Non-uniqueness of cohesive-crack stress-separation law of human and bovine bones and remedy by size effect tests

Abstract

It is shown that if the bilinear stress-separation law of the cohesive crack model is identified from the complete softening load-deflection curve of a notched human bone specimen of only one size, the problem is ill-conditioned and the result is non-unique. The same measured load-deflection curve can be fitted with values of initial fracture energy and tensile strength differing, respectively, by up to 100 and 72.4 % (of the lower value). The material parameters, however, give very different load-deflection curves when the specimen is scaled up or down significantly. This implies that the aforementioned non-uniqueness could be avoided by testing human bone specimens of different sizes. To demonstrate it, tests of notched bovine bone beams of sizes in the ratio of 1:$\sqrt{6}$:6 are conducted. To minimize random scatter, all the specimens are cut from one and the same bovine bone, even though this limits the number of specimens to 8. A strong size effect is found, but an anomaly in the size effect data trend is obtained, probably due to random scatter and too small a number of specimens. Further it is shown that the optimum range of size effect testing based on Bažant’s size effect law approximately coincides with the size range of beams that can be cut from one bovine bone. By size effect fitting of previously published data on human bone, it is shown that the optimum size range calls for beam depths under 10 mm, which is too small for the standard equipment of mechanics of materials labs and would require a special miniaturized precision equipment.

Appendices

Appendix I: Treatment of bone orthotropy and numerical simulation procedures

The bovine femur, as well as all bone, is an orthotropic material. The orthotropy has easily been taken into account in finite element computations of the dimensionless energy release rate function $g(\alpha )$ and stress intensity factor $K_{I}$.

For analytical formulation, the effective modulus based on Bao et al. (1992) was adopted. Assuming orthotropic transversely isotropic properties, one has five independent elastic constants, with the following values for the bovine femur: $E_1 = 30$ GPa, $E_2 = 15$ GPa, $G_{12}= 12.5$ GPa, $\nu _{12} = 0.206, \nu _{21} = 0.103$. The subscripts 1 and 2 refer to the longitudinal directions of the beam specimens and of the bone fibers (Fig. 2). According to Bao et al., the stress intensity factor can be written as:

$$\begin{aligned} K_{I}&= \sigma \sqrt{\pi {D}\alpha }Y(\rho )F(\alpha ) \end{aligned}$$

(16)

where

$$\begin{aligned}&\rho = \sqrt{E_1 E_2}\ /2G_{12} - \sqrt{ \nu _{12} \nu _{21} } \quad \lambda = E_2 / E_1\end{aligned}$$

(17)

$$\begin{aligned}&Y(\rho ) = \left[ 1 + 0.1(\rho -1) - 0.015(\rho -1)^{2} \right. \nonumber \\&\quad \left. + 0.002(\rho -1)^{3}\right] \left[ (1+\rho ) /2 \right] ^{-1/4} \end{aligned}$$

(18)

$F(\alpha )$ is the same function of the relative crack length $\alpha = a/D$ as for isotropic materials. The energy release rate for an orthotropic material is:

$$\begin{aligned} G(\alpha ) = \sqrt{\frac{1+\rho }{2E_{1}E_{2}\sqrt{\lambda }}}K_{I}^{2} \end{aligned}$$

(19)

Substituting Eq. (16) into (19), and changing the form as for the isotropic materials, we have

$$\begin{aligned} G(\alpha )&= \frac{K_{I}^{2}}{E_{ef}} = \frac{\sigma ^{2}D\pi \alpha }{E_{ef}}[F(\alpha )]^{2} = \frac{\sigma ^{2}D}{E_{ef}}g(\alpha ) \end{aligned}$$

(20)

where

$$\begin{aligned} E_{ef}&= [Y(\rho )]^{-2} \sqrt{ 2E_1 E_2 \sqrt{\lambda }/(1+\rho )} \end{aligned}$$

(21)

$$\begin{aligned} g(\alpha )&= \pi \alpha [F(\alpha )]^{2} = [k(\alpha )]^2 \end{aligned}$$

(22)

Energy release rate $g(\alpha )$ is the dimensionless energy release function. $E_{ef}$ is the effective modulus which is needed for formulating the size effect law (Eq. 6). For the three-point bending specimen (Tada et al. 1985):

$$\begin{aligned} k(\alpha ) = \sqrt{\alpha }\frac{\begin{array}{c}1.900 - \alpha [-0.089+0.603(1-\alpha )\\ -0.441(1-\alpha )^{2} +1.223(1-\alpha )^{3}]\end{array}}{(1+2\alpha )(1-\alpha )^{3/2}} \end{aligned}$$

(23)

Note that, $K_{I}^2 = GE_{ef}$ where $G =$ energy release rate and $E_{ef} =$ effective Young’s modulus. Then we have $g(\alpha ) = \pi \alpha [F(\alpha )]^{2}$. The equations for $g(\alpha )$ and $g^{\prime }(\alpha )$ for energy release rate at initial notch depth $0.2D$ are as follows:

$$\begin{aligned} g(\alpha )&= \frac{\alpha }{(1+2\alpha )^2(1-\alpha )^{3}}\{1.900-\alpha [-0.089\nonumber \\&+0.603(1-\alpha )-0.441(1-\alpha )^{2}\nonumber \\&+1.223(1-\alpha )^{3}]\}^{2} \end{aligned}$$

(24)

Finally $g^{\prime }(\alpha _0)$ can be obtained through numerical differentiation.

Appendix II: Comparison with human bone examples Yang et al. (2006a, b)

Using finite elements, the softening load-deflection curves of human bone, reported in Yang et al. (2006a, b), are reproduced in Figs. 8 and 11 using the following parameters:

$$\begin{aligned}&f^{\prime }_t = p_c = 60~\text{ MPa },~~w_c = u_c = 36~\upmu \text{ m },\nonumber \\&\sigma _1 = p_0 = 30~\text{ MPa },~~w_1 = u_0 = 6~\upmu \text{ m }\end{aligned}$$

(25)

$$\begin{aligned}&W_{tip} = 0.544~\text{ kJ/m }^2,~~W_{brid} = 0.90~\text{ kJ/m }^2 \end{aligned}$$

(26)

Yang’s bilinear values were varied to fit data scatter. For example, they assumed that $p_c$ has the mean, 60 MPa and standard error, 10 MPa ($60 \pm 10$ MPa).

According to Fig. 1, our definition of $G_f$ is the area of the triangle under the initial straight softening segment. On the other hand, $W_{tip}$ is area of trapezoids according to Yang’s definition. To interpret results in terms of $G_f$, $W_{tip}$ must be converted to $G_f$ or otherwise the initial fracture energy could be underestimated since $G_f > W_{tip}$. After conversion, the initial fracture energy of Yang’s bilinear law $G_f$ becomes 0.72 $\mathrm{kJ/m}^2$.

The total fracture energy, $G_F$, for Yang et al.’s tests can be calculated by adding $W_{tip}$ to $W_{brid}$,

$$\begin{aligned} G_F = W_{tip} + W_{brid} = 0.544 + 0.90 = 1.44~\text{ kJ/m }^2 \nonumber \\ \end{aligned}$$

(27)