Civil Engineering Reference
In-Depth Information
Also
7
.
5)
2
y
D
=
0
.
0311
×
(
−
=
1
.
75m
so that
h
D
=
7
−
1
.
75
=
5
.
25m
Taking moments about A for the overall equilibrium of the arch we have
R
B,V
×
25
+
R
B,H
×
3
.
89
−
10
×
15
×
7
.
5
=
0
which simplifies to
R
B,V
+
0
.
16
R
B,H
−
45
.
0
=
0
(iii)
Now taking moments about C for the forces to the right of C we obtain
R
B,V
×
10
−
R
B,H
×
3
.
11
=
0
which gives
R
B,V
−
0
.
311
R
B,H
=
0
(iv)
The simultaneous solution of Eqs (iii) and (iv) gives
R
B,V
=
29
.
7kN
R
B,H
=
95
.
5kN
From the horizontal equilibrium of the arch we have
R
A,H
=
R
B,H
=
95
.
5kN
and from the vertical equilibrium
R
A,V
+
R
B,V
−
10
×
15
=
0
which gives
R
A,V
=
120
.
3kN
To calculate the normal force and shear force at the point D we require the slope of
the arch at D. From Eq. (ii)
d
y
d
x
D
=
2
×
0
.
0311
×
(
−
7
.
5)
=−
0
.
4665
=−
tan
α
Hence
25
.
0
◦
α
=
