Skip to main content

Thin copper shell: pressure from extra 80 cc liquid

Short answer

Volume V = 212058 cm³, volumetric strain = 80/V. Using ε_v = pd(5−4ν)/(4tE) with ν=0.36, E=10⁶ kg/cm², solve p = 19.78 kg/cm² (1.94 MPa).

A thin walled cylindrical shell made up of copper plate has been filled with a liquid at atmospheric pressure. An additional 80cc of liquid is then pumped into 3 m cylindrical shell whose internal diameter is 300 mm and wall thickness 14 mm. Find the values of pressure developed on the wall of cylinder due to this extra liquid. Take Poisson ratio = 0.36 and Modulus of elasticity E=106 kg/cm2E = 10^6\ \text{kg/cm}^2.

The extra liquid fills the space created by the expansion of the shell. The liquid is assumed incompressible (no bulk modulus is given), so δVshell=80 cm3\delta V_{shell} = 80\ \text{cm}^3.

Data: L=300L = 300 cm, d=30d = 30 cm, t=1.4t = 1.4 cm, ν=0.36\nu = 0.36, E=106 kg/cm2E = 10^6\ \text{kg/cm}^2.

Volumetric strain of a thin cylinder

εv=2εc+εl=pd2tE(52−2ν)=pd4tE(5−4ν)\varepsilon_v = 2\varepsilon_c + \varepsilon_l = \frac{pd}{2tE}\left(\frac{5}{2} - 2\nu\right) = \frac{pd}{4tE}(5-4\nu)

Volume and pressure

V=π4(30)2(300)=212057.50 cm3V = \frac{\pi}{4}(30)^2(300) = 212057.50\ \text{cm}^3 δVV=80212057.50=3.7726×10−4=p(30)4(1.4)(106)(5−4×0.36)=1.9071×10−5 p\frac{\delta V}{V} = \frac{80}{212057.50} = 3.7726\times10^{-4} = \frac{p(30)}{4(1.4)(10^6)}(5 - 4\times0.36) = 1.9071\times10^{-5}\,p p=19.781 kg/cm2≈1.940 N/mm2p = 19.781\ \text{kg/cm}^2 \approx 1.940\ \text{N/mm}^2

Answer: pressure developed ≈19.78 kg/cm2\approx 19.78\ \text{kg/cm}^2 (1.94 MPa).

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Strength of Materials chapter 5 questions →