Solution to Problem 625 | Moment Diagram by Parts | Strength of Materials Review at MATHalino

Problem 625
For the beam loaded as shown in Fig. P-625, compute the moment of area of the M diagrams between the reactions about both the left and the right reaction. (Hint: Draw the moment diagram by parts from right to left.)

Uniform load over 3/4 of span and concentrated load at midspan of simple beam

Solution 625

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$\Sigma M_{R2} = 0$

$4R_1 = 400(3)(2.5) + 500(2)$

$R_1 = 1000 \, \text{N}$

$\Sigma M_{R1} = 0$

$4R_2 = 400(3)(1.5) + 500(2)$

$R_2 = 700 \, \text{N}$

$(Area_{AB})\,\bar{X}_A = \frac{1}{2}(4)(2800)(\frac{4}{3}) - \frac{1}{2}(2)(1000)(\frac{2}{3}) - \frac{1}{3}(3)(1800)(\frac{3}{4})$

$(Area_{AB})\,\bar{X}_A = 5\,450 \, \text{N}\cdot\text{m}^3$ answer

$(Area_{AB})\,\bar{X}_B = \frac{1}{2}(4)(2800)(\frac{8}{3}) - \frac{1}{2}(2)(1000)(\frac{10}{3}) - \frac{1}{3}(3)(1800)(\frac{13}{4})$

$(Area_{AB})\,\bar{X}_B = 5\,750 \, \text{N}\cdot\text{m}^3$ answer

Solution to Problem 625 | Moment Diagram by Parts

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