For a general isotropic medium, the stress tensor pij and strain tensors eij are related by pij = σE/(1 + σ)(1 − 2σ) ekkδij + E/1 + σ eij, where E is Young’s modulus and σ is Poisson’s ratio. By considering an isotropic body subjected to a uniform hydrostatic pressure (no shearing stress), show

Question

For a general isotropic medium, the stress tensor [latex]{p}_{ij}[/latex] and strain tensors [latex]{e}_{ij}[/latex] are related by

[latex]{ p }_{ ij }=\frac { \sigma E }{ (1+\sigma )(1-2\sigma ) } { e }_{ kk }{ \delta }_{ ij }+\frac { E }{ 1+\sigma } { e }_{ ij } [/latex]

where E is Young’s modulus and σ is Poisson’s ratio.

By considering an isotropic body subjected to a uniform hydrostatic pressure (no shearing stress), show that the bulk modulus k, defined by the ratio of the pressure to the fractional decrease in volume, is given by k = E/[3(1 − 2σ)].

Accepted Answer

Consider a small rectangular parallelepiped, with one corner at the origin and the opposite one at [latex]\left(a_{1}, a_{2}, a_{3} ight)[/latex], subjected to a uniform hydrostatic pressure. The isotropy of the pressure means that all forces are normal to the surfaces on which they act and that the stress and strain tensor components [latex]p_{i j}=e_{i j}=0[/latex] for i ≠ j. Furthermore, because of the symmetry of the situation, when i ≠ j, not only is [latex]e_{i j}[/latex] zero, but so are the individual [latex]\partial u_{i} / \partial x_{j}[/latex] that are its constituents.

In the current situation, [latex]p_{11}=p_{22}=p_{33}=-p[/latex] and so, writing [latex]\sum_{k} e_{k k}[/latex] as [latex] heta[/latex], we have, for each [latex]i(i=1,2,3)[/latex] with no summation over i implied, that

[latex]-p=p_{i i}=\frac{E}{(1+\sigma)(1-2 \sigma)}\left[\sigma heta+(1-2 \sigma) e_{i i} ight] .[/latex]

Adding the three equations together gives

[latex]-3 p=\frac{E}{(1+\sigma)(1-2 \sigma)}[3 \sigma heta+(1-2 \sigma) heta]=\frac{E heta}{1-2 \sigma} ext {. }[/latex]

Now the fractional increase f in the volume of the parallelepiped is given by

[latex]\frac{1}{a_{1} a_{2} a_{3}}\left(a_{1}+\frac{\partial u_{1}}{\partial x_{i}} a_{i}+\cdots ight)\left(a_{2}+\frac{\partial u_{2}}{\partial x_{i}} a_{i}+\cdots ight)\left(a_{3}+\frac{\partial u_{3}}{\partial x_{i}} a_{i}+\cdots ight)-1 .[/latex]

Since [latex]\partial u_{i} / \partial x_{j}=0[/latex] for i ≠ j, the only three non-zero first-order terms are

[latex]f=\frac{\partial u_{1}}{\partial x_{1}}+\frac{\partial u_{2}}{\partial x_{2}}+\frac{\partial u_{3}}{\partial x_{3}}=e_{11}+e_{22}+e_{33}= heta .[/latex]

We conclude that the bulk modulus, k, is given by

[latex]k=\frac{p}{-f}=\frac{-E heta}{3(1-2 \sigma)} \frac{1}{(- heta)}=\frac{E}{3(1-2 \sigma)} .[/latex]

Question 26.21: For a general isotropic medium, the stress tensor pij and st...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions