Solution to Problem 427 | Relationship Between Load, Shear, and Moment | Strength of Materials Review at MATHalino

Problem 427
Beam loaded as shown in Fig. P-427.

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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. (Note to instructor: Problems 403 to 420 may also be assigned for solution by semi-graphical method describes in this article.)

Solution 427

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$\Sigma M_C = 0$

$12R_1 = 100(12)(6) + 800(3)$

$R_1 = 800 \, \text{lb}$

$\Sigma M_A = 0$

$12R_2 = 100(12)(6) + 800(9)$

$R_2 = 1200 \, \text{lb}$

To draw the Shear Diagram

V_A = R₁ = 800 lb
V_B = VA + Area in load diagram
V_B = 800 - 100(9)
V_B = -100 lb
V_B2 = -100 - 800 = -900 lb
V_C = V_B2 + Area in load diagram
V_C = -900 - 100(3)
V_C = -1200 lb
Solving for x:
x / 800 = (9 - x) / 100
100x = 7200 - 800x
x = 8 ft

To draw the Moment Diagram

M_A = 0
M_x = M_A + Area in shear diagram
M_x = 0 + ½(8)(800) = 3200 lb·ft;
M_B = M_x + Area in shear diagram
M_B = 3200 - ½(1)(100) = 3150 lb·ft
M_C = M_B + Area in shear diagram
M_C = 3150 - ½(900 + 1200)(3) = 0
The moment curve BC is downward parabola with vertex at A'. A' is the location of zero shear for segment BC.

Solution to Problem 427 | Relationship Between Load, Shear, and Moment

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