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 (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. 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.

 

DARK PORTION IN FIGURE IS THE ELEMENT

Fig. Longitudinal failure of a cylinder

Equate     resisting force = 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

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

σh L t + σh L t =2 σh L t

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.

Fig. Circumferential failure of a cylinder

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 minimum d/t ratio 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 = p Di/4tE – μ p Di/2tE

δL/L = (p Di/4tE) (1-2μ) =h /2E)(1-2μ)

δL = (p Di 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 = p Di/2tE – μ p Di/4tE

δD/D = (p Di/4tE) (2-μ) =

δD = (p Di Di /4tE) (2-μ)= (pDi2 /4tE) (2-μ)

(c) Change in volume, δV

V = (π/4) D2L

Taking log on both sides

log V= log(π/4) +log D2 +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 Di/4tE) (2-μ)+ (p Di/4tE) (1-2μ)

δV/V =  (p Di/4tE) (5–4μ) =h /2E) (5–4μ)

δV = (p Di/4tE) (5–4μ) (π/4)D2L

δV =(πpD3L/16tE) (5–4μ)

SPHERICAL SHELL

Stresses

Hoop stress

hoop stress

radial stress

Derivation for hoop stress

Bursting force=(π/4)Di2 pi

Resisting force =Di t σh)

Bursting force=Resisting force

(π/4)Di2 pi=Di t σh)

σh = 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(σh)

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

  • Hoop stress or Tangential stress or Circumferential stress(σh)

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

  • Radial stress(σr)

  • 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) σh = 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 = σh πD t

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

= p (π/4)Di2

Therefore  σh πD t = p (π/4)Di2

σh = pD/4t

If the efficiency of the circumferential joint is considered then  σh = pD/2t ηcircum

(b) Radial stress = σr = inside fluid pressure

σr = pi

 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= σh/E -μ σh/E = p Di/4tE – μ p D/4tE

δD/D = (p Di/4tE) (1-μ)

δD = (p Di Di /4tE) (2-μ)= (pDi2 /4tE) (2-μ)

δD = (pDi2 /4tE) (1-μ)

(ii)Change in volume, δV

 Volumetric strain = 3 δD/D

                               = 3 (p Di/4tE) (1-μ)

                       δV/V=  (¾)(p Di/t E) (1–μ)

V = (4π/3) R3 = (π/6) D3

δV = (p Di/4tE) (1–μ) (π/6) D3

δV =(π/6)(pD4/t E) (1–μ)=(πpD4/6tE) (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               σh = pi Di / 2t
Longitudinal stress   σL = pi Di / 4t
Radial stress               σr = pi
Minimum value of Di / t is 20.
Hoop stress                σh = 10 pi
Longitudinal stress    σL =5 pi
Radial stress               σr = pi
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 = σh = pi Di/2t = 1.5 x 200/ (2 x1.5) = 100 MPa
Longitudinal stress=pi Di/4t = 1.5 x 200/ (4 x1.5) = 50 MPa
Radial stress=pi Di/2t = 1.5  = 1.5 MPa
Hoop strain=(1/2E)[σh-μσl) =(1/2×200 x 1000)[100- 0.25 x 50)
=87.5/400000 = 0.000021875
Longitudinal strain=(1/2E)[σl-μσh) =(1/2×200 x 1000)[50- 0.25 x 100)  =25/400000 = 00000625

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