# The Rayleigh Ritz Method Computer Science Essay

✅ Paper Type: Free Essay |
✅ Subject: Computer Science |

✅ Wordcount: 3869 words |
✅ Published: 1st Jan 2015 |

The given assignment is done with the soul purpose of developing an intense knowledge and understanding of vibrational behaviour and dynamic response of structures.

The assignment aims to apply up to date methods of structural dynamics in aerospace and aerospace system engineering. Here we use Rayleigh-Ritz method and Finite Element method to obtain the natural frequency and mode shape of the given cantilever beam.

## 1. Rayleigh-Ritz Method

Rayleigh-Ritz method is an extension of the Rayleigh method which was developed by the Swiss mathematician and physicist Walter Ritz. Its one of the widely used method to calculate more accurate value of fundamental frequency, further it also gives approximations to the higher frequencies and mode shapes.

In the Ritz method the single shape function is replaced by a series of shape functions multiplied by constant coefficients, that is the single function of deflection choose in Rayleigh method is assumed to be a sum of several functions multiplied by constant coefficients. The coefficients values are modified by reducing the frequency with respect to each of the coefficients, which result in n algebraic equations in. The solution of these equations will give the value of natural frequency and mode shapes of the system. It should be taken into account that the success of the method is only possible so long as the shape function taken satisfies the geometric boundary conditions of the problem. The method should also be differentiable to the order of the derivatives of the equations. Here the function can ignore discontinuities like shear due to concentrated masses that involve third derivatives in beam.

The Rayleigh-Ritz method is done by assuming the deflection curve of the beam by

The function are the assumed displacement functions that satisfy geometrical boundary conditions.

For a cantilever beam the boundary conditions are

They are selected such that it is possible to obtain a good approximation to each of the required natural modes by superposition.

The quantities are generalized coordinates representing contributions of each assumed functions.

For a beam divided into â€›n’ span wise stations the total differential equation can be formulate using Lagrange equation as

Putting as a solution , where the amplitude of the displacement is, is the frequency and is the phase angle.

This set of characteristics equations can be solved for n discrete values of . This equation can easily be put into a matrix form for numerical calculation as

For a beam divided into n span wise station the mass and stiffness terms can be formulated into matrices as

Where = matrix of assumed modes

= mass matrix

= matrix of weighting coefficients

= rigidity matrix

Hence we write as

The above equation is considered to be convenient for computation, but has limitations in the manner of expressing the strain energy.

Given Data

Length L=1.5

Modulus of Elasticity E=74 GPa

Poisson’s Ratio P=0.33

Material density

The depth of the beam tapers uniformly from 0.3 at the fixed end to 0.1 at the free end.

The breadth of the beam tapers uniformly from 0.02 at the fixed end to 0.005 at the free end.

The assumed modes are given by the polynomial function:

## MATLAB Operation

>> L=1.5

L =

1.5000

>>x=[0,0.15,0.3,0.45,0.6,0.75,0.9,1.05,1.2,1.35,1.5]

x =

0 0.1500 0.3000 0.4500 0.6000 0.7500 0.9000 1.0500 1.2000 1.3500 1.5000

>> s=x/L

s =

0 0.1000 0.2000 0.3000 0.4000 0.5000 0.6000 0.7000 0.8000 0.9000 1.0000

>> V1= 2*s.^2-(4/3)*s.^3+(1/3)*s.^4

V1 =

0 0.0187 0.0699 0.1467 0.2432 0.3542 0.4752 0.6027 0.7339 0.8667 1.0000

>> V2=(10/3)*s.^3-(10/3)*s.^4+s.^5

V2 =

0 0.0030 0.0217 0.0654 0.1382 0.2396 0.3658 0.5111 0.6690 0.8335 1.0000

>> V=[V1;V2]

V =

0 0.0187 0.0699 0.1467 0.2432 0.3542 0.4752 0.6027 0.7339 0.8667 1.0000

0 0.0030 0.0217 0.0654 0.1382 0.2396 0.3658 0.5111 0.6690 0.8335 1.0000

>> dV1=(1/(L.^2))*(4-8*s+4*(s.^2))

dV1 =

1.7778 1.4400 1.1378 0.8711 0.6400 0.4444 0.2844 0.1600 0.0711 0.0178 0

>> dV2= (1/(L.^2))*(20*s-40*(s.^2)+20*(s.^3))

dV2 =

0 0.7200 1.1378 1.3067 1.2800 1.1111 0.8533 0.5600 0.2844 0.0800 0

>> dV=[dV1;dV2]

dV =

1.7778 1.4400 1.1378 0.8711 0.6400 0.4444 0.2844 0.1600 0.0711 0.0178 0

0 0 .7200 1.1378 1.3067 1.2800 1.1111 0.8533 0.5600 0.2844 0.0800 0

Weighting matrix can be formulated using Trapezoidal rule, Simpson’s rule and Lagrange’s Interpolation formula.

By Lagrange’s interpolation formula if the beam is divided into 10 equal elements with spacing â€›d ‘ then weighting matrix is computed as:

## MATLAB Operation

>> d=0.15

d =

0.1500

>> W1=(d/3.7266)*[1,6.616,-3.020,16.954,-16.216,26.599,-16.216,16.954, -3.020, 6.616,1]

W1 =

0.0403 0.2663 -0.1216 0.6824 -0.6527 1.0706 -0.6527 0.6824 -0.1216 0.2663 0.0403

>> W=diag(W1)

W =

0.0403 0 0 0 0 0 0 0 0 0 0

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Essay Writing Service0 0.2663 0 0 0 0 0 0 0 0 0

0 0 -0.1216 0 0 0 0 0 0 0 0

0 0 0 0.6824 0 0 0 0 0 0 0

0 0 0 0 -0.6527 0 0 0 0 0 0

0 0 0 0 0 1.0706 0 0 0 0 0

0 0 0 0 0 0 -0.6527 0 0 0 0

0 0 0 0 0 0 0 0.6824 0 0 0

0 0 0 0 0 0 0 0 -0.1216 0 0

0 0 0 0 0 0 0 0 0 0.2663 0

0 0 0 0 0 0 0 0 0 0 0.0403

Mass matrix is a diagonal matrix representing the mass per unit length at the eleven span wise stations.

The matrix can be calculated by

Material density = 2700

The depth of the beam at a station with a distance x from the fixed end is given by

Depth

Similarly the breadth of the beam at a station with a distance x from the fixed end is given by

Breadth

## MATLAB Operation

>> h=0.3-(s*0.2)

h =

0.3000 0.2800 0.2600 0.2400 0.2200 0.2000 0.1800 0.1600 0.1400 0.1200 0.1000

>> b=0.02-(s*0.015)

b =

0.0200 0.0185 0.0170 0.0155 0.0140 0.0125 0.0110 0.0095 0.0080 0.0065 0.0050

>> m=2700*diag(b)*diag(h)

m =

16.2000 0 0 0 0 0 0 0 0 0 0

0 13.9860 0 0 0 0 0 0 0 0 0

0 0 11.9340 0 0 0 0 0 0 0 0

0 0 0 10.0440 0 0 0 0 0 0 0

0 0 0 0 8.3160 0 0 0 0 0 0

0 0 0 0 0 6.7500 0 0 0 0 0

0 0 0 0 0 0 5.3460 0 0 0 0

0 0 0 0 0 0 0 4.1040 0 0 0

0 0 0 0 0 0 0 0 3.0240 0 0

0 0 0 0 0 0 0 0 0 2.1060 0

0 0 0 0 0 0 0 0 0 0 1.3500

The Second moment of area of the beam is given by

## MATLAB Operation

>> I=diag(h)*(diag(b).^3)/12

I =

1.0e-006 *

0.2000 0 0 0 0 0 0 0 0 0 0

0 0.1477 0 0 0 0 0 0 0 0 0

0 0 0.1064 0 0 0 0 0 0 0 0

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View our services0 0 0 0.0745 0 0 0 0 0 0 0

0 0 0 0 0.0503 0 0 0 0 0 0

0 0 0 0 0 0.0326 0 0 0 0 0

0 0 0 0 0 0 0.0200 0 0 0 0

0 0 0 0 0 0 0 0.0114 0 0 0

0 0 0 0 0 0 0 0 0.0060 0 0

0 0 0 0 0 0 0 0 0 0.0027 0

0 0 0 0 0 0 0 0 0 0 0.0010

Rigidity matrix is the diagonal matrix that gives the product of modulus of elasticity and the second moment of area of the beam about the neutral axis.

EI=74000000000*I

EI =

1.0e+004 *

1.4800 0 0 0 0 0 0 0 0 0 0

0 1.0933 0 0 0 0 0 0 0 0 0

0 0 0.7877 0 0 0 0 0 0 0 0

0 0 0 0.5511 0 0 0 0 0 0 0

0 0 0 0 0.3723 0 0 0 0 0 0

0 0 0 0 0 0.2409 0 0 0 0 0

0 0 0 0 0 0 0.1477 0 0 0 0

0 0 0 0 0 0 0 0.0846 0 0 0

0 0 0 0 0 0 0 0 0.0442 0 0

0 0 0 0 0 0 0 0 0 0.0203 0

0 0 0 0 0 0 0 0 0 0 0.0077

Substituting in Rayleigh-Ritz equation:

This gives

Simplifying

The above equation is a quadratic in , which can be solved

## =

Result: The approximate values of the first and second natural frequencies of the given beam under flexural vibrations, by the use of Rayleigh- Ritz method, was found to be

## 2. Mode shapes

Consider the equation

Substituting the values of in the above equation and simplifying

The column matrix that represents the mode shape at the eleven stations is obtained by putting,

## =

0.0578

Substituting the value of in the above equation and simplifying

The column matrix that represents the mode shape at the eleven stations is obtained by putting,

= 0.0693

## 3. Finite Element Method

Finite Element Method (FEM) is considered to be one of the profound developments in the static and dynamics analysis of continuous systems. It provides a discrete approximation to vibration of continuous systems. The finite element method can be developed as a special case of the Rayleigh -Ritz method. The method was originally developed for the static- stress analysis of complex distributed parameter structures. Now a days FEM is widely applied to disciplines of heat transfer, electro magnetics, fluid flow and vibrations.

In finite element method the structure is divided into a large number of small but finite parts called elements which are interconnected at points called nodes. For each element a displacement function is assumed which satisfies the geometric boundary condition so that continuity is achieved between the elements. The variations in displacement of each element( which can be linear, quadratic etc.), are assumed over the length of the element. This method allows the displacement of any point in the element to be expressed in terms of the displacement at the end of the element. These displacements by finite element terminology are called nodal variables.

Unlike Rayleigh-Ritz in finite element method the global coordinate is replaced by a local coordinate where is the length of the element. The kinetic and strain energy of the element is obtained by integrating along the element’s length, in terms of the nodal variables. By superposing the energies contributed by the individual elements into which the structure is divided, we can obtain the kinetic and strain energy of the structure or system in terms of the nodal variables of the whole structure. The finite element method is mainly based on variational principles.

The method is considered very much versatile and can be used to physical problems with arbitrary shapes, loads and support conditions. The finite element model has a close resemblance to the actual structure.

Many general finite element code packages have been written over the years with user friendly windows and menus (GUI) which allow for easy geometry setup, boundary condition manipulation and evaluation/post processing of common structural problems. Some of the most popular codes in the industry are ANSYS, MSC Nastran and MARC. ANSYS will be the code used for this assignment.

## ANSYS Operation

## Define Material

Step 1: Set preferences

Preferences are set in order to filter quantities that pertain to this discipline.

Step 2: Define constant material properties.

Modulus of elasticity, Poisson’s ratio and Density are defined.

Step 3:- Modeling

Create the beam with required geometry.

## Generating Mesh

Step 4: Define element type

Two element types are defined: a 2-D element and a 3-D element. The beam cross-sectional area is meshed with 2-D elements, and then the area is to be extruded to create a 3-D volume. The mesh will be extruded along with the geometry so 3-D elements will automatically be created in the volume.

Step 5: Mesh the area

Mesh control are specified in order to obtain a particular mesh density.

Element edge length is set at 0.01

Note: – Mesh density is very important. If the mesh is too coarse your result can contain serious errors. If the mesh is too fine, would cause waste of computer resources, experience excessively long run time, the model may be too large to be run on the computer system. Unfortunately it cannot be definitively specified how fine the mesh density should be. But one way to find out is to perform the analysis with what seems to be a reasonable mesh. Then reanalyse the problem with twice as many elements in the critical region and compare the results. If the two mesh give the same result then the mesh probably be adequate. If there is substantial difference between the two results then further refinement of the mesh is required.

Step 6:- Extrude the meshed area into a meshed volume.

The 3-D volume is generated by first changing the element type to SOLID 45, which is defined as element type 2, and then extruding the area into a volume.

The number of element divisions is set as 10

Offsets of extrusion are set as 0, 0, 1.5

Tapering ratio is given as 0.33333, 0.25, 0

## Apply loads

Step 7:- Unselect 2-D elements

Before applying constraints to the fixed end of the wing, unselect all the PLANE42 elements used in the 2-D area mesh since they will not be used for the analysis.

Step 8:- Apply constraints to the model

Constraints will be applied to all nodes located where the wing is fixed to the body. Select all nodes at z = 0, then apply the displacement constraints.

## Obtain Solution

Step 9:- Specify analysis type and options

Specify a modal analysis type.

Number of modes to be extracted is given as 5

Number of modes to be expanded is given as 5

Step 10:- Solve

## Observe results

Step 11:-List natural frequencies

***** INDEX OF DATA SETS ON RESULTS FILE *****

SET TIME/FREQ LOAD STEP SUBSTEP CUMULATIVE

1 11.964 1 1 1

2 40.840 1 2 2

3 100.05 1 3 3

4 144.08 1 4 4

5 182.70 1 5 5

Step 12:- Animate the two mode shapes.

Set the results for the first mode to be animated. Observe the first mode shape. Animate the next mode shape. Observe the second mode shape. Repeat the same procedure to obtain the other three mode shapes.

## First Mode Shape

## Second Mode Shape

## Third Mode Shape

## Fourth Mode Shape

## Fifth Mode Shape

## 4. Comparison of Rayleigh-Ritz Method and Finite Element Method

## Rayleigh-Ritz method

Rayleigh-Ritz method uses the principle of conservation of energy to formulate the matrix equation.one major advantage of this method is that it allows us to neglect the non-applied forces like forces at a point of rolling contact, forces at frictionless guides etc.

Considering the method to be an extension of the Rayleigh method it has an improved accuracy by assuming the deflection curve of the beam to be

Nevertheless the assumed function should satisfy all the boundary conditions and should be linearly independent. This can be easily achieved by using polynomial expression to derive the deflection function.

Disadvantages and limitations

The selection of assumed deflection function requires a good knowledge and expertise of the method

Good approximation of the true natural modes are only possible as the assumed function are limited in numbers and nature

All n modal solutions will not give a good approximation to the true mode, so it is necessary to discard some higher frequency modes.

The approximations are only good for lower modes and it becomes worse for higher modes.

The major limitation of this method is in the manner in which the strain energy can be expressed.

## Finite Element Method

The finite element method (its practical application is often as Finite Element Analysis)is a powerful technique developed in the analysis of complex structural mechanics. In this method the structure is divided into large number of finite parts or elements which are interconnected at points called nodes. The elements will have properties like thickness, Youngs modulus, Poisson’s ratio etc. An assumption is made over the variation over the length of the element.

This allows to find the displacement at any point in the given structure by introducing

Disadvantages and limitations

The method is not considered convenient for simple structures.

It’s a time consuming operation.

Its accuracy depends on the number of elements the structure is divided.

It does not provide a closed-form solution, denying analytical study of the effects of changing parameters.

It needs a reliable program for support.

Creating a good model requires experience.

A good amount of data are required and voluminous output must be sorted and studied.

## Comparison of results

## Obtained by Rayleigh-Ritz Method

## Obtained by Finite Element Method

## % Difference

0.77%

26.82%

There is difference of 26.82% for the second frequency of the system. The value obtained by the Rayleigh-Ritz method can be brought closer to accuracy by increasing the number of assumed functions and by improving their nature. Normally the approximation becomes worse as we move to higher modes .With use of only two assumed functions, the solutions obtained are considered to be satisfactory.

## Errors in Rayleigh-Ritz method

To use the method with ease the assumed functions are kept as simple as possible by using simple polynomial functions and at fewer times only the functions of sine and cosine are used. There is no exact answer as to which function the good approximate value can be obtained.

There are always some terms omitted in the function which results to an ineffective solution.

This method is considered to be inflexible as the actual displacement of the structure is restricted to only the shape generated by superposing the finite number of functions selected by the analyst.

## Recommendation to improve Rayleigh-Ritz method

The iterative process can be carried out with each time adding the term in the assumed function until it gives the exact value.

## Errors in Finite Element method

Distorted mesh can result in flawed stiffness and mass terms

Errors are always presented at joints and constrained boundaries due to uncertainty.

## Recommendation to improve Finite Element method

Iterative method has to be applied to see the number of elements actually required to break down the structure so as to obtain more accurate value.

Appropriate methods should be employed for remeshing like HYPERMESH, Mesh++ based on a posterior error.

## CONCLUSION

The first and second natural frequencies of the given beam are found out by using both Rayleigh-Ritz method and Finite element method, and mode shapes for these frequencies are drawn.

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