Introduction: The Direct Stiffness Method - 2 GATE Notes | EduRev

Structural Analysis

GATE : Introduction: The Direct Stiffness Method - 2 GATE Notes | EduRev

 Page 1


B must be 
B
?
which is unknown .The relation between  and 
B
M -
B
?
is 
established as follows. Apply a unit rotation at B and calculate the 
moment. ( caused by it. That is given by the relation  
)
BB
k
 
L
EI
k
BB
4
=
     (23.6) 
 
where  is the stiffness coefficient and is defined as the force at joint B due to 
unit displacement at joint B. Now, moment caused by 
BB
k
B
?
rotation is 
 
B BB B
k M ? =
     (23.7) 
 
3. Now, write the equilibrium equation for joint B. The total moment at B is 
B BB B
k M ? +
, but in the actual structure the moment at B is zero as 
support B is hinged. Hence, 
 
0 = +
B BB B
k M ?
    (23.8) 
 
BB
B
B
k
M
- = ?
 
 
EI
wl
B
48
3
= ?
     (23.9) 
 
The relation B B
L
EI
M ?
4
=
 has already been derived in slope –deflection method 
in lesson 14. Please note that exactly the same steps are followed in slope-
deflection method. 
 
 
 
                                                         
Page 2


B must be 
B
?
which is unknown .The relation between  and 
B
M -
B
?
is 
established as follows. Apply a unit rotation at B and calculate the 
moment. ( caused by it. That is given by the relation  
)
BB
k
 
L
EI
k
BB
4
=
     (23.6) 
 
where  is the stiffness coefficient and is defined as the force at joint B due to 
unit displacement at joint B. Now, moment caused by 
BB
k
B
?
rotation is 
 
B BB B
k M ? =
     (23.7) 
 
3. Now, write the equilibrium equation for joint B. The total moment at B is 
B BB B
k M ? +
, but in the actual structure the moment at B is zero as 
support B is hinged. Hence, 
 
0 = +
B BB B
k M ?
    (23.8) 
 
BB
B
B
k
M
- = ?
 
 
EI
wl
B
48
3
= ?
     (23.9) 
 
The relation B B
L
EI
M ?
4
=
 has already been derived in slope –deflection method 
in lesson 14. Please note that exactly the same steps are followed in slope-
deflection method. 
 
 
 
                                                         
 
 
 
 
 
 
 
 
 
                                                         
Page 3


B must be 
B
?
which is unknown .The relation between  and 
B
M -
B
?
is 
established as follows. Apply a unit rotation at B and calculate the 
moment. ( caused by it. That is given by the relation  
)
BB
k
 
L
EI
k
BB
4
=
     (23.6) 
 
where  is the stiffness coefficient and is defined as the force at joint B due to 
unit displacement at joint B. Now, moment caused by 
BB
k
B
?
rotation is 
 
B BB B
k M ? =
     (23.7) 
 
3. Now, write the equilibrium equation for joint B. The total moment at B is 
B BB B
k M ? +
, but in the actual structure the moment at B is zero as 
support B is hinged. Hence, 
 
0 = +
B BB B
k M ?
    (23.8) 
 
BB
B
B
k
M
- = ?
 
 
EI
wl
B
48
3
= ?
     (23.9) 
 
The relation B B
L
EI
M ?
4
=
 has already been derived in slope –deflection method 
in lesson 14. Please note that exactly the same steps are followed in slope-
deflection method. 
 
 
 
                                                         
 
 
 
 
 
 
 
 
 
                                                         
23.3 Two degrees of freedom structure 
Consider a plane truss as shown in Fig.23.4a.There is four members in the truss 
and they meet at the common point at E. The truss is subjected to external loads 
and acting at E. In the analysis, neglect the self weight of members. There 
are two unknown displacements at joint E and are denoted by  and .Thus 
the structure is kinematically indeterminate to second degree. The applied forces 
and unknown joint displacements are shown in the positive directions. The 
members are numbered from (1), (2), (3) and (4) as shown in the figure. The 
length and axial rigidity of i-th member is and respectively. Now it is sought 
to evaluate  and by stiffness method. This is done in following steps: 
1
P
2
P
1
u
2
u
i
l
i
EA
1
u
2
u
 
1. In the first step, make all the unknown displacements equal to zero by 
altering the boundary conditions as shown in Fig.23.4b. On this restrained 
/kinematically determinate structure, apply all the external loads except 
the joint loads and calculate the reactions corresponding to unknown joint 
displacements  and . Since, in the present case, there are no 
external loads other than the joint loads, the reactions and  
will be equal to zero. Thus, 
1
u
2
u
1
) (
L
R
2
) (
L
R
 
?
?
?
?
?
?
=
?
?
?
?
?
?
0
0
) (
) (
2
1
L
L
R
R
         (23.10) 
 
2. In the next step, calculate stiffness coefficients and .This is 
done as follows. First give a unit displacement along holding 
displacement along to zero and calculate reactions at E corresponding 
to unknown displacements and in the kinematically determinate 
structure. They are denoted by . The joint stiffness of the 
whole truss is composed of individual member stiffness of the truss. This 
is shown in Fig.23.4c. Now consider the member
12 21 11
, , k k k
22
k
1
u
2
u
1
u
2
u
21 11
,k k
21 11
,k k
AE . Under the action of 
unit displacement along , the joint 
1
u
E displaces toE '. Obviously the new 
length is not equal to lengthAE . Let us denote the new length of the 
members by , where
1 1
l l ? + l ? , is the change in length of the 
member E A '. The member E A ' also makes an angle 
1
?
with the 
horizontal. This is justified as 
1
l ?
is small. From the geometry, the change 
in length of the members E A ' is 
 
1 1
cos ? = ?l
     (23.11a) 
 
                                                         
Page 4


B must be 
B
?
which is unknown .The relation between  and 
B
M -
B
?
is 
established as follows. Apply a unit rotation at B and calculate the 
moment. ( caused by it. That is given by the relation  
)
BB
k
 
L
EI
k
BB
4
=
     (23.6) 
 
where  is the stiffness coefficient and is defined as the force at joint B due to 
unit displacement at joint B. Now, moment caused by 
BB
k
B
?
rotation is 
 
B BB B
k M ? =
     (23.7) 
 
3. Now, write the equilibrium equation for joint B. The total moment at B is 
B BB B
k M ? +
, but in the actual structure the moment at B is zero as 
support B is hinged. Hence, 
 
0 = +
B BB B
k M ?
    (23.8) 
 
BB
B
B
k
M
- = ?
 
 
EI
wl
B
48
3
= ?
     (23.9) 
 
The relation B B
L
EI
M ?
4
=
 has already been derived in slope –deflection method 
in lesson 14. Please note that exactly the same steps are followed in slope-
deflection method. 
 
 
 
                                                         
 
 
 
 
 
 
 
 
 
                                                         
23.3 Two degrees of freedom structure 
Consider a plane truss as shown in Fig.23.4a.There is four members in the truss 
and they meet at the common point at E. The truss is subjected to external loads 
and acting at E. In the analysis, neglect the self weight of members. There 
are two unknown displacements at joint E and are denoted by  and .Thus 
the structure is kinematically indeterminate to second degree. The applied forces 
and unknown joint displacements are shown in the positive directions. The 
members are numbered from (1), (2), (3) and (4) as shown in the figure. The 
length and axial rigidity of i-th member is and respectively. Now it is sought 
to evaluate  and by stiffness method. This is done in following steps: 
1
P
2
P
1
u
2
u
i
l
i
EA
1
u
2
u
 
1. In the first step, make all the unknown displacements equal to zero by 
altering the boundary conditions as shown in Fig.23.4b. On this restrained 
/kinematically determinate structure, apply all the external loads except 
the joint loads and calculate the reactions corresponding to unknown joint 
displacements  and . Since, in the present case, there are no 
external loads other than the joint loads, the reactions and  
will be equal to zero. Thus, 
1
u
2
u
1
) (
L
R
2
) (
L
R
 
?
?
?
?
?
?
=
?
?
?
?
?
?
0
0
) (
) (
2
1
L
L
R
R
         (23.10) 
 
2. In the next step, calculate stiffness coefficients and .This is 
done as follows. First give a unit displacement along holding 
displacement along to zero and calculate reactions at E corresponding 
to unknown displacements and in the kinematically determinate 
structure. They are denoted by . The joint stiffness of the 
whole truss is composed of individual member stiffness of the truss. This 
is shown in Fig.23.4c. Now consider the member
12 21 11
, , k k k
22
k
1
u
2
u
1
u
2
u
21 11
,k k
21 11
,k k
AE . Under the action of 
unit displacement along , the joint 
1
u
E displaces toE '. Obviously the new 
length is not equal to lengthAE . Let us denote the new length of the 
members by , where
1 1
l l ? + l ? , is the change in length of the 
member E A '. The member E A ' also makes an angle 
1
?
with the 
horizontal. This is justified as 
1
l ?
is small. From the geometry, the change 
in length of the members E A ' is 
 
1 1
cos ? = ?l
     (23.11a) 
 
                                                         
The elongation is related to the force in the member
1
l ?
E A ' , by '
AE
F
 
E A
l F
l
AE
1
1
1
'
= ?
     (23.11b) 
 
Thus from (23.11a) and (23.11b), the force in the members E A ' is  
 
1
1
1
cos ?
l
EA
F
AE
= '
    (23.11c) 
 
This force acts along the member axis. This force may be resolved along and 
directions. Thus, horizontal component of force 
1
u
2
u
AE
F ' is 1
2
1
1
cos ?
l
EA
 (23.11d) 
and vertical component of force 
AE
F ' is 1 1
1
1
sin cos ? ?
l
EA
   (23.11e) 
 
 
 
                                                         
Page 5


B must be 
B
?
which is unknown .The relation between  and 
B
M -
B
?
is 
established as follows. Apply a unit rotation at B and calculate the 
moment. ( caused by it. That is given by the relation  
)
BB
k
 
L
EI
k
BB
4
=
     (23.6) 
 
where  is the stiffness coefficient and is defined as the force at joint B due to 
unit displacement at joint B. Now, moment caused by 
BB
k
B
?
rotation is 
 
B BB B
k M ? =
     (23.7) 
 
3. Now, write the equilibrium equation for joint B. The total moment at B is 
B BB B
k M ? +
, but in the actual structure the moment at B is zero as 
support B is hinged. Hence, 
 
0 = +
B BB B
k M ?
    (23.8) 
 
BB
B
B
k
M
- = ?
 
 
EI
wl
B
48
3
= ?
     (23.9) 
 
The relation B B
L
EI
M ?
4
=
 has already been derived in slope –deflection method 
in lesson 14. Please note that exactly the same steps are followed in slope-
deflection method. 
 
 
 
                                                         
 
 
 
 
 
 
 
 
 
                                                         
23.3 Two degrees of freedom structure 
Consider a plane truss as shown in Fig.23.4a.There is four members in the truss 
and they meet at the common point at E. The truss is subjected to external loads 
and acting at E. In the analysis, neglect the self weight of members. There 
are two unknown displacements at joint E and are denoted by  and .Thus 
the structure is kinematically indeterminate to second degree. The applied forces 
and unknown joint displacements are shown in the positive directions. The 
members are numbered from (1), (2), (3) and (4) as shown in the figure. The 
length and axial rigidity of i-th member is and respectively. Now it is sought 
to evaluate  and by stiffness method. This is done in following steps: 
1
P
2
P
1
u
2
u
i
l
i
EA
1
u
2
u
 
1. In the first step, make all the unknown displacements equal to zero by 
altering the boundary conditions as shown in Fig.23.4b. On this restrained 
/kinematically determinate structure, apply all the external loads except 
the joint loads and calculate the reactions corresponding to unknown joint 
displacements  and . Since, in the present case, there are no 
external loads other than the joint loads, the reactions and  
will be equal to zero. Thus, 
1
u
2
u
1
) (
L
R
2
) (
L
R
 
?
?
?
?
?
?
=
?
?
?
?
?
?
0
0
) (
) (
2
1
L
L
R
R
         (23.10) 
 
2. In the next step, calculate stiffness coefficients and .This is 
done as follows. First give a unit displacement along holding 
displacement along to zero and calculate reactions at E corresponding 
to unknown displacements and in the kinematically determinate 
structure. They are denoted by . The joint stiffness of the 
whole truss is composed of individual member stiffness of the truss. This 
is shown in Fig.23.4c. Now consider the member
12 21 11
, , k k k
22
k
1
u
2
u
1
u
2
u
21 11
,k k
21 11
,k k
AE . Under the action of 
unit displacement along , the joint 
1
u
E displaces toE '. Obviously the new 
length is not equal to lengthAE . Let us denote the new length of the 
members by , where
1 1
l l ? + l ? , is the change in length of the 
member E A '. The member E A ' also makes an angle 
1
?
with the 
horizontal. This is justified as 
1
l ?
is small. From the geometry, the change 
in length of the members E A ' is 
 
1 1
cos ? = ?l
     (23.11a) 
 
                                                         
The elongation is related to the force in the member
1
l ?
E A ' , by '
AE
F
 
E A
l F
l
AE
1
1
1
'
= ?
     (23.11b) 
 
Thus from (23.11a) and (23.11b), the force in the members E A ' is  
 
1
1
1
cos ?
l
EA
F
AE
= '
    (23.11c) 
 
This force acts along the member axis. This force may be resolved along and 
directions. Thus, horizontal component of force 
1
u
2
u
AE
F ' is 1
2
1
1
cos ?
l
EA
 (23.11d) 
and vertical component of force 
AE
F ' is 1 1
1
1
sin cos ? ?
l
EA
   (23.11e) 
 
 
 
                                                         
 
 
 
 
                                                         
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