direct stiffness - trusses

8/7/2019 Direct Stiffness - Trusses

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March 20, 2003March 20, 2003

9:35 AM9:35 AM

Little 109Little 109

CES 4141CES 4141

Forrest MastersForrest Masters

A Recap ofStiffness by

Definition andthe Direct

Stiffness Method


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Farther Down the Yellow BrickFarther Down the Yellow Brick

Road..Road..

V i t r u a l W

F o r c e M

S l o p e D e

M o m e n t -

C l a s s i c a l

S t i f f n e s s b

T r u s

B e a

D i r e c t S t i

M a t r i x M

S t r u c t u r a l


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Our Emphasis This Week: Trusses..Our Emphasis This Week: Trusses..

s Composed of slender,Composed of slender,

lightweight memberslightweight members

s All loading occurs onAll loading occurs on

jointsjoints

s No moments orNo moments or

rotations in the jointsrotations in the jointss Axial Force MembersAxial Force Members

s Tension (+)Tension (+)

s

Compression (-)Compression (-)


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StiffnessStiffness

s KKijij = the amount of force required at i= the amount of force required at i

to cause a unit displacement at j, withto cause a unit displacement at j, withdisplacements at all other DOF = zerodisplacements at all other DOF = zero

s A function of:A function of: System geometrySystem geometry

Material properties (E, I)Material properties (E, I)

Boundary conditions (Pinned, Roller orBoundary conditions (Pinned, Roller orFree for a truss)Free for a truss)

s NOT a function of external loadsNOT a function of external loads

K = AE/LK = AE/L


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From Strength of Materials..From Strength of Materials..

Combine two equations to get aCombine two equations to get a stiffnessstiffness

elementelement

F LA E

F = k *F = k *

k F

F

A E

L

SpringSpring

AxialAxial

DeformationDeformation

kA E

LUnits ofUnits of

Force perForce per

LengthLength


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Go to the Board..Go to the Board..

Lets take aLets take a

look at lastlook at last

weeksweekshomework tohomework to

shed someshed some

light on thelight on the

Stiffness byStiffness by

DefinitionDefinitionProcedureProcedure

DOFDOF


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From Stiffness byFrom Stiffness by

DefinitionDefinitions We can create aWe can create a stiffness matrixstiffness matrix thatthataccounts for the material and geometricaccounts for the material and geometricproperties of the structureproperties of the structure

s A square, symmetric matrix KA square, symmetric matrix Kijij

= K= Kjiji

s Diagonal terms always positiveDiagonal terms always positive

s The stiffness matrix is independent ofThe stiffness matrix is independent ofthe loads acting on the structure. Manythe loads acting on the structure. Manyloading casesloading cases can be tested withoutcan be tested without

recalculating the stiffness matrixrecalculating the stiffness matrix

Stiffness by Definition only uses a small part of theStiffness by Definition only uses a small part of the

information available to tackle the probleminformation available to tackle the problem

However ..


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Stiffness by DefinitionStiffness by Definition OnlyOnlyConsiders..Considers..

s Stiffnesses fromStiffnesses from

ImposedImposed

DisplacementsDisplacementss UnknownUnknown

DisplacementsDisplacements

s

Known LoadingsKnown Loadings

For each released DOF, we get one equationFor each released DOF, we get one equation

that adds to the stiffness, displacement andthat adds to the stiffness, displacement and

loading matricesloading matrices

K * r = RStiffness

Matrix

Unknown

Displacements

Known

External

Forces


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A Better Method:A Better Method: Direct StiffnessDirect Stiffness

Consider all DOFs Stiffness ByConsider all DOFs Stiffness By DirectDirect

DefinitionDefinition

StiffnessStiffness

..now we have..now we have more equationsmore equations to work withto work with

PINPIN

ROLLERROLLER

00

11

22

22


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A Simple ComparisonA Simple Comparison

66

55

44

33

22

11

Stiffness by DefinitionStiffness by Definition

s 2 Degrees of2 Degrees of

FreedomFreedomDirect StiffnessDirect Stiffness

s 6 Degrees of6 Degrees of

FreedomFreedom

s DOFs 3,4,5,6 = 0DOFs 3,4,5,6 = 0

s Unknown ReactionsUnknown Reactions

(to be solved)(to be solved)

included in Loadingincluded in LoadingRemember..Remember.. More DOFs = More EquationsMore DOFs = More Equations


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Node Naming ConventionNode Naming Convention

s Unknown orUnknown or

Unfrozen DegreesUnfrozen Degrees

of Freedom areof Freedom are

numbered firstnumbered first

r1, r2r1, r2

s Unknown orUnknown or

Unfrozen DegreesUnfrozen Degreesof Freedom followof Freedom follow

r3, r4, r5, r6r3, r4, r5, r6

66

55

44

33

22

11

If Possible..If Possible.. X-direction before Y-directionX-direction before Y-direction


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Stiffness by Definition vsStiffness by Definition vs

Direct StiffnessDirect Stiffness

KK1111 KK1212 KK1313 KK1414 KK1515 KK1616

KK2121 KK2222 KK2323 KK2424 KK2525 KK2626

KK3131 KK3232 KK3333 KK3434 KK3535 KK3636

KK4141 KK4242 KK4343 KK4444 KK4545 KK4646

KK5151 KK5252 KK5353 KK5454 KK5555 KK5656

KK6161 KK6262 KK6363 KK6464 KK6565 KK6666

rr11

rr22

rr33

rr44

rr55

rr66

RR11

RR22

RR33

RR44

RR55

RR66

==

6655

4433

2211

Stiffness by Definition Solution inStiffness by Definition Solution in

REDREDDirect Stiffness Solution inDirect Stiffness Solution in

REDRED/YELLOW/YELLOW


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The FundamentalThe Fundamental

ProcedureProcedure

s Calculate theCalculate the Stiffness MatrixStiffness Matrix

s Determine Local Stiffness Matrix, KeDetermine Local Stiffness Matrix, Ke

sTransform it into Global Coordinates, KGTransform it into Global Coordinates, KG

s Assemble all matricesAssemble all matrices

s Solve for theSolve for the Unknown DisplacementsUnknown Displacements

s Use unknown displacements to solve for theUse unknown displacements to solve for the

Unknown ReactionsUnknown Reactions

s Calculate theCalculate the Internal ForcesInternal Forces


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To continue..To continue..

s You need yourYou need your Direct Stiffness TrussDirect Stiffness Truss

Application HandoutApplication Handout to follow theto follow the

remaining lecture. If you forgot it, lookremaining lecture. If you forgot it, lookon your neighbors, pleaseon your neighbors, please

s I have yourI have your new homeworknew homework (if you(if you

dont have it already)dont have it already)

Go toGo to http://www.ce.ufl.edu/~kgurlhttp://www.ce.ufl.edu/~kgurl for the handoutfor the handout

FOR MORE INFO ..
http://www.ce.ufl.edu/~kgurlhttp://www.ce.ufl.edu/~kgurlhttp://www.ce.ufl.edu/~kgurl


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OverviewOverview

First, we willFirst, we will

decompose the entiredecompose the entire

structure into a set ofstructure into a set of

finite elementsfinite elements

Next, we will build aNext, we will build a

stiffness matrixstiffness matrix forfor

each element (6each element (6Here)Here)

Later, we will combineLater, we will combine

all of the localall of the local

stiffness matrices intostiffness matrices into

Node 1Node 1

Node 2Node 2

11

4422

33 55


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Element Stiffness Matrix in LocalElement Stiffness Matrix in Local

CoordinatesCoordinates

s Remember KRemember Kijij = the amount of force required at i= the amount of force required at i

to cause a unit displacement at j, withto cause a unit displacement at j, withdisplacements at all other DOF = zerodisplacements at all other DOF = zero

s

For a truss element (which has 2 DOF)..For a truss element (which has 2 DOF)..

K11*v1 + K12*v2 = S1K11*v1 + K12*v2 = S1

K21*v1 + K22*v2 = S2K21*v1 + K22*v2 = S2

S1S1

S2S2

v2v2v1v1

K11K11 K12K12

K21K21 K22K22

v1v1

v2v2

==S1S1

S2S2

Gurley refers to theGurley refers to the axial displacementaxial displacement as as vvand theand the internal forceinternal force as as SS in the local in the local


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s Use Stiffness by Definition to finding Ks of LocalUse Stiffness by Definition to finding Ks of LocalSystemSystem

Node 1Node 1

Node 2Node 2

AEAELL KK2222

KK1212

AEAE

LL

KK1111KK2121

KK1212 = - AE / L= - AE / L

KK2222 = AE / L= AE / L

KK1111 = AE / L= AE / L

KK2121 = - AE / L= - AE / L


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Coordinates Cont..Coordinates Cont..

Put the local stiffness elements inPut the local stiffness elements in matrixmatrix

formform

Simplified..Simplified..

For a truss elementFor a truss element


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Displacement TransformationDisplacement Transformation

MatrixMatrix

s Structures are composed of many members inStructures are composed of many members in

many orientationsmany orientations

s We must move the stiffness matrix from aWe must move the stiffness matrix from a

locallocal to ato a globalglobal coordinate systemcoordinate system

S1S1

S2S2

v2v2v1v1

r1r1

r2r2

r4r4r3r3

xx

yy

LOCALLOCAL

GLOBALGLOBAL


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How do we do that?How do we do that?

s Meaning if I give you a point (x,y) in CoordinateMeaning if I give you a point (x,y) in Coordinate

System Z, how do I find the coordinates (x,y) inSystem Z, how do I find the coordinates (x,y) in

Coordinate System ZCoordinate System Z

xxyy

xx

yyUse aUse a

DisplacementDisplacementTransformationTransformation

MatrixMatrix


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To change the coordinates of a truss..To change the coordinates of a truss..

s Each node has one displacementEach node has one displacementin the local system concurrent toin the local system concurrent tothe element (v1 and v2)the element (v1 and v2)

s In the global system, every nodeIn the global system, every node

has two displacements in the xhas two displacements in the xand y directionand y direction

r1r1

r2r2

r4r4

r3r3

xx

yy

v1v1

v2v2

v1 will be expressed by r1 and r2v1 will be expressed by r1 and r2

v2 will be expressed by r3 and r4v2 will be expressed by r3 and r4


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Displacement Transformation MatrixDisplacement Transformation Matrix

Cont..Cont..

s The relationship between v and r is theThe relationship between v and r is thevector sum:vector sum:

v1 = r1*cosv1 = r1*cos xx + r2*cos+ r2*cos YY

v2 = r3*cosv2 = r3*cos xx + r4*cos+ r4*cos YY

x

Y

v1

r1

r2

We can simplify the cosine terms:We can simplify the cosine terms:

Lx = cosLx = cos xx

Ly = cosLy = cos yyv1 = r1*Lx +v1 = r1*Lx +r2*Lyr2*Ly

v2 = r3*Lx +v2 = r3*Lx +r4*Lyr4*Ly

Put in matrix formPut in matrix form


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Displacement Transformation MatrixDisplacement Transformation Matrix

Cont..Cont..

v1 = r1*Lx + r2*Lyv1 = r1*Lx + r2*Ly

v2 = r3*Lx + r4*Lyv2 = r3*Lx + r4*Ly

v1

v2

Lx

0

Ly

0

0

Lx

0

Ly

r1

r2

r3

r4

aLx

0

Ly

0

0

Lx

0

Ly

Transformation matrix, aTransformation matrix, a gives us thegives us the

relationship we soughtrelationship we sought

So..So.. v = a*rv = a*r


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Force Transformation MatrixForce Transformation Matrix

Similarly, we can perform aSimilarly, we can perform a

transformation on the internal forcestransformation on the internal forces

R1

R2

R3

R4

Lx

Ly

0

0

0

0

Lx

Ly

S1

S2 S1S1

S2S2

R1R1R2R2

R3R3

R4R4


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Element Stiffness Matrix in GlobalElement Stiffness Matrix in Global


Letsput itall together.. We knowthat theLetsput itall together.. We knowthat the

Internalforce =stiffness* local displacementInternalforce =stiffness* localdisplacement ( S=k*v )( S=k*v )

Units:Force =(Force/Length) * LengthUnits:Force =(Force/Length) *Length

localdisp =transformmatrix* globaldisplocaldisp =transformmatrix* globaldisp ( v =a*r )( v =a*r )

Substitutelocal displacementSubstitutelocaldisplacement

Internalforce =stiffness* transformmatrix* globaldispInternalforce =stiffness* transformmatrix* globaldisp

( S=k* a*r )( S=k* a*r )

PremultiplybythetransposeofaPremultiplyby thetransposeofa

aaTT * S=a* S=aTT * k*a* r* k*a* r

and substitute R= aand substitute R= aTT * Stoget* Stoget R= aR= aTT * k * a * r* k * a * r


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is an important relationshipis an important relationship

between the loading, stiffnessbetween the loading, stiffness

and displacements of the structureand displacements of the structure

in terms of the global systemin terms of the global system

R =R = aaTT * k * a* k * a * r* r

StiffnessStiffness

termterm

sWe have a stiffness term,We have a stiffness term, KeKe, for each element, for each element

in the structurein the structure

sWe use them to build the global stiffnessWe use them to build the global stiffness

matrix,matrix, KGKG

Ke =Ke = aaTT * k * a* k * a


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Lets expand all of terms to getLets expand all of terms to get

a Ke that we can use.a Ke that we can use.Ke =Ke = aaTT * k * a* k * a

KeA E

L

Lx

Ly

0

0

0

0

Lx

Ly

1

1

1

1

Lx

0

Ly

0

0

Lx

0

Ly

KeA E

L

Lx2

LxLy

Lx2

Lx Ly

LxLy

Ly2

Lx Ly

Ly2

Lx2

Lx Ly

Lx2

LxLy

Lx Ly

Ly2

LxLy

Ly2

(14) From notes(14) From notes

Great formula toGreat formula to

plug into yourplug into your

calculatorcalculator


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sLets use aLets use a

problem to illustrateproblem to illustrate

the rest of thethe rest of the

procedureprocedure

sWe will start byWe will start by

calculating KEs forcalculating KEs for

the two elementsthe two elements

66

55

44

33

22

11

3 ft3 ft

4 ft4 ft

NodeNode

11

NodeNode 33

NodeNode 22

ElementElement

22

ElementElement

11

bl f h Gl b l S iff i


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Assembly of the Global Stiffness MatrixAssembly of the Global Stiffness Matrix

(KG)(KG)

r1r1

r2r2

r3r3

r4r43 ft3 ft

Element 1Element 1

LL = 3= 3

LxLx == x / L = (3-0) / 3 = 1x / L = (3-0) / 3 = 1

LyLy == y / L = (0-0) / 3 = 0y / L = (0-0) / 3 = 0

NearNear FarFar

r1r1 r2r2 r3r3 r4r4

r1r1r2r2

r3r3

r4r4

Ke1 A E

0.333

0

0.333

0

0

0

0

0

0.3330

0.333

0

0

0

0

0

Pick aPick a NearNear and aand a

FarFar

Plug Lx, Ly and LPlug Lx, Ly and L

into equation 14 tointo equation 14 to

etget

A bl f h Gl b l S iff M i


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Assembly of the Global Stiffness MatrixAssembly of the Global Stiffness Matrix

(KG)(KG)

r1r1

r2r2

r5r5

r6r6

3 ft3 ft

4 ft4 ft

5 ft5 ft

Element 2Element 2

LL = 5= 5

LxLx == x / L = (3-0) / 5 =x / L = (3-0) / 5 =

0.60.6LyLy == y / L = (4-0) / 5 =y / L = (4-0) / 5 =

0.80.8

NearNear

FarFar

Ke2 A E

0.072

0.096

0.072

0.096

0.096

0.128

0.096

0.128

0.0720.096

0.072

0.096

0.0960.128

0.096

0.128

r1r1 r2r2 r5r5 r6r6

r1r1r2r2

r5r5

r6r6


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The Entire Local StiffnessThe Entire Local Stiffness

Matrix in Global TermsMatrix in Global Terms

Ke2 A E

0.072

0.096

0.0720.096

0.096

0.128

0.0960.128

0.072

0.096

0.072

0.096

0.096

0.128

0.096

0.128

r1r1 r2r2 r5r5 r6r6

r1r1

r2r2

r5r5r6r6

0.072

0.096

0

0

0.072

0.096

0.096

0.128

0

0

0.096

0.128

0

0

0

0

0

0

0

0

0

0

0

0

0.072

0.096

0

0

0.072

0.096

0.096

0.128

0

0

0.096

0.128

r1r1

r2r2

r3r3

r4r4

r5r5

r6r6

r1 r2 r3 r4 r5r1 r2 r3 r4 r5

r6r6

Notice that thereNotice that therearent any termsarent any terms

in the localin the local

matrix formatrix for r3r3 andand

r4r4

ShorthandShorthand

RealReal

MatrixMatrix


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Assembly of the GlobalAssembly of the Global

Stiffness Matrix (KG)Stiffness Matrix (KG)

Summing Ke1 and Ke2Summing Ke1 and Ke2

r1r1

KG A E

0.405

0.096

0.333

0.000

0.072

0.096

0.096

0.128

0.000

0.000

0.096

0.128

0.333

0.000

0.333

0.000

0.000

0.000

0.000

0.000

0.000

0.000

0.000

0.000

0.072

0.096

0.000

0.000

0.072

0.096

0.096

0.128

0.000

0.000

0.096

0.128

r2r2 r3r3 r4r4 r5r5 r6r6

r1r1

r2r2

r3r3

r4r4r5r5

r6r6

==KK rr RR

How does this relate to Stiffness by Definition?How does this relate to Stiffness by Definition?


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Solution ProcedureSolution Procedure

Now, we can examine the full systemNow, we can examine the full system

ReactionsReactions Known displacementsKnown displacements

@ reactions ( = 0 )@ reactions ( = 0 )

Unknown DeflectionsUnknown DeflectionsLoads acting on the nodesLoads acting on the nodesR1 0.405 0.096 -0.333 0.000 -0.072 -0.096 r1

R2 0.096 0.128 0.000 0.000 -0.096 0.128 r2

R3 -0.333 0.000 0.333 0.000 0.000 0.000 r3

R4 0.000 0.000 0.000 0.000 0.000 0.000 r4

R5 -0.072 -0.096 0.000 0.000 0.072 0.096 r5

R6 -0.096 -0.128 0.000 0.000 0.096 0.128 r6

= X


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Solution Procedure cont..Solution Procedure cont..

o find the unknowns, we must subtend the matrico find the unknowns, we must subtend the matrice

Rk

Ru

AE K11

K21

K12

K22

rurk

K11K11

K22K22

K12K12

K21K21==

TwoTwo

ImportantImportant

EquationsEquations

Rk = AE ( K11*ru +Rk = AE ( K11*ru +K12*rk )K12*rk )

Ru = AE ( K21*ru +Ru = AE ( K21*ru +

K22*rk )K22*rk )

(24)(24)

(25)(25)

Going to be ZERO.Going to be ZERO.


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66

55

44

33

22

11

3 ft3 ft

4 ft4 ft

10 kips10 kips

We will apply a load at DOFWe will apply a load at DOF

22

Then use equation (24)Then use equation (24)Rk = AE ( K11*ru + K12*rk )Rk = AE ( K11*ru + K12*rk )

0

10

AE0.405

0.096

0.096

0.128

r1

r2

AE K12

0

0

0

0

+

0 = AE ( 0.405*r1 + 0.096*r2)0 = AE ( 0.405*r1 + 0.096*r2)

-10 = AE ( 0.096*r1 + 0.128*r2)-10 = AE ( 0.096*r1 + 0.128*r2)

r1 = 22.52/AEr1 = 22.52/AE

r2 = -95.02/AEr2 = -95.02/AE

solvedsolved

00


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With the displacements, we can use equation (25)With the displacements, we can use equation (25)

to find the reactions at the pinned endsto find the reactions at the pinned ends

Ru = AE ( K21*ru + K22*rk )Ru = AE ( K21*ru + K22*rk )R3

R4

R5

R6

AE

0.333

0

0.072

0.096

0

0

0.096

0.128

22.52

AE

95.02

AE

AE K22

0

0

0

0

+

00

R3 = -7.5 kipsR3 = -7.5 kips R4 = 0 kipsR4 = 0 kips

R5 = 7.5 kipsR5 = 7.5 kips

R6 = 10 kipsR6 = 10 kips


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Internal Member Force RecoveryInternal Member Force Recovery

sTo find the internal force inside of anTo find the internal force inside of an

element, we must return to the localelement, we must return to the local

coordinate systemcoordinate system

sRemember the equationRemember the equation S = k * a * rS = k * a * r ??

S1

S2

AE

L

1

1

1

1

Lx

0

Ly

0

0

Lx

0

Ly

r1

r2

r3

r4

But S1 alwaysBut S1 always

Equals S2Equals S2

soso S AEL

Lx Ly Lx Ly( )

r1

r2

r3

r4

I t l M b F RInternal Member Force Reco er


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Internal Member Force RecoveryInternal Member Force Recovery

Cont..Cont..

sFor Element 1For Element 1

sFor Element 2For Element 2

S1AE

31

0 1 0( )

22.52

AE

95.02

AE

0

0

r1r1

r2r2

r3r3

r4r4

= -7.5 kips= -7.5 kips

r1r1

r2r2

r5r5

r6r6

= 12.5 kips= 12.5 kipsS2 AE5

0.6 0.8 0.6 0.8( )

22.52

AE

95.02AE

0

0


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ConclusionConclusion

We solvedWe solved

s Element StiffnessesElement Stiffnesses

s

UnknownUnknownDisplacementsDisplacements

s ReactionsReactions

s Internal ForcesInternal Forces

I will cover another example in theI will cover another example in the

laboratorylaboratory

MatricesMatrices


40/56

Matrices.Matrices.

..

a x b y+ c z+ dStart with a basic equation

In order to solve x,y,z ..You must have three

equations

a 1

a 2

a 3

b 1

b 2

b 3

c 1

b 2

b 3

x

y

z

a1 x b1 y+ c1 z+ d1=

a2 x b2 y+ b2 z+ d2=a3 x b3 y+ b3 z+ d3=

But you must put these

equations in matrix

form

d 1

d 2

d 3

=

41 A Sample Problem solved with Stiffness by Definition


41/56

12

3

B

CA

10 kips

5 kips

A Sample Problem solved with Stiffness by Definition

and Direct Stiffness

42


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ForStiffness by Definition, we are only concerned with

the three DOFs that are free to move:

r1

r2

r3

43


43/56

ForColumn 1, we set r1 = 1 and r2 = r3 = 0

A

B

C

BElement Change in LengthElement Change in Length

1 6/10 Long

2 8/10 Short

3 0Unit DisplacementUnit Displacement

44


44/56


A

B

C

B

Element Change in LengthElement Change in Length

1 8/10 Short

2 6/10 Short

3 0

Unit DisplacementUnit Displacement

45


45/56


A

B

C

Element Change in LengthElement Change in Length

1 0

2 4/5 Long

3 1 Long

C

Unit DisplacementUnit Displacement

46


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K

7

50

1

50

2

25

1

50

91

600

3

50

2

25

3

50

9

50

r1 r2 r3

r1

r2

r3

The final stiffness matrix is as follows..

r1 r2 r3

0.14 -0.02 -0.08 r1

-0.02 0.152 -0.06 r2

-0.08 -0.06 0.18 r3

47


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ForDirect Stiffness, we are concerned with all six

DOFs in the structural system:

r1

r2

r3

r4

r5

r6

48


48/56

In the Direct Stiffness Method, we will use this equation

for each elements 1, 2 and 3:

Ke

A E

L

Lx2

LxLy

Lx2

Lx Ly

LxLy

Ly2

Lx Ly

Ly2

Lx2

Lx Ly

Lx2

LxLy

Lx Ly

Ly2

LxLy

Ly2

Near X Near Y Far X Far Y

Near X

Near Y

Far X

Far Y

DOF

Location

49


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Element 1

r5 r6 r1 r2

r5

r6

r1

r2

L = 6Lx = 0.6

Ly = -0.8

Ke1

AE

3

50

2

25

3

50

2

25

2

25

8

75

2

25

8

75

3

50

2

25

3

50

2

25

2

25

8

75

2

25

8

75

50


50/56

Element 1 Another View

r1 r2 r3 r4 r5 r6

r1

Ke1 AE

3

502

25

0

03

50

2

25

2

25

8

75

0

02

25

8

75

0

0

0

0

0

0

0

0

0

0

0

0

3

50

2

25

0

03

50

2

25

2

258

75

0

02

25

8

75

r2

r3

r4

r5

r6

51


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Element 2

r1 r2 r3 r4

r1

r2

r3

r4

L = 8Lx = 0.8

Ly = 0.6

Ke 2 AE

2

25

3

50

225

3

50

2

50

9

200

350

9

200

2

253

50

225

3

50

3

509

200

350

9

200

52


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Element 3

r5 r6 r3 r4

r5

r6

r3

r4

L = 10Lx = 1

Ly = 0

Ke 3 AE

1

10

0

1

10

0

0

0

0

0

1

10

0

1

10

0

0

0

0

0

53


53/56

Summing Elements 1 through 3

Ke 1 AE

3

50

2

25

3

50

2

25

2

25

8

75

2

25

8

75

3

50

2

25

3

50

2

25

2

25

8

75

2

25

8

75

Ke 2 AE

2

25

3

50

2

25

3

50

2

50

9

200

3

50

9

200

2

25

3

50

2

25

3

50

3

50

9

200

3

50

9

200

Ke 3 AE

1

10

0

1

10

0

0

0

0

0

1

10

0

1

10

0

0

0

0

0

+ +

Remember: We must take care to add the correct elements

from the local stiffness matrix to the global stiffness matrix.

54


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KG AE

3

50

2

25+

2

25 3

50+

2

25

3

50

3

50

2

25

2

25

3

50+

8

75

9

200+

3

50

9

200

2

25

8

75

2

25

3

50

2

25

1

10+

3

500+

1

10

0

3

50

9

200

3

500+

9

2000+

0

0

3

50

2

25

1

10

0

3

50

1

10+

2

25 0+

2

25

8

75

0

0

2

25 0+

8

750+

r1 r2 r3 r4 r5 r6

r1

r2

r3

r4

r5

r6

55


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r1 r2 r3 r4 r5 r6

0.14 -0.02 -0.08 -0.06 -0.06 0.08 r1

-0.02 0.15 -0.06 -0.05 0.08 -0.11 r2

-0.08 -0.06 0.18 0.06 -0.10 0.00 r3

-0.06 -0.05 0.06 0.05 0.00 0.00 r4

-0.06 0.08 -0.10 0.00 0.16 -0.08 r5

0.08 -0.11 0.00 0.00 -0.08 0.11 r6

Look Familiar? We found the yellow portion

in the Stiffness by Definition Method

S iff b D fi i i DiStiff b D fi iti Di t


56/56

X

Stiffness by Definition vs DirectStiffness by Definition vs Direct

StiffnessStiffness

KK

KK

completedcompleted

rrunknownunknown

RRunknownunknownrrknownknown

RRknownknown=

=X

ReactionsReactions

Zero UnlessZero Unless

SettlementSettlement

OccursOccurs

direct stiffness - trusses

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