March 20, 2003
9:35 AM
Little 109
CES 4141
Forrest Masters
A Recap of
Stiffness by
Definition and
the Direct
Stiffness Method
Farther Down the Yellow Brick Road..
Structural Analysis
Classical Methods
Matrix Methods
Vitrual Work
Stiffness by Definition
Force Method
Direct Stiffness
Slope Deflection
Moment-Area
Trusses
Beams
Our Emphasis This Week: Trusses..






Composed of slender,
lightweight members
All loading occurs on joints
No moments or rotations
in the joints
Axial Force Members
Tension (+)
Compression (-)
Stiffness


Kij = the amount of force
required at i to cause a unit
displacement at j, with
displacements at all other DOF
= zero
A function of:
– System geometry
– Material properties (E, I)
– Boundary conditions (Pinned,
Roller or Free for a truss)

NOT a function of external loads
K = AE/L
From Strength of Materials..
Combine two equations to get a stiffness element
F=k*
Spring
k
F

k

F L
F
AE
A E

L
Axial Deformation
A E
L
Units of
Force per
Length
Go to the Board..
Let’s take a
look at last
week’s
homework to
shed some light
on the Stiffness
by Definition
Procedure
DOF
From Stiffness by Definition

We can create a stiffness matrix that
accounts for the material and geometric
properties of the structure

A square, symmetric matrix Kij = Kji
Diagonal terms always positive


The stiffness matrix is independent of the
loads acting on the structure. Many
loading cases can be tested without
recalculating the stiffness matrix
However ..
Stiffness by Definition only uses a small part of
the information available to tackle the problem
Stiffness by Definition Only Considers..



Stiffnesses from
Imposed
Displacements
Unknown
Displacements
Known Loadings
Stiffness
Matrix
Unknown
Displacements
K*r=R
Known
External
Forces
For each released DOF, we get one equation that adds
to the stiffness, displacement and loading matrices
But what about Reactions and Known Displacements?
A Better Method: Direct Stiffness
Consider all DOFs
PIN
ROLLER
Stiffness By Direct
Definition Stiffness
0
2
1
2
..now we have more equations to work with
A Simple Comparison
6
2
1
5
4
3
Stiffness by Definition
 2 Degrees of Freedom
Direct Stiffness
 6 Degrees of Freedom
 DOFs 3,4,5,6 = 0
 Unknown Reactions (to
be solved) included in
Loading Matrix
Remember.. More DOFs = More Equations
Node Naming Convention
6
2

5

1
4
3
Unknown or “Unfrozen”
Degrees of Freedom are
numbered first…
r1, r2
Unknown or “Unfrozen”
Degrees of Freedom
follow
r3, r4, r5, r6
If Possible.. X-direction before Y-direction
6
Stiffness by Definition vs
Direct Stiffness
2
Stiffness by Definition Solution in RED
Direct Stiffness Solution in RED/YELLOW
K11 K12 K13 K14 K15 K16
r1
R1
K21 K22 K23 K24 K25 K26
r2
R2
K31 K32 K33 K34 K35 K36
r3
R3
K41 K42 K43 K44 K45 K46
r4
K51 K52 K53 K54 K55 K56
r5
R5
K61 K62 K63 K64 K65 K66
r6
R6
=
R4
1
5
4
3
The Fundamental Procedure




Calculate the Stiffness Matrix
 Determine Local Stiffness Matrix, Ke
 Transform it into Global Coordinates, KG
 Assemble all matrices
Solve for the Unknown Displacements
Use unknown displacements to solve for the
Unknown Reactions
Calculate the Internal Forces
To continue..


You need your Direct Stiffness – Truss
Application Handout to follow the
remaining lecture. If you forgot it,
look on your neighbor’s, please
I have your new homework (if you
don’t have it already)
FOR MORE INFO ..
Go to http://www.ce.ufl.edu/~kgurl for the
handout
Overview
Node 2
Node 1
1
2
4
3
5
First, we will decompose
the entire structure into
a set of finite elements
Next, we will build a
stiffness matrix for each
element (6 Here)
Later, we will combine all
of the local stiffness
matrices into ONE global
stiffness matrix
Element Stiffness Matrix in Local Coordinates


Remember Kij = the amount of force required at i to
cause a unit displacement at j, with displacements at all
other DOF = zero
For a truss element (which has 2 DOF)..
S2
K11*v1 + K12*v2 = S1
K21*v1 + K22*v2 = S2
K11 K12
v1
K21 K22
v2
=
S1
S2
v1
v2
S1
Gurley refers to the axial displacement as “v” and the
internal force as “S” in the local coordinate system
Element Stiffness Matrix in Local Coordinates

Use Stiffness by Definition to finding Ks of Local System
K21
Node 2
Node 1
K11
AE
L
AE
L
K22
K12
K11 = AE / L
K12 = - AE / L
K21 = - AE / L
K22 = AE / L
Element Stiffness Matrix in Local Coordinates
Cont..
Put the local stiffness elements in matrix form
Simplified..
For a truss element
Displacement Transformation Matrix


Structures are composed of many members in many
orientations
We must move the stiffness matrix from a local to a
global coordinate system
GLOBAL
S2
r4
v1
S1
r3
v2
LOCAL
r2
r1
y
x
How do we do that?

Meaning if I give you a point (x,y) in
Coordinate System Z, how do I find the
coordinates (x’,y’) in Coordinate System Z’
y’
y
x
x’
Use a
Displacement
Transformation
Matrix
To change the coordinates of a truss..


Each node has one
displacement in the local
system concurrent to
the element (v1 and v2)
In the global system,
every node has two
displacements in the x
and y direction
v2
v1
r4
r3
r2
r1
v1 will be expressed by r1 and r2
v2 will be expressed by r3 and r4
y
x
Displacement Transformation Matrix Cont..
r2
v1
QY
Qx
r1

The relationship between v and r is
the vector sum:
v1 = r1*cos Qx + r2*cos QY
v2 = r3*cos Qx + r4*cos QY
We can simplify the cosine terms:
v1 = r1*Lx + r2*Ly
v2 = r3*Lx + r4*Ly
Lx = cos Qx
Ly = cos Qy
Put in matrix form
Displacement Transformation Matrix Cont..
 r1 
 v1   Lx Ly 0 0    r2 
  
 
 v2   0 0 Lx Ly  r3
 r4 
 
v1 = r1*Lx + r2*Ly
v2 = r3*Lx + r4*Ly
Lx Ly 0 0 

a 

 0 0 Lx Ly 
Transformation matrix, a gives us the
relationship we sought
So..
v = a*r
Force Transformation Matrix
Similarly, we can perform a transformation
on the internal forces
 R1 

  

 R2   Ly 0    S1 
 R3   0 Lx   S2 
 R4   0 Ly 
  

S2
 Lx 0
R3
S1
R1
R2
R4
Element Stiffness Matrix in Global Coordinates
Let’s put it all together.. We know that the
Internal force = stiffness * local displacement (S = k * v)
Units: Force = (Force/Length) * Length
local disp = transform matrix * global disp
(v = a * r)
Substitute local displacement
Internal force = stiffness * transform matrix * global disp
(S = k * a * r)
Premultiply by the transpose of “a”
aT * S= aT * k * a * r
and substitute R = aT * S to get
R = aT * k * a * r
Element Stiffness Matrix in Global Coordinates
Cont..
R = aT * k * a * r
Stiffness
term
is an important relationship
between the loading, stiffness
and displacements of the structure
in terms of the global system
We
have a stiffness term, Ke, for each element in the
structure
Ke = aT * k * a
We
use them to build the global stiffness matrix, KG
Element Stiffness Matrix in Global Coordinates
Cont..
Ke =
Ke
Ke
aT
*k*a
Let’s expand all of terms to get
a Ke that we can use.
Lx 0 



0 
A  E  Ly 0   1 1   Lx Ly 0



 

L  0 Lx   1 1   0
0 Lx Ly 
 0 Ly 


 Lx2 Lx Ly Lx2 Lx Ly 

2
2 
Lx

Ly
Ly

Lx

Ly

Ly
A E 



L 
2
2
Lx Lx Ly Lx
Lx Ly


 Lx Ly Ly 2 Lx Ly Ly 2 


(14) From notes
Great formula to plug
into your calculator
Element Stiffness Matrix in Global Coordinates
Cont..
Node
Let’s
use a problem
to illustrate the rest of
the procedure
2
Node 2
Element
will start by
calculating KE’s for the
two elements
6
2
1
5
4 ft
1
We
4
3
3 ft
Element
1
Node
3
Assembly of the Global Stiffness Matrix (KG)
Near
Element 1
L
=3
Lx
= Dx / L = (3-0) / 3 = 1
Ly
= Dy / L = (0-0) / 3 = 0
Far
r2
3 ft
r1
r4
r3
Pick a Near and a Far
Plug Lx, Ly and L into
equation 14 to get
r1 r2
 0.333
0

Ke1 A  E
 0.333
 0

r3 r4
0 0.333 0  r1

0
0
0  r2
0 0.333 0  r3

0
0
0  r4
Assembly of the Global Stiffness Matrix (KG)
Far
r6
r5
5 ft
Element 2
L
=5
Lx
= Dx / L = (3-0) / 5 = 0.6
Ly
= Dy / L = (4-0) / 5 = 0.8
4 ft
r2
r1
Near
3 ft
r1
 0.072
0.096

Ke2 A  E
 0.072
 0.096

r2
r5
r6
0.096 0.072 0.096  r1

0.128 0.096 0.128  r2
0.096 0.072 0.096  r5

0.128 0.096 0.128  r6
The Entire Local Stiffness
Matrix in Global Terms
r1
Ke2
 0.072
0.096
A  E 
 0.072
 0.096

r2
0.096
r5
r6
0.072 0.096 

0.128 0.096 0.128 
0.096 0.072 0.096 

0.128 0.096 0.128 
r1
r2
r5
r6
r1
Notice that there
aren’t any terms in
the local matrix for
r3 and r4
Shorthand
 0.072
 0.096

 0
 0
 0.072

 0.096
Real Matrix
r2
r3 r4
r5
r6
0.096 0 0 0.072 0.096 
r1

0
0 
0
0 

0.072 0.096

0.096 0.128 
r2
0.128 0 0 0.096 0.128 
0
0 0
0
0 0
0.096 0 0
0.128 0 0
r3
r4
r5
r6
Assembly of the Global
Stiffness Matrix (KG)
Summing Ke1 and Ke2
r1
 0.405
 0.096

0.333

KG A  E
 0.000
 0.072

 0.096
K
r2
r3
r
r4
= R
r5
r6
0.096 0.333 0.000 0.072 0.096  r1
0.128
0.000 0.000 0.096 0.128
0.000
0.333 0.000 0.000
0.000
0.000
0.000 0.000 0.000
0.000
0.096 0.000 0.000 0.072
0.096
0.128 0.000 0.000 0.096
0.128







How does this relate to Stiffness by Definition?
r2
r3
r4
r5
r6
Solution Procedure
Now, we can examine the full system
Loads acting on the nodes
R1
R2
R3
R4
R5
R6
=
Reactions
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
Unknown Deflections
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
r1
0.128
r2
0.000
r3
X
0.000
r4
0.096
r5
0.128
r6
Known displacements
@ reactions ( = 0 )
Solution Procedure cont..
To find the unknowns, we must subtend the matrices
=
K11 K12
K21 K22
Two Important
Equations
 Rk  AE  K11 K12    ru 
 

 
 Ru 
 K21 K22   rk 
Rk = AE ( K11*ru + K12*rk ) (24)
Ru = AE ( K21*ru + K22*rk ) (25)
Going to be ZERO. Why?
Solution Procedure cont..
6
We will apply a load at DOF 2
Then use equation (24)
5
Rk = AE ( K11*ru + K12*rk )
2
4 ft
1
4
3 ft
10 kips
3
0
 0 
 0  AE  0.405 0.096    r1   AE K12  0 



 
0

10
0.096
0.128



  r2 
0
 
0 = AE ( 0.405*r1 + 0.096*r2)
-10 = AE ( 0.096*r1 + 0.128*r2)
solved r1 = 22.52/AE
r2 = -95.02/AE
Solution Procedure cont..
With the displacements, we can use equation (25) to find
the reactions at the pinned ends
Ru = AE ( K21*ru + K22*rk )
 R3 
 0.333 0  
0 
 R4  AE  0

 R5 
 0.072 0.096  
 R6 
 0.096 0.128  
 


R3 = -7.5 kips
R5 = 7.5 kips
 0 

AE 
0

  AE K22  
95.02
0

0
AE 
 
22.52
R4 = 0 kips
R6 = 10 kips
0
Internal Member Force Recovery
To
find the internal force inside of an element, we
must return to the local coordinate system
Remember the equation S = k * a * r ?
 S1 
 
 S2 
so
 r1 
AE  1 1   Lx Ly 0 0   r2 


 
L  1 1   0 0 Lx Ly  r3
 r4 
 
S
 r1 
r2 
AE

 ( Lx Ly Lx Ly ) 
 r3 
L
 r4 
 
But S1 always
Equals –S2
Internal Member Force Recovery Cont..
For
Element 1



AE
 ( 1 0 1 0 )  

3




AE 

95.02 
AE 
0 

0 
r1



AE
 ( 0.6 0.8 0.6 0.8 )  

5




AE 

95.02 
AE 
0 

0 
r1
S1
For
22.52
r2
r3
r4
= -7.5 kips
Element 2
S2
22.52
r2
r5
r6
= 12.5 kips
Conclusion
We solved
 Element Stiffnesses
 Unknown
Displacements
 Reactions
 Internal Forces
I will cover another
example in the
laboratory
Matrices..
Start with a basic equation
a1 x  b1  y  c1 z  d1
a2 x  b2  y  b2 z  d2
a3 x  b3  y  b3 z  d3
a x  b  y  c z
d
In order to solve x,y,z ..
You must have three
equations
a1 b1 c1  x
 
But you must put these 
 a 2 b 2 b 2   y 
equations in matrix
a b b  z
form
 3 3 3
=
d1
 
d2
d 
 3
4
1
A Sample Problem solved with Stiffness by Definition
and Direct Stiffness
3
A
C
1
2
B
10 kips
5 kips
4
2
For Stiffness by Definition, we are only concerned with
the three DOF’s that are free to move:
r3
r2
r1
4
3
For Column 1, we set
r1 = 1 and r2 = r3 = 0
A
C
B
B’
Unit Displacement
Element Change in Length
1
2
3
6/10
8/10
0
Long
Short
4
4
For Column 2, we set
r2 = 1 and r1 = r3 = 0
A
C
B’
Unit Displacement
B
Element Change in Length
1
2
3
8/10
6/10
0
Short
Short
4
5
For Column 3, we set
r3 = 1 and r1 = r2 = 0
C
C’
A
Unit Displacement
B
Element Change in Length
1
2
3
0
4/5
1
Long
Long
4
6
The final stiffness matrix is as follows..
r1
 7
 50

1


K
 50
 2

 25
r2

1
50
91
600

3
50
r3

25 

3 

50 
9 

50 

2
r1
r2
r3
r1
r2
r3
0.14 -0.02 -0.08 r1
-0.02 0.152 -0.06 r2
-0.08 -0.06 0.18 r3
4
7
For Direct Stiffness, we are concerned with all six
DOF’s in the structural system:
r6
r4
r5
r3
r2
r1
4
8
In the Direct Stiffness Method, we will use this equation
for each elements 1, 2 and 3:
DOF
Location
Near X
Ke
Near Y
Far X
Far Y
 Lx2 Lx Ly Lx2 Lx Ly 

2
2 
Ly
Lx Ly Ly 
A  E  Lx Ly


L 
2
2
Lx Lx Ly Lx
Lx Ly


 Lx Ly Ly 2 Lx Ly Ly 2 


Near X
Near Y
Far X
Far Y
4
9
Element 1
L=6
Lx = 0.6
Ly = -0.8
r5
Ke 1
 3
 50
 2
 25
AE 
 3
 50
 2
 25

r6
r1
2
3

25

50
8
2
75
25
2
3
25
50
8
2

75

25
r2

25

8 

75 

2
 
25 
8 

75 
2
r5
r6
r1
r2
50
Element 1
– Another View
r1
 3
 50

 2
 25
 0
Ke1 AE 
 0
 3
 50

 2
 25
r2 r3 r4 r5

2
25
8
75
0 0
50
2
25
0
0 0
0
0
0 0
0
2
25

0 0 
3
8
75
0 0
3
50
0 0 
2
25
r6

25 

8
 
75 
0 

0

2 

25 

8

75 
2
r1
r2
r3
r4
r5
r6
51
Element 2
L=8
Lx = 0.8
Ly = 0.6
r1
Ke 2
r2
2
 2
 25 50
9
 3
 50 200
AE 
 2  3
 25 50
 3
9
  50  200

r3


2
25
3
50
2
25
3
50
r4

50

9 

200 

3

50 
9 

200 

3
r1
r2
r3
r4
52
Element 3
L = 10
Lx = 1
Ly = 0
r5
Ke 3
 1
 10

0

AE
 1
  10

 0
r6
0
0
0
0
r3
r4

0 
10

0 0

1
0
10

0 0
1
r5
r6
r3
r4
53
Summing Elements 1
Ke 1
 3
 50
 2
 25
AE 
 3
 50
 2
 25


2
25

3
50
8
2
75
25
2
3
25
50
8
2

75

25


8 

75 

2
 
25 
8 

75 
through 3
2
25
+
Ke 2
2
 2
25
50

9
 3
 50 200
AE 
 2  3
 25 50
 3
9
  50  200



2
25
3
50
2
25
3
50


9 

200 

3

50 
9 

200 

3
50
+
Ke 3
 1
 10

0
AE 
 1
  10

 0
0 
0
0
0
10
0 
0
0
1
1


10
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
r1
through 3
r2
r3
r4
r5
r6
2
3
3
2
 3  2 2  3




 50 25 25 50

25
50
50
25


3
9
2
8 
 2  3 8  9



 25 50 75 200
50
200
25
75 


2
3
2
1
3
1
 


0

0

25
50
25
10 50
10

KG AE 


3
9
3
9



0

0
0
0


50
200
50
200


3
2
1
3
1
2
 

0


 0
50
25
10
50
10
25




2
8
2
8

0
0

0
0 

25
75
25
75


r1
r2
r3
r4
r5
r6
55
Summing Elements 1
through 3
Look Familiar? We found the yellow portion
in the Stiffness by Definition Method
r1
0.14
-0.02
-0.08
-0.06
-0.06
0.08
r2
-0.02
0.15
-0.06
-0.05
0.08
-0.11
r3
-0.08
-0.06
0.18
0.06
-0.10
0.00
r4
-0.06
-0.05
0.06
0.05
0.00
0.00
r5
-0.06
0.08
-0.10
0.00
0.16
-0.08
r6
0.08
-0.11
0.00
0.00
-0.08
0.11
r1
r2
r3
r4
r5
r6
Stiffness by Definition vs Direct Stiffness
K
runknown
X
Zero Unless
Settlement Occurs
Rknown
=
X
K completed
=
rknown
Reactions
Runknown