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Shape functions, eight-noded rectangular element

Short answer

For an eight-noded rectangle in natural coordinates (xi,eta), mid-side shape functions are N5=0.5(1-xi^2)(1-eta), N6=0.5(1+xi)(1-eta^2), N7=0.5(1-xi^2)(1+eta), N8=0.5(1-xi)(1-eta^2). Corner shape functions are Ni=0.25(1+xi*xi_i)(1+eta*eta_i)(xi*xi_i+eta*eta_i-1).

Derive the shape functions for the eight-noded rectangular element given in the figure.

[Figure: eight-noded 2-D rectangular element in natural coordinates ξ\xi, η\eta with origin at the centre and half-sides of 1 (or aa and bb); corner nodes 1, 2, 3, 4 taken anticlockwise from the bottom-left corner, and mid-side nodes 5 (bottom), 6 (right), 7 (top), 8 (left)]

Natural coordinates and nodes

Use ξ=x/a\xi=x/a, η=y/b\eta=y/b so the element is the square −1≤ξ,η≤1-1\le\xi,\eta\le1 (for half-sides a,ba,b; if the figure has unit half-sides, ξ=x\xi=x, η=y\eta=y). Nodes:

Node12345678
(ξ,η)(\xi,\eta)(−1,−1)(-1,-1)(1,−1)(1,-1)(1,1)(1,1)(−1,1)(-1,1)(0,−1)(0,-1)(1,0)(1,0)(0,1)(0,1)(−1,0)(-1,0)
   4-----7-----3
   |           |
   8     +     6      eta ^
   |           |          |
   1-----5-----2          +--> xi

Assumed displacement

Eight nodal values need an 8-term polynomial (the serendipity family: complete quadratic plus two cubic terms):

u=α1+α2ξ+α3η+α4ξ2+α5ξη+α6η2+α7ξ2η+α8ξη2u=\alpha_1+\alpha_2\xi+\alpha_3\eta+\alpha_4\xi^2+\alpha_5\xi\eta+\alpha_6\eta^2+\alpha_7\xi^2\eta+\alpha_8\xi\eta^2

Writing it at the 8 nodes gives {u}=[C]{α}\{u\}=[C]\{\alpha\} and N=[1 ξ η … ][C]−1N=[1\ \xi\ \eta\ \dots][C]^{-1}. The same result is obtained more quickly by building each NiN_i as a product of straight-line factors that vanish at all the other nodes, as below.

Mid-side node 5 (0,−1)(0,-1)

N5N_5 must vanish on the sides ξ=±1\xi=\pm1 and η=1\eta=1, and have a parabola along the bottom side: N5=C(1−ξ2)(1−η)N_5=C(1-\xi^2)(1-\eta). At node 5, C(1)(2)=1⇒C=12C(1)(2)=1\Rightarrow C=\tfrac12. Likewise for nodes 6, 7, 8:

N5=12(1−ξ2)(1−η)N6=12(1+ξ)(1−η2)N7=12(1−ξ2)(1+η)N8=12(1−ξ)(1−η2)\begin{aligned} N_5&=\tfrac12(1-\xi^2)(1-\eta)\\ N_6&=\tfrac12(1+\xi)(1-\eta^2)\\ N_7&=\tfrac12(1-\xi^2)(1+\eta)\\ N_8&=\tfrac12(1-\xi)(1-\eta^2) \end{aligned}

Corner node 1 (−1,−1)(-1,-1)

N1N_1 must vanish on the sides ξ=1\xi=1 and η=1\eta=1 (nodes 2, 6, 3, 7, 4) and at the mid-side nodes 5 and 8, which lie on the line 1+ξ+η=01+\xi+\eta=0: N1=C(1−ξ)(1−η)(1+ξ+η)N_1=C(1-\xi)(1-\eta)(1+\xi+\eta). At node 1: C(2)(2)(−1)=1⇒C=−14C(2)(2)(-1)=1\Rightarrow C=-\tfrac14. Hence

N1=−14(1−ξ)(1−η)(1+ξ+η)N_1=-\tfrac14(1-\xi)(1-\eta)(1+\xi+\eta)

and for the other corners the general form is

Ni=14(1+ξξi)(1+ηηi)(ξξi+ηηi−1),i=1,2,3,4\boxed{N_i=\tfrac14(1+\xi\xi_i)(1+\eta\eta_i)(\xi\xi_i+\eta\eta_i-1)},\quad i=1,2,3,4

that is

N2=14(1+ξ)(1−η)(ξ−η−1)N3=14(1+ξ)(1+η)(ξ+η−1)N4=14(1−ξ)(1+η)(−ξ+η−1)\begin{aligned} N_2&=\tfrac14(1+\xi)(1-\eta)(\xi-\eta-1)\\ N_3&=\tfrac14(1+\xi)(1+\eta)(\xi+\eta-1)\\ N_4&=\tfrac14(1-\xi)(1+\eta)(-\xi+\eta-1) \end{aligned}

Checks

At the centre (0,0)(0,0): corners give 4×14(−1)=−14\times\tfrac14(-1)=-1, mid-sides give 4×12=24\times\tfrac12=2; the sum is 11. In general ∑Ni=1\sum N_i=1 and each Ni=1N_i=1 at its own node and 00 at the other seven. Each side carries a quadratic variation through three nodes, so adjacent elements are compatible.

✓ Answer checked by Claude Opus 5.5 (high effort)

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