- •Copyright Notice
- •Table of Contents
- •Chapter 1. Introduction
- •I. Basic Concepts
- •Examples:
- •Why Finite Element Method?
- •Applications of FEM in Engineering
- •Examples:
- •A Brief History of the FEM
- •FEM in Structural Analysis (The Procedure)
- •Example:
- •Available Commercial FEM Software Packages
- •Objectives of This FEM Course
- •II. Review of Matrix Algebra
- •Linear System of Algebraic Equations
- •Matrix Addition and Subtraction
- •Scalar Multiplication
- •Matrix Multiplication
- •Transpose of a Matrix
- •Symmetric Matrix
- •Unit (Identity) Matrix
- •Determinant of a Matrix
- •Singular Matrix
- •Matrix Inversion
- •Examples:
- •Solution Techniques for Linear Systems of Equations
- •Positive Definite Matrix
- •Differentiation and Integration of a Matrix
- •Types of Finite Elements
- •III. Spring Element
- •One Spring Element
- •Spring System
- •Checking the Results
- •Notes About the Spring Elements
- •Example 1.1
- •Chapter 2. Bar and Beam Elements
- •I. Linear Static Analysis
- •II. Bar Element
- •Stiffness Matrix --- Direct Method
- •Stiffness Matrix --- A Formal Approach
- •Example 2.1
- •Example 2.2
- •Distributed Load
- •Bar Elements in 2-D and 3-D Space
- •2-D Case
- •Transformation
- •Stiffness Matrix in the 2-D Space
- •Element Stress
- •Example 2.3
- •Example 2.4 (Multipoint Constraint)
- •3-D Case
- •III. Beam Element
- •Simple Plane Beam Element
- •Direct Method
- •Formal Approach
- •3-D Beam Element
- •Example 2.5
- •Equivalent Nodal Loads of Distributed Transverse Load
- •Example 2.6
- •Example 2.7
- •FE Analysis of Frame Structures
- •Example 2.8
- •Chapter 3. Two-Dimensional Problems
- •I. Review of the Basic Theory
- •Plane (2-D) Problems
- •Stress-Strain-Temperature (Constitutive) Relations
- •Strain and Displacement Relations
- •Equilibrium Equations
- •Exact Elasticity Solution
- •Example 3.1
- •II. Finite Elements for 2-D Problems
- •A General Formula for the Stiffness Matrix
- •Constant Strain Triangle (CST or T3)
- •Linear Strain Triangle (LST or T6)
- •Linear Quadrilateral Element (Q4)
- •Quadratic Quadrilateral Element (Q8)
- •Example 3.2
- •Transformation of Loads
- •Stress Calculation
- •I. Symmetry
- •Types of Symmetry:
- •Examples:
- •Applications of the symmetry properties:
- •Examples:
- •Cautions:
- •II. Substructures (Superelements)
- •Physical Meaning:
- •Mathematical Meaning:
- •Advantages of Using Substructures/Superelements:
- •Disadvantages:
- •III. Equation Solving
- •Direct Methods (Gauss Elimination):
- •Iterative Methods:
- •Gauss Elimination - Example:
- •Iterative Method - Example:
- •IV. Nature of Finite Element Solutions
- •Stiffening Effect:
- •V. Numerical Error
- •VI. Convergence of FE Solutions
- •Type of Refinements:
- •Examples:
- •VII. Adaptivity (h-, p-, and hp-Methods)
- •Error Indicators:
- •Examples:
- •Chapter 5. Plate and Shell Elements
- •Applications:
- •Forces and Moments Acting on the Plate:
- •Stresses:
- •Relations Between Forces and Stresses
- •Thin Plate Theory ( Kirchhoff Plate Theory)
- •Examples:
- •Under uniform load q
- •Thick Plate Theory (Mindlin Plate Theory)
- •II. Plate Elements
- •Kirchhoff Plate Elements:
- •Mindlin Plate Elements:
- •Discrete Kirchhoff Element:
- •Test Problem:
- •Mesh
- •III. Shells and Shell Elements
- •Example: A Cylindrical Container.
- •Shell Theory:
- •Shell Elements:
- •Curved shell elements:
- •Test Cases:
- •Chapter 6. Solid Elements for 3-D Problems
- •I. 3-D Elasticity Theory
- •Stress State:
- •Strains:
- •Stress-strain relation:
- •Displacement:
- •Strain-Displacement Relation:
- •Equilibrium Equations:
- •Stress Analysis:
- •II. Finite Element Formulation
- •Displacement Field:
- •Stiffness Matrix:
- •III. Typical 3-D Solid Elements
- •Tetrahedron:
- •Hexahedron (brick):
- •Penta:
- •Element Formulation:
- •Solids of Revolution (Axisymmetric Solids)
- •Axisymmetric Elements
- •Applications
- •Chapter 7. Structural Vibration and Dynamics
- •I. Basic Equations
- •A. Single DOF System
- •B. Multiple DOF System
- •Example
- •II. Free Vibration
- •III. Damping
- •IV. Modal Equations
- •V. Frequency Response Analysis
- •VI. Transient Response Analysis
- •B. Modal Method
- •Cautions in Dynamic Analysis
- •Examples
- •Chapter 8. Thermal Analysis
- •Further Reading
Lecture Notes: Introduction to Finite Element Method |
Chapter 2. Bar and Beam Elements |
Example 2.7
Y |
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E,I |
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k |
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L |
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4 |
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Given: P = 50 kN, k = 200 kN/m, L = 3 m,
E = 210 GPa, I = 2×10-4 m4.
Find: Deflections, rotations and reaction forces.
Solution:
The beam has a roller (or hinge) support at node 2 and a spring support at node 3. We use two beam elements and one spring element to solve this problem.
The spring stiffness matrix is given by,
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v3 |
v4 |
ks |
k |
− k |
= |
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− k |
k |
Adding this stiffness matrix to the global FE equation (see
Example 2.5), we have
© 1997-2002 Yijun Liu, University of Cincinnati |
65 |
Lecture Notes: Introduction to Finite Element Method |
Chapter 2. Bar and Beam Elements |
v1 |
θ1 |
v2 |
θ2 |
v3 |
θ3 |
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12 |
6L |
−12 |
6L |
0 |
0 |
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2 |
−6L |
2 |
0 |
0 |
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4L |
2L |
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24 |
0 |
−12 |
6L |
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EI |
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2 |
−6L |
2 |
L3 |
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8L |
2L |
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12 + k' |
−6L |
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4L2 |
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Symmetry |
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in which |
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k'= |
L3 |
k |
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EI |
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is used to simply the notation. |
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We now apply the boundary conditions, |
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v1 =θ1 = v2 = v4 = 0, |
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M2 = M3 = 0, |
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F3Y = −P |
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v4 |
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0 |
v1 |
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F1Y |
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0 |
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θ1 |
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M1 |
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0 |
v |
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0 |
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2Y |
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= M |
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2 |
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− k' v3 |
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F3Y |
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0 |
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θ3 |
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M3 |
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k' v4 |
F4Y |
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‘Deleting’ the first three and seventh equations (rows and columns), we have the following reduced equation,
8L2 |
−6L 2L2 |
θ2 |
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EI |
−6L |
12 + k' −6L |
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L3 |
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v3 |
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= − P |
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2 |
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−6L |
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2L |
4L |
θ3 |
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0 |
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Solving this equation, we obtain the deflection and rotations at node 2 and node 3,
© 1997-2002 Yijun Liu, University of Cincinnati |
66 |
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
θ2 |
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3 |
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v3 |
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= − |
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7L |
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EI (12 + |
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7k') |
9 |
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θ3 |
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The influence of the spring k is easily seen from this result. Plugging in the given numbers, we can calculate
θ2 −0.002492 rad
v3 = −0.01744 mθ3 −0.007475 rad
From the global FE equation, we obtain the nodal reaction forces as,
F1Y |
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−69.78 kN |
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M1 |
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−69.78 kN m |
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= |
116.2 kN |
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2Y |
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F |
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3.488 kN |
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4Y |
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Checking the results: Draw free body diagram of the beam
69.78 kN |
50 kN |
1 |
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69.78 kN m |
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116.2 kN |
3.488 kN |
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© 1997-2002 Yijun Liu, University of Cincinnati |
67 |
Lecture Notes: Introduction to Finite Element Method |
Chapter 2. Bar and Beam Elements |
FE Analysis of Frame Structures
Members in a frame are considered to be rigidly connected. Both forces and moments can be transmitted through their joints. We need the general beam element (combinations of bar and simple beam elements) to model frames.
Example 2.8 |
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500 lb/ft |
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3000 lb |
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E, I, A |
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8 ft |
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4 |
X |
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12 ft |
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Given: |
E = 30 ×106 psi, |
I = 65 in.4 , |
A = 68. in.2 |
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Find: Displacements and rotations of the two joints 1 and 2.
Solution:
For this example, we first convert the distributed load to its equivalent nodal loads.
© 1997-2002 Yijun Liu, University of Cincinnati |
68 |
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
3000 lb |
3000 lb |
72000 lb-in. |
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3000 lb |
1 |
1 |
2 |
72000 lb-in. |
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4 |
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In local coordinate system, the stiffness matrix for a general 2-D beam element is
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ui |
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vi |
θi |
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uj |
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vj |
θj |
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EA |
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0 |
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− |
EA |
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0 |
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0 |
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L |
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0 |
12EI |
6EI |
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− |
12EI |
6EI |
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L3 |
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L2 |
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L3 |
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L2 |
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0 |
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6EI |
4EI |
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− |
6EI |
2EI |
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k = |
EA |
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L2 |
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L |
EA |
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− |
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12EI |
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6EI |
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6EI |
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0 |
− |
− |
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− |
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L3 |
L2 |
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L3 |
L2 |
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0 |
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6EI |
2EI |
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− |
6EI |
4EI |
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2 |
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L |
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© 1997-2002 Yijun Liu, University of Cincinnati |
69 |
Lecture Notes: Introduction to Finite Element Method |
Chapter 2. Bar and Beam Elements |
Element Connectivity Table
Element |
Node i (1) |
Node j (2) |
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1 |
1 |
2 |
2 |
3 |
1 |
3 |
4 |
2 |
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For element 1, we have
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u1 |
v1 |
θ1 |
u2 |
v2 |
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1417. |
0 |
0 |
−1417. |
0 |
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0 |
0.784 |
56.4 |
0 |
−0.784 |
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0 |
56.4 |
5417 |
0 |
−56.4 |
k1 |
= k1 |
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' = 104 × |
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0 |
0 |
1417. |
0 |
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−1417. |
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0 |
−0.784 |
−56.4 |
0 |
0.784 |
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0 |
56.4 |
2708 |
0 |
−56.4 |
For elements 2 and 3, the stiffness matrix in local system is,
θ2
0
56.4
2708
0 −56.4 5417
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ui ' |
vi ' |
θi ' |
uj ' |
v j ' |
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212.5 |
0 |
0 |
−212.5 |
0 |
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0 |
2.65 |
127 |
0 |
−2.65 |
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0 |
127 |
8125 |
0 |
−127 |
k2 |
' = k3 |
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' = 104 × |
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0 |
0 |
212.5 |
0 |
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−212.5 |
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0 |
−2.65 |
−127 |
0 |
2.65 |
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0 |
127 |
4063 |
0 |
−127 |
where i=3, j=1 for element 2 and i=4, j=2 for element 3.
θj '
0
127
4063
0 −127 8125
© 1997-2002 Yijun Liu, University of Cincinnati |
70 |
Lecture Notes: Introduction to Finite Element Method |
Chapter 2. Bar and Beam Elements |
In general, the transformation matrix T is,
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l |
m |
0 |
0 |
0 |
0 |
−m |
l |
0 |
0 |
0 |
0 |
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0 |
0 |
1 |
0 |
0 |
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0 |
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T = |
0 |
0 |
0 |
l |
m |
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0 |
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0 |
0 |
0 |
−m |
l |
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0 |
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0 |
0 |
0 |
0 |
0 |
1 |
We have
l = 0, m = 1
for both elements 2 and 3. Thus,
0 |
1 |
0 |
0 |
0 |
0 |
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−1 |
0 |
0 |
0 |
0 |
0 |
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0 |
1 |
0 |
0 |
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0 |
0 |
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T = |
0 |
0 |
0 |
0 |
1 |
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0 |
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0 |
0 |
0 |
−1 |
0 |
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0 |
0 |
0 |
0 |
0 |
1 |
Using the transformation relation,
k = TT k'T
we obtain the stiffness matrices in the global coordinate system for elements 2 and 3,
© 1997-2002 Yijun Liu, University of Cincinnati |
71 |
Lecture Notes: Introduction to Finite Element Method |
Chapter 2. Bar and Beam Elements |
k2 = 104
and
k3 = 104
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u3 |
2.65 |
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0 |
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−127 |
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× |
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−2.65 |
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0 |
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−127 |
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u4 |
2.65 |
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−127 |
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× |
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−2.65 |
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0 |
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−127 |
v3 |
θ3 |
u1 |
v1 |
θ1 |
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0 |
−127 |
−2.65 |
0 |
−127 |
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212.5 |
0 |
0 |
−212.5 |
0 |
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0 |
8125 |
127 |
0 |
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4063 |
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0 |
127 |
2.65 |
0 |
127 |
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−212.5 |
0 |
0 |
212.5 |
0 |
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0 |
4063 |
127 |
0 |
8125 |
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v4 |
θ4 |
u2 |
v2 |
θ2 |
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0 |
−127 |
−2.65 |
0 |
−127 |
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212.5 |
0 |
0 |
−212.5 |
0 |
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0 |
8125 |
127 |
0 |
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4063 |
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0 |
127 |
2.65 |
0 |
127 |
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−212.5 |
0 |
0 |
212.5 |
0 |
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0 |
4063 |
127 |
0 |
8125 |
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Assembling the global FE equation and noticing the following boundary conditions,
u3 = v3 = θ3 = u4 = v4 =θ4 = 0
F1X |
= 3000lb, F2 X = 0, |
F1Y = F2Y = −3000lb, |
M1 |
= −72000lb in., M2 |
= 72000lb in. |
we obtain the condensed FE equation,
© 1997-2002 Yijun Liu, University of Cincinnati |
72 |
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
144.3 |
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127 |
−1417. |
0 |
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0 |
u |
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213.3 |
56.4 |
0 |
−0.784 |
56.4 |
v1 |
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127 |
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56.4 |
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13542 |
0 |
−56.4 |
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2708 θ |
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104 × |
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144.3 |
0 |
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127 |
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−1417. |
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u2 |
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−0.784 |
−56.4 |
0 |
213.3 |
−56.4 |
v2 |
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56.4 |
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2708 |
127 |
−56.4 |
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13542 θ2 |
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3000 |
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−3000 |
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−72000 |
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−3000 |
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72000 |
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Solving this, we get |
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u1 |
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0.092 in. |
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v |
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0.00104in. |
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1 |
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0.00139 rad |
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θ |
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0.0901in. |
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u2 |
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v2 |
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−0.0018in. |
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θ |
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−388. ×10−5 rad |
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To calculate the reaction forces and moments at the two ends, we employ the element FE equations for element 2 and element 3. We obtain,
© 1997-2002 Yijun Liu, University of Cincinnati |
73 |
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
F |
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−672.7lb |
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3 X |
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2210 lb |
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F3Y |
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= |
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M3 |
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60364 lb in. |
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and |
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F |
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−2338lb |
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4 X |
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3825 lb |
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F4Y |
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= |
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|
|
M4 |
|
112641 lb in. |
||
Check the results:
Draw the free-body diagram of the frame. Equilibrium is maintained with the calculated forces and moments.
3000 lb |
3000 lb |
72000 lb-in. |
|
3000 lb |
|
|
60364 lb-in. |
|
672.7 lb |
2210 lb |
3825 lb |
72000 lb-in.
112641 lb-in.
2338 lb
© 1997-2002 Yijun Liu, University of Cincinnati |
74 |
