Civil Engineering Reference
In-Depth Information
Resolving forces parallel to AB
τ
AB
=
σ
x
BC sin
θ
−
σ
y
AC cos
θ
or
BC
AB
sin
θ
AC
AB
cos
θ
τ
=
σ
x
−
σ
y
which gives
σ
x
−
sin 2
θ
σ
y
τ
=
(14.4)
2
Again we see that although the applied loads produce only direct stresses on planes
perpendicular and parallel to the axis of the cylinder, both direct and shear stresses
exist on inclined planes. Furthermore, for given values of
σ
x
and
σ
y
(i.e.
p
) the shear
stress
τ
is a maximum on planes inclined at 45
◦
to the axis of the cylinder.
E
XAMPLE
14.1
A cylindrical pressure vessel has an internal diameter of 2m and
is fabricated from plates 20mm thick. If the pressure inside the vessel is 1
.
5N
/
mm
2
and, in addition, the vessel is subjected to an axial tensile load of 2500 kN, calculate
the direct and shear stresses on a plane inclined at an angle of 60
◦
to the axis of the
vessel. Calculate also the maximum shear stress.
From Eq. (7.63) the circumferential stress is
10
3
pd
2
t
1
.
5
×
2
×
75N
/
mm
2
=
=
2
×
20
From Eq. (7.62) the longitudinal stress is
pd
4
t
=
37
.
5N
/
mm
2
The direct stress due to axial load is, from Eq. (7.1)
10
3
2500
×
19
.
9N
/
mm
2
20
=
π
×
2000
×
Therefore on a rectangular element at any point in the wall of the vessel there is a
biaxial stress system as shown in Fig. 14.6. Now considering the equilibrium of the
triangular element ABC we have, resolving forces perpendicular to AB
20 cos 30
◦
+
20 cos 60
◦
σ
n
AB
×
20
=
57
.
4BC
×
75AC
×
Since the walls of the vessel are thin the thickness of the two-dimensional element may
be taken as 20mm. However, as can be seen, the thickness cancels out of the above
equation so that it is simpler to assume unit thickness for two-dimensional elements
in all cases. Then
57
.
4 cos
2
30
◦
+
75 cos
2
60
◦
σ
n
=
