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DistributedLoadsinBarElement

Uniformlydistributedaxialloadq(N/mm,N/m,lb/in)canbeconvertedtotwoequivalentnodalforcesofmagnitudeqL/2.

This

can

be

verified

by

considering

the

work

done

by

the

load

q

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SincetheU=Wconceptfortheelement,ityields:

Then,

In

an

assembly

of

bars,

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BarElementsin2D(Truss)

: v oe o o u e o e e o ,w thelineartheory.

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BarElementsin2D

Transformation:

Inmatrixform,with

Forthetwonodesofthebarelement,wehave

Thenodalforcesaretransformedin

,

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StiffnessMatrixinthe2DSpace

Inthelocalcoordinatesystem:

multiplyingbothsidesby

Thus,theelementstiffnessmatrixkintheglobalcoordinatesystemis

whichis

a4x4

symmetric

matrix.

Inexplicitform,itcanbewrittenas

Calculationofthedirectionalcosineslandm:

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ElementStress

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Example1

s mp e

p ane

truss

sma e

o

two

ent ca

bars(withE,A,andL),andloadedasshowninthefigure.Find1)dis lacementofnode2;

2)stressineachbar.

Thissimplestructureisusedheretodemonstrate

theassembly

and

solution

process

using

the

bar

e ement n space.

Inlocalcoordinatesystems,

Thesetwomatricescannotbeassembledtogether,becausetheyareindifferent

coordinatesystems.WeneedtoconvertthemtoglobalcoordinatesystemOXY.

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Element1:

Element2:

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AssemblethestructureFEequation:

Loadand

boundary

conditions

(BC):

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Thestressesinthetwobars:

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Example2

Determinethedisplacementsandreactionforces.

Wehave

an

inclined

roller

at

node

3,

which

needs

special

attention

in

the

FE

solution.

We

.

Element1:

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ement :

Element3:

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TheglobalFEequationis:

Loadandboundaryconditions(BC):

From the transformation relation and the BC we have

This is a multi oint constraint MPC .

Similarly,wehavearelationfortheforceatnode3,

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ApplyingtheloadandBCsinthestructureFEequationby

eliminating1st,2ndand4throwsandcolumns,wehave

Further,fromtheMPCandtheforcerelationatnode3,the

equationbecomes,

The3rdequationyields

Thusit

can

be

rearranged

to

yield

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FromtheglobalFEequation,wecancalculatethereactionforces,