Civil Engineering Reference
In-Depth Information
Now resolving forces vertically
R
B,V
+
R
A,V
−
60
−
100
=
0
which, on substituting for
R
A,V
, gives
R
B,V
=
29
.
0kN
Since no horizontal loads are present, we see by inspection that
R
A,H
=
R
B,H
Finally, taking moments of forces to the right of C about C (this is a little simpler than
considering forces to the left of C) we have
R
B,H
×
6
−
R
B,V
×
6
=
0
from which
R
B,H
=
29
.
0kN
=
R
A,H
The normal force at the point X is obtained by resolving the forces to one side of X
in a direction tangential to the arch at X. Thus, considering forces to the left of X and
taking tensile forces as positive
R
A,V
cos 45
◦
−
R
A,H
sin 45
◦
+
60 cos 45
◦
N
X
=−
so that
N
X
=−
70
.
7kN
and is compressive.
The shear force at X is found by resolving the forces to one side of X in a direction
perpendicular to the tangent at X. We shall take a positive shear force as acting radially
inwards when it is to the left of a section. So, considering forces to the left of X
R
A,V
sin 45
◦
+
R
A,H
cos 45
◦
+
60 sin 45
◦
S
X
=−
which gives
S
X
=−
29
.
7kN
Now taking moments about X for forces to the left of X and regarding a positive
moment as causing tension on the underside of the arch, we have
6 cos 45
◦
)
6 sin 45
◦
−
60 (6 cos 30
◦
−
6 cos 45
◦
)
M
X
=
R
A,V
(6
−
−
R
A,H
×
from which
M
X
=+
50
.
0kNm
Note that in Ex. 6.1 the sign conventions adopted for normal force, shear force and
bending moment are the same as those specified in Chapter 3.
