Problem 6.1Determine the axial forces in the mem-bers of the truss and indicate whether they are in tension(T) or compression (C).Strategy:Draw free-body diagram of jointA.Bywriting the equilibrium equations for the joint, you candetermine the axial forces in the two members.BAC30°60°800 NSolution:30°FABFAC60°A800 NyxAssume the forces are in the directions shown (both in tension). If aforce turns up negative, that force will be in compression.Equilibrium Eqns.XFx:−FABcos 30◦+FACcos 60◦+800=0XFy:−FABsin 30◦−FACsin 60◦=0BAC30°60°800 NSolving: We getFAB= 693N (tension)FAC=−400N (compression)Problem 6.2The truss supports a 10-kN load atC.(a) Draw the free-body diagram of the entire truss, anddetermine the reactions at its supports.(b) Determine the axial forces in the members. Indicatewhether they are in tension (T) or compression (C).10 kNCBA3 m4 mSolution:(a) The free-body diagram of the system is shown. Thesum of the moments aboutBis:MB=3Ax−4(10) = 0, fromwhichAx=13.33kN. The sums of the forces:XFx=Ax+Bx,from whichBx=−Ax=−13.33kN.XFy=By−10=0,from whichBy=10kN. (b) The interior angleACBisα=tan−1(0.75)=36.87◦. (b) Assume that the unknown forces actaway from the joint. Denote the axial force in the memberI,KbyIK. The axial forces areFCB=BC(−icosα+jsinα), andFCA=−ACi. Summing the forces:XFy=sinα−,from which=16.67kN(T).XFx=−cosα−AC,from whichAC=−13.33kN(C). For the jointA,10 kNCBA3 m4 mBXAXBY3 m4 mWXFy=AB,from whichAB
Problem 6.3In Example 6.1, suppose that the 2-kNload is applied atDin the horizontal direction, point-ing fromDtowardB. What are the axial forces in themembers?Solution:First, solve for the support forces and then use themethod of joints at each joint to solve for the forces.BXAXAYBθ6 mtan =10 mDxy2 kN610=30.96XFx:Bx+Ax−2kN=0XFy:AyXMB:−6AxSolving, we getAx=Ay,Bx=2kNJoint A:AX=0AY=0ABAC=30.96θ=30.96◦XFx:ACcosθXFy:−AB−ACsinθSolving, we getAB=ACJoint C:BCCDCAC = 0 fromabove =30.96°XFx:0AC/cosθ−BCcosθ+CDcosθ−+C5 m5 mADB2 kN3 m3 mXFy:0AC/sinθ−sinθ−sinθ−−Solving, we get=Joint B:ABBCBDBXxWe already knowAB=andBxkNXFx:0/cosθ+BD+Bx+2kN=−2kN(compression)XFy:(all forces zero)=0we haveAB=AC===−2kN(c)Note that we did not have to use jointDas we had already solved for the forcesthere. TheFBDatDis, with the (−) sign, is opposite the direction shown.CD=0BD=−2 kN (c)D2 kN
Problem 6.4Determine the axial forces in the mem-bers of the truss.0.6 m0.4 m0.3 m1.2 m2 kNCBASolution:First, solve for the support reactions atBandC, andthen use the method of joints to solve for the forces in the members.
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