Civil Engineering Reference
In-Depth Information
60 mm diameter
1.5 mm
1200 N m
F
IGURE
14.8
Cantilever beam of
Ex. 14.2
50 000 N
21.2 N/mm
2
28.3 N/mm
2
B
28.3 N/mm
2
s
n
21.2 N/mm
2
21.2 N/mm
2
17.7
3.5
F
IGURE
14.9
Two-dimensional
stress system in
cantilever beam of
Ex. 14.2
τ
28.3 N/mm
2
C
A
60
°
28.3 N/mm
2
28.3 N/mm
2
the axis of the beam and are given, respectively, by Eqs (7.1) and (9.9). Therefore
50 000
π
×
17
.
7N
/
mm
2
σ
x
(axial load)
=
60
2
/
4
=
75 000
×
30
3
.
5N
/
mm
2
σ
x
(bending moment)
=
60
2
/
64
=
π
×
The shear stress
τ
xy
at the same point due to the torque is obtained from Eq. (11.4)
and is
10
3
1200
×
×
30
28
.
3N
/
mm
2
τ
xy
=
=
60
4
/
32
π
×
The stress system acting on a two-dimensional rectangular element at the point is
as shown in Fig. 14.9. Note that, in this case, the element is at the bottom of the
cylinder so that the shear stress is opposite in direction to that in Fig. 14.7. Considering
the equilibrium of the triangular element ABC and resolving forces in a direction
perpendicular to AB we have
21
.
2 BC cos 30
◦
+
28
.
3 BC sin 30
◦
+
28
.
3AC cos 30
◦
σ
n
AB
=−
Dividing through by AB we obtain
21
.
2 cos
2
30
◦
+
28
.
3 cos 30
◦
sin 30
◦
+
28
.
3 sin 30
◦
cos 30
◦
σ
n
=−
which gives
8
.
6N
/
mm
2
σ
n
=
