Computation of Full-field Strains Using Principal Component Analysis

Abstract

The primary output from several full-field deformation measurement techniques, e.g., Digital Image Correlation (DIC), is the displacement vector at a dense grid of points covering the area of interest. Since such displacement data inherently contain noise, they are usually smoothed first and then differentiated to obtain strains. Another common approach is to use finite-element shape functions for the strains and compute them by treating the measured displacements as nodal displacements. In this paper, we propose a novel method for strain calculation from full-field data, based on the multivariate analysis technique of Principal Component Analysis (PCA) using which we first obtain the singular values and singular vectors for each component of the displacement field. By choosing only the dominant singular values and their corresponding singular vectors, we show that the dimensionality of the displacement data is sharply reduced and a significant portion of the noise is eliminated. Moreover, the shapes of the dominant singular vectors offer physical insight into dominant deformation patterns. We demonstrate the accuracy of the proposed technique by applying it to two cases each of homogeneous and inhomogeneous strain fields and show that in all cases the proposed method yields excellent results.

Appendices

Appendix A: Finite Displacement and Strain in an Incompressible Hollow Cylinder Undergoing Inflation

An extensive discussion of a similar problem (Fig. 11) is found in Saravanan and Rajagopal [52]. Working in cylindrical polar coordinates, and using upper-case letters for the coordinates in the reference configuration and lower-case for those in the current configuration respectively, we have

$$ \begin{aligned} r & = R + f(R) \\ \theta & = \mathit{\Theta} \\ z & = Z \\ \end{aligned} $$

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where f(R) is the radial displacement function to be determined. The deformation gradient tensor, F, for this deformation is

$$ \mathbf{F} = \begin{aligned} \frac{df}{dR} &\; 0 & 0 \\ 0 &\; \frac{r}{R} & 0 \\ 0 &\; 0 & 1 \end{aligned} $$

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As the material is incompressible, J = detF = 1, thus yielding function f in terms of R.

$$ f(R) = \sqrt{R^2+r_{o}^2-R_{o}^{2}}, $$

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where R _o and r _o are the outer radius of the cylinder before and after inflation respectively.

Thus, the displacement in the radial direction can be written as

$$ U_{R} = \sqrt{R^2+r_{o}^2-R_{o}^{2}} - R $$

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The displacement in X and Y directions can be expressed as

$$ \begin{aligned} U_{X} = U_{R} \cos \phi \\ U_{Y} = U_{R} \sin \phi \\ \end{aligned} $$

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where, ϕ is the angle between radial direction and X direction. The displacement matrices U and V are assembled by computing U _X and U _Y from equation (16) at grid points inside the square region of interest (Fig. 11). The Green-Lagrange strain tensor E, defined as

$$ \mathbf{E} = \frac{1}{2}[\mathbf{F}^{T}\mathbf{F}-\mathbf{I}] $$

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is then obtained as

$$ \mathbf{E} = \frac{1}{2} \left[\begin{aligned} \left(\frac{R_{o}^2-r_{o}^{2}}{R^2+r_{o}^2-R_{o}^{2}}\right)^{2} & 0 & 0 \\ 0 & \left(\frac{r_{o}^2-R_{o}^{2}}{R^{2}}\right)^{2} & 0 \\ 0 & 0 & 0 \end{aligned} \right]$$

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The strain components in Cartesian basis are then readily obtained as

$$ \begin{aligned} E_{XX} &= E_{RR}\cos^{2}\phi + E_{\mathit{\Theta \Theta}}\sin^{2}\phi \\ E_{YY} &= E_{RR}\sin^{2}\phi + E_{\mathit{\Theta \Theta}}\cos^{2}\phi \\ E_{XY} &= E_{RR}\cos \phi \sin \phi - E_{\mathit{\Theta \Theta}}\sin \phi \cos \phi \\ \end{aligned} $$

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These equations are used to generate the ground-truth strain components over the region of interest.

Appendix B: Infinitesimal Displacement and Strain in a Spherical Isotropic Inclusion -Eshelby’s Solution [57]

Solution Inside Spherical Inclusion

The displacement, strain and stress inside the spherical inclusion are as follows [57, 58]:

$$ u_{i} = S_{ijkl}^{*}{\epsilon}^{T}_{kl}x_j $$

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$$ \epsilon_{ij} = S_{ijkl}^{*} \epsilon_{kl}^T $$

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$$ \sigma_{ij} \!=\! \frac{E}{(1\!+\!\nu)}\left[S_{ijkl}^* \epsilon_{kl}^T \!+\! \frac{\nu}{(1\!-\!2\nu)}\delta_{ij}S_{ppkl}^*\epsilon_{kl}^T\right] \!-\! p_{ij}^T $$

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where $S_{ijkl}^{*}$ is the Eshelby tensor. For a spherical inclusion, we obtain

$$ S_{1111}^{*} = S_{2222}^{*} = S_{3333}^{*} = \frac{7-5\nu}{15(1-\nu)} $$

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$$ S_{1212}^{*} = S_{2323}^{*} = S_{3131}^{*} = \frac{4-5\nu}{15(1-\nu)} $$

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$$ S_{1122}^{*} \,=\, S_{2233}^{*} \,=\, S_{3311}^{*} \,=\, S_{1133}^{*} \,=\, S_{3322}^{*} \,=\, \frac{5\nu-1}{15(1-\nu)}\\ $$

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Additional terms of $S_{ijkl}^{*}$ follow from symmetry i.e. $S_{ijkl}^{*} = S_{jikl}^{*} = S_{ijlk}^{*} = S_{jilk}^{*}$, while the remaining terms are zero.

Solution Outside Spherical Inclusion

The displacement field outside the spherical inclusion in Cartesian basis is given by

$$\begin{array}{rcl} u_i \!=\! \frac{(1\!+\!\nu)a^3}{2(1\!-\!{\nu})E}\left[\frac{(2p_{ik}^T x_k \!+\! p_{kk}^T x_i)}{15R^5} \left(3a^2-5R^2\right)\right.\nonumber\\ &&\left.{\kern5pc} \!+\! \frac{p_{jk}^T x_j x_k x_i}{R^7} \left(R^2\!-\!a^2\right) \!+\! \frac{4(1\!-\!{\nu})p_{ik}^Tx_k}{3R^3}\right] \end{array}$$

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where $R = \sqrt {x_kx_{k}}$ and a is the radius of spherical inclusion. The stress fields in Cartesian basis are

$$\begin{array}{@{}rcl@{}} &&\sigma_{ij} = \frac{{a}^{3}}{2(1-\nu)R^{3}}\left[\frac{p_{ij}^{T}}{15}\left(10(1-2{\nu}) + \frac{6{a}^{2}}{R^{2}}\right)\right.\\ &&{\kern6.8pc}+ \frac{p_{ik}^Tx_kx_{j} + p_{jk}^Tx_kx_{i}}{R^{2}}\left(2\nu - 2\frac{{a}^{2}}{R^{2}}\right) \\ &&{\kern6.8pc} + \frac{{\delta}_{ij}p_{kk}^{T}}{15} + \left(\frac{3{a}^{2}}{R^{2}} - 5(1-2{\nu}) \right) \\ &&{\kern6.8pc}+ \frac{{\delta}_{ij}p_{kl}^Tx_{k} x_{l}}{R^{2}}\left((1-2\nu) - \frac{{a}^{2}}{R^{2}}\right) \\ &&{\kern6.8pc}- \frac{x_ix_jp_{kl}^Tx_kx_{l}}{R^{4}}\left(5-\frac{7{a}^{2}}{R^{2}}\right)\\ &&{\kern6.8pc}+ \left. \frac{x_ix_jp_{kk}^{T}}{R^{2}}\left(1-\frac{{a}^{2}}{R^{2}}\right)\right]\\ \end{array} $$

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