Civil Engineering Reference
In-Depth Information
or
[
C
][
t
]
=
[
F
]
where [
C
] is the coordinate matrix, [
t
] the tension coefficient matrix and [
F
] the force
matrix. Then
[
C
]
−
1
[
F
]
[
t
]
=
Computer programs exist which will carry out the inversion of [
C
] so that the tension
coefficients are easily obtained.
In the above thematrix equationonly represents the equilibriumof joint F. There are, in
fact, sixmembers in the truss so that a total of six equations are required. The additional
equations are Eqs (vii), (viii) and (ix) in Ex. 4.6. Therefore, to obtain a complete
solution, these equations would be incorporated giving a 6
×
6 square matrix for [
C
].
In practice plane and space frame programs exist which, after the relevant data have
been input, give the member forces directly. It is, however, important that the funda-
mentals on which these programs are based are understood. We shall return to matrix
methods later.
PROBLEMS
P.4.1
Determine the forces in the members of the truss shown in Fig. P.4.1 using the
method of joints and check the forces in the members JK, JD and DE by the method
of sections.
Ans.
AG
=+
37
.
5, AB
=−
22
.
5, BG
=−
20
.
0, BC
=−
22
.
5, GC
=−
12
.
5, GH
=+
30
.
0,
HC
=
0, HJ
=+
30
.
0, CJ
=+
12
.
5, CD
=−
37
.
5, JD
=−
10
.
0, JK
=+
37
.
5, DK
=+
12
.
5,
DE
=−
45
.
0, EK
=−
70
.
0, FE
=−
45
.
0, FK
=+
75
.
0. All in kN.
G
H
J
K
8m
A
B
C
D
E
F
30 kN
60 kN
5
6m
F
IGURE
P.4.1
P.4.2
Calculate the forces in the members of the truss shown in Fig. P.4.2.
Ans
.
AC
=−
30
.
0, AP
=+
26
.
0, CP
=−
8
.
7, CE
=−
25
.
0, PE
=+
8
.
7, PF
=+
17
.
3,
FE
=−
17
.
3, GE
=−
20
.
0, HE
=+
8
.
7, FH
=+
17
.
3, GH
=−
8
.
7,
JG
=−
15
.
0,
HJ
=+
26
.
0, FB
=
0, BJ
=−
15
.
0. All in kN.
