# A Recap of Stiffness by Definition and the Direct

Document Sample

```					                  A Recap of
Stiffness by
Definition and
the Direct
March 20, 2003
9:35 AM
Stiffness Method
Little 109
CES 4141
Forrest Masters
Farther Down the Yellow Brick Road..

Structural Analysis

Classical Methods        Matrix Methods

Vitrual Work        Stiffness by Definition

Force Method            Direct Stiffness

Slope Deflection
Trusses
Moment-Area
Beams
Our Emphasis This Week: Trusses..

   Composed of slender,
lightweight members
   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      K = AE/L
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
From Strength of Materials..

Combine two equations to get a stiffness element

F=k*                   F
k
Spring                                     A E
k
L
F L           F       AE
                                          Units of
A E                   L             Force per
Length
Axial Deformation
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
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               Stiffness
K*r=R
Displacements          Matrix
   Unknown                              Known
Unknown
Displacements                        External
Displacements
Forces

For each released DOF, we get one equation that adds
But what about Reactions and Known Displacements?
A Better Method: Direct Stiffness

Consider all DOFs          Stiffness By Direct
Definition Stiffness

PIN                            0          2

ROLLER                          1          2

..now we have more equations to work with
A Simple Comparison
Stiffness by Definition
6
5            2 Degrees of Freedom
Direct Stiffness
 6 Degrees of Freedom
2                            DOFs 3,4,5,6 = 0
1
4
 Unknown Reactions (to
3
be solved) included in

Remember.. More DOFs = More Equations
Node Naming Convention
   Unknown or “Unfrozen”
6
5                  Degrees of Freedom are
numbered first…
r1, r2
2                                  Unknown or “Unfrozen”
1
4              Degrees of Freedom
3          follow
r3, r4, r5, r6

If Possible.. X-direction before Y-direction
6
5
Stiffness by Definition vs
Direct Stiffness                              2
1           4
Stiffness by Definition Solution in RED                            3
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           R4
K51 K52 K53 K54 K55 K56    r5           R5
K61 K62 K63 K64 K65 K66    r6           R6
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
remaining lecture. If you forgot it,
   I have your new homework (if you

Go to http://www.ce.ufl.edu/~kgurl for the
handout
Overview
First, we will decompose
the entire structure into
Node 2      a set of finite elements
Next, we will build a
Node 1               stiffness matrix for each
element (6 Here)
2      4       Later, we will combine all
of the local stiffness
matrices into ONE global
stiffness matrix
1      3      5
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                         v1
v2

K11 K12       v1       S1
=                          S1
K21 K22       v2       S2
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       AE         K22
L
K11
Node 2                           AE
L              K12

Node 1            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                                          r3
v2
r2             y
S1           LOCAL                     r1                      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’
Use a
y                       x                Displacement
Transformation
Matrix
x’
To change the coordinates of a truss..

   Each node has one
v2
displacement in the local                    r4
system concurrent to
r3
the element (v1 and v2)     v1
   In the global system,              r2        y
every node has two            r1
x
displacements in the x
and y direction
v1 will be expressed by r1 and r2
v2 will be expressed by r3 and r4
Displacement Transformation Matrix Cont..

r2          v1      The relationship between v and r is
the vector sum:
QY                     v1 = r1*cos Qx + r2*cos QY
v2 = r3*cos Qx + r4*cos QY
Qx
r1
Lx = cos Qx
We can simplify the cosine terms:
Ly = cos Qy
v1 = r1*Lx + r2*Ly
Put in matrix form
v2 = r3*Lx + r4*Ly
Displacement Transformation Matrix Cont..

 r1 
 
v1 = r1*Lx + r2*Ly               v1   Lx Ly 0 0    r2 
                
v2 = r3*Lx + r4*Ly               v2   0 0 Lx Ly   r3 
 r4 
 

 Lx Ly 0 0    Transformation matrix, a gives us the
a            
 0 0 Lx Ly    relationship we sought

So..           v = a*r
Force Transformation Matrix

Similarly, we can perform a transformation
on the internal forces
S2
 R1         
 Lx 0
           
 R2   Ly 0    S1 
 
R4
 R3   0 Lx   S2                     R3
 R4   0 Ly                       R2
                       S1   R1
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    is an important relationship
Stiffness       and displacements of the structure
term          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..

Let’s expand all of terms to get
Ke =    aT   *k*a
a Ke that we can use.

 Lx   0  
         
A  E  Ly     0   1 1   Lx Ly 0 0 
Ke                                       
L  0        Lx   1 1   0  0 Lx Ly 
 0       
      Ly 

 Lx2 Lx Ly Lx2 Lx Ly 
                             
           2               2    (14) From notes
AE   Lx Ly  Ly   Lx Ly Ly 

Ke
L         2           2             Great formula to plug
Lx Lx Ly Lx      Lx Ly
 Lx Ly Ly2 Lx Ly Ly2 
                             
Element Stiffness Matrix in Global Coordinates
Cont..
Node           1
Let’s  use a problem                         6
5
to illustrate the rest of
the procedure
Element    2                                 4 ft
2
Node   2        1
4
3
We  will start by                        3 ft
calculating KE’s for the
two elements                    Element   1           Node           3
Assembly of the Global Stiffness Matrix (KG)

Element 1
Near               Far
L          =3
r2        3 ft
r4
Lx   = Dx / L = (3-0) / 3 = 1
Ly   = Dy / L = (0-0) / 3 = 0
r1                r3
r1 r2     r3 r4
Pick a Near and a Far
 0.333 0 0.333 0    r1
                  
 0 0 0 0             r2
Plug Lx, Ly and L into       Ke1 A  E
 0.333 0 0.333 0    r3
equation 14 to get
 0 0 0 0             r4
                  
Assembly of the Global Stiffness Matrix (KG)

Element 2
r6             L          =5
Far                      Lx   = Dx / L = (3-0) / 5 = 0.6
r5
5 ft                       Ly   = Dy / L = (4-0) / 5 = 0.8
4 ft
r2                                   r1     r2      r5     r6
3 ft
 0.072 0.096 0.072 0.096    r1
r1                                                       
 0.096 0.128 0.096 0.128    r2
Ke2 A  E
Near                                0.072 0.096 0.072 0.096    r5
 0.096 0.128 0.096 0.128    r6
                           
The Entire Local Stiffness
Matrix in Global Terms
r1        r2      r5    r6
 0.072    0.096 0.072 0.096
r1
Shorthand
                             
0.128 0.096 0.128     r2
A  E 
0.096
Ke2
 0.072   0.096 0.072 0.096 
Real Matrix
r5
 0.096                      
          0.128 0.096 0.128      r6
r1        r2   r3 r4   r5   r6

 0.072 0.096 0       0 0.072 0.096
Notice that there                        0.096 0.128 0
r1
0 0.096 0.128   r2
aren’t any terms in                                                        
 0        0    0     0   0     0       r3
the local matrix for                     0        0    0     0   0     0       r4
r3 and r4                                0.072 0.096 0                   
0 0.072 0.096      r5
                                   
 0.096 0.128 0     0 0.096 0.128     r6
Assembly of the Global
Stiffness Matrix (KG)
Summing Ke1 and Ke2
K    r   = R

r1    r2     r3       r4    r5    r6
 0.405 0.096 0.333 0.000 0.072 0.096   r1
 0.096 0.128 0.000 0.000 0.096 0.128     r2
                                       
 0.333 0.000 0.333 0.000 0.000 0.000     r3
KG A  E
 0.000 0.000 0.000 0.000 0.000 0.000      r4
 0.072 0.096 0.000 0.000 0.072 0.096    r5
                                       
 0.096 0.128 0.000 0.000 0.096 0.128    r6
How does this relate to Stiffness by Definition?
Solution Procedure

Now, we can examine the full system

Loads acting on the nodes                 Unknown Deflections
R1         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
=                                                       X
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
Reactions                                Known displacements
@ reactions ( = 0 )
Solution Procedure cont..

To find the unknowns, we must subtend the matrices

K11 K12
 Rk  AE  K11 K12   ru 
=                                        
K21 K22             Ru       K21 K22  rk 

Two Important    Rk = AE ( K11*ru + K12*rk ) (24)
Equations        Ru = AE ( K21*ru + K22*rk ) (25)

Going to be ZERO. Why?
Solution Procedure cont..

6                       We will apply a load at DOF 2
5                  Then use equation (24)
Rk = AE ( K11*ru + K12*rk )                       0
0
4 ft                                                  
2                                  0  AE  0.405 0.096   r1   AE K1 2  0 
                     
1                              10     0.096 0.128  r2                0
4                                                            0
 
3
0 = AE ( 0.405*r1 + 0.096*r2)
3 ft
-10 = AE ( 0.096*r1 + 0.128*r2)

10 kips                                      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 )                              0
 R3       0.333 0   22.52                  0
                                             
 R4  AE  0        0 

AE
  AE K2 2 0
 R5       0.072 0.096  95.02             0
 R6       0.096 0.128  AE                 0
                                               

R3 = -7.5 kips        R4 = 0 kips
R5 = 7.5 kips         R6 = 10 kips
Internal Member Force Recovery

Tofind the internal force inside of an element, we
Remember the equation S = k * a * r ?
 r1 
 
 S1 
 
AE  1 1   Lx Ly 0 0   r2 
                
But S1 always
 S2     L  1 1   0 0 Lx Ly   r3 
 r4        Equals –S2
 

 r1 
 
so     S
AE
 ( Lx Ly Lx Ly )   r2 
L                        r3 
 r4 
 
Internal Member Force Recovery Cont..

For   Element 1                                22.52 
 AE        r1
        
95.02 
 ( 1 0 1 0 )              r2   = -7.5 kips
AE
S1
3                   AE 
 0         r3
        
 0         r4

For   Element 2
 22.52     r1
 AE 
        
95.02 
S2
AE
 ( 0.6 0.8 0.6 0.8)              r2   = 12.5 kips
5                           AE 
 0         r5
        
 0         r6
Conclusion
We solved
 Element Stiffnesses
 Unknown
Displacements
 Reactions
 Internal Forces

I will cover another
example in the
laboratory
Matrices..

Start with a basic equation       a x  b  y  c z   d
a1 x  b1  y  c1 z  d1     In order to solve x,y,z ..
a2 x  b2  y  b2 z  d2     You must have three
equations
a3 x  b3  y  b3 z  d3

a1 b1 c1  x               d1
But you must put these                            
equations in matrix     a 2 b 2 b 2   y     =   d2
form                   a b b  z                  d 
 3 3 3                       3
4
1
A Sample Problem solved with Stiffness by Definition
and Direct Stiffness

A                 3             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’
Element Change in Length

1       6/10   Long
Unit Displacement       2       8/10   Short
3       0
4
4

For Column 2, we set   r2 = 1 and r1 = r3 = 0

A                              C
B’
Unit Displacement

B
Element Change in Length

1       8/10   Short
2       6/10   Short
3       0
4
5

For Column 3, we set   r3 = 1 and r1 = r2 = 0
C        C’

A

Unit Displacement

B
Element Change in Length

1     0
2     4/5    Long
3     1      Long
4
6

The final stiffness matrix is as follows..

r1     r2          r3

 7         
1

2   r1
 50            50     25 
                         
K  1         91

3    r2           r1   r2   r3
 50        600        50            0.14 -0.02 -0.08 r1
 2             3      9             -0.02 0.152 -0.06 r2
                          r3
 25            50    50             -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   Near Y   Far X     Far Y

 Lx2 Lx Ly Lx2 Lx Ly 
                                 Near X
           2               2 
A  E  Lx Ly   Ly   Lx Ly Ly        Near Y
Ke         
L           2           2          
Lx Lx Ly Lx      Lx Ly
                                 Far X
 Lx Ly Ly2 Lx Ly Ly2 
                                 Far Y
4
9

Element 1

L=6
Lx = 0.6
Ly = -0.8               r5    r6           r1   r2

 3        2        3     2      r5
                           
 50       25       50    25

 2       8        2

8    r6
 25   75       25         75 
Ke 1   AE                              
 3       2        3
 
2
r1
 50   25       50         25 
 2        8        2      8 
 25                           r2
          75       25    75 
50

Element 1   – Another View

r1    r2 r3 r4 r5 r6
 3      
2
0 0 
3     2      r1
 50         25           50    25 
                                   
 2         8
0 0
2
 
8
r2
 25     75             25       75 
 0          0    0 0     0      0     r3
Ke1 AE                                       r4
 0          0    0 0     0      0

 3          2            3       2 
  50                         
0 0                   r5
25             50       25 
                                   
 2      
8
0 0 
2      8
   r6
 25         75           25    75 
51

Element 2

L=8
Lx = 0.8
Ly = 0.6                r1    r2         r3    r4
 2      2         2        3 
                              r1
 25 50            25      50

 3      9

3

9 
r2
 50 200           50     200 
Ke 2   AE                              
 2  3           2      3
   r3
 25 50        25         50 
 3       9        3      9 
  50  200                  
              50        200     r4
52

Element 3

L = 10
Lx = 1
Ly = 0                  r5     r6    r3   r4

 1           
1
0
r5
 10     0
10 
                       r6
Ke 3   AE  0      0      0 0
 1             1   
  10   0
10
0    r3

                   
 0      0      0 0     r4
53

Summing Elements 1                                  through 3
 3                                              2                                              1                 0
2        3     2                               2         2        3                                   1
                                                                                                   0 
 50       25       50   25
                    25   50         25       50
                   10           10    
 2       8        2

8                    3      9

3

9                                        
 25                       75                    50 200                  200 
AE 
75       25                                                  50                               0     0     0    0
Ke 1   AE 
 3       2        3
 
2
   +   Ke 2   AE 
 2  3           2      3


+   Ke 3
 1            1     
 50   25       50         25                    25 50        25         50                      10   0
10
0
 2        8        2      8                     3       9        3      9                                         
 25                                            50  200                  
          75       25    75                                  50        200                     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

r1       r2       r3      r4       r5      r6
 3  2 2  3    
2

3

3        2    
 50 25 25 50       25      50        50       25
   r1
                                                    
 2  3 8  9    
3

9        2

8 
 25 50 75 200      50      200      25         75      r2
      2      3 2      1  3            1

                         0                0       r3
KG AE                                                     
25     50 25 10 50              10
      3      9  3        9                          
                  0       0      0         0       r4
50    200  50      200
                                                    
     3     2

1
0
3

1

2
 0
     50    25      10           50 10 25               r5
     2       8                     2        8       
                  0       0          0       0 
r6
    25      75                    25       75       
55

Summing Elements 1   through 3
Look Familiar? We found the yellow portion
in the Stiffness by Definition Method

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
Stiffness by Definition vs Direct Stiffness

K         X       runknown   =    Rknown
X              =
K completed        rknown         Runknown

Zero Unless
Settlement Occurs       Reactions

```
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