Determine the force in members: (a) BC, (b) CD, (c) AD, (d) BD, (e) AB of the loaded truss. (Note: the compression force must be negative.)Problem 6-7Determine the force in each member of the truss and state if the members are in tension orcompression.Units Used:kN 1000N:=Given:F13kN:=F28kN:=F34kN:=F410kN:=a 2m:=b 1.5m:=Solution:ΣMA0=;F1b()− F3a()− F42a()− Ey2a()+ 0= For the exclusive use of adopters of the Hibbeler series of books.© 2006 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected under all copyright laws as they currently exist. No portion of this material maybe reproduced, in any form or by any means, without permission in writing from the publisher.EyF1b()⋅ F3a()⋅+ F42a()⋅+2a:=Ey13.125 kN=+↑ΣFy = 0;AyF2− F3− F4− Ey+ 0=AyF2F3+ F4+ Ey−:=Ay8.875 kN=+→ΣFx = 0;AxF1− 0= AxF1:=Ax3kN=Joint B:+→F1FBC− 0= FBCF1:=ΣFx = 0;FBC3kN= C()+↑ΣFy = 0;FBAF2− 0= FBAF2:=FBA8kN= C()Joint A:+↑ΣFy = 0;AyFBA−bb2a2+FAC− 0=FACAyFBA−bb2a2+:=FAC1.46 kN= C()+→Ax−ab2a2+FAC− FAF+ 0=ΣFx = 0;FAFAxFACa⋅b2a2++:=FAF4.17 kN= T()Joint C:+→FBCab2a2+FAC⋅+ FCD− 0=ΣFx = 0;FCDFBCFACa⋅b2a2++:=FCD4.17 kN= C()+↑ΣFy = 0;FCFF3−FACb⋅b2a2++ 0=FCFF3FACb⋅b2a2+−:=FCF3.13 kN= C()Joint E:+→ΣFx = 0;FEF0kN:= FEF0.00 kN=+↑ΣFy = 0;EyFED− 0=FEDEy:= FED13.1 kN= C()Joint D:+↑ΣFy = 0;FEDF4−FDFb⋅b2a2+− 0=FDFFEDF4−()b2a2+⋅b:=FDF5.21 kN= T()Sample Problem 4/1Compute the force in each member of the loaded cantilever truss by themethod of joints.Solution. If it were not desired to calculate the external reactions at D and E,the analysis for a cantilever truss could begin with the joint at the loaded end.However, this truss will be analyzed completely, so the first step will be to com-pute the external forces at D and E from the free-body diagram of the truss as awhole. The equations of equilibrium giveNext we draw free-body diagrams showing the forces acting on each of theconnecting pins. The correctness of the assigned directions of the forces is veri-fied when each joint is considered in sequence. There should be no questionabout the correct direction of the forces on joint A. Equilibrium requiresAns.Ans.where T stands for tension and C stands for compression.Joint B must be analyzed next, since there are more than two unknownforces on joint C. The force BC must provide an upward component, in whichcase BD must balance the force to the left. Again the forces are obtained fromAns.Ans.Joint C now contains only two unknowns, and these are found in the sameway as before:Ans.Ans.Finally, from joint E there resultsAns.and the equation ΣFx 0 checks.DE  11.55 kN C0.866DE  10[ΣFy  0] CE  63.5 kN C CE 17.32 0.5(34.6) 0.5(57.7)  0[ΣFx  0] CD  57.7 kN T 0.866CD 0.866(34.6) 20  0[ΣFy  0] BD  34.6 kN T BD 2(0.5)(34.6)  0[ΣFx  0] BC  34.6 kN C 0.866BC 0.866(34.6)  0[ΣFy  0]AC  17.32 kN C AC 0.5(34.6)  0[ΣFx  0]AB  34.6 kN T 0.866AB 30  0[ΣFy  0] Ey  10 kN 80 sin 30  Ey 20 30  0[ΣFy  0] Ex  69.3 kN 80 cos 30 Ex  0[ΣFx  0] T  80 kN 5T 20(5) 30(10)  0[ΣME  0]180 Chapter 4 Structures5 m30 kN 20 kN5 m5 m 5 m 5 m 5 m5 mCEBDAyxTExEy30°60°30 kN 20 kN5 m5 m5 m5 m30 kNACBCBDAB =34.6 kNABJoint A Joint Bxy60°60°60°20 kN 10 kNDECDCEBC =34.6 kNAC =17.32 kNCE =63.5 kN69.3 kNJoint C Joint E60°60° 60°Helpful Hint It should be stressed that the ten-sion/compression designation refersto the member, not the joint. Notethat we draw the force arrow on thesame side of the joint as the memberwhich exerts the force. In this waytension (arrow away from the joint)is distinguished from compression(arrow toward the joint).c04.qxd 1/26/06 1:25 PM Page 180Sample Problem 4/5The space truss consists of the rigid tetrahedron ABCD anchored by a ball-and-socket connection at A and prevented from any rotation about the x-, y-, orz-axes by the respective links 1, 2, and 3. The load L is applied to joint E, whichis rigidly fixed to the tetrahedron by the three additional links. Solve for theforces in the members at joint E and indicate the procedure for the determina-tion of the forces in the remaining members of the truss.Solution. We note first that the truss is supported with six properly placedconstraints, which are the three at A and the links 1, 2, and 3. Also, with m  9members and j  5 joints, the condition m  6  3j for a sufficiency of membersto provide a noncollapsible structure is satisfied.The external reactions at A, B, and D can be calculated easily as a first step,although their values will be determined from the solution of all forces on eachof the joints in succession.We start with a joint on which at least one known force and not more thanthree unknown forces act, which in this case is joint E. The free-body diagram ofjoint E is shown with all force vectors arbitrarily assumed in their positive ten-sion directions (away from the joint). The vector expressions for the three un-known forces areEquilibrium of joint E requiresRearranging terms givesEquating the coefficients of the i-, j-, and k-unit vectors to zero gives the threeequationsSolving the equations gives usAns.Thus, we conclude that FEBand FECare compressive forces and FEDis tension.Unless we have computed the external reactions first, we must next analyzejoint C with the known value of FECand the three unknowns FCB, FCA, and FCD.The procedure is identical with that used for joint E. Joints B, D, and A are thenanalyzed in the same way and in that order, which limits the scalar unknowns tothree for each joint. The external reactions computed from these analyses must,of course, agree with the values which can be determined initially from an analy-sis of the truss as a whole.FEB  L/ 2 FEC  5L/6 FED  5L/6FEC  FED  0FEB 2  3FEC5  L FEB 2  3FED5  0 L FEB 2 3FEC5i   FEB 2 3FED5j   4FEC5 4FED5k  0 Li  FEB 2 ( i j)  FEC5 ( 3i  4k)  FED5 ( 3j 4k)  0[ΣF  0] L  FEB  FEC  FED  0 orFEB  FEB 2 ( i j), FEC  FEC5 ( 3i 4k), FED  FED5 ( 3j 4k)Article 4/5 Space Trusses 199xyAECDLB214 m3 m4 m3z3 mxyAEFEDFECFEBLB4 m3 m3 m4 mzHelpful Hints Suggestion: Draw a free-body dia-gram of the truss as a whole and ver-ify that the external forces acting onthe truss are Ax Li, Ay