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Finite Stretching and Shearing of an Internally Balanced Elastic Solid


When seeking to separately describe elastic and inelastic effects in solids undergoing finite strain, it is typical to factor the deformation gradient \(\bf F\) into parts, say \(\mathbf{F}={\hat {\mathbf{F}}}{\mathbf{F}}^{*}\). Such a factorization can also be used to describe a notion of internal elastic balance. For equilibrium deformations this balance emerges naturally from an appropriately formulated energy minimization by considering the variation with respect to the multiplicative decomposition itself. The internal balance requirement is then a tensor relation between F, \({\hat {\mathbf{F}}}\) and F . Such theories holds promise for describing various substructural reconfigurations in solids. This includes the development of surfaces that localize slip once a threshold value of load is obtained. We consider an incompressible internally balanced elastic material with a constitutive law that is motivated by neo-Hookean behavior in the standard hyperelastic theory. The role of the internal balance relation is examined in uniaxial loading and also in simple shearing. In uniaxial loading the principle directions remain fixed. In simple shearing the principle directions change as the deformation proceeds. To address the latter, we establish a specific sense in which the internal balance relation can be solved analytically for general isochoric \(\bf F\).

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Helpful discussions with H. Tsai when early aspects of this treatment were first formulated are gratefully acknowledged. This publication was made possible by NPRP grant #4-1333-1-214 from the Qatar National Research Fund (a member of the Qatar Foundation). The statements made herein are solely the responsibility of the authors.

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Correspondence to Thomas J. Pence.


Appendix A: Theorem Demonstration

We prove the theorem from Sect. 5 which is as follows:


Let c be a real number and \(\bf A\) and \(\bf Y\) be symmetric tensors from \(\mathcal{R}^{3} \to \mathcal{R}^{3}\) such that

$$ {\mathbf{Y}}^2+c\mathbf{Y}=\mathbf{A}. $$

Then c is a root of the sixth order polynomial equation

$$\begin{aligned} &4AY^4 c^6+ 2 \bigl( A a_1^2 Y^3- A a_2 Y^3- a_1^2 Y^5+ a_2 Y^5 \bigr) c^5 \\ &\quad{} + \bigl( 4 A^2 a_1 Y^2+ 8Aa_1 Y^4 - a_1^4 Y^4 + 2 a_1^2 a_2 Y^4-a_2^2 Y^4+4 a_1 Y^6 \bigr)c^4 \\ &\quad{} + 4 \bigl( A^3 Y + 5 A^2 Y^3-5 A Y^5-Y^7 \bigr)c^3 \\ &\quad{} + 4 \bigl( A^2 a_1^2 Y^2 - A^2 a_2 Y^2-2A a_1^2 Y^4 +2 A a_2 Y^4+a_1^2 Y^6 - a_2 Y^6 \bigr) c^2 \\ &\quad{} -4 A^4 + 16 A^3 Y^2 - 24 A^2 Y^4+ 16 A Y^6-4 Y^8 =0, \end{aligned}$$


$$ a_1=\operatorname{tr}(\mathbf{A}), \qquad a_2=\operatorname{tr}\bigl( \mathbf{A}^2\bigr), \qquad A= \det{\mathbf{A}}, \qquad Y=\det{ \mathbf{Y}}. $$


For any symmetric tensor \(\mathbf{M}: \mathcal{R}^{3}\to \mathcal{R}^{3}\) let the determinant of \(\bf M\) be denoted by M, and the trace of M n for arbitrary positive integer n be denoted by m n , i.e.,

$$ \det{\mathbf{M}}=M, \qquad \operatorname{tr}\bigl(\mathbf{M}^n \bigr)=m_n. $$

Use a corresponding notational scheme for symmetric tensors \(\bf A\) and \(\bf Y\). This notation gives: A=detA, \(a_{n}=\operatorname{tr}(\mathbf{A}^{n})\), Y=detY, \(y_{n}=\operatorname{tr}(\mathbf{Y}^{n})\) and so is consistent with (A.3). Also, in this notational scheme the Cayley–Hamilton identity is written

$$ {\mathbf{M}}^3 - m_1 {\mathbf{M}}^2 + \frac{1}{2}\bigl(m_1^2-m_2\bigr) \mathbf{M} - M \mathbf{I}=\mathbf{0}. $$

Taking the trace of (A.5) yields

$$ m_3=-\frac{1}{2}m_1^3 + \frac{3}{2}m_1 m_2 + 3M. $$

Multiplying (A.5) by \(\bf M\), taking the trace, and using (A.6) gives

$$ m_4=-\frac{1}{2}m_1^4 + m_1^2 m_2 +\frac{1}{2} m_2^2 + 4m_1M. $$

We now apply the results (A.5)–(A.7) to (A.1).

First, take the trace of (A.1) so as to obtain \(\operatorname{tr}(\mathbf{Y}^{2}) + c \operatorname{tr}(\mathbf{Y})=\operatorname{tr}(\mathbf{A})\), i.e.,

$$ y_1 c + y_2=a_1. $$

Second, square (A.1) and take the resulting trace so as to obtain \(\operatorname{tr}(\mathbf{Y}^{4})+2c\operatorname{tr}(\mathbf{Y}^{3}) +c^{2}\operatorname{tr}(\mathbf{Y}^{2})=\operatorname{tr}(\mathbf{A}^{2})\). Eliminate \(\operatorname{tr}(\mathbf{Y}^{4})=y_{4}\) and \(\operatorname{tr}(\mathbf{Y}^{3})=y_{3}\) by means of (A.6) and (A.7); this gives

$$ y_2 c^2 + \bigl(-y_1^3+3 y_1 y_2+6Y \bigr) c -\frac{1}{2}y_1^4+y_1^2 y_2+ \frac{1}{2}y_2^2+ 4 y_1 Y =a_2. $$

Third, consider the determinant of (A.1) in the form detA=detYdet(Y+c I). Set M=Y+c I in (A.5), take the trace of this result, and perform elementary manipulations so as to obtain

$$ 3c^3+ 3 y_1 c^2 + \frac{3}{2} \bigl(y_1^2-y_2\bigr) c +y_3 - \frac{3}{2} y_1 y_2 + \frac{1}{2} y_1^3 - 3 \det({\mathbf{Y}+ c \mathbf{I}})=0. $$

Using the form for y 3 that follows from (A.6) substitute from (A.10) into detA=detYdet(Y+c I) so as to obtain

$$ c^3+ y_1 c^2 + \frac{1}{2} \bigl(y_1^2-y_2\bigr) c +Y = \frac{A}{Y}. $$

Three Eqs. (A.8), (A.9) and (A.11) relate c to the values A, a 1, a 2, Y, y 1, y 2, of which only c, A, a 1, a 2, Y appear in (A.2). The rest of the demonstration provides an elimination so as to obtain a single equation that does not include y 1 and y 2.

Eliminating y 2 between (A.8) and (A.11) gives a quadratic equation in y 1

$$ y_1^2+3c y_1+ \biggl( 2c^2-a_1+ \frac{2 Y}{c}- \frac{2 A}{c Y} \biggr)=0, $$

that in turn yields possible y 1-values

$$ y_1=-\frac{3 c}{2} \pm \frac{1}{2} \sqrt{c^2+ 4 \bigl(a_1 - 2 Y/c+2A/(c Y ) \bigr)}. $$

Eliminating y 1 and y 2 from (A.9) by means of (A.13) and (A.8) gives an equation that can be manipulated into the form

$$\begin{aligned} &\pm \biggl( Y+\frac{A}{Y} \biggr) \sqrt{c^2+ 4 a_1 -\frac{8 Y}{c}+ \frac{8A}{c Y } } \\ &\quad{}= a_2 -a_1^2 -c Y + \frac{2 Y^4-4 A Y^2+ A c^3 Y + 2 A^2}{c^2 Y^2} \end{aligned}$$

where we have retained both root possibilities for y 1. Squaring each side and performing elementary operations now yields (A.2). □

Appendix B: Spurious Roots

The polynomial equation (5.17) may yield multiple roots, only one of which makes \(\det{\hat {\mathbf{B}}}=1\). For example, if \(\hat{\alpha}= \alpha^{*}\) then β=1 and (5.17) is simply

$$ c^6+ \bigl(4 I_1- I_2^2\bigr) c^4 =0. $$

One root of this equation is c=0 which gives the correct solution q=0 as described previously for this special case (see the discussion after (5.13)). However if \(I_{2}^{2} - 4 I_{1} > 0\) then there exist two additional roots c= \(\pm \sqrt{I_{2}^{2} - 4 I_{1}}\) to (B.1). In examining the possibilities for the sign of \(I_{2}^{2} - 4 I_{1} \) the analysis of [20] shows how the constraint detF=1 restricts I 1 and I 2 to lie in a cusp-shaped portion of the (I 1,I 2)-parameter plane (Fig. 1 of [20]). This cusp-shaped region is within the larger region I 1≥3, I 2≥3 such that the cusp occurs at (I 1,I 2)=(3,3). In particular, the ray I 1=I 2=3+γ 2 which emanates from the cusp is in this region; points on this ray correspond to a simple shearing deformation with shear amount γ. On the γ-parametrized ray I 1=(3+γ 2), I 2=(3+γ 2) it follows that \(I_{2}^{2} - 4 I_{1} =(3+\gamma^{2}) (-1+\gamma^{2})\) which is positive for |γ|>1. However in this case the two additional roots \(c= \pm \sqrt{(3+\gamma^{2}) (-1+\gamma^{2})}\) to (B.1) give a \({\hat{\mathbf{B}}}^{(q)}\) via (5.1) that makes \(\det {\hat {\mathbf{B}}}^{(q)} \ne 1\). Hence these additional roots are spurious.

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Demirkoparan, H., Pence, T.J. Finite Stretching and Shearing of an Internally Balanced Elastic Solid. J Elast 121, 1–23 (2015).

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  • Hyperelasticity
  • Internal balance
  • Neo-Hookean
  • Substructural reconfiguration

Mathematics Subject Classification

  • 74A20
  • 74A60
  • 74B20