Question

What is the stiffness matrix for the beam shown in the figure (fixed at one end, guided roller at the other)?

• A. $\begin{bmatrix} \frac{AE}{L} & \frac{4EI}{L} \\ 0 & 0 \end{bmatrix}$
• B. $\begin{bmatrix} \frac{AE}{L} & 0 \\ \frac{3EI}{L} & 0 \end{bmatrix}$
• C. $\begin{bmatrix} \frac{AE}{L} & 0 \\ 0 & \frac{4EI}{L} \end{bmatrix}$
• D. $\begin{bmatrix} 0 & \frac{4EI}{L} \\ 0 & \frac{AE}{L} \end{bmatrix}$

Hint:

For standard 1D beam elements, axial and flexural effects are uncoupled. This means the off-diagonal terms in the stiffness matrix related to axial force vs. rotation will always be zero.

Answer 1

The Correct Option is C

Concept: The stiffness matrix $[K]$ relates displacements to forces. The element $k_{ij}$ represents the force required at coordinate $i$ due to a unit displacement at coordinate $j$, while all other displacements are zero.

Step 1: Analyze the axial stiffness (Coordinate 1).
A unit axial displacement ($\delta = 1$) at the free end requires an axial force: \[ F = \frac{AE}{L} \delta \Rightarrow k_{11} = \frac{AE}{L} \] Since axial displacement doesn't produce rotation in a linear beam, $k_{21} = 0$.

Step 2: Analyze the rotational stiffness (Coordinate 2).
For a beam fixed at one end, the moment required to produce a unit rotation ($\theta = 1$) at the other end is: \[ M = \frac{4EI}{L} \theta \Rightarrow k_{22} = \frac{4EI}{L} \] Since rotation doesn't produce axial force in this idealized case, $k_{12} = 0$.

Step 3: Assemble the matrix.
\[ [K] = \begin{bmatrix} k_{11} & k_{12} \\ k_{21} & k_{22} \end{bmatrix} = \begin{bmatrix} \frac{AE}{L} & 0 \\ 0 & \frac{4EI}{L} \end{bmatrix} \]