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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.

Structural Analysis
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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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
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
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