THIN CYLINDRICAL SHELL

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THIN CYLINDRICAL SHELL

 

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.

Various Stresses in a Thin Cylindrical Shell

(i) Hoop stress or Tangential stress or Circumferential stress(σh)–It acts in the tangential direction at the point of consideration

(ii) Longitudinal stress (σl)--It acts in the longitudinal direction at the point of consideration

(iii) Radial stress(σr)--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 but radial stress is always compressive.

Examples of thin shells

  1. Steam boilers

  2. P.G. Cylinder

  3. Air compressor

  4. Gas storage shell in a refinery

  5. Pressure cooker

  6. Reaction vessels

Shapes of thin shells

  1. Cylindrical

  2. Spherical

  3. Modified

Derivation of formulas for various stresses

(i) σh = hoop stress

It will break the cylinder length wise as shown in Fig.

Equating the resistance of the material to the breaking force.

Material resistance = σh L t + σh L t = 2 σh L t

Breaking force due to internal fluid pressure = p x projected area

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

Therefore equating the resisting force to bursting force, we get

2 σh L t = p DL

σh = p D/2t

If the efficiency of the longitudinal joint is considered then  σh = p D/2t ηlong

(ii) σl = Longitudinal stress

If the cylinder fails at the circumference joint due to fluid pressure as shown in Fig.

 

Equating the resistance of the material to the breaking force.

Material resistance = σl (π/4)(Do2 –Di2)

= σl (π/4)(Di +2t)2 –Di2)= σl π Di t

Breaking force due to internal fluid pressure = p x internal area(on which pressure is acting)

= p (π/4)Di2)

Therefore  σl π Di t = p(π/4)Di2)

σl = p Di/4t

If the efficiency of the circumferential joint is considered then σl = pD/4t ηcircum

(iii) Radial stress = σr = inside fluid pressure

σr = pi

From the equations, it can be easily concluded that the radial stress is negligible because of d/t ratio minimum value is 20.

 

Built-up shells

 

Shells with joints are called built-up shells.

 

Change in the dimensions of the shell

 

(a) Change in length δL

δL/L = σl/E -μ σh/E = pDi/4tE – μ pD/2tE

δL/L = (pDi/4tE) (1-2μ)

δL = (pDi L/4tE) (1-2μ)

(b) Change in diameter of the shell, δD

δD/D = Change in circumference / Original circumference= π δD/πD

Therefore δD/D= σh/E -μ σL/E = pDi/2tE – μ pD/4tE

δD/D = (pDi/4tE) (2-μ)

δD = (pDiDi /4tE) (2-μ)= (pDi2 /4tE) (2-μ)=

(c) Change in volume, δV

δV/V = Volumetric strain = 2 δD/D + δL/L

δV/V = 2 (pDi/4tE) (2-μ)+ (pDi/4tE) (1-2μ)

δV/V =  (pDi/4tE) (5–4μ)

V = (π/4) D2L

δV = (pDi/4tE) (5–4μ) (π/4)D2L

δV =(πpD3L/16tE) (5–4μ)

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.