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- 1. Method of Joints If a truss is in equilibrium, then each of its jointsmust be in equilibrium. The method of joints consists of satisfying theequilibrium equations for forces acting on eachjoint.0yF 0xF Method of Joints Recall, that the line of action of a force acting ona joint is determined by the geometry of the trussmember. The line of action is formed by connecting thetwo ends of each member with a straight line. Since direction of the force is known, theremaining unknown is the magnitude of theforce.Method of JointsTension ForceCompression ForceJoint A Joint BJoint BJoint AMethod of JointsVerticalsUpper chord membersLower chord membersDiagonalsMethod of Jointsgusset plateweldIdealized joint –members connected bya frictionless pinMethod of JointsLower chord in tensionUpper chord in compressionThis is a Pratt trussCIVL 3121 Trusses - Method of Joints 1/5
- 2. Method of JointsLower chord in tensionUpper chord in compressionThis is a Howe trussMethod of Joints Procedure for analysis - the following is aprocedure for analyzing a truss using themethod of joints:1. If possible, determine the support reactions2. Draw the free body diagram for each joint. Ingeneral, assume all the force member reactions aretension (this is not a rule, however, it is helpful inkeeping track of tension and compressionmembers).Method of Joints Procedure for analysis - the following is aprocedure for analyzing a truss using themethod of joints:3. Write the equations of equilibrium for each joint,0yF 0xF Method of Joints4. If possible, begin solving the equilibrium equationsat a joint where only two unknown reactions exist.Work your way from joint to joint, selecting the newjoint using the criterion of two unknown reactions.5. Solve the joint equations of equilibriumsimultaneously, typically using a computer or anadvanced calculator. Procedure for analysis - the following is aprocedure for analyzing a truss using themethod of joints:500 lb10 ft10 ftMethod of JointsExample - Consider the following trussFirst, determine the support reactions for the truss500 lbAyAxCy500 lb10 ft10 ftMethod of JointsExample - Consider the following trussFirst, determine the support reactions for the truss0AM 500 (10 ) (10 )ylb ft C ft 500 lbAyAxCyCy = 500 lb0yF y yA C Ay = -500 lb0xF 500xA lb Ax = -500 lbCIVL 3121 Trusses - Method of Joints 2/5
- 3. Method of JointsThe equations of equilibrium for Joint AFAC0xF 500ACF lb FAC = 500 lb0yF 500ABF lb 500 lb500 lbFABFAB = 500 lbMethod of JointsThe equations of equilibrium for Joint B0xF cos45 500BCF lb FBC = -707.2 lbFAB500 lbFBCFAC = 500 lb (T)FAB = 500 lb (T) FBC = 707.2 lb (C)The forces in the truss can be summarized as:Method of JointsProblem – Determine the force in each member of the trussshown belowMethod of JointsProblem – Determine the force in each member of the trussshown below800 lb4 ft 4 ft4 ftAB CDE6060Method of JointsProblem – Determine the force in each member of the trussshown belowZero Force Members Truss analysis may be simplified by determiningmembers with no loading or zero–force. These members may provide stability or be useful if theloading changes. Zero–force members may be determined by inspectionof the jointsCIVL 3121 Trusses - Method of Joints 3/5
- 4. Zero Force MembersCase 1: If two members are connected at a joint and there isno external force applied to the jointF1F2xy0yF 1 sinF F1 = 00xF 1 2cosF F F2 = 0Zero Force MembersCase 2: If three members are connected at a joint and thereis no external force applied to the joint and two of themembers are colinearF1F2xy0yF 1 sinF F1 = 0F3Zero Force MembersDetermine the force in each member of the truss shownbelow:ABCDEFG8 ft 8 ft800 lbUsing Case 2 FBG and FDFare zero-force membersZero Force MembersDetermine the force in each member of the truss shownbelow:ABCDEFG8 ft 8 ft800 lbUsing Case 1 FAG and FCGare zero-force membersUsing Case 1 FEF and FCFare zero-force membersZero Force MembersDetermine the force in each member of the truss shownbelow:ABCDEFG8 ft 8 ft800 lbThe remaining non-zeroforces can be found usingthe method of jointsMethod of JointsThe equations of equilibrium for Joint C0xF 4 45 5BC CDF F FBC = FCDFBC800 lbFCD43430yF 3 38005 5BC CDF F lb FBC = -666.7 lbFBC = 666.7 lb (C)CIVL 3121 Trusses - Method of Joints 4/5
- 5. End of Trusses - Part 2Any questions?CIVL 3121 Trusses - Method of Joints 5/5

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