PRESSURE VESSELS
Applications of plane stress
(Pressure vessels, Beams and Combined loadings)
• Plane stress is a common stress condition that exists in all ordinary structures,
including buildings, machines, vehicles and aircraft
• In this chapter we will investigate some practical applications involving plane
stress
• We will start with pressure vessels by determining the stresses and strains in
the walls of these structures caused because of the internal pressures from the
compressed gases or liquids
Spherical pressure vessels
• Pressure vessels are closed structures containing liquids or gases under
pressure
• Examples include tanks, pipes, beverage containers and pressurized cabins in
aircraft and space vehicles
• If pressure vessels have walls that are thin in comparison to their overall
dimensions they are known as shell structures
• In this section we consider thin – walled (r/t>10) pressure vessels of spherical
shape
FIG. 8-2
Cross section of spherical pressure
vessel showing inner radius r, wall
thickness t, and internal pressure p
Fig. 8.1) Spherical pressure vessel
THIN AND THICK CYLINDERS
INTRODUCTION:
In many engineering applications, cylinders are frequently
used for transporting or storing of liquids, gases or fluids.
Eg: Pipes, Boilers, storage tanks etc.
These cylinders are subjected to fluid pressures. When a
cylinder is subjected to a internal pressure, at any point on the
cylinder wall, three types of stresses are induced on three
mutually perpendicular planes.
They are,
1. Hoop or Circumferential Stress (σC) – This is directed along the
tangent to the circumference and tensile in nature. Thus, there
will be increase in diameter.
2. Longitudinal Stress (σL ) – This stress is directed along the
length of the cylinder. This is also tensile in nature and tends
to increase the length.
3. Radial pressure ( pr ) – It is compressive in nature.
Its magnitude is equal to fluid pressure on the inside wall and
zero on the outer wall if it is open to atmosphere.
σC
σC
p
σC
σL
p
σC
σL
σL
pr
p
σL
1. Hoop Stress (C) 2. Longitudinal Stress (L)
pr
Element on the cylinder
wall subjected to these
three stresses
3. Radial Stress (pr)
σC
σL
σL
σC
pr
THIN CYLINDERS
INTRODUCTION:
A cylinder or spherical shell is considered to be thin when the
metal thickness is small compared to internal diameter.
i. e., when the wall thickness, ‘t’ is equal to or less than
‘d/20’, where ‘d’ is the internal diameter of the cylinder or shell,
we consider the cylinder or shell to be thin, otherwise thick.
Magnitude of radial pressure is very small compared to other
two stresses in case of thin cylinders and hence neglected.
t
Circumferential stress
Longitudinal
Longitudinal stress
axis
The stress acting along the circumference of the cylinder is called
circumferential stresses whereas the stress acting along the length of
the cylinder (i.e., in the longitudinal direction ) is known as
longitudinal stress
The bursting will take place if the force due to internal (fluid)
pressure (acting vertically upwards and downwards) is more than the
resisting force due to circumferential stress set up in the material.
P - internal pressure (stress)
σc –circumferential stress
p
σc
σc
σc
t
p
dL
P - internal pressure (stress)
σc – circumferential stress
EVALUATION OF CIRCUMFERENTIAL or HOOP STRESS (σC):
t
p
d
p
dl
t
d
σc
σc
Consider a thin cylinder closed at both ends and subjected to internal
pressure ‘p’ as shown in the figure.
Let d=Internal diameter,
L = Length of the cylinder.
t = Thickness of the wall
To determine the Bursting force across the diameter:
Consider a small length ‘dl’ of the cylinder and an elementary
area ‘dA’ as shown in the figure.
Force on the elementary area,
dF  p  dA  p  r  dl  dθ
d
 p   dl  dθ
2
Horizontal component of this force
d
dFx  p   dl  cos θ  dθ
2
Vertical component of this force
d
dFy  p   dl  sin θ  dθ
2
dA
dθ
p
θ
t
dl
d
σc
σc
dA
The horizontal components cancel out
when integrated over semi-circular
portion as there will be another equal
and opposite horizontal component on
the other side of the vertical axis.
dθ
p
θ
t
dl
d
σc
σc

d
 Total diametrica l bursting force   p   dl  sin   dθ
2
0
d
 p   dl   cos   0  p  d  dl
2
 p  projected area of the curved surface.
 Resisting force (due to circumfere ntial stress σ c )  2  σ c  t  dl
Under equillibri um, Resisting force  Bursting force
i.e., 2  σ c  t  dl  p  d  dl
pd
 Circumfere ntial stress, σ c 
........................(1)
2 t
t
σc
p
dL
Assumed as rectangular
Force due to fluid pressure = p × area on which p is acting = p ×(d ×L)
(bursting force)
Force due to circumferential stress = σc × area on which σc is acting
(resisting force) = σc × ( L × t + L ×t ) = σc × 2 L × t
Under equilibrium bursting force = resisting force
p ×(d ×L) = σc × 2 L × t
pd
 Circumfere ntial stress, σ c 
........................(1)
2 t
LONGITUDINAL STRESS (σL):
A
The bursting of the cylinder takes
place along the section AB
P
B
σL
p
The force, due to pressure of the fluid, acting at the ends of the
thin cylinder, tends to burst the cylinder as shown in figure
EVALUATION OF LONGITUDINAL STRESS (σL):
t
σL
p
π 2
Longitudin al bursting force (on the end of cylinder)  p   d
4
Area of cross section resisting this force  π  d  t
Let σ L  Longitudin al stress of the material of the cylinder.
 Resisting force  σ L  π  d  t
Under equillibri um, bursting force  resisting force
π
i.e., p   d 2  σ L  π  d  t
4
pd
 Longitudin al stress, σ L 
...................( 2)
4 t
From eqs (1) & (2),
σC  2  σL
Force due to fluid pressure  p  area on which p is acting
π 2
 p d
4
Re sisting force  σ L  area on which σ L is acting
 σL  π  d  t
circumference
Under equillibri um, bursting force  resisting force
π 2
pπd d  t
i.e.,
p


d

σ
 Longitudin al stress, σ L L
...................( 2)
4
4 t
EVALUATION OF STRAINS
σL=(pd)/(4t)
σ C=(pd)/(2t)
σ C=(pd)/(2t)
σ L=(pd)/(4t)
A point on the surface of thin cylinder is subjected to biaxial
stress system, (Hoop stress and Longitudinal stress) mutually
perpendicular to each other, as shown in the figure. The strains due
to these stresses i.e., circumferential and longitudinal are obtained
by applying Hooke’s law and Poisson’s theory for elastic materials.
Circumfere ntial strain, ε C :
σC
σL
εC 
μ
E
E
σL
σL
 2
μ
E
E
σL

 (2  μ)
E
i.e.,
σ L=(pd)/(4t)
σ C=(pd)/(2t)
σC=(pd)/(2t)
σ L=(pd)/(4t)
δd
pd
εC 

 (2  μ)................................(3)
d 4 t  E
Note: Let δd be the change in diameter. Then
final circumfere nce  original circumfere nce
c 
original circumfere nce
  d  d   d  d



d
d


Longitudin al strain, ε L :
σL
σC
εL 
μ
E
E
σL
(2  σ L ) σ L

μ

 (1  2  μ)
E
E
E
i.e.,
δl
pd
εL  
 (1  2  μ)................................(4)
L 4 t  E
VOLUMETRIC STRAIN,
v
V
Change in volume = δV = final volume – original volume
original volume = V = area of cylindrical shell × length

 d2
4
L
final volume = final area of cross section × final length




4
d   d  2 L   L 

2

2

d
4

d
4

 ( d ) 2  2 d  d  L   L 
L  ( d ) 2 L  2 L d  d  d 2  L  ( d ) 2  L  2 d  d  L

neglecting the smaller quantities such as ( d ) 2 L, ( d ) 2  L and 2 d  d  L
Final volume 


d
4
2
L  2 L d  d  d 2 L
change in volume V 
V 



d
4
2


L  2 L d  d  d 2 L 

2 L d  d  d  L
4
2

4
d  2 L

π
2 d L d  L d 2
dv 4

π 2
V
d L
4

L
dV
V
L
 2

d
d
= ε L + 2 × εC
pd
pd

(1  2  μ)  2 
(2  μ)
4 t  E
4 t  E
i.e.,
dv
pd

(5  4  μ).................(5)
V 4 t  E
Maximum Shear stress :
There are two principal stresses at any point,
viz., Circumfere ntial and longitudin al. Both
these stresses are normal and act perpendicu lar
to each other.
 Maximum Shear stress, τ max
pd pd

 2t 4t
2
i.e.,
τ max
σC - σL

2
σ L=(pd)/(4t)
σC=(pd)/(2t)
pd
 .....................(5)
8t
σ C=(pd)/(2t)
σ L=(pd)/(4t)
Maximum Shear stress :
 Maximum Shear stress, τ max
σC - σL

2
pd pd

 2t 4t
2
i.e.,
τ max
pd
 .....................(5)
8t