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Thin copper shell: pressure from extra 50 cc liquid

Short answer

Cylinder volume 98175 cm³, volumetric strain 50/V. From ε_v = pd(5−4ν)/(4tE) with ν=0.34, E=10⁶ kg/cm², pressure p = 26.86 kg/cm² (2.63 MPa).

A thin walled cylindrical shell made up of copper plate has been filled with a liquid at atmospheric pressure. An additional 50 c.c. of liquid is then pumped in to 2m cylinder whose internal diameter is 25 cm and wall thickness is 12 mm. Find the values of pressure developed on the wall of cylinder due to this extra liquid. Take poission ratio = 0.34 and modules of Elasticity = 106 kg/cm210^6\ \text{kg/cm}^2.

The extra liquid occupies the additional space made by the expansion of the shell (the liquid is taken as incompressible), so δV=50 cm3\delta V = 50\ \text{cm}^3.

Data: L=200L = 200 cm, d=25d = 25 cm, t=1.2t = 1.2 cm, ν=0.34\nu = 0.34, E=106 kg/cm2E = 10^6\ \text{kg/cm}^2.

Relation between pressure and volume change

For a thin cylinder, σc=pd2t\sigma_c = \dfrac{pd}{2t}, σl=pd4t\sigma_l = \dfrac{pd}{4t}, and

εv=2εc+εl=pd4tE(5−4ν)\varepsilon_v = 2\varepsilon_c + \varepsilon_l = \frac{pd}{4tE}(5-4\nu) V=π4(25)2(200)=98174.77 cm3V = \frac{\pi}{4}(25)^2(200) = 98174.77\ \text{cm}^3 δVV=5098174.77=5.0930×10−4=p(25)4(1.2)(106)(5−4×0.34)=1.8958×10−5 p\frac{\delta V}{V} = \frac{50}{98174.77} = 5.0930\times10^{-4} = \frac{p(25)}{4(1.2)(10^6)}(5-4\times0.34) = 1.8958\times10^{-5}\,p p=26.864 kg/cm2≈2.634 N/mm2p = 26.864\ \text{kg/cm}^2 \approx 2.634\ \text{N/mm}^2

Answer: pressure developed ≈26.86 kg/cm2\approx 26.86\ \text{kg/cm}^2 (2.63 MPa).

✓ Answer checked by Claude Opus 5.5 (high effort)

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