Matrix Methods of Structural Analysis

Matrix Methods of Structural Analysis
Lecture Notes
Dr. Atteshamuddin S. Sayyad, SRES’s Sanjivani College of Engineering, Kopargaon-423603
Matrix Methods of
Structural Analysis
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

Lecture Notes
Unit-I
Computational Techniques
Gauss Elimination Method
Gauss Elimination method can be adopted to find the solution of linear
simultaneous equations arising in engineering problems. In the method, equations
are solved by elimination procedure of the unknowns successively. In this method
the unknowns are gradually eliminated by combining the equations Basically the
method involves the reduction of ‘n’ equations in ‘n’ unknowns into one unknown,
which is then solved by back substitution.
Example: Solve by Gauss-elimination method
1 2 3
1 2 3
1 2 3
2 4 4
3 5
3 2 2 1
x x x
x x x
x x x
  
   
   
Solution: Let assume
1 2 3
2 4 4
x x x
   (1)
1 2 3
3 5
x x x
    (2)
1 2 3
3 2 2 1
x x x
    (3)
From Eq. (1) find value of x1
 
1 2 3
1
4 4
2
x x x
   (4)
Put into Eqs. (2) and (3)
Eq. (2) becomes
 
2 3 2 3
1
4 4 3 5
2
x x x x
     
2 3
3 5 3 7
. x x
     (5)
Eq. (3) becomes
 
2 3 2 3
3
4 4 2 2 1
2
x x x x
     
2 3
3 5 4 7
. x x
     (6)
From Eq. (5) find value of x2
 
2 3 3
1
7 3 2 0 0 867
3 5
x x . . x
.
       (7)
Put into Eq. (6)
 
3 3
3 5 2 0 0 867 4 7
. . . x x
     

Lecture Notes
3 0
x 
Find value of x2 from eq. (7) and x1 from eq. (4)
2 1
2 0 and 1 0
x . x .
 

Lecture Notes

Lecture Notes
Gauss Jordon Method
In this method matrix is diagonalized by row operations so that the solution is
directly obtained.
Note: We can perform only row operations in this method.
Example: Solve by Gauss-Jordon method
1 2 3
1 2 3
1 2 3
2 4 4
3 5
3 2 2 1
x x x
x x x
x x x
  
   
   
Solution: Let write given set of equations into matrix form
[A]{x} = {B}
1
2
3
2 1 4 4
1 3 1 5
3 2 2 1
x
x
x
 
   
   
 
   
   
 
   
 
 
   
 
Write augmented matrix (A, B)
 
2 1 4 4
1 3 1 5
3 2 2 1
A,B
 
 
   
 
 
 
 
1 2
R R

1 3 1 5
2 1 4 4
3 2 2 1
  
 
 
 
 
 
 

Lecture Notes
2 2 3 1
2 3
R R , R R
 
1 3 1 5
0 7 6 14
0 7 5 14
  
 
 
 
 
 
3 2
R R

1 3 1 5
0 7 6 14
0 0 1 0
  
 
 
 
 

 
2 3
6
R R

1 3 1 5
0 7 0 14
0 0 1 0
  
 
 
 
 

 
 
2 3
7 1
R , R
  
1 3 1 5
0 1 0 2
0 0 1 0
  
 
 
 
 
 
1 3
3
R R

1 0 1 1
0 1 0 2
0 0 1 0

 
 
 
 
 
1 3
R R

1 0 0 1
0 1 0 2
0 0 1 0
 
 
 
 
 
Therefore unknowns are
1
2
3
1 0 0 1
0 1 0 2
0 0 1 0
x
x
x
 
   
   
  
   
 
   
 
   
 
1 2 3
1 0 2 0 and 0
x . , x . x
   

Lecture Notes
Gauss-Siedel Method (Iterative Method)
Example 1: Solve by Gauss-Siedel method
1 2 3
1 2 3
1 2 3
2 5 12
5 4
4 2 15
x x x
x x x
x x x
  
  
  
Solution:
Make the given system diagonally predominant
1 2 3
5 4
x x x
   (1)
1 2 3
4 2 15
x x x
   (2)
1 2 3
2 5 12
x x x
   (3)
Find value of x1 from Eq. (1)
 
1 2 3
1
4
5
x x x
   (4)
 
2 1 3
1
15 2
4
x x x
   (5)
 
3 1 2
1
12 2
5
x x x
   (6)
Let assume initial values of 2 3
0 0
x , x
 
Iteration 1:
2 3
0 0
x , x
  x1 = 0.8 from eq. (4)
x1 = 0.8, x3= 0 x2 = 3.55 from eq. (5)
x1 = 0.8, x2 = 3.55 x3 = 3.66 from eq. (6)
Iteration 2:
x2 = 3.55, x3 = 3.66 x1 = 0.822 from Eq. (4)
x1 = 0.822, x3 = 3.66 x2 = 1.714 from Eq. (5)
x1 = 0.822, x2 = 1.714 x3 = 2.921 from Eq. (6)
Iteration 3:
x2 = 1.714, x3 = 2.921 x1 = 1.041 from Eq. (4)
x1 = 1.041, x3 = 2.921 x2 = 2.029 from Eq. (5)

Lecture Notes
x1 = 1.041, x2 = 2.029 x3 = 3.003 from Eq. (6)
Iteration 4:
x2 = 2.029, x3 = 3.003 x1 = 0.995 from Eq. (4)
x1 = 0.995, x3 = 3.003 x2 = 1.999 from Eq. (5)
x1 = 0.995, x2 = 1.999 x3 = 3.000 from Eq. (6)
Iteration 5:
x2 = 1.999, x3 = 3.000 x1 = 1.000 from Eq. (4)
x1 = 1.000, x3 = 3.000 x2 = 2.000 from Eq. (5)
x1 = 1.000, x2 = 2.000 x3 = 3.000 from Eq. (6)
Iteration 6:
x2 = 2.000, x3 = 3.000 x1 = 1.000 from Eq. (4)
x1 = 1.000, x3 = 3.000 x2 = 2.000 from Eq. (5)
x1 = 1.000, x2 = 2.000 x3 = 3.000 from Eq. (6)
1 2 3
1 0 2 0 and 0
x . , x . x
   
Example 2: Solve by Gauss-Siedel method
1 2 3
1 2 3
1 2 3
5 2 3 1
3 9 2
2 7 3
x x x
x x x
x x x
   
   
  
Solution:
Make the given system diagonally predominant
1 2 3
5 2 3 1
x x x
    (1)
1 2 3
3 9 2
x x x
    (2)
1 2 3
2 7 3
x x x
   (3)
 
1 2 3
1
1 2 3
5
x x x
    (4)
 
2 1 3
1
2 3
9
x x x
   (5)
 
3 1 2
1
3 2
7
x x x

   (6)

Lecture Notes
Let assume initial values of 2 3
0 0
x , x
 
Iterations we can represent in the tabular form also
Iteration No. (n) 1 2 3 4 5 6
x1 0 -0.2 0.167 0.191 0.186 0.186
x2 0 0.156 0.334 0.333 0.331 0.331
x3 0 -0.508 -0.429 -0.422 -0.423 -0.423

Lecture Notes
Exercise
 Solve using Gauss-Siedel Method
i) ii)
1 2 3
1 2 3
1 3
4 7
7 2 2
3 4 11
x x x
x x x
x x
  
   
 
1 2 3
1 2 3
1 2 3
4 2 2 0
3 4 5
3 7
x x x
x x x
x x x
 
  
  
 Solve using Gauss-Elimination or Gauss-Jordon Method
i) ii)
10
2 2
3 5 8 66
x y z
x y z
x y z
   
  
   
3 1
2 5 2 0
3 2 3
x y z
x y z
x y z
  
  
  
iii) iv)
6 3 2 1
5 4 3 0
0
x y z
x y z
x y z
  
  
  
3 2 14
3 2 8
2 6 4 30
x y z
x y z
x y z
  
   
    

Lecture Notes
Unit-II
Stiffness matrix method for bars and trusses
Stiffness matrix method for analysis of bar structures
Bars are 1D structures subjected to axial force only. The degree of freedom at each
node is one i.e. axial displacement. Total degrees of freedom are two. Therefore
size of stiffness matrix is 2×2.
Stiffness matrix for bar element having axial stiffness (AE/L) can be obtained by
giving unit displacement one by one at each node.
Let consider a two noded bar 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
AE
K
u
L

 
  

 
Procedure for the solution of numerical examples
1) Divide the given bar structures into number of members
2) Calculate total degrees of freedom
3) Determine stiffness matrix of each bar 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

Lecture Notes
Example 1: Two bars one of aluminum and other of steel are joint together and
subjected to load as shown in figure. Determine displacements at common joints
and the member forces. Take AAl=200 mm2
, EAl=70 kN/mm2
, Ast=400 mm2
,
Est=200 kN/mm2
Solution:
Let assume u1, u2, u3 are the displacements at three nodes.
Step 1: Divide given bar structure into number of members/elements
Member AE/L (kN/mm) Nodes Displacements (mm) Boundary conditions
1 100 1-2 u1-u2 u1 = 0
2 80 2-3 u2-u3 u3 = 0
Step 2: Element stiffness matrices
 
1 2
1
1
2
1 1 1 1
2000 70
100
1 1 1 1
1400
u u
u
K
u
 
   

 
   
 
   
 
2 3
2
2
3
1 1 1 1
400 200
80
1 1 1 1
1000
u u
u
K
u
 
   

 
   
 
   
Step 3: Global stiffness matrix
   
1 2 3
1
2
3
100 100 0
100 100 80 80
0 80 80
u u u
u
K u
u
 
 
 
   
 
 
 
 
 
Step 4: Reduced stiffness matrix
Imposing boundary conditions i.e. u1 = 0, u3 = 0 eliminate first row, first column
and third row, third column.
Therefore reduced stiffness matrix is

Lecture Notes
   
2
2
180
u
K u

Step 5: Determine unknown joint displacements
Applying Equation of Equilibrium
    
K f
 
    
2
180 9
u 
 
2 0 05
u . mm
  
Step 6: Calculation of member forces
Member 1:
     
1 1
1
K f
 
1 1
2 2
1 1
100
1 1
u f
u f
    
 

   
 

    
1
2
1 1 0
100
1 1 0 05
f
f
.
  
  

   
 

    
( 1 2
0 and 0 05
u u .
 
)
1 2
5 and 5
f kN f kN
  
Similarly, Member 2: 3 4
4 and 4
f kN f kN
   ( 2 3
0 05 and 0
u . u
  )
Example 2: A circular rod ABCD of different c/s is loaded as shown in figure.
Find displacements at all joints using stiffness matrix method. Take E = 200 GPa.
Solution:
Let assume u1, u2, u3, u4 are the displacements at four nodes.
1 200 1-2 u1-u2 u1 = 0
2 70 2-3 u2-u3 ---
3 100 3-4 u3-u4 ---

Lecture Notes
 
1 2
1
1
2
1 1 1 1
1000 200
200
1 1 1 1
1000
u u
u
K
u
 
   

 
   
 
   
 
2 3
2
2
3
1 1 1 1
700 200
70
1 1 1 1
2000
u u
u
K
u
 
   

 
   
 
   
 
3 4
3
3
4
1 1 1 1
500 200
100
1 1 1 1
1000
u u
u
K
u
 
   

 
   
 
   
Imposing boundary conditions i.e. u1 = 0. Size of reduced stiffness matrix is 3×3.
 
 
 
2 3 4
2
3
4
200 70 70 0
70 100 70 100
0 100 100
u u u
u
K u
u
 
 
 
   
 
 

 
    
K f
 
 
 
2
3
4
200 70 70 0 100
70 100 70 100 50
0 100 100 25
u
u
u
 
    
   
 
    
   
 
   
 
    
 
     
2 3 4
0 375 0 0178 0 267
u . mm ,u . mm ,u . mm
      
Example 3: Bars of three different areas are connected together as shown in
figure. Determine the displacement at each joint. Take E=200 GPa, A1=3000 mm2
,
A2=2000 mm2
, A3=1000 mm2
.

Lecture Notes
1 600 1-2 u1-u2 u1 = 0
2 500 2-3 u2-u3 ---
3 333.33 3-4 u3-u4 u4 = 0
 
1 2
1
1
2
1 1 1 1
3000 200
600
1 1 1 1
1000
u u
u
K
u
 
   

 
   
 
   
 
2 3
2
2
3
1 1 1 1
2000 200
500
1 1 1 1
800
u u
u
K
u
 
   

 
   
 
   
 
3 4
3
3
4
1 1 1 1
1000 200
333 33
1 1 1 1
600
u u
u
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
600 600 0 0
600 600 500 500 0
0 500 500 333 33 333 33
0 0 333 33 333 33
u u u u
u
u
K
u
. .
u
. .
 
 
 
  
 

 
  
 
 
 
 
Imposing boundary conditions i.e. u1 = 0, u4 = 0. Size of reduced stiffness matrix is
2×3.
 
2 3
2
1
3
1100 500
500 833 33
u u
u
K
u
.

 
  

 
    
K f
 

Lecture Notes
2
3
1100 500 100
500 833 33 50
u
u
.
  
   

   
 
 
   
 
   
2 3
0 0875 0 0075
u . mm , u . mm
    
Example 4: A circular steel rod ABCD of different c/s is loaded as shown in
figure. Find displacements at each joint using stiffness matrix method. Take E =
200 GPa.
 
 
 
2 2
1
2 2
2
2 2 2
3
70 3848 45
4
50 1963 49
4
50 30 1256 64
4
A . mm
A . mm
A . mm



 
 
  
1 769.69 1-2 u1-u2 u1 = 0
2 196.35 2-3 u2-u3 ---
3 251.33 3-4 u3-u4 ---

Lecture Notes
 
1 2
1
1
2
1 1 1 1
3848 45 200
769 69
1 1 1 1
1000
u u
u
.
K .
u
 
   

 
   
 
   
 
2 3
2
2
3
1 1 1 1
1963 49 200
196 35
1 1 1 1
2000
u u
u
.
K .
u
 
   

 
   
 
   
 
3 4
3
3
4
1 1 1 1
1256 64 200
251 33
1 1 1 1
1000
u u
u
.
K .
u
 
   

 
   
 
   
 
 
 
2 3 4
2
3
4
769 69 196 35 196 35 0
196 35 196 35 251 33 251 33
0 251 33 251 33
u u u
. . . u
K . . . . u
. . u
 
 
 
   
 
 

 
    
K f
 
2
3
4
966 04 196 35 0 100
196 35 447 68 251 33 50
0 251 33 251 33 25
. . u
. . . u
. . u

    
   
 
   
   
 
   
 

    
     
2 3 4
0 097 0 0298 0 0695
u . mm ,u . mm ,u . mm
      
Example 5: Determine the support reaction forces at the two ends of the bar
loaded as shown in figure. The c/s area of the bar is 300 mm2
. Take E=200 GPa
and 0 2
. mm
 

Lecture Notes
Let assume u1, u2, u3 are the displacements at three nodes.
1 300 1-2 u1-u2 u1 = 0
2 300 2-3 u2-u3 u3 = 0.2
 
1 2
1
1
2
1 1 1 1
300 200
300
1 1 1 1
200
u u
u
K
u
 
   

 
   
 
   
 
2 3
2
2
3
1 1 1 1
300 200
300
1 1 1 1
200
u u
u
K
u
 
   

 
   
 
   
 
 
2 3
2
3
300 300 300
300 300
u u
u
K
u
 
 
  

 
    
K f
 
2
3
80
600 300
300 300 0 2
u
R
.
  
  

   
 

    
   
3 2
9 9 0 233
R . kN ,u . mm
    
Reaction at joint 3 is  
9 9
.  kN.
Reaction at joint 1 is  
70 1
. kN  kN.

Lecture Notes
Example 6: Bars of three different c/s area are connected together as shown in figure.
Determine displacements at joints using stiffness matrix method. Take E=210 GPa.
Example 7: A rod is composed of an aluminum section rigidly attached between steel and
bronze as shown in Figure. If the cross-section area of rod is 800 mm2
determine nodal
displacements. Take Est = 210 GPa, EAl = 70 GPa and Ebr = 110 GPa.
Example 8: A circular steel rod ABCD of different cross-section is loaded as shown in Figure
3. Find the displacements at all joints using stiffness matrix method. Take E=200 GPa.

Lecture Notes
Stiffness matrix method for analysis of trusses
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. 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
 
1
1
2
2
e
u
v
x
u
v
 
 
 
  
 
 
 

Lecture Notes
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
 
  
  
 
   
  
  
 
 
 
0
0
0
0
l
m l m l m
AE
K
l l m l m
L
m
 
   
 
 
  
   
 
 
 

Lecture Notes
 
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 )
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) B.C.
AB 4 0 4 1 0 50 uA = 0
A
v 
BC -2 -1.5 2.5 -0.8 -0.6 64 0
B
v 
CA -2 1.5 2.5 -0.8 0.6 64 ---
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:

Lecture Notes
 
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 90 96 30 72 40 96 30 72
0 0 30 72 23 04 30 72 23 04
40 96 30 72 40 96 30 72 81 92 0
30 72 23 04 30 72 23 04 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
     
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.

Lecture Notes
Solution:
Step 1: Degrees of freedom: 08 ( A A B B c c D D
u ,v ,u ,v ,u ,v ,u ,v )
Assume origin support A (0, 0). The coordinates of other nodes B (1000, 0),
C(2000, 0) and D(1500, 1000)
AD 1500 1000 1802.8 0.832 0.555 88.75 0
A A
u v
 
BD 500 1000 1118 0.447 0.894 143.112 0
B B
u v
 
CD -500 1000 1118 -0.447 0.894 143.112 0
c c
u v
 
Stiffness matrix of element AD:
 
61 43 40 98 61 43 40 98
40 98 27 34 40 98 27 34
61 43 40 98 61 43 40 98
40 98 27 34 40 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
  )
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:

Lecture Notes
 
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
  )
 
118 61 40 98
40 98 256 10
D D
D
D
u v
u
. .
K
v
. .
 
  
 
    
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
  
Example 3: for the truss as shown in figure using stiffness matrix 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 ). Assume
origin point B. The coordinates of points areA (-4, 3), B (0,0), C (4,-3), D (-4, -3)
AB 4000 -3000 5000 0.8 -0.6 40 0
A A
u v
 
DB 4000 3000 5000 0.8 0.6 40 0
D D
u v
 
CB -4000 3000 5000 -0.8 0.6 40 0
c c
u v
 

Lecture Notes
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
  )
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
  )
 
1 92 0 48
40
0 48 1 08
B B
B
B
u v
u
. .
K
v
. .

 
  

 
    
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.

Lecture Notes
Solution:
Step 1: Degrees of freedom: 06 ( 1 1 2 2 3 3
u ,v ,u ,v ,u ,v )
Take origin node 1. The coordinates of nodes are
1(0, 0), 2(-3.535, 3.535), 3(-10, 0)
1-2 -3.535 3.535 5 -0.707 0.707 200×102
2 2 0
u v
 
1-3 -10 0 10 -1 0 100×102
3 3 0
u v
 
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

Lecture Notes
 
1 4
1
3
4
1 1
2000
1 1
v v
v
K
v

 
  
 
 

 
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.

Lecture Notes
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.
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
stiffness matrix method.

Lecture Notes
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.
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

Lecture Notes
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 stiffness matrix method.
Example: Analyze the two member truss shown in figure using stiffness matrix
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 stiffness matrix method. Assume A =
2000 mm2
and E = 200 GPa.

Lecture Notes
Unit-IV
Stiffness Matrix Method for Beams
Stiffness Matrix Method for 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 among all
these methods Stiffness Matrix Method is program oriented method.
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
, ,
 
 )

Lecture Notes
Structure Approach
Equation of equilibrium
      
D DL
A A K D
 
where
 
D
A =Action corresponding to unknown displacements in the original structure
 
DL
A  Action corresponding to unknown displacements in the restrained
structure (If sinking is given, add sinking moments also in this vector)
 
K  Stiffness matrix
 
D Vector of unknown displacements
Steps for the solution of continuous (Indeterminate) beams using stiffness
matrix method:
1. Identify total nonzero degrees of freedom
2. Write  
D
A matrix
3. Restrained the structure, determine fixed end moments and reactions (reactions
are due to external load + reaction due to moments)
4. write  
DL
A matrix
5. Derive stiffness matrix [K] by giving unit displacements one by one
6. Apply equation of equilibrium and determine unknown joint displacements.
7. Determine unknown moments and reactions using following equation
      
M ML MD
A A A D
 
where
 
M
A =Unknown moments and reactions in the original structure
 
ML
A  Values of unknown moments and reactions in the restrained structure (If
sinking is given, add sinking moments and reactions also)
 
MD
A  Values of unknown moments and reactions in the unit displacement
figures
 
D Vector of unknown displacements which are calculated in previous step
Note: 1) Action corresponding to translation is reaction
2) Action corresponding to rotation is moment

Lecture Notes
Restrained Structures (Fixed end moments and reactions)
In stiffness matrix 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.
Uniformly distributed load
Central point load
Eccentric point load
Example 1: Analyse the beam as shown in figure using structure approach of
stiffness matrix method. Take EI = constant.
Solution:
I) Dki=02 ( B C
,
  )

Lecture Notes
II)  
0
0
B
D
C
A


 
  
 
(No moments acting at joint B and joint C)
III) Restrained Structure
 
75 50 25
50 50
B
DL
C
A


  
   
 
   
 
   
(Values of moments at B and C)
IV) Derivation of Stiffness matrix
1
B
Let  
Moment at B = 1.33 EI + EI = 2.333 EI
Moment at C = 0.5 EI
1
C
Let  
Moment at B = 0.5 EI
 
1 1
2 333 0
K = EI
5
0 5 1 0
B C
B
C
. .
. .
 


 
 
 
 
V) Equation of Equilibrium
      
D DL
A A K D
 

Lecture Notes
0 25 2 333 0 5
0 50 0 5 1 0
. .
. .


  
     
 
     
 

      
B
C
EI
B C
50
0 0 and
EI
.
 
 
VI) Moment Calculations
      
M ML MD
A A A D
 
75 0 67 0 75
75 1 33 0 0 75
1
50 1 0 0 5 50 75
50 0 5 1 0 0
AB
BA
BC
CB
M .
M .
EI
M . . EI
M . .
       
       
 
 
     
 
  
       
  
     
 
     

     
 
Example 2: Analyse the beam using structure approach of stiffness matrix method
if support B sink by 25mm. Take EI = 3800 kN.m2
Solution:
I) Dki=02 ( B C
,
  )
II)  
0
0
B
D
C
A


 
  
 
(No moments acting at joint B and joint C)
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.

Lecture Notes
2 2
6 6 3800 0.025
Sinking moments 15.833 .
6
EI
kN m
L
  
  
 
30 26 67 15 833 15 833 3 333
13 38 15 833 29 213
B
DL
C
. . . .
A
. . .


    
   
 
   
  
   
1
B
Let  
Moment at B = 0.67 EI + 0.67 EI = 1.333 EI
1
C
Let  
 
1 1
1 333 0 333
0 333 0
= EI
7
K
6
B C
B
C
. .
. .
 


 
 
 
 
      
D DL
A A K D
 
0 3 333 1 333 0 333
0 29 213 0 333 0 67
. . .
. . .


  
     
 
     
 

      
B
C
EI

Lecture Notes
B C
9 45 48 189
EI EI
. .
 

 
      
M ML MD
A A A D
 
30 15 833 0 33 0 42 714
75 15 833 0 67 0 9 45 20 491
1
26 67 15 833 0 67 0 33 48 189 20 491
13 38 15 833 0 33 0 67 0
AB
BA
BC
CB
M . . .
M . . . .
EI
M . . . . . .
EI
M . . . .

       
       
   
 
     
 
  
       
 
  
     
 
     
 
     
 
Example 3: 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 structure approach of stiffness matrix method.
Solution:
I) Dki=02 ( B C
,
  )
II)  
0
0
B
D
C
A

 
  

 
(No moments acting at joint B and point load at C)
 
20
10
B
DL
C
Moment at B
A
Reaction at C


 
  

 

Lecture Notes
1
B
Let  
Reaction at C = -0.09375 EI
1
C
Let  
Moment at B = -0.0937 EI
Reaction at C = 0.0234 EI
 
1 1
1 0 0 0937
0 0937 0 02
K =
4
EI
3
B C
B
C
. .
. .


  

 
  

 
      
D DL
A A K D
 
0 20 1 0 0 0937
0 10 0 0937 0 0234
. .
. .

   
     
 
     
  

      
B
C
EI
B C
32 078 555 8
and
EI EI
. .

 
  
      
M ML MD
A A A D
 
40 0 25 0 31 98
40 0 5 0 32 078 56 04
1
20 0 5 0 0937 555 8 56 04
20 0 25 0 0937 24 05
AB
BA
BC
CB
M . .
M . . .
EI
M . . . .
EI
M . . .
       
       
  
 
     
 
  
       
 
 
 
     
 
     
 
     
 

Lecture Notes
Example 4: Analyze the continuous beam using structure approach of stiffness
matrix method. Take EI constant
Solution:
I) Dki=03 ( B C C
, ,
   )
(Clockwise moments acting at joint B, no moment and point load at C)
II)  
30
0
0
B
D C
C
A



 
 
  
 
 
 
20
0
0
B
DL C
C
Moment at B
A Moment at C
Reaction at C



 
 
  
 
 
1
B
Let  
1
C
Let  

Lecture Notes
1
C
Let  
Moment at B = -1.5 EI
Moment at C = -1.5 EI
Reaction at C = 2.5 EI
 
3 0 1 0 1 5
K = EI 1 0 2 0 1 5
1 5 1 5 2
1
5
1 1
B C C
B
C
C
. . .
. . .
. . .
 



 
 







 
 
 

      
D DL
A A K D
 
3 0 1 0 1 5
1 0 2
30 20
0 0 0 1 5
1 5 1 5 2 5
0 0
. . .
. . .
. . .


   
   
     
 


 
 
    
 

   

     
 
 
 
 
B
C
C
EI
B C C
4 782 2 608 0 434
and
EI EI EI
. . .
,
 
 
   
      
M ML MD
A A A D
 
20 0 5 0 0 17 60
4 782
20 1 0 0 0 24 782
1
0 434
0 2 0 1 0 1 5 5 218
2 608
0 1 0 2 0 1 5 0
AB
BA
BC
CB
M . .
.
M . .
EI .
M . . . .
EI
.
M . . .
       

 
       
 
       
 
  
       
 
 
       

  
     

     
 
Note: Spring support is provided at support C, spring force developed due to
deformation of spring is added in the reaction at C

Lecture Notes
Example 5: Analyze the indeterminate beam as shown in figure using structure
approach of stiffness matrix method. The beam is fixed at A, C and has internal
hinge at B. Take EI constant.
Solution:
I) Dki=03 ( B BA BC
, ,
 
 )
II)  
0
0
0
B
D BA
BC
A 


 
 
  
 
 
 
90
5
20
B
DL BA
BC
Reaction B
A Moment BA
Moment BC



 
 
 
 
 
 
1
B
Let  
Reaction at B = 12 EI + 1.5 EI = 13.5 EI
Moment BA = -6.0 EI
Moment BC = 1.5 EI

Lecture Notes
1
BA
Let  
Reaction at B = -6.0 EI
Moment BA = 4.0 EI
Moment BC = 0
1
BC
Let  
Reaction at B = 1.5 EI
Moment BA = 0
Moment BC = 2 EI
 
13 5 6 0 1 5
K = EI 6 0 4 0 0
1 5 0
1
0
1
2
1
B BA BC
B
BA
BC
. . .
. .
. .
 


  

 
 









      
D DL
A A K D
 
13 5 6 0 1 5
6 0 4 0 0
1 5 0 2 0
0 90
0 5
0 20
. . .
. .
. .



 
   
     
  
     
     
   

 
 

 
 



B
BA
BC
EI
BA BC
20 28 75 5
and
EI EI EI
B
.
,  
 
   

Lecture Notes
Exercise
Example: For the following beam, find the vertical deflection and rotation at joint
B using structure approach of Stiffness Matrix Method. Take EI = 12×103
kN.m2
Example: Determine the unknown joint displacements of the beam as shown in
figure using structure approach of Stiffness Matrix Method. Take EI constant.
Example: Analyse the beam using structure approach of Stiffness Matrix 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 structure approach of Stiffness Matrix
Method and draw SFD and BMD. Take E = 200 GPa and I=4×106
mm4
.
Example: Analyze the continuous beam ABC as shown in Figure using structure
approach of Stiffness Matrix Method. Take EI constant.

Lecture Notes
Example: Analyse the prismatic beam ABC loaded and supported as shown in
Figure using structure approach of stiffness matrix method. Support B is sink by 25
mm. Draw SFD and BMD. Take EI =3500 kN.m2
.
Example: Determine support reactions of continuous beam ABC if support B sink
by 10 mm. Take EI = 6000 kN.m2
. Use structure approach of Stiffness Matrix
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 structure
approach of Stiffness Matrix Method.

Lecture Notes
Member Approach
Steps for the solution of continuous (Indeterminate) beams using member
approach:
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
K
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

Lecture Notes
Example 1: Analyse the beam as shown in figure using member approach of
stiffness matrix 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

Lecture Notes
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)
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.

Lecture Notes
 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
       
       
       
       
  
       

       
       

       

     
 

Lecture Notes
Example 2: Analyse the continuous beam as shown in figure using member
approach of stiffness matrix 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
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

Lecture Notes
 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
. . . .
. . . .
. . . .
. . . .

 
 

 
 
  
 

 
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
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

Lecture Notes
     
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
.
.
.
&
. .

 
  
     
      
   

     
     
     
  
   
 

  
 
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.
 
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
   
   
   
   

   
   
   
   
 
 

Lecture Notes
Example 3: Analyse the beam using member approach of stiffness matrix 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)
Discretization
 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

Lecture Notes
 
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




   
 
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
  
  

Lecture Notes
   
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
.
.
.
.
.
.

 
  
 
  
 

 
  
 

 
 
F q
 
3 33 4
F
29 163 6
.
.
 
  
 
    
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
.
.
.
. .
.
. . . .
     
     
     
     
   

     
 
     
     
     

     
     

Lecture Notes
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 member approach of stiffness matrix 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
Discretization
2 2-3 3,4,5,6 3=zero (simple support)
6=zero (guided support)
 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
. . . .
. . . .
. . . .
. . . .

 
 

 
 
  
 
 

Lecture Notes
 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
. . . .
. . . .
. . . .
. . . .

 
 

 
 
  
 
 
 
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


   
1 = 2 = zero (Fixed support), 3 = zero (simple supports), 6 = zero (guided support)
 
4 5
1 0 0 0937 4
0 0937 0 0234 5
K = EI
. . ,
. . ,


 
  

 
B
C
 AB
20 1
40 2
20 3
40 4
q

 
  
 
 

 
 

 
 &  BC
10 3
20 4
10 5
20 6
q

 
  
 
 
 
 
 
 


Lecture Notes
 
F q
 
20 4
10 5



 

 
f
    
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 member approach of stiffness
matrix method. Take EI constant
Solution: Step 1: Degrees of Freedom: 06
DOF at point C are 02 (rotation and translation due to spring).
Discretization
2 2-3 3,4,5,6 3=zero (simple support)
3 3-4 5, 8 8=zero (spring fixed at bottom)

Lecture Notes
 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

 
 

 
 
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
. . . .
. . . .
. . . . . .
. . . . . .
. . . .
. . . .

 
 

 
 
  
 

 
 
  


 

 

  
1 = 2 = zero (Fixed support), 3 = zero (simple supports)
 
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
. . . ,
. . . ,
. . . ,



 
 
  
 
 

 

Lecture Notes
 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

 
  
 
  
 

 
 
 
 

 
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
    
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
. . .
,
 
 
   
Example 5: Analyze the indeterminate beam as shown in figure using member
approach of stiffness matrix method. The beam is fixed at A, C and has internal
hinge at B. Take EI constant.

Lecture Notes
Solution: Step 1: Degrees of Freedom: 07
DOF at point B are 03 (two rotations and translation).
Discretization
1 1-2 1,2,3,4 1=2= zero (Fixed support)
2 2-3 3,5,6,7 6=7=zero (Fixed support)
 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
. . . .
. . . .
. . . .
. . . .

 
 

 
 
   
 
 
 
 

Lecture Notes
 
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
. . .
. .
. ,
.
,
,


 
 

 



 
 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
 
 
 
 
 
 
 
F q
 
90 3
f 5 4
20 5

 
 
  
 

 
    
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
.
,  
 
   

Lecture Notes
Example: For the following beam, find the vertical deflection and rotation at joint
B using member approach of stiffness matrix method. Take EI = 12×103
kN.m2
Example: Determine the unknown joint displacements of the beam as shown in
figure using member approach of stiffness matrix method. Take EI constant.
Example: Analyse the beam using member approach of stiffness matrix 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 member approach of stiffness matrix
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 member approach
of stiffness matrix 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 member

Lecture Notes
Example: Analyse the beam ABC shown in Figure 1 using member approach of
stiffness matrix 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 member approach of stiffness matrix method. Take EI
= constant
Example: Obtain fixed end moment at support A using member approach of
stiffness matrix 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 member approach of stiffness matrix
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 member
approach of stiffness matrix method.

Lecture Notes
Unit-V
Stiffness Matrix Method for Frame
Structure Approach
 The plane frame is a combination of plane truss and beam. All members are
connected by rigid joints in case of frame.
 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
 .
Example 1: Analyze the portal frame as shown in figure using Structure approach
of stiffness matrix method. Take EI constant.
Solution:
I) Dki=03 ( B C
, ,
  )
II)  
0
0
0
B
D C
A


 
 
  
  
 
(No moments and horizontal force acting at joint B and C)
10.4 Moment at B
{ } 1.6 Moment at C
4.72 Sum of Horizontal Reaction at B and C
B
DL C
A



 
 
 
 
 
 
 

Lecture Notes
1
B
Let   1
C
Let  
Moment at B = 1.6EI + 4.0EI = 5.6 EI Moment at B = 2.0 EI
Moment at C = 2.0 EI Moment at C = 4.0EI+0.8EI=4.8EI
Reaction at B = 0.24 EI Reaction at C = 0.24 EI

Lecture Notes
1
Let  
Reaction at B+C=0.048EI+0.096EI=0.144EI
 
5 6 2 0 0 24
K = EI 2 0 4 8 0 24
0 24 0 24 0 144
1 1 1
B C
B
C
. . .
. . .
. . .
 


 
 
 
  




 
      
D DL
A A K D
 
5 6 2 0 0 2
0 10 4
0 1 6
0 4
4
2 0 4 8 0 24
0 24 0 24 0 4
2 4
7 1
B
C
. . .
EI .
.
. . .
.
. . .



     
     
  
     
     
 
    
 
 

 
  

1.0419 1.803 34.046
; ;
B C
rad rad m
EI EI EI
 
    
9 6 0 8 0 0 24 18 604
14 4 1 6 0 0 24 4 562
1 0419
4 4 2 0 4 562
1
1 803
4 2 4 0 9 128
34 046
2 4 0 0 8 0 24 9 128
3 6 0 0 4 0 24 3 849
AB
BA
BC
CB
CD
DC
M . . . .
M . . . .
.
M .
EI .
M .
EI
.
M . . . .
M . . . .
      
      
 
      
 
      
  
      
 
 
     
 
 
     
     

   
 



 

 
 
 
 

Lecture Notes
Example 2: Analyze the rigid frame by using Structure approach of stiffness
matrix method. Take EI constant.
Solution:
I) Dki=03 ( B C
, ,
  )
II)  
0
0
50
B
D C
A


 
 
  
  
 
(No moments acting at joint B and C, horizontal force at B)
75 Moment at B
{ } 75 Moment at C
0 Sum of Horizontal Reaction at B and C
B
DL C
A


 
 
 
 
  
 

Lecture Notes
1
B
Let   1
C
Let  
Moment at B = 2.0 EI Moment at B = 0.5 EI
Moment at C = 0.5 EI Moment at C = 2.0EI
Reaction at B = 0.375 EI Reaction at C = 0.375 EI
1
Let  
Reaction at B+C=0.375EI
 
2 0 0 5 0 375
1 1
K = EI 0 5 2 0 0 375
0 375 0 375 0 37
1
5
B C
B
C
. . .
. . .
. . .
 


 
 
 
 
 
   


Lecture Notes
      
D DL
A A K D
 
2 0 0 5 0 375
0 5 2 0 0 375
0
0 75
3
0 75
50 0 75 0 375 0 375
B
C
. . .
EI . . .
. . .


     
     
  
     
   
 
 
 
 

 

  
   
𝜽𝑩 =
−𝟕𝟖.𝟓𝟕𝟏
𝑬𝑰
𝒓𝒂𝒅 𝜽𝑪 =
𝟐𝟏.𝟒𝟐𝟖
𝑬𝑰
𝒓𝒂𝒅 ∆=
𝟏𝟗𝟎.𝟒𝟕𝟔
𝑬𝑰
𝒎
0 0 5 0 0 375 32 143
0 1 0 0 0 375 7 143
78 571
75 1 0 0 5 0 7 143
1
21 428
75 0 5 1 0 0 92 857
190 476
0 0 0 5 0 375 92 857
0 0 1 0 0 375 8
AB
BA
BC
CB
CD
DC
M . . .
M . . .
.
M . . .
EI .
M . . .
EI
.
M . . .
M . .
     
      
      
 
      
  
     
 
 
     
 
 
     
     
   
  2 143
.
 
 
 
 
 
 
 
 
 
Example 3: Determine the rotation of joint B, and the horizontal displacements of
joints B and C. Take EI constant.
Solution:
I) Dki=03 ( B C
, ,
  )
II)  
0
0
0
B
D C
A


 
 
  
  
 

Lecture Notes
13.33 Moment at B
{ } 40 Moment at C
40 Horizontal Reaction at B
B
DL C
A


 
 
 
 
 
 
 
1
B
Let  

Lecture Notes
1
C
Let  
1
Let  
 
2 0 0 5 0 375
K = EI 0 5 1 0 0
0 375 0 0 187
1 1 1
B C
B
C
. . .
. .
. .
 


 
 
 
 



  

      
D DL
A A K D
 
2 0 0 5 0 375
0
0 13 33
0 40
0 40
5 1 0 0
0 375 0 0 187
B
C
. . .
EI .
.
.
. .


     
     
  
     
     
 
 
 
 
 
 
     
113.77 96.88 442.05
; ;
B C
rad rad m
EI EI EI
 

   
26 67 0 5 0 0 375 135 55
113 77
26 67 1 0 0 0 375 25 33
1
96 88
40 1 0 0 5 0 25 33
442 05
40 0 5 1 0 0 0
AB
BA
BC
CB
M . . . .
.
M . . . .
EI .
M . . .
EI
.
M . .
       

 
       

       
 
  
       
  
       
  
     

     
 

Lecture Notes
Example 4: Analyze the rigid jointed portal frame shown in Figure using Structure
Solution:
I) Dki=03 ( B C
,
  )
II)  
0
0
B
D
C
A


 
  
 
 
4 0 Moment at B
20 Moment at C
B
DL
C
.
A



 
  

 

Lecture Notes
1
B
Let  
1
C
Let  
 
1 1
1 8 0 5
0 5 1 0
B C
B
C
. .
K EI
. .
 


 
 
  
 
      
D DL
A A K D
 
0 4 0 1 8 0 5
0 20 0 5 1 0
B
C
. . .
EI
. .


  
     
 
     
 

      
3.870 21.935
;
B C
rad rad
EI EI
 

 
36 0 4 0 34 45
24 0 8 0 3 870 27 09
1
20 1 0 0 5 21 935 27 09
20 0 5 1 0 0
AB
BA
BC
CB
M . .
M . . .
EI
M . . . .
EI
M . .
       
       
  
 
     
 
  
       
  
     
 
     

     
 

Lecture Notes
Exercise
Example 5: Determine global stiffness matrix of the frame ABC shown in figure
using Structure approach of stiffness matrix method. Take EI constant. Neglect
axial deformation.
Example 6: Analyse the frame shown in Figure using Structure approach of
stiffness matrix method and draw bending moment diagram. Neglect axial
deformation.
(Ans. B AB BA BC CB
1.2/ EI, M 0.8kN.m,M 1.6kN.m, M 18.4kN.m, M 14.8kN.m
      )
Example 7: Determine the unknown joint displacements of the portal frame as
shown in Figure using Structure approach of stiffness matrix method. Take EI
constant. Neglect axial deformation.

Lecture Notes
Example 8: Analyse the portal frame as shown in Figure using Structure approach
of stiffness matrix method. Neglect axial deformation.
shown in Figure using Structure approach of stiffness matrix method. Take EI
Example 11: Analyze the rigid jointed portal frame shown in Figure 6 using
Structure approach of stiffness matrix method. Take EI constant. Draw BMD.
Neglect axial deformation.

Lecture Notes
Member Approach
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

Lecture Notes
 
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
D
K'
D
AE / L AE / L
D
D

 
 

 
 

  

 
 
  
 

 
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.

Lecture Notes
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

Lecture Notes
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
AE / L AE / L
K
AE / L AE / L
 
  
 

 
 

  

 
 

 

 
 

Lecture Notes
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 member approach
of stiffness matrix 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
K
 

 

 

 
  
 

 
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
K
 
  
 

 

 

 

 

Lecture Notes
Example 1: Analyze the portal frame as shown in figure using member approach
of stiffness matrix 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
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)

Lecture Notes
 
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
 
 
 

 

 
   
 

 
 
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
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

Lecture Notes
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
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

Lecture Notes
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 member approach of stiffness matrix
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
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

Lecture Notes
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
   
 
 
 
 

 

 

 
 
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 


 
 

 
 
 

Lecture Notes
Step 3: Element Nodal Load Vector
0 1 37.5 5
0 3 6
{ } { }
4 37.5 8
0 6 9
0 10
0 12
{ }
7
9
75
0
75
0
0
AB BC
DC
q q
q
   
   
   
 
   
   
   
   
 
 
 
  
 
 



Reduced element nodal load vector
 
0 4
75 6
75 9
q
 
 
  
 

 
Step 4: Equivalent Load Vector
F q
  
 
0 50 50 4
75 0 75 6
75 0 75 9
F
     
     
    
     
     
     
0.375 0.375 0.375 50
0.375 2 0.5 75
0.375 0.5 2 75
B
C
EI 


 
   
   
   
   
 
   
 
   
 
𝜽𝑩 =
−𝟕𝟖.𝟓𝟕𝟏
𝑬𝑰
𝒓𝒂𝒅 𝜽𝑪 =
𝟐𝟏.𝟒𝟐𝟖
𝑬𝑰
𝒓𝒂𝒅 ∆=
𝟏𝟗𝟎.𝟒𝟕𝟔
𝑬𝑰
𝒎
Step 6: Moment Calculations
{f} = [K]{Δ}+{q}

Lecture Notes
Member AB
0
0.375 1 0.375 0.5 0 0 32.143
1
0.375 0.5 0.375 1 190.476 0 7.143
78.571
AB
BA
M
EI
M EI
 
 

       
 
  
       
 
 
     
   
 

 
Member BC
1 0.5 6 78.571 75 7.143
1
0.5 1 9 21.428 75 92.857
BC
CB
M
EI
M EI

         
  
       
   
       
 
Member DC
0
0.375 1 0.375 0.5 0 0 82.143
1
0.375 0.5 0.375 1 190.476 0 92.857
21.428
DC
CD
M
EI
M EI
 
 

       
 
  
       
 

     
   
 
 
Example 4: Determine global stiffness matrix of the frame ABC shown in figure
using member approach of stiffness matrix method. Take EI constant. Neglect
axial deformation.
Example 5: Analyse the frame shown in Figure using member approach of
stiffness matrix method and draw bending moment diagram. Neglect axial
deformation.
(Ans. B AB BA BC CB
1.2/ EI, M 0.8kN.m,M 1.6kN.m, M 18.4kN.m, M 14.8kN.m
      )

Lecture Notes
Example 6: Determine the rotation of joint B, and the horizontal displacements of
joints B and C. Take EI = 10 ×103
KN.m2
. Neglect axial deformations.
Example 7: Analyze the rigid jointed portal frame shown in Figure using member
approach of stiffness matrix method. Take EI constant. Neglect axial
deformation.
shown in Figure using member approach of stiffness matrix method. Take EI

Lecture Notes
Example 9: Analyse the portal frame as shown in Figure using member approach
of stiffness matrix method. Neglect axial deformation.
shown in Figure using member approach of stiffness matrix method. Take EI
Example 11: Derive the stiffness matrix of portal frame ABC as shown in figure
using member approach of stiffness matrix method. Neglect axial deformation.
Example 12: Analyze the rigid jointed portal frame shown in Figure 6 using
member approach of stiffness matrix method. Take EI constant. Draw BMD.
Neglect axial deformation.

Lecture Notes
Unit-VI
Stiffness Matrix Method for Grid
The property of grid member is basically a combination of 2D beam with twisting
effect. The plane frame is subjected to tangential load (loaded in its own plane)
whereas grid is subjected to load perpendicular to its plane. As a result twisting
effects are included in the grid analysis. Thus grid structures withstand bending
moment, shear force and twisting moment.
Degree Kinematic Indeterminacy (Degrees of Freedom)
Grid has three degrees of freedom at each joint i.e. one translation and two
rotations (bending and twisting)
Twisting stiffness
A force (torque) required to twist the member by unit radian is called as twisting
stiffness.
T G
J L


When 1
T ? 
 
GJ
T
L


Lecture Notes
Example 1: Analyze the grid structure as shown in figure using structure approach
of stiffness matrix method. Take GJ = 0.4 EI
Solution:
I) Dki = 03 ( Bz Bx By
, ,
 
 )
II)
 
70
0
0
Bz
D Bx
By
A 

 
 
 
  
 
 
III) Restrained structure
No member forces on members. Therefore all fixed end moments and reactions are
zero
 
0
0
0
Bz
DL Bx
By
A 


 
 
  
 
 
IV) Stiffness matrix derivation
1
 
Bz
Perpendicular direction
should approach the
observer

Lecture Notes
1

Bx

All moments are in x-direction
1

By

All moments are in y-direction
 
1 1 1
0.540 0.67 0.24
0.67 1.413 0
0.24 0 0.933
Bz Bx By
Bz
Bx
By
K EI
 


   
  
 
 
 
 
 

 
V) Equation of equilibrium
      
D DL
A A K D
 
70 0 0.540 0.67 0.24
0 0 0.67 1.413 0
0 0 0.24 0 0.933
Bz
Bx
By
EI 

 
   
     
     
 
  
     
 
     
 

       
Bz Bx By
EI EI EI
428.371 202.21 110.192
; ;
 
  
   
      
M ML MD
A A A D
 

Lecture Notes
0 0 0 08 0
0 0 24 0 0 4
0 0 0 08 0
428 371
0 0 24 0 0 8
202
0 0 67 1 33 0
0 0 0 0 133
0 0 67 0 67 0
0 0 0 0 133
ABx
ABy
BAx
BAy
BCx
BCy
CBx
CBy
M .
M . .
M .
.
M . .
EI
M . .
M .
M . .
M .

     
     

     
     

     

     
  
     

     
     
     

     
    
 
   
 
16 176
58 732
16 176
14 655
1
21
16 176
110 192
14 655
150 84
14 655
.
.
.
.
.
.
EI
.
.
.
.
 
 
 
 

   
   

   
   

   

 
 
 
 
Example 2: Analyze the balcony grid as shown in figure using structure approach
of stiffness matrix method. Take EI = 1600 kN.m2
and GJ = 800 kN.m2
Solution:
, ,
 
 )
II)
 
0
0
0
Bz
D Bx
By
A 


 
 
  
 
 
 
80
20
40
Bz
DL Bx
By
A 


 
 
  
 
 

Lecture Notes
For BA member moments are in y-direction (member is along x-direction)
For BC member moments are in x-direction (member is along y-direction)
1
 
Bz
1

Bx


Lecture Notes
1

By

 
1 1 1
600 600 600
600 1800 0
600 0 1800
Bz Bx By
Bz
Bx
By
K
 


   

 
 
  
 
 
      
D DL
A A K D
 
0 80 600 600 600
0 20 600 1800 0
0 40 600 0 1800
Bz
Bx
By


 

     
     
 
 
     
 
     
 
      
0.3 ; 0.0888 ; 0.0777
Bz Bx By
m rad rad
 
    
      
M ML MD
A A A D
 
0 0 200 0
40 600 0 800
0 0 200 0
0 3
40 600 0 1600
0 0888
20 600 1600 0
0 0777
0 0 0 200
20 600 800 0
0 0 0 200
ABx
ABy
BAx
BAy
BCx
BCy
CBx
CBy
M
M
M
.
M
.
M
.
M
M
M

     
     

     
     

 
     
    
 
 
    
 
    
 
     
     

     
    
 
   
 
17 76
157 78
17 76
15 56
17 76
15 56
128 89
15 56
.
.
.
.
.
.
.
.

 
 

 
 
 

  

  

  
  
 

 
 

 

Lecture Notes
Example 3: Using structure approach of stiffness matrix method, determine
unknown joint displacements of the grid as shown in figure. E=2×105
MPa, I =
20×105
mm4
, G = 0.8×105
MPa, J =50×105
mm4
Solution:
5 5 2
2 10 20 10 400
EI kN.m
    
5 5 2
0 8 10 50 10 400
GJ . kN.m
    
, ,
 
 )
II)
 
0
0
0
Bz
D Bx
By
A 


 
 
  
 
 
 
6
0
3
Bz
DL Bx
By
A 


 
 
  
 

 

Lecture Notes
1
 
Bz
1

Bx

1

By

 
1 1 1
777.78 600 266.67
600 933.33 0
266.67 0 733.33
Bz Bx By
Bz
Bx
By
K
 


   
 
 
 
  
 

 
      
D DL
A A K D
 

Lecture Notes
 
 
     
 
     
 
     
 
     
 
 
      
0 6 777.78 600 266.67
0 0 600 933.33 0
0 3 266.67 0 733.33
Bz
Bx
By


0.0166 ; 0.0106 ; 0.00195
Bz Bx By
m rad rad
 
    
      
M ML MD
A A A D
 
0 600 400 0
0 0 0 200
0 600 800 0
0 0166
0 0 0 200
0
0 0 133 33 0
3 266 67 0 533 33
0 0 133 33 0
3 266 67 0 266 67
ABx
ABy
BAx
BAy
BCx
BCy
CBx
CBy
M
M
M
.
M
.
M .
M . .
M .
M . .
     
     

     
     

     
     
 
     
     
   
 
 
     

     
    
 
   
 
5 72
0 39
1 48
0 39
0106
1 41
0 00195
0 39
1 41
6 90
.
.
.
.
.
.
.
.
.

 
 
 
 

   

   

   
   

   
 

 
 
 
Example 5: Analyze the grid structure ABC as shown in Figure using Stiffness
matrix method. Take EI=2×105
kN.m2
and GJ = 1.2×105
kN.m2
.

Lecture Notes
Example 6: Orthogonal grid ABC is in x-z plane. It consists of two prismatic
members having same EI and GJ as well as length ‘L’. Develop stiffness matrix for
grid.
Example 7: Analyze the grid structure ABC as shown in Figure using structure
approach of stiffness matrix method. Take E = 210 GPa, G = 84 GPa, I = 16.6 X10-
5
m4
, J = 4.6 X10-5
m4
for all elements.

Lecture Notes
Member Approach
Stiffness matrix for grid element
The total degrees of freedom at each node of the grid are 03 (vertical displacement,
bending rotation and twisting rotation). Therefore size of stiffness matrix is 6×6.
D1 = Displacement in y-direction at node 1
D2= Rotation in x-direction at node 1
D3 = Rotation in z-direction at node 1
D4 = Displacement in y-direction at node 2
D5= Rotation in x-direction at node 2
D6 = Rotation in z-direction at node 2
To derive the stiffness matrix, give the unit displacements at each node one by one
Unit rotation in x-direction at node 1
Moments in this figure are about
z-axis, no moment about x-axis.
Reactions are along y direction
Moments in this figure are about x-
axis i.e. twisting moment. Reactions
along y direction and moments along
z-direction are zero

Lecture Notes
Unit rotation in x-direction at node 2
Stiffness matrix for grid element in local coordinate system from above six figures
 
EI L EI L EI L EI L
GJ L GJ L
EI L EI L EI L EI L
K
EI L EI L EI L EI L
GJ L GJ L
EI L
1 2 3 4 5 6
3 2 3 2
2 2
3 2 3 2
2
D D D D D D
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 /




  

EI L EI L EI L
2
0 2 / 6 / 0 4 /
 
 
 
 
 
 
 
 

 
 
Transformation Matrix of grid Element
In grid the members are orthogonally 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.
Moments in this figure are about x-
axis i.e. twisting moment. Reactions
along y direction and moments along
z-direction are zero

Lecture Notes
Let consider a grid element at an angle θ with respect to positive x-axis in
clockwise sense.
𝐷1
′
, 𝐷2
′
and 𝐷3
′
= D.O.F. at each node for local co-ordinate system
D1, D2 and D3 = D.O.F. at each node for global co-ordinate system
Let the local DOF be expressed into global DOF
At Node 1
'
1 1
'
2 2 3
'
3 2 3
cos sin
sin cos
D D
D D D
D D D
 
 

 
  
At Node 2
'
4 4
'
5 5 6
'
6 5 6
cos sin
sin cos
D D
D D D
D D D
 
 

 
  
In matrix form: (l = cosθ and m = sinθ are direction cosines)

Lecture Notes
 
'
1
1
'
2
2
'
3
3
'
4
4
'
5
5
'
6
6
1 0 0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 0 0 0
0 1 0 0 0 0 1 0 0 0
0 0 0 1 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
D
D
D
l m l m
D
D
m m
D
L
D
D
D
l m l m
D
D
m l m l
D
   
   
   
   
   
   
   
   
 
 
  
   
   
   
   
   
   
   
   
 
     
 
 
    
'
x L x

[L] = Transformation Matrix
 
'
x = Local Displacement Vector
 
x =Global Displacement Vector
The stiffness matrix of member is global coordinate system is obtained by
using relation (θ=900
, l = 0, m = 1)
[𝐾] = [𝐿]𝑇[𝐾′][𝐿]
 
3 2 3 2
2 2
1
3 2 3 2
2 2
12 6 12 6
0 0
0 0 0 0
1 0 0 0 0 0
0 0 1 0 0 0 6 4 6 2
0 0
0 1 0 0 0 0
0 0 0 1 0 0 12 6 12 6
0 0
0 0 0 0 0 1
0 0 0 0 1 0 0 0 0 0
6 2 6 4
0 0
T
EI EI EI EI
L L L L
GJ GJ
L L
EI EI EI EI
L L L L
L K
EI EI EI EI
L L L L
GJ GJ
L L
EI EI EI EI
L L L L




 
  
  

   
  
   
 
  
    
 
 
  
 
  




















 

3 2 3 2
2 2
3 2 3 2
2 2
12 6 12 6
0 0
6 4 6 2
0 0 1 0 0 0 0 0
0 0 1 0 0 0
0 0 0 0
0 1 0 0 0 0
12 6 12 6 0 0 0 1 0 0
0 0
0 0 0 0 0 1
6 2 6 4 0 0 0 0 1 0
0 0
0 0 0 0
EI EI EI EI
L L L L
EI EI EI EI
L L L L
GJ GJ
L L
EI EI EI EI
L L L L
EI EI EI EI
L L L L
GJ GJ
L L
 

 
 
 
   
 

  

 
 
 
 
 
 
  

 
 

  
 
 
 

 
 




 
 
 
 
 


Lecture Notes
 
EI L EI L EI L EI L
EI L EI L EI L EI L
GJ L GJ L
K
EI L EI L EI L EI L
EI L EI L EI L EI L
GJ L GJ L
3 2 3 2
2 2
3 2 3 2
2 2
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 / 0
0 0 / 0 0 /
 
  
 

 
 

  

 
 

 

 
 
Example 1: Orthogonal grid is in xz plane. It consists of two prismatic members
having same EI and GJ as well as length L. Develop stiffness matrix for grid using
member approach of stiffness matrix method.
Degrees of freedom and boundary conditions
Solution: Total DOF: 09 (03 at each node)
Step 1: Discretization
1 AB 1,2,3,4,5,6 1=2=3=zero
2 BC 4,5,6,7,8,9 7=8=9=zero
Note: First select the element to which standard stiffness matrix is applicable.
a) Direction of element must be towards right
b) Perpendicular direction must approach the observer
Stiffness matrix for member AB: (Since standard element is along x-axis,
assume twisting rotation along x and bending rotation along z)

Lecture Notes
 
3 2 3
2 2
3 2 3 2
2
1 2 3 4 5 6
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
AB
EI L EI L EI L EI L
GJ L GJ L
EI L EI L EI L
EI L E
EI
I L
L
K
EI L EI L
G L GJ
J




 


2 2
1
2
3
4
5
6
6 / 0 2 /
/ 0
6 4
0 /
/
L
EI L E
EI L EI L I L

 
 

 
  
 
 
 
 
 
 
 


Stiffness matrix for member CB:
Rotate member CB in clockwise
direction about joint C at an angle of 900
w. r. t to positive x-axis. (θ=900
, l=0,
m=1). Therefore take ‘C’ as a first node
and ‘B’ as a second node.
 
3
3
2 3 2
2 2
3 2 2
7 8 9 4 5 6
12 / 6 / 0 12 / 6 / 0
6 / 4 / 0 6 / 2 / 0
0 0 / 0
1
0 /
12 2 / 6 / 0
/ 6 / 0
CB
EI L EI L EI L EI L
EI L EI L EI L EI L
GJ L GJ L
K
EI EI L
L EI EI
L L
  




2 2
7
8
9
4
5
6 / 2 / 0
6
0 0 /
6 / 4 / 0
0 /
0
EI L EI L
GJ L
EI L EI L
GJ L

 
 

 
  
 
 
 

 

 
 
  
III) Reduced stiffness matrix
     
   
3 2 2
2
2
4 5 6
24 / 6 / 6 / 4,
6 / 4 / / 0 5,
6 / 0 4 / / 6,
Bz
Bx
By
EI L EI L EI L
K EI L EI L GJ L
EI L EI L GJ L


 
 
 
 
 
 
 
 

Lecture Notes
Example 2: Analyze the grid structure as shown in figure using member approach
of stiffness matrix method. Take GJ = 0.4 EI
Solution: Total DOF: 09 (03 at each node)
No. of elements: 02 (AB and CB)
1 AB 1,2,3,4,5,6 1=2=3=zero
2 CB 4,7,8,9,5,6 7=8=9=zero
Note: First select the element to which standard stiffness matrix is applicable.
a) Direction of element must be towards right
b) Perpendicular direction must approach the observer
Standard stiffness matrix for member AB: (Since standard element AB is along
x-axis, assume twisting rotation along x and bending rotation along y)
Stiffness matrix of element AB (Using standard stiffness matrix)
 
1 2 3 4 5 6
0.096 0 0.24 0.096 0 0.24
0 0.08 0 0 0.08 0
0.24 0 0.8 0.24 0 0.4
0.096 0 0.24 0.096 0 0.24
0 0.08 0 0 0.08 0
0
AB
K EI




  

1
2
3
4
5
.24 0 0.4 0.24 0 0.8 6

 
  
 
  
 
 
 
 

 
  

Lecture Notes
Stiffness matrix for member BC:
Rotate member BC in clockwise
direction about joint C at an angle of 900
w. r. t to positive x-axis. (θ=900
, l=0,
m=1). Therefore take ‘B’ as a first node
and ‘C’ as a second node.
 
4 5 6 7 8 9
0.444 0.667 0 0.444 0.667 0
0.667 1.333 0 0.667 0.667 0
0 0 0.133 0 0 0.133
0.444 0.667 0 0.444 0.667 0
0.667 0.6
BC
K EI
  





4
5
6
7
67 0 0.667 1.333 0 8
0 0 0.133 0 0 0.133 9
 
 
 
 
 

 
  
 
 
 
  
Reduced stiffness matrix is
 
4 5 6
0.540 0.667 0.24 4,
0.667 1.413 0 5,
0.24 0 0.933 6,
Bz
Bx
By
K EI 

  
 
 
 
 
 

 
Step 3: Element nodal load vector
Since there is no load on the members, fixed end moments and corresponding
reactions are zero. Therefore element nodal load vector is null matrix.
     
0 1 0 4
0 2 0 5
0 4
0 3 0 6
0 5
0 4 0 7
0 6
0 5 0 8
0 6 0 9
AB BC
q and q q
   
   
     
     
   
     
     
 
   
   
   

Lecture Notes
f q
  
 
0 70 70 4
0 0 0 5
0 0 0 6
f
 
     
     
  
     
     
     
[K]{Δ} = {f}
0.540 0.667 0.24 70
0.667 1.413 0 0
0.24 0 0.933 0
Bz
Bx
By
EI 

 
   
   
   
 
 
   
 
   
 

   
 
Bz Bx By
EI EI EI
428.371 202.21 110.192
; ;
 
  
   
Element AB
       
AB AB AB
AB
K q f
  
0
0 0.08 0 0 0.08 0 0 0
0.24 0 0.8 0.24 0 0.4 0 0
1
0 0.08 0 0 0.08 0 428.371 0
0.24 0 0.4 0.24 0 0.8 202.21 0
110.192
Ax
Ay
Bx
By
M
M
EI
M
EI
M
 
 
  
   
   
   
 
    
   
     
 
 
     
      
 
     
 

 
Element BC
428.371
0.667 1.333 0 0.667 0.667 0 202.21 0
0 0 0.133 0 0 0.133 110.192 0
1
0.667 0.667 0 0.667 1.333 0 0 0
0 0 0.133 0 0 0.133 0 0
0
Bx
By
Cx
Cy
M
M
EI
M
EI
M

 
 
   
   
   
   
 
     
   
     
 
      
      

     
 
 

Lecture Notes
Ax
Ay
Bx
By
M
M
M
M
16.176
58.732
16.176
14.655
   
   
   

   

   
   
 
 
and
Bx
By
Cx
Cy
M
M
M
M
16.176
14.655
150.84
14.655
   
   

   

   
   
   
 
 
Example 3: Analyze the balcony grid as shown in figure using member approach
of stiffness matrix method. Take EI = 1600 kN.m2
and GJ = 800 kN.m2
Solution:
Step 1: Total DOF = 09 (03 at each node)
No. of elements: 02 (AB and BC)
1 AB 1,2,3,4,5,6 1=2=3=zero
2 BC 4,7,8,9,5,6 7=8=9=zero
Note:
1) According to standard derivation, element BC is satisfying conditions of
standard derivation (direction along the member is along right and
perpendicular direction approaching the observer). Therefore, standard
stiffness matrix is applicable to BC and 900
matrix is applicable to member
AB.
2) Since BC is along y-direction assume rotation in y-direction ( By
 ) as second
unknown and x-direction rotation ( Bx
 ) as third unknown.
3) While measuring angle for second member, measure w.r.t. positive y-axis.

Lecture Notes
Stiffness matrix of element BC
 
4 5 6 7 8 9
300 0 600 300 0 600 4
0 200 0 0 200 0 5
600 0 1600 600 0 800 6
300 0 600 300 0 600 7
0 200 0 0 200 0 8
600 0 800 600 0 1600 9
BC
K

 
 

 
 

  
   
 
 
 
 
 
 
  
Stiffness matrix for member AB:
Rotate member AB in clockwise
direction about joint A at an angle of 900
w. r. t to positive y-axis. (θ=900
, l=0,
m=1). Therefore take ‘A’ as a first node
and ‘B’ as a second node.
 
1 2 3 4 5 6
300 600 0 300 600 0 1
600 1600 0 600 800 0 2
0 0 200 0 0 200 3
300 600 0 300 600 0 4
600 800 0 600 1600 0 5
0 0 200 0 0 200 6
AB
K
   
 
 
 
 
 
 
  

 
 

 

 
  
 
4 5 6
600 600 600 4,
600 1800 0 5,
600 0 1800 6,
Bz
By
Bx
K 


 
 
  
 
 

Lecture Notes
Step 3: Element nodal load vector: (Fixed end moments and reactions)
   
 
20 4 60 1
0 5 40 2
20 6 0 3
and
20 7 60 4
0 8 40 5
20 9 0 6
80 4,
40 5,
20 6,
BC AB
Bz
By
Bx
q q
R
q M
M

   
   
 
   
    
 
   

   
   

   
 
   
 
 
  
 
 
f q
  
 
80 4
40 5
20 6
f

 
 
 
 
 

 
[K]{Δ} = {f}
600 600 600 80
600 1800 0 40
600 0 1800 20
Bz
By
Bx


 
 
   
   
   
   
 
   
  
   
 
0.3 ; 0.0888 ; 0.0777
Bz Bx By
m rad rad
 
    
      
K q f
  
Element AB
0
600 1600 0 600 800 0 0 0
0 0 200 0 0 200 0 40
600 800 0 600 1600 0 0.3 0
0 0 200 0 0 200 0.0888 40
0.0777
Ay
Ax
By
Bx
M
M
M
M
 
 
  
   
 
 
   
 
    
   
     
 
 
     
      
 
     
 
 

Lecture Notes
0.3
0 200 0 0 200 0 0.0888 20
600 0 1600 600 0 800 0.0777 0
1
0 200 0 0 200 0 0 20
600 0 800 600 0 1600 0 0
0
By
Bx
Cy
Cx
M
M
EI
M
EI
M

 
 
  
   
 
 
   
 
    
   
     
 
 
     
      

     
 
 
Element BC
Ay
Ax
By
Bx
M
M
M
M
17.76
157.84
17.76
15.68
  
 
   

   

   
   
   

 
 
and
By
Bx
Cy
Cx
M
M
M
M
17.76
15.68
128.96
15.68
  
 
   

   

   

   
   

 
 
Example 4: Using member approach of stiffness matrix method, determine
unknown joint displacements of the grid as shown in figure. E=2×105
MPa, I =
20×105
mm4
, G = 0.8×105
MPa, J =50×105
mm4
Solution:
Step 1: Total DOF = 09 (03 at each node)
No. of elements: 02 (BA and BC)
EI = 2×105
× 20×105
= 40×1010
N.mm2
= 400 kN.m2
GJ = 0.8×105
× 50×105
= 40×1010
N.mm2
= 400 kN.m2

Lecture Notes
1 BA 4,5,6,1,2,3 1=2=3=zero
2 BC 4,7,8,9,5,6 7=8=9=zero
Note:
1) According to standard derivation, element BA is satisfying conditions of
standard derivation (direction along the member is along right and
perpendicular direction approaching the observer). Therefore, standard
stiffness matrix is applicable to BA and 900
matrix is applicable to member
BC.
2) Since BA is along y-direction assume rotation in y-direction ( By
 ) as second
unknown and x-direction rotation ( Bx
 ) as third unknown.
3) While measuring angle for second member, measure w.r.t. positive y-axis.
 
4 5 6 1 2 3
600 0 600 600 0 600 4
0 200 0 0 200 0 5
600 0 800 600 0 400 6
600 0 600 600 0 600 1
0 200 0 0 200 0 2
600 0 400 600 0 800 3
BA
K

 
 

 
 

  
   
 
 
 
 
 
 
  
Stiffness matrix for member BC:
Rotate member BC in clockwise
direction about joint B at an angle of 900
w. r. t to positive y-axis. (θ=900
, l=0,
m=1). Therefore take ‘B’ as a first node
and ‘C’ as a second node.

Lecture Notes
 
4 5 6 7 8 9
177.77 266.67 0 177.77 266.67 0
266.67 533.33 0 266.67 266.67 0
0 0 133.33 0 0 133.33
177.77 266.67 0 0.177 266.67 0
BC
K
  




4
5
6
7
266.6 266.67 0 266.67 533.33 0 8
0 0 133.3 0 0 133.33 9
 
 
 
 
 

 
 
  
 
 
 
  
 
 
 
 
 
 
 
 
Bz
By
BC
Bx
K
4 5 6
777.77 266.67 600 4
266.67 733.33 0 5
600 0 933.33 6


Step 3: Element nodal load vector: (Fixed end moments and reactions)
     
0 4 6 4
0 5 3 5
6 4
0 6 0 6
and 3 5
0 1 6 7
0 6
0 2 3 8
0 3 0 9
BA BC
q q q
   
   

     
     
    
     
     
 
   
   
   
f q
  

Lecture Notes
 
6 4
3 5
0 6
f

 
 
  
 
 
[K]{Δ} = {f}
 
  
   
   
 
 
   
 
   
 
   
 
Bz
By
Bx
777.77 266.67 600 6
266.67 733.33 0 3
600 0 933.33 0


0.0166 ; 0.00195 ; 0.0106
Bz By Bx
m rad rad
 
     
Example 6: Derive the stiffness matrix for the grid elements as shown in Figure
using member approach of stiffness matrix method. Take flexural rigidity EI and
torsional rigidity GJ same for both the elements
Example 7: Analyze the grid structure ABC as shown in Figure using member
approach of stiffness matrix method. Take EI=2×105
kN.m2
and GJ = 1.2×105
kN.m2
.

Lecture Notes
Example 8: Orthogonal grid ABC is in x-z plane. It consists of two prismatic
members having same EI and GJ as well as length ‘L’. Develop stiffness matrix for
grid using member approach of stiffness matrix method.
Example 9: Analyze the grid structure ABC as shown in Figure using member
approach of stiffness matrix method. Take E = 210 GPa, G = 84 GPa, I = 16.6 X10-
5
m4
, J = 4.6 X10-5
m4
for all elements.

Matrix Methods of Structural Analysis

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