Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

6.3Stiffness Matrix for Two Elastic Springs in Line 175

ThefirststepistorewriteEq.(6.13)inpartitionedformas

⎧

⎨

⎩

Fx,1

Fx,2

Fx,3

⎫

⎬

⎭

=

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

ka

..

. −ka 0
························
−ka

..

. ka+kb −kb

0

..

. −kb kb

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

⎧

⎨

⎩

u 1 = 0

u 2

u 3

⎫

⎬

⎭

(6.18)

InEq.(6.18),Fx,1istheunknownreactionatnode1,u 1 andu 2 areunknownnodaldisplacements,while
Fx,2andFx,3areknownappliedloads.ExpandingEq.(6.18)bymatrixmultiplication,weobtain

{Fx,1}=[−ka 0]

{

u 2

u 3

}{

Fx,2

Fx,3

}

=

[

ka+kb −kb

−kb kb

]{

u 2

u 3

}

(6.19)

InversionofthesecondofEqs.(6.19)givesu 2 andu 3 intermsofFx,2andFx,3.Substitutionofthese
valuesinthefirstequationthenyieldsFx,1.
Thus,
{
u 2
u 3

}

=

[

ka+kb −kb

−kb kb

]− 1 {

Fx,2

Fx,3

}

or
{
u 2
u 3

}

=

[

1 /ka 1 /ka

1 /ka 1 /kb+ 1 /ka

]{

Fx,2

Fx,3

}

Hence,

{Fx,1}=[−ka 0]

[

1 /ka 1 /ka

1 /ka 1 /kb+ 1 /ka

]{

Fx,2

Fx,3

}

whichgives

Fx,1=−Fx,2−Fx,3

aswouldbeexpectedfromequilibriumconsiderations.Inproblemswherereactionsarenotrequired,
equationsrelatingknownappliedforcestounknownnodaldisplacementsmaybeobtainedbydeleting
the rows and columns of [K] corresponding to zero displacements. This procedure eliminates the
necessityofrearrangingrowsandcolumnsintheoriginalstiffnessmatrixwhenthefixednodesarenot
convenientlygroupedtogether.
Finally,theinternalforcesinthespringsmaybedeterminedfromtheforce–displacementrelationship
ofeachspring.Thus,ifSaistheforceinthespringjoiningnodes1and2,then

Sa=ka(u 2 −u 1 )

Similarly,forthespringbetweennodes2and3

Sb=kb(u 3 −u 2 )