Distributed Forces - College of Engineering | University ...
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SOLUTION
R, = (200lblft)(4 ft) = Sgg rO
1
R n = :([50 lb/ft)(3 ft) = zz5 tb2
+J:r, = g'
{xzo = 6'
A - 800Lb +2251b =0
MA - (8001bX2 f0 + Q25lb)(5 f0 = 0
A=575lbl<
Me = 475b .rt *) 1
t+/al,!
PROBLEM 5.70
Determine the reactions at the beam supports for the given loading.
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PROBLEM 5,71
Determine the reactions at the beam supports for the given
loading.
SOLUTION1
nr=i@kN/mX6 m)
=12kNRo =(2kN/m)(10m)
=20kN
*1 >r, =o' A-12kN-2okN=o
(xMr\
{>lro = g, M A- $2t<NX2m)-(20 kNX5 m) =0
L=32.0kN1<
Mo=r\ .OtN'm) {
pRopRrErARy MArERtaL.o 20r3 rhe Mccrrw-Hirr,C;Ylil:';# ,ll,i,X};;;):['i iil!";illii:I;i!,,:i!,!f;;:i i':i,:il:i'r;;;;;;;
"' rlistributed in nnt fitrtn or b1' Ltnv means'
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635
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