Shape Functions of beam element with 3 nodes (quadratic element)

Question

For a two node beam element there are four shape functions for four degree of freedom:

For a straight three node beam element how shape functions are?

Please note that beam is straight and not curved.

Answer 1

If the nodes are at $\xi = -1, 0, +1$ you can find the shape functions using Lagrangian polynomial interpolation.

In fact you don't need to work through the general procedure, since you can write down the general form the shape functions must take with only a few unknown parameters, and then solve for the unknown values.

Consider the shape functions that are non-zero, and have non-zero derivative, at $\xi = 1$. They must be zero, and have zero derivatives, at $\xi = -1, 0$. Therefore they must have the general form $$N(\xi) = (\xi+1)^2 \xi^2 (a\xi + b)$$ for some values of $a$ and $b$. Using the product rule, the derivative is $$N'(\xi) = 2(\xi+1)\xi(2\xi + 1)(a\xi + b) + (\xi+1)^2 \xi^2 a.$$ At $\xi = 1$ we therefore have $$\begin{align}N(1) &= 4a + 4b \\ N'(1) &= 16a + 12b\end{align}$$

For one shape function we want $N(1) = 1, N'(1) = 0$ and for the other, $N(1) = 0, N'(1) = 1$.

Solving the simultaneous equations for $a$ and $b$ gives the two shape functions as $$\begin{gather}(\xi+1)^2 \xi^2 (-3\xi+4)/4 \\ (\xi+1)^2 \xi^2 (\xi-1)/4 \end{gather}$$

You can get the shape functions that are nonzero at $\xi = -1$ simply by changing $\xi$ to $-\xi$ (be careful with the sign of the non-zero slope shape function!)

The shape functions for the middle node can be found in the same way, but if you see what is going on you can just write them down by inspection: $$\begin{gather}(\xi-1)^2(\xi+1)^2 \\ (\xi-1)^2 \xi (\xi+1)^2 \end{gather}$$