Lecture 28-29
MECH3361
Finite Element Stiffness Matrices for Equivalent: ku  f
Recap
 k   ui   f i 
  
k  u j   f j 
k
 k

1D Single Spring element
EA  1  1
L  1 1
1D Bar Elements:
'
 l 2  lm   ui   f i ' 
 '   
 lm  m 2   vi   0 
 '  
l2
lm  u j   f j' 
 '
lm
m 2  v j   0 
 l2
lm

m2
EA  lm
L   l 2  lm

2
 lm  m
2D Bar Element
  ui   f i 
 u    f 
 j   j 
Combined spring and bar system
Step 1: Elemental equilibrium equations (stiffness matrix)
 k
Spring Element 1:  1
  k1
k
Bar Element 2:  2
 k2
 k1   u1   f1(1) 
 

k1  u2   f 2(1) 
k1
1
 k2  u2   f1( 2) 
 

k2  u3   f 2( 2) 
L
P
2
k2 
EA
L
L
P
3
Step 2: Form global equilibrium equation: An alternative way of assembling the whole stiffness
matrix or global equilibrium equation is to “Enlarging” the elemental stiffness matrices (and
equilibrium equations), as
(1)
 k1  k1 0  u1   f1 





 k
k1 0 u2    f 2(1) 
 1

 0
0 0 u3   0 


0
0
0 k
2

0  k2
0   u1   0 

  
 k2  u2    f1( 2) 

k2  u3   f 2( 2) 
Add these two expanded equations:
  k1  k1 0 0
0



  k1 k1 0  0 k 2
 0
0 0 0  k 2

0   u1   k1
 k1
  

 k 2  u2    k1 k1  k 2

k 2  
 k2
u3   0
0   0   f1(1)   0 
 
  

 k 2  u2    f 2(1)    f1( 2) 
 
  ( 2) 
k 2  
u3   0   f 2 
Step 3: Apply B.C. & load: cross the row and column corresponding to u1  0 , external forces:
F2  F3  P
k  k
(1) 1st equation: F1  k1u2 and (2) the 2nd and 3rd equations:  1 2
  k2
 k2  u2   P 
  
k2  u3   P 
Step 4: Solve the “reduced” global equilibrium equation for unknowns U  u2 u3 T as
k1  k2
 k
2

 k2  u2   P 
     , which leads to displacements:
k2  u3   P 
2 P / k1
u  

U   2  

u3  2 P / k1  P / k 2 
Step 5: Solve for the reaction force F1  k1u2  k1 2P / k1   2P
1
Lecture 28-29
MECH3361
8.3 Beam Element
Simple Plane Beam Element
L
I
E
v = v(x)
length
2nd moment of inertia of the cross-sectional area
elastic modulus
deflection (lateral displacement) of the neutral axis
rotation about the z-axis
  dv / dx
F = F(x) shear force
M = M(x) moment about z-axis
As per Beam Theory (Mechanics of Solids I): EI
My
d 2v
 M ( x ) and   
2
I
dx
Formal Approach
Step 1: Introduce the shape functions:
3
2
2
1
N1 ( x )  1  2 x 2  3 x 3 ,
N 2 ( x)  x  x 2  2 x 3
L
L
L
L
3
2
1
1
N 3 ( x)  2 x 2  3 x 3 ,
N 4 ( x)   x 2  2 x 3
L
L
L
L
 vi 
 
 i
Step 2: Calculate the deflection: v( x)  Nu  N1 ( x) N 2 ( x) N 3 ( x) N 4 ( x) 
v j 
 j 
 
which is a cubic function w.r.t. the coordinate of x and u is nodal displacement vector.
d 2v d 2
Nu  Bu
Curvature of the beam is,

dx 2 dx 2
where the strain-displacement matrix B is given by,
d2
4 6 x 6 12 x
2 6x 
 6 12 x
B  2 N  N1'' ( x) N 2'' ( x) N3'' ( x) N 4'' ( x)   2  3   2
 3   2
2
L L
L L 
dx
L
L
L
 L
Step 3: Strain energy stored in the beam element is


T
1
1  My  1  My 
1
T 1
U    T dV    
Mdx


dAdx   M
2V
2V  I  E  I 
2L
EI
T

 d 2v 
1  d 2v 
1
1 1
   2  EI  2 dx   Bu T EI Bu dx  uT   BT EIBdx u
 dx 

2 L  dx 
2L
2  2 L



We conclude that the stiffness matrix for the simple beam element is
1
k   BT EIBdx
2L
After integrating, equilibrium eqn is:
Combining the bar element, we obtain the stiffness matrix of a general 2-D beam element,
2
Lecture 28-29
MECH3361
8.4 Distributed Load – Equivalent nodal load
Uniformly distributed axial load q (N/mm, N/m, lb/in) can be converted to two equivalent
nodal forces of magnitude qL/2. We verify this by considering the work done by the load q,
Wq  
L1
0

2
uqdx 
 ui 
1 1
qL 1
qL 1
u( )q( Ld )   u( )d   N i ( ) N j ( )  d

2 0
2 0
2 0
u j 

 ui  1  qL
qL  1
 0 1    d     
 u j  2  2
2 
qL   ui  1
   ui
2  u j  2



qL / 2 
uj 

qL / 2 
qL / 2 
1
That is Wq  uT f q where f q  

2
qL / 2 
Thus, from U=W principle for element, we should have
1
1
1
U  uT ku  Wq  uT f  uT f q

2
2 other 2
nodal
force
distribute d
load
which yields ku  f  f q . Thus the new nodal force vector is
 f i  qL / 2   f i  qL / 2 
f  fq     


 f j  qL / 2   f j  qL / 2 
in an assemble bar system
Multiple bar element system
For a multiple bar element system, e.g. two elements (as figure next page), one can expand
the force vector in each element and then add the expanded force vectors, as
 f1(1)  qL / 2  
0
  f1  qL / 2  qL / 2 
 (1)
  ( 2)
 
 

f  f q element 1  f  f q element 1   f 2  qL / 2    f1  qL / 2   f 2  qL    qL 

  f (1)  qL / 2   f  qL / 2 qL / 2 
0
 

  3

  2




Equivalent nodal load for transverse force
3
Lecture 28-29
MECH3361
This can be verified by considering the work done by the distributed load q. For the 2 beam
elements, the equivalent loading can be calculated as follows:
Example 8.5
Given: A cantilever beam with distributed lateral load p as shown above.
Find: The deflection and rotation at the right end, the reaction force and moment at the left
end.
Solution:
Step 1: Calculate the quivalent nodal load: The work-equivalent nodal loads are shown on
right, in which the equivalent nodal loads are: f  pL / 2, m  pL2 / 12 .
Step 2: Applying the FE equilibrium equation, we have
6 L  12 6 L   v1   F1Y 
 12
 6 L 4 L2  6L 2 L2     M 
EI 
  1    1 
3   12  6 L
12  6 L v2   F2Y 
L


2
 6L 4 L2   2   M 2 
 6L 2 L
Step 3: Load and constrains (BCs) are: F2Y   f , M 2  m, v1  1  0
Reduce the FE equation into
6 L  12 6 L   0   F1Y 
 12
 6 L 4 L2  6 L 2 L2   0   M 
EI 
  1

 

3   12  6 L
12  6 L v2   f 
L

2
2
m 

2 
 
 6L 2 L  6L 4 L  
EI  12  6L v2   f 
Thus:

2    

L3  6L 4 L   2   m 
Step 4: Solve for the reduced FE equation
4
Lecture 28-29
MECH3361
v2 
L  2 L2 f  3Lm   pL4 /(8EI ) 

 



3

6
EI

3
Lf

6
m
 2

  pL /(6 EI ) 
These nodal values are the same as the engineering beam theorey solution. Note that the
deflection v(x) (for 0 < x< 0) in the beam by the FEM is, however, different from that by the
exact solution. The exact solution by the engineering beam theory is a 4th order polynomial of
x, while the FE solution of v is only a 3rd order polynomial of x. If more elements are used,
such an approximate error can be reduced.
Step 5: Reaction force and moments
From the eliminated FE equation, we can calculate the reaction force and moment as,
 F1Y  L3   12 6L  v2   pL / 2 
 

2    
2

 M1  EI  6L 2 L   2  5 pL / 12
Example 8.6
Given: P = 50 kN, k = 200 kN/m, L = 3 m, E = 210 GPa, I = 210-4 m4.
Find: Deflections, rotations and reaction forces.
Solution: This is a combined problem of beam and spring elements
Step 1: The system has 4 nodes as well as 2 beam elements and 1 spring element.
Step 2: The spring element has stiffness matrix as,
which is related to nodes #3 and #4 with displacement v3 and v4
6 L  12 6 L   v1   f1Y 
 12
 6 L 4 L2  6L 2 L2     m 
EI
  1    1 
Beam element 1: 3 
L  12  6 L 12  6 L v2   f 2Y 


2
 6L 4 L2   2   m2 
 6L 2 L
6 L  12 6 L  v2   f 2Y 
 12
2

 6L 2 L2   2   m2 
EI 6 L 4 L
   
Beam element 2: 3 

L  12  6 L 12  6 L  v3   f 3Y 


2
 6L 4 L2   3   m3 
 6L 2 L
Step 3: The global FE equations can be assembled as (where k '  L3k /( EI ) )
5
Lecture 28-29
MECH3361
Beam element 1
Beam element 2
spring element
Step 4: Apply the boundary conditions
v1  1  v2  v4  0, M 2  M 3  0,
F3Y   P
 8L2
 6L 2 L2   2   0 

   
EI
The reduced FE equation becomes: 3  6L 12  k '  6L  v3    P 
L  2
2L
 6L 4 L2   3   0 


Step 5 Solve for the reduced FE equation:
 2 
 3   0.002492rad 
PL2
 
  

 v3   
7 L     0.01744m 
EI (12  7k ' )   
 

 3
 9    0.007475rad 
Step 6: Reaction force and moment can be found from the eliminated equations:
 F1Y    69.78kN 
 M   69.78kN  m 
 1 




 F2Y   116.2kN 
 F4Y   3.488kN 
Thus the free-body diagram can be drawn as below:
6
Lecture 28-29
MECH3361
8.5 Two-Dimensional Problems
General Formula for the Stiffness Matrix
Recall the displacement in terms of “shape function”
 ui 
u  [ N i N j ]   Nu
u j 
We can extend it to 2D, where the displacements (u, v) in a plane element are interpolated
from nodal displacements (ui, vi) using shape functions Ni as follows,
u  u( x, y )  N1 ( x, y )u1  N 2 ( x, y )u2  

v  v( x, y )  N1 ( x, y )v1  N 2 ( x, y )v2  
u1 
v 
N
0
N
0

u
  1 
   1
2
In the matrix form: u     
 u2   Nd
v   0 N1 0 N 2   
v2 
  
where N is the shape function matrix, u the displacement vector and d the elemental nodal
displacement vector. Here we assume that u depends on the nodal values of u only, and v on
nodal values of v only.
From strain-displacement relation, the strain vector can be derived as:
  Du  DNd
  Bd
Or
B
where  DN is the strain-displacement matrix
Consider the strain energy stored in an element,
1
1
U   [T ]dV   ( xx xx   yy yy   xy xy )dV
2V
2V

1
1
1
 T dV   T dV    Βd)T Βd)dV

2V
2V
2V
Since nodal displacement vector d is independent on the elemental coordinate dV.
1

U  dT   ΒT ΒdV d  dT kd
2

 V

From this, we obtain the general formula for the element stiffness matrix,
Remarks:
 Note that unlike 1-D cases, E here is a matrix that is given by the stress-strain relation
 The stiffness matrix k is symmetric since E is symmetric.
 Also note that given the material property, the behavior of k depends on the B matrix
only, which in turn on the shape functions. Thus, the quality of finite elements in
representing the behaviour of a structure is entirely determined by the choice of shape
functions.
7
Lecture 28-29
MECH3361
Constant Strain Triangle (CST or T3)
This is the simplest 2-D element, which is also called linear triangular element.
For this element, we have three nodes at the vertices of
the triangle, which are numbered around the element in
the counterclockwise direction. Each node has two
degrees of freedom (can move in the x and y directions).
The displacements u and v are assumed to be linear
functions within the element, that is,
u  b1  b2 x  b3 y
v  b4  b5 x  b6 y
where bi (i = 1, 2, ..., 6) are constants. From these, the
strains are found to be,
u 
b1  b2 x  b3 y   b2

x x
v

b4  b5 x  b6 y   b6


y y
 xx 
 yy
 u v  

2 xy      b4  b5 x  b6 y   b1  b2 x  b3 y   b3  b5
y
 y x  x
which are constant throughout the element. Thus, we have the name “constant strain
triangle” (CST) element.
Displacements should satisfy the following six equations,
u1  b1  b2 x1  b3 y1
v1  b4  b5 x1  b6 y1
u2  b1  b2 x2  b3 y 2
v2  b4  b5 x2  b6 y 2
u3  b1  b2 x3  b3 y3
v3  b4  b5 x3  b6 y3
Solving these equations, we can find the coefficients b1, b2, ..., b6 in terms of nodal
displacements and coordinates. Substituting these coefficients displacement equation and
rearranging the terms, we obtain:
 u1 
v 
 1
N
0
N
0
N
0
u
   1
 u2 
2
3
(linear distribution)
 
 
v   0 N1 0 N 2 0 N 3  v2 
 u3 
 
 v3 
where the shape functions (linear functions in x and y) are
1
( x2 y3  x3 y2 )  ( y2  y3 ) x  ( x3  x2 ) y 
N1 
2A
1
( x3 y1  x1 y3 )  ( y3  y1 ) x  ( x1  x3 ) y 
N2 
2A
1
( x1 y2  x2 y1 )  ( y1  y2 ) x  ( x2  x1 ) y 
N3 
2A
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Lecture 28-29
MECH3361
1 x1 y1 
1 
where A  det 1 x2 y2  is the area of the triangle

2 
1 x3 y3 
Using the strain-displacement relation, we have,
where xij = xi - xj and yij = yi - yj (i, j = 1, 2, 3). Again, we see constant strains within the
element. From stress-strain relation (Hooke’s law), we see that stresses obtained using the
CST element are also constant.
Applying general formula for stiffness matrix, we obtain it for the CST element,
in which t is the thickness of the element. Notice that k for CST is a 6 by 6 symmetric matrix.
The matrix multiplication can be carried out by a computer program.
Applications of the CST Element:
 Use in areas where the strain gradient is small.
 Use in mesh transition areas (fine mesh to coarse mesh).
 Avoid using CST in stress concentration or other crucial areas in the structure, such as
edges of holes and corners.
 Recommended for quick and preliminary FE analysis of 2-D problems.
Linear Strain Triangle (LST or T6)
This element is also called quadratic triangular element.
There are six nodes on this element: three corner nodes
and three midside nodes. Each node has two degrees of
freedom (DOF) as before. The displacements (u, v) are
assumed to be quadratic functions of (x, y),
u  b1  b2 x  b3 y  b4 x 2  b5 xy  b6 y 2
v  b7  b8 x  b9 y  b10 x 2  b11xy  b12 y 2
where bi (i=1,2,…12) are constants to be determined.
Thus the strains are found to be:
u 
 xx 

b1  b2 x  b3 y  b4 x 2  b5 xy  b6 y 2  b2  2b4 x  b5 y
x x
v 
 yy  
b7  b8 x  b9 y  b10 x 2  b11 xy  b12 y 2  b9  b11 x  2b12 y
y y




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Lecture 28-29
MECH3361


 u v  
2 xy     
b1  b2 x  b3 y  b4 x 2  b5 xy  b6 y 2
 y x  x


b7  b8 x  b9 y  b10 x 2  b11xy  b12 y 2
y
 b3  b5   b3  2b10 x  2b6  2b11  y
which are linear functions. Thus, we have the “linear strain triangle” (LST), which provides
better results than the CST.


In a natural coordinate system, the six shape functions for the LST element are:
N1   (2  1), N 2   (2  1),
N 3  (1     )2(1     )  1
N 4  4,
N 5  4 (1     ),
N 6  4(1     )
6
Displacement can be written as:
u   N i ui ,
i 1
6
v   N i vi (quadratic distribution)
i 1
which means that for given nodal displacements, one can know the displacement at any point
within the element. In other words, we use shape functions and nodal displacement to express
displacement field (function)
Remarks: Note that the element stiffness matrix is still given by k e   BT E BdV , but
V
T
B EB is quadratic in x and y. In general the integral has to be computed numerically.
Bi-Linear Quadrilateral Element (Q4)
There are four nodes at the corners of the
quadrilateral shape. In the natural coordinate system
(, ), the four shape functions are,
1
1
N1  (1   )(1   ) , N 2  (1   )(1   ) ,
4
4
1
1
N 3  (1   )(1   ) , N 4  (1   )(1   )
4
4
Mapping from Cartesian coordinate to natural
coordinate systems can be done as follows:
4
x   N i xi ,
i 1
Displacement can be written as:
4
y   N i yi ,
i 1
4
u   N i ui ,
i 1
Strain:
i 1
u  

   N i ui 
x x  i 1

6
v  


   N i vi 
y y  i 1

 xx 
 yy
4
v   N i vi (bilinear distribution)
6
 u v    6
   6

2 xy        N i vi     N i ui 
 y  i 1

 y x  x  i 1
We cannot directly obtain the results because u and v are defined as functions of natural
coordinate  and  rather than as a function of x and y. We can however use chain rule:
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Lecture 28-29
MECH3361
 u   x y   u 
 u   x y 
        x 
 x     
or



 u 
 u    x y 
x y   u 

  
 
  
 y     
        y 



1
 u 
  
 u 
 
  
[ J ]1
[J ]
where [J] is called Jacobina matrix.
Coefficients in [J] can be obtained as:
 N 
 N

x
y
   i xi ,
   i yi 


  
 

u
* u
* u
Thus  x 
 J 11
 J 12
x


*
*
, J 12
where J 11
are the coefficients in the first row of inverse Jacobian matrix [J]-1.
 N 
u
   i ui ,

  
 N 
u
   i ui 

  
Thus we change the integration from Cartesian coordinate to natural coordinate as
k   B E BdV 
e
T
V
1 1
 B
T
EBt J dd
11
which needs to use numerical integration method.
According to Gauss rule of integration, one can have:
I
    , dd  WiW j i , j 
1 1
n
m
i 1 j 1
11
Gaussian integration points:

3
 
3

3
3

3
3


y
3
3
x
Gaussian points and weights:
m,n
11
22
Wi
, 
0.0
1
 3 / 3  0.5774
0
0.8888
33
0.5555
 0.6  0.7746
The stress, strain results at the Gaussian point should be most accurate in the field.
11
Lecture 28-29
MECH3361
Quadratic Quadrilateral Element (Q8)


3
7
4
4
y
8
6
1
x
5
7
3
mapping

6
8
2
1
1
N1   (1   )(  1)(    1)
4
1
N 2   (1   )(  1)(    1)
4
1
N 3   (1   )(  1)(1     )
4
1
N 4   (1   )(  1)(1     )
4
5
8
i 1
The displacement field is given by: u   Niu,
i 1
2
1
N 5  (1   2 )(1   )
2
1
N 6  (1   )(1   2 )
2
1
N 7  (1   2 )(1   )
2
1
N 8  (1   )(1   2 )
2
Mapping from Cartesian coordinate to natural coordinate systems: x   Ni xi ,
8

8
y   Ni yi ,
i 1
8
v   Nivi (quadratic distributions)
i 1
The stiffness matrix:
k e   BT E BdV 
V
1 1
 B
T
EBt J dd
11
which needs to use numerical integration method as
I
    , dd  WiW j i , j 
1 1
11
n
m
i 1 j 1
Remarks:
 Q4 and T3 are usually used together in a mesh with linear elements.
 Q8 and T6 are usually applied in a mesh composed of quadratic elements.
 Quadratic elements are preferred for stress analysis, because of their high accuracy
and the flexibility in modeling complex geometry, such as curved boundaries.
12