A Beam
Beam AB, which is simply supported at Points A & B, is subjected to a distributed load as shown. If the beam’s weight is negligible, what is the reaction force at Point B?
Expand Hint
Draw the free body diagram:
Hint 2
Take the moment about Point A to reduce the amount of unknown forces.
$$$\sum M_A=0=Force \times Distance$$$
First, draw the free body diagram:
The triangular force distribution can be replaced with a concentrated force
$$F$$
, which is located through the triangle’s centroid. The force’s magnitude is equal to the triangle’s area:
$$$F=\frac{1}{2}bh=\frac{1}{2}(12m)(200N/m)=1,200\:N$$$
where
$$b$$
is the triangle’s base, and
$$h$$
is the triangle’s height.
We can take the moment about Point A to reduce our unknown variables down to 1 (
$$A_x$$
,
$$B_x$$
and
$$R_A$$
will zero out since their radius vector goes through Point A). Remember,
$$Moment=Force \times Distance$$
:
$$$\sum M_A=0=(18m)R_B-(10m)F$$$
$$$(18m)R_B=(10m)(1,200N)$$$
$$$R_B=\frac{(10m)(1,200N)}{18m}=\frac{12,000N}{18}=667\:N$$$
667 N
Time Analysis
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