1.
TOPICS
 INTRODUCTION
 STRESSES IN THIN CYLINDRICAL SHEELS
HOOP STRESS(CIRCUMFERENTIAL STRESS)
LONGITUDINAL STRESS
 C (IN) D (OF) (A) TCS (DUE) (TO) IP
 C (IN) V (OF) (A) TCS (DUE) (TO) IP
 THIN SPHERICAL SHEELS

2.
THIN CYLINDER
If the wall thickness is less than about 7% of the inner
diameter then the cylinder may be treated as a thin one.
Thin walled cylinders are used as boiler shells,pressure
tanks, pipes and in other low pressure pocessing
equipments.
Ingeneral three types of stresses are developed in pressure
cylinders viz.
Circumferential or Hoop stress, Longitudinal stress in
closed end cylinders and Radial stresses.

3.
In a thin walled cylinder the circumferential
stresses may be assumed to be constant over the
wall thickness and stress in the radial direction
may be neglected for the analysis. Considering
the equilibrium of a cut out section the
circumferential stress and longitudinal stress can
be found.

4.
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.
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.

5.
σ C σ L
1. Hoop Stress (C) 2. Longitudinal Stress (L) 3. Radial Stress (pr)
Element on the cylinder
wall subjected to these
three stresses
σ C
σ C
σC
p
σ L
σ L
σ L
p p
pr
σ Lσ L
σ C
σ C
pr
pr

6.
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/10
to d/15’, where ‘d’ is the internal diameter of the cylinder or shell,
we consider the cylinder or shell to be thin, otherwise thick. In
thin cylindrical stress hoop stress and longitudinal stresses are
constant over the thickness and radial stresses are negligible.
Magnitude of radial pressure is very small compared to other
two stresses in case of thin cylinders and hence neglected.
THIN CYLINDERS

7.
Longitudinal
axisLongitudinal stress
Circumferential stress
t
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

8.
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
σc σc
P - internal pressure (stress)
σc –circumferential stress

9.
P - internal pressure (stress)
σc – circumferential stress
dL
σc
p
t

10.
EVALUATION OF CIRCUMFERENTIAL or HOOP STRESS (σC):
Consider a thin cylinder closed at both ends and subjected to internal
pressure ‘p’ as shown in the figure.
Let d=Internal diameter, t = Thickness of the wall
L = Length of the cylinder.
p d
t
σcσc
dlt
p
d

11.
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.
rpp dθdldAdF 
dθdldFx  θcos
2
d
p
dA
σcσc
dlt
p
d
dθ
θ
Force on the elementary area,
Horizontal component of this force
dθdl 
2
d
p
dθdldFy  θsin
2
d
p
Vertical component of this force

12.
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.
sin
2
d
pforceburstingldiametricaTotal
0
 

 dθdl
dA
σcσc
dlt
p
θ
d
dθ
 
surface.curvedtheofareaprojectedp
dpcosdl
2
d
p 0

 dl


13.
dlcc  tσ2)σstressntialcircumfereto(dueforceResisting
dldl  dptσ2i.e., c
dL
σc
p
t
forceBurstingforceResistingum,equillibriUnder 
Circumferential stress 𝜎𝑐 =
𝑝𝑑
2𝑡

14.
LONGITUDINAL STRESS (σL):
p
σL
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
P
A
B
The bursting of the cylinder takes
place along the section AB

15.
EVALUATION OF LONGITUDINAL STRESS (σL):
d
4
π
pcylinder)ofend(on theforceburstingalLongitudin 2

p
t
σL
tdπσforceResisting L 
tdπforcethisresistingsectioncrossofArea 
cylinder.theofmaterialtheofstressalLongitudinσLet L 

16.
tdπσd
4
π
pi.e., L
2

UNDER EQULIBIRIUM
BURSTING FORCE = RESISTING FORCE
∴ 𝑙𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠 𝜎𝐿 =
𝑝𝑑
4𝑡

17.
CHANGE IN DIMENSIONS OF A THIN CYLINDRICAL
SHELL DUE TO INTERNAL PRESSURE

18.
Consider a thin cylindrical shell subjected to internal pressure
p = internal pressure
d = internal diameter of shell
t = thickness of shell
l = length of the shell
WE KNOW THAT,
σc =
pd
2t and σL =
pd
4t
Let, δd= change in diameter of shell
δl = change in length of shell
∴ circumferential strain,
ε 1 =
𝛿𝑑
𝑑
=
𝜎 𝑐
𝐸
−
𝜎 𝑙
𝑚𝐸
=
𝑝𝑑
2𝑡𝐸
−
𝑝𝑑
4𝑡𝑚𝐸
∈1=
𝑝𝑑
2𝑡𝐸
1 −
1
2𝑚

19.
∈2=
𝛿𝑙
𝑙
=
𝜎𝑙
𝐸
−
𝜎𝑐
𝑚𝐸
=
𝑝𝑑
4𝑡𝐸
−
𝑝𝑑
2𝑡𝑚𝐸
∈2=
𝑝𝑑
2𝑡𝐸
1
2
−
1
𝑚
∴ 𝑙𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛
Change in diameter = 𝛿𝑑 =∈1 𝑑
Change in length = 𝛿𝑙 =∈2 𝑙
When the shell is subjected to an internal pressure, there will be an increase in the
Diameter as well as the length of the shell

20.
CHANGE IN VOLUME OF A THIN CYLINDRICAL
SHELL DUE TO INTERNAL PRESSURE

21.
Voloume of shell ,
𝑉 =
𝜋
4
𝑑2
𝑙
∴ Final volume
v+𝛿𝑣 =
𝜋
4
(𝑑 + 𝛿𝑑)2
× 𝑙 + 𝛿𝑙
∴ Change in voloume ,
𝛿𝑣 = 𝑣 + 𝛿𝑣 − 𝑣
=
𝜋
4
(𝑑 + 𝛿𝑑)2
(𝑙 + 𝛿𝑙) −
𝜋
4
𝑑2
𝑙
=
𝜋
4
𝑑2
+ 2. 𝛿𝑑. 𝑑𝑙 + 𝛿𝑑2
. 𝑙 + 𝑑2
. 𝛿𝑙 + 2𝛿𝑑. 𝑑. 𝛿𝑙 + 𝛿𝑑2
. 𝛿𝑙 −
𝜋
4
𝑑2
. 𝑙
=
𝜋
4
𝑑2. 𝛿𝑙 + 2𝛿𝑑. 𝑑. 𝑙 𝑛𝑒𝑔𝑙𝑒𝑐𝑡𝑖𝑛𝑔 𝑠𝑚𝑎𝑙𝑙 𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑖𝑒𝑠

22.
𝛿𝑣
𝑉
=
𝜋
4
(𝑑2
. 𝛿𝑙 + 2. 𝛿𝑑. 𝑑. 𝑙)
𝜋
4
𝑑2 𝑙
=
𝛿𝑙
𝑙
+
2𝛿𝑑
𝑑
𝛿𝑣
𝑉
= 𝜖2 + 2. 𝜖1
∴ 𝛿𝑣 = 𝑉(𝜀2 + 2. 𝜖1)
∵ 𝜖1 =
𝛿𝑑
𝑑
𝜀2 =
𝛿𝑙
𝑙
∈1= 𝑐𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛
∈2= 𝑙𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛

23.
THIN SPHERICAL SHELL

24.
Consider a thin spherical shell subjected to internal pressure p
p= internal pressure
d= internal dia of shell
t= thickness of shell
𝜎 = stress in the shell material
Total force,
P=
𝜋
4
× 𝑑2
× 𝑝
Resisting section = 𝜋𝑑𝑡
∴ Stress in the shell,
𝜎 =
𝑡𝑜𝑡𝑎𝑙 𝑓𝑜𝑟𝑐𝑒
𝑟𝑒𝑠𝑖𝑠𝑡𝑖𝑛𝑔 𝑠𝑒𝑐𝑡𝑖𝑜𝑛
Type equation here.
=
𝜋
4
𝑑2×𝑝
𝜋𝑑𝑡
𝜎 =
𝑝𝑑
4𝑡

25.
CHANGE IN DIAMETER AND VOLUME OF A THIN SPHERICAL
SHELL DUE TO INTERNAL PRESSURE

26.
Consider a thin spherical shell subjected to internal pressure
d=internal dia of shell
p=internal pressure
t=thickness of the shell
𝜎= stress in the shell material
We know that the thin spherical shell,
𝜎 =
𝑝𝑑
4𝑡
Strain in any direction,
𝜖 =
𝜎
𝐸
−
𝜎
𝑚𝐸
=
𝑝𝑑
4𝑡𝐸
−
𝑝𝑑
4𝑡.𝑚𝐸
=
𝑝𝑑
4𝑡𝐸
(1−
1
𝑚
)

27.
2
4t
pd
2t
pd
2
σ-σ
τstress,ShearMaximum
other.eachto
larperpendicuactandnormalarestressesthese
Bothal.longitudinandntialCircumfereviz.,
point,anyatstressesprincipaltwoareThere
:stressShearMaximum
LC
max



)5.(....................
8t
pd
τi.e., max 
σ C=(pd)/(2t)
σC=(pd)/(2t)
σ L=(pd)/(4t)
σ L=(pd)/(4t)

28.
2
4t
pd
2t
pd
2
σ-σ
τstress,ShearMaximum
:stressShearMaximum
LC
max



i.e. 𝜏 𝑚𝑎𝑥 =
𝑝𝑑
8𝑡