Civil Engineering Reference
In-Depth Information
so that
R
A,H
=
27
.
6kN
Alternatively we could have obtained
R
A,H
by using the fact that the resultant reaction,
R
A
, at A is in line with the cable at A, i.e.
R
A,V
/
R
A,H
=
tan 18
.
4
◦
, which gives
tan
α
=
R
A,H
=
27
.
6 kN as before. Having obtained
R
A,V
and
R
A,H
,
T
CA
follows. Thus
R
2
A,H
+
27
.
6
2
R
2
A,V
=
9
.
2
2
T
CA
=
R
A
=
+
i.e.
T
CA
=
29
.
1kN
From a consideration of the vertical equilibrium of the forces acting at C we have
29
.
1 sin 18
.
4
◦
−
T
CD
sin
β
+
T
CA
sin
α
−
10
=
T
CD
sin
β
+
10
=
0
which gives
T
CD
sin
β
=
0
.
815
(i)
From the horizontal equilibrium of the forces at C
29
.
1 cos 18
.
4
◦
=
T
CD
cos
β
−
T
CA
cos
α
=
T
CD
cos
β
−
0
so that
T
CD
cos
β
=
27
.
612
(ii)
Dividing Eq. (i) by Eq. (ii) yields
tan
β
=
0
.
0295
from which
1
.
69
◦
=
β
Therefore from either of Eq. (i) or (ii)
T
CD
=
27
.
6kN
We can obtain the tension in DB in a similar manner. Thus, from the vertical
equilibrium of the forces at D, we have
27
.
6 sin 1
.
69
◦
−
T
DB
sin
γ
−
T
DC
sin
β
−
6
=
T
DB
sin
γ
−
6
=
0
from which
T
DB
sin
γ
=
6
.
815
(iii)
