An analysis strategy for swept solids

Rakesh Vemulapally; Krishnan Suresh

doi:10.1007/s00366-008-0117-y

Engineering with Computers

March 2009, 25:155 | Cite as

An analysis strategy for swept solids

Authors
Authors and affiliations

Rakesh Vemulapally
Krishnan SureshEmail author

Original Article

First Online: 26 November 2008

Received: 04 October 2007

Accepted: 21 October 2008

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Abstract

Numerous engineering components such as gears, turbine-blades, splined-shafts, and MEMS devices have a common underlying geometric structure in that they may be represented as a sweep of a 2D cross-section along a 3D trajectory. We consider here boundary value problems over such solids and propose an efficient strategy for solving such problems. The proposed strategy avoids the construction of a 3D finite element mesh, and therefore does not require a 3D mesh generator/solver. Instead a 2D mesh of the cross-section is coupled with an implicit 1D ‘hierarchical mesh’ over the trajectory. The latter is symbolically eliminated prior to computational analysis. This, we show, leads to the reduction of 3D scalar boundary value problems to simpler 2D vector problems over the cross-section. The proposed method is particularly well suited for the design exploration of swept solids, where numerous combinations of trajectories and/or cross-sections must be examined. Its computational superiority over extant 3D strategies is established through theory and numerical experiments.

Keywords

Swept solids Engineering analysis Extrusions CAD/CAE

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Notes

Acknowledgments

The authors wish to acknowledge the reviewers, and the support of the National Science Foundation under grants OCI-0636206, and CMMI-0726635, CMMI-0745398.

Appendix A

Proof of Proposition 1:

Consider substituting (10) in (8). First consider the sub-space:

$$ {\text{U}}(\Upomega ) = \left\{ {U \in {\mathbf{H}}^{1} (\Upomega )\quad U = r_{i} \;{\text{on}}\;\Upgamma_{i} \quad \forall i \in D} \right\} $$

From (10), we have:

$$ {\mathbf{N}}(\xi )^{T} {\mathbf{u}}(x_{\omega } ,y_{\omega } ) = r_{i} \;{\text{on}}\;\Upgamma_{i} \quad \forall ~ i \in D $$

Consider the following cases:

If Dirichlet conditions are prescribed over the starting cross-section, i.e., if 1 ∈ D⇒N(−1)^T u(x _ω, y _ω) = r ₁.
If Dirichlet conditions are prescribed over the terminal cross-section, i.e., if 2 ∈ D⇒N(1)^T u(x _ω, y _ω) = r ₂.
If Dirichlet conditions are prescribed over the cylindrical surfaces i.e., if i > 2 ∈ D⇒N(ξ)^T u(x _ω, y _ω) = r _i. Multiply both sides by N(ξ) and integrate out ξ, resulting in a set of boundary conditions for u(x _ω, y _ω) over γ _i−2:

$$ \left[ {\int\limits_{ - 1}^{1} {{\mathbf{N}}(\xi ){\mathbf{N}}^{T} (\xi )d\xi } } \right]{\mathbf{u}}(x_{\omega } ,y_{\omega } ) = \left\{ {\int\limits_{ - 1}^{1} {{\mathbf{N}}(\xi )r_{i} d\xi } } \right\}\;{\text{on}}\;\gamma_{i - 2} \;{\text{if}}\;i > 2 \in D $$

We write this in a compact form:

$$ {\mathbf{Bu}} = {\mathbf{r}}_{i} \;{\text{on}}\;\gamma_{i - 2} \;{\text{if}}\;i \ge 3 \in D $$

where $ {\mathbf{B}} = \int\limits_{ - 1}^{1} {{\mathbf{NN}}^{T} d\xi } ,\quad {\mathbf{r}}_{i} = \int\limits_{ - 1}^{1} {{\mathbf{N}}(\xi )r_{i} d\xi } $

Thus, the trial sub-space is reduced to:

$$ U(\omega _{1} ) = \left\{ {\begin{array}{*{20}c} {{\mathbf{u}} = \left\{ {\begin{array}{*{20}c} {u_{1} (x_{\omega } ,y_{\omega } )} & \ldots & \ldots & {u_{p} (x_{\omega } ,y_{\omega } )} \\ \end{array} } \right\} \in {\mathbf{H}}_{p}^{1} (\omega _{1} )} \hfill \\ {{\mathbf{N}}( - 1)^{T} {\mathbf{u}}(x_{\omega } ,y_{\omega } ) = r_{1} \;{\text{on}}\;\omega _{1} \;{\text{if}}\;1 \in D} \hfill \\ {{\mathbf{N}}( + 1)^{T} {\mathbf{u}}(x_{\omega } ,y_{\omega } ) = r_{2} \;{\text{on}}\;\omega _{1} \;{\text{if}}\;{\text{2}} \in D} \hfill \\ {{\mathbf{Bu}} = {\mathbf{r}}_{i} \;{\text{on}}\;\gamma _{{i - 2}} \;{\text{if}}\;i \ge 3 \in D} \hfill \\ \end{array} } \right\} $$

(14)

Similarly the test sub-space is reduced to:

$$ V(\omega _{1} ) = \left\{ {\begin{array}{*{20}c} {{\mathbf{v}} = \left\{ {\begin{array}{*{20}c} {v_{1} (x_{\omega } ,y_{\omega } )} & \ldots & \ldots & {v_{p} (x_{\omega } ,y_{\omega } )} \\ \end{array} } \right\} \in {\mathbf{H}}_{p}^{1} (\omega _{1} )} \hfill \\ {{\mathbf{N}}( - 1)^{T} {\mathbf{v}}(x_{\omega } ,y_{\omega } ) = 0\;{\text{on}}\;\omega _{1} \;{\text{if}}\;1 \in D} \hfill \\ {{\mathbf{N}}( + 1)^{T} {\mathbf{v}}(x_{\omega } ,y_{\omega } ) = 0\;{\text{on}}\;\omega _{1} \;{\text{if}}\;{\text{2}} \in D} \hfill \\ {{\mathbf{Bv}} = 0\;{\text{on}}\;\gamma _{{i - 2}} \;{\text{if}}\;i \ge 3 \in D} \hfill \\ \end{array} } \right\} $$

(15)

Now consider the first term of Equation (8). Observe that:

$$ \nabla U = {\mathbf{J}}^{ - 1} \nabla^{*} U = {\mathbf{J}}^{ - 1} \left\{ {\begin{array}{*{20}c} {\frac{\partial U}{{\partial x_{\omega } }}} \\ {\frac{\partial U}{{\partial y_{\omega } }}} \\ {\frac{\partial U}{\partial s}} \\ \end{array} } \right\} = {\mathbf{J}}^{ - 1} \left\{ {\begin{array}{*{20}c} {{\mathbf{N}}^{T} {\mathbf{u}}_{{,x_{\omega } }} } \\ {{\mathbf{N}}^{T} {\mathbf{u}}_{{,y_{\omega } }} } \\ {\frac{2}{L}{\mathbf{N}}_{,\xi }^{T} {\mathbf{u}}} \\ \end{array} } \right\} $$

Substituting the above into Equation (8) results in:

$$ \int\limits_{\omega } {\left[ {\left\{ {\begin{array}{*{20}c} {{\mathbf{v}}_{{,x_{\omega } }}^{T} } & {{\mathbf{v}}_{{,y_{\omega } }}^{T} } & {{\mathbf{v}}_{{}}^{T} } \\ \end{array} } \right\}{\mathbf{C}}\left\{ {\begin{array}{*{20}c} {{\mathbf{u}}_{{,x_{\omega } }} } \\ {{\mathbf{u}}_{{,y_{\omega } }} } \\ {\mathbf{u}} \\ \end{array} } \right\}} \right]{\text{d}}\omega } $$

where

$$ \begin{gathered} \left[ {\mathbf{C}} \right]_{(3p,3p)} = \int\limits_{ - 1}^{1} {\frac{\left( L \right)}{2}\left[ {\mathbf{M}} \right]\left[ {{\mathbf{J}}^{ - T} c{\mathbf{J}}^{ - 1} } \right]\left[ {\mathbf{M}} \right]^{T} \left| {\mathbf{J}} \right|} {\text{d}}\xi \hfill \\ \left[ {\mathbf{M}} \right] = \left[ {\begin{array}{*{20}c} {\mathbf{N}} & 0 & 0 \\ 0 & {\mathbf{N}} & 0 \\ 0 & 0 & {\frac{2}{L}{\mathbf{N}}_{,\xi } } \\ \end{array} } \right]_{(3p,3)} \hfill \\ \end{gathered} $$

Similarly the right hand side of Equation (8) may be simplified to:

$$ \int\limits_{\omega } {{\mathbf{v}}^{T} {\mathbf{f}}d\omega } \quad {\text{where}}\quad {\mathbf{f}} = \int\limits_{{ - 1}}^{1} {\frac{L}{2}{\mathbf{N}}f\left| {\mathbf{J}} \right|} d\xi $$

Finally consider the boundary integrals. Direct substitution of the assumed functional form into the first term yields:

$$ \int\limits_{{\omega _{1} \;{\text{if}}\;1 \in N}} {\left( {{\mathbf{v}}^{T} {\mathbf{Q}}_{1} {\mathbf{u}} - {\mathbf{v}}^{T} {\mathbf{g}}_{1} } \right){\text{d}}\omega } \quad {\text{where}}\quad \begin{array}{*{20}c} {{\mathbf{Q}}_{1} = {\mathbf{N}}( - 1)q_{1} {\mathbf{N}}^{T} ( - 1)} \hfill \\ {{\mathbf{g}}_{1} = {\mathbf{N}}( - 1)g_{1} } \hfill \\ \end{array} $$

Similarly, the second term reduces to:

$$ \int\limits_{{\omega _{1} \;{\text{if}}\;2 \in N}} {\left( {{\mathbf{v}}^{T} {\mathbf{Q}}_{2} {\mathbf{u}} - {\mathbf{v}}^{T} {\mathbf{g}}_{2} } \right){\text{d}}\omega } \quad {\text{where}}\quad \begin{array}{*{20}c} {{\mathbf{Q}}_{2} = {\mathbf{N}}( + 1)q_{2} {\mathbf{N}}^{T} ( + 1)\lambda ^{2} (L)} \hfill \\ {{\mathbf{g}}_{2} = {\mathbf{N}}( + 1)g_{2} \lambda ^{2} (L)} \hfill \\ \end{array} $$

And the last term yields

$$ \int\limits_{{\gamma _{{i - 2}} \;{\text{if}}\;i \in N}} {\left( {{\mathbf{v}}^{T} {\mathbf{Q}}_{i} {\mathbf{u}} - {\mathbf{v}}^{T} {\mathbf{g}}_{i} } \right){\text{d}}\omega } \quad {\text{where}}\quad \begin{array}{*{20}c} {{\mathbf{Q}}_{i} = \int\limits_{0}^{L} {{\mathbf{N}}q_{i} {\mathbf{N}}^{T} \left| {{\mathbf{G}}_{i} } \right|{\text{d}}s} } \hfill \\ {{\mathbf{g}}_{i} = \int\limits_{0}^{L} {{\mathbf{N}}g_{i} \left| {{\mathbf{G}}_{i} } \right|{\text{d}}s} } \hfill \\ \end{array} $$

□

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Authors and Affiliations

Rakesh Vemulapally
- 1
Krishnan Suresh
- 1
Email author

1.Department of Mechanical EngineeringUniversity of WisconsinMadisonUSA

An analysis strategy for swept solids

Abstract

Keywords

Notes

Acknowledgments

References

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