Solution to Problem 441 | Relationship Between Load, Shear, and Moment | Strength of Materials Review at MATHalino

Problem 441
A beam ABCD is supported by a roller at A and a hinge at D. It is subjected to the loads shown in Fig. P-441, which act at the ends of the vertical members BE and CF. These vertical members are rigidly attached to the beam at B and C. (Draw shear and moment diagrams for the beam ABCD only.)

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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 441

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$F_{BH} = 14 \, \text{kN to the right}$

$M_B = 14(2)$
$M_B = 28 \, \text{kN}\cdot\text{m}$ counterclockwise

$F_{CH} = \frac{3}{5} (10)$
$F_{CH} = 6 \, \text{ kN}$ to the right

$F_{CV} = 4/5 (10)$
$F_{CV} = 8 \, \text{ kN}$ upward

$M_C = F_{CH} (2) = 6(2)$
$M_C = 12 \, \text{ kN}\cdot\text{m}$ clockwise

$\Sigma M_D = 0$
$6R_A + 12 + 8(2) = 28$
$R_A = 0$

$\Sigma M_A = 0$
$6R_{DV} + 12 = 28 + 8(4)$
$R_{DV} = 8 \, \text{kN}$

$\Sigma F_H = 0$
$R_{DH} = 14 + 6$
$R_{DH} = 20 \, \text{kN}$

To draw the Shear Diagram

Shear in segments AB and BC is zero.
V_C = 8
V_D = V_C + Area in load diagram
V_D = 8 + 0 = 8 kN
V_D2 = V_D - R_DV
V_D2 = 8 - 8 = 0

To draw the Moment Diagram

Moment in segment AB is zero
M_B = -28 kN·m
M_C = M_B + Area in shear diagram
M_C = -28 + 0 = -28 kN·m
M_C2 = M_C + 12 = -28 + 12
M_C2 = -16 kN·m
M_D = M_C2 + Area in shear diagram
MD = -16 + 8(2)
MD = 0

Solution to Problem 441 | Relationship Between Load, Shear, and Moment

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