Civil Engineering Reference
In-Depth Information
E
XAMPLE
2.2
Calculate the support reactions in the simply supported beamABCD
shown in Fig. 2.20.
The different types of support have been discussed in Section 1.4. In Fig. 2.20 the sup-
port at A is a pinned support which allows rotation but no translation in any direction,
while the support at D allows rotation and translation in a horizontal direction but
not in a vertical direction. Therefore there will be no moment reactions at A or D and
only a vertical reaction at D,
R
D
. It follows that the horizontal component of the 5 kN
load can only be resisted by the support at A,
R
A,H
, which, in addition, will provide a
vertical reaction,
R
A,V
.
Since the forces acting on the beam are coplanar, Eqs. (2.10) are used. From the first
of these, i.e.
F
x
=
0, we have
5 cos 60
◦
=
R
A,H
−
0
which gives
R
A,H
=
2
.
5kN
The use of the second equation,
F
y
=
0, at this stage would not lead directly to either
R
A,V
or
R
D
since both would be included in the single equation. A better approach is
to use the moment equation,
M
z
=
0, and take moments about either A or D (it is
immaterial which), thereby eliminating one of the vertical reactions. Taking moments,
say, about D, we have
(5 sin 60
◦
)
R
A,V
×
−
×
−
×
=
1
.
2
3
0
.
9
0
.
4
0
(i)
Note that in Eq. (i) the moment of the 5 kN force about D may be obtained either
by calculating the perpendicular distance of its line of action from D (0
.
4 sin 60
◦
)
or by resolving it into vertical and horizontal components (5 sin 60
◦
and 5 cos 60
◦
,
respectively) where only the vertical component exerts a moment about D. From
3kN
y
5kN
60
°
A
B
C
D
R
A,H
x
R
A,V
R
D
F
IGURE
2.20
Beam of Ex. 2.2
0.3 m
0.5 m
0.4 m
