V_B = V_A + Area in load diagram
V_B = 0 - 200(2) = -400 lb
V_B2 = V_B + R₁ = -400 + 900 = 500 lb
V_H = V_B2 + Area in load diagram
V_H = 500 - 200(4) = -300 lb
V_C = V_H + Area in load diagram
V_C = -300 - 200(2) = -700 lb
Location of zero shear:
x / 500 = (4 - x) / 300
300x = 2000 - 500x
x = 2.5 ft

438-load-shear-and-moment-diagrams-hinged-cantilever-beam.gif

To draw the Moment Diagram

M_A = 0
M_B = M_A + Area in shear diagram
M_B = 0 - ½ (400)(2) = -400 lb·ft
M_x = M_B + Area in load diagram
M_x = -400 + ½ (500)(2.5)
M_x = 225 lb·ft
M_H = M_x + Area in load diagram
M_H = 225 - ½(300)(4 - 2.5) = 0 ok!
M_C = M_H + Area in load diagram
M_C = 0 - ½ (300 + 700)(2)
M_C = -1000 lb·ft
The location of zero moment in segment BH can easily be found by symmetry.

Solution to Problem 438 | Relationship Between Load, Shear, and Moment

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