The Force Method of Analysis: Trusses - 2 Civil Engineering (CE) Notes | EduRev

Structural Analysis

Civil Engineering (CE) : The Force Method of Analysis: Trusses - 2 Civil Engineering (CE) Notes | EduRev

 Page 2


 
 
The plane truss shown in Fg.10.2a is externally indeterminate to degree one. 
Truss is internally determinate. Select the horizontal reaction atE , as the 
redundant. Releasing the redundant (replacing the hinge at 
Ex
R
E by a roller support) 
a stable determinate truss is obtained as shown in Fig. 10.2b. The member axial 
forces and reactions of the released truss are shown in Fig. 10.2b. 
 
Now calculate the displacement 
L
? corresponding to redundant reaction in 
the released structure. This can be conveniently done in a table (see Figs. 10.2b, 
10.2c and the table). Hence from the table, 
Ex
R
 
()
4
15 10 m
i
Liv
i
ii
L
PP
AE
-
?=
=×
?
    (1) 
 
 
 
Page 3


 
 
The plane truss shown in Fg.10.2a is externally indeterminate to degree one. 
Truss is internally determinate. Select the horizontal reaction atE , as the 
redundant. Releasing the redundant (replacing the hinge at 
Ex
R
E by a roller support) 
a stable determinate truss is obtained as shown in Fig. 10.2b. The member axial 
forces and reactions of the released truss are shown in Fig. 10.2b. 
 
Now calculate the displacement 
L
? corresponding to redundant reaction in 
the released structure. This can be conveniently done in a table (see Figs. 10.2b, 
10.2c and the table). Hence from the table, 
Ex
R
 
()
4
15 10 m
i
Liv
i
ii
L
PP
AE
-
?=
=×
?
    (1) 
 
 
 
In the next step apply a unit load, along the redundant reaction  and calculate 
the displacement using unit load method. 
Ex
R
11
a
 
()
2
11
5
410 m
i
v
i
ii
L
aP
AE
-
=
=×
?
      (2) 
 
The support at E is hinged. Hence the total displacement at Emust vanish. 
Thus, 
      0
11
= + ?
AD L
F a    (3) 
 
0 10 4 10 15
5 4
= × + ×
- -
Ex
R 
 
4
5
15 10
410
37.5 kN(towards left)
Ex
R
-
-
×
=-
×
=-
 
 
The actual member forces and reactions are shown in Fig. 10.2d. 
 
Table 10.2 Numerical computation for example 10.2 
 
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 
 
()
AE
L
P P
i
i v i
 
 
()
i i
i
i v
E A
L
P
2
 
 
()
i v AD i
i
P F P
F
+
=
 
 m 
( )
5
10 kN 
kN kN 
( )
4
10
-
m ( )
5
10
-
m/Kn
kN 
AB 3 3 33.75 +1 3.375 1 -3.75 
BC 3 3 33.75 +1 3.375 1 -3.75 
CD 3 3 41.25 +1 4.125 1 3.75 
DE 3 3 41.25 +1 4.125 1 3.75 
FG 6 3 -7.50 0 0 0 -7.5 
FB 4 2 0.00 0 0 0 0 
GD 4 2 0.00 0 0 0 0 
AF 5 5 -6.25 0 0 0 -6.25 
FC 5 5 6.25 0 0 0 6.25 
CG 5 5 -6.25 0 0 0 -6.25 
GE 5 5 -68.75 0 0 0 -68.75 
    Total 15 4  
 
Page 4


 
 
The plane truss shown in Fg.10.2a is externally indeterminate to degree one. 
Truss is internally determinate. Select the horizontal reaction atE , as the 
redundant. Releasing the redundant (replacing the hinge at 
Ex
R
E by a roller support) 
a stable determinate truss is obtained as shown in Fig. 10.2b. The member axial 
forces and reactions of the released truss are shown in Fig. 10.2b. 
 
Now calculate the displacement 
L
? corresponding to redundant reaction in 
the released structure. This can be conveniently done in a table (see Figs. 10.2b, 
10.2c and the table). Hence from the table, 
Ex
R
 
()
4
15 10 m
i
Liv
i
ii
L
PP
AE
-
?=
=×
?
    (1) 
 
 
 
In the next step apply a unit load, along the redundant reaction  and calculate 
the displacement using unit load method. 
Ex
R
11
a
 
()
2
11
5
410 m
i
v
i
ii
L
aP
AE
-
=
=×
?
      (2) 
 
The support at E is hinged. Hence the total displacement at Emust vanish. 
Thus, 
      0
11
= + ?
AD L
F a    (3) 
 
0 10 4 10 15
5 4
= × + ×
- -
Ex
R 
 
4
5
15 10
410
37.5 kN(towards left)
Ex
R
-
-
×
=-
×
=-
 
 
The actual member forces and reactions are shown in Fig. 10.2d. 
 
Table 10.2 Numerical computation for example 10.2 
 
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 
 
()
AE
L
P P
i
i v i
 
 
()
i i
i
i v
E A
L
P
2
 
 
()
i v AD i
i
P F P
F
+
=
 
 m 
( )
5
10 kN 
kN kN 
( )
4
10
-
m ( )
5
10
-
m/Kn
kN 
AB 3 3 33.75 +1 3.375 1 -3.75 
BC 3 3 33.75 +1 3.375 1 -3.75 
CD 3 3 41.25 +1 4.125 1 3.75 
DE 3 3 41.25 +1 4.125 1 3.75 
FG 6 3 -7.50 0 0 0 -7.5 
FB 4 2 0.00 0 0 0 0 
GD 4 2 0.00 0 0 0 0 
AF 5 5 -6.25 0 0 0 -6.25 
FC 5 5 6.25 0 0 0 6.25 
CG 5 5 -6.25 0 0 0 -6.25 
GE 5 5 -68.75 0 0 0 -68.75 
    Total 15 4  
 
Example 10.3 
Determine the reactions and the member axial forces of the truss shown in 
Fig.10.3a by force method due to external load and rise in temperature of 
member  by . The cross sectional areas of the members in square 
centimeters are shown in parenthesis. Assume 
and
FB C ° 40
52
2.0 10 N/mm E=×
1
per °C
75000
a = . 
 
 
 
The given truss is indeterminate to second degree. The truss has both internal 
and external indeterminacy. Choose horizontal reaction at and the axial 
force in member as redundant actions. Releasing the restraint against 
redundant actions, a stable determinate truss is obtained as shown in Fig. 10.3b. 
D()
1
R
EC()
2
R
 
 
Page 5


 
 
The plane truss shown in Fg.10.2a is externally indeterminate to degree one. 
Truss is internally determinate. Select the horizontal reaction atE , as the 
redundant. Releasing the redundant (replacing the hinge at 
Ex
R
E by a roller support) 
a stable determinate truss is obtained as shown in Fig. 10.2b. The member axial 
forces and reactions of the released truss are shown in Fig. 10.2b. 
 
Now calculate the displacement 
L
? corresponding to redundant reaction in 
the released structure. This can be conveniently done in a table (see Figs. 10.2b, 
10.2c and the table). Hence from the table, 
Ex
R
 
()
4
15 10 m
i
Liv
i
ii
L
PP
AE
-
?=
=×
?
    (1) 
 
 
 
In the next step apply a unit load, along the redundant reaction  and calculate 
the displacement using unit load method. 
Ex
R
11
a
 
()
2
11
5
410 m
i
v
i
ii
L
aP
AE
-
=
=×
?
      (2) 
 
The support at E is hinged. Hence the total displacement at Emust vanish. 
Thus, 
      0
11
= + ?
AD L
F a    (3) 
 
0 10 4 10 15
5 4
= × + ×
- -
Ex
R 
 
4
5
15 10
410
37.5 kN(towards left)
Ex
R
-
-
×
=-
×
=-
 
 
The actual member forces and reactions are shown in Fig. 10.2d. 
 
Table 10.2 Numerical computation for example 10.2 
 
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 
 
()
AE
L
P P
i
i v i
 
 
()
i i
i
i v
E A
L
P
2
 
 
()
i v AD i
i
P F P
F
+
=
 
 m 
( )
5
10 kN 
kN kN 
( )
4
10
-
m ( )
5
10
-
m/Kn
kN 
AB 3 3 33.75 +1 3.375 1 -3.75 
BC 3 3 33.75 +1 3.375 1 -3.75 
CD 3 3 41.25 +1 4.125 1 3.75 
DE 3 3 41.25 +1 4.125 1 3.75 
FG 6 3 -7.50 0 0 0 -7.5 
FB 4 2 0.00 0 0 0 0 
GD 4 2 0.00 0 0 0 0 
AF 5 5 -6.25 0 0 0 -6.25 
FC 5 5 6.25 0 0 0 6.25 
CG 5 5 -6.25 0 0 0 -6.25 
GE 5 5 -68.75 0 0 0 -68.75 
    Total 15 4  
 
Example 10.3 
Determine the reactions and the member axial forces of the truss shown in 
Fig.10.3a by force method due to external load and rise in temperature of 
member  by . The cross sectional areas of the members in square 
centimeters are shown in parenthesis. Assume 
and
FB C ° 40
52
2.0 10 N/mm E=×
1
per °C
75000
a = . 
 
 
 
The given truss is indeterminate to second degree. The truss has both internal 
and external indeterminacy. Choose horizontal reaction at and the axial 
force in member as redundant actions. Releasing the restraint against 
redundant actions, a stable determinate truss is obtained as shown in Fig. 10.3b. 
D()
1
R
EC()
2
R
 
 
 
 
 
 
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