Study Notes for Spring and Bar Elements

Review for the Final Exam

Spring Element Overview

  • Components of a spring system:   - Nodes: Two nodes labeled as i and j.   - Nodal Displacements (U):     - Denoted as ( u_i ) and ( u_j ) which can be in units such as m or mm.   - Nodal Forces (F):     - Denoted as ( f_i ) and ( f_j ) which can be in lb or Newtons.   - Spring Constant (Stiffness):     - Noted as k with units (lb/in, N/m, N/mm).   - Spring Force-Displacement Relationship:     - Represented by the equation ( F = k \lambda )     - Where ( riangle = u_i - u_j );

Forces and Equilibrium

  • At node i:   - Forces acting can be described as ( F_i = k(u_j - u_i) )

  • At node j:   - Forces can also be expressed similarly.

Spring System Configurations

  • Elements configurations:   - Considering multiple springs.

  • For element 1 and 2, the forces can be related through their nodal displacements, indicated as follows:   - Node 1: ( F_1 ) and Node 2: ( F_2 ) with respective displacements ( u_1 ) and ( u_2 ).

Stiffness Matrix Assembly

  • The stiffness matrix for the system can be represented in matrix form:   - General form:
      (k1amp;−k1 −k1amp;k1+k2 −k2amp;k2 ext… ext… ext… ext… ext… ext… ext… ext… )\begin{pmatrix} k_1 & -k_1 \ -k_1 & k_1 + k_2 \ -k_2 & k_2 \ ext{…} \ ext{…} \ ext{…} \ ext{…} \ ext{…} \ ext{…} \ ext{…} \ ext{…} \ \end{pmatrix}   - Where each ( k ) is the stiffness of the respective nodes.

Boundary Conditions and Load Conditions

  • Assuming conditions for simple calculations:   - ( u_1 = 0 )   - If both ends are fixed, it can lead to a system of equations to solve for unknown forces like ( F_i ).

Example 1.1: Spring System Analysis

  • Given Spring Constants and Displacement:   - ( k_1 = 100 N/mm ), ( k_2 = 200 N/mm ), ( k_3 = 100 N/mm )   - Force ( P = 500 N )   - Assumed condition ( u_1 = 0 )

Part (a): Global Stiffness Matrix

  • The element stiffness matrices for springs can be summarized as:   - For Element 1:   k1=(100amp;−100 −100amp;100)k_1 = \begin{pmatrix} 100 & -100 \ -100 & 100 \end{pmatrix}   - For Element 2:   k2=(200amp;−200 −200amp;200)k_2 = \begin{pmatrix} 200 & -200 \ -200 & 200 \end{pmatrix}

Part (b): Calculating Node Displacements

  • Using the global stiffness matrix to find ( u_2 ) and ( u_3 ).   - Boundary conditions to remove corresponding rows and columns in matrix manipulation.

Part (c): Reaction Forces Calculation

  • Using overall equilibrium principles and equations yielding forces at nodes such as:   - ( F_1 = -100u_1 = -200 N )   - ( F_3 = -100u_3 = -300 N )

Part (d): Assessing Spring Forces

  • Evaluated spring force calculation for each element, such as:   - ( F = f_i = -f_j ) yields to numerically concluded forces based on found displacements.

Bar Elements

Uniform Prismatic Bar Overview

  • Configurations of bars include parameters:   - Length ( L ), cross-sectional area ( A ), and elastic modulus ( E ).

  • Displacement: Given as ( u ), as it varies along bar length.

  • Stress and Strain Relations:   - Stress: ( \sigma = E \cdot \varepsilon )   - Strain Relation: ( \varepsilon = \frac{du}{dx} )

Element Stiffness Matrix Development

  • Assumed linear variations in displacements lead to the calculation of:   - Stiffness ( k = \frac{EA}{L} ) based on previous stress-strain relations.

Global FE Equation Formulation

  • General equilibrium and stiffness formulation can derive in a structured multi-variable iterative solution method depicted as:   - EA(1amp;−1 −1amp;1)EA\begin{pmatrix} 1 & -1 \ -1 & 1 \end{pmatrix}

Beam Elements

  • Characteristics:   - Elements must account for bending moments, transverse shear forces, defining degree of freedom at each joint.   - Stiffness matrix in local coordinate representation relates to:   (12EIL3amp;6EIL2 6EIL2amp;4EIL)\begin{pmatrix} \frac{12EI}{L^3} & \frac{6EI}{L^2} \ \frac{6EI}{L^2} & \frac{4EI}{L} \end{pmatrix}

  • Utilization follows similar methodologies as springs and bars, based on determined stiffness with axial loads accounted separately.