The parametric Bézier curve and surface can be expressed as Equation (2) and Equation (3) respectively,

p ( t ) = ∑ i = 0 n b i B i , n ( t ) ,   0 ≤ t ≤ 1 (2)

r ( u , v ) = ∑ i = 0 m ∑ j = 0 n b i , j B i , m ( u ) B j , n ( v ) ,   0 ≤ u , v ≤ 1 (3)

where B i , m ( t ) = C m i t i ( 1 − t ) m − i , i = 0 , 1 , ⋯ , m , j = 0 , 1 , ⋯ , n , C m i = m ! i ! ( m − i ) ! . t, u and v are inner variables, and b i are vertex vectors.

De Casteljau subdivision algorithm is constructed from a control polygon or grid, inserting new vertices according to certain subdivision rules, and connecting these new vertices to obtain new control polygons or control grids. The new control polygon or grid obtained from once subdivision process is used as the initial control geometry, and repeat the subdivision process above. Finally, the control polygons or grids generated by recursive subdivision converge to the curve or surface corresponding to the initial control polygon or grid.

For Bézier curves, the arithmetic is expressed as,

b j k = { b j ,   k = 0 ( 1 − t ) b j k − 1 + t b j + 1 k − 1 ,   k = 1 , 2 , ⋯ , n ,     j = 0 , 1 , ⋯ , n − k (4)

where 0 ≤ t ≤ 1 and b j k are middle control points. Take n = 3 , t = 0.5 as an example, two new control polygons are expressed as { b 0 k } and { b k 3 − k } respectively and the subdivision process can be shown as Figure 1.

Similarly, for Bézier surface, the corresponding subdivision (dichotomy) arithmetic is expressed as,

b i , j k = { b i , j ,   k = 0 ( 1 − t ) b i , j k − 1 + t b i , j + 1 k − 1 ,   k = 1 , 2 , ⋯ , n ,     i = 0 , 1 , ⋯ , n ;   j = 0 , 1 , ⋯ , n − k (5)

where 0 ≤ t ≤ 1 and b i , j k are middle control points. Take m = n = 3 , t = 0.5 as an example, two new control grids are expressed as { b i , 0 k } and { b i , 3 − k k } respectively and the subdivision process can be shown as Figure 2.

Lemma [19] For a m × n Bézier grid, if the control vertices satisfy the following conditions:

{ ( Δ 1 , 0 b i , j , Δ 0 , 1 b p , q , Δ 2 , 0 b s , t ) ≥ 0 ( ≤ 0 ) ( Δ 1 , 0 b i , j , Δ 0 , 1 b p , q , Δ 0 , 2 b α , β ) ≥ 0 ( ≤ 0 ) (6)

i = 0 , 1 , ⋯ , m − 1 ;   j , t = 0 , 1 , ⋯ , n ;   p , α = 0 , 1 , ⋯ , m ;

q = 0 , 1 , ⋯ , n − 1 ;   s = 0 , 1 , ⋯ , m − 2 ;   β = 0 , 1 , ⋯ , n − 2

then the Bézier grid is downward (upward) convex.

In this expression, Δ 1 , 0 b i , j , Δ 0 , 1 b i , j and Δ 2 , 0 b i , j , Δ 0 , 2 b i , j represent the first-order difference and second-order difference of the control grid, respectively. It reflects the positive and negative consistency of all possible mixed products with two first-order differences and one second-order difference. In fact, the necessary and sufficient conditions of the convex Bézier surface proposed by Floater [15] and Koras [18] include three inequalities, and Equation (6) essentially corresponds to the first two inequalities in the necessary and sufficient conditions (Equation (1)). The following is Yao’s proof of the lemma.

Definition [19] For a m × n Bézier grid, if the control vertices satisfy the following conditions:

{ l i , j ⋅ ( b i + k 1 , j − b i , j ) ≥ 0 ( ≤ 0 ) l i , j ⋅ ( b i , j + k 2 − b i , j ) ≥ 0 ( ≤ 0 ) (7)

where

l i , j = Δ 1 , 0 b i , j + k 2 × Δ 0 , 1 b i + k 1 , j ,

i = 0 , 1 , ⋯ , m ;   j = 0 , 1 , ⋯ , n ;   k 1 = 1 , 2 , ⋯ , m − i ;   k 2 = 1 , 2 , ⋯ , n − j ,

then the Bézier grid is downward (upward) convex.

Proof. [19] If the vertices of the Bézier grid satisfy the Equation (6), it can be deduced that

{ ( Δ 1 , 0 b i , j , Δ 0 , 1 b p , q , Δ 2 , 0 b i + p , j ) ≥ 0 ( ≤ 0 ) ( Δ 1 , 0 b i , j , Δ 0 , 1 b p , q , Δ 0 , 2 b i , j + q ) ≥ 0 ( ≤ 0 )

i = 0 , 1 , ⋯ , m − i − 2 ;   j = 0 , 1 , ⋯ , n − j − 2.

Then,

{ ( Δ 1 , 0 b i , j , Δ 0 , 1 b p , q , Δ 1 , 0 b i + p + 1 , j − Δ 1 , 0 b i + p , j ) ≥ 0 ( ≤ 0 ) ( Δ 1 , 0 b i , j , Δ 0 , 1 b p , q , Δ 0 , 1 b i , j + q + 1 − Δ 0 , 1 b i , j + q ) ≥ 0 ( ≤ 0 )

Then,

{ ( Δ 1 , 0 b i , j , Δ 0 , 1 b p , q , ∑ r = 1 k 1 ∑ p = 0 r − 2 ( Δ 1 , 0 b i + p + 1 , j − Δ 1 , 0 b i + p , j ) ) ≥ 0 ( ≤ 0 ) ( Δ 1 , 0 b i , j , Δ 0 , 1 b p , q , ∑ r = 1 k 2 ∑ q = 0 r − 2 ( Δ 0 , 1 b i , j + q + 1 − Δ 0 , 1 b i , j + q ) ) ≥ 0 ( ≤ 0 )

k 1 = 1 , 2 , ⋯ , m − i ;   k 2 = 1 , 2 , ⋯ , n − j .

Then,

{ ( Δ 1 , 0 b i , j , Δ 0 , 1 b p , q , b i + k 1 , j − b i , j ) ≥ 0 ( ≤ 0 ) ( Δ 1 , 0 b i , j , Δ 0 , 1 b p , q , b k , j + k 2 − b i , j ) ≥ 0 ( ≤ 0 )

and it leads to Equation (7).

For a 2 × 2 Bézier patch (see Figure 3), the external control vertices have been given by experimental data, and only the internal vertex b 1 , 1 is unknown. Specifically, four corner vertices b 0 , 0 , b 0 , 2 , b 2 , 0 and b 2 , 2 represent the experimental points and they are on the surface; b 0 , 1 , b 1 , 0 , b 2 , 1 and b 1 , 2 are determined by the normal data of these corner vertices, and they are not on the surface. If these mixed products unrelated to b 1 , 1 have satisfied the convexity condition, b 1 , 1 is the only point to determine the shape of the surface patch. Therefore, the difference terms associated with b 1 , 1 need to be substituted into the convexity formulas to determine the values of the mixed product as shown below:

{ ( Δ 1 , 0 b i , j , Δ 0 , 1 b p , q , Δ 2 , 0 b 0 , 1 ) ( Δ 1 , 0 b i , j , Δ 0 , 1 b p , q , Δ 0 , 2 b 1 , 0 ) ( Δ 1 , 0 b i , 1 , Δ 0 , 1 b p , q , Δ 2 , 0 b s , t ) ( Δ 1 , 0 b i , j , Δ 0 , 1 b 1 , q , Δ 0 , 2 b α , β ) (15)

i , q = 0 , 1 ;   j , t , p , α = 0 , 1 , 2 ;   s , β = 0

For example, the control grid of a Bézier patch is represented as

b i , j = ( ( 0 , 0 , 0 ) ( 0 , 1 , 1 ) ( 0 , 2 , 0 ) ( 1 , 0 , 1 ) b 1 , 1 ( 1 , 2 , 1 ) ( 2 , 0 , 0 ) ( 2 , 1 , 1 ) ( 2 , 2 , 0 ) ) ,

where b 1 , 1 = ( x , y , z ) .

Then,

Δ 1 , 0 b i , j = [ ( 1 , 0 , 1 ) ( x , y − 1 , z − 1 ) ( 1 , 0 , 1 ) ( 1 , 0 , − 1 ) ( 2 − x , 1 − y , 1 − z ) ( 1 , 0 , − 1 ) ]

Δ 0 , 1 b i , j = [ ( 0 , 1 , 1 ) ( 0 , 1 , − 1 ) ( x − 1 , y , z − 1 ) ( 1 − x , 2 − y , 1 − z ) ( 0 , 1 , 1 ) ( 0 , 1 , − 1 ) ]

Δ 2 , 0 b i , j = [ ( 0 , 0 , − 2 ) ( 2 − 2 x , 2 − 2 y , 2 − 2 z ) ( 0 , 0 , − 2 ) ]

Δ 0 , 2 b i , j = [ ( 0 , 0 − 2 ) ( 2 − 2 x , 2 − 2 y , 2 − 2 z ) ( 0 , 0 , − 2 ) ]

Suppose there are two alternatives for b 1 , 1 , one is ( 1 , 1 , 2 ) , and the other is ( 1 , 1 , 0 ) . For b i , j = ( 1 , 1 , 2 ) ,

{ ( Δ 1 , 0 b i , j , Δ 0 , 1 b p , q , Δ 2 , 0 b s , t ) = − 2 ≤ 0 ( Δ 1 , 0 b i , j , Δ 0 , 1 b p , q , Δ 0 , 2 b α , β ) = − 2 ≤ 0

So, it satisfies the convexity condition. For b i , j = ( 1 , 1 , 0 ) ,

{ ( Δ 1 , 0 b i , j , Δ 0 , 1 b p , q , Δ 2 , 0 b s , t ) = { − 2 , 2 } ( Δ 1 , 0 b i , j , Δ 0 , 1 b p , q , Δ 0 , 2 b α , β ) = { − 2 , 2 }

Therefore, it does not satisfy the convexity condition.

In some cases, a degenerate surface may be used to construct a surface with poles or sharp points, such as quasi-ellipsoid or quasi-cone, which causes the vertices of Bézier patch on one edge converging to a point. In this case, the number of interpolation nodes becomes three. As shown in Figure 4, a 3 × 3 planar Béziergrid is constructed, of which the control grid is composed of plane polygons, i.e., first row of triangles, second row of trapezoids and third row of rectangles.

Denote A D = h 1 , A F = h 2 , D B = F C = v , B E = h 3 and E C = h 4 , take A 1 D = a 1 h 1 , A 2 F = a 1 h 2 , D B 2 = F C 2 = a 2 v , B 1 E = a 3 h 3 and E C 1 = a 3 h 4 , where 0 ≤ a i ≤ 1 .

Denote 1 − j a i ≜ a i j . Obviously, a i , a i 1 ≥ 0 . Then, the matrixes composed of the differential vectors can be expressed as:

Δ 1 , 0 b i , j = [ a 1 1 h 1 a 1 1 ( h 1 + a 3 1 h 3 ) a 1 1 ( h 2 − a 4 1 h 4 ) a 1 1 h 2 a 2 v + a 1 h 1 a 2 v + a 1 ( h 1 + a 3 1 h 3 ) a 2 v + a 1 ( h 2 − a 4 1 h 4 ) a 2 v + a 1 h 2 a 2 1 v a 2 1 v a 2 1 v a 2 1 v ]

Δ 0 , 1 b i , j = [ 0 0 0 a 1 1 a 3 1 h 3 a 1 1 ( a 3 h 3 + a 4 h 4 ) a 1 1 a 4 1 h 4 a 3 1 h 3 a 3 h 3 + a 4 h 4 a 4 1 h 4 a 3 1 h 3 a 3 h 3 + a 4 h 4 a 4 1 h 4 ]

Δ 2 , 0 b i , j = [ a 2 v − a 1 2 h 1 a 2 v − a 1 2 ( h 1 + a 3 1 h 3 ) a 2 v − a 1 2 ( h 2 − a 4 1 h 4 ) a 2 v − a 1 2 h 2 a 2 2 v − a 1 h 1 a 2 2 v − a 1 ( h 1 + a 3 1 h 3 ) a 2 2 v − a 1 ( h 2 − a 4 1 h 4 ) a 2 2 v − a 1 h 2 ]

Δ 0 , 2 b i , j = [ 0 0 a 1 1 ( a 4 h 4 − a 3 2 h 3 ) a 1 1 ( a 4 2 h 4 − a 3 h 3 ) a 4 h 4 − a 3 2 h 3 a 4 2 h 4 − a 3 h 3 a 4 h 4 − a 3 2 h 3 a 4 2 h 4 − a 3 h 3 ]

Then,

Δ 1 , 0 b i , j × Δ 0 , 1 b p , q = ∑ λ l 1 ( h m × h n ) + ∑ λ l 2 ( v × h n ) ,   m = 1 , 2 , 3 ;   n = 3 , 4 (16)

In the expression, the parameters λ l 1 and λ l 2 are the multiply-add combinations of a i 1 and a i , which leads to λ l 1 , λ l 2 ≥ 0 . For the two sets of mixed products in Equation (6), the corresponding values are only determined by a i 2 . It is convenient to control the surface convexity by choosing the values of a i .

In the field of solid mechanics, the yield surface corresponding to the plastic deformation of materials can be considered to be constructed with Bézier patches. Based on plasticity theory, the yield surface is required to be convex. When constructing a yield surface under plane stress condition in a general stress space, the form of the yield surface is represented by a quasi-ellipsoid. In this case, point A in Figure 4 is the pole along z direction, and B, C and their normals are on the XOY plane, which makes h m × h n ∥ v and ( h m , h n , h k ) = 0 .

Consider ( Δ 1 , 0 b i , j , Δ 0 , 1 b p , q , Δ 2 , 0 b s , t ) , if ( h m , h n , v ) < 0 is guaranteed, compare the first and second row of the matrix Δ 2 , 0 b s , t , and based on the positive and negative consistency, they should have a same structure. We get:

{ a i ≥ 0 a j 1 ≥ 0 a i 2 ≥ 0 ,   i = 1 , 2 ;   j = 3 , 4 (17)

Then, 0 ≤ a 1 , a 2 ≤ 1 / 2 .

Similarly, consider ( Δ 1 , 0 b i , j , Δ 0 , 1 b p , q , Δ 0 , 2 b α , β ) , we get:

{ a i ≥ 0 a 1 1 ≥ 0 a i 2 ≥ 0 ,   i = 3 , 4 (18)

Then, 0 ≤ a 3 , a 4 ≤ 1 / 2 . In summary, 0 ≤ a i ≤ 1 / 2 , i = 1 , 2 , 3 , 4 .

In order to illustrate applications of the modeling method in constructing plastic yield surface, the yield data of IF steel (see Table 1, data after [21]) is chosen to give modeling.

In Table 1, θ 1 = − arctan ( r 0 r 0 + 1 ) , θ 2 = π 2 + arctan ( r 90 + 1 r 90 ) , r 0 = 1.85 , r 90 = 2.51 , θ 3 = arctan 0.77 , and r 0 , r 90 represent the anisotropy coefficients along 0˚ and 90˚, respectively.

As shown in Figure 5, The data above can construct two interpolated surfaces, one with f U N , 0 , f B I and f S H , 0 as the interpolation points and the other with f U N , 90 , f B I and f S H , 0 as the interpolation points. The corresponding tangential directions (perpendicular to the normals) of these interpolation points represent the direction of the edge of the control polygon at the interpolation point. The choice for a i can be obtained based on existing conditions. For example, the yield points of f U N , 45 are on the surface which can be used to determine the value of a 1 or a 2 ; the C1 continuous splicing condition of the two surface patches requires that the sides on both sides of the common point be collinear and of equal length; if there is no limit conditions, take a i = 1 / 3 which is obtained when the Bézier curve is raised from the second order to the third order. The value scheme of a i is shown in Table 2.

The final results are shown in Figure 6 and Figure 7.