1. ENGR-1100 Introduction to Engineering Analysis
Lecture 17

2. Today Lecture Outline
Trusses analysis- method of section.

3. 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

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

5. Example 7-16
Determine the forces in members BC, CD, and DF of the truss shown in fig. P7-16

6. Solution
M=9
j=6
M=2j-3
From a free-body diagram for the complete truss:
MA = Ey (8) - 10 (4) - 8 (6) = 0
Ey = 11 kN = 11 kN
From a free-body joint E:
Fy = TDE sin 30 + 11 = 0
TDE = -22kN = 22 kN (C)

7. From a free-body joint D:
Fy = -TDF cos 30 - 8 cos 30 = 0
TDF = -8 kN = 8 kN (C)
Fx = TDE - TCD + 8 sin 30 - TDF sin 30
= (-22) - TCD + 8 sin 30 - (-8) sin 30=0
TCD = -14 kN = 14 kN (C)
From a free-body joint C:
Fx = -TBC cos 30 + TCD cos 30
= - TBC cos 30 + (-14.00) cos 30 = 0
TBC = -14 kN = 14 kN (C)

8. Class Assignment: Exercise set 7-46
please submit to TA at the end of the lecture
Determine the forces in members BC, BF, and EF of the truss shown in Fig. P7-46 using the joint method.
Answer:
TBF= 3.75 kN (T)
TBC= 2.25 kN (T)
TEF= 4.5 kN (C)

9. Solution (by Heather Alexander)

10. Method of Sections

11. Example 7-45
Determine the in members CD, CF, and FG of the bridge shown in Fig. P7-45 using the method of sections.

12. Solution
M=11
j=7
M=2j-3 x y
From a free-body diagram for the complete truss:
MA = Ey (45) - 10 (15) - 20 (30) = 0
Ey = kip  kip

13. From a free-body diagram of the part of the truss to the right of member CG:
ME = -TCF sin 60 (15) + 20 (15)= 0
TCF = kip  23.1 kip (T)
MF = TCD (15 sin 30) (15) = 0
TCD = kip  33.3 kip (C) x y
MC = -TFG (22.5 tan 30) - 20(7.5) (22.5) = 0
TFG = kip  kip (T)

14. Class Assignment: Exercise set P7-46
please submit to TA at the end of the lecture
Determine the in members BC, BF, and EF of the truss shown in Fig. P7-46 using the method sections.
Answer:
TBF= 3.75 kN (T)
TBC= 2.25 kN (T)
TEF= 4.5 kN (C)

15. 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

16. 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

17. Class Assignment: Exercise set
7-35 to 7-38
please submit to TA at the end of the lecture
Identify the zero-force members present when the trusses shown in the following problems are subjected to the loadings indicated.