Saint-Venant End Effects in the Plane Problem for Linearly Elastic Functionally Graded Materials

Abstract

An isotropic elastic strip, with a continuous inhomogeneity profile for Young’s modulus is considered, subject to a self-equilibrated load on one of its axial ends and free of traction on the remainder of its surface boundaries. By taking advantage of the analytical flexibility of an exponential inhomogeneity profile, the full equations of linear theory of elasticity are employed to find the two-dimensional Saint-Venant decay rate in terms of an inhomogeneity parameter that measures gradation steepness in a Functionally Graded Material (FGM). The results show that softening axial inhomogeneity may be introduced to engineered FGM strips and beams to accelerate the decay of stresses. For a hardening axial inhomogeneity on the other hand, the end effects extend beyond a one-width-size distance from the loaded end. The plane problem end effects are shown to decay faster compared to the anti-plane shear counterparts, consistent with what is observed in homogeneous media. For inhomogeneity in the lateral direction from the core toward the outer edges, the qualitative behaviour changes with the degree of inhomogeneity. Whether the core is softer or harder relative to the outer edges, a steep lateral gradation of elastic modulus can significantly increase the decay length of end effects.

Appendices

Appendix A: Derivation of Compatibility Equation for an Inhomogeneous Isotropic Material

The two-dimensional compatibility equation requires strains to satisfy

$$ \mathscr{E}_{22,11}+\mathscr{E}_{11,22}-2\mathscr{E}_{12,12}=0. $$

(43)

In view of (3), the plane-stress constitutive assumptions in terms of the Airy stress function $\phi$, for an isotropic material with inhomogeneity in the form $E=E(x_{1},x_{2})$, is given by

$$ \begin{bmatrix} \mathscr{E}_{11} \\ \mathscr{E}_{22} \\ \mathscr{E}_{12}\end{bmatrix} =\frac{1}{E(x_{1},x_{2})} \begin{bmatrix} 1 & -\nu & 0 \\ -\nu & 1 & 0 \\ 0 & 0 & -(1+\nu ) \end{bmatrix} \begin{bmatrix} \phi _{,22} \\ \phi _{,11} \\ \phi _{,12} \end{bmatrix} . $$

(44)

Inserting (44) in the compatibility equation (43) yields

$$\begin{aligned} &\frac{1}{E}(\phi _{,1111}+\phi _{,2222}+2\phi _{,1122})+\big( \frac{1}{E}\big)_{,22}(\phi _{,22}-\nu \phi _{,11})+\big(\frac{1}{E} \big)_{,11}(\phi _{,11}-\nu \phi _{,22}) \\ &\quad {}+2(1+\nu )\big(\frac{1}{E}\big)_{,12}\phi _{,12} +2\big(\frac{1}{E} \big)_{,1}(\phi _{,111}+\phi _{,221})+2\big(\frac{1}{E}\big)_{,2}( \phi _{,112}+\phi _{,222})=0, \end{aligned}$$

(45)

which upon introducing the two-dimensional Laplace operator $\Delta = \partial _{11}+\partial _{22}$ simplifies to

$$ \Delta \big(\frac{\Delta \phi }{E}\big)-(\nu +1)\big[\big(\frac{1}{E} \big)_{,11} \phi _{,22}+\big(\frac{1}{E}\big)_{,22} \phi _{,11}\big]+2(1+ \nu )\big(\frac{1}{E}\big)_{,12}\phi _{,12}=0. $$

(46)

This is the governing equation (4).

Appendix B: Determining the Sign of $p$ from $p^{2}-4q>0$ in (18)

Recall the definitions of $p$ and $q$ given by (15):

$$ p=2m(m+\gamma )-\gamma ^{2}\nu \qquad \text{and} \qquad q=m^{2}(m+ \gamma )^{2}, $$

(47)

by which

$$ p^{2}-4q=\gamma ^{2}\nu (\gamma ^{2}\nu -4m(m+\gamma )), $$

(48)

and since $\nu$ is nonnegative by definition, if $p^{2}-4q>0$, it follows that

$$ 4m(m+\gamma )< \gamma ^{2}\nu . $$

(49)

Now if $m(m+\gamma )>0$, then $2m(m+\gamma )<4m(m+\gamma )$, so that

$$ 2m(m+\gamma )< 4m(m+\gamma )< \gamma ^{2}\nu , $$

(50)

by which

$$ p=2m(m+\gamma )-\gamma ^{2}\nu < 0. $$

(51)

If on the other hand $m(m+\gamma )\le 0$, $p$ is negative or zero on account of being the sum of two nonpositive terms by its definition (47)₁, irrespective of the sign of ${p^{2}-4q}$. As a result, ${p^{2}-4q>0}$ always implies that $p$ is either negative or equal to zero.

Appendix C: Analysis of Equation (22)

a) If the roots $\alpha _{1}$ and $\alpha _{2}$ given by (18) are complex conjugate ($p^{2}-4q<0$):

Define the roots in terms of real numbers $a$ and $b$ as $\alpha _{1}=a+ib$ and $\alpha _{2}=a-ib$. Then, with the expansions

$$ \begin{aligned} &\sinh (a\pm ib)=\sinh (a)\cos (b)\pm i \cosh (a)\sinh (b) , \\ &\cosh (a\pm ib)=\cosh (a)\cos (b)\pm i \sinh (a)\sinh (b), \end{aligned} $$

(52)

equation (22) yields

$$ aw\sin (2bw)+bw\sinh (2aw)=0, $$

(53)

by which either $a$ or $b$ should vanish, leading to $\alpha _{1}=\pm \alpha _{2}$.

b) If the roots $\alpha _{1}$ and $\alpha _{2}$ given by (18) are real ($p^{2}-4q>0$):

Consider the function $\bar{f}(x)=x\tanh (x)$. It can be easily verified that it satisfies the following:

• $\bar{f}'(x)>0$ for $x>0$, so $\bar{f}$ is increasing for $x>0$

• $\bar{f}'(x)<0$ for $x<0$, so $\bar{f}$ is decreasing for $x<0$

• $\lim _{x \to \pm \infty } \bar{f}(x)=\infty$

• $\bar{f}(x)>0$ for $x\ne 0$.

From these properties and the intermediate value theorem, we have that for any $L>0$, there are unique numbers $a>0$ and $b<0$ such that $L=\bar{f}(a)=\bar{f}(b)$. Moreover, for any $x$, $\bar{f}(x)=\bar{f}(-x)$. It follows that equation (22), or its equivalent $\bar{f}(w\alpha _{1})=\bar{f}(w\alpha _{2})$, leads to $\alpha _{1}=\pm \alpha _{2}$.