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.

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