1. ANALYSIS OF STRUCTURES
CHAPTER 6
ANALYSIS OF STRUCTURES

2. Trusses Analysis of Structures Frames Machines Method of Sections
Method of Joints
Trusses
Analysis of
Structures
Frames
Machines

3. 6.1 Introduction
Trusses
6.2 Definition of truss
6.3 Simple trusses
6.4 Analysis of trusses by the method of joints
6.5 Joints under special loading conditions
6.7 Analysis of trusses by the method of sections
6.10 Analysis of frame
6.11 Frames which cease to be rigid when detached from
their supports
6.12 Machines

4. Three categories of engineering structures are considered:
6.1 Introduction
For the equilibrium of structures made of several connected parts, the internal forces as well the external forces are considered.
In the interaction between connected parts, Newton’s 3rd Law states that the forces of action and reaction between bodies in contact have the same magnitude, same line of action, and opposite sense.
Three categories of engineering structures are considered:
Frames: contain at least one one multi-force member, i.e., member acted upon by 3 or more forces.
Trusses: formed from two-force members, i.e., straight members with end point connections
Machines: structures containing moving parts designed to transmit and modify forces.

5. 6.2 Definition of a Truss
A truss consists of straight members connected at joints. No member is continuous through a joint.
Most structures are made of several trusses joined together to form a space framework. Each truss carries those loads which act in its plane and may be treated as a two-dimensional structure.
Bolted or welded connections are assumed to be pinned together. Forces acting at the member ends reduce to a single force and no couple. Only two-force members are considered.
When forces tend to pull the member apart, it is in tension. When the forces tend to compress the member, it is in compression.

6. Force 500 N pulling the structure to the right, therefore results in tension in member AB, compression in member BC B
500 N
Beam in tension
Beam compressed A C

7. At joint B 500 N 500 N B B FBC (comp) 45 45 FBC (comp) FBA(tens)

8. Members of a truss are slender and not capable of supporting large lateral loads. Loads must be applied at the joints.

9. A number of typical trusses

10. A rigid truss will not collapse under the application of a load.
6.3 Simple Trusses
A rigid truss will not collapse under the application of a load.
A simple truss is constructed by successively adding two members and one connection to the basic triangular truss.
In a simple truss, m = 2n - 3 where m is the total number of members and n is the number of joints.

11. 6.4 Analysis of trusses by the method of joints
Dismember the truss and create a freebody diagram for each member and pin.
The two forces exerted on each member are equal, have the same line of action, and opposite sense.
Forces exerted by a member on the pins or joints at its ends are directed along the member and equal and opposite.
Conditions of equilibrium on the pins provide 2n equations for 2n unknowns. For a simple truss, 2n = m May solve for m member forces and 3 reaction forces at the supports.
Conditions for equilibrium for the entire truss provide 3 additional equations which are not independent of the pin equations.

12. 6.5 Joints under Special Loading Conditions
Forces in opposite members intersecting in two straight lines at a joint are equal.
The forces in two opposite members are equal when a load is aligned with a third member. The third member force is equal to the load (including zero load).
The forces in two members connected at a joint are equal if the members are aligned and zero otherwise.
Recognition of joints under special loading conditions simplifies a truss analysis.

13. Example (a) (b) Fx = + (300N) cos α Fy = - (300N) sin α
A man pulls with a force of 300N on a rope attached
to a building as shown in the picture.
What are the horizontal and vertical components of the
force exerted by the rope at point A?
Fx = + (300N) cos α
Fy = - (300N) sin α
Observing that AB = 10m
cos α = 8m = 8m = 4
AB m
sin α = 6m = 6m = 3
(a)
We thus obtain
Fx = + (300N) = +240 N 4 5
Fy = - (300N) = -180 N 3 5
We write
F = + (240N) i - (180 N) j
tan θ= Fx Fy
F =
(b)

14. Problems 2.23
Determine the x and y components of each of the forces shown.
1.2 m
1.4 m
2.3 m
1.5 m
2.0 m
1800 N
950N
900N

15. Solution
2.59m
2.69m
2.5 m

16. Solution
2.59m
2.69m
2.5 m

17. Using the method of joints, determine the force in each member of the
Sample problem 6.1
9.0 kN
4.5 kN
3.6 m
2.4 m
1.8 m
Using the method of joints, determine the force in each member of the
truss shown.

18. 9.0 kN 4.5 kN E = 45 kN Cy = 31.5 kN Solution 3.6 m 2.4 m 1.8 m
Free body : Entire truss . A free-body diagram of the entire truss is drawn. external forces acting on this free body
consist of the applied loads and the reactions at C and E. We write the following equilibrium equations.
E = 45 kN
Cy = 31.5 kN

19. 9.0 kN
Free body : Joint A . This joint is subjected to only 2 unknown forces, namely, the forces exerted by members AB and AD.
A force triangle is used to determine FAB and FAD. We note that member AB pulls on the joint and thus is in tension and
that member AD pushes on the joint and thus is in compression.
The magnitudes of the 2 forces are obtained from the proportion
FAB = 6.75 kN
FAD = kN

20. FDB = FDA FDB = 11.25 kN FDE = 2 FDA 3 5 FDE = 13.5 kN 11.25 kN
Free body : Joint D . Since the force exerted by member AD has been determine, only 2 unknown forces are now
involved at this joint. Again, a force triangle is used to determine the unknown forces in members DB and DE
FDB = FDA
FDE = FDA 3 5
FDB = kN
FDE = 13.5 kN

23. Summing y components,
we obtain a check of our computations:

25. PROBLEMS

26. Problem 6.1
Using the method of joints, determine the force in each member of the
trust shown. State whether each member is in tension or compression.

27. Joint FBD’S By Bx Cy
315 N A C 4m
1.25m

28. C = Compression T = Tension 315 N A Bx C By Cy FAB B FBC By = 75N FAC
Joint B
Joint C
FAC
Cy = 240 N C
240 N
100 N 3 4
C = Compression
T = Tension

29. Using the method of joints, determine the force in each member of the
Problem 6.2
1.2 m
4.2 m
2.25 m
25 kN
Using the method of joints, determine the force in each member of the
trust shown. State whether each member is in tension or compression.

30. B A C
4.2 m Ay Ax Cx
2.25 m
1.2 m
25 kN
1.2 m
4.2 m
2.25 m
Roller

31. FAC FBC FBC FAC CX = 13.39 kN Ay Ax A FAB FAB FAC 2.25 m Ay Ax A 1.2 m
Joint C
FAC
FBC
FBC 4
FAC 3 4 C 3
CX = kN
13.39 kN
Joint A Ay
= 25 kN Ax A
7.5 kN 4 4
= kN
7.5
FAB
FAB
13.39 kN
FAC
= kN

32. Using the method of joints, determine the force in each member of the
Problem 6.3
2.4 m
27 kN
1.8 m
2.7 m
Using the method of joints, determine the force in each member of the
trust shown. State whether each member is in tension or compression.

33. 2.4 m
1.8 m A B C
27 kN
2.7 m By Cy Cx
2.4 m
1.8 m
2.7 m
27kN

34. 2.4 m 1.8 m 2.7 m 27kN Joint C 2.4 m A B C 27 kN By Cy Cx 18kN FAC FAC15
FAC 8 8 C
18kN
FBC 15
FBC
18kN

35. 2.4 m A B C 27 kN By Cy Cx 18kN 2.4 m 1.8 m 2.7 m Joint B
FBC = kN
FAB
45 kN B
FAB = kN
45 kN 4 3
33.75 kN

36. Problem 6.9
Determine the force in each member of the Gambrel roof truss shown.
State whether each member is in tension or compression.

37. 3 kN
6 kN Hy Hx A
3 kN
FAB 3 A 4
FAC
12 kN

38. 3 kN
6 kN Hy Hx A
FBD 3 B 10 3 3 4 4
FBE
15kN

39. 3 kN
6 kN Hy Hx A
6kN D
FDF
11.93 kN
FDE

40. Determine the force in each member of the Howe roof truss shown.
Problem 6.10
6 kN
6 kN
6 kN
3 kN
3 kN
Determine the force in each member of the Howe roof truss shown.
State whether each member is in tension or compression.

41. Hy Hx
3 kN
FAB A
FAC
12 kN
FAC
9 kN
FAB

42. Hy Hx
FBC
FCE
12 kN c

43. Hy Hx
6 kN α
FBD B β
FBE
15 kN

44. Hy Hx
6 kN
10 kN 3 3
10 kN 4 4
FDE

45. Problem 6.15
Determine the force in each member of the Pratt bridge truss shown.
State whether each member is in tension or compression.

46. Ax Ay B D F C E G H A 3m 3m 3m 3m

47. Ax Ay
FAB 4 3 A
FAC
9 kN

48. Ax Ay
FBC C
6.75 kN
FCE
6 kN

49. Ax Ay
FBD
FBE B
6 kN
11.25 kN 4 3

50. THE END