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Trusses

2. COMSATS University Islamabad,Abbottabad
CEE 207:Basic StructureAnalysis
Module04
Determinate TrussAnalysis
Analysis of pin-connected determinate
frames (Trusses) and to measure the Axial forces.
By:
Engr. Muhammad Jamal Butt
CivilEngineeringDepartment
CUI, Abbottabad
[email protected]

3. TopicstobeCovered
•Trusses
•CommonTypesofTrusses
•TrussAnalysisAssumptions
•ClassificationofCoplanarTrusses
•DeterminacyofCoplanarTrusses
•StabilityofCoplanarTrusses
•MethodofJoints
•Zero-ForceMembers
•MethodofSections

4. Trusses:
Atrussisastructurecomposedofslendermembersjoinedtogetherat
theirendpoints.
Planartrusseslieinasingleplane.
Typically,thejointconnectionsareformedbyboltingorweldingthe
endmemberstogethertoa commonplate,calleda gussetplate.
Analysisof StaticallyDeterminateTrusses
gussetplate

5. Trusses:
Analysisof StaticallyDeterminateTrusses

6. Trusses:
Analysisof StaticallyDeterminateTrusses

7. TypesofTrusses:
1.Rooftrusses:
Ingeneral,theroofloadistransmittedtothetrussbyaseriesof
purlins.
The roof truss along with its supporting columns is termed a bent.
Thespacebetweenbentsiscalleda bay.
Analysisof StaticallyDeterminateTrusses
topcord
roof
purlins
knee
brace
bottom
cord
span bay

8. TypesofTrusses:
1.Rooftrusses:
Analysisof StaticallyDeterminateTrusses
Howetruss Pratttruss
ModifiedHowetrussWarrentruss
saw-toothtruss
Finktruss
three-hingedarch

9. TypesofTrusses:
2.Bridgetrusses:
Theloadistransmittedbythedecktoaseriesofstringersandthento
asetoffloor beams.
Thefloorbeamsaresupportedby twoparalleltrusses.
Thesupportingtrussesareconnectedtopandbottombylateral
bracing.
Additionalstabilitymaybeprovidedbyportaland sway bracing
Analysisof StaticallyDeterminateTrusses

10. TypesofTrusses:
2.Bridgetrusses:
Analysisof StaticallyDeterminateTrusses
topcord
deck
stringers
portal
endpost
swaybracing
toplateralbracing
Portalbracing
panel
floorbeam
bottomcord

11. TypesofTrusses:
2.Bridgetrusses:
Analysisof StaticallyDeterminateTrusses

12. TypesofTrusses:
2.Bridgetrusses:
Analysisof StaticallyDeterminateTrusses
troughPratttruss WarrentrussHowetruss
deckPratttruss parkertruss
(Pratttrusswithcurvedchord)
BaltimoretrussKtruss

13. TrussAnalysisAssumptions:
Twoimportantassumptionswillbemadeinordertoidealizethetruss.
1.Themembersarejoinedtogetherbysmoothpins.
Incaseswhereboltedorweldedjointconnectionsareused,this
assumptionisgenerallysatisfactoryprovidedthecenterlinesofthe
joiningmembersareconcurrentatapoint,asinFigure.
Analysisof StaticallyDeterminateTrusses
gussetplate

14. TrussAnalysisAssumptions:
Twoimportantassumptionswillbemadeinordertoidealizethetruss.
1.Themembersarejoinedtogetherbysmoothpins.
Theactualconnectionsdogivesomerigiditytothejointandthisin
turnintroducesbendingoftheconnectedmemberswhenthetrussis
subjectedtoaload.
The bending stress developed in the members is called secondary
stress,whereasthestressinthemembersoftheidealizedtruss,having
pin-connectedjoints,iscalledprimarystress.
Asecondarystressanalysisofatrusscanbeperformed usinga
computer.
Analysisof StaticallyDeterminateTrusses

15. TrussAnalysisAssumptions:
Twoimportantassumptionswill bemadeinordertoidealizethetruss.
2.Allloadingsareappliedatthejoints.
Inmostsituations,suchasforbridgeandrooftrusses,thisassumption
istrue.
Frequentlyintheforceanalysis,theweightofthemembersis
neglected,sincetheforcesupportedbythemembersislargein
comparisonwiththeirweight.
Iftheweightistobe includedin theanalysis,itisgenerallysatisfactory
to apply it as a vertical force, half of its magnitude applied at each end
ofthemember.
Analysisof StaticallyDeterminateTrusses

16. TrussAnalysisAssumptions:
Becauseofthesetwoassumptions,eachtrussmemberactsasanaxial
forcemember,andthereforetheforcesactingattheendsofthemember
mustbedirectedalongtheaxisofthemember.
Iftheforcetendstoelongatethemember,itisatensileforce(T),
whereasiftheforcetendstoshortenthemember,itisacompressive
force(C).
Intheactualdesignofatrussitisimportanttostatewhethertheforceis
tensileorcompressive.
Analysisof StaticallyDeterminateTrusses

17. ClassificationofCoplanarTrusses:
1.SimpleTruss
Thesimplestframeworkthatisrigidorstableisatriangle.
Therefore,asimpletrussisconstructedstartingwithabasictriangular
elementandconnectingtwomemberstoformadditionalelements.
Aseachadditionalelementoftwomembersisplacedonatruss,the
numberofjointsisincreasedbyone.
Mostoften,compressionmembersmustbemadethickerthantension
members,becauseofthebucklingorsuddeninstabilitythatmayoccur
incompressionmembers.
Analysisof StaticallyDeterminateTrusses

18. ClassificationofCoplanarTrusses:
1.SimpleTruss
Thesimplestframeworkthatisrigidorstableisa triangle.
Analysisof StaticallyDeterminateTrusses

19. ClassificationofCoplanarTrusses:
2.CompoundTruss
Thistrussisformedby connectingtwoormoresimpletrussestogether.
Thistypeof trussisoftenusedforlargespans.
Therearethreewaysinwhich simpletrussesmaybe connectedto form
acompoundtruss:
i.Trussesmaybeconnectedby acommonjointandbar.
Analysisof StaticallyDeterminateTrusses

20. ClassificationofCoplanarTrusses:
2.CompoundTruss
Therearethreewaysinwhich simpletrussesmaybe connectedto form
acompoundtruss:
ii.Trussesmaybejoinedbythreebars.
Analysisof StaticallyDeterminateTrusses

21. ClassificationofCoplanarTrusses:
2.CompoundTruss
Therearethreewaysinwhich simpletrussesmaybe connectedto form
acompoundtruss:
iii.Trussesmaybe joinedwherebarsof a largesimple truss, calledthe
main truss, have been substituted by simple trusses, called secondary
trusses
Analysisof StaticallyDeterminateTrusses

22. ClassificationofCoplanarTrusses:
3.ComplexTruss
Thisis atrussthatcannotbeclassifiedasbeingeithersimple or
compound.
Analysisof StaticallyDeterminateTrusses

23. ClassificationofCoplanarTrusses:
Analysisof StaticallyDeterminateTrusses

24. DeterminacyofCoplanarTrusses:
•Since all the elements of a truss are two-force members, the moment
equilibriumisautomaticallysatisfied.
•Thereforetherearetwo equationsofequilibriumforeachjoint,j, in
a truss. If r is the number of reactions and b is the number of
members
Analysisof StaticallyDeterminateTrusses
Inparticular,thedegreeofindeterminacyisspecifiedby thedifferenceinthe
numbers(b+r) -2j.

25. Analysisof StaticallyDeterminateTrusses
DeterminacyofCoplanarTrusses:
Example01:

26. Analysisof StaticallyDeterminateTrusses
DeterminacyofCoplanarTrusses:
Example02:

27. Analysisof StaticallyDeterminateTrusses
DeterminacyofCoplanarTrusses:
Example03:
S.I=b+r–2j
S.I=14+4–16=2°
S.I=21+3–20=4°

28. StabilityofCoplanarTrusses:
•Ifb+r<2j,atrusswillbeunstable,whichmeansthestructurewill
collapsesincetherearenotenoughreactionstoconstrainallthe
joints.
•Atrussmayalsobeunstableifb+r>2j.Inthiscase,stabilitywill
bedeterminedbyinspection.
Analysisof StaticallyDeterminateTrusses

29. Analysisof StaticallyDeterminateTrusses
StabilityofCoplanarTrusses:
Example04:

30. StabilityofCoplanarTrusses:
Externalstability–
Astructure(truss)isexternallyunstableifitsreactionsareconcurrent
orparallel.
Analysisof StaticallyDeterminateTrusses

31. StabilityofCoplanarTrusses:
Internalstability–
•Maybedeterminedbyinspectionofthearrangementofthetruss
members.
•Asimpletrusswillalwaysbeinternallystable.
•Thestabilityofacompoundtrussisdeterminedbyexamininghow
thesimple trussesareconnected
•Thestabilityofacomplextrusscanoftenbedifficulttodetermineby
inspection.
Analysisof StaticallyDeterminateTrusses

32. StabilityofCoplanarTrusses:
Internalstability–
•Ingeneral,thestabilityofanytrussmaybecheckedbyperforminga
completeanalysisofthestructure.Ifauniquesolutioncanbefound
forthesetofequilibriumequations,thenthetrussisstable.
Analysisof StaticallyDeterminateTrusses

33. Analysisof StaticallyDeterminateTrusses
StabilityofCoplanarTrusses:
Internalstability–

34. Example05:
Classifyeachofthetrussesinthefigurebelowasstable,unstable,
staticallydeterminate,orstaticallyindeterminate.Thetrussesare
subjectedtoarbitraryexternalloadingsthatareassumedtobeknown
andcanactanywhereonthetrusses.
Analysisof StaticallyDeterminateTrusses

35. Example05:
Solution:
Analysisof StaticallyDeterminateTrusses
Externallystable,sincethereactions
arenotconcurrentorparallel.Since
b=19,r=3,j=11,thenb+r=2j
or22=22.Therefore,thetrussis
staticallydeterminate.
Byinspectionthetrussisinternally
stable.
Externallystable.Sinceb=15,r=4,
j = 9, then b + r > 2j or 19 > 18. The
truss is statically indeterminate to the
firstdegree.By inspectionthetrussis
internallystable.

36. Example05:
Solution:
Analysisof StaticallyDeterminateTrusses
Externallystable.Sinceb=9,r= 3,
j=6,thenb+r=2jor12=12.The
truss is statically determinate. By
inspection the truss is internally
stable.
Externallystable.Sinceb=12,r =3,
j = 8, then b + r < 2j or 15 < 16. The
trussisinternallyunstable.

37. Analysis ofDeterminateTrusses
Analysisof StaticallyDeterminateTrusses

38. MethodofJoints:
Ifatrussisinequilibrium,theneachofitsjointsmustbein
equilibrium.
Themethodofjointsconsistsofsatisfyingtheequilibriumequations
for forcesactingon eachjoint.
Analysisof StaticallyDeterminateTrusses

39. MethodofJoints:
Recall,thatthelineofactionof aforceactingon a jointisdetermined
bythegeometryof thetrussmember.
Thelineofactionisformedbyconnectingthetwoendsofeach
memberwitha straightline.
Sincedirectionof theforceisknown, theremainingunknown isthe
magnitudeoftheforce.
Analysisof StaticallyDeterminateTrusses

40. MethodofJoints:
Procedurefor analysis:
Thefollowingisaprocedureforanalyzingatrussusingthemethodof
joints:
1.Determinethesupportreactions
2.Drawthefreebodydiagramforeachjoint.
Ingeneral,assumealltheforcememberreactionsaretension(thisis
notarule,however,itishelpfulinkeepingtrackoftensionand
compressionmembers).
3.Writetheequationsofequilibriumforeachjoint,
Analysisof StaticallyDeterminateTrusses

41. MethodofJoints:
Procedureforanalysis:
Thefollowingisaprocedureforanalyzingatrussusingthemethodof
joints:
4.Ifpossible,beginsolvingtheequilibriumequationsatajointwhere
onlytwounknownreactionsexist.Workyourwayfromjointtojoint,
selectingthenewjointusingthecriterionoftwounknownreactions.
5.Solvethejointequationsofequilibriumsimultaneously,typically
usingacomputeroranadvancedcalculator.
Analysisof StaticallyDeterminateTrusses

42. Analysisof StaticallyDeterminateTrusses
r=3
b=3
j=3
r+b=2xj
3+3=2x3
6=6,StaticallyDeterminate
2m
2m
45
o
A
MethodofJoints:
Example06:Analyzethefollowingtruss.
B
500N
C

43. Analysisof StaticallyDeterminateTrusses
Method of Joints:
Example 06:
Solution:Reactions

44. Analysisof StaticallyDeterminateTrusses
Method of Joints:
Example 06:
Solution:Reactions
2m
45
o
B
500N
2m
C
C
y=500N
A
A
x=500N
A
y=500N

45. MethodofJoints:
Example06:
Solution:Equilibriumatjoint“A”
Analysisof StaticallyDeterminateTrusses
2m
45
o
B
500N
2m
A
C
C
y=500N
A
x=500N
A
y=500N

46. MethodofJoints:
Example06:
Solution:Equilibriumatjoint“B”
Analysisof StaticallyDeterminateTrusses
2m
45
o
B
500N
2m
A
C
C
y=500N
A
x=500N
A
y=500N

47. Analysisof StaticallyDeterminateTrusses
F
ac=500lb
F
ab=500lb
F
bc=707.2lb
45
o
A
MethodofJoints:
Example06:
Solution:FinalAnalyzedStructure
B
500N
C
C
y=500N
A
x=500N
A
y=500N

48. MethodofJoints:
Example 07: Determine the force in each member of the truss shown
inthefigure.
Analysisof StaticallyDeterminateTrusses

49. MethodofJoints:
Example07:
Solution:Determinacy
Analysisof StaticallyDeterminateTrusses
r=3
b=9
j=6
r+b=2xj
3+9=2x6
12=12,
StaticallyDeterminate

50. Analysisof StaticallyDeterminateTrusses
MethodofJoints:
Example07:
Solution:Reactions
∑F
x=0=A
x
A
x= 0k
∑M
A=0= –12x20 –24x40+D
yx60
D
y=20k
∑M
D= 0 = –A
yx60+12x40+24x20
A
y=16k

51. Analysisof StaticallyDeterminateTrusses
F
AB
AB
Fcos(36.87)
F
ABsin(36.87)
MethodofJoints:
Example07:
Solution:Equilibriumatjoint“A”
∑F
y=0 =A
y+F
ABsin(36.87)
F
AB=–26.67k
∑F
x=0=A
x+F
AF+F
ABcos(36.87)
F
AF=+21.34k
F
AF

52. Analysisof StaticallyDeterminateTrusses
F
AF=21.34k F
FE
F
FB
MethodofJoints:
Example07:
Solution:Equilibriumatjoint“F”
∑F
y=0=–12+F
FB
F
FB =+12k
∑F
x=0= –F
AF +F
FE
F
FE =+21.34k

53. MethodofJoints:
Example07:
Solution:Equilibriumatjoint“B”
Analysisof StaticallyDeterminateTrusses
F
BC
F
BF
F
BE
36.87
o
53.13
o 36.87
o
BE
Fcos(36.87)
=12 k
F
BEsin(36.87)
F
BAsin(53.13)
F
BA=26.67k
F
BAcos(53.13)
∑F
y=0
0=–12–F
BEsin(36.87)+F
ABcos(53.13)
F
BE=+6.7k
∑F
x=0
0 = F
BE + F
BEcos(36.87) + F
AB sin(53.13)
F
BC=–26.67k

54. Analysisof StaticallyDeterminateTrusses
F
DCcos(36.87)
F
DC=33.33k
F
DC sin(36.87)
F
CB
F
CE
MethodofJoints:
Example07:
Solution:Equilibriumatjoint“C”
∑F
x=0 = F
CB –F
DCcos(36.87)
F
DC=33.33k
∑F
y=0=–F
C E+F
DCsin(36.87)
F
CE=19.99k

55. MethodofJoints:
Example07:
Solution:Equilibriumatjoint“C”
Analysisof StaticallyDeterminateTrusses
ED
F=26.67k
EC
F=19.99k
EF
F=21.34k
F
EBsin(36.87)
F
EB
F
EBcos(36.87)
∑F
y=0 =–24 +F
EC+F
EBsin(36.87)
F
EB=6.66k
∑F
x=0 =–F
EF +F
ED–F
EBcos(36.87)
EB
F=6.66k

56. MethodofJoints:
Example07:
Solution:FinalAnalyzedStructure
Analysisof StaticallyDeterminateTrusses
26.67k
21.33k
12k
21.34k
6.67k
26.67k
33.34k
19.99k
26.67k
16k
20k
0k

57. Zero-ForceMembers:
Trussanalysisusingthemethodofjointsisgreatlysimplifiedifoneis
abletofirstdeterminethosemembersthatsupportnoloading.
Thesezero-forcemembersmaybenecessaryforthestabilityofthe
trussduringconstructionandtoprovidesupportiftheappliedloading
ischanged.
Thezero-forcemembersofatrusscangenerallybedeterminedby
inspectionofthejoints.
Analysisof StaticallyDeterminateTrusses

58. Zero-ForceMembers:
Thezero-forcemembersofatrusscangenerallybedeterminedby
inspectionof thejointsand theyareoccurintwocases,
Case1:
Iftwo membersareconnectedatajointandthereisno externalforce
appliedto thejoint
Analysisof StaticallyDeterminateTrusses

59. Zero-ForceMembers:
Thezero-forcemembersofatrusscangenerallybedeterminedby
inspectionof thejointsand theyareoccurintwocases,
Case2:
Ifthreemembersareconnectedatajointandthereisnoexternalforce
appliedto thejointand twoofthemembersarecollinear.
Analysisof StaticallyDeterminateTrusses

60. Zero-ForceMembers:
Example08:Indicateallthemembersofthetrussshowninthefigure
belowthathavezeroforce.
Analysisof StaticallyDeterminateTrusses
P
A
B
C
DE
D
yE
y

61. Zero-ForceMembers:
Example08:
Solution:
Analysisof StaticallyDeterminateTrusses
F
CD
F
CB
JointC:
C
F
x= 0:F
CB=0
F
y= 0:F
CD=0
+
+
F
AEA

JointA:
F
AB
y
F=0:
AB AB
Fsin= 0,F=0+
F
x=0:F
AE+0=0,F
AE= 0
+
P
A
B
C
DE
D
yE
y
Zeroforcemember
areshownwithdash
lines

62. Zero-ForceMembers:
Example09:Indicateallthemembersofthetrussshowninthefigure
belowthathavezeroforce.
Analysisof StaticallyDeterminateTrusses
P
A B
C
D
EFG
H

63. Zero-ForceMembers:
Example09:
Solution:
Analysisof StaticallyDeterminateTrusses
Zeroforcemember
areshownwithdash
lines
A
x
A
x
G
x
P
A B
C
D
EFG
H
0
F
EC
F
EF
E
F
y=0:+
F
DCsin=0,F
DC=0
F
x=0:
+
F
DE+0=0,F
DE=0
y
D

x
F
DC
F
DE
JointD
F
EF=0
P
F
x=0:
+
JointE

64. Zero-ForceMembers:
Example09:
Solution:
Analysisof StaticallyDeterminateTrusses
Zeroforcemember
areshownwithdash
lines
A
x
A
x
G
x
P
A B
C
D
EFG
H
JointH
F
HF
F
HA
F
HB
y
H
F
y=0:
+
x
F
HB=0
JointG
G
G
x F
GF
F
GA
F
y=0:+
F
GA=0

65. MethodofSections:
Iftheforcesinonlyafewmembersofatrussaretobedetermined,the
methodofsectionsisgenerallythemostappropriateanalysis
procedure.
Themethodofsectionsconsistsofpassinganimaginarylinethrough
thetruss,cuttingitintosections.
Eachimaginarysectionmustbeinequilibriumiftheentiretrussisin
equilibrium.
Analysisof StaticallyDeterminateTrusses

66. MethodofSections:
Procedureforanalysis:
Thefollowingisaprocedureforanalyzingatrussusingthemethodof
sections:
1.First,ifnecessary,determinethesupportreactionsfortheentire
truss.
2.Next,makeadecisiononhowthetrussshouldbe“cut”intosections
anddrawthecorrespondingfree-bodydiagrams.Ingeneral,thecut
passesthroughnotmorethanthreemembersinwhichtheforcesare
unknown.
Analysisof StaticallyDeterminateTrusses

67. MethodofSections:
Procedureforanalysis:
Thefollowingisaprocedureforanalyzingatrussusingthemethodof
sections:
3.Trytoapplythethreeequationsofequilibriumsuchthat
simultaneoussolutionisnotrequired.
Momentsshouldbesummedaboutpointsthatlieattheintersectionof
thelinesofactionoftwounknownforces,sothattheremainingforce
maybedetermined.
Eachimaginarysectionmustbeinequilibriumiftheentiretrussisin
equilibrium.
Analysisof StaticallyDeterminateTrusses

68. Analysisof StaticallyDeterminateTrusses
r=3
b=3
j=3
r+b=2xj
3+3=2x3
6=6,
StaticallyDeterminate
100N
A
B C D
EFG
2m
2m 2m 2m
E
x
D
x
MethodofSections:
Example 10: Determine the force in members BC, GC and GF of the
trussshowninfig below usingmethodofsections.
D
y

69. MethodofSections:
Example10:
Solution:Typicallythesectionwiththefewestforcesorwithsection
withthemostconvenientgeometryisselected.
•Inthisexampletheleft-handside.
•Applythethreeequationsof
equilibriumtothesection.
•Ifpossible,attempttodevelop
anequationinjustoneunknown.
•Lookforpointswherethelines
of action of several forces are
concurrent.
Analysisof StaticallyDeterminateTrusses
100N
A
B C D
EFG
2m
2m 2m 2m
E
x
D
x
D
y
a
a

70. MethodofSections:
Example10:
Solution:Typicallythesectionwiththefewestforcesorwithsection
withthemostconvenientgeometryisselected.
Analysisof StaticallyDeterminateTrusses
100N
A
G
2m
45
o
F
GF
F
GC
B F
BC
C
+
+M
G=
100(2)–F
BC(2) =0,F
BC=100N(T)
F
y=0:
-100+F
GCsin45
o
=0,F
GC=141.42N(T)
+M
C=
100(4)–F
GF(2)=0,F
GF=200N (C)

71. Analysisof StaticallyDeterminateTrusses
A
x=0
A
y=9kN
E
y=7kN
A
B C D
E
F
MethodofSections:
Example 11: Determine the force in members GF and GD of the truss
shown in the figure below. State whether the members are in tension or
compression.Thereactionsatthesupportshavebeencalculated.
G
H
6kN8kN 2kN
3 m3m 3m 3m
3m
4.5m

72. MethodofSections:
Example11:
Solution:Typicallythesectionwiththefewestforcesorwithsection
withthemostconvenientgeometryisselected.
Analysisof StaticallyDeterminateTrusses
a
a
A
x=0
A
y=9kN
E
y=7kN
A
B C D
E
F
G
H
6kN8kN 2kN
3 m3m 3m 3m
3m
4.5m

73. MethodofSections:
Example11:
Solution:Equilibriumatsectionaa.
Analysisof StaticallyDeterminateTrusses
E=7kN
E
F
Sectiona-a
F
FG
F
DG
F
DC
D
2 kN
y
3m 3m
26.6
o
O
26.6
o
56.3
o
+∑M
D=0:
FG
Fsin26.6
o
(6)+7(3)=0,
F
FG=-7.82kN(C)
+
O
∑M=0:
-7(3)+2(6)+F
DGsin56.3
o
(6)=0,
F
DG=1.80kN(C)

74. References
•StructuralAnalysisby R.C. Hibbeler
•OnlineCivilEngineeringblogs