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# The Direct Stiffness Method: Truss Analysis - 4 GATE Notes | EduRev

## GATE : The Direct Stiffness Method: Truss Analysis - 4 GATE Notes | EduRev

``` Page 1

Instructional Objectives
After reading this chapter the student will be able to
1. Transform member stiffness matrix from local to global co-ordinate system.
2. Assemble member stiffness matrices to obtain the global stiffness matrix.
3. Analyse plane truss by the direct stiffness matrix.
4. Analyse plane truss supported on inclined roller supports.

25.1 Introduction
In the previous lesson, the direct stiffness method as applied to trusses was
discussed. The transformation of force and displacement from local co-ordinate
system to global co-ordinate system were accomplished by single transformation
matrix. Also assembly of the member stiffness matrices was discussed. In this
lesson few plane trusses are analysed using the direct stiffness method. Also the
problem of inclined support will be discussed.

Example 25.1
Analyse the truss shown in Fig. 25.1a and evaluate reactions. Assume EA to be
constant for all the members.

Page 2

Instructional Objectives
After reading this chapter the student will be able to
1. Transform member stiffness matrix from local to global co-ordinate system.
2. Assemble member stiffness matrices to obtain the global stiffness matrix.
3. Analyse plane truss by the direct stiffness matrix.
4. Analyse plane truss supported on inclined roller supports.

25.1 Introduction
In the previous lesson, the direct stiffness method as applied to trusses was
discussed. The transformation of force and displacement from local co-ordinate
system to global co-ordinate system were accomplished by single transformation
matrix. Also assembly of the member stiffness matrices was discussed. In this
lesson few plane trusses are analysed using the direct stiffness method. Also the
problem of inclined support will be discussed.

Example 25.1
Analyse the truss shown in Fig. 25.1a and evaluate reactions. Assume EA to be
constant for all the members.

The numbering of joints and members are shown in Fig. 25.1b. Also, the possible
displacements (degrees of freedom) at each node are indicated. Here lower
numbers are used to indicate unconstrained degrees of freedom and higher
numbers are used for constrained degrees of freedom. Thus displacements 6,7
and 8 are zero due to boundary conditions.

First write down stiffness matrix of each member in global co-ordinate system
and assemble them to obtain global stiffness matrix.

Element 1: . 619 . 4 , 60 m L = ° = ? Nodal points 4-1

[]
?
?
?
?
?
?
?
?
?
?
?
?
- -
- -
- -
- -
=
75 . 0 433 . 0 75 . 0 433 . 0
433 . 0 25 . 0 433 . 0 25 . 0
75 . 0 433 . 0 75 . 0 433 . 0
433 . 0 25 . 0 433 . 0 25 . 0
619 . 4
1
EA
k   (1)

Element 2: . 00 . 4 , 90 m L = ° = ? Nodal points 2-1

Page 3

Instructional Objectives
After reading this chapter the student will be able to
1. Transform member stiffness matrix from local to global co-ordinate system.
2. Assemble member stiffness matrices to obtain the global stiffness matrix.
3. Analyse plane truss by the direct stiffness matrix.
4. Analyse plane truss supported on inclined roller supports.

25.1 Introduction
In the previous lesson, the direct stiffness method as applied to trusses was
discussed. The transformation of force and displacement from local co-ordinate
system to global co-ordinate system were accomplished by single transformation
matrix. Also assembly of the member stiffness matrices was discussed. In this
lesson few plane trusses are analysed using the direct stiffness method. Also the
problem of inclined support will be discussed.

Example 25.1
Analyse the truss shown in Fig. 25.1a and evaluate reactions. Assume EA to be
constant for all the members.

The numbering of joints and members are shown in Fig. 25.1b. Also, the possible
displacements (degrees of freedom) at each node are indicated. Here lower
numbers are used to indicate unconstrained degrees of freedom and higher
numbers are used for constrained degrees of freedom. Thus displacements 6,7
and 8 are zero due to boundary conditions.

First write down stiffness matrix of each member in global co-ordinate system
and assemble them to obtain global stiffness matrix.

Element 1: . 619 . 4 , 60 m L = ° = ? Nodal points 4-1

[]
?
?
?
?
?
?
?
?
?
?
?
?
- -
- -
- -
- -
=
75 . 0 433 . 0 75 . 0 433 . 0
433 . 0 25 . 0 433 . 0 25 . 0
75 . 0 433 . 0 75 . 0 433 . 0
433 . 0 25 . 0 433 . 0 25 . 0
619 . 4
1
EA
k   (1)

Element 2: . 00 . 4 , 90 m L = ° = ? Nodal points 2-1

[]
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
0 . 4
2
EA
k       (2)

Element 3: . 619 . 4 , 120 m L = ° = ? Nodal points 3-1

[]
?
?
?
?
?
?
?
?
?
?
?
?
- -
- -
- -
- -
=
75 . 0 433 . 0 75 . 0 433 . 0
433 . 0 25 . 0 433 . 0 25 . 0
75 . 0 433 . 0 75 . 0 433 . 0
433 . 0 25 . 0 433 . 0 25 . 0
619 . 4
3
EA
k   (3)

Element 4: . 31 . 2 , 0 m L = ° = ? Nodal points 4-2

[]
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
31 . 2
4
EA
k      (4)

Element 5: . 31 . 2 , 0 m L = ° = ? Nodal points 2-3

[]
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
31 . 2
5
EA
k      (5)

The assembled global stiffness matrix of the truss is of the order 8 8 × . Now
assemble the global stiffness matrix. Note that the element
1
11
k of the member
stiffness matrix of truss member 1 goes to location ( ) 7 , 7 of global stiffness matrix.
On the member stiffness matrix the corresponding global degrees of freedom are
indicated to facilitate assembling. Thus,

Page 4

Instructional Objectives
After reading this chapter the student will be able to
1. Transform member stiffness matrix from local to global co-ordinate system.
2. Assemble member stiffness matrices to obtain the global stiffness matrix.
3. Analyse plane truss by the direct stiffness matrix.
4. Analyse plane truss supported on inclined roller supports.

25.1 Introduction
In the previous lesson, the direct stiffness method as applied to trusses was
discussed. The transformation of force and displacement from local co-ordinate
system to global co-ordinate system were accomplished by single transformation
matrix. Also assembly of the member stiffness matrices was discussed. In this
lesson few plane trusses are analysed using the direct stiffness method. Also the
problem of inclined support will be discussed.

Example 25.1
Analyse the truss shown in Fig. 25.1a and evaluate reactions. Assume EA to be
constant for all the members.

The numbering of joints and members are shown in Fig. 25.1b. Also, the possible
displacements (degrees of freedom) at each node are indicated. Here lower
numbers are used to indicate unconstrained degrees of freedom and higher
numbers are used for constrained degrees of freedom. Thus displacements 6,7
and 8 are zero due to boundary conditions.

First write down stiffness matrix of each member in global co-ordinate system
and assemble them to obtain global stiffness matrix.

Element 1: . 619 . 4 , 60 m L = ° = ? Nodal points 4-1

[]
?
?
?
?
?
?
?
?
?
?
?
?
- -
- -
- -
- -
=
75 . 0 433 . 0 75 . 0 433 . 0
433 . 0 25 . 0 433 . 0 25 . 0
75 . 0 433 . 0 75 . 0 433 . 0
433 . 0 25 . 0 433 . 0 25 . 0
619 . 4
1
EA
k   (1)

Element 2: . 00 . 4 , 90 m L = ° = ? Nodal points 2-1

[]
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
0 . 4
2
EA
k       (2)

Element 3: . 619 . 4 , 120 m L = ° = ? Nodal points 3-1

[]
?
?
?
?
?
?
?
?
?
?
?
?
- -
- -
- -
- -
=
75 . 0 433 . 0 75 . 0 433 . 0
433 . 0 25 . 0 433 . 0 25 . 0
75 . 0 433 . 0 75 . 0 433 . 0
433 . 0 25 . 0 433 . 0 25 . 0
619 . 4
3
EA
k   (3)

Element 4: . 31 . 2 , 0 m L = ° = ? Nodal points 4-2

[]
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
31 . 2
4
EA
k      (4)

Element 5: . 31 . 2 , 0 m L = ° = ? Nodal points 2-3

[]
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
31 . 2
5
EA
k      (5)

The assembled global stiffness matrix of the truss is of the order 8 8 × . Now
assemble the global stiffness matrix. Note that the element
1
11
k of the member
stiffness matrix of truss member 1 goes to location ( ) 7 , 7 of global stiffness matrix.
On the member stiffness matrix the corresponding global degrees of freedom are
indicated to facilitate assembling. Thus,

[]
8
7
6
5
4
3
2
1
162 . 0 0934 . 0 0 0 0 0 162 . 0 094 . 0
0934 . 0 487 . 0 0 0 0 433 . 0 094 . 0 054 . 0
0 0 162 . 0 094 . 0 0 0 162 . 0 094 . 0
0 0 094 . 0 487 . 0 0 433 . 0 094 . 0 054 . 0
0 0 0 0 25 . 0 0 25 . 0 0
0 433 . 0 0 433 . 0 0 866 . 0 0 0
162 . 0 094 . 0 162 . 0 094 . 0 25 . 0 0 575 . 0 0
094 . 0 054 . 0 094 . 0 054 . 0 0 0 0 108 . 0
8 7 6 5 4 3 2 1
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- -
- - -
- -
- - -
-
- -
- - - -
- - -
= EA K

(6)

Writing the load-displacement relation for the truss, yields

?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- -
- - -
- -
- - -
-
- -
- - - -
- - -
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
8
7
6
5
4
3
2
1
8
7
6
5
4
3
2
1
162 . 0 0934 . 0 0 0 0 0 162 . 0 094 . 0
0934 . 0 487 . 0 0 0 0 433 . 0 094 . 0 054 . 0
0 0 162 . 0 094 . 0 0 0 162 . 0 094 . 0
0 0 094 . 0 487 . 0 0 433 . 0 094 . 0 054 . 0
0 0 0 0 25 . 0 0 25 . 0 0
0 433 . 0 0 433 . 0 0 866 . 0 0 0
162 . 0 094 . 0 162 . 0 094 . 0 25 . 0 0 575 . 0 0
094 . 0 054 . 0 094 . 0 054 . 0 0 0 0 108 . 0
u
u
u
u
u
u
u
u
EA
p
p
p
p
p
p
p
p
(7)

The displacements
1
u to
5
u are unknown. The displacements 0
8 7 6
= = = u u u .

Also  0
5 3 2 1
= = = = p p p p . But
4
10 kN p = - .

?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- -
-
-
-
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
5
4
3
2
1
487 . 0 0 433 . 0 094 . 0 054 . 0
0 25 . 0 0 25 . 0 0
433 . 0 0 866 . 0 0 0
094 . 0 25 . 0 0 575 . 0 0
054 . 0 0 0 0 108 . 0
0
10
0
0
0
u
u
u
u
u
EA
(8)

Solving which, the unknown displacements are evaluated. Thus,

Page 5

Instructional Objectives
After reading this chapter the student will be able to
1. Transform member stiffness matrix from local to global co-ordinate system.
2. Assemble member stiffness matrices to obtain the global stiffness matrix.
3. Analyse plane truss by the direct stiffness matrix.
4. Analyse plane truss supported on inclined roller supports.

25.1 Introduction
In the previous lesson, the direct stiffness method as applied to trusses was
discussed. The transformation of force and displacement from local co-ordinate
system to global co-ordinate system were accomplished by single transformation
matrix. Also assembly of the member stiffness matrices was discussed. In this
lesson few plane trusses are analysed using the direct stiffness method. Also the
problem of inclined support will be discussed.

Example 25.1
Analyse the truss shown in Fig. 25.1a and evaluate reactions. Assume EA to be
constant for all the members.

The numbering of joints and members are shown in Fig. 25.1b. Also, the possible
displacements (degrees of freedom) at each node are indicated. Here lower
numbers are used to indicate unconstrained degrees of freedom and higher
numbers are used for constrained degrees of freedom. Thus displacements 6,7
and 8 are zero due to boundary conditions.

First write down stiffness matrix of each member in global co-ordinate system
and assemble them to obtain global stiffness matrix.

Element 1: . 619 . 4 , 60 m L = ° = ? Nodal points 4-1

[]
?
?
?
?
?
?
?
?
?
?
?
?
- -
- -
- -
- -
=
75 . 0 433 . 0 75 . 0 433 . 0
433 . 0 25 . 0 433 . 0 25 . 0
75 . 0 433 . 0 75 . 0 433 . 0
433 . 0 25 . 0 433 . 0 25 . 0
619 . 4
1
EA
k   (1)

Element 2: . 00 . 4 , 90 m L = ° = ? Nodal points 2-1

[]
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
0 . 4
2
EA
k       (2)

Element 3: . 619 . 4 , 120 m L = ° = ? Nodal points 3-1

[]
?
?
?
?
?
?
?
?
?
?
?
?
- -
- -
- -
- -
=
75 . 0 433 . 0 75 . 0 433 . 0
433 . 0 25 . 0 433 . 0 25 . 0
75 . 0 433 . 0 75 . 0 433 . 0
433 . 0 25 . 0 433 . 0 25 . 0
619 . 4
3
EA
k   (3)

Element 4: . 31 . 2 , 0 m L = ° = ? Nodal points 4-2

[]
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
31 . 2
4
EA
k      (4)

Element 5: . 31 . 2 , 0 m L = ° = ? Nodal points 2-3

[]
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
31 . 2
5
EA
k      (5)

The assembled global stiffness matrix of the truss is of the order 8 8 × . Now
assemble the global stiffness matrix. Note that the element
1
11
k of the member
stiffness matrix of truss member 1 goes to location ( ) 7 , 7 of global stiffness matrix.
On the member stiffness matrix the corresponding global degrees of freedom are
indicated to facilitate assembling. Thus,

[]
8
7
6
5
4
3
2
1
162 . 0 0934 . 0 0 0 0 0 162 . 0 094 . 0
0934 . 0 487 . 0 0 0 0 433 . 0 094 . 0 054 . 0
0 0 162 . 0 094 . 0 0 0 162 . 0 094 . 0
0 0 094 . 0 487 . 0 0 433 . 0 094 . 0 054 . 0
0 0 0 0 25 . 0 0 25 . 0 0
0 433 . 0 0 433 . 0 0 866 . 0 0 0
162 . 0 094 . 0 162 . 0 094 . 0 25 . 0 0 575 . 0 0
094 . 0 054 . 0 094 . 0 054 . 0 0 0 0 108 . 0
8 7 6 5 4 3 2 1
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- -
- - -
- -
- - -
-
- -
- - - -
- - -
= EA K

(6)

Writing the load-displacement relation for the truss, yields

?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- -
- - -
- -
- - -
-
- -
- - - -
- - -
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
8
7
6
5
4
3
2
1
8
7
6
5
4
3
2
1
162 . 0 0934 . 0 0 0 0 0 162 . 0 094 . 0
0934 . 0 487 . 0 0 0 0 433 . 0 094 . 0 054 . 0
0 0 162 . 0 094 . 0 0 0 162 . 0 094 . 0
0 0 094 . 0 487 . 0 0 433 . 0 094 . 0 054 . 0
0 0 0 0 25 . 0 0 25 . 0 0
0 433 . 0 0 433 . 0 0 866 . 0 0 0
162 . 0 094 . 0 162 . 0 094 . 0 25 . 0 0 575 . 0 0
094 . 0 054 . 0 094 . 0 054 . 0 0 0 0 108 . 0
u
u
u
u
u
u
u
u
EA
p
p
p
p
p
p
p
p
(7)

The displacements
1
u to
5
u are unknown. The displacements 0
8 7 6
= = = u u u .

Also  0
5 3 2 1
= = = = p p p p . But
4
10 kN p = - .

?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- -
-
-
-
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
5
4
3
2
1
487 . 0 0 433 . 0 094 . 0 054 . 0
0 25 . 0 0 25 . 0 0
433 . 0 0 866 . 0 0 0
094 . 0 25 . 0 0 575 . 0 0
054 . 0 0 0 0 108 . 0
0
10
0
0
0
u
u
u
u
u
EA
(8)

Solving which, the unknown displacements are evaluated. Thus,

AE
u
AE
u
AE
u
AE
u
AE
u
334 . 13
;
642 . 74
;
668 . 6
;
64 . 34
;
668 . 6
5 4 3 2 1
=
-
= =
-
= =  (9)
Now reactions are evaluated from equation,

?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
?
?
?
?
?
?
?
?
?
?
- -
- - -
- -
=
?
?
?
?
?
?
?
?
?
?
334 . 13
642 . 74
668 . 6
64 . 34
668 . 6
1
0 0 0 162 . 0 094 . 0
0 0 433 . 0 094 . 0 054 . 0
094 . 0 0 0 162 . 0 094 . 0
8
7
6
EA
EA
p
p
p
(10)

Thus,

67 8
5.00 kN ; 0 ; 5.00 kN pp p == = .    (11)

Now calculate individual member forces.

Member 1:  m L m l 619 . 4 ; 866 . 0 ; 50 . 0 = = = .

{} []
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- - =
2
1
8
7
1
619 . 4
'
u
u
u
u
m l m l
AE
p

{} []
1
6.667
1
' 0.5 0.866 5.77 kN
34.64 4.619
AE
p
AE
??
=- - =
??
-
??
(12)

Member 2:  m L m l 0 . 4 ; 0 . 1 ; 0 = = = .

{}[]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- - =
2
1
4
3
1
0 . 4
'
u
u
u
u
m l m l
AE
p

{} []
1
74.642
1
' 1 1 10.0 kN
34.64 4.619
AE
p
AE
-??
=- =-
??
-
??
(13)

Member 3:  m L m l 619 . 4 ; 866 . 0 ; 50 . 0 = = - = .

```
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## Structural Analysis

30 videos|122 docs|28 tests

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