# 统计代写|主成分分析代写Principal Component Analysis代考|Simplifying Finite Element Formulations with Area Coordinates

The passage explains how area coordinates, also known as natural coordinates, can greatly simplify the interpolation functions and integration steps involved in developing finite element formulations for triangular elements. Unlike the Cartesian coordinates (x, y), which lead to algebraically complex interpolation functions (as seen in Equation 6.36), area coordinates L1,L2,L_1, L_2,L 1 ​ ,L 2 ​ , and L3L_3L 3 ​ are based on the fractional areas subtended by a point P(x,y)P(x, y)P(x,y) inside a triangle relative to the total area AAA of the triangle.

In the context of a three-node linear triangular element, the area coordinates satisfy certain properties that align them naturally with the shape functions needed for interpolating the field variable. They are non-negative and sum to unity (Equation 6.41), and each area coordinate assumes a specific value (unity) at one of the nodes while being zero at the others (Equation 6.42). Therefore, the field variable ϕ(x,y)\phi(x, y)ϕ(x,y) can be represented simply as a weighted sum of the nodal values using the area coordinates:

ϕ(x,y)=L1ϕ1+L2ϕ2+L3ϕ3\phi(x, y) = L_1\phi_1 + L_2\phi_2 + L_3\phi_3ϕ(x,y)=L 1 ​ ϕ 1 ​ +L 2 ​ ϕ 2 ​ +L 3 ​ ϕ 3 ​

Although this looks different from the Cartesian-based representation in Equation 6.36, once the area coordinates are expressed in terms of nodal coordinates, the two representations are indeed equivalent.

Moving on to higher-order elements like the six-node quadratic triangular element, the passage illustrates how the use of area coordinates becomes even more beneficial. With six nodes, a quadratic polynomial is used to represent the field variable (Equation 6.44), and this is rewritten using six interpolation functions Ni(x,y)N_i(x, y)N i ​ (x,y) that are quadratic and satisfy specific nodal conditions. By examining the geometry and behavior of lines of constant area coordinates, one can construct the interpolation functions systematically.

For instance, the interpolation function N1N_1N 1 ​ is designed to be unity at node 1 and zero at all other nodes. By expressing N1N_1N 1 ​ in terms of area coordinates and enforcing these conditions, the authors arrive at a simplified quadratic form N1=L1(2L1−1)N_1 = L_1(2L_1 – 1)N 1 ​ =L 1 ​ (2L 1 ​ −1).

Similarly, the rest of the interpolation functions are derived to meet the nodal requirements, resulting in expressions such as N2=L2(2L2−1)N_2 = L_2(2L_2 – 1)N 2 ​ =L 2 ​ (2L 2 ​ −1), N3=L3(2L3−1)N_3 = L_3(2L_3 – 1)N 3 ​ =L 3 ​ (2L 3 ​ −1), etc., with bilinear products N4,N5,N_4, N_5,N 4 ​ ,N 5 ​ , and N6N_6N 6 ​ corresponding to the midside nodes.

The passage suggests that this method can be extended to derive interpolation functions for even higher-order elements, such as the 10-node cubic element, with similar procedures. The advantage of using area coordinates lies in their simplicity and ease of manipulation when constructing higher-order shape functions, as well as in the reduced complexity of integrating these functions over the element domain. This approach streamlines the process of deriving element characteristic matrices and ultimately speeds up the solution process in finite element analysis. As seen in Chapter 5 and encountered again in later chapters, integration of various forms of the interpolation functions over the domain of an element are required in formulating element characteristic matrices and load vectors. When expressed in area coordinates, integrals of the form A La 1Lb 2Lc 3 dA (6.48) (where A is the total area of a triangle defined by nodes 1, 2, 3) must often be evaluated. The relation A La 1Lb 2Lc 3 dA = (2A) a!b!c! (a + b + c + 2)! (6.49) has been shown [7] to be valid for all exponents a, b, c that are positive integers (recall 0! forward. = 1). Therefore, integration in area coordinates is quite straightAs will be shown in Chapter 7, the convection terms of the stiffness matrix for a 2-D heat transfer element are of the form ki j = A h Ni Nj dA where h is the convection coefficient and A is element area. Use the interpolation functions for a six-node triangular element given by Equation 6.47 to compute k24 . ■ Solution Using Equation 6.47b and 6.47d, we have N2 = L2(2L2 − 1) N4 = 4L1 L2 so (assuming h is a constant) k24 = h A L2(2L2 − 1)4L1 L2 dA = h A 8L1 L3 2 − 4L1 L2 2 dA Applying Equation 6.49, we have h A 8L1 L3 2 dA = 8h(2A) (1!)(3!)(0!) (1 + 3 + 0 + 2)! = 96h A 720 = 2h A 15 h A 4L1 L2 2 dA = 4h(2A) (1!)(2!)(0!) (1 + 2 + 0 + 2)! = 16h A 120 = 2h A 15 Therefore, k24 = 2h A 15 − 2h A 15 = 0

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