U = C 1 / 2 ${\mathbf{U}} = \mathbf{C}^{1/2}$ and Its Invariants in Terms of C $\mathbf{C}$ and Its Invariants
- PDF / 776,355 Bytes
- 17 Pages / 439.37 x 666.142 pts Page_size
- 62 Downloads / 207 Views
U = C1/2 and Its Invariants in Terms of C and Its Invariants N.H. Scott1
Received: 6 August 2019 © Springer Nature B.V. 2020
Abstract We consider N × N tensors for N = 3, 4, 5, 6. In the case N = 3, it is desired to find the three principal invariants i1 , i2 , i3 of U in terms of the three principal invariants I1 , I2 , I3 of C = U2 . Equations connecting the iα and Iα are obtained by taking determinants of the factorisation λ2 I − C = (λI − U)(λI + U), and comparing coefficients. On eliminating i2 we obtain a quartic equation with coefficients depending solely on the Iα whose largest root is i1 . Similarly, we may obtain a quartic equation whose largest root is i2 . For N = 4 we find that i2 is once again the largest root of a quartic equation and so all the iα are expressed in terms of the Iα . Then U and U−1 are expressed solely in terms of C, as for N = 3. For N = 5 we find, but do not exhibit, a twentieth degree polynomial of which i1 is the largest root and which has four spurious zeros. We are unable to express the iα in terms of the Iα for N = 5. Nevertheless, U and U−1 are expressed in terms of powers of C with coefficients now depending on the iα . For N = 6 we find, but do not exhibit, a 32 degree polynomial which has largest root i12 . Sixteen of these roots are relevant, which we exhibit, but the other 16 are spurious. U and U−1 are expressed in terms of powers of C. The cases N > 6 are discussed. Keywords Continuum mechanics · Polar decomposition · Tensor square roots · Principal invariants · Cubic equations · Quartic equations · Equations of degree 16 Mathematics Subject Classification 15A16 · 74B20
1 Introduction All tensors occurring in this paper are square of dimension N × N . In Sects. 1–3 we consider only the case N = 3 except at the end of this section we briefly discuss the case N = 2. In Sects. 4–6 we discuss the cases N = 4–6, respectively.
B N.H. Scott
[email protected]
1
School of Mathematics, University of East Anglia, Norwich Research Park, Norwich NR4 7TJ, UK
N.H. Scott
In terms of the deformation gradient F the right and left Cauchy-Green strain tensors are defined by the symmetric positive definite tensors C = FT F, B = FFT , respectively, where T denotes the transpose. The polar decomposition theorem is F = RU = VR,
RRT = RT R = I,
in which R is a proper orthogonal, or rotation, tensor, U and V are respectively the right and left stretch tensors and I is the identity. We see that C = U2 , so that U = C1/2 , where U is the unique symmetric positive definite tensor square root of C. Assuming F is known then C is easy to calculate but U less so as it is a square root. Denote the (necessarily positive) eigenvalues of C by λ2i with corresponding eigenvectors ei so that the eigenvalues of U are λi with the same eigenvectors. The λi are the principal stretches. From the spectral representations C=
3 i=1
λ2i ei ⊗ ei ,
U=
3
λi e i ⊗ e i ,
i=1
we see that one method would be to find the eigenvalues and eigenvectors of C numerically, square root the eigenvalues, and the
Data Loading...
