INTRODUCTION, 1 Fundamentals of Vibration, 2 Causes of Vibrations, 3 Types of Vibrations, EQUATION OF MOTION, CONSISTENT MASS MATRICES, 1 Single DOF System, 2.Multiple DOF System, VECTOR ITERATION METHODS, MODELLING OF DAMPING, 1 Proportional Damping (Rayleigh Damping), 2 Frequency Response Analysis, TRANSIENT RESPONSE ANALYSIS, 1Cautions in Dynamic Analysis.
DYNAMIC ANALYSIS USING ELEMENT METHOD
It provides the basic equations necessary for structural dynamical analysis and developed both the lumped and the consistent mass matrix involved in the analysis of bar beam and spring elements.
1 Fundamentals of Vibration
Any motion which repeats itself after an interval of time is called vibration or oscillation or periodic motion
All bodies possessing mass and elasticity are capable of producing vibration.
2 Causes of Vibrations
o Unbalanced forces in the machine. These force are produced from within the machine itself
o Elastic nature of the system.
o Self excitations produced by the dry friction between the two mating surfaces. o External excitations applied on the system.
o Wind may causes vibrations
o Earthquakes may causes vibrations
3.Types of Vibrations
1.According to the actuating force
Free or natural vibrations
2.According to motion of system with respect to axis
EQUATION OF MOTION
There is two types of equation of motion
Longitudinal vibration of beam or axial vibration of a rod
Transverse vibration of a beam
where Ni, Nxi and Nyi are shape functions. This is an incompatible element! The stiffness matrix is still of the form
k = BTEBdV ,
where B is the strain-displacement matrix, and E the stress- strain matrix.
Minding Plate Elements:
Three independent fields.
Deflection w(x,y) is linear for Q4, and quadratic for Q8.
Discrete Kirchhoff Element:
Triangular plate element (not available in ANSYS). Start with a 6-node riangular element,
Total DOF = 9 (DKT Element).
Incompatible w(x,y); convergence is faster (w is cubic along each edge) and it is efficient.
L/t = 10, = 0.3
ANSYS 4-node quadrilateral plate element.
ANSYS Result for wc
Mesh wc ( PL2/D)
2 2 0.00593
4 4 0.00598
8 8 0.00574
16 16 0.00565
Exact Solution 0.00560
Question:Converges from “above”? Contradiction to what we learnt about the nature of the FEA solution?
Reason: This is an incompatible element ( See comments on p. 177).
Shells and Shell Elements
Shells – Thin structures witch span over curved surfaces.
Sea shell, egg shell (the wonder of the nature); Containers, pipes, tanks; Car bodies;
Roofs, buildings (the Superdome), etc. Forces in shells:
Membrane forces + Bending Moments
(cf. plates: bending only)
Example: A Cylindrical Container.
Thin shell theory
Shell theories are the most complicated ones to formulate and analyze in mechanics (Russian’s contributions).
Engineering Craftsmanship Demand strong analytical skill
cf.: bar + simple beam element => general beam element.
DOF at each node:
Based on shell theori es;
Most general shell el ements (flat shell and plate elements are subsets);
Complicated in form ulation.
CONSISTENT MASS MAT RICES
Natural frequencies a nd modes
Frequency response ( F(t)=Fo sinwt) Transient
response (F(t) arbitrary)
1 Single DOF System
f(t) = 0 and no damping (c = 0) Eq. (1) becomes
(meaning: inertia force + stiffness force = 0) Assume:
This is the circular natural frequency of the single DOF system (rad/s). The cyclic frequency (1/s = Hz) is
2.Multiple DOF System
Equation of Motion
Equation of motion for the whole structure is
Mu Cu Ku f (t) , (8)
in which: u nodal displacement vector,
M mass matrix,
C damping matrix,
K stiffness matrix, f forcing vector.
Physical meaning of Eq. (8):
Inertia forces + Damping forces + Elastic forces
= Applied forces
Lumped mass matrix (1-D bar element):
VECTOR ITERATION METHODS
Study of the dynamic characteristics of a structure: natural frequencies normal modes shapes)
Let f(t) = 0 and C = 0 (ignore damping) in the dynamic equation (8) and obtain
Mu Ku 0
Assume that displacements vary harmonically with time, that is,
where u i
s the vector of nodal displacement
Eq. (12) yields,
This is a generalized eigenvalue problem (EVP).
This is an n-th order polynomial of from which we can find n solutions (roots) or eigenvalues
i (i = 1, 2, …, n) are the natural frequencies (or characteristic frequencies) of the structure (the smallest one) is called the fundamental frequency. For each gives one
solution (or eigen) vector
if wi wj . That is, modes are orthogonal (or independent) to each other with respect to K and
Magnitudes of displacements (modes) or stresses in normal mode analysis have no physical meaning.
For normal mode analysis, no support of the structure is necessary.
i = 0 there are rigid body motions of the whole or a part of the structure. apply this to check
the FEA model (check for mechanism or free elements in the models).
Lower modes are more accurate than higher modes in the FE calculations (less spatial variations in the lower modes fewer elements/wave length are needed).
MODELLING OF DAMPING
Two commonly used models for viscous damping.
1 Proportional Damping (Rayleigh Damping)
C M K
where the constants & are found from
1,2 , 1 & 2 (damping ratio) being selected.
Incorporate the viscous damping in modal equations.
Use the normal modes (modal matrix) to transform the coupled system of dynamic equations to uncoupled system of equations.
Equations in (22) or (24) are called modal equations. These are uncoupled, second-order differential equations, which are much easier to solve than the original dynamic equation (coupled system).
To recover u from z, apply transformation (21) again, once z is obtained from (24).
Only the first few m odes may be needed in constructing the mod al matrix (i.e., could be an n m rectangular matrix with m<n). Thus, significant reduction in the
size of the system ca n be achieved.
Modal equations are best suited for problems in which higher mode s are not important (i.e., structural vibrations, but not shock loading).
2.Frequency Response Analysis
(Harmonic Response Analysis)
Ku E u
Modal method: Apply th e modal equations,
TRANSIENT RESPONSE ANALYSIS
(Dynamic Response/Time-History Analysis)
Structure response to arbitrary, time-dependent loading.
Compute responses by integrating through time:
B. Modal Method
First, do the transformation of the dynamic equations using the modal matrix before the time marching:
Then, solve the uncoupled equations using an integration method. Can use, e.g., 10%, of the total modes (m= n/10).
Uncoupled system, Fewer equations,
No inverse of matrices,
More efficient for large problems.
1Cautions in Dynamic Analysis
Symmetry: It should not be used in the dynamic analysis (normal modes, etc.) because symmetric structures can have antisymmetric modes.
Mechanism, rigid body motion means = 0. Can use this to check FEA models to see if they are properly connected and/or supported.
Input for FEA: loading F(t) or F( ) can be very complicated in real applications and often needs to be filtered first before used as input for FEA.
Impact, drop test, etc.
In the spring structure shown k1 = 10 lb./in., k2 = 15 lb./in., k3 = 20 lb./in., P= 5 lb. Determine the deflection at nodes 2 and 3.
Again apply the three steps outlined previously.
Step 1: Find the Element Stiffness Equations
Step 2: Find the Global stiffness matrix
Now the global structural equation can be written as above.
Step 3: Solve for Deflections
The known boundary conditions are: u1 = u4 = 0, F3 = P = 3lb. Thus, rows and columns 1 and 4 will drop out, resulting in t following matrix equation,
Solving, we get u2 = 0.0692 & u3 = 0.1154
In the spring structure shown, k1 = 10 N/mm, k2 = 15 N/mm, k3 = 20 N/mm, k4 = 25 N/mm, k5 = 30 N/mm, k6 = 35 N/mm. F2 = 100 N. Find the deflections in all springs.
Here again, we follow the three-step approach described earlier, without specifically mentioning at each step.
Now, apply the boundary conditions, u1 = u4 = 0, F2 = 100 N. This is carried out by deleting the rows 1 and 4, columns 1 and 4, and replacing F2 by 100N. The final matrix equation is,
Spring 1: u4 – u1 = 0
Spring 2: u2 – u1 = 1.54590
Spring 3: u3 – u2 = -0.6763
Spring 4: u3 – u2 = -0.6763
Spring 5: u4 – u2 = -1.5459
Spring 6: u4 – u3 = -0.8696