Civil Engineering Reference
In-Depth Information
T
ABLE
16.2
F
1,
j
F
1,
j
Member
L
j
F
0,
j
F
1,
j
(
X
1
)
F
1,
j
(
R
2
)
F
0,
j
F
1,
j
(
X
1
)
L
j
F
0,
j
F
1,
j
(
R
2
)
L
j
(
X
1
)
L
j
(
R
2
)
L
j
F
1,
j
(X
1
)
F
1,
j
(
R
2
)
L
j
F
a,
j
AB
1
10.0
−
0.71
−
2.0
−
7.1
−
20.0
0.5
4.0
1.41
0.67
BC
1.41
0
0
−
1.41
0
0
0
2.81
0
−
4.45
CD
1
0
0
1.0
0
0
0
1.0
0
3.15
DE
1
0
−
0.71
1.0
0
0
0.5
1.0
−
0.71
0.12
AD
1.41
0
1.0
0
0
0
1.41
0
0
4.28
BE
1.41
−
14.14
1.0
1.41
−
20.0
−
28.11
1.41
2.81
2.0
−
5.4
BD
1
0
−
0.71
0
0
0
0.5
0
0
−
3.03
=−
=−
=
4.32
=
11.62
=
27.1
48.11
2.7
Ex. 16.7). The calculations are similar to those carried out in Ex. 16.8 and are shown
in Table 16.2.
From Table 16.2
n
F
0,
j
F
1,
j
(
X
1
)
L
j
AE
=
−
27
.
1
AE
AD
=
(i.e. AD increases in length)
j
=
1
n
F
0,
j
F
1,
j
(
R
2
)
L
j
AE
=
−
48
.
11
AE
v
C
=
(i.e. C displaced downwards)
j
=
1
F
1,
j
(
X
1
)
L
j
AE
n
4
.
32
AE
a
11
=
=
j
=
1
n
F
1,
j
(
R
2
)
L
j
AE
11
.
62
AE
a
22
=
=
j
=
1
n
F
1,
j
(
X
1
)
F
1,
j
(
R
2
)
L
j
AE
2
.
7
AE
a
12
=
a
21
=
j
=
1
Substituting in Eqs (i) and (ii) and multiplying through by AE we have
−
27
.
1
+
4
.
32
X
1
+
2
.
7
R
2
=
0
(iii)
−
48
.
11
+
2
.
7
X
1
+
11
.
62
R
2
=
0
(iv)
Solving Eqs (iii) and (iv) we obtain
X
1
=
4
.
28 kN
R
2
=
3
.
15 kN
The actual forces,
F
a,
j
, in the members of the complete truss are now calculated by
the method of joints and are listed in the final column of Table 16.2.
