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
2Common Types of Trussesgusset plate• Roof Trussestop cordroofpurlinsknee bracebottomcord gusset platespan, 18 - 30 m, typicalbay, 5-6 m typical
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
7Classification of Coplanar Trusses• Simple Trussesab cd (new joint)new membersA BC DP C DA BP PA BC
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• 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
11A BCD EFOInternal Unstable8 + 3 = 11 < 2(6)AD, BE, and CF are concurrent at point O
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.
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.
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.
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
162 m2 m500 N45oABCCy = 500 NAx = 500 NAy = 500 NFACΣFx = 0:+500 - FAC = 0FAC = 500 N (T)Joint A500 N500 N500 N
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
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
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
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
36Compound TrussesProcedure for AnalysisStep 1. Identify the simple trussesStep 2. Obtain external loadingStep 3. Solve for simple trusses separately
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
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
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
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
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
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
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
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
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
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
51Space TrussesPb + r < 3j unstable trussb + r = 3j statically determinate-check stability>b + r 3j statically determinate-check stability• Determinacy and Stability
52xyzyzxyzxzxyxyzFyFzFzFxFxFzFyshort linkyxzrollerzxyslotted rollerconstrainedin a cylinderyzx ball-and -socket
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
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