1. Truss: Method of Joints and Sections
Theory of Structure - I

2. Lecture Outlines Method of Joints Zero Force Members
Method of Sections
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

3. Method of Joints
If a truss is in equilibrium, then each of its joints must also be in equilibrium
The method of joints consists of satisfying the equilibrium conditions for the forces exerted “on the pin” at each joint of the truss
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

4. Truss members are all straight two-force members lying in the same plane
The force system acting at each pin is coplanar and concurrent (intersecting)
Rotational or moment equilibrium is automatically satisfied at the joint, only need to satisfy ∑ Fx = 0, ∑ Fy = 0
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

5. Procedure for Analysis
Draw the free-body diagram of a joint having at least one known force and at most two unknown forces (may need to first determine external reactions at the truss supports)
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

6. Establish the sense of the unknown forces
Always assume the unknown member forces acting on the joint’s free-body diagram to be in tension (pulling on the “pin”)
Assume what is believed to be the correct sense of an unknown member force
In both cases a negative value indicates that the sense chosen must be reversed
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

7. Orient the x and y axes such that the forces can be easily resolved into their x and y components
Apply ∑ Fx = 0 and ∑ Fy = 0 and solve for the unknown member forces and verify their correct sense
Continue to analyze each of the other joints, choosing ones having at most two unknowns and at least one known force
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

8. Members in compression “push” on the joint and members in tension “pull” on the joint
Mechanics of Materials and building codes are used to size the members once the forces are known
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

9. The Method of Joints 2 m 500 N 45o A B C Ax = 500 N Ay = 500 N
Cy = 500 N
500 N B
45o
Joint B
SFx = 0: + x y
500 - FBCsin45o = 0
FBC = N (C)
FBA
FBC
SFy = 0: +
- FBA + FBCcos45o = 0
FBA = 500 N (T)
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

10. B 500 N 2 m 45o A C Ax = 500 N Ay = 500 N Cy = 500 N Joint A +
SFx = 0: +
500 - FAC = 0
FAC = 500 N (T)
FAC
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

11. Zero Force Members
Truss analysis using the method of joints is greatly simplified if one is able to determine those members which support no loading (zero-force members)
These zero-force members are used to increase stability of the truss during construction and to provide support if the applied loading is changed
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

12. If only two members form a truss joint and no external load or support reaction is applied to the joint, the members must be zero-force members.
If three members form a truss for which two of the members are collinear, the third member is a zero-force member provided no external force or support reaction is applied.
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

13. Zero-force members
Frequently the analysis can be simplified by identifying members that carry no load
two typical cases are found
When only two members form a non-collinear joint and there is no external force or reaction at that joint, then both members must be zero-force
If either TCB or TCD ≠ 0, then C cannot be in equilibrium, since there is no restoring force towards the right.
Hence both BC and CD are zero-load members here. P A B C D E
TCB
TCD C
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

14. When three members form a truss joint for which two members are collinear and the third is at an angle to these, then this third member must be zero-force
in the absence of an external force or reaction from a support P
Here, joint B has only one force in the vertical direction.
Hence, this force must be zero or B would move (provided there are no external loads/reactions)
Also TAB = TBC A B C D
TBC
TAB
TBD B E
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

15. While zero-force members can be removed in this configuration, care should be taken
any change in the loading can lead to the member carrying a load
the stability of the truss can be degraded by removing the zero-force member P
You may think that we can remove AD and BD to make a triangle …
This satisfies the statics requirements
However, this leaves a long CE member to carry a compressive load. This long member is highly susceptible to failure by buckling. A B C D E
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

16. P A B C D E Dx Ey Dy C FCD FCB SFx = 0: FCB = 0 + SFy = 0: FCD = 0 +
FAE
FAB
SFy = 0: FABsinq = 0, FAB = 0 + A q
SFx = 0: FAE + 0 = 0, FAE = 0 +
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

17. Example 3-4
Using the method of joints, indicate all the members of the truss shown in the figure below that have zero force. P A B C D E F G H
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

18. SOLUTION P A B C D E F G H Ax Gx Joint D SFy = 0: + x y FDC FDE
Joint D
SFy = 0: + x y
FDC
FDE
FDCsinq = 0, FDC = 0 D q
SFx = 0: +
FDE + 0 = 0, FDE = 0
FEC
FEF
SFx = 0: +
Joint E P E
FEF = 0
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

19. P A B C D E F G H Ax Gx x y Joint H SFy = 0: + FHF FHA FHB H FHB = 0Gx x y
Joint H
SFy = 0: +
FHF
FHA
FHB H
FHB = 0
FGF
FGA
Joint G
SFy = 0: + G Gx
FGA = 0
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

20. Method of Sections
Based on the principle that if a body is in equilibrium, then any part of the body is also in equilibrium
Procedure for analysis
Section or “cut” the truss through the members where the forces are to be determined
Before isolating the appropriate section, it may be necessary to determine the truss’s external reactions (then 3 equs. of equilibrium can be used to solve for unknown member forces in the section).
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

21. Establish the sense of the unknown member forces
Draw the free-body diagram of that part of the sectioned truss that has the least number of forces acting on it
Establish the sense of the unknown member forces
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

22. Apply 3 equations of equilibrium trying to avoid equations that need to be solved simultaneously
Moments should be summed about a point that lies at the intersection of the lines of action of two unknown forces
If two unknown forces are parallel – sum forces perpendicular to the direction of these unknowns
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

23. The Method of Sections Ex Dx Dy a 100 N A B C D E F G 2 m + SMG = 0:
FGF
FGC
FBC
100 N A B G
2 m
45o
100(2) - FBC(2) = 0
FBC = 100 N (T) C
SFy = 0: +
FGCsin45o = 0
FGC = N (T) +
SMC = 0:
100(4) - FGF(2) = 0
FGF = 200 N (C)
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

24. Example 3-6
Determine the force in members GF and GD of the truss shown in the figure below. State whether the members are in tension or compression. The reactions at the supports have been calculated.
6 kN
8 kN
2 kN
Ax = 0
Ay = 9 kN
Ey = 7 kN A B C D E F G H
3 m
4.5 m
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

25. 6 kN 8 kN 2 kN Ax = 0 Ay = 9 kN Ey = 7 kN A B C D E F G H 3 m 4.5 m
SOLUTION a
2 kN
Ey = 7 kN D E F
Section a-a
3 m
FFG
FDG
FDC
26.6o
56.3o +
SMD = 0:
3 m
26.6o O
FFGsin26.6o(3.6) + 7(3) = 0,
FFG = kN (C) +
SMO = 0:
- 7(3) + 2(6) + FDGsin56.3o(6) = 0,
FDG = 1.80 kN (C)
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

26. Example 3-7
Determine the force in members BC and MC of the K-truss shown in the figure below. State whether the members are in tension or compression. The reactions at the supports have been calculated. A B C D E F G H I J K L M N O P
5.34 kN
6.67 kN
8 kN
7.11 kN
12.9 kN
3 m
4.6 m
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

27. a SOLUTION A B C D E F G H I J K L M N O P 5.34 kN 6.67 kN 8 kN
3 m
4.6 m
Section a-a A B L
5.34 kN
12.9 kN
4.6 m
6 m
FLK
FBM
FLM
FBC +
SML = 0:
FBC(6) (4.6) = 0,
FBC = 9.89 kN (T)
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

28. -FCMcos33.1o(6) - 9.89(6) - 8(4.6) + 7.11(18.4) = 0 FCM = 6.90 kN (T)b A B C D E F G H I J K L M N O P
5.34 kN
6.67 kN
8 kN
7.11 kN
12.9 kN
3 m
4.6 m
FKL
FKM
FCM C D E F G H I J K N O P
6.67 kN
8 kN
7.11 kN
3 m
4.6 m
9.89 kN
33.1o +
SMK = 0:
-FCMcos33.1o(6) (6) - 8(4.6) (18.4) = 0
FCM = 6.90 kN (T)
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

29. Practice Problems Book Examples and exercise questions related to
Structural Analysis by R.C. Hibbeler
Chapter no. 03
Examples and exercise questions related to
Joint Method
Zero Force Members
Section Method
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

30. Thank You Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan