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

## 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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## Structural Analysis

30 videos|122 docs|28 tests

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