written 6.4 years ago by | modified 5.4 years ago by |
Derive the expression for the volumetric strain for thin cylindrical shells.
written 6.4 years ago by | modified 5.4 years ago by |
Derive the expression for the volumetric strain for thin cylindrical shells.
written 6.4 years ago by | • modified 6.4 years ago |
p is the internal pressure, t is the thickness and d is diameter.
written 5.4 years ago by |
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 tE−pd4tE×μ
EL=pd2tE(12−μ)
Circumferential/hoop strain=d2tE−2dμ4tE
En=pd2tE(1−12−μ)
En=pd4t E(2−μ)
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(12−μ)+2pd2tE(1−12 μ)
δVV=Pd2tE[12−μ+2−μ]=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).