89589540-Strength-of-Materials-by-S-K-Mondal-10.pdf

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10. Thin Cylinder

Theory at a Glance (for IES, GATE, PSU)1. Thin Rings

Uniformly distributed loading (radial) may be due to

either

• Internal pressure or external pressure

• Centrifugal force as in the case of a rotating ring

Case-I: Internal pressure or external pressure

• s = qr Where q = Intensity of loading in kg/cm of Oce

r = Mean centreline of radius

s = circumferential tension or hoop’s

tension

(Radial loading ducted outward)

• Unit stress, σ = =s qr

A A

• Circumferential strain,E

c

qr

AE

σ ∈ = =

• Diametral strain, (∈d ) = Circumferential strain, (∈c

)

Case-II: Centrifugal force

• Hoop's Tension,

2 2ω

= w r

sg

Where w = wt. per unit length of circumferential element

ω = Angular velocity

• Radial loading, q =

2ω

= w r s

r g

• Hoop's stress,2 2.σ ω = =

s wr

A Ag

2. Thin Walled Pressure Vessels

For thin cylinders whose thickness may be considered small compared to their diameter.

iInner dia of the cylinder (d )

15 or 20wall thickness(t)

>

Page 308 of 429



Chapter-10 Thin Cylinder S K Mondal’s

3. General Formula

1 2

1 2

σ σ + =

p

r r t

Where 1σ =Meridional stress at A

2σ =Circumferential / Hoop's stress

P = Intensity of internal gas pressure/ fluid pressure

t = Thickness of pressure vessel.

4. Some cases:

• Cylindrical vessel

1 2=

2 4 2σ σ = = =

pr pD pr pD

t t t t 1 2

,r r r ⎡ ⎤→ ∞ =⎣ ⎦

1 2max

2 4 8

σ σ τ

−= = =

pr pD

t t

• Spherical vessel

1 22 4

σ σ = = = pr pD

t t [r1 = r2 = r]

• Conical vessel

1 1

tan[ ]

2 cos

pyr

t

α σ

α = → ∞ and

2

tan

cos

py

t

α σ

α =

Notes:

• Volume 'V' of the spherical shell,3V=

6

π

i D

1/36

π

⎛ ⎞⇒ = ⎜ ⎟

⎝ ⎠i

V D

• Design of thin cylindrical shells is based on hoop's stress

5. Volumetric Strain (Dilation)

• Rectangular block,

0

x y z

V

V

Δ=∈ + ∈ + ∈

• Cylindrical pressure vessel

∈1=Longitudinal strain = [ ]1 2 1 22

pr

E E Et

σ σ μ μ − = −

2∈ =Circumferential strain = [ ]2 1 1 22

σ σ μ μ − = −

pr

E E Et

Volumetric Strain,1 2

2 [5 4μ] [5 4μ]2 4

Δ=∈ + ∈ = − = −

o

V pr pD

V Et Et

i.e. ( ) ( ) ( )1 2, 2

vVolumetric strain longitudinal strain circumferential strain∈ = ∈ + × ∈

• Spherical vessels

1 2 [1 ]2

pr

Et μ ∈=∈ =∈ = −

α

α α

α

1σ

2σ

2σ

Page 309 of 429



Chapter-10 Thin Cylinder S K Mondal’s

0

33 [1 ]

2

V pr

V Et μ

Δ= ∈= −

6. Thin cylindrical shell with hemispherical end

Condition for no distortion at the junction of cylindrical and hemispherical portion

2

1

1

2

t

t

μ

μ

−=

− Where, t1= wall thickness of cylindrical portion

t2 = wall thickness of hemispherical portion

7. Alternative method

Consider the equilibrium of forces in the z-direction acting on the part

cylinder shown in figure.

Force due to internal pressure p acting on area π D2/4 = p. π D2/4

Force due to longitudinal stress sL acting on area π Dt =1

σ π Dt

Equating: p. π D2/4 =1

σ π Dt

or 14 2

pd pr

t tσ = =

Now consider the equilibrium of forces in the x-direction acting on the

sectioned cylinder shown in figure. It is assumed that the

circumferential stress2

σ is constant through the thickness of the

cylinder.

Force due to internal pressure p acting on area Dz = pDz

Force due to circumferential stress2

σ acting on area 2tz =2

σ 2tz

Equating: pDz =2

σ 2tz

or 2

2

pD pr

t tσ = =

Page 310 of 429

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