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# The Force Method of Analysis: Trusses - 3 Civil Engineering (CE) Notes | EduRev

## Civil Engineering (CE) : The Force Method of Analysis: Trusses - 3 Civil Engineering (CE) Notes | EduRev

``` Page 1

Member
i
L
i i
E A Forces in
the
released
truss due
to applied
i
P
Forces in
the
released
truss due
to unit
i v
P
Forces in
the
released
truss due
to unit
i v
Q

()
AE
L
P P
i
i v i

()
AE
L
Q P
i
i v i

m
( )
5
10 kN
kN kN kN
( )
4
10
-
m ( )
4
10
-
m
AB 4 3 40 +1 0 5.333 0.000
BC 4 3 60 +1 -0.8 8.000 -6.400
CD 4 3 60 +1 0 8.000 0.000
EF 4 3 -20 0 -0.8 0.000 2.133
EB 3 2 15 0 -0.6 0.000 -1.350
FC 3 2 0 0 -0.6 0.000 0.000
AE 5 4 -25 0 0 0.000 0.000
BF 5 4 -25 0 +1 0.000 -3.125
FD 5 4 -75 0 0 0.000 0.000
EC 5 4 0 0 +1 0.000 0.000
Total  21.333 -8.742

Deflection of the released structure along redundant and respectively are,
1
R
2
R

()
4
1
21.33 10 m (towards right)
L
-
?= ×

()
4
2
8.742 10 m (shortening)
L
-
?=- ×   (1)

In the next step, compute the flexibility coefficients (ref. Fig. 10.3c and Fig. 10.3d
and the accompanying table)

Page 2

Member
i
L
i i
E A Forces in
the
released
truss due
to applied
i
P
Forces in
the
released
truss due
to unit
i v
P
Forces in
the
released
truss due
to unit
i v
Q

()
AE
L
P P
i
i v i

()
AE
L
Q P
i
i v i

m
( )
5
10 kN
kN kN kN
( )
4
10
-
m ( )
4
10
-
m
AB 4 3 40 +1 0 5.333 0.000
BC 4 3 60 +1 -0.8 8.000 -6.400
CD 4 3 60 +1 0 8.000 0.000
EF 4 3 -20 0 -0.8 0.000 2.133
EB 3 2 15 0 -0.6 0.000 -1.350
FC 3 2 0 0 -0.6 0.000 0.000
AE 5 4 -25 0 0 0.000 0.000
BF 5 4 -25 0 +1 0.000 -3.125
FD 5 4 -75 0 0 0.000 0.000
EC 5 4 0 0 +1 0.000 0.000
Total  21.333 -8.742

Deflection of the released structure along redundant and respectively are,
1
R
2
R

()
4
1
21.33 10 m (towards right)
L
-
?= ×

()
4
2
8.742 10 m (shortening)
L
-
?=- ×   (1)

In the next step, compute the flexibility coefficients (ref. Fig. 10.3c and Fig. 10.3d
and the accompanying table)

Table 10.3b Computation of flexibility coefficients

Member
i
L
i i
E A ()
i v
P
()
i i
i
i v
E A
L
P
2

( )
i v
Q
()
i i
i
i v
E A
L
Q
2
()( )
AE
L
Q P
i
i v i v

m
( )
5
10 kN
kN
( )
5
10
-
m/kN
kN
( )
5
10
-
m/kN ( )
5
10
-
m/kN
AB 4 3 +1.00 1.333 0.000 0.000 0.000
BC 4 3 +1.00 1.333 -0.800 0.853 -1.067
CD 4 3 +1.00 1.333 0.000 0.000 0.000
EF 4 3 0 0.000 -0.800 0.853 0.000
EB 3 2 0 0.000 -0.600 0.540 0.000
FC 3 2 0 0.000 -0.600 0.540 0.000
AE 5 4 0 0.000 0.000 0.000 0.000
BF 5 4 0 0.000 1.000 1.250 0.000
FD 5 4 0 0.000 0.000 0.000 0.000
EC 5 4 0 0.000 1.000 1.250 0.000
Total 4.000   5.286 -1.064

Thus,

5
22
5
21 12
5
11
10 286 . 5
10 064 . 1
10 4
-
-
-
× =
× - = =
× =
a
a a
a
(2)

Analysis of truss when only external loads are acting
The compatibility conditions of the problem may be written as,

( ) 0
2 12 1 11 1
= + + ? R a R a
L

( ) 0
2 22 1 21 2
= + + ? R a R a
L
(3)

Solving     and
1
51.73 kN (towards left) R =-
2
6.136 kN (tensile) R =

The actual member forces and reactions in the truss are shown in Fig 10.3c.
Now, compute deflections corresponding to redundants due to rise in
temperature in the member .  Due to rise in temperature, the change in length
of member is,
FB
FB

Page 3

Member
i
L
i i
E A Forces in
the
released
truss due
to applied
i
P
Forces in
the
released
truss due
to unit
i v
P
Forces in
the
released
truss due
to unit
i v
Q

()
AE
L
P P
i
i v i

()
AE
L
Q P
i
i v i

m
( )
5
10 kN
kN kN kN
( )
4
10
-
m ( )
4
10
-
m
AB 4 3 40 +1 0 5.333 0.000
BC 4 3 60 +1 -0.8 8.000 -6.400
CD 4 3 60 +1 0 8.000 0.000
EF 4 3 -20 0 -0.8 0.000 2.133
EB 3 2 15 0 -0.6 0.000 -1.350
FC 3 2 0 0 -0.6 0.000 0.000
AE 5 4 -25 0 0 0.000 0.000
BF 5 4 -25 0 +1 0.000 -3.125
FD 5 4 -75 0 0 0.000 0.000
EC 5 4 0 0 +1 0.000 0.000
Total  21.333 -8.742

Deflection of the released structure along redundant and respectively are,
1
R
2
R

()
4
1
21.33 10 m (towards right)
L
-
?= ×

()
4
2
8.742 10 m (shortening)
L
-
?=- ×   (1)

In the next step, compute the flexibility coefficients (ref. Fig. 10.3c and Fig. 10.3d
and the accompanying table)

Table 10.3b Computation of flexibility coefficients

Member
i
L
i i
E A ()
i v
P
()
i i
i
i v
E A
L
P
2

( )
i v
Q
()
i i
i
i v
E A
L
Q
2
()( )
AE
L
Q P
i
i v i v

m
( )
5
10 kN
kN
( )
5
10
-
m/kN
kN
( )
5
10
-
m/kN ( )
5
10
-
m/kN
AB 4 3 +1.00 1.333 0.000 0.000 0.000
BC 4 3 +1.00 1.333 -0.800 0.853 -1.067
CD 4 3 +1.00 1.333 0.000 0.000 0.000
EF 4 3 0 0.000 -0.800 0.853 0.000
EB 3 2 0 0.000 -0.600 0.540 0.000
FC 3 2 0 0.000 -0.600 0.540 0.000
AE 5 4 0 0.000 0.000 0.000 0.000
BF 5 4 0 0.000 1.000 1.250 0.000
FD 5 4 0 0.000 0.000 0.000 0.000
EC 5 4 0 0.000 1.000 1.250 0.000
Total 4.000   5.286 -1.064

Thus,

5
22
5
21 12
5
11
10 286 . 5
10 064 . 1
10 4
-
-
-
× =
× - = =
× =
a
a a
a
(2)

Analysis of truss when only external loads are acting
The compatibility conditions of the problem may be written as,

( ) 0
2 12 1 11 1
= + + ? R a R a
L

( ) 0
2 22 1 21 2
= + + ? R a R a
L
(3)

Solving     and
1
51.73 kN (towards left) R =-
2
6.136 kN (tensile) R =

The actual member forces and reactions in the truss are shown in Fig 10.3c.
Now, compute deflections corresponding to redundants due to rise in
temperature in the member .  Due to rise in temperature, the change in length
of member is,
FB
FB

m
L T
T
3
10 67 . 2 5 40
75000
1
-
× = × × =
= ? a
(4)

Due to change in temperature, the deflections corresponding to redundants
and  are
1
R
2
R

() ( ) ( )
() ( )() m Q
P
i
T i v T
i
T i v T
3
2
1
10 67 . 2
0
-
× = ? = ?
= ? = ?
?
?
(5)

compatibility equations can be written as,

()( ) 0
2 12 1 11 1 1
= + + ? + ? R a R a
T L

() ( ) 0
2 22 1 21 2 2
= + + ? + ? R a R a
T L
(6)

Solving     and
1
65.92 kN(towards left) R =-
2
47.26 kN (compressive) R = -

The actual member forces and reactions are shown in Fig. 10.3f

Page 4

Member
i
L
i i
E A Forces in
the
released
truss due
to applied
i
P
Forces in
the
released
truss due
to unit
i v
P
Forces in
the
released
truss due
to unit
i v
Q

()
AE
L
P P
i
i v i

()
AE
L
Q P
i
i v i

m
( )
5
10 kN
kN kN kN
( )
4
10
-
m ( )
4
10
-
m
AB 4 3 40 +1 0 5.333 0.000
BC 4 3 60 +1 -0.8 8.000 -6.400
CD 4 3 60 +1 0 8.000 0.000
EF 4 3 -20 0 -0.8 0.000 2.133
EB 3 2 15 0 -0.6 0.000 -1.350
FC 3 2 0 0 -0.6 0.000 0.000
AE 5 4 -25 0 0 0.000 0.000
BF 5 4 -25 0 +1 0.000 -3.125
FD 5 4 -75 0 0 0.000 0.000
EC 5 4 0 0 +1 0.000 0.000
Total  21.333 -8.742

Deflection of the released structure along redundant and respectively are,
1
R
2
R

()
4
1
21.33 10 m (towards right)
L
-
?= ×

()
4
2
8.742 10 m (shortening)
L
-
?=- ×   (1)

In the next step, compute the flexibility coefficients (ref. Fig. 10.3c and Fig. 10.3d
and the accompanying table)

Table 10.3b Computation of flexibility coefficients

Member
i
L
i i
E A ()
i v
P
()
i i
i
i v
E A
L
P
2

( )
i v
Q
()
i i
i
i v
E A
L
Q
2
()( )
AE
L
Q P
i
i v i v

m
( )
5
10 kN
kN
( )
5
10
-
m/kN
kN
( )
5
10
-
m/kN ( )
5
10
-
m/kN
AB 4 3 +1.00 1.333 0.000 0.000 0.000
BC 4 3 +1.00 1.333 -0.800 0.853 -1.067
CD 4 3 +1.00 1.333 0.000 0.000 0.000
EF 4 3 0 0.000 -0.800 0.853 0.000
EB 3 2 0 0.000 -0.600 0.540 0.000
FC 3 2 0 0.000 -0.600 0.540 0.000
AE 5 4 0 0.000 0.000 0.000 0.000
BF 5 4 0 0.000 1.000 1.250 0.000
FD 5 4 0 0.000 0.000 0.000 0.000
EC 5 4 0 0.000 1.000 1.250 0.000
Total 4.000   5.286 -1.064

Thus,

5
22
5
21 12
5
11
10 286 . 5
10 064 . 1
10 4
-
-
-
× =
× - = =
× =
a
a a
a
(2)

Analysis of truss when only external loads are acting
The compatibility conditions of the problem may be written as,

( ) 0
2 12 1 11 1
= + + ? R a R a
L

( ) 0
2 22 1 21 2
= + + ? R a R a
L
(3)

Solving     and
1
51.73 kN (towards left) R =-
2
6.136 kN (tensile) R =

The actual member forces and reactions in the truss are shown in Fig 10.3c.
Now, compute deflections corresponding to redundants due to rise in
temperature in the member .  Due to rise in temperature, the change in length
of member is,
FB
FB

m
L T
T
3
10 67 . 2 5 40
75000
1
-
× = × × =
= ? a
(4)

Due to change in temperature, the deflections corresponding to redundants
and  are
1
R
2
R

() ( ) ( )
() ( )() m Q
P
i
T i v T
i
T i v T
3
2
1
10 67 . 2
0
-
× = ? = ?
= ? = ?
?
?
(5)

compatibility equations can be written as,

()( ) 0
2 12 1 11 1 1
= + + ? + ? R a R a
T L

() ( ) 0
2 22 1 21 2 2
= + + ? + ? R a R a
T L
(6)

Solving     and
1
65.92 kN(towards left) R =-
2
47.26 kN (compressive) R = -

The actual member forces and reactions are shown in Fig. 10.3f

Page 5

Member
i
L
i i
E A Forces in
the
released
truss due
to applied
i
P
Forces in
the
released
truss due
to unit
i v
P
Forces in
the
released
truss due
to unit
i v
Q

()
AE
L
P P
i
i v i

()
AE
L
Q P
i
i v i

m
( )
5
10 kN
kN kN kN
( )
4
10
-
m ( )
4
10
-
m
AB 4 3 40 +1 0 5.333 0.000
BC 4 3 60 +1 -0.8 8.000 -6.400
CD 4 3 60 +1 0 8.000 0.000
EF 4 3 -20 0 -0.8 0.000 2.133
EB 3 2 15 0 -0.6 0.000 -1.350
FC 3 2 0 0 -0.6 0.000 0.000
AE 5 4 -25 0 0 0.000 0.000
BF 5 4 -25 0 +1 0.000 -3.125
FD 5 4 -75 0 0 0.000 0.000
EC 5 4 0 0 +1 0.000 0.000
Total  21.333 -8.742

Deflection of the released structure along redundant and respectively are,
1
R
2
R

()
4
1
21.33 10 m (towards right)
L
-
?= ×

()
4
2
8.742 10 m (shortening)
L
-
?=- ×   (1)

In the next step, compute the flexibility coefficients (ref. Fig. 10.3c and Fig. 10.3d
and the accompanying table)

Table 10.3b Computation of flexibility coefficients

Member
i
L
i i
E A ()
i v
P
()
i i
i
i v
E A
L
P
2

( )
i v
Q
()
i i
i
i v
E A
L
Q
2
()( )
AE
L
Q P
i
i v i v

m
( )
5
10 kN
kN
( )
5
10
-
m/kN
kN
( )
5
10
-
m/kN ( )
5
10
-
m/kN
AB 4 3 +1.00 1.333 0.000 0.000 0.000
BC 4 3 +1.00 1.333 -0.800 0.853 -1.067
CD 4 3 +1.00 1.333 0.000 0.000 0.000
EF 4 3 0 0.000 -0.800 0.853 0.000
EB 3 2 0 0.000 -0.600 0.540 0.000
FC 3 2 0 0.000 -0.600 0.540 0.000
AE 5 4 0 0.000 0.000 0.000 0.000
BF 5 4 0 0.000 1.000 1.250 0.000
FD 5 4 0 0.000 0.000 0.000 0.000
EC 5 4 0 0.000 1.000 1.250 0.000
Total 4.000   5.286 -1.064

Thus,

5
22
5
21 12
5
11
10 286 . 5
10 064 . 1
10 4
-
-
-
× =
× - = =
× =
a
a a
a
(2)

Analysis of truss when only external loads are acting
The compatibility conditions of the problem may be written as,

( ) 0
2 12 1 11 1
= + + ? R a R a
L

( ) 0
2 22 1 21 2
= + + ? R a R a
L
(3)

Solving     and
1
51.73 kN (towards left) R =-
2
6.136 kN (tensile) R =

The actual member forces and reactions in the truss are shown in Fig 10.3c.
Now, compute deflections corresponding to redundants due to rise in
temperature in the member .  Due to rise in temperature, the change in length
of member is,
FB
FB

m
L T
T
3
10 67 . 2 5 40
75000
1
-
× = × × =
= ? a
(4)

Due to change in temperature, the deflections corresponding to redundants
and  are
1
R
2
R

() ( ) ( )
() ( )() m Q
P
i
T i v T
i
T i v T
3
2
1
10 67 . 2
0
-
× = ? = ?
= ? = ?
?
?
(5)

compatibility equations can be written as,

()( ) 0
2 12 1 11 1 1
= + + ? + ? R a R a
T L

() ( ) 0
2 22 1 21 2 2
= + + ? + ? R a R a
T L
(6)

Solving     and
1
65.92 kN(towards left) R =-
2
47.26 kN (compressive) R = -

The actual member forces and reactions are shown in Fig. 10.3f

Summary
In this lesson, the flexibility matrix method is used to analyse statically
indeterminate planar trusses. The equation to calculate the degree of statical
indeterminacy of a planar truss is derived. The forces induced in the members
due to temperature loading and member lack of fit is also discussed in this
lesson.  Few examples are solved to illustrate the force method of analysis as
applied to statically indeterminate planar trusses.

```
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## Structural Analysis

30 videos|122 docs|28 tests

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