Structural Analysis
6th Edition
ISBN: 9781337630931
Author: KASSIMALI, Aslam.
Publisher: Cengage,
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Why is there two values of 1.5? Where did the other one came from?
Transcribed Image Text:153.7
RAx ↑
RAY
152,09N T
划g N
12036N
Now usirg Fore and Momert
equilibruum equations
At suffert A end B t
EFX =0 RAX +4ax3 -120.36_132.ng TRey
B
EFyzo RAy + Rey -159.72+ 111.09
fis
EMA =° (momenf at A=o)
RBY X 30 + 141•09X7 – 4aX 3X(1s+1-5)– 159.79X4
RB4X {o =71.63 + 378 F 630.88
RBY X 1° = 239.25
RBy = 23.925N
%3D
how using equation Li) →
RA
152.09 + 120-36 – 126
%3D
= 147.26
using equntion lüv -
Ray + Roy = 159.72 – 311. 69
%3D
Ray
= 4P.63- RB
Ray = 48.63-23, §25
KBY= 23-9 5N)
%3D
(Ray = 24r705 N
Transcribed Image Text:Through your solutions, prove that the support reactions are similar to the values shown below.
Rb = 23.927 N, Upward
Ray = 24.709 N, Upward
Rax = 147.267 N, Left
200 N
53
36
1.5
189 N
42 kN/m
3
B
4
3
3
Note: Uniform distributed load = 42 N/m, if you believe that the correct answer/s is/are not included
above, justify your answer.
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