W= 9kN/m P= 60 kN a= 4m b=2m ++ 4b 4P P D E 9 P 25 2b F + a 2b Question 1 W = 9 kN/m P= 60.KN a=4m b = 2m i) Reactions at support A and B. Vertical force: Fy ot Ray + Roy 4P-46W= 0 RAY+RBY = 4P + 46W RAY + 46y = 4×60 + 4×2×9 RAY+RBY = 312 KN EM@AOA (4Pxa) + 4bWx (a + b/2) - 2a RBy = 0 960 + 72 x (4 + 2/2) - 8RBY = 0 -8RBY =- 1320 RBY = 165 kN upwards ↑ EM@BOA 4pa + 45W x (a + b/2) - 2a RBy = 0 2aRBY = 4bwx (a+b/2) + 4 Pa RBY = 4bwx (a+b/2) + 4 pa 29 RBy = bw (1+ b/a)+2p -Sub into RAY + RBY = 312 RAY + bw (1 +5/a) +2p₁ = 4P + 4bW RAY = 2P - bw x1-b/a) = RAY 111 KN

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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For the three-pinned frame shown in picture i am trying to detemine the: 

1. Determine the reactions at supports A and B. 

2. Draw the shear force diagram for member AC. 

3. Draw the bending moment diagram for member BFE. 

4. Draw the axial force diagram for member CD. 

5. Determine the maximum bending moment in member AC. 

Detail answer with actual  calculations and drawings please for better comprehension. I have tried to attempt this question multiple times, however if seems to be incorrect. I will upvote, if answer is detailed Thank you 

W= 9kN/m
P= 60 kN
a= 4m
b=2m
++
4b
4P
P
D
E
9
P
25
2b
F +
a
2b
Transcribed Image Text:W= 9kN/m P= 60 kN a= 4m b=2m ++ 4b 4P P D E 9 P 25 2b F + a 2b
Question 1
W = 9 kN/m
P= 60.KN
a=4m b = 2m
i) Reactions at support A and B.
Vertical force:
Fy ot
Ray + Roy
4P-46W= 0
RAY+RBY = 4P + 46W
RAY
+ 46y = 4×60 + 4×2×9
RAY+RBY
=
312 KN
EM@AOA
(4Pxa) + 4bWx (a + b/2) - 2a RBy = 0
960 + 72 x (4 + 2/2) - 8RBY = 0
-8RBY
=- 1320
RBY =
165 kN upwards ↑
EM@BOA
4pa + 45W x (a + b/2) - 2a RBy = 0
2aRBY
= 4bwx (a+b/2) + 4 Pa
RBY
=
4bwx (a+b/2) + 4 pa
29
RBy = bw (1+ b/a)+2p
-Sub into RAY + RBY = 312
RAY + bw (1 +5/a) +2p₁ = 4P + 4bW
RAY
=
2P
-
bw x1-b/a)
=
RAY 111 KN
Transcribed Image Text:Question 1 W = 9 kN/m P= 60.KN a=4m b = 2m i) Reactions at support A and B. Vertical force: Fy ot Ray + Roy 4P-46W= 0 RAY+RBY = 4P + 46W RAY + 46y = 4×60 + 4×2×9 RAY+RBY = 312 KN EM@AOA (4Pxa) + 4bWx (a + b/2) - 2a RBy = 0 960 + 72 x (4 + 2/2) - 8RBY = 0 -8RBY =- 1320 RBY = 165 kN upwards ↑ EM@BOA 4pa + 45W x (a + b/2) - 2a RBy = 0 2aRBY = 4bwx (a+b/2) + 4 Pa RBY = 4bwx (a+b/2) + 4 pa 29 RBy = bw (1+ b/a)+2p -Sub into RAY + RBY = 312 RAY + bw (1 +5/a) +2p₁ = 4P + 4bW RAY = 2P - bw x1-b/a) = RAY 111 KN
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