Expansion of a thin-walled spherical shell

A thin-walled spherical shell made of a perfectly plastic material of yield stress, Y, original radius, $r_{o}$ , and thickness, $t_{o}$ , is being expanded by internal pressure. (a) Calculate the work done in expanding this shell to a radius of $r_{f}$ . (b) If the diameter expands at a constant rate, what changes take place in the power consumed as the radius increases?

Question

Expansion of a thin-walled spherical shell

A thin-walled spherical shell made of a perfectly plastic material of yield stress, Y, original radius, [latex]r_{o}[/latex], and thickness, [latex]t_{o}[/latex], is being expanded by internal pressure. (a) Calculate the work done in expanding this shell to a radius of [latex]r_{f}[/latex]. (b) If the diameter expands at a constant rate, what changes take place in the power consumed as the radius increases?

Accepted Answer

The membrane stresses are given by

[latex]\sigma_{1}[/latex]= [latex]\sigma_{2}[/latex]= Y

(from Example 2.6), where r and t are instantaneous dimensions. The true strains in the membrane are given by

[latex]\epsilon_{1}=\epsilon_{2}=\ln \left(\frac{2 \pi r_{f}}{2 \pi r_{o}} ight)=\ln \left(\frac{r_{f}}{r_{o}} ight)[/latex]

Because an element in this shell is subjected to equal biaxial stretching, the specific energy is

[latex]u=\int_{0}^{\epsilon_{1}} \sigma_{1} d \epsilon_{1}+\int_{0}^{\epsilon_{2}} \sigma_{2} d \epsilon_{2}=2 \sigma_{1} \epsilon_{1}=2 Y \ln \left(\frac{r_{f}}{f_{o}} ight)[/latex]

Since the volume of the shell material is [latex]4 \pi r_{o}^{2} t_{o} [/latex]the work done is

[latex]W=(u)( ext { Volume })=8 \pi Y r_{o}^{2} t_{o} \ln \left(\frac{r_{f}}{r_{o}} ight)[/latex]

The specific energy can also be calculated from the effective stresses and strains. Thus, according to the distortion-energy criterion

[latex]\bar{\sigma}=\frac{1}{\sqrt{2}}\left[(0)^{2}+\left(\sigma_{2} ight)^{2}+\left(-\sigma_{1} ight)^{2} ight]^{-1 / 2}=\sigma_{1}=\sigma_{2}[/latex]

and

[latex]\bar{\epsilon}=\frac{\sqrt{2}}{3}\left[(0)^{2}+\left(\epsilon_{2}+2 \epsilon_{2} ight)^{2}+\left(-2 \epsilon_{2}-\epsilon_{2} ight)^{2} ight]^{1 / 2}=2 \epsilon_{2}=2 \epsilon_{1}[/latex]

(The thickness strain [latex]\epsilon_{3}[/latex]= -2[latex]\epsilon_{2}[/latex]= -2[latex]\epsilon_{1}[/latex]because of volume constancy in plastic deformation, where [latex]\epsilon_{1}[/latex]+ [latex]\epsilon_{2}[/latex]+ [latex]\epsilon_{3}[/latex]= 0.) Hence

[latex]u=\int_{0}^{\epsilon} \bar{\sigma} \bar{\epsilon}=\int_{0}^{2 \epsilon_{1}} \sigma_{1} d \epsilon_{1}=2 \sigma_{1} \epsilon_{1}[/latex]

Thus, the answer is the same.
Power is defined as the rate of work; thus

Power=[latex]\frac{d W}{d t}[/latex]

Since all other factors in the expression are constant, the expression for work can be written as being proportional to strain

[latex]W \propto \ln \left(\frac{r}{r_{o}} ight) \propto\left(\ln r-\ln r_{o} ight)[/latex]

Hence

Power[latex]\propto \frac{1}{r} \frac{d r}{d t}[/latex]

Because the shell is expanding at a constant rate, [latex]d_{r}[/latex]/[latex]d_{t}[/latex] = constant. Hence the power is related to the instantaneous radius r by

Power[latex]\propto \frac{1}{r}[/latex]

Question 2.7: Expansion of a thin-walled spherical shell A thin-walled sph...

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