Solution to Problem 425 | Relationship Between Load, Shear, and Moment | Strength of Materials Review at MATHalino

Problem 425
Beam loaded as shown in Fig. P-425.

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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 425

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$\Sigma M_A = 0$

$6R_2 = 2(60) + 7(30)$

$R_2 = 55 \, \text{kN}$

$\Sigma M_C = 0$

$6R_1 + 1(30) = 4(60)$

$R_1 = 35 \, \text{kN}$

To draw the Shear Diagram:

V_A = R₁ = 35 kN
V_B = V_A + Area in load diagram - 60 kN
V_B = 35 + 0 - 60 = -25 kN
V_C = V_B + area in load diagram + R₂
V_C = -25 + 0 + 55 = 30 kN
V_D = V_C + Area in load diagram - 30 kN
V_D = 30 + 0 - 30 = 0

To draw the Moment Diagram:

M_A = 0
M_B = M_A + Area in shear diagram
M_B = 0 + 35(2) = 70 kN·m
M_C = M_B + Area in shear diagram
M_C = 70 - 25(4) = -30 kN·m
M_D = M_C + Area in shear diagram
M_D = -30 + 30(1) = 0

Solution to Problem 425 | Relationship Between Load, Shear, and Moment

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