Identifiability of material parameters in solid mechanics

Abstract

Material parameter identification using constitutive models of elasticity, viscoelasticity, rate-independent plasticity and viscoplasticity has a long history with regard to homogeneous and inhomogeneous deformations. For example, uniaxial tensile tests, pure shear tests, torsion experiments of thin-walled tubes or biaxial tensile tests are used to obtain the material parameters by solving the inverse problem. Frequently, the parameters are determined by numerical optimization tools. In this paper, we investigate some very basic single- and two-layered examples regarding identifiability, because these tests are the basis for more complex geometrical and physical nonlinear problems. These simple examples are the uniaxial tensile/compression case, biaxial tensile tests of a cruciform specimen, torsion of a thin-walled tube, a thick-walled tube under internal pressure and the indentation test. For the thick-walled tube under internal and external pressure with an axial pre-strain with several layers, an analytical solution is provided directly suitable for programming. The aim is to get an understanding whether some problems lead to non-identifiable parameters.

Appendices

Appendix

Thick-walled tube under internal and external pressure and axial strains

The analytical solution of the thick-walled tube under both agencies, internal and external pressure and axial strains for isotropic, linear elasticity in the small-strain regime, is recapped, since it is required for the more general multi-layered results in 8, see references cited in [80] as well. We assume a displacement field

$$\begin{aligned} \mathbf {u} (r,z) = u(r) \mathbf {e} _r+ w(z) \mathbf {e} _z, \end{aligned}$$

(49)

leading to the linearized Green strain tensor ${\mathbf {E}} = u,_r \mathbf {e} _r\otimes \mathbf {e} _r+ (u/r) \mathbf {e} _\vartheta \otimes \mathbf {e} _\vartheta + w,_z \mathbf {e} _z\otimes \mathbf {e} _z$. $r,\vartheta ,z$ are cylindrical coordinates, $\mathbf {e} _r, \mathbf {e} _\vartheta$ and $\mathbf {e} _z$ the normalized tangent vectors in radial, circumferential and axial direction. The notation $u,_r$ symbolizes the partial derivative of the radial displacement u with respect to r, $u,_r = \partial u/ \partial r$. u(r) is unknown and $w(0) = 0$ and $w(H) = w_{\text {max}}$ are prescribed (H is the thickness of the circular plate), see Fig. 6a. The radial and circumferential strains are $\varepsilon _r = u,_r$ and $\varepsilon _\vartheta = u/r$, respectively, and the axial strain $\varepsilon _z = w,_z = \text {const.}$ is assumed to be constant over z. Using the elasticity relation (24), we have

$$\begin{aligned} \begin{aligned} \sigma _r&= \frac{E}{(1+\nu )(1-2\nu )} \big ( (1-\nu ) \varepsilon _r + \nu \varepsilon _\vartheta + \nu \varepsilon _z \big ), \\ \sigma _\vartheta&= \frac{E}{(1+\nu )(1-2\nu )} \big ( \nu \varepsilon _r + (1-\nu ) \varepsilon _\vartheta + \nu \varepsilon _z \big ), \\ \sigma _z&= \frac{E}{(1+\nu )(1-2\nu )} \big ( \nu \varepsilon _r + \nu \varepsilon _\vartheta + (1-\nu ) \varepsilon _z \big ), \end{aligned} \end{aligned}$$

(50)

or the inverse relation

$$\begin{aligned} \begin{aligned} \varepsilon _r = u,_r&= \frac{\displaystyle 1}{\displaystyle E} \big ( \sigma _r - \nu \sigma _\vartheta - \nu \sigma _z \big ), \\ \varepsilon _\vartheta = \frac{u}{r}&= \frac{\displaystyle 1}{\displaystyle E} \big ( -\nu \sigma _r + \sigma _\vartheta - \nu \sigma _z \big ), \\ \varepsilon _z = w,_z&= \frac{\displaystyle 1}{\displaystyle E} \big ( -\nu \sigma _r - \nu \sigma _\vartheta + \sigma _z \big ). \end{aligned} \end{aligned}$$

(51)

Additionally, we have to fulfill the equilibrium conditions ${{\mathrm{div}}}{\mathbf {T}} = \mathbf {0}$ in the absence of specific body forces, leading to

$$\begin{aligned} \begin{aligned} \sigma _r,_r + \frac{\sigma _r - \sigma _\vartheta }{r} = 0 \quad&\rightarrow \quad (r \sigma _r),_r - \sigma _\vartheta = 0 \\ \sigma _\vartheta ,_\vartheta = 0 \quad&\rightarrow \quad \sigma _\vartheta \text { is constant over } \vartheta \\ \sigma _z,_z = 0 \quad&\rightarrow \quad \sigma _z \text { is constant over } z \end{aligned} \end{aligned}$$

(52)

see, for example, [54].

We insert Eqs. (50)$_1$ and (50)$_2$ into Eq. (51)$_3$ and obtain

$$\begin{aligned} \sigma _z = E \varepsilon _z + \nu (\sigma _r + \sigma _\vartheta ) \end{aligned}$$

(53)

leading to

$$\begin{aligned} \sigma _z,_r = \nu (\sigma _r,_r + \sigma _\vartheta ,_r), \end{aligned}$$

(54)

because of a constant axial strain $\varepsilon _z$. Now, we multiply Eq. (51)$_2$ with the Young modulus E and the radial coordinate r—evaluating $\varepsilon _\vartheta = u/r$—leading to

$$\begin{aligned} E u = r (\sigma _\vartheta - \nu \sigma _r - \nu \sigma _z). \end{aligned}$$

(55)

The derivative with respect to the radial coordinate employing $u,_r = \varepsilon _r$ and Eq. (51)$_1$ leads to

$$\begin{aligned} E u,_r = \sigma _\vartheta - \nu \sigma _r - \nu \sigma _z + r \big ( \sigma _\vartheta ,_r - \nu \sigma _r,_r - \nu \sigma _z,_r \big ) = E \varepsilon _r = \sigma _r - \nu \sigma _\vartheta - \nu \sigma _z. \end{aligned}$$

After inserting Eq. (54), we arrive at

$$\begin{aligned} \sigma _\vartheta - \sigma _r + r (1-\nu ) \sigma _\vartheta ,_r - r \nu \sigma _r,_r = 0, \end{aligned}$$

(56)

where, additionally, the equilibrium conditions (52)$_1$ have to be considered. This leads to

$$\begin{aligned} \sigma _r,_r + \sigma _\vartheta ,_r = 0, \end{aligned}$$

(57)

i.e., in view of Eq. (54), the axial stresses are constant over r, $\sigma _z,_r = 0$. If we integrate Eq.(57)

$$\begin{aligned} \sigma _r + \sigma _\vartheta = 2 C_1, \end{aligned}$$

(58)

and again insert the circumferential stress from Eq. (52)$_1$ and multiply the relation with r, we arrive at

$$\begin{aligned} \big (r^2 \sigma _r\big ),_r = 2 C_1 r, \end{aligned}$$

or, if we integrate this relation, the radial stress

$$\begin{aligned} r^2 \sigma _r = C_1 r^2 + C_2 \quad \Rightarrow \quad \sigma _r = C_1 + \frac{C_2}{r^2} \end{aligned}$$

(59)

can be derived. Since $\sigma _z,_r = 0$ vanishes, we obtain the circumferential stress from Eq. (58),

$$\begin{aligned} \sigma _\vartheta = C_1 - \frac{C_2}{r^2}. \end{aligned}$$

(60)

The constants are determined by the boundary conditions

$$\begin{aligned} \begin{aligned} \sigma _r(r_i) = -p_i&= C_1 + \frac{C_2}{r_i^2} \\ \sigma _r(r_o) = -p_o&= C_1 + \frac{C_2}{r_o^2} \\ \end{aligned} \end{aligned}$$

(61)

leading to the coefficients

$$\begin{aligned} C_1 = \frac{p_i r_i^2 - p_o r_o^2}{r_o^2 - r_i^2}, \qquad C_2 = \frac{r_i^2 r_o^2}{r_o^2 - r_i^2} (p_o - p_i). \end{aligned}$$

(62)

$p_i$ and $p_o$ are the internal and external pressure, respectively. $r_i$ and $r_o$ are the internal and external radius.

In view of experimental results, we can measure the axial force by integrating the axial stresses (53) using Eqs. (59) and (60),

$$\begin{aligned} F_z = 2 \pi \int _{r_i}^{r_o} \sigma _z(r) r {\mathrm {d}}r = \sigma _z a, \quad \text {with} \quad \sigma _z = E \varepsilon _z + 2 \nu C_1, \end{aligned}$$

(63)

with the cross section $a = \pi (r_o^2 - r_i^2)$. Additionally, the radial displacements (or the circumferential strains) on the outside are measurable,

$$\begin{aligned} u(r_o) = \frac{2 r_o r_i^2 (1-\nu ^2)}{E (r_o^2 - r_i^2)} p_i - \frac{r_o \big (r_i^2 (1+\nu ) + r_o^2 (1-\nu +2 \nu ^2)\big )}{E (r_o^2 - r_i^2)} p_o - \nu r_o \varepsilon _z. \end{aligned}$$

(64)

Analytical solution of thick-walled cylinder with several layers

Multi-layered thick-walled tubes occur, for example, in drilling pipes, pressure vessels, or arteries. In the following, each layer k, $k=1, \ldots , N$, is assumed to be of an isotropic, linear elastic material behavior. The layers are defined by the radius $r^{(k)}$, $k=0,\ldots ,N$, with the internal and external radius $r^{(0)} = r_i$ and $r^{(N)} = r_o$, respectively, see Fig. 6b. As in 7, we assume that the lower cut of the tube is fixed in axial direction and the upper part is stretched by $w_{\text {max}}$. The pressures $p_i$ and $p_o$ are applied internally and externally. Thus, we have the boundary conditions

$$\begin{aligned} \sigma _r^{(1)}(r_i) = \sigma _r^{(1)}(r^{(0)}) = -p_i, \quad \sigma _r^{(N)}(r_o) = \sigma _r^{(1)}(r^{(N)}) = -p_0. \end{aligned}$$

(65)

Additionally, the internal constraints in between the layers are given,

$$\begin{aligned} \begin{aligned} u^{(k-1)}(r^{(k-1)})&= u^{(k)}(r^{(k-1)}), \\ \sigma _r^{(k-1)}(r^{(k-1)})&= \sigma _r^{(k)}(r^{(k-1)}), \end{aligned} \end{aligned}$$

(66)

for $k=2,\ldots ,N$. The right superscript (k), $k=1,\ldots ,N$, in the stress and displacement field, defines the layer number. In view of Eq. (59), and for the radial displacements (55) the two unknowns $C_1^{(k)}$ and $C_2^{(k)}$ have to be specified in each layer, i.e., we have a total of 2N unknowns. On the other hand, we have with Eqs. (65) and (66) $n_{\text {eq}} = 2 (N-1) + 2 = 2 N$ equations. From the conditions (65), the two equations

$$\begin{aligned} \sigma _r^{(1)}(r^{(0)})&= C_1^{(1)} + \frac{C_2^{(1)}}{{r^{(0)}}^2} = - p_i \end{aligned}$$

(67)

$$\begin{aligned} \sigma _r^{(N)}(r^{(N)})&= C_1^{(N)} + \frac{C_2^{(N)}}{{r^{(N)}}^2} = - p_o \end{aligned}$$

(68)

are obtained. Applying Eq. (59) for the internal constraint (66)$_2$, we arrive at

$$\begin{aligned} C_1^{(k-1)} + \frac{C_2^{(k-1)}}{{r^{(k-1)}}^2} = C_1^{(k)} + \frac{C_2^{(k)}}{{r^{(k-1)}}^2} \end{aligned}$$

(69)

Next, we insert the radial, circumferential and axial stresses (59), (60) and (63) depending on the constants $C_1$ and $C_2$ into the radial displacements (55). This leads to

$$\begin{aligned} u = \frac{r}{E} \left( (1-\nu - 2 \nu ^2) C_1 - \frac{1+\nu }{r^2} C_2\right) - \nu r \varepsilon _z. \end{aligned}$$

(70)

For each layer, the constraint (66)$_1$ yields

$$\begin{aligned} A_1^{(k-1)} C_1^{(k-1)} - \frac{A_2^{(k-1)}}{{r^{(k-1)}}^2} C_2^{(k-1)} - \left( A_1^{(k)} C_1^{(k)} - \frac{A_2^{(k)}}{{r^{(k-1)}}^2} C_2^{(k)} \right) = \big (\nu ^{(k-1)} - \nu ^{(k)}\big ) \varepsilon _z, \end{aligned}$$

(71)

with

$$\begin{aligned} A_1^{(k)} = \frac{1 - \nu ^{(k)} - 2 {\nu ^{(k)}}^2}{E^{(k)}}, \quad A_2^{(k)} = \frac{1 + \nu ^{(k)}}{E^{(k)}}, \end{aligned}$$

(72)

The combination of Eqs. (67), (68), (69) and (71) leads to the system of linear equations

(73)

with

(74)

and

(75)

After solving the linear system (73), we can determine the stress distribution in each layer , , and , and the displacements. In view of a sensitivity analysis of an experiment, the resulting force

(76)

with the partial cross sections of layers, $a_k = \pi \big ({r^{(k)}}^2 - {r^{(k-1)}}^2\big )$, $k=1,\ldots ,N$, can be calculated. Additionally, we can compute the outer surface displacement

$$\begin{aligned} u(r_o) = u^{(N)}(r^{(N)}) = r^{(N)} \big (A_1^{(N)} C_1^{(N)} - A_2^{(N)} C_2^{(N)}\big ) - \nu ^{(N)} r^{(N)} \varepsilon _z. \end{aligned}$$

(77)