Derive beam element stiffness matrix

A beam element has 4 degrees of freedom: $\{d\} = \{v_1,\ \theta_1,\ v_2,\ \theta_2\}^T$ with corresponding nodal forces $\{f_1,\ m_1,\ f_2,\ m_2\}^T$.

   v1,f1               v2,f2
    |  (th1,m1)         |  (th2,m2)
    o===================o
    1         L         2

Displacement function

Since there are 4 nodal values, a cubic is chosen: $v(x) = a_1 + a_2x + a_3x^2 + a_4x^3$.

Applying $v(0) = v_1$, $v'(0) = \theta_1$, $v(L) = v_2$, $v'(L) = \theta_2$ and solving gives the shape functions (Hermite cubics):

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

Strain energy and stiffness

Curvature is $\kappa = \dfrac{d^2v}{dx^2} = [B]\{d\}$, where

$$[B] = \left[\frac{-6}{L^2} + \frac{12x}{L^3},\ \ \frac{-4}{L} + \frac{6x}{L^2},\ \ \frac{6}{L^2} - \frac{12x}{L^3},\ \ \frac{-2}{L} + \frac{6x}{L^2}\right]$$

The strain energy is $U = \dfrac{1}{2}\displaystyle\int_0^L EI\,\kappa^2\,dx$, which gives

$$[k] = EI\int_0^L [B]^T[B]\,dx$$

Evaluating the integrals term by term (for example $k_{11} = EI\int_0^L \left(\frac{12x}{L^3} - \frac{6}{L^2}\right)^2dx = \frac{12EI}{L^3}$):

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

Meaning of a column

Column 1 is the set of forces and moments needed to give a unit deflection $v_1 = 1$ at node 1 while $\theta_1 = v_2 = \theta_2 = 0$: shear $12EI/L^3$ and moment $6EI/L^2$ at node 1, and the opposite shear $-12EI/L^3$ and moment $6EI/L^2$ at node 2.

The matrix is symmetric and singular (rigid-body motion is possible until supports are applied).