Solution to Problem 627 | Moment Diagram by Parts | Strength of Materials Review at MATHalino

Problem 627
For the beam loaded as shown in Fig. P-627compute the moment of area of the M diagrams between the reactions about both the left and the right reaction. (Hint: Resolve the trapezoidal loading into a uniformly distributed load and a uniformly varying load.)

Solution 627

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$\Sigma M_{R2} = 0$

$4R_1 = 200(4)(2) + \frac{1}{2}(3)(400)(1)$

$R_1 = 550 \, \text{N}$

$\Sigma M_{R1} = 0$

$4R_2 = 200(4)(2) + \frac{1}{2}(3)(400)(3)$

$R_2 = 850 \, \text{N}$

$(Area_{AB})\, \bar{X}_A = \frac{1}{2}(4)(2200)(\frac{8}{3}) - \frac{1}{3}(4)(1600)(3) - \frac{1}{4}(3)(600)(\frac{17}{5})$

$(Area_{AB})\, \bar{X}_A = 3 \, 803.33 \, \text{N}\cdot\text{m}^3$ answer

$(Area_{AB})\, \bar{X}_B = \frac{1}{2}(4)(2200)(\frac{4}{3}) - \frac{1}{3}(4)(1600)(1) - \frac{1}{4}(3)(600)(\frac{3}{5})$

$(Area_{AB})\, \bar{X}_B = 3 \, 463.33 \, \text{N}\cdot\text{m}^3$ answer

Solution to Problem 627 | Moment Diagram by Parts

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