Dynamics cheat sheet

Dynamics cheat sheet
Nasser M. Abbasi
September 27, 2015
page compiled on September 27, 2015 at 5:28pm
Contents
1 Modal analysis for two degrees of freedom system 3
1.1 Step one, finding the equations of motion in normal coordinates space . . . . . . . . . . . . . . . 3
1.2 Step 2, solving the eigenvalue problem, finding the natural frequencies . . . . . . . . . . . . . . . 4
1.3 Step 3, finding the non-normalized eigenvectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.4 Step 4. Mass normalization of the shape vectors (or the eigenvectors) . . . . . . . . . . . . . . . . 6
1.5 Step 5, obtain the modal transformation matrix Φ . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.6 Step 6, applying modal transformation to decouple the original equations of motion . . . . . . . . 8
1.7 Step 7. Converting modal solution to normal coordinates solution . . . . . . . . . . . . . . . . . . 9
1.8 Numerical solution using modal analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
2 Fourier series representation of a periodic function 14
3 Transfer functions for different vibration systems 16
3.1 Force transmissibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
3.2 vibration isolation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
3.3 accelerometer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
3.4 seismometer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
3.5 Summary of vibration transfer functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
4 Steady state solutions to single degree freedom system 20
4.1 Summary: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
4.2 Details . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
4.3 Undamped case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
4.4 damped cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
5 Summary table of solutions 22
5.1 Harmonic loading mu ′′ + cu ′ + ku = F sin ϖt . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
5.2 constant loading mu ′′ + cu ′ + ku = F . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
5.3 no loading (free) mu ′′ + cu ′ + ku = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
5.4 impulse F 0 δ (t) loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
5.5 Impulse sin function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
6 Tree view look at the different cases 28
7 Some common formulas 31
8 Derivation of the solutions of the equations of motion 33
8.1 Free vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
8.1.1 Undamped free vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
1

8.1.2 under-damped free vibration c < c r , ξ < 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
8.1.3 critically damped free vibration ξ = c
c r
= 1 . . . . . . . . . . . . . . . . . . . . . . . . . . 34
8.1.4 over-damped free vibration ξ = c
c r
> 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
8.2 Constant input F . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
8.2.1 Undamped case ζ = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
8.2.2 under-damped ζ < 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
8.2.3 Critical damping ζ = 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
8.2.4 Over-damped ζ > 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
8.3 Harmonic forced input F sin (ϖt) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
8.3.1 Undamped forced vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
8.3.2 Under-damped forced vibration c < c r , ξ < 1 . . . . . . . . . . . . . . . . . . . . . . . . . 38
8.3.3 critically damping forced vibration ξ = c
c r
= 1 . . . . . . . . . . . . . . . . . . . . . . . . . 40
8.3.4 over-damped forced vibration ξ = c
c r
> 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
8.4 Response to impulsive loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
8.4.1 impulse input . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
8.4.2 sin impulse input . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
8.5 references . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
9 Derivation of rotation formula 49
10 Miscellaneous hints 50
11 Formulas 51
12 Velocity and acceleration diagrams 53
12.1 Spring pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
12.2 pendulum with blob moving in slot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
12.3 spring pendulum with block moving in slot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
12.4 double pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
13 Velocity and acceleration of rigid body 2D 57
14 Velocity and acceleration of rigid body 3D 58
14.1 Using Vehicle dynamics notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
14.2 3D Not Using vehicle dynamics notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
14.2.1 Derivation for F = ma in 3D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
14.2.2 Derivation for τ = Iω in 3D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
14.2.3 Derivation for τ = Iω in 3D using principle axes . . . . . . . . . . . . . . . . . . . . . . . 62
15 Wheel spinning precession 63
16 References 63
17 Misc. items 64
18 Astrodynamics 65
18.1 Ellipse main parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
18.2 Table of common equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
18.3 Flight path angle for ellipse γ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
18.4 Parabolic trajectory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
18.5 Hyperbolic trajectory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
18.6 showing that energy is constant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
18.7 Earth satellite Transfer orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
18.7.1 Hohmann transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
2

18.7.2 Bi-Elliptic transfer orbit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
18.7.3 semi-tangential elliptical transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
18.8 Rocket engines, Hohmann transfer, plane change at equator . . . . . . . . . . . . . . . . . . . . . 75
18.8.1 Rocket equation with plane change not at equator . . . . . . . . . . . . . . . . . . . . . . 77
18.9 Spherical coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
18.10interplanetary transfer orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
18.10.1 interplanetary hohmann transfer orbit, case one . . . . . . . . . . . . . . . . . . . . . . . . 79
18.11rendezvous orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
18.11.1 Two satellite, walking rendezvous using Hohmann transfer . . . . . . . . . . . . . . . . . . 81
18.11.2 Two satellite, separate orbits, rendezvous using Hohmann transfer, coplaner . . . . . . . . 83
18.11.3 Two satellite, separate orbits, rendezvous using bi-elliptic transfer, coplaner . . . . . . . . 85
18.12Semi-tangential transfers, elliptical, parabolic and hyperbolic . . . . . . . . . . . . . . . . . . . . 88
18.13Lagrange points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
18.14Orbit changing by low contiuous thrust . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
18.15References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
19 Dynamic of flights 92
19.1 Wing geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
19.2 Summary of main equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94
19.2.1 Writing the equations in linear form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96
19.3 definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98
19.4 images and plots collected . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
19.5 Some strange shaped airplanes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
19.6 links . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119
19.7 references . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119
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
x 1
x 2
k 1 k 2
m 1 m 2
f 1 t
f 2 t
In the above the generalized coordinates are x 1 and x 2 . Hence the system requires two equations of motion
(EOM’s).
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 that both masses are moving in the positive direction and
that x 2 is larger than x 1 when these equations of equilibrium are written
3

x 1
x 2
k 1 x 1
k 2 x 2 x 1
m
k 2 x 2 x 1
m 2
1
f 1 t
f 2 t
F m 1 x 1
k 1 x 1 k 2 x 2 x 1 f 1 t m 1 x 1
F m 2 x 2
k 2 x 2 x 1 f 2 t m 2 x 2
Hence, from the above the equations of motion are
or
In Matrix form
⎡ ⎤ ⎧
⎣ m 1 0 ⎨
⎦
0 m
⎩
2
m 1 x ′′
1 + k 1 x 1 − k 2 (x 2 − x 1 ) = f 1 (t)
m 2 x ′′
2 + k 2 (x 2 − x 1 ) = f 2 (t)
m 1 x ′′
1 + x 1 (k 1 + k 2 ) − k 2 x 2 = f 1 (t)
m 2 x ′′
2 + k 2 x 2 − k 2 x 1 = f 2 (t)
x ′′
1
x ′′
2
⎫ ⎡
⎤ ⎧ ⎫ ⎧ ⎫
⎬
⎭ + ⎣ k 1 + k 2 −k 2
⎨x ⎦ 1 ⎬ ⎨
−k 2 k
⎩
2 x
⎭ = f 1 (t) ⎬
⎩
2 f 2 (t)
⎭
The above two EOM are coupled in stiffness, but not mass coupled. Using short notations, the above is
written as
[M]{x ′′ } + [K]{x} = {f}
Modal analysis now starts with the goal to decouple the EOM and obtain the fundamental shape functions that
the system can vibrate in. To make these derivations more general, the mass matrix and the stiffness matrix are
written in general notations as follows
⎡ ⎤ ⎧
⎣ m 11 m 12
⎨
⎦
m 21 m
⎩
22
x ′′
1
x ′′
2
⎫ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫
⎬
⎭ + ⎣ k 11 k 12
⎨x ⎦ 1 ⎬ ⎨
k
⎩
22 x
⎭ = f 1 (t) ⎬
⎩
2 f 2 (t)
⎭
k 21
The mass matrix [M] and the stiffness matrix [K] must always come out to be symmetric. If they are not
symmetric, then a mistake was made in obtaining them. As a general rule, the 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.
1.2 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
1 http://en.wikipedia.org/wiki/Fundamental_equation_of_constrained_motion
4

are the natural frequencies of the system. Since there are two degrees of freedom, there will be two natural
frequencies ω 1 , ω 2 for the system.
det ( [K] − ω 2 [M] ) = 0
⎛⎡
⎤ ⎡ ⎤⎞
det ⎝⎣ k 11 k 12
⎦ − ω 2 ⎣ m 11 m 12
⎦⎠ = 0
k 21 k 22 m 21 m 22
⎡
⎤
det ⎣ k 11 − ω 2 m 11 k 12 − ω 2 m 12
⎦ = 0
k 21 − ω 2 m 21 k 22 − ω 2 m 22
(
k11 − ω 2 ) (
m 11 k22 − ω 2 ) (
m 22 − k12 − ω 2 ) (
m 12 k21 − ω 2 )
m 21 = 0
ω 4 (m 11 m 22 − m 12 m 21 ) + ω 2 (−k 11 m 22 + k 12 m 21 + k 21 m 12 − k 22 m 11 ) + k 11 k 22 − k 12 k 21 = 0
The above is a polynomial in ω 4 . Let ω 2 = λ it becomes
λ 2 (m 11 m 22 − m 12 m 21 ) + λ (−k 11 m 22 + k 12 m 21 + k 21 m 12 − k 22 m 11 ) + k 11 k 22 − k 12 k 21 = 0
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 ϕ 1 , ϕ 2 also
called the eigenvectors associated with ω 1 and ω 2
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 , ϕ 2
⎡
⎡
⎧
⎫
⎣ k 11 k 12
k 21
⎤
⎦ − ω1
2
k 22
⎣ m 11 m 12
m 21
⎤ ⎫ ⎧
⎨ϕ ⎦ 11 ⎬ ⎨
m
⎩
22 ϕ
⎭ = 0⎬
⎩
21 0
⎭
and
⎡ ⎤
⎣ k 11 k 12
⎦ − ω2
2
k 21 k 22
⎡
⎣ m 11 m 12
m 21
⎤ ⎧ ⎫ ⎧ ⎫
⎨ϕ ⎦ 12 ⎬ ⎨
m
⎩
22 ϕ
⎭ = 0⎬
⎩
22 0
⎭
For ω 1 , let ϕ 11 = 1 and solve for
⎡ ⎤ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫
⎣ k 11 k 12
⎦ − ω1
2 ⎣ m 11 m 12
⎨ 1 ⎬ ⎨
⎦
k 21 k 22 m 21 m
⎩
22 ϕ
⎭ = 0⎬
⎩
21 0
⎭
⎡
⎤ ⎧ ⎫ ⎧ ⎫
⎣ k 11 − ω1 2m 11 k 12 − ω1 2m ⎨
12 1 ⎬ ⎨
⎦
k 21 − ω1 2m 21 k 22 − ω1 2m ⎩
22 ϕ
⎭ = 0⎬
⎩
21 0
⎭
Which gives one equation now to solve for ϕ 21 (the first row equation is only used)
(
k11 − ω 2 1m 11
)
+ ϕ21
(
k12 − ω 2 1m 12
)
= 0
Hence
ϕ 21 = − ( k 11 − ω1 2m )
11
(
k12 − ω1 2m 12)
5

Therefore the first shape vector is
⎧ ⎫ ⎧
⎨ϕ 11 ⎬ ⎪⎨
ϕ 1 =
⎩
ϕ
⎭ = 1
⎪
21
⎩
− ( k 11 −ω1 2m )
( 11
k12 −ω1 2m 12)
⎫
⎪⎬
⎪⎭
Similarly the second shape function is obtained. For ω 2 , let ϕ 12 = 1 and solve for
⎡ ⎤ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫
⎣ k 11 k 12
⎦ − ω2
2 ⎣ m 11 m 12
⎨ 1 ⎬ ⎨
⎦
k 21 k 22 m 21 m
⎩
22 ϕ
⎭ = 0⎬
⎩
22 0
⎭
⎡
⎤ ⎧ ⎫ ⎧ ⎫
⎣ k 11 − ω2 2m 11 k 12 − ω2 2m ⎨
12 1 ⎬ ⎨
⎦
k 21 − ω2 2m 21 k 22 − ω2 2m ⎩
22 ϕ
⎭ = 0⎬
⎩
22 0
⎭
Which gives one equation now to solve for ϕ 22 (the first row equation is only used)
(
k11 − ω 2 2m 11
)
+ ϕ22
(
k12 − ω 2 2m 12
)
= 0
Hence
ϕ 22 = − ( k 11 − ω2 2m )
11
(
k12 − ω2 2m 12)
Therefore the second shape vector is
⎧ ⎫ ⎧
⎨ϕ 12 ⎬ ⎪⎨
ϕ 2 =
⎩
ϕ
⎭ = 1
⎪
22
⎩
− ( k 11 −ω2 2m )
( 11
k12 −ω2 2m 12)
Now that the two non-normalized shape vectors are found, the next step is to perform mass normalization
1.4 Step 4. Mass normalization of the shape vectors (or the eigenvectors)
Let
µ 1 = ϕ T 1 [M] ϕ 1
This results in a scalar value µ 1 , which is later used to normalize ϕ 1 . Similarly
µ 2 = ϕ T 2 [M] ϕ 2
⎫
⎪⎬
⎪⎭
For example, to find µ 1
⎧ ⎫T ⎡ ⎤ ⎧ ⎫
⎨ϕ 11 ⎬
µ 1 = ⎣ m 11 m 12
⎨ϕ ⎦ 11 ⎬
⎩
ϕ
⎭
21 m 21 m
⎩
22 ϕ
⎭
21
{ } ⎡ ⎤ ⎧ ⎫
= ϕ 11 ϕ ⎣ m 11 m 12
⎨ϕ ⎦ 11 ⎬
21
m
⎩
22 ϕ
⎭
21
m 21
}
=
{ϕ ⎧ ⎫
⎨ϕ 11 ⎬
11 m 11 + ϕ 21 m 21 ϕ 11 m 12 + ϕ 21 m 22
⎩
ϕ
⎭
21
= ϕ 11 (ϕ 11 m 11 + ϕ 21 m 21 ) + ϕ 21 (ϕ 11 m 12 + ϕ 21 m 22 )
6

Similarly, µ 2 is found
⎧ ⎫T ⎡ ⎤ ⎧ ⎫
⎨ϕ 12 ⎬
µ 2 = ⎣ m 11 m 12
⎨ϕ ⎦ 12 ⎬
⎩
ϕ
⎭
22 m 21 m
⎩
22 ϕ
⎭
22
{ } ⎡ ⎤ ⎧ ⎫
= ϕ 12 ϕ ⎣ m 11 m 12
⎨ϕ ⎦ 12 ⎬
22
m
⎩
22 ϕ
⎭
22
m 21
}
=
{ϕ ⎧ ⎫
⎨ϕ 12 ⎬
12 m 11 + ϕ 22 m 21 ϕ 12 m 12 + ϕ 22 m 22
⎩
ϕ
⎭
22
= ϕ 12 (ϕ 12 m 11 + ϕ 22 m 21 ) + ϕ 22 (ϕ 12 m 12 + ϕ 22 m 22 )
Now that µ 1 , µ 2 are obtained, the mass normalized shape vectors are found. They are called Φ 1 , Φ 2
⎧ ⎫
⎨ϕ 11 ⎬
⎧ ⎫
Φ 1 = ϕ ⎩
ϕ
⎭
1 21
⎨ √µ1
ϕ 11 ⎬
√ = √ =
µ1 µ1 ⎩ √µ1
ϕ 21 ⎭
Similarly
Φ 2 = ϕ 2
√
µ2
=
⎧
⎨
⎩
ϕ 12
ϕ 22
⎫
⎬
⎭
√
µ2
=
⎧
⎨
õ2
ϕ 12
⎫
⎬
⎩ √µ2
ϕ 22 ⎭
1.5 Step 5, obtain the modal transformation matrix Φ
The modal transformation matrix is the 2 × 2 matrix made of of Φ 1 , Φ 2 in each of its columns
[Φ] = [Φ 1 Φ 2 ]
⎡
= ⎣
⎤
õ1
ϕ 11 √µ2
ϕ 12
⎦
õ1
ϕ 21 √µ2
ϕ 22
Now the [Φ] is found, the transformation from the normal coordinates {x} to modal coordinates, which is
called {η} is found
{x} = [Φ] {η}
⎧ ⎫ ⎡
⎨x 1 (t) ⎬ ϕ 11
⎩
x 2 (t)
⎭ = ⎣
⎤
õ1 õ2
ϕ 12
⎦
õ1
ϕ 21 √µ2
ϕ 22
⎧ ⎫
⎨η 1 (t) ⎬
⎩
η 2 (t)
⎭
The transformation from modal coordinates back to normal coordinates is
{η} = [Φ] −1 {x}
⎧ ⎫ ⎡
⎨η 1 (t) ⎬ ϕ 11
⎩
η 2 (t)
⎭ = ⎣
⎤
õ1 õ2
ϕ 12
⎦
õ1
ϕ 21 √µ2
ϕ 22
−1 ⎧ ⎫
⎨ x 1 (t) ⎬
⎩
x 2 (t)
⎭
7

However, [Φ] −1 = [Φ] T [M] therefore
{η} = [Φ] T [M] {x}
⎧ ⎫ ⎡ ⎤
⎨η 1 (t) ⎬ √µ1
ϕ 11 √µ2
ϕ 12
⎩
η 2 (t)
⎭ = ⎣ ⎦
õ1
ϕ 21 √µ2
ϕ 22
⎤ ⎧ ⎫
⎣ m 11 m 12
⎨x ⎦ 1 (t) ⎬
m 21 m
⎩
22 x 2 (t)
⎭
T ⎡
The next step is to apply this transformation to the original equations of motion in order to decouple them
1.6 Step 6, applying modal transformation to decouple the original equations of motion
The EOM in normal coordinates is
⎡
⎣ m 11 m 12
m 21
⎤ ⎧
⎨
⎦
m
⎩
22
x ′′
1
x ′′
2
⎫ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫
⎬
⎭ + ⎣ k 11 k 12
⎨x ⎦ 1 ⎬ ⎨
k
⎩
22 x
⎭ = f 1 (t) ⎬
⎩
2 f 2 (t)
⎭
Applying the above modal transformation {x} = [Φ] {η} on the above results in
⎡ ⎤ ⎧ ⎫ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫
⎣ m 11 m 12
⎨η 1
⎦ ′′ ⎬
[Φ]
m 21 m
⎩
22 η ′′ ⎭ + ⎣ k 11 k 12
⎨η ⎦ 1 ⎬ ⎨
[Φ]
k 21 k
⎩
22 η
⎭ = f 1 (t) ⎬
⎩
2 f 2 (t)
⎭
pre-multiplying by [Φ] T results in
⎡ ⎤ ⎧
[Φ] T ⎣ m 11 m 12
⎨
⎦ [Φ]
m
⎩
22
m 21
η 1
′′
η 2
′′
2
k 21
⎫ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫
⎬
⎭ + [Φ]T ⎣ k 11 k 12
⎨η ⎦ 1 ⎬ ⎨
[Φ]
k
⎩
22 η
⎭ = f 1 (t) ⎬
[Φ]T ⎩
2 f 2 (t)
⎭
⎡ ⎤
⎡ ⎤
The result of [Φ] T ⎣ m 11 m 12
⎦ [Φ] will always be ⎣ 1 0 ⎦. This is because mass normalized shape vectors
m 21 m 22 0 1
are used. If the shape functions were not mass normalized, then the diagonal values will not be 1 as shown.
⎡ ⎤
⎡ ⎤
The result of [Φ] T ⎣ k 11 k 12
⎦ [Φ] will be ⎣ ω2 1 0
⎦.
k 21 k 22 0 ω2
2
⎧ ⎫ ⎧ ⎫
⎨
Let the result of [Φ] T f 1 (t) ⎬ ⎨ ˜f
⎩
f 2 (t)
⎭ be 1 (t) ⎬
,Therefore, in modal coordinates the original EOM becomes
⎩ ˜f 2 (t)
⎭
⎡ ⎤ ⎧
⎣ 1 0 ⎨
⎦
0 1
⎩
η 1
′′
η 2
′′
k 21
⎫ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫
⎬
⎭ + ⎣ ω2 1 0 ⎨η ⎦ 1 ⎬ ⎨ ˜f
0 ω2
2 ⎩
η
⎭ = 1 (t) ⎬
⎩
2
˜f 2 (t)
⎭
The EOM are now decouples and each can be solved as follows
η ′′
1 (t) + ω 2 1η 1 (t) = ˜f 1 (t)
η ′′
2 (t) + ω 2 2η 2 (t) = ˜f 2 (t)
To solve these EOM’s, the initial conditions in normal coordinates must be transformed to modal coordinates
using the above transformation rules
{η (0)} = [Φ] T [M] {x (0)}
{
η ′ (0) } = [Φ] T [M] { x ′ (0) }
8

Or in full form
and
⎧ ⎫ ⎡
⎨η 1 (0) ⎬
⎩
η 2 (0)
⎭ = ⎣
⎧ ⎫ ⎡
⎨η 1 ′ (0) ⎬
⎩
η 2 ′ (0) ⎭ = ⎣
⎤T ⎡
õ1
ϕ 11 √µ2
ϕ 12
⎦
õ1
ϕ 21 √µ2
ϕ 22
⎤T ⎡
õ1
ϕ 11 √µ2
ϕ 12
⎦
õ1
ϕ 21 √µ2
ϕ 22
⎣ m 11 m 12
m 21
⎣ m 11 m 12
m 21
⎤ ⎧ ⎫
⎨x ⎦ 1 (0) ⎬
m
⎩
22 x 2 (0)
⎭
⎤ ⎧ ⎫
⎨x ⎦
′ 1 (0) ⎬
m
⎩
22 x ′ 2 (0) ⎭
Each of these EOM are solved using any of the standard methods. This will result is solutions η 1 (t) and
η 2 (t)
1.7 Step 7. Converting modal solution to normal coordinates solution
The solutions found above are in modal coordinates η 1 (t) , η 2 (t). The solution needed is x 1 (t) , x 2 (t). Therefore,
the transformation {x} = [Φ] {η} is now applied to convert the solution to normal coordinates
⎧ ⎫ ⎡ ⎤ ⎧ ⎫
⎨x 1 (t) ⎬ √µ1
ϕ 11 √µ2
ϕ 12 ⎨
⎩
x 2 (t)
⎭ = η
⎣ ⎦ 1 (t) ⎬
õ1
ϕ 21 √µ2
ϕ 22 ⎩
η 2 (t)
⎭
⎧
⎫
⎨ √µ1
ϕ 11
η 1 (t) + √ ϕ 12
µ2
η 2 (t) ⎬
=
⎩ √µ1
ϕ 21
η 1 (t) + √ ϕ 22
µ2
η 2 (t)
⎭
Hence
and
x 1 (t) = ϕ 11
√
µ1
η 1 (t) + ϕ 12
√
µ2
η 2 (t)
x 2 (t) = ϕ 21
√
µ1
η 1 (t) + ϕ 22
√
µ2
η 2 (t)
Notice that the solution in normal coordinates is a linear combination of the modal solutions. The terms
ϕ
√ ij
µ
are just scaling factors that represent the contribution of each modal solution to the final solution. This
completes modal analysis
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 = 1, k 2 = 2, m 1 = 1, m 2 = 3 and let f 1 (t) = 0 and f 2 (t) = sin (5t). Let initial conditions be
x 1 (0) = 0, x ′ 1 (0) = 1, x 2 (0) = 1.5, x ′ 2 (0) = 3, hence
⎧ ⎫ ⎫
⎨x 1 (0) ⎬ 0⎬
⎧
⎨
⎩
x 2 (0)
⎭ = ⎩
1
⎭
and
⎧ ⎫ ⎧ ⎫
⎨x ′ 1 (0) ⎬ ⎨
⎩
x ′ 2 (0) ⎭ = 1.5⎬
⎩
3
⎭
In normal coordinates, the EOM are
9

⎡ ⎤ ⎧ ⎫ ⎡
⎤ ⎧ ⎫ ⎧ ⎫
⎣ m 1 0 ⎨x ⎦
′′ ⎬
1
0 m
⎩
2 x ′′ ⎭ + ⎣ k 1 + k 2 −k 2
⎨x ⎦ 1 ⎬ ⎨
2 −k 2 k
⎩
2 x
⎭ = f 1 (t) ⎬
⎩
2 f 2 (t)
⎭
⎡ ⎤ ⎧ ⎫ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫
⎣ 1 0 ⎨x ⎦
′′ ⎬
1
0 3
⎩
x ′′ ⎭ + ⎣ 3 −2 ⎨x ⎦ 1 ⎬ ⎨
−2 2
⎩
x
⎭ = 0 ⎬
⎩
2 sin (5t)
⎭
2
In this example m 11 = 1, m 12 = 0, m 21 = 0, m 22 = 3 and k 11 = 3, k 12 = −2, k 21 = −2, k 22 = 2 and f 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
Let ω 2 = λ hence
The solution is λ 1 = 3. 475 and λ 2 = 0.192, therefore
And
det ( [K] − ω 2 [M] ) = 0
⎛⎡
⎤ ⎡ ⎤⎞
det ⎝⎣ 3 −2 ⎦ − ω 2 ⎣ 1 0 ⎦⎠ = 0
−2 2 0 3
⎡
⎤
det ⎣ 3 − ω2 −2
⎦ = 0
−2 2 − 3ω 2
(
3 − ω
2 ) ( 2 − 3ω 2) − (−2) (−2) = 0
3ω 4 − 11ω 2 + 2 = 0
3λ 2 − 11λ + 2 = 0
ω 1 = √ 3.475 = 1.864
ω 2 = √ 0.192 = 0.438
step 3 is now applied which finds 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 eigen vectors ϕ 1 , ϕ 2
⎛⎡
⎤ ⎡ ⎤⎞
⎧ ⎫ ⎧ ⎫
⎝⎣ 3 −2 ⎦ − ω1
2 ⎣ 1 0 ⎨ϕ ⎦⎠
11 ⎬ ⎨
−2 2 0 3
⎩
ϕ
⎭ = 0⎬
⎩
21 0
⎭
For ω 1 = 1. 864
⎛⎡
⎤ ⎡ ⎤⎞
⎧ ⎫ ⎧ ⎫
⎝⎣ 3 −2 ⎦ − 1.864 2 ⎣ 1 0 ⎨ϕ ⎦⎠
11 ⎬ ⎨
−2 2
0 3
⎩
ϕ
⎭ = 0⎬
⎩
21 0
⎭
⎡
⎤ ⎧ ⎫ ⎧ ⎫
⎣ −0.475 −2 ⎨ 1 ⎬ ⎨
⎦
−2 −8.424
⎩
ϕ
⎭ = 0⎬
⎩
21 0
⎭
This gives one equation to solve for ϕ 21 (the first row equation is only used)
−0.475 − 2ϕ 21 = 0
10

Hence
The first eigen vector is
Similarly for ω 2 = 0.438
ϕ 21 = 0.475
−2 = −0.237
⎧ ⎫ ⎧ ⎫
⎨ϕ 11 ⎬ ⎨
ϕ 1 =
⎩
ϕ
⎭ = 1 ⎬
⎩
21 −0.237
⎭
⎛⎡
⎤ ⎡ ⎤⎞
⎧ ⎫ ⎧ ⎫
⎝⎣ 3 −2 ⎦ − 0.438 2 ⎣ 1 0 ⎨ϕ ⎦⎠
12 ⎬ ⎨
−2 2
0 3
⎩
ϕ
⎭ = 0⎬
⎩
22 0
⎭
⎡
⎤ ⎧ ⎫ ⎧ ⎫
⎣ 2.808 −2 ⎨ 1 ⎬ ⎨
⎦
−2 1.425
⎩
ϕ
⎭ = 0⎬
⎩
22 0
⎭
This gives one equation to solve for ϕ 22 (the first row equation is only used)
2.808 − 2ϕ 22 = 0
Hence
The second eigen vector is
ϕ 22 = −2.808
−2
= 1.404
⎧ ⎫ ⎧ ⎫
⎨ϕ 12 ⎬ ⎨
ϕ 2 =
⎩
ϕ
⎭ = 1 ⎬
⎩
22 1.404
⎭
Now step 4 is applied, which is mass normalization of the shape vectors (or the eigenvectors)
µ 1 = ϕ T 1 [M] ϕ 1
Hence
Similarly, µ 2 is found
Hence
⎧ ⎫
⎨ 1 ⎬
µ 1 =
⎩
−0.237
⎭
= 1. 169
⎧ ⎫
⎨ 1 ⎬
µ 2 =
⎩
1.404
⎭
= 6.914
⎤ ⎧ ⎫
⎣ 1 0 ⎨ 1 ⎬
⎦
0 3
⎩
−0.237
⎭
T ⎡
µ 2 = ϕ T 2 [M] ϕ 2
⎤ ⎧ ⎫
⎣ 1 0 ⎨ 1 ⎬
⎦
0 3
⎩
1.404
⎭
T ⎡
11

Now that µ 1 , µ 2 are found, the mass normalized eigen vectors are found. They are called Φ 1 , Φ 2
⎧ ⎧ ⎫
⎨ ⎨ 1 ⎬
Φ 1 = ϕ 1
√
µ1
=
⎩
ϕ 11
⎧ ⎫
⎨ 0.925 ⎬
=
⎩
−0.219
⎭
ϕ 21
⎫
⎬
⎭
√
µ1
=
⎩
−0.237
⎭
√
1.169
Similarly
Φ 2 = ϕ 2
√
µ2
=
⎧
⎨
⎩
⎧ ⎫
⎨0.380⎬
=
⎩
0.534
⎭
ϕ 12
ϕ 22
⎫
⎬
⎭
√
µ2
=
⎧
⎨
1
⎫
⎬
⎩
1.404
⎭
√
6.914
Therefore, the modal transformation matrix is
[Φ] = [Φ 1 Φ 2 ]
⎡
⎤
0.925
= ⎣
0.380
⎦
−0.219 0.534
This result can be verified using Matlab’s eig function as follows
EDU>> K=[3 -2;-2 2]; M=[1 0;0 3];
EDU>> [phi,lam]=eig(K,M)
phi =
-0.3803 -0.9249
-0.5340 0.2196
EDU>> diag(sqrt(lam))
0.4380
1.8641
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 coordinates, called
{η} , is obtained
{x} = [Φ] {η}
⎧ ⎫ ⎡
⎤ ⎧ ⎫
⎨x 1 (t) ⎬
⎩
x 2 (t)
⎭ = 0.925 0.380 ⎨η ⎣ ⎦ 1 (t) ⎬
−0.219 0.534
⎩
η 2 (t)
⎭
12

The transformation from modal coordinates back to normal coordinates is
However, [Φ] −1 = [Φ] T [M] therefore
{η} = [Φ] −1 {x}
⎧ ⎫ ⎡
⎤−1 ⎧ ⎫
⎨η 1 (t) ⎬
⎩
η 2 (t)
⎭ = 0.925 0.380 ⎨ x
⎣ ⎦ 1 (t) ⎬
−0.219 0.534
⎩
x 2 (t)
⎭
{η} = [Φ] T [M] {x}
⎧ ⎫ ⎡
⎤T ⎡ ⎤ ⎧ ⎫
⎨η 1 (t) ⎬
⎩
η 2 (t)
⎭ = 0.925 0.380
⎣ ⎦ ⎣ 1 0 ⎨x ⎦ 1 (t) ⎬
−0.219 0.534 0 3
⎩
x 2 (t)
⎭
⎡
⎤ ⎧ ⎫
0.925 −0.657 ⎨x = ⎣ ⎦ 1 (t) ⎬
0.38 1.6
⎩
x 2 (t)
⎭
The next step is to apply this transformation to the original equations of motion in order to decouple them.
Applying step 6 results in
⎡ ⎤ ⎧ ⎫ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫
⎣ 1 0 ⎨η 1
⎦
′′ ⎬
0 1
⎩
η 2
′′ ⎭ + ⎣ ω2 1 0 ⎨η ⎦ 1 ⎬ ⎨
0 ω2
2 ⎩
η
⎭ = 0 ⎬
[Φ]T ⎩
2 sin (5t)
⎭
⎡ ⎤ ⎧ ⎫ ⎡
⎤ ⎧ ⎫ ⎡
⎤T ⎧ ⎫
⎣ 1 0 ⎨η 1
⎦
′′ ⎬
0 1
⎩
η 2
′′ ⎭ + ⎣ 1.8642 0 ⎨η ⎦ 1 ⎬
0 0.438 2 ⎩
η
⎭ = 0.925 0.380 ⎨ 0 ⎬
⎣ ⎦
2 −0.219 0.534
⎩
sin (5t)
⎭
⎡ ⎤ ⎧ ⎫ ⎡
⎤ ⎧ ⎫ ⎧ ⎫
⎣ 1 0 ⎨η 1
⎦
′′ ⎬
0 1
⎩
η ′′ ⎭ + 3. 47 0 ⎨η ⎣ ⎦ 1 ⎬ ⎨
0 0.192
⎩
η
⎭ = −0.219 sin (5t) ⎬
⎩
2 0.534 sin (5t)
⎭
2
The EOM are now decoupled and each EOM can be solved easily as follows
η ′′
1 (t) + 3.47η 1 (t) = −0.219 sin (5t)
η ′′
2 (t) + 0.192η 2 (t) = 0.534 sin (5t)
To solve these EOM’s, the initial conditions in normal coordinates must be transformed to modal coordinates
using the above transformation rules
⎧ ⎫ ⎡
⎤ ⎧ ⎫
⎨η 1 (0) ⎬
⎩
η 2 (0)
⎭ = 0.925 −0.657 ⎨x ⎣ ⎦ 1 (0) ⎬
0.38 1.6
⎩
x 2 (0)
⎭
⎧ ⎫ ⎡
⎤ ⎧ ⎫
⎨η 1 (0) ⎬
⎩
η 2 (0)
⎭ = 0.925 −0.657 ⎨0⎬
⎣ ⎦
0.38 1.6
⎩
1
⎭
⎧ ⎫
⎨−0.657⎬
=
⎩
1.6
⎭
13

and
⎧ ⎫ ⎡
⎤ ⎧ ⎫
⎨η 1 ′ (0) ⎬
⎩
η 2 ′ (0) ⎭ = 0.925 −0.657 ⎨x ⎣ ⎦
′ 1 (0) ⎬
0.38 1.6
⎩
x ′ 2 (0) ⎭
⎡
⎤ ⎧ ⎫
0.925 −0.657 ⎨1.5⎬
= ⎣ ⎦
0.38 1.6
⎩
3
⎭
⎧ ⎫
⎨−0.584⎬
=
⎩
5.37
⎭
Each of these EOM are solved using any of the standard methods. This results in solutions η 1 (t) and η 2 (t) .
Hence the following EOM’s are solved
and also
η ′′
1 (t) + 3.47η 1 (t) = −0.219 sin (5t)
η 1 (0) = −0.657
η ′ 1 (0) = −0.584
η ′′
2 (t) + 0.192η 2 (t) = 0.534 sin (5t)
η 2 (0) = 1.6
η ′ 2 (0) = 5.37
The solutions η 1 (t) , η 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
⎧ ⎫ ⎡
⎤ ⎧ ⎫
⎨x 1 (t) ⎬
⎩
x 2 (t)
⎭ = 0.925 0.380 ⎨η ⎣ ⎦ 1 (t) ⎬
−0.219 0.534
⎩
η 2 (t)
⎭
⎧
⎫
⎨ 0.925 η 1 + 0.38η 2 ⎬
=
⎩
0.534 η 2 − 0.219 η
⎭
1
Hence
and
x 1 (t) = 0.925η 1 (t) + 0.38η 2 (t)
x 2 (t) = 0.534η 1 (t) − 0.219η 2 (t)
The above shows that the solution x 1 (t) and x 2 (t) has contributions from both nodal solutions.
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
(
˜f (t) = 1 N
)
2 F ∑
2π
in
0 + Re F n e T t
= 1 2 F 0 + 1 2
= 1 2
N∑
n=−N
n=1
N∑
F n e
n=1
2π
in
F n e T t
in
2π
T t + F ∗ ne
2π
−in T t
14

Where
F n = 2 T
F 0 = 2 T
∫ T
0
∫ T
0
2π
−in
f (t) e T t dt
f (t) dt
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.
˜f (t) = a 0 +
a 0 = 1 T
n=1
∫ T
0
a n = 1
T/2
b n = 1
T/2
N∑
a n cos n 2π N T t + ∑
b n sin n 2π
T t
f (t) dt
∫ T
0
∫ T
0
n=1
(
f (t) cos n 2π )
T t dt
(
f (t) sin n 2π )
T t dt
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.
15

3 Transfer functions for different vibration systems
y
c
m
k
z
u y z
3.1 Force transmissibility
Let steady state
x ss = Re
{ ˆF
k D (r, ζ) eiϖt }
Then
f tr (t) = f spring + f damper
= kx + cx ′
{
= Re k ˆF
} {
k D (r, ζ) eiϖt + Re ciϖ ˆF
}
k D (r, ζ) eiϖt
{(
= Re ˆF + ciϖ ˆF
) }
D (r, ζ) e iϖt
k
Hence
|f tr (t)| max
= ∣ ˆF
√
∣ √
∣ |D| 1 + c 2 ϖ2 ∣∣
k 2 = ˆF ∣ |D| 1 + (2ζr) 2
So TR or force transmissibility is
T R = |f √
tr (t)| ∣ max
∣ ˆF
= |D| 1 + (2ζr) 2
∣
If r > √ 2 then we want small ζ to reduce force transmitted to base. For r < √ 2, it is the other way round.
3.2 vibration isolation
We need transfer function between y and z.
Equation of motion
my ′′ = −c ( y ′ − z ′) − k (y − z)
my ′′ + cy ′ + ky = cz ′ + kz
16

Let z = Re { Ze iωt} , z ′ = Re { iωZe iωt} and let y = Re { Y e iωt} , y ′ = Re { iωY e iωt} , y ′′ = Re { −ω 2 Y e iωt} ,
hence the above becomes
Hence |D (r, ζ)| =
m Re { −ω 2 Y e iωt} + c Re { iωY e iωt} + k Re { Y e iωt} = c Re { iωZe iωt} + k Re { Ze iωt}
√
1+(2ζr) 2
√
(1−r 2 ) 2 +(2ζr) 2
ciω + k
Y =
−ω 2 m + ciω + k Z
i2ζω n mω + k
=
−ω 2 m + i2ζω n mω + k Z
i2ζω n ω + ωn
2 =
−ω 2 + i2ζω n ω + ωn
2 Z
i2ζr + 1
=
(1 − r 2 ) + i2ζr Z
( )
and arg (D) = tan−1 (2ζr) − tan −1 2ζr
1−r 2
where r = ω ω n
√
Hence for good vibration isolation we need |Y | max
|Z|
to be small. i.e. |D| 1 + (2ζr) 2 to be small. This is the
same TR as for force isolation above.
For small |D|, we need small ζ and small k (the small k is to make r > √ 2) see plot
In Matlab, the above can be plotted using
17

close all;
zeta=linspace(0.1, 0.7, 10);
r=linspace(0, 3, 10);
D=@(r,z) (sqrt(1+(2*z*r).^2)./sqrt((1-r.^2).^2+(2*z*r).^2));
figure;
hold on;
for i=1:length(zeta)
plot(r,D(r,zeta(i)));
end
grid on;
legend
3.3 accelerometer
We need transfer function between u and z a where now z a is the amplitude of the ground acceleration.
This device is used to measure base acceleration by relating it linearly to relative displacement of m to base.
Equation of motion. We use relative distance now.
m ( u ′′ + z ′′) + cu ′ + ku = 0
mu ′′ + cu ′ + ku = −mz ′′
Let z ′′ = Re { Z a e iωt} . Notice we here jumped right away to the z ′′ itself and wrote it as Re { Z a e iωt} and we
did not go through the steps as above starting from base motion. This is because we want the transfer function
between relative motion u and acceleration of base.
Now, u = Re { Ue iωt} , u ′ = Re { iωUe iωt} , u ′′ = Re { −ω 2 Ue iωt} , hence the above becomes
Hence |D (r, ζ)| =
m Re { −ω 2 Ue iωt} + c Re { iωUe iωt} + k Re { Ue iωt} = −m Re { Z a e iωt}
−m
U =
−ω 2 m + iωc + k Z a
−1
=
−ω 2 + iω2ζω n + ωn
2 Z a
−1
=
(ωn 2 − ω 2 Z a
) + iω2ζω n
( )
−1
√(ω and arg (D) = −1800 − tan −1 2ωζωn
n−ω 2 2 ) 2 +(2ωζω n) 2 ωn−ω 2 2
When system is very stiff, which means ω n very large compared to ω , then D (r, ζ) ≈ −1
ω 2 n
Z a , hence by
measuring u we estimate Z a the amplitude of the ground acceleration since ω 2 n is known. For accuracy, need
ω n > 5ω at least.
3.4 seismometer
Now we need to measure the base motion (not base acceleration like above). But we still use the relative
displacement. Now the transfer function is between u and z where now z is the base motion amplitude.
Equation of motion. We use relative distance now.
m ( u ′′ + z ′′) + cu ′ + ku = 0
mu ′′ + cu ′ + ku = −mz ′′
Let z = Re { Ze iωt} , z ′ = Re { iωZe iωt} ,z ′′ = Re { −ω 2 Ze iωt} ,and let u = Re { Ue iωt} , u ′ = Re { iωUe iωt} , u ′′ =
Re { −ω 2 Ue iωt} , hence the above becomes
18

Now, u = Re { Ue iωt} , u ′ = Re { iωUe iωt} , u ′′ = Re { −ω 2 Ue iωt} , hence the above becomes
m Re { −ω 2 Ue iωt} + c Re { iωUe iωt} + k Re { Ue iωt} = −m Re { −ω 2 Ze iωt}
U =
mω 2
−ω 2 m + iωc + k Z
ω 2
=
−ω 2 + iω2ζω n + ωn
2 Z
r 2
=
(1 − r 2 ) + i2ζr Z
( )
r
Hence |D (r, ζ)| = √ 2
and arg (D) = − 2ζr
(1−r 2 )+i2ζr tan−1 1−r 2 1
Now if r is very large, which happens when ω n ≪ ω, then
(1−r 2 )+i2ζr ⇒ 1 since r 2 is the dominant factor.
−r 2
r
Therefore U = 2
(1−r 2 )+i2ζr Z a now becomes U ≃ −Z a therefore measuring the relative displacement U gives
linear estimate of the ground motion. However, this device requires that ω n be much smaller than ω, which
means that m has to be massive. So this device is heavy compared to accelerometer.
3.5 Summary of vibration transfer functions
For good isolation of mass from ground motion, rule of thumb: Make damping low, and stiffness low (soft spring).
system equation used to derive transfer function
isolate base from force
transmitted by machine
isolate machine
from motion of base
accelerometer: Measure
base acc. using relative
displacement
f tr (t) = f spring + f damper
Use absolute mass position
my ′′ + cy ′ + ky = cz ′ + kz
|f tr(t)| ∣ max
∣ ˆF
∣
|Y | max
|Z|
√
= |D| 1 + (2ζr) 2
√
= |D| 1 + (2ζr) 2
−1
U =
Use relative mass position (ωn 2 − ω 2 Z a ⇒ |D (r, ζ)|
) + iω2ζω n
−1
mu ′′ + cu ′ + ku = −mz ′′ = √
(ωn 2 − ω 2 ) 2 + (2ωζω n ) 2
seismometer: Measure
base motion using
relative displacement
Use relative mass position
mu ′′ + cu ′ + ku = −mz ′′ U =
r 2
(1−r 2 )+i2ζr Z ⇒ |D (r, ζ)| = r 2 √
(1−r 2 )+i2ζr
19

4 Steady state solutions to single degree freedom system
4.1 Summary:
( )
my ′′ + cy ′ + ky = Re ˆF e
iϖt
{ }
x = Re ˆXe
iϖt
ˆX = ˆF D (r, ζ)
k
1
D (r, ζ) =
(1 − r 2 ) + 2iζr
{ }
ˆF
x = Re
k |D (r, ζ)| ei(ϖt−θ)
θ = tan −1
2ζr
1 − r 2
4.2 Details
( )
Let load be harmonic and represented in general as Re ˆF e
iϖt
where ˆF is the complex amplitude of the force.
Hence system is represented by
( )
my ′′ + cy ′ + ky = Re ˆF e
iϖt
( ) ˆF
y ′′ + 2ζω n y ′ + ωny 2 = Re
m eiϖt
)
)
)
Let y = Re
(Ŷ e
iϖt
Hence y ′ = Re
(iϖŶ eiϖt , y ′′ = Re
(−ϖ 2 Ŷ e iϖt , therefore the differential equation
becomes
(
Re −ϖ 2 Ŷ e iϖt) (
+ 2ζω n Re iϖŶ eiϖt) + ωn 2 Re
(Ŷ ) ( )
e
iϖt ˆF
= Re
m eiϖt
Dividing numerator and denominator ω 2 n gives
Where r = ϖ ω n
, hence the response is
Therefore, the phase of the response is
y = Re
( )
arg (y) = arg ˆF
( ˆF
−ϖ 2 + 2ζω n iϖ + ωn) 2 Ŷ =
m
ˆF
m
Ŷ =
(−ϖ 2 + 2ζω n iϖ + ωn)
2
Ŷ = ˆF 1
k (1 − r 2 ) + i2ζr
( )
ˆF 1
k (1 − r 2 ) + i2ζr eiϖt
( 2ζr
− tan −1 (1 − r 2 )
20
)
+ ϖt

Hence at t = 0 the phase of the response will be
( ) ( ) 2ζr
arg (y) = arg ˆF − tan −1 (1 − r 2 )
So when ˆF
( )
is real, the phase of the response is simply − tan −1 2ζr
(1−r 2 )
4.3 Undamped case
When ζ = 0 the above becomes
For real force this becomes
∣
The magnitude ∣Ŷ ∣ = F k
4.4 damped cases
ζ > 0
1
(1−r 2 )
( ˆF
y = Re
k
∣ ˆF
∣
=
k
)
1
(1 − r 2 ) eiϖt
1
(
(1 − r 2 ) cos
y = F k
and phase zero.
∣
∣Ŷ
y = Re
( ))
ϖt + arg ˆF
1
(1 − r 2 cos (ϖt)
)
( )
ˆF 1
k (1 − r 2 ) + i2ζr eiϖt
∣ ˆF
∣ 1
∣ = √
k
(1 − r 2 ) 2 + (2ζr) 2
) (
arg
(Ŷ = φ = arg ˆF
Hence for real force and at t = 0 the phase of displacement is
( ) 2ζr
− tan −1 1 − r 2
)
− tan −1 ( 2ζr
1 − r 2 )
+ ϖt
lag behind the load.
When r < 1 then φ goes from 0 to −90 0 Therefore phase of displacement is 0 to −90 0 behind force. The
minus sign at the front was added since the complex number is in the denominator. Hence the response will
always be lagging in phase relative for load.
For r > 1
Now 1 − r 2 is negative, hence the phase will be from −90 ◦ to −180 ◦
When r = 1
( ) ˆF 1
y = Re
k i2ζ eiϖt
∣ ∣
∣Ŷ
)
arg
(Ŷ
∣ ˆF
∣
∣ =
k
= −90 ◦
1
2ζ
Now phase is −90 ◦ 21

Phase of response complex
amplitude for underdamped
and when r1
Phase will be from -90 to -
180 degrees
Phase of response complex
amplitude for
underdamped and when
r=1
Phase will -90 degrees
2r
2r
1 r 2
1 r 2
Examples. System has ζ =
)
0.1 and
√
m = 1, k = 1 subjected for force 3 cos (0.5t) find the steady state solution.
Answer y (t) = Re
(Ŷ e
iϖt
, ω n = = 1 rad/sec, hence r = 0.5 under the response is
k
m
(∣ ∣∣ y (t) = Re Ŷ ∣ e iϖt)
∣
= ∣Ŷ ∣ cos (ϖt)
= F k
1
√(1 − r 2 ) 2 + (2ζr) 2 cos
(
( ))
2 (0.1) 0.5
.5t − tan −1 1 − 0.5 2
1
= 3
cos (.5t − 7.59 √(1 ◦ )
− 0.5 2 ) 2 + (2 (0.1) 0.5) 2
= 3. 964 9 cos (.5t − 7.59 ◦ )
5 Summary table of solutions
5.1 Harmonic loading mu ′′ + cu ′ + ku = F sin ϖt
The equation of motion can also be written as u ′′ + 2ζωu ′ + ω 2 u = F m
sin ϖt.
22

The following table gives the solutions for initial conditions are u (0) and u ′ (0) under all damping conditions.
The roots shown are the roots of the quadratic characteristic equation λ 2 + 2ζωλ + ω 2 λ = 0. Special handling is
needed to obtain the solution of the differential equation for the case of ζ = 0 and ϖ = ω as described in the
detailed section below.
⎧
⎨ −iω
roots
ζ = 0
⎩
⎧
⎪⎨
u (t)
⎪⎩
⎧
⎨
roots
⎩
+iω
ϖ = ω → u (0) cos ϖt + u′ (0)
ϖ
sin ϖt − F K
ϖ ≠ ω → u (0) cos ωt +
−ξω + iω n
√
1 − ξ 2
−ξω − iω n
√
1 − ξ 2
(
u ′ (0)
ω
− F K
ϖt
2
cos (ϖt)
)
r
sin ωt + F 1
1−r 2 K
sin ϖt
1−r 2
ζ < 1
u (t) = e −ξωt (A cos ω d t + B sin ω d t) + F K
A = u 0 + F K
1 √(1−r 2 ) 2 +(2ξr) 2 sin θ
1
sin (ϖt − θ)
√(1−r 2 ) 2 2
+(2ξr)
B = v 0
ω d
+ u 0ξω
ω d
⎧
⎨ −ω
roots
⎩
−ω
+ F √ 1
K
ω d (1−r 2 ) 2 +(2ξr)
2
(ξω sin θ − ϖ cos θ)
ζ = 1
ζ > 1
u (t) = (A + Bt) e −ωt + F K
A = u 0 + F K
1 √(1−r 2 ) 2 +(2r) 2 sin θ
1
sin (ϖt − θ)
√(1−r 2 ) 2 2
+(2r)
F/k
B = v 0 + u 0 ω +
(ω sin θ − ϖ cos θ)
√(1−r 2 ) 2 2
+(2r)
⎧
√
⎨ −ω n ξ + ω n ξ 2 − 1
roots
⎩
√
−ω n ξ − ω n ξ 2 − 1
(
−ξ+ √ (
ξ
u (t) = Ae
2 −1
)ω nt −ξ− √ )
ξ + Be 2 −1 ω nt +
F
K
β sin (ϖt − θ)
A = v 0+u 0 ωξ+u 0 ω √ ((
ξ 2 −1+ F K β ξ+ √ )
)
ξ 2 −1 ω sin θ−ϖ cos θ
2ω √ ξ 2 −1
B = − v 0+u 0 ωξ−u 0 ω √ ((
ξ 2 −1+ F K β ξ− √ )
)
ξ 2 −1 ω sin θ−ϖ cos θ
2ω √ ξ 2 −1
23

5.2 constant loading mu ′′ + cu ′ + ku = F
⎧
⎨ −iω
roots
ζ = 0 ⎩
+iω
ζ < 1
ζ = 1
u (t) = ( u 0 − F )
k cos ωt +
v 0
ω
sin ωt + F k
⎧
√
⎨ −ξω + iω n 1 − ξ 2
roots
⎩
√
−ξω − iω n 1 − ξ 2
( (u0
u (t) = e −ξωt − F )
k cos ωd t +
⎧
⎨ −ω
roots
⎩
−ω
(
) )
v 0 +u 0 ξω− F k ξω
ω d
sin ω d t + F k
u (t) = (( u 0 − F ) (
k + v0 + u 0 ω − F k ω) t ) e −ωt + F k
⎧
√
⎨ −ω n ξ + ω n ξ 2 − 1
roots
⎩
√
−ω n ξ − ω n ξ 2 − 1
ζ > 1
B = F k p 1−u 0 p 1 +v 0
(p 2 −p 1 )
A = u 0 − F k − B
p 1 = − c
2m + √ ( c
2m
p 2 = − c
2m − √ ( c
2m
u (t) = Ae p 1t + Be p 2t + F k
) 2 −
k
m = −ω √
nξ + ω n ξ 2 − 1
) 2 −
k
m = −ω √
nξ − ω n ξ 2 − 1
24

5.3 no loading (free) mu ′′ + cu ′ + ku = 0
⎧
⎨ −iω
roots
⎩
+iω
ζ = 0
u ′′ + ω 2 u = 0
ζ < 1
ζ = 1
u (t) = u(0) cos ωt + u′ (0)
ω
sin ωt
⎧
u (t) = A cos (ωt − φ)
⎪⎨ √
( )
A = u(0) + u ′ 2
(0)
ω
( )
⎪⎩ φ = tan −1 u ′ (0)/ω
u(0)
⎧
⎨ −ξω + iω √ 1 − ξ 2
roots
⎩
−ξω − iω √ 1 − ξ 2
(
u (t) = e −ξωt u(0) cos ω d t + u′ (0)+u(0)ξω
ω
⎧
d
⎨ −ω
roots
⎩
−ω
)
sin ω d t
ζ > 1
u (t) = (u(0)(1 + ωt) + u ′ (0)t) e −ωt
⎧
⎨ λ 1 = −ωξ + ω √ ξ 2 − 1
roots
⎩
λ 2 = −ωξ − ω √ ξ 2 − 1
⎧
u (t) = Ae
⎪⎨
λ1ωt + Be λ 2ωt
A = u′ (0)−u(0)λ 2
2ω √ ξ 2 −1
⎪⎩
B = −u′ (0)+u(0)λ 1
2ω √ ξ 2 −1
25

5.4 impulse F 0 δ (t) loading
⎧
⎨ −iω
roots
ζ = 0
⎩
+iω
u ′′ + ω 2 u = 0
ζ < 1
ζ = 1
ζ > 1
transient
{ }} {
u (t) = u(0) cos ωt + u′ (0)
⎧
ω
⎨ −ξω + iω √ 1 − ξ 2
roots
⎩
−ξω − iω √ 1 − ξ 2
steady state
{ }} {
sin ωt + F 0
sin ωt
mω
transient
steady state
{ (
}} ){
{ ( }} ){
u (t) = e −ξωt u(0) cos ω d t + u′ (0) + u(0)ξω
sin ω d t + e −ξωt F0
sin ω d t
⎧
ω d
mω d
⎨ −ω
roots
⎩
−ω
u (t) = (u (0) (1 + ωt) + u ′ (0) t) e −ωt + F 0t
⎧
⎨ λ 1 = −ωξ + ω √ ξ 2 − 1
roots
⎩
λ 2 = −ωξ − ω √ ξ 2 − 1
⎧
u (t) = Ae
⎪⎨
λ1ωt + Be λ2ωt +
⎪⎩
A = u′ (0)−u(0)λ 2
2ω √ ξ 2 −1
B = −u′ (0)+u(0)λ 1
2ω √ ξ 2 −1
m e−ωt
F 0
m
2ω √ ξ 2 −1 eλ 1ωt −
The impulse response can be implemented in Mathematica as
parms = {m -> 10, c -> 1.2, k -> 4.3, a -> 1};
tf = TransferFunctionModel[a/(m s^2 + c s + k) /. parms, s]
sol = OutputResponse[tf, DiracDelta[t], t];
F 0
m
2ω √ ξ 2 −1 eλ 2ωt
Plot[sol, {t, 0, 60}, PlotRange -> All, Frame -> True,
FrameLabel -> {{z[t], None}, {Row[{t, " (sec)"}], eq}},GridLines -> Automatic]
26

5.5 Impulse sin function
Input is given by F (t) = F 0 sin (ϖt) where ϖ = 2π
2t 1
= π t 1
ζ = 0
ζ < 1
ζ = 1
⎧
⎨ −iω
roots
⎩
+iω
⎧
⎧
⎪⎨
ϖ = ω →
(
⎪⎨
⎪⎩ −u (0) +
F 0 π
k 2
⎧
(
u (t)
⎪⎨ u (0) cos ωt +
ϖ ≠ ω →
⎪⎩ ⎪⎩
⎧
√
⎨ −ξω + iω n 1 − ξ 2
roots
⎩
√
−ξω − iω n 1 − ξ 2
⎧
⎪⎨
u (t) =
⎪⎩
u (0) cos ωt + u′ (0)
ω sin ωt − F 0
k
ωt
)
cos ωt +
−u ′ (0)+ F 0
k
u ′ (0)
ϖ
( −
(F π/t1
0/k) ω
π/t1
1−(
ω
2 cos (ωt) 0 ≤ t ≤ t 1
π
2t 1
))
) 2
ω
sin ωt t > t 1
sin ωt + F 0/k
1−( π/t1
ω
( )
) 2 sin πt
t 1
0 ≤ t ≤ t 1
u (t 1 ) cos ωt + u′ (t 1 )
ω
sin ωt t > t 1
time constant τ = 1
ζω n
e −ξωt (A cos ω d t + B sin ω d t) + F 0
√ 1
k
(
e −ξωt u (t 1 ) cos ω d t + u′ (t 1 )+u(t 1 )ξω
ω d
sin ω d t
A = u (0) + F 0
k
1 √(1−r 2 ) 2 +(2ξr) 2 sin θ
B = u′ (0)
ω d
⎧
⎨
roots
+ u(0)ξω
ω d
+ F 0
−ω
√ 1
k
ω d (1−r 2 ) 2 +(2ξr)
(
sin
(1−r 2 ) 2 +(2ξr)
)
2
2
(ξω sin θ − ϖ cos θ)
)
t 1
− θ
⎩
−ω
⎧
( )
⎪⎨ (A + Bt) e −ωt + F 0 √ 1
sin k
π t
u (t) =
(1−r 2 ) 2 +(2r) 2 t 1
− θ 0 ≤ t ≤ t 1
⎪⎩
u (t) = (u (t 1 ) + (u ′ (t 1 ) + u (t 1 ) ω) t) e −ωt t > t 1
π t
0 ≤ t ≤ t 1
t > t 1
A = u (0) + F 0
k
1 √(1−r 2 ) 2 +(2r) 2 sin θ
B = u ′ (0) + u (0) ω + F 0
(
√ 1
k
(1−r 2 ) 2 +(2r) 2
)
ω sin θ − π t 1
cos θ
27

ζ > 1
⎧
√
⎨ −ω n ξ + ω n ξ 2 − 1
roots
⎩
√
−ω n ξ − ω n ξ 2 − 1
⎧ (
⎪⎨
−ξ+ √ (
ξ
Ae
2 −1
)ωt + Be
u (t) =
(
⎪⎩
−ξ+ √ )
ξ 2 −1
ω
u (t) = A 1 e
nt + B1 e
A = u′ (0)+u(0)ωξ+u(0)ω √ ((
ξ 2 −1+ F k β ξ+ √ )
ξ 2 −1
2ω √ ξ 2 −1
B = − u′ (0)+u(0)ωξ−u(0)ω √ ξ 2 −1+ F k β ((
ξ− √ ξ 2 −1
1
β = √
(1−r 2 ) 2 +(2ξr) 2
(
A 1 = − ˙u(t 1)+u(t 1 )ω n ξ− √ )
ξ 2 −1
√
2ω n ( ξ 2 −1
B 1 = ˙u(t 1)+u(t 1 )ξω n ξ+ √ )
ξ 2 −1
2ω n
√
ξ 2 −1
2ω √ ξ 2 −1
−ξ− √ )
ξ 2 −1 ωt +
F
π t
(
−ξ− √ )
ξ 2 −1 ω nt
k β sin (
)
ω sin θ− π cos θ t 1
)
)
ω sin θ− π cos θ t 1
)
t 1
− θ
0 ≤ t ≤ t 1
t > t 1
6 Tree view look at the different cases
This tree illustrates the different cases that needs to be considered for the solution of single degree of freedom
system with harmonic loading.
There are 12 cases to consider. Resonance needs to be handled as special case when damping is absent due to
the singularity in the standard solution when the forcing frequency is the same as the natural frequency. When
damping is present, there is no resonance, however, there is what is called practical response which occur when
the forcing frequency is almost the same as the natural frequency.
28

Solution to single degree of freedom system, second order, linear, time invariant
By Nasser M. Abbasi
December 11, 2012
Single_degree_system_tree.vsd
undamped
damped
r
c
c cr
free
mü ku 0
forced
mü ku Fsint
free
forced
mü cu ku 0
mü cu ku Fsint
1 1 1
r 1 r 1
r 1 r 1
1
1
1 1 1
1
r 1 resonance
1 underdamped, roots a ib with a real and negative
1 critical damping roots a,a
1 overdamped roots both real and negative a,b
x
x
x
xx
The following is another diagram made sometime ago which contains more useful information and is kept
here for reference.
29

One degree freedom
linear system unforced
response
y t c m y t k m yt 0
y t 2 n y t n2 yt 0
Some
defintions
k
n m rad/sec
damping ratio
c c c
r
c
2 km 2 n m
Solution roots
s 1,2 n n 2 1
3 cases
d n 1 2 30
Defined only
for under
damped
case(else zero)
1 1 1
Underdamped
2 roots,
complex
conjugate
Critical damped
(one real root,
double
multiplicity)
Over damped,
2 real roots,
distinct
yt e nt Acos d t Bsin d t
A y0, B y 0A n
d
yt A Bte nt
A y0, B y 0 A n
T d 2
d
1
n
Nasser M. Abbasi
May 25, 2011
One_DOF_system.vsd
Damped period of oscillation
Time constant
yt Ae 2 1 n t Be 2 1 n t
A y 0y0 n
2 n
B y 0y0 n
2 n
2 1
2 1
2 1
2 1

7 Some common formulas
These definitions are used throughout the derivations below.
ξ = c c r
=
c
2 √ km = c
2ω n m
u st = F static deflection
√
k
k
ω n =
m
ω D = ω n
√
1 − ξ 2 note: not defined for ξ > 1 since becomes complex
r = ϖ ω n
T d = 2π damped period of oscillation
ω
{ d
−1
τ = , −1 }
time constants where λ i are roots of characteristic equation
λ 1 λ2
1
β =
magnification factor
√(1 − r 2 ) 2 + (2rξ) 2
β max when r = √ 1 − 2ξ 2
1
β max =
2ξ √ 1 − ξ 2
y n
= e ζωn2π
ω D
y
( n+1
)
yn
ln = ζ2π
y n+1
( )
1
M ln yn
= ζ2π
y n+M
small damping
⇒
√
ζ2π
1−ζ 2
e
⇒ e ζ2π
This table shows many cycles it takes for the peak to decay by half its original value as a function of the
damping ζ. For example, we see that when ζ = 2.7% then it takes 4 cycles for the peak (i.e. displacement) to
reduce to half its value.
data = Table[{i, (1/i Log[2]/(2*Pi)*100)}, {i, 1, 20}];
TableForm[N@data,
TableHeadings -> {None, {Column[{"number of cycles",
"needed for peak", "to decay by half"}], "\[Zeta] (%)"}}]
31

The roots of the characteristic equation for u ′′ + 2ξωu ′ + ω 2 u = 0 are given in this table
ξ < 1
roots
{ √ √ }
−ξω + jω n 1 − ξ 2 , −ξω − iω n 1 − ξ 2
time constant τ
1
ξω
1
ξ = 1 {−ω, −ω}
ω
{ √ √ }
ξ > 1 −ω n ξ + ω n ξ 2 − 1, −ω n ξ − ω n ξ 2 1
− 1
√ , 1√ (which to use? the bigger?)
ω nξ−ω n ξ 2 −1 ω nξ+ω n ξ 2 −1
32

8 Derivation of the solutions of the equations of motion
8.1 Free vibration
8.1.1 Undamped free vibration
u
K
M
mu ′′ + ku = 0
u ′′ + ω 2 u = 0
The solution is
u (t) = u (0) cos ωt + u′ (0)
ω
sin ωt
8.1.2 under-damped free vibration c < c r , ξ < 1
u
K
M
C
The solution is
mu ′′ + cu ′ + ku = 0
u ′′ + 2ξωu ′ + ω 2 u = 0
u = e −ξωt (A cos ω d t + B sin ω d t)
Applying initial conditions gives A = u (0) and B = u′ (0)+u(0)ξω
ω d
. Therefore the solution becomes
(
)
u (t) = e −ξωt u (0) cos ω d t + u′ (0) + u (0) ξω
sin ω d t
ω d
33

8.1.3 critically damped free vibration ξ = c
c r
= 1
The solution is
u (t) = (A + Bt) e −( cr
2m
)
t
= (A + Bt) e −ωt
where A, B are found from initial conditions A = u (0),B = u ′ (0) + u (0) ω, hence
8.1.4 over-damped free vibration ξ = c
c r
> 1
The solution is
where A, B are found from initial conditions.
u (t) = ( u (0) + ( u ′ (0) + u (0) ω ) t ) e −ωt
u (t) = Ae λ 1t + Be λ 2t
A = u′ (0) − u (0) λ 2
2ω √ ξ 2 − 1
B = −u′ (0) + u (0) λ 1
2ω √ ξ 2 − 1
where λ 1 and λ 2 are the roots of the characteristic equation
8.2 Constant input F
8.2.1 Undamped case ζ = 0
λ 1 = − c
√ ( c
2m + 2m
λ 2 = − c
2m − √ ( c
2m
) 2 k −
m = −ξω + ω√ ξ 2 − 1
) 2 k −
m = −ξω − ω√ ξ 2 − 1
mu ′′ + ku = F
u ′′ + ω 2 u = F
u (t) = u h + u p
Where u h = A cos ωt + B sin ωt and u p = F k
, the solution is
Applying initial conditions gives
u (t) = A cos ωt + B sin ωt + F k
And complete solution is
A = u (0) − F k
B = u′ (0)
ω
u (t) = F (
k + u (0) − F )
cos ωt + u′ (0)
sin ωt
k
ω
34

8.2.2 under-damped ζ < 1
The general solution is
u (t) = e −ξωt (A cos ω d t + B sin ω d t) + F k
From initial conditions
Hence the solution is
u (t) = e −ξωt ( (
u (0) − F k
A = u (0) − F k
B = u′ (0) + u (0) ξω − F k ξω
ω d
) (
u ′ (0) + u (0) ξω − F k
cos ω d t +
ξω
) )
sin ω d t + F ω d k
8.2.3 Critical damping ζ = 1
The general solution is
u(t) = (A + Bt)e −ωt + F k
Where from initial conditions
A = u(0) − F k
B = u ′ (0) + u(0)ω − F k ω
8.2.4 Over-damped ζ > 0
The solution is
u (t) = Ae p1t + Be p2t + F k
Where now
B =
F
k p 1 − u 0 p 1 + u ′ (0)
(p 2 − p 1 )
A = u (0) − F k − B
Hence the solution is
u (t) = Ae p 1t + Be p 2t + F k
Where
p 1 = − c
√ ( c
2m + 2m
p 2 = − c
2m − √ ( c
2m
) 2 k −
m = −ωξ + ω √
n ξ 2 − 1
) 2 k −
m = −ωξ − ω √
n ξ 2 − 1
35

8.3 Harmonic forced input F sin (ϖt)
8.3.1 Undamped forced vibration
u
K
M
ReF e it
mu ′′ + ku = F sin ϖt
Since there is no damping in the system, then there is no steady state solution. In other words, the particular
solution is not the same as the steady state solution in this case. We need to find the particular solution using
method on undetermined coefficients.
Let u = u h + u p . By guessing that u p = c 1 sin ϖt then we find the solution to be
u = A cos ω n t + B sin ω n t + F 1
sin ϖt
k 1 − r2 Applying initial conditions is always done on the full solution. Applying initial conditions gives
u (0) = A
u ′ (t) = −Aω sin ω n t + Bω cos ω n t + ϖ F 1
cos ϖt
k 1 − r2 u ′ (0) = Bω n + ϖ F 1
k 1 − r 2
B = u′ (0)
− F r
ω n k 1 − r 2
Where r = ϖ ω n
The complete solution is
( u ′ (0)
u (t) = u (0) cos ω n t + − F k
ω n
r
1 − r 2 )
sin ω n t + F k
1
sin ϖt (1)
1 − r2 Example: Given force f (t) = 3 sin (5t) then ϖ = 5 rad/sec, and ˆF = 3. Let m = 1, k = 1, then ω n = 1
rad/sec. Hence r = 5, Let initial conditions be zero, then
( )
5
1
u = −3
1 − 5 2 sin t + 3 sin 5t
1 − 52 = 0.625 sin t − 0.125 sin 5.0t
Resonance forced vibration When ϖ ≈ ω we obtain resonance since r → 1 in the solution given in Eq (1)
above and as written the solution can not be used for analysis. To obtain a solution for resonance some calculus
is needed. Eq (1) is written as
( u ′ (0)
u (t) = u (0) cos ωt +
ω
− F k
ωϖ
ω 2 − ϖ 2 )
sin ωt + F k
ω 2
ω 2 sin ϖt
− ϖ2 (1A)
36

When ϖ ≈ ω but less than ω, letting
where ∆ is very small positive quantity. And since ϖ ≈ ω let
Multiplying Eq (2) and (3) gives
Eq (1A) can now be written in terms of Eqs (2,3) as
( u ′ (0)
u (t) = u (0) cos ωt +
ω
(
v0
= u (0) cos ωt +
ω − F k
ω − ϖ = 2∆ (2)
ω + ϖ ≈ 2ϖ (3)
ω 2 − ϖ 2 = 4∆ϖ (4)
− F ωϖ
k 4∆ϖ
ω
4∆
Since ϖ ≈ ω the above becomes
( u ′ (0)
u (t) = u (0) cos ωt +
ω
− F k
= u (0) cos ωt + u′ (0)
ω
sin ωt + F k
)
sin ωt + F k
)
sin ωt + F k
)
ω
sin ωt + F 4∆ k
Using sin ϖt − sin ωt = 2 sin ( ϖ−ω
2
t ) cos ( ϖ+ω
2
t ) the above becomes
u (t) = u (0) cos ωt + u′ (0)
ω
From Eqs (2,3) the above can be written as
sin ωt + F k
u (t) = u (0) cos ωt + u′ (0)
ω
ω
2∆
sin ωt + F k
(
sin
ω 2
sin ϖt
4∆ϖ
ω 2
sin ϖt
4∆ϖ
ω
sin ϖt
4∆
ω
(sin ϖt − sin ωt)
4∆
( ϖ − ω
2
) ( )) ϖ + ω
t cos t
2
ω
(sin (−∆t) cos (ϖt))
2∆
Since lim ∆→0
sin(∆t)
∆
= t the above becomes
This is the solution to use for resonance.
u (t) = u (0) cos ωt + u′ (0)
ω
sin ωt − F k
ωt
2
cos (ωt)
37

8.3.2 Under-damped forced vibration c < c r , ξ < 1
u
K
C
M
Fsin t
mu ′′ + cu ′ + ku = F sin ϖt
u ′′ + 2ξωu ′ + ω 2 u = F sin ϖt
m
The solution is
where
and
u (t) = u h + u p
u h (t) = e −ξωt (A cos ω d t + B sin ω d t)
u p (t) =
F
√(k − mϖ) 2 + (cϖ) 2 sin (ϖt − θ)
where
tan θ =
cϖ
k − mϖ 2 =
2ξr
1 − r 2
Very important note here in the calculations of tan θ above, one should be careful on the sign of the
denominator. When the forcing frequency ϖ > ω the denominator will become negative (the case of ϖ = ω is
resonance and is handled separately). Therefore, one should use arctan that takes care of which quadrant the
angle is. For example, in Mathematica use
ArcTan[1 - r^2, 2 Zeta r]]
and in Matlab use
atan2(2 Zeta r,1 - r^2)
Otherwise, wrong solution will result when ϖ > ω The full solution is
u (t) = e −ξωt (A cos ω d t + B sin ω d t) + F k
1
√(1 − r 2 ) 2 + (2ξr) 2 sin (ϖt − θ) (1)
Applying initial conditions gives
A = u (0) + F k
B = u′ (0)
ω d
+
1
√(1 − r 2 ) 2 + (2ξr) 2 sin θ
u (0) ξω
ω d
+ F k
Another form of these equations is given as follows
1
ω d
√
(1 − r 2 ) 2 + (2ξr) 2 (ξω sin θ − ϖ cos θ)
38

Hence the full solution is
u p = p 0 1 ((
1 − r
2 )
k (1 − r 2 ) 2 + (2ζr) 2 sin ϖt − 2ζr cos ϖt )
u (t) = e −ξωnt (A cos ω d t + B sin ω d t) + F k
Applying initial conditions now gives
1
(1 − r 2 ) 2 + (2ζr) 2 ((
1 − r
2 ) sin ϖt − 2ζr cos ϖt ) (1)
A = u (0) +
B = u′ (0)
ω d
2F rξ
k
1
(1 − r 2 ) 2 + (2ξr) 2
+ u (0) ξω n
− F ( 1 − r 2) ϖ 2F rζ
ω d kω d (1 − r 2 ) 2 +
2
+ (2ζr) kω d
ω n
(1 − r 2 ) 2 + (2ζr) 2
The above 2 sets of equations are equivalent. One uses the phase angle explicitly and the second ones do not.
Also, the above assume the force is F sin ϖt and not F cos ϖt. If the force is F cos ϖt then in Eq 1 above, the
term reverse places as in
u (t) = e −ξωnt (A cos ω d t + B sin ω d t) + F k
Applying initial conditions now gives
(
1 − r
2 )
A = u (0) + F k (1 − r 2 ) 2 + (2ξr) 2
B = u′ (0)
ω d
+ u (0) ξω n
ω d
+
2F rζ
kω d
1 ((
1 − r
2 )
(1 − r 2 ) 2 + (2ζr) 2 cos ϖt − 2ζr sin ϖt )
ϖ
(1 − r 2 ) 2 + (2ζr) 2 − F ( 1 − r 2)
kω d
ω n
(1 − r 2 ) 2 + (2ζr) 2
When a system is damped, the problem with the divide by zero when r = 1 does not occur here as was the
case with undamped system, since when when ϖ ≈ ω or r = 1, the solution in Eq (1) becomes
u (t) = e −ξωt ((
u (0) + F k
) (
1
u ′
2ξ sin θ (0) u (0) ξω
cos ω d t + + + F ω d ω d k
) )
1
2ω d ξ (ξω sin θ − ϖ cos θ) sin ω d t
+ F k
1
sin (ϖt − θ)
2
and the problem with the denominator going to zero does not show up here. The amplitude when steady state
response is maximum can be found as follows. The amplitude of steady state motion is F 1
k
√(1−r . This
2 ) 2 +(2ξr)
√
2
1
is maximum when the magnification factor β = √
is maximum or when (1 − r 2 ) 2 + (2ξr) 2 or
(1−r 2 ) 2 +(2ξr)
√ 2
(
1 − ( ) )
ϖ 2 2 ( )
ω + 2ξ
ϖ 2
ω is minimum. Taking derivative w.r.t. ϖ and equating the result to zero and solving
for ϖ gives
ϖ = ω √ 1 − 2ξ 2
We are looking for positive ϖ, hence when ϖ = ω √ 1 − 2ξ 2 the under-damped response is maximum.
39

8.3.3 critically damping forced vibration ξ = c
c r
= 1
The solution is
Where u h = (A + Bt) e −ωt and u p = F k
arctan definition). Hence
u (t) = u h + u p
1
2r
sin (ϖt − θ) where tan θ = (making sure to use correct
√(1−r 2 ) 2 2 1−r
+(2r) 2
u (t) = (A + Bt) e −ωt + F k
1
√(1 − r 2 ) 2 + (2r) 2 sin (ϖt − θ)
where A, B are found from initial conditions
A = u (0) + F k
B = u ′ (0) + u (0) ω + F k
1
√(1 − r 2 ) 2 + (2r) 2 sin θ
1
√(1 − r 2 ) 2 + (2r) 2 (ω sin θ − ϖ cos θ)
8.3.4 over-damped forced vibration ξ = c
c r
> 1
The solution is
where
and
u p (t) = F k
u (t) = u h + u p
u h (t) = Ae p 1t + Be p 2t
1
√(1 − r 2 ) 2 + (2ξr) 2 sin (ϖt − θ)
hence
u = Ae p 1t + Be p 2t + F k
1
√(1 − r 2 ) 2 + (2ξr) 2 sin (ϖt − θ)
where tan θ = 2ξr
1−r 2
Hence the solution is
and
p 1 = − c
√ ( c
2m + 2m
p 2 = − c
2m − √ ( c
2m
) 2 k −
m = −ωξ + ω √
n ξ 2 − 1
) 2 k −
m = −ωξ − ω √
n ξ 2 − 1
(
−ξ+ √ (
ξ
u (t) = Ae
2 −1
)ωt −ξ− √ )
ξ + Be 2 −1 ωt F + β sin (ϖt − θ)
k
u ′ (0) + u (0) ωξ + u (0) ω √ ξ 2 − 1 + F k
((ξ β + √ )
ξ 2 − 1
A =
2ω √ ξ 2 − 1
u ′ (0) + u (0) ωξ − u (0) ω √ ξ 2 − 1 + F k
((ξ β − √ )
ξ 2 − 1
B = −
2ω √ ξ 2 − 1
)
ω sin θ − ϖ cos θ
)
ω sin θ − ϖ cos θ
40

8.4 Response to impulsive loading
8.4.1 impulse input
Undamped system with impulse
mü + ku = F 0 δ(t)
with initial conditions u (0) = 0 and u ′ (0) = 0.Assuming the impulse acts for a very short time period from
0 to t 1 seconds, where t 1 is small amount. Integrating the above differential equation gives
∫ t1
0
müdt +
∫ t1
0
kudt =
∫ t1
0
F 0 δ(t)
Since t 1 is very small, it can be assumed that u changes is negligible, hence the above reduces to
since we assumed u ′ (0) = 0 and since ∫ t 1
0
∫ t1
0
∫ t1
müdt =
0
( ) d ˙u
m dt =
dt
∫ ˙u(t1 )
˙u(0)
∫ t1
0
∫ t1
0
d ˙u = F 0
m
˙u (t 1 ) − ˙u (0) = F 0
m
˙u (t 1 ) = F 0
m
F 0 δ(t)
F 0 δ(t)
∫ t1
0
∫ t1
0
∫ t1
0
δ(t)
δ(t)
δ(t)
δ(t) = 1 then the above reduces to
˙u (t 1 ) = F 0
m
Therefore, the effect of the impulse is the same as if the system was a free system but with initial velocity given
by F 0
m
and zero initial position. Hence the system is now solved as follows
With u (0) = 0 and u ′ (0) = F 0
m
. The solution is
mü + ku = 0
u impulse (t) = F 0
sin ωt
mω
If the initial conditions were not zero, then the solution for these are added to the above. From earlier, it was
found that the solution is u (t) = u(0) cos ωt + u′ (0)
ω
sin ωt, therefore, the full solution is
u (t) =
under-damped with impulse c < c r , ξ < 1
due to IC only
due to impulse
{ }} { { }} {
u(0) cos ωt + u′ (0)
ω
sin ωt + F 0 sin ωt
mω
mü + c ˙u + ku = δ(t)
ü + 2ξω ˙u + ω 2 u = δ(t)
with initial conditions u (0) = 0 and u ′ (0) = 0.Integrating gives
∫ t1
0
müdt +
∫ t1
0
c ˙udt +
41
∫ t1
0
kudt =
∫ t1
0
F 0 δ(t)

Since t 1 is very small, it can be assumed that u changes is negligible as well as the change in velocity, hence the
above reduces to the same result as in the case of undamped. Therefore, the system is solved as free system, but
with initial velocity u ′ (0) = F 0 /m and zero initial position.
Initial conditions are u (0) = 0 and u ′ (0) = 0 then the solution is
applying initial conditions gives A = 0 and B =
u impulse = e −ξωt (A cos ω d t + B sin ω d t)
( F0
)
m
ω d
, hence
u impulse (t) = e −ξωt (
F0
mω d
sin ω d t
If the initial conditions were not zero, then ( the solution for these are added ) to the above. From earlier, it
was found that the solution is u (t) = e −ξωt u(0) cos ω d t + u′ (0)+u(0)ξω
ω d
sin ω d t , therefore, the full solution is
due to IC only
{ (
}} ){
{ ( }} ){
u (t) = e −ξωt u(0) cos ω d t + u′ (0) + u(0)ξω
sin ω d t + e −ξωt F0
sin ω d t
ω d
mω d
)
due to impulse
critically damped with impulse input ξ = c
c r
= 1 with initial conditions u (0) = 0 and u ′ (0) = 0 then the
solution is
u (t) = (A + Bt) e −( cr
2m
)
t
= (A + Bt) e −ωt
where A, B are found from initial conditions A = u (0) = 0 and B = u ′ (0) + u (0) ω = F 0
m
, hence the solution
is
u impulse (t) = F 0t
m e−ωt
If the initial conditions were not zero, then the solution for these are added to the above. From earlier, it
was found that the solution is u (t) = (u 0 (1 + ωt) + u ′ (0) t) e −ωt , therefore, the full solution is
u (t) =
due to IC only
{ }} {
(
u (0) (1 + ωt) + u ′ (0) t ) e −ωt +
due to impulse
{ }} {
F 0 t
m e−ωt
over-damped with impulse input ξ = c
c r
> 1 With initial conditions are u (0) = 0 and u ′ (0) = 0 the
solution is
u impulse (t) = Ae λ 1ωt + Be λ 2ωt
where A, B are found from initial conditions and
λ 1 = −ωξ + ω √ ξ 2 − 1
λ 2 = −ωξ − ω √ ξ 2 − 1
A = u′ (0) − u (0) λ 2
2ω √ ξ 2 − 1
B = −u′ (0) + u (0) λ 1
2ω √ ξ 2 − 1
Hence the solution is
(
−ξ+ √ (
ξ
u impulse (t) = Ae
2 −1
)ωt −ξ− √ )
ξ + Be 2 −1 ωt
42

where
A =
B =
F 0
m
2ω √ ξ 2 − 1
− F 0
m
2ω √ ξ 2 − 1
Hence
u impulse (t) =
F 0
(
m
2ω √ ξ 2 − 1 e −ξ+ √ )
ξ 2 F 0
(
−1 ωt −
m
2ω √ ξ 2 − 1 e −ξ− √ )
ξ 2 −1 ωt
If the initial conditions were not zero, then the solution for these are added to the above. From earlier, it
was found that the solution is u (t) = Ae p 1t + Be p 2t , therefore, the full solution is
u (t) = Ae λ 1ωt + Be λ 2ωt +
F 0
F 0
m
2ω √ ξ 2 − 1 eλ 1ωt m
−
2ω √ ξ 2 − 1 eλ 2ωt
8.4.2 sin impulse input
Now assume the input is as follows
A = u′ (0) − u (0) λ 2
2ω √ ξ 2 − 1
B = −u′ (0) + u (0) λ 1
2ω √ ξ 2 − 1
given by F (t) = F 0 sin (ϖt) where ϖ = 2π
2t 1
= π t 1
undamped system with sin impulse
⎧
⎨ F 0 sin (ϖt) 0 ≤ t ≤ t 1
mü + ku =
⎩
0 t > t 1
with u (0) = u 0 and ˙u (0) = v 0 . For 0 ≤ t ≤ t 1 the solution is
( )
( )
v0
u (t) = u 0 cos ωt +
ω − u r
1 π
st
1 − r 2 sin ωt + u st
1 − r 2 sin t
t 1
43

where r = ϖ ω = π/t 1
ω
= T
2t 1
where T is the natural period of the system. u st = F 0
k
, hence the above becomes
⎛
⎜
u (t) = u 0 cos ωt + ⎝ v 0
ω − F 0
k
When u 0 = 0 and v 0 = 0 then
u (t) = − F 0
k
u (t) = F 0
k
( )
π/t1
ω
1 −
(
π/t1
ω
1
1 −
(
π/t1
ω
( )
π/t1
ω
1 −
(
π/t1
ω
) 2
sin ωt + F 0
k
) 2
(sin
) 2
⎞
⎟
⎠ sin ωt + F 0
k
1
1 −
(
π/t1
ω
1
1 −
(
π/t1
ω
(
π t
t 1
)
− π/t 1
ω sin ωt )
(
) 2
sin π t )
t 1
The above Eq (1) gives solution during the time 0 ≤ t ≤ t 1
Now after t = t 1 the force will disappear, the differential equation becomes
mü + ku = 0 t > t 1
(
) 2
sin π t )
t 1
but with the initial conditions evaluate at t = t 1 . From (1)
( )
v0
u (t 1 ) = u 0 cos ωt 1 +
ω − u r
1
st
1 − r 2 sin ωt 1 + u st
1 − r 2 sin ϖt 1
( )
v0
= u 0 cos ωt 1 +
ω − u r r
st
1 − r 2 u st
1 − r 2 sin ωt 1 (2)
since sin ϖt 1 = 0. taking derivative of Eq (1)
˙u (t) = −ωu 0 sin ωt + ω
and at t = t 1 the above becomes
˙u (t 1 ) = −ωu 0 sin ωt 1 + ω
= −ωu 0 sin ωt 1 + ω
(
v0
)
ω − u r
st
1 − r 2 cos ωt + ϖ 1 cos ϖt
1 − r2 (
v0
)
ω − u r
st
1 − r 2 )
ω − u r
st
1 − r 2
(
v0
cos ωt 1 + ϖ 1
1 − r 2 cos ϖt 1
(1)
cos ωt 1 − ϖ 1
1 − r 2 (3)
since cos ϖt 1 = −1. Now (2) and (3) are used as initial conditions to solve mü + ku = 0 . The solution for
t > t 1 is
u (t) = u (t 1 ) cos ωt + ˙u (t 1)
ω
sin ωt
Resonance with undamped sin impulse When ϖ ≈ ω and t ≤ t 1 we obtain resonance since r → 1 in
the solution shown up and as written the solution can’t be used for analysis in this case. To obtain a solution
for resonance some calculus is needed. Eq (1) is written as
(
)
ϖ
v 0
u (t) = u 0 cos ωt +
ω − u ω
st
1 − ( 1
)
ϖ 2
sin ωt + u st
ω
1 − ( )
ϖ 2
sin ϖt
ω
( )
v0
= u 0 cos ωt +
ω − u ωϖ
ω 2
st
ω 2 − ϖ 2 sin ωt + u st
ω 2 sin ϖt
− ϖ2 (1A)
Now looking at case when ϖ ≈ ω but less than ω, hence let
ω − ϖ = 2∆ (2)
44

where ∆ is very small positive quantity. and we also have
Multiplying Eq (2) and (3) with each others gives
Going back to Eq (1A) and rewriting it as
ω + ϖ ≈ 2ϖ (3)
ω 2 − ϖ 2 = 4∆ϖ (4)
( v0
u (t) = u 0 cos ωt +
ω − u ωϖ
)
ω 2
st sin ωt + u st sin ϖt
4∆ϖ
4∆ϖ
( v0
= u 0 cos ωt +
ω − u ω
)
ω 2
st sin ωt + u st sin ϖt
4∆
4∆ϖ
Since ϖ ≈ ω the above becomes
( v0
u (t) = u 0 cos ωt +
ω − u ω
)
ω
st sin ωt + u st sin ϖt
4∆
4∆
= u 0 cos ωt + v 0
ω sin ωt + u ω
st (sin ϖt − sin ωt)
4∆
now using sin ϖt − sin ωt = 2 sin ( ϖ−ω
2
t ) cos ( ϖ+ω
2
t ) the above becomes
u (t) = u 0 cos ωt + v ( ( )
0
ω sin ωt + u ω ϖ − ω
st sin t cos
2∆ 2
From Eq(2) ϖ − ω = −2∆ and ω + ϖ ≈ 2ϖ hence the above becomes
or since ϖ ≈ ω
Now lim ∆→0
sin(∆t)
∆
This can also be written as
( ϖ + ω
u (t) = u 0 cos ωt + v 0
ω sin ωt + u ω
st (sin (−∆t) cos (ϖt))
2∆
u (t) = u 0 cos ωt + v 0
ω sin ωt − u ω
st (sin (∆t) cos (ωt))
2∆
= t hence the above becomes
u (t) = u 0 cos ωt + v 0
ω sin ωt − u ωt
st cos (ωt)
2
2
))
t
u (t) = u 0 cos ϖt + v 0
ϖ sin ϖt − u ϖt
st cos (ϖt) (1)
( )
2
π
= u 0 cos t + v ( ) ( ) ( )
0 π π π
t 1 ϖ sin t − u st t cos t
t 1 2t 1 t 1
since ϖ ≈ ω in this case. This is the solution to use for resonance and for t ≤ t 1
Hence for t > t 1 , the above equations is used to determine initial conditions at t = t 1
u (t 1 ) = u 0 cos ϖt 1 + v 0
ϖ sin ϖt 1 − u st
ϖt 1
2 cos (ϖt 1)
but cos ϖt 1 = cos π t 1
t 1 = −1 and sin ϖt 1 = 0 and ϖt 1
2
= π 2
, hence the above becomes
Taking derivative of Eq (1) gives
u (t 1 ) = −u 0 + u st
π
2
ϖ 2 t
˙u (t) = −ϖu 0 sin ϖt + v 0 cos ϖt + u st
2 sin (ϖt) − u ϖ
st cos (ϖt)
2
45

and at t = t 1
ϖ 2 t 1
ϖ
˙u (t 1 ) = −ϖu 0 sin ϖt 1 + v 0 cos ϖt 1 + u st sin (ϖt 1 ) − u st
2
2 cos (ϖt 1)
ϖ
= −v 0 + u st
2
Now the solution for t > t 1 is
u (t) = u (t 1 ) cos ωt + ˙u (t 1)
ω
(
π
= −u (0) + u st
2
under-damped with sin impulse c < c r , ξ < 1
sin ωt
)
cos ωt + −u′ (0) + u st
π
2t 1
ω
⎧
⎨ F 0 sin (ϖ) 0 ≤ t ≤ t 1
mü + c ˙u + ku =
⎩
0 t > t 1
sin ωt
or
⎧
⎨
ü + 2ξω ˙u + ω 2 F 0 sin (ϖ) 0 ≤ t ≤ t 1
u =
⎩
0 t > t 1
mü + c ˙u + ku = F sin ϖt
ü + 2ξω ˙u + ω 2 u = F sin ϖt
m
For t ≤ t 1 Initial conditions are u (0) = u 0 and ˙u (0) = v 0 and u st = F k
u (t) = e −ξωt (A cos ω d t + B sin ω d t) +
then the solution from above is
u st
√(1 − r 2 ) 2 + (2ξr) 2 sin (ϖt − θ) (1)
Applying initial conditions gives
For t > t 1 . From (1)
A = u 0 +
B = v 0
ω d
+ u 0ξω
ω d
+
u st
√(1 − r 2 ) 2 + (2ξr) 2 sin θ
u st
u (t 1 ) = e −ξωt 1
(A cos ω d t 1 + B sin ω d t 1 ) +
ω d
√
(1 − r 2 ) 2 + (2ξr) 2 (ξω sin θ − ϖ cos θ)
u st
√(1 − r 2 ) 2 + (2ξr) 2 sin (ϖt 1 − θ) (2)
Taking derivative of (1) gives
˙u (t) = −ξωe −ξωt (A cos ω d t + B sin ω d t) + e −ξωt (−Aω d sin ω d t + ω d B cos ω d t)
u st
+ ϖ
cos (ϖt − θ)
√(1 − r 2 ) 2 + (2ξr) 2
46

at t = t 1
˙u (t 1 ) = −ξωe −ξωt 1
(A cos ω d t 1 + B sin ω d t 1 ) + e −ξωt 1
(−Aω d sin ω d t 1 + ω d B cos ω d t 1 )
u st
+ ϖ
cos (ϖt 1 − θ) (3)
√(1 − r 2 ) 2 + (2ξr) 2
Now for t > t 1 the equation becomes
which has the solution
where A = u (t 1 ) and B = ˙u(t 1)+u(t 1 )ξω
ω d
mü + c ˙u + ku = 0
u = e −ξωt (A cos ω d t + B sin ω d t)
critically damped with sin impulse ξ = c
c r
= 1 For t ≤ t 1 Initial conditions are u (0) = u 0 and ˙u (0) = v 0
then the solution is from above
u (t) = (A + Bt) e −ωt u st
+
sin (ϖt − θ) (1)
√(1 − r 2 ) 2 + (2r) 2
Where tan θ =
cϖ
k−mϖ 2
= 2ξr
1−r 2 . A, B are found from initial conditions
A = u 0 +
u st
√(1 − r 2 ) 2 + (2r) 2 sin θ
B = v 0 + u 0 ω +
u st
√(1 − r 2 ) 2 + (2r) 2 (ω sin θ − ϖ cos θ)
For t > t 1 the solution is
To find u (t 1 ) , from Eq(1)
u (t) = (u (t 1 ) + ( ˙u (t 1 ) + u (t 1 ) ω) t) e −ωt (2)
u (t 1 ) = (A + Bt) e −ωt 1
+
u st
√(1 − r 2 ) 2 + (2r) 2 sin (ϖt 1 − θ)
taking derivative of (1) gives
˙u (t) = −ω (A + Bt) e −ωt + Be −ωt + ϖ
sin (ϖt − θ) (3)
√(1 − r 2 ) 2 + (2r) 2
u st
at t = t 1
˙u (t 1 ) = −ω (A + Bt 1 ) e −ωt 1
+ Be −ωt 1
+ ϖ
sin (ϖt 1 − θ) (4)
√(1 − r 2 ) 2 + (2r) 2
u st
Hence Eq (2) can now be evaluated using Eq(3,4)
over-damped with sin impulse ξ = c
c r
> 1 For t ≤ t 1 Initial conditions are u (0) = u 0 and ˙u (0) = v 0 then
the solution is
u = Ae p1t + Be p2t u st
+
sin (ϖt − θ)
√(1 − r 2 ) 2 + (2ξr) 2
where tan θ = 2ξr
1−r 2 (make sure you use correct quadrant, see not above on arctan) and
p 1 = − c
√ ( c
) 2
2m + k −
2m m
= −ωξ + ω √ ξ 2 − 1
47

and
leading to the solution where tan θ = 2ξr
1−r 2
p 2 = − c
√ ( c
) 2
2m − k −
2m m
= −ωξ − ω √ ξ 2 − 1
and
is
p 1 = − c
√ ( c
2m + 2m
p 2 = − c
2m − √ ( c
2m
) 2 k −
m = −ωξ + ω √
n ξ 2 − 1
) 2 k −
m = −ωξ − ω √
n ξ 2 − 1
(
−ξ+ √ (
ξ
u (t) = Ae
2 −1
)ωt −ξ− √ )
ξ + Be 2 −1 ωt F + β sin (ϖt − θ)
k
u ′ (0) + u (0) ωξ + u (0) ω √ ξ 2 − 1 + F k
((ξ β + √ )
ξ 2 − 1
A =
2ω √ ξ 2 − 1
u ′ (0) + u (0) ωξ − u (0) ω √ ξ 2 − 1 + F k
((ξ β − √ )
ξ 2 − 1
B = −
2ω √ ξ 2 − 1
1
β = √
(1 − r 2 ) 2 + (2ξr) 2
)
ω sin θ − ϖ cos θ
)
ω sin θ − ϖ cos θ
For t > t 1 . From Eq(1) and at t = t 1
Taking derivative of Eq (1)
(
−ξ+ √ )
(
ξ
u (t 1 ) = Ae
2 −1 ωt 1 −ξ− √ )
ξ
+ Be
2 −1 ωt 1
+ D sin (ϖt 1 − θ) (2)
(
−ξ+ √ (
ξ
˙u (t) = ωAe
2 −1
)ωt −ξ− √ )
ξ + ωBe 2 −1 ωt + ϖD cos (ϖt − θ)
At t = t 1
−ξ+
˙u (t 1 ) = ωAe( √ )
(
ξ 2 −1 ωt 1 −ξ− √ )
ξ
+ ωBe
2 −1 ωt 1
+ ϖD cos (ϖt 1 − θ) (3)
Equation of motion now is
which has solution for over-damped given by
ü + 2ξω ˙u + ω 2 u = 0
(
−ξ+ √ (
ξ
u (t) = Ae
2 −1
)ω nt −ξ− √ )
ξ + Be 2 −1 ω nt
where
˙u (t 1 ) + u (t 1 ) ω n
(ξ − √ )
ξ 2 − 1
A = −
√
2ω n ξ 2 − 1
˙u (t 1 ) + u (t 1 ) ξω n
(ξ + √ )
ξ 2 − 1
B =
√
2ω n ξ 2 − 1
48

8.5 references
1. Vibration analysis by Robert K. Vierck
2. Structural dynamics theory and computation, 5th edition by Mario Paz, William Leigh
3. Dynamic of structures, Ray W. Clough and Joseph Penzien
4. Theory of vibration,volume 1, by A.A.Shabana
5. Notes on Diffy Qs, Differential equations for engineers, by Jiri Lebl, online PDF book, chapter 2.6, oct
1,2012 http://www.jirka.org/diffyqs/
9 Derivation of rotation formula
This formula is very important. Will show its derivation now in details. It is how to express vectors in rotating
frames.
Consider this diagram
P
r
Y
X
r p
y
r o
o
x
Moving frame of
reference, attached
to body of interest
Absolute (or inertial frame of reference)
In the above, the small axis x, y is a frame attached to some body which rotate around this axis with angular
velocity ω (measured by the inertial frame of course). All laws derived below are based on the following one rule
d
dt r ∣
∣∣∣absolute
= d dt r ∣
∣∣∣relative
+ ω × r (1)
Lets us see how to apply this rule. Let us express the position vector of the particle r p . We can see by
normal vector additions that the position vector of particle is
r p = r o + r (2)
Notice that nothing special is needed here, since we have not yet looked at rate of change with time. The
complexity (i.e. using rule (1)) appears only when we want to look at velocities and accelerations. This is when
we need to use the above rule (1). Let us now find the velocity of the particle. From above
ṙ p = ṙ o + ṙ
49

Every time we take derivatives, we stop and look. For any vector that originates from the moving frame,
we must apply rule (1) to it. That is all. In the above, only r needs rule (1) applied to it, since that is the
only vector measure from the moving frame. Replacing ṙ p by V p and ṙ o by V o , meaning the velocity of P and o,
Hence the above becomes
V p = V o + ṙ
and now we apply rule (1) to expand ṙ
V p = V o + (V rel + ω × r) (3)
where V rel is just d dt r∣ ∣
relative
The above is the final expression for the velocity of the particle V p using its velocity as measured by the
moving frame in order to complete the expression.
So the above says that the absolute velocity of the particle is equal to the absolute velocity of the base of the
moving frame + something else and this something else was (V rel + ω × r)
Now we will find the absolute acceleration of P . Taking time derivatives of (3) gives
˙V p = ˙V
( )
o + ˙Vrel + ˙ω × r + ω × ṙ
(4)
As we said above, each time we take time derivatives, we stop and look for vectors which are based on the
moving frame, and apply rule (1) to them. In the above, ˙Vrel and ṙ qualify. Apply rule (1) to ˙V rel gives
˙V rel = a rel + ω × V rel (5)
where a rel just means the acceleration relative to moving frame. And applying rule (1) to ṙ gives
Replacing (5) and (6) into (4) gives
ṙ = V rel + ω × r (6)
a p = a o + (a rel + ω × V rel + ˙ω × r + ω × (V rel + ω × r))
= a o + a rel + (ω × V rel ) + ( ˙ω × r) + (ω × V rel ) + (ω × (ω × r))
= a o + a rel + 2 (ω × V rel ) + ( ˙ω × r) + (ω × (ω × r)) (7)
Eq (7) says that the absolute acceleration a p of P is the sum of the acceleration of the base a o of the
moving frame plus the relative acceleration a rel of the particle to the moving frame plus 2 (ω × V rel ) + ( ˙ω × r) +
(ω × (ω × r))
Hence, using Eq(3) and Eq(7) gives us the expressions we wanted for velocity and acceleration.
10 Miscellaneous hints
1. When finding the generalized force for the user with the Lagrangian method (the hardest step), using the
virtual work method, if the force (or virtual work by the force) ADDS energy to the system, then make
the sign of the force positive otherwise the sign is negative.
2. For damping force, the sign is always negative.
3. External forces such as linear forces applied, torque applied, in general, are positive.
4. Friction force is negative (in general) as friction takes energy from the system like damping.
50

11 Formulas
sin 2 x 1cos2x
2
cos 2 x 1cos2x
2
sin 2x 2 sin x cosx
cos2x cos 2 x sin 2 x
1 2 sin 2 x
V p V o r V rel
Velocity of p
relative to o
a p a 0 r r 2 V rel a rel
p
IF rigid body is rotating AND translation at the same time THEN
Take moments around its cg only
ELSE
can also take moments about any other points on it (but
change I)
END IF
o
r
Y
R Y
X
R y
R X
R
x
y
R x
BODY FIXED
COORDINATES
This gives the relation
between the rate of
change with respect to
time of a vector expressed
in a frame of reference
which is body fixed (y,x)
here, to the rate of change
with respect to time of the
same vector expressed
using an inertial frame
coordinates XY.
The omega here is the
angular velocity of the
rigid body rotation, or the
body fixed coordinates,
with respect to the inertial
frame.
d
dt R X,Y d dt R x,y R X,Y
R
r
R r
r
The small disk rotates
at angular speed of
The condition of no slip can
be seen to be
R r r
For the sign of generalized force:
If work done by force takes away
energy from system, then the
sign is negative, else positive. So
Friction will always have negative
sign, so will damping force.
d
dt
cosxt sinxtx t
d
dt
df
dxt
d
dxt
d
cosxt
dt
df
dxt
d
d
cos xt
dxt
d
xt dt
xt dt
51

F ma
linear momentum p mv
F d p dt
particle kinetic energy T 1 2 mv2
I
angular momentum H I cg
d
dt H
rigid body T 1 Mv 2 cg
2 1 I 2
2
cg
my 2 n y n 2 y fy,t
y cy k m y fy,t
n
For small
epsilon
1
1
1
1
1
1
k
m ,c 2 n
m
conservation of angular momentum d dt
H constant
x n2 x 0
If n 0 then sinusodial solution, ok
if n 0 then solution blows up, exponential
52

12 Velocity and acceleration diagrams
12.1 Spring pendulum
Length of pendulum fixed
Length of pendulum changes with time
2L
L
L
L
Velocity diagram
acceleration
diagram
L 2 L Lt L 2
L
L
Velocity diagram
acceleration
diagram
y
Length of pendulum changes with time and base of pendulum moves
ÿ
x
x
y
L
x
L 2
ÿ
L
x
2L
Velocity diagram
L
acceleration
diagram
r
Nasser M. Abbasi
Oct 10, 2013 (drawing2.vsd)
53

12.2 pendulum with blob moving in slot
r
y
y
r
y
System. Pendulum with
blob which moves in slot
perpendicular to axis
Velocity diagram
2y
r 2
y
ÿ
r
acceleration diagram
y 2
54

12.3 spring pendulum with block moving in slot
r
y
y
r
System. Pendulum with blob which
moves in slot perpendicular to axis. In
addition, length of the axis of pendulum
itself changes in time. y and r are
measured from static equilibrium of
springs
y
Velocity diagram
r
2y
r 2
2r
y
ÿ
r
acceleration diagram
y 2
r
55

12.4 double pendulum
Double
pendulum
L 1
L 2
m 1
m 1
L 1
• 2
m 2
m 2
L 1
L 1
L 2
m 2
••
acceleration diagram
L 1
m 1
L 2
L 2
• 2
L 1
• 2
L 2
••
L 1
•
L 1
••
Velocity diagram
L 2
•
L 1
•
56

13 Velocity and acceleration of rigid body 2D
V
y
x
U
V U
y
x
U V
Y
X
Linear
Velocity
diagram
Rigid body,
rotating at
angular
speed
Linear
acceleration
diagram
Notice the
sign
difference
Nasser M. Abbasi
Drawing_rigid_body_rotati
on_1.vsd
May 23, 2011
a
a x
a y
U V
V U
Finding linear acceleration of center of mass of a rigid body under pure rotation using fixed body coordinates.
In the above U is the speed of the center of mass in the direction of the x axis, where this axis is fixed on
the body itself. Similarly, V is the speed of the center of mass in the direction of the y axis, where the y axis is
attached to the body itself.
Just remember that all these speeds (i.e. U,V ) and accelerations (a x , a y ) are still being measured by an
observer in the inertial frame. It is only that the directions of the velocity components of the center of mass
is along an axis fixed on the body. Only the direction. But actual speed measurements are still done by a
stationary observer. Since clearly if the observer was sitting on the body itself, then they will measure the speeds
to be zero in that case.
57

14 Velocity and acceleration of rigid body 3D
14.1 Using Vehicle dynamics notations
F x
Nasser M. Abbasi
3d_1.vsd
May 26, 2011
L
I
N
F z
W
z
r
Linear force
torque
Linear velocity
angular velocity
v
y
p q V
x
M
F y
U
F
Forces, Torques, linear
velocities and angular velocities
for a general 3D rigid body
I xx I xy I xz
I yx I yy I yz
linear momentum
U qW rV
F d dt p d dt mvm d dt v m a
V rU pW
W pV qU
U
V
W
p
q
r
F x
F y
F z
L
M
N
d dt H d dt I
Angular momentum
Derivation of this is much more complicated than with
the case of linear motion (F=ma), since m is scalar
there, but for rotation, I is matrix. See next page for the
derivation
58

14.2 3D Not Using vehicle dynamics notations
Nasser M. Abbasi
3d_1.vsd
May 26, 2011
z
F z
Linear force
torque
Linear velocity
v
v x
v y
v z
angular velocity
v z
F x
x
I
v x
x
x
z
z
y
y
F y
F
Forces, Torques, linear
velocities and angular
velocities for a general 3D
rigid body
I xx I xy I xz
I yx I yy I yz
F d p dt d mvm d v m dt dt
a y z v x x v z
a
a x y v z z v y
v y
y
x
y
z
F x
F y
F z
x
y
z
linear momentum
a z x v y y v z
The derivation of the above is given next, but it uses the standard formula given by
d
A d
A A
dt dt resolved
This is in the
inertial frame of
reference
This is the same A, but its components
are with respect to the body fixed
coordinates system,
Cross product
59

14.2.1 Derivation for F = ma in 3D
F = d dt p
= d dt (mv)
= m d dt v
⎡⎛
⎞ ⎛ ⎞ ⎛ ⎞⎤
a x
ω x
v x
= m
⎢⎜a y ⎟
⎣⎝
⎠ + ⎜ω y ⎟
⎝ ⎠ ⊗ ⎜v y ⎟⎥
⎝ ⎠⎦
a z ω z v z
⎡⎛
⎞ ∣ ∣⎤
∣∣∣∣∣∣∣∣∣ ∣∣∣∣∣∣∣∣∣ a x
i j k
= m
⎢⎜a y ⎟
⎣⎝
⎠ + det ω x ω y ω z ⎥
⎦
a z v x v y v z
⎡⎛
⎞ ⎛
⎞⎤
a x
ω y v z − ω z v y
= m
⎢⎜a y ⎟
⎣⎝
⎠ + ⎜− (ω x v z − ω z v x )
⎟⎥
⎝
⎠⎦
a z ω x v y − ω y v x
⎛
⎞
a x + ω y v z − ω z v y
= m
⎜a y − ω x v z + ω z v x ⎟
⎝
⎠
a z + ω x v y − ω y v x
60

14.2.2 Derivation for τ = Iω in 3D
Let A = Iω then using the rule
( ) d
τ =
dt A ( ) d
=
dt A + ω × A
resolved
Then τ = Iω can be found for the general case
⎡
⎤
A
A
{ ⎛ }} ⎞ ⎛ ⎞{
⎛ ⎞ { ⎛ }} ⎞ ⎛ ⎞{
τ = d I xx I xy I xz
ω x
ω x
I xx I xy I xz
ω x
dt
⎜I yx I yy I yz ⎟ ⎜ω y ⎟
+
⎝
⎠ ⎝ ⎠
⎜ω y ⎟
⎝ ⎠ ×
⎜I yx I yy I yz ⎟ ⎜ω y ⎟
⎝
⎠ ⎝ ⎠
⎢
⎣ I zx I yz I zz ω z
⎥
⎦ ω z I zx I yz I zz ω z
⎛
⎞ ⎛ ⎞ ⎛ ⎞ ⎛
⎞
I xx I xy I xz
α x
ω x
I xx ω x + I xy ω y + I xz ω z
=
⎜I yx I yy I yz ⎟ ⎜α y ⎟
⎝
⎠ ⎝ ⎠ + ⎜ω y ⎟
⎝ ⎠ × ⎜I yx ω x + I yy ω y + I yz ω z ⎟
⎝
⎠
I zx I yz I zz α z ω z I zx ω x + I yz ω y + I zz ω z
⎛
⎞ ⎛ ⎞ ∣ ∣∣∣∣∣∣∣∣∣ I xx I xy I xz
α x
i j k
=
⎜I yx I yy I yz ⎟ ⎜α y ⎟
⎝
⎠ ⎝ ⎠ + det ω x ω y ω z
I zx I yz I zz α z (I xx ω x + I xy ω y + I xz ω z ) (I yx ω x + I yy ω y + I yz ω z ) (I zx ω x + I yz ω y + I zz ω z ) ∣
⎛
⎞ ⎛ ⎞ ⎛
⎞
I xx I xy I xz
α x
ω y (I zx ω x + I yz ω y + I zz ω z ) − ω z (I yx ω x + I yy ω y + I yz ω z )
=
⎜I yx I yy I yz ⎟ ⎜α y ⎟
⎝
⎠ ⎝ ⎠ + ⎜ω x (I zx ω x + I yz ω y + I zz ω z ) − ω z (I xx ω x + I xy ω y + I xz ω z )
⎟
⎝
⎠
I zx I yz I zz α z ω x (I yx ω x + I yy ω y + I yz ω z ) − ω y (I xx ω x + I xy ω y + I xz ω z )
61

14.2.3 Derivation for τ = Iω in 3D using principle axes
The above derivation simplifies now since we will be using principle axes. In this case, all cross products of
moments of inertia vanish.
⎛
⎞
I xx 0 0
I =
⎜ 0 I yy 0
⎟
⎝
⎠
0 0 I zz
Hence
⎡
⎤
A
A
{ ⎛ }} ⎞ ⎛ ⎞{
⎛ ⎞ { ⎛ }} ⎞ ⎛ ⎞{
τ = d I xx 0 0
ω x
ω x
I xx 0 0
ω x
dt
⎜ 0 I yy 0
⎟ ⎜ω y ⎟
+
⎝
⎠ ⎝ ⎠
⎜ω y ⎟
⎝ ⎠ ×
⎜ 0 I yy 0
⎟ ⎜ω y ⎟
⎝
⎠ ⎝ ⎠
⎢
⎣ 0 0 I zz ω z
⎥
⎦ ω z 0 0 I zz ω z
⎛
⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
I xx 0 0
α x
ω x
I xx ω x
=
⎜ 0 I yy 0
⎟ ⎜α y ⎟
⎝
⎠ ⎝ ⎠ + ⎜ω y ⎟
⎝ ⎠ × ⎜I yy ω y ⎟
⎝ ⎠
0 0 I zz α z ω z I zz ω z
⎛ ⎞ ∣ ∣ ∣∣∣∣∣∣∣∣∣ ∣∣∣∣∣∣∣∣∣
I xx α x
i j k
=
⎜I yy α y ⎟
⎝ ⎠ + det ω x ω y ω z
I zz α z I xx ω x I yy ω y I zz ω z
⎛ ⎞ ⎛
⎞
I xx α x
ω y (I zz ω z ) − ω z (I yy ω y )
=
⎜I yy α y ⎟
⎝ ⎠ + ⎜−ω x (I zz ω z ) + ω z (I xx ω x )
⎟
⎝
⎠
I zz α z ω x (I yy ω y ) − ω y (I xx ω x )
⎛ ⎞ ⎛
⎞
I xx α x
ω y ω z (I zz − I yy )
=
⎜I yy α y ⎟
⎝ ⎠ + ⎜ω x ω z (I xx − I zz )
⎟
⎝
⎠
I zz α z ω x ω y (I yy − I xx )
So, we can see how much simpler it became when using principle axes. Compare the above to
⎛
⎞ ⎛ ⎞ ⎛
⎞
I xx I xy I xz
α x
ω y (I zx ω x + I yz ω y + I zz ω z ) − ω z (I yx ω x + I yy ω y + I yz ω z )
⎜I yx I yy I yz ⎟ ⎜α y ⎟
⎝
⎠ ⎝ ⎠ + ⎜ω x (I zx ω x + I yz ω y + I zz ω z ) − ω z (I xx ω x + I xy ω y + I xz ω z )
⎟
⎝
⎠
I zx I yz I zz α z ω x (I yx ω x + I yy ω y + I yz ω z ) − ω y (I xx ω x + I xy ω y + I xz ω z )
So, always use principle axes for the body fixed coordinates system!
62

15 Wheel spinning precession
x
MgL
L
y
z
Mg
Wheel spinning around y-axis,
hanging from ceiling
Weight create a torque which
changes the angular
momentum
Torque vector
representation
y
Important: This
analysis is valid
only for LARGE
y
H Angular momentum
x
z
y
H y I yy y
p
y
t
Angular momentum
change due to the
torque applied
After torque applied for t
z
This is what wheel will
look like after del T
time due to
precesses, notice that
SPIN angular
momentum changes
in the direction of the
torque
y
Time it takes
for the wheel
to precesses
one full cycle
T p 2
p
H
z
Angular momentum
after
t
t
H aftert
x
H I yy y t
x
I yy y
p
y
H t H
t
But
H
t
I yy y p , hence I yy y p hence p
I yy y
MgL
I yy y
Nasser M.
Abbasi
Precesses.vsd
6/1/2011
Therefore, precession velocity p is MgL
I yy y
16 References
1. Structural Dynamics 5th edition. Mario Paz, William Leigh
63

17 Misc. items
The Jacobian matrix for a system of differential equations, such as
is given by
x ′ (t) = f (x, y, z)
y ′ (t) = g (x, y, z)
z ′ (t) = h (x, y, z)
⎛
J =
⎜
⎝
For example, for the given the following 3 set of coupled differential equations in n 3
df
dx
dg
dx
dh
dx
df
dy
dg
dy
dh
dy
df
dz
dg
dz
dh
dz
x ′ (t) = −y (t) − z (t)
y ′ (t) = x (t) + ay (t)
⎞
⎟
⎠
z ′ (t) = b + z (t) (x (t) − c)
then the Jacobian matrix is
⎛
⎞
0 −1 −1
J =
⎜ 1 a 0
⎟
⎝
⎠
z (t) 0 x (t) − c
Now to find stability of this system, we evaluate this matrix at t = t 0 where x (t 0 ) , y (t 0 ) , z (t 0 ) is a point in
this space (may be stable point or initial conditions, etc...) and then J become all numerical now. Then we can
evaluate the eigenvalues of the resulting matrix and look to see if all eigenvalues are negative. If so, this tells us
that the point is a stable point. I.e. the system is stable.
If X is N(0, 1) distributed then mu + sigma ∗ X is N(mu, sigma 2 ) distributed.
64

18 Astrodynamics
18.1 Ellipse main parameters
18.2 Table of common equations
The following table contains the common relations to use for elliptic motion. Equation of ellipse is x2
a 2 + y2
b 2 = 1
term to find
conversion between E and θ
position of satellite at time t
Solve for E, then find θ. τ
here is time at perigee and n
is mean satellite speed.
eccentricity e
relation
( θ
tan =
2)
√ ( )
1 + e E
1 − e tan 2
cos E = e + cos θ
1 + e cos θ
cos θ = e − cos E
e cos E − 1
E − e sin E = n (t − τ)
√
e = c a = ra−rp
r a+r p
= 1 + 2Eh2
µ 2
65

Maor axes a
a = r p(1 + e)
1 − e 2
= r a(1 − e)
1 − e 2
= − µ
2E
= √ b 2 + c 2
= p
1 − e 2
Minor axes b b = a √ 1 − e 2
r p = a(1 − e2 )
r p 1 + e
= a(1 − e)
r a = a(1 − e2 )
1 − e
r a = a (1 + e)
p
specific angular momentum h
Total Energy E
= h2
µ
1
1 − e
r p
r a
= 1−e
1+e
p = a ( 1 − e 2) = h2
µ = r p (1 + e) = r a (1 − e)
h = r p v p = r a v a = ⃗r × ⃗v = √ pµ
h = √ µr
(circular orbit)
E = v2
2 − µ r = − µ 2a
√ ( 2
v = µ
r a)
− 1
(vis-viva)
velocity v
√
2µ
v escape = (escape velocity for parabola)
r
√ µ
v radial =
p e sin θ
√ µ
v normal = (1 + e cos θ)
√
p
( )
µ 1 + e
v p =
a 1 − e
v perigee (closest)
v apogee (furthest)
v a =
√ µ
= (1 + e)
p
√ ( 2
= µ − 1 r p a)
√
µ
a
( ) 1 − e
1 + e
√ µ
= (1 − e)
p
√ ( 2
= µ − 1 r a a)
66

magnitude of ⃗r
period T
mean satellite speed n
r = a ( 1 − e 2)
1 + e cos θ = h2 1
µ 1 + e cos θ
r cos θ = a (cos E − e)
r = a (1 − e cos E)
√
T = 2 h πab = 2π a 3
µ
n = 2π
T = √
µ
a 3
eccentric anomaly E tan θ 2 = √
1+e
1−e tan E 2
area sweep rate
dA
dt = h 2
equation of motion ¨⃗r +
µ
r 3 ⃗r = 0
(eq 4.2-14 Bate book)
spherical coordinates relation cos(i) = sin(A z ) cos(φ) where i is the inclination
and A z is the azimuth and φ is latitude
2
Notice in the above, that the period T of satellite depends only on a (for same µ)
In the above, µ = GM where M is the mass of the body at the focus of the ellipse and G is the gravitational
constant. h is the specific mass angular momentum (moment of linear momentum) of the satellite. Hence the
units of h2
µ
is length.
To draw the locus of the satellite (the small body moving around the ellipse, all what we need is the
eccentricity e and a, the major axes length. Then by changing the angle θ the path of the satellite is drawn. I
have a demo on this here
See http://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html for earth facts
This table below is from my class EMA 550 handouts (astrodynamics, spring 2014)
2 Image is from my class notes, page 7-9, EMA 550, Univ. Of Wisconsin, Madison by professor S. Sandrik
67

68

18.3 Flight path angle for ellipse γ
69

√
v = |⃗v| = |⃗v r | 2 + |⃗v n | 2
√ ( 2
= µ
r a)
− 1
To find γ, if r is given then use
If θ is given, then use
√
cos γ =
√
ap
r (2a − r) =
a 2 (1 − e 2 )
r (2a − r)
tan γ =
e sin θ
1 + e cos θ
18.4 Parabolic trajectory
This diagram shows the parabolic trajectory
18.5 Hyperbolic trajectory
This diagram shows the hyperbolic trajectory
70

This diagram below from Orbital mechanics for Engineering students, second edition, by Howard D. Curtis,
page 109
71

18.6 showing that energy is constant
Showing that energy E = v2
2 − µ r
is constant.
Most of such relations starts from the same place. The equation of motion of satellite under the assumption
that its mass is much smaller than the mass of the large body (say earth) it is rotating around. Hence we can
use ν = GM and the equation of motion reduces to
¨⃗r + µ r 3⃗r = 0
In the above equation, the vector ⃗r is the relative vector from the center of the earth to the center of the satellite.
The reason the center of earth is used as the origin of the inertial frame of reference is due to the assumption
that M ≫ m where M is the mass of earth (or the body at the focal of the ellipse) and m is the mass of the
satellite. Hence the median center of mass between the earth and the satellite is taken to be the center of earth.
This is an approximation, but a very good approximation.
The first step is to dot product the above equation with ˙⃗r giving
˙⃗r · µ
r 3 ⃗r = µ ṙ
r 2
˙⃗r · ¨⃗r + ˙⃗r · µ
r3⃗r = 0 (1)
( )
ṙ2
2
and we also see that
And there is the main trick. We look ahead and see that ˙⃗r · ¨⃗r = ṙ¨r but ṙ¨r = d dt
but µ ṙ = d ( −µ
)
r 2 dt r Hence equation 1 above can be written as
Hence
(
d v
2
dt 2 − r )
= 0
µ
E = v2
2 − r µ
Where E is a constant, which is the total energy of the satellite.
18.7 Earth satellite Transfer orbits
18.7.1 Hohmann transfer
This diagram shows the Hohmann transfer
72

73

18.7.2 Bi-Elliptic transfer orbit
74

18.7.3 semi-tangential elliptical transfer
18.8 Rocket engines, Hohmann transfer, plane change at equator
two cases: Hohmann transfer, 2 burns, or semi-tangential. All burns at equator.
75

76

18.8.1 Rocket equation with plane change not at equator
77

18.9 Spherical coordinates
78

18.10 interplanetary transfer orbits
18.10.1 interplanetary hohmann transfer orbit, case one
The following are the steps to accomplish the above. The first stage is getting into the Hohmann orbit from
planet 1, then reaching the sphere of influence of the second planet. Then we either do a fly-by or do a parking
orbit around the second planet. These steps below show how to reach the second planet and do a parking orbit
around it.
79

The input is the following.
1. µ 1 planet one standard gravitational parameter
2. µ 2 planet two standard gravitational parameter
3. µ sun standard gravitational parameter for the sun 1.327 × 10 8 km
4. r 1 planet one radius
5. r 2 planet two radius
6. alt 1 original satellite altitude above planet one. For example, for LEO use 300 km
7. alt 2 satellite altitude above second planet. (since goal is to send satellite for circular orbit around second
planet)
8. R 1 mean distance of center of first planet from the sun. For earth use AU = 1.495978 × 10 8 km
9. R 2 mean distance of center of second planet from the sun. For Mars use 1.524 AU
10. SOI 1 sphere of influence for first planet. For earth use 9.24 × 10 8 km
11. SOI 2 sphere of influence for second planet.
Given the above input, there are the steps to achieve the above maneuver
1. Find the burn out distance of the satellite r bo = r 1 + alt 1
2. Find satellite speed around planet earth (relative to planet) V sat =
√
µ1
r bo
3. Find Hohmann ellipse a = R 1+R 2
2
4. Find speed of satellite at perigee relative to sun V perigee =
√
µ sun
(
2
R 1
− 1 a
)
5. Find speed of earth (first planet) relative to sun V 1 =
√
µsun
R 1
6. Find escape velocity from first planet V ∞,out = V perigee − V 1
7. Find burn out speed at first planet by solving the energy equation V 2
bo
2 − µ 1
r bo
8. Find ∆V 1 needed at planet one ∆V 1 = V bo − V sat
9. Find e the eccentricity of the escape hyperbola e =
√
1 + V 2 ∞V 2
bo r2 bo
µ 2 1
10. Find the angle with the path of planet one velocity vector η = arccos ( − 1 e
11. Find the dusk-line angle θ = 180 0 − η
)
= V 2 ∞,out
2
− µ 1
SOI 1
for V bo
The above completes the first stage, now the satellite is in the Hohmann transfer orbit. Assuming it reached
the orbit of the second planet ahead of it as shown in the diagram above. Now we start the second stage to land
the satellite on a parking orbit around the second planet at altitude alt 2 above the surface of the second planet.
These are the steps needed.
1. Find the apogee speed of the satellite V apogree =
√
µ sun
(
2
R 2
− 1 a
)
80

2. Find speed of second planet V 2 =
√
µsun
R 2
3. Find V ∞ entering the second planet sphere of influence V ∞,in = V 2 − V apogree
4. Find burn in radius where the satellite will be closest to the second planet. r bo = r 1 + alt 2
5. Find burn out speed at second planet by solving the energy equation V 2
bo
2 − µ 2
r bo
6. Find impact parameter b on entry to second planet SOI b = r boV bo
V ∞,in
7. Find the required satellite speed around the second planet V sat =
√
µ2
r bo
8. Find ∆V 2 needed at planet two ∆V 2 = V sat − V bo
9. Find e the eccentricity of the approaching hyperbola on second planet e =
10. Find the angle with the path of planet two velocity vector η = arccos ( − 1 e
11. Find the dusk-line angle for second planet θ = 2η − 180 0
18.11 rendezvous orbits
18.11.1 Two satellite, walking rendezvous using Hohmann transfer
)
√
= V 2 ∞,in
2
− µ 2
SOI 2
for V bo
1 + V 2 ∞ V 2
bo r2 bo
µ 2 2
81

Algorithm 1 Hohmann Walking Rendezvous Orbit, case 1
1: function hohmann walking rendezvous(θ 0 , r, altitude, µ)
2: θ 0 := θ 0
π
180
⊲ convert from degrees to radian
3: r a := r + altitude
4: T := 2π√
r 3 aµ
⊲ period of circular orbit
5: n := 1
6: done:=false
7: while not(done) do
(
8: TOF := N − θ 0
9: a := solve
10: r p := 2a − r a
11: if rp < r then
12: N = N + 1
13: else
14: done:=true
15: end if
16: end while
2π
)
T
(
T OF = 2π
√
17: V befor := µ
h
√
18: V after := µ ( 2
h − 1 )
a
19: ∆V := 2(V after − V before )
20: return (TOF, ∆V )
21: end function
√
An example implementation is below
)
a 3
µ
for a
1 hohmannRendezvousSameOrbit[\[Theta]00_, r_, alt_, mu_] :=
2 Module[{\[Theta]0 = \[Theta]00*Pi/180, n = 1, delT, v1, v2, period, a,
3 rp, ra, done = False, vBefore, vAfter},
4 ra = r + alt;
5 period = 2 Pi Sqrt[ra^3/mu];
6 While[Not[done],
7 delT = (n - \[Theta]0 /(2 Pi)) period ;
8 a = First@Select[a /. NSolve[delT == 2 Pi Sqrt[a^3/mu], a], Element[#, Reals] &];
9 rp = 2 a - ra;
10 If[rp < r,(*we hit the earth, try again*)
11 n = n + 1,
12 done = True
13 ]
14 ];
15 vBefore = Sqrt[mu/h];
16 vAfter = Sqrt[mu (2/h - 1/a)];
17 {delT, 2 (vAfter - vBefore)}
18 ]
And calling the above
82

1 mu = 324859;
2 alt = 1475.776;
3 r = 6052;
4 \[Theta]0 = 3.80562; (*degree*)
5 hohmannRendezvousSameOrbit[\[Theta]0, r, alt, mu]
6 {7123.89, -0.0467913}
18.11.2 Two satellite, separate orbits, rendezvous using Hohmann transfer, coplaner
83

Algorithm 2 Hohmann rendezvous algorithm, case 1
1: function hohmann rendezvous 1(θ 0 , r a , r b , µ)
π
2: θ 0 := θ 0 180
⊲ convert from degrees to radian
3: a := ra+r b
2
⊲ Hohmann orbit semi-major axes
√
4: TOF := π
5: θ H := π
6: ω a :=
a 3
µ
⊲ time of flight on Hohmann orbit
( ( ) ) 3/2
1 − ra+r b
2r b
√ µ
r 3 a
⊲ required phase angle before starting Hohmann transfer
⊲ angular speed of lower rad/sec
7: ω b := √ µ
⊲ angular speed of higher satellite rad/sec
rb
3
8: if θ 0 ≤ θ H then ⊲ adjust initial angle if needed
9: θ 0 := θ 0 + 2π
10: end if
11: wait time := θ 0−θ H
ω a−ω b
⊲ how long to wait before starting Hohmann transfer
12: wait time := wait time + TOF ⊲ now ready to go, add Hohmann transfer time
13: return wait time
14: end function
An example implementation is below
1 hohmann_rendezvous_1:= proc({
2 theta::numeric:=0,
3 r1::numeric:=0,
4 r2::numeric:=0,
5 mu::numeric:=3.986*10^5})
6 local theta0,thetaH,TOF,a,omega1,omega2,wait_time;
7 theta0 := evalf(theta*Pi/180);
8 a := (r1+r2)/2;
9 TOF := Pi*(sqrt(a^3/mu));
10 omega1 := sqrt(mu/r1^3);
11 omega2 := sqrt(mu/r2^3);
12 thetaH := evalf(Pi*(1-((r1+r2)/(2*r2))^(3/2)));
13 if theta0

18.11.3 Two satellite, separate orbits, rendezvous using bi-elliptic transfer, coplaner
In this transfer, the lower (fast satellite) does not have to wait for phase lock as in the case with Hohmann
transfer. The transfer can starts immediately. There is a free parameter N that one select depending on fuel
cost requiments or any limitiation on the first transfer orbit semi-major axes distance required. One can start
with N = 0 and adjust as needed.
85

Algorithm 3 Hohmann rendezvous algorithm, case 2
1: function hohmann rendezvous 2(θ 0 , r 1 , r 2 , N, µ)
π
2: θ 0 := θ 0 180 (
⊲ convert from degrees to radian
( ) ) 3/2
3: θ H := π 1 − r1 +r 2
2r 2
⊲ Find Hohmann ideal phase angle before transfer
4: if θ 0 = θ H and N = 0 then ⊲ adjust for special case
5: a := r 2+r 1
2 (√
6: TOF := π
)
a 3
µ
7: else
8: ω 2 := √ µ
r2
3
9: t 2 := (2π−θ 0)+2πN
ω 2
10: a 1 := rt+r 1
2
11: a 2 := rt+r 2
2
12: TOF := π
(√
a 3
1
µ
+ a3 2
µ
)
⊲ angular speed of slower satellite in rad/sec
⊲ find time of light of the slower satellite
⊲ time of flight for the fast satellite
13: r t := solve (t 2 = T OF ) for r t ⊲ Solve numerically for r t
14: end if
15: return TOF
16: end function
An example implementation is below
1 hohmann_rendezvous_2:= proc({
2 theta::numeric:=0,
3 r1::numeric:=0,
4 r2::numeric:=0,
5 N::nonnegint:=0,
6 mu::numeric:=3.986*10^5})
7 local theta0,thetaH,TOF;
8 theta0 := theta*Pi/180;
9 thetaH := Pi*(1-((r1+r2)/(2*r2))^(3/2));
10 if theta0 = thetaH and N = 0 then
11 proc()
12 local a:=(r1+r2)/2;
13 TOF:= Pi*(sqrt(a^3/mu));
14 end proc()
15 else
16 proc()
17 local t2,a1,a2,rt,omega2;
18 omega2 := sqrt(mu/r2^3);
19 t2 := ((2*Pi-theta0)+2*Pi*N)/omega2;
20 a1 := (rt+r1)/2;
21 a2 := (rt+r2)/2;
22 TOF := Pi*(sqrt(a1^3/mu)+sqrt(a2^3/mu));
23 rt := op(select(is, [solve(t2=TOF,rt)], real));
24 end proc()
25 fi;
26 eval(TOF);
27 end proc:
And calling the above for two different cases gives
86

1 TOF:=hohmann_rendezvous_2(theta=0,r1=6678,r2=6878,N=0):
2 evalf(TOF/(60*60)); #in hrs
3 1.576892101
4 TOF:=hohmann_rendezvous_2(theta=160,r1=6678,r2=6878,N=1):
5 evalf(TOF/(60*60)); #in hrs
6 2.452943266
87

18.12 Semi-tangential transfers, elliptical, parabolic and hyperbolic
88

89

18.13 Lagrange points
90

18.14 Orbit changing by low contiuous thrust
18.15 References
1. Orbital mechanics for Engineering students, second edition, by Howard D. Curtis
2. Orbital Mechanics, Vladimir Chobotov, second edition, AIAA
3. Fundamentals of Astrodynamics, Bates, Muller and White. Dover1971
91

19 Dynamic of flights
19.1 Wing geometry
length of mean
aerodynamic chord
projected on the
root chord
Leading edge
sweep angle
root chord
centroid of area of
one half of wing
located on mean
aerodynamic chord
local chord
wing tip chord
wing1.vsdx
Nasser M. Abbasi
020914
half wing span
92

C r below is the core chord of the wing.
A.C.
tail
control-fixed
static margin
apex of
wing
wing
A.C.
mac
c.g
positive for
static
stability
wing_4_generic.vsdx
Nasser M. Abbasi
020914
This is a diagram to use to generate equations of longitudinal equilibrium.
This distance is called the stick-fixed static margin k m = (h n − h) ¯c Must be positive for static stability
93

tail mean
aerodynamic
chord
The whole Airplane
neutral point (aft c.g.
for static stability)
Wing mean
aerodynamic chord
km
forces.vsd
updated 2/3/
14
Nasser M.
Abbasi
19.2 Summary of main equations
This table contain some definitions and equations that can be useful.
# equation meaning/use
94

1
C L = ∂C L
∂α α
= C Lα α
= aα
C L is lift coefficient. α is angle
of attack. a is slope ∂C L
∂α
which
is the same as C Lα
2 C Lw = C Lwα α wing lift coefficient
3 C D = C Dmin + kC 2 L
drag coefficient
4 C mw = C macw + (C Lw + C Dmin α w ) (h − h nw ) + (C L α w − C Dw ) z¯c pitching moment coefficient
due to wing only about the
C.G. of the airplane assuming
small α w . This is simplified
more by assuming C Dw α w ≪
C Lw and (C L α w − C Dw ) ≪ 1
5 C mw = C macw + C Lw (h − h nw ) simplified wing Pitching moment
C mwb = C macwb + C Lwb (h − h nw )
6
= C macwb + ∂C L wb
∂α wb
α wb (h − h nw )
= C macwb + a wb α wb (h − h nw )
simplified pitching moment coefficient
due to wing and body
about the C.G. of the airplane.
α wb is the angle of attack
7 C Lt = Lt
1
2 ρV 2 S t
C Lt is the lift coefficient generated
by tail. S t is the tail area.
V is airplane air speed
8 L = L wb + L t total lift of airplane. L wb is lift
due to body and wing and L t
is lift due to tail
9 C L = C Lwb + St
S C L t
coefficient of total lift of airplane.
C Lwb is coefficient of lift
due to wing and body. C Lt is
lift coefficient due to tail. S is
the total wing area. S t is tail
area
10
1
M t = −l t L t = −l t C Lt 2 ρV 2 S t pitching moment due to tail
about C.G. of airplane
11 C mt = Mt = − lt¯c
S t
1
2 ρV 2 S t¯c S C L t
= −V H C Lt pitching moment coefficient
due to tail. V H = lt¯c
S t
S
is called
tail volume
12
V H = lt¯c
S t
S
¯V H = ¯l t¯c
S t
S
introducing ¯V H bar tail volume
which is V H but uses ¯l t instead
of l t . Important note. V H
depends on location of C.G.,
but ¯V H does not. ¯lt = l t +
(h − h nwb ) ¯c
95

13 C mt = − ¯V H C Lt + C Lt
S t
S (h − h n wb
) pitching moment coefficient
due to tail expressed using ¯V H .
This is the one to use.
14 C mp pitching moment coefficient
due to propulsion about airplane
C.G.
15 C m = C mwb + C mt + C mp total airplane pitching moment
coefficient about airplane C.G.
16
C m = C mwb + C mt + C mp
[
]
= C macwb + C Lwb (h − h nw )
C L
+ [ − ¯V H C Lt + C Lt
S t
S (h − h n wb
) ] + C mp
{ ( }} ) {
simplified total Pitching moment
coefficient about airplane
S t
= C macwb + C Lwb + C Lt (h − h nw ) −
S
¯V H C Lt + C mp
C.G.
= C macwb + C L (h − h nw ) − ¯V H C Lt + C mp
17
∂C m
∂α
= ∂Cmac wb
∂α
C mα = ∂Cmac wb
∂α
18 h n = h nwb − 1 ∂C L
∂α
+ ∂C L
∂α
(h − h n w
) − ¯V H
∂C Lt
∂α
+ C Lα (h − h nw ) − ¯V H
∂C Lt
∂α
( ∂Cmacwb
∂α
− ¯V H
∂C Lt
∂α
)
+
∂Cmp
∂α
+ ∂Cmp
∂α
+
∂Cmp
∂α
derivative of total pitching moment
coefficient C m w.r.t airplane
angle of attack α
location of airplane neutral
point of airplane found by setting
C mα = 0 in the above
equation
19
∂C m
∂α
= ∂C L
∂α (h − h n)
C mα = C Lα (h − h n )
rewrite of C mα in terms of h n .
Derived using the above two
equations.
20 k n = h n − h static margin. Must be Positive
for static stability
19.2.1 Writing the equations in linear form
The following equations are derived from the above set of equation using what is called the linear form. The main
point is to bring into the equations the expression for C Lt written in term of α wb . This is done by expressing the
∂C Lwb
tail angle of attack α t in terms of α wb via the downwash angle and the i t angle.
∂α wb
in the above equations are
replaced by a wb and ∂C L t
∂α t
is replaced by a t . This replacement says that it is a linear relation between C L and
the corresponding angle of attack. The main of this rewrite is to obtain an expression for C m in terms of α wb
where α t is expressed in terms of α wb , hence α t do not show explicitly. The linear form of the equations is what
from now on.
# equation meaning/use
96

1
2
3
C Lwb = ∂C L wb
∂α wb
α wb
= a wb α wb
C Lt = a t α t
C mp = C m0 p + ∂C mp
∂α α
α t = α wb − i t − ɛ
ɛ = ɛ 0 + ∂ɛ
∂α α wb
C Lt = a t α t
[ (
= a t α wb 1 − ∂ɛ ) ]
− i t − ɛ 0
∂α
a wb is constant, represents
∂C L wb
∂α wb
and C m0p
is propulsion pitching
moment coeff. at zero
angle of attack α
main relation that associates
α wb with α t . α wb
is the wing-body angle
of attack, ɛ is downwash
angle at tail, and
i t is tail angle with horizontal
reference (see diagram)
Lift due to tail expressed
using α wb and
ɛ (notice that α t do not
show explicitly)
(
4 a = a wb
[1 + at S t
a wb S 1 −
∂ɛ
∂α) ] a defined for use with
overall lift coefficient
5
C L =
a wb α { }}
wb
{
C Lwb
+ St
S C L t
= a wb α wb + St
S a t
= aα
= (C L ) αwb =0 + aα wb
[
αwb
(
1 −
∂ɛ
∂α
)
− it − ɛ 0
]
overall airplane lift using
linear relations
6 α = α wb − at S t
a S (i t + ɛ 0 )
overall angle of attack
α as function of the
wing and body angle of
attack α wb and tail angles
cl_apha.vsdx
Nasser M. Abbasi
021 214
7
C m = C m0 + ∂Cm
∂α α = C m0 + C mα α
C m = ¯C m0 + ∂Cm
∂α α wb = ¯C m0 + C mα α wb
overall airplane pitch
moment. Two versions
one uses α wb and one
uses α
97

8
9
( )
C mα = a (h − h nwb ) − a t ¯VH 1 −
∂ɛ ∂C
∂α + mp
∂α
C mα = a wb (h − h nwb ) − a t V H
(
1 −
∂ɛ
∂α
)
+
∂C mp
∂α
C m0 = C macwb + C mop + a t ¯VH (ɛ 0 + i t ) [ 1 − at S t
a S
¯C m0 = C macwb + ¯C mop + a t V H (ɛ 0 + i t )
( )]
1 −
∂ɛ
∂α
∂C m
∂α
Two versions of
one for α wb and one one
uses α
C m0 is total pitching
moment coef. at zero
lift (does not depend on
C.G. location) but ¯C m0
is total pitching moment
coef. at α wb = 0
(not at zero lift). This
depends on location of
C.G.
10
¯Cm0p = C m0p + (α − α wb ) ∂Cmp
∂α
h n = h nwb + at ¯V
( )
a H 1 −
∂ɛ
∂α −
1 ∂C mp
a ∂α
11
= h nwb +
a t (
a wb
[1+ a t S t
a wb S
1− ∂ɛ
∂α
)] ¯VH
(
1 −
∂ɛ
∂α
)
−
1
a wb
[1+ a t S t
a wb S
(
)] ∂Cmp
1− ∂ɛ ∂α
∂α
Used to determine h n
19.3 definitions
1. Remember that for symmetric airfoil, when the chord is parallel to velocity vector, then the angle of attack
is zero, and also the left coefficient is zero. But this is only for symmetric airfoil. For the common campbell
airfoil shape, when the chord is parallel to the velocity vector, which means the angle of attack is zero,
there will still be lift (small lift, but it is there). What this means, is that the chord line has to tilt down
more to get zero lift. This extra tilting down makes the angle of attack negative. If we now draw a line
from the right edge of the airfoil parallel to the velocity vector, this line is called the zero lift line (ZLL)
see diagram below.
Just remember, that angle of attack (which is always the angle between the chord and the velocity vector,
the book below calls it the geometrical angle of attack) is negative for zero lift. This is when the airfoil is
not symmetric. For symmetric airfoil, ZLL and the chord line are the same. This angle is small, −3 0 or so.
Depending on shape. See Foundations of Aerodynamics, 5th ed, by Chow and Kuethe, here is the diagram.
98

2. stall from http://en.wikipedia.org/wiki/Stall_(flight)
In fluid dynamics, a stall is a reduction in the lift coefficient
generated by a foil as angle of attack increases.[1] This occurs when
the critical angle of attack of the foil is exceeded. The critical
angle of attack is typically about 15 degrees, but it
may vary significantly depending on the fluid, foil, and Reynolds number.
3. Aerodynamics in road vehicle wiki page
4. some demos relating to airplane control http://demonstrations.wolfram.com/ControllingAirplaneFlight/
http://demonstrations.wolfram.com/ThePhysicsOfFlight/
5. http://www.americanflyers.net/aviationlibrary/pilots_handbook/chapter_3.htm
6. Lectures Helicopter Aerodynamics and Dynamics by Prof. C. Venkatesan, Department of Aerospace
Engineering, IIT Kanpur http://www.youtube.com/watch?v=DKWj2WzYXtQ&list=PLAE677E56C97A7C7D
7. http://avstop.com/ac/apgeneral/terminology.html has easy to understand definitions airplane geometry.
”The MAC is the mean average chord of the wing”
8. http://www.tdmsoftware.com/afd/afd.html airfoil design software
99

19.4 images and plots collected
These are diagrams and images collected from different places. References is given next to each image.
Perpendicular to c.g. velocity vector V
Body x-axes
(chord of
airplane
Zero left line
for airplane
Current
velocity
vector of
airplane c.g.
Parallel to V
vector originate
from A.C. and not
C.G.
Angle of attack of thrust
Absolute angle of attack
Weight of
airplane.
Vertically down
By Nasser M. Abbasi
my_drawing.vsd
1/24/14
100

http://en.wikipedia.org/wiki/Flight_dynamics_%28aircraft%29
Dihedral angle
Reference: http://en.wikipedia.org/wiki/Dihedral_%28aircraft%29
101

definitions
From Performance, Stability, Dynamics, and Control of Airplanes
By Baudu N. Pamadi
I do not see viscosity here?
http://en.wikipedia.org/wiki/Drag_coefficient
This below from http://www.grc.nasa.gov/WWW/k-12/UEET/StudentSite/dynamicsofflight.html
102

http://www.grc.nasa.gov/WWW/k-12/UEET/StudentSite/dynamicsofflight.html
http://www.grc.nasa.gov/WWW/k-12/airplane/alr.html
103

http://www.grc.nasa.gov/WWW/k-12/airplane/alr.html
http://www.grc.nasa.gov/WWW/k-12/airplane/alr.html
From http://en.wikipedia.org/wiki/Lift_coefficient and http://en.wikipedia.org/wiki/File:Aeroforces
104

http://en.wikipedia.org/wiki/Lift_coefficient
http://en.wikipedia.org/wiki/File:Aeroforces.svg
svg
from http://adg.stanford.edu/aa241/drag/sweepncdc.html
http://en.wikipedia.org/wiki/Lift_%28force%29
105

http://adg.stanford.edu/aa241/drag/sweepncdc.html
Images from http://adamone.rchomepage.com/cg_calc.htm and Flight dynamics principles by Cook, 1997.
106

http://adamone.rchomepage.com/cg_calc.htm
From http://chrusion.com/BJ7/SuperCalc7.html
107

From http://www.willingtons.com/aircraft_center_of_gravity_calcu.html
108

From http://www.solar-city.net/2010/06/airplane-control-surfaces.html nice diagram that shows
clearly how the elevator causes the pitching motion (nose up/down). From same page, it says ”The purpose of
109

the flaps is to generate more lift at slower airspeed, which enables the airplane to fly at a greatly reduced speed
with a lower risk of stalling.”
Images from flight dynamics principles, by Cook, 1997.
110

Page 184, flight dynamics principles, by Cook
Page 41, flight dynamics principles, by Cook
Images from Performance, stability, dynamics and control of Airplanes. By Pamadi, AIAA press. Page 169.
and http://www.americanflyers.net/aviationlibrary/pilots_handbook/chapter_3.htm
111

From: Performance, stability, dynamics and control of
Airplanes. By Pamadi, AIAA press. Page 169
http://www.americanflyers.net/aviationlibrary/pilots_handbook/chapter_3.htm
Image from http://www.americanflyers.net/aviationlibrary/pilots_handbook/chapter_3.htm
112

Image from http://www.americanflyers.net/aviationlibrary/pilots_handbook/chapter_3.htm
113

The profile drag of a streamlined object held
in a fixed position relative to the airflow increases
approximately as the square of the velocity;
thus, doubling the airspeed increases the drag four times,
and tripling the airspeed increases the drag nine times.
The amount of induced drag varies inversely
as the square of the airspeed.
From the foregoing discussion, it can be noted that parasite drag increases
as the square of the airspeed, and induced drag varies inversely as
the square of the airspeed.
lift varies directly with the wing area
a wing with a planform area of 200 square feet lifts twice as
much at the same angle of attack as a wing with an area of 100
square feet.
http://www.americanflyers.net/aviationlibrary/pilots_handbook/chapter_3.htm
Image from FAA pilot handbook and http://www.youtube.com/watch?v=8uT55aei1NI
114

From FAA pilot’s handbook
I do not understand the above
From FAA pilot’s handbook
http://www.youtube.com/watch?v=8uT55aei1NI
Image http://www.youtube.com/watch?v=8uT55aei1NI and http://www.youtube.com/user/DAMSQAZ?
feature=watch
115

http://www.youtube.com/watch?v=8uT55aei1NI
http://www.youtube.com/user/DAMSQAZ?feature=watch
A=area of wing
D=density of air
V=wind speed relative to
wing
116

zero-lift angle
The angle of attack at which an airfoil does not produce any lift. Its value is generally less than zero unless
the airfoil is symmetrical.
This from Foundations of aerodynamics, 5 th ed. By Chow and Kuethe
http://www.answers.com/topic/zero-lift-angle
These are from text :foundation of aerodynamics by Kuethe and Chow
1. the center of pressure is at ¼ chord for all values of lift coeff.
2. center of pressure (c.p.) of a force is defined as the point about which the moment vanishes.
3. The geometric angle of attack is defined as the angle between the flight path and the chord line of the airfoil.
When the geometric angle of attack is zero, the lift coeff. Is zero.
4. The point about which the moment coeff. Is independent of the angle of attack is called the aerodynamic
center of the section. (a.c.)
5. a.c. is at the ¼ chord line point.
6. The value of the angle of attack that makes the lift coeff. Zero is called the angle of zero lift (Z.L.L.)
There are from Foundations
of aerodynamics, 5 th ed. By
Chow and Kuethe
19.5 Some strange shaped airplanes
Image http://edition.cnn.com/2014/01/16/travel/inside-airbus-beluga/index.html?hpt=ibu_c2
117

Image from http://edition.cnn.com/2014/01/16/travel/inside-airbus-beluga/index.html?hpt=ibu_
c2
Image from http://www.nasa.gov/centers/dryden/Features/super_guppy.html
118

Image from http://www.aerospaceweb.org/question/aerodynamics/q0130.shtml ”Boeing Pelican ground
effect vehicle”
19.6 links
1. https://3dwarehouse.sketchup.com/search.html?redirect=1&tags=airplane
19.7 references
1. Etkin and Reid, Dynamics of flight, 3rd edition.
2. Cook, Flight Dynamics principles, third edition.
3. Lecture notes, EMA 523 flight dynamics and control, University of Wisconsin, Madison by Professor
Riccardo Bonazza
4. Kuethe and Chow, Foundations of Aerodynamics, 4th edition
119