1. What is a Truss?
A structure composed of members connected together to form a rigid framework.
Usually composed of interconnected triangles.
Members carry load in tension or compression.

2. Component Parts
Support (Abutment)

3. Standard Truss Configurations

4. Types of Structural Members
These shapes are called
cross-sections.

5. Types of Truss Connections
Pinned
Connection
Gusset Plate
Connection
Most modern bridges use gusset plate connections

6. Forces, Loads, & Reactions
Force – A push or pull.
Load – A force applied to a structure.
Reaction – A force developed at the support of a structure to keep that structure in equilibrium.
Self-weight of structure, weight of vehicles, pedestrians, snow, wind, etc.
Forces are represented mathematically as
VECTORS.

7. Equilibrium A Load... ...and Reactions Newton’s First Law:
An object at rest will remain at rest, provided it is not acted upon by an unbalanced force.
A Load...
...and Reactions

8. Tension and Compression
An unloaded member experiences no deformation
Tension causes a member to get longer
Compression causes a member to shorten

9. Tension and Compression
EXTERNAL FORCES and INTERNAL FORCES
Must be in equilibrium with each other.

10. Structural Analysis
For a given load, find the internal forces (tension and compression) in all members.
Why?
Procedure:
Model the structure:
Define supports
Define loads
Draw a free body diagram.
Calculate reactions.
Calculate internal forces using “Method of Joints.”

11. Model the Structure
15 cm D A B C
mass=5 kg
=2.5 kg per truss

12. Draw a Free Body Diagram
15 cm D A B C x y RA RC
24.5N
mass=2.5 kg

13. Calculate Reactions Total downward force is 24.5 N.
Total upward force must be 24.5 N.
Loads, structure, and reactions are all symmetrical.
RA and RC must be equal.

14. Calculate Reactions
15 cm
15 cm D
15 cm A B C y
12.25 N RA RC
24.5 N x

15. Method of Joints Isolate a Joint. 15 cm C B D RC 24.5 N 12.25 N A yx

16. Method of Joints Isolate a Joint.
Draw a free body diagram of the joint.
Include any external loads of reactions applied at the joint.
Include unknown internal forces at every point where a member was cut.
Assume unknown forces in tension.
Solve the Equations of Equilibrium for the Joint.
FAD A x y
FAB
12.25 N
EXTERNAL FORCES and INTERNAL FORCES
Must be in equilibrium with each other.

17. Equations of Equilibrium
The sum of all forces acting in the x-direction must equal zero.
The sum of all forces acting in the y-direction must equal zero.
For forces that act in a diagonal direction, we must consider both the x-component and the y-component of the force.
12.25 N A x y
FAD
FAB

18. Components of Force A (FAD)y (FAD)x q FAD q A
If magnitude of FAD is represented as the hypotenuse of a right triangle...
Then the magnitudes of (FAD)x and (FAD)y are represented by the lengths of the sides.

19. Trigonometry Review x y q
Definitions: H
Therefore:

20. Components of Force y
(FAD)y
FAD
q=?
q=?
45o x A A
(FAD)x
Therefore:

21. Equations of Equilibrium
0.707 FAD
FAD A x y
FAB
12.3 N ?
FAB=12.25 N (tension)
FAD=17.3 N (compression)

22. Method of Joints...Again Isolate another Joint. 12.25 N A 15 cm C D RCB
24.5 N y x

23. Equations of Equilibrium
FBD
FBC=12.25 N (tension)
FAB
FBC B x y
24.5 N
FBD=24.5 N (tension)

24. Results of Structural AnalysisD B
24.5 N
12.25 N (T)
24.5 N (T)
17.3 N (C)
Do these results make sense?

25. Results of Structural AnalysisD B
24.5 N
12.25 N (T)
24.5 N (T)
17.3 N (C)
In our model, what kind of members are used for tension? for compression?