A novel space-constrained vehicle suspension mechanism synthesized by a systematic design process employing topology optimization

Abstract

Demands on larger motors and battery packs in electric vehicles cause a suspension layout issue not appearing in gas-powered vehicles. Motivated by this need, our research aims to design a new-concept suspension applicable to electric vehicles, where given space-related constraints are satisfied without scarifying their kinematic performance. Here, we propose a three-phase design process for the synthesis of rear suspensions of an electric vehicle: concept topology design, kinematic feature identification, and detailed design. For the concept design to determine the mechanism topology, we employ the topology optimization method developed for mechanism synthesis subjected to a reduced suspension design space as well as a tighter condition on the camber rate—known as yielding better vehicle’s dynamic performance. The next phase is to extract the underlying kinematic features of the synthesized suspension obtained by the topology optimization method as it may be difficult to directly figure out how the synthesized mechanism functions kinematically. For the extraction, we propose a connectivity-mapping technique followed by the wrench calculation. This phase is followed by the final detailed design to meet the specific requirements imposed on the target suspensions. The new suspensions designed by the proposed three-phase design approach will be shown to successfully resolve the suspension layout issue typically encountered in electric vehicles.

Appendix. Vector calculus for the wrench analysis

Equations (A1a-e) can be calculated based on (28) and (31). For the transpose and dot product of the ordered pair of vectors, the rules defined in (22) and (23) are applied.

$$ {\$}_r^{\mathrm{T}}\bullet {\$}_1=\left({\mathbf{s}}_{\mathbf{F1}}\times {\mathbf{S}}_N\right)\bullet {\mathbf{S}}_1+{\mathbf{S}}_N\bullet \left({\mathbf{s}}_{\mathbf{M24}}\times {\mathbf{S}}_1\right) $$

(A1a)

$$ {\$}_r^{\mathrm{T}}\bullet {\$}_2=\left({\mathbf{s}}_{\mathbf{F1}}\times {\mathbf{S}}_N\right)\bullet {\mathbf{S}}_2+{\mathbf{S}}_N\bullet \left({\mathbf{s}}_{\mathbf{F1}}\times {\mathbf{S}}_2\right) $$

(A1b)

$$ {\$}_r^{\mathrm{T}}\bullet {\$}_3=\left({\mathbf{s}}_{\mathbf{F1}}\times {\mathbf{S}}_N\right)\bullet {\mathbf{S}}_3+{\mathbf{S}}_N\bullet \left({\mathbf{s}}_{\mathbf{F1}}\times {\mathbf{S}}_3\right) $$

(A1c)

$$ {\$}_r^{\mathrm{T}}\bullet {\$}_4=\left({\mathbf{s}}_{\mathbf{F1}}\times {\mathbf{S}}_N\right)\bullet {\mathbf{S}}_4+{\mathbf{S}}_N\bullet \left({\mathbf{s}}_{\mathbf{F1}}\times {\mathbf{S}}_4\right) $$

(A1d)

$$ {\$}_r^{\mathrm{T}}\bullet {\$}_5=\left({\mathbf{s}}_{\mathbf{F1}}\times {\mathbf{S}}_N\right)\bullet {\mathbf{S}}_5+{\mathbf{S}}_N\bullet \left({\mathbf{s}}_{\mathbf{K34}}\times {\mathbf{S}}_5\right) $$

(A1e)

By using the identities in (A2), it is possible to check if the wrench in (31–33) satisfies the relation (30). Although the identities in (A2) are well-known from the vector calculus, they can be checked from the simple calculation after assuming three-dimensional vector components, e.g., A = {A₁, A₂, A₃}^T.

$$ \mathbf{A}\times \left(\mathbf{B}\times \mathbf{C}\right)=\mathbf{B}\left(\mathbf{A}\bullet \mathbf{C}\right)-\mathbf{C}\left(\mathbf{A}\bullet \mathbf{B}\right) $$

(A2a)

$$ \left(\mathbf{A}\times \mathbf{B}\right)\bullet \left(\mathbf{C}\times \mathbf{D}\right)=\left(\mathbf{A}\bullet \mathbf{C}\right)\left(\mathbf{B}\bullet \mathbf{D}\right)-\left(\mathbf{A}\bullet \mathbf{D}\right)\left(\mathbf{B}\bullet \mathbf{C}\right) $$

(A2b)

In the case of (A1b), one can obtain (A3) by using the identity expressed by (A2) and (32). Likewise, one can see that (A1c) and (A1d) are equal to zero from the result of (A3).

$$ {\displaystyle \begin{array}{c}{\$}_r^{\mathrm{T}}\bullet {\$}_2=\left({\mathbf{s}}_{\mathbf{F1}}\times \left(c{\mathbf{N}}_1\times {\mathbf{N}}_2\right)\right)\bullet {\mathbf{S}}_2+\left(c{\mathbf{N}}_1\times {\mathbf{N}}_2\right)\bullet \left({\mathbf{s}}_{\mathbf{F1}}\times {\mathbf{S}}_2\right)\\ {}=c\left\{\left({\mathbf{s}}_{\mathbf{F1}}\bullet {\mathbf{N}}_2\right)\left({\mathbf{N}}_1\bullet {\mathbf{S}}_2\right)-\left({\mathbf{s}}_{\mathbf{F1}}\bullet {\mathbf{N}}_1\right)\left({\mathbf{N}}_2\bullet {\mathbf{S}}_2\right)+\left({\mathbf{N}}_1\bullet {\mathbf{s}}_{\mathbf{F1}}\right)\left({\mathbf{N}}_2\bullet {\mathbf{S}}_2\right)-\left({\mathbf{N}}_1\bullet {\mathbf{S}}_2\right)\left({\mathbf{N}}_2\bullet {\mathbf{s}}_{\mathbf{F1}}\right)\right\}=0\end{array}} $$

(A3)

Next, (A4) can also be obtained by applying the identities (A2) into (A1a).

$$ {\displaystyle \begin{array}{c}{\$}_r^{\mathrm{T}}\bullet {\$}_1=\left({\mathbf{s}}_{\mathbf{F1}}\times \left(c{\mathbf{N}}_1\times {\mathbf{N}}_2\right)\right)\bullet {\mathbf{S}}_1+\left(c{\mathbf{N}}_1\times {\mathbf{N}}_2\right)\bullet \left({\mathbf{s}}_{\mathbf{M24}}\times {\mathbf{S}}_1\right)\\ {}\begin{array}{c}=c\left\{\left({\mathbf{s}}_{\mathbf{F1}}\bullet {\mathbf{N}}_2\right)\left({\mathbf{N}}_1\bullet {\mathbf{S}}_1\right)-\left({\mathbf{s}}_{\mathbf{F1}}\bullet {\mathbf{N}}_1\right)\left({\mathbf{N}}_2\bullet {\mathbf{S}}_1\right)+\left({\mathbf{N}}_1\bullet {\mathbf{s}}_{\mathbf{M24}}\right)\left({\mathbf{N}}_2\bullet {\mathbf{S}}_2\right)-\left({\mathbf{N}}_1\bullet {\mathbf{S}}_2\right)\left({\mathbf{N}}_2\bullet {\mathbf{s}}_{\mathbf{M24}}\right)\right\}\\ {}=c\left({\mathbf{N}}_1\bullet {\mathbf{S}}_1\right)\left\{{\mathbf{N}}_2\bullet \left({\mathbf{s}}_{\mathbf{F1}}-{\mathbf{s}}_{\mathbf{M24}}\right)\right\}+c\left({\mathbf{N}}_2\bullet {\mathbf{S}}_1\right)\left\{-{\mathbf{N}}_1\bullet \left({\mathbf{s}}_{\mathbf{F1}}-{\mathbf{s}}_{\mathbf{M24}}\right)\right\}=0\end{array}\end{array}} $$

(A4)

Because the vectors N₁ and N₂ are perpendicular to the vector s_F1 − s_M24, (A4) becomes zero. In the same manner, one can show that (A1e) is equal to zero. Thus, all equations in (A1) are zero. Therefore, the linear combination of (A1), i.e., (30), becomes zero.