Problem 30 - Concepts and Applications of Finite Element Analysis

Question

(a) A bar element of axial stiffness \(k=A E / L\) is permitted only axial nodal displacements \(u_{1}\) and \(u_{2} .\) Write \(\left[\mathrm{k}_{E E}\right]\), and from it determine \([\mathrm{k}]\). (b) Repeat part (a), but let there be four d.o.f. \(\\{d\\}=\left\lfloor\begin{array}{llll}u_{1} & v_{1} & u_{2} & \left.v_{2}\right\\}^{T} \text {, as }\end{array}\right.\) in Eq. \(2.4-3\), so that plane motion is possible.

Step-by-Step Solution

Answer

The 1D stiffness matrix for the bar is \(\left[\mathrm{k}_{E E}\right] = \frac{A E}{L} \left[\begin{array}{cc}1 & -1 \ -1 & 1\end{array}\right]\). In the 2D case, the stiffness matrix \([\mathrm{k}]\) is given by \(\left[\mathrm{k}\right] = \left[T\right]^T \left[\mathrm{k}_{E E}\right] \left[T\right]\), where the transformation matrix \(\left[T\right]\) depends on the orientation of the bar with respect to the global axes.

Step 1: Formulate the 1D Stiffness Matrix

For a bar element only permitted axial nodal displacements (\(u_{1}\) and \(u_{2}\)), the element stiffness matrix \(\left[\mathrm{k}_{E E}\right]\) in the local coordinate system can be written as: \[ \left[\mathrm{k}_{E E}\right] = \frac{A E}{L} \left[\begin{array}{cc}1 & -1 \ -1 & 1 \end{array}\right] \] where \(A\), \(E\), and \(L\) are the cross-sectional area, Young's modulus, and length of the bar respectively. The global stiffness matrix \([\mathrm{k}]\) is the same as \(\left[\mathrm{k}_{E E}\right]\) in 1D case.

Step 2: Formulate the 2D Stiffness Matrix

For a bar element allowed plane motion (i.e., has four degrees of freedom as given by the vector \(\{d\}=\left\{ {u_{1}, v_{1}, u_{2}, v_{2}}\right\}\)), the element stiffness matrix cannot be directly written like in the 1D case. It requires transforming the 1D stiffness matrix \(\left[\mathrm{k}_{E E}\right]\) to the 2D case. This involves multiplying \(\left[\mathrm{k}_{E E}\right]\) by a transformation matrix \(\left[T\right]\) and its transpose: \[\left[\mathrm{k}\right] = \left[T\right]^T \left[\mathrm{k}_{E E}\right] \left[T\right]\] The transformation matrix \(\left[T\right]\) is computed based on the angle the bar makes with respect to the global axes.

Key Concepts

Stiffness Matrix

The stiffness matrix is a key component in Finite Element Analysis (FEA), which describes how a structure resists deformational forces.
For a simple bar element with only axial displacements, the stiffness matrix \([\mathrm{k}_{E E}]\) is formulated using the properties of the material and its geometry. These properties include:

• \(A\) - the cross-sectional area
• \(E\) - Young's modulus, which is a measure of the material's stiffness
• \(L\) - the length of the bar

The basic 1D stiffness matrix for axial deformations only is:\[[\mathrm{k}_{E E}] = \frac{A E}{L} \begin{bmatrix} 1 & -1 \ -1 & 1 \end{bmatrix}\]This matrix reflects the relationship between forces and displacements at the nodal points of the element. It helps in calculating the response under applied loads. In this context, it essentially measures the resistance to axial stretching or compressing of the bar.

Degrees of Freedom

Degrees of Freedom (DOF) refer to the number of independent movements allowed in a system. They specify the possible motions at nodes in FEA models.
For a simple bar element, each node initially has only one DOF (axial displacement), denoted by \(u_1\) and \(u_2\). This 1D case is straightforward as the movement is along the length of the element.
When transitioning to a more complex system, such as a 2D or plane motion scenario, each node may have additional DOF. Here, the DOF \[\{d\} = \begin{bmatrix} u_1 & v_1 & u_2 & v_2 \end{bmatrix}^T\]\ increases to include both \(u\) (axial) and \(v\) (transverse) movements.
This complexity introduces the need for a larger stiffness matrix to account for movements in multiple dimensions, requiring transformation techniques for better accuracy in simulations.

Axial Displacement

Axial displacement refers to the change in position along the axis of a structural element. For a bar, it is the movement of nodes \(u_1\) and \(u_2\) in response to applied loads.
In Finite Element Analysis, accurately calculating axial displacement is crucial for determining how forces affect an object's shape and stability.

Axial displacement is determined using the stiffness matrix and is directly proportional to the applied force and inversely proportional to the stiffness.
Mathematically, the relationship can be expressed as: \[F = k imes u\]\ where \(F\) is the force applied, \(k\) is the stiffness derived from the stiffness matrix, and \(u\) represents the displacement.
Understanding axial displacement helps predict not only the immediate effects of loading but also any potential long-term issues like fatigue or failure.

Transformation Matrix

The transformation matrix is an essential tool for translating the stiffness matrix of an element from its local coordinate system to a global one.
When elements in a structure are not aligned along a global axis, their stiffness calculations need to be adjusted to account for their orientation.
In our problem, the transformation matrix \[T\]\ helps convert the 1D stiffness matrix \[\mathrm{k}_{E E}\]\ to a 2D system for elements capable of plane motion. This is achieved through:

• Multiplying the local stiffness matrix by the transformation matrix and its transpose
• Aligning the results with the global coordinate system, taking into account the element's angular position

This process ensures that the calculations for displacement and force consider the actual positional setup of the element within the overall structure, leading to more accurate and relevant results.