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When writing the momentum equation in a lagrangian configuration is the the stress tensor used the first Piola-Kirchhoff stress tensor or the nominal stress tensor (which is the transpose of the 1st P-K)? In other words, which of the two equations below is correct:

$$\nabla_0\cdot P + \rho_0 f_0 = \rho_0 a$$

or

$$\nabla_0\cdot N + \rho_0 f_0 = \rho_0 a$$

I have seen it written in terms of the 1st P-K here (section 4.5.2) but in terms of the nominal stress here and here (page 224). The 1st P-K is not symmetric so presumably this does make a difference which will impact the acceleration components.

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All of these resources are saying the same thing, but you have to pay extremely close attention to the definitions of their differential operators. Specifically, in the brown.edu link, they define the divergence of a tensor $\mathbf{A}$ as $$ \nabla \cdot \mathbf{A} = \frac{\partial A_{ij}}{\partial x_i} $$ with summation over the first index of $\mathbf{A}$.

In most other texts (and in my own experience), the divergence of a tensor is defined as $$ div(\mathbf{A}) = \frac{\partial A_{ij}}{\partial x_j} $$ with summation over the second index of $\mathbf{A}$. When I do my work, I deliberately use the "$div$" notation (or "$Div$" if it's a divergence wrt a reference configuration coordinate) rather than the "$\nabla \cdot$" notation to avoid confusion with any sort of inner product ideas that the dot would seem to imply.

Keeping this in mind, you can see how different authors can come to seemingly different conclusions for something as trivial as momentum balance. However, they are really arriving at the same result. At the end of the day, the balance of linear momentum is given by

$$ Div \mathbf{P} + \rho_0 \mathbf{f}_0 = \rho_0 \mathbf{a}_0 $$

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  • $\begingroup$ I missed the difference in the differential operator definitions. I understand that this definition can result in the transpose of what might be expected. Yet in the brown.edu link even though they use the summation over the first index: $$\nabla \cdot \boldsymbol{A} = \frac{\partial A_{ij}}{\partial x_i}$$ They still end up with: $$ \frac{\partial \sigma_{11}}{\partial x_1}+\frac{\partial \sigma_{21}}{\partial x_2}+\frac{\partial \sigma_{31}}{\partial x_3}+\rho_0 f_1 = \rho_0 a_1......$$ Which is the same as conventional summation over the second index. EDIT - Being Daft. $\endgroup$ Commented May 16, 2016 at 15:45
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    $\begingroup$ Well, if you sum over the second index, then you'd end up with $$ \frac{\partial P_{11}}{\partial X_1} + \frac{\partial P_{12}}{\partial X_2} + \frac{\partial P_{13}}{\partial X_3} + ... $$ As you noted, in your question, the first Piola-Kirchhoff stress tensor is not symmetric, so it's not the same as summing over the first index. Since $N = P^T$, using the Brown convention would result in the same expansion in indices as I have written out here. $\endgroup$ Commented May 16, 2016 at 15:48
  • $\begingroup$ My understanding of the summation notation must be flawed. I thought the 'i'th subscript was held constant whilst the 'j'th was iterated and summed to the previous result until completion. In this case the summation over the second subscript makes perfect sense. Yet in the case of the summation over the first index it is obviously incorrect as the index in the denominator iterates within each sum. It seems to me that the summation iteration must be with respect to the denominator subscript. Could you please comment? $\endgroup$ Commented May 16, 2016 at 16:28
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    $\begingroup$ In index notation, the summation is performed over the repeated index. In the brown.edu's notation, $$(\nabla \cdot \mathbf{A})_j = \frac{\partial A_{ij}}{\partial x_i}$$ This implies summation over the first index of $A$ and the (only) index of $x$. In the convention followed by many other authors, the divergence of a tensor is given by $$ (div\mathbf{A})_i = \frac{\partial A_{ij}}{\partial x_j}$$ This convention implies summation over the second index of $A$ and the (only) index of $x$. In each case, the result of the operation is a first-order tensor (vector), but they will not be equal. $\endgroup$ Commented May 16, 2016 at 16:35
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    $\begingroup$ Yup you've got it. No problem at all, this is an important concept to get right, so I'm happy to help. $\endgroup$ Commented May 16, 2016 at 18:01

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