Solution to Problem 624 | Moment Diagram by Parts | Strength of Materials Review at MATHalino

Problem 624
For the beam loaded as shown in Fig. P-624, compute the moment of area of the M diagrams between the reactions about both the left and the right reaction.

Solution 624
$\Sigma M_{R2} = 0 $

$6R_1 = 400 + 1000(2)$

$R_1 = 400 \, \text{N}$

$\Sigma M_{R1} = 0$

$6R_2 + 400 = 1000(2)$

$R_2 = 600 \, \text{N}$

Moment diagram by parts can be drawn in different ways; three are shown below.

1st Solution: Click here to read or hide

$(Area_{AB})\bar{X}_A = \frac{1}{2}(2)(800)(\frac{4}{3}) + \frac{1}{2}(4)(2400)(\frac{10}{3}) - \frac{1}{2}(2)(2000)(\frac{8}{3})$

$(Area_{AB})\bar{X}_A = 11\,733.33 \, \text{N}\cdot\text{m}^3$

$(Area_{AB})\bar{X}_B = \frac{1}{2}(2)(800)(\frac{14}{3}) + \frac{1}{2}(4)(2400)(\frac{8}{3}) - \frac{1}{2}(2)(2000)(\frac{10}{3})$

$(Area_{AB})\bar{X}_B = 9\,866.67 \, \text{N}\cdot\text{m}^3$

2nd Solution: Click here to read or hide

$(Area_{AB}) \bar{X}_A = 400(2)(1) + \frac{1}{2}(6)(3600)(2) - \frac{1}{2}(4)(4000)(\frac{4}{3})$

$(Area_{AB})\bar{X}_A = 11\,733.33 \, \text{N}\cdot\text{m}^3$

$(Area_{AB}) \bar{X}_B = 400(2)(5) + \frac{1}{2}(6)(3600)(4) - \frac{1}{2}(4)(4000)(\frac{14}{3})$

$(Area_{AB})\bar{X}_B = 9\,866.67 \, \text{N}\cdot\text{m}^3$

3rd Solution: Click here to read or hide

$(Area_{AB}) \, \bar{X}_A = \frac{1}{2}(6)(2400)(4) - 400(4)(4) - \frac{1}{2}(2)(2000)(\frac{16}{3})$

$(Area_{AB})\bar{X}_A = 11\,733.33 \, \text{N}\cdot\text{m}^3$

$(Area_{AB}) \bar{X}_B = \frac{1}{2}(6)(2400)(2) - 400(4)(2) - \frac{1}{2}(2)(2000)(\frac{2}{3})$

$(Area_{AB})\bar{X}_B = 9\,866.67 \, \text{N}\cdot\text{m}^3$

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