Derive bar element stiffness matrix

A bar element carries only axial force. Take nodes 1 and 2 with axial displacements $u_1$ and $u_2$ and nodal forces $f_1$ and $f_2$.

  f1 -->  o===================o  --> f2
         node 1     L       node 2
         (u1)               (u2)

Direct method

The bar behaves like a spring of stiffness $k = \dfrac{AE}{L}$. The elongation is $\delta = u_2 - u_1$ and the axial force is $F = k\delta$.

Equilibrium of the element requires $f_1 = -F$ and $f_2 = +F$:

$$\begin{aligned} f_1 &= \frac{AE}{L}(u_1 - u_2) \\ f_2 &= \frac{AE}{L}(-u_1 + u_2) \end{aligned}$$

In matrix form:

$$\begin{Bmatrix} f_1 \\ f_2 \end{Bmatrix} = \frac{AE}{L}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}\begin{Bmatrix} u_1 \\ u_2 \end{Bmatrix}$$

Check by energy method

With linear shape functions $N_1 = 1 - x/L$, $N_2 = x/L$, the strain is $\varepsilon = \dfrac{du}{dx} = \dfrac{u_2 - u_1}{L}$, so $[B] = \dfrac{1}{L}[-1\ \ 1]$. Then

$$[k] = \int_0^L [B]^T AE [B]\,dx = AE L\cdot\frac{1}{L^2}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} = \frac{AE}{L}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$$

Both methods give the same result.

Properties

• Symmetric, since $k_{12} = k_{21}$.
• Singular ($\det = 0$): the element can move as a rigid body until supports are applied.
• Each row sums to zero, which expresses equilibrium for rigid-body motion.