Mathematical design and preliminary mechanical analysis of the new lattice structure: “GE-SEZ*” structure processed by ABS polymer and FDM technology

Abstract

In this study, the authors propose a new design of a novel class of lattice structures. The new design is based on two main geometrical properties, the “volume” and the “surface to volume ratio”. It takes advantage of the strongest column designed against buckling proposed by Keller (Arch Ration Mech Anal 5:275–285, 1960) and Seiranyan (J Appl Math Mech 51(2):272–275, 1987). The Schoen minimal gyroid is used as a reference to establish the necessary lightweight property of the proposed design. This is to say, the surface to volume ratio and the volume of the new class of structures and their gyroid equivalent are equal. Models are built using CAD software and printed with UP mini 2.0 using acrylonitrile butadiene styrene (ABS) copolymer. Moreover, compression tests are conducted, using “MTS Criterion—Model 45”. The results show that after the phases of elasticity, relaxation and plasticity, the structure (three samples) is stable in the sense that it did not buckle nor collapse. Furthermore, reaching 7.35 mm of platen displacement (14.7% of strain) and 7.5 kN of resistance (3 MPa of equivalent stress), an additional progressive hardening is observed due to material densification and friction phenomena. A normalized comparison between the proposed structure and several lattice structures is conducted, showing a higher competitive behavior of the new design. The results of this study could possibly be a major contribution in the fields of biomechanics, aeronautic, and mechanical parts design.

Appendices

Appendix A1: Symmetry proof of the pinned–pinned KT column

Symmetries definition:

• (a) Symmetry between $\left( {x,y_{1} \left( x \right)} \right)$ and $\left( {x,f_{2} \left( x \right)} \right)$ according to the $\left( {\Delta_{1} :x = {\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 4}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$4$}}} \right)$ axis;

• (b) Symmetry of $\left( {x,f_{2} \left( x \right)} \right)$ itself around the $\left( {\Delta_{2} :x = {\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 2}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$2$}}} \right)$ axis;

• (c) Symmetry between $\left( {x,f_{1} \left( x \right)} \right)$ and $\left( {x,f_{3} \left( x \right)} \right)$ according to the $\left( {\Delta_{2} :x = {\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 2}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$2$}}} \right)$ axis;

Proof

Symmetry (a)

To prove the symmetry between $\left( {x,f_{1} \left( x \right)} \right)$ and $\left( {x,f_{2} \left( x \right)} \right)$ relatively to the axis $\left( {\Delta_{1} :x = {\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 4}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$4$}}} \right)$, we prove that:

$$f_{1} \left( x \right) = f_{2} \left( {x + 2\alpha } \right)\quad {\text{where}} \,\alpha = \frac{1}{4} - x \forall x \in \left[ {0,\frac{1}{4} } \right].$$

(24)

We calculate the right member of the Eq. (24):

$$f_{2} \left( {x + 2\alpha } \right) = f_{2} \left( {x + 2\left( {\frac{1}{4} - x} \right)} \right) = f_{2} \left( { - x + \frac{1}{2}} \right)$$

$$f_{2} \left( {x + 2\alpha } \right) = f_{2} \left( { - x + \frac{1}{2}} \right) = \frac{3}{2}\left( { - 3 + 16\left( { - x + \frac{1}{2}} \right) - 16\left( { - x + \frac{1}{2}} \right)^{2} } \right).$$

Thus $y_{2} \left( {x + 2\alpha } \right) = \frac{3}{2}\left( {1 - 16x^{2} } \right) = y_{1} \left( x \right)$

$$\forall x \in \left[ {0,\frac{1}{4} } \right].$$

Symmetry (b)

To prove the symmetry between $\left( {x,f_{2} \left( x \right)} \right)$ to itself around the axis $\left( {\Delta_{1} :x = {\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 2}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$2$}}} \right)$, we prove that:

$$f_{2} \left( x \right) = f_{2} \left( {x + 2\alpha } \right)\quad {\text{where}}\,\alpha = \frac{1}{2} - x\quad \forall x \in \left[ {\frac{1}{4},\frac{1}{2} } \right]$$

(24a)

We calculate the right member of the Eq. (24a):

$$f_{2} \left( {x + 2\alpha } \right) = f_{2} \left( {x + 2\left( {\frac{1}{2} - x} \right)} \right) = f_{2} \left( { - x + 1} \right)$$

$$f_{2} \left( {x + 2\alpha } \right) = f_{2} \left( { - x + 1} \right) = \frac{3}{2}\left( { - 3 - 16x^{2} + 16x} \right)$$

Thus $f_{2} \left( {x + 2\alpha } \right) = f_{2} \left( x \right)\quad \forall x \in \left[ {0,\frac{1}{2} } \right].$

Symmetry (c)

To prove the symmetry between $\left( {x,f_{1} \left( x \right)} \right)$ and $\left( {x,f_{2} \left( x \right)} \right)$ relatively to the axis $\left( {\Delta_{1} :x = {\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 4}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$4$}}} \right)$, we prove that:

$$f_{1} \left( x \right) = f_{3} \left( {x + 2\alpha } \right)\quad {\text{where}}\,\alpha = \frac{1}{2} - x\quad \forall x \in \left[ {0,\frac{1}{4} } \right]$$

(24b)

We calculate the right member of the Eq. (24b):

$$f_{3} \left( {x + 2\alpha } \right) = f_{3} \left( {x + 2\left( {\frac{1}{2} - x} \right)} \right) = f_{3} \left( { - x + 1} \right)$$

$$f_{3} \left( {x + 2\alpha } \right) = f_{3} \left( { - x + 1} \right) = \frac{3}{2}\left( { - 15 - 16\left( { - x + 1} \right)^{2} + 32\left( { - x + 1} \right)} \right)$$

$$f_{3} \left( {x + 2\alpha } \right) = \frac{3}{2}\left( {1 - 16x^{2} } \right)$$

Thus $f_{3} \left( {x + 2\alpha } \right) = f_{1} \left( x \right)\quad \forall x \in \left[ {0,\frac{1}{2} } \right].$

Appendix A2: SVR and volume of the GE-SEZ structure calculation

The external and internal lateral surfaces of the SEZ structure, are calculated using the Gauss formula (25):

$$S_{{{\text{lateral}}}} = 2\pi \mathop \int \limits_{a}^{b} g\left( x \right)\sqrt {1 + \left( {g^{\prime}\left( x \right)} \right)^{2} } {\text{d}}x$$

(25)

We apply Eq. (15) on the curve $C_{{{\text{ext}}}} \left( {x,\tilde{y}_{{{\text{ext}}}} } \right)$ presented in the Eq. (12.2):

$$S_{{{\text{lateral}} \left[ {0,\frac{L}{4m}} \right]}} = 2\pi \mathop \int \limits_{0}^{\frac{L}{4m}} \tilde{y}_{{{\text{ext}}}} \left( x \right)\sqrt {1 + \left( {\tilde{y}_{{{\text{ext}}}}^{\prime } \left( x \right)} \right)^{2} } {\text{d}}x.$$

(26)

This equation describes a quart of the surface of a SEZ. In other words, since the authors consider the different symmetries of the SEZ structure, as proved in the Sect. 3.1.1, the lateral revolution surface of the SEZ will be equal to four times the surface calculated by the Eq. (26). Thus, the total external surface is given by Eq. (17):

$$S_{{{\text{SEZ}} \left[ {0,\frac{L}{m}} \right]}} = 4 \times \left( {2\pi \mathop \int \limits_{0}^{\frac{L}{4m}} \tilde{y}_{{{\text{ext}}}} (x)\sqrt {1 + \left( {\tilde{y}_{{{\text{ext}}}}^{\prime } \left( x \right)} \right)^{2} } {\text{d}}x} \right).$$

(27)

It should be noted that in the Eq. (15), the index of $S_{{{\text{lateral}} \left[ {0,\frac{L}{4m}} \right]}}$ specifies that the calculation is made on the interval $\left[ {0,\frac{L}{4m}} \right]$. The index of $S_{{{\text{SEZ}} \left[ {0,\frac{L}{m}} \right]}}$ mentions that the calculation of this surface is performed on the whole length of the SEZ in the corresponding interval of $\left[ {0,\frac{L}{m}} \right]$ (see Fig. 5). An equivalent calculus could be performed by Pappus theorems [52], using the orthogonal distance between the barycenter of the function that will be spinning.

From the mapping (4), we recall that:

$$\tilde{y}_{{{\text{ext}}}} \left( x \right) = c\left( {a\left( {1 - 16\left( {d x} \right)^{2} } \right) + b} \right).$$

The derivative of $\tilde{y}_{{{\text{ext}}}} \left( x \right)$ is equal to:

$$\tilde{y}_{{{\text{ext}}}}^{\prime } \left( x \right) = \frac{{d\left( {\tilde{y}_{{{\text{ext}}}} \left( x \right)} \right)}}{{{\text{d}}x}} = 32acd^{2} x.$$

(28)

The Eq. (27) becomes:

$${\text{SEZ}}_{{ \left[ {0,\frac{L}{m}} \right]}} = 8\pi \mathop \int \limits_{0}^{\frac{L}{4m}} c\left( {a\left( {1 - 16\left( {{\text{d}}x} \right)^{2} } \right) + b} \right)\sqrt {1 + \left( {32acd^{2} x} \right)^{2} } {\text{d}}x,$$

which leads to the expression (29):

$$\begin{array}{c} \begin{array}{c} P_{n,m}\left(a,b\right)=\frac{8\ \pi \ \left(A\times B-C\times D\right)}{E}-\gamma \\ \left\{ \begin{array}{c} \ \\ A=A_{n,m}\left(a,b\right)=256a\left(a+b\right)d^2c^2+1 \\ B=B_{n,m}\left(a,b\right)={\mathrm{log} \left(\sqrt{32a*c*d^2+1}+32acd^2x_1\right)\ } \\ C=C_{n,m}\left(a,b\right)=32acd^2x_1\sqrt{{\left(32acd^2x_1\right)}^2+1} \\ D=D_{n,m}\left(a,b\right)=256ac^2d^2\left(a\left(8d^2x^2_1-1\right)-b\right)+1 \\ E_{n,m}\left(a,b\right)=16{\left(32acd^2\right)}^2 \end{array} \right. \end{array} \\ s.t\ \left\{ \begin{array}{c} c=\frac{L}{2n\left(a+b\right)} \\ d=\frac{m}{L} \\ x_1=\frac{L}{4m} \\ \gamma ={\left({S}/{V_{Gy}}\right)}_{Gyroid} \\ V_{Gy}:the\ effective\ volume\ of\ the\ Gyroid \end{array} \right. \end{array}$$

(29)

Since the cubic volume, for $n^{2} \times m$ SEZ structures confined into the is equation to L³, the ${\text{SVR}}_{{\left( {\text{GE } - \text{ SEZ}} \right)_{{n^{2} \times m}} }}$ is obtained by the following equation:

$${\text{SVR}}_{{\left( {\text{GE } - \text{ SEZ}} \right)_{{n^{2} \times m}} }} = \frac{{n^{2} {\text{mS}}_{{{\text{SEZ}} \left[ {0,\frac{L}{m}} \right]}} }}{{L^{3} }}.$$

(30)

We calculate the volume of the GE-SEZ assembly as a sum of SEZs volumes Eq. (31).

The volume of a SEZ unit if given by the Gauss formula presented in the Eq. (32):

$$V_{{\text{GE } - \text{ SEZ}}} = n^{2} \times m V_{{{\text{SEZ}}}}$$

(31)

$$V_{{{\text{SEZ}}}} = \pi \mathop \int \limits_{0}^{\frac{L}{m}} \left( {y_{{{\text{ext}}}}^{2} - y_{{\text{int}}}^{2} } \right){\text{d}}x$$

(32)

Since the authors consider the symmetries introduced in the Sect. 3.1.1 (see “Appendix A1”), the volume of the SEZ can be calculated by:

$$V_{{{\text{SEZ}}}} = \pi 4 \times \left( {\mathop \int \limits_{0}^{\frac{L}{4m}} \left( {y_{{{\text{ext}}}}^{2} - y_{{\text{int}}}^{2} } \right){\text{d}}x} \right).$$

(33)

where

$$y_{{{\text{ext}}}}^{2} - y_{{\text{int}}}^{2} = \left( {y_{{{\text{ext}}}} - y_{{\text{int}}} } \right)\left( {y_{{{\text{ext}}}} + y_{{\text{int}}} } \right)$$

according to the Eq. (11.2):

$$y_{{{\text{ext}}}} - y_{{\text{int}}} = e.$$

And according to the expression of the functions $y_{{{\text{ext}}}}$ and $y_{{\text{int}}}$:

$$V_{{{\text{SEZ}}}} = \frac{4\pi e}{{nm}}\left[ {Ln + enm - \frac{{aL^{2} }}{{12\left( {a + b} \right)}}} \right].$$

(34)

Hence, the volume of the ${\text{GE } - \text{ SEZ}}_{n \times n \times m}$ assembly is equal to:

$$V_{{\text{GE } - \text{ SEZ}}} = n^{2} m V_{{{\text{SEZ}}}} = 4\pi en\left[ {Ln + enm - \frac{{aL^{2} }}{{12\left( {a + b} \right)}}} \right].$$

(35)

The volume equivalence is ensured by the equality between the GE-SEZ volume and the volume of the Gyroïd as shown below:

$$\frac{{V_{{\text{GE } - \text{ SEZ}}} }}{{V_{{\text{G}}} }} = 1.$$

(36)

Reporting the expression (35) in (38), the second characteristic equation corresponds to the Eq. (37):

$$4\pi en\left[ {Ln + enm - \frac{{aL^{2} }}{{12\left( {a + b} \right)}}} \right] - V_{{\text{G}}} = 0$$

(37)

$$\begin{gathered} A_{a,b} e^{2} + B_{a,b} e + C_{a,b} = 0 \hfill \\ \left\{ {\begin{array}{*{20}l} {A_{a,b} = \frac{{4\pi n^{2} m}}{{V_{{\text{g}}} }}} \hfill \\ {B_{a,b} = \frac{4\pi n}{{V_{{\text{g}}} }}\left( {Ln - \frac{{aL^{2} }}{{12\left( {a + b} \right)}}} \right)} \hfill \\ {C = - 1} \hfill \\ \end{array} } \right.. \hfill \\ \end{gathered}$$

(38)

The characteristic Eq. (37) permits to calculate the thickness e, that permits to ensure the second characteristic equation, that is to say, the equality of the volumes of the GE-SEZ and the gyroid.

The discriminant of the equation is given by:

$$\Delta = B_{a,b}^{2} - 4 A_{a,b} C_{a,b} .$$

Hence

$$\Delta = \left( {\frac{4\pi n}{{V_{{\text{g}}} }}\left( {Ln - \frac{{aL^{2} }}{{12\left( {a + b} \right)}}} \right)} \right)^{2} + 4\left( {\frac{{4\pi n^{2} m}}{{V_{{\text{g}}} }}} \right).$$

Since the parameters n, m and V_g are strictly positive, the discriminant is also strictly positive.

So the roots of the Eq. (37) are defined as:

$$e = \left\{ {\begin{array}{*{20}c} {e^{ + } = \frac{{ - B_{a,b} + \sqrt \Delta }}{{2A_{a,b} }}} \\ {e^{ - } = \frac{{ - B_{a,b} - \sqrt \Delta }}{{2A_{a,b} }}} \\ \end{array} } \right..$$

(39)

Appendix A3: Statistics on dimensions measurement of the $\left( {{\text{GE-SEZ}}} \right)_{{20 \times 20 \times 6}}$ and the cylindrical specimens

Table 8 groups the statistics of the dimension measurement related to the biggest diameter, neck diameter (smallest diameter), the width and height of the $\left( {{\mathbf{GE}}{\text{-}}{\mathbf{SEZ}}} \right)_{{{\mathbf{20}} \times {\mathbf{20}} \times {\mathbf{6}}}}$ manufactured.

Table 8 Statistics related to $\left( {\text{GE-} \text{ SEZ}} \right)_{20 \times 20 \times 6}$ samples dimensions

Table 9 groups the statistics of the dimension measurement related to the diameter and height of the cylindrical specimens manufactured.

Table 9 Statistics of dimensions of the cylindrical specimens