Civil Engineering Reference
In-Depth Information
Now resolving forces to the left (or right) of D in a direction parallel to the tangent at
D we obtain the normal force at D. Hence
R
A,V
sin 25
.
0
◦
−
R
A,H
cos 25
.
0
◦
+
7
.
5 sin 25
.
0
◦
N
D
=−
10
×
which gives
N
D
=−
105
.
7 kN (compression)
The shear force at D is then
R
A,V
cos 25
.
0
◦
+
R
A,H
sin 25
.
0
◦
+
7
.
5 cos 25
.
0
◦
S
D
=−
10
×
so that
S
D
=−
0
.
7kN
Finally the bending moment at D is
7
.
5
2
M
D
=
R
A,V
×
7
.
5
−
R
A,H
×
5
.
25
−
10
×
7
.
5
×
from which
M
D
=+
119
.
6kNm
6.3 A T
HREE-PINNED
P
ARABOLIC
A
RCH
C
ARRYING A
U
NIFORM
H
ORIZONTALLY
D
ISTRIBUTED LOAD
In Section 5.2 we saw that a flexible cable carrying a uniform horizontally distributed
load took up the shape of a parabola. It follows that a three-pinned parabolic arch
carrying the same loading would experience zero shear force and bending moment
at all sections. We shall now investigate the bending moment in the symmetrical
three-pinned arch shown in Fig. 6.7.
w
C
P(
x
,
y
)
y
h
B
A
R
A,H
x
R
B,H
F
IGURE
6.7
Parabolic arch
carrying a uniform
horizontally
distributed load
R
A,V
R
B,V
L
/2
L
/2
