1 / 17

Lecture 17

ENGR-1100 Introduction to Engineering Analysis. Lecture 17. Today Lecture Outline. Trusses analysis- method of section. Stability Criteria. 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.

Download Presentation

Lecture 17

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. ENGR-1100 Introduction to Engineering Analysis Lecture 17

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

  3. Stability Criteria 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. M=9 j=6 M=2j-3 From a free-body diagramfor the complete truss: MA = Ey (8) - 10 (4) - 8 (6) = 0 Ey = 11 kN = 11 kN From a free-body joint E: Solution 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. y x Solution M=11 j=7 M=2j-3 From a free-body diagramfor the complete truss: MA = Ey (45) - 10 (15) - 20 (30) = 0 Ey = 16.699 kip  16.67 kip

  13. y x 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 = 23.09 kip  23.1 kip (T) MF =TCD (15 sin 30) + 16.667 (15) = 0 TCD = -33.33 kip  33.3 kip (C) MC = -TFG (22.5 tan 30) - 20(7.5) +16.667(22.5) = 0 TFG = 17.321 kip  17.32 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. Free body diagram on joint C y FBC x FCD FCD=0 FBC=0 SFx=0 Member BC and DC are zero force members Zero Force Members SFy=0

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

More Related