Shape functions for a beam element

Beam element

A two-node beam element of length $L$ (Euler-Bernoulli) has at each node a transverse deflection $v$ and a rotation $\theta=dv/dx$: $\{d\}=\{v_1,\theta_1,v_2,\theta_2\}^T$. Because the bending moment $EI\,d^2v/dx^2$ must be finite, both $v$ and $dv/dx$ must be continuous at the nodes, so the displacement function must contain a cubic, the lowest polynomial with four constants:

$$v(x)=\alpha_1+\alpha_2x+\alpha_3x^2+\alpha_4x^3=[1\ \ x\ \ x^2\ \ x^3]\{\alpha\}$$ $$\theta(x)=\alpha_2+2\alpha_3x+3\alpha_4x^2$$

Nodal conditions

At $x=0$: $v=v_1$, $\theta=\theta_1$. At $x=L$: $v=v_2$, $\theta=\theta_2$.

$$\begin{Bmatrix}v_1\\\theta_1\\v_2\\\theta_2\end{Bmatrix}=\begin{bmatrix}1&0&0&0\\0&1&0&0\\1&L&L^2&L^3\\0&1&2L&3L^2\end{bmatrix}\begin{Bmatrix}\alpha_1\\\alpha_2\\\alpha_3\\\alpha_4\end{Bmatrix}\ \Rightarrow\ \{d\}=[C]\{\alpha\}$$ $$\{\alpha\}=[C]^{-1}\{d\}=\begin{bmatrix}1&0&0&0\\0&1&0&0\\-\dfrac{3}{L^2}&-\dfrac{2}{L}&\dfrac{3}{L^2}&-\dfrac{1}{L}\\\dfrac{2}{L^3}&\dfrac{1}{L^2}&-\dfrac{2}{L^3}&\dfrac{1}{L^2}\end{bmatrix}\{d\}$$

Shape functions

$$v(x)=[1\ x\ x^2\ x^3][C]^{-1}\{d\}=[N_1\ N_2\ N_3\ N_4]\{d\}$$ $$\begin{aligned} N_1&=1-\frac{3x^2}{L^2}+\frac{2x^3}{L^3}\\ N_2&=x-\frac{2x^2}{L}+\frac{x^3}{L^2}\\ N_3&=\frac{3x^2}{L^2}-\frac{2x^3}{L^3}\\ N_4&=-\frac{x^2}{L}+\frac{x^3}{L^2} \end{aligned}$$

These are the Hermite cubic functions. With $s=x/L$: $N_1=1-3s^2+2s^3$, $N_2=L(s-2s^2+s^3)$, $N_3=3s^2-2s^3$, $N_4=L(-s^2+s^3)$.

Properties

Also $N_1+N_3=1$, so a rigid-body translation is represented.

Use: stiffness matrix

Curvature $\kappa=\dfrac{d^2v}{dx^2}=[B]\{d\}$ with $[B]=\left[-\dfrac{6}{L^2}+\dfrac{12x}{L^3},\ -\dfrac{4}{L}+\dfrac{6x}{L^2},\ \dfrac{6}{L^2}-\dfrac{12x}{L^3},\ -\dfrac{2}{L}+\dfrac{6x}{L^2}\right]$, and

$$[k]=\int_0^L EI\,[B]^T[B]\,dx=\frac{EI}{L^3}\begin{bmatrix}12&6L&-12&6L\\6L&4L^2&-6L&2L^2\\-12&-6L&12&-6L\\6L&2L^2&-6L&4L^2\end{bmatrix}$$