B W D 2a a За 4a W = 30 KN. RAX. 4 P: 30 KN. a = um 30KN P с 4m; : . B 30 KN 4m 6m 2m -RAY ≤ FX = 0 + - RAX +30=0 RAX SFY = O = - 30 KN RBY Diagonal √√(2)²+(4)2 = 4.47 m =2√5 Angles of diagonal member Slope ya = 2 7 2a tan 0 = 2 ≤ M @ c = 0 + VASO tan-1 (2) RAY (6) (6) + 30 (4)"= RAX 30 (4) GRAY 6 Total SFy = 30 + + W 30 - RAY = Roy 30 ( 4 ) = 0 120 = = 20 kN a + 9V- a 30 7 20 = RCY AONE RcY EM @ A = D = X4.02 -RBY (6) - 30 (2) + 30 (4) 99207 729N8 MU 1-8 = 0.02 mp.s - RBY (6) = -60 •BBY = PRO KN
B W D 2a a За 4a W = 30 KN. RAX. 4 P: 30 KN. a = um 30KN P с 4m; : . B 30 KN 4m 6m 2m -RAY ≤ FX = 0 + - RAX +30=0 RAX SFY = O = - 30 KN RBY Diagonal √√(2)²+(4)2 = 4.47 m =2√5 Angles of diagonal member Slope ya = 2 7 2a tan 0 = 2 ≤ M @ c = 0 + VASO tan-1 (2) RAY (6) (6) + 30 (4)"= RAX 30 (4) GRAY 6 Total SFy = 30 + + W 30 - RAY = Roy 30 ( 4 ) = 0 120 = = 20 kN a + 9V- a 30 7 20 = RCY AONE RcY EM @ A = D = X4.02 -RBY (6) - 30 (2) + 30 (4) 99207 729N8 MU 1-8 = 0.02 mp.s - RBY (6) = -60 •BBY = PRO KN
Residential Construction Academy: House Wiring (MindTap Course List)
4th Edition
ISBN:9781285852225
Author:Gregory W Fletcher
Publisher:Gregory W Fletcher
Chapter19: Service Panel Trim-out
Section: Chapter Questions
Problem 8RQ
Related questions
Question
The pin-jointed plane frame (truss) in the picture below use the method of virtual work to determine the vertical deflection at joint D and the horizontal deflection at joint B. All members of the truss have a constant EA of 1×105kN.
I have attempted the first half of the question however struggling at the second part. Detail answer with actual working out of the joint section AC,BD, CD and the deflection table of each member for better comprehension. I do not need written do list but the calculation pls!
I will upvote, if answer is detailed (calculations) Thank you
random picks for loads and dimensions:W = 30 kN P= 30 kN a= 1m
Transcribed Image Text:B
W
D
2a
a
За
4a
Transcribed Image Text:W = 30 KN.
RAX.
4
P: 30 KN.
a = um
30KN P
с
4m;
:
.
B
30 KN
4m
6m
2m
-RAY
≤ FX = 0 +
-
RAX +30=0
RAX
SFY
= O
=
- 30 KN
RBY
Diagonal
√√(2)²+(4)2
= 4.47 m
=2√5
Angles of diagonal member
Slope ya = 2
7
2a
tan 0 = 2
≤ M @ c = 0 + VASO tan-1 (2)
RAY (6)
(6) + 30 (4)"=
RAX
30 (4)
GRAY
6
Total SFy = 30 + +
W
30 - RAY = Roy
30 ( 4 ) = 0
120
=
= 20 kN a +
9V-
a
30
7
20 = RCY
AONE RcY
EM @ A = D
=
X4.02
-RBY (6) - 30 (2) + 30 (4)
99207 729N8 MU
1-8
= 0.02
mp.s - RBY (6) = -60
•BBY = PRO KN
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