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  • can anybody suggest how to solve truss when loading is at a midpoint of a member ? joint equilibrium won't work because nowhere the load will come in analysis of joints and section equilibrium shows forces even in zero force members
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  • 1. 1! Common Types of Trusses! Classification of Coplanar Trusses! The Method of Joints! Zero-Force Members! The Method of Sections! Compound Trusses! Complex Trusses! Space TrussesAnalysis of Statically DeterminateTrusses
  • 2. 2Common Types of Trussesgusset plate• Roof Trussestop cordroofpurlinsknee bracebottomcord gusset platespan, 18 - 30 m, typicalbay, 5-6 m typical
  • 3. 3Howe truss18 - 30 mPratt truss18 - 30 mHowe trussflat roofWarren trussflat roofsaw-tooth trussskylight Fink truss> 30 mthree-hinged archhangar, gymnasium
  • 4. 4• Bridge Trussestop corddeckswaybracingtop lateralbracingstringersportalend postportalbracingpanelfloor beambottom cord
  • 5. 5trough Pratt trussdeck Pratt trussWarren trussparker truss(pratt truss with curved chord)Howe truss baltimore trussK truss
  • 6. 61. All members are connected at both ends by smooth frictionless pins.2. All loads are applied at joints (member weight is negligible).Notes: Centroids of all joint members coincide at the joint.All members are straight.All load conditions satisfy Hooke’s law.Assumptions for Design
  • 7. 7Classification of Coplanar Trusses• Simple Trussesab cd (new joint)new membersA BC DP C DA BP PA BC
  • 8. 8• Compound Trussessimple trusssimple trussType 2simple trusssimple trussType 1secondarysimple trusssecondarysimple trusssecondarysimple trusssecondarysimple trussmain simple trussType 3
  • 9. 9• Complex Trusses• Determinacyb + r = 2j statically determinateb + r > 2j statically indeterminateIn particular, the degree of indeterminacy is specified by the difference in thenumbers (b + r) - 2j.
  • 10. 10• Stabilityb + r < 2j unstableb + r 2j unstable if truss support reactions are concurrent or parallelor if some of the components of the truss form a collapsiblemechanism>External UnstableUnstable-parallel reactions Unstable-concurrent reactions
  • 11. 11A BCD EFOInternal Unstable8 + 3 = 11 < 2(6)AD, BE, and CF are concurrent at point O
  • 12. 12Example 3-1Classify each of the trusses in the figure below as stable, unstable, staticallydeterminate, or statically indeterminate. The trusses are subjected to arbitraryexternal loadings that are assumed to be known and can act anywhere on thetrusses.
  • 13. 13SOLUTIONExternally stable, since the reactions are not concurrent or parallel. Since b = 19,r = 3, j = 11, then b + r = 2j or 22 = 22. Therefore, the truss is statically determinate.By inspection the truss is internally stable.Externally stable. Since b = 15, r = 4, j = 9, then b + r > 2j or 19 > 18. The trussis statically indeterminate to the first degree. By inspection the truss is internallystable.
  • 14. 14Externally stable. Since b = 9, r = 3, j = 6, then b + r = 2j or 12 = 12. The truss isstatically determinate. By inspection the truss is internally stable.Externally stable. Since b = 12, r = 3, j = 8, then b + r < 2j or 15 < 16. The trussis internally unstable.
  • 15. 152 m2 m500 N45oABCThe Method of JointsCy = 500 NFBAFBC500 NB45oJoint BxyΣFx = 0:+500 - FBCsin45o = 0FBC = 707 N (C)ΣFy = 0:+- FBA + FBCcos45o = 0FBA = 500 N (T)Ax = 500 NAy = 500 N
  • 16. 162 m2 m500 N45oABCCy = 500 NAx = 500 NAy = 500 NFACΣFx = 0:+500 - FAC = 0FAC = 500 N (T)Joint A500 N500 N500 N
  • 17. 17Zero-Force MembersDxDyEyCFCDFCB ΣFx = 0: FCB = 0+ΣFy = 0: FCD = 0+FAEFABAθΣFy = 0: FABsinθ = 0, FAB = 0+ΣFx = 0: FAE + 0 = 0, FAE = 0+0000PAB CDE
  • 18. 18Example 3-4Using the method of joints, indicate all the members of the truss shown in thefigure below that have zero force.PA BCDEFGH
  • 19. 19PA BCDEFGHSOLUTIONJoint DΣFy = 0:+ FDCsinθ = 0, FDC = 0ΣFx = 0:+ FDE + 0 = 0, FDE = 00FECFEFPEFEF = 0ΣFx = 0:+Joint EAxAxGx000xyDθFDCFDE
  • 20. 20Joint HΣFy = 0:+FHB = 0AxAxGxJoint GΣFy = 0:+ FGA = 0000 00FHFFHAFHBxyHGGx FGFFGAPA BCDEFGH
  • 21. 212 kN2 kN2 kNA BCDEF1 kNGHIJK5@3m = 15 m5 mExample 3-5• Determine all the member forces• Identify zero-force members
  • 22. 222 kN2 kN2 kNA BCDEF1 kNGHIJK5@3m = 15 m5 mSOLUTIONr + b = 2j,4 18 2(11)• Determinate• Stable+ ΣMA = 0: 0)12(1)9(2)6(2)3(2)5( =−−−−xKKx = 7.6 kN,ΣFx = 0:+ ,06.7 =+− xA Ax = 7.6 kN,ΣFy = 0:+ ,01222 =−−−−yA Ay = 7 kN,KyKxAyAx0Use method of joints
  • 23. 232 kN2 kN2 kNA BCDEF1 kNGHIJK5@3m = 15 m5 m07 kN7.6 kN7.6 kNF x´y´θ00• Joint FFFEFFGΣF x´ = 0:+ FFG = 0ΣF y´ = 0:+ 0sin =θFEFFFE = 0Use method of joint
  • 24. 242 kN2 kN2 kNA BCDEF1 kNGHIJK5@3m = 15 m5 m07 kN7.6 kN7.6 kN00ΣF x´ = 0:+ -FED = 0ΣF y´ = 0:+ 0cos =θEGFFEG = 0y´Ex´θFED0FEG• Joint E
  • 25. 252 kN2 kN2 kNA BCDEF1 kNGHIJK5@3m = 15 m5 m07 kN7.6 kN7.6 kN 00ΣF x´ = 0:+FHG = 1.5 kN (T)ΣF y´ = 0:+ 0169.33sin =−oDGFFDG = 1.803 kN (C)• Joint GFHGFDG 01 kNGx´y´033.69oo69.33)32(tan 1== −θθ069.33cos803.1 =+− HGF
  • 26. 262 kN2 kN2 kNA BCDEF1 kNGHIJK5@3m = 15 m5 m07 kN7.6 kN7.6 kN0FHI• Joint HHx´y´1.5 kNFHDΣF y´ = 0:+ FHD = 0ΣF x´ = 0:+FHI = 1.5 kN (T)05.1 =+− HIF000
  • 27. 2733.69oI2 kNDEF1 kNGH3 m 3 m3 mUse method of sectionsFHIFDCFDI18.44o+ ΣMD = 0: 0)3(1)2( =−HIFFHI = 1.5 kN (T)+ ΣMF = 0: 0)6(2)3(1)9(69.33sin =++− DIFFDI = 3 kN (T)+ ΣMI = 0: 0)3(2)6(1)9(44.18sin =−−− DCFFDC = -4.25 kN (C)Check : 01244.18sin69.33sin =−−− DCDI FF O.K.ΣF y = 0:+3 -4.25
  • 28. 28CThe Method of SectionsExDxDyaa100 NABG2 m45oFGFFGCFBC+ ΣMG = 0:100(2) - FBC(2) = 0FBC = 100 N (T)ΣFy = 0:+-100 + FGCsin45o = 0FGC = 141.42 N (T)+ ΣMC = 0:100(4) - FGF(2) = 0FGF = 200 N (C)100 NAB C DEFG2 m2 m 2 m 2 m
  • 29. 292 kN2 kN2 kNA BCDEF1 kNGHIJK5@3m = 15 m5 mExample 3-6• Determine member force CD, ID, and IH
  • 30. 3033.69oI2 kNDEF1 kNGH3 m 3 m3 mUse method of sectionsFHIFDCFDI18.44o+ ΣMD = 0: 0)3(1)2( =−HIFFHI = 1.5 kN (T)+ ΣMF = 0: 0)6(2)3(1)9(69.33sin =++− DIFFDI = 3 kN (T)+ ΣMI = 0: 0)3(2)6(1)9(44.18sin =−−− DCFFDC = -4.25 kN (C)Check : 01244.18sin69.33sin =−−− DCDI FF O.K.ΣF y = 0:+3 -4.25SOLUTION+1.50E+00-4.22E+003.00E+00
  • 31. 31Example 3-7Determine the force in members GF and GD of the truss shown in the figurebelow. State whether the members are in tension or compression. The reactions atthe supports have been calculated.6 kN 8 kN 2 kNAx = 0Ay = 9 kN Ey = 7 kNAB C DEFGH3 m 3 m 3 m 3 m3 m4.5 m
  • 32. 32Ax = 0Ay = 9 kNSOLUTION aa+ ΣMD = 0:FFG sin26.6o(3.6) + 7(3) = 0,FFG = -17.83 kN (C)+ ΣMO = 0:- 7(3) + 2(6) + FDG sin56.3o(6) = 0,FDG = 1.80 kN (C)6 kN 8 kN 2 kN Ey = 7 kNAB C DEFGH3 m 3 m 3 m 3 m3 m4.5 m2 kN Ey = 7 kND EFSection a-a3 mFFGFDGFDC3 m26.6oO26.6o56.3o
  • 33. 33Example 3-8Determine the force in members BC and MC of the K-truss shown in the figurebelow. State whether the members are in tension or compression. The reactions atthe supports have been calculated.AB C D E FGHIJKLM N O P6 kN 6 kN 8 kN 7 kN13 kN03 m3 m5 m 5 m 5 m 5 m 5 m 5 m
  • 34. 34SOLUTIONaaSection a-aABL6 kN13 kN5 m6 mFLKFBMFLMFBC+ ΣML = 0:FBC(6) - 13(5) = 0,FBC = 10 kN (T)AB C D E FGHIJKLM N O P6 kN 6 kN 8 kN 7 kN13 kN03 m3 m5 m 5 m 5 m 5 m 5 m 5 m
  • 35. 35bbFKLFKMFCM10 kN+ ΣMK = 0: -FCMcos31o(6) - 10(6) - 8(5) + 7(20) = 0FCM = 7.77 kN (T)31oAB C D E FGHIJKLM N O P6 kN 6 kN 8 kN 7 kN13 kN03 m3 m5 m 5 m 5 m 5 m 5 m 5 mC D E FGHIJKN O P6 kN 8 kN 7 kN3 m3 m5 m 5 m 5 m 5 m
  • 36. 36Compound TrussesProcedure for AnalysisStep 1. Identify the simple trussesStep 2. Obtain external loadingStep 3. Solve for simple trusses separately
  • 37. 37Example 3-9Indicate how to analyze the compound truss shown in the figure below. Thereactions at the supports have been calculated.AB4 kN 2 kN 4 kNC D4 m2 m2 mEy = 5 kNAy = 5 kNAy = 0IHJ K FEG2 m 2 m 2 m 2 m
  • 38. 38SOLUTIONaaCFHGFBCFJC4 sin60o mAB4 kNAy = 5 kNIHJ2 m 2 m+ ΣMC = 0:-5(4) + 4(2) + FHG(4sin60o) = 0FHG = 3.46 kN (C)AB4 kN 2 kN 4 kNC D4 m2 m2 mEy = 5 kNAy = 5 kNAy = 0IHJ K FEG2 m 2 m 2 m 2 m
  • 39. 39AB4 kN 2 kNCAy = 5 kNIHJ2 m 2 m4 sin60o m3.46 kN+ ΣMA = 0:3.46(4sin60o) + FCKsin60o(4) - 4(2) - 2(4) = 0FCK = 1.16 kN (T)FCKFCD60oΣFx = 0:+-3.46 + 1.16cos60o + FCD = 0FCK = 2.88 kN (T)AB4 kN 2 kN 4 kNC D4 m2 m2 mEy = 5 kNAy = 5 kNAy = 0IHJ K FEG2 m 2 m 2 m 2 m
  • 40. 40AB4 kN 2 kNCAy = 5 kNIHJ2 m 2 m4 sin60o m3.46 kNFCK = 1.162.88 kN60oUsing the method of joints.Joint A : Determine FAB and FAIJoint H : Determine FHI and FHJJoint I : Determine FIJ and FIBJoint B : Determine FBC and FBJJoint J : Determine FJCAB4 kN 2 kN 4 kNC D4 m2 m2 mEy = 5 kNAy = 5 kNAy = 0IHJ K FEG2 m 2 m 2 m 2 m
  • 41. 41Example 3-10Indicate how to analyze the compound truss shown in the fugure below. Thereactions at the supports have been calculated.A BCDEFGH45o45o 45o2 m 2 m 2 m 2 m 2 m2 m15 kN 15 kNAy = 15 kNAx = 0 kNFy = 15 kN4 m
  • 42. 42SOLUTIONaa+ ΣMB = 0:-15(2) - FDG(2 sin45o) - FCEcos45o(4)- FCEsin45o(2) = 0 -----(1)ΣFy = 0:+15 - 15 + FBHsin45o - FCEsin45o = 0FBH = FCE-----(2)ΣFx = 0:+FCEFBHFDGA BCD45o45o2 m 2 m2 m15 kNAy = 15 kN4 m45o2 sin 45o mFBHcos45o + FDG + FCEcos45o = 0 -----(3)From eq.(1)-(3): FBH = FCE = -13.38 kN (C)FDG = 18.92 kN (T)A BCDEFGH45o45o 45o2 m 2 m 2 m 2 m 2 m2 m15 kN 15 kNAy = 15 kNAx = 0 kNFy = 15 kN4 m
  • 43. 43From eq.(1)-(3): FBH = FCE = -13.38 kN (C)FDG = 18.92 kN (T)Analysis of each connected simple trusscan now be performed using the method ofjoints.Joint A : Determine FAB and FADJoint D : Determine FDC and FDBJoint C : Determine FCBaaA BCD45o45o2 m 2 m2 m15 kNAy = 15 kN4 m45oA BCDEFGH45o45o 45o2 m 2 m 2 m 2 m 2 m2 m15 kN 15 kNAy = 15 kNAx = 0 kNFy = 15 kN4 mFCEFBHFDG
  • 44. 44Example 3-11Indicate how to analyze the symmetrical compound truss shown in the figurebelow. The reactions at the supports have been calculated.5 kN3 kN3 kNA CEBF DG H45o 45oAy = 4.62 kNAx = 0 kNFy = 4.62 kN6 m 6 m5o5o5o5o
  • 45. 453 kNCEDH1.5 kN1.5 kNFECFEC1.5 kN1.5 kN3 kNEFGAFAEFAE5 kN3 kN3 kNA CEBF DG H45o 45oAy = 4.62 kNAx = 0 kNFy = 4.62 kN6 m 6 m5o5o5o5o
  • 46. 461.5 kN5 kN45o 45o4.62 kN 4.62 kN1.5 kN1.5 kNAB C45o45o1.5 kNEΣFy = 0:+4.62 - 1.5sin45o - FAEsin45o = 0FAE = 5.03 kN (C)ΣFx = 0:+1.5cos45o - 5.03cos45o + FAB = 0FAB = 2.50 kN (T)45o4.62 kNA1.5 kN45oFAEFAB5 kN3 kN3 kNA CEBF DG H45o 45oAy = 4.62 kNAx = 0 kNFy = 4.62 kN6 m 6 m5o5o5o5o
  • 47. 4711Complex Trusses+=X xF´EC + x f´EC = 0x =F´ECf´ECr + b = 2j,3 9 2(6)• Determinate• StableF´EC= FADFADf´ECFi = F´i + x f´iPABCDFEPABCDFEPABCDFE
  • 48. 48Example 3-12Determine the force in each member of the complex truss shown in the figurebelow. Assume joints B, F, and D are on the same horizontal line. State whetherthe members are in tension or compression.45o 45o20 kNABCDEF1 m1 m2.5 m0.25 m
  • 49. 49SOLUTION+ =xFBD+ x f´BD = 0x =FBDf´BDFi = Fi + x f´i45o 45o20 kNABCDEF45o 45oABCDEF 1 kN45o 45o20 kNABCDEF1 m1 m2.5 m0.25 m
  • 50. 50x45o 45o20 kNABCDEF+45o 45oABCDEF 1 kN-0.707 -0.7070.707-0.539-0.3-0.5391-0.30.70714.14 -14.140 00-1821.54-10+10100)1(100==+−=+xxxfF BDBDFirst determine reactions and next use the method of joint, start at join C, F, E, D, and B.20 kN(20x2.25)/2.5 = 18 kN18 kN00045o 45o20 kNABCDEF1 m1 m2.5 m0.25 m7.077.07 -21.217.07-5.39-2116.1571020 kN18 kN18 kN1 m1 m2.5 m0.25 m
  • 51. 51Space TrussesPb + r < 3j unstable trussb + r = 3j statically determinate-check stability>b + r 3j statically determinate-check stability• Determinacy and Stability
  • 52. 52xyzyzxyzxzxyxyzFyFzFzFxFxFzFyshort linkyxzrollerzxyslotted rollerconstrainedin a cylinderyzx ball-and -socket
  • 53. 53xyzlFBA zFxFyFzxy222zyxl ++=)(lxFFx = )(lyFFy = )(lzFFz =222zyx FFFF ++=• x, y, z, Force Components.• Zero-Force MembersΣFz = 0 , FD = 0FDFA = 0FBFC = 0ABBDxyzCase 2Case 1FDFAFBFCABCDxyzΣFz = 0 , FB = 0ΣFy = 0 , FD = 0
  • 54. 54Example 3-13Determine the force in each member of the space truss shown in the figure below.The truss is supported by a ball-and-socket joint at A, a slotted roller joint at B,and a cable at C.2.67 kN1.22 m1.22 m2.44 m2.44 mABCDExyz
  • 55. 55SOLUTIONCyByBxAzAxAyThe truss is statically determinate since b + r = 3j or 9 + 6 = 3(5)ΣMz = 0: Cy = 0 kNΣMx = 0: By(2.44) - 2.67(2.44) = 0 By = 2.67 kNΣMy = 0: -2.67(1.22) + Bx(2.44) = 0 Bx = 1.34 kNΣFx = 0: -Ax + 1.34 = 0 Ax = 1.34 kNΣFz = 0: Az - 2.67 = 0 Az = 2.67 kNΣFy = 0: Ay - 2.67 = 0 Ay = 2.67 kN2.67 kN1.22 m2.44 m2.44 mBCDExyz
  • 56. 56ByBxAzAxAy2.67 kN1.22 m2.44 m2.44 mBCDExyzJoint D.ΣFZ= 0: FDC = 02.73 m3.66 mΣFY = 0: FDE = 00zxy0FDEFDCDJoint C.zxy0000 FCEC0000000ΣFx = 0: FDA = 0ΣFy = 0: FCE = 0ΣFx = 0: FCB = 0ΣFz = 0: FCA = 0
  • 57. 57ByBxAzAxAy2.67 kN1.22 m2.44 m2.44 mBCDExyz2.73 m3.66 m02.67 kN1.34 kNBzFBAFBEFBCxyJoint B.ΣFy = 0: - 2.67 + FBE(2.44/3.66) = 0 FBE = 4 kN (T)ΣFz = 0: FBA - 4(2.44/3.66) = 0 FBA = 2.67 kN (C)ΣFx = 0: 1.34 - FBC -4(1.22/3.66) = 0 FBC = 00000000
  • 58. 58ByBxAzAxAy2.67 kN1.22 m2.44 m2.44 mBCDExyz2.73 m3.66 m02.67 kNAzxy2.67 kN1.34 kN2.67 kNFACFADFAE45o2 1ΣFy = 0: - FAE(52) + 2.67 = 0 FAE = 2.99 kN (C)ΣFz = 0: - 1.34 + FAD + 2.99( 51) = 0 FAD = 0, OKJoint A.ΣFz = 0: 2.67 - 2.67 - FACsin45o = 0 FAC = 0, OK0000000