CIVE1187 Statics
Week 8 Tutorial
Truss Analysis - Section Method
Jie Li
Jie Li 1
The Method of Sections
• Used to determine the loadings within a body
• If a body is in equilibrium, any part of the body is in
equilibrium
• To find forces within members, an imaginary section is
used to cut each member into 2 and expose each
internal force as external
The Method of Sections
Procedure for Analysis
Free-Body Diagram
• Decide the section of the truss
• Determine the truss’s external reactions
• Use equilibrium equations to solve member forces at
the cut session
• Draw FBD of the sectioned truss which has the least
number of forces acting on it
• Find the sense of an unknown member force
Equations of Equilibrium
• Summed moments about a point
• Find the 3rd unknown force from moment equation
Q1
Determine the force in members GE, GC, and BC of the truss.
Indicate whether the members are in tension or compression.
Ax
Ay Dy
Draw FBD of the entire truss
Fx 0; 400 N Ax 0 Ax 400 N
M A 0; 1200 N (8m) 400 N (3m) D y (12m) 0 D y 900 N
Fy 0; Ay 1200 N 900 N 0 Ay 300 N
Choose section a-a since it cuts through the 3 members
5m
sin = 3/5
Draw FBD for the section portion
M G 0; 300 N (4m ) 400 N (3m) FBC (3m) 0 FBC 800 N (T )
M C 0; 300 N (8m ) FGE (3m ) 0 FGE 800 N (C )
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Fy 0; 300 N FGC 0 FGC 500 N (T )
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Q2 Determine the force in members BC, HC, and HG of
the bridge truss, and indicate whether the members are in
tension or compression.
FBD
Jie Li 6
Support Reactions :
E
ME = 0
18(3) + 14(6) + 12(9) – Ay(12) = 0
Ay = 20.5 kN
MC = 0 -FHG(3) + 12(3) - 20.5(6) = 0
FHG = - 29.0 kN (C) H FHG
MH = 0 3m
FBC(3) – 20.5(3) = 0 45o
FHC
FBC = 20.5 kN (T)
FBC C
Fy = 0; 20.5 – 12 – FHC sin 45° = 0 3m 3m
12 kN
FHC = 12.0 kN (T) Ay=20.5 kN
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Q3 Determine the force in members EI and JI of the truss
which serves to support the deck of a bridge. State if these
members are in tension or compression.
Jie Li 8
FEF
FEI
Gy
FJI
+ MA = 0 - 40(3) - 80(3x2)- 50(3x5) + Gy(3x6) = 0
Gy = 75 kN
+ ME = 0 –50(3) + 75(6) – FJI(4) = 0
FJI = 75 kN (T)
+↑ Fy = 0; 75 - 50 + FEI = 0
FEI = - 25 kN (C ) 9
Q4 Determine the force in members BC, CG, and GF of
the Warren truss. Indicate if the members are in tension or
compression.
Ax
Ay Ey
+ Fx = 0; Ax= 0
+ ME = 0 6(6) + 8(3) – Ay(9) = 0, Ay = 6.667 kN
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+ MC = 0
FGF(3 sin 60°) + 6(1.5) – 6.667(4.5) = 0
FGF = 8.08 kN (T)
+ MG = 0
- FBC(3 sin 60°) – 6.667(3) = 0
FBC = - 7.70 kN (C)
+↑ Fy = 0; 6.667 – 6 + FCG sin 60° = 0
FCG = - 0.770 kN (C)
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Q5 Determine the force in members IC and CG of the truss
and state if these members are in tension or compression.
Also, indicate all zero-force members.
AB, BC, CD, DE, HI,
and GI are all zero-
force members.
+ Fx = 0; Ax= 0
Ax
+ ME = 0
6(1.5) + 6(3) – Ay(1.5x4) = 0
Ay
Ay = 4.5 kN
Jie Li 12
3m FIC cos()
FBC = 0
sin = 3/5 B FIC sin()
5m cos = 4/5 C
FIC
4m I
FIG=0
A
FHG G
Ay= 4.5 kN 6 kN
+ MG = 0 - 4.5 (3) - FIC (3/5)(4) = 0 FIC = - 5.62 kN (C)
Joint C + Fx = 0; 5.62 sin + FCJ sin = 0
FCJ = - 5.62 kN (C)
0 0
+↑ Fy = 0; (4/5)5.625 + (4/5)5.625 - FCG = 0
5.62 kN FCJ
FCG FCG = 9 kN (T)
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