March 20, 2003March 20, 2003
9:35 AM9:35 AM
Little 109Little 109
CES 4141CES 4141
Forrest MastersForrest Masters
A Recap of Stiffness by Definition and the Direct Stiffness Method
Farther Down the Yellow Brick Farther Down the Yellow Brick Road..Road..
Vitrual W ork
Force M ethod
S lope Deflection
M om ent-A rea
C lassical M ethods
Stiffness by Definition
Trusses
Beams
Direct Stiffness
M atrix Methods
Structural Analysis
Our Emphasis This Week: Trusses..Our Emphasis This Week: Trusses..
Composed of slender, Composed of slender, lightweight members lightweight members
All loading occurs on All loading occurs on jointsjoints
No moments or No moments or rotations in the jointsrotations in the joints
Axial Force MembersAxial Force Members Tension (+) Tension (+) Compression (-)Compression (-)
StiffnessStiffness
KKijij = the amount of force = the amount of force required at i to cause a unit required at i to cause a unit displacement at j, with displacement at j, with displacements at all other DOF displacements at all other DOF = zero= zero
A function of:A function of:– System geometrySystem geometry– Material properties (E, I)Material properties (E, I)– Boundary conditions (Pinned, Boundary conditions (Pinned,
Roller or Free for a truss)Roller or Free for a truss) NOT a function of external loadsNOT a function of external loads
K = AE/LK = AE/L
From Strength of Materials..From Strength of Materials..
Combine two equations to get a Combine two equations to get a stiffness stiffness elementelement
F LA E
F = k * F = k * k
F
F
A EL
SpringSpring
Axial Axial DeformationDeformation
kA E
L
Units ofUnits ofForce per Force per
LengthLength
Go to the Board..Go to the Board..
Let’s take a Let’s take a look at last look at last week’s week’s homework to homework to shed some shed some light on the light on the Stiffness by Stiffness by DefinitionDefinition ProcedureProcedure
DOFDOF
From Stiffness by From Stiffness by DefinitionDefinition
We can create aWe can create a stiffness matrixstiffness matrix that that accounts for the material and accounts for the material and geometric properties of the structuregeometric properties of the structure
A square, symmetric matrix KA square, symmetric matrix Kijij = K = Kjiji
Diagonal terms always positiveDiagonal terms always positive The stiffness matrix is independent The stiffness matrix is independent
of the loads acting on the structure. of the loads acting on the structure. ManyMany loading casesloading cases can be tested can be tested without recalculating the stiffness without recalculating the stiffness matrixmatrix
Stiffness by Definition only uses a small part of the Stiffness by Definition only uses a small part of the information available to tackle the probleminformation available to tackle the problem
However ..
Stiffness by Definition Stiffness by Definition OnlyOnly Considers..Considers..
Stiffnesses from Stiffnesses from Imposed Imposed DisplacementsDisplacements
Unknown Unknown Displacements Displacements
Known LoadingsKnown Loadings
For each released DOF, we get one equation For each released DOF, we get one equation that adds to the stiffness, displacement and that adds to the stiffness, displacement and loading matricesloading matrices
But what aboutBut what about ReactionsReactions andand Known Known Displacements?Displacements?
K * r = RStiffness Matrix
Unknown Displacements
Known External Forces
A Better Method: A Better Method: Direct StiffnessDirect Stiffness
Consider all DOFs Stiffness ByConsider all DOFs Stiffness By Direct Direct Definition Stiffness Definition Stiffness
..now we have ..now we have more equationsmore equations to work with to work with
PINPIN
ROLLERROLLER
00
11
22
22
A Simple ComparisonA Simple Comparison
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Stiffness by DefinitionStiffness by Definition 2 Degrees of 2 Degrees of
FreedomFreedom
Direct StiffnessDirect Stiffness 6 Degrees of 6 Degrees of
FreedomFreedom DOFs 3,4,5,6 = 0DOFs 3,4,5,6 = 0 Unknown Reactions Unknown Reactions
(to be solved) (to be solved) included in Loading included in Loading MatrixMatrix
Remember.. Remember.. More DOFs = More EquationsMore DOFs = More Equations
Node Naming ConventionNode Naming Convention
Unknown or Unknown or “Unfrozen” Degrees “Unfrozen” Degrees of Freedom are of Freedom are numbered first…numbered first…
r1, r2r1, r2 Unknown or Unknown or
“Unfrozen” Degrees “Unfrozen” Degrees of Freedom followof Freedom follow
r3, r4, r5, r6r3, r4, r5, r6
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If Possible.. If Possible.. X-direction before Y-directionX-direction before Y-direction
Stiffness by Definition vs Stiffness 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
==
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Stiffness by Definition Solution in Stiffness by Definition Solution in REDREDDirect Stiffness Solution in Direct Stiffness Solution in REDRED/YELLOW/YELLOW
The Fundamental The Fundamental ProcedureProcedure
Calculate the Calculate the Stiffness MatrixStiffness Matrix Determine Local Stiffness Matrix, KeDetermine Local Stiffness Matrix, Ke Transform it into Global Coordinates, KGTransform it into Global Coordinates, KG Assemble all matricesAssemble all matrices
Solve for the Solve for the Unknown Displacements Unknown Displacements Use unknown displacements to solve for the Use unknown displacements to solve for the
Unknown ReactionsUnknown Reactions Calculate the Calculate the Internal ForcesInternal Forces
To continue..To continue..
You need your You need your Direct Stiffness – Direct Stiffness – Truss Application HandoutTruss Application Handout to follow to follow the remaining lecture. If you forgot the remaining lecture. If you forgot it, look on your neighbor’s, pleaseit, look on your neighbor’s, please
I have your I have your new homework new homework (if you (if you don’t have it already)don’t have it already)
Go to Go to http://www.ce.ufl.edu/~kgurlhttp://www.ce.ufl.edu/~kgurl for the handout for the handout
FOR MORE INFO ..
OverviewOverviewFirst, we will First, we will
decompose the entire decompose the entire structure into a set of structure into a set of finite elementsfinite elements
Next, we will build a Next, we will build a stiffness matrixstiffness matrix for for each element (6 each element (6 Here)Here)
Later, we will combine Later, we will combine all of the local all of the local stiffness matrices into stiffness matrices into ONE ONE globalglobal stiffness stiffness matrixmatrix
Node 1Node 1
Node 2Node 2
11
4422
33 55
Element Stiffness Matrix in Local Element Stiffness Matrix in Local CoordinatesCoordinates
Remember KRemember Kijij = the amount of force required at i = the amount of force required at i to cause a unit displacement at j, with to cause a unit displacement at j, with displacements at all other DOF = zerodisplacements at all other DOF = zero
For a truss element (which has 2 DOF)..For a truss element (which has 2 DOF)..
K11*v1 + K12*v2 = S1K11*v1 + K12*v2 = S1K21*v1 + K22*v2 = S2K21*v1 + K22*v2 = S2
S1S1
S2S2
v2v2v1v1
K1K111
K1K122
K2K211
K2K222
vv11
vv22
==SS11
SS22
Gurley refers to the Gurley refers to the axial displacementaxial displacement as “ as “vv” ” and the and the internal forceinternal force as “ as “SS” in the local ” in the local
coordinate systemcoordinate system
Element Stiffness Matrix in Local Element Stiffness Matrix in Local CoordinatesCoordinates
Use Stiffness by Definition to finding Ks of Local Use Stiffness by Definition to finding Ks of Local SystemSystem
Node 1Node 1
Node 2Node 2
AEAE LL
KK2222
KK1212
AEAE LL
KK1111
KK2121
KK1212 = - AE / L = - AE / L
KK2222 = AE / L = AE / L
KK1111 = AE / L = AE / L
KK2121 = - AE / L = - AE / L
Element Stiffness Matrix in Local Element Stiffness Matrix in Local Coordinates Cont..Coordinates Cont..
Put the local stiffness elements in Put the local stiffness elements in matrix matrix formform
Simplified..Simplified..
For a truss elementFor a truss element
Displacement Transformation Displacement Transformation MatrixMatrix
Structures are composed of many members in Structures are composed of many members in many orientationsmany orientations
We must move the stiffness matrix from a We must move the stiffness matrix from a locallocal to a to a globalglobal coordinate system coordinate system
S1S1
S2S2
v2v2v1v1
r1r1r2r2
r4r4r3r3
xxyy
LOCALLOCAL
GLOBALGLOBAL
How do we do that?How do we do that?
Meaning if I give you a point (x,y) in Meaning if I give you a point (x,y) in Coordinate System Z, how do I find the Coordinate System Z, how do I find the coordinates (x’,y’) in Coordinate System Z’coordinates (x’,y’) in Coordinate System Z’
xxyy
x’x’
y’y’Use aUse a
Displacement Displacement Transformation Transformation
MatrixMatrix
To change the coordinates of a truss..To change the coordinates of a truss..
Each node has one Each node has one displacement in the local displacement in the local system concurrent to the system concurrent to the element (v1 and v2)element (v1 and v2)
In the global system, every In the global system, every node has two displacements node has two displacements in the x and y directionin the x and y direction
r1r1r2r2
r4r4r3r3
xxyy
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
Displacement Transformation Matrix Displacement Transformation Matrix Cont..Cont..
The relationship between v The relationship between v and r is the vector sum:and r is the vector sum:
v1 = r1*cos v1 = r1*cos xx + r2*cos + r2*cos YY
v2 = r3*cos v2 = r3*cos xx + r4*cos + r4*cos YYx
Y
v1
r1
r2
We can simplify the cosine terms:We can simplify the cosine terms:Lx = cos Lx = cos xx Ly = cos Ly = cos yy v1 = r1*Lx + v1 = r1*Lx +
r2*Ly r2*Ly
v2 = r3*Lx + v2 = r3*Lx + r4*Lyr4*Ly
Put in matrix formPut in matrix form
Displacement Transformation Matrix Displacement Transformation Matrix Cont..Cont..
v1 = r1*Lx + r2*Ly v1 = 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 the gives us therelationship we soughtrelationship we sought
So.. So.. v = a*rv = a*r
Force Transformation MatrixForce Transformation Matrix
Similarly, we can perform a Similarly, 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
R3R3R4R4
Element Stiffness Matrix in Global Element Stiffness Matrix in Global CoordinatesCoordinates
Let’s put it all together.. We know that the Let’s put it all together.. We know that the
Internal force = stiffness * local displacementInternal force = stiffness * local displacement (S = k * v)(S = k * v)Units: Force = (Force/Length) * LengthUnits: Force = (Force/Length) * Lengthlocal disp = transform matrix * global displocal disp = transform matrix * global disp (v = a * r)(v = a * r)Substitute local displacementSubstitute local displacementInternal force = stiffness * transform matrix * global dispInternal force = stiffness * transform matrix * global disp
(S = k * a * r)(S = k * a * r)Premultiply by the transpose of “a”Premultiply by the transpose of “a”aaTT * S= a * S= aTT * k * a * r * k * a * r
and substitute R = aand substitute R = aTT * S to get * S to get R = aR = aTT * k * a * r * k * a * r
Element Stiffness Matrix in Global Element Stiffness Matrix in Global Coordinates Cont..Coordinates Cont..
is an important relationship is an important relationship
between the loading, stiffness between the loading, stiffness
and displacements of the and displacements of the structurestructure
in terms of the global systemin terms of the global system
RR = = aaTT * k * a * k * a * * rr
StiffnessStiffness
termterm
We have a stiffness term, We have a stiffness term, KeKe, for each element , for each element in the structure in the structure
We use them to build the global stiffness We use them to build the global stiffness matrix, matrix, KGKG
KeKe = = aaTT * k * a * k * a
Element Stiffness Matrix in Global Element Stiffness Matrix in Global Coordinates Cont..Coordinates Cont..
Let’s expand all of terms to get Let’s expand all of terms to get
a Ke that we can use.a Ke that we can use.KeKe = = 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 to Great formula to plug into your plug into your calculatorcalculator
Element Stiffness Matrix in Global Element Stiffness Matrix in Global Coordinates Cont..Coordinates Cont..
Let’s use a Let’s use a problem to illustrate problem to illustrate the rest of the the rest of the procedureprocedure
We will start by We will start by calculating KE’s for calculating KE’s for the two elementsthe two elements
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3 ft3 ft
4 ft4 ft
Node Node 11
Node Node 33
Node Node 22
Element Element
22
Element Element
11
Assembly of the Global Stiffness Matrix Assembly of the Global Stiffness Matrix (KG)(KG)
r1r1
r2r2
r3r3
r4r43 ft3 ft
Element 1Element 1
LL = 3= 3
Lx Lx = = x / L = (3-0) / 3 = 1x / L = (3-0) / 3 = 1
Ly Ly = = y / L = (0-0) / 3 = 0y / L = (0-0) / 3 = 0
NearNear FarFar
r1r1 r2r2 r3r3 r4r4
r1r1
r2r2
r3r3
r4r4
Ke1 A E
0.333
0
0.333
0
0
0
0
0
0.333
0
0.333
0
0
0
0
0
Pick a Pick a NearNear and a and a FarFar
Plug Lx, Ly and L Plug Lx, Ly and L into equation 14 to into equation 14 to getget
Assembly of the Global Stiffness Matrix Assembly of the Global Stiffness Matrix (KG)(KG)
r1r1
r2r2
r5r5
r6r6
3 ft3 ft
4 ft4 ft5 ft5 ft
Element 2Element 2
LL = 5= 5
Lx Lx = = x / L = (3-0) / 5 = x / L = (3-0) / 5 = 0.60.6
Ly Ly = = 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
0.072
0.096
0.072
0.096
0.096
0.128
0.096
0.128
r1r1 r2r2 r5r5 r6r6
r1r1
r2r2
r5r5
r6r6
The Entire Local Stiffness The Entire Local Stiffness Matrix in Global TermsMatrix in Global Terms
Ke2 A E
0.072
0.096
0.072
0.096
0.096
0.128
0.096
0.128
0.072
0.096
0.072
0.096
0.096
0.128
0.096
0.128
r1r1 r2r2 r5r5 r6r6r1r1r2r2r5r5r6r6
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 r5 r1 r2 r3 r4 r5 r6r6
Notice that there Notice that there aren’t any terms aren’t any terms in the local in the local matrix for matrix for r3r3 and and r4r4
ShorthandShorthand
Real Real MatrixMatrix
Assembly of the Global Assembly 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
r4r4
r5r5
r6r6
==KK rr RR
How does this relate to Stiffness by Definition?How does this relate to Stiffness by Definition?
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 r1R2 0.096 0.128 0.000 0.000 -0.096 0.128 r2R3 -0.333 0.000 0.333 0.000 0.000 0.000 r3R4 0.000 0.000 0.000 0.000 0.000 0.000 r4R5 -0.072 -0.096 0.000 0.000 0.072 0.096 r5R6 -0.096 -0.128 0.000 0.000 0.096 0.128 r6
= X
Solution Procedure cont..Solution Procedure cont..
To find the unknowns, we must subtend the matricesTo find the unknowns, we must subtend the matrices
Rk
Ru
AEK11
K21
K12
K22
ru
rk
K11K11
K22K22
K12K12
K21K21==
Two Two Important Important 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. Why?Why?
Solution Procedure cont..Solution Procedure cont..
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4 ft4 ft
10 kips10 kips
We will apply a load at DOF We 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/AEr2 = -95.02/AEr2 = -95.02/AE
solvedsolved
00
Solution Procedure cont..Solution Procedure cont..
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
95.02AE
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
Internal Member Force RecoveryInternal Member Force Recovery
To find the internal force inside of an To find the internal force inside of an element, we must return to the local element, we must return to the local coordinate systemcoordinate systemRemember the equation Remember 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 always But S1 always
Equals –S2Equals –S2
soso SAE
LLx Ly Lx Ly( )
r1
r2
r3
r4
Internal Member Force Recovery Internal Member Force Recovery Cont..Cont..
For Element 1For Element 1
For Element 2For Element 2
S1AE
31 0 1 0( )
22.52
AE
95.02AE
0
0
r1r1
r2r2
r3r3r4r4
= -7.5 kips= -7.5 kips
r1r1
r2r2
r5r5r6r6
= 12.5 kips= 12.5 kipsS2AE
50.6 0.8 0.6 0.8( )
22.52
AE
95.02AE
0
0
ConclusionConclusion
We solvedWe solved Element StiffnessesElement Stiffnesses Unknown Unknown
DisplacementsDisplacements ReactionsReactions Internal ForcesInternal Forces
I will cover another example I will cover another example in the laboratoryin the laboratory
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
12
3
B
CA
10 kips
5 kips
A Sample Problem solved with Stiffness by Definition and Direct Stiffness
42
For Stiffness by Definition, we are only concerned with the three DOF’s that are free to move:
r1
r2
r3
43
For Column 1, we set r1 = 1 and r2 = r3 = 0
A
B
C
B’Element Change in LengthElement Change in Length
1 6/10 Long2 8/10 Short3 0
Unit DisplacementUnit Displacement
44
For Column 2, we set r2 = 1 and r1 = r3 = 0
A
B
C
B’
Element Change in LengthElement Change in Length
1 8/10 Short2 6/10 Short3 0
Unit DisplacementUnit Displacement
45
For Column 3, we set r3 = 1 and r1 = r2 = 0
A
B
C
Element Change in LengthElement Change in Length
1 02 4/5 Long3 1 Long
C’
Unit DisplacementUnit Displacement
46
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 r30.14 -0.02 -0.08 r1-0.02 0.152 -0.06 r2-0.08 -0.06 0.18 r3
47
For Direct Stiffness, we are concerned with all six DOF’s in the structural system:
r1
r2
r3
r4
r5
r6
48
In the Direct Stiffness Method, we will use this equation for each elements 1, 2 and 3:
KeA 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
DOFLocation
49
Element 1
r5 r6 r1 r2
r5
r6
r1
r2
L = 6Lx = 0.6Ly = -0.8
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
50
Element 1 – Another View
r1 r2 r3 r4 r5 r6
r1
Ke1 AE
3
50
2
25
0
0
3
50
2
25
2
25
8
75
0
0
2
25
8
75
0
0
0
0
0
0
0
0
0
0
0
0
3
50
2
25
0
0
3
50
2
25
2
25
8
75
0
0
2
25
8
75
r2
r3r4
r5
r6
51
Element 2
r1 r2 r3 r4
r1
r2
r3
r4
L = 8Lx = 0.8Ly = 0.6
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
52
Element 3
r5 r6 r3 r4
r5
r6
r3
r4
L = 10Lx = 1Ly = 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
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
Summing Elements 1 through 3
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
Summing Elements 1 through 3
r1 r2 r3 r4 r5 r60.14 -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 r50.08 -0.11 0.00 0.00 -0.08 0.11 r6
Look Familiar? We found the yellow portion in the Stiffness by Definition Method
X
Stiffness by Definition vs Direct Stiffness by Definition vs Direct StiffnessStiffness
KK
K K completedcompleted
rrunknownunknown
RRunknownunknownrrknownknown
RRknownknown=
=X
ReactionsReactions
Zero Unless Zero Unless Settlement Settlement
OccursOccurs