Iso/super/sub-parametric; Hermite shape function

Types of parametric elements

• Isoparametric: the same shape functions (same number of nodes) are used for the geometry and for the displacement, $x=\sum N_ix_i$, $u=\sum N_iu_i$.
• Superparametric: the geometry is described by a higher-order interpolation (more nodes) than the displacement, e.g. quadratic geometry and linear displacement.
• Subparametric: the geometry is described by a lower-order interpolation than the displacement, e.g. straight sides with a quadratic displacement.

Hermite shape functions for a beam element

A beam element of length $L$ has two nodes with two DOFs each: deflection $v$ and slope $\theta=dv/dx$. Both $v$ and $\theta$ must be continuous between elements ($C^1$ continuity), so a cubic (4 constants) is assumed:

$$v(x)=\alpha_1+\alpha_2x+\alpha_3x^2+\alpha_4x^3$$ $$\theta(x)=\frac{dv}{dx}=\alpha_2+2\alpha_3x+3\alpha_4x^2$$

Apply the four nodal conditions: $v(0)=v_1$, $\theta(0)=\theta_1$, $v(L)=v_2$, $\theta(L)=\theta_2$:

$$\begin{Bmatrix}v_1\\\theta_1\\v_2\\\theta_2\end{Bmatrix}=\begin{bmatrix}1&0&0&0\\0&1&0&0\\1&L&L^2&L^3\\0&1&2L&3L^2\end{bmatrix}\begin{Bmatrix}\alpha_1\\\alpha_2\\\alpha_3\\\alpha_4\end{Bmatrix}$$

Solving: $\alpha_1=v_1$, $\alpha_2=\theta_1$,

$$\alpha_3=\frac{3(v_2-v_1)}{L^2}-\frac{2\theta_1+\theta_2}{L},\qquad \alpha_4=\frac{2(v_1-v_2)}{L^3}+\frac{\theta_1+\theta_2}{L^2}$$

Substituting and collecting the coefficients of $v_1,\theta_1,v_2,\theta_2$ gives $v=[N_1\ N_2\ N_3\ N_4]\{d\}$ with

$$\begin{aligned} N_1&=1-\frac{3x^2}{L^2}+\frac{2x^3}{L^3}\\ N_2&=x-\frac{2x^2}{L}+\frac{x^3}{L^2}\\ N_3&=\frac{3x^2}{L^2}-\frac{2x^3}{L^3}\\ N_4&=-\frac{x^2}{L}+\frac{x^3}{L^2} \end{aligned}$$

In terms of $s=x/L$: $N_1=1-3s^2+2s^3$, $N_2=L(s-2s^2+s^3)$, $N_3=3s^2-2s^3$, $N_4=L(-s^2+s^3)$.

Check: $N_1(0)=1$, $N_1(L)=0$, $N_1'(0)=N_1'(L)=0$; $N_2'(0)=1$, $N_2(0)=N_2(L)=N_2'(L)=0$, and so on for $N_3,N_4$. Also $N_1+N_3=1$ (rigid-body translation is represented).