Civil Engineering Reference
In-Depth Information
from which
W
L
(
L
R
A,V
=
−
a
)
Substituting in the expressions for
M
0
gives
W
L
(
L
M
0
=
−
a
)
x
(0
x
a
)
(ii)
W
L
(
L
M
0
=
−
x
)
a
x
L
)
(iii)
The denominator in Eq. (16.15) may be evaluated separately. Thus, from Eq. (i)
L
4
h
L
2
2
8
h
2
L
15
y
2
d
x
x
2
)
2
d
x
=
(
Lx
−
=
Profile
0
Then, from Eq. (16.15) and Eqs (ii) and (iii)
a
x
2
)d
x
L
15
8
h
2
L
W
L
(
L
a
)
x
4
h
Wa
L
(
L
x
)
4
h
x
2
)d
x
R
1
=
−
L
2
(
Lx
−
+
−
L
2
(
Lx
−
0
a
which gives
5
Wa
8
hL
3
(
L
3
a
3
2
La
2
)
R
1
=
+
−
(iv)
The remaining support reactions follow froma considerationof the statical equilibrium
of the arch.
If, in Ex. 16.14, we had expressed the load position in terms of the span of the arch,
say
a
=
kL
, Eq. (iv) in Ex. 16.14 becomes
5
WL
8
h
k
4
2
k
3
)
R
1
=
(
k
+
−
(16.16)
Therefore, for a series of concentrated loads positioned at distances
k
1
L
,
k
2
L
,
k
3
L
,
etc., from A, the reaction,
R
1
, may be calculated for each load acting separately using
Eq. (16.16) and the total reaction due to all the loads obtained by superposition.
The result expressed in Eq. (16.16) may be used to determine the reaction,
R
1
, due to a
part-span uniformly distributed load. Consider the arch shown in Fig. 16.30. The arch
profile is parabolic and its second moment of area varies as the secant assumption. An
elemental length,
x
is
very small, we may regard this load as a concentrated load. This will then produce an
increment,
δ
x
, of the load produces a load
w
δ
x
on the arch. Thus, since
δ
δ
R
1
, in the horizontal support reaction which, from Eq. (16.16), is given by
5
8
w
x
L
k
4
2
k
3
)
δ
R
1
=
δ
h
(
k
+
−
in which
k
=
x
/
L
. Therefore, substituting for
k
in the expression for
δ
R
1
and then
integrating over the length of the load we obtain
x
L
+
d
x
x
2
x
4
L
4
−
2
x
3
L
3
5
wL
8
h
R
1
=
x
1
