Techniques for integration within Bezier domains and approximation of circular arcs

Abstract

This work presents techniques to perform integration of arbitrary functions within Bezier domains in two dimensions. Techniques that are considered for general nth order Bezier domains are the use of Green’s theorem and method of integration over parametrized surface. Derivation and steps to implement Green’s theorem for the integrations are provided. New expressions for calculation of integration of functions within quadratic and cubic Bezier domains are then derived based on the concept of bijective mapping, with following advantages:

• •Comparison with alternate method (integration over parametrized surface) for integration within a quadratic Bezier domain shows that the proposed formula is simpler and easier to use.
• •Method proposed for the cubic Bezier domains can be used to approximate arbitrary circular arcs and subsequently perform integration of functions within the circular arc domains.

The new methods are then implemented within the context of Extended Finite Element Method (X-FEM) and Strain-based Finite Element Method (SB-FEM) to solve problems with circular arc boundaries.

• •Simulation results show improvement in the solution accuracy.

Graphical abstract

Specifications table

Subject area	Engineering
More specific subject area	Computational Method
Name of your method	1. Integration within Bezier domains by using Green’s theorem, method of integration over parametrized surface and bijective mapping.
	2. Approximation of circular arcs by using bijective mapping of Bezier curve.
Name and reference of original method	Marsden, J.E., & Tromba, A. (2012). Vector calculus (6th ed., International ed.). W.H. Freeman, New York.
Resource availability	The methods are described in this paper.

Introduction

Integration of functions within a domain is particularly of interest within the context of Finite Element Analysis (FEA). FEA discretizes the problem domain by using finite number of elements and integration of functions within the individual elements is performed to obtain the element matrices, which are later merged to form global matrices for the entire domain. Among the factors that affect accuracy of the solution are the accuracy of the domain’s geometry representation (through discretization) and the technique used for the integration within the individual elements. Various discretization methods can be seen within different types of FEA. Each discretization method is facilitated with a suitable integration technique, which can be interchangeably used or merged with other techniques. Three most common discretization methods are the area discretization (used in Finite Element Method (FEM) and Strain-based FEM), boundary discretization (used in Boundary Element Method (BEM)) and uniform background mesh (used in Extended FEM (XFEM), Immersed Boundary Method (or CutFEM), and Finite Cell Method).

In case of area discretization, the entire active area of the domain is discretized by using finite number of elements that conform to the shape of the domain. Discretization of the inner area of the domain can be performed easily by using conventional linear finite elements (such as triangular or quadrilateral elements), but care must be taken when regions near the curved boundaries of the domain are to be discretised. Curvatures of the domain’s boundaries need to be represented correctly for better solution accuracy. Several techniques are available to approximate curvatures of the boundaries. Those are the use of adaptive mesh [1], isoparametric elements [2], higher order elements (HO) [[3], [4], [5]], isogeometric analysis [6], use of curved elements [7,8] and etc. Frequently, integration within the elements in FEM are carried out by first transforming the curved elements in actual problem domain (physical space) to a reference domain (parent space) by using suitable coordinate transformation (Jacobian transformation) followed by use of standard numerical integration schemes onto the reference domain (that involve integration points and weights), such as Newton-Cotes integration, Gauss-Legendre integration, classical Gaussian Quadrature, Generalized Gaussian Quadrature [9], trapezoidal quadrature [10], Double Exponential (DE) formula [11], etc. Use of adaptive integration techniques for arbitrary domains and to reduce computational time can be seen in [12].

Boundary discretization such as in BEM involves division of boundary of the problem domain into several sub elements for analysis, instead of partitioning entire area of the problem domain. Integration of functions is then carried out along the discretised boundaries. BEM produces integral equations that need to be treated for singularity. Analytical and numerical regularizations can be used to treat the singularity, which include analytical line integral, coordinate transformations followed by the use of Gaussian Quadrature, use of Taylor series expansion or power series expansion, radial integration method, semi-analytical approaches, integration by parts, hybrid approach and etc [13].

In case of uniform background mesh, the problem domain lies within the regular and structured grids that do not conform to the actual shape of the problem domain. The grids are categorised into three; grids that are cut/intersected by the domain’s boundaries, grids that are outside the problem domain and grids that represent voids within the problem domain. Integration within the first category requires special techniques to capture the curvature of the problem domain. Various techniques can be implemented, such as subdivision of the grid (that is intersected by the domain’s boundary) by using high number of linear elements, quadtree/octree meshes or by using polygonal elements followed by integration in parent space through mapping [14], use of equivalent functions [14], use of moment–fitting, local quasi-parametrization and Monte-Carlo techniques [15]. Fubini’s theorem can be used to perform the integration when equations of the boundary curves are known [16]. Another method employs homogeneous numerical integration scheme to convert area integration to boundary integration (dimension reduction) along the parametric curves (domain’s boundary) that cut the grids [17]. Use of Stokes or divergence theorem and Green’s formula for the dimension reduction can also be seen in the literature [12].

Some of the limitations found in the existing techniques are:

• 1.Some of the existing techniques rely on Jacobian transformation (to perform integration) which could lead to matrix singularities for high element distortions.
• 2.Uniform background meshing techniques involve high number of elements to approximate the irregular boundaries and use of computationally complex methods such as Monte Carlo.
• 3.Fubini’s theorem can be used to avoid the above-mentioned drawbacks. However, this method requires explicit equations of the boundary curves, which might not be available especially when the boundaries are represented by parametric curves.
• 4.Dimension reduction by using Stokes or divergence theorem and Green’s formula requires the integrals to be in a specific vector form, or in the form of $Mdx+Ndy$.

In this work, integration of arbitrary functions, f (x, y) within Bezier domains is studied and implemented in Finite Element Analysis (FEA). Three methods are considered. Those are the use of Green’s theorem, method of integration over parametrized surface and bijective mapping. The second method (method of integration over parametrized surface) is a standard existing method classified as Method 2 and described in the section: Method 2: Integration over parametrized surface. New approaches are then introduced in the implementation of these methods, which address limitations of the existing techniques. The modifications/new approaches are described below:

• 1.Green’s theorem is often used to calculate area of a domain and to solve line integrals that are in vector form, or in the form of $Mdx+Ndy$. However, implementation of Green’s theorem for integration of arbitrary functions f (x, y) within a Bezier domain is not readily available. In this work, function M is taken as zero and the Green’s theorem is implemented in the order $dxdy$ to perform integration of arbitrary functions within a Bezier domain. Steps for the implementation of the technique are provided. Limitations of this method due to selection of M = 0 are described and alternate approach to overcome these limitations (by letting N = 0) are provided in the section: Limitations.
• 2.Bijective mapping is used to form new integration formulas for quadratic Bezier domains that are simpler and easier to compute. Jacobian of the transformation is a constant. Three different approaches are presented, and numerical integrations are performed by transforming the integration limits with the use of original formula [18].
• 3.Method of integration over parametrized surface and bijective mapping are combined to form integration formulas for cubic Bezier domains. This combination is shown to be suitable for approximation of circular arcs and subsequently perform integration of arbitrary functions within the (Bezier) circular arc domain with the following advantages:

• -The new approach can be used to approximate circular arcs by using nth order Bezier curve.
• -Calculation of cross products is avoided during integration of the arbitrary functions.

These new techniques and formulas are applicable in various fields that require calculation of area or integration of functions within Bezier domains.

The methods

Method 1: Green’s theorem

Green’s theorem is given by [19]:

$$\int \int \left(\frac{\partial N\left(x,y\right)}{\partial x}-\frac{\partial M\left(x,y\right)}{\partial y}\right)d\Omega =\underset{C}{\oint }M\left(x,y\right)dx+N\left(x,y\right)dy$$ (1)

Let:

$$\int \int \left(\frac{\partial N\left(x,y\right)}{\partial x}-\frac{\partial M\left(x,y\right)}{\partial y}\right)d\Omega =\int \int f\left(x,y\right)d\Omega$$ (2)

where $f\left(x,y\right)$ is the arbitrary function to be integrated within the domain, $\Omega$. This function can be represented in differential form as:

$$f\left(x,y\right)=\frac{\partial V\left(x,y\right)}{\partial x}$$ (3)

and the integration in right hand side of Eq. (2) can be carried out according to the sequence dxdy:

$$\int \int f\left(x,y\right)d\Omega =\int \int \frac{\partial V\left(x,y\right)}{\partial x}dxdy$$ (4)

From Eq. (3):

$$V\left(x,y\right)=\int f\left(x,y\right)dx+C\left(y\right)$$ (5)

where $C\left(y\right)$ is the unknown arbitrary function that is taken as zero due to the following reasons: i) to obtain a single valid function and ii) the unknown arbitrary function $C\left(y\right)$ will cancel itself out when the integration is performed within the closed domain (closed integration bounds or closed path) [20]. Substituting Eq. (3) into Eq. (2) and direct comparison yields:

$$\frac{\partial N\left(x,y\right)}{\partial x}-\frac{\partial M\left(x,y\right)}{\partial y}=\frac{\partial V\left(x,y\right)}{\partial x}$$ (6)

In order to satisfy Eq. (6), let $N=V$ and $M=0.$ Eq. (1) can now be written as:

$$\int \int \left(\frac{\partial V\left(x,y\right)}{\partial x}\right)d\Omega =\underset{C}{\oint }V\left(x,y\right)dy$$ (7)

Substituting Eq. (7) into Eq. (4) gives:

$$\int \int f\left(x,y\right)d\Omega =\underset{C}{\oint }V\left(x,y\right)dy$$ (8)

Integration within a general nth order Bezier domain enclosed by k number of Bezier curves can then be carried out along the domain’s enclosed boundaries by the following substitutions into Eq. (8):

$$V\left(x,y\right)=V\left(B_{c,x}\left(t\right),B_{c,y}\left(t\right)\right)=V\left({\overrightarrow{B}}_c\left(t\right)\right)$$ (9)

$$dy=\left(\frac{d{B}_{c,y}\left(t\right)}{dt}\right)dt$$ (10)

that lead to:

$$\int \int f\left(x,y\right)d\Omega =\underset{C}{\oint }V\left({\overrightarrow{B}}_c\left(t\right)\right)\left(\frac{d{B}_{c,y}\left(t\right)}{dt}\right)dt=\sum _{i=1}^k\underset{0}{\overset{1}{\int }}V\left({\overrightarrow{B}}_{ci}\left(t\right)\right)\left(\frac{d{B}_{ci,y}\left(t\right)}{dt}\right)dt$$ (11)

where $B_{c,x}\left(t\right)$ and $B_{c,y}\left(t\right)$ are the parametric coordinates of a Bezier curve that is defined as [21]:

$${\overrightarrow{B}}_c\left(t\right)=\sum _{i=1}^n{\overrightarrow{P}}_i{B}_{i,n}\left(t\right),0<t<1$$ (12)

where

$${\overrightarrow{B}}_c\left(t\right)=\left\{B_{c,x}\left(t\right),B_{c,y}\left(t\right)\right\}$$

$$B_{i,n}\left(t\right)=\left(\begin{array}{c}n\\ i\end{array}\right)t^i{\left(1-t\right)}^{n-i},$$

$${\overrightarrow{P}}_i=\{P_{i,x},P_{i,y}\}$$

$$\left(\begin{array}{c}n\\ i\end{array}\right)=\frac{n!}{i!\left(n-i\right)!}$$

The term n represents degree of the Bezier curve (n = 3 for quadratic curve, n = 4 for cubic curve, etc.), ${\overrightarrow{P}}_i$ represents coordinates of the Bezier control points and $B_{i,n}\left(t\right)$ represents Bernstein basis polynomials. Steps to implement Green’s theorem for integration of arbitrary function $f\left(x,y\right)$ within a Bezier domain are:

• 1.Get $V\left(x,y\right)$ by symbolic integration of $f\left(x,y\right)$ with respect to x according to Eq. (5). Let $C\left(y\right)=0.$
• 2.Obtain ${\overrightarrow{B}}_c\left(t\right)=\left\{B_{c,x}\left(t\right),B_{c,y}\left(t\right)\right\}$ for the curve that encloses the domain by using Eq. (12).
• 3.Differentiate $B_{c,y}\left(t\right)$ with respect to t to get $\frac{d{B}_{c,y}\left(t\right)}{dt}$.
• 4.Obtain $V\left({\overrightarrow{B}}_c\left(t\right)\right)$ by substituting ${\overrightarrow{B}}_c\left(t\right)$ from step 2 into $V\left(x,y\right)$ obtained from step 1, according to Eq. (9).
• 5.Multiply $V\left({\overrightarrow{B}}_c\left(t\right)\right)$ from step 4 with $\frac{d{B}_{c,y}\left(t\right)}{dt}$ from step 3.
• 6.Perform integration with respect to t with interval of 0 to 1 onto the resultant equation from step 5. This integration can either be solved analytically or numerically.
• 7.Repeat steps 2-6 for each Bezier curve that encloses the domain and sum up the results of the integration (from step 6) for all the curves.

Method 2: Integration over parametrized surface

Integration of functions within a Bezier domain (domain enclosed by the Bezier curve and a line connecting the start and end points of the curve) can be performed by following two steps. The first step is to generate a surface equation for the Bezier domain in two parametric dimensions u and v by using the extended version of Eq. (12) as shown below [19]:

$${\overrightarrow{B}}_s\left(u,v\right)=\sum _{i=0}^m\sum _{j=0}^n{\overrightarrow{P}}_{i,j}{B}_{i,m}\left(u\right)B_{j,n}\left(v\right),0<u<1,0<v<1$$ (13)

where

$$B_{i,m}\left(u\right)=\left(\begin{array}{c}m\\ i\end{array}\right)u^i{\left(1-u\right)}^{m-i},$$

$$B_{j,n}\left(v\right)=\left(\begin{array}{c}n\\ j\end{array}\right)v^j{\left(1-v\right)}^{n-j},$$

$$\left(\begin{array}{c}m\\ i\end{array}\right)=\frac{m!}{i!\left(m-i\right)!},$$

$$\left(\begin{array}{c}n\\ j\end{array}\right)=\frac{n!}{j!\left(n-j\right)!}$$

$${\overrightarrow{P}}_{i,j}=\{P_{i,j,x},P_{i,j,y}\}$$

$${\overrightarrow{B}}_s\left(u,v\right)=\left\{B_{s,x}\left(u,v\right),B_{s,y}\left(u,v\right)\right\}$$

${\overrightarrow{P}}_{i,j}$ represents coordinates of the Bezier surface control points while n and m represent degrees of the surface along u and v directions. Second step is to perform the surface integration by using the following equation involving cross product [22]:

$$\int \underset{s}{\int }f\left(x,y\right)ds=\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}f\left({\overrightarrow{B}}_s\left(u,v\right)\right)\parallel {\overrightarrow{B}}_{s,u}\times {\overrightarrow{B}}_{s,v}\parallel dudv$$ (14)

where

$${\overrightarrow{B}}_{s,u}=\left\{\frac{\partial {\overrightarrow{B}}_s\left(u,v\right)}{\partial u},0\right\},$$

$${\overrightarrow{B}}_{s,v}=\left\{\frac{\partial {\overrightarrow{B}}_s\left(u,v\right)}{\partial v},0\right\}$$

Method 3: Bijective mapping

Formula for integration of functions within a Bezier domain can be derived by using proposition below:

Let P_i be the Bezier control points in the physical coordinate system corresponding to predetermined p_i in reference coordinate system. Then, bijective mapping of P_i to p_i will generate unique curve and plane in the reference coordinate system.

Integrations and other operations can then be performed onto the unique curve and plane in the reference coordinate system, hence simplifies the calculation.

Quadratic Bezier domain

Since Bezier curve lies within convex hull of the control points, then the parametrized Bezier curve in physical coordinate system (x and y) can be traced within the unit triangle in reference coordinate system (r and s) by using mapping (Fig. 1) given by the following equation [23]:

$$r=\frac{1}{\left(2*Area\right)}\left[\left(P_{3,y}-P_{1,y}\right)\left(x-P_{1,x}\right)-\left(P_{3,x}-P_{1,x}\right)\left(y-P_{1,y}\right)\right]$$ (15a)

$$s=\frac{1}{\left(2*Area\right)}\left[\left(P_{1,y}-P_{2,y}\right)\left(x-P_{1,x}\right)+\left(P_{2,x}-P_{1,x}\right)\left(y-P_{1,y}\right)\right]$$ (15b)

Fig. 1.

Mapping of a quadratic Bezier curve (a) random Bezier curve in physical coordinate system with convex hull that connects the control points $P_i$: $P_1,P_2$ and $P_3$(b) unit triangle in the reference coordinate system with the unique Bezier curve associated with points $p_i:$$p_1,p_2$ and $p_3$.

$P_{i,x}$ and $P_{i,y}$ represent x- and y- coordinates respectively, of the Bezier control points in physical coordinate system. Area represents area of the convex hull of the Bezier control points in physical coordinate system that can be calculated by using the formula:

$$Area=\left|\frac{1}{2}\left[\left(P_{2,x}-P_{1,x}\right)\left(P_{3,y}-P_{2,y}\right)-\left(P_{2,y}-P_{1,y}\right)\left(P_{3,x}-P_{2,x}\right)\right]\right|$$ (16)

Eq. (15) represents inverse transformation by using shape functions of a 3-nodes triangular element. Prove of bijectivity of the original transformation (reference coordinate system to physical coordinate system) can be found in [24]. Since the original transformation is bijective, therefore the inverse transformation given by Eq. (15) is also bijective. Closed form integration formula for a quadratic Bezier domain is derived by referring to the area enclosed by the mapped Bezier curve (Fig. 1 (b)) and the line that connects the first and last control points (base of the unit triangle). Parametric equation of the unique Bezier curve in the reference coordinate system can be obtained through substitution of Eq. (12) (with n = 3) into Eq. (15). Parameter t can be eliminated by solving Eq. (15a) for t and then substituting the resultant solution into Eq. (15b). Equation of the curve that results is:

$$s=2\sqrt{r}-2r$$ (17)

Integration formula for a quadratic Bezier domain, ${\Omega }^2$ is then given by:

$$\int \underset{{\Omega }^2}{\int }f\left(x,y\right)d\Omega =\left(2*Area\right)*\underset{0}{\overset{1}{\int }}\underset{0}{\overset{\left(2\sqrt{r}-2r\right)}{\int }}f\left(\overrightarrow{P}\right)dsdr$$ (18)

where

$$\overrightarrow{P}=\sum _{i=1}^j{N}_i{\overrightarrow{P}}_i$$ (19)

$$j=3$$

$$N_1=1-r-s$$

$$N_2=r$$

$$N_3=s$$

$${\overrightarrow{P}}_i=\{P_{i,x},P_{i,y}\}$$

$$\overrightarrow{P}=\{P_x,P_y\}$$

$P_x$ and $P_y$ represent x- and y- coordinates respectively, of an arbitrary point within convex hull of the Bezier curve in physical coordinate system. $P_{i,x}$ and $P_{i,y}$ represent x- and y- coordinates respectively, of the Bezier control points in physical coordinate system. $N_1$, $N_2$ and $N_3$ are the shape functions corresponding to the three nodes of unit triangle in the reference coordinate system. Mapping the start and end points of the Bezier curve in physical coordinate system to nodes 1 ($N_1$) and 3 ($N_3$) of the reference triangle (Fig. 1(b)) respectively, will yield the following formula for the integration:

$$\underset{{\Omega }^2}{\int }f\left(x,y\right)d\Omega =\left(2*Area\right)*\left(\underset{0}{\overset{0.5}{\int }}\underset{\frac{1}{4}{\left(-1+\sqrt{1-2r}\right)}^2}{\overset{0.25}{\int }}f\left(\overrightarrow{P}\right)\mathrm{d}s\mathrm{d}r+\underset{0}{\overset{0.5}{\int }}\underset{0.25}{\overset{\frac{1}{4}{\left(1+\sqrt{1-2r}\right)}^2}{\int }}f\left(\overrightarrow{P}\right)\mathrm{d}s\mathrm{d}r\right)$$ (20)

Plane equation for the unique curve in the reference coordinate system (Fig. 1(b)) and norm of the cross product are:

$${\overrightarrow{B}}_s\left(u,v\right)=\left[2\left(u-1\right)\left(v-1\right)v\right]\widehat{i}+\left[v\left(u\left(1-v\right)+v\right)\right]\widehat{j}+0\widehat{k}$$ (21)

$$\parallel \stackrel{⟶}{B_{s,u}}\times \stackrel{⟶}{B_{s,v}}\parallel =2v\left(1-v\right)$$ (22)

The reference coordinate system (r and s) can be converted to the two parametric dimensions (u and v) by using the following substitutions that are taken from Eq. (21):

$$r=2\left(u-1\right)\left(v-1\right)v$$ (23a)

$$s=v\left(u\left(1-v\right)+v\right)$$ (23b)

The physical coordinates x and y can be replaced with $P_x$ and $P_y$ (Eq. (19) respectively, which are now in terms of u and v (due to the conversion given by Eq. (23)):

$$x=P_x\left(u,v\right)$$ (24a)

$$y=P_y\left(u,v\right)$$ (24b)

Integration within the quadratic Bezier domain is then given by (merged with Method 2):

$$\int \underset{{\Omega }^2}{\int }f\left(x,y\right)d\Omega =(2*Area)*\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}\left[f\left(P_x\left(u,v\right),P_y\left(u,v\right)\right)*\left(2v\left(1-v\right)\right)\right]dudv$$ (25)

Numerical integrations can be carried out by transforming the integration limits according to the quadrature rule of choice. Transformation of the integration limits to [0,1] would enable use of generalized gaussian quadrature which permits integration of wider class of functions. Transformation of the integration limits can be accomplished by using the formula [18]:

$$\begin{array}{c}{I}_1=\underset{a}{\overset{b}{\int }}\underset{r\left(x\right)}{\overset{s\left(x\right)}{\int }}f\left(x,y\right)dydx\\ or\\ I_2=\underset{a}{\overset{b}{\int }}\underset{r\left(y\right)}{\overset{s\left(y\right)}{\int }}f\left(x,y\right)dxdy\end{array}\}=\underset{L}{\overset{U}{\int }}\underset{L}{\overset{U}{\int }}f\left(m_{x}u+c_x,m_{y}v+c_y\right)m_x{m}_{y}dvdu=\sum _j^n\sum _i^n{w}_i{w}_j{m}_x{m}_{y}f\left(m_x{u}_i+c_x,m_y{v}_i+c_y\right)$$

$$where$$

$$Uisupperlimit$$

$$Lislowerlimit$$

$$w_{i}and{w}_{j}areintegrationweights$$

$$u_{i}and{v}_{j}areintegrationpoints$$

$$i=1,2,3,\dots ,n$$

$$nisintegrationorder$$

$for{I}_1:$

$$\begin{array}{cc}\hfill & m_x=\frac{a-b}{L-U};m_y=\frac{r\left(m_{x}u+c_x\right)-s\left(m_{x}u+c_x\right)}{L-U};\\ \hfill & c_x=\frac{\left(b\times L\right)-\left(a\times U\right)}{L-U};c_y=\frac{\left(s\left(m_{x}u+c_x\right)\times L\right)-\left(r\left(m_{x}u+c_x\right)\times U\right)}{L-U}\end{array}$$

$for{I}_2:$

$$\begin{array}{cc}\hfill & m_x=\frac{r\left(m_{y}v+c_y\right)-s\left(m_{y}v+c_y\right)}{L-U};m_y=\frac{a-b}{L-U};\\ \hfill & c_x=\frac{\left(s\left(m_{y}v+c_y\right)\times L\right)-\left(r\left(m_{y}v+c_y\right)\times U\right)}{L-U};c_y=\frac{\left(b\times L\right)-\left(a\times U\right)}{L-U};\end{array}$$ (26)

Cubic Bezier domain

Convex hull of a Bezier curve with four control points will be either a triangular or a quadrilateral as shown in Fig. 2. In the case of triangular convex hull (Fig. 2(a) and 2(b)), the previous bijective mapping can be implemented. However, for the latter case, the convex hull needs to be converted to a triangular domain by extending two non-parallel sides of the quadrilaterals to form a triangular domain that encloses the Bezier curve. Extension of the sides of the quadrilaterals to form triangular domain is shown in Fig. 2 (c-e). Intersection of the extended lines is denoted as V. This intersecting point V is mapped to the origin of the reference element (with coordinates (0,0)) and the rest of the points are mapped in counterclockwise direction. Two of the Bezier points are mapped onto the edges as shown in Fig. 3. Eq. (13) can be implemented to generate plane equation for the mapped curve in the reference element followed by the use of Eq. (14) to perform the integration. Fig. 4 shows flow chart for the case-based integration of cubic Bezier domains with four control points.

Fig. 2.

Examples of convex hulls for Bezier curves with four control points (P₁, P₂, P₃ and P₄). (a, b) triangular convex hulls with one internal control point. (c-e) triangulation of rectangular convex hulls through formation of vertex V.

Fig. 3.

Mapping of a cubic Bezier curve with vertex V. (a) random Bezier curve in physical coordinate system with convex hull that connects the control points $P_i$: $P_1,P_2,$$P_3$ and $P_4$ (b) unit triangle in the reference coordinate system with the mapped Bezier curve associated with points $p_i:$$p_1,p_2$, $p_3$ and $p_4$. $p_2$ and $p_3$ act as edge nodes.

Fig. 4.

Flow chart for the case-based integration of cubic Bezier domains with four control points.

The concept of formation of triangular convex hull with the creation of virtual vertex, V for the cubic Bezier curves can be used to generate Bezier control points for arbitrary circular arcs, as detailed below.

Approximation of circular arcs

Bezier control points for approximation of quarter of a unit circle with high accuracy are available in [25] and shown in Fig. 5(a). Bezier control points are labelled as P_i (i = 1, 2, 3, 4) and midpoint of the quarter circle is labelled as M. Convex hull of the Bezier control points for this case is a quadrilateral and therefore the two non-parallel sides of the quadrilateral (P₁ P₂ and P₄ P₃) are extended to form a triangular domain for the purpose of mapping. Intersection of the extended lines is denoted as V and control points P₁, P₄ and vertex V act as corner nodes of the triangle. Coordinates (in physical x-y coordinate system) of the points shown in Fig. 5(a) are obtained as:

$$P_1=\left(0,1\right)$$

$$P_2=\left(0.5534,0.99871\right)$$

$$P_3=\left(0.99871,0.5534\right)$$

$$P_4=\left(1,0\right)$$

$$M=\left(\sqrt{2}/2,\sqrt{2}/2\right)$$ (27)

$$V=\left(V_x,V_y\right)=\left(0.9977,0.9977\right)$$

Fig. 5.

Approximation of arbitrary circular arc through bijective mapping (a) Bezier approximation of quarter of a unit circle (b) Mapping of quarter of circle to the reference triangle (c) Approximation of arbitrary circular arc by mapping of the reference triangle.

These control points are mapped to a unit reference triangle (in reference r-s coordinate system), as shown in Fig. 5(b). Vertex V is mapped to the origin of the reference element as v (with coordinates (0,0)) and the rest of the control points are mapped in counterclockwise direction as p_i (i = 1, 2, 3, 4). Coordinates of midpoint m, p₂ and p₃ are obtained by using Eq. (15):

$$r=\frac{\left(V_y-V_{y}x-y+V_{x}y\right)}{\left(V_x+V_y-1\right)}$$ (28a)

$$s=\frac{\left(V_x+V_{y}x-x-V_{x}y\right)}{\left(V_x+V_y-1\right)}$$ (28b)

x and y are the coordinates of the points in physical coordinate system (Fig. 5(a)). A random circular arc can be approximated by a cubic Bezier curve through bijective mapping of the control points in reference coordinate system as shown in Fig. 5(c). The mapping can be achieved by using Eq. (19):

$$P=N_{1}V+N_2{P}_1+N_3{P}_4$$ (29)

P is an arbitrary point on the circular arc domain in physical coordinate system. $P_1$ and $P_4$ are the start and end points of the arbitrary circular arc in physical coordinate system. Bezier control points P₂ and P₃ for the arbitrary circular arc can be determined once location of vertex V for the arbitrary arc has been identified. This process requires start and end points of the arbitrary circular arc ($P_1$ and $P_4$) and arc’s midpoint coordinates (P).

Location of V for the arc in physical coordinate system is then determined by substituting location of the arc’s midpoint in reference coordinate system (obtain r and s coordinates for the arc’s midpoint from Eq. (28) by substituting x and y values with coordinates of M given in Eq. (27)) and in physical coordinate system (P) into Eq. (29) and solving for V. P₂ and P₃ can then be determined by substituting the V and corresponding r and s values (for P₂ and P₃, respectively) into Eq. (29). r and s values for P₂ and P₃ can be respectively obtained by substituting x and y values with coordinates of P₂ and P₃ given in Eq. (27). The general equation for approximation of arbitrary circular arc by using cubic Bezier curve through mapping as described in this work is given as:

$$P_1=Arcstartpoint$$

$$P_2=\left(1.3330826\times ArcMidpoint\right)+\left(0.0560985P_1\right)-\left(0.3891810P_4\right)$$

$$P_3=\left(1.3330826\times ArcMidpoint\right)-\left(0.3891810P_1\right)+\left(0.0560985P_4\right)$$ (30)

$$P_4=Arcendpoint$$

Once Bezier control points for an arbitrary circular arc are found, integration within the domain enclosed by the circular arc can be carried out by using the Eq. (14), with the following substitutions:

$$\overrightarrow{B_s}\left(u,v\right)=\left[1.+\left(-1.6642+0.1642u\right)v+\left(0.3283+1.1717u\right)v^2+\left(0.3358-1.3358u\right)v^3\right]\widehat{i}+\left[1.3358v^2-0.3358v^3+uv\left(1.5-2.8358v+1.3358v^2\right)\right]\widehat{j}$$ (31)

$$\parallel {\overrightarrow{B}}_{s,u}\times {\overrightarrow{B}}_{s,v}\parallel =2.4962v+\left(-5.2657-2.2231u\right)$$ (32)

This technique can be used with general nth order Bezier, Non-Uniform Rational B-Splines (NURBS) or any other established Bezier curves for the approximation of the circular arc by simply extending two edges of the two non-parallel sides of the convex hull. However, complexity of the integrand will differ, according to the type of approximation used. In case of approximation by using NURBS, the integrand will consist of rational functions which require special or more complex integration techniques. Accuracy of approximation of the circular arc might be higher when NURBS is used, but the integration of rational functions within the domain needs to be handled with care. Furthermore, the final simulation results for some cases might not show significant improvement when NURBS is used, since the current technique has shown very good results (for instance, smallest error of 0.05% for example 2 in the section: Example 2: analysis of thick curved beam).

Method validation

Validation of methods 1 and 2

Integration of several functions within a domain enclosed by two curves as shown in Fig. 6 is carried out in Mathematica by using Methods 1 and 2. Bezier curves of orders 2, 3 and 4 are used to approximate the curves. Control points for the 2^nd order Bezier curves are:

$$P_1=\left\{0,0\right\};P_2=\left\{0.02485,0\right\};P_3=\left\{0.0497,0.1235\right\}$$

for curve 1 and followings for curve 2:

$$P_1=\left\{0.0497,0.1235\right\};P_2=\left\{0,0.111467\right\};P_3=\left\{0,0\right\}$$

Fig. 6.

A domain enclosed by Bezier curves.

Control points for the higher order approximations are obtained through degree elevation of the 2^nd order control points. Known equations of the curves are used to determine the reference solution given by:

$$\underset{0}{\overset{0.0497}{\int }}\underset{\left(50x^2\right)}{\overset{\left(\sqrt{x}-2x\right)}{\int }}f\left(x,y\right)dydx$$ (33)

Table 1 shows the simulation results that are obtained by using symbolic computation in Mathematica. Percentage of error is obtained by using the following equation:

$$Percentageerror(\%)=\left(\frac{Referencesolution-Simulationresult}{Referencesolution}\right)\times 100\%$$ (34)

Table 1.

Simulation results (symbolic computations) by using Methods 1 and 2 for the Bezier domain in Fig. 6.

Integrand $f\left(x,y\right)$	Reference solution	Simulation result (percentage error, %)
		Method 1: Green’s Theorem				Method 2: Integration over parametric surface
		Order of Bezier domain				Order of Bezier domain
		2nd order	3rd order	4th order	2nd order		3rd order	4th order
$2xy+\left(x^2{y}^2\right)$	$9.478313\times {10}^{-6}$	$9.474038\times {10}^{-6}$ ($4.51\times {10}^{-2}$)	$9.474038\times {10}^{-6}$ ($4.51\times {10}^{-2}$)	$9.474038\times {10}^{-6}$ ($4.51\times {10}^{-2}$)	$9.474038\times {10}^{-6}$ ($4.51\times {10}^{-2}$)		$9.474038\times {10}^{-6}$ ($4.51\times {10}^{-2}$)	$9.474038\times {10}^{-6}$ ($4.51\times {10}^{-2}$)
$\sqrt{1+x}$	$2.901296\times {10}^{-3}$	$2.900735\times {10}^{-3}$ ($1.94\times {10}^{-2}$)	-	-	-		$2.900735\times {10}^{-3}$ ($1.94\times {10}^{-2}$)	$2.900734\times {10}^{-3}$ ($1.94\times {10}^{-2}$)
$e^{\left(1+x\right)}$	$7.973990\times {10}^{-3}$	$7.972443\times {10}^{-3}$ ($1.94\times {10}^{-2}$)	$7.972443\times {10}^{-3}$ ($1.94\times {10}^{-2}$)	-	-		$7.972443\times {10}^{-3}$ ($1.94\times {10}^{-2}$)	-
$1+\frac{x}{y}$	$3.973870\times {10}^{-3}$	$3.973291\times {10}^{-3}$ ($1.46\times {10}^{-2}$)	$3.973291\times {10}^{-3}$ ($1.46\times {10}^{-2}$)	-	-		-	-
$\left(\ln y\right)\left(x+y\right)$	$-6.464878\times {10}^{-4}$	$-6.463095\times {10}^{-4}$ ($2.76\times {10}^{-2}$)	$-$	-	-		-	-

It is seen that Green’s theorem is able to provide solutions with good accuracy for all the integrands under 2^nd order approximation, while method of integration over parametric surface failed to provide solutions for some of the cases within the same order of approximation. Both methods are found to be unable to generate solutions for some of the integrands under 3^rd and 4^th order approximations. These are caused by complexity of the integrands that resulted after plane transformation and due to highly demanding symbolic manipulations. Results in Table 1 show that the integration error is independent of order of the Bezier curve, since it employs analytical approach through symbolic manipulation. Results for all the integrands can be obtained by implementing numerical integration technique, given by Eq. (26). Table 2 shows simulation results by using the semi-analytical and numerical integration techniques (generalized gaussian quadrature are implemented). Unlike the results in Table 1, results in Table 2 are now dependent on order of integration, due to the numerical approach. Fig. 7 shows convergence of the simulation results (for first and last integrands from Table 2) towards the exact solution with increasing integration order. Initial error is higher for higher-order curves due to the presence of higher-degree terms in the equations. This initial error is then reduced dramatically when integration order is increased.

Table 2.

Simulation results by using semi-analytical and numerical integration techniques for the Bezier domain in Fig. 6.

Integrand$f\left(x,y\right)$	Number of integration points and weights, n	Simulation result (percentage error, %)
		Method 1: Green’s Theorem (semi analytical)	Method 2: Integration over parametric surface (numerical)
		Order of Bezier domain	Order of Bezier domain
		2nd , 3rd and 4th orders	2nd order	3rd order	4th order
$2xy+\left(x^2{y}^2\right)$	n = 5	$9.472069\times {10}^{-6}$ ($6.59\times {10}^{-2}$)	$9.473137\times {10}^{-6}$ ($5.46\times {10}^{-2}$)	$9.505534\times {10}^{-6}$ ($2.87\times {10}^{-1}$)	$9.554411\times {10}^{-6}$ ($8.03\times {10}^{-1}$)
	n = 10	$9.474038\times {10}^{-6}$ ($4.51\times {10}^{-2}$)	$9.474038\times {10}^{-6}$$\left(4.51\times {10}^{-2}\right)$	$9.474038\times {10}^{-6}$$\left(4.51\times {10}^{-2}\right)$	$9.474038\times {10}^{-6}$$\left(4.51\times {10}^{-2}\right)$
	n = 15	$\text{same}\text{as}\text{above}$	$\text{same}\text{as}\text{above}$	$\text{same}\text{as}\text{above}$	$\text{same}\text{as}\text{above}$
	n = 20	$\text{same}\text{as}\text{above}$	$\text{same}\text{as}\text{above}$	$\text{same}\text{as}\text{above}$	$\text{same}\text{as}\text{above}$
$\sqrt{1+x}$	n = 5	$2.900735\times {10}^{-3}$ ($1.93\times {10}^{-2}$)	$2.900735\times {10}^{-3}$ ($1.94\times {10}^{-2}$)	$2.900752\times {10}^{-3}$ ($1.88\times {10}^{-2}$)	$2.900784\times {10}^{-3}$ ($1.76\times {10}^{-2}$)
	n = 10	$\text{same}\text{as}\text{above}$	$\text{same}\text{as}\text{above}$	$2.900735\times {10}^{-3}$ ($1.94\times {10}^{-2}$)	$2.900735\times {10}^{-3}$ ($1.94\times {10}^{-2}$)
	n = 15	$\text{same}\text{as}\text{above}$	$\text{same}\text{as}\text{above}$	$\text{same}\text{as}\text{above}$	$\text{same}\text{as}\text{above}$
	n = 20	$\text{same}\text{as}\text{above}$	$\text{same}\text{as}\text{above}$	$\text{same}\text{as}\text{above}$	$\text{same}\text{as}\text{above}$
$e^{\left(1+x\right)}$	n = 5	$7.972446\times {10}^{-3}$ ($1.94\times {10}^{-2}$)	$7.972444\times {10}^{-3}$ ($1.94\times {10}^{-2}$)	$7.972562\times {10}^{-3}$ ($1.79\times {10}^{-2}$)	$7.972762\times {10}^{-3}$ ($1.54\times {10}^{-2}$)
	n = 10	$7.972443\times {10}^{-3}$ ($1.94\times {10}^{-2}$)	$7.972443\times {10}^{-3}$ ($1.94\times {10}^{-2}$)	$7.972443\times {10}^{-3}$ ($1.94\times {10}^{-2}$)	$7.972443\times {10}^{-3}$ ($1.94\times {10}^{-2}$)
	n = 15	$\text{same}\text{as}\text{above}$	$\text{same}\text{as}\text{above}$	$\text{same}\text{as}\text{above}$	$\text{same}\text{as}\text{above}$
	n = 20	$\text{same}\text{as}\text{above}$	$\text{same}\text{as}\text{above}$	$\text{same}\text{as}\text{above}$	$\text{same}\text{as}\text{above}$
$1+\frac{x}{y}$	n = 5	$3.973288\times {10}^{-3}$ ($1.46\times {10}^{-2}$)	$3.973307\times {10}^{-3}$ ($1.42\times {10}^{-2}$)	$3.976200\times {10}^{-3}$ ($5.86\times {10}^{-2}$)	$3.976351\times {10}^{-3}$ ($6.24\times {10}^{-2}$)
	n = 10	$3.973291\times {10}^{-3}$ ($1.46\times {10}^{-2}$)	$3.973291\times {10}^{-3}$ ($1.46\times {10}^{-2}$)	$3.973488\times {10}^{-3}$ ($0.96\times {10}^{-2}$)	$3.973490\times {10}^{-3}$ ($0.96\times {10}^{-2}$)
	n = 15	$\text{same}\text{as}\text{above}$	$\text{same}\text{as}\text{above}$	$3.973332\times {10}^{-3}$ ($1.35\times {10}^{-2}$)	$3.973332\times {10}^{-3}$ ($1.35\times {10}^{-2}$)
	n = 20	$\text{same}\text{as}\text{above}$	$\text{same}\text{as}\text{above}$	$3.973304\times {10}^{-3}$ ($1.42\times {10}^{-2}$)	$3.973304\times {10}^{-3}$ ($1.42\times {10}^{-2}$)
$\left(\ln y\right)\left(x+y\right)$	n = 5	$-6.462024\times {10}^{-4}$ ($4.41\times {10}^{-2}$)	$-6.463106\times {10}^{-4}$ ($2.74\times {10}^{-2}$)	$-6.460463\times {10}^{-4}$ ($6.83\times {10}^{-2}$)	$-6.459156\times {10}^{-4}$ ($8.85\times {10}^{-2}$)
	n = 10	$-6.463092\times {10}^{-4}$ ($2.76\times {10}^{-2}$)	$-6.463095\times {10}^{-4}$ ($2.76\times {10}^{-2}$)	$-6.463063\times {10}^{-4}$ ($2.81\times {10}^{-2}$)	$-6.463036\times {10}^{-4}$ ($2.85\times {10}^{-2}$)
	n = 15	$-6.463095\times {10}^{-4}$ ($2.76\times {10}^{-2}$)	$\text{same}\text{as}\text{above}$	$-6.463092\times {10}^{-4}$ ($2.76\times {10}^{-2}$)	$-6.463090\times {10}^{-4}$ ($2.77\times {10}^{-2}$)
	n = 20	$\text{same}\text{as}\text{above}$	$\text{same}\text{as}\text{above}$	$-6.463094\times {10}^{-4}$ ($2.76\times {10}^{-2}$)	$-6.463094\times {10}^{-4}$ ($2.76\times {10}^{-2}$)

Fig. 7.

Convergence of simulation results from Table 2: (a) convergence of the first integrand: $2xy+\left(x^2{y}^2\right)$ and (b) convergence of the last integrand: $\left(\ln y\right)\left(x+y\right)$.

Validation of method 3

Quadratic Bezier domain

Simulation results for quadratic Bezier domain by using Method 3 are obtained by considering the domain in Fig. 6 with second order approximation for the curves. Results for all the three approaches described in the section: Quadratic Bezier domain (under Method 3: Bijective mapping) are obtained and shown in Table 3 for symbolic computation and in Table 4 for numerical computation. It is seen that simulation results using symbolic computations are not available for some of the integrands when approaches 2 and 3 are used (Table 3). Unlike symbolic computation, the numerical integration technique is able to produce results for all the integrands by using all the three approaches. Approach 1 shows continual convergence when integration order is increased, for all the integrands. Similar continual convergence is not seen in the other approaches. Approach 2 produces the smallest error among all the methods for all integrands, except for the function with square root (second integrand in Table 4). Approach 3 in Table 4 and Method 2 in Table 2 produce similar results. Fig. 8 shows convergence of the simulation results (for first and last integrands from Table 4) towards the exact solution with increasing integration order for all three approaches. Convergence of approaches 1 and 3 is identical (Fig. 8(a)) or nearly the same (Fig. 8(b)).

Table 3.

Simulation results (symbolic computations) by using Method 3 for the Bezier domain in Fig. 6 with quadratic approximations for the curves.

Integrand $f\left(x,y\right)$	Reference solution	Simulation result (percentage error, %)
		Approach 1: Using Eq. (18)	Approach 2: Using Eq. (20)	Approach 3: Using Eq. (25)
$2xy+\left(x^2{y}^2\right)$	$9.478313\times {10}^{-6}$	$9.474038\times {10}^{-6}$ ($4.51\times {10}^{-2}$)	$9.474038\times {10}^{-6}$ ($4.51\times {10}^{-2}$)	$9.474038\times {10}^{-6}$ ($4.51\times {10}^{-2}$)
$\sqrt{1+x}$	$2.901296\times {10}^{-3}$	$2.900455\times {10}^{-3}$ ($2.90\times {10}^{-2}$)	$2.902102\times {10}^{-3}$ ($2.78\times {10}^{-2}$)	-
$e^{\left(1+x\right)}$	$7.973990\times {10}^{-3}$	$7.972443\times {10}^{-3}$ ($1.94\times {10}^{-2}$)	$7.972443\times {10}^{-3}$ ($1.94\times {10}^{-2}$)	-
$1+\frac{x}{y}$	$3.973870\times {10}^{-3}$	$3.973291\times {10}^{-3}$ ($1.46\times {10}^{-2}$)	-	-
$\left(\ln y\right)\left(x+y\right)$	$-6.464878\times {10}^{-4}$	$-6.463095\times {10}^{-4}$ ($2.76\times {10}^{-2}$)	-	-

Table 4.

Simulation results (numerical integrations) by using Method 3 for the Bezier domain in Fig. 6 with quadratic approximations for the curves.

Integrand $f\left(x,y\right)$	Simulation results (percentage error, %)
	Number of integration points and weights, n	Approach 1: Using Eqs. (18) and (26)	Approach 2: Using Eqs. (20) and (26)	Approach 3: Using Eqs. (25) and (26)
$2xy+\left(x^2{y}^2\right)$	n = 5	$9.473038\times {10}^{-6}$($5.57\times {10}^{-2}$)	$9.486441\times {10}^{-6}$($8.58\times {10}^{-2}$)	$9.473028\times {10}^{-6}$($5.57\times {10}^{-2}$)
	n = 10	$9.474016\times {10}^{-6}$$\left(4.53\times {10}^{-2}\right)$	$9.475978\times {10}^{-6}$($2.46\times {10}^{-2}$)	$9.474038\times {10}^{-6}$$\left(4.51\times {10}^{-2}\right)$
	n = 15	$9.474035\times {10}^{-6}$($4.51\times {10}^{-2}$)	$9.474656\times {10}^{-6}$($3.86\times {10}^{-2}$)	$9.474038\times {10}^{-6}$($4.51\times {10}^{-2}$)
	n = 20	$9.474037\times {10}^{-6}$($4.51\times {10}^{-2}$)	$9.474309\times {10}^{-6}$($4.22\times {10}^{-2}$)	$9.474038\times {10}^{-6}$($4.51\times {10}^{-2}$)
$\sqrt{1+x}$	n = 5	$2.900510\times {10}^{-3}$($2.71\times {10}^{-2}$)	$2.898598\times {10}^{-3}$($9.30\times {10}^{-2}$)	$2.900735\times {10}^{-3}$($1.94\times {10}^{-2}$)
	n = 10	$2.900730\times {10}^{-3}$($1.95\times {10}^{-2}$)	$2.892610\times {10}^{-3}$($2.99\times {10}^{-1}$)	$2.900735\times {10}^{-3}$($1.94\times {10}^{-2}$)
	n = 15	$2.900734\times {10}^{-3}$($1.94\times {10}^{-2}$)	$2.891900\times {10}^{-3}$($3.24\times {10}^{-1}$)	$2.900735\times {10}^{-3}$($1.94\times {10}^{-2}$)
	n = 20	$2.900734\times {10}^{-3}$($1.94\times {10}^{-2}$)	$2.891716\times {10}^{-3}$($3.30\times {10}^{-1}$)	$2.900735\times {10}^{-3}$($1.93\times {10}^{-2}$)
$e^{\left(1+x\right)}$	n = 5	$7.971826\times {10}^{-3}$($2.71\times {10}^{-2}$)	$7.991609\times {10}^{-3}$($2.21\times {10}^{-1}$)	$7.972443\times {10}^{-3}$($1.94\times {10}^{-2}$)
	n = 10	$7.972430\times {10}^{-3}$($1.96\times {10}^{-2}$)	$7.975272\times {10}^{-3}$($1.61\times {10}^{-2}$)	$7.972443\times {10}^{-3}$($1.94\times {10}^{-2}$)
	n = 15	$7.972442\times {10}^{-3}$($1.94\times {10}^{-2}$)	$7.973336\times {10}^{-3}$($8.20\times {10}^{-3}$)	$7.972443\times {10}^{-3}$($1.94\times {10}^{-2}$)
	n = 20	$7.972443\times {10}^{-3}$($1.94\times {10}^{-2}$)	$7.972832\times {10}^{-3}$($1.45\times {10}^{-2}$)	$7.972443\times {10}^{-3}$($1.94\times {10}^{-2}$)
$1+\frac{x}{y}$	n = 5	$3.967419\times {10}^{-3}$($1.62\times {10}^{-1}$)	$3.983786\times {10}^{-3}$($2.50\times {10}^{-1}$)	$3.973294\times {10}^{-3}$($1.45\times {10}^{-2}$)
	n = 10	$3.972658\times {10}^{-3}$($3.05\times {10}^{-2}$)	$3.974801\times {10}^{-3}$($2.34\times {10}^{-2}$)	$3.973291\times {10}^{-3}$($1.46\times {10}^{-2}$)
	n = 15	$3.973143\times {10}^{-3}$($1.83\times {10}^{-2}$)	$3.973758\times {10}^{-3}$($2.82\times {10}^{-3}$)	$3.973291\times {10}^{-3}$($1.46\times {10}^{-2}$)
	n = 20	$3.973240\times {10}^{-3}$($1.59\times {10}^{-2}$)	$3.973492\times {10}^{-3}$($9.51\times {10}^{-3}$)	$3.973291\times {10}^{-3}$($1.46\times {10}^{-2}$)
$\left(\ln y\right)\left(x+y\right)$	n = 5	$-6.462471\times {10}^{-4}$($3.72\times {10}^{-2}$)	$-6.478897\times {10}^{-4}$($2.17\times {10}^{-1}$)	$-6.460301\times {10}^{-4}$($7.08\times {10}^{-2}$)
	n = 10	$-6.463082\times {10}^{-4}$($2.78\times {10}^{-2}$)	$-6.465402\times {10}^{-4}$($8.10\times {10}^{-3}$)	$-6.463052\times {10}^{-4}$($1.94\times {10}^{-2}$)
	n = 15	$-6.463093\times {10}^{-4}$($2.76\times {10}^{-2}$)	$-6.463822\times {10}^{-4}$($1.63\times {10}^{-2}$)	$-6.463091\times {10}^{-4}$($2.76\times {10}^{-2}$)
	n = 20	$-6.463094\times {10}^{-4}$($2.76\times {10}^{-2}$)	$-6.463412\times {10}^{-4}$($2.27\times {10}^{-2}$)	$-6.463094\times {10}^{-4}$($2.76\times {10}^{-2}$)

Fig. 8.

Convergence of simulation results for quadratic domain from Table 4: (a) convergence of the first integrand: $2xy+\left(x^2{y}^2\right)$ and (b) convergence of the last integrand: $\left(\ln y\right)\left(x+y\right)$.

Cubic Bezier domain

A domain enclosed by a quadrant and a sine-based curve as shown in Fig. 9 is approximated by cubic Bezier curves with control points that form both triangular and quadrilateral convex hulls. Bezier control points that are used for the curves are:

Fig. 9.

A domain enclosed by two cubic Bezier curves.

Sine-based curve = $\left(0,3\right),\left(0.8374,1.717\right),\left(1.837,0.019\right),\left(3,0\right)$.

Quadrant = $\left(0,3.00017\right),\left(1.66028,2.99621\right),\left(2.99621,1.66028\right),\left(3.00017,0\right)$.

Known equations of the curves are used to determine the reference solutions given by:

$$\underset{0}{\overset{3}{\int }}\underset{\left(3-3Sin\left(0.52x\right)\right)}{\overset{\left(\sqrt{\left(9-x^2\right)}\right)}{\int }}f\left(x,y\right)dydx$$ (35)

Results are shown in Table 5. Results for all the integrands can be obtained by implementing numerical integration technique (Eq. (26)), which showed continual convergence when integration order is increased as shown in Fig. 10.

Table 5.

Results for the Bezier domain in Fig. 9 with cubic approximations for the curves.

Integrand $f\left(x,y\right)$	Reference solution	Simulation results (percentage error, %)
		Symbolic integration	Numerical integration (n = Number of integration points and weights)
$2xy+\left(x^2{y}^2\right)$	39.4053	39.4159 ($2.695\times {10}^{-2}$)	n = 5	39.4959 ($2.3\times {10}^{-1}$)
			n = 10	39.4159 ($2.7\times {10}^{-2}$)
			n = 15	$\text{same}\text{as}\text{above}$
			n = 20	$\text{same}\text{as}\text{above}$
$\sqrt{1+x}$	6.1448	-	n = 5	6.1483 ($5.62\times {10}^{-2}$)
			n = 10	6.1480 ($5.13\times {10}^{-2}$)
			n = 15	$\text{same}\text{as}\text{above}$
			n = 20	$\text{same}\text{as}\text{above}$
$e^{\left(1+x\right)}$	70.0488	70.0735 ($3.530\times {10}^{-2}$)	n = 5	70.1994 ($2.15\times {10}^{-1}$)
			n = 10	70.0735 ($3.53\times {10}^{-2}$)
			n = 15	$\text{same}\text{as}\text{above}$
			n = 20	$\text{same}\text{as}\text{above}$
$1+\frac{x}{y}$	13.6779	-	n = 5	13.7619 ($6.14\times {10}^{-1}$)
			n = 10	13.7602 ($6.02\times {10}^{-1}$)
			n = 15	13.7601 ($6.01\times {10}^{-1}$)
			n = 20	$\text{same}\text{as}\text{above}$
$\left(\ln y\right)\left(x+y\right)$	3.84045	-	n = 5	3.7538 $(2.26\times {10}^{-1}$)
			n = 10	3.8318 ($2.25\times {10}^{-1}$)
			n = 15	3.8361 ($1.14\times {10}^{-1}$)
			n = 20	3.8368 ($9.42\times {10}^{-2}$)

Fig. 10.

Convergence of simulation results for cubic domain from Table 5: (a) convergence of the first integrand: $2xy+\left(x^2{y}^2\right)$ and (b) convergence of the last integrand: $\left(\ln y\right)\left(x+y\right)$.

Approximation of circular arcs

Problems from XFEM and SB-FEM are used as examples to validate the new methods. Example 1 is based on XFEM, which is analysis of a plate with circular hole. Examples 2 and 3 are based on SB-FEM, which are analysis of thick circular beam and thick-walled cylinder, respectively. Bezier control points for the circular arcs in all the examples are obtained by using Eq. (30). Midpoint of a curve is found based on the start and end points of the curve as well as the radius.

Example 1: Analysis of a plate with circular hole

Circular hole in the plate (with 7 by 7 mesh) is represented by four cubic Bezier curves as shown in Fig. 11. The enriched elements (highlighted in grey) are determined based on intersection of the Bezier curves with the boundaries of the square elements. Integration within the enriched elements is carried out by referring to the 12 possible combinations. These 12 possible combinations are further divided into smaller subdomains as shown in Fig. 12, to facilitate integration within the circular arc domain by using Method 3. This further division is not required when Green’s theorem is used but it requires specific counterclockwise orientation for the curves and lines enclosing the domain. For combinations 1a to 4a, the required domain, Ω is R₁ (Fig. 12(a)). Integration within R₁ is accomplished by subtracting R₂ and R₃ from the entire square element:

$$\int \underset{\Omega =R_1}{\int }f\left(r,s\right)ds=\int \underset{Square}{\int }f\left(r,s\right)ds-\int \underset{R_3={\Delta }_{123}}{\int }f\left(r,s\right)ds-\int \underset{R_2}{\int }f\left(r,s\right)ds$$ (36)

Fig. 11.

Representation of a circular hole with 4 cubic Bezier curves and enriched elements (greyed) under 7 by 7 mesh.

Fig. 12.

Subdivision of the 12 possible combinations (a) subdivision of combinations 1a to 4a (b) subdivision of combination 5a (c) subdivision of combination 6a (d) subdivision of combinations 1b to 4b (e) subdivision of combination 5b (f) subdivision of combination 6b.

For combinations 5a and 6a, integration within the required domain R₁ is accomplished by subtracting R₂ from the quadrilateral 1-2-3-4 (Fig. 12(b) and 12(c)):

$$\int \underset{\Omega =R_1}{\int }f\left(r,s\right)ds=\int \underset{Quad}{\int }f\left(r,s\right)ds-\int \underset{R_2}{\int }f\left(r,s\right)ds$$ (37)

For combinations 1b to 4b, the required domain, Ω is R₁ (Fig. 12(d)). Integration within R₁ is accomplished by subtracting R₂ from the triangle 1-2-3:

$$\int \underset{\Omega =R_1}{\int }f\left(r,s\right)ds=\int \underset{{\Delta }_{123}}{\int }f\left(r,s\right)ds-\int \underset{R_2}{\int }f\left(r,s\right)ds$$ (38)

Comparison of L2 error norm between classical method and current proposed method for different mesh sizes is shown in Fig. 13. The classical method uses linear approximation for the curves. The proposed method yields more accurate results.

Fig. 13.

Comparison of L2 errors between classical and proposed methods.

Example 2: Analysis of thick curved beam

Analysis of a thick curved beam by using SB-FEM with quadrilateral elements is shown in Fig. 14. Bottom of the thick curved beam is fixed, and the free end is subjected to shear load, F. Linear approximation of the beam’s boundary (used in the conventional SB-FEM) is shown by the dashed lines. Vertical displacement of point A is calculated and compared with the reference solution. Curvatures of the thick beam are defined by circular arcs. Integration of functions within the problem domain is now accomplished by using curved quadrilateral elements to achieve better boundary representation. Modification of linear quadrilateral elements to curved ones require removal and addition of material stiffness (Ω₁ and Ω₂, respectively) as shown in Fig. 14.

Fig. 14.

Analysis of a thick curved beam with four linear quadrilateral elements. Addition and removal of material stiffness are required to achieve better boundary representation.

Stiffness matrix, k for the curved quadrilateral element is then given by:

$$k=\int \underset{\begin{array}{c}Curved\\ Quad\end{array}}{\int }f\left(x,y\right)ds=\int \underset{\begin{array}{c}Linear\\ Quad\end{array}}{\int }f\left(x,y\right)ds-\int \underset{{\Omega }_1}{\int }f\left(x,y\right)ds+\int \underset{{\Omega }_2}{\int }f\left(x,y\right)ds$$ (39)

Three different cases are studied, which are exact boundary representation by addition and removal of material stiffness according to Eq. (39) (case 1), considering inner curvature only through removal of material stiffness according to Eq. (40) (case 2):

$$k=\int \underset{\begin{array}{c}Curved\\ Quad\end{array}}{\int }f\left(x,y\right)ds=\int \underset{\begin{array}{c}Linear\\ Quad\end{array}}{\int }f\left(x,y\right)ds-\int \underset{{\Omega }_1}{\int }f\left(x,y\right)ds$$ (40)

and considering outer curvature only through addition of material stiffness according to Eq. (41) (case 3):

$$k=\int \underset{\begin{array}{c}Curved\\ Quad\end{array}}{\int }f\left(x,y\right)ds=\int \underset{\begin{array}{c}Linear\\ Quad\end{array}}{\int }f\left(x,y\right)ds+\int \underset{{\Omega }_2}{\int }f\left(x,y\right)ds$$ (41)

Integrands within the element level stiffness matrices ($f\left(x,y\right)$) remain the same, but the domain of integration changes.

Table 6 shows that displacement of point A based on addition of material alone is always the lowest among all the cases considered. Conversely, removal of material alone yields the highest displacement among all the cases considered. These are because when material is added, the curved beam becomes stiffer and resistance towards deformation is increased. Likewise, removal of material causes the curved beam to lose some of its stiffness and therefore deforms more easily. Fig. 15 shows that all the three cases considered in this example converge towards the reference solution when the number of elements in the mesh is increased. It is also seen that convergence of solutions obtained by using exact boundary representation is almost similar to the convergence of solutions using linear approximation for mesh size of 64 elements and onwards. Lowest percentage of error is recorded by case 2 (addition of material only) which is 0.05% with mesh size of 256 elements.

Table 6.

Simulation results for thick curved beam.

Number of elements in the mesh	Vertical displacement of point A (analytical solution: 90.1)
	Linear approximation by using linear quadrilaterals	Case 1: Exact boundary approximation (involves both addition and removal of material)	Case 2: Addition of material only	Case 3: Removal of material only
4	87.1234	82.7176	77.1005	93.7679
16	87.2196	86.9251	85.0739	89.1345
64	89.4775	89.4533	88.9415	89.9935
256	90.2711	90.2728	90.1413	90.4078

Fig. 15.

Percentage of error in the calculation of displacement of point A for the thick curved beam.

Example 3: Analysis of thick-walled cylinder

Analysis of quarter of a thick-walled cylinder by using SB-FEM with linear quadrilateral mesh is shown in Fig. 16. The internal curved surface is subjected to internal pressure, and the radial displacement of point A is calculated and compared with the reference solution. Hybrid mesh is used to further improve the solution accuracy by meshing the inner curved surface with pentagonal elements, while leaving the rest of the elements as linear quadrilaterals. Similar to Example 2, the elements at the inner and outer surfaces can be modified to curved ones by removal and addition of material stiffness (Ω₁ and Ω₂, respectively) as shown in Fig. 16.

Fig. 16.

Analysis of quarter of a thick-walled cylinder subjected to internal pressure with linear quadrilateral mesh.

Stiffness matrix, k for the curved quadrilateral elements representing the inner surface is obtained by subtraction of material stiffness according to Eq. (42):

$$k_{\begin{array}{c}inner\\ surface\end{array}}=\int \underset{\begin{array}{c}Curved\\ Quad\end{array}}{\int }f\left(x,y\right)ds=\int \underset{\begin{array}{c}Linear\\ Quad\end{array}}{\int }f\left(x,y\right)ds-\int \underset{{\Omega }_1}{\int }f\left(x,y\right)ds$$ (42)

Addition of material stiffness for curved quadrilateral elements representing the outer surface is given by Eq. (43):

$$k_{\begin{array}{c}outer\\ surface\end{array}}=\int \underset{\begin{array}{c}Curved\\ Quad\end{array}}{\int }f\left(x,y\right)ds=\int \underset{\begin{array}{c}Linear\\ Quad\end{array}}{\int }f\left(x,y\right)ds+\int \underset{{\Omega }_2}{\int }f\left(x,y\right)ds$$ (43)

Simulation results for the vertical displacement of point A are shown in Table 7.

Table 7.

Simulation results for thick-walled cylinder.

Number of elements in the mesh			Vertical displacement of point A (Percentage of error, %)
	Reference solution	Linear approximation by using linear quadrilaterals		Linear approximation by using pentagonal elements for the inner surface (Hybrid mesh)	Case 1: Exact boundary approximation (involves both addition and removal of material)	Case 2: Removal of material only	Case 3: Addition of material only
40	0.004583	0.00451033 (1.5856)		0.00452535 (1.2579)	0.00453809 (0.9798)	0.00454284 (0.8762)	0.00450563 (1.6882)

Three different cases are studied, which are exact boundary representation by both addition and removal of material stiffness according to both Eqs. (42) and (43) (case 1), considering inner curvature only through removal of material stiffness according to Eq. (42) (case 2) and considering outer curvature only through addition of material stiffness according to Eq. (43) (case 3). Table 7 shows that displacement of point A based on addition of material alone is always the lowest among all the cases considered. Conversely, removal of material alone yields the highest displacement among all the cases considered. The exact boundary approximation by using the proposed method (case 1) yielded better result compared to the linear approximations by the quadrilateral and pentagonal elements. Lowest percentage of error is recorded by case 2 (removal of material only), which is 0.88%.

Limitations

Limitation of the method 1 and alternate approach (letting N = 0)

The Method 1 in the section: Method 1: Green's theorem is described for the case where M is set to be zero. This causes the integration to be performed with respect to x as shown by Eq. (5) and imposes restriction that, the function $f\left(x,y\right)$ should have closed-form antiderivatives with respect to x (an inherent dependence on x-integration). This limitation can be overcome by letting N = 0 to allow integration to be performed with respect to y instead. Hence, selection of M or N to be zero can be made based on the availability of closed-form antiderivatives of the function $f\left(x,y\right)$. Table 8 shows several integrands with inherent dependence and their solutions. It is seen that solutions for all the integrands can be obtained by alternating between the two integration orders. However, solutions for the integration might not exist when the function $f\left(x,y\right)$ has discontinuities or singularities that prevent convergence.

Table 8.

Simulation results (symbolic computations) by using Method 1 for the cubic Bezier domain in Fig. 6.

Integrand $f\left(x,y\right)$	Selecting M = 0 (Integration order dxdy)	Selecting N = 0 (Integration order dydx)
$y{x}^x$	-	$1.669856\times {10}^{-4}$
$x{y}^y$	$5.144681\times {10}^{-5}$	-
$y^{2}tan\left(x^2\right)$	-	$1.335285\times {10}^{-8}$
$x^{2}tan\left(y^2\right)$	$1.335331\times {10}^{-8}$	-
$\frac{y{e}^x}{x}$	-	$1.082208\times {10}^{-2}$
$\frac{x{e}^y}{y}$	$1.167995\times {10}^{-3}$	-

Limitations and mitigations of method 3 for cubic Bezier domains

The method described in the section: Cubic Bezier domain (under Method 3: Bijective mapping) (for cubic Bezier domains) is limited to convex hulls with non-parallel sides that can be extended to form a virtual vertex V. However, convex hull of the control points that are in the form of parallelogram, square and other shapes with parallel sides can still be solved by using the method presented in this work but requires splitting of the original Bezier curve into two or more smaller segments. An example is shown in Fig. 17 for a convex hull of control points in the form of a parallelogram. The original curve is split into two segments (Fig. 17(b)). Convex hulls of control points of these two segments have non-parallel sides which satisfy the requirement of the method. Integration can then be carried out within three subdomains labelled as R1 to R3 as shown in the Fig. 17(c).

Fig. 17.

Integration within a cubic Bezier domain with convex hull of control points in the form of a parallelogram (a) original convex hull with parallel sides (b) splitting of the curve into two segments (c) formation of virtual vertex V and subdomains R1 to R3.

CRediT author statement

Logah Perumal: Sole contributor.

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The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.

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No data was used for the research described in the article.