#### Transcript ppt

```MANE 4240 & CIVL 4240
Introduction to Finite Elements
Prof. Suvranu De
Development of Truss
Equations
Chapter 3: Sections 3.1-3.9 + Lecture notes
Summary:
• Stiffness matrix of a bar/truss element
• Coordinate transformation
• Stiffness matrix of a truss element in 2D space
•Problems in 2D truss analysis (including multipoint
constraints)
•3D Truss element
Trusses: Engineering structures that are composed only
of two-force members. e.g., bridges, roof supports
Actual trusses: Airy structures composed of slender
members (I-beams, channels, angles, bars etc) joined
together at their ends by welding, riveted connections or
large bolts and pins
A typical truss structure
Gusset plate
Ideal trusses:
Assumptions
• Ideal truss members are connected only at their ends.
• Ideal truss members are connected by frictionless pins (no
moments)
• The truss structure is loaded only at the pins
• Weights of the members are neglected
A typical truss structure
Frictionless pin
These assumptions allow us to idealize each truss
member as a two-force member (members loaded only
at their extremities by equal opposite and collinear
forces)
member in
compression
member in
tension
Connecting pin
FEM analysis scheme
Step 1: Divide the truss into bar/truss elements connected to
each other through special points (“nodes”)
Step 2: Describe the behavior of each bar element (i.e. derive its
stiffness matrix and load vector in local AND global coordinate
system)
Step 3: Describe the behavior of the entire truss by putting
together the behavior of each of the bar elements (by assembling
their stiffness matrices and load vectors)
Step 4: Apply appropriate boundary conditions and solve
Stiffness matrix of bar element
E, A
© 2002 Brooks/Cole Publishing / Thomson Learning™
L: Length of bar
A: Cross sectional area of bar
E: Elastic (Young’s) modulus of bar
uˆ ( xˆ ) :displacement of bar as a function of local coordinate
The strain in the bar at xˆ
d uˆ
ε( xˆ ) 
d xˆ
The stress in the bar (Hooke’s law)
 ( xˆ )  E ε( xˆ )
xˆ
of bar
dˆ 2x
Tension in the bar
T( xˆ )  EA ε
dˆ 1x
xˆ
Assume that the displacement
uˆ ( xˆ )
xˆ
L
is varying linearly along the bar
xˆ  ˆ
xˆ ˆ

uˆ ( xˆ )   1   d 1x  d 2x
L
L

Then, strain is constant along the bar:
Stress is also constant along the bar:
Tension is constant along the bar:
xˆ 
xˆ

uˆ ( xˆ )   1   dˆ 1x  dˆ 2x
L
L

ε
d uˆ
d xˆ

dˆ 2x  dˆ 1x
L

E ˆ
  Eε 
d 2x  dˆ 1x
L


EA ˆ
T  EA ε 
d 2x  dˆ 1x
L

k
The bar is acting like a spring with stiffness
k 
EA
L

Recall the lecture on springs
E, A
© 2002 Brooks/Cole Publishing / Thomson Learning™
Two nodes: 1, 2
Nodal displacements: dˆ 1x dˆ 2x
Nodal forces: fˆ1x fˆ 2x
Spring constant: k  EA
L
Element stiffness matrix in local coordinates
Element force
vector

- k
 fˆ1x 
 k
 dˆ 1x 

ˆf  kˆ dˆ
ˆ   
 ˆ 
k 

 - k
d 2x 
2x 
f






Element nodal 
Element
stiffness
matrix
displacement
vector
fˆ
kˆ
dˆ
What if we have 2 bars?
E1, A1
E2, A2
L2
L1
This is equivalent to the following system of springs
k1 
E 1A 1
L1
k2 
E 2A 2
L2
x
Element 1 2 Element 23
1
d1x
PROBLEM
d2x
d3x
Problem 1: Find the stresses in the two-bar assembly loaded as
shown below
E, 2A
E, A
P
1
2
3
L
L
Solution: This is equivalent to the following system of springs
k1 
1
2E A
L
k2 
EA
L
x
Element 1 2 Element 23
d1x
d2x
d3x
We will first compute the displacement at node 2 and then the
stresses within each element
The global set of equations can be generated using the technique
developed in the lecture on “springs”
 k1

k
 1
 0
here
 k1
k1  k 2
k2
d1 x  d 3 x  0
0   d 1x   F1x 
 


 k 2  d 2 x    F2 x 

 F 
k 2  
d
 3x   3x 
and F2 x  P
Hence, the above set of equations may be explicitly written as
 k 1 d 2 x  F1 x
(1)
( k1  k 2 ) d 2 x  P
(2 )
 k 2 d 2 x  F3 x
(3)
From equation (2) d 2 x 
P
k1  k 2

PL
3EA
To calculate the stresses:
For element #1 first compute the element strain

(1)

d 2 x  d1x
d2x

L

L
P
3EA
and then the stress as

(1)
 E
P

(1)
(element in tension)
3A
Similarly, in element # 2

(2)

d3x  d2x

L

(2)
 E
(2)
d2x
L

P
3A

P
3EA
(element in compression)
© 2002 Brooks/Cole Publishing / Thomson Learning™
Inter-element continuity of a two-bar structure
Bars in a truss have various orientations
member in
compression
member in
tension
Connecting pin
d 2y , f 2y
y
dˆ 2y , fˆ2y  0
yˆ
dˆ 2x , fˆ 2x
d 2x , f 2x
d1y , f1y
θ
dˆ 1y , fˆ1y  0
dˆ 1x , fˆ1x
xˆ
d1x , f1x
x
At node 1:
θ
dˆ 1x
d1x
θ
f1x
dˆ 1y
d1y
fˆ1y  0
fˆ1 x
f1y
At node 2:
θ
dˆ 2 x
f 2x
d 2y
fˆ 2y  0
d 2x
θ
dˆ 2y
fˆ 2 x
f 2y
In the global coordinate system, the vector of nodal
 d 1x 


d
 1y 
d 
;
d 2x


 d 2y 
 f 1x 
 
 f 1y 
f  
f 2x
 
 f 2y 
Our objective is to obtain a relation of the form
f  k d
4 1
4  4 4 1
Where k is the 4x4 element stiffness matrix in global coordinate
system
The key is to look at the local coordinates
xˆ
y
dˆ 2y , fˆ2y  0
yˆ
dˆ 2x , fˆ 2x
dˆ 1y , fˆ1y  0

 fˆ1x 
 k
ˆ   

 f 2x 
 - k
θ
- k
 dˆ 1x 

 ˆ 
k 
 d 2x 

dˆ 1x , fˆ1x
k 
x
Rewrite as  fˆ 
1x
k
 

0
 fˆ1y 

ˆ  
 f 2x   - k
ˆ   0

 f 2y 
0
-k
0
0
0
k
0
0
EA
L
ˆ
0   d 1x 


 ˆ
0  d 1y 


ˆ
0   d 2x 

0  dˆ 
 2y 
fˆ  kˆ dˆ
NOTES
1. Assume that there is no stiffness in the local ^y direction.
2. If you consider the displacement at a point along the local x
direction as a vector, then the components of that vector along the
global x and y directions are the global x and y displacements.
3. The expanded stiffness matrix in the local coordinates is
symmetric and singular.
NOTES
5. In local coordinates we have
fˆ  kˆ dˆ
4 1
4  4 4 1
But or goal is to obtain the following relationship
f  k d
4 1
4  4 4 1
Hence, need a relationship between dˆ and d
dˆ
and between fˆ and f
1y
ˆ 

d
1x
 d 1x 




ˆ
 d 1y  ˆ  d 1y 
d
 d  ˆ 
d 2x


 d 2x 
 d 2y 
ˆ 
 d 2y 
θ
dˆ 1x
d1x
θ
d 2x
d1y
dˆ 2y
dˆ 2 x
d 2y
Need to understand
how the components
of a vector change
with coordinate
transformation
Transformation of a vector in two dimensions
yˆ
y
v y cos θ
vx
ˆx
v
θ
v
vˆ y
vy
v x sin θ
xˆ
Angle q is
measured positive
in the counter
clockwise direction
from the +x axis)
θ
v x cos θ
v y sin θ
x
The vector v has components (vx, vy) in the global coordinate system
and (v^x, v^y) in the local coordinate system. From geometry
ˆ x  v x cos θ  v y sin θ
v
ˆ y   v x sin θ  v y cos θ
v
In matrix form
ˆ x   cos θ
v
 
ˆ
 v y    sin θ
sin θ   v x 
 v 
cos θ   y 
Direction cosines
Or
ˆx  l
v
 
ˆ
 v y   m
m vx 
 
l  vy 
where
l  cos q
m  sin q
Transformation matrix for a single vector in 2D
 l
T 
 m
*
where
m

l
relates
*
ˆ T v
v
ˆx
v
vx 
ˆ    and v   
v
ˆ
vy 
vy 
are components of the same
vector in local and global
coordinates, respectively.
Relationship between dˆ and d for the truss element
dˆ 1y
At node 1
At node 2

d
dˆ 1x 

*  1x 
ˆ   T  

d1y 
d1y 


d
dˆ 2x 

*  2x 
ˆ   T 

d
d

 2y 
 2y 

Putting these together
dˆ  T d
dˆ 1x
θ
dˆ 2y
d1x
θ
dˆ 2 x
d 2y
d 2x
 dˆ 1x 
m
0
0   d 1x 
 l





m l
0
0  d 1y 
 dˆ 1y 

ˆ   

d


0
0
l
m  2x 
 d 2x 


ˆ   0
0  m l   d 2y 
d
2y 
       


d
T
dˆ
d1y
T *
T  
4 4
0
0 
*
T 
Relationship between fˆ and f for the truss element
At node 1
At node 2

f
fˆ1x 

*  1x 
ˆ   T  

f1y 
f1y 


f
fˆ2x 

*  2x 
ˆ   T  

f 2y 
f 2y 

Putting these together
fˆ  T f
 fˆ1x 
m
0
0   f 1x 
 l
 

 
m
l
0
0  f 1y 
 fˆ1y 
 
ˆ   
0
l
m   f 2x 
 f 2x   0


ˆ   0
0 m
l   f 2y 
f
2y 
       


f
T
fˆ
fˆ1 y
θ
f1y
fˆ1 x
fˆ 2 y
f1x
θ
fˆ 2 x
f 2y
f 2x
T *
T  
4 4
0
0 
*
T 
Important property of the transformation matrix T
The transformation matrix is orthogonal, i.e. its inverse is its
transpose
T
1
T
T
Use the property that l2+m2=1
Putting all the pieces together
xˆ
y
dˆ 2y , fˆ 2y
yˆ
dˆ  T d
dˆ 2x , fˆ 2x
dˆ 1y , fˆ1y
fˆ  kˆ dˆ
θ
dˆ 1x , fˆ1x
 T f  kˆ T d
x
The desired relationship is
Where
k T
4 4
fˆ  T f
T
kˆ T
4 4 4 4 4 4
f  k d
4 1


 f  T kˆ T d



1
k
4  4 4 1
is the element stiffness matrix in the
global coordinate system
 l

m

T 
 0

 0
m
0
l
0
0
l
0
m
0 

0

m 

l 
 l2

EA  lm
T ˆ
k  T kT 
L   l2

  lm
k

0
ˆk  
- k

 0
l
lm
m
2
 lm
2
 lm
l
m
lm
2
2
0
-k
0
0
0
k
0
0
 lm
m
lm
m
2
2






0

0

0

0
Computation of the direction cosines
l  cos q 
x2  x1
L
2 (x2,y2)
L
m  sin q 
y2  y1
θ
1
L
(x1,y1)
What happens if I reverse the node numbers?
l '  cos q 
x1  x2
L
 l
L
m'  sin q 
y1  y2
L
 m
θ
2 (x ,y )
2 2
Question: Does the stiffness matrix change?
1 (x1,y1)
Example Bar element for stiffness matrix evaluation
E  30  10
© 2002 Brooks/Cole Publishing / Thomson Learning™
A  2 in
6
psi
2
L  60 in
q  30

l  cos 30 
3
2
 3

 4
 3
6

30  10  2   4
k 

60
3
 
 4

3


 4
3
4
1

4



3
3
4
3
4
3
4
1
4
3
4
4
3 


4 
1 

4  lb
 in
3

4


1

4

m  sin 30 
1
2
Computation of element strains
© 2002 Brooks/Cole Publishing / Thomson Learning™
Recall that the element strain is
ε 
dˆ 2x  dˆ 1x

L

1
L

1
L
1
L
 1
 1
0
1
0  dˆ
 1
0
1
0 T d
0
1
 dˆ 1x

 dˆ 1y
0 
ˆ
d
 2x
ˆ
 d 2y







ε 
1
L

1
L

1
L
 1
 l
 l
0
 l

m

0
 0

 0
1
m
m
l
l
m
0
l
0
0
l
0
m
m d
 d 1x

 d 1y
m 
d
 2x
 d 2y








0
d
m 

l 
0
Computation of element stresses stress and tension
Recall that the element stress is
  Eε 


E ˆ
E
d 2x  dˆ 1x   l
L
L
Recall that the element tension is
T  EAε 
EA
L
l
m
l
md
m
l
m d
Steps in solving a problem
Step 1: Write down the node-element connectivity table
linking local and global nodes; also form the table of
direction cosines (l, m)
Step 2: Write down the stiffness matrix of each element in
global coordinate system with global numbering
Step 3: Assemble the element stiffness matrices to form the
global stiffness matrix for the entire structure using the
node element connectivity table
Step 4: Incorporate appropriate boundary conditions
Step 5: Solve resulting set of reduced equations for the unknown
displacements
Step 6: Compute the unknown nodal forces
Node element connectivity table
ELEMENT Node 1 Node 2
1
1
2
2
2
3
3
3
1
1
El 1
2
60
60
L
El 3
60
El 2
θ
3
1 (x ,y )
1 1
2 (x2,y2)
Stiffness matrix of element 1
d1x d1y d2x d2y

 d1x


d1y
(1 )


k 

 d2x



 d2y
Stiffness matrix of element 3
d3x d3y d1x d1y

 d3x


d3y
(3)


k


 d1x



 d1y
Stiffness matrix of element 2
d2x d2y d3x d3y

 d2x


d2y
(2)


k


 d3x



 d3y
There are 4 degrees of
freedom (dof) per
element (2 per node)
k
(1 )
Global stiffness matrix
d1x d1y d2x d2y d3x d3y

 d1x



 d1y

 d2x
K  
 d
2y



 d3x

 d3y

 6  6
How do you incorporate boundary conditions?
k
k
(2)
(3)
Example 2
The length of bars 12 and 23 are equal (L)
E: Young’s modulus
A: Cross sectional area of each bar
Solve for
P1 (1) d and d
2x
2y
(2) Stresses in each bar
y
3
El#2 P2
El#1
2
45o
1
x
Solution
Step 1: Node element connectivity table
ELEMENT Node 1 Node 2
1
1
2
2
2
3
Table of nodal coordinates
Node
x
y
1
0
0
2
Lcos45
Lsin45
3
0
2Lsin45
Table of direction cosines
ELEMENT
Length
l
x2  x1
length
m
y2  y1
length
1
L
cos45
sin45
2
L
-cos45
sin45
Step 2: Stiffness matrix of each element in global coordinates
with global numbering
Stiffness matrix of element 1
k
(1)
 l2

E A  lm

L  l 2

  lm
d1x
1

EA  1

2L   1

1
l
lm
m
2
 lm
2
 lm
l
m
lm
2
2
d1y d2x
1
1
1
1
1
1
1
1
 lm 

2
m 
lm 

2
m

d2y
1 

1

1 

1 
d1x
d1y
d2x
d2y
Stiffness matrix of element 2
k
(2)
d2x d2y
d3x
d3y
1
1
1
1
1
1
1
1


1

1 

1 
1

EA 1

2L   1

1
1
d2x
d2y
d3x
d3y
Step 3: Assemble the global stiffness matrix
1

1

EA 1
K 

2L   1
0

0
1
1
1
0
1
1
1
0
1
2
0
1
1
0
2
1
0
1
1
1
0
1
1
1
0 

0

1 

 1
 1

1 
The final set of equations is K d  F
Step 4: Incorporate boundary conditions
 0

0

d2x
d  
d2y
 0

 0









Hence reduced set of equations to solve for unknown
displacements at node 2
EA 2

2 L 0
0 d2x 
 P1 

 d   
2  2y 
 P2 
Step 5: Solve for unknown displacements
d2x

d2y




  




P1 L 

EA 

P2 L 
EA 

Step 6: Obtain stresses in the elements
For element #1:

(1)
E  1
 
L
2

E
2L

1
2
(d 2 x  d 2 y ) 
1
2
P1  P2
A 2
 d1x 


d
1   1y 
 d 
2   2x 
d2 y 


0
0
For element #2:

(2)


E 1
L  2
E
2L

1
2
(d 2 x  d 2 y ) 

1
2
P1  P2
A 2
d2x 


d
1   2y 
 d 
2   3x 
d3y 


0
0
Multi-point constraints
© 2002 Brooks/Cole Publishing / Thomson Learning™
Figure 3-19 Plane truss with inclined boundary
conditions at node 3 (see problem worked out in class)
Problem 3: For the plane truss
y
P
3
El#2
2
El#1
P=1000 kN,
L=length of elements 1 and 2 = 1m
E=210 GPa
A = 6×10-4m2 for elements 1 and 2
= 6 2 ×10-4 m2 for element 3
El#3
45o
1
x
Determine the unknown displacements
and reaction forces.
Solution
Step 1: Node element connectivity table
ELEMENT Node 1 Node 2
1
2
3
1
2
1
2
3
3
Table of nodal coordinates
Node
x
y
1
0
0
2
0
L
3
L
L
Table of direction cosines
ELEMENT
Length
l
x2  x1
m
y2  y1
length
length
1
L
0
1
2
L
1
0
3
L
2
1/
2
1/
2
Step 2: Stiffness matrix of each element in global coordinates
with global numbering
Stiffness matrix of element 1
k
(1)
 l2

E A  lm

L  l 2

  lm
 lm
l
m
lm
 lm 

2
m 
lm 

2
m

d1x
d1y d2x
l
lm
m
2
 lm
2
2
2
0
9
-4 
(210  10 )(6  10 )  0

0
1

0
0
0
1
0
0
0
1
0
d2y


1

0 

1 
0
d1x
d1y
d2x
d2y
Stiffness matrix of element 2 d2x d2y
d3x
d3y
1
9
-4 
(210  10 )(6  10 )  0

1
1

 0
0
1
0
0
0
1
0
0
0 

0

0 

0 
k
(2)
d2x
d2y
d3x
d3y
Stiffness matrix of element 3
k
(3)
d1x
d1y
d3x
 0.5
9
-4 
(210  10 )(6 2  10 )  0.5

  0.5
2

  0.5
0.5
 0.5
0.5
 0.5
 0.5
0.5
 0.5
0.5
d3y
 0.5 

 0.5

0.5 

0.5 
d1x
d1y
d3x
d3y
Step 3: Assemble the global stiffness matrix
 0.5

0.5

 0
5
K  1260  10 
 0
  0.5

  0.5
0.5
0
0
 0.5
1.5
0
1
 0.5
0
1
0
1
1
0
1
0
 0.5
1
0
1.5
 0.5
0
0
0.5
The final set of equations is K d  F
 0.5 

 0.5

0 

0 
0.5 

0.5 
Eq(1)
N/m
Step 4: Incorporate boundary conditions
y
 0 

0


d2x
d  
0

 d3x


d3y
Also,










d 3y  0
P
y
x
3
El#2
2
El#1
El#3
45o
1
x
in the local coordinate system of element 3
How do I convert this to a boundary condition in the global (x,y)
coordinates?
y
 F1 x

F
 1y

 P
F  
F
 2y
 F3 x


 F3 y
Also,











F 3x  0
x
y
P
3
El#2
2
El#1
El#3
45o
1
x
in the local coordinate system of element 3
How do I convert this to a boundary condition in the global (x,y)
coordinates?
Using coordinate transformations

 d 3x

d 3y


 l

  
m



 d 3x
 

d 3y
m   d3x 
 d  l  m 
l   3y 
 1



2

 
1





2

d 3y  0
 d

2
1 
 1


2  d3x 
 2

  
1  d3y 
 1

 2
2 

(Multi-point constraint)
3y
1
1
2
d
 d3 y  d3x  0
3y
 d3x

Eq (2)
0

d

d
 3x
3 y 


 d 3 y  d 3 x  

Similarly for the forces at node 3

F


F
3x
3y

F
 

F

 l

  
m


3x
3y
m   F3 x 
 F  l  m 
n   3y 
 1



2

 
1





2

3x

2
1 
 1


2  F3 x 
 2

  
1   F3 y 
 1

 2
2 

F 3x  0
 F
1
1
2
F
 F3 y  F3 x  0
3y
 F3 x

Eq (3)
0

F

F
 3x
3 y 


 F3 y  F3 x  

Therefore we need to solve the following equations simultaneously
Kd  F
Eq(1)
Eq(2)
d3 y  d3x  0
Eq(3)
F3 y  F3 x  0
Incorporate boundary conditions and reduce Eq(1) to
 1
5 
1260  10
1


 0
1
1 .5
0 .5
0 

0 .5

0 .5 

d2x

 d3x
d
 3y

 P




 F3 x

F

 3y





Write these equations out explicitly
Eq(4)
1260  10 (d 2 x  d 3x )  P
5
1 2 6 0  1 0 (  d 2 x  1 .5 d 3 x  0 .5 d 3 y )  F3 x
Eq(5)
1 2 6 0  1 0 (0 .5 d 3 x  0 .5 d 3 y )  F3 y
Eq(6)
5
5
1 2 6 0  1 0 (  d 2 x  2 d 3 x  d 3 y )  F3 x  F3 y  0 using
5
 1260  10 ( d 2 x  3d 3 x )  0
5
using Eq(2)
 d 2 x  3 d 3 x Eq(7)
 1 2 6 0  1 0 (3 d 3 x  d 3 x )  P
5
Plug this into Eq(4)
 2520  10 d3x  10
5
6
Eq(3)
 d 3 x  0 .0 0 3 9 6 8 m
d 2 x  3 d 3 x  0 .0 1 1 9 m
Compute the reaction forces
 F1 x

F
 1y

 F2 y
F
 3x

 F3 y




5

1
2
6
0

1
0





500 


500




  0  kN
500 



 500 

 0

0

 0

1

 0
 0 .5
 0 .5
0
1 .5
0 .5
 0 .5 

 0 .5

0 

0 .5 
0 .5 

d2x

 d3x
d
 3y





Physical significance of the stiffness matrix
In general, we will have a stiffness matrix of the form
 k 11

K  k 21

 k 31
k 12
k 22
k 32
k 13 

k 23

k 33 
And the finite element force-displacement relation
 k 11

k
 21
 k 31
k 12
k 22
k 32
k 13   d 1   F1 
   
k 23  d 2    F2 

k 33   d 3   F3 
Physical significance of the stiffness matrix
The first equation is
k 11 d 1  k 12 d 2  k 13 d 3  F1
Force equilibrium
equation at node 1
Columns of the global stiffness matrix
What if d1=1, d2=0, d3=0 ?
F1  k 11
F2  k 21
F3  k 31
While d.o.f 2 and 3 are held fixed
Force along d.o.f 1 due to unit displacement at d.o.f 1
Force along d.o.f 2 due to unit displacement at d.o.f 1
Force along d.o.f 3 due to unit displacement at d.o.f 1
Similarly we obtain the physical significance of the other
entries of the global stiffness matrix
In general
k ij
= Force at d.o.f ‘i’ due to unit displacement at d.o.f ‘j’
keeping all the other d.o.fs fixed
Example
The length of bars 12 and 23 are equal (L)
E: Young’s modulus
A: Cross sectional area of each bar
Solve for d2x and d2y using the “physical
P1 interpretation” approach
y
3
El#2 P2
El#1
2
45o
1
x
Solution
Notice that the final set of equations will be of the form
 k 11

 k 21
k 12   d 2 x   P1 
 d    
k 22   2 y   P2 
Where k11, k12, k21 and k22 will be determined using the
“physical interpretation” approach
To obtain the first column
 2  1.cos(45) 
y
3
F2y=k21
El#2
2
2
x
 1  1.cos(45) 
d2x=1
Force equilibrium

F
F x  k11  T1 cos(45)  T 2 cos(45)  0
y
d2x  1
d2y  0
F2y=k21
T2
F2x=k11
2’
El#1
1
1
 k1 1 

 apply
 k 21  y
 k 21  T1 sin(45)  T 2 sin(45)  0
F2x=k11
T1
2
x
1
2
Force-deformation relations
T1 
EA
T2 
EA
L
L
1
2
Combining force equilibrium and force-deformation relations
k11 
k 21 
 T1  T2 

2
EA
2L
 T1  T2 

2
EA
2L
 1   2 
 1   2 
Now use the geometric (compatibility) conditions (see figure)
 1  1.cos(45) 
1
 2  1.cos(45) 
1
2
2
Finally
k 11 
EA
k 21 
EA
2L
2L
 1   2  
EA
 1   2   0
2L
(
2
2
)
EA
L
k
To obtain the second column  1 2  apply d 2 x  0


y
d2y  1
 k 2 2 y
3
1
   1.cos(45)  
F2y=k22
2
2’
T2
El#2
d2y=1
F2x=k12
2
El#1
T1
2
x
1
x
  1.cos(45) 
1
2
1
Force equilibrium

F
F x  k12  T1 cos(45)  T 2 cos(45)  0
y
 k 22  T1 sin(45)  T 2 sin(45)  0
2
Force-deformation relations
T1 
EA
T2 
EA
L
L
1
2
Combining force equilibrium and force-deformation relations
k12 
k 22 
 T1  T2 

EA
2
 T1  T2 
2L

EA
2
2L
 1   2 
 1   2 
Now use the geometric (compatibility) conditions (see figure)
 1  1.cos(45) 
1
2
 2   1.cos(45)  
1
This negative is due to compression
2
Finally
k 12 
EA
k 22 
EA
2L
2L
 1   2   0
 1   2  
EA
2L
(
2
2
)
EA
L
© 2002 Brooks/Cole Publishing / Thomson Learning™
3D Truss (space truss)
fˆ  kˆ dˆ
In local coordinate system
 fˆ1x

 fˆ1y

 fˆ1z
ˆ
 f 2x
ˆ
f
 2y
ˆ
 f 2z

  k
  0
 
  0
 
  k
  0
 
  0

0
0
k
0
0
0
0
0
0
0
0
0
0
0
k
0
0
0
0
0
0
0
0
0
 dˆ 1x
0
  dˆ
0
1y

0   dˆ 1z
 ˆ
0   d 2x
0   dˆ
  2y
0   ˆ
 d 2z











The transformation matrix for a single vector in 3D
*
dˆ  T d
 l1
*

T  l2

l 3
m1
m2
m3
n1 

n2

n3 
© 2002 Brooks/Cole Publishing / Thomson Learning™
l1, m1 and n1 are the direction cosines of x^
l1  cos q x
m 1  cos q y
n 1  cos q z
Transformation matrix T relating the local and global
displacement and load vectors of the truss element
dˆ  T d
T *
T  
6 6
 0
fˆ  T f
0 
*
T 
Element stiffness matrix in global coordinates
k  T
6 6
T
kˆ T
6 6 6 6 6 6
 l1 2

 l1 m 1
EA  l1 n 1
T ˆ
k  T kT 

2
L   l1
 l m
 1 1
  l1 n 1
l1 m 1
m1
2
m 1 n1
 l1 m 1
 m1
2
 m 1 n1
2
l1 n 1
 l1
m 1 n1
 l1 m 1
 m1
l1 n 1
m 1 n1
n1
2
 l1 n 1
 m 1 n1
 n1
2
l1
2
 l1 m 1
2
l1 m 1
2
l1 m 1
m1
l1 n 1
m 1 n1
 l1 n 1 

 m 1 n1 
2
 n1 

l1 n 1 
m 1 n1 

2
n 1 
Notice that the direction cosines of only the local ^x axis enter the
k matrix
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