6161103 6.4 the method of sections

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6161103 6.4 the method of sections

  1. 1. 6.4 The Method of SectionsUsed to determine the loadingswithin a bodyIf a body is in equilibrium, any partof the body is in equilibriumTo determine the forces within themembers, an imaginary sectionindicated by the blue line, can beused to cut each member into twoand expose each internal force asexternal
  2. 2. 6.4 The Method of SectionsIt can be seen that equilibriumrequires the member in tension (T) besubjected to a pull and the member incompression (C) be subjected to apushMethod of section can be used to cutor section members of an entire trussApply equations of equilibrium on thatpart to determine the members
  3. 3. 6.4 The Method of SectionsConsider the truss shownTo determine the force in the memberGC, section aa would be considered
  4. 4. 6.4 The Method of SectionsConsider the FBDNote the line of action of each member force isspecified from the geometry of the trussMember forces acting on one part of the trussare equal and opposite to those acting on theother part – Newton’s Law
  5. 5. 6.4 The Method of SectionsMembers assumed to be in tension (BCand GC) are subjected to a pull whereasthe member in compression (GF) issubjected to a pushApply equations of equilibrium
  6. 6. 6.4 The Method of SectionsDetermining the Correct Sense of the Unknown Member Always assume the unknown member forces in the cut section are in tension - The numerical solution of the equilibrium will yield positive scalars for members in tension and negative scalars for members in compression
  7. 7. 6.4 The Method of SectionsDetermining the Correct Sense of the Unknown Member The correct sense of a direction of an unknown force can be determined by inspection - In more complicated problems, the sense of the member can be assumed - A positive answer indicates that the assumed sense is correct and a negative answer indicates that the assumed sense must be reversed
  8. 8. 6.4 The Method of SectionsProcedure for Analysis Draw the FBD of a joint having at least one known force and at most two unknown forces If this joint is at one of the supports, it ma be necessary to know the external reactions at the truss support Use one of two methods for determining the correct sense of the member Orient the x and y axes so that the forces on the FBD can be easily resolved into x and y components
  9. 9. 6.4 The Method of SectionsProcedure for AnalysisFree-Body Diagram Decide how to cut or session the truss through the members where forces are to be determined Before isolating the appropriate section, determine the truss’s external reactions Use the equilibrium equations to solve for member forces at the cut session
  10. 10. 6.4 The Method of SectionsProcedure for AnalysisFree-Body Diagram Draw the FBD of that part of the sectioned truss which has the least number of forces acting on it Use one of the two methods for establishing the sense of an unknown member force
  11. 11. 6.4 The Method of SectionsProcedure for AnalysisEquations of Equilibrium Moments are summed about a point that lies at the intersection of lines of action of the two unknown forces The third unknown force is determined directly from moment equation If two of the unknown forces are parallel, forces may be summed perpendicular to the direction of these unknowns to determine the third unknown force
  12. 12. 6.4 The Method of SectionsExample 6.5Determine the force in members GE, GC,and BC of the truss. Indicate whether themembers are in tension or compression.
  13. 13. 6.4 The Method of SectionsSolution Choose section aa since it cuts through the three members FBD of the entire truss
  14. 14. 6.2 The Method of JointsSolution+ → ∑ Fx = 0;400 N − Ax = 0Ax = 400 N∑ M A = 0;− 1200 N (8m) − 400 N (3m) + D y (12m) = 0D y = 900 N+ ↑ ∑ Fy = 0;Ay − 1200 N + 900 N = 0Ay = 300 N
  15. 15. 6.4 The Method of SectionsSolutionFBD of the sectioned truss
  16. 16. 6.2 The Method of JointsSolution∑ M G = 0;− 300 N (4m) − 400 N (3m) + FBC (3m) = 0FBC = 800 N (T )∑ M C = 0;− 300 N (8m) + FGE (3m) = 0FGE = 800 N (C )+ ↑ ∑ Fy = 0; 3300 N − FGC = 0 5FGC = 500 N (T )
  17. 17. 6.4 The Method of SectionsExample 6.6Determine the force in member CF of the bridgetruss. Indicate whether the member are in tensionor compression. Assume each member is pin connected.
  18. 18. 6.4 The Method of SectionsSolutionFBD of the entire truss
  19. 19. 6.4 The Method of SectionsSolutionFBD of the sectioned truss Three unknown FFG, FCF, FCD
  20. 20. 6.4 The Method of SectionsSolutionEquations of Equilibrium For location of O measured from E 4 / (4 + x) = 6 / (8 + x) x = 4m Principle of Transmissibility ∑ M O = 0; − FCF sin 45o (12m) + (3kN )(8m) − (4.75kN )(4m) = 0 FCF = 0.589kN (C )
  21. 21. 6.4 The Method of SectionsExample 6.7Determine the force in member EB of the rooftruss. Indicate whether the member are intension or compression.
  22. 22. 6.4 The Method of SectionsSolutionFBD of the sectioned truss
  23. 23. 6.4 The Method of SectionsSolution Force system is concurrent Sectioned FBD is same as the FBD for the pin at E (method of joints)
  24. 24. 6.4 The Method of SectionsSolution∑ M B = 0;1000 N (4m) − 3000 N (2m) − 4000 N ( 4m) + FED sin 30o (4m) = 0FED = 3000 N (C )+ → ∑ Fx = 0;FEF coso − 3000 cos 30o N = 0FEF = 3000 N (C )+ ↑ ∑ Fy = 0;2(3000 sin 30o N ) − 1000 N − FEB = 0FEB = 2000 N (T )

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