Fiber-reinforced materials: finite elements for the treatment of the inextensibility constraint

Abstract

The present paper proposes a numerical framework for the analysis of problems involving fiber-reinforced anisotropic materials. Specifically, isotropic linear elastic solids, reinforced by a single family of inextensible fibers, are considered. The kinematic constraint equation of inextensibility in the fiber direction leads to the presence of an undetermined fiber stress in the constitutive equations. To avoid locking-phenomena in the numerical solution due to the presence of the constraint, mixed finite elements based on the Lagrange multiplier, perturbed Lagrangian, and penalty method are proposed. Several boundary-value problems under plane strain conditions are solved and numerical results are compared to analytical solutions, whenever the derivation is possible. The performed simulations allow to assess the performance of the proposed finite elements and to discuss several features of the developed formulations concerning the effective approximation for the displacement and fiber stress fields, mesh convergence, and sensitivity to penalty parameters.

Analytical solutions

In the following sections, we report the derivation of the analytical solutions of some performed numerical tests.

1.1 Traction test

The solution of the traction test carried out in Sect. 5.1 is computed by considering that the pure uniaxial traction state implies:

$$\begin{aligned} \begin{aligned}&\sigma _{xx}=q\\&\sigma _{yy}= 0 \\&\sigma _{xy}= 0 \end{aligned} \end{aligned}$$

(52)

It is possible to compute the displacement field as follows:

$$\begin{aligned} \begin{aligned} u(x,y)&= \varepsilon _{xx} \ x + \varepsilon _{xy} \ y \\ v(x,y)&= \varepsilon _{xy} \ x + \varepsilon _{yy} \ y \end{aligned} \end{aligned}$$

(53)

We then consider the stress-strain relationship (refer to system (16)):

$$\begin{aligned} \left[ \begin{array}{l} \varepsilon _{xx}\\ \varepsilon _{yy}\\ 2\varepsilon _{xy}\\ \sigma _f \end{array} \right] = \left[ \begin{array}{llll} \mathbb {S}_{11} &{}\quad \mathbb {S}_{12} &{}\quad \mathbb {S}_{13} &{}\quad \mathbb {S}_{14}\\ \mathbb {S}_{21} &{}\quad \mathbb {S}_{22} &{}\quad \mathbb {S}_{23} &{}\quad \mathbb {S}_{24} \\ \mathbb {S}_{31} &{}\quad \mathbb {S}_{32} &{}\quad \mathbb {S}_{33} &{}\quad \mathbb {S}_{34}\\ \mathbb {S}_{41} &{}\quad \mathbb {S}_{42} &{}\quad \mathbb {S}_{43} &{}\quad \mathbb {S}_{44} \end{array} \right] \left[ \begin{array}{l} \sigma _{xx}\\ \sigma _{yy}\\ \sigma _{xy}\\ 0 \end{array} \right] \end{aligned}$$

(54)

where:

$$\begin{aligned} \mathbb {S}_{11}= & {} \dfrac{a_y^4 \mu + a_x^2a_y^2 (\lambda + 2 \mu )}{(a_x^2+a_y^2)^2\mu (\lambda + 2 \mu )}\nonumber \\ \mathbb {S}_{22}= & {} \dfrac{ a_x^4 \mu + a_x^2a_y^2 (\lambda + 2 \mu )}{(a_x^2+a_y^2)^2\mu (\lambda + 2 \mu )}\nonumber \\ \mathbb {S}_{33}= & {} \dfrac{-2a_x^2a_y^2 \lambda + (a_x^4+a_y^4) (\lambda + 2 \mu )}{(a_x^2+a_y^2)^2\mu (\lambda + 2 \mu )}\nonumber \\ \mathbb {S}_{44}= & {} -\dfrac{4\mu (\lambda +\mu )}{(a_x^2+a_y^2)^2(\lambda + 2 \mu )}\nonumber \\ \mathbb {S}_{12}= & {} \mathbb {S}_{21} = -\dfrac{a_x^2 a_y^2 (\lambda + \mu )}{(a_x^2+a_y^2)^2\mu (\lambda + 2 \mu )}\\ \mathbb {S}_{13}= & {} \mathbb {S}_{31} = \dfrac{a_x a_y^3 \lambda - a_x^3 a_y (\lambda + 2 \mu )}{(a_x^2+a_y^2)^2\mu (\lambda + 2 \mu )}\nonumber \\ \mathbb {S}_{14}= & {} \mathbb {S}_{41} =\dfrac{-a_y^2 \lambda + a_x^2 (\lambda + 2 \mu )}{(a_x^2+a_y^2)^2(\lambda + 2 \mu )}\nonumber \\ \mathbb {S}_{23}= & {} \mathbb {S}_{32} = \dfrac{a_x^3 a_y \lambda - a_x a_y^3 (\lambda + 2 \mu )}{(a_x^2+a_y^2)^2\mu (\lambda + 2 \mu )} \nonumber \\ \mathbb {S}_{24}= & {} \mathbb {S}_{42} =\dfrac{-a_x^2 \lambda + a_y^2 (\lambda + 2 \mu )}{(a_x^2+a_y^2)^2(\lambda + 2 \mu )}\nonumber \\ \mathbb {S}_{34}= & {} \mathbb {S}_{43} =\dfrac{4a_x a_y(\lambda +\mu )}{(a_x^2+a_y^2)^2(\lambda + 2 \mu )}\nonumber \end{aligned}$$

(55)

After applying system (52), we can provide the following analytical solution in terms of displacement components:

$$\begin{aligned} \begin{aligned} u(x,y)&= \left( \mathbb {S}_{11}x+\dfrac{\mathbb {S}_{31}}{2}y\right) \sigma _{xx} \\ v(x,y)&= \left( \dfrac{\mathbb {S}_{31}}{2}x+\mathbb {S}_{21}y\right) \sigma _{xx} \end{aligned} \end{aligned}$$

(56)

and Lagrange multiplier:

$$\begin{aligned} \sigma _f=\mathbb {S}_{41}\sigma _{xx} \end{aligned}$$

(57)

where components $\mathbb {S}_{11}$, $\mathbb {S}_{21}$, $\mathbb {S}_{31}$, and $\mathbb {S}_{41}$ are reported in Eq. (55).

1.2 Pure bending test

The solution of the bending test carried out in Sect. 5.2 can be obtained considering the pure bending of a beam under applied moment M (see Fig. 12). The elementary beam theory predicts that stress component $\sigma _{xx}$ varies linearly with y. Therefore, we consider the following Airy stress function $\phi$:

$$\begin{aligned} \phi =Ay^3 \end{aligned}$$

(58)

where A is a constant to be determined.

Fig. 12

Pure bending of a beam subject to moment M

The Airy stress function enable us to determine the stress components by applying the following relations:

$$\begin{aligned} \begin{aligned} \sigma _{xx}&= \dfrac{\partial ^2\phi }{\partial y^2} = 6Ay \\ \sigma _{yy}&= \dfrac{\partial ^2\phi }{\partial x^2} = 0 \\ \sigma _{xy}&= -\dfrac{\partial ^2\phi }{\partial x \partial y} = 0 \\ \end{aligned} \end{aligned}$$

(59)

Now, we establish a relation between the moment M and the stress distribution at the beam ends in integral form to fully define the stress field in terms of problem parameters. Particularly, after setting $h=H/2$, we obtain:

$$\begin{aligned} M = \int _{-h}^{h} \sigma _{xx} y\ dy = 6A \int _{-h}^{h} y^2 dy = 4A h^3 \ \Rightarrow \ A = \dfrac{M}{4 h^3} \end{aligned}$$

(60)

Therefore, the stress component $\sigma _{xx}$ becomes:

$$\begin{aligned} \sigma _{xx} = \dfrac{3M}{2h^3}y \end{aligned}$$

(61)

Now, we apply the strain-stress relationship (54) and we obtain:

$$\begin{aligned} \begin{aligned} \varepsilon _{xx}&= \dfrac{\partial u}{\partial x} =\mathbb {S}_{11} \sigma _{xx} = \mathbb {S}_{11}\dfrac{3M}{2h^3} y\\ \varepsilon _{yy}&= \dfrac{\partial v}{\partial y} =\mathbb {S}_{21} \sigma _{xx} = \mathbb {S}_{21}\dfrac{3M}{2h^3} y\\ 2\varepsilon _{xy}&= \dfrac{\partial u}{\partial y} + \dfrac{\partial v}{\partial x} = \mathbb {S}_{31}\sigma _{xx} = \mathbb {S}_{31}\dfrac{3M}{2h^3}y \end{aligned} \end{aligned}$$

(62)

where $\mathbb {S}_{11}$, $\mathbb {S}_{21}$, and $\mathbb {S}_{31}$ are given in Eq. (55).

After integrating relations (62)$_1$ and (62)$_2$, we obtain the following displacement field:

$$\begin{aligned} \begin{aligned} u(x,y)&= \mathbb {S}_{11}\dfrac{3M}{2h^3}xy + g(y)\\ v(x,y)&= \mathbb {S}_{21}\dfrac{3M}{4h^3}y^2 + h(x) \end{aligned} \end{aligned}$$

(63)

After introducing the displacement components of Eq. (63) into (62)$_3$, we obtain the following equation:

$$\begin{aligned} \mathbb {S}_{11}\dfrac{3M}{2h^3}x+g'(y)+h'(x) - \mathbb {S}_{31}\dfrac{3M}{2h^3}y=0 \end{aligned}$$

which can be separated into two independent relations in x and y:

$$\begin{aligned} \left\{ \begin{aligned}&h'(x) + \mathbb {S}_{11}\dfrac{3M}{2h^3}x =C_1 \\&g'(y) -\mathbb {S}_{31}\dfrac{3M}{2h^3}y = -C_1 \end{aligned} \right. \end{aligned}$$

which leads to:

$$\begin{aligned} h(x)=-\mathbb {S}_{11}\dfrac{3M}{4h^3} x^2+C_1 x + C_2 \end{aligned}$$

(64)

and

$$\begin{aligned} g(y) = \mathbb {S}_{31}\dfrac{3M}{4h^3}y^2 -C_1 y + C_3 \end{aligned}$$

(65)

By replacing Eqs. (65) and (64) into Eqs. (63), we obtain:

$$\begin{aligned} \begin{aligned} u(x,y)&= \dfrac{3M}{2h^3}\left( \mathbb {S}_{11}xy+\dfrac{1}{2}\mathbb {S}_{31}y^2\right) -C_1 y + C_3 \\ v(x,y)&= \dfrac{3M}{4h^3}\left( \mathbb {S}_{21}y^2-\mathbb {S}_{11}x^2\right) +C_1 x + C_2 \end{aligned} \end{aligned}$$

(66)

where $C_1$, $C_2$, and $C_3$ are constants to be determined by applying boundary conditions.

If we apply the boundary conditions reported in Fig. 6:

$$\begin{aligned} \begin{aligned} u(0,-h)&= \mathbb {S}_{31}\dfrac{3M}{4h} + C_1 h + C_3 = 0\\ v(0,-h)&= \mathbb {S}_{21}\dfrac{3M}{4h} + C_2 = 0\\ u(0,h)&= \mathbb {S}_{31}\dfrac{3M}{4h} -C_1 h + C_3= 0 \end{aligned} \end{aligned}$$

(67)

we can compute constants $C_1$, $C_2$, and $C_3$, as follows:

$$\begin{aligned} \begin{aligned} C_1&= 0\\ C_2&= -\mathbb {S}_{21}\dfrac{3M}{4h}\\ C_3&= -\mathbb {S}_{31}\dfrac{3M}{4h} \end{aligned} \end{aligned}$$

(68)

Now, setting $M=\dfrac{fH^2}{6}$ and replacing constants $C_1$, $C_2$, and $C_3$ in Eqs. (66), we obtain the following analytical solutions:

$$\begin{aligned} \begin{aligned} u(x,y)&= \dfrac{2f}{H}\left[ \mathbb {S}_{11}xy+\dfrac{1}{2}\mathbb {S}_{31}\left( y^2-\dfrac{H^2}{4}\right) \right] \\ v(x,y)&= \dfrac{f}{H}\left[ \mathbb {S}_{21}\left( y^2 -\dfrac{H^2}{4}\right) -\mathbb {S}_{11}x^2 \right] \end{aligned} \end{aligned}$$

(69)

We recall that the analytical solution under plane strain for a purely isotropic material is given by [28]:

$$\begin{aligned} \left\{ \begin{aligned}&u(x,y) = \dfrac{2f(1-\nu ^2)}{EH}x\left( \dfrac{H}{2}-y\right) \\&v(x,y) = \dfrac{f(1-\nu ^2)}{EH}\left[ x^2+\dfrac{\nu }{1-\nu }y(y-H)\right] \end{aligned} \right. \end{aligned}$$

(70)