1. Math for Truss Calculation
Principles of EngineeringTM
Lesson Statics
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Copyright 2007

2. Math for Truss Calculation

3. Trusses
Trusses are structures made up of beams joined together at their endpoints.
Solve for the force in each member of the truss in this example to find whether the members are in tension or compression.
Start be replacing the supports with reaction forces:
500 N AX CY AY
500 N B A C 2m
45

4. Trusses This truss consists of: 500 N AX CY AY 500 N AX CY AY
2 unknown member forces and 1 unknown external force at joint B
2 unknown member forces and 2 unknown reaction forces at joint A
2 unknown member forces and 1 unknown reaction force at joint C
Draw the free-body diagrams for each pin:
500 N AX CY AY
500 N AX CY AY

5. Trusses These three separate free-body diagrams assume that:
Member AB is in tension
Member BC is in compression
Member AC is in tension
If these assumptions are incorrect, then our solution will show a negative quantity for force.
500 N AX CY AY
500 N AX CY AY
FBC
FAB
FAC

6. Trusses To really “see” that
Member AB is in tension
Member BC is in compression
Member AC is in tension
We must look at the free-body diagrams of the beams, which show the effects of the pins on the beams.

7. Trusses
Before we can solve for the forces, we must break FBC into its x and y components:
FBC
FBCX
FBCY
45
FBCX = FBC * cos (45)
FBCY = FBC * sin (45)

8. Trusses + FX = 0 ; 500N – FBCcos(45) = 0
FBCcos(45) = 500N
cos(45) cos(45)
FBC = N (C)
+FY = 0 ;
FBCsin(45) – FAB = 0
707.1*sin(45) – FAB = 0
FAB = 707.1*sin(45)
FAB = 500 N (T)
The following is the free-body diagram of joint B. The force FBC has been replaced with its x and y components:
500 N
FBC cos(45)
FAB
FBC sin(45)

9. Trusses + FX = 0 ; 500N – FBCcos(45) = 0
FBCcos(45) = 500N
cos(45) cos(45)
FBC = N (C)
+FY = 0 ;
FBCsin(45) – FAB = 0
707.1*sin(45) – FAB = 0
FAB = 707.1*sin(45)
FAB = 500 N (T)
The (T) indicates tension and the (C) indicates compression. Both solutions were positive therefore our initially assumptions were correct. (If the solution had been a negative number, then we would simply reverse our assumption from tension to compression or vice versa)

10. Trusses + FX = 0 ; – FAC+ FBC cos (45) = 0
– FAC cos (45) = 0
FAC = cos(45)
FAC = 500 N (T)
+FY = 0 ;
CY – FBC *sin(45) = 0
CY – *sin(45) = 0
CY = 500 N
The following is the free-body diagram of joint C. The force FBC has been replaced with its x and y components: CY
FAC
FBC
45

11. Trusses + FX = 0 ; FAC - Ax = 0 500 N - Ax = 0 AX = 500 N
+FY = 0 ;
FAB - AY = 0
500 N - AY = 0
AY = 500 N
The following is the free-body diagram of joint A: AX AY
FAB
FAC

12. Trusses
Solved!!!
500 N B A C 2m
45
500 N
707.1 N (C)
500 N (T)