Linear shape functions for one-dimensional element

Displacement function

Two nodal values $u_1$ (at $x = 0$) and $u_2$ (at $x = L$) allow a linear polynomial:

$$u(x) = a_1 + a_2x$$

   u1                u2
   o-----------------o
  x=0               x=L

Apply the nodal conditions:

$$u(0) = a_1 = u_1,\qquad u(L) = a_1 + a_2L = u_2 \Rightarrow a_2 = \frac{u_2 - u_1}{L}$$

Substitute back:

$$u = u_1 + \frac{u_2 - u_1}{L}x = \left(1 - \frac{x}{L}\right)u_1 + \frac{x}{L}\,u_2$$

So $u = N_1u_1 + N_2u_2$ with

$$N_1 = 1 - \frac{x}{L},\qquad N_2 = \frac{x}{L}$$

Properties

• $N_i = 1$ at node $i$ and $0$ at the other node (Kronecker delta property).
• $N_1 + N_2 = 1$ everywhere (partition of unity); this allows rigid-body motion.
• Each $N_i$ is linear, so the displacement is continuous between elements ($C^0$), but the strain is constant per element.
• In natural coordinate $\xi = 2x/L - 1$: $N_1 = \dfrac{1 - \xi}{2}$ and $N_2 = \dfrac{1 + \xi}{2}$.

Strain-displacement matrix

$$\varepsilon = \frac{du}{dx} = \frac{dN_1}{dx}u_1 + \frac{dN_2}{dx}u_2 = \frac{1}{L}\left[-1\ \ 1\right]\begin{Bmatrix} u_1 \\ u_2 \end{Bmatrix}$$ $$[B] = \frac{1}{L}\left[-1\ \ 1\right]$$