Solution to Problem 407 | Shear and Moment Diagrams

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Problem 407
Beam loaded as shown in Fig. P-407.

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Write shear and moment equations for the beams in the following problems. In each problem, let x be the distance measured from left end of the beam. Also, draw shear and moment diagrams, specifying values at all change of loading positions and at points of zero shear. Neglect the mass of the beam in each problem.

Solution 407

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$\Sigma M_A = 0$

$6R_D = 4[2(30)]$

$R_D = 40 \, \text{kN}$

$\Sigma M_D = 0$

$6R_A = 2[2(30)]$

$R_A = 20 \text{kN}$

Segment AB:
$V_{AB} = 20 \, \text{kN}$
$M_{AB} = 20x \, \text{kN}\cdot\text{m}$

Segment BC:
$V_{BC} = 20 - 30(x - 3)$
$V_{BC} = 110 - 30x \, \text{kN}$
$M_{BC} = 20x - 30(x - 3)(x - 3)/2$
$M_{BC} = 20x - 15(x - 3)^2 \, \text{kN}\cdot\text{m}$

Segment CD:
$V_{CD} = 20 - 30(2)$
$V_{CD} = -40 \, \text{kN}$
$M_{CD} = 20x - 30(2)(x - 4)$
$M_{CD} = 20x - 60(x - 4) \, \text{kN}\cdot\text{m}$

To draw the Shear Diagram:

For segment AB, the shear is uniformly distributed at 20 kN.
V_BC = 110 - 30x for segment BC; at x = 3 m, V_BC = 20 kN; at x = 5 m, V_BC = -40 kN. For V_BC = 0, x = 3.67 m or 0.67 m from B.
The shear for segment CD is uniformly distributed at -40 kN.

To draw the Moment Diagram:

For AB, M_AB = 20x; at x = 0, M_AB = 0; at x = 3 m, M_AB = 60 kN·m.
M_BC = 20x - 15(x - 3)² for segment BC is second degree curve; at x = 3 m, M_BC = 60 kN·m; at x = 5 m, M_BC = 40 kN·m. Note that maximum moment occurred at zero shear; at x = 3.67 m, M_BC = 66.67 kN·m.
M_CD = 20x - 60(x - 4) for segment BC is linear; at x = 5 m, M_CD = 40 kN·m; at x = 6 m, M_CD = 0.

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