Engineering mechanics dynamics formula sheet, Cheat Sheet of Mechanical Engineering

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General report

Dynamics cheat sheet

Nasser M. Abbasi

December 2019

Contents CONTENTS

iv

Chapter 1

Vibration

1.1 Modal analysis for two degrees of freedom system

Detailed steps to perform modal analysis are given below for a standard undamped two

degrees of freedom system. The main advantage of solving a multidegree system using

modal analysis is that it decouples the equations of motion (assuming they are coupled)

making solving them much simpler.

In addition it shows the fundamental shapes that the system can vibrate in, which gives

more insight into the system. Starting with standard 2 degrees of freedom system

m 1

m 2

k 1

k 2

x 1

f 1

 t 

f 2  t 

x (^) 2

Figure 1.1: 2 degrees of freedom system

In the above the generalized coordinates are 𝑥

1

and 𝑥

2

. Hence the system requires two

equations of motion (EOM’s).

1.1.1 Step one. Finding the equations of motion in normal

coordinates space

The two EOM’s are found using any method such as Newton’s method or Lagrangian

method. Using Newton’s method, free body diagram is made of each mass and then 𝐹 = 𝑚𝑎

is written for each mass resulting in the equations of motion. In the following it is assumed

mass matrix [𝑀] is PSD (positive definite matrix) and the [𝐾] matrix is positive semi-definite

matrix. The reason the [𝑀] is PSD is that 𝑥

𝑇

[𝑀]{𝑥} represents the kinetic energy of the

system, which is typically positive and not zero. But reading some other references

1

it is

possible that [𝑀] can be positive semi-definite. It depends on the application being modeled.

1.1.2 Step 2. Solving the eigenvalue problem, finding the natural

frequencies

The first step in modal analysis is to solve the eigenvalue problem det�[𝐾] − 𝜔

2 [𝑀]� = 0

in order to determine the natural frequencies of the system. This equations leads to a

polynomial in 𝜔 and the roots of this polynomial are the natural frequencies of the system.

Since there are two degrees of freedom, there will be two natural frequencies 𝜔

1

2

for the

system.

det�[𝐾] − 𝜔

2 [𝑀]� = 0

det

11

12

21

22

2

11

12

21

22

det

11

2 𝑚 11

12

2 𝑚 12

21

2 𝑚 21

22

2 𝑚 22

11

2 𝑚 11

22

2 𝑚 22

12

2 𝑚 12

21

2 𝑚 21

4 (𝑚 11

22

12

21

2 (−𝑘 11

22

12

21

21

12

22

11

11

22

12

21

The above is a polynomial in 𝜔

4

. Let 𝜔

2

= 𝜆 it becomes

2 (𝑚 11

22

12

21

11

22

12

21

21

12

22

11

11

22

12

21

This quadratic polynomial in 𝜆 which is now solved using the quadratic formula. Then the

positive square root of each 𝜆 root to obtain 𝜔

1

and 𝜔

2

which are the roots of the original

eigenvalue problem. Assuming from now that these roots are 𝜔

1

and 𝜔

2

the next step is to

obtain the non-normalized shape vectors 𝜑⃗

1

2

also called the eigenvectors associated with

1

and 𝜔

2

1.1.3 Step 3. Finding the non-normalized eigenvectors

For each natural frequency 𝜔

1

and 𝜔

2

the corresponding shape function is found by solving

the following two sets of equations for the vectors 𝜑

1

2

11

12

21

22

2

1

11

12

21

22

11

21

1 http://en.wikipedia.org/wiki/Fundamental_equation_of_constrained_motion

and

11

12

21

22

2

2

11

12

21

22

12

22

For 𝜔

1

, let 𝜑

11

= 1 and solve for

11

12

21

22

2

1

11

12

21

22

21

11

2

1

11

12

2

1

12

21

2

1

21

22

2

1

22

21

Which gives one equation now to solve for 𝜑

21

(the first row equation is only used)

11

2

1

11

21

12

2

1

12

Hence

21

11

2

1

11

12

2

1

12

Therefore the first shape vector is

1

11

21

−�𝑘 11

−𝜔

2

1

𝑚 11

�

�𝑘 12

−𝜔

2

1

𝑚 12

�

Similarly the second shape function is obtained. For 𝜔

2

, let 𝜑

12

= 1 and solve for

11

12

21

22

2

2

11

12

21

22

22

11

2

2

11

12

2

2

12

21

2

2

21

22

2

2

22

22

Which gives one equation now to solve for 𝜑

22

(the first row equation is only used)

11

2

2

11

22

12

2

2

12

Hence

22

11

2

2

11

12

2

2

12

Similarly, 𝜇

2

is found

2

12

22

11

12

21

22

12

22

12

22

11

12

21

22

12

22

12

11

22

21

12

12

22

22

12

22

12

12

11

22

21

22

12

12

22

22

Now that 𝜇

1

2

are obtained, the mass normalized shape vectors are found. They are called

1

2

1

1

1

11

21

1

𝜑 11

√

𝜇 1

𝜑 21

√

𝜇 1

Similarly

2

2

2

12

22

2

𝜑 12

√

𝜇 2

𝜑 22

√

𝜇 2

1.1.5 Step 5, obtain the modal transformation matrix Φ

The modal transformation matrix is the 2 × 2 matrix made of of

1

2

in each of its columns

[Φ] = �⃗Φ

1

2

𝜑 11

√

𝜇 1

𝜑 12

√

𝜇 2

𝜑 21

√

𝜇 1

𝜑 22

√

𝜇 2

Now the [Φ] is found, the transformation from the normal coordinates {𝑥} to modal coordi-

nates, which is called �𝜂�^ is found

{𝑥} = [Φ]�𝜂�

1

2

𝜑 11

√

𝜇 1

𝜑 12

√

𝜇 2

𝜑 21

√

𝜇 1

𝜑 22

√

𝜇 2

1

2

The transformation from modal coordinates back to normal coordinates is

�𝜂� = [Φ]

−

{𝑥}

1

2

𝜑 11

√

𝜇 1

𝜑 12

√

𝜇 2

𝜑 21

√

𝜇 1

𝜑 22

√

𝜇 2

1

2

However, [Φ]

−

= [Φ]

𝑇

[𝑀] therefore

�𝜂� = [Φ]

𝑇

[𝑀]{𝑥}

1

2

𝜑 11

√

𝜇 1

𝜑 12

√

𝜇 2

𝜑 21

√

𝜇 1

𝜑 22

√

𝜇 2

11

12

21

22

1

2

The next step is to apply this transformation to the original equations of motion in order to

decouple them

1.1.6 Step 6. Applying modal transformation to decouple the

original equations of motion

The EOM in normal coordinates is

11

12

21

22

′′

1

′′

2

11

12

21

22

1

2

1

2

Applying the above modal transformation {𝑥} = [Φ]�𝜂� on the above results in

11

12

21

22

[Φ]

′′

1

′′

2

11

12

21

22

[Φ]

1

2

1

2

pre-multiplying by [Φ]

𝑇

results in

[Φ]

𝑇

11

12

21

22

[Φ]

′′

1

′′

2

+ [Φ]

𝑇

11

12

21

22

[Φ]

1

2

= [Φ]

𝑇

1

2

1.1.7 Step 7. Converting modal solution to normal coordinates

solution

The solutions found above are in modal coordinates 𝜂

1

2

(𝑡). The solution needed is

1

2

(𝑡). Therefore, the transformation {𝑥} = [Φ]�𝜂� is now applied to convert the solution

to normal coordinates

1

2

𝜑 11

√

𝜇 1

𝜑 12

√

𝜇 2

𝜑 21

√

𝜇 1

𝜑 22

√

𝜇 2

1

2

𝜑 11

√

𝜇 1

1

𝜑 12

√

𝜇 2

2

𝜑 21

√

𝜇 1

1

𝜑 22

√

𝜇 2

2

Hence

1

11

1

1

12

2

2

and

2

21

1

1

22

2

2

Notice that the solution in normal coordinates is a linear combination of the modal solutions.

The terms

𝜑𝑖𝑗

√

𝜇

are just scaling factors that represent the contribution of each modal solution

to the final solution. This completes modal analysis

1.1.8 Numerical solution using modal analysis

This is a numerical example that implements the above steps using a numerical values for

[𝐾] and [𝑀]. Let 𝑘

1

2

1

2

= 3 and let 𝑓

1

(𝑡) = 0 and 𝑓

2

(𝑡) = sin(5𝑡). Let initial

conditions be 𝑥

1

′

1

2

′

2

(0) = 3, hence

1

2

and

′

1

′

2

In normal coordinates, the EOM are

1

2

′′

1

′′

2

1

2

2

2

2

1

2

1

2

′′

1

′′

2

1

2

sin(5𝑡)

In this example 𝑚

11

12

21

22

= 3 and 𝑘

11

12

21

22

= 2 and

1

(𝑡) = 0 and 𝑓

2

(𝑡) = sin(5𝑡)

step 2 is now applied which solves the eigenvalue problem in order to find the two natural

frequencies

det�[𝐾] − 𝜔

2 [𝑀]� = 0

det

2

det

2 −

2

2 ��2 − 3𝜔

2 � − (−2)(−2) = 0

4 − 11𝜔

2

• 2 = 0

Let 𝜔

2

= 𝜆 hence

2 − 11𝜆 + 2 = 0

The solution is 𝜆

1

= 3. 475 and 𝜆

2

= 0.192, therefore

1

And

2

step 3 is now applied which finds the non-normalized eigenvectors. For each natural fre-

quency 𝜔

1

and 𝜔

2

the corresponding shape function is found by solving the following two

sets of equations for the eigen vectors 𝜑

1

2

2

1

11

21

Now step 4 is applied, which is mass normalization of the shape vectors (or the eigenvectors)

1

𝑇

1

[𝑀]⃗𝜑

1

Hence

1

Similarly, 𝜇

2

is found

2

𝑇

2

[𝑀]⃗𝜑

2

Hence

2

Now that 𝜇

1

2

are found, the mass normalized eigen vectors are found. They are called

1

2

1

1

1

11

21

1

Similarly

2

2

2

12

22

2

Therefore, the modal transformation matrix is

[Φ] = �⃗Φ

1

2

This result can be verified using Matlab’s eig function as follows

K=[3 -2;-2 2]; M=[1 0;0 3];

[phi,lam]= eig (K,M)

phi =

diag ( sqrt (lam))

Matlab result agrees with the result obtained above. The sign difference is not important.

Now step 5 is applied. Matlab generates mass normalized eigenvectors by default.

Now that [Φ] is found, the transformation from the normal coordinates {𝑥} to modal coordi-

nates, called �𝜂�, is obtained

{𝑥} = [Φ]�𝜂�

1

2

1

2

The transformation from modal coordinates back to normal coordinates is

�𝜂� = [Φ]

−

{𝑥}

1

2

1

2

and

′

1

′

2

′

1

′

2

Each of these EOM are solved using any of the standard methods. This results in solutions

1

(𝑡) and 𝜂

2

(𝑡). Hence the following EOM’s are solved

′′

1

1

(𝑡) = −0.219 sin(5𝑡)

1

′

1

and also

′′

2

2

(𝑡) = 0.534 sin(5𝑡)

2

′

2

The solutions 𝜂

1

2

(𝑡) are found using basic methods shown in other parts of these notes.

The last step is to transform back to normal coordinates by applying step 7

1

2

1

2

1

2

2

1

Hence

1

1

2

and

2

1

2

The above shows that the solution 𝑥

1

(𝑡) and 𝑥

2

(𝑡) has contributions from both nodal solutions.

1.2. Fourier series representation of a … CHAPTER 1. VIBRATION

1.2 Fourier series representation of a periodic function

Given a periodic function 𝑓(𝑡) with period 𝑇 then its Fourier series approximation

𝑓(𝑡) using

𝑁 terms is

0

+ Re

𝑁

𝑛=

𝑛

𝑖𝑛

2𝜋

𝑇

𝑡

0

𝑁

𝑛=

𝑛

𝑖𝑛

2𝜋

𝑇

𝑡

• 𝐹

∗

𝑛

−𝑖𝑛

2𝜋

𝑇

𝑡

𝑁

𝑛=−𝑁

𝑛

𝑖𝑛

2𝜋

𝑇

𝑡

Where

𝑛

𝑇

0

−𝑖𝑛

2𝜋

𝑇

𝑡

𝑑𝑡

0

𝑇

0

Another way to write the above is to use the classical representation using cos and sin. The

same coefficients (i.e. the same series) will result.

0

𝑁

𝑛=

𝑛

cos 𝑛

𝑁

𝑛=

𝑛

sin 𝑛

0

𝑇

0

𝑛

𝑇

0

𝑓(𝑡) cos

𝑛

𝑇

0

𝑓(𝑡)^ sin�𝑛

Just watch out in the above, that we divide by the full period when finding 𝑎

0

and divide by

half the period for all the other coefficients. In the end, when we find

𝑓(𝑡) we can convert

that to complex form. The complex form seems easier to use.