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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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FAQs on The Force Method of Analysis: Trusses - 3 - Structural Analysis - Civil Engineering (CE)
| 1. What is the Force Method of Analysis for trusses? |  |
Ans. The Force Method of Analysis is a structural analysis technique used to determine the internal forces and displacements of truss structures. It involves breaking down the truss into individual members and analyzing each member separately, considering the equilibrium of forces at each joint.
| 2. How does the Force Method of Analysis work for trusses? | |
Ans. The Force Method of Analysis works by assuming unknown forces in the truss members and then applying the equations of equilibrium to determine these forces. The method involves solving a series of simultaneous equations based on the equilibrium conditions at each joint, considering both the external loads and the internal member forces.
| 3. What are the advantages of using the Force Method of Analysis for trusses? | |
Ans. Some advantages of using the Force Method of Analysis for trusses include:
- It provides a detailed understanding of the internal forces and displacements within the truss.
- It allows for the analysis of complex truss structures with multiple supports and loads.
- It is a systematic and practical approach that can be easily applied to real-world engineering problems.
- It can account for the effects of member stiffness, material properties, and boundary conditions on the truss behavior.
- It can be used to assess the stability and strength of the truss structure.
| 4. Are there any limitations or assumptions of the Force Method of Analysis for trusses? | |
Ans. Yes, there are some limitations and assumptions associated with the Force Method of Analysis for trusses. These include:
- It assumes that the truss members are connected by frictionless pins or hinges, allowing for rotation but no moment transfer.
- It assumes that the truss is statically determinate, meaning that the number of unknown forces can be determined solely from equilibrium conditions.
- It neglects the effects of member deformations due to axial, bending, or torsional stiffness, assuming that the members remain linearly elastic.
- It assumes that the joints are perfectly rigid, without any deformation or rotation.
- It assumes that the truss is loaded within the elastic range of the materials used.
| 5. How can the Force Method of Analysis be applied in practical civil engineering projects? | |
Ans. The Force Method of Analysis can be applied in practical civil engineering projects by following these steps:
1. Identify the truss structure to be analyzed and gather information about its geometry, member properties, and support conditions.
2. Determine the external loads acting on the truss, including point loads, distributed loads, and moments.
3. Break down the truss into individual members and assign unknown forces to each member.
4. Apply the equations of equilibrium at each joint to develop a system of simultaneous equations.
5. Solve the simultaneous equations to determine the unknown forces in the truss members.
6. Calculate the displacements and deformations of the truss members using the known forces and member properties.
7. Evaluate the stability, strength, and performance of the truss structure based on the calculated internal forces and displacements.
8. Make any necessary adjustments or modifications to the truss design to ensure its safety and functionality.
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The Force Method of Analysis: Trusses - 3 Notes
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