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Derive the expression for the volumetric strain for thin cylindrical shells.

Derive the expression for the volumetric strain for thin cylindrical shells.

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p is the internal pressure, t is the thickness and d is diameter. 1 2 3 4 5

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Let El and Eh be the longitudinal strain and hoop strain.

And μ is the poison ration

And E is the modulus of elasticity.

Now, Longitudinal strain EL=Pd2 tEpd4tE×μ

EL=pd2tE(12μ)

Circumferential/hoop strain=d2tE2dμ4tE

En=pd2tE(112μ)

En=pd4t E(2μ)

Hoop strain or circumferential strain=change in circumferenceoriginal circumference

En=πδdπd=δdd (δd is change in diameter.)

But δdd=strain of diameter.

 Change in diameter=Eh×Original diameter

And also, Longitudinal strain=δll=EL

 Change in length=El×Original length

Now, capacity of shell= V =π4×d2 l

Change in capacity of shell=δV=π4d2×δl+π4×2d δd l

(i.e. taking differential.)

Now,

δVV=δlL+2 δdd [1]

δVV=El+En=Pd2tE(12μ)+2pd2tE(112 μ)

δVV=Pd2tE[12μ+2μ]=pd2tE(52 2μ)

\frac{\delta V}{V} = \frac{pd}{4 t E} [5 – 4 \mu] \rightarrow [2]

Equation [1] and [2] are the volumetric strain for thin cylindrical shell.

Note:

If the compressibility of the fluid in cylinder is also considered then the change in capacity of shell.

\delta V = \frac{pd}{4tE} (5 – 4 \mu) V + \frac{p}{k} \ V

i.e.

Volumetric \ Strain = \frac{\delta V}{V} = \frac{pd}{4tE} (5 – 4 \mu) V + \frac{p}{k}

( Where K is bulk modulus of fluid).

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