UNIS Template
Finite Element (FE) equili
ium equation: ku=f
1D spring element
1D bar element
Beam element
1
Finite Element Method (FEM) formulation
2
Finite Element Method (FEM) formulation
Nodal force: In FE formulation, loading has only been applied on nodes.
Distributed load: In some cases, distributed loading is applied along elements, where the distributed loads can be converted to the nodes by using equivalent nodal forces.
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Classroom activity #14
1
2
3
(2)
(1)
EA
2EA
L
L
A step-bar system with uniform material density uniform Young’s modulus cross-section areas of for bar AB and for bar BC. The gravity () is considered as a distributed force. Determine (1) the displacement at the middle node B and (2) the reaction forces from the wall A (Exam 2014, 2017, Quiz 2015).
A
B
C
Solution
Step 0: Analysis of the problem:
Type of elements: Bar (loading follows axial direction of structural components)
Number of elements (one thick bar and one thin bar elements) , : 2
Number of degree of freedom:
Number of nodes (1, 2, 3) = 3
Step 1: uniformly distributed gravitational load:
Element AB (1):
Element BC (2):
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Classroom Activity #14 – Cont’d
Step 3: Expanded elemental equili
ium equations for global equili
ium equation:
Step 2: Elemental equili
ium equations:
Element A:
Element B:
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Classroom Activity #14 – Cont’d
Step 3: Apply boundary and loading conditions:
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Classroom Activity #15
1
2
3
(2)
(1)
EA
2EA
L
L
A
B
C
If there is a “stopper” is placed in the bottom, please solve the problem again.
which has are two unknowns: ,
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Classroom Activity #15 – Cont’d
1
2
3
(2)
(1)
EA
2EA
L
L
A
B
C
From the second equation above:
From the “dropped equation” in the global equili
ium
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Chapter 7 Finite Element Method
7.5 Vi
ation Analysis
MECH3361/9361 Mechanics of Solids II
Qing Li
School of Aerospace, Mechanical and Mechatronic Engineering
9
7.5 FE methods for vi
ational analysis
Vi
ational analysis is often performed on mechanical systems to evaluate their dynamic behaviour and stability during vi
ations
Single degree-of-freedom systems: A system consisting of one mass, one spring, and one damper vi
ating in one direction is a typical single DOF system, A single mass vi
ating in a single direction is a single DOF system
We are concerned with the motion of the mass m, i.e. x(t). The mass m is considered to be a rigid body.
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7.5 FEM for vi
ational analysis – Multiple degree of freedom
A 3D single mass can have a maximum of 6 DOFs, i.e. 3 DOFs in translation (straight line motion) and 3 DOFs in rotation.
Multiple degree of freedom: A system that must be described by two or more independent co-ordinates is called a multi-DOF system.
A single mass vi
ating in different directions constitutes a multi DOF system, as well as multiple masses vi
ating in a single (or multiple) direction(s).
General Multiple DOF: In an n-DOF system, the number of co-ordinates required to describe the motion of the masses is N. An N-DOF system has N number of natural frequencies (ωn). For each ωn, there is a co
esponding vi
ation mode (i.e. mode shape).
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7.5.3 FEM for vi
ational analysis - Equations of motion
Displacement of each mass (i.e. x(t)) is measured from its static equili
ium position.
Thus we can ignore the gravity force of each mass and the initial deformation of the springs when we derive the vi
ation differential equations using Newton’s second law.
Dynamic equili
ium positions
Static equili
ium
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7.5.3 FEM for vi
ational analysis - Equations of motion
Equations of motion: consider the free body diagrams of the 2 DOF system. Newton’s second law () can be applied to each mass.
For mass 2:
For mass 1:
Express these two equations in matrix form
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7.5.3 FEM for vi
ational analysis - Equations of motion
is the mass matrix
is the stiffness matrix;
is the unknown displacement vecto
Equations of motion:
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7.5.4. Characteristic free vi
ation (synchronous)
Synchronous solutions: In the above example, each mass undergoes harmonic motion at the same frequency (synchronous) and exhibits a certain vi
ation mode. Therefore, we assume the following synchronous solutions
Substituting the synchronous solutions into the equation of motion:
Characteristic equation: For a non-trivial solution (at least one of the A’s is non-zero):
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7.5.5. Natural frequencies
Natural frequencies can be found from the solutions to the characteristic equation. For simplicity, let’s look at a special system where and . The characteristic equation becomes:
Divide by , and put , where is an assumed reference frequency.
This is a quadratic equation, in terms of . Therefore, solving the quadratic equation
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7.5.5. Natural frequencies
Two natural frequencies (two DOF) are obtained, and .
1st natural circular frequency:
2nd natural circular frequency:
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7.5.6. Mode Shape
For each natural frequency , there is a co
esponding mode shape.
For a multi DOF spring-mass system, a mode shape is simply an amplitude ratio.
Recall the synchronous solutions, we provided two synchronous solutions with different amplitudes and .
Hence, the mode shape can be expressed mathematically as: .
Mode shape expressions can be derived from either equations
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7.5.6. Mode Shape – Cont’d
Mode shape expressions can be derived from either equations
Both ratios are equivalent for a given . For our special case in section 7.5.5, this can be confirmed by substituting the natural circular frequency : into both equations in
F
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7.5.6. Mode Shape – Cont’d
Similarly, for
For convenience, the amplitude is taken to be a unit (i.e. ). As a result, the mode shapes can also be expressed in terms of a vector form as
1st mode shape:
2nd mode shape:
Remarks: these are actually the eigenvectors co
esponding to the eigenvalues ( found from the characteristic equation
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7.5.6. Mode Shape – Cont’d
For and , the first order vi
ation mode (if A2=1, then A1=0.618) can be drawn
For and , the second order vi
ation mode (if A2=1, then A1= XXXXXXXXXXcan be drawn. Notably, when vi
ating at the 2nd order natural frequency, the two masses are out of phase, i.e. they are vi
ating in different directions.
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UNIS Template
Finite Element (FE) equili
ium equation: ku=f
1D spring element
1D bar element
Beam element
1
Solution
Step 0: Analysis of the problem:
Number of elements (one bar and one spring elements) : 2
Classroom Activity #12
A spring – bar system with the spring constant and . It is assumed that an axial force is applied in node 2 and a stopper is placed next to node 2 with a gap of . Determine (1) the displacement at the middle node and (2) the reaction force from the stoppe
walls.
Number of degrees of freedom:
Step 1: Elemental equili
ium equations:
Number of nodes (1, 2, 3) = 3
Bar Element
Spring Element
2
Classroom Activity #12 – Cont’d
Step 2: Expanded elemental equili
ium equations for global equili
ium equation:
3
Classroom Activity #12 – Cont’d
Step 3: Apply boundary and loading conditions:
Note that includes all the forces applied at node 2, which comprises external force and reaction from the stopper .
Equili
ium
Thus
4
Assignment 3 – Q1 1D finite element method (10 marks)
The two horizontal bars are connected by a linear spring. The system is fully clamped at ends A and D, as illustrated in Fig. 1.
1m
1m
1m
EA=2
k
s
=
3
EA=1
x
2P
4P
A
B
C
D
0.5m
0.5m
O
Discretise the system into two bar elements and one spring element. Write each elemental equili
ium equation, and compile the global equili
ium equation.
Calculate the displacements at B and C, and the reaction forces at A and D.
Calculate the displacement at O (the midpoint of spring element) by using shape functions.
Plot the distributions of displacement and strain in the system.
If a stopper is placed to the left of node C with a distance of separation =0.1P as shown in