1. ENGR-1100 Introduction to Engineering Analysis
Lecture 16

2. Today Lecture Outline Structures: Trusses Frames
Trusses analysis- method of joints
Stability criteria.

3. Two important structures types
Trusses: Structures composed entirely of two force members.
Frames: Structures containing at least one member acted on by forces at three or more points.

4. Plane Trusses Plane trusses: lie in a single plane.
Space trusses: not contained in a single plane and/or loaded out of the structure plane.

5. Assumptions
1) Truss members are connected together at their ends only.
2) Truss members are connected together by frictionless pins.
3) The truss structure is loaded only at the joints.
4) The weight of the member may be neglected.

6. Truss members are two-force members
The Four Assumptions
Truss members are two-force members F F
Truss member F F

7. Straight Members Forces act along the axis of the member
Compressive forces tend to shorten the member. F
Tensile forces tend to elongate the member.

8. “Rigid” trusses
“Rigid”- the truss will retain its shape when removed from its support
Simple truss- constructed by attaching several triangles together.
Allows a simple way to check rigidity.

9. How many equations are available? How many unknowns?
What are we looking for?
The support reaction .
The force in each member.
How many equations are available? How many unknowns?
Each joint- 2 equations
Unknowns- number of members+ support reaction.

10. Stability Criteria m=2j-3 m<2j-3 m>2j-3
2j- number of equations to be solved.
m- number of members.
3- number of support reaction
m<2j-3
Truss unstable
m>2j-3
Statically indeterminate

11. Example m (Number of members) = 13 j (Number of joints) = 8 m=2j-3
Number of supports= 3

12. Method of Joints Separate free-body diagrams for: each member each pin
Equilibrium equations for each pin:
SF=0
no moment equation

13. Example 7-3
Use the method of joints to determine the force in each member of the truss shown in Fig. P7-3. State whether each member is in tension or compression.

14. Solution Fy = TBD - 3000 = 0 TBD = 3000 lb = 3000 lb (T)
TAB = lb = 1500 lb (c)
TBC = lb  2600 lb (c)
Solution
TCD
TAD
TBD
Joint D y x
From a free-body diagram on joint D:
TAB
TBD
Joint B
TBC y x
From a free-body diagram on joint B:
Fx = TBC sin 30º - TAB sin 60º = 0
Fy = - TBC cos 30º - TAB cos 60º - TBD = 0

15. known
TCD Cy
Joint C
TBC y x
From a free-body diagram on joint C:
Fx = - TCD - TBCsin 30
= - TCD -(2598) sin 30 = 0
TCD = 1299 lb = 1299 lb (T)
From a free-body diagram on joint D:
TCD
TAD
TBD
Joint D y x
Fx = -TAD + TCD = - TAD = 0
TAD = 1299 lb = 1299 lb (T)

16. Class Assignment: Exercise set 7-4
please submit to TA at the end of the lecture
Answer:
TBC=14.1 kN (T)
TAC=5.13 kN (T)
TAB=28.2 kN (C)

17. Class Assignment: Exercise set 7-6
please submit to TA at the end of the lecture
Answer:
TBD=3.23 kN (C)
TCD=4.62 kN (T)
TAD=0.567 kN (T)
TBC=4 kN (C)
TAB=2.38kN (C)

18. Zero Force Members SFy=0 FCD=0 FBC=0 SFx=0
Free body diagram on joint C
FBC
FCD y x
SFy=0
FCD=0
SFx=0
FBC=0
Member BC and DC are zero force members

19. SFy=0 FBD=0 FAD=0 Free body diagram on joint B y FCB FAB x FBD
Free body diagram on joint D
FAD
FED y x
FDC
FAD=0

20. Example 7-19
The truss shown in figure P7-19 support one side of a bridge; an identical truss supports the other side. Floor beams carry vehicle loads to the truss joints. Calculate the forces in members BC, BG, and CG when a truck weighing 7500 lb is stopped in the middle of the bridge as shown. The center of gravity of the truck is midway between the front and rear wheels.

21. Solution From symmetry: FH = FF Free body diagram on floor beams GF
MG = 2FF (10) – (6) = 0
FF = 1125 lb
Free body diagram on floor beams FH
Fy = 2FG - 2(3750) + 4(1125) = 0
FG= 1500 lb

22. f          = tan-1 8/5 = 57.99
= tan-1 5/10 =26.57
= tan-1 8/10 =38.66
Free body diagram on complete truss
Fx = Ax = 0
Ax = 0
ME= Ay (30) (25) (15) (5) = 0
Ay = 1875 lb = 1875 lb

23. TAB = lb  2210 lb (c)
TBH = 1125 lb = 1125 lb (T)
Free body diagram on joint A
Fx = TAB cos 57.99 + TAH = 0
Fy = TAB sin 57.99 = 0
Fy = TBH = 0
Free body diagram on joint H

24. TBC = lb  1451 lb (C)
TBG = lb  lb (T)
Free body diagram on joint B
Fx = TBC cos 26.57 + TBG cos 38.66
+ 2211cos 57.99 = 0
Fy = TBC cos 26.57 - TBG sin 38.66
sin 57.99 = 0
Free body diagram on joint C
Fx = TCD cos 26.57 cos 26.57 = 0
Fy = TCD sin 26.57 - TCG sin 26.57 = 0
TCD = lb  1451 lb (C)
TCG = lb  1298 lb (T)

25. Class Assignment: Exercise set 7-23
please submit to TA at the end of the lecture