METHODS OF JOINTS
Problem Find the force acting in all members of the truss shown in Figure T-01.
Summary
FAB=5.56 kN tension
FAE=75.56 kN tension
FBC=4.45 kN tension
FBE=3.34 kN compression
FCD=88.87 kN compression
FCE=5.57 kN tension
FCF=50 kN tension
FDF=71.11 kN tension
FFE=71.11 kN tension
Problem 002-mj
The structure in Fig. T-02 is a truss which is pinned to the floor at point A, and supported by a roller
at point D. Determine the force to all members of the truss.
Summary
FAB = 8.73 kN tension
FAG = 21.82 kN compression
FBC = 15.71 kN tension
FBF = 8.73 kN compression
FBG = 8.73 kN tension
FCD = 5.24 kN tension
FCE = 13.09 kN tension
FCF = 13.09 kN compression
FDE = 13.09 kN compression
FEF = 10.48 kN compression
FFG = 12.22 kN compression
Problem 003-mj
Find the force in each member of the truss shown in Fig. T-04.
Summary
Top chords
FDE = 64 kN tension
FEF = 176 kN tension
Bottom chords
FAB = 120 kN compression
FBC = 32 kN compression
Web members
FAF = 140 kN tension
FAE = 150.78 kN compression
FBE = 150.78 kN tension
FBD = 86.16 kN compression
FCD = 86.16 kN tension
Problem 004-mj
The truss pinned to the floor at D, and supported by a roller at point A is loaded as shown in Fig. T06. Determine the force in member CG.
FCG=8.01 kN tensionFCG=8.01 kN tension
answer
FCG=8.02 kN tensionFCG=8.02 kN tension
answer
Problem 005-mj
Compute the force in all members of the truss shown in Fig. T-08.
Summary
Problem 403
Determine the force in each bar of the truss shown in Fig. P-403.
Summary
Problem 404
Determine the forces in the members of the roof truss shown in Fig. P-404.
Summary
AB = 450 N compression
AC = 389.71 N tension
BC = 450 N tension
BD = 900 N compression
CD = 389.71 N tension
Problem 405
Determine the force in each bar of the truss shown in Fig. P-405 caused by lifting the 120 kN load at a
constant velocity of 8 m per sec. What change in these forces, if any, results from placing the roller
support at D and the hinge support at A?
a. Summary
AB = 160 kN compression
AC = 128 kN tension
BC = 192 kN tension
CD = 32 kN tension
BD = 160 kN compression
b. Summary
AB = 160 kN compression
AC = 224 kN tension
BC = 192 kN tension
CD = 128 kN tension
BD = 160 kN compression
Problem 406
The cantilever truss in Fig. P-406 is hinged at D and E. Find the force in each member.
Summary
AB = 2000 N tension
AC = 1732.05 N compression
BC = 866.02 N compression
BD = 2500 N tension
CD = 2020.72 N tension
CE = 3175.42 N compression
Problem 407
In the cantilever truss shown in Fig. P-407, compute the force in members AB, BE, and DE.
We may check FDE by Method of Sections (Optional)
Summary (Answers)
AB = 1732.05 lb tension
BE = 1154.70 lb compression
DE = 1154.70 lb compression
Problem 408
Compute the force in each member of the Warren truss shown in Fig. P-408.
Summary
AB = 4041.45 lb compression
AC = 2020.72 lb tension
BC = 2886.75 lb tension
BD = 3464.10 lb compression
CD = 1732.05 lb tension
CE = 2598.07 lb tension
DE = 5196.15 lb compression
Problem 409
Determine the force in members AB, BD, BE, and DE of the Howe roof truss shown in Fig. P-409.
FAB=9.45kN compression
answer
FBE=2.7kN compression
answer
FBD=6.75kN compression
answer
FDF=6.75kN compression
FDE=6.75kN tension
answer
Problem 410
Determine the force in each member of the Pratt roof truss shown in Fig. P-410.
Summary
AB = FH = 20 kN compression
AC = GH = 16 kN tension
BC = FG = 8 kN compression
BD = DF = 20 kN compression
CD = DG = 9.6148 kN tension
CE = EG = 10.6667 kN tension
DE = 0
Problem 411
Determine the force in members AB, AC, BD, CD, and CE of the cantilever truss shown in Fig. P-411. If
the loads were applied at C and E instead of at B, specify which members would have their internal
force changed.
Summary of Answers
AB = 180.2776 kN tension
AC = 150 kN compression
BD = 180.2776 kN tension
CD = 250 kN tension
CE = 300 kN compression
FDF=360.5551kN tension
FEF=335.4102kN tension
FEG=450kN compression
FDE=300kN compression
With 200 kN loads moved from B and D to C and E, respectively
Summary of Changes in Internal Forces
BC from 200 kN compression to 0
DE from 300 kN compression to 100 kN compression
Problem 412
Compute the force in each member of the truss shown in Fig. P-412. If the loads at B and D are
shifted vertically downward to add to the loads at C and E, would there be any change in the
reactions? Which members, if any, would undergo a change in internal force?
Summary
AB = 983.87 kN compression
AC = 440 kN tension
BD = 491.94 kN compression
BC = 700 kN tension
CD = 1118.03 kN compression
CE = 1440 kN tension
DE = 200 kN tension
EF = 1440 kN tension
DF = 1609.97 kN compression
With Loads at B and D moved and added to loads at C and E, respectively
RA and RF will not change, thus, internal forces of AB, AC, DF, and EF will not change.
By inspection at joint E, CE will not change because EF did not change but DE changed from 200 kN
tension to 1000 kN tension.
Thus, only BC and DE changed
BC; from 700 kN tension to 1100 kN tension
DE; from 200 kN tension to 1000 kN tension
Problem 413
Determine the force in each member of the crane shown in Fig. P-413.
Summary
AC = 131 kN tension
AB = 120 kN compression
BC = 60 kN compression
BD = 104 kN compression
CD = 0
CE = 90 kN tension
Problem 414
Determine the force in members AB, BD, and CD of the truss shown in Fig. P-414. Also solve for the
force on members FH, DF, and DG.
Summary
AB = 84.85 kN compression
BD = 63.24 kN compression
CD = 12.5 kN compression
Solving for force in members FH, DF, and DG
Summary
FH = 127.28 kN compression
DF = 94.87 kN compression
DG = 37.5 kN tension
METHODS OF SECTIONS
Problem 001-ms
From the truss in Fig. T-01, determine the force in mebers BC, CE, and
EF.
FBC=4.45 kN tension
FEF=71.11 kN tension
FCE=5.55 kN tension
answer
answer
answer
Problem 002-ms
The roof truss shown in Fig. T-03 is pinned at point A, and supported by a
roller at point H. Determine the force in member DG.
FDG=18.14 kN tension
answer
Problem 003-ms
The truss in Fig. T-04 is pinned to the wall at point F, and supported by a
roller at point C. Calculate the force (tension or compression) in
members BC, BE, and DE.
Figure T-04
FBE=150.78 kN tension
answer
FBC=120 kN compression
answer
FDE=64 kN tension
answer
Problem 004-ms
For the truss shown in Fig. T-05, find the internal fore in member BE.
FBE=0
answer
Problem 005-ms
The structure shown in Figure T-07 is pinned to the floor at A and H.
Determine the magnitude of all the support forces acting on the structure
and find the force in member BF.
RH=400 kN upward
AV=300 kN downward
AH=80 kN to the left
answer
answer
answer
If you wish to check the reaction forces at A and H, pass a cutting plane through members AB, AG,
and AH. As shown in the figure above, the cutting plane b-b cut these members. Take the section
above b-b and replace the cut members by internal force to maintain equilibrium. Solve for the
internal force in members AB, AG, and AH. With these forces known, you can then check the
reactions through joints A and H. The free body diagram of the section above b-b is similar to the
free body diagram of section above a-a. See the solution of section a-a below for easy reference of
doing the section b-b.
From section above a-a
ΣFH=0ΣFH=0
441√FBF=80441FBF=80
FBF=128.06 kN tension
answer
Problem 417
Using the method of sections, determine the force in members BD, CD,
and CE of the roof truss shown in Fig. P-417.
FBD=160kN compression
answer
FCD=200kN compression
answer
FCE=320kN tension
answer
Problem 418
The Warren truss loaded as shown in Fig. P-418 is supported by a roller at C and a hinge at G. By the
method of sections, compute the force in the members BC, DF, and CE.
FBC=44.721 kN compression
answer
FCE=10 kN tension
answer
FDF=80 kN compression
answer
Problem 419
Use the method of sections to determine the force in members BD, CD,
and CE of the Warren truss described in Problem 408.
FBD=3464.10 lb compression
answer
FCD=1732.05 lb tension
answer
FCE=2598.08 lb tension
answer
Problem 420
Determine the force in members DF, DG, and EG of the Howe truss
shown in Fig. P-420.
FDF=280 kN compression
answer
FDG=150 kN compression
answer
FEG=400 kN tension
answer
Problem 421
Use the method of sections to compute for the force in members DF, EF, and EG of the cantilever
truss described in Problem 411 and Fig. P-411.
FDF=360.5551 kN tension
FEF=335.4102 kN tension
FEG=450 kN compression
answer
answer
answer
Problem 422
Refer to the truss described in Problem 412 and compute the force in members BD, CD, and CE by
the method of sections.
FBD=491.93 kN compression
answer
FCD=1118.03 kN compression
answer
FCE=1440 kN tension
answer
Problem 423
Use the method of sections to determine the force acting in members DF, EF, and EG of the Howe
truss described in Problem 409.
FDF=6.75 kN compression
answer
FEF=1.8 kN compression
answer
FEG=7.404 kN tension
answer
Problem 424 - Method of Joints Checked by Method of Sections
Problem 424
For the truss shown in Fig. P-424, determine the force in BF by the method of joints and then check
this result using the method of sections. Hint: To apply the method of sections, first obtain the value
of BE by inspection.
FBF=2.5 kN compression
answer
Problem 425
In the Fink truss shown in Fig. P-425, the web members BC and EF are perpendicular to the inclined
members at their midpoints. Use the method of sections to determine the force in members DF, DE,
and CE.
FCE=40 kN tension
answer
FDE=513−−√ kN=18.03 kN tension
answer
FDF=255–√ kN=55.90 kN compression
answer
Problem 426
Using the method of sections, compute the force in bars FH, GH, and EK.
FFH=110 kN compression
answer
FGH=303–√ kN=51.96 kN tension
answer
FEK=403–√ kN=69.28 kN tension
answer