Question

A cylindrical container of radius $R=1m$, wall thickness $1mm$ is filled with water up to a depth of $2m$ and suspended along its upper rim. The density of water is $1000kg/m^3$ and acceleration due to gravity is $10m/s^2$. The self-weight of the cylinder is negligible. The formula for hoop stress in a thin-walled cylinder can be used at all points along the height of the cylindrical container.



If the Young's modulus and Poisson's ratio of the container material are 100 Gpa and 0.3 respectively, the axial strain in the cylinder wall at mid - depth is

• A. $2\times {10}^{-5}$
• B. $6\times {10}^{-5}$
• C. $7\times {10}^{-5}$
• D. $1.2\times {10}^{-4}$

Answer 1

The correct option is C $7\times {10}^{-5}$

Note : why $\frac{pd}{2t}and\frac{pd}{4t}$ not applicable for the given case?
⦁ In case of closed end cylinder when pressure acts along all direction the resisting force developed in longitudinal as well as circumferential direction that's why the both stresses come into picture.
⦁ This is only applicable for thin cylinder with internal gas pressure and applicable for thin cylinder with internal gas pressure and both ends closed.
⦁ If ends of the cylinder are open then longitudinal stress will be zero. For example in pipe flow case. In pipes flow the fluid pressure acts in circumferential direction because of this the resisting force develops along circumference and that's why the circumferential stresses come into picture.
In our problem
⦁ When this container which is hold at the bottom side then longitudinal stress will not exist at any point along longitudinal direction. Because the weight of water which is hold by bottom support and the net resisting force along longitudinal direction is zero. But circumferential stress comes because of fluid pressure which is dependent from height of water column.
⦁ when this container which is hold at the top side then longitudinal stress comes into picture because the weight of water develops the resisting force along the longitudinal direction which is equal in magnitude at every point along longitudinal direction and circumferential stresses varying point to point if the height of water column varying.
⦁ So this is the special case in thin cylinder.
Special Case : When thin cylinder containing liquid with open end.
As per given data

${\sigma }_1=\frac{pgHr}{2t}=\frac{1\times 2\times 1000\times 10}{2\times 0.001}=10MPa$

${\sigma }_c=\frac{pgHr}{t}=\frac{1000\times 10\times 1\times 1}{0.001}=10MPa$

$\because$ Axial strain is given by,

$=\frac{{\sigma }_1}{E}-\mu \frac{{\sigma }_c}{E}$

Axial strain at mid - depth
$=\frac{10-0.3\times 10}{100\times {10}^3}=7\times {10}^{-5}$