TOPICS INTRODUCTION STRESSES IN THIN CYLINDRICAL SHEELSHOOP 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

THIN CYLINDERIf 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.

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.

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.

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 Cσ C

σC

p

σ L

σ L

σ L

p ppr

σ Lσ L

σ C

σ C

pr

pr

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

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

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

P - internal pressure (stress)

σc – circumferential stressdL

σc

p

t

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

dltp

d

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 θcos2dp

dA

σcσc

dlt

p

d

dθ

θ

Force on the elementary area,

Horizontal component of this force

dθdl 2dp

dθdldFy θsin2dp

Vertical component of this force

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.

sin2dpforce bursting ldiametrica Total

0

dθdl

dA

σcσc

dlt

pθ

d

dθ

surface. curved theof area projectedp

dp cosdl2dp 0

dl

dlcc tσ 2)σ stress ntialcircumfere to(due force Resisting

dldl dptσ2 i.e., c

dL

σc

p

t

force Burstingforce Resisting um,equillibriUnder

Circumferential stress

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

EVALUATION OF LONGITUDINAL STRESS (σL):

d4πp cylinder) of end (on the force bursting alLongitudin 2

p

t

σL

tdπσ force Resisting L

td π force thisresistingsection cross of Area

cylinder. theof material theof stress alLongitudinσLet L

tdπσd4πp i.e., L

2

UNDER EQULIBIRIUM

BURSTING FORCE = RESISTING FORCE

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

Consider a thin cylindrical shell subjected to internal pressure

p = internal pressured = internal diameter of shellt = thickness of shelll = length of the shell

WE KNOW THAT,

σc =

pd2t and σL =

pd4t

Let, δd= change in diameter of shell δl = change in length of shell

circumferential strain,

ε 1 =

∈1=𝑝𝑑

2 𝑡𝐸 (1 − 12𝑚 )

∈2=𝛿𝑙𝑙 =

𝜎 𝑙

𝐸 −𝜎 𝑐

𝑚𝐸= 𝑝𝑑4 𝑡𝐸 − 𝑝𝑑

2𝑡𝑚𝐸

∈2=𝑝𝑑2 𝑡𝐸 ( 1

2− 1𝑚 )

∴𝑙𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛

Change in diameter Change in length

When the shell is subjected to an internal pressure, there will be an increase in theDiameter as well as the length of the shell

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

Voloume of shell ,

v

(

¿𝜋4

(𝑑2+2. 𝛿𝑑 .𝑑𝑙+𝛿𝑑2 . 𝑙+𝑑2 . 𝛿𝑙+2𝛿𝑑 . 𝑑 .𝛿 𝑙+𝛿𝑑2 . 𝛿𝑙 ) − 𝜋4 𝑑2 . 𝑙

¿𝜋4

(𝑑2 .𝛿 𝑙+2 𝛿𝑑 .𝑑 .𝑙 )𝑛𝑒𝑔𝑙𝑒𝑐𝑡𝑖𝑛𝑔 𝑠𝑚𝑎𝑙𝑙𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑖𝑒𝑠

𝛿𝑣𝑉 =

𝜋4 (𝑑2 .𝛿𝑙+2.𝛿𝑑 .𝑑 . 𝑙)

𝜋4𝑑2𝑙

¿𝛿𝑙𝑙 +

2𝛿𝑑𝑑

𝛿𝑣𝑉 =𝜖2+2.𝜖1

∴𝛿𝑣=𝑉 (𝜀2+2.𝜖1)

∵𝜖1=𝛿𝑑𝑑

𝜀2=𝛿 𝑙𝑙

THIN SPHERICAL SHELL

Consider a thin spherical shell subjected to internal pressure p

p= internal pressured= internal dia of shellt= thickness of shell

= stress in the shell material

Total force,

P

Resisting section

Stress in the shell,

=

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

Consider a thin spherical shell subjected to internal pressure

d=internal dia of shell p=internal pressuret=thickness of the shell= stress in the shell material

We know that the thin spherical shell,

Strain in any direction,

(1)

24tpd

2tpd

2σ-σ τstress,Shear Maximum

other.each tolar perpendicuact and normal are stresses these

Both al.longitudin and ntialCircumfere viz.,point,any at stresses principal twoare There

:stressShear Maximum

LCmax

)5.(....................8tpdτ i.e., max

σ C=(pd)/(2t)σC=(pd)/(2t)

σ L=(pd)/(4t)

σ L=(pd)/(4t)

24tpd

2tpd

2σ-σ τstress,Shear Maximum

:stressShear Maximum

LCmax

i.e.