Calculate the force in members BC, CH, and GH by method of sections 8000 kN A₂ = 0 A B H 6m A,= 7500KN FBC 6m FCH FGH G 8m = 0 ΣFx = 6 FGH + FCH+ FBC = 0 10 = 0 ΣFy = Ay + 8 10 FCH 80000 8 10 FCH = 625 kN (T) 7500+ FCH - 8000 = 0 + Mc = 0 -Ay(12)+8000 (6)+FGH (8) = 0 = 0 -7500 (12)+8000 (6)+FGH (8) FGH = 5250 kN (T) 6 5250 + 625+ FBC=0 10 FBC = -5625 kN FBC = 5625 kN (C)

Structural Analysis
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Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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QUESTION: Can you please explain where did the "10" from 8/10 and 6/10 came from?

Calculate the force in members BC, CH, and GH by method of sections
8000 KN
Ax = 0
6m
A, = 7500KN
FBC
6m
FCH
FGH
G
8m
ΣFx
= 0
FGH + FCH+ FBC = 0
10
ΣF, = 0
8
10
Ay +
FCH-8000=0
8
10
FCH = 625 kN (T)
7500+ FCH-8000=0
+G ΣMc = 0
-Ay(12)+8000 (6)+FGH (8)
-7500 (12)+8000 (6)+FGH (8)
FGH
5250 +
=
10
FBC
FBC
5250 kN (T)
=
= -5625 kN
= 0
625+ FBC = 0
5625 kN (C)
= 0
Transcribed Image Text:Calculate the force in members BC, CH, and GH by method of sections 8000 KN Ax = 0 6m A, = 7500KN FBC 6m FCH FGH G 8m ΣFx = 0 FGH + FCH+ FBC = 0 10 ΣF, = 0 8 10 Ay + FCH-8000=0 8 10 FCH = 625 kN (T) 7500+ FCH-8000=0 +G ΣMc = 0 -Ay(12)+8000 (6)+FGH (8) -7500 (12)+8000 (6)+FGH (8) FGH 5250 + = 10 FBC FBC 5250 kN (T) = = -5625 kN = 0 625+ FBC = 0 5625 kN (C) = 0
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