Courses

# The Direct Stiffness Method: Temperature Changes & Fabrication Errors in Truss Analysis - 3 GATE Notes | EduRev

## GATE : The Direct Stiffness Method: Temperature Changes & Fabrication Errors in Truss Analysis - 3 GATE Notes | EduRev

``` Page 1

4
300
'
2
= p
[0 -1 0 1]
L
L AE
u
u
u
u
?
-
?
?
?
?
?
?
?
?
?
?
?
?
?
?
2
1
4
3

()
( )
4
01 . 0 300
10 3478 . 4
4
300
3
-
- × - =
-

kN 424 . 0 4239 . 0 ? =      (7)

Example 26.3
Evaluate the member forces of truss shown in Fig.26.4a.The temperature of the
member BC is raised by C
o
40 and member BD is raised by C
o
50 .Assume
AE=300KN for all members and
75000
1
= a per
o
C.

Page 2

4
300
'
2
= p
[0 -1 0 1]
L
L AE
u
u
u
u
?
-
?
?
?
?
?
?
?
?
?
?
?
?
?
?
2
1
4
3

()
( )
4
01 . 0 300
10 3478 . 4
4
300
3
-
- × - =
-

kN 424 . 0 4239 . 0 ? =      (7)

Example 26.3
Evaluate the member forces of truss shown in Fig.26.4a.The temperature of the
member BC is raised by C
o
40 and member BD is raised by C
o
50 .Assume
AE=300KN for all members and
75000
1
= a per
o
C.

Solution
For this problem assembled stiffness matrix is available in Fig.26.4b.The joints
and members are numbered as shown in Fig.26.4b. In the given problem
4 3 2 1
, , , u u u u and
5
u represent unconstrained degrees of freedom. Due to support
conditions, 0
8 7 6
= = = u u u .

The temperature of the member (2) is raised by
o
50 C.Thus,

m T L L
3 2
10 333 . 3 50 5
75000
1
-
× = × × = ? = ? a
(1)

The forces are developed in member (2), as it was prevented from expansion.

()
()
()
()
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
× × =
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
?
?
?
?
sin
cos
sin
cos
10 333 . 3 300
3
2
1
8
7
f
f
f
f
p
p
p
p

Page 3

4
300
'
2
= p
[0 -1 0 1]
L
L AE
u
u
u
u
?
-
?
?
?
?
?
?
?
?
?
?
?
?
?
?
2
1
4
3

()
( )
4
01 . 0 300
10 3478 . 4
4
300
3
-
- × - =
-

kN 424 . 0 4239 . 0 ? =      (7)

Example 26.3
Evaluate the member forces of truss shown in Fig.26.4a.The temperature of the
member BC is raised by C
o
40 and member BD is raised by C
o
50 .Assume
AE=300KN for all members and
75000
1
= a per
o
C.

Solution
For this problem assembled stiffness matrix is available in Fig.26.4b.The joints
and members are numbered as shown in Fig.26.4b. In the given problem
4 3 2 1
, , , u u u u and
5
u represent unconstrained degrees of freedom. Due to support
conditions, 0
8 7 6
= = = u u u .

The temperature of the member (2) is raised by
o
50 C.Thus,

m T L L
3 2
10 333 . 3 50 5
75000
1
-
× = × × = ? = ? a
(1)

The forces are developed in member (2), as it was prevented from expansion.

()
()
()
()
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
× × =
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
?
?
?
?
sin
cos
sin
cos
10 333 . 3 300
3
2
1
8
7
f
f
f
f
p
p
p
p

?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
=
1
0
1
0
(2)

The displacement of the member (5) was raised by C
o
40 . Thus,

m T L L
3 5
10 771 . 3 40 2 5
000 , 75
1
-
× = × × = ? = ? a

The forces developed in member (5) as it was not allowed to expand is

()
()
()
()
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
× × =
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
707 . 0
707 . 0
707 . 0
707 . 0
10 771 . 3 300
3
8
7
6
5
t
t
t
t
p
p
p
p

?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
1
1
1
1
8 . 0
(3)

The global force vector due to thermal load is

()
()
()
()
()
()
()
()
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
1
0
8 . 0
8 . 0
0
0
8 . 1
8 . 0
8
7
6
5
4
3
2
1
t
t
t
t
t
t
t
t
p
p
p
p
p
p
p
p
(4)

Writing the load-displacement relation for the entire truss is given below.

Page 4

4
300
'
2
= p
[0 -1 0 1]
L
L AE
u
u
u
u
?
-
?
?
?
?
?
?
?
?
?
?
?
?
?
?
2
1
4
3

()
( )
4
01 . 0 300
10 3478 . 4
4
300
3
-
- × - =
-

kN 424 . 0 4239 . 0 ? =      (7)

Example 26.3
Evaluate the member forces of truss shown in Fig.26.4a.The temperature of the
member BC is raised by C
o
40 and member BD is raised by C
o
50 .Assume
AE=300KN for all members and
75000
1
= a per
o
C.

Solution
For this problem assembled stiffness matrix is available in Fig.26.4b.The joints
and members are numbered as shown in Fig.26.4b. In the given problem
4 3 2 1
, , , u u u u and
5
u represent unconstrained degrees of freedom. Due to support
conditions, 0
8 7 6
= = = u u u .

The temperature of the member (2) is raised by
o
50 C.Thus,

m T L L
3 2
10 333 . 3 50 5
75000
1
-
× = × × = ? = ? a
(1)

The forces are developed in member (2), as it was prevented from expansion.

()
()
()
()
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
× × =
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
?
?
?
?
sin
cos
sin
cos
10 333 . 3 300
3
2
1
8
7
f
f
f
f
p
p
p
p

?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
=
1
0
1
0
(2)

The displacement of the member (5) was raised by C
o
40 . Thus,

m T L L
3 5
10 771 . 3 40 2 5
000 , 75
1
-
× = × × = ? = ? a

The forces developed in member (5) as it was not allowed to expand is

()
()
()
()
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
× × =
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
707 . 0
707 . 0
707 . 0
707 . 0
10 771 . 3 300
3
8
7
6
5
t
t
t
t
p
p
p
p

?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
1
1
1
1
8 . 0
(3)

The global force vector due to thermal load is

()
()
()
()
()
()
()
()
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
1
0
8 . 0
8 . 0
0
0
8 . 1
8 . 0
8
7
6
5
4
3
2
1
t
t
t
t
t
t
t
t
p
p
p
p
p
p
p
p
(4)

Writing the load-displacement relation for the entire truss is given below.

?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- - -
- - -
- - -
- - -
- -
- - -
- - -
- - -
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
1
0
8 . 0
8 . 0
0
0
8 . 1
8 . 0
271 . 0 071 . 0 0 0 071 . 0 071 . 0 2 . 0 0
071 . 0 271 . 0 0 2 . 0 071 . 0 071 . 0 0 0
0 0 271 . 0 071 . 0 2 . 0 0 071 . 0 071 . 0
0 2 . 0 071 . 0 271 . 0 0 0 071 . 0 071 . 0
071 . 0 071 . 0 2 . 0 0 129 . 0 071 . 0 0 0
071 . 0 071 . 0 0 0 071 . 0 271 . 0 0 20 . 0
2 . 0 0 071 . 0 071 . 0 0 0 271 . 0 071 . 0
0 0 071 . 0 071 . 0 0 20 . 0 071 . 0 271 . 0
8
7
6
5
4
3
2
1
8
7
6
5
4
3
2
1
u
u
u
u
u
u
u
u
AE
p
p
p
p
p
p
p
p
(5)

In the above problem 0
8 7 6 5 4 3 2 1
= = = = = = = = p p p p p p p p and
0
8 7 6
= = = u u u .

Thus solving for unknown displacements,

?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
-
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- -
-
- -
-
- -
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
8 . 0
0
0
8 . 1
8 . 0
0
0
0
0
0
271 . 0 0 0 071 . 0 071 . 0
0 129 . 0 071 . 0 0 0
0 071 . 0 271 . 0 0 20 . 0
071 . 0 0 0 271 . 0 071 . 0
071 . 0 0 2 . 0 071 . 0 271 . 0
1
5
4
3
2
1
AE
u
u
u
u
u
(5)

Solving equation (5), the unknown displacements are calculated as

m u
u m u m u m u
0013 . 0
0 , 0005 . 0 , 0020 . 0 , 0013 . 0
5
4 3 2 1
- =
= - = = =
(6)

Now, reactions are computed as,
?
?
?
?
?
?
?
?
?
?
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- -
- -
- - -
=
?
?
?
?
?
?
?
?
?
?
1
0
8 . 0
0 071 . 0 071 . 0 2 . 0 0
2 . 0 071 . 0 071 . 0 0 0
071 . 0 2 . 0 0 071 . 0 071 . 0
5
4
3
2
1
8
7
6
u
u
u
u
u
p
p
p
(7)

All reactions are zero as truss is externally determinate and hence change in
temperature does not induce any reaction. Now member forces are calculated by
using equation (26.10b)

Member (1): L=5m,
o
0 = ?

Page 5

4
300
'
2
= p
[0 -1 0 1]
L
L AE
u
u
u
u
?
-
?
?
?
?
?
?
?
?
?
?
?
?
?
?
2
1
4
3

()
( )
4
01 . 0 300
10 3478 . 4
4
300
3
-
- × - =
-

kN 424 . 0 4239 . 0 ? =      (7)

Example 26.3
Evaluate the member forces of truss shown in Fig.26.4a.The temperature of the
member BC is raised by C
o
40 and member BD is raised by C
o
50 .Assume
AE=300KN for all members and
75000
1
= a per
o
C.

Solution
For this problem assembled stiffness matrix is available in Fig.26.4b.The joints
and members are numbered as shown in Fig.26.4b. In the given problem
4 3 2 1
, , , u u u u and
5
u represent unconstrained degrees of freedom. Due to support
conditions, 0
8 7 6
= = = u u u .

The temperature of the member (2) is raised by
o
50 C.Thus,

m T L L
3 2
10 333 . 3 50 5
75000
1
-
× = × × = ? = ? a
(1)

The forces are developed in member (2), as it was prevented from expansion.

()
()
()
()
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
× × =
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
?
?
?
?
sin
cos
sin
cos
10 333 . 3 300
3
2
1
8
7
f
f
f
f
p
p
p
p

?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
=
1
0
1
0
(2)

The displacement of the member (5) was raised by C
o
40 . Thus,

m T L L
3 5
10 771 . 3 40 2 5
000 , 75
1
-
× = × × = ? = ? a

The forces developed in member (5) as it was not allowed to expand is

()
()
()
()
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
× × =
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
707 . 0
707 . 0
707 . 0
707 . 0
10 771 . 3 300
3
8
7
6
5
t
t
t
t
p
p
p
p

?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
1
1
1
1
8 . 0
(3)

The global force vector due to thermal load is

()
()
()
()
()
()
()
()
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
1
0
8 . 0
8 . 0
0
0
8 . 1
8 . 0
8
7
6
5
4
3
2
1
t
t
t
t
t
t
t
t
p
p
p
p
p
p
p
p
(4)

Writing the load-displacement relation for the entire truss is given below.

?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- - -
- - -
- - -
- - -
- -
- - -
- - -
- - -
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
1
0
8 . 0
8 . 0
0
0
8 . 1
8 . 0
271 . 0 071 . 0 0 0 071 . 0 071 . 0 2 . 0 0
071 . 0 271 . 0 0 2 . 0 071 . 0 071 . 0 0 0
0 0 271 . 0 071 . 0 2 . 0 0 071 . 0 071 . 0
0 2 . 0 071 . 0 271 . 0 0 0 071 . 0 071 . 0
071 . 0 071 . 0 2 . 0 0 129 . 0 071 . 0 0 0
071 . 0 071 . 0 0 0 071 . 0 271 . 0 0 20 . 0
2 . 0 0 071 . 0 071 . 0 0 0 271 . 0 071 . 0
0 0 071 . 0 071 . 0 0 20 . 0 071 . 0 271 . 0
8
7
6
5
4
3
2
1
8
7
6
5
4
3
2
1
u
u
u
u
u
u
u
u
AE
p
p
p
p
p
p
p
p
(5)

In the above problem 0
8 7 6 5 4 3 2 1
= = = = = = = = p p p p p p p p and
0
8 7 6
= = = u u u .

Thus solving for unknown displacements,

?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
-
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- -
-
- -
-
- -
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
8 . 0
0
0
8 . 1
8 . 0
0
0
0
0
0
271 . 0 0 0 071 . 0 071 . 0
0 129 . 0 071 . 0 0 0
0 071 . 0 271 . 0 0 20 . 0
071 . 0 0 0 271 . 0 071 . 0
071 . 0 0 2 . 0 071 . 0 271 . 0
1
5
4
3
2
1
AE
u
u
u
u
u
(5)

Solving equation (5), the unknown displacements are calculated as

m u
u m u m u m u
0013 . 0
0 , 0005 . 0 , 0020 . 0 , 0013 . 0
5
4 3 2 1
- =
= - = = =
(6)

Now, reactions are computed as,
?
?
?
?
?
?
?
?
?
?
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- -
- -
- - -
=
?
?
?
?
?
?
?
?
?
?
1
0
8 . 0
0 071 . 0 071 . 0 2 . 0 0
2 . 0 071 . 0 071 . 0 0 0
071 . 0 2 . 0 0 071 . 0 071 . 0
5
4
3
2
1
8
7
6
u
u
u
u
u
p
p
p
(7)

All reactions are zero as truss is externally determinate and hence change in
temperature does not induce any reaction. Now member forces are calculated by
using equation (26.10b)

Member (1): L=5m,
o
0 = ?

5
'
2
AE
p = [-1 0 1 0]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
2
1
4
3
u
u
u
u
(8)

=
'
2
p 0.1080 Kn

Member 2:  L=5m,
o
90 = ? ,nodal points 4-1

5
'
2
AE
p = [0 -1 0 1]
5
2
1
8
7
10 771 . 3 300
-
× × -
?
?
?
?
?
?
?
?
?
?
?
?
?
?
u
u
u
u
(9)

=0.1087 kN

Member (3): L=5m,
o
0 = ? ,nodal points 3-4

5
300
'
2
= p [-1 0 1 0]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
8
7
6
5
u
u
u
u
(10)

=0.0780kN

Member (4): , 5 , 90 m L = =
o
? nodal points 3-2

5
300
'
2
= p [0 -1 0 1]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
4
3
6
5
u
u
u
u
=0      (11)

Member (5): 2 5 , 45 = = L
o
? ,nodal points 3-1

2 5
300
'
2
= p [-0.707 -0.707 0.707 0.707]
3
2
1
6
5
10 333 . 3 300
-
× × -
?
?
?
?
?
?
?
?
?
?
?
?
?
?
u
u
u
u
(12)

=-0.8619 kN

Member (6) : 2 5 , 135 = = L
o
? ,nodal points 4-2
```
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

## Structural Analysis

30 videos|122 docs|28 tests

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;