THIN SHELLS CLASS NOTES FOR MECHANICAL ENGINEERING
THIN CYLINDRICAL SHELL
CLASS NOTES FOR
MECHANICAL ENGINEERING
A vessel is a thin if its thickness is less as
compared to its diameter. Mathematically
it is expressed as a thin shell if D/t. A
vessel is a thin shell where stresses are
assumed to be uniform. Uniform stresses
mean stress at the inner to outer radius is
of same magnitude.
Thin sell
where D/t > 20 and where stress are uniform from the inner surface to the outer surface of the shell.
Various Stresses in a Thin Cylindrical Shell
(i) Hoop stress or Tangential stress or Circumferential stress (σ_{h})
It acts in the tangential direction at the point of consideration
(ii) Longitudinal stress (σ_{l})
It acts in the longitudinal direction at the point of consideration
(iii) Radial stress(σ_{r})
It acts in the radial direction at the point of consideration
NOTE: All the three stresses are tensile or compressive and are at right angles to each other. Therefore, these are principal stresses. Hoop stress is tensile. Longitudinal stress is tensile. Radial stress is always compressive.
Examples of thin shells

Steam boilers

P.G. Cylinder

Air compressor

Gas storage shell in a refinery

Pressure cooker

Reaction vessels
Shapes of thin shells

Cylindrical

Spherical

Modified
Derivation of formulas for various stresses
(i) σ_{h} = hoop stress
It will break the cylinder length wise as shown in Fig.
DARK PORTION IN FIGURE IS THE ELEMENT
Fig. Longitudinal failure of a cylinder
Equate resisting force = breaking force.
Material resistance = σ_{h} L t + σ_{h} L t = 2 σ_{h} L t
Breaking force due to internal fluid pressure = p x projected area
Bursting force on element with dimensions rdθ and L
dF= p Sinθ rdθ L
Integrating θ from 0 to 180 OR it will be twice of when θ is taken from 0 to 90
F = 2p∫Sinθ rdθ L . limits from 0 to 90
F = 2prL = p D L
RESISTING FORCE
σ_{h} L t + σ_{h} L t =2 σ_{h} L t
Therefore equating the resisting force to bursting force, we get
2 σ_{h} L t = p DL
σ_{h} = p D/2t
If the efficiency of the longitudinal joint is considered then σ_{h} = p D/2t η_{long}
(ii) σ_{l} = Longitudinal stress
If the cylinder fails at the circumference joint due to fluid pressure as shown in Fig.
Fig. Circumferential failure of a cylinder
Equating the resistance of the material to the breaking force.
Material resistance = σ_{l} (π/4)(D_{o}^{2} –D_{i}^{2})
= σ_{l} (π/4)(D_{i} +2t)^{2} –D_{i}^{2})= σ_{l} π D_{i} t
Breaking force due to internal fluid pressure = p x internal area(on which pressure is acting)
= p (π/4)D_{i}^{2})
Therefore σ_{l} π D_{i} t = p(π/4)D_{i}^{2})
σ_{l} = p D_{i}/4t
If the efficiency of the circumferential joint is considered then σ_{l} = pD/4t η_{circum}
(iii) Radial stress = σ_{r }= inside fluid pressure
σ_{r }= p_{i}
From the equations, it can be easily concluded that the radial stress is negligible because of minimum d/t ratio is 20.
Builtup shells
Shells with joints are called builtup shells.
Change in the dimensions of the shell
(a) Change in length δL
δL/L = σ_{l}/E μ σ_{h}/E = p D_{i}/4tE – μ p D_{i}/2tE
δL/L = (p D_{i}/4tE) (12μ) =(σ_{h} /2E)(12μ)
δL = (p D_{i} L/4tE) (12μ)
(b) Change in diameter of the shell, δD
δD/D = Change in circumference / Original circumference
= π δD/πD
Therefore δD/D= σ_{h}/E μ σ_{L}/E = p D_{i}/2tE – μ p D_{i}/4tE
δD/D = (p D_{i}/4tE) (2μ) =
δD = (p D_{i }D_{i} /4tE) (2μ)= (pD_{i}^{2} /4tE) (2μ)
(c) Change in volume, δV
V = (π/4) D^{2}L
Taking log on both sides
log V= log(π/4) +log D^{2 }+log L
log V= log(π/4) + 2 log D^{ }+log L
Differentiate
δV/V= 2 δD/D + δL/L
Volumetric strain
δV/V = 2 (p D_{i}/4tE) (2μ)+ (p D_{i}/4tE) (12μ)
δV/V = (p D_{i}/4tE) (5–4μ) =(σ_{h} /2E) (5–4μ)
δV = (p D_{i}/4tE) (5–4μ) (π/4)D^{2}L
δV =(πpD^{3}L/16tE) (5–4μ)
SPHERICAL SHELL
Stresses
Hoop stress
hoop stress
radial stress
Derivation for hoop stress
Bursting force=(π/4)D_{i}^{2} p_{i}
Resisting force =(πD_{i }t σ_{h})
Bursting force=Resisting force
(π/4)D_{i}^{2} p_{i}=(πD_{i }t σ_{h})
σ_{h} = p D/4t
SHELL HEADS OR SHELL ENDS
All cylindrical shells under internal fluid pressure require ends on both sides to close the shell. These covers are called HEADS. These heads are flat type, flanged dished type, conical head type and hemispherical head type. The use of a particular type of head will depend upon the fluid pressure. Flat heads for minimum and hemispherical heads are suitable for maximum fluid pressure.
THIN SPHERICAL SHELL
Stresses in a thin spherical shell

Hoop stress or Tangential stress or Circumferential stress(σ_{h})

It acts in the tangential direction at the point of consideration.

Hoop stress or Tangential stress or Circumferential stress(σ_{h})

It acts in the tangential direction at the point of consideration (say in a vertical circle)

Radial stress(σ_{r})

It acts in the radial direction at the point of consideration
NOTE: All the three stresses are tensile or compression and are at right angles to each other. Therefore, these are principal stresses.
Hoop stress is tensile.
Hoop stress is tensile.
Radial stress is compression stress.
Derivation of formula of various stresses

(a) σ_{h} = hoop stress
It will break the sphere at the circumference as shown in Fig.
Refer to Fig. and equating the resistance of the material to the breaking force.
Fig. Circumferential failure of a sphere
Material resistance = σ_{h} πD t
Breaking force due to internal fluid pressure = p x projected area
= p (π/4)D_{i}^{2}
Therefore σ_{h} πD t = p (π/4)D_{i}^{2}
σ_{h} = pD/4t
If the efficiency of the circumferential joint is considered then σ_{h} = pD/2t η_{circum}
(b) Radial stress = σ_{r }= inside fluid pressure
σ_{r }= p_{i}
Radial stress is negligible as compared to hoop stress. Because of d/t minimum ratio is 20.
Change in the dimensions of the shell
(i) Change in diameter of the shell, δD
δD/D = Change in circumference / Original circumference= π δD/πD
Therefore δD/D= σ_{h}/E μ σ_{h}/E = p D_{i}/4tE – μ p D/4tE
δD/D = (p D_{i}/4tE) (1μ)
δD = (p D_{i }D_{i} /4tE) (2μ)= (pD_{i}^{2} /4tE) (2μ)
δD = (pD_{i}^{2} /4tE) (1μ)
(ii)Change in volume, δV
Volumetric strain = 3 δD/D
= 3 (p D_{i}/4tE) (1μ)
δV/V= (¾)(p D_{i}/t E) (1–μ)
V = (4π/3) R^{3 }= (π/6) D^{3}
δV = (p D_{i}/4tE) (1–μ) (π/6) D^{3}
δV =(π/6)(pD^{4}/t E) (1–μ)=(πpD^{4}/6tE) (1–μ)
Effect of joint efficiencies on the stress induced
Joint efficiencies increase the stresses in the cylinder.
Efficiency of longitudinal joint increase the hoop stress.
Efficiency of circumferential joint increase the longitudinal stress.
Advantages of thin spherical shells over thin cylindrical shells
(i) Thickness requirement is less for the same pressure, same diameter and same material.
(ii) Higher pressure can be with the same thickness of a cylindrical shell of same material and same diameter
(iii) It can store or process more volume in the same space i.e. it is compact.
Disadvantages of thin spherical shells over thin cylindrical shells
(i) It is difficult to manufacture and a costly affair.
(ii) It is difficult to transport from the industry to the point of use.
(iii) It is difficult to support at the site of use.
(iv)It is difficult to repair and maintain.
Thus invariably cylindrical thin shells are in use. Only sparingly a spherical shell is used for storage in a refinery.
Negligible Radial stress
Hoop stress σ_{h} = p_{i} D_{i} / 2t
Longitudinal stress σ_{L} = p_{i} D_{i} / 4t
Radial stress σ_{r} = p_{i}
Minimum value of Di / t is 20.
Hoop stress σ_{h} = 10 p_{i}
Longitudinal stress σ_{L} =5 p_{i}
Radial stress σ_{r} = p_{i}
Radial stress is equal to fluid pressure and is very small as compared to hoop stress as well as longitudinal stress.
A boiler shell 200 cm diameter and 1.5 cm thickness is subjected to an internal fluid pressure of 1.5 M Pa. What is the hoop, longitudinal and radial stresses induced in the vessel? Also calculate the hoop and longitudinal strains.
Boiler is always a cylindrical vessel.
Hoop stress = σ_{h} = p_{i} D_{i}/2t = 1.5 x 200/ (2 x1.5) = 100 MPa
Longitudinal stress=p_{i} D_{i}/4t = 1.5 x 200/ (4 x1.5) = 50 MPa
Radial stress=p_{i} D_{i}/2t = 1.5 = 1.5 MPa
Hoop strain=(1/2E)[σ_{h}μσ_{l}) =(1/2×200 x 1000)[100 0.25 x 50)
=87.5/400000 = 0.000021875
Longitudinal strain=(1/2E)[σ_{l}μσ_{h}) =(1/2×200 x 1000)[50 0.25 x 100) =25/400000 = 00000625
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