Truss- Problems With Solution (Structural)

7/30/2019 Truss- Problems With Solution (Structural)

1/10

QUESTION1:

Determinetheforcesactinginallthemembersofthetrussshowninthefigure.

Answer:First,weshoulddeterminethesupportreactionsbyconsideringthe

FBDofthewholetruss.Then,wecanstarttosolvetheproblemfrom

aconvenientjointbyapplyingthemethodofjoints.

EquationsofEquilibriumforthewholetruss:

(4) 3(2) 0 ; C 1.5 k

0 ; 1.5 k

3 0 ; 3 k

A

y y y

x x x

M C N

F C A A N

F A A N

We start applying theMOJ from thejoint C as it has 2 unknowns

which can be solved via eqn`s of eq`m. Joint A could also be a

candidatebut the equationswouldbemore complicateddue to 2

componentsofreaction.

JointC:

0 cos 30 sin 45 0

0 sin 30 cos 45 1.5x CD CB

y CD CB

F F F

F F F

CoupledsolutionofthesetyieldsFCB=5.02kN (C)andFCD=4.10kN

(T). The same solutioncanbe foundwithoutacoupled solution if

anothercoordinatesystemsisusedas infigure(d)and(e).Seeyour

bookforthedetails.

JointD:

0 cos 30 4.10 cos 30 0 ; 4.10 kN (T)

0 2(4.10 sin 30) 0 ; 4.10 kN (T)x DA DA

y DB DB

F F F

F F F

TheforceinBAcanbecalculatedeitherbyusingjointAorB.Weuse

pointBasthereisnodifferenceatallinbetweentwosolutions.

0 sin 45 sin 45 0 ; 0.776 kN (y BD BC BA BAF F F F F

TheFreeBodyDiagrams:


2/10

QUESTION2:

Determinetheforceineachmemberofthetrussandstateifthemembersare intensionorcompression.SetP1=500

kNandP2=100kN

Answer:We can start solving the problem directly with MOJ

without the need of calculating support reactions. Ifwe

start fromjointB, thenumberofunknowns is2andcan

directlybesolvedviaequationsofequilibrium.

JointB:

30 cos 45 100

54

0 sin 45 5005

x BC BA

y BC BA

F F F

F F F

- - - - - - - -

CoupledsolutionofthesetyieldsFBC=384kN(T)andFBA=

286kN(T).

JointC:

0 384 cos 45 ; 271 kN (C)

0 384 sin 45 ; 271 kNx CA CA

y y y

F F F

F C C

Ifitisnecessary,thesupportreactionsatA(Ax,Ay)canbe

calculated by analyzing the joint A using the calculated

values.


JointB:

JointC:


3/10

QUESTION3:

Determine the force inmember CF of the truss shown.Indicatethememberisintensionorcompression.

Answer:AlthoughtheproblemcanalsobesolvedbyMOJ,Method

of Sections (MOS) provides a shorter solution for the

problembycreatingasectionaa`.Ifthesupportreactions

areknown,theproblemcandirectlybesolvedintheright

handpartofthetrusssincethereexist3unknownswhich

canbesolvedbyequationsofequilibrium.


(16) 5(8) 3(12) 0 ; 4.75

0 ; 0

5 3 4.74 0 ; 3.25

A

x x x

y y y

M E E kN

F A A kN

F A A kN

Sectionaa`:

Now, we can start analyzing the righthand part of the

truss which separated by section aa`.We selected this

partbecauseitinvolveslessunknowns.

The most straightforward way of solving FCF is the

application of moment equation about a point O thateliminatestwounknownforces.However,inanycase,the

systemofequationscanbesolvedthoughwithsomemore

effort. We select the point O, as the point where the

moment equation eliminates 2 unknowns.However, the

position of point O is not known but can be calculated

using some geometry. Considering the similar triangles

FDOandGCO:

4 44 m

8 6

xx

x

Then,weusetheprincipleoftransmissibilitytomoveFCF

tothepointCandresolveitintorectangularcomponents.

Thus,wecanwriteourmomentequation:0 ( sin 45)(12) 3(8) 4.75(4)

0.589 kN (C)

O CF

CF

M F

F


WholeTruss:

Sectionaa`:


4/10

QUESTION4:

Determinethe forces inmembersBC,HCandHGof the bridge truss and indicate whether the

members are in tension or compression. Repeat

theprocedureformembersGF,CFandCD.

Answer:We apply theMOS to solve the problem as the

MOJwillbequite timeconsuming.First,westart

with determination of support reactions. Then

applytheMOSonsectionsaa`andbb`.


(12) 18(9) 14(6) 12(3) 0

23.5

0 0

12 14 18 23.5 0

20.5

A y

y

x x x

y y

y

M E

E kN

F E E kN

F A

A kN

Sectionaa`:

(3) 12(3) 20.5(6) 0

29.0 (C)

(3) 20.5(3) 0

20.5 (T)

20.5 12 sin 45 0

12.0 (T)

c HG

HG

H BC

BC

y HC

HC

M F

F kN

M F

F kN

F F

F kN

Sectionbb`:

(3) 23.5(6) 18(3) 0

29.0 (C)

(3) 23.5(3) 0

23.5 (T)

23.5 18 sin 45 0

7.78 (T)

C GF

GF

F CD

CD

y CF

CF

M F

F kN

M F

F kN

F F

F kN


WholeTruss:

Sectionaa`:

Sectionbb`:


5/10

QUESTION5:

Determine thehorizontaland vertical componentsof forcewhichthepinatCexertsonmemberCBoftheframeinthe

figure.

Answer:The inspection of freebody diagrams of the members

reveals thatAB isa twoforcememberso the forcesacting

onthejointsshouldbeactingonitsdirectionandhaveequal

magnitudeswithreversedirections.

Theunknowns in the freebodydiagram (Cx,Cy,FAB)canbe

determinedfrom3equationsofequilibriumonthemember

CB.

2000(2) ( sin60)(4) 0 1154.7

1154.7 cos60 C 577

1154.7 sin 60 2000 0 1000

C AB AB

x x x

y y y

M F F N

F C N

F C C N

Notice that if don`t recognize AB as a twoforcemember,

then the work involved becomes quite complicated and

requires the solution of 6 equations using 6 equilibrium

equations(3onmemberABand3onmemberBC).



6/10

QUESTION6:

Determine thehorizontaland vertical componentsof forceatCwhichmemberABCexertsonmemberCEF.

Answer:First,we disassemble the structure and sketch the FBD of

eachmember.Wehave6unknowns(Bx,By,Cx,Cy,Ex,Ey)and9

(3 member x 3 equilibrium equations) equations to

determinetheunknowns.

MemberBED:

30(6) (3) 0 E 60 k

30 0 (1)

60 30 0 1000

B y y

x x x

y y y

M E N

F B E

F B B N

MemberFEC:

30(3) (4) 0 22.5

Using equation (1) 7.5

30 22.5 0 7.5

C x x

x

x x x

M E E kN

B kN

F C C kN

MemberABC:

7.5(8) (6) 7.5(4) 30(3) 0 10A y yM C C kN



7/10

QUESTION7:

Determinetheforceneededtosupport20kgmassusingtheSpanishBurtonrig.Whatarethereactionsatthesupporting

hooksA,BandC?

Answer:First,we disassemble the structure and sketch the FBD of

eachmember.We start sketching the FBDs from pulley H

whichhastheunknownforce(P)forsimplicityandproceed

with the neighboring pulleys by expressing the other

unknown reactions in terms of P using Newton`s 3rd law.

Thus,whenwereachtopulleyD,wecansolvetheequations

ofequilibriumforPbyusingtheknownweightattheendof

pulleyD.

ForpulleyD:

9 20(9.81) 0 21.8yF P P N

Supportreactions:

SupportA:

2 43.6AR P N

SupportB:

2 43.6BR P N

SupportC:

6 131CR P N

ThesupportCcarriesthelargestload,soapossiblefailureis

moreprobablethere.



8/10

QUESTION8:

Thecompoundbeam ispinsupportedatCandsupportedbyaroller at A and B. There is a hinge (pin) atD.Determine the

reactionsatthesupports.Neglectthethicknessofthebeam.

Answer:First,wedisassemblethestructureandsketchtheFBDofeach

member. Then, we start the solution from the FBD of first

member (FBD a) as it posses only 3 unknownswhich can be

solvedfromequationsofequilibriumwhiletheotherparthas4

unknownswhichcannotbeindependentlysolved.

EquationsofEquilibriumforFBD(a):

(6) 4 cos 30(12) 8(2) 0 ; 9.59

4 sin 30 0 ; 1.87

9.59 4 cos 30 0 ; 2.0

D y y

x x y

y y x

M A A kN

F D D kN

F D D kN

EquationsofEquilibriumforFBD(b):

4(16) 12 1.87(24) 15 0 ; 8.54

53

2.00 12 0 ; C 2.9358.54 1.87 0 ; C 9.20

C y y

x x y

y y x

M B B kN

F C kN

F C kN

- - - - - - - -



9/10

QUESTION9:

ThepistonCmovesverticallybetweenthetwosmoothwalls.Ifthespringhasastiffnessofk=15N/cm,and isnotstretched

when=0o,determinethecoupleMthatmustbeappliedto

ABtoholdthemechanisminequilibriumwhen=30o.

Answer:Westartthesolutionbycalculatingtheanglesofthedeformed

geometryand stretched lengthof the springby trigonometry.

Wefirstapplythesine lawtocalculatetheotherangles(,)

ofthetriangleABCandproceedwiththecalculationoftheedge

ACusinglawofcosine.

2 2

sin sin 3019.47 180 130.53

8 12

8 12 2(8)(12)cos130.53 18.242 cm

(8 12) 18.242 1.758 cm

15(1.758) 26.37 N

o o

AC

oAC AC AC

sp

l

l l l

F

ZZ G R Z

%

The solution of the problem can be simplified if one realizes

thatmember CB is a two forcemember. Thus one does notneedtocalculatethedirectionofreactionatCandcandirectly

usetheanglecalculatedfrompurelygeometricapproach.

EquationsofEquilibrium:

UsingtheMOJatC:

cos 0 ; F 27.97y CB sp CB F F F kN Z UsingtheFBDofthememberAB;

27.97 cos 40.53 8 0 ; 1.70AM M M kN



10/10

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Truss- Problems With Solution (Structural)