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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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FAQs on Introduction: The Direct Stiffness Method - 3 - Structural Analysis - Civil Engineering (CE)
| 1. What is the Direct Stiffness Method? |  |
Ans. The Direct Stiffness Method is a structural analysis technique used to determine the displacements and internal forces in a structure. It is based on the principles of linear elasticity and uses the stiffness matrix of the structure to solve for these quantities.
| 2. How does the Direct Stiffness Method work? | |
Ans. The Direct Stiffness Method breaks down the structure into smaller elements, such as beams or trusses, and represents each element with a stiffness matrix. These stiffness matrices are then assembled to form the global stiffness matrix of the entire structure. By applying appropriate boundary conditions and loading, the displacements and internal forces can be determined by solving the resulting system of equations.
| 3. What are the advantages of using the Direct Stiffness Method? | |
Ans. The Direct Stiffness Method offers several advantages in structural analysis. It allows for the analysis of complex structures, including those with irregular geometries or varying material properties. It can handle both linear and nonlinear problems, making it a versatile technique. Additionally, it provides accurate and reliable results when compared to experimental testing.
| 4. Are there any limitations to the Direct Stiffness Method? | |
Ans. Yes, the Direct Stiffness Method has some limitations. It assumes that the structure behaves linearly under the applied loads, which may not always be the case. It can also be computationally intensive and time-consuming for large-scale structures. Additionally, the accuracy of the results depends on the accuracy of the input data, including material properties and boundary conditions.
| 5. How is the Direct Stiffness Method used in engineering practice? | |
Ans. The Direct Stiffness Method is commonly used in engineering practice for structural analysis and design. It is implemented in computer software and used to analyze various types of structures, such as buildings, bridges, and aerospace components. By accurately determining the displacements and internal forces, engineers can ensure the structural integrity and safety of the design.
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Nov 22, 2024 Last updated |
Document Description: Introduction: The Direct Stiffness Method - 3 for Civil Engineering (CE) 2024 is part of Structural Analysis preparation.
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