Short Notes: Matrix Method of Analysis | Short Notes for Civil Engineering - Civil Engineering (CE) PDF Download

 Page 1


 
 
MATRIX METHOD OF ANALYSIS 
1. DISPLACEMENT METHOD/STIFFNESS METHOD: 
In this method displacements at the joints are taken as unknowns and equation are expressed 
in terms of these unknown displacement. Additional joint equilibrium equations are developed 
to find the unknown displacement. This method is suitable when the Kinematic indeterminacy 
is less than the static indeterminacy.  
1.1. Stiffness (k) 
It is the load required to produce unit displacement. Stiffness for various cases are as 
follows. 
 
(1) Axial stiffness (k11) = 
AE
l
 
(2) Transverse stiffness (k22) = 
3
12EI
l
 
(3) Flexural stiffness (k33) = 
4EI
l
 
(4) Torsional stiffness (k44) = 
GJ
l
 
1.2. Procedure to Construct Stiffness Matrix 
 To get first column of stiffness matrix, fix all the coordinates and give unit displacement 
at the 1
st
 coordinate and find forces developed at all other coordinates similarly to get the 
second column of stiffness matric apply unit displacement at coordinate 2 and find forces 
at all coordinates. 
 
The cantilever beam shown in the figure above will be subjected to three displacements 
(1), (2) and (3). 
When the unit displacement is given in direction of (1) i.e., horizontal deflection only, 
K11 = Force at (1) due to unit displacement at (1) = 
????
??  
Page 2


 
 
MATRIX METHOD OF ANALYSIS 
1. DISPLACEMENT METHOD/STIFFNESS METHOD: 
In this method displacements at the joints are taken as unknowns and equation are expressed 
in terms of these unknown displacement. Additional joint equilibrium equations are developed 
to find the unknown displacement. This method is suitable when the Kinematic indeterminacy 
is less than the static indeterminacy.  
1.1. Stiffness (k) 
It is the load required to produce unit displacement. Stiffness for various cases are as 
follows. 
 
(1) Axial stiffness (k11) = 
AE
l
 
(2) Transverse stiffness (k22) = 
3
12EI
l
 
(3) Flexural stiffness (k33) = 
4EI
l
 
(4) Torsional stiffness (k44) = 
GJ
l
 
1.2. Procedure to Construct Stiffness Matrix 
 To get first column of stiffness matrix, fix all the coordinates and give unit displacement 
at the 1
st
 coordinate and find forces developed at all other coordinates similarly to get the 
second column of stiffness matric apply unit displacement at coordinate 2 and find forces 
at all coordinates. 
 
The cantilever beam shown in the figure above will be subjected to three displacements 
(1), (2) and (3). 
When the unit displacement is given in direction of (1) i.e., horizontal deflection only, 
K11 = Force at (1) due to unit displacement at (1) = 
????
??  
 
 
K21 = Force at (2) due to unit displacement at (1) = 0 
K31 = Force at (3) due to unit displacement at (1) = 0 
When the unit displacement is given in direction of (2) i.e., vertical deflection only, 
 
K12 = Force at (1) due to unit displacement at (2) = 0 
K22 = Force at (2) due to unit displacement at (2) = 
2?? ?? ?? 3
 
K32 = Force at (3) due to unit displacement at (2) = -
6????
?? 2
 
When the unit displacement is given in direction of (3) i.e., rotation only, 
 
K13 = Force at (1) due to unit displacement at (3) = 0 
K23 = Force at (2) due to unit displacement at (3) = -
6????
?? 2
 
K33 = Force at (3) due to unit displacement at (3) = 
4????
?? 
So, the stiffness matrix is 
?? =
[
 
 
 
 
 
????
?? 0 0
0
2????
?? 3
-
6????
?? 2
0 -
6????
?? 2
4????
?? ]
 
 
 
 
 
 
2. FLEXIBILITY MATRIX METHOD: 
In this method, forces are taken as unknown and equations are expressed in terms of these 
forces. Additional equation called compatibility condition are developed to find all the unknown 
forces. This method is suitable when the static indeterminacy is less than kinematic 
indeterminacy. 
2.1. Flexibility (d) 
Flexibility is defined as the displacement produced due to unit force. It is the inverse of 
stiffness. Flexibility for various cases are as follows 
(a) Axial flexibility =
1
????
?? =
?? ????
 
(b) Transverse flexibility =
1
????
?? 3
=
?? 3
12????
 
Page 3


 
 
MATRIX METHOD OF ANALYSIS 
1. DISPLACEMENT METHOD/STIFFNESS METHOD: 
In this method displacements at the joints are taken as unknowns and equation are expressed 
in terms of these unknown displacement. Additional joint equilibrium equations are developed 
to find the unknown displacement. This method is suitable when the Kinematic indeterminacy 
is less than the static indeterminacy.  
1.1. Stiffness (k) 
It is the load required to produce unit displacement. Stiffness for various cases are as 
follows. 
 
(1) Axial stiffness (k11) = 
AE
l
 
(2) Transverse stiffness (k22) = 
3
12EI
l
 
(3) Flexural stiffness (k33) = 
4EI
l
 
(4) Torsional stiffness (k44) = 
GJ
l
 
1.2. Procedure to Construct Stiffness Matrix 
 To get first column of stiffness matrix, fix all the coordinates and give unit displacement 
at the 1
st
 coordinate and find forces developed at all other coordinates similarly to get the 
second column of stiffness matric apply unit displacement at coordinate 2 and find forces 
at all coordinates. 
 
The cantilever beam shown in the figure above will be subjected to three displacements 
(1), (2) and (3). 
When the unit displacement is given in direction of (1) i.e., horizontal deflection only, 
K11 = Force at (1) due to unit displacement at (1) = 
????
??  
 
 
K21 = Force at (2) due to unit displacement at (1) = 0 
K31 = Force at (3) due to unit displacement at (1) = 0 
When the unit displacement is given in direction of (2) i.e., vertical deflection only, 
 
K12 = Force at (1) due to unit displacement at (2) = 0 
K22 = Force at (2) due to unit displacement at (2) = 
2?? ?? ?? 3
 
K32 = Force at (3) due to unit displacement at (2) = -
6????
?? 2
 
When the unit displacement is given in direction of (3) i.e., rotation only, 
 
K13 = Force at (1) due to unit displacement at (3) = 0 
K23 = Force at (2) due to unit displacement at (3) = -
6????
?? 2
 
K33 = Force at (3) due to unit displacement at (3) = 
4????
?? 
So, the stiffness matrix is 
?? =
[
 
 
 
 
 
????
?? 0 0
0
2????
?? 3
-
6????
?? 2
0 -
6????
?? 2
4????
?? ]
 
 
 
 
 
 
2. FLEXIBILITY MATRIX METHOD: 
In this method, forces are taken as unknown and equations are expressed in terms of these 
forces. Additional equation called compatibility condition are developed to find all the unknown 
forces. This method is suitable when the static indeterminacy is less than kinematic 
indeterminacy. 
2.1. Flexibility (d) 
Flexibility is defined as the displacement produced due to unit force. It is the inverse of 
stiffness. Flexibility for various cases are as follows 
(a) Axial flexibility =
1
????
?? =
?? ????
 
(b) Transverse flexibility =
1
????
?? 3
=
?? 3
12????
 
 
 
(c) Flexural flexibility =
1
4????
?? =
?? 4????
 
(d) Torsional flexibility =
1
????
?? =
?? ????
 
2.2. Procedure to construct Flexibility Matrix 
To get the first column of flexibility matrix, apply unit force at coordinate (1) and find 
displacement at all coordinates in the released structure. Similarly, to get II column of 
the flexibility matrix apply unit force at coordinate (2) and find displacement at all 
coordinates in the released structure. 
 
The cantilever beam shown in the figure above is subjected unit forces in three 
directions. 
When the unit force is applied in direction of (1) 
d11 = displacement at coordinate (1) due to unit load at coordinate (1) =
?? ????
 
d21 = displacement at coordinate (2) due to unit load at coordinate (1) = 0 
d31 = displacement at coordinate (3) due to unit load at coordinate (1) = 0 
When the unit load is applied in the direction of (2) 
 
d12 = displacement at coordinate (1) due to unit load at coordinate (2) = 0 
d22 = displacement at coordinate (2) due to unit load at coordinate (2) = 
?? 3
3????
 
d32 = displacement at coordinate (3) due to unit load at coordinate (2) = -
?? 2
2????
 
When the unit load is applied in the direction of (3) 
 
d13 = displacement at coordinate (1) due to unit load at coordinate (3) = 0 
d23 = displacement at coordinate (2) due to unit load at coordinate (3) = -
?? 2
2????
 
d33 = displacement at coordinate (3) due to unit load at coordinate (3) = 
?? ????
 
Page 4


 
 
MATRIX METHOD OF ANALYSIS 
1. DISPLACEMENT METHOD/STIFFNESS METHOD: 
In this method displacements at the joints are taken as unknowns and equation are expressed 
in terms of these unknown displacement. Additional joint equilibrium equations are developed 
to find the unknown displacement. This method is suitable when the Kinematic indeterminacy 
is less than the static indeterminacy.  
1.1. Stiffness (k) 
It is the load required to produce unit displacement. Stiffness for various cases are as 
follows. 
 
(1) Axial stiffness (k11) = 
AE
l
 
(2) Transverse stiffness (k22) = 
3
12EI
l
 
(3) Flexural stiffness (k33) = 
4EI
l
 
(4) Torsional stiffness (k44) = 
GJ
l
 
1.2. Procedure to Construct Stiffness Matrix 
 To get first column of stiffness matrix, fix all the coordinates and give unit displacement 
at the 1
st
 coordinate and find forces developed at all other coordinates similarly to get the 
second column of stiffness matric apply unit displacement at coordinate 2 and find forces 
at all coordinates. 
 
The cantilever beam shown in the figure above will be subjected to three displacements 
(1), (2) and (3). 
When the unit displacement is given in direction of (1) i.e., horizontal deflection only, 
K11 = Force at (1) due to unit displacement at (1) = 
????
??  
 
 
K21 = Force at (2) due to unit displacement at (1) = 0 
K31 = Force at (3) due to unit displacement at (1) = 0 
When the unit displacement is given in direction of (2) i.e., vertical deflection only, 
 
K12 = Force at (1) due to unit displacement at (2) = 0 
K22 = Force at (2) due to unit displacement at (2) = 
2?? ?? ?? 3
 
K32 = Force at (3) due to unit displacement at (2) = -
6????
?? 2
 
When the unit displacement is given in direction of (3) i.e., rotation only, 
 
K13 = Force at (1) due to unit displacement at (3) = 0 
K23 = Force at (2) due to unit displacement at (3) = -
6????
?? 2
 
K33 = Force at (3) due to unit displacement at (3) = 
4????
?? 
So, the stiffness matrix is 
?? =
[
 
 
 
 
 
????
?? 0 0
0
2????
?? 3
-
6????
?? 2
0 -
6????
?? 2
4????
?? ]
 
 
 
 
 
 
2. FLEXIBILITY MATRIX METHOD: 
In this method, forces are taken as unknown and equations are expressed in terms of these 
forces. Additional equation called compatibility condition are developed to find all the unknown 
forces. This method is suitable when the static indeterminacy is less than kinematic 
indeterminacy. 
2.1. Flexibility (d) 
Flexibility is defined as the displacement produced due to unit force. It is the inverse of 
stiffness. Flexibility for various cases are as follows 
(a) Axial flexibility =
1
????
?? =
?? ????
 
(b) Transverse flexibility =
1
????
?? 3
=
?? 3
12????
 
 
 
(c) Flexural flexibility =
1
4????
?? =
?? 4????
 
(d) Torsional flexibility =
1
????
?? =
?? ????
 
2.2. Procedure to construct Flexibility Matrix 
To get the first column of flexibility matrix, apply unit force at coordinate (1) and find 
displacement at all coordinates in the released structure. Similarly, to get II column of 
the flexibility matrix apply unit force at coordinate (2) and find displacement at all 
coordinates in the released structure. 
 
The cantilever beam shown in the figure above is subjected unit forces in three 
directions. 
When the unit force is applied in direction of (1) 
d11 = displacement at coordinate (1) due to unit load at coordinate (1) =
?? ????
 
d21 = displacement at coordinate (2) due to unit load at coordinate (1) = 0 
d31 = displacement at coordinate (3) due to unit load at coordinate (1) = 0 
When the unit load is applied in the direction of (2) 
 
d12 = displacement at coordinate (1) due to unit load at coordinate (2) = 0 
d22 = displacement at coordinate (2) due to unit load at coordinate (2) = 
?? 3
3????
 
d32 = displacement at coordinate (3) due to unit load at coordinate (2) = -
?? 2
2????
 
When the unit load is applied in the direction of (3) 
 
d13 = displacement at coordinate (1) due to unit load at coordinate (3) = 0 
d23 = displacement at coordinate (2) due to unit load at coordinate (3) = -
?? 2
2????
 
d33 = displacement at coordinate (3) due to unit load at coordinate (3) = 
?? ????
 
 
 
So, the flexibility matrix is 
[?? ] =
[
 
 
 
 
 
 
?? ????
0 0
0
?? 3
3????
-
?? 2
2????
0 -
?? 2
2????
?? ????
]