Eccentric Rheometry for Viscoelastic Characterization of Small, Soft, Anisotropic, and Irregularly Shaped Biopolymer Gels and Tissue Biopsies

Abstract

Quantification of the physical properties of tissue biopsies and cell-remodeled hydrogels is critical for understanding tissue development and pathophysiological tissue remodeling. However, due to the low modulus, small size, irregular shape, and anisotropy of samples from these materials, accurate viscoelastic characterization using standard rheometric methods is problematic. The goal of this work is to utilize image analysis to extend rotational rheometry to these samples. In this method, the sample is offset to increase the torque generated; a custom clear glass geometry, right angle prism, and camera are used to determine the exact shape and location of the sample relative to the axis of rotation for calculation of the sample shear modulus, G′. Values of G′ for standard polydimethylsiloxane gels tested in centered and eccentric configurations were not statistically different (respectively 137 ± 37 kPa and 126 ± 8 kPa, p = 0.58), indicating accuracy of the method. Additionally, G′ values from circular and irregularly shaped collagen gels yielded equivalent results (31 ± 1.8 Pa and 31 ± 5.1 Pa, p = 0.29). A blood clot and a lipid plaque sample recovered from human patients (G′ ~ 4 kPa) were successfully tested with this method demonstrating applicability to clinical diagnostics.

Appendices

Appendix A: Integral Derivation for a Circular Sample

The following is a derivation for torque for an offset circular sample. It is represent by a small circle of radius r _s, inside a larger circle of radius R _g offset a distance, d, from the axis of rotation, so that it lies along the edge of the larger circle. In the following equations, ρ is the distance from the axis of rotation to an incremental area dA, G is the elastic modulus, T is the torque, h is the gap height between the peltier plate and bottom surface of the geometry. Integrating over domain D which is a circle of radius r _s centered at x = R _g − r _s (=d) and y = 0

$$ T = \iint\limits_{D} {\rho \tau \left( \rho \right) dA} $$

(A1)

Integration in Cartesian coordinates gives:

$$ \rho \tau \left( \rho \right) = \frac{G\theta }{h}\rho^{2} = \frac{G\theta }{h}\left( {x^{2} + y^{2} } \right). $$

(A2)

Equation (1) from text becomes

$$ T = \frac{G\theta }{h}\iint\limits_{D} {(x^{2} + y^{2} ){{d}}y{{d}}x} $$

(A3)

Symmetry about x-axis allows for doubling of the integration over top of disc:

$$ T = 2\frac{G\theta }{h}\int\limits_{{R_{\text{g}} - 2r_{\text{s}} }}^{{R_{\text{g}} }} {\int\limits_{0}^{{y(x)^{ + } }} {(x^{2} + y^{2} ){{d}}y{{d}}x} } . $$

(A4)

Integration of Eq. (A4) with respect to y gives:

$$ T = 2\frac{G\theta }{h}\int\limits_{{R_{\text{g}} - 2r_{\text{s}} }}^{{R_{\text{g}} }} {x^{2} y + \frac{1}{3}y^{3} \left| {_{0}^{y \left( x \right) + } } \right.{{d}}x.} $$

(A5)

$$ T = 2\frac{G\theta }{h}\int\limits_{{R_{\text{g}} - 2r_{\text{s}} }}^{{R_{\text{g}} }} {(x^{2} + \frac{1}{3}(r_{\text{s}}^{2} - \left( {x - \left( {R_{\text{g}} - r_{\text{s}} ))^{2} } \right)} \right)\sqrt {r_{\text{s}}^{2} - \left( {x - \left( {R_{\text{g}} - r_{\text{s}} } \right)} \right)^{2} } } {{d}}x. $$

(A6)

Let $V = \frac{x}{{r_{\rm s} }} - \left( {\frac{{R_{\rm g} }}{{r_{\rm s} }} - 1} \right)$ and $dV = \frac{1}{{r_{\rm s} }}{{d}}x.$

Substitution of terms in Eq. (A6) gives:

$$ T\,=\, 2\frac{G\theta }{h}r_{\text{s}}^{4} \int\limits_{ - 1}^{1} {\left[ {\left( {V + \left( {\frac{{R_{\text{g}} }}{{r_{\text{s}} }}} \right) - 1} \right)^{2} + \frac{1}{3}\left( {1 - V^{2} } \right)} \right]\sqrt {1 - V^{2} } {{d}}V.} $$

(A7)

Expanding and collecting terms gives:

$$ T = 2\frac{G\theta }{h}r_{\rm s}^{4} \int\limits_{ - 1}^{1} {\left[ {\frac{2}{3}V^{2} + 2\left( {\frac{{R_{\rm g} }}{{r_{\rm s} }} - 1} \right)V + \left( {\frac{{R_{\rm g} }}{{r_{\rm s} }} - 1} \right)^{2} + \frac{1}{3}} \right]\sqrt {1 - V^{2} } dV} . $$

(A8)

With further expansion, Eq. (A8) becomes:

$$ T = 2\frac{G\theta }{h}r_{\rm s}^{4} \left[ {\frac{2}{3}\int\limits_{ - 1}^{1} {V^{2} \sqrt {1 - V^{2} } dV + 2\left( {\frac{{R_{\rm g} }}{{r_{\rm s} }} - 1} \right)\int_{ - 1}^{1} {V\sqrt {1 - V^{2} } dV + \left( {\left( {\frac{{R_{\rm g} }}{{r_{\rm s} }} - 1} \right)^{2} + \frac{1}{3}} \right)\int_{ - 1}^{1} {\sqrt {1 - V^{2} } dV} } } } \right] $$

(A9)

Which, upon integration yields:

$$ T = 2\frac{G\theta }{h}r_{\text{s}}^{4} \left[ {\frac{2}{3}\frac{\pi }{8} + 2\left( {\frac{{R_{\text{g}} }}{{r_{\text{s}} }} - 1} \right)0 + \left( {\left( {\frac{{R_{\text{g}} }}{{r_{\text{s}} }} - 1} \right)^{2} + \frac{1}{3}} \right)\frac{\pi }{2}} \right]. $$

(A10)

Let r _s = r, d = R _g − r _s, then $\frac{{R_{\rm g} }}{{r_{\rm s} }} - 1 = \frac{{R_{\rm g} - r_{\rm s} }}{{r_{\rm s} }} = \frac{d}{r}.$

Rewriting Eq. (A10) with the following substitutions gives:

$$ T = \frac{2G\theta }{h}r^{4} \left[ {\frac{2}{3}\frac{\pi }{8} + \frac{{d^{2} }}{{r^{2} }}\frac{\pi }{2} + \frac{1}{3}\frac{\pi }{2}} \right]. $$

(A11)

Multiplying and combining constants in Eq. (A11) gives:

$$ T = \frac{G\theta }{h}r^{4} \left[ {\frac{\pi }{2} + \frac{{d^{2} }}{{r^{2} }}\pi } \right]. $$

(A12)

Simplification of Eq. (A12) gives:

$$ T = \frac{G\theta }{h}\left[ {\frac{\pi }{2}r^{4} + d^{2} \pi r^{2} } \right]. $$

(A13)

Equation (A13) is equivalent to Eq. (8) in text.

Appendix B: Error Analysis

To estimate the error associated with image analysis, the derivative of the equation for b(=J _rheo/J _sample) was used:

$$ \delta b = \sqrt {\left( {\frac{\partial b}{{\partial J_{\text{rheo}} }}\delta J_{\text{rheo}} } \right)^{2} + \left( {\frac{\partial b}{{\partial J_{\text{s}} }}\delta J_{\text{s}} } \right)^{2} }. $$

(B1)

For a theoretical perfectly circular sample of radius r _s at an offset, d, with J _s = πr ⁴_s  + d ²πr ²_s (Eq. (8)), and J _rheo given by Eq. (6) with R = R _g, the relative error is given by:

$$ \frac{\delta b}{b} = \sqrt {\left( {4\frac{{\delta R_{\rm g} }}{{R_{\rm g} }}} \right)^{2} + \left( {4\frac{{\left( {r_{\rm s}^{2} + d^{2} } \right)}}{{\left( {r_{\rm s}^{2} + 2d^{2} } \right)}}\frac{{\delta r_{\rm s} }}{{r_{\rm s} }}} \right)^{2} + \left( {\frac{{4d^{2} }}{{\left( {r_{\rm s}^{2} + 2d^{2} } \right)}}\frac{\delta d}{d}} \right)^{2} } . $$

(B2)

Repeated manual measurements on actual images of circular samples yielded 1% error for δR _g/R _g, δr _s/r _s, and δd/d (e.g., ±1 pixel for a 100 pixel distance). Eq. (B2) shows that the error associated with the image analysis depends upon the size of the sample and the degree of offset; it also highlights the impact of having the polar moment of inertia dependent upon the square of the offset and 4th power of the radii. For example, for a sample of radius 7.5 mm and offset 24.5 mm (at the edge), with 1% error for R _g, r _s, and d, the total error is 4.90%, whereas if it were offset to only 12.5 mm, the total error is 4.92%. For the smallest sample tested (radius 2 mm with offset 30 mm (at the edge)), the error is 4.9%. However, for the first case (r _s = 7.5 mm and d = 24.5 mm) with δr _s/r _s = 2%, the total error in the rises to 6.09%. These hypothetical cases indicate that the error is ~5% for representative circular samples, and that the sample dimensions and location have very little impact on the overall error, whereas the measurement error for each variable (i.e., δr _s/r _s) has a large impact on the total error.

In the actual implementation, J _rheo is calculated from R _g which is estimated from 5 points and fit to equation of a circle in MATLAB, and J _s is numerically calculated (r _s and d are not applicable for non-circular samples), thus the following equation is used:

$$ \frac{\delta b}{b} = \sqrt {\left( {4\frac{{\delta R_{\rm g} }}{{R_{\rm g} }}} \right)^{2} + \left( {\frac{{\delta J_{\rm s} }}{{J_{\rm s} }}} \right)^{2} } . $$

(B3)

When analyzed repeatedly for different images by two separate users, we found δR _g/R _g = 1.6%. Although J _s is calculated in an automated fashion, variability associated with measurement of J _s could arise from camera focus and aperture, lighting, prism alignment, and the filtering and threshold used to determine the sample boundaries. With proper focus and light levels, even lighting using a ring flash, appropriate filtering and threshold setting, and automated perimeter selection, analysis of multiple independent images taken of the same sample after systematic readjustment over the widest applicable range of camera settings and prism placement yielded 0.99% error in J _s. The combined estimated error due to image analysis using δR _g/R _g = 1.6% and δJ _s/J _s = 0.99% is 3.3% for the correction multiplier, b.