Effect of geometrical parameters on the effective elastic modulus for an X-type lattice truss panel structure

2.1 Homogenization theory

The X-type LTPS consists of upper face sheet, lower face sheet, and a truss core, as shown in Figure 1. The basic material used in this work is P92 steel, and its elastic modulus and Poisson’s ratio are 203 GPa and 0.3, respectively. The X-type LTPS is a kind of periodic cellular structure whose mechanical properties can be analyzed by a representative unit cell, as shown in Figure 2. The geometrical parameters of X-type LTPS: the truss length L, truss width w, truss thickness t, face sheet thickness T, node length S, stamping angle α and shearing angle β are listed in Table 1. The length L_X̅,width L_Y̅ and height H of the unit cell are 21, 28 and 14 mm, respectively.

Figure 1:

Schematic of X-type lattice truss panel structure.

Figure 2:

Sketching of the unit cell.

According to homogenization method, the X-type LTPS is deemed as a homogeneous solid at the macroscopic scale, while it is considered as discrete structural elements at microscopic scale [26]. Considering the continuum mechanics and homogenized theory, the relationship between macroscopic scale and microscopic scale for strain and stress can be described as

where Ξ and Σ are the equivalent strain and stress tensors at macroscopic scale, respectively; ε and σ are microscopic strain and stress tensors, respectively. Ω denotes the volume of the unit cell.

The volume averaged strain energy density of an inhomogeneous material is determined by multiplying the separate volume averages of microscopic stresses and strains [26]:

where Σ·Ξ is the macroscopic strain energy density and ∫Ωσ⋅εdΩ is the total strain energy of the admissible microscopic field.

Eight beams of the unit cell would be analyzed to study the mechanical properties of the X-type LTPS. According to the deformation analysis of the unit cell, the displacements of the end nodes of the beam are

where Ξ is the three-order equivalent strain tensor, L₀ means the length of the beam, and n₀ denotes the direction vector of the beam in global coordinate system:

Because each sub-geometry (the unit cell) of X-type LTPS contains equal truss members and the length of the trusses is equal. Then, using the Euler-Bernoulli beam model derived from mechanics of materials, the axial stretching of the beam prevails over the bending deformation [26]. Therefore, the bending deformation and associated rotations are neglected here. The nodal displacement vector u⁽ⁱ⁾ for the ith beam can be characterized by the two ends of the beam. For u⁽ⁱ⁾, The previous six values represent displacement of the one end, and the following six values denotes that of the other end:

where the ω means the rotation:

If the unit cell is composed of N Euler-Bernoulli beam members, its strain energy density can be defined as

where Ω is the volume of the unit cell, u⁽ⁱ⁾ is the nodal displacement vector of the ith beam, and K⁽ⁱ⁾ is the global stiffness matrix that satisfies the transformation between local and global coordinates; it can be calculated by

where T′ is the transformation matrix, which is related with the global coordinate system and the local coordinate system of the beam element. K^e(i) is the element stiffness matrix:

The macroscopic strain vector acting on the unit cell is defined as

The effective stiffness of the unit cell can be calculated as [26]

where H denotes the homogenized effective stiffness.

Combining Equations (10) and (14), the effective stiffness matrix of the X-type LTPS is calculated by a program compiled by Matlab code:

It is obvious that the stiffness matrix has nine independent parameters, namely, C₁₁₁₁, C₁₁₂₂, C₁₁₃₃, C₂₂₂₂, C₂₂₃₃, C₃₃₃₃, C₁₂₁₂, C₁₃₁₃ and C₂₃₂₃, indicating that the core is regarded as an orthotropic material after the homogenization.

2.2 Calculation of the effective elastic modulus

The homogenous calculation in Section 2.1 just gets the effective stiffness matrix of the X-type core of LTPS. In order to calculate the EEM of the whole panel structure, a three-dimensional equivalent model is established by finite element software ABAQUS, as shown in Figure 3. The X-type core is assumed to be an equivalent homogenous-solid plate with the same length, width, and height of X-type core. The effective elastic constants of the equivalent homogenous-solid plate are specified by the effective stiffness matrix obtained in Section 2.1 (Equation (15)). The equivalent solid plate is bonded tightly to the upper face sheet and lower face sheet to form a sandwich panel with three layers.

Figure 3:

Equivalent homogenous solid plate model.

A finite element analysis of the uniaxial tensile test was performed to calculate the EEM of the X-type core of LTPS. In order to get a uniform deformation of the model, a rigid plate was added and bonded tightly to the top surface of the upper face sheet. A tensile load was applied on the rigid plate, and the effective strain is calculated by the displacement of rigid plate divided by the original height of the composite plate. A plot of the obtained elastic stage of the equivalent stress-strain curve is shown in Figure 4, and the curve slope is the EEM.

Figure 4:

The elastic stage of the equivalent stress-strain curves by homogenization method and FEM.

4.1 Comparison between homogeneous method and FEM

Figure 4 shows the elastic stage of the stress-strain curve of the X-type LTPS obtained by homogeneous method and FEM. The two slopes by the two methods show a good agreement, indicating that the homogeneous method is right. The obtained EEM by the two methods are 393 MPa and 420 MPa, respectively, and the error is only 6.43%. It is a pity that we did not present an experimental validation, which still needs further study in the future.

4.2 Effect of the geometrical parameters

As shown in Figure 2, the X-type LTPS is determined by seven parameters including the thickness of face sheet T, stamping angle α, shearing angle β, truss length L, truss width w, truss thickness t and node length S. Basing on the single-factor analysis, how these parameters affect the EEM is fully discussed here by the developed homogeneous method. When one of the parameters is discussed, its value is changed, and the rest of the parameters are kept constant. The initial values are listed in Table 1.

Figure 6 shows the effects of truss length, truss width, truss thickness, stamping angle, shearing angle, thickness of face sheet and node length on EEM by homogenization method. Obviously, as the truss length, stamping angle and shearing angle increase, the EEM decreases. The EEM decreases slightly as the node length increases, while it increases with the increase of truss width, truss thickness and thickness of face sheet. The effect of the geometrical parameters can be verified by previous work [30]. As the truss length increases from 14 to 22 mm, the EEM decreases from 1500 to 432 MPa. However, as the truss width and thickness increase from 1.5 to 3.5 mm and 1.0 to 1.8 mm, the EEM increases from 432 to 700 MPa and from 432 to 880 MPa, respectively. As the stamping and shearing angle increase from 10 to 70° and 20 to 60°, the EEM decreases from 1040 to 502 MPa and from 2024 to 494 MPa, respectively. And their decrease rates are 8.97 MPa and 38.25 MPa per each degree, respectively. As the thickness of face sheet increases from 1.2 to 3.2 mm, the EEM increases from 450 to 455 MPa slightly, while it decreases slightly from 620 to 460 MPa as the node length increases from 1 to 5 mm. As found above, the most significant important factors effecting the EEM are shearing angle, stamping angle, truss width, truss length and truss thickness, while the effects of the node length and face sheet thickness are not obvious.

Figure 6:

Effects of truss length (A), truss width (B), truss thickness (C), stamping angle (D), shearing angle (E), thickness of face sheet (F) and node length (G) on EEM.

4.3 Discussion of influence mechanism

To find out the reason why the geometric parameters have such significant effect on the EEM of X-type LTPS, we perform a stress analysis to a single truss as shown in Figure 7. P is a tensile load at the truss end in Z̅ direction. It is decomposed into three forces P_X, P_Y and P_Z in X, Y and Z directions, respectively, which produce axial tensile deformation D_X, bending deformation D_Y and D_Z in Y and Z directions, respectively.

Figure 7:

The force diagram of the beam.

P_X, P_Y and P_Z are calculated by

D_X, D_Y and D_Z are calculated by

where A is the cross-sectional area of truss, and I_Y and I_Z are the inertia moments of the cross-sectional area of truss about the Y and Z axes, respectively. We decompose D_X, D_Y and D_Z in Z̅, directions, respectively:

d_X, d_Y and d_Z are the deformation components induced by D_X, D_Y and D_Z in Z̅ direction.

The total displacement d′ in Z̅ direction is the sum of d_X, d_Y and d_Z:

The truss height H shown in Figure 2 is calculated by

The equivalent strain ε′ in Z̅ direction is

The EEM (E_eq) is calculated by

Having a derivation to L, w, t, α and β by Equation (21), respectively, we obtain

After a calculation of Equations (23)–(27), it is found that ∂ε′∂L>0,∂ε′∂α>0,∂ε′∂β>0, which means that ε′ increases with the increase of L, α and β (10°≤α≤70°, 10°≤β≤60°). As a result, EEM decreases according to Equation (22), whist ∂ε′∂w<0 and ∂ε′∂t<0, which proves that the EEM increases as w and t increase.

The reason of node length S effect on EEM is discussed as follows:

As shown in Figure 2, the length L_X̅ and width L_Y̅ of the unit cell are calculated, respectively, by

As the panel structure bears a distributed load p, the equivalent load P of each truss in Z̅ direction as shown in Figure 7 is calculated by

Substituting Equation (30) into Equation (21), we obtain

Having a derivation to S by Equation (31), we obtain

It shows that ∂ε′∂S>0, which proves that the ε′ increases with the increase of S, and correspondingly the EEM decreases according to Equation (22).

As calculated in Section 2, it reveals that the effective modulus of lattice truss core is far smaller than that of face sheet. Therefore, the deformation of face sheet can be ignored compared with the core. As the thickness of face sheet increases, the total height of the panel structure increases, while the vertical deformation of core is constant at a constant stress. As a result, the effective strain in Z̅ direction decreases, which leads to a slight increase of EEM.

4.4 Formula fitted

Based on the above analysis, it shows that each parameter has a different effect on the EEM. It is essential to develop a theory formula to calculate the EEM by the seven variables. For foam materials, it has been proved that the EEM is related with the relative density ρ¯ [23]:

For low density foam, Equation (33) agrees well with experiment, but for large density foam they have no clear correlation [23].

In order to illustrate whether the LTPS obeys Equation (33), we plot the EEM as a function of relative density with different geometrical dimensions, as shown in Figure 8. It should be noted that when one parameter is discussed the other parameters are kept constant. Here the relative density of the X-type LTPS is calculated by

Figure 8:

EEM of X-type LTPS versus relative density with different geometrical parameters.

Figure 8 demonstrates that EEM of the X-type LTPS does not exhibit a clear correlation with the relative density. This conclusion is consistent with that of the pyramidal LTPS [30]. The EEM totally increases as the relative density increases. As the relative density is smaller than 2%, the effects of different parameters are similar. As the relative density is larger than 2%, the most significant impact factor is the shearing angle, and then followed by the truss width, length and thickness with the same effect; the smallest influencing factor is the stamping angle. Therefore, EEM cannot be simply expressed by a formula with relative density.

Gibson and Ashby [31] have developed a theoretical formula to calculate the EEM for honeycomb and foam structures which include the geometrical parameters of unit cells, respectively (see Equations (35) and (36)). Figures 9 and 10 show the sketching of the unit cells of honeycomb and foam structures, respectively. For honeycombs, there are four geometrical parameters including length of side l and h, wall thickness d and inclination angle of side θ. The foams have two geometrical parameters: length a and thickness b.

Figure 9:

Unit cell of honeycomb structures.

Figure 10:

Unit cell of foams.

where Eeqh and Eeqf denote the EEM of honeycomb and foam structures, respectively. E₀ is the Young’s modulus of the parent material. ρ¯ is the relative density. C₁ and I are the constant of proportionality and the second moment of area.

However, the honeycomb structures belong to the ordered two-dimensional cellular solid, and foams belong to unordered cellular solid, whereas X-type LTPS belongs to ordered three-dimensional cellular solid. Besides, the honeycomb and the foam structures include four and two geometrical parameters, respectively, while the X-type LTPS has seven geometrical parameters. The EEM proposed by Gibson and Ashby is not suitable for X-type LTPS. Therefore, it is essential to establish the relationship between the EEM and geometrical parameters of X-type LTPS. The structure of LTPS is very complex, and it is very difficult to get an analytical formula to calculate the EEM. In order to get a formula, we performed a large number of FEM calculations with different geometric dimensions and then fitted a formula by Origin software as follows

where E_eq is the EEM, and E₀ is the elastic modulus of basic metal. Their unit is megapascal. The units of the geometrical parameters L, w, t, T, and S are millimeters, and the units of α, β are degrees. But only the values of the geometrical parameters are put into Equation (37), and their units are not considered in the calculation.

The EEM of the X-type LTPS with different geometrical parameters derived from the present fitted formula Equation (37) is also plotted in Figure 6. It obviously shows that there is a good consistence between the homogenization and formula results, which proves that the present formula can be used to calculate the EEM. Kawashima et al. [32] and Mizokami et al. [33] studied the effective strength of plate-fin structure. The porous structure is treated as the equivalent homogenous solid plate, and the tensile and creep strength are predicted successfully. In their method, the calculation of EEM is an important issue, and they predict it by FEM. But how the geometrical parameters influence the EEM is still unclear. The effective mechanical strength of LTPS, such as tensile, creep and fatigue, still needs to be studied further by the equivalent homogenous solid method, and the formula can be used to calculate the EEM for future work, proving that this work plays a key role for the development of structural design procedure of compact heat exchanger by LTPS. It should be noted that the flexural deformation has not been considered in this model [8], because for heat exchanger it is forbidden to have such obvious deformation in order to ensure safety. The effect of flexural deformation on strength still needs further study in the future.