Civil Engineering Reference
In-Depth Information
Finally from the third of Eqs. (2.10),
M
z
=
0, and taking moments about A, thereby
eliminating
R
A,H
and
R
A,V
(2 sin 45
◦
)
M
A
−
5
×
0
.
4
−
×
1
.
0
=
0
from which
M
A
=
3
.
4kNm
InExs 2.2 and 2.3, the directions or sense of the support reactions is reasonably obvious.
However, where this is not the case, a direction or sense is assumed which, if incorrect,
will result in a negative value.
Occasionally the resultant reaction at a support is of interest. In Ex. 2.2 the resultant
reaction at A is found using the first of Eqs. (2.4), i.e.
R
2
A
=
R
2
A,H
+
R
2
A,V
which gives
R
2
A
=
2
.
5
2
3
.
7
2
+
so that
R
A
=
4
.
5kN
The inclination of
R
A
to, say, the vertical is found from the second of Eqs. (2.4). Thus
R
A,H
R
A,V
=
2
.
5
3
.
7
=
tan
θ
=
0
.
676
from which
34
.
0
◦
θ
=
E
XAMPLE
2.4
Calculate the reactions at the supports in the plane truss shown in
Fig. 2.22.
The truss is supported in the same manner as the beam in Ex. 2.2 so that there will be
horizontal and vertical reactions at A and only a vertical reaction at B.
The angle of the truss,
α
, is given by
tan
−
1
2
.
4
3
38
.
7
◦
α
=
=
From the first of Eqs. (2.10) we have
5 sin 38
.
7
◦
−
10 sin 38
.
7
◦
=
R
A,H
−
0
