Introduction: The Direct Stiffness Method - 3 GATE Notes | EduRev

Structural Analysis

GATE : Introduction: The Direct Stiffness Method - 3 GATE Notes | EduRev

 Page 1


Expressions of similar form as above may be obtained for all  members. The sum 
of all horizontal components of individual forces gives us the stiffness coefficient 
 and sum of all vertical component of forces give us the required stiffness 
coefficient . 
11
k
21
k
 
4
2
4
4
3
2
3
3
2
2
2
2
1
2
1
1
11
cos cos cos cos ? ? ? ?
l
EA
l
EA
l
EA
l
EA
k + + + =
 
 
i
i
i
i
l
EA
k ?
2
4
1
11
cos
?
=
=
      (23.12) 
 
i i
i
i
l
EA
k ? ? sin cos
21
?
=
     (23.13) 
 
In the next step, give a unit displacement along holding displacement along 
equal to zero and calculate reactions at 
2
u
1
u
E corresponding to unknown 
displacements and in the kinematically determinate structure. The 
corresponding reactions are denoted by and  as shown in Fig.23.4d. The 
joint 
1
u
2
u
12
k
22
k
E gets displaced to E ' when a unit vertical displacement is given to the joint 
as shown in the figure. Thus, the new length of the member E A ' is 
1 1
l l ? +
. 
From the geometry, the elongation 
1
l ?
is given by 
 
1 1
sin ? = ?l
         (23.14a) 
 
Thus axial force in the member along its centroidal axis  is 1
1
1
sin ?
l
EA
 (23.14b) 
 
Resolve the axial force in the member along and directions. Thus, 
horizontal component of force in the member 
1
u
2
u
E A ' is 1 1
1
1
cos sin ? ?
l
EA
      (23.14c)  
and vertical component of force in the member E A ' is  1
2
1
1
sin ?
l
EA
       (23.14d) 
 
In order to evaluate , we need to sum vertical components of forces in all the 
members meeting at joint 
22
k
E .Thus, 
                                                         
Page 2


Expressions of similar form as above may be obtained for all  members. The sum 
of all horizontal components of individual forces gives us the stiffness coefficient 
 and sum of all vertical component of forces give us the required stiffness 
coefficient . 
11
k
21
k
 
4
2
4
4
3
2
3
3
2
2
2
2
1
2
1
1
11
cos cos cos cos ? ? ? ?
l
EA
l
EA
l
EA
l
EA
k + + + =
 
 
i
i
i
i
l
EA
k ?
2
4
1
11
cos
?
=
=
      (23.12) 
 
i i
i
i
l
EA
k ? ? sin cos
21
?
=
     (23.13) 
 
In the next step, give a unit displacement along holding displacement along 
equal to zero and calculate reactions at 
2
u
1
u
E corresponding to unknown 
displacements and in the kinematically determinate structure. The 
corresponding reactions are denoted by and  as shown in Fig.23.4d. The 
joint 
1
u
2
u
12
k
22
k
E gets displaced to E ' when a unit vertical displacement is given to the joint 
as shown in the figure. Thus, the new length of the member E A ' is 
1 1
l l ? +
. 
From the geometry, the elongation 
1
l ?
is given by 
 
1 1
sin ? = ?l
         (23.14a) 
 
Thus axial force in the member along its centroidal axis  is 1
1
1
sin ?
l
EA
 (23.14b) 
 
Resolve the axial force in the member along and directions. Thus, 
horizontal component of force in the member 
1
u
2
u
E A ' is 1 1
1
1
cos sin ? ?
l
EA
      (23.14c)  
and vertical component of force in the member E A ' is  1
2
1
1
sin ?
l
EA
       (23.14d) 
 
In order to evaluate , we need to sum vertical components of forces in all the 
members meeting at joint 
22
k
E .Thus, 
                                                         
i
i
i
i
l
EA
k ?
2
4
1
22
sin
?
=
=
    (23.15) 
 
Similarly, i i
i
i
i
l
EA
k ? ? cos sin
4
1
12 ?
=
=
     (23.16) 
 
3. Joint forces in the original structure corresponding to unknown 
displacements and  are  
1
u
2
u
 
?
?
?
?
?
?
=
?
?
?
?
?
?
2
1
2
1
P
P
F
F
      (23.17) 
Now the equilibrium equations at joint E states that the forces in the original 
structure are equal to the superposition of (i) reactions in the kinematically 
restrained structure corresponding to unknown joint displacements and (ii) 
reactions in the restrained structure due to unknown displacements themselves. 
This may be expressed as, 
 
()
2 12 1 11 1 1
u k u k R F
L
+ + =
 
()
2 22 1 21 2 2
u k u k R F
L
+ + =
   (23.18) 
 
This may be written compactly as 
 
{} { } [ ] { } u k R F
i
+ =
     (23.19) 
 
where, 
 
{}
?
?
?
?
?
?
=
2
1
F
F
F
; 
 
{}
()
()
?
?
?
?
?
?
=
2
1
L
L
L
R
R
R
 
 
[]
?
?
?
?
?
?
=
22 21
12 11
k k
k k
k 
 
{}
?
?
?
?
?
?
=
2
1
u
u
u
      (23.20) 
                                                         
Page 3


Expressions of similar form as above may be obtained for all  members. The sum 
of all horizontal components of individual forces gives us the stiffness coefficient 
 and sum of all vertical component of forces give us the required stiffness 
coefficient . 
11
k
21
k
 
4
2
4
4
3
2
3
3
2
2
2
2
1
2
1
1
11
cos cos cos cos ? ? ? ?
l
EA
l
EA
l
EA
l
EA
k + + + =
 
 
i
i
i
i
l
EA
k ?
2
4
1
11
cos
?
=
=
      (23.12) 
 
i i
i
i
l
EA
k ? ? sin cos
21
?
=
     (23.13) 
 
In the next step, give a unit displacement along holding displacement along 
equal to zero and calculate reactions at 
2
u
1
u
E corresponding to unknown 
displacements and in the kinematically determinate structure. The 
corresponding reactions are denoted by and  as shown in Fig.23.4d. The 
joint 
1
u
2
u
12
k
22
k
E gets displaced to E ' when a unit vertical displacement is given to the joint 
as shown in the figure. Thus, the new length of the member E A ' is 
1 1
l l ? +
. 
From the geometry, the elongation 
1
l ?
is given by 
 
1 1
sin ? = ?l
         (23.14a) 
 
Thus axial force in the member along its centroidal axis  is 1
1
1
sin ?
l
EA
 (23.14b) 
 
Resolve the axial force in the member along and directions. Thus, 
horizontal component of force in the member 
1
u
2
u
E A ' is 1 1
1
1
cos sin ? ?
l
EA
      (23.14c)  
and vertical component of force in the member E A ' is  1
2
1
1
sin ?
l
EA
       (23.14d) 
 
In order to evaluate , we need to sum vertical components of forces in all the 
members meeting at joint 
22
k
E .Thus, 
                                                         
i
i
i
i
l
EA
k ?
2
4
1
22
sin
?
=
=
    (23.15) 
 
Similarly, i i
i
i
i
l
EA
k ? ? cos sin
4
1
12 ?
=
=
     (23.16) 
 
3. Joint forces in the original structure corresponding to unknown 
displacements and  are  
1
u
2
u
 
?
?
?
?
?
?
=
?
?
?
?
?
?
2
1
2
1
P
P
F
F
      (23.17) 
Now the equilibrium equations at joint E states that the forces in the original 
structure are equal to the superposition of (i) reactions in the kinematically 
restrained structure corresponding to unknown joint displacements and (ii) 
reactions in the restrained structure due to unknown displacements themselves. 
This may be expressed as, 
 
()
2 12 1 11 1 1
u k u k R F
L
+ + =
 
()
2 22 1 21 2 2
u k u k R F
L
+ + =
   (23.18) 
 
This may be written compactly as 
 
{} { } [ ] { } u k R F
i
+ =
     (23.19) 
 
where, 
 
{}
?
?
?
?
?
?
=
2
1
F
F
F
; 
 
{}
()
()
?
?
?
?
?
?
=
2
1
L
L
L
R
R
R
 
 
[]
?
?
?
?
?
?
=
22 21
12 11
k k
k k
k 
 
{}
?
?
?
?
?
?
=
2
1
u
u
u
      (23.20) 
                                                         
For example take
P P P = =
2 1
,
i
i
L
L
? sin
=
,
A A A A A = = = =
4 3 2 1
 and 
° = 35
1
?
,
° = 70
2
?
,
° = 105
3
?
and 
° = 140
4
?
 
 
Then. 
 
{}
?
?
?
?
?
?
=
P
P
F
     (23.21) 
 
{}
?
?
?
?
?
?
=
0
0
L
R
 
 
L
EA
L
EA
k
i i
9367 . 0 sin cos
2
11
= =
?
? ?
 
 
L
EA
L
EA
k
i i
0135 . 0 cos sin
2
12
= =
?
? ?
 
 
L
EA
L
EA
k
i i
0135 . 0 cos sin
2
21
= =
?
? ?
 
 
L
EA
L
EA
k
i
1853 . 2 sin
3
22
= =
?
?
   (23.22) 
 
1
2
0.9367 0.0135
0.0135 2.1853
u P
EA
u P L
? ? ?? ? ?
=
?? ? ?
??
?? ? ???
 
 
Solving which, yields 
 
EA
L
u 0611 . 1
1
=
 
EA
L
u 451 . 0
2
=
 
 
 
 
 
 
                                                         
Page 4


Expressions of similar form as above may be obtained for all  members. The sum 
of all horizontal components of individual forces gives us the stiffness coefficient 
 and sum of all vertical component of forces give us the required stiffness 
coefficient . 
11
k
21
k
 
4
2
4
4
3
2
3
3
2
2
2
2
1
2
1
1
11
cos cos cos cos ? ? ? ?
l
EA
l
EA
l
EA
l
EA
k + + + =
 
 
i
i
i
i
l
EA
k ?
2
4
1
11
cos
?
=
=
      (23.12) 
 
i i
i
i
l
EA
k ? ? sin cos
21
?
=
     (23.13) 
 
In the next step, give a unit displacement along holding displacement along 
equal to zero and calculate reactions at 
2
u
1
u
E corresponding to unknown 
displacements and in the kinematically determinate structure. The 
corresponding reactions are denoted by and  as shown in Fig.23.4d. The 
joint 
1
u
2
u
12
k
22
k
E gets displaced to E ' when a unit vertical displacement is given to the joint 
as shown in the figure. Thus, the new length of the member E A ' is 
1 1
l l ? +
. 
From the geometry, the elongation 
1
l ?
is given by 
 
1 1
sin ? = ?l
         (23.14a) 
 
Thus axial force in the member along its centroidal axis  is 1
1
1
sin ?
l
EA
 (23.14b) 
 
Resolve the axial force in the member along and directions. Thus, 
horizontal component of force in the member 
1
u
2
u
E A ' is 1 1
1
1
cos sin ? ?
l
EA
      (23.14c)  
and vertical component of force in the member E A ' is  1
2
1
1
sin ?
l
EA
       (23.14d) 
 
In order to evaluate , we need to sum vertical components of forces in all the 
members meeting at joint 
22
k
E .Thus, 
                                                         
i
i
i
i
l
EA
k ?
2
4
1
22
sin
?
=
=
    (23.15) 
 
Similarly, i i
i
i
i
l
EA
k ? ? cos sin
4
1
12 ?
=
=
     (23.16) 
 
3. Joint forces in the original structure corresponding to unknown 
displacements and  are  
1
u
2
u
 
?
?
?
?
?
?
=
?
?
?
?
?
?
2
1
2
1
P
P
F
F
      (23.17) 
Now the equilibrium equations at joint E states that the forces in the original 
structure are equal to the superposition of (i) reactions in the kinematically 
restrained structure corresponding to unknown joint displacements and (ii) 
reactions in the restrained structure due to unknown displacements themselves. 
This may be expressed as, 
 
()
2 12 1 11 1 1
u k u k R F
L
+ + =
 
()
2 22 1 21 2 2
u k u k R F
L
+ + =
   (23.18) 
 
This may be written compactly as 
 
{} { } [ ] { } u k R F
i
+ =
     (23.19) 
 
where, 
 
{}
?
?
?
?
?
?
=
2
1
F
F
F
; 
 
{}
()
()
?
?
?
?
?
?
=
2
1
L
L
L
R
R
R
 
 
[]
?
?
?
?
?
?
=
22 21
12 11
k k
k k
k 
 
{}
?
?
?
?
?
?
=
2
1
u
u
u
      (23.20) 
                                                         
For example take
P P P = =
2 1
,
i
i
L
L
? sin
=
,
A A A A A = = = =
4 3 2 1
 and 
° = 35
1
?
,
° = 70
2
?
,
° = 105
3
?
and 
° = 140
4
?
 
 
Then. 
 
{}
?
?
?
?
?
?
=
P
P
F
     (23.21) 
 
{}
?
?
?
?
?
?
=
0
0
L
R
 
 
L
EA
L
EA
k
i i
9367 . 0 sin cos
2
11
= =
?
? ?
 
 
L
EA
L
EA
k
i i
0135 . 0 cos sin
2
12
= =
?
? ?
 
 
L
EA
L
EA
k
i i
0135 . 0 cos sin
2
21
= =
?
? ?
 
 
L
EA
L
EA
k
i
1853 . 2 sin
3
22
= =
?
?
   (23.22) 
 
1
2
0.9367 0.0135
0.0135 2.1853
u P
EA
u P L
? ? ?? ? ?
=
?? ? ?
??
?? ? ???
 
 
Solving which, yields 
 
EA
L
u 0611 . 1
1
=
 
EA
L
u 451 . 0
2
=
 
 
 
 
 
 
                                                         
Example 23.1 
Analyze the plane frame shown in Fig.23.5a by the direct stiffness method. 
Assume that the flexural rigidity for all members is the same .Neglect axial 
displacements. 
 
 
 
Solution 
In the first step identify the degrees of freedom of the frame .The given frame has 
three degrees of freedom (see Fig.23.5b): 
 
(i) Two rotations as indicated by and and 
1
u
2
u
(ii) One horizontal displacement of joint B and C as indicated by . 
3
u
 
In the next step make all the displacements equal to zero by fixing joints B and C 
as shown in Fig.23.5c. On this kinematically determinate structure apply all the 
external loads and calculate reactions corresponding to unknown joint 
displacements .Thus, 
 
                                                         
Page 5


Expressions of similar form as above may be obtained for all  members. The sum 
of all horizontal components of individual forces gives us the stiffness coefficient 
 and sum of all vertical component of forces give us the required stiffness 
coefficient . 
11
k
21
k
 
4
2
4
4
3
2
3
3
2
2
2
2
1
2
1
1
11
cos cos cos cos ? ? ? ?
l
EA
l
EA
l
EA
l
EA
k + + + =
 
 
i
i
i
i
l
EA
k ?
2
4
1
11
cos
?
=
=
      (23.12) 
 
i i
i
i
l
EA
k ? ? sin cos
21
?
=
     (23.13) 
 
In the next step, give a unit displacement along holding displacement along 
equal to zero and calculate reactions at 
2
u
1
u
E corresponding to unknown 
displacements and in the kinematically determinate structure. The 
corresponding reactions are denoted by and  as shown in Fig.23.4d. The 
joint 
1
u
2
u
12
k
22
k
E gets displaced to E ' when a unit vertical displacement is given to the joint 
as shown in the figure. Thus, the new length of the member E A ' is 
1 1
l l ? +
. 
From the geometry, the elongation 
1
l ?
is given by 
 
1 1
sin ? = ?l
         (23.14a) 
 
Thus axial force in the member along its centroidal axis  is 1
1
1
sin ?
l
EA
 (23.14b) 
 
Resolve the axial force in the member along and directions. Thus, 
horizontal component of force in the member 
1
u
2
u
E A ' is 1 1
1
1
cos sin ? ?
l
EA
      (23.14c)  
and vertical component of force in the member E A ' is  1
2
1
1
sin ?
l
EA
       (23.14d) 
 
In order to evaluate , we need to sum vertical components of forces in all the 
members meeting at joint 
22
k
E .Thus, 
                                                         
i
i
i
i
l
EA
k ?
2
4
1
22
sin
?
=
=
    (23.15) 
 
Similarly, i i
i
i
i
l
EA
k ? ? cos sin
4
1
12 ?
=
=
     (23.16) 
 
3. Joint forces in the original structure corresponding to unknown 
displacements and  are  
1
u
2
u
 
?
?
?
?
?
?
=
?
?
?
?
?
?
2
1
2
1
P
P
F
F
      (23.17) 
Now the equilibrium equations at joint E states that the forces in the original 
structure are equal to the superposition of (i) reactions in the kinematically 
restrained structure corresponding to unknown joint displacements and (ii) 
reactions in the restrained structure due to unknown displacements themselves. 
This may be expressed as, 
 
()
2 12 1 11 1 1
u k u k R F
L
+ + =
 
()
2 22 1 21 2 2
u k u k R F
L
+ + =
   (23.18) 
 
This may be written compactly as 
 
{} { } [ ] { } u k R F
i
+ =
     (23.19) 
 
where, 
 
{}
?
?
?
?
?
?
=
2
1
F
F
F
; 
 
{}
()
()
?
?
?
?
?
?
=
2
1
L
L
L
R
R
R
 
 
[]
?
?
?
?
?
?
=
22 21
12 11
k k
k k
k 
 
{}
?
?
?
?
?
?
=
2
1
u
u
u
      (23.20) 
                                                         
For example take
P P P = =
2 1
,
i
i
L
L
? sin
=
,
A A A A A = = = =
4 3 2 1
 and 
° = 35
1
?
,
° = 70
2
?
,
° = 105
3
?
and 
° = 140
4
?
 
 
Then. 
 
{}
?
?
?
?
?
?
=
P
P
F
     (23.21) 
 
{}
?
?
?
?
?
?
=
0
0
L
R
 
 
L
EA
L
EA
k
i i
9367 . 0 sin cos
2
11
= =
?
? ?
 
 
L
EA
L
EA
k
i i
0135 . 0 cos sin
2
12
= =
?
? ?
 
 
L
EA
L
EA
k
i i
0135 . 0 cos sin
2
21
= =
?
? ?
 
 
L
EA
L
EA
k
i
1853 . 2 sin
3
22
= =
?
?
   (23.22) 
 
1
2
0.9367 0.0135
0.0135 2.1853
u P
EA
u P L
? ? ?? ? ?
=
?? ? ?
??
?? ? ???
 
 
Solving which, yields 
 
EA
L
u 0611 . 1
1
=
 
EA
L
u 451 . 0
2
=
 
 
 
 
 
 
                                                         
Example 23.1 
Analyze the plane frame shown in Fig.23.5a by the direct stiffness method. 
Assume that the flexural rigidity for all members is the same .Neglect axial 
displacements. 
 
 
 
Solution 
In the first step identify the degrees of freedom of the frame .The given frame has 
three degrees of freedom (see Fig.23.5b): 
 
(i) Two rotations as indicated by and and 
1
u
2
u
(ii) One horizontal displacement of joint B and C as indicated by . 
3
u
 
In the next step make all the displacements equal to zero by fixing joints B and C 
as shown in Fig.23.5c. On this kinematically determinate structure apply all the 
external loads and calculate reactions corresponding to unknown joint 
displacements .Thus, 
 
                                                         
() ?
?
?
?
?
? × ×
- +
× ×
=
36
9 3 24
16
4 2 48
1
F
D
R
      
 (1) 
 
24 18 6 kN.m =- =       
  
  
()
2
24 kN.m
F
D
R =-
 
 
( )
3
12 kN.m
F
D
R =
       (2) 
 
Thus, 
 
()
()
()
1
2
3
6
24
12
F
D
F
D
F
D
R
R
R
??
??
??
???
=-
???
???
??
??
??
?
?
?
      (3) 
Next calculate stiffness coefficients. Apply unit rotation along and calculate 
reactions corresponding to the unknown joint displacements in the kinematically 
determinate structure (vide Fig.23.5d) 
1
u
 
                                                         
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