why in the slope deflection method of step 1 is ic from (3) for MDC negative shouldn't it be positive? im just trying to understand the concept.
why in the slope deflection method of step 1 is ic from (3) for MDC negative shouldn't it be positive? im just trying to understand the concept.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Question
why in the slope deflection method of step 1 is ic from (3) for MDC negative shouldn't it be positive?
im just trying to understand the concept.
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how does this relate to the original question, even in the answer it shows in this case for MAB the theta B is positive. but in the question I asked earlier it showed this to be a negative theta in the base equation.
![To draw a free body diagram of the given figure:
40 kN
Slope deflection method:
- 40×4
8
MAC
=
WR
8
=
A
MCA
McD = MDc=0
wl
MDF
8
MFD = 30 kN
MAC =0
MCA = -MCA-
20 3EI
MCA = 20+ + .(id)
20 kN
33
4
MDc=0+ -
-60 × 4
8
(Eli)
2EI
MCA = 30 +
McD = MCD
2EI
McD = 0+ -(2ic+id)
4
2EI
2 m
B
- 20 kN
MAC 3EI
+ -(ic
2
2EI
MDC=MDC + (2id-ic)
I
(2id-ic)
- 30 kN
-후)
(2ic+ig-30)
2 m
....(a)
.....(b)
.....(c)
4 m
D
2m
60 KN
E
2 m
F](https://content.bartleby.com/qna-images/question/2fcb0ac2-c365-43e1-ab5c-a271b4af4a10/9afa4581-c752-409e-a0de-ae4fb092cacb/dikn4fi_thumbnail.png)
Transcribed Image Text:To draw a free body diagram of the given figure:
40 kN
Slope deflection method:
- 40×4
8
MAC
=
WR
8
=
A
MCA
McD = MDc=0
wl
MDF
8
MFD = 30 kN
MAC =0
MCA = -MCA-
20 3EI
MCA = 20+ + .(id)
20 kN
33
4
MDc=0+ -
-60 × 4
8
(Eli)
2EI
MCA = 30 +
McD = MCD
2EI
McD = 0+ -(2ic+id)
4
2EI
2 m
B
- 20 kN
MAC 3EI
+ -(ic
2
2EI
MDC=MDC + (2id-ic)
I
(2id-ic)
- 30 kN
-후)
(2ic+ig-30)
2 m
....(a)
.....(b)
.....(c)
4 m
D
2m
60 KN
E
2 m
F
Solution
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