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

Structural Analysis

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

 Page 1


 
Table 10.3a Deflection due to external loading 
 
Member 
i
L 
i i
E A Forces in 
the 
released 
truss due 
to applied 
loading 
i
P 
Forces in 
the 
released 
truss due 
to unit 
load ( )
i v
P 
Forces in 
the 
released 
truss due 
to unit 
load ( )
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


 
Table 10.3a Deflection due to external loading 
 
Member 
i
L 
i i
E A Forces in 
the 
released 
truss due 
to applied 
loading 
i
P 
Forces in 
the 
released 
truss due 
to unit 
load ( )
i v
P 
Forces in 
the 
released 
truss due 
to unit 
load ( )
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


 
Table 10.3a Deflection due to external loading 
 
Member 
i
L 
i i
E A Forces in 
the 
released 
truss due 
to applied 
loading 
i
P 
Forces in 
the 
released 
truss due 
to unit 
load ( )
i v
P 
Forces in 
the 
released 
truss due 
to unit 
load ( )
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) 
 
When both external loading and temperature loading are acting  
When both temperature loading and the external loading are considered, the 
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


 
Table 10.3a Deflection due to external loading 
 
Member 
i
L 
i i
E A Forces in 
the 
released 
truss due 
to applied 
loading 
i
P 
Forces in 
the 
released 
truss due 
to unit 
load ( )
i v
P 
Forces in 
the 
released 
truss due 
to unit 
load ( )
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) 
 
When both external loading and temperature loading are acting  
When both temperature loading and the external loading are considered, the 
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


 
Table 10.3a Deflection due to external loading 
 
Member 
i
L 
i i
E A Forces in 
the 
released 
truss due 
to applied 
loading 
i
P 
Forces in 
the 
released 
truss due 
to unit 
load ( )
i v
P 
Forces in 
the 
released 
truss due 
to unit 
load ( )
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) 
 
When both external loading and temperature loading are acting  
When both temperature loading and the external loading are considered, the 
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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