Finite Element Method/Analysis Lecture Notes

1. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Finite Element Method
in Civil Engineering
Lecture Notes
Dr. A. S. Sayyad
Professor
Department of Civil Engineering
SRES’s Sanjivani College of Engineering,
Savitribai Phule Pune University,
Kopargaon-423603
Email: attu_sayyad@yahoo.co.in
Ph. No.: (+91) 9763567881
Year-2017

2. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Unit- I
Theory of Elasticity
1.1 Fundamentals of theory of elasticity
Assumptions made in theory of elasticity:
1) Material of elastic body is continuous i.e. no sudden discontinuity such as
cracks, flaws, deep notches etc.
2) Material is homogenous, isotropic and elastic
3) strains and displacements are small
4) stress strain relationship is linear
5) higher order differential terms are neglected
6) small angle assumptions are valid such that
1
sin ; cos ; tan
    
  
1.2 Basic terms:
External forces: 1) Surface force 2) body force
1) Surface force: The force distributed over the surface of the body is called as
surface force. It is denoted by x,y & z and resolved into three components.
2) Body force: The forces distributed over the volume of the body are called as
body forces and are caused by gravity, magnetism and acceleration. Body forces
can be resolved into three components (X, Y & Z).
Internal forces: Internal forces are the stress resultants existing on the cut faces of
the isolated part of the body. Internal force distributed over an internal face may be
resolved into two components.
1) Normal component perpendicular to face (Normal stress)
2) Shear component tangential to face (Shear stress)
Normal stress: The normal force per unit area is called as normal stress. It is
denoted by ij
 where i represent the plane in which it acts and j represents the
direction of the stress.
Example:
plane and direction
plane and direction
plane and direction
xx
yy
zz
' yz' ' x'
' xz' ' y'
' xy' ' z'







3. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Sign conventions: Tensile Normal stress (Positive)
Compressive Normal stress (Negative)
Shear stress: The shear force components per unit area is called as shear stress
and denoted by ij
 .
plane and direction
xy ' yz' ' y'
 
Displacements: As a result of a deformation of elastic body subjected to external
forces, point of a body are displaced. The displacements of a point are represented
by u, v, w in x, y, z directions respectively.
Strains: Measure of deformation of a body.
1) Linear strain: Change in dimension
2) Lateral strain: Change in lateral dimensions
3) Shear strain: Change in angle between originally perpendicular elements.
Considered as positive, if original angle is reduced and considered as negative, if
original angle is increased.
1.3 State of stress at a point:
In one plane there are three stress components. One normal stress ( ) and two
shear stresses ( ). At every point there are three mutually perpendicular planes.
i.e. orthogonal planes. Therefore, at every point there are nine stress components
(three normal stresses and six shear stresses). But since shear stresses are
complimentary ( xy yx xz zx yz zy
, ,
     
   ), nine stress components are reduced to
six.
   
3 3 3 3
xx xy xz xx xy xz
yx yy yz xy yy yz
zx zy zz xz yz zz
     
       
     
 
   
   
  
   
   
   
1.4 State of strain at a point:
In one plane there are three strain components. One normal strain ( ) and two
shear strains ( ). At every point there are three mutually perpendicular planes. i.e.
orthogonal planes. Therefore, at every point there are nine strain components (three
normal strains and six shear strains). But since shear strains are complimentary (
xy yx xz zx yz zy
, ,
     
   ), nine strain components are reduced to six.

4. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
   
3 3 3 3
xx xy xz xx xy xz
yx yy yz xy yy yz
zx zy zz xz yz zz
     
       
     
 
   
   
  
   
   
   
1.5 Equilibrium equations for 3D elasticity problem
Let consider as infinitesimal element of sides dx, dy and dz as shown in figure.
The stresses acting on the element due to external forces and body forces. These
stresses can be represented by nine components as given below.
   
T
xx yy zz xy yx xz zx yz zy
, , , , , , , ,
         

where, xx yy zz
, ,
   are the normal stresses and xy yx xz zx yz zy
, , , , ,
      are the shear
stresses.
Plane Stresses on Positive faces Stresses on Negative faces Area of Plane
yz ' xx
xx xx dx
x

 

 

xy
'
xy xy dx
x

 

 

' xz
xz xz dx
x

 

 

xx

xy

xz

dy dz

5. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
xz yy
'
yy yy dy
y

 

 

yx
'
yx yx dy
y

 

 

yz
'
yz yz dy
y

 

 

yy

yx

yz

dx dz
xy ' zz
zz zz dz
z

 

 

' zx
zx zx dz
z

 

 

zy
'
zy zy dz
z

 

 

zz

zx

zy

dx dy
X = Component of body force in x-direction,
Y= Component of body force in y-direction and
Z= Component of body force in z-direction
Appling the conditions for forces and moments of static equilibrium,
0
x
F 

0
yx
xx
xx xx yx
zx
yx zx zx
dx dydz dydz dy dxdz
x y
dxdz dz dxdy dxdy X dxdydz
z


  

  

 

 
   
   
 
   

 
     
 

 
(1)
Simplifying the equation (1) we get
0
yx
xx zx
dxdydz dxdydz dxdydz X dxdydz
x y z

 

 
   
  
(2)
Dividing by dxdydz to the equation (2), we will get
0
yx zx
xx
X
x y z
 
  

   
  
(First governing equation)
Similarly from 0
y
F 

0
xy yy zy
Y
x y z
  
  
   
  
(Second governing equation)

6. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
and from 0
z
F 

0
yz
xz zz
Z
x y z

 

 
   
  
(Third governing equation)
Appling conditions of moment equilibrium to prove shear stresses are
complimentary
Now, taking moment about x-axis through the centroid of the element. The
coordinates of centroid of an element are
2 2 2
dx dy dz
, ,
 
 
 
.
Notes:
1) Moment of 0
xx yy zz
  
   because they are passing through centroid.
2) Moment of 0
xy yx xz zx
   
   
3) Only moment due to yz zy
 
 will present about x-axis.
4) Anticlockwise moment positive and clockwise moment negative
2 2
0
2 2
yz
x yz yz
zy
zy zy
dy dy
M dy dxdz dxdz
y
dz dz
dz dxdy dxdy
z

 

 

 
  
 

 

 
   
 

 

(3)
Neglecting higher power of differential coefficients    
2 2
dy , dz
 
 
in equation (3),
we get
2 0
2 2
yz zy
dy dz
dxdz dxdy
 
 
yz zy
 
 (Shear stress is complimentary)
Similarly, from 0
y xz zx
M  
  

and from 0
z xy yx
M  
  


7. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
1.6 Strain-Displacement Relationship
The six strain components  
x y z xy xz yz
, , , , ,
      are related to three displacement
component (u, v, w) at a point. Let consider elements AB and AC initially
perpendicular to each other in xy plane as shown in figure. ' '
A B and ' '
A C are the
deformed lengths of AB and AC respectively.
AB = Original line element in x-direction
' '
A B = Deformed line element in x-direction
AC = Original line element in y-direction
' '
A C = Deformed line element in y-direction
u = Displacement of point A in x-direction
u
u dx
x



= Displacement of point B in x-direction
v = Displacement of point A in y-direction
v
v dy
y



= Displacement of point C in y-direction
Change in length of the line element AB =
u
u dx u
x

 
 
 

 
=
u
dx
x


Linear/Normal strain of the element AB is x

Change in length
Original length
x
u
dx
u
x
dx x




  

Similarly

8. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
y
v
y




and z
w
z




The angular displacement of line element AB = 
v
dx
v
x
tan
dx x




  

Similarly the angular displacement of line element AC = 
u
dy
u
y
tan
dy y




  

Total shear strain xy
  
  =
v u
x y
 

 
Similarly
xz
w u
x z

 
 
 
and yz
w v
y z

 
 
 
Therefore, strain displacement relationship for the 3D elasticity problem is
x
y
z
xy
xz
yz
u / x
v / y
w / z
u / y v / x
u / z w / x
v / z w / y






 
   
   
 
   
   
 
 

   
    
   
   
    
   
    
   
 
Example 1: An elastic body under the action of external forces has the
displacement field given by,      
2 2 2
2 5 3
D x y i z y j x y k
     
Evaluate components of strain at point (3, 1, 2)
Solution:
 
31 2
4 12
1 1
0 0
2 2
3 3
5 2 7
x
y
z
xy
xz
yz , , mm/ mm
u / x x
v / y
w / z
u / y v / x y
u / z w / x
v / z w / y y 






 
       
       
   
       
       
 
 
  
       
    
       
       
    
       
     
       
 

9. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Example 2: The general displacement field in Cartesian coordinates for stressed
body is,
2 2
2
0 015 0 03 0 005 0 03
0 03 0 001 0 005
u . x y . , v . y . xy
w . x . yz .
   
  
Where displacement coordinates u, v, w and x, y, z are in constant units. Find
Cartesian strain components at point (1, 0, 2)
Solution:
 
2
1 0 2
0 03 0
0 01 0 03 0 03
0 001 0
0 015 0 03 0 015
0 006 0 006
0 001 0 002
x
y
z
xy
xz
yz , ,
u / x . xy
v / y . y . x .
w / z . y
u / y v / x . x . y .
u / z w / x . x .
v / z w / y . z .






 
      
     
  
     
     
 
 
  
      
     
     
     
    
     
    
     
  mm/ mm


 
 
 

 
 
 
 
Example 3: In an strained elastic body under the action of external forces has the
displacement field given by
     
3 2 2
3 2 4 2 4
D x y i z y j y z k
     
Evaluate components of strain at point (2, 4, 1)
Example 4: In an strained elastic body under the action of external forces has the
displacement field given by
     
2 2 2
2 3
D x y i z y j x y k
     
Evaluate components of strain at point (3, 1, 2)
1.7 Strain-Compatibility Conditions or Saint Venant’s Strain
compatibility conditions
In general, the elastic problem is a statically indeterminate and hence the
solution of elastic problem needs the concept of compatibility and stress-strain
relationship in addition to equilibrium equations.
The displacements are taken to be continuous single valued function and hence
the strains must be such that they do not cause any dislocations cracks or overlaps.
This means that continuity of structure must be maintained. This is the physical
interpretation of compatibility is that six components of strains must satisfy six

10. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
compatibility equations. These equations are derived from strain-displacement
relations.
2 3
2 2
x
x
u u
x y x y


  
  
   
(Differentiating x
 twice w.r.t y) (1)
2 3
2 2
y
y
v v
y x x y



 
  
   
(Differentiating y
 twice w.r.t x) (2)
2 3 3
2 2
xy
xy
v u v u
x y x y y x x y



   
    
       
(Differentiating xy
 w.r.t x & y) (3)
Therefore, from equations (1)-(3) one can write
2 2
2
2 2
y xy
x
y x x y
 
  

 
   
(I)
Similarly
2 2
2
2 2
z xz
x
z x x z
 
  

 
   
(II)
2 2
2
2 2
y yz
z
z y y z
 

 

 
   
(III)
2 2
xy
xy
u v u v
y x z y z x z



   
    
      
(Differentiating xy
 w.r.t z) (4)
2 2
xz
xz
u w u w
z x y y z x y


    
    
      
(Differentiating xz
 w.r.t y) (5)
2 2
yz
yz
v w v w
z y x x z x y



   
    
      
(Differentiating yz
 w.r.t x) (6)
Solving equations (4) – (6)
2
2
xy yz
xz u
z y x y z
 

 
 
  
    
(7)
Differentiating equation (7) w.r.t x
3
2
xy yz
xz u
x z y x x y z
 

 
 
  
  
 
      
 
2
2
xy yz
xz x
x z y x y z
 
 
 
 
 

   
 
     
 
(IV)
Similarly

11. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
2
2
xy yz y
xz
y z x y x z
  

  
 


  
 
     
 
(V)
2
2
yz xy
xz x
z y x z x y
 
 
 
 
 

  
 
     
 
(VI)
These are the six strain-compatibility conditions for the 3D elasticity problems.
Example 1: The general displacement field in Cartesian coordinates for stressed
body is
2
2
2
0 015 0 03
0 005 0 03
0 03 0 001 0 005
u . x y .
v . y . xy
w . x . yz .
 
 
  
Check whether the strain field given by above displacement field is compatible.
Solution:
2 2
2 2 2 2
2 2 2 2
2 2
2
2 2
0 0 0 0 0 0
0 0 0
y xy
x x z xz
y yz
z
, , , , , ,
y x x y z x x z
, ,
z y y z
 
   
 

 
   
     
       
 

  
   
Above displacement field satisfy all strain displacement relationship. Therefore, it
gives possible/compatible state of strain.
Example 2: Check whether the following system of strain is possible.
2 3
2 2 2
2 3
3 3 2
4 6 4
3 34 12 9 2
x
y
xy
xy y x
x y x y
x y xy x y



    
   
    

12. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
1.8 Stress-strain relationship
3D Hooke’s law:
 
 
 
1 0 0 0
1 0 0 0
1 0 0 0
1
0 0 0 2 1
0 0 0 2 1
0 0 0 2 1
x x
y y
z z
xy xy
xz xz
yz yz
E
 
 
 
 
 
 
 

 

 

 
   
 
   
 
 
   
 
   
 
 
   
  
   

 
   
 
   

 
   

 
   
 
   
It is also written in the form
  
 
 
 
1 0 0 0
1 0 0 0
1 0 0 0
1 2
0 0 0
2
1 1 2
1 2
0 0 0
2
1 2
0 0 0
2
x x
y y
z z
xy xy
xz xz
yz yz
E
  
  
 
  
 

 
 
 

 
 


 
 

   
 
   
 

   
 

   
 
   
  
   
   
   

 
   
 
   
   
 
   

 
 
 
where E is Young’s modulus and  is Poisson’s ratio.
2D Hooke’s law:
Plane stress problem
2
1 0
1 0
1
0 0 1 2
x x
y y
xy xy
E
/
  
  

  
   
 
   
 

   
 

   
 

 
   
Plane strain problem
  
1 0
1 0
1 1 2
0 0 1 2
x x
y y
xy xy
E
/
   
   
 
  
   

 
   
 
 
   
 
 
   
 

 
   
1D Hooke’s law: E
 


13. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Example: The state of strain at a point is given by
  4
18 0 36
0 54 5 4 10
36 5 4 0
.
.
 

 
 
  
 
 

 
mm/mm
Determine the stress matrix if E = 210 GPa and 0 3
.
 
Solution:
3
4
0 7 0 3 0 3 0 0 0 18
0 3 0 7 0 3 0 0 0 54
0 3 0 3 0 7 0 0 0 0
210 10
10
0 0 0 0 2 0 0 0
1 3 0 4
0 0 0 0 0 2 0 36
0 0 0 0 0 0 2 5 4
x
y
z
xy
xz
yz
. . .
. . .
. . .
.
. .
.
. .







    
    

    
    

 
 
   
 

   
 
   
  
   
 
    
 
145 38
1308 04
436 6
0
290 71
43 6
x
y
z
xy
xz
yz
.
.
.
MPa
.
.







   
   

   
   

 

   
   
   

   
   
 
or  
145 38 0 290 71
0 1308 04 43 6
290 71 43 6 436 6
. .
. .
. . .

 
 
 
 
 
 
 
 
1.9 2D Elasticity Problems
1) Plane stress problem
2) Plane strain problem
3) Axisymmetric problem
Plane stress problem: Two dimensional elasticity problems under the following
conditions are considered as plane stress problem.
1) One dimension is very small as compared to other two dimensions
e.g. Rectangular plate (Thickness is very small as compared to length and
width)
2) The loads acting on the body are in the plane perpendicular to the thickness
of the body i.e. z-axis. The loads are uniformly distributed over the
thickness.
3) The stresses in the small direction (normally z) must be zero.
0
zz xz yz
  
   but, 0
z
 

14. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Equations of Equilibrium
0
xy
xx
X
x y

 

  
 
and 0
xy yy
Y
x y
 
 
  
 
State of stress at a point
 
xx
yy
xy

 

 
 
 
  
 
 
 
State of strain at a point
 
xx
yy
xy

 

 
 
 
  
 
 
 
and 0
z
 
Strain-Displacement relation
x
u
x




, xy
v u
x y

 
 
 
Strain compatibility condition
2 2
2
2 2
y xy
x
y x x y
 
  

 
   
Stress-strain relation
 
1 0
1
1 0
0 0 2 1
x x
y y
xy xy
E
  
  
  
   

 
   
 
 
   
 
   
 

 
   
or 2
1 0
1 0
1
0 0 1 2
x x
y y
xy xy
E
/
  
  

  
   
 
   
 

   
 

   
 

 
   

15. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Compatibility conditions in-terms of stresses for plane stress
problem:
As discussed in the Saint Venant’s strain compatibility conditions, substituting
plane stress conditions in the three dimensional Saint venant’s compatibility
conditions. Non-zero strain components in plane stress problems are
x y z xy
, , &
    which functions of x and y. Substitute these components of strain
in compatibility equation of plane stress problem.
2 2
2
2 2
y xy
x
y x x y
 
  

 
   
Other strain compatibility conditions are not imposed in plane stress problem. The
stress compatibility can be obtained either by substituting plane stress condition in
the Beltrami-Michell compatibility or directly substituting strains in-terms of
stress.
     
2 2 2
2 2 2 2
2 1
1 1
x y y x xy
E E
y x x y E

    
 

 
   
  
     
   
     
     
(1)
 
2 2 2
2 2
2 2 2 2
2 1
y y xy
x x
y x x y x y
  
 
 
   
  
 
    
   
     
   
(2)
Now, from equilibrium equations of plane stress problem neglecting body forces
2
2
2
0
xy xy
x x
x y x x y
 
 
 
 
    
    
(3)
2 2
2
0
xy y y xy
x y y x y
   
   
    
    
(4)
Adding equations (3) and (4)
2 2
2
2 2
2
y xy
x
x y x y
 
  

  
   
(5)
Put equation (5) into the equation (2)
 
2 2 2
2 2 2
2 2 2 2 2 2
1
y y y
x x x
y x x y x y
  
  
 
     
  
  
      
     
     
     
2 2
2 2
2 2 2 2
0
y y
x x
x x y y
 
 
 
 
   
   
 
2 2
2 2
0
x y
y x
 
 
 
  
 
 
 
(6)

16. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
 
2
0
x y
 
   where
2 2
2
2 2
0
y x
 
 
  
 
 
 
This is called as stress compatibility conditions in-terms of stress. It is also written
in the form (putting Eq. (5) into Eq. (2) one can wirte)
 
2 2 2
2
2 2
2 2 1
y xy xy
x
y x x y x y
  

 
 
  

   
 
     
 
2 2
2
2 2
2
y xy
x
y x x y
 
  

 
   
(7)
This is also called as compatibility conditions in-terms of stresses. This is the plane
stress idealization of Beltrami-Michell compatibility conditions. The stress
compatibility equation is valid only for an isotropic body with constant body force
under equilibrium.
Example 1: Show that the following state of stress is in equilibrium
2 2
2 2
2
2
3 4 8 4
2 3
6 2
2
x
y
xy
x xy y
x xy y
x
xy y



   
  
 
   
 
 
Solution:
Differential equation of equilibrium of 2D elasticity problem is given by
0
xy
x
x y

 

 
 
and 0
xy y
x y
 
 
 
 
(1)
6 4 6 4
6 6
xy
x
xy y
x y, x y
x y
x y, x y
x y


 


    
 
 
    
 
Putting these differential quantities in equations of equilibrium (1) and satisfying
those. This proves that the given state of stress is in equilibrium.

17. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Example 2: A 2D stress distribution at a point in xy coordinate system is given
below. Find constants A, B & C if system of stress is in equilibrium. The body
force is zero.
2 3
2
3 2
10
1 5
x
y
xy
xy Ax
. B xy
By Cx y



  
 
  
Solution:
2 2 2 2
10 3 3
xy
x
y Ax , By Cx
x y

 

     
 
2 3
xy y
Cxy, Bxy
x y
 
 
   
 
Putting these derivatives in following two governing equations
   
2 2
3 10 3 0
xy
x
A C x B y
x y

 

      
 
and
2 3 0
xy y
Cxy Bxy
x y
 
 
    
 
3 2
C B /
   Putting this in above equation
L.H.S. = 0 only when
3 0 and 10+3B=0
A C
 
10 3 and A=5/3
B /
  
Plane strain problem: Two dimensional elasticity problems under the following
conditions are considered as plane strain problem.
1) One dimension is infinitely long as compared to other two dimensions
e.g. Retaining wall, Dam, Bridge (Length is infinitely long as compared to
depth and width)
2) External force is perpendicular to the z-axis.
3) The strains in the long direction (normally z) must be zero.
0
zz xz yz
  
   but 0
z
 

18. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Equations of Equilibrium
0
xy
xx
X
x y

 

  
 
and 0
xy yy
Y
x y
 
 
  
 
State of stress at a point
 
xx
yy
xy

 

 
 
 
  
 
 
 
0
z
 
State of strain at a point
 
xx
yy
xy

 

 
 
 
  
 
 
 
Strain-Displacement relation
x
u
x




, xy
v u
x y

 
 
 
Strain compatibility condition
2 2
2
2 2
y xy
x
y x x y
 
  

 
   
Stress-strain relation
1 0
1
1 0
0 0 2
x x
y y
xy xy
E
   

   
 
   
 
 

   
 
  
   
 
   
 
 
   
or
  
1 0
1 0
1 1 2
0 0 1 2
x x
y y
xy xy
E
/
   
   
 
  
   

 
   
 
 
   
 
 
   
 

 
   

19. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Compatibility conditions in-terms of stresses for plane strain problem:
2 2
2
2 2
y xy
x
y x x y
 
  

 
   
(1)
 
 
   
 
 
 
2 2
2 2
2
1 1
1 1
2 1
x y y x
xy
y E x E
x y E
 
     


 
   
 
    
   
 
   

 

  
   
(2)
Now, from equilibrium equations of plane stress problem neglecting body forces
2
2
2
0
xy xy
x x
x y x x y
 
 
 
 
    
    
(3)
2 2
2
0
xy y y xy
x y y x y
   
   
    
    
(4)
Adding equations (3) and (4)
2 2
2
2 2
2
y xy
x
x y x y
 
  

  
   
(5)
Put equation (5) into equation (2)
 
2 2
2 2
0
x y
y x
 
 
 
  
 
 
 
 
2
0
x y
 
  

20. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Summary of elasticity problems
Elasticity problem Equations No. of Equations No. of Unknowns
3D Equilibrium equations
0
yx zx
xx
X
x y z
 
  

   
  
0
xy yy zy
Y
x y z
  
  
   
  
0
yz
xz zz
Z
x y z

 

 
   
  
03 06 stresses
Stress-strain relations
 
 
 
x x y z
y y x z
z z x y
xy xy
xz xz
yz yz
/ E
/ E
/ E
/ G
/ G
/ G
   
   
   
 
 
 
  
  
  



06 06 strains
Strain-displacement relations
x
y
z
xy
xz
yz
u / x
v / y
w / z
u / y v / x
u / z w / x
v / z w / y






  
  
  
     
     
     
06 03 displacements
Compatibility conditions
2 2
2
2 2
y xy
x
y x x y
 
  

 
   
2 2
2
2 2
z xz
x
z x x z
 
  

 
   
2 2
2
2 2
y yz
z
z y y z
 

 

 
   
03 ---
2D
Plane stress
Equilibrium equations
0
yx
xx
X
x y

 

  
 
0
xy yy
Y
x y
 
 
  
 
02 03 stresses

21. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Stress-strain relations
 
 
 
1
x x y
y y x
z x y
xy xy
/ E
/ E
/ E
G
  
  
  
 
 
 
  

03 03 strains
Strain-displacement relations
x
y
xy
u / x
v / y
u / y v / x



  
  
     
03 02 displacements
Compatibility conditions
2 2
2
2 2
y xy
x
y x x y
 
  

 
   
01 ---
2D
Plane strain
Equilibrium equations
0
yx
xx
X
x y

 

  
 
0
xy yy
Y
x y
 
 
  
 
02 03 stresses
Stress-strain relations
 
 
 
1
x x y z
y y x z
z x y
xy xy
/ E
/ E
G
   
   
  
 
  
  
  

03 03 strains
Strain-displacement relations
x
y
xy
u / x
v / y
u / y v / x



  
  
     
03 02 displacements
Compatibility conditions
2 2
2
2 2
y xy
x
y x x y
 
  

 
   
01 ---

22. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Unit-II
Finite Element Analysis of Spring Assembly
Springs are 1D structures subjected to axial force only. The degree of freedom at
each node is one i.e. axial displacement. Stiffness matrix for spring element having
stiffness constant k is given below which can be obtained by giving unit
displacement one by one at each node.
Let consider a two noded spring element with ui and uj displacements at each
nodes.
Let unit displacement at node i
Let unit displacement at node j  
1 1
1 1
i j
i
j
u u
u
K k
u

 
  

 
Procedure for the solution of numerical examples
1) Divide the spring assembly into number of members
2) Calculate total degrees of freedom
3) Determine stiffness matrix of each spring element
4) Assemble the global stiffness matrix
5) Impose the boundary conditions
6) Determine reduced stiffness matrix
7) Apply governing equation to determine unknown joint displacements.
    
K f
 
where  
f = Nodal load vector
Example 1: Determine elongations at each node and hence the forces in springs.
Solution:
Step 1: Discretization
Element k (N/m) Nodes Displacements (m) Boundary conditions
1 500 1-2 u1-u2 u1 = 0
2 100 2-3 u2-u3 ---

23. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Step 2: Element stiffness matrices
 
1 2
1
1 1
2
1 1 1 1
500
1 1 1 1
u u
u
K k
u
 
   
 
   
 
   
 
2 3
2
2 2
3
1 1 1 1
100
1 1 1 1
u u
u
K k
u
 
   
 
   
 
   
Step 3: Global stiffness matrix
Assemble the element stiffness matrices to get the global stiffness matrix
   
1 2 3
1
2
3
500 500 0
500 500 100 100
0 100 100
u u u
u
K u
u
 
 
 
   
 
 

 

Step 4: Reduced stiffness matrix
Imposing boundary conditions i.e. u1 = 0 eliminate first row and first column.
Therefore reduced stiffness matrix is
 
2 3
2
3
600 100
100 100
u u
u
K
u

 
  

 
Step 5: Determine unknown joint displacements
Applying Equation of Equilibrium
    
K f
 
2
3
600 100 0
100 100 5
u
u
  
   

   
 

   
 
   
2 3
0 01 and 0 06
u . m u . m
    
Step 6: Calculation of spring force
Spring 1:
     
1 1
1
K f
 
1 1
2 2
1 1
500
1 1
u f
u f
    
 

   
 

    
1
2
1 1 0
500
1 1 0 01
f
f
.
  
  

   
 

    
( 1 2
0 and 0 01
u u .
  )
   
1 2
5 T and 5 T
f N f N
 
Spring 2:    
3 4
5 T and 5 T
f N f N
  ( 2 3
0 01 and 0 06
u . u .
  )

24. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Example 2: Determine elongations at each node and hence the forces in springs.
Take F3 = 5000 N.
Solution:
Step 1: Discretization
Element k (N/m) Nodes Displacements (m) Boundary conditions
1 1000 1-2 u1-u2 u1 = 0
2 2000 2-3 u2-u3 ---
3 3000 3-4 u3-u4 u4 = 0
Step 2: Element stiffness matrices
 
1 2
1
1 1
2
1 1 1 1
1000
1 1 1 1
u u
u
K k
u
 
   
 
   
 
   
 
2 3
2
2 2
3
1 1 1 1
2000
1 1 1 1
u u
u
K k
u
 
   
 
   
 
   
 
3 4
3
3 3
4
1 1 1 1
3000
1 1 1 1
u u
u
K k
u
 
   
 
   
 
   
Step 3: Global stiffness matrix
Assemble the element stiffness matrices to get the global stiffness matrix
 
 
 
1 2 3 4
1
2
3
4
1000 1000 0 0
1000 1000 2000 2000 0
0 2000 2000 3000 3000
0 0 3000 3000
u u u u
u
u
K
u
u
 
 
 
  
 

 
  
 
 
 
 
Step 4: Reduced stiffness matrix
Imposing boundary conditions i.e. u1 = 0 and u4 = 0
Eliminate first row, first column and fourth row and fourth column.
Therefore reduced stiffness matrix is

25. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
 
2 3
2
3
3000 2000
2000 5000
u u
u
K
u

 
  

 
Step 5: Determine unknown joint displacements
Applying Equation of Equilibrium
    
K f
 
2
3
3000 2000 0
2000 5000 5000
u
u
  
   

   
 

   
 
Ans.    
2 3
0 909 and 1 363
u . m u . m
    
Step 6: Calculation of spring force
Spring 1:
     
1 1
1
K f
 
1 1
2 2
1 1
1000
1 1
u f
u f
    
 

   
 

    
1
2
1 1 0
1000
1 1 0 909
f
f
.
  
  

   
 

    
( 1 2
0 and 0 909
u u .
  )
   
1 2
909 T and 909 T
f N f N
 
Spring 2:    
3 4
909 T and 909 T
f N f N
  ( 2 3
0 909 and 1 363
u . u .
  )
Spring 3:    
5 6
4091 C and 4091 C
f N f N
  ( 3 4
1 363 and 0 0
u . u .
  )
Example 3: Determine elongations at nodes 2 and 4 if u3 = 0.02 m and hence the
force at node 3.
Solution:
Step 1: Discretization
Element k (N/m) Nodes Displacements (m) Boundary conditions
1 500 1-2 u1-u2 u1 = 0
2 100 2-3 u2-u3 u3 = 0.02
3 200 3-4 u3-u4 ---
Step 2: Element stiffness matrices

26. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
 
1 2
1
1 1
2
1 1 1 1
500
1 1 1 1
u u
u
K k
u
 
   
 
   
 
   
 
2 3
2
2 2
3
1 1 1 1
100
1 1 1 1
u u
u
K k
u
 
   
 
   
 
   
 
3 4
3
3 3
4
1 1 1 1
200
1 1 1 1
u u
u
K k
u
 
   
 
   
 
   
Step 3: Global stiffness matrix
Assemble the element stiffness matrices to get the global stiffness matrix
 
 
 
1 2 3 4
1
2
3
4
500 500 0 0
500 500 100 100 0
0 100 100 200 200
0 0 200 200
u u u u
u
u
K
u
u
 
 
 
  
 

 
  
 

 

Step 4: Reduced stiffness matrix
Imposing boundary conditions i.e. u1 = 0 eliminate first row and first column.
Therefore reduced stiffness matrix is
 
2 3 4
2
3
4
600 100 0
100 300 200
0 200 200
u u u
u
K u
u

 
 
  
 
 

 
Step 5: Determine unknown joint displacements
Applying Equation of Equilibrium
    
K f
 
2
2
4
600 100 0 0
100 300 200 0 02
0 200 200 100
u
. f
u

    
   
 
  
   
 
   
 

    
     
2 4 2
0 00333 0 52 and 98 33
u . m , u . m f . N
      

27. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Example 4: Determine elongation at node 2 and pulling force (F) at node 3 for the
spring assembly given below. Take pull at node 3, 0.06m.
Solution
Step 1: Discretization
Element k (N/m) Nodes Displacements (m) Boundary conditions
1 500 1-2 u1-u2 u1 = 0
2 100 2-3 u2-u3 u1 = 0.06m
Step 2: Element stiffness matrices
 
1 2
1
1 1
2
1 1 1 1
500
1 1 1 1
u u
u
K k
u
 
   
 
   
 
   
 
2 3
2
2 2
3
1 1 1 1
100
1 1 1 1
u u
u
K k
u
 
   
 
   
 
   
Step 3: Global stiffness matrix
Assemble the element stiffness matrices to get the global stiffness matrix
   
1 2 3
1
2
3
500 500 0
500 500 100 100
0 100 100
u u u
u
K u
u
 
 
 
   
 
 

 

Step 4: Reduced stiffness matrix
Imposing boundary conditions i.e. u1 = 0 eliminate first row and first column.
Therefore reduced stiffness matrix is
 
2 3
2
3
600 100
100 100
u u
u
K
u

 
  

 
Step 5: Determine unknown joint displacements
Applying Equation of Equilibrium
    
K f
 
2
600 100 0
100 100 0 06
u
. F

    

   
 

    
   
2 0 01 and 5
u . m F N
    

28. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Example 5: Determine spring elongations and force at node 5 for the spring
assembly as shown in figure. Take stiffness of all spring elements 200 kN/m.
Solution:
Step 1: Discretization
Element k (N/m) Nodes Displacements (m) Boundary conditions
1 200 1-2 u1-u2 u1 = 0
2 200 2-3 u2-u3 ---
3 200 3-4 u3-u4 ---
4 200 4-5 u4-u5 u5 = 0.02 m
Step 2: Element stiffness matrices
 
1 2
1
1 1
2
1 1 1 1
200
1 1 1 1
u u
u
K k
u
 
   
 
   
 
   
 
2 3
2
2 2
3
1 1 1 1
200
1 1 1 1
u u
u
K k
u
 
   
 
   
 
   
 
3 4
3
3 3
4
1 1 1 1
200
1 1 1 1
u u
u
K k
u
 
   
 
   
 
   
 
4 5
4
4 4
5
1 1 1 1
200
1 1 1 1
u u
u
K k
u
 
   
 
   
 
   
Step 3: Global stiffness matrix
Assemble the element stiffness matrices to get the global stiffness matrix
 
1 2 3 4 5
1
2
3
4
5
200 200 0 0 0
200 400 200 0 0
0 200 400 200 0
0 0 200 400 200
0 0 0 200 200
u u u u u
u
u
K u
u
u
 
 
 
 
 
 
  
 
 
 
 

 


29. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Step 4: Reduced stiffness matrix
Imposing boundary conditions i.e. u1 = 0 eliminate first row and first column.
Therefore reduced stiffness matrix is
 
2 3 4 5
2
3
4
5
400 200 0 0
200 400 200 0
0 200 400 200
0 0 200 200
u u u u
u
u
K
u
u

 
 
 
 

 
 
 

 
Step 5: Determine unknown joint displacements
Applying Equation of Equilibrium
    
K f
 
2
3
4
400 200 0 0 0
200 400 200 0 0
0 200 400 200 0
0 0 200 200 0 02
u
u
u
. F

    
    
     
  
   
 
     
    

    
Ans.
       
2 3 4
0 005 0 01 0 015 and 1
u . m ,u . m ,u . m F kN
       
Example 6: Figure shows three springs connected parallel. Using finite element
method determines the deflections of individual springs.
Solution: Step 1: Discretization
Element k (N/mm) Nodes Displacements (mm) Boundary conditions
1 10 1-2 u1-u2 u1 = 0, u2 =?
2 20 3-4 u3-u4 u3 = 0, u2 = u4,
3 40 5-6 u5-u6 u5 = 0, u2 = u6,
Step 2: Element stiffness matrices

30. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
 
1 2
1
1 1
2
1 1 1 1
10
1 1 1 1
u u
u
K k
u
  
   
 
   
 
   

 
3 4
3
2 2
4
1 1 1 1
20
1 1 1 1
u u
u
K k
u
  
   
 
   
 
   

 
5 6
5
3 3
6
1 1 1 1
40
1 1 1 1
u u
u
K k
u
  
   
 
   
 
   

Step 3: Reduced stiffness matrix
Imposing boundary conditions i.e. u1 = 0, u3 = 0, u5 = 0, u2 = u4, u2 = u6.
Therefore reduced stiffness matrix is
10 20 40 70
K    
Step 4: Determine unknown joint displacements
Applying Equation of Equilibrium
    
K f
 
2
70 700
u 
 
2 4 6 10
u u u mm
   
Example 7: Figure shows cluster of four springs. One end of the spring assembly
is fixed and a force of 1000 N is applied at the other end. Using the finite element
method, determine deflection of each spring.

31. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Solution:
Step 1: Discretization
Element k (N/mm) Nodes Displacements (mm) Boundary conditions
1 4 1-2 u1-u2 u1 = 0, u2 =?
2 8 3-4 u3-u4 u3 = 0, u4 = u2
3 20 5-6 u5-u6 u5 = 0, u6 =?
4 10 7-8 u7-u8 u7 = u2, u8 = u6
Step 2: Element stiffness matrices
 
1 2
1
1 1
2
1 1 1 1
4
1 1 1 1
u u
u
K k
u
  
   
 
   
 
   

 
3 4
3
2 2
4
1 1 1 1
8
1 1 1 1
u u
u
K k
u
  
   
 
   
 
   

 
5 6
5
3 3
6
1 1 1 1
20
1 1 1 1
u u
u
K k
u
  
   
 
   
 
   

 
7 8
7
3 3
8
1 1 1 1
10
1 1 1 1
u u
u
K k
u
 
   
 
   
 
   
Step 3: Reduced stiffness matrix
Imposing boundary conditions i.e. u1 = 0, u3 = 0, u5 = 0, u2 = u4, u2 = u7, u8 = u6
Therefore reduced stiffness matrix is
 
2 6
2
6
22 10
10 30
u u
u
K
u

 
  

 
Step 4: Determine unknown joint displacements
Applying Equation of Equilibrium
    
K f
 
2
6
22 10 0
10 30 1000
u
u
  
   

   
 

   
 
   
2 4 7 8 6
17 857 39 286
u u u . mm u u . mm
      

32. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Example 8: A three spring system shown in figure has stiffnesses k1 = 40 N/mm,
k2 = 50 N/mm and k3 = 80 N/mm. The loads applied are F1 = 100 N and F2 = 50 N.
Calculate displacements at nodal points.
Example 9: The system of springs, subjected to a load of 20 kN is shown in figure.
Find the deflection of each spring.
Example 10: The figure shows cluster of five springs. One end of the assembly is
fixed while a force of 1 kN is applied at the other end. Using finite element method
determines the deflection of each spring. (Ans. 24. 39mm, 24.39mm, 34.146mm,
14.634mm)

33. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Finite Element Analysis of trusses
Stiffness matrix of a truss element
The truss may be statically determinate or indeterminate. All members are
subjected to only direct stresses (tensile or compressive). Joint displacements are
selected as unknown variables. Since there is no bending of the members we have
to ensure only displacement continuity (C0
continuity) and there is no need to
worry about slope continuity (C1
continuity).
Here we select two noded bar element for the formulation of stiffness matrix of
truss element. Since the members are subjected to only axial forces, the
displacements are only in the axial directions of the members. Therefore, the nodal
displacement vector for the bar element is
  1
2
e
u'
x'
u'
 
  
 
where, 1 2
and
' '
u u are the displacements in axial direction of the element. The
stiffness matrix of a bar element is
1 2
1
2
1 1
1 1
'
u' u'
u'
AE
K
u'
L

 
    
  
 
Transformation matrix for the truss:
' '
x y = Local coordinate
systems
x, y = global coordinate
system
1 2
u' ,u' = Displacements in
local coordinate system
1 1 2 2
u , v ,u , v = Displacements
in global coordinate system
 =Angle measured in
anticlockwise sense w.r.t.
positive x-axis.
Since axial directions of all members of truss are not same, hence in global
coordinate system (x-y) there are two displacement components at every node.
Hence the nodal displacement vector for typical truss element is

34. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
 
1
1
2
2
e
u
v
x
u
v
 
 
 
  
 
 
 
Refereeing above figure,
At Node 1, At Node 2,
1 1 1
cos sin
'
u u v
 
  2 2 2
cos sin
'
u u v
 
 
Therefore, in matrix form above relation are
1
1
1
2
2
2
cos sin 0 0
0 0 cos sin
'
'
u
v
u
u
u
v
 
 
 
 
    

   
 
 
   
 
 
    
'
e
x L x

where,
 
'
e
x = vector of local unknowns
 
x = vector of global unknowns
 
L = Transformation matrix =  
0 0
0 0
l m
L
l m
 
  
 
where, 2 1 2 1
cos or sin or
x x y y
l l m m
L L
 
 
   
Stiffness matrix of truss element in global coordinate system
      
'
T
K L K L

 
0
0 1 1 0 0
0 1 1 0 0
0
l
m l m
AE
K
l l m
L
m
 
  
  
 
   
  
  
 
 

35. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
 
0
0
0
0
l
m l m l m
AE
K
l l m l m
L
m
 
   
 
 
  
   
 
 
 
 
1 1 2 2
2 2
1
2 2
1
2 2
2
2 2
2
u v u v
u
l lm l lm
v
lm m lm m
AE
K
u
L l lm l lm
v
lm m lm m
 
 
 
 
 

 
 
 
 
 
Example 1: Analyze the truss as shown in figure. Cross-sectional area of members
are AB=1000 mm2
, BC=800 mm2
, CA= 800 mm2
. Take E = 2 × 105
MPa
Solution: Step 1: Degrees of freedom: 06 ( A A B B c c
u ,v ,u ,v ,u ,v )
Discretization
Element Nodes Displacements (mm) Boundary conditions
1 AB uA, vA, uB, vB uA = 0
A
v 
2 BC uB, vB, uC, vC 0
B
v 
3 CA uC, vC, uA, vA ---
Assume x-axis horizontal through point c and vertical through point A. The
coordinate of node A(0, 1.5), B(4, 1.5) and C (2, 0). Take E in GPa
Member x2-x1 y2-y1 L l m AE/L (kN/mm)
AB 4 0 4 1 0 50
BC -2 -1.5 2.5 -0.8 -0.6 64
CA -2 1.5 2.5 -0.8 0.6 64

36. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Step 2: Element stiffness matrices
Stiffness matrix of element AB: Stiffness matrix of element BC:
 
1 0 1 0
0 0 0 0
50
1 0 1 0
0 0 0 0
A A B B
A
A
AB
B
B
u v u v
u
v
K
u
v

 
 
 

 

 
 
 
0 64 0 48 0 64 0 48
0 48 0 36 0 48 0 36
64
0 64 0 48 0 64 0 48
0 48 0 36 0 48 0 36
B B c c
B
B
BC
c
c
u v u v
u
. . . .
v
. . . .
K
u
. . . .
v
. . . .
 
 
 
 
 

 
 
 
 
 
Stiffness matrix of element CA:
 
0 64 0 48 0 64 0 48
0 48 0 36 0 48 0 36
64
0 64 0 48 0 64 0 48
0 48 0 36 0 48 0 36
c c A A
c
c
CA
A
A
u v u v
u
. . . .
v
. . . .
K
u
. . . .
v
. . . .
 
 
 
 
 

 
 
 
 
 
Step 3: Global stiffness matrix (Total DOF are 06, size of stiffness matrix 6×6)
 
90 96 30 72 50 0 40 96 30 72
30 72 23 04 0 0 30 72 23 04
50 0 30 72
0 0 30 72 23 0
90 96 40 96 30 72
4
4 30 72 23 04
40 96 30 72 30
0 96 81 92 0
72
30 72 23 04 23 04
30 72 0 46 08
A A B B c c
A
A
B
B
c
u v u v u v
u
. . . .
v
. . . .
u
.
K
v
. .
. . .
. .
.
. .
u
. .
. .
.
. .
  
 
 
 
 
 

  
 
 
 
 
 
 
 



 c
v



  
Step 4: Reduced stiffness matrix (Since uA= 0
A
v  , 0
B
v  eliminate corresponding
rows and columns from global stiffness matrix)
 
90 96 40 96 30 72
40 96 81 9 0
30 72 0 46 08
B c c
B
c
c
. . .
. .
. .
u u v
u
K u
v
 
 

 
 






Step 5: Equation of equilibrium
    
K f
 
90 96 40 96 30 72 0
40 96 81 9 0 0
30 72 0 46 08 120
B
c
c
. . . u
. . u
. . v
   
   
   
 
 
   
 
   
 
 
   
 
1 6 0 8 3 67
B c c
u . mm, u . mm, v . mm
     

37. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Example 2: Figure shows a plane truss with three members. Cross-sectional area
of all members 800 mm2
Young modulus is 200 KN/mm2
. Determine
deflection at loaded joint.
Solution:
Step 1: Degrees of freedom: 08 ( A A B B c c D D
u ,v ,u ,v ,u ,v ,u ,v )
Discretization
Element Nodes Displacements (mm) Boundary conditions
1 AD uA, vA, uD, vD 0
A A
u v
 
2 BD uB, vB, uD, vD 0
B B
u v
 
3 CD uC, vC, uD, vD 0
c c
u v
 
Assume origin support A (0, 0). The coordinates of other nodes B (1000, 0),
C(2000, 0) and D(1500, 1000)
Member x2-x1 y2-y1 L l m AE/L (kN/mm)
AD 1500 1000 1802.8 0.832 0.555 88.75
BD 500 1000 1118 0.447 0.894 143.112
CD -500 1000 1118 -0.447 0.894 143.112
Step 2: Element stiffness matrices
Stiffness matrix of element AD:
 
61 43 40 98 61 43 40 98
40 98 27 34 40 98 27 34
6 61 43 40 98
1 43 4
40
0 98
40 98 27 34 98 27 34
A A D D
A
A
AD
D
D
u v u v
u
. . . .
v
. . . .
K
u
. . .
.
.
. . . v
  
 
 
  
 

 
 
 
 
 
 
( 0
A A
u v
  )

38. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Stiffness matrix of element BD:
 
28 59 57 19 28 59 57 19
57 19 114 38 57 19 114 38
28 59 57 19 28 59 57 19
57 19 114 38 57 19 114 38
B B D D
B
B
BD
D
D
u v u v
u
. . . .
v
. . . .
K
u
. . . .
v
. . . .
  
 
 
  
 

 
 
 
 
 
 
( 0
B B
u v
  )
Stiffness matrix of element CD:
 
28 59 57 19 28 59 57 19
57 19 114 38 57 19 114 38
28 59 57 19 28 59 57 19
57 19 114 38 57 19 114 38
C C D D
C
C
CD
D
D
u v u v
u
. . . .
v
. . . .
K
u
. . . .
v
. . . .
  
 
 
  
 

 
 
 
 
 
 
( 0
c c
u v
  )
Step 3: Reduced stiffness matrix
 
118 61 40 98
40 98 256 10
D D
D
D
u v
u
. .
K
v
. .
 
  
 
Step 4: Equation of equilibrium
    
K f
 
118 61 40 98 200
40 98 256 10 0
D
D
u
. .
v
. .
 
   

   
 
   
 
1 785 0 286
D D
u . mm, v . mm
  

39. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Example 3: for the truss as shown in figure using finite element method,
determines deflections at loaded joints. The joint B is subjected to 50 kN
horizontal force towards left and 80 kN force vertically downward. Take cross-
sectional area of all members 1000 mm2
Young modulus is 200 GPa.
Solution: Step 1: Degrees of freedom: 06 ( A A B B c c D D
u ,v ,u ,v ,u ,v ,u ,v ).
Discretization
Element Nodes Displacements (mm) Boundary conditions
1 AB uA, vA, uB, vB 0
A A
u v
 
2 DB uD, vD, uB, vB 0
D D
u v
 
3 CB uC, vC, uB, vB 0
c c
u v
 
Assume origin point B. The coordinates of points areA (-4, 3), B (0,0), C (4,-3), D
(-4, -3)
Member x2-x1 y2-y1 L l m AE/L (kN/mm)
AB 4000 -3000 5000 0.8 -0.6 40
DB 4000 3000 5000 0.8 0.6 40
CB -4000 3000 5000 -0.8 0.6 40
Step 2: Stiffness matrix of element AB:
 
0 64 0 48 0 64 0 48
0 48 0 36 0 48 0 36
40
0 64 0 48 0 64 0 48
0 48 0 36 0 48 0 36
A A B B
A
A
AB
B
B
u v u v
u
. . . .
v
. . . .
K
u
. . . .
v
. . . .
  
 
 
  
 

 
 
 
 
 
 
( 0
A A
u v
  )

40. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Stiffness matrix of element DB:
 
0 64 0 48 0 64 0 48
0 48 0 36 0 48 0 36
40
0 64 0 48 0 64 0 48
0 48 0 36 0 48 0 36
D D B B
D
D
DB
B
B
u v u v
u
. . . .
v
. . . .
K
u
. . . .
v
. . . .
  
 
 
  
 

 
 
 
 
 
 
( 0
D D
u v
  )
Stiffness matrix of element CB:
 
0 64 0 48 0 64 0 48
0 48 0 36 0 48 0 36
40
0 64 0 48 0 64 0 48
0 48 0 36 0 48 0 36
C C B B
C
C
CB
B
B
u v u v
u
. . . .
v
. . . .
K
u
. . . .
v
. . . .
  
 
 
  
 

 
 
 
 
 
 
( 0
c c
u v
  )
Step 3: Reduced stiffness matrix
 
1 92 0 48
40
0 48 1 08
B B
B
B
u v
u
. .
K
v
. .

 
  

 
Step 4: Equation of equilibrium
    
K f
 
1 92 0 48 50
40
0 48 1 08 80
B
B
u
. .
v
. .
 
 
   

   
 
 
   
 
1 25 2 4
B B
u . mm, v . mm
   
Example 3: Determine the deflections at loaded joint in two bar truss supported by
spring as shown in figure. Bar one has length of 5m and bar two a length of 10m.
The stiffness of spring is 2000 kN/m. Take A = 5×10-4
m2
and E = 200 GPa.

41. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Solution:
Step 1: Degrees of freedom: 06 ( 1 1 2 2 3 3
u ,v ,u ,v ,u ,v )
Discretization
Element Nodes Displacements (mm) Boundary conditions
1 1-2 u1, v1, u2, v2 2 2 0
u v
 
2 1-3 u1, v1, u3, v3 3 3 0
u v
 
3 1-4 v1, v4 v4 = 0
Take origin node 1. The coordinates of nodes are
1(0, 0), 2(-3.535, 3.535), 3(-10, 0)
Member x2-x1 y2-y1 L l m AE/L (kN/mm)
1-2 -3.535 3.535 5 -0.707 0.707 200×102
1-3 -10 0 10 -1 0 100×102
1-4 --- --- --- --- --- ---
Step 2: Stiffness matrix of element 1-2:
 
1 1 2 2
1
1
2
1 2
2
2
0 5 0 5 0 5 0 5
0 5 0 5 0 5 0 5
200 10
0 5 0 5 0 5 0 5
0 5 0 5 0 5 0 5
u v u v
u
. . . .
v
. . . .
K
u
. . . .
v
. . . .

 
 
 
 
 
 
 
  
 
  
 
 
( 2 2 0
u v
  )
Stiffness matrix of element 1-3:
 
1 1 3 3
1
1
2
1 3
3
3
1 0 1 0
0 0 0 0
100 10
1 0 1 0
0 0 0 0
u v u v
u
v
K
u
v


 
 
 
 
 
 
 

 
 
( 3 3 0
u v
  )
Stiffness matrix of Spring element
 
1 4
1
3
4
1 1
2000
1 1
v v
v
K
v

 
  
 
 

Step 3: Reduced stiffness matrix

42. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
 
1 1
1
1
20000 10000
10000 12000
u v
u
K
v

 
  

 
Step 4: Equation of equilibrium:     
K f
 
1
1
20000 10000 0
10000 12000 40
u
v
  
   

   
 
 
   
 
1 1
2 857 5 714
u . mm, v . mm
   
Example: For the plane truss shown in figure, determine the x and y components
of displacements at node 1. Take E = 70 GPa and A = 500 mm2
for all elements.
Length of member 1-3 is 2500mm.
Example: For the plane truss composed of three elements shown in figure
subjected to a downward force of 50 kN applied at node 1, determine the x and y
components of displacements at node 1. Take E = 200 GPa and A = 1000 mm2
for
all elements.
Example: Figure shows a plane truss with two members. Both the members are of
cross-sectional area 70.71 mm2
. Young’s modulus is 200 kN/mm2
. Determine
deflections of loaded joint and hence the member forces.

43. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Example: A steel truss as shown in figure. The modulus of elasticity is 210 GPa.
The cross sectional area of member AB is 300 mm2
, BC is 400 mm2
and AC is 500
mm2
. Calculate the horizontal and vertical displacements at point ‘A’ using finite
element method.
Example: Figure shows a plane truss with three members. All members are of
length 1000 mm and cross-sectional area 600 mm2
. Young’s modulus is 150
kN/mm2
. Determine unknown joint displacements of the truss.
Example: For the two bar truss shown in figure determine the displacements at the
loaded joint using stiffness matrix method. Take A = 200 mm2
and E = 70
GPa.

44. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Example: Find the vertical and horizontal deflection at point C for the two
member truss as shown in figure. Area of inclined member is 2000 mm2
whereas horizontal member is 1600 mm2
. Take E = 200 GPa
Example: Figure shows plane truss with three members. All members are of
length 1000mm and c/s area 600mm2. E=150 KN/mm2. Determine forces in
members of truss using finite element method.
Example: Analyze the two member truss shown in figure using finite element
method. Take c/s area of each member 1000 mm2
and E = 200 GPa. The length
of each member is 5m.
Example: For the plane truss structure shown in figure, determine the
displacements at the loaded joint using finite element method. Assume A =
2000 mm2
and E = 200 GPa.

45. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Finite Element Analysis of Continuous Beams
A beam is a structural member which is subjected to bending deformation. There
are several methods available in the literature for the analysis of continuous beams
such as slope deflection method, moment distribution method, flexibility matrix
method, stiffness matrix method, three moment theorem etc. However all these
methods have limitations if either geometry, loading material properties or
boundary conditions. Finite element method can well handle such problems easily.
Element nodal load vector/ Equivalent load vector
In finite element method, the external forces are necessary to act at the joints
corresponding to joint displacements, which do not happen always. Beams are
often subjected to member forces, therefore these member forces we have to
convert into nodal forces. Vector of these forces is called as element nodal load
vector and apposite vector is called as equivalent load vector.
Degree of Kinematic Indeterminacy/Degrees of Freedom
Beam has two degrees of freedom at each point i.e. vertical translation and
rotation. Whereas frame has three degrees of freedom at each point i.e. two
displacements and one rotation.
Type of Support Kinematic
Unknowns for Beam
Kinematic Unknowns
for Frame
Hinge 1 ( ) 1 ( )
Roller 1 ( ) 2 ( ,
 )
Fixed 0 0
Spring 2 ( ,
 ) 2 ( ,
 )
Guided/Slider 1 ( ) 1 ( )
Internal Hinge 3 ( 1 2
, ,
 
 ) 3 ( 1 2
, ,
 
 )

46. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Steps for the solution of continuous (Indeterminate) beams using finite
element method:
1. Divide the beam into number of elements (Take one member as one element)
2. Identify total degrees of freedom (Two D.O.F. at each node, translation and
rotation)
3. Determine stiffness matrices of all elements ([K]1, [K]2………)
4. Assemble the global stiffness matrix [K]
5. Impose the boundary conditions and determine reduced stiffness matrix
6. Determine element nodal load vector [q] (Restrained structure)
7. Determine equivalent load vector [f]
8. Apply equation of equilibrium [K]{Δ}={f} and determine unknown joint
displacements.
9. Apply equation [K]{Δ}+[q] ={f} to determine reactions and moments
Stiffness matrix of beam
1 = Translation at node A
2= Rotation at node A
3 = Translation at node B
4= Rotation at node B
 
EI / L EI / L EI / L EI / L
EI / L EI / L EI / L EI / L
K
EI / L EI / L EI / L EI / L
EI / L EI / L EI / L EI / L
3 2 3 2
2 2
3 2 3 2
2 2
1 2 3 4
1 Reaction
12 6 12 6
2 Moment
6 4 6 2
3 Reaction
12 6 12 6
4 Moment
6 2 6 4
Reaction Moment Reaction Moment

 

 


 

  
  
 


 
   
Note: 1) Action corresponding to translation is reaction
2) Action corresponding to rotation is moment

47. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Example 1: Analyse the beam as shown in figure using finite element method.
Take EI = constant.
Solution:
Step 1: Degrees of freedom: 06 (02 DOF at each node, translation and rotation)
No. of elements: 02 (AB, BC)
Discretization
Element Nodes Displacements Boundary conditions
1 1-2 1,2,3,4 1=2=3= zero (Fixed support)
2 2-3 3,4,5,6 3=5=zero (simple supports)
Step 2: Element stiffness matrices: Using standard stiffness matrix of beam
element, obtain local stiffness matrix of each element separately. (Note that the
moment of inertia of AB is 2I and BC is I).
The local stiffness matrix of element AB is:
 
1 2 3 4
0 111 0 333 0 111 0 333 1
0 333 1 333 0 333 0 667 2
0 111 0 333 0 111 0 333 3
0 333 0 667 0 333 1 33
K = EI
3 4
AB
. . . .
. . . .
. . . .
. . . .

 
 

 
 
  
 

 
Similarly the local stiffness matrix of element BC is:
 
3 4 5 6
0 1875 0 375 0 1875 0 375 3
0 375 1 0 0 375 0 5 4
0 1875 0 375 0 1875 0 375 5
0 375 0 5 0 375 1 0 6
K = EI
BC
. . . .
. . . .
. . . .
. . . .

 
 

 
 
  
 

 
Step 3: Assemble global stiffness matrix

48. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Size of global stiffness matrix will be 6×6, because total DOF are 6. Joint B is
common in both the elements; therefore elements corresponding to unknown at
joint B (3 and 4) will be added together.
 
1 2 3 4 5 6
0 111 0 333 0 111 0 333 0 0 1
0 333 1 333 0 333 0 667 0 0 2
0 111 0 333 0 2985 0 042 0 1875 0 375 3
0 333 0 667 0 042 0 375 4
0 0 0 1875 0 375 0 1875 0 375 5
0 0 0 37
K = E
5 0 375 6
I
. . . .
. . . .
. . . . . .
. . . .
. . . .
. .
 
 
 
 
 
 
   
 

 
 
   
 

 
   
2.333 0.5
0.5 1.0
Step 4: Impose the boundary conditions
1 = 2 = zero (Fixed support), 3 = 5 = zero (simple supports)
Step 5: Reduced stiffness matrix
The nonzero joint displacements are 4 and 6. Therefore collect the elements
corresponding to 4 and 6 from global stiffness matrix.
 
4 6
2 333 0 5 4
0 5 1 0 6
K = EI
. . ,
. . ,


 
 
 
B
C
Step 6: Element Nodal Load Vector:
The element nodal load vector is obtained by restraining the beam at all supports.
Determine fixed end moments, reactions due to external load and reactions due to
moments. Write down element nodal load vector for both the elements and then
determine reduced element nodal load vector.

49. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
 AB
75 1
75 2
75 3
75 4
 
 
 
 
 
 

 

q  AB
50 3
50 4
50 5
50 6
 
 
 
  
 
 

 
q  
75 1
75 2
125 3
25 4
50 5
50 6

 
  
 
  
  

 
  
 

 
q
Step 7: Equivalent load vector
Equivalent load vector is opposite to element nodal load vector.
    Joint forces
 
F q
 
25 0 25 4
50 0 50 6
     
  
     
     
F
Step 8: Equation of equilibrium:
    
B
C
K F
2 333 0 5 25
EI
0 5 1 0 50
. .
. .


 
 
   

   
 
   
 
B C
50
0 0 and
EI
.
 
 
Step 9: Reactions and Moments:
      
A
A
B
B
C
C
f K q
R 0 111 0 333 0 111 0 333 0 0
M 0 333 1 333 0 333 0 667 0 0
R 0 111 0 333 0 2985 0 042 0 1875 0 375
M 0 333 0 667 0 042 2 333 0 375 0 5
R 0 0 0 1875 0 375 0 1875 0 375
M 0 0 0 375 0 5 0 375 1 0
. . . .
. . . .
. . . . . .
EI
. . . . . .
. . . .
. . . .
  

  
  
 
    
 

 

 
    
 

 
 
0 75
0 75
0 125
1
0 25
EI
0 50
50 50
    
     
     
     

   
 

   
 
   
 
   
 

     
A
A
B
B
C
C
R 0 75 75
M 0 75 75
R 18 75 125 106 25
M 25 25 0
R 18 75 50 31 25
M 50 50 0
kN
kN.m
. . kN
kN.m
. . kN
kN.m
       
       
       
       
  
       

       
       

       

     
 

50. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Example 2: Analyse the continuous beam as shown in figure using finite element
method. Take EI constant.
Solution:
Step 1: Degrees of Freedom: 06
No. of elements: 02 (AB, BC)
For simplicity, convert overhang portion into moment.
Discretization
Element Nodes Displacements Boundary conditions
1 1-2 1,2,3,4 1=2=3= zero (Fixed support)
2 2-3 3,4,5,6 3=5=zero (simple supports)
Step 2: Element stiffness matrix
Using standard stiffness matrix of beam element, obtain stiffness matrix of each
element separately.
Stiffness matrix element AB
 AB
1 2 3 4
0 096 0 24 0 096 0 24 1
0 24 0 8 0 24 0 4 2
0 096 0 24 0 096 0 24 3
0 24 0 4 0 24 0 8
E
4
K = I
. . . .
. . . .
. . . .
. . . .

 
 

 
 
  
 

 
Stiffness matrix of element BC
 BC
1 2 3 4
0 1875 0 375 0 1875 0 375 1
0 375 1 0 0 375 0 5 2
0 1875 0 375 0 1875 0 375 3
0 375 0 5 0 375 1 0
K = E
4
I
. . . .
. . . .
. . . .
. . . .

 
 

 
 
  
 

 

51. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Step 3: Global Stiffness matrix:
Size of global stiffness matrix will be 6×6, because total DOF are 6. Joint B is
common in both the elements; therefore elements corresponding to unknown at
joint B (3 and 4) will be added together.
 
1 2 3 4 5 6
0 096 0 24 0 096 0 24 0 0 1
0 24 0 8 0 24 0 4 0 0 2
0 096 0 24 0 2835 0 135 0 1875 0 375 3
0 24 0 4 0 135 0 375 4
0 0 0 1875 0 375 0 1875 0 375 5
0 0 0 375
K = EI
0 375 6
. . . .
. . . .
. . . . . .
. . . .
. . . .
. .
 
 
 
 
 
 
   
 

 
 
   
 

 
   
1.8 0.5
0.5 1.0
Step 4: Impose the boundary conditions
1 = 2 = zero (Fixed support), 3 = 5 = zero (simple supports)
Step 5: Reduced stiffness matrix
The nonzero joint displacements are 4 and 6. Therefore collect the elements
corresponding to 4 and 6 from global stiffness matrix.
 
. . ,
. . ,
4 6
1 8 0 5 4
0 5 1
K =
0
EI
6
B
C


 
 
 
Step 6: Element nodal load vector
     
AB BC
17 6 1
17 6 1 60 3 24 2
24 2 40 4 92 4 3
32 4 3 60 5 4 0 4
36 4 40 6 60 5
4
q q
0
q
6
.
.
.
&
. .

 
  
     
      
   

     
     
     
  
   
 

  
 
Step 7: Equivalent load vector

52. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Equivalent load vector is opposite to element nodal load vector. For simplicity
convert overhang portion into moment. (20×1.5=30kN.m clockwise) acting at joint
c. This joint moment will be considered in equivalent load vector directly.
    Joint forces
 
F q
 
4 0 4 0 4
40 30 10 6
.
 
     
 
     

     

f
Step 8: Equation of Equilibrium:
    
B
C
K F
1 8 0 5 4 0
EI
0 5 1 0 10
. . .
. .


 

 
   

   
 
   
 
B C
5 806 12 903
and
EI EI
. .
 

 
Step 9: Moments and Reaction Calculation
      
f K q
  
A
A
B
B
C
C
R 0 096 0 24 0 096 0 24 0 0
M 0 24 0 8 0 24 0 4 0 0
R 0 096 0 24 0 2835 0 135 0 1875 0 375 1
M 0 24 0 4 0 135 1 8 0 375 0 5 EI
R 0 0 0 1875 0 375 0 1875 0 375
M 0 0 0 375 0 5 0 375 1 0
. . . .
. . . .
. . . . . .
. . . . . .
. . . .
. . . .

   
   

   
   
  
 

   

   
   
  
   

   
 
EI
0 17 6
0 24
0 92 4
5 806 4 0
0 60
12 903 40
.
.
. .
.
   
   
   
   

   

   
   
   

   
A
A
B
B
C
C
R 16 207
M 21 68
R 96 46
M 0
R 57 337
M 30
. kN
. kN.m
. kN
kN.m
. kN
kN.m
   
   
   
   

   
   
   
   
 
 
Example 3: Analyse the beam using finite element method if support B sink by
25mm. Take EI = 3800 kN.m2
Solution: Step 1: Degrees of Freedom: 06 and No. of elements: 02 (AB, BC)

53. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Discretization
Element Nodes Displacements Boundary conditions
1 1-2 1,2,3,4 1=2=3= zero (Fixed support)
2 2-3 3,4,5,6 3=5=zero (simple supports)
Step 2: Element stiffness matrices
 AB
1 2 3 4
0 0555 0 1667 0 0555 0 1667 1
0 1667 0 667 0 1667 0 333 2
0 0555 0 1667 0 0555 0 1667 3
0 1667 0 333 0 1667 0 66
K = EI
7 4
. . . .
. . . .
. . . .
. . . .

 
 

 
 
  
 

 
 BC
3 4 5 6
0 0555 0 1667 0 0555 0 1667 3
0 1667 0 667 0 1667 0 333 4
0 0555 0 1667 0 0555 0 1667 5
0 1667 0 333 0 1667 0 66
K = EI
7 6
. . . .
. . . .
. . . .
. . . .

 
 

 
 
  
 

 
Step 3: Global Stiffness matrix
 
1 2 3 4 5 6
0 0555 0 1667 0 0555 0 1667 0 0 1
0 1667 0 667 0 1667 0 333 0 0 2
0 0555 0 1667 0 111 0 0 0555 0 1667 3
0 1667 0 333 0 0 1667 4
0 0 0 0555 0 1667 0 0555 0 1667
0 0 0 1667 0 166
K
7
= EI
. . . .
. . . .
. . . . .
. . .
. . . .
. .

 
 

 
 
  
 

 
 
  
 

 
1.33 0.333
0.333 0.667
5
6




   
Step 4: Impose the boundary conditions
1 = 2 = zero (Fixed support), 3 = 5 = zero (simple supports)

54. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Step 5: Reduced stiffness matrix
 
4 6
1 333 0 333 4
0 333 0 667
K = EI
6
B
C
. . ,
. . ,


 
 
 
Step 6: Element nodal load vector:
   
1 1
AB BC
30 1 22 22 3
30 2 26 67 4
q q
30 3 7 78 5
30 4 13 33 6
.
.
.
.
   
   
   
 
   
   
   
 
   
Sinking Moments: Sinking is given in mm. Put this in m while calculating sinking
moments. Since both the element are having same length. Sinking moment of both
the elements will be same.
2 2
6 6 3800 0.025
Sinking moments 15.833 .
6
EI
kN m
L
  
  
   
AB BC
5 28 1 5 28 3
15 833 2 15 833 4
q q
5 28 3 5 28 5
15 833 4 15 833 6
. .
. .
. .
. .

   
   

   
 
   

   
   

   
 
35 28 1
45 833 2
41 66 3
q
3 33 4
13 06 5
29 163 6
.
.
.
.
.
.

 
  
 
  
 

 
  
 

 
Step 7: Equivalent load vector
    Joint forces
 
F q
 
3 33 4
F
29 163 6
.
.
 
  
 

55. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Step 8: Equation of Equilibrium:
    
K F
 
B
C
1 333 0 333 3 333
EI
0 333 0 667 29 163
. . .
. . .


 
   

   
 
   
 
B C
9 45 48 189
EI EI
. .
 

 
Step 9: Moments and Reaction Calculation
      
A
A
B
B
C
C
f K q
R 0 0555 0 1667 0 0555 0 1667 0 0
M 0 1667 0 667 0 1667 0 333 0 0
R 0 0555 0 1667 0 111 0 0 0555 0 1667
M 0 1667 0 333 0 1 333 0 1667 0 333
R 0 0 0 0555 0 1667 0 0555 0 1667
M 0 0 0 1667 0 333
. . . .
. . . .
. . . . .
. . . . .
. . . .
. .
  

 
  
 
    
 

 

 
    
 

 
 
EI
0 35 28
0 45 833
0 41 66
1
9 45 3 33
EI
0 13 06
0 1667 0 667 48 189 29 163
.
.
.
. .
.
. . . .
     
     
     
     
   

     
 
     
     
     

     
     
A
A
B
B
C
C
R 33 704
M 42 686
R 49 693
M 0 00
R 6 602
M 0 00
. kN
. kN.m
. kN
. kN.m
. kN
. kN.m
   
   
   
   
 

   
   
   
   
   
 
Example 4: A continuous beam ABC is loaded as shown in fig. It has constant
flexural rigidity. Fixed support at A, roller support at B and guided support at C.
Analyze the beam using finite element method.
Step 1: Degrees of Freedom: 06
Note: Guided support is having only vertical displacement. (Rotation is always
zero.) Therefore reaction at guided support is zero

56. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Discretization
Element Nodes Displacements Boundary conditions
1 1-2 1,2,3,4 1=2=3= zero (Fixed support)
2 2-3 3,4,5,6 3=zero (simple support)
6=zero (guided support)
Step 2: Element stiffness matrices
 AB
1 2 3 4
0 0234 0 0937 0 0234 0 0937 1
0 0937 0 5 0 0937 0 25 2
0 0234 0 0937 0 0234 0 093
K =
7 3
0 0937 0 25 0 0937 0 5
E
4
I
. . . .
. . . .
. . . .
. . . .

 
 

 
 
  
 
 
 BC
3 4 5 6
0 0234 0 0937 0 0234 0 0937 3
0 0937 0 5 0 0937 0 25 4
0 0234 0 0937 0 0234 0 093
K =
7 5
0 0937 0 25 0 0937 0 5
E
6
I
. . . .
. . . .
. . . .
. . . .

 
 

 
 
  
 
 
Step 3: Global Stiffness matrix
 
1 2 3 4 5 6
0 0234 0 0937 0 0234 0 0937 0 0 1
0 0937 0 5 0 0937 0 25 0 0 2
0 0234 0 0937 0 0468 0 0 0234 0 0937 3
0 0937 0 25 0 1874 0 25 4
0 0 0 0234 0 0937 5
0 0 0 09
K =
37 0 25 0 0937 0 5 6
EI
. . . .
. . . .
. . . . .
. . . .
. .
. . . .
 
 
 
 
 
 
  
 
 
 
 
 
 
1.0 -0.0937
-0.0937 0.0234


   
Step 4: Impose the boundary conditions
1 = 2 = zero (Fixed support), 3 = zero (simple supports), 6 = zero (guided support)
Step 5: Reduced stiffness matrix
 
4 5
1 0 0 0937 4
0 0937 0 0234 5
K = EI
. . ,
. . ,


 
  

 
B
C

57. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Step 6: Element nodal load vector:
 AB
20 1
40 2
20 3
40 4
q

 
  
 
 

 
 

 
 &  BC
10 3
20 4
10 5
20 6
q

 
  
 
 
 
 
 
 

Step 7: Equivalent load vector
    Joint forces
 
F q
 
20 4
10 5



 

 
f
Step 8: Equation of Equilibrium:
    
K F
 
B
C
20 1 0 0 0937
EI
10 0 0937 0 0234
. .
. .

  
   

   
  
 
    
B C
32 078 555 8
and
EI EI
. .

 
  
Example 4: Analyze the continuous beam using finite element method. Take EI
constant
Solution: Step 1: Degrees of Freedom: 06
DOF at point C are 02 (rotation and translation due to spring).

58. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Discretization
Element Nodes Displacements Boundary conditions
1 1-2 1,2,3,4 1=2=3= zero (Fixed support)
2 2-3 3,4,5,6 3=zero (simple support)
3 3-4 5, 8 8=zero (spring fixed at bottom)
Step 2: Element stiffness matrix
 AB
1 2 3 4
0 1875 0 375 0 1875 0 375 1
0 375 1 0 0 375 0 5 2
0 1875 0 375 0 1875 0 375 3
0 375 0 5 0 375 1 0
K = E
4
I
. . . .
. . . .
. . . .
. . . .

 
 

 
 
  
 

 
 BC
3 4 5 6
1 5 1 5 1 5 1 5 3
1 5 2 0 1 5 1 0 4
1 5 1
K =
5 1 5 1 5 5
1 5 1 0 1 5 2 0
EI
6
. . . .
. . . .
. . . .
. . . .

 
 

 
 
  
 

 
Stiffness matrix of spring element
 
1 1
5
5
K = EI
8
1 1 8
CD

 
 

 
Step 3: Global Stiffness matrix
 
0 1875 0 375 0 1875 0 375 0 0
0 375 1 0 0 375 0 5 0 0
0 1875 0 375 1 6875 1 125 1 5 1 0
K = EI
0 375 0 5 1 125 3 0 1 5 1 0
0 0 1 5 1 5 2 5 1 5
0 0 1 5 1 0 1 5 2 0
1 2 3 4 5 6
1
2
3
4
5
6
. . . .
. . . .
. . . . . .
. . . . . .
. . . .
. . . .

 
 

 
 
  
 

 
 
  


 

 

  
Step 4: Impose the boundary conditions
1 = 2 = zero (Fixed support), 3 = zero (simple supports)

59. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Step 5: Reduced stiffness matrix
 
4 5 6
3 0 1 5 1 0 4
1 5 2 5 1 5 5
1 0 1 5 2
E
6
K I
0
=
B
C
C
. . . ,
. . . ,
. . . ,



 
 
  
 
 

 
Step 6: Element nodal load vector:
 AB
20 1
20 2
20 3
20
q
4
 
 
 
 
 
 

 
 &  BC
0 3
0 4
0 5
0 6
q
 
 
 
 
 
 
 
  
20 1
20 2
20 3
20 4
0 5
0
q
6

 
  
 
  
 

 
 
 
 

Step 7: Equivalent load vector
    Joint forces
 
F q
Note- External moment 30 kN.m clockwise is acting at B, it is accounted in the
element corresponding to rotation at B i.e. 4
 
20 30 10 4
0 0 0 5
0 0 0 6
 
     
     
  
     
     
     
f
Step 8: Equation of Equilibrium:
    
K F
 
B
C
C
10 3 0 1 5 1 0
0 EI 1 5 2 5 1 5
0 1 0 1 5 2 0
. . .
. . .
. . .


   
   
   
 
   
   
 
   
 

    
B C C
4 782 2 608 0 434
and
EI EI EI
. . .
,
 
 
   

60. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Example 5: Analyze the indeterminate beam as shown in figure using finite
element method. The beam is fixed at A, C and has internal hinge at B. Take EI
constant.
Solution: Step 1: Degrees of Freedom: 07
DOF at point B are 03 (two rotations and translation).
Discretization
Element Nodes Displacements Boundary conditions
1 1-2 1,2,3,4 1=2= zero (Fixed support)
2 2-3 3,5,6,7 6=7=zero (Fixed support)
Step 2: Element stiffness matrix
 AB
1 2 3 4
12 6 12 6 1
6 4 6 2 2
12 6 12 6 3
6
K = EI
2 6 4 4
 
 
 
 
 
 
  
 

 
 
 BC
3 5 6 7
1 5 1 5 1 5 1 5 3
1 5 2 0 1 5 1 0 5
1 5 1
K =
5 1 5 1 5 6
1 5 1 0 1 5
EI
2 0 7
. . . .
. . . .
. . . .
. . . .

 
 

 
 
   
 
 
 
 

61. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Step 5: Reduced stiffness matrix
 
13 5 6 0 1 5
K = EI 6 0 4 0 0
1 5 0 2 0
3 4 5
3
4
5
B
BA
BC
. . .
. .
. ,
.
,
,


 
 

 



 
Step 6: Element nodal load vector:
 AB
30 1
5 2
30 3
5 4
q
 
 
 
 
 
 

 
 &  BC
60 3
20 5
60 6
20
q
7
 
 
 
 
 
 

 
  
90 3
q 5 4
20 5
 
 
 
 
 
 
Step 7: Equivalent load vector
    Joint forces
 
F q
 
90 3
f 5 4
20 5

 
 
  
 

 
Step 8: Equation of Equilibrium:
    
K F
 
13 5 6 0 1 5
6 0 4 0 0
90
5
1 2
5 0 0
2 0
. . .
. .
. .



 
 

 
 
 
   
   

 
   
   

 
 
B
BA
BC
EI
BA BC
20 28 75 5
and
EI EI EI
B
.
,  
 
   

62. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Example: For the following beam, find the vertical deflection and rotation at joint
B using finite element method. Take EI = 12×103
kN.m2
Example: Determine the unknown joint displacements of the beam as shown in
figure using finite element method. Take EI constant.
Example: Analyse the beam using finite element method if support B is sink by
25mm. Take EI = 3800 kN.m2
Example: A continuous beam has fixed support at node 1 and roller supports at
nodes 2 and 3. Analyse the beam using finite element method and draw SFD and
BMD. Take E = 200 GPa and I=4×106
mm4
.
Example: Obtain rotation at B for the beam shown below using finite element
method. Consider given beam as one element. Take E = 2×108
kN/m2
and I =
4×10-6
m4
.
Example: Analyze the continuous beam ABC as shown in Figure using finite
element method. Take EI constant.

63. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Example: Analyse the beam ABC shown in Figure 1 using finite element method.
AB = 3 m and BC = 6 m. Take EI = constant
Example: Analyse the prismatic beam ABC loaded and supported as shown in
Figure using finite element approach. Support B is sink by 25 mm. Draw SFD and
BMD. Take EI constant.
Example: Determined the prop reaction of the propped cantilever beam AB as
shown in Figure 1 using finite element method. Take EI = constant
Example: Obtain fixed end moment at support A using finite element method.
Take E = 2×108
kN/m2
and I = 4×10-6
m4
.
Example: Determine support reactions of continuous beam ABC if support B sink
by 10 mm. Take EI = 6000 kN.m2
. Use finite element method.
Example: Determine support reactions of continuous beam ABC as shown in
Figure 1 if support B sink by 10 mm. Take EI = 6000 kN.m2
. Use finite element
method.

64. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Unit-III
Finite Element Analysis of Plane Frames
The plane frame is a combination of plane truss and beam. All members are
connected by rigid joints in case of frame.
Stiffness matrix of frame element in local coordinate system
Let consider a frame element of length L, flexural rigidity EI and axial rigidity AE.
A frame is having three degrees of freedom at each node i.e. displacement in x-
direction, displacement in y-direction and rotation. Therefore the size of stiffness
matrix of frame element is 6 6
 .
D1 = Displacement in x-direction at node A
D2= Displacement in y-direction at node A
D3 = Rotation at node A
D4 = Displacement in x-direction at node B
D5= Displacement in y-direction at node B
D6 = Rotation at node B
To derive the stiffness matrix, give the unit displacements at each node one by one
Unit displacement in x-direction at node 1
Unit displacement in y-direction at node 1
Unit rotation in z-direction at node 1
Unit displacement in x-direction at node 2

65. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Unit displacement in y-direction at node 2
Unit rotation in z-direction at node 2
 
1 2 3 4 5 6
1
3 2 3 2
2
2 2
3
4
3 2 3 2
5
2 2
6
0 0 0 0
0 12 6 0 12 6
0 6 4 0 6 2
0 0 0 0
0 12 6 0 12 6
0 6 2 0 6 4
D D D D D D
D
AE / L AE / L
D
EI / L EI / L EI / L EI / L
D
EI / L EI / L EI / L EI / L
K'
D
AE / L AE / L
D
EI / L EI / L EI / L EI / L
D
EI / L EI / L EI / L EI / L

 
 

 
 

  

 
 
  
 

 
Transformation Matrix of Frame Element
In plane frame the members are oriented in different directions and hence it is
necessary to transfer stiffness matrix of individual member from local coordinate
system to global coordinate system. This is performed by using transformation
matrix.

66. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Let consider a frame element at an angle θ with respect to positive x-axis.
D1, D2 and D3 D.O.F. at each node for global co-ordinate system
𝐷1
′
, 𝐷2
′
and 𝐷3
′
D.O.F. at each node for local co-ordinate system
Let the local DOF be expressed into global DOF
At Node 1
'
1 1 2
'
2 1 2
'
3 3
D Dl D m
D D m D l
D D
 
  

At Node 2
'
4 4 5
'
5 4 5
'
6 6
D D l D m
D D m D l
D D
 
  

In matrix form:
l = cosθ and m = sinθ are direction cosines.
1 2 3 4 5 6
'
1
1
'
2
2
'
3
3
'
4
4
'
5
5
'
6
6
0 0 0 0
0 0 0 0
0 0 1 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0 0 1
D D D D D D
D
l m
D
D
m l
D
D
D
D
l m
D
D
m l
D
D
D
   
 
   
 

   
 
   
 
 

   
 
   
 
   
 

   
 
    
 
where  
0 0 0 0
0 0 0 0
0 0 1 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0 0 1
l m
m l
L
l m
m l
 
 

 
 
  
 
 

 
 
    
'
x L x

[L] = Transformation Matrix
 
'
x = Local Displacement Vector
 
x =Global Displacement Vector

67. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
The stiffness matrix of member is global coordinate system is obtained by
using relation (Taking θ = 900
, l=0, m=1)
[𝐾] = [𝐿]𝑇[𝐾′][𝐿]
    
3 2 3 2
2 2
3 2 3 2
2 2
'
0 1 0 0 0 0 / 0 0 / 0 0
1 0 0 0 0 0 0 12 / 6 / 0 12 / 6 /
0 0 1 0 0 0 0 6 / 4 / 0 6 / 2 /
0 0 0 0 1 0 / 0 0 / 0 0
0 0 0 1 0 0 0 12 / 6 / 0 12 / 6 /
0 0 0 0 0 1 0 6 / 2 / 0 6 / 4 /
T
L K L
AE L AE L
EI L EI L EI L EI L
EI L EI L EI L EI L
AE L AE L
EI L EI L EI L EI L
EI L EI L EI L EI L

 
  
  
 
  

 
 
 
    
 

 
0 1 0 0 0 0
1 0 0 0 0 0
0 0 1 0 0 0
0 0 0 0 1 0
0 0 0 1 0 0
0 0 0 0 0 1

 
 
 
 
 
 
 
 
 
 

 
 
 
 
 

 
 
Therefore, the stiffness matrix of any member which is perpendicular
(θ = 900
) to reference member
 
3 2 3 2
2 2
3 2 3 2
2 2
12 0 6 12 0 6
0 0 0 0
6 0 4 6 0 2
12 0 6 12 0 6
0 0 0 0
6 0 2 6 0 4
EI / L EI / L EI / L EI / L
AE / L AE / L
EI / L EI / L EI / L EI / L
K
EI / L EI / L EI / L EI / L
AE / L AE / L
EI / L EI / L EI / L EI / L
 
  
 

 
 

  

 
 

 

 
 

68. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Note:
If we neglect the axial deformation these two matrices reduced to order 4×4.
(The columns and row corresponding to axial stiffness AE/L are neglected)
Steps for the solution of Indeterminate plane frames using finite element
method:
1. Divide the frame into number of elements (Take one member as one element)
2. Identify total degrees of freedom (Three D.O.F. at each node, two
displacements and rotation)
3. Determine stiffness matrices of all elements ([K]1, [K]2………)
4. Assemble the global stiffness matrix [K]
5. Impose the boundary conditions and determine reduced stiffness matrix
6. Determine element nodal load vector [q] (Restrained structure)
7. Determine equivalent load vector [f]
8. Apply equation of equilibrium [K]{Δ}={f} and determine unknown joint
displacements.
9. Apply equation [K]{Δ}+[q] ={f} to determine reactions and moments
Stiffness matrix for Beam Member neglecting axial deformation
(Neglect first and fourth row and columns)
 
3 2 3 2
2 2
3 2 3 2
2 2
12 6 12 6
6 4 6 2
12 6 12 6
6 2 6 4
EI / L EI / L EI / L EI / L
EI / L EI / L EI / L EI / L
K
EI / L EI / L EI / L EI / L
EI / L EI / L EI / L EI / L
 

 

 

 
  
 

 
Stiffness matrix for Column Member (θ=900
, l = 0, m = 1) always take bottom
of column as a first node. (Neglect second and fifth row and columns)
 
3 2 3 2
2 2
3 2 3 2
2 2
12 6 12 6
6 4 6 2
12 6 12 6
6 2 6 4
EI / L EI / L EI / L EI / L
EI / L EI / L EI / L EI / L
K
EI / L EI / L EI / L EI / L
EI / L EI / L EI / L EI / L
 
  
 

 

 

 

 

69. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Example 1: Analyze the portal frame as shown in figure using finite element
method. Take EI constant. Neglect axial deformation.
Solution: Step 1: Total DOF = 12
(Three DOF at each node, two displacements and one rotation)
No. of elements: 03 (AB, BC, DC)
Discretization
Element Nodes Displacements Boundary conditions
1 AB 1,2,3,4,5,6 1=2=3=zero
2 BC 4,5,6,7,8,9 5=8=zero, 4=7
3 DC 10,11,12,7,8,9 10=11=12=zero
Step 2: Element Stiffness matrices
Element stiffness matrix for AB (Column member)
 
   
 
 
 
 

 

 

 
 
1 3 4 6
0.048 0.24 0.048 0.24 1
0.24 1.6 0.24 0.8 3
0.04 0.048 0
8 0.24 4
0.2
.24
0.2
4 0 4 1.6
.8 6
AB
K EI
Imposing Boundary Conditions
1=3=0
Element Stiffness Matrix for BC: (Beam member)

70. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
 
5 6 8 9
0.75 1.5 0.75 1.5 5
1.5 1.5 6
0.75 1.5 0.75 1.5 8
1.5
4 2
2 4
1.5 9
BC
K EI
 
 
 

 

 
   
 

 
 
Imposing Boundary Conditions
5=8=0
Element Stiffness Matrix for DC: (Column member)
 DC
K EI
   
 
 
 
 

 

 

 
 
10 12 7 9
0.096 0.24 0.096 0.24 10
0.24 0.8 0.24 0.4 12
0.09 0.096 0.24
0.24
6 0.24 7
0.24 0. 9
0.8
4
Imposing Boundary Conditions
10=12=0
Step 3: Reduced Stiffness Matrix:
Since horizontal sway at B and C are same (4=7), we can modify the above
stiffness matrix as
4 6 9
0.144 0.24 0.24 4,
[ ] 0.24 5.6 2 6,
0.24 2 4.8 9,
B
C
K EI 


 
 
  
 
 
Step 4: Element Nodal Load Vector

71. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
3.52 1
9.6 3
{ }
4
6
6 5
6
{ }
6 8
9
3.24 10
3.6 12
6.48
14.4
4
4
1.76
}
2 9
.
{
4
7
AB
BC
DC
q
q
q

 
 
 
  
 
 
 
 
 
 
  
 
 
 
 
 

 
  
 
 





Reduced element nodal load vector
4.72 4
{ } 10.4 6
1.6 9
q

 
 
 
 
 

 
Step 4: Equivalent Load Vector
    Joint forces
F q
  
 
4.72 4
10.4 6
1.6 9
F
 
 
  
 
 
Step 5: Equation of Equilibrium
[ ]{ } { }
0.144 0.24 0.24 4.72
0.24 5.6 2 10.4
0.24 2 4.8 1.6
B
C
K F
EI 

 

 
   
   
  
   
 
   
 
   
 
34.046 1.0419 1.803
; ;
B C
m rad rad
EI EI EI
 
    
Step 6: Moment Calculations {f} = [K]{Δ}+{q}
Member AB

72. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
0
0.24 1.6 0.24 0.8 0 9.6 18.604
1
0.24 0.8 0.24 1.6 34.046 14.4 4.562
1.0419
AB
BA
M
EI
M EI
 
 

       
 
  
       
 
  
     
   
 
 
Member BC
0
1.5 4 1.5 2 1.0419 4 4.562
1
1.5 2 1.5 4 0 4 9.128
1.803
BC
CB
M
EI
M EI
 
 

       
 
  
       
 
  
     
   
 

 
Member DC
0
0.24 0.8 0.24 0.4 0 3.6 3.849
1
0.24 0.4 0.24 0.8 34.046 2.4 9.128
1.803
DC
CD
M
EI
M EI
 
 
 
       
 
  
       
 

     
   
 

 
Example 2: Analyze the rigid frame by using finite element method. Take EI
constant. Neglect axial deformation.
Solution: Step 1: Total DOF = 12
(Three DOF at each node, two displacements and one rotation)
No. of elements: 03 (AB, BC, DC)
Discretization
Element Nodes Displacements Boundary conditions
1 AB 1,2,3,4,5,6 1=2=3=zero
2 BC 4,5,6,7,8,9 5=8=zero, 4=7
3 DC 10,11,12,7,8,9 10=11=12=zero

73. Finite Element Method
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Step 2: Element stiffness matrix for Column AB:
 
1 3 4 6
0.1875 0.375 0.1875 0.375 1
0.375 1 0.375 0.5 3
0.1875 0.375 4
0.37
0.1875 0.375
0.375 1
5 0.5 6
AB
K EI
   
 
 
 
 

 

 

 
 
Element Stiffness Matrix of beam BC
 
5 6 8 9
0.0469 0.1875 0.0469 0.1875 5
0.1875 0.1875 6
0.0469 0.1875 0.0469 0.1875 8
0.1875 0.187
1 0.5
0.5 1
5 9
BC
K EI
 
 
 

 

 
   
 

 
 
Element Stiffness Matrix for column DC
 
10 12 7 9
0.1875 0.375 0.1875 0.375 10
0.375 1 0.375 0.5 12
0.1875 0.375 7
0.375 0
0.1875 0.375
0.375 1
.5 9
DC
K EI
   
 
 
 
 

 

 

 
 
Imposing Boundary Conditions
1=2=3=5=8=10=11=12=0
Step 2: Reduced Stiffness Matrix
Since horizontal sway at B and C are same (4=7), we can modify the above
stiffness matrix as
4 6 9
0.375 0.375 0.375 4,
[ ] 0.375 2 0.5 6,
0.375 0.5 2 9,
B
C
K EI 


 
 

 
 
 
Step 3: Element Nodal Load Vector