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Derive shape functions of CST element

Short answer

For a linear triangle, shape functions are N_i=(a_i+b_i x+c_i y)/(2A), derived by solving nodal conditions. They satisfy Kronecker delta, sum to unity, and equal area coordinates. Strains are constant because b_i and c_i do not depend on x or y.

Derive the shape functions of a three-node linear triangular (constant strain triangle) element in terms of nodal coordinates. State their properties and explain why the element is called a constant strain triangle.

Consider a triangle with nodes 1, 2, 3 at (x1,y1)(x_1, y_1), (x2,y2)(x_2, y_2), (x3,y3)(x_3, y_3) numbered anticlockwise. Each node has two displacements (ui,vi)(u_i, v_i), so the element has 6 degrees of freedom.

         3 (x3,y3)
        / \
       /   \
      /     \
     1-------2

Displacement field

Choose a complete linear polynomial (from Pascal's triangle):

u=α1+α2x+α3y,v=α4+α5x+α6yu = \alpha_1 + \alpha_2x + \alpha_3y,\qquad v = \alpha_4 + \alpha_5x + \alpha_6y

Apply the nodal conditions for uu:

{u1u2u3}=[1x1y11x2y21x3y3]{α1α2α3}\begin{Bmatrix} u_1 \\ u_2 \\ u_3 \end{Bmatrix} = \begin{bmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 1 & x_3 & y_3 \end{bmatrix}\begin{Bmatrix} \alpha_1 \\ \alpha_2 \\ \alpha_3 \end{Bmatrix}

Solving by Cramer's rule, where the determinant of the matrix equals 2A2A (AA = area of triangle):

u=N1u1+N2u2+N3u3,Ni=12A(ai+bix+ciy)u = N_1u_1 + N_2u_2 + N_3u_3,\qquad N_i = \frac{1}{2A}\left(a_i + b_ix + c_iy\right)

with

a1=x2y3−x3y2,b1=y2−y3,c1=x3−x2a2=x3y1−x1y3,b2=y3−y1,c2=x1−x3a3=x1y2−x2y1,b3=y1−y2,c3=x2−x1\begin{aligned} a_1 &= x_2y_3 - x_3y_2, & b_1 &= y_2 - y_3, & c_1 &= x_3 - x_2 \\ a_2 &= x_3y_1 - x_1y_3, & b_2 &= y_3 - y_1, & c_2 &= x_1 - x_3 \\ a_3 &= x_1y_2 - x_2y_1, & b_3 &= y_1 - y_2, & c_3 &= x_2 - x_1 \end{aligned} 2A=x1(y2−y3)+x2(y3−y1)+x3(y1−y2)2A = x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)

The same NiN_i apply to vv, so

{uv}=[N10N20N300N10N20N3]{d}\begin{Bmatrix} u \\ v \end{Bmatrix} = \begin{bmatrix} N_1 & 0 & N_2 & 0 & N_3 & 0 \\ 0 & N_1 & 0 & N_2 & 0 & N_3 \end{bmatrix}\{d\}

Properties

  • Ni=1N_i = 1 at node ii and 00 at the other two nodes.
  • N1+N2+N3=1N_1 + N_2 + N_3 = 1 at every point (and ∑Nixi=x\sum N_ix_i = x, ∑Niyi=y\sum N_iy_i = y).
  • NiN_i varies linearly along each edge and depends only on the two nodes of that edge, so displacements are continuous between adjacent elements.
  • NiN_i equal the area coordinates Li=Ai/AL_i = A_i/A.

Why constant strain

The strains are

εx=∂u∂x=12A∑biui,εy=12A∑civi,γxy=12A∑(ciui+bivi)\varepsilon_x = \frac{\partial u}{\partial x} = \frac{1}{2A}\sum b_iu_i,\quad \varepsilon_y = \frac{1}{2A}\sum c_iv_i,\quad \gamma_{xy} = \frac{1}{2A}\sum (c_iu_i + b_iv_i)

bib_i and cic_i depend only on the nodal coordinates, not on xx or yy. So the strains (and stresses) are the same everywhere inside the element.

✓ Answer checked by Claude Opus 5.5 (high effort)

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