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# The Direct Stiffness Method: Truss Analysis - 5 GATE Notes | EduRev

## : The Direct Stiffness Method: Truss Analysis - 5 GATE Notes | EduRev

``` Page 2

The joint and member numbers are indicated in Fig. 25.2b. The possible degree
of freedom are also shown in Fig. 25.2b. In the given problem
2 1
,u u and
3
u
represent unconstrained degrees of freedom and  0
8 7 6 5 4
= = = = = u u u u u due
to boundary condition. First let us generate stiffness matrix for each of the six
members in global co-ordinate system.

Element 1: . 00 . 5 , 0 m L = ° = ? Nodal points 2-1

[]
2
1
4
3
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
0 . 5
2 1 4 3
1
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
(1)

Element 2: . 00 . 5 , 90 m L = ° = ? Nodal points 4-1

Page 3

The joint and member numbers are indicated in Fig. 25.2b. The possible degree
of freedom are also shown in Fig. 25.2b. In the given problem
2 1
,u u and
3
u
represent unconstrained degrees of freedom and  0
8 7 6 5 4
= = = = = u u u u u due
to boundary condition. First let us generate stiffness matrix for each of the six
members in global co-ordinate system.

Element 1: . 00 . 5 , 0 m L = ° = ? Nodal points 2-1

[]
2
1
4
3
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
0 . 5
2 1 4 3
1
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
(1)

Element 2: . 00 . 5 , 90 m L = ° = ? Nodal points 4-1

[]
2
1
8
7
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
0 . 5
2 1 8 7
2
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
(2)

Element 3: . 00 . 5 , 0 m L = ° = ? Nodal points 3-4

[]
8
7
6
5
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
0 . 5
8 7 6 5
3
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
(3)

Element 4: . 00 . 5 , 90 m L = ° = ? Nodal points 3-2

[]
4
3
6
5
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
0 . 5
4 3 6 5
4
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
(4)

Element 5: . 07 . 7 , 45 m L = ° = ? Nodal points 3-1

[]
2
1
6
5
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
07 . 7
2 1 6 5
5
?
?
?
?
?
?
?
?
?
?
?
?
- -
- -
- -
- -
=
EA
k
(5)

Element 6: . 07 . 7 , 135 m L = ° = ? Nodal points 4-2

Page 4

The joint and member numbers are indicated in Fig. 25.2b. The possible degree
of freedom are also shown in Fig. 25.2b. In the given problem
2 1
,u u and
3
u
represent unconstrained degrees of freedom and  0
8 7 6 5 4
= = = = = u u u u u due
to boundary condition. First let us generate stiffness matrix for each of the six
members in global co-ordinate system.

Element 1: . 00 . 5 , 0 m L = ° = ? Nodal points 2-1

[]
2
1
4
3
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
0 . 5
2 1 4 3
1
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
(1)

Element 2: . 00 . 5 , 90 m L = ° = ? Nodal points 4-1

[]
2
1
8
7
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
0 . 5
2 1 8 7
2
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
(2)

Element 3: . 00 . 5 , 0 m L = ° = ? Nodal points 3-4

[]
8
7
6
5
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
0 . 5
8 7 6 5
3
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
(3)

Element 4: . 00 . 5 , 90 m L = ° = ? Nodal points 3-2

[]
4
3
6
5
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
0 . 5
4 3 6 5
4
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
(4)

Element 5: . 07 . 7 , 45 m L = ° = ? Nodal points 3-1

[]
2
1
6
5
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
07 . 7
2 1 6 5
5
?
?
?
?
?
?
?
?
?
?
?
?
- -
- -
- -
- -
=
EA
k
(5)

Element 6: . 07 . 7 , 135 m L = ° = ? Nodal points 4-2

[]
4
3
8
7
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
07 . 7
4 3 8 7
6
?
?
?
?
?
?
?
?
?
?
?
?
- -
- -
- -
- -
=
EA
k
(6)

There are eight possible global degrees of freedom for the truss shown in the
figure. Hence the global stiffness matrix is of the order ( 8 8 × ). On the member
stiffness matrix, the corresponding global degrees of freedom are indicated to
facilitate assembly. Thus the global stiffness matrix is,

[]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- - -
- - -
- - -
- - -
- - -
- - -
- - -
- - -
=
271 . 0 071 . 0 0 0 071 . 0 071 . 0 20 . 0 0
071 . 0 271 . 0 0 20 . 0 071 . 0 071 . 0 0 0
0 0 271 . 0 071 . 0 20 . 0 0 071 . 0 071 . 0
0 20 . 0 071 . 0 271 . 0 0 0 071 . 0 071 . 0
071 . 0 071 . 0 20 . 0 071 . 0 271 . 0 071 . 0 0 0
071 . 0 071 . 0 0 0 071 . 0 271 . 0 0 20 . 0
20 . 0 0 071 . 0 071 . 0 0 0 271 . 0 071 . 0
0 0 071 . 0 071 . 0 0 2 . 0 071 . 0 271 . 0
AE K     (7)

The force-displacement relation for the truss is,

?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- - -
- - -
- - -
- - -
- - -
- - -
- - -
- - -
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
8
7
6
5
4
3
2
1
8
7
6
5
4
3
2
1
271 . 0 071 . 0 0 0 071 . 0 071 . 0 20 . 0 0
071 . 0 271 . 0 0 20 . 0 071 . 0 071 . 0 0 0
0 0 271 . 0 071 . 0 20 . 0 0 071 . 0 071 . 0
0 20 . 0 071 . 0 271 . 0 0 0 071 . 0 071 . 0
071 . 0 071 . 0 20 . 0 071 . 0 271 . 0 071 . 0 0 0
071 . 0 071 . 0 0 0 071 . 0 271 . 0 0 20 . 0
20 . 0 0 071 . 0 071 . 0 0 0 271 . 0 071 . 0
0 0 071 . 0 071 . 0 0 2 . 0 071 . 0 271 . 0
u
u
u
u
u
u
u
u
EA
p
p
p
p
p
p
p
p

(8)

The displacements
2 1
,u u
and
3
u
are unknowns.
Here,
12 3
5kN ; 10 ; 0 pp p ==- =
and
0
8 7 6 5 4
= = = = = u u u u u
.
Page 5

The joint and member numbers are indicated in Fig. 25.2b. The possible degree
of freedom are also shown in Fig. 25.2b. In the given problem
2 1
,u u and
3
u
represent unconstrained degrees of freedom and  0
8 7 6 5 4
= = = = = u u u u u due
to boundary condition. First let us generate stiffness matrix for each of the six
members in global co-ordinate system.

Element 1: . 00 . 5 , 0 m L = ° = ? Nodal points 2-1

[]
2
1
4
3
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
0 . 5
2 1 4 3
1
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
(1)

Element 2: . 00 . 5 , 90 m L = ° = ? Nodal points 4-1

[]
2
1
8
7
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
0 . 5
2 1 8 7
2
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
(2)

Element 3: . 00 . 5 , 0 m L = ° = ? Nodal points 3-4

[]
8
7
6
5
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
0 . 5
8 7 6 5
3
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
(3)

Element 4: . 00 . 5 , 90 m L = ° = ? Nodal points 3-2

[]
4
3
6
5
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
0 . 5
4 3 6 5
4
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
EA
k
(4)

Element 5: . 07 . 7 , 45 m L = ° = ? Nodal points 3-1

[]
2
1
6
5
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
07 . 7
2 1 6 5
5
?
?
?
?
?
?
?
?
?
?
?
?
- -
- -
- -
- -
=
EA
k
(5)

Element 6: . 07 . 7 , 135 m L = ° = ? Nodal points 4-2

[]
4
3
8
7
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
5 . 0 5 . 0 5 . 0 5 . 0
07 . 7
4 3 8 7
6
?
?
?
?
?
?
?
?
?
?
?
?
- -
- -
- -
- -
=
EA
k
(6)

There are eight possible global degrees of freedom for the truss shown in the
figure. Hence the global stiffness matrix is of the order ( 8 8 × ). On the member
stiffness matrix, the corresponding global degrees of freedom are indicated to
facilitate assembly. Thus the global stiffness matrix is,

[]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- - -
- - -
- - -
- - -
- - -
- - -
- - -
- - -
=
271 . 0 071 . 0 0 0 071 . 0 071 . 0 20 . 0 0
071 . 0 271 . 0 0 20 . 0 071 . 0 071 . 0 0 0
0 0 271 . 0 071 . 0 20 . 0 0 071 . 0 071 . 0
0 20 . 0 071 . 0 271 . 0 0 0 071 . 0 071 . 0
071 . 0 071 . 0 20 . 0 071 . 0 271 . 0 071 . 0 0 0
071 . 0 071 . 0 0 0 071 . 0 271 . 0 0 20 . 0
20 . 0 0 071 . 0 071 . 0 0 0 271 . 0 071 . 0
0 0 071 . 0 071 . 0 0 2 . 0 071 . 0 271 . 0
AE K     (7)

The force-displacement relation for the truss is,

?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- - -
- - -
- - -
- - -
- - -
- - -
- - -
- - -
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
8
7
6
5
4
3
2
1
8
7
6
5
4
3
2
1
271 . 0 071 . 0 0 0 071 . 0 071 . 0 20 . 0 0
071 . 0 271 . 0 0 20 . 0 071 . 0 071 . 0 0 0
0 0 271 . 0 071 . 0 20 . 0 0 071 . 0 071 . 0
0 20 . 0 071 . 0 271 . 0 0 0 071 . 0 071 . 0
071 . 0 071 . 0 20 . 0 071 . 0 271 . 0 071 . 0 0 0
071 . 0 071 . 0 0 0 071 . 0 271 . 0 0 20 . 0
20 . 0 0 071 . 0 071 . 0 0 0 271 . 0 071 . 0
0 0 071 . 0 071 . 0 0 2 . 0 071 . 0 271 . 0
u
u
u
u
u
u
u
u
EA
p
p
p
p
p
p
p
p

(8)

The displacements
2 1
,u u
and
3
u
are unknowns.
Here,
12 3
5kN ; 10 ; 0 pp p ==- =
and
0
8 7 6 5 4
= = = = = u u u u u
.

?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- - -
- - -
- - -
- - -
- - -
- - -
- - -
- - -
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
0
0
0
0
0
271 . 0 071 . 0 0 0 071 . 0 071 . 0 20 . 0 0
071 . 0 271 . 0 0 20 . 0 071 . 0 071 . 0 0 0
0 0 271 . 0 071 . 0 20 . 0 0 071 . 0 071 . 0
0 20 . 0 071 . 0 271 . 0 0 0 071 . 0 071 . 0
071 . 0 071 . 0 20 . 0 071 . 0 271 . 0 071 . 0 0 0
071 . 0 071 . 0 0 0 071 . 0 271 . 0 0 20 . 0
20 . 0 0 071 . 0 071 . 0 0 0 271 . 0 071 . 0
0 0 071 . 0 071 . 0 0 2 . 0 071 . 0 271 . 0
0
10
5
3
2
1
8
7
6
5
4
u
u
u
EA
p
p
p
p
p
(9)

Thus,

?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
?
?
?
?
?
?
?
?
?
?
-
3
2
1
271 . 0 0 20 . 0
0 271 . 0 071 . 0
20 . 0 071 . 0 271 . 0
0
10
5
u
u
u
(10)

Solving which, yields

AE
u
AE
u
AE
u
825 . 53
;
97 . 55
;
855 . 72
3 2 1
=
-
= =

Now reactions are evaluated from the equation,

?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
- -
- -
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
3
2
1
8
7
6
5
4
071 . 0 20 . 0 0
071 . 0 0 0
0 071 . 0 071 . 0
0 071 . 0 071 . 0
071 . 0 0 0
u
u
u
p
p
p
p
p
(11)

45 6 7 8
3.80 kN ; 1.19 kN ; 1.19 kN ; 3.8 0 ; 15.00 kN pp p p kNp =- =- =- = =

In the next step evaluate forces in members.

Element 1: . 00 . 5 , 0 m L = ° = ? Nodal points 2-1

```
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