By utilizing the local support properties of NURBS, the geometry of NURBS meshes can be defined as (1)xe=∑i=1p+1Pieφi, pe(ζ)   ζ∈[ζ1e, ζ2e],where φi, pe represents the i-th NURBS basis function on element e. Pie is the control point.

The physical field u on the element is approximated using uniform B-spline interpolation [20], and a transformation matrix is introduced to represent it as (2)ue=N(η)ue   ξ∈[0,1],where ξ is the parameter coordinate that defines the field variable, and the transformation formula with geometric parameter coordinate ζ is defined as (3)ζ=ζ1e+(ζ2e−ζ1e)ξ,

ue={u1, u2,⋯,um}T is the column vector composed of the physical degrees of freedom (DOFs) of element nodes; N is the row vector composed of shape functions, represented as (4)N=ΦpT={ψ1, ψ2,⋯,ψm}, ψi(ξ)=∑j=1mϕj, p(ξ)Tjiwhere ψi is the i-th shape function; Φp and T are row vectors composed of uniform B-spline basis functions and C⁰ type transformation matrix, respectively, with their expressions given as (5)Φp={ϕ1, p, ϕ2, p,⋯, ϕm, p} (6)T={ΦpT(ξ1),ΦpT(ξ2),⋯,ΦpT(ξm)}−T,where m is the number of element nodes; ϕi, p represents the i-th p-degree uniform B-spline basis function; ξi represents the parameter coordinate of node i, expressed as (i−1)/(m−1). The subscript T refers to the transpose of the matrix.

The differential equation for the axial force bar is described as [38] (7)EAd2dx2u(x)=q(x),where the solution domain range is x∈[a, b]; u represents the displacement in the x-coordinate direction; E, A, and q(x) denote the elastic modulus, cross-sectional area, and continuous load, respectively; The end conditions of the bar are u=u¯ and Aσx=P, respectively. σx and P represent stress and prescribed load, respectively. Then the functional equation of the axial force bar [38] is given as (8)Π(u)=∫abEA2(dudx)2dx−∫abq(x)udx−∑j=1N1Pju(xj),where Pj refers to the concentrated force at point j and N₁ is the number of concentrated forces. Then, by employing element discretization and parameter coordinate transformation, Eq. (8) is transformed into the field parameter domain. Under the condition of δΠ=0, the expression of the IGFEM equation is derived as (9)Keue=Pe,where the element in the bar element matrix Ke is represented as (10)kij=EA∫011J∑a=1mdϕa, p(ξ)dξTai∑a=1mdϕa, p(ξ)dξTajdξwhere J=dxdζdζdξ is Jacobian transformation, and the column vector of the load Pe is given as (11)Pe=Pfe+Pje,where elements in Pfe and Pje are respectively represented as (12)p1i=∫01∑j=1mϕj, p(ξ)Tjif(ξ)Jdξ (13)p2i=∑jePje∑j=1mϕj, p(ξje)Tji.

The physical field w on the element is approximated using uniform B-spline and Hermite interpolation, represented as (14)we=Nb(ξ)we   ξ∈[0, 1],where we={w1, θ1, w2,⋯,wm−1, wm, θm}T is the column vector composed of the physical DOFs of element nodes; θi=(dw/dx)ii=1, m represents the first derivative of w at point i; The number of DOFs within the element is m + 2; The row vector Nb composed of shape functions is represented as (15)Nb=ΦpbTbe={ψ1b, ψ2b,⋯, ψm+2b},ψib=∑j=1m+2ϕj, p(ξ)Tjiwhere ψi is the i-th shape function; Φpb is a row vector composed of p-degree uniform B-spline basis functions and their first-order derivatives, represented as (16)Φpb={ϕ1, p, dϕ1, pdη,ϕ2, p,⋯,ϕm−1, p, ϕm, p, dϕm, pdη},

Tbe is C¹ type element transformation matrix, deduced as (17)Tbe={(Φpb(η1))T, (dΦpbdη(ξ1))T, (Φpb(ξ2))T,⋯,(Φpb(ξm−1))T, (Φpb(ξm))T,(dΦpbdη(ξm))T}−T.

The basic equation for the Euler beam [38] is given as (18)EId4wdx4=q(x)   x∈[a, b],where w is the deflection; E and I represent the elastic modulus and moment of inertia, respectively; q(x) denotes the continuous load. The end conditions of the beam are w=w¯, dw/dx=θ¯ and M=M¯, respectively. θ and M refer to the angle of rotation and bending moment, respectively.

The functional equation [38] that is equivalent to the basic equation of Euler beam is given as (19)Π(u)=∫abEI2(d2wdx2)2dx−∫abq(x)wdx−∑j=1N1Pjw(xj)+∑k=1N2Mk(dwdx)k,where Pj and Mk denote the concentrated force of point j and the concentrated bending moment at point k, respectively. N₁ and N₂ are the number of concentrated forces and the number of concentrated bending moments, respectively.

Similar to the derivation of axial force bars, Eq. (19) is transformed into the field parameter domain. Under the condition of δΠ=0, the expression of the IGFEM equation is derived as (20)Kewe=Fe,where the element k_ij of Ke and Fe are respectively deduced as (21)kij=∫01EIJ3∑a=1m+2d2ϕa, p(ξ)dξ2Tai∑a=1m+2d2ϕa, p(ξ)dξ2Tajdξ (22)Fe=Fqe+Fje−Fke,where elements in Fqe, Fje and Fke are respectively given as (23)f1i=∫01∑j=1m+2ϕj, p(ξ)Tjiq(ξ)Jdξ (24)f2i=∑je∑a=1m+2ϕa, p(ξje)TaiPje (25)f3i=∑ke1J(∑a=1m+2dϕa, p(ξke)dξTai)TMke.

The expression at each point x(x, y) on element e can be defined as (26)xe=∑i=1nPieφi, de(ζ, ϑ)   (ζ, ϑ)∈[ζ1e, ζ2e]×[ϑ1e, ϑ2e],where n = (p + 1) × (q + 1), with p and q referring to degree of parameter coordinate (ζ, ϑ) direction. Pie is the control point. φi, de(ζ, ϑ)=φa, p(ζ)×φb, q(ϑ) is i-th two-dimensional NURBS basis function, with d = p × q representing the degree of NURBS.

The displacement is described as (27)u=N(ξ, η)ue   (ξ, η)∈[0, 1]×[0, 1],where u={u1, u2}T, ue={u1(ξ1, η1), u2(ξ2, η1),⋯,uM(ξm1, ηm2)}T, with M = m₁ × m₂, denoting the number of element nodes. N(ξ, η) in Eq. (27) is expressed as (28)N(ξ, η)=ΦT,where (29)Φ={ϕ1, d(ξ, η)I, ϕ2, d(ξ, η)I,⋯,ϕM, d(ξ, η)I} (30)T={ΦT(ξ1,η1),ΦT(ξ2,η1),⋯,ΦT(ξm2,ηm1)}−T,where ϕi, d is i-th d-degree two-dimensional B-spline basis function, I is 2 × 2 identity matrix.

The system of equation for linear elasticity occupying a domain Ω with boundary Γ is given as [38] (31)∇⋅σ+f=0 in Ωσn=t¯ on Γtu=u¯ on Γu,where σ and f are stress tensor and body force, respectively. n represents an outward normal vector. The traction t¯ and displacement u¯ are applied on the boundary Γt and Γu respectively with Γ=Γt∪Γu.

According to the principle of potential energy, the generalized function for the plane stress in the element is defined as (32)Πe(u)=12∫ΩeεTDεdΩ−∫ΩeuTfdΩ−∫ΓeuTtdΓ,where D and ε are the elasticity matrix of the plane stress case and strain vector, respectively. Applying the variation principle, the IGFEM equation is expressed as (33)Keue=Fe.

The 2 × 2 matrix K_ij of K^e is given as (34)Kij=∫01∫01(Bi)TDBjJdξdη,with (35)Bi=[∂/∂x00∂/∂y∂/∂y∂/∂x]Ni, J=det(∂(x, y)∂(ζ,ϑ))×det(∂(ζ, ϑ)∂(ξ, η)). (36){∂Ni∂x, ∂Ni∂y}T=(∂(ζ, ϑ)∂(ξ, η)×∂(x, y)∂(ζ, ϑ))−1{∂Ni∂ξ, ∂Ni∂η}T.

The 2 × 1 vector F_i of F^e at the right end of the Eq. (33) is given as (37)Fi=∫01∫01NifJdξdη+∫01NitJ1dξ,where J1=(dx/dζ)2+(dy/dζ)2×(dζ/dξ).

Consider an axial force bar with a length of L = 10 and a cross-sectional area of A = 1, as shown in Fig. 1.

The elastic modulus is E = 1. The essential boundary condition is given as u₀ (x = 0) = 0.5. The continuously distributed load is q(x)=q1+(q2−q1)x/L, where q1 = −3 and q2 = 5. The end of the bar (x = L) is subjected to a concentrated force P = −1. The analytical solution to the problem is given as (38)u=(q1−q2)x36EAL−q1x22EA+(3L2(q1+q2)+6LP)x6EAL+u0.

18 points are equally distributed within [0, L] for evaluation, and displacement and stress are calculated using a 3-degree PFEM1 (MM = 4) element. Relative errors (%) with the corresponding exact solution are shown in Fig. 2. In Fig. 2, the PFEM1 (p = 3, MM = 4) element achieves high-precision displacement and stress solutions, which indicates the effectiveness and accuracy of the method. From Fig. 2b, it can also be observed that the numerical error is relatively large near the free end. This may be because the node is at the free end, and the element only has C⁰ continuity at that point, resulting in a higher stress error. However, the error results show that the impact is relatively small.

Consider an axial force bar with a length of L = 1 and a cross-sectional area of A = 1. The geometric model is the same as Fig. 1. The elastic modulus is E = 1. The essential boundary conditions are given as u (x = 0) = 0 and u (x = L) = 1. Within the range of [0.48, 0.52], the bar bears a locally distributed load, expressed as (39)q(x)=2α2−4(α(x−0.5))2e−(α(x−0.5))2   x∈[0.42,0.58],where α=50. The analytical solution for displacement is given as (40)u=x+e−(α(x−0.5))2   x∈[0, L],

A cantilever beam in Fig. 11 is analyzed, with a length of L = 8 and an equal cross-section.

The moment of inertia and elastic modulus are I = bh³/12 and E = 2.1e11, respectively. The load conditions are as follows: (1) The entire beam is subjected to a uniformly distributed load q = −1; (2) The free end is subjected to a concentrated force P = −1e5; (3) The free end is subjected to a concentrated bending moment M = −1e7. The corresponding analytical solutions for the boundary conditions (1)–(3) are given as (41)w=−q24EI(x4−4Lx+6L2), (42)w=−P6EI(3Lx2−x3), (43)w=−M2EIx2.

A PFEM1 (p = 4, MM = 5) element is used. 20 evaluation points are equally spaced along the beam. The angle of rotation θ and bending moment M are calculated under boundary condition (1), and relative errors between their results and analytical results are depicted in Fig. 12.

When the beam is subjected to boundary conditions (2) and (3), respectively, relative errors of angle of rotation θ and bending moment M are calculated. Relative errors between their results and analytical results are depicted in Figs. 13, 14.

In Figs. 12–14, the calculated results closely match analytical solutions, demonstrating the high-precision capabilities of the PFEM1 (p = 4, MM = 5) element in solving this problem.

A variable cross-section cantilever beam is analyzed in Fig. 15. The length is L = 1, while the width and height of the rectangular section are given as b = 0.1 and h = (1 + 0.9x/L − 1.8 (x/L)2)/10, respectively.

The free end is subjected to a concentrated force P = −1. The elastic modulus is E = 1.2e6. The analytical solution for the bending moment is provided in [39]. The numerical solution obtained by FEM and the proposed method is rounded to six decimal places. The equation for calculating the overall relative error (%) is given as

(44)e=1n∑i=1n|uih−uiauia|×100,where n is the number of evaluation points; uih and uia represent numerical and analytical solutions, respectively. The value of the last evaluation point in the reference solution is 0. To satisfy Eq. (44), the first eight points in [39] are taken as evaluation points. The solution results of the FEM and PFEM1 element with third-degree and two nodes are compared. Uniform h-refinement is used in this example, with the mesh and element nodes shown in Fig. 16.

Fig. 17a displays the error results with increasing DOFs. As can be seen, the results are similar. When MM = 11, the errors obtained by PFEM1 with p = 3, 4, and 5 are compared with those of FEM at the same degree, as depicted in Fig. 17b–d. Fig. 17a,b shows a non-monotonic convergence. The possible reason is that elements based on cubic polynomials are significantly affected by changes in cross-sectional dimensions, leading to unstable numerical results on coarse grids. Increasing the polynomial degree and obtaining a smoother approximation function can help reduce this effect. In Fig. 17b–d, the PFEM1 element can achieve lower errors with fewer DOFs compared to the same degree FEM, indicating good convergence. Among them, the results obtained by the 4- and 5-degree PFEM1 elements are similar, with lower computational errors compared to the 3-degree PFEM1 element. This indicates that the 4-degree PFEM1 element performs better in solving the problem.

Table 2 shows the overall running time of PFEM1 and FEM at different DOFs when p = 3 and MM = 2. In Table 2, t₀ and t₁ respectively refer to the time of PFEM1 and FEM. It can be seen that under the same DOFs, the proposed method has less running time.

Table 3 shows the overall running time of PFEM1 (p = 4, MM = 11) and FEM (p = 4, MM = 3) at different DOFs, with DOFs_0 and DOFs_1 referring to the DOFs of PFEM1 and FEM, respectively. From Tables 2 and 3, it can be seen that the calculation time of PFEM1 (p = 4, MM = 11) element is relatively less than that of (p = 3, MM = 2) and (p = 4, MM = 3) FEM elements, and is similar to PFEM1 (p = 3, MM = 2) element.

Similarly, a comparison is made between the overall errors of PFEM1 and FEM, with Fig. 18 showing the L₂ norm errors of the bending moment M. It can be seen that the proposed method can achieve higher computational accuracy with similar degrees of freedom. In addition, both elements exhibit similar convergence rates as degrees of freedom increase.

In the final example, the plane stress case of Lame’s problem is studied.

A thick hollow cylinder with internal radius a = 8 and external radius b = 10 under constant pressure p = 1 is simulated. Due to the symmetry of the problem, a quarter of the geometric model is established. The geometry and loading conditions are specified in Fig. 19. The elastic modulus and Poisson’s ratio are set to E = 1 and v = 0.3, respectively. For this problem, the analysis solution of the stress field is obtained as [40]

(45)σrr=a2(r2−b2)(b2−a2)r2σrθ=a2p(r2+b2)(b2−a2)r2where r and θ are the polar coordinates. To evaluate PFEM1, the energy norm is given as 0.5∫Ω(σh−σe)TD−1(σh−σe)dΩ.

When the degree is 4 × 4, PFEM1 (MM = 8 × 8) element is constructed and compared with FEM (MM = 5 × 5). Fig. 20 shows the energy norm errors for the two types of elements, abbreviated as PFEM1 (p = 4, MM = 8) and FEM (p = 4, MM = 5).

From the results, the PFEM1 shows significantly higher computational accuracy with similar degrees of freedom. This is because the geometry of the proposed method is precise, which means there is no geometric discretization error in coarse grids, resulting in lower errors. In addition, the convergence rate is faster with fewer degrees of freedom. However, with grid refinement, the convergence rate remains similar.