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<strong>Dynamics</strong> <strong>cheat</strong> <strong>sheet</strong><br />
Nasser M. Abbasi<br />
September 27, 2015<br />
page compiled on September 27, 2015 at 5:28pm<br />
Contents<br />
1 Modal analysis for two degrees of freedom system 3<br />
1.1 Step one, finding the equations of motion in normal coordinates space . . . . . . . . . . . . . . . 3<br />
1.2 Step 2, solving the eigenvalue problem, finding the natural frequencies . . . . . . . . . . . . . . . 4<br />
1.3 Step 3, finding the non-normalized eigenvectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5<br />
1.4 Step 4. Mass normalization of the shape vectors (or the eigenvectors) . . . . . . . . . . . . . . . . 6<br />
1.5 Step 5, obtain the modal transformation matrix Φ . . . . . . . . . . . . . . . . . . . . . . . . . . 7<br />
1.6 Step 6, applying modal transformation to decouple the original equations of motion . . . . . . . . 8<br />
1.7 Step 7. Converting modal solution to normal coordinates solution . . . . . . . . . . . . . . . . . . 9<br />
1.8 Numerical solution using modal analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9<br />
2 Fourier series representation of a periodic function 14<br />
3 Transfer functions for different vibration systems 16<br />
3.1 Force transmissibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16<br />
3.2 vibration isolation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16<br />
3.3 accelerometer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18<br />
3.4 seismometer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18<br />
3.5 Summary of vibration transfer functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19<br />
4 Steady state solutions to single degree freedom system 20<br />
4.1 Summary: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20<br />
4.2 Details . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20<br />
4.3 Undamped case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21<br />
4.4 damped cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21<br />
5 Summary table of solutions 22<br />
5.1 Harmonic loading mu ′′ + cu ′ + ku = F sin ϖt . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22<br />
5.2 constant loading mu ′′ + cu ′ + ku = F . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24<br />
5.3 no loading (free) mu ′′ + cu ′ + ku = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25<br />
5.4 impulse F 0 δ (t) loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26<br />
5.5 Impulse sin function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27<br />
6 Tree view look at the different cases 28<br />
7 Some common formulas 31<br />
8 Derivation of the solutions of the equations of motion 33<br />
8.1 Free vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33<br />
8.1.1 Undamped free vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33<br />
1
8.1.2 under-damped free vibration c < c r , ξ < 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 33<br />
8.1.3 critically damped free vibration ξ = c<br />
c r<br />
= 1 . . . . . . . . . . . . . . . . . . . . . . . . . . 34<br />
8.1.4 over-damped free vibration ξ = c<br />
c r<br />
> 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34<br />
8.2 Constant input F . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34<br />
8.2.1 Undamped case ζ = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34<br />
8.2.2 under-damped ζ < 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35<br />
8.2.3 Critical damping ζ = 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35<br />
8.2.4 Over-damped ζ > 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35<br />
8.3 Harmonic forced input F sin (ϖt) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36<br />
8.3.1 Undamped forced vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36<br />
8.3.2 Under-damped forced vibration c < c r , ξ < 1 . . . . . . . . . . . . . . . . . . . . . . . . . 38<br />
8.3.3 critically damping forced vibration ξ = c<br />
c r<br />
= 1 . . . . . . . . . . . . . . . . . . . . . . . . . 40<br />
8.3.4 over-damped forced vibration ξ = c<br />
c r<br />
> 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40<br />
8.4 Response to impulsive loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41<br />
8.4.1 impulse input . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41<br />
8.4.2 sin impulse input . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43<br />
8.5 references . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49<br />
9 Derivation of rotation formula 49<br />
10 Miscellaneous hints 50<br />
11 Formulas 51<br />
12 Velocity and acceleration diagrams 53<br />
12.1 Spring pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53<br />
12.2 pendulum with blob moving in slot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54<br />
12.3 spring pendulum with block moving in slot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55<br />
12.4 double pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56<br />
13 Velocity and acceleration of rigid body 2D 57<br />
14 Velocity and acceleration of rigid body 3D 58<br />
14.1 Using Vehicle dynamics notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58<br />
14.2 3D Not Using vehicle dynamics notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59<br />
14.2.1 Derivation for F = ma in 3D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60<br />
14.2.2 Derivation for τ = Iω in 3D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61<br />
14.2.3 Derivation for τ = Iω in 3D using principle axes . . . . . . . . . . . . . . . . . . . . . . . 62<br />
15 Wheel spinning precession 63<br />
16 References 63<br />
17 Misc. items 64<br />
18 Astrodynamics 65<br />
18.1 Ellipse main parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65<br />
18.2 Table of common equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65<br />
18.3 Flight path angle for ellipse γ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69<br />
18.4 Parabolic trajectory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70<br />
18.5 Hyperbolic trajectory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70<br />
18.6 showing that energy is constant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72<br />
18.7 Earth satellite Transfer orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72<br />
18.7.1 Hohmann transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72<br />
2
18.7.2 Bi-Elliptic transfer orbit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74<br />
18.7.3 semi-tangential elliptical transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75<br />
18.8 Rocket engines, Hohmann transfer, plane change at equator . . . . . . . . . . . . . . . . . . . . . 75<br />
18.8.1 Rocket equation with plane change not at equator . . . . . . . . . . . . . . . . . . . . . . 77<br />
18.9 Spherical coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78<br />
18.10interplanetary transfer orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79<br />
18.10.1 interplanetary hohmann transfer orbit, case one . . . . . . . . . . . . . . . . . . . . . . . . 79<br />
18.11rendezvous orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81<br />
18.11.1 Two satellite, walking rendezvous using Hohmann transfer . . . . . . . . . . . . . . . . . . 81<br />
18.11.2 Two satellite, separate orbits, rendezvous using Hohmann transfer, coplaner . . . . . . . . 83<br />
18.11.3 Two satellite, separate orbits, rendezvous using bi-elliptic transfer, coplaner . . . . . . . . 85<br />
18.12Semi-tangential transfers, elliptical, parabolic and hyperbolic . . . . . . . . . . . . . . . . . . . . 88<br />
18.13Lagrange points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90<br />
18.14Orbit changing by low contiuous thrust . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91<br />
18.15References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91<br />
19 Dynamic of flights 92<br />
19.1 Wing geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92<br />
19.2 Summary of main equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94<br />
19.2.1 Writing the equations in linear form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96<br />
19.3 definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98<br />
19.4 images and plots collected . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100<br />
19.5 Some strange shaped airplanes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117<br />
19.6 links . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119<br />
19.7 references . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119<br />
1 Modal analysis for two degrees of freedom system<br />
Detailed steps to perform modal analysis are given below for a standard undamped two degrees of freedom<br />
system. The main advantage of solving a multidegree system using modal analysis is that it decouples the<br />
equations of motion (assuming they are coupled) making solving them much simpler.<br />
In addition it shows the fundamental shapes that the system can vibrate in, which gives more insight into<br />
the system. Starting with standard 2 degrees of freedom system<br />
x 1<br />
x 2<br />
k 1 k 2<br />
m 1 m 2<br />
f 1 t<br />
f 2 t<br />
In the above the generalized coordinates are x 1 and x 2 . Hence the system requires two equations of motion<br />
(EOM’s).<br />
1.1 Step one, finding the equations of motion in normal coordinates space<br />
The two EOM’s are found using any method such as Newton’s method or Lagrangian method. Using Newton’s<br />
method, free body diagram is made of each mass and then F = ma is written for each mass resulting in the<br />
equations of motion. In the following it is assumed that both masses are moving in the positive direction and<br />
that x 2 is larger than x 1 when these equations of equilibrium are written<br />
3
x 1<br />
x 2<br />
k 1 x 1<br />
k 2 x 2 x 1 <br />
m<br />
k 2 x 2 x 1 <br />
m 2<br />
1<br />
f 1 t<br />
f 2 t<br />
F m 1 x 1<br />
<br />
k 1 x 1 k 2 x 2 x 1 f 1 t m 1 x 1<br />
<br />
F m 2 x 2<br />
<br />
k 2 x 2 x 1 f 2 t m 2 x 2<br />
<br />
Hence, from the above the equations of motion are<br />
or<br />
In Matrix form<br />
⎡ ⎤ ⎧<br />
⎣ m 1 0 ⎨<br />
⎦<br />
0 m<br />
⎩<br />
2<br />
m 1 x ′′<br />
1 + k 1 x 1 − k 2 (x 2 − x 1 ) = f 1 (t)<br />
m 2 x ′′<br />
2 + k 2 (x 2 − x 1 ) = f 2 (t)<br />
m 1 x ′′<br />
1 + x 1 (k 1 + k 2 ) − k 2 x 2 = f 1 (t)<br />
m 2 x ′′<br />
2 + k 2 x 2 − k 2 x 1 = f 2 (t)<br />
x ′′<br />
1<br />
x ′′<br />
2<br />
⎫ ⎡<br />
⎤ ⎧ ⎫ ⎧ ⎫<br />
⎬<br />
⎭ + ⎣ k 1 + k 2 −k 2<br />
⎨x ⎦ 1 ⎬ ⎨<br />
−k 2 k<br />
⎩<br />
2 x<br />
⎭ = f 1 (t) ⎬<br />
⎩<br />
2 f 2 (t)<br />
⎭<br />
The above two EOM are coupled in stiffness, but not mass coupled. Using short notations, the above is<br />
written as<br />
[M]{x ′′ } + [K]{x} = {f}<br />
Modal analysis now starts with the goal to decouple the EOM and obtain the fundamental shape functions that<br />
the system can vibrate in. To make these derivations more general, the mass matrix and the stiffness matrix are<br />
written in general notations as follows<br />
⎡ ⎤ ⎧<br />
⎣ m 11 m 12<br />
⎨<br />
⎦<br />
m 21 m<br />
⎩<br />
22<br />
x ′′<br />
1<br />
x ′′<br />
2<br />
⎫ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫<br />
⎬<br />
⎭ + ⎣ k 11 k 12<br />
⎨x ⎦ 1 ⎬ ⎨<br />
k<br />
⎩<br />
22 x<br />
⎭ = f 1 (t) ⎬<br />
⎩<br />
2 f 2 (t)<br />
⎭<br />
k 21<br />
The mass matrix [M] and the stiffness matrix [K] must always come out to be symmetric. If they are not<br />
symmetric, then a mistake was made in obtaining them. As a general rule, the mass matrix [M] is PSD (positive<br />
definite matrix) and the [K] matrix is positive semi-definite matrix. The reason the [M] is PSD is that x T [M]{x}<br />
represents the kinetic energy of the system, which is typically positive and not zero. But reading some other<br />
references 1 it is possible that [M] can be positive semi-definite. It depends on the application being modeled.<br />
1.2 Step 2, solving the eigenvalue problem, finding the natural frequencies<br />
The first step in modal analysis is to solve the eigenvalue problem det ( [K] − ω 2 [M] ) = 0 in order to determine<br />
the natural frequencies of the system. This equations leads to a polynomial in ω and the roots of this polynomial<br />
1 http://en.wikipedia.org/wiki/Fundamental_equation_of_constrained_motion<br />
4
are the natural frequencies of the system. Since there are two degrees of freedom, there will be two natural<br />
frequencies ω 1 , ω 2 for the system.<br />
det ( [K] − ω 2 [M] ) = 0<br />
⎛⎡<br />
⎤ ⎡ ⎤⎞<br />
det ⎝⎣ k 11 k 12<br />
⎦ − ω 2 ⎣ m 11 m 12<br />
⎦⎠ = 0<br />
k 21 k 22 m 21 m 22<br />
⎡<br />
⎤<br />
det ⎣ k 11 − ω 2 m 11 k 12 − ω 2 m 12<br />
⎦ = 0<br />
k 21 − ω 2 m 21 k 22 − ω 2 m 22<br />
(<br />
k11 − ω 2 ) (<br />
m 11 k22 − ω 2 ) (<br />
m 22 − k12 − ω 2 ) (<br />
m 12 k21 − ω 2 )<br />
m 21 = 0<br />
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<br />
The above is a polynomial in ω 4 . Let ω 2 = λ it becomes<br />
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<br />
This quadratic polynomial in λ which is now solved using the quadratic formula. Then the positive square<br />
root of each λ root to obtain ω 1 and ω 2 which are the roots of the original eigenvalue problem. Assuming from<br />
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<br />
called the eigenvectors associated with ω 1 and ω 2<br />
1.3 Step 3, finding the non-normalized eigenvectors<br />
For each natural frequency ω 1 and ω 2 the corresponding shape function is found by solving the following two<br />
sets of equations for the vectors ϕ 1 , ϕ 2<br />
⎡<br />
⎡<br />
⎧<br />
⎫<br />
⎣ k 11 k 12<br />
k 21<br />
⎤<br />
⎦ − ω1<br />
2<br />
k 22<br />
⎣ m 11 m 12<br />
m 21<br />
⎤ ⎫ ⎧<br />
⎨ϕ ⎦ 11 ⎬ ⎨<br />
m<br />
⎩<br />
22 ϕ<br />
⎭ = 0⎬<br />
⎩<br />
21 0<br />
⎭<br />
and<br />
⎡ ⎤<br />
⎣ k 11 k 12<br />
⎦ − ω2<br />
2<br />
k 21 k 22<br />
⎡<br />
⎣ m 11 m 12<br />
m 21<br />
⎤ ⎧ ⎫ ⎧ ⎫<br />
⎨ϕ ⎦ 12 ⎬ ⎨<br />
m<br />
⎩<br />
22 ϕ<br />
⎭ = 0⎬<br />
⎩<br />
22 0<br />
⎭<br />
For ω 1 , let ϕ 11 = 1 and solve for<br />
⎡ ⎤ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫<br />
⎣ k 11 k 12<br />
⎦ − ω1<br />
2 ⎣ m 11 m 12<br />
⎨ 1 ⎬ ⎨<br />
⎦<br />
k 21 k 22 m 21 m<br />
⎩<br />
22 ϕ<br />
⎭ = 0⎬<br />
⎩<br />
21 0<br />
⎭<br />
⎡<br />
⎤ ⎧ ⎫ ⎧ ⎫<br />
⎣ k 11 − ω1 2m 11 k 12 − ω1 2m ⎨<br />
12 1 ⎬ ⎨<br />
⎦<br />
k 21 − ω1 2m 21 k 22 − ω1 2m ⎩<br />
22 ϕ<br />
⎭ = 0⎬<br />
⎩<br />
21 0<br />
⎭<br />
Which gives one equation now to solve for ϕ 21 (the first row equation is only used)<br />
(<br />
k11 − ω 2 1m 11<br />
)<br />
+ ϕ21<br />
(<br />
k12 − ω 2 1m 12<br />
)<br />
= 0<br />
Hence<br />
ϕ 21 = − ( k 11 − ω1 2m )<br />
11<br />
(<br />
k12 − ω1 2m 12)<br />
5
Therefore the first shape vector is<br />
⎧ ⎫ ⎧<br />
⎨ϕ 11 ⎬ ⎪⎨<br />
ϕ 1 =<br />
⎩<br />
ϕ<br />
⎭ = 1<br />
⎪<br />
21<br />
⎩<br />
− ( k 11 −ω1 2m )<br />
( 11<br />
k12 −ω1 2m 12)<br />
⎫<br />
⎪⎬<br />
⎪⎭<br />
Similarly the second shape function is obtained. For ω 2 , let ϕ 12 = 1 and solve for<br />
⎡ ⎤ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫<br />
⎣ k 11 k 12<br />
⎦ − ω2<br />
2 ⎣ m 11 m 12<br />
⎨ 1 ⎬ ⎨<br />
⎦<br />
k 21 k 22 m 21 m<br />
⎩<br />
22 ϕ<br />
⎭ = 0⎬<br />
⎩<br />
22 0<br />
⎭<br />
⎡<br />
⎤ ⎧ ⎫ ⎧ ⎫<br />
⎣ k 11 − ω2 2m 11 k 12 − ω2 2m ⎨<br />
12 1 ⎬ ⎨<br />
⎦<br />
k 21 − ω2 2m 21 k 22 − ω2 2m ⎩<br />
22 ϕ<br />
⎭ = 0⎬<br />
⎩<br />
22 0<br />
⎭<br />
Which gives one equation now to solve for ϕ 22 (the first row equation is only used)<br />
(<br />
k11 − ω 2 2m 11<br />
)<br />
+ ϕ22<br />
(<br />
k12 − ω 2 2m 12<br />
)<br />
= 0<br />
Hence<br />
ϕ 22 = − ( k 11 − ω2 2m )<br />
11<br />
(<br />
k12 − ω2 2m 12)<br />
Therefore the second shape vector is<br />
⎧ ⎫ ⎧<br />
⎨ϕ 12 ⎬ ⎪⎨<br />
ϕ 2 =<br />
⎩<br />
ϕ<br />
⎭ = 1<br />
⎪<br />
22<br />
⎩<br />
− ( k 11 −ω2 2m )<br />
( 11<br />
k12 −ω2 2m 12)<br />
Now that the two non-normalized shape vectors are found, the next step is to perform mass normalization<br />
1.4 Step 4. Mass normalization of the shape vectors (or the eigenvectors)<br />
Let<br />
µ 1 = ϕ T 1 [M] ϕ 1<br />
This results in a scalar value µ 1 , which is later used to normalize ϕ 1 . Similarly<br />
µ 2 = ϕ T 2 [M] ϕ 2<br />
⎫<br />
⎪⎬<br />
⎪⎭<br />
For example, to find µ 1<br />
⎧ ⎫T ⎡ ⎤ ⎧ ⎫<br />
⎨ϕ 11 ⎬<br />
µ 1 = ⎣ m 11 m 12<br />
⎨ϕ ⎦ 11 ⎬<br />
⎩<br />
ϕ<br />
⎭<br />
21 m 21 m<br />
⎩<br />
22 ϕ<br />
⎭<br />
21<br />
{ } ⎡ ⎤ ⎧ ⎫<br />
= ϕ 11 ϕ ⎣ m 11 m 12<br />
⎨ϕ ⎦ 11 ⎬<br />
21<br />
m<br />
⎩<br />
22 ϕ<br />
⎭<br />
21<br />
m 21<br />
}<br />
=<br />
{ϕ ⎧ ⎫<br />
⎨ϕ 11 ⎬<br />
11 m 11 + ϕ 21 m 21 ϕ 11 m 12 + ϕ 21 m 22<br />
⎩<br />
ϕ<br />
⎭<br />
21<br />
= ϕ 11 (ϕ 11 m 11 + ϕ 21 m 21 ) + ϕ 21 (ϕ 11 m 12 + ϕ 21 m 22 )<br />
6
Similarly, µ 2 is found<br />
⎧ ⎫T ⎡ ⎤ ⎧ ⎫<br />
⎨ϕ 12 ⎬<br />
µ 2 = ⎣ m 11 m 12<br />
⎨ϕ ⎦ 12 ⎬<br />
⎩<br />
ϕ<br />
⎭<br />
22 m 21 m<br />
⎩<br />
22 ϕ<br />
⎭<br />
22<br />
{ } ⎡ ⎤ ⎧ ⎫<br />
= ϕ 12 ϕ ⎣ m 11 m 12<br />
⎨ϕ ⎦ 12 ⎬<br />
22<br />
m<br />
⎩<br />
22 ϕ<br />
⎭<br />
22<br />
m 21<br />
}<br />
=<br />
{ϕ ⎧ ⎫<br />
⎨ϕ 12 ⎬<br />
12 m 11 + ϕ 22 m 21 ϕ 12 m 12 + ϕ 22 m 22<br />
⎩<br />
ϕ<br />
⎭<br />
22<br />
= ϕ 12 (ϕ 12 m 11 + ϕ 22 m 21 ) + ϕ 22 (ϕ 12 m 12 + ϕ 22 m 22 )<br />
Now that µ 1 , µ 2 are obtained, the mass normalized shape vectors are found. They are called Φ 1 , Φ 2<br />
⎧ ⎫<br />
⎨ϕ 11 ⎬<br />
⎧ ⎫<br />
Φ 1 = ϕ ⎩<br />
ϕ<br />
⎭<br />
1 21<br />
⎨ √µ1<br />
ϕ 11 ⎬<br />
√ = √ =<br />
µ1 µ1 ⎩ √µ1<br />
ϕ 21 ⎭<br />
Similarly<br />
Φ 2 = ϕ 2<br />
√<br />
µ2<br />
=<br />
⎧<br />
⎨<br />
⎩<br />
ϕ 12<br />
ϕ 22<br />
⎫<br />
⎬<br />
⎭<br />
√<br />
µ2<br />
=<br />
⎧<br />
⎨<br />
õ2<br />
ϕ 12<br />
⎫<br />
⎬<br />
⎩ √µ2<br />
ϕ 22 ⎭<br />
1.5 Step 5, obtain the modal transformation matrix Φ<br />
The modal transformation matrix is the 2 × 2 matrix made of of Φ 1 , Φ 2 in each of its columns<br />
[Φ] = [Φ 1 Φ 2 ]<br />
⎡<br />
= ⎣<br />
⎤<br />
õ1<br />
ϕ 11 √µ2<br />
ϕ 12<br />
⎦<br />
õ1<br />
ϕ 21 √µ2<br />
ϕ 22<br />
Now the [Φ] is found, the transformation from the normal coordinates {x} to modal coordinates, which is<br />
called {η} is found<br />
{x} = [Φ] {η}<br />
⎧ ⎫ ⎡<br />
⎨x 1 (t) ⎬ ϕ 11<br />
⎩<br />
x 2 (t)<br />
⎭ = ⎣<br />
⎤<br />
õ1 õ2<br />
ϕ 12<br />
⎦<br />
õ1<br />
ϕ 21 √µ2<br />
ϕ 22<br />
⎧ ⎫<br />
⎨η 1 (t) ⎬<br />
⎩<br />
η 2 (t)<br />
⎭<br />
The transformation from modal coordinates back to normal coordinates is<br />
{η} = [Φ] −1 {x}<br />
⎧ ⎫ ⎡<br />
⎨η 1 (t) ⎬ ϕ 11<br />
⎩<br />
η 2 (t)<br />
⎭ = ⎣<br />
⎤<br />
õ1 õ2<br />
ϕ 12<br />
⎦<br />
õ1<br />
ϕ 21 √µ2<br />
ϕ 22<br />
−1 ⎧ ⎫<br />
⎨ x 1 (t) ⎬<br />
⎩<br />
x 2 (t)<br />
⎭<br />
7
However, [Φ] −1 = [Φ] T [M] therefore<br />
{η} = [Φ] T [M] {x}<br />
⎧ ⎫ ⎡ ⎤<br />
⎨η 1 (t) ⎬ √µ1<br />
ϕ 11 √µ2<br />
ϕ 12<br />
⎩<br />
η 2 (t)<br />
⎭ = ⎣ ⎦<br />
õ1<br />
ϕ 21 √µ2<br />
ϕ 22<br />
⎤ ⎧ ⎫<br />
⎣ m 11 m 12<br />
⎨x ⎦ 1 (t) ⎬<br />
m 21 m<br />
⎩<br />
22 x 2 (t)<br />
⎭<br />
T ⎡<br />
The next step is to apply this transformation to the original equations of motion in order to decouple them<br />
1.6 Step 6, applying modal transformation to decouple the original equations of motion<br />
The EOM in normal coordinates is<br />
⎡<br />
⎣ m 11 m 12<br />
m 21<br />
⎤ ⎧<br />
⎨<br />
⎦<br />
m<br />
⎩<br />
22<br />
x ′′<br />
1<br />
x ′′<br />
2<br />
⎫ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫<br />
⎬<br />
⎭ + ⎣ k 11 k 12<br />
⎨x ⎦ 1 ⎬ ⎨<br />
k<br />
⎩<br />
22 x<br />
⎭ = f 1 (t) ⎬<br />
⎩<br />
2 f 2 (t)<br />
⎭<br />
Applying the above modal transformation {x} = [Φ] {η} on the above results in<br />
⎡ ⎤ ⎧ ⎫ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫<br />
⎣ m 11 m 12<br />
⎨η 1<br />
⎦ ′′ ⎬<br />
[Φ]<br />
m 21 m<br />
⎩<br />
22 η ′′ ⎭ + ⎣ k 11 k 12<br />
⎨η ⎦ 1 ⎬ ⎨<br />
[Φ]<br />
k 21 k<br />
⎩<br />
22 η<br />
⎭ = f 1 (t) ⎬<br />
⎩<br />
2 f 2 (t)<br />
⎭<br />
pre-multiplying by [Φ] T results in<br />
⎡ ⎤ ⎧<br />
[Φ] T ⎣ m 11 m 12<br />
⎨<br />
⎦ [Φ]<br />
m<br />
⎩<br />
22<br />
m 21<br />
η 1<br />
′′<br />
η 2<br />
′′<br />
2<br />
k 21<br />
⎫ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫<br />
⎬<br />
⎭ + [Φ]T ⎣ k 11 k 12<br />
⎨η ⎦ 1 ⎬ ⎨<br />
[Φ]<br />
k<br />
⎩<br />
22 η<br />
⎭ = f 1 (t) ⎬<br />
[Φ]T ⎩<br />
2 f 2 (t)<br />
⎭<br />
⎡ ⎤<br />
⎡ ⎤<br />
The result of [Φ] T ⎣ m 11 m 12<br />
⎦ [Φ] will always be ⎣ 1 0 ⎦. This is because mass normalized shape vectors<br />
m 21 m 22 0 1<br />
are used. If the shape functions were not mass normalized, then the diagonal values will not be 1 as shown.<br />
⎡ ⎤<br />
⎡ ⎤<br />
The result of [Φ] T ⎣ k 11 k 12<br />
⎦ [Φ] will be ⎣ ω2 1 0<br />
⎦.<br />
k 21 k 22 0 ω2<br />
2<br />
⎧ ⎫ ⎧ ⎫<br />
⎨<br />
Let the result of [Φ] T f 1 (t) ⎬ ⎨ ˜f<br />
⎩<br />
f 2 (t)<br />
⎭ be 1 (t) ⎬<br />
,Therefore, in modal coordinates the original EOM becomes<br />
⎩ ˜f 2 (t)<br />
⎭<br />
⎡ ⎤ ⎧<br />
⎣ 1 0 ⎨<br />
⎦<br />
0 1<br />
⎩<br />
η 1<br />
′′<br />
η 2<br />
′′<br />
k 21<br />
⎫ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫<br />
⎬<br />
⎭ + ⎣ ω2 1 0 ⎨η ⎦ 1 ⎬ ⎨ ˜f<br />
0 ω2<br />
2 ⎩<br />
η<br />
⎭ = 1 (t) ⎬<br />
⎩<br />
2<br />
˜f 2 (t)<br />
⎭<br />
The EOM are now decouples and each can be solved as follows<br />
η ′′<br />
1 (t) + ω 2 1η 1 (t) = ˜f 1 (t)<br />
η ′′<br />
2 (t) + ω 2 2η 2 (t) = ˜f 2 (t)<br />
To solve these EOM’s, the initial conditions in normal coordinates must be transformed to modal coordinates<br />
using the above transformation rules<br />
{η (0)} = [Φ] T [M] {x (0)}<br />
{<br />
η ′ (0) } = [Φ] T [M] { x ′ (0) }<br />
8
Or in full form<br />
and<br />
⎧ ⎫ ⎡<br />
⎨η 1 (0) ⎬<br />
⎩<br />
η 2 (0)<br />
⎭ = ⎣<br />
⎧ ⎫ ⎡<br />
⎨η 1 ′ (0) ⎬<br />
⎩<br />
η 2 ′ (0) ⎭ = ⎣<br />
⎤T ⎡<br />
õ1<br />
ϕ 11 √µ2<br />
ϕ 12<br />
⎦<br />
õ1<br />
ϕ 21 √µ2<br />
ϕ 22<br />
⎤T ⎡<br />
õ1<br />
ϕ 11 √µ2<br />
ϕ 12<br />
⎦<br />
õ1<br />
ϕ 21 √µ2<br />
ϕ 22<br />
⎣ m 11 m 12<br />
m 21<br />
⎣ m 11 m 12<br />
m 21<br />
⎤ ⎧ ⎫<br />
⎨x ⎦ 1 (0) ⎬<br />
m<br />
⎩<br />
22 x 2 (0)<br />
⎭<br />
⎤ ⎧ ⎫<br />
⎨x ⎦<br />
′ 1 (0) ⎬<br />
m<br />
⎩<br />
22 x ′ 2 (0) ⎭<br />
Each of these EOM are solved using any of the standard methods. This will result is solutions η 1 (t) and<br />
η 2 (t)<br />
1.7 Step 7. Converting modal solution to normal coordinates solution<br />
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,<br />
the transformation {x} = [Φ] {η} is now applied to convert the solution to normal coordinates<br />
⎧ ⎫ ⎡ ⎤ ⎧ ⎫<br />
⎨x 1 (t) ⎬ √µ1<br />
ϕ 11 √µ2<br />
ϕ 12 ⎨<br />
⎩<br />
x 2 (t)<br />
⎭ = η<br />
⎣ ⎦ 1 (t) ⎬<br />
õ1<br />
ϕ 21 √µ2<br />
ϕ 22 ⎩<br />
η 2 (t)<br />
⎭<br />
⎧<br />
⎫<br />
⎨ √µ1<br />
ϕ 11<br />
η 1 (t) + √ ϕ 12<br />
µ2<br />
η 2 (t) ⎬<br />
=<br />
⎩ √µ1<br />
ϕ 21<br />
η 1 (t) + √ ϕ 22<br />
µ2<br />
η 2 (t)<br />
⎭<br />
Hence<br />
and<br />
x 1 (t) = ϕ 11<br />
√<br />
µ1<br />
η 1 (t) + ϕ 12<br />
√<br />
µ2<br />
η 2 (t)<br />
x 2 (t) = ϕ 21<br />
√<br />
µ1<br />
η 1 (t) + ϕ 22<br />
√<br />
µ2<br />
η 2 (t)<br />
Notice that the solution in normal coordinates is a linear combination of the modal solutions. The terms<br />
ϕ<br />
√ ij<br />
µ<br />
are just scaling factors that represent the contribution of each modal solution to the final solution. This<br />
completes modal analysis<br />
1.8 Numerical solution using modal analysis<br />
This is a numerical example that implements the above steps using a numerical values for [K] and [M].<br />
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<br />
x 1 (0) = 0, x ′ 1 (0) = 1, x 2 (0) = 1.5, x ′ 2 (0) = 3, hence<br />
⎧ ⎫ ⎫<br />
⎨x 1 (0) ⎬ 0⎬<br />
⎧<br />
⎨<br />
⎩<br />
x 2 (0)<br />
⎭ = ⎩<br />
1<br />
⎭<br />
and<br />
⎧ ⎫ ⎧ ⎫<br />
⎨x ′ 1 (0) ⎬ ⎨<br />
⎩<br />
x ′ 2 (0) ⎭ = 1.5⎬<br />
⎩<br />
3<br />
⎭<br />
In normal coordinates, the EOM are<br />
9
⎡ ⎤ ⎧ ⎫ ⎡<br />
⎤ ⎧ ⎫ ⎧ ⎫<br />
⎣ m 1 0 ⎨x ⎦<br />
′′ ⎬<br />
1<br />
0 m<br />
⎩<br />
2 x ′′ ⎭ + ⎣ k 1 + k 2 −k 2<br />
⎨x ⎦ 1 ⎬ ⎨<br />
2 −k 2 k<br />
⎩<br />
2 x<br />
⎭ = f 1 (t) ⎬<br />
⎩<br />
2 f 2 (t)<br />
⎭<br />
⎡ ⎤ ⎧ ⎫ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫<br />
⎣ 1 0 ⎨x ⎦<br />
′′ ⎬<br />
1<br />
0 3<br />
⎩<br />
x ′′ ⎭ + ⎣ 3 −2 ⎨x ⎦ 1 ⎬ ⎨<br />
−2 2<br />
⎩<br />
x<br />
⎭ = 0 ⎬<br />
⎩<br />
2 sin (5t)<br />
⎭<br />
2<br />
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<br />
and f 2 (t) = sin (5t)<br />
step 2 is now applied which solves the eigenvalue problem in order to find the two natural frequencies<br />
Let ω 2 = λ hence<br />
The solution is λ 1 = 3. 475 and λ 2 = 0.192, therefore<br />
And<br />
det ( [K] − ω 2 [M] ) = 0<br />
⎛⎡<br />
⎤ ⎡ ⎤⎞<br />
det ⎝⎣ 3 −2 ⎦ − ω 2 ⎣ 1 0 ⎦⎠ = 0<br />
−2 2 0 3<br />
⎡<br />
⎤<br />
det ⎣ 3 − ω2 −2<br />
⎦ = 0<br />
−2 2 − 3ω 2<br />
(<br />
3 − ω<br />
2 ) ( 2 − 3ω 2) − (−2) (−2) = 0<br />
3ω 4 − 11ω 2 + 2 = 0<br />
3λ 2 − 11λ + 2 = 0<br />
ω 1 = √ 3.475 = 1.864<br />
ω 2 = √ 0.192 = 0.438<br />
step 3 is now applied which finds the non-normalized eigenvectors. For each natural frequency ω 1 and ω 2 the<br />
corresponding shape function is found by solving the following two sets of equations for the eigen vectors ϕ 1 , ϕ 2<br />
⎛⎡<br />
⎤ ⎡ ⎤⎞<br />
⎧ ⎫ ⎧ ⎫<br />
⎝⎣ 3 −2 ⎦ − ω1<br />
2 ⎣ 1 0 ⎨ϕ ⎦⎠<br />
11 ⎬ ⎨<br />
−2 2 0 3<br />
⎩<br />
ϕ<br />
⎭ = 0⎬<br />
⎩<br />
21 0<br />
⎭<br />
For ω 1 = 1. 864<br />
⎛⎡<br />
⎤ ⎡ ⎤⎞<br />
⎧ ⎫ ⎧ ⎫<br />
⎝⎣ 3 −2 ⎦ − 1.864 2 ⎣ 1 0 ⎨ϕ ⎦⎠<br />
11 ⎬ ⎨<br />
−2 2<br />
0 3<br />
⎩<br />
ϕ<br />
⎭ = 0⎬<br />
⎩<br />
21 0<br />
⎭<br />
⎡<br />
⎤ ⎧ ⎫ ⎧ ⎫<br />
⎣ −0.475 −2 ⎨ 1 ⎬ ⎨<br />
⎦<br />
−2 −8.424<br />
⎩<br />
ϕ<br />
⎭ = 0⎬<br />
⎩<br />
21 0<br />
⎭<br />
This gives one equation to solve for ϕ 21 (the first row equation is only used)<br />
−0.475 − 2ϕ 21 = 0<br />
10
Hence<br />
The first eigen vector is<br />
Similarly for ω 2 = 0.438<br />
ϕ 21 = 0.475<br />
−2 = −0.237<br />
⎧ ⎫ ⎧ ⎫<br />
⎨ϕ 11 ⎬ ⎨<br />
ϕ 1 =<br />
⎩<br />
ϕ<br />
⎭ = 1 ⎬<br />
⎩<br />
21 −0.237<br />
⎭<br />
⎛⎡<br />
⎤ ⎡ ⎤⎞<br />
⎧ ⎫ ⎧ ⎫<br />
⎝⎣ 3 −2 ⎦ − 0.438 2 ⎣ 1 0 ⎨ϕ ⎦⎠<br />
12 ⎬ ⎨<br />
−2 2<br />
0 3<br />
⎩<br />
ϕ<br />
⎭ = 0⎬<br />
⎩<br />
22 0<br />
⎭<br />
⎡<br />
⎤ ⎧ ⎫ ⎧ ⎫<br />
⎣ 2.808 −2 ⎨ 1 ⎬ ⎨<br />
⎦<br />
−2 1.425<br />
⎩<br />
ϕ<br />
⎭ = 0⎬<br />
⎩<br />
22 0<br />
⎭<br />
This gives one equation to solve for ϕ 22 (the first row equation is only used)<br />
2.808 − 2ϕ 22 = 0<br />
Hence<br />
The second eigen vector is<br />
ϕ 22 = −2.808<br />
−2<br />
= 1.404<br />
⎧ ⎫ ⎧ ⎫<br />
⎨ϕ 12 ⎬ ⎨<br />
ϕ 2 =<br />
⎩<br />
ϕ<br />
⎭ = 1 ⎬<br />
⎩<br />
22 1.404<br />
⎭<br />
Now step 4 is applied, which is mass normalization of the shape vectors (or the eigenvectors)<br />
µ 1 = ϕ T 1 [M] ϕ 1<br />
Hence<br />
Similarly, µ 2 is found<br />
Hence<br />
⎧ ⎫<br />
⎨ 1 ⎬<br />
µ 1 =<br />
⎩<br />
−0.237<br />
⎭<br />
= 1. 169<br />
⎧ ⎫<br />
⎨ 1 ⎬<br />
µ 2 =<br />
⎩<br />
1.404<br />
⎭<br />
= 6.914<br />
⎤ ⎧ ⎫<br />
⎣ 1 0 ⎨ 1 ⎬<br />
⎦<br />
0 3<br />
⎩<br />
−0.237<br />
⎭<br />
T ⎡<br />
µ 2 = ϕ T 2 [M] ϕ 2<br />
⎤ ⎧ ⎫<br />
⎣ 1 0 ⎨ 1 ⎬<br />
⎦<br />
0 3<br />
⎩<br />
1.404<br />
⎭<br />
T ⎡<br />
11
Now that µ 1 , µ 2 are found, the mass normalized eigen vectors are found. They are called Φ 1 , Φ 2<br />
⎧ ⎧ ⎫<br />
⎨ ⎨ 1 ⎬<br />
Φ 1 = ϕ 1<br />
√<br />
µ1<br />
=<br />
⎩<br />
ϕ 11<br />
⎧ ⎫<br />
⎨ 0.925 ⎬<br />
=<br />
⎩<br />
−0.219<br />
⎭<br />
ϕ 21<br />
⎫<br />
⎬<br />
⎭<br />
√<br />
µ1<br />
=<br />
⎩<br />
−0.237<br />
⎭<br />
√<br />
1.169<br />
Similarly<br />
Φ 2 = ϕ 2<br />
√<br />
µ2<br />
=<br />
⎧<br />
⎨<br />
⎩<br />
⎧ ⎫<br />
⎨0.380⎬<br />
=<br />
⎩<br />
0.534<br />
⎭<br />
ϕ 12<br />
ϕ 22<br />
⎫<br />
⎬<br />
⎭<br />
√<br />
µ2<br />
=<br />
⎧<br />
⎨<br />
1<br />
⎫<br />
⎬<br />
⎩<br />
1.404<br />
⎭<br />
√<br />
6.914<br />
Therefore, the modal transformation matrix is<br />
[Φ] = [Φ 1 Φ 2 ]<br />
⎡<br />
⎤<br />
0.925<br />
= ⎣<br />
0.380<br />
⎦<br />
−0.219 0.534<br />
This result can be verified using Matlab’s eig function as follows<br />
EDU>> K=[3 -2;-2 2]; M=[1 0;0 3];<br />
EDU>> [phi,lam]=eig(K,M)<br />
phi =<br />
-0.3803 -0.9249<br />
-0.5340 0.2196<br />
EDU>> diag(sqrt(lam))<br />
0.4380<br />
1.8641<br />
Matlab result agrees with the result obtained above. The sign difference is not important. Now step 5 is<br />
applied. Matlab generates mass normalized eigenvectors by default.<br />
Now that [Φ] is found, the transformation from the normal coordinates {x} to modal coordinates, called<br />
{η} , is obtained<br />
{x} = [Φ] {η}<br />
⎧ ⎫ ⎡<br />
⎤ ⎧ ⎫<br />
⎨x 1 (t) ⎬<br />
⎩<br />
x 2 (t)<br />
⎭ = 0.925 0.380 ⎨η ⎣ ⎦ 1 (t) ⎬<br />
−0.219 0.534<br />
⎩<br />
η 2 (t)<br />
⎭<br />
12
The transformation from modal coordinates back to normal coordinates is<br />
However, [Φ] −1 = [Φ] T [M] therefore<br />
{η} = [Φ] −1 {x}<br />
⎧ ⎫ ⎡<br />
⎤−1 ⎧ ⎫<br />
⎨η 1 (t) ⎬<br />
⎩<br />
η 2 (t)<br />
⎭ = 0.925 0.380 ⎨ x<br />
⎣ ⎦ 1 (t) ⎬<br />
−0.219 0.534<br />
⎩<br />
x 2 (t)<br />
⎭<br />
{η} = [Φ] T [M] {x}<br />
⎧ ⎫ ⎡<br />
⎤T ⎡ ⎤ ⎧ ⎫<br />
⎨η 1 (t) ⎬<br />
⎩<br />
η 2 (t)<br />
⎭ = 0.925 0.380<br />
⎣ ⎦ ⎣ 1 0 ⎨x ⎦ 1 (t) ⎬<br />
−0.219 0.534 0 3<br />
⎩<br />
x 2 (t)<br />
⎭<br />
⎡<br />
⎤ ⎧ ⎫<br />
0.925 −0.657 ⎨x = ⎣ ⎦ 1 (t) ⎬<br />
0.38 1.6<br />
⎩<br />
x 2 (t)<br />
⎭<br />
The next step is to apply this transformation to the original equations of motion in order to decouple them.<br />
Applying step 6 results in<br />
⎡ ⎤ ⎧ ⎫ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫<br />
⎣ 1 0 ⎨η 1<br />
⎦<br />
′′ ⎬<br />
0 1<br />
⎩<br />
η 2<br />
′′ ⎭ + ⎣ ω2 1 0 ⎨η ⎦ 1 ⎬ ⎨<br />
0 ω2<br />
2 ⎩<br />
η<br />
⎭ = 0 ⎬<br />
[Φ]T ⎩<br />
2 sin (5t)<br />
⎭<br />
⎡ ⎤ ⎧ ⎫ ⎡<br />
⎤ ⎧ ⎫ ⎡<br />
⎤T ⎧ ⎫<br />
⎣ 1 0 ⎨η 1<br />
⎦<br />
′′ ⎬<br />
0 1<br />
⎩<br />
η 2<br />
′′ ⎭ + ⎣ 1.8642 0 ⎨η ⎦ 1 ⎬<br />
0 0.438 2 ⎩<br />
η<br />
⎭ = 0.925 0.380 ⎨ 0 ⎬<br />
⎣ ⎦<br />
2 −0.219 0.534<br />
⎩<br />
sin (5t)<br />
⎭<br />
⎡ ⎤ ⎧ ⎫ ⎡<br />
⎤ ⎧ ⎫ ⎧ ⎫<br />
⎣ 1 0 ⎨η 1<br />
⎦<br />
′′ ⎬<br />
0 1<br />
⎩<br />
η ′′ ⎭ + 3. 47 0 ⎨η ⎣ ⎦ 1 ⎬ ⎨<br />
0 0.192<br />
⎩<br />
η<br />
⎭ = −0.219 sin (5t) ⎬<br />
⎩<br />
2 0.534 sin (5t)<br />
⎭<br />
2<br />
The EOM are now decoupled and each EOM can be solved easily as follows<br />
η ′′<br />
1 (t) + 3.47η 1 (t) = −0.219 sin (5t)<br />
η ′′<br />
2 (t) + 0.192η 2 (t) = 0.534 sin (5t)<br />
To solve these EOM’s, the initial conditions in normal coordinates must be transformed to modal coordinates<br />
using the above transformation rules<br />
⎧ ⎫ ⎡<br />
⎤ ⎧ ⎫<br />
⎨η 1 (0) ⎬<br />
⎩<br />
η 2 (0)<br />
⎭ = 0.925 −0.657 ⎨x ⎣ ⎦ 1 (0) ⎬<br />
0.38 1.6<br />
⎩<br />
x 2 (0)<br />
⎭<br />
⎧ ⎫ ⎡<br />
⎤ ⎧ ⎫<br />
⎨η 1 (0) ⎬<br />
⎩<br />
η 2 (0)<br />
⎭ = 0.925 −0.657 ⎨0⎬<br />
⎣ ⎦<br />
0.38 1.6<br />
⎩<br />
1<br />
⎭<br />
⎧ ⎫<br />
⎨−0.657⎬<br />
=<br />
⎩<br />
1.6<br />
⎭<br />
13
and<br />
⎧ ⎫ ⎡<br />
⎤ ⎧ ⎫<br />
⎨η 1 ′ (0) ⎬<br />
⎩<br />
η 2 ′ (0) ⎭ = 0.925 −0.657 ⎨x ⎣ ⎦<br />
′ 1 (0) ⎬<br />
0.38 1.6<br />
⎩<br />
x ′ 2 (0) ⎭<br />
⎡<br />
⎤ ⎧ ⎫<br />
0.925 −0.657 ⎨1.5⎬<br />
= ⎣ ⎦<br />
0.38 1.6<br />
⎩<br />
3<br />
⎭<br />
⎧ ⎫<br />
⎨−0.584⎬<br />
=<br />
⎩<br />
5.37<br />
⎭<br />
Each of these EOM are solved using any of the standard methods. This results in solutions η 1 (t) and η 2 (t) .<br />
Hence the following EOM’s are solved<br />
and also<br />
η ′′<br />
1 (t) + 3.47η 1 (t) = −0.219 sin (5t)<br />
η 1 (0) = −0.657<br />
η ′ 1 (0) = −0.584<br />
η ′′<br />
2 (t) + 0.192η 2 (t) = 0.534 sin (5t)<br />
η 2 (0) = 1.6<br />
η ′ 2 (0) = 5.37<br />
The solutions η 1 (t) , η 2 (t) are found using basic methods shown in other parts of these notes. The last step<br />
is to transform back to normal coordinates by applying step 7<br />
⎧ ⎫ ⎡<br />
⎤ ⎧ ⎫<br />
⎨x 1 (t) ⎬<br />
⎩<br />
x 2 (t)<br />
⎭ = 0.925 0.380 ⎨η ⎣ ⎦ 1 (t) ⎬<br />
−0.219 0.534<br />
⎩<br />
η 2 (t)<br />
⎭<br />
⎧<br />
⎫<br />
⎨ 0.925 η 1 + 0.38η 2 ⎬<br />
=<br />
⎩<br />
0.534 η 2 − 0.219 η<br />
⎭<br />
1<br />
Hence<br />
and<br />
x 1 (t) = 0.925η 1 (t) + 0.38η 2 (t)<br />
x 2 (t) = 0.534η 1 (t) − 0.219η 2 (t)<br />
The above shows that the solution x 1 (t) and x 2 (t) has contributions from both nodal solutions.<br />
2 Fourier series representation of a periodic function<br />
Given a periodic function f (t) with period T then its Fourier series approximation ˜f (t) using N terms is<br />
(<br />
˜f (t) = 1 N<br />
)<br />
2 F ∑<br />
2π<br />
in<br />
0 + Re F n e T t<br />
= 1 2 F 0 + 1 2<br />
= 1 2<br />
N∑<br />
n=−N<br />
n=1<br />
N∑<br />
F n e<br />
n=1<br />
2π<br />
in<br />
F n e T t<br />
in<br />
2π<br />
T t + F ∗ ne<br />
2π<br />
−in T t<br />
14
Where<br />
F n = 2 T<br />
F 0 = 2 T<br />
∫ T<br />
0<br />
∫ T<br />
0<br />
2π<br />
−in<br />
f (t) e T t dt<br />
f (t) dt<br />
Another way to write the above is to use the classical representation using cos and sin. The same coefficients<br />
(i.e. the same series) will result.<br />
˜f (t) = a 0 +<br />
a 0 = 1 T<br />
n=1<br />
∫ T<br />
0<br />
a n = 1<br />
T/2<br />
b n = 1<br />
T/2<br />
N∑<br />
a n cos n 2π N T t + ∑<br />
b n sin n 2π<br />
T t<br />
f (t) dt<br />
∫ T<br />
0<br />
∫ T<br />
0<br />
n=1<br />
(<br />
f (t) cos n 2π )<br />
T t dt<br />
(<br />
f (t) sin n 2π )<br />
T t dt<br />
Just watch out in the above, that we divide by the full period when finding a 0 and divide by half the period<br />
for all the other coefficients. In the end, when we find ˜f (t) we can convert that to complex form. The complex<br />
form seems easier to use.<br />
15
3 Transfer functions for different vibration systems<br />
y<br />
c<br />
m<br />
k<br />
z<br />
u y z<br />
3.1 Force transmissibility<br />
Let steady state<br />
x ss = Re<br />
{ ˆF<br />
k D (r, ζ) eiϖt }<br />
Then<br />
f tr (t) = f spring + f damper<br />
= kx + cx ′<br />
{<br />
= Re k ˆF<br />
} {<br />
k D (r, ζ) eiϖt + Re ciϖ ˆF<br />
}<br />
k D (r, ζ) eiϖt<br />
{(<br />
= Re ˆF + ciϖ ˆF<br />
) }<br />
D (r, ζ) e iϖt<br />
k<br />
Hence<br />
|f tr (t)| max<br />
= ∣ ˆF<br />
√<br />
∣ √<br />
∣ |D| 1 + c 2 ϖ2 ∣∣<br />
k 2 = ˆF ∣ |D| 1 + (2ζr) 2<br />
So TR or force transmissibility is<br />
T R = |f √<br />
tr (t)| ∣ max<br />
∣ ˆF<br />
= |D| 1 + (2ζr) 2<br />
∣<br />
If r > √ 2 then we want small ζ to reduce force transmitted to base. For r < √ 2, it is the other way round.<br />
3.2 vibration isolation<br />
We need transfer function between y and z.<br />
Equation of motion<br />
my ′′ = −c ( y ′ − z ′) − k (y − z)<br />
my ′′ + cy ′ + ky = cz ′ + kz<br />
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} ,<br />
hence the above becomes<br />
Hence |D (r, ζ)| =<br />
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}<br />
√<br />
1+(2ζr) 2<br />
√<br />
(1−r 2 ) 2 +(2ζr) 2<br />
ciω + k<br />
Y =<br />
−ω 2 m + ciω + k Z<br />
i2ζω n mω + k<br />
=<br />
−ω 2 m + i2ζω n mω + k Z<br />
i2ζω n ω + ωn<br />
2 =<br />
−ω 2 + i2ζω n ω + ωn<br />
2 Z<br />
i2ζr + 1<br />
=<br />
(1 − r 2 ) + i2ζr Z<br />
( )<br />
and arg (D) = tan−1 (2ζr) − tan −1 2ζr<br />
1−r 2<br />
where r = ω ω n<br />
√<br />
Hence for good vibration isolation we need |Y | max<br />
|Z|<br />
to be small. i.e. |D| 1 + (2ζr) 2 to be small. This is the<br />
same TR as for force isolation above.<br />
For small |D|, we need small ζ and small k (the small k is to make r > √ 2) see plot<br />
In Matlab, the above can be plotted using<br />
17
close all;<br />
zeta=linspace(0.1, 0.7, 10);<br />
r=linspace(0, 3, 10);<br />
D=@(r,z) (sqrt(1+(2*z*r).^2)./sqrt((1-r.^2).^2+(2*z*r).^2));<br />
figure;<br />
hold on;<br />
for i=1:length(zeta)<br />
plot(r,D(r,zeta(i)));<br />
end<br />
grid on;<br />
legend<br />
3.3 accelerometer<br />
We need transfer function between u and z a where now z a is the amplitude of the ground acceleration.<br />
This device is used to measure base acceleration by relating it linearly to relative displacement of m to base.<br />
Equation of motion. We use relative distance now.<br />
m ( u ′′ + z ′′) + cu ′ + ku = 0<br />
mu ′′ + cu ′ + ku = −mz ′′<br />
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<br />
did not go through the steps as above starting from base motion. This is because we want the transfer function<br />
between relative motion u and acceleration of base.<br />
Now, u = Re { Ue iωt} , u ′ = Re { iωUe iωt} , u ′′ = Re { −ω 2 Ue iωt} , hence the above becomes<br />
Hence |D (r, ζ)| =<br />
m Re { −ω 2 Ue iωt} + c Re { iωUe iωt} + k Re { Ue iωt} = −m Re { Z a e iωt}<br />
−m<br />
U =<br />
−ω 2 m + iωc + k Z a<br />
−1<br />
=<br />
−ω 2 + iω2ζω n + ωn<br />
2 Z a<br />
−1<br />
=<br />
(ωn 2 − ω 2 Z a<br />
) + iω2ζω n<br />
( )<br />
−1<br />
√(ω and arg (D) = −1800 − tan −1 2ωζωn<br />
n−ω 2 2 ) 2 +(2ωζω n) 2 ωn−ω 2 2<br />
When system is very stiff, which means ω n very large compared to ω , then D (r, ζ) ≈ −1<br />
ω 2 n<br />
Z a , hence by<br />
measuring u we estimate Z a the amplitude of the ground acceleration since ω 2 n is known. For accuracy, need<br />
ω n > 5ω at least.<br />
3.4 seismometer<br />
Now we need to measure the base motion (not base acceleration like above). But we still use the relative<br />
displacement. Now the transfer function is between u and z where now z is the base motion amplitude.<br />
Equation of motion. We use relative distance now.<br />
m ( u ′′ + z ′′) + cu ′ + ku = 0<br />
mu ′′ + cu ′ + ku = −mz ′′<br />
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 ′′ =<br />
Re { −ω 2 Ue iωt} , hence the above becomes<br />
18
Now, u = Re { Ue iωt} , u ′ = Re { iωUe iωt} , u ′′ = Re { −ω 2 Ue iωt} , hence the above becomes<br />
m Re { −ω 2 Ue iωt} + c Re { iωUe iωt} + k Re { Ue iωt} = −m Re { −ω 2 Ze iωt}<br />
U =<br />
mω 2<br />
−ω 2 m + iωc + k Z<br />
ω 2<br />
=<br />
−ω 2 + iω2ζω n + ωn<br />
2 Z<br />
r 2<br />
=<br />
(1 − r 2 ) + i2ζr Z<br />
( )<br />
r<br />
Hence |D (r, ζ)| = √ 2<br />
and arg (D) = − 2ζr<br />
(1−r 2 )+i2ζr tan−1 1−r 2 1<br />
Now if r is very large, which happens when ω n ≪ ω, then<br />
(1−r 2 )+i2ζr ⇒ 1 since r 2 is the dominant factor.<br />
−r 2<br />
r<br />
Therefore U = 2<br />
(1−r 2 )+i2ζr Z a now becomes U ≃ −Z a therefore measuring the relative displacement U gives<br />
linear estimate of the ground motion. However, this device requires that ω n be much smaller than ω, which<br />
means that m has to be massive. So this device is heavy compared to accelerometer.<br />
3.5 Summary of vibration transfer functions<br />
For good isolation of mass from ground motion, rule of thumb: Make damping low, and stiffness low (soft spring).<br />
system equation used to derive transfer function<br />
isolate base from force<br />
transmitted by machine<br />
isolate machine<br />
from motion of base<br />
accelerometer: Measure<br />
base acc. using relative<br />
displacement<br />
f tr (t) = f spring + f damper<br />
Use absolute mass position<br />
my ′′ + cy ′ + ky = cz ′ + kz<br />
|f tr(t)| ∣ max<br />
∣ ˆF<br />
∣<br />
|Y | max<br />
|Z|<br />
√<br />
= |D| 1 + (2ζr) 2<br />
√<br />
= |D| 1 + (2ζr) 2<br />
−1<br />
U =<br />
Use relative mass position (ωn 2 − ω 2 Z a ⇒ |D (r, ζ)|<br />
) + iω2ζω n<br />
−1<br />
mu ′′ + cu ′ + ku = −mz ′′ = √<br />
(ωn 2 − ω 2 ) 2 + (2ωζω n ) 2<br />
seismometer: Measure<br />
base motion using<br />
relative displacement<br />
Use relative mass position<br />
mu ′′ + cu ′ + ku = −mz ′′ U =<br />
r 2<br />
(1−r 2 )+i2ζr Z ⇒ |D (r, ζ)| = r 2 √<br />
(1−r 2 )+i2ζr<br />
19
4 Steady state solutions to single degree freedom system<br />
4.1 Summary:<br />
( )<br />
my ′′ + cy ′ + ky = Re ˆF e<br />
iϖt<br />
{ }<br />
x = Re ˆXe<br />
iϖt<br />
ˆX = ˆF D (r, ζ)<br />
k<br />
1<br />
D (r, ζ) =<br />
(1 − r 2 ) + 2iζr<br />
{ }<br />
ˆF<br />
x = Re<br />
k |D (r, ζ)| ei(ϖt−θ)<br />
θ = tan −1<br />
2ζr<br />
1 − r 2<br />
4.2 Details<br />
( )<br />
Let load be harmonic and represented in general as Re ˆF e<br />
iϖt<br />
where ˆF is the complex amplitude of the force.<br />
Hence system is represented by<br />
( )<br />
my ′′ + cy ′ + ky = Re ˆF e<br />
iϖt<br />
( ) ˆF<br />
y ′′ + 2ζω n y ′ + ωny 2 = Re<br />
m eiϖt<br />
)<br />
)<br />
)<br />
Let y = Re<br />
(Ŷ e<br />
iϖt<br />
Hence y ′ = Re<br />
(iϖŶ eiϖt , y ′′ = Re<br />
(−ϖ 2 Ŷ e iϖt , therefore the differential equation<br />
becomes<br />
(<br />
Re −ϖ 2 Ŷ e iϖt) (<br />
+ 2ζω n Re iϖŶ eiϖt) + ωn 2 Re<br />
(Ŷ ) ( )<br />
e<br />
iϖt ˆF<br />
= Re<br />
m eiϖt<br />
Dividing numerator and denominator ω 2 n gives<br />
Where r = ϖ ω n<br />
, hence the response is<br />
Therefore, the phase of the response is<br />
y = Re<br />
( )<br />
arg (y) = arg ˆF<br />
( ˆF<br />
−ϖ 2 + 2ζω n iϖ + ωn) 2 Ŷ =<br />
m<br />
ˆF<br />
m<br />
Ŷ =<br />
(−ϖ 2 + 2ζω n iϖ + ωn)<br />
2<br />
Ŷ = ˆF 1<br />
k (1 − r 2 ) + i2ζr<br />
( )<br />
ˆF 1<br />
k (1 − r 2 ) + i2ζr eiϖt<br />
( 2ζr<br />
− tan −1 (1 − r 2 )<br />
20<br />
)<br />
+ ϖt
Hence at t = 0 the phase of the response will be<br />
( ) ( ) 2ζr<br />
arg (y) = arg ˆF − tan −1 (1 − r 2 )<br />
So when ˆF<br />
( )<br />
is real, the phase of the response is simply − tan −1 2ζr<br />
(1−r 2 )<br />
4.3 Undamped case<br />
When ζ = 0 the above becomes<br />
For real force this becomes<br />
∣<br />
The magnitude ∣Ŷ ∣ = F k<br />
4.4 damped cases<br />
ζ > 0<br />
1<br />
(1−r 2 )<br />
( ˆF<br />
y = Re<br />
k<br />
∣ ˆF<br />
∣<br />
=<br />
k<br />
)<br />
1<br />
(1 − r 2 ) eiϖt<br />
1<br />
(<br />
(1 − r 2 ) cos<br />
y = F k<br />
and phase zero.<br />
∣<br />
∣Ŷ<br />
y = Re<br />
( ))<br />
ϖt + arg ˆF<br />
1<br />
(1 − r 2 cos (ϖt)<br />
)<br />
( )<br />
ˆF 1<br />
k (1 − r 2 ) + i2ζr eiϖt<br />
∣ ˆF<br />
∣ 1<br />
∣ = √<br />
k<br />
(1 − r 2 ) 2 + (2ζr) 2<br />
) (<br />
arg<br />
(Ŷ = φ = arg ˆF<br />
Hence for real force and at t = 0 the phase of displacement is<br />
( ) 2ζr<br />
− tan −1 1 − r 2<br />
)<br />
− tan −1 ( 2ζr<br />
1 − r 2 )<br />
+ ϖt<br />
lag behind the load.<br />
When r < 1 then φ goes from 0 to −90 0 Therefore phase of displacement is 0 to −90 0 behind force. The<br />
minus sign at the front was added since the complex number is in the denominator. Hence the response will<br />
always be lagging in phase relative for load.<br />
For r > 1<br />
Now 1 − r 2 is negative, hence the phase will be from −90 ◦ to −180 ◦<br />
When r = 1<br />
( ) ˆF 1<br />
y = Re<br />
k i2ζ eiϖt<br />
∣ ∣<br />
∣Ŷ<br />
)<br />
arg<br />
(Ŷ<br />
∣ ˆF<br />
∣<br />
∣ =<br />
k<br />
= −90 ◦<br />
1<br />
2ζ<br />
Now phase is −90 ◦ 21
Phase of response complex<br />
amplitude for underdamped<br />
and when r1<br />
Phase will be from -90 to -<br />
180 degrees<br />
Phase of response complex<br />
amplitude for<br />
underdamped and when<br />
r=1<br />
Phase will -90 degrees<br />
2r<br />
2r<br />
1 r 2<br />
1 r 2<br />
Examples. System has ζ =<br />
)<br />
0.1 and<br />
√<br />
m = 1, k = 1 subjected for force 3 cos (0.5t) find the steady state solution.<br />
Answer y (t) = Re<br />
(Ŷ e<br />
iϖt<br />
, ω n = = 1 rad/sec, hence r = 0.5 under the response is<br />
k<br />
m<br />
(∣ ∣∣ y (t) = Re Ŷ ∣ e iϖt)<br />
∣<br />
= ∣Ŷ ∣ cos (ϖt)<br />
= F k<br />
1<br />
√(1 − r 2 ) 2 + (2ζr) 2 cos<br />
(<br />
( ))<br />
2 (0.1) 0.5<br />
.5t − tan −1 1 − 0.5 2<br />
1<br />
= 3<br />
cos (.5t − 7.59 √(1 ◦ )<br />
− 0.5 2 ) 2 + (2 (0.1) 0.5) 2<br />
= 3. 964 9 cos (.5t − 7.59 ◦ )<br />
5 Summary table of solutions<br />
5.1 Harmonic loading mu ′′ + cu ′ + ku = F sin ϖt<br />
The equation of motion can also be written as u ′′ + 2ζωu ′ + ω 2 u = F m<br />
sin ϖt.<br />
22
The following table gives the solutions for initial conditions are u (0) and u ′ (0) under all damping conditions.<br />
The roots shown are the roots of the quadratic characteristic equation λ 2 + 2ζωλ + ω 2 λ = 0. Special handling is<br />
needed to obtain the solution of the differential equation for the case of ζ = 0 and ϖ = ω as described in the<br />
detailed section below.<br />
⎧<br />
⎨ −iω<br />
roots<br />
ζ = 0<br />
⎩<br />
⎧<br />
⎪⎨<br />
u (t)<br />
⎪⎩<br />
⎧<br />
⎨<br />
roots<br />
⎩<br />
+iω<br />
ϖ = ω → u (0) cos ϖt + u′ (0)<br />
ϖ<br />
sin ϖt − F K<br />
ϖ ≠ ω → u (0) cos ωt +<br />
−ξω + iω n<br />
√<br />
1 − ξ 2<br />
−ξω − iω n<br />
√<br />
1 − ξ 2<br />
(<br />
u ′ (0)<br />
ω<br />
− F K<br />
ϖt<br />
2<br />
cos (ϖt)<br />
)<br />
r<br />
sin ωt + F 1<br />
1−r 2 K<br />
sin ϖt<br />
1−r 2<br />
ζ < 1<br />
u (t) = e −ξωt (A cos ω d t + B sin ω d t) + F K<br />
A = u 0 + F K<br />
1 √(1−r 2 ) 2 +(2ξr) 2 sin θ<br />
1<br />
sin (ϖt − θ)<br />
√(1−r 2 ) 2 2<br />
+(2ξr)<br />
B = v 0<br />
ω d<br />
+ u 0ξω<br />
ω d<br />
⎧<br />
⎨ −ω<br />
roots<br />
⎩<br />
−ω<br />
+ F √ 1<br />
K<br />
ω d (1−r 2 ) 2 +(2ξr)<br />
2<br />
(ξω sin θ − ϖ cos θ)<br />
ζ = 1<br />
ζ > 1<br />
u (t) = (A + Bt) e −ωt + F K<br />
A = u 0 + F K<br />
1 √(1−r 2 ) 2 +(2r) 2 sin θ<br />
1<br />
sin (ϖt − θ)<br />
√(1−r 2 ) 2 2<br />
+(2r)<br />
F/k<br />
B = v 0 + u 0 ω +<br />
(ω sin θ − ϖ cos θ)<br />
√(1−r 2 ) 2 2<br />
+(2r)<br />
⎧<br />
√<br />
⎨ −ω n ξ + ω n ξ 2 − 1<br />
roots<br />
⎩<br />
√<br />
−ω n ξ − ω n ξ 2 − 1<br />
(<br />
−ξ+ √ (<br />
ξ<br />
u (t) = Ae<br />
2 −1<br />
)ω nt −ξ− √ )<br />
ξ + Be 2 −1 ω nt +<br />
F<br />
K<br />
β sin (ϖt − θ)<br />
A = v 0+u 0 ωξ+u 0 ω √ ((<br />
ξ 2 −1+ F K β ξ+ √ )<br />
)<br />
ξ 2 −1 ω sin θ−ϖ cos θ<br />
2ω √ ξ 2 −1<br />
B = − v 0+u 0 ωξ−u 0 ω √ ((<br />
ξ 2 −1+ F K β ξ− √ )<br />
)<br />
ξ 2 −1 ω sin θ−ϖ cos θ<br />
2ω √ ξ 2 −1<br />
23
5.2 constant loading mu ′′ + cu ′ + ku = F<br />
⎧<br />
⎨ −iω<br />
roots<br />
ζ = 0 ⎩<br />
+iω<br />
ζ < 1<br />
ζ = 1<br />
u (t) = ( u 0 − F )<br />
k cos ωt +<br />
v 0<br />
ω<br />
sin ωt + F k<br />
⎧<br />
√<br />
⎨ −ξω + iω n 1 − ξ 2<br />
roots<br />
⎩<br />
√<br />
−ξω − iω n 1 − ξ 2<br />
( (u0<br />
u (t) = e −ξωt − F )<br />
k cos ωd t +<br />
⎧<br />
⎨ −ω<br />
roots<br />
⎩<br />
−ω<br />
(<br />
) )<br />
v 0 +u 0 ξω− F k ξω<br />
ω d<br />
sin ω d t + F k<br />
u (t) = (( u 0 − F ) (<br />
k + v0 + u 0 ω − F k ω) t ) e −ωt + F k<br />
⎧<br />
√<br />
⎨ −ω n ξ + ω n ξ 2 − 1<br />
roots<br />
⎩<br />
√<br />
−ω n ξ − ω n ξ 2 − 1<br />
ζ > 1<br />
B = F k p 1−u 0 p 1 +v 0<br />
(p 2 −p 1 )<br />
A = u 0 − F k − B<br />
p 1 = − c<br />
2m + √ ( c<br />
2m<br />
p 2 = − c<br />
2m − √ ( c<br />
2m<br />
u (t) = Ae p 1t + Be p 2t + F k<br />
) 2 −<br />
k<br />
m = −ω √<br />
nξ + ω n ξ 2 − 1<br />
) 2 −<br />
k<br />
m = −ω √<br />
nξ − ω n ξ 2 − 1<br />
24
5.3 no loading (free) mu ′′ + cu ′ + ku = 0<br />
⎧<br />
⎨ −iω<br />
roots<br />
⎩<br />
+iω<br />
ζ = 0<br />
u ′′ + ω 2 u = 0<br />
ζ < 1<br />
ζ = 1<br />
u (t) = u(0) cos ωt + u′ (0)<br />
ω<br />
sin ωt<br />
⎧<br />
u (t) = A cos (ωt − φ)<br />
⎪⎨ √<br />
( )<br />
A = u(0) + u ′ 2<br />
(0)<br />
ω<br />
( )<br />
⎪⎩ φ = tan −1 u ′ (0)/ω<br />
u(0)<br />
⎧<br />
⎨ −ξω + iω √ 1 − ξ 2<br />
roots<br />
⎩<br />
−ξω − iω √ 1 − ξ 2<br />
(<br />
u (t) = e −ξωt u(0) cos ω d t + u′ (0)+u(0)ξω<br />
ω<br />
⎧<br />
d<br />
⎨ −ω<br />
roots<br />
⎩<br />
−ω<br />
)<br />
sin ω d t<br />
ζ > 1<br />
u (t) = (u(0)(1 + ωt) + u ′ (0)t) e −ωt<br />
⎧<br />
⎨ λ 1 = −ωξ + ω √ ξ 2 − 1<br />
roots<br />
⎩<br />
λ 2 = −ωξ − ω √ ξ 2 − 1<br />
⎧<br />
u (t) = Ae<br />
⎪⎨<br />
λ1ωt + Be λ 2ωt<br />
A = u′ (0)−u(0)λ 2<br />
2ω √ ξ 2 −1<br />
⎪⎩<br />
B = −u′ (0)+u(0)λ 1<br />
2ω √ ξ 2 −1<br />
25
5.4 impulse F 0 δ (t) loading<br />
⎧<br />
⎨ −iω<br />
roots<br />
ζ = 0<br />
⎩<br />
+iω<br />
u ′′ + ω 2 u = 0<br />
ζ < 1<br />
ζ = 1<br />
ζ > 1<br />
transient<br />
{ }} {<br />
u (t) = u(0) cos ωt + u′ (0)<br />
⎧<br />
ω<br />
⎨ −ξω + iω √ 1 − ξ 2<br />
roots<br />
⎩<br />
−ξω − iω √ 1 − ξ 2<br />
steady state<br />
{ }} {<br />
sin ωt + F 0<br />
sin ωt<br />
mω<br />
transient<br />
steady state<br />
{ (<br />
}} ){<br />
{ ( }} ){<br />
u (t) = e −ξωt u(0) cos ω d t + u′ (0) + u(0)ξω<br />
sin ω d t + e −ξωt F0<br />
sin ω d t<br />
⎧<br />
ω d<br />
mω d<br />
⎨ −ω<br />
roots<br />
⎩<br />
−ω<br />
u (t) = (u (0) (1 + ωt) + u ′ (0) t) e −ωt + F 0t<br />
⎧<br />
⎨ λ 1 = −ωξ + ω √ ξ 2 − 1<br />
roots<br />
⎩<br />
λ 2 = −ωξ − ω √ ξ 2 − 1<br />
⎧<br />
u (t) = Ae<br />
⎪⎨<br />
λ1ωt + Be λ2ωt +<br />
⎪⎩<br />
A = u′ (0)−u(0)λ 2<br />
2ω √ ξ 2 −1<br />
B = −u′ (0)+u(0)λ 1<br />
2ω √ ξ 2 −1<br />
m e−ωt<br />
F 0<br />
m<br />
2ω √ ξ 2 −1 eλ 1ωt −<br />
The impulse response can be implemented in Mathematica as<br />
parms = {m -> 10, c -> 1.2, k -> 4.3, a -> 1};<br />
tf = TransferFunctionModel[a/(m s^2 + c s + k) /. parms, s]<br />
sol = OutputResponse[tf, DiracDelta[t], t];<br />
F 0<br />
m<br />
2ω √ ξ 2 −1 eλ 2ωt<br />
Plot[sol, {t, 0, 60}, PlotRange -> All, Frame -> True,<br />
FrameLabel -> {{z[t], None}, {Row[{t, " (sec)"}], eq}},GridLines -> Automatic]<br />
26
5.5 Impulse sin function<br />
Input is given by F (t) = F 0 sin (ϖt) where ϖ = 2π<br />
2t 1<br />
= π t 1<br />
ζ = 0<br />
ζ < 1<br />
ζ = 1<br />
⎧<br />
⎨ −iω<br />
roots<br />
⎩<br />
+iω<br />
⎧<br />
⎧<br />
⎪⎨<br />
ϖ = ω →<br />
(<br />
⎪⎨<br />
⎪⎩ −u (0) +<br />
F 0 π<br />
k 2<br />
⎧<br />
(<br />
u (t)<br />
⎪⎨ u (0) cos ωt +<br />
ϖ ≠ ω →<br />
⎪⎩ ⎪⎩<br />
⎧<br />
√<br />
⎨ −ξω + iω n 1 − ξ 2<br />
roots<br />
⎩<br />
√<br />
−ξω − iω n 1 − ξ 2<br />
⎧<br />
⎪⎨<br />
u (t) =<br />
⎪⎩<br />
u (0) cos ωt + u′ (0)<br />
ω sin ωt − F 0<br />
k<br />
ωt<br />
)<br />
cos ωt +<br />
−u ′ (0)+ F 0<br />
k<br />
u ′ (0)<br />
ϖ<br />
( −<br />
(F π/t1<br />
0/k) ω<br />
π/t1<br />
1−(<br />
ω<br />
2 cos (ωt) 0 ≤ t ≤ t 1<br />
π<br />
2t 1<br />
))<br />
) 2<br />
ω<br />
sin ωt t > t 1<br />
sin ωt + F 0/k<br />
1−( π/t1<br />
ω<br />
( )<br />
) 2 sin πt<br />
t 1<br />
0 ≤ t ≤ t 1<br />
u (t 1 ) cos ωt + u′ (t 1 )<br />
ω<br />
sin ωt t > t 1<br />
time constant τ = 1<br />
ζω n<br />
e −ξωt (A cos ω d t + B sin ω d t) + F 0<br />
√ 1<br />
k<br />
(<br />
e −ξωt u (t 1 ) cos ω d t + u′ (t 1 )+u(t 1 )ξω<br />
ω d<br />
sin ω d t<br />
A = u (0) + F 0<br />
k<br />
1 √(1−r 2 ) 2 +(2ξr) 2 sin θ<br />
B = u′ (0)<br />
ω d<br />
⎧<br />
⎨<br />
roots<br />
+ u(0)ξω<br />
ω d<br />
+ F 0<br />
−ω<br />
√ 1<br />
k<br />
ω d (1−r 2 ) 2 +(2ξr)<br />
(<br />
sin<br />
(1−r 2 ) 2 +(2ξr)<br />
)<br />
2<br />
2<br />
(ξω sin θ − ϖ cos θ)<br />
)<br />
t 1<br />
− θ<br />
⎩<br />
−ω<br />
⎧<br />
( )<br />
⎪⎨ (A + Bt) e −ωt + F 0 √ 1<br />
sin k<br />
π t<br />
u (t) =<br />
(1−r 2 ) 2 +(2r) 2 t 1<br />
− θ 0 ≤ t ≤ t 1<br />
⎪⎩<br />
u (t) = (u (t 1 ) + (u ′ (t 1 ) + u (t 1 ) ω) t) e −ωt t > t 1<br />
π t<br />
0 ≤ t ≤ t 1<br />
t > t 1<br />
A = u (0) + F 0<br />
k<br />
1 √(1−r 2 ) 2 +(2r) 2 sin θ<br />
B = u ′ (0) + u (0) ω + F 0<br />
(<br />
√ 1<br />
k<br />
(1−r 2 ) 2 +(2r) 2<br />
)<br />
ω sin θ − π t 1<br />
cos θ<br />
27
ζ > 1<br />
⎧<br />
√<br />
⎨ −ω n ξ + ω n ξ 2 − 1<br />
roots<br />
⎩<br />
√<br />
−ω n ξ − ω n ξ 2 − 1<br />
⎧ (<br />
⎪⎨<br />
−ξ+ √ (<br />
ξ<br />
Ae<br />
2 −1<br />
)ωt + Be<br />
u (t) =<br />
(<br />
⎪⎩<br />
−ξ+ √ )<br />
ξ 2 −1<br />
ω<br />
u (t) = A 1 e<br />
nt + B1 e<br />
A = u′ (0)+u(0)ωξ+u(0)ω √ ((<br />
ξ 2 −1+ F k β ξ+ √ )<br />
ξ 2 −1<br />
2ω √ ξ 2 −1<br />
B = − u′ (0)+u(0)ωξ−u(0)ω √ ξ 2 −1+ F k β ((<br />
ξ− √ ξ 2 −1<br />
1<br />
β = √<br />
(1−r 2 ) 2 +(2ξr) 2<br />
(<br />
A 1 = − ˙u(t 1)+u(t 1 )ω n ξ− √ )<br />
ξ 2 −1<br />
√<br />
2ω n ( ξ 2 −1<br />
B 1 = ˙u(t 1)+u(t 1 )ξω n ξ+ √ )<br />
ξ 2 −1<br />
2ω n<br />
√<br />
ξ 2 −1<br />
2ω √ ξ 2 −1<br />
−ξ− √ )<br />
ξ 2 −1 ωt +<br />
F<br />
π t<br />
(<br />
−ξ− √ )<br />
ξ 2 −1 ω nt<br />
k β sin (<br />
)<br />
ω sin θ− π cos θ t 1<br />
)<br />
)<br />
ω sin θ− π cos θ t 1<br />
)<br />
t 1<br />
− θ<br />
0 ≤ t ≤ t 1<br />
t > t 1<br />
6 Tree view look at the different cases<br />
This tree illustrates the different cases that needs to be considered for the solution of single degree of freedom<br />
system with harmonic loading.<br />
There are 12 cases to consider. Resonance needs to be handled as special case when damping is absent due to<br />
the singularity in the standard solution when the forcing frequency is the same as the natural frequency. When<br />
damping is present, there is no resonance, however, there is what is called practical response which occur when<br />
the forcing frequency is almost the same as the natural frequency.<br />
28
Solution to single degree of freedom system, second order, linear, time invariant<br />
By Nasser M. Abbasi<br />
December 11, 2012<br />
Single_degree_system_tree.vsd<br />
undamped<br />
damped<br />
r <br />
<br />
c<br />
c cr<br />
free<br />
mü ku 0<br />
forced<br />
mü ku Fsint<br />
free<br />
forced<br />
mü cu ku 0<br />
mü cu ku Fsint<br />
1 1 1<br />
r 1 r 1<br />
r 1 r 1<br />
1<br />
1<br />
1 1 1<br />
1<br />
r 1 resonance<br />
1 underdamped, roots a ib with a real and negative<br />
1 critical damping roots a,a<br />
1 overdamped roots both real and negative a,b<br />
x<br />
x<br />
x<br />
xx<br />
The following is another diagram made sometime ago which contains more useful information and is kept<br />
here for reference.<br />
29
One degree freedom<br />
linear system unforced<br />
response<br />
y t c m y t k m yt 0<br />
y t 2 n y t n2 yt 0<br />
Some<br />
defintions<br />
k<br />
n m rad/sec<br />
damping ratio<br />
c c c<br />
r<br />
c<br />
2 km 2 n m<br />
Solution roots<br />
s 1,2 n n 2 1<br />
3 cases<br />
d n 1 2 30<br />
Defined only<br />
for under<br />
damped<br />
case(else zero)<br />
1 1 1<br />
Underdamped<br />
2 roots,<br />
complex<br />
conjugate<br />
Critical damped<br />
(one real root,<br />
double<br />
multiplicity)<br />
Over damped,<br />
2 real roots,<br />
distinct<br />
yt e nt Acos d t Bsin d t<br />
A y0, B y 0A n<br />
d<br />
yt A Bte nt<br />
A y0, B y 0 A n<br />
T d 2<br />
d<br />
1<br />
n<br />
Nasser M. Abbasi<br />
May 25, 2011<br />
One_DOF_system.vsd<br />
Damped period of oscillation<br />
Time constant<br />
yt Ae 2 1 n t Be 2 1 n t<br />
A y 0y0 n<br />
2 n<br />
B y 0y0 n<br />
2 n<br />
2 1<br />
2 1<br />
2 1<br />
2 1
7 Some common formulas<br />
These definitions are used throughout the derivations below.<br />
ξ = c c r<br />
=<br />
c<br />
2 √ km = c<br />
2ω n m<br />
u st = F static deflection<br />
√<br />
k<br />
k<br />
ω n =<br />
m<br />
ω D = ω n<br />
√<br />
1 − ξ 2 note: not defined for ξ > 1 since becomes complex<br />
r = ϖ ω n<br />
T d = 2π damped period of oscillation<br />
ω<br />
{ d<br />
−1<br />
τ = , −1 }<br />
time constants where λ i are roots of characteristic equation<br />
λ 1 λ2<br />
1<br />
β =<br />
magnification factor<br />
√(1 − r 2 ) 2 + (2rξ) 2<br />
β max when r = √ 1 − 2ξ 2<br />
1<br />
β max =<br />
2ξ √ 1 − ξ 2<br />
y n<br />
= e ζωn2π<br />
ω D<br />
y<br />
( n+1<br />
)<br />
yn<br />
ln = ζ2π<br />
y n+1<br />
( )<br />
1<br />
M ln yn<br />
= ζ2π<br />
y n+M<br />
small damping<br />
⇒<br />
√<br />
ζ2π<br />
1−ζ 2<br />
e<br />
⇒ e ζ2π<br />
This table shows many cycles it takes for the peak to decay by half its original value as a function of the<br />
damping ζ. For example, we see that when ζ = 2.7% then it takes 4 cycles for the peak (i.e. displacement) to<br />
reduce to half its value.<br />
data = Table[{i, (1/i Log[2]/(2*Pi)*100)}, {i, 1, 20}];<br />
TableForm[N@data,<br />
TableHeadings -> {None, {Column[{"number of cycles",<br />
"needed for peak", "to decay by half"}], "\[Zeta] (%)"}}]<br />
31
The roots of the characteristic equation for u ′′ + 2ξωu ′ + ω 2 u = 0 are given in this table<br />
ξ < 1<br />
roots<br />
{ √ √ }<br />
−ξω + jω n 1 − ξ 2 , −ξω − iω n 1 − ξ 2<br />
time constant τ<br />
1<br />
ξω<br />
1<br />
ξ = 1 {−ω, −ω}<br />
ω<br />
{ √ √ }<br />
ξ > 1 −ω n ξ + ω n ξ 2 − 1, −ω n ξ − ω n ξ 2 1<br />
− 1<br />
√ , 1√ (which to use? the bigger?)<br />
ω nξ−ω n ξ 2 −1 ω nξ+ω n ξ 2 −1<br />
32
8 Derivation of the solutions of the equations of motion<br />
8.1 Free vibration<br />
8.1.1 Undamped free vibration<br />
u<br />
K<br />
M<br />
mu ′′ + ku = 0<br />
u ′′ + ω 2 u = 0<br />
The solution is<br />
u (t) = u (0) cos ωt + u′ (0)<br />
ω<br />
sin ωt<br />
8.1.2 under-damped free vibration c < c r , ξ < 1<br />
u<br />
K<br />
M<br />
C<br />
The solution is<br />
mu ′′ + cu ′ + ku = 0<br />
u ′′ + 2ξωu ′ + ω 2 u = 0<br />
u = e −ξωt (A cos ω d t + B sin ω d t)<br />
Applying initial conditions gives A = u (0) and B = u′ (0)+u(0)ξω<br />
ω d<br />
. Therefore the solution becomes<br />
(<br />
)<br />
u (t) = e −ξωt u (0) cos ω d t + u′ (0) + u (0) ξω<br />
sin ω d t<br />
ω d<br />
33
8.1.3 critically damped free vibration ξ = c<br />
c r<br />
= 1<br />
The solution is<br />
u (t) = (A + Bt) e −( cr<br />
2m<br />
)<br />
t<br />
= (A + Bt) e −ωt<br />
where A, B are found from initial conditions A = u (0),B = u ′ (0) + u (0) ω, hence<br />
8.1.4 over-damped free vibration ξ = c<br />
c r<br />
> 1<br />
The solution is<br />
where A, B are found from initial conditions.<br />
u (t) = ( u (0) + ( u ′ (0) + u (0) ω ) t ) e −ωt<br />
u (t) = Ae λ 1t + Be λ 2t<br />
A = u′ (0) − u (0) λ 2<br />
2ω √ ξ 2 − 1<br />
B = −u′ (0) + u (0) λ 1<br />
2ω √ ξ 2 − 1<br />
where λ 1 and λ 2 are the roots of the characteristic equation<br />
8.2 Constant input F<br />
8.2.1 Undamped case ζ = 0<br />
λ 1 = − c<br />
√ ( c<br />
2m + 2m<br />
λ 2 = − c<br />
2m − √ ( c<br />
2m<br />
) 2 k −<br />
m = −ξω + ω√ ξ 2 − 1<br />
) 2 k −<br />
m = −ξω − ω√ ξ 2 − 1<br />
mu ′′ + ku = F<br />
u ′′ + ω 2 u = F<br />
u (t) = u h + u p<br />
Where u h = A cos ωt + B sin ωt and u p = F k<br />
, the solution is<br />
Applying initial conditions gives<br />
u (t) = A cos ωt + B sin ωt + F k<br />
And complete solution is<br />
A = u (0) − F k<br />
B = u′ (0)<br />
ω<br />
u (t) = F (<br />
k + u (0) − F )<br />
cos ωt + u′ (0)<br />
sin ωt<br />
k<br />
ω<br />
34
8.2.2 under-damped ζ < 1<br />
The general solution is<br />
u (t) = e −ξωt (A cos ω d t + B sin ω d t) + F k<br />
From initial conditions<br />
Hence the solution is<br />
u (t) = e −ξωt ( (<br />
u (0) − F k<br />
A = u (0) − F k<br />
B = u′ (0) + u (0) ξω − F k ξω<br />
ω d<br />
) (<br />
u ′ (0) + u (0) ξω − F k<br />
cos ω d t +<br />
ξω<br />
) )<br />
sin ω d t + F ω d k<br />
8.2.3 Critical damping ζ = 1<br />
The general solution is<br />
u(t) = (A + Bt)e −ωt + F k<br />
Where from initial conditions<br />
A = u(0) − F k<br />
B = u ′ (0) + u(0)ω − F k ω<br />
8.2.4 Over-damped ζ > 0<br />
The solution is<br />
u (t) = Ae p1t + Be p2t + F k<br />
Where now<br />
B =<br />
F<br />
k p 1 − u 0 p 1 + u ′ (0)<br />
(p 2 − p 1 )<br />
A = u (0) − F k − B<br />
Hence the solution is<br />
u (t) = Ae p 1t + Be p 2t + F k<br />
Where<br />
p 1 = − c<br />
√ ( c<br />
2m + 2m<br />
p 2 = − c<br />
2m − √ ( c<br />
2m<br />
) 2 k −<br />
m = −ωξ + ω √<br />
n ξ 2 − 1<br />
) 2 k −<br />
m = −ωξ − ω √<br />
n ξ 2 − 1<br />
35
8.3 Harmonic forced input F sin (ϖt)<br />
8.3.1 Undamped forced vibration<br />
u<br />
K<br />
M<br />
ReF e it <br />
mu ′′ + ku = F sin ϖt<br />
Since there is no damping in the system, then there is no steady state solution. In other words, the particular<br />
solution is not the same as the steady state solution in this case. We need to find the particular solution using<br />
method on undetermined coefficients.<br />
Let u = u h + u p . By guessing that u p = c 1 sin ϖt then we find the solution to be<br />
u = A cos ω n t + B sin ω n t + F 1<br />
sin ϖt<br />
k 1 − r2 Applying initial conditions is always done on the full solution. Applying initial conditions gives<br />
u (0) = A<br />
u ′ (t) = −Aω sin ω n t + Bω cos ω n t + ϖ F 1<br />
cos ϖt<br />
k 1 − r2 u ′ (0) = Bω n + ϖ F 1<br />
k 1 − r 2<br />
B = u′ (0)<br />
− F r<br />
ω n k 1 − r 2<br />
Where r = ϖ ω n<br />
The complete solution is<br />
( u ′ (0)<br />
u (t) = u (0) cos ω n t + − F k<br />
ω n<br />
r<br />
1 − r 2 )<br />
sin ω n t + F k<br />
1<br />
sin ϖt (1)<br />
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<br />
rad/sec. Hence r = 5, Let initial conditions be zero, then<br />
( )<br />
5<br />
1<br />
u = −3<br />
1 − 5 2 sin t + 3 sin 5t<br />
1 − 52 = 0.625 sin t − 0.125 sin 5.0t<br />
Resonance forced vibration When ϖ ≈ ω we obtain resonance since r → 1 in the solution given in Eq (1)<br />
above and as written the solution can not be used for analysis. To obtain a solution for resonance some calculus<br />
is needed. Eq (1) is written as<br />
( u ′ (0)<br />
u (t) = u (0) cos ωt +<br />
ω<br />
− F k<br />
ωϖ<br />
ω 2 − ϖ 2 )<br />
sin ωt + F k<br />
ω 2<br />
ω 2 sin ϖt<br />
− ϖ2 (1A)<br />
36
When ϖ ≈ ω but less than ω, letting<br />
where ∆ is very small positive quantity. And since ϖ ≈ ω let<br />
Multiplying Eq (2) and (3) gives<br />
Eq (1A) can now be written in terms of Eqs (2,3) as<br />
( u ′ (0)<br />
u (t) = u (0) cos ωt +<br />
ω<br />
(<br />
v0<br />
= u (0) cos ωt +<br />
ω − F k<br />
ω − ϖ = 2∆ (2)<br />
ω + ϖ ≈ 2ϖ (3)<br />
ω 2 − ϖ 2 = 4∆ϖ (4)<br />
− F ωϖ<br />
k 4∆ϖ<br />
ω<br />
4∆<br />
Since ϖ ≈ ω the above becomes<br />
( u ′ (0)<br />
u (t) = u (0) cos ωt +<br />
ω<br />
− F k<br />
= u (0) cos ωt + u′ (0)<br />
ω<br />
sin ωt + F k<br />
)<br />
sin ωt + F k<br />
)<br />
sin ωt + F k<br />
)<br />
ω<br />
sin ωt + F 4∆ k<br />
Using sin ϖt − sin ωt = 2 sin ( ϖ−ω<br />
2<br />
t ) cos ( ϖ+ω<br />
2<br />
t ) the above becomes<br />
u (t) = u (0) cos ωt + u′ (0)<br />
ω<br />
From Eqs (2,3) the above can be written as<br />
sin ωt + F k<br />
u (t) = u (0) cos ωt + u′ (0)<br />
ω<br />
ω<br />
2∆<br />
sin ωt + F k<br />
(<br />
sin<br />
ω 2<br />
sin ϖt<br />
4∆ϖ<br />
ω 2<br />
sin ϖt<br />
4∆ϖ<br />
ω<br />
sin ϖt<br />
4∆<br />
ω<br />
(sin ϖt − sin ωt)<br />
4∆<br />
( ϖ − ω<br />
2<br />
) ( )) ϖ + ω<br />
t cos t<br />
2<br />
ω<br />
(sin (−∆t) cos (ϖt))<br />
2∆<br />
Since lim ∆→0<br />
sin(∆t)<br />
∆<br />
= t the above becomes<br />
This is the solution to use for resonance.<br />
u (t) = u (0) cos ωt + u′ (0)<br />
ω<br />
sin ωt − F k<br />
ωt<br />
2<br />
cos (ωt)<br />
37
8.3.2 Under-damped forced vibration c < c r , ξ < 1<br />
u<br />
K<br />
C<br />
M<br />
Fsin t<br />
mu ′′ + cu ′ + ku = F sin ϖt<br />
u ′′ + 2ξωu ′ + ω 2 u = F sin ϖt<br />
m<br />
The solution is<br />
where<br />
and<br />
u (t) = u h + u p<br />
u h (t) = e −ξωt (A cos ω d t + B sin ω d t)<br />
u p (t) =<br />
F<br />
√(k − mϖ) 2 + (cϖ) 2 sin (ϖt − θ)<br />
where<br />
tan θ =<br />
cϖ<br />
k − mϖ 2 =<br />
2ξr<br />
1 − r 2<br />
Very important note here in the calculations of tan θ above, one should be careful on the sign of the<br />
denominator. When the forcing frequency ϖ > ω the denominator will become negative (the case of ϖ = ω is<br />
resonance and is handled separately). Therefore, one should use arctan that takes care of which quadrant the<br />
angle is. For example, in Mathematica use<br />
ArcTan[1 - r^2, 2 Zeta r]]<br />
and in Matlab use<br />
atan2(2 Zeta r,1 - r^2)<br />
Otherwise, wrong solution will result when ϖ > ω The full solution is<br />
u (t) = e −ξωt (A cos ω d t + B sin ω d t) + F k<br />
1<br />
√(1 − r 2 ) 2 + (2ξr) 2 sin (ϖt − θ) (1)<br />
Applying initial conditions gives<br />
A = u (0) + F k<br />
B = u′ (0)<br />
ω d<br />
+<br />
1<br />
√(1 − r 2 ) 2 + (2ξr) 2 sin θ<br />
u (0) ξω<br />
ω d<br />
+ F k<br />
Another form of these equations is given as follows<br />
1<br />
ω d<br />
√<br />
(1 − r 2 ) 2 + (2ξr) 2 (ξω sin θ − ϖ cos θ)<br />
38
Hence the full solution is<br />
u p = p 0 1 ((<br />
1 − r<br />
2 )<br />
k (1 − r 2 ) 2 + (2ζr) 2 sin ϖt − 2ζr cos ϖt )<br />
u (t) = e −ξωnt (A cos ω d t + B sin ω d t) + F k<br />
Applying initial conditions now gives<br />
1<br />
(1 − r 2 ) 2 + (2ζr) 2 ((<br />
1 − r<br />
2 ) sin ϖt − 2ζr cos ϖt ) (1)<br />
A = u (0) +<br />
B = u′ (0)<br />
ω d<br />
2F rξ<br />
k<br />
1<br />
(1 − r 2 ) 2 + (2ξr) 2<br />
+ u (0) ξω n<br />
− F ( 1 − r 2) ϖ 2F rζ<br />
ω d kω d (1 − r 2 ) 2 +<br />
2<br />
+ (2ζr) kω d<br />
ω n<br />
(1 − r 2 ) 2 + (2ζr) 2<br />
The above 2 sets of equations are equivalent. One uses the phase angle explicitly and the second ones do not.<br />
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<br />
term reverse places as in<br />
u (t) = e −ξωnt (A cos ω d t + B sin ω d t) + F k<br />
Applying initial conditions now gives<br />
(<br />
1 − r<br />
2 )<br />
A = u (0) + F k (1 − r 2 ) 2 + (2ξr) 2<br />
B = u′ (0)<br />
ω d<br />
+ u (0) ξω n<br />
ω d<br />
+<br />
2F rζ<br />
kω d<br />
1 ((<br />
1 − r<br />
2 )<br />
(1 − r 2 ) 2 + (2ζr) 2 cos ϖt − 2ζr sin ϖt )<br />
ϖ<br />
(1 − r 2 ) 2 + (2ζr) 2 − F ( 1 − r 2)<br />
kω d<br />
ω n<br />
(1 − r 2 ) 2 + (2ζr) 2<br />
When a system is damped, the problem with the divide by zero when r = 1 does not occur here as was the<br />
case with undamped system, since when when ϖ ≈ ω or r = 1, the solution in Eq (1) becomes<br />
u (t) = e −ξωt ((<br />
u (0) + F k<br />
) (<br />
1<br />
u ′<br />
2ξ sin θ (0) u (0) ξω<br />
cos ω d t + + + F ω d ω d k<br />
) )<br />
1<br />
2ω d ξ (ξω sin θ − ϖ cos θ) sin ω d t<br />
+ F k<br />
1<br />
sin (ϖt − θ)<br />
2<br />
and the problem with the denominator going to zero does not show up here. The amplitude when steady state<br />
response is maximum can be found as follows. The amplitude of steady state motion is F 1<br />
k<br />
√(1−r . This<br />
2 ) 2 +(2ξr)<br />
√<br />
2<br />
1<br />
is maximum when the magnification factor β = √<br />
is maximum or when (1 − r 2 ) 2 + (2ξr) 2 or<br />
(1−r 2 ) 2 +(2ξr)<br />
√ 2<br />
(<br />
1 − ( ) )<br />
ϖ 2 2 ( )<br />
ω + 2ξ<br />
ϖ 2<br />
ω is minimum. Taking derivative w.r.t. ϖ and equating the result to zero and solving<br />
for ϖ gives<br />
ϖ = ω √ 1 − 2ξ 2<br />
We are looking for positive ϖ, hence when ϖ = ω √ 1 − 2ξ 2 the under-damped response is maximum.<br />
39
8.3.3 critically damping forced vibration ξ = c<br />
c r<br />
= 1<br />
The solution is<br />
Where u h = (A + Bt) e −ωt and u p = F k<br />
arctan definition). Hence<br />
u (t) = u h + u p<br />
1<br />
2r<br />
sin (ϖt − θ) where tan θ = (making sure to use correct<br />
√(1−r 2 ) 2 2 1−r<br />
+(2r) 2<br />
u (t) = (A + Bt) e −ωt + F k<br />
1<br />
√(1 − r 2 ) 2 + (2r) 2 sin (ϖt − θ)<br />
where A, B are found from initial conditions<br />
A = u (0) + F k<br />
B = u ′ (0) + u (0) ω + F k<br />
1<br />
√(1 − r 2 ) 2 + (2r) 2 sin θ<br />
1<br />
√(1 − r 2 ) 2 + (2r) 2 (ω sin θ − ϖ cos θ)<br />
8.3.4 over-damped forced vibration ξ = c<br />
c r<br />
> 1<br />
The solution is<br />
where<br />
and<br />
u p (t) = F k<br />
u (t) = u h + u p<br />
u h (t) = Ae p 1t + Be p 2t<br />
1<br />
√(1 − r 2 ) 2 + (2ξr) 2 sin (ϖt − θ)<br />
hence<br />
u = Ae p 1t + Be p 2t + F k<br />
1<br />
√(1 − r 2 ) 2 + (2ξr) 2 sin (ϖt − θ)<br />
where tan θ = 2ξr<br />
1−r 2<br />
Hence the solution is<br />
and<br />
p 1 = − c<br />
√ ( c<br />
2m + 2m<br />
p 2 = − c<br />
2m − √ ( c<br />
2m<br />
) 2 k −<br />
m = −ωξ + ω √<br />
n ξ 2 − 1<br />
) 2 k −<br />
m = −ωξ − ω √<br />
n ξ 2 − 1<br />
(<br />
−ξ+ √ (<br />
ξ<br />
u (t) = Ae<br />
2 −1<br />
)ωt −ξ− √ )<br />
ξ + Be 2 −1 ωt F + β sin (ϖt − θ)<br />
k<br />
u ′ (0) + u (0) ωξ + u (0) ω √ ξ 2 − 1 + F k<br />
((ξ β + √ )<br />
ξ 2 − 1<br />
A =<br />
2ω √ ξ 2 − 1<br />
u ′ (0) + u (0) ωξ − u (0) ω √ ξ 2 − 1 + F k<br />
((ξ β − √ )<br />
ξ 2 − 1<br />
B = −<br />
2ω √ ξ 2 − 1<br />
)<br />
ω sin θ − ϖ cos θ<br />
)<br />
ω sin θ − ϖ cos θ<br />
40
8.4 Response to impulsive loading<br />
8.4.1 impulse input<br />
Undamped system with impulse<br />
mü + ku = F 0 δ(t)<br />
with initial conditions u (0) = 0 and u ′ (0) = 0.Assuming the impulse acts for a very short time period from<br />
0 to t 1 seconds, where t 1 is small amount. Integrating the above differential equation gives<br />
∫ t1<br />
0<br />
müdt +<br />
∫ t1<br />
0<br />
kudt =<br />
∫ t1<br />
0<br />
F 0 δ(t)<br />
Since t 1 is very small, it can be assumed that u changes is negligible, hence the above reduces to<br />
since we assumed u ′ (0) = 0 and since ∫ t 1<br />
0<br />
∫ t1<br />
0<br />
∫ t1<br />
müdt =<br />
0<br />
( ) d ˙u<br />
m dt =<br />
dt<br />
∫ ˙u(t1 )<br />
˙u(0)<br />
∫ t1<br />
0<br />
∫ t1<br />
0<br />
d ˙u = F 0<br />
m<br />
˙u (t 1 ) − ˙u (0) = F 0<br />
m<br />
˙u (t 1 ) = F 0<br />
m<br />
F 0 δ(t)<br />
F 0 δ(t)<br />
∫ t1<br />
0<br />
∫ t1<br />
0<br />
∫ t1<br />
0<br />
δ(t)<br />
δ(t)<br />
δ(t)<br />
δ(t) = 1 then the above reduces to<br />
˙u (t 1 ) = F 0<br />
m<br />
Therefore, the effect of the impulse is the same as if the system was a free system but with initial velocity given<br />
by F 0<br />
m<br />
and zero initial position. Hence the system is now solved as follows<br />
With u (0) = 0 and u ′ (0) = F 0<br />
m<br />
. The solution is<br />
mü + ku = 0<br />
u impulse (t) = F 0<br />
sin ωt<br />
mω<br />
If the initial conditions were not zero, then the solution for these are added to the above. From earlier, it was<br />
found that the solution is u (t) = u(0) cos ωt + u′ (0)<br />
ω<br />
sin ωt, therefore, the full solution is<br />
u (t) =<br />
under-damped with impulse c < c r , ξ < 1<br />
due to IC only<br />
due to impulse<br />
{ }} { { }} {<br />
u(0) cos ωt + u′ (0)<br />
ω<br />
sin ωt + F 0 sin ωt<br />
mω<br />
mü + c ˙u + ku = δ(t)<br />
ü + 2ξω ˙u + ω 2 u = δ(t)<br />
with initial conditions u (0) = 0 and u ′ (0) = 0.Integrating gives<br />
∫ t1<br />
0<br />
müdt +<br />
∫ t1<br />
0<br />
c ˙udt +<br />
41<br />
∫ t1<br />
0<br />
kudt =<br />
∫ t1<br />
0<br />
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<br />
above reduces to the same result as in the case of undamped. Therefore, the system is solved as free system, but<br />
with initial velocity u ′ (0) = F 0 /m and zero initial position.<br />
Initial conditions are u (0) = 0 and u ′ (0) = 0 then the solution is<br />
applying initial conditions gives A = 0 and B =<br />
u impulse = e −ξωt (A cos ω d t + B sin ω d t)<br />
( F0<br />
)<br />
m<br />
ω d<br />
, hence<br />
u impulse (t) = e −ξωt (<br />
F0<br />
mω d<br />
sin ω d t<br />
If the initial conditions were not zero, then ( the solution for these are added ) to the above. From earlier, it<br />
was found that the solution is u (t) = e −ξωt u(0) cos ω d t + u′ (0)+u(0)ξω<br />
ω d<br />
sin ω d t , therefore, the full solution is<br />
due to IC only<br />
{ (<br />
}} ){<br />
{ ( }} ){<br />
u (t) = e −ξωt u(0) cos ω d t + u′ (0) + u(0)ξω<br />
sin ω d t + e −ξωt F0<br />
sin ω d t<br />
ω d<br />
mω d<br />
)<br />
due to impulse<br />
critically damped with impulse input ξ = c<br />
c r<br />
= 1 with initial conditions u (0) = 0 and u ′ (0) = 0 then the<br />
solution is<br />
u (t) = (A + Bt) e −( cr<br />
2m<br />
)<br />
t<br />
= (A + Bt) e −ωt<br />
where A, B are found from initial conditions A = u (0) = 0 and B = u ′ (0) + u (0) ω = F 0<br />
m<br />
, hence the solution<br />
is<br />
u impulse (t) = F 0t<br />
m e−ωt<br />
If the initial conditions were not zero, then the solution for these are added to the above. From earlier, it<br />
was found that the solution is u (t) = (u 0 (1 + ωt) + u ′ (0) t) e −ωt , therefore, the full solution is<br />
u (t) =<br />
due to IC only<br />
{ }} {<br />
(<br />
u (0) (1 + ωt) + u ′ (0) t ) e −ωt +<br />
due to impulse<br />
{ }} {<br />
F 0 t<br />
m e−ωt<br />
over-damped with impulse input ξ = c<br />
c r<br />
> 1 With initial conditions are u (0) = 0 and u ′ (0) = 0 the<br />
solution is<br />
u impulse (t) = Ae λ 1ωt + Be λ 2ωt<br />
where A, B are found from initial conditions and<br />
λ 1 = −ωξ + ω √ ξ 2 − 1<br />
λ 2 = −ωξ − ω √ ξ 2 − 1<br />
A = u′ (0) − u (0) λ 2<br />
2ω √ ξ 2 − 1<br />
B = −u′ (0) + u (0) λ 1<br />
2ω √ ξ 2 − 1<br />
Hence the solution is<br />
(<br />
−ξ+ √ (<br />
ξ<br />
u impulse (t) = Ae<br />
2 −1<br />
)ωt −ξ− √ )<br />
ξ + Be 2 −1 ωt<br />
42
where<br />
A =<br />
B =<br />
F 0<br />
m<br />
2ω √ ξ 2 − 1<br />
− F 0<br />
m<br />
2ω √ ξ 2 − 1<br />
Hence<br />
u impulse (t) =<br />
F 0<br />
(<br />
m<br />
2ω √ ξ 2 − 1 e −ξ+ √ )<br />
ξ 2 F 0<br />
(<br />
−1 ωt −<br />
m<br />
2ω √ ξ 2 − 1 e −ξ− √ )<br />
ξ 2 −1 ωt<br />
If the initial conditions were not zero, then the solution for these are added to the above. From earlier, it<br />
was found that the solution is u (t) = Ae p 1t + Be p 2t , therefore, the full solution is<br />
u (t) = Ae λ 1ωt + Be λ 2ωt +<br />
F 0<br />
F 0<br />
m<br />
2ω √ ξ 2 − 1 eλ 1ωt m<br />
−<br />
2ω √ ξ 2 − 1 eλ 2ωt<br />
8.4.2 sin impulse input<br />
Now assume the input is as follows<br />
A = u′ (0) − u (0) λ 2<br />
2ω √ ξ 2 − 1<br />
B = −u′ (0) + u (0) λ 1<br />
2ω √ ξ 2 − 1<br />
given by F (t) = F 0 sin (ϖt) where ϖ = 2π<br />
2t 1<br />
= π t 1<br />
undamped system with sin impulse<br />
⎧<br />
⎨ F 0 sin (ϖt) 0 ≤ t ≤ t 1<br />
mü + ku =<br />
⎩<br />
0 t > t 1<br />
with u (0) = u 0 and ˙u (0) = v 0 . For 0 ≤ t ≤ t 1 the solution is<br />
( )<br />
( )<br />
v0<br />
u (t) = u 0 cos ωt +<br />
ω − u r<br />
1 π<br />
st<br />
1 − r 2 sin ωt + u st<br />
1 − r 2 sin t<br />
t 1<br />
43
where r = ϖ ω = π/t 1<br />
ω<br />
= T<br />
2t 1<br />
where T is the natural period of the system. u st = F 0<br />
k<br />
, hence the above becomes<br />
⎛<br />
⎜<br />
u (t) = u 0 cos ωt + ⎝ v 0<br />
ω − F 0<br />
k<br />
When u 0 = 0 and v 0 = 0 then<br />
u (t) = − F 0<br />
k<br />
u (t) = F 0<br />
k<br />
( )<br />
π/t1<br />
ω<br />
1 −<br />
(<br />
π/t1<br />
ω<br />
1<br />
1 −<br />
(<br />
π/t1<br />
ω<br />
( )<br />
π/t1<br />
ω<br />
1 −<br />
(<br />
π/t1<br />
ω<br />
) 2<br />
sin ωt + F 0<br />
k<br />
) 2<br />
(sin<br />
) 2<br />
⎞<br />
⎟<br />
⎠ sin ωt + F 0<br />
k<br />
1<br />
1 −<br />
(<br />
π/t1<br />
ω<br />
1<br />
1 −<br />
(<br />
π/t1<br />
ω<br />
(<br />
π t<br />
t 1<br />
)<br />
− π/t 1<br />
ω sin ωt )<br />
(<br />
) 2<br />
sin π t )<br />
t 1<br />
The above Eq (1) gives solution during the time 0 ≤ t ≤ t 1<br />
Now after t = t 1 the force will disappear, the differential equation becomes<br />
mü + ku = 0 t > t 1<br />
(<br />
) 2<br />
sin π t )<br />
t 1<br />
but with the initial conditions evaluate at t = t 1 . From (1)<br />
( )<br />
v0<br />
u (t 1 ) = u 0 cos ωt 1 +<br />
ω − u r<br />
1<br />
st<br />
1 − r 2 sin ωt 1 + u st<br />
1 − r 2 sin ϖt 1<br />
( )<br />
v0<br />
= u 0 cos ωt 1 +<br />
ω − u r r<br />
st<br />
1 − r 2 u st<br />
1 − r 2 sin ωt 1 (2)<br />
since sin ϖt 1 = 0. taking derivative of Eq (1)<br />
˙u (t) = −ωu 0 sin ωt + ω<br />
and at t = t 1 the above becomes<br />
˙u (t 1 ) = −ωu 0 sin ωt 1 + ω<br />
= −ωu 0 sin ωt 1 + ω<br />
(<br />
v0<br />
)<br />
ω − u r<br />
st<br />
1 − r 2 cos ωt + ϖ 1 cos ϖt<br />
1 − r2 (<br />
v0<br />
)<br />
ω − u r<br />
st<br />
1 − r 2 )<br />
ω − u r<br />
st<br />
1 − r 2<br />
(<br />
v0<br />
cos ωt 1 + ϖ 1<br />
1 − r 2 cos ϖt 1<br />
(1)<br />
cos ωt 1 − ϖ 1<br />
1 − r 2 (3)<br />
since cos ϖt 1 = −1. Now (2) and (3) are used as initial conditions to solve mü + ku = 0 . The solution for<br />
t > t 1 is<br />
u (t) = u (t 1 ) cos ωt + ˙u (t 1)<br />
ω<br />
sin ωt<br />
Resonance with undamped sin impulse When ϖ ≈ ω and t ≤ t 1 we obtain resonance since r → 1 in<br />
the solution shown up and as written the solution can’t be used for analysis in this case. To obtain a solution<br />
for resonance some calculus is needed. Eq (1) is written as<br />
(<br />
)<br />
ϖ<br />
v 0<br />
u (t) = u 0 cos ωt +<br />
ω − u ω<br />
st<br />
1 − ( 1<br />
)<br />
ϖ 2<br />
sin ωt + u st<br />
ω<br />
1 − ( )<br />
ϖ 2<br />
sin ϖt<br />
ω<br />
( )<br />
v0<br />
= u 0 cos ωt +<br />
ω − u ωϖ<br />
ω 2<br />
st<br />
ω 2 − ϖ 2 sin ωt + u st<br />
ω 2 sin ϖt<br />
− ϖ2 (1A)<br />
Now looking at case when ϖ ≈ ω but less than ω, hence let<br />
ω − ϖ = 2∆ (2)<br />
44
where ∆ is very small positive quantity. and we also have<br />
Multiplying Eq (2) and (3) with each others gives<br />
Going back to Eq (1A) and rewriting it as<br />
ω + ϖ ≈ 2ϖ (3)<br />
ω 2 − ϖ 2 = 4∆ϖ (4)<br />
( v0<br />
u (t) = u 0 cos ωt +<br />
ω − u ωϖ<br />
)<br />
ω 2<br />
st sin ωt + u st sin ϖt<br />
4∆ϖ<br />
4∆ϖ<br />
( v0<br />
= u 0 cos ωt +<br />
ω − u ω<br />
)<br />
ω 2<br />
st sin ωt + u st sin ϖt<br />
4∆<br />
4∆ϖ<br />
Since ϖ ≈ ω the above becomes<br />
( v0<br />
u (t) = u 0 cos ωt +<br />
ω − u ω<br />
)<br />
ω<br />
st sin ωt + u st sin ϖt<br />
4∆<br />
4∆<br />
= u 0 cos ωt + v 0<br />
ω sin ωt + u ω<br />
st (sin ϖt − sin ωt)<br />
4∆<br />
now using sin ϖt − sin ωt = 2 sin ( ϖ−ω<br />
2<br />
t ) cos ( ϖ+ω<br />
2<br />
t ) the above becomes<br />
u (t) = u 0 cos ωt + v ( ( )<br />
0<br />
ω sin ωt + u ω ϖ − ω<br />
st sin t cos<br />
2∆ 2<br />
From Eq(2) ϖ − ω = −2∆ and ω + ϖ ≈ 2ϖ hence the above becomes<br />
or since ϖ ≈ ω<br />
Now lim ∆→0<br />
sin(∆t)<br />
∆<br />
This can also be written as<br />
( ϖ + ω<br />
u (t) = u 0 cos ωt + v 0<br />
ω sin ωt + u ω<br />
st (sin (−∆t) cos (ϖt))<br />
2∆<br />
u (t) = u 0 cos ωt + v 0<br />
ω sin ωt − u ω<br />
st (sin (∆t) cos (ωt))<br />
2∆<br />
= t hence the above becomes<br />
u (t) = u 0 cos ωt + v 0<br />
ω sin ωt − u ωt<br />
st cos (ωt)<br />
2<br />
2<br />
))<br />
t<br />
u (t) = u 0 cos ϖt + v 0<br />
ϖ sin ϖt − u ϖt<br />
st cos (ϖt) (1)<br />
( )<br />
2<br />
π<br />
= u 0 cos t + v ( ) ( ) ( )<br />
0 π π π<br />
t 1 ϖ sin t − u st t cos t<br />
t 1 2t 1 t 1<br />
since ϖ ≈ ω in this case. This is the solution to use for resonance and for t ≤ t 1<br />
Hence for t > t 1 , the above equations is used to determine initial conditions at t = t 1<br />
u (t 1 ) = u 0 cos ϖt 1 + v 0<br />
ϖ sin ϖt 1 − u st<br />
ϖt 1<br />
2 cos (ϖt 1)<br />
but cos ϖt 1 = cos π t 1<br />
t 1 = −1 and sin ϖt 1 = 0 and ϖt 1<br />
2<br />
= π 2<br />
, hence the above becomes<br />
Taking derivative of Eq (1) gives<br />
u (t 1 ) = −u 0 + u st<br />
π<br />
2<br />
ϖ 2 t<br />
˙u (t) = −ϖu 0 sin ϖt + v 0 cos ϖt + u st<br />
2 sin (ϖt) − u ϖ<br />
st cos (ϖt)<br />
2<br />
45
and at t = t 1<br />
ϖ 2 t 1<br />
ϖ<br />
˙u (t 1 ) = −ϖu 0 sin ϖt 1 + v 0 cos ϖt 1 + u st sin (ϖt 1 ) − u st<br />
2<br />
2 cos (ϖt 1)<br />
ϖ<br />
= −v 0 + u st<br />
2<br />
Now the solution for t > t 1 is<br />
u (t) = u (t 1 ) cos ωt + ˙u (t 1)<br />
ω<br />
(<br />
π<br />
= −u (0) + u st<br />
2<br />
under-damped with sin impulse c < c r , ξ < 1<br />
sin ωt<br />
)<br />
cos ωt + −u′ (0) + u st<br />
π<br />
2t 1<br />
ω<br />
⎧<br />
⎨ F 0 sin (ϖ) 0 ≤ t ≤ t 1<br />
mü + c ˙u + ku =<br />
⎩<br />
0 t > t 1<br />
sin ωt<br />
or<br />
⎧<br />
⎨<br />
ü + 2ξω ˙u + ω 2 F 0 sin (ϖ) 0 ≤ t ≤ t 1<br />
u =<br />
⎩<br />
0 t > t 1<br />
mü + c ˙u + ku = F sin ϖt<br />
ü + 2ξω ˙u + ω 2 u = F sin ϖt<br />
m<br />
For t ≤ t 1 Initial conditions are u (0) = u 0 and ˙u (0) = v 0 and u st = F k<br />
u (t) = e −ξωt (A cos ω d t + B sin ω d t) +<br />
then the solution from above is<br />
u st<br />
√(1 − r 2 ) 2 + (2ξr) 2 sin (ϖt − θ) (1)<br />
Applying initial conditions gives<br />
For t > t 1 . From (1)<br />
A = u 0 +<br />
B = v 0<br />
ω d<br />
+ u 0ξω<br />
ω d<br />
+<br />
u st<br />
√(1 − r 2 ) 2 + (2ξr) 2 sin θ<br />
u st<br />
u (t 1 ) = e −ξωt 1<br />
(A cos ω d t 1 + B sin ω d t 1 ) +<br />
ω d<br />
√<br />
(1 − r 2 ) 2 + (2ξr) 2 (ξω sin θ − ϖ cos θ)<br />
u st<br />
√(1 − r 2 ) 2 + (2ξr) 2 sin (ϖt 1 − θ) (2)<br />
Taking derivative of (1) gives<br />
˙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)<br />
u st<br />
+ ϖ<br />
cos (ϖt − θ)<br />
√(1 − r 2 ) 2 + (2ξr) 2<br />
46
at t = t 1<br />
˙u (t 1 ) = −ξωe −ξωt 1<br />
(A cos ω d t 1 + B sin ω d t 1 ) + e −ξωt 1<br />
(−Aω d sin ω d t 1 + ω d B cos ω d t 1 )<br />
u st<br />
+ ϖ<br />
cos (ϖt 1 − θ) (3)<br />
√(1 − r 2 ) 2 + (2ξr) 2<br />
Now for t > t 1 the equation becomes<br />
which has the solution<br />
where A = u (t 1 ) and B = ˙u(t 1)+u(t 1 )ξω<br />
ω d<br />
mü + c ˙u + ku = 0<br />
u = e −ξωt (A cos ω d t + B sin ω d t)<br />
critically damped with sin impulse ξ = c<br />
c r<br />
= 1 For t ≤ t 1 Initial conditions are u (0) = u 0 and ˙u (0) = v 0<br />
then the solution is from above<br />
u (t) = (A + Bt) e −ωt u st<br />
+<br />
sin (ϖt − θ) (1)<br />
√(1 − r 2 ) 2 + (2r) 2<br />
Where tan θ =<br />
cϖ<br />
k−mϖ 2<br />
= 2ξr<br />
1−r 2 . A, B are found from initial conditions<br />
A = u 0 +<br />
u st<br />
√(1 − r 2 ) 2 + (2r) 2 sin θ<br />
B = v 0 + u 0 ω +<br />
u st<br />
√(1 − r 2 ) 2 + (2r) 2 (ω sin θ − ϖ cos θ)<br />
For t > t 1 the solution is<br />
To find u (t 1 ) , from Eq(1)<br />
u (t) = (u (t 1 ) + ( ˙u (t 1 ) + u (t 1 ) ω) t) e −ωt (2)<br />
u (t 1 ) = (A + Bt) e −ωt 1<br />
+<br />
u st<br />
√(1 − r 2 ) 2 + (2r) 2 sin (ϖt 1 − θ)<br />
taking derivative of (1) gives<br />
˙u (t) = −ω (A + Bt) e −ωt + Be −ωt + ϖ<br />
sin (ϖt − θ) (3)<br />
√(1 − r 2 ) 2 + (2r) 2<br />
u st<br />
at t = t 1<br />
˙u (t 1 ) = −ω (A + Bt 1 ) e −ωt 1<br />
+ Be −ωt 1<br />
+ ϖ<br />
sin (ϖt 1 − θ) (4)<br />
√(1 − r 2 ) 2 + (2r) 2<br />
u st<br />
Hence Eq (2) can now be evaluated using Eq(3,4)<br />
over-damped with sin impulse ξ = c<br />
c r<br />
> 1 For t ≤ t 1 Initial conditions are u (0) = u 0 and ˙u (0) = v 0 then<br />
the solution is<br />
u = Ae p1t + Be p2t u st<br />
+<br />
sin (ϖt − θ)<br />
√(1 − r 2 ) 2 + (2ξr) 2<br />
where tan θ = 2ξr<br />
1−r 2 (make sure you use correct quadrant, see not above on arctan) and<br />
p 1 = − c<br />
√ ( c<br />
) 2<br />
2m + k −<br />
2m m<br />
= −ωξ + ω √ ξ 2 − 1<br />
47
and<br />
leading to the solution where tan θ = 2ξr<br />
1−r 2<br />
p 2 = − c<br />
√ ( c<br />
) 2<br />
2m − k −<br />
2m m<br />
= −ωξ − ω √ ξ 2 − 1<br />
and<br />
is<br />
p 1 = − c<br />
√ ( c<br />
2m + 2m<br />
p 2 = − c<br />
2m − √ ( c<br />
2m<br />
) 2 k −<br />
m = −ωξ + ω √<br />
n ξ 2 − 1<br />
) 2 k −<br />
m = −ωξ − ω √<br />
n ξ 2 − 1<br />
(<br />
−ξ+ √ (<br />
ξ<br />
u (t) = Ae<br />
2 −1<br />
)ωt −ξ− √ )<br />
ξ + Be 2 −1 ωt F + β sin (ϖt − θ)<br />
k<br />
u ′ (0) + u (0) ωξ + u (0) ω √ ξ 2 − 1 + F k<br />
((ξ β + √ )<br />
ξ 2 − 1<br />
A =<br />
2ω √ ξ 2 − 1<br />
u ′ (0) + u (0) ωξ − u (0) ω √ ξ 2 − 1 + F k<br />
((ξ β − √ )<br />
ξ 2 − 1<br />
B = −<br />
2ω √ ξ 2 − 1<br />
1<br />
β = √<br />
(1 − r 2 ) 2 + (2ξr) 2<br />
)<br />
ω sin θ − ϖ cos θ<br />
)<br />
ω sin θ − ϖ cos θ<br />
For t > t 1 . From Eq(1) and at t = t 1<br />
Taking derivative of Eq (1)<br />
(<br />
−ξ+ √ )<br />
(<br />
ξ<br />
u (t 1 ) = Ae<br />
2 −1 ωt 1 −ξ− √ )<br />
ξ<br />
+ Be<br />
2 −1 ωt 1<br />
+ D sin (ϖt 1 − θ) (2)<br />
(<br />
−ξ+ √ (<br />
ξ<br />
˙u (t) = ωAe<br />
2 −1<br />
)ωt −ξ− √ )<br />
ξ + ωBe 2 −1 ωt + ϖD cos (ϖt − θ)<br />
At t = t 1<br />
−ξ+<br />
˙u (t 1 ) = ωAe( √ )<br />
(<br />
ξ 2 −1 ωt 1 −ξ− √ )<br />
ξ<br />
+ ωBe<br />
2 −1 ωt 1<br />
+ ϖD cos (ϖt 1 − θ) (3)<br />
Equation of motion now is<br />
which has solution for over-damped given by<br />
ü + 2ξω ˙u + ω 2 u = 0<br />
(<br />
−ξ+ √ (<br />
ξ<br />
u (t) = Ae<br />
2 −1<br />
)ω nt −ξ− √ )<br />
ξ + Be 2 −1 ω nt<br />
where<br />
˙u (t 1 ) + u (t 1 ) ω n<br />
(ξ − √ )<br />
ξ 2 − 1<br />
A = −<br />
√<br />
2ω n ξ 2 − 1<br />
˙u (t 1 ) + u (t 1 ) ξω n<br />
(ξ + √ )<br />
ξ 2 − 1<br />
B =<br />
√<br />
2ω n ξ 2 − 1<br />
48
8.5 references<br />
1. Vibration analysis by Robert K. Vierck<br />
2. Structural dynamics theory and computation, 5th edition by Mario Paz, William Leigh<br />
3. Dynamic of structures, Ray W. Clough and Joseph Penzien<br />
4. Theory of vibration,volume 1, by A.A.Shabana<br />
5. Notes on Diffy Qs, Differential equations for engineers, by Jiri Lebl, online PDF book, chapter 2.6, oct<br />
1,2012 http://www.jirka.org/diffyqs/<br />
9 Derivation of rotation formula<br />
This formula is very important. Will show its derivation now in details. It is how to express vectors in rotating<br />
frames.<br />
Consider this diagram<br />
P<br />
r<br />
Y<br />
X<br />
r p<br />
y<br />
r o<br />
o<br />
<br />
x<br />
Moving frame of<br />
reference, attached<br />
to body of interest<br />
Absolute (or inertial frame of reference)<br />
In the above, the small axis x, y is a frame attached to some body which rotate around this axis with angular<br />
velocity ω (measured by the inertial frame of course). All laws derived below are based on the following one rule<br />
d<br />
dt r ∣<br />
∣∣∣absolute<br />
= d dt r ∣<br />
∣∣∣relative<br />
+ ω × r (1)<br />
Lets us see how to apply this rule. Let us express the position vector of the particle r p . We can see by<br />
normal vector additions that the position vector of particle is<br />
r p = r o + r (2)<br />
Notice that nothing special is needed here, since we have not yet looked at rate of change with time. The<br />
complexity (i.e. using rule (1)) appears only when we want to look at velocities and accelerations. This is when<br />
we need to use the above rule (1). Let us now find the velocity of the particle. From above<br />
ṙ p = ṙ o + ṙ<br />
49
Every time we take derivatives, we stop and look. For any vector that originates from the moving frame,<br />
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<br />
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,<br />
Hence the above becomes<br />
V p = V o + ṙ<br />
and now we apply rule (1) to expand ṙ<br />
V p = V o + (V rel + ω × r) (3)<br />
where V rel is just d dt r∣ ∣<br />
relative<br />
The above is the final expression for the velocity of the particle V p using its velocity as measured by the<br />
moving frame in order to complete the expression.<br />
So the above says that the absolute velocity of the particle is equal to the absolute velocity of the base of the<br />
moving frame + something else and this something else was (V rel + ω × r)<br />
Now we will find the absolute acceleration of P . Taking time derivatives of (3) gives<br />
˙V p = ˙V<br />
( )<br />
o + ˙Vrel + ˙ω × r + ω × ṙ<br />
(4)<br />
As we said above, each time we take time derivatives, we stop and look for vectors which are based on the<br />
moving frame, and apply rule (1) to them. In the above, ˙Vrel and ṙ qualify. Apply rule (1) to ˙V rel gives<br />
˙V rel = a rel + ω × V rel (5)<br />
where a rel just means the acceleration relative to moving frame. And applying rule (1) to ṙ gives<br />
Replacing (5) and (6) into (4) gives<br />
ṙ = V rel + ω × r (6)<br />
a p = a o + (a rel + ω × V rel + ˙ω × r + ω × (V rel + ω × r))<br />
= a o + a rel + (ω × V rel ) + ( ˙ω × r) + (ω × V rel ) + (ω × (ω × r))<br />
= a o + a rel + 2 (ω × V rel ) + ( ˙ω × r) + (ω × (ω × r)) (7)<br />
Eq (7) says that the absolute acceleration a p of P is the sum of the acceleration of the base a o of the<br />
moving frame plus the relative acceleration a rel of the particle to the moving frame plus 2 (ω × V rel ) + ( ˙ω × r) +<br />
(ω × (ω × r))<br />
Hence, using Eq(3) and Eq(7) gives us the expressions we wanted for velocity and acceleration.<br />
10 Miscellaneous hints<br />
1. When finding the generalized force for the user with the Lagrangian method (the hardest step), using the<br />
virtual work method, if the force (or virtual work by the force) ADDS energy to the system, then make<br />
the sign of the force positive otherwise the sign is negative.<br />
2. For damping force, the sign is always negative.<br />
3. External forces such as linear forces applied, torque applied, in general, are positive.<br />
4. Friction force is negative (in general) as friction takes energy from the system like damping.<br />
50
11 Formulas<br />
sin 2 x 1cos2x<br />
2<br />
cos 2 x 1cos2x<br />
2<br />
sin 2x 2 sin x cosx<br />
cos2x cos 2 x sin 2 x<br />
1 2 sin 2 x<br />
V p V o r V rel<br />
Velocity of p<br />
relative to o<br />
a p a 0 r r 2 V rel a rel<br />
p<br />
IF rigid body is rotating AND translation at the same time THEN<br />
Take moments around its cg only<br />
ELSE<br />
can also take moments about any other points on it (but<br />
change I)<br />
END IF<br />
o<br />
r<br />
<br />
Y<br />
R Y<br />
X<br />
R y<br />
R X<br />
R<br />
x<br />
y<br />
<br />
R x<br />
BODY FIXED<br />
COORDINATES<br />
This gives the relation<br />
between the rate of<br />
change with respect to<br />
time of a vector expressed<br />
in a frame of reference<br />
which is body fixed (y,x)<br />
here, to the rate of change<br />
with respect to time of the<br />
same vector expressed<br />
using an inertial frame<br />
coordinates XY.<br />
The omega here is the<br />
angular velocity of the<br />
rigid body rotation, or the<br />
body fixed coordinates,<br />
with respect to the inertial<br />
frame.<br />
d<br />
dt R X,Y d dt R x,y R X,Y<br />
<br />
R<br />
r<br />
<br />
R r<br />
r<br />
The small disk rotates<br />
at angular speed of <br />
The condition of no slip can<br />
be seen to be<br />
R r r<br />
For the sign of generalized force:<br />
If work done by force takes away<br />
energy from system, then the<br />
sign is negative, else positive. So<br />
Friction will always have negative<br />
sign, so will damping force.<br />
d<br />
dt<br />
cosxt sinxtx t<br />
d<br />
dt<br />
df<br />
dxt<br />
<br />
d<br />
dxt<br />
d<br />
cosxt <br />
dt<br />
df<br />
dxt<br />
d<br />
d<br />
cos xt<br />
dxt<br />
d<br />
xt dt<br />
xt dt<br />
51
F ma<br />
linear momentum p mv<br />
F d p dt<br />
particle kinetic energy T 1 2 mv2<br />
I<br />
angular momentum H I cg <br />
d<br />
dt H<br />
rigid body T 1 Mv 2 cg<br />
2 1 I 2<br />
2<br />
cg<br />
my 2 n y n 2 y fy,t<br />
y cy k m y fy,t<br />
n <br />
For small<br />
epsilon<br />
1<br />
1<br />
1<br />
1<br />
1 <br />
1 <br />
k<br />
m ,c 2 n<br />
m<br />
conservation of angular momentum d dt<br />
H constant<br />
x n2 x 0<br />
If n 0 then sinusodial solution, ok<br />
if n 0 then solution blows up, exponential<br />
52
12 Velocity and acceleration diagrams<br />
12.1 Spring pendulum<br />
<br />
Length of pendulum fixed<br />
Length of pendulum changes with time<br />
2L <br />
L<br />
L<br />
L<br />
Velocity diagram<br />
acceleration<br />
diagram<br />
L 2 L Lt L 2<br />
<br />
L<br />
L<br />
Velocity diagram<br />
acceleration<br />
diagram<br />
y<br />
Length of pendulum changes with time and base of pendulum moves<br />
ÿ<br />
x<br />
x<br />
y<br />
L<br />
x<br />
L 2<br />
ÿ<br />
L<br />
x<br />
2L <br />
Velocity diagram<br />
<br />
L<br />
acceleration<br />
diagram<br />
r<br />
Nasser M. Abbasi<br />
Oct 10, 2013 (drawing2.vsd)<br />
53
12.2 pendulum with blob moving in slot<br />
<br />
r<br />
<br />
y<br />
y<br />
r<br />
y<br />
System. Pendulum with<br />
blob which moves in slot<br />
perpendicular to axis<br />
Velocity diagram<br />
2y <br />
r 2<br />
<br />
y<br />
ÿ<br />
r<br />
acceleration diagram<br />
y 2<br />
54
12.3 spring pendulum with block moving in slot<br />
<br />
r<br />
<br />
y<br />
y<br />
r<br />
System. Pendulum with blob which<br />
moves in slot perpendicular to axis. In<br />
addition, length of the axis of pendulum<br />
itself changes in time. y and r are<br />
measured from static equilibrium of<br />
springs<br />
y<br />
Velocity diagram<br />
r<br />
2y <br />
r 2<br />
2r<br />
<br />
y<br />
ÿ<br />
r<br />
acceleration diagram<br />
y 2<br />
r<br />
55
12.4 double pendulum<br />
<br />
<br />
Double<br />
pendulum<br />
L 1<br />
L 2<br />
<br />
<br />
<br />
m 1<br />
m 1<br />
L 1<br />
• 2<br />
m 2<br />
m 2<br />
L 1<br />
<br />
<br />
L 1<br />
L 2<br />
m 2<br />
••<br />
acceleration diagram<br />
L 1<br />
<br />
<br />
m 1<br />
L 2<br />
L 2<br />
• 2<br />
L 1<br />
• 2<br />
<br />
L 2<br />
••<br />
<br />
L 1<br />
•<br />
L 1<br />
••<br />
Velocity diagram<br />
L 2<br />
•<br />
<br />
<br />
L 1<br />
•<br />
56
13 Velocity and acceleration of rigid body 2D<br />
<br />
V<br />
y<br />
x<br />
U<br />
V U<br />
y<br />
x<br />
U V<br />
Y<br />
X<br />
Linear<br />
Velocity<br />
diagram<br />
Rigid body,<br />
rotating at<br />
angular <br />
speed<br />
Linear<br />
acceleration<br />
diagram<br />
Notice the<br />
sign<br />
difference<br />
Nasser M. Abbasi<br />
Drawing_rigid_body_rotati<br />
on_1.vsd<br />
May 23, 2011<br />
a <br />
a x<br />
a y<br />
<br />
U V<br />
V U<br />
Finding linear acceleration of center of mass of a rigid body under pure rotation using fixed body coordinates.<br />
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<br />
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<br />
attached to the body itself.<br />
Just remember that all these speeds (i.e. U,V ) and accelerations (a x , a y ) are still being measured by an<br />
observer in the inertial frame. It is only that the directions of the velocity components of the center of mass<br />
is along an axis fixed on the body. Only the direction. But actual speed measurements are still done by a<br />
stationary observer. Since clearly if the observer was sitting on the body itself, then they will measure the speeds<br />
to be zero in that case.<br />
57
14 Velocity and acceleration of rigid body 3D<br />
14.1 Using Vehicle dynamics notations<br />
F x<br />
Nasser M. Abbasi<br />
3d_1.vsd<br />
May 26, 2011<br />
L<br />
I <br />
N<br />
F z<br />
W<br />
z<br />
r<br />
Linear force<br />
torque<br />
Linear velocity<br />
angular velocity<br />
v <br />
y<br />
<br />
p q V<br />
x<br />
M<br />
F y<br />
U<br />
F <br />
Forces, Torques, linear<br />
velocities and angular velocities<br />
for a general 3D rigid body<br />
I xx I xy I xz<br />
<br />
I yx I yy I yz<br />
linear momentum<br />
U qW rV<br />
F d dt p d dt mvm d dt v m a<br />
V rU pW<br />
W pV qU<br />
U<br />
V<br />
W<br />
p<br />
q<br />
r<br />
F x<br />
F y<br />
F z<br />
L<br />
M<br />
N<br />
d dt H d dt I<br />
Angular momentum<br />
Derivation of this is much more complicated than with<br />
the case of linear motion (F=ma), since m is scalar<br />
there, but for rotation, I is matrix. See next page for the<br />
derivation<br />
58
14.2 3D Not Using vehicle dynamics notations<br />
Nasser M. Abbasi<br />
3d_1.vsd<br />
May 26, 2011<br />
z<br />
F z<br />
Linear force<br />
torque<br />
Linear velocity<br />
v <br />
v x<br />
v y<br />
v z<br />
angular velocity<br />
v z<br />
F x<br />
x<br />
I <br />
v x<br />
x<br />
x<br />
z<br />
z<br />
y<br />
y<br />
F y<br />
<br />
F <br />
Forces, Torques, linear<br />
velocities and angular<br />
velocities for a general 3D<br />
rigid body<br />
I xx I xy I xz<br />
<br />
I yx I yy I yz<br />
F d p dt d mvm d v m dt dt<br />
a y z v x x v z<br />
a<br />
a x y v z z v y<br />
v y<br />
y<br />
x<br />
y<br />
z<br />
F x<br />
F y<br />
F z<br />
x<br />
y<br />
z<br />
linear momentum<br />
a z x v y y v z<br />
The derivation of the above is given next, but it uses the standard formula given by<br />
d<br />
A d<br />
A A<br />
dt dt resolved<br />
This is in the<br />
inertial frame of<br />
reference<br />
This is the same A, but its components<br />
are with respect to the body fixed<br />
coordinates system,<br />
Cross product<br />
59
14.2.1 Derivation for F = ma in 3D<br />
F = d dt p<br />
= d dt (mv)<br />
= m d dt v<br />
⎡⎛<br />
⎞ ⎛ ⎞ ⎛ ⎞⎤<br />
a x<br />
ω x<br />
v x<br />
= m<br />
⎢⎜a y ⎟<br />
⎣⎝<br />
⎠ + ⎜ω y ⎟<br />
⎝ ⎠ ⊗ ⎜v y ⎟⎥<br />
⎝ ⎠⎦<br />
a z ω z v z<br />
⎡⎛<br />
⎞ ∣ ∣⎤<br />
∣∣∣∣∣∣∣∣∣ ∣∣∣∣∣∣∣∣∣ a x<br />
i j k<br />
= m<br />
⎢⎜a y ⎟<br />
⎣⎝<br />
⎠ + det ω x ω y ω z ⎥<br />
⎦<br />
a z v x v y v z<br />
⎡⎛<br />
⎞ ⎛<br />
⎞⎤<br />
a x<br />
ω y v z − ω z v y<br />
= m<br />
⎢⎜a y ⎟<br />
⎣⎝<br />
⎠ + ⎜− (ω x v z − ω z v x )<br />
⎟⎥<br />
⎝<br />
⎠⎦<br />
a z ω x v y − ω y v x<br />
⎛<br />
⎞<br />
a x + ω y v z − ω z v y<br />
= m<br />
⎜a y − ω x v z + ω z v x ⎟<br />
⎝<br />
⎠<br />
a z + ω x v y − ω y v x<br />
60
14.2.2 Derivation for τ = Iω in 3D<br />
Let A = Iω then using the rule<br />
( ) d<br />
τ =<br />
dt A ( ) d<br />
=<br />
dt A + ω × A<br />
resolved<br />
Then τ = Iω can be found for the general case<br />
⎡<br />
⎤<br />
A<br />
A<br />
{ ⎛ }} ⎞ ⎛ ⎞{<br />
⎛ ⎞ { ⎛ }} ⎞ ⎛ ⎞{<br />
τ = d I xx I xy I xz<br />
ω x<br />
ω x<br />
I xx I xy I xz<br />
ω x<br />
dt<br />
⎜I yx I yy I yz ⎟ ⎜ω y ⎟<br />
+<br />
⎝<br />
⎠ ⎝ ⎠<br />
⎜ω y ⎟<br />
⎝ ⎠ ×<br />
⎜I yx I yy I yz ⎟ ⎜ω y ⎟<br />
⎝<br />
⎠ ⎝ ⎠<br />
⎢<br />
⎣ I zx I yz I zz ω z<br />
⎥<br />
⎦ ω z I zx I yz I zz ω z<br />
⎛<br />
⎞ ⎛ ⎞ ⎛ ⎞ ⎛<br />
⎞<br />
I xx I xy I xz<br />
α x<br />
ω x<br />
I xx ω x + I xy ω y + I xz ω z<br />
=<br />
⎜I yx I yy I yz ⎟ ⎜α y ⎟<br />
⎝<br />
⎠ ⎝ ⎠ + ⎜ω y ⎟<br />
⎝ ⎠ × ⎜I yx ω x + I yy ω y + I yz ω z ⎟<br />
⎝<br />
⎠<br />
I zx I yz I zz α z ω z I zx ω x + I yz ω y + I zz ω z<br />
⎛<br />
⎞ ⎛ ⎞ ∣ ∣∣∣∣∣∣∣∣∣ I xx I xy I xz<br />
α x<br />
i j k<br />
=<br />
⎜I yx I yy I yz ⎟ ⎜α y ⎟<br />
⎝<br />
⎠ ⎝ ⎠ + det ω x ω y ω z<br />
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 ) ∣<br />
⎛<br />
⎞ ⎛ ⎞ ⎛<br />
⎞<br />
I xx I xy I xz<br />
α x<br />
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 )<br />
=<br />
⎜I yx I yy I yz ⎟ ⎜α y ⎟<br />
⎝<br />
⎠ ⎝ ⎠ + ⎜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 )<br />
⎟<br />
⎝<br />
⎠<br />
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 )<br />
61
14.2.3 Derivation for τ = Iω in 3D using principle axes<br />
The above derivation simplifies now since we will be using principle axes. In this case, all cross products of<br />
moments of inertia vanish.<br />
⎛<br />
⎞<br />
I xx 0 0<br />
I =<br />
⎜ 0 I yy 0<br />
⎟<br />
⎝<br />
⎠<br />
0 0 I zz<br />
Hence<br />
⎡<br />
⎤<br />
A<br />
A<br />
{ ⎛ }} ⎞ ⎛ ⎞{<br />
⎛ ⎞ { ⎛ }} ⎞ ⎛ ⎞{<br />
τ = d I xx 0 0<br />
ω x<br />
ω x<br />
I xx 0 0<br />
ω x<br />
dt<br />
⎜ 0 I yy 0<br />
⎟ ⎜ω y ⎟<br />
+<br />
⎝<br />
⎠ ⎝ ⎠<br />
⎜ω y ⎟<br />
⎝ ⎠ ×<br />
⎜ 0 I yy 0<br />
⎟ ⎜ω y ⎟<br />
⎝<br />
⎠ ⎝ ⎠<br />
⎢<br />
⎣ 0 0 I zz ω z<br />
⎥<br />
⎦ ω z 0 0 I zz ω z<br />
⎛<br />
⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞<br />
I xx 0 0<br />
α x<br />
ω x<br />
I xx ω x<br />
=<br />
⎜ 0 I yy 0<br />
⎟ ⎜α y ⎟<br />
⎝<br />
⎠ ⎝ ⎠ + ⎜ω y ⎟<br />
⎝ ⎠ × ⎜I yy ω y ⎟<br />
⎝ ⎠<br />
0 0 I zz α z ω z I zz ω z<br />
⎛ ⎞ ∣ ∣ ∣∣∣∣∣∣∣∣∣ ∣∣∣∣∣∣∣∣∣<br />
I xx α x<br />
i j k<br />
=<br />
⎜I yy α y ⎟<br />
⎝ ⎠ + det ω x ω y ω z<br />
I zz α z I xx ω x I yy ω y I zz ω z<br />
⎛ ⎞ ⎛<br />
⎞<br />
I xx α x<br />
ω y (I zz ω z ) − ω z (I yy ω y )<br />
=<br />
⎜I yy α y ⎟<br />
⎝ ⎠ + ⎜−ω x (I zz ω z ) + ω z (I xx ω x )<br />
⎟<br />
⎝<br />
⎠<br />
I zz α z ω x (I yy ω y ) − ω y (I xx ω x )<br />
⎛ ⎞ ⎛<br />
⎞<br />
I xx α x<br />
ω y ω z (I zz − I yy )<br />
=<br />
⎜I yy α y ⎟<br />
⎝ ⎠ + ⎜ω x ω z (I xx − I zz )<br />
⎟<br />
⎝<br />
⎠<br />
I zz α z ω x ω y (I yy − I xx )<br />
So, we can see how much simpler it became when using principle axes. Compare the above to<br />
⎛<br />
⎞ ⎛ ⎞ ⎛<br />
⎞<br />
I xx I xy I xz<br />
α x<br />
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 )<br />
⎜I yx I yy I yz ⎟ ⎜α y ⎟<br />
⎝<br />
⎠ ⎝ ⎠ + ⎜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 )<br />
⎟<br />
⎝<br />
⎠<br />
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 )<br />
So, always use principle axes for the body fixed coordinates system!<br />
62
15 Wheel spinning precession<br />
x<br />
MgL<br />
L<br />
y<br />
z<br />
Mg<br />
Wheel spinning around y-axis,<br />
hanging from ceiling<br />
Weight create a torque which<br />
changes the angular<br />
momentum<br />
Torque vector<br />
representation<br />
y<br />
<br />
Important: This<br />
analysis is valid<br />
only for LARGE<br />
y<br />
H Angular momentum<br />
x<br />
z<br />
y<br />
H y I yy y<br />
p<br />
y<br />
t<br />
Angular momentum<br />
change due to the<br />
torque applied<br />
After torque applied for t<br />
z<br />
This is what wheel will<br />
look like after del T<br />
time due to<br />
precesses, notice that<br />
SPIN angular<br />
momentum changes<br />
in the direction of the<br />
torque<br />
y<br />
Time it takes<br />
for the wheel<br />
to precesses<br />
one full cycle<br />
T p 2<br />
p<br />
H<br />
z<br />
Angular momentum<br />
after<br />
t<br />
t<br />
H aftert<br />
x<br />
H I yy y t<br />
x<br />
I yy y<br />
p<br />
y<br />
H t H<br />
t<br />
<br />
But<br />
H<br />
t<br />
I yy y p , hence I yy y p hence p <br />
<br />
I yy y<br />
MgL<br />
I yy y<br />
Nasser M.<br />
Abbasi<br />
Precesses.vsd<br />
6/1/2011<br />
Therefore, precession velocity p is MgL<br />
I yy y<br />
16 References<br />
1. Structural <strong>Dynamics</strong> 5th edition. Mario Paz, William Leigh<br />
63
17 Misc. items<br />
The Jacobian matrix for a system of differential equations, such as<br />
is given by<br />
x ′ (t) = f (x, y, z)<br />
y ′ (t) = g (x, y, z)<br />
z ′ (t) = h (x, y, z)<br />
⎛<br />
J =<br />
⎜<br />
⎝<br />
For example, for the given the following 3 set of coupled differential equations in n 3<br />
df<br />
dx<br />
dg<br />
dx<br />
dh<br />
dx<br />
df<br />
dy<br />
dg<br />
dy<br />
dh<br />
dy<br />
df<br />
dz<br />
dg<br />
dz<br />
dh<br />
dz<br />
x ′ (t) = −y (t) − z (t)<br />
y ′ (t) = x (t) + ay (t)<br />
⎞<br />
⎟<br />
⎠<br />
z ′ (t) = b + z (t) (x (t) − c)<br />
then the Jacobian matrix is<br />
⎛<br />
⎞<br />
0 −1 −1<br />
J =<br />
⎜ 1 a 0<br />
⎟<br />
⎝<br />
⎠<br />
z (t) 0 x (t) − c<br />
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<br />
this space (may be stable point or initial conditions, etc...) and then J become all numerical now. Then we can<br />
evaluate the eigenvalues of the resulting matrix and look to see if all eigenvalues are negative. If so, this tells us<br />
that the point is a stable point. I.e. the system is stable.<br />
If X is N(0, 1) distributed then mu + sigma ∗ X is N(mu, sigma 2 ) distributed.<br />
64
18 Astrodynamics<br />
18.1 Ellipse main parameters<br />
18.2 Table of common equations<br />
The following table contains the common relations to use for elliptic motion. Equation of ellipse is x2<br />
a 2 + y2<br />
b 2 = 1<br />
term to find<br />
conversion between E and θ<br />
position of satellite at time t<br />
Solve for E, then find θ. τ<br />
here is time at perigee and n<br />
is mean satellite speed.<br />
eccentricity e<br />
relation<br />
( θ<br />
tan =<br />
2)<br />
√ ( )<br />
1 + e E<br />
1 − e tan 2<br />
cos E = e + cos θ<br />
1 + e cos θ<br />
cos θ = e − cos E<br />
e cos E − 1<br />
E − e sin E = n (t − τ)<br />
√<br />
e = c a = ra−rp<br />
r a+r p<br />
= 1 + 2Eh2<br />
µ 2<br />
65
Maor axes a<br />
a = r p(1 + e)<br />
1 − e 2<br />
= r a(1 − e)<br />
1 − e 2<br />
= − µ<br />
2E<br />
= √ b 2 + c 2<br />
= p<br />
1 − e 2<br />
Minor axes b b = a √ 1 − e 2<br />
r p = a(1 − e2 )<br />
r p 1 + e<br />
= a(1 − e)<br />
r a = a(1 − e2 )<br />
1 − e<br />
r a = a (1 + e)<br />
p<br />
specific angular momentum h<br />
Total Energy E<br />
= h2<br />
µ<br />
1<br />
1 − e<br />
r p<br />
r a<br />
= 1−e<br />
1+e<br />
p = a ( 1 − e 2) = h2<br />
µ = r p (1 + e) = r a (1 − e)<br />
h = r p v p = r a v a = ⃗r × ⃗v = √ pµ<br />
h = √ µr<br />
(circular orbit)<br />
E = v2<br />
2 − µ r = − µ 2a<br />
√ ( 2<br />
v = µ<br />
r a)<br />
− 1<br />
(vis-viva)<br />
velocity v<br />
√<br />
2µ<br />
v escape = (escape velocity for parabola)<br />
r<br />
√ µ<br />
v radial =<br />
p e sin θ<br />
√ µ<br />
v normal = (1 + e cos θ)<br />
√<br />
p<br />
( )<br />
µ 1 + e<br />
v p =<br />
a 1 − e<br />
v perigee (closest)<br />
v apogee (furthest)<br />
v a =<br />
√ µ<br />
= (1 + e)<br />
p<br />
√ ( 2<br />
= µ − 1 r p a)<br />
√<br />
µ<br />
a<br />
( ) 1 − e<br />
1 + e<br />
√ µ<br />
= (1 − e)<br />
p<br />
√ ( 2<br />
= µ − 1 r a a)<br />
66
magnitude of ⃗r<br />
period T<br />
mean satellite speed n<br />
r = a ( 1 − e 2)<br />
1 + e cos θ = h2 1<br />
µ 1 + e cos θ<br />
r cos θ = a (cos E − e)<br />
r = a (1 − e cos E)<br />
√<br />
T = 2 h πab = 2π a 3<br />
µ<br />
n = 2π<br />
T = √<br />
µ<br />
a 3<br />
eccentric anomaly E tan θ 2 = √<br />
1+e<br />
1−e tan E 2<br />
area sweep rate<br />
dA<br />
dt = h 2<br />
equation of motion ¨⃗r +<br />
µ<br />
r 3 ⃗r = 0<br />
(eq 4.2-14 Bate book)<br />
spherical coordinates relation cos(i) = sin(A z ) cos(φ) where i is the inclination<br />
and A z is the azimuth and φ is latitude<br />
2<br />
Notice in the above, that the period T of satellite depends only on a (for same µ)<br />
In the above, µ = GM where M is the mass of the body at the focus of the ellipse and G is the gravitational<br />
constant. h is the specific mass angular momentum (moment of linear momentum) of the satellite. Hence the<br />
units of h2<br />
µ<br />
is length.<br />
To draw the locus of the satellite (the small body moving around the ellipse, all what we need is the<br />
eccentricity e and a, the major axes length. Then by changing the angle θ the path of the satellite is drawn. I<br />
have a demo on this here<br />
See http://nssdc.gsfc.nasa.gov/planetary/fact<strong>sheet</strong>/earthfact.html for earth facts<br />
This table below is from my class EMA 550 handouts (astrodynamics, spring 2014)<br />
2 Image is from my class notes, page 7-9, EMA 550, Univ. Of Wisconsin, Madison by professor S. Sandrik<br />
67
68
18.3 Flight path angle for ellipse γ<br />
69
√<br />
v = |⃗v| = |⃗v r | 2 + |⃗v n | 2<br />
√ ( 2<br />
= µ<br />
r a)<br />
− 1<br />
To find γ, if r is given then use<br />
If θ is given, then use<br />
√<br />
cos γ =<br />
√<br />
ap<br />
r (2a − r) =<br />
a 2 (1 − e 2 )<br />
r (2a − r)<br />
tan γ =<br />
e sin θ<br />
1 + e cos θ<br />
18.4 Parabolic trajectory<br />
This diagram shows the parabolic trajectory<br />
18.5 Hyperbolic trajectory<br />
This diagram shows the hyperbolic trajectory<br />
70
This diagram below from Orbital mechanics for Engineering students, second edition, by Howard D. Curtis,<br />
page 109<br />
71
18.6 showing that energy is constant<br />
Showing that energy E = v2<br />
2 − µ r<br />
is constant.<br />
Most of such relations starts from the same place. The equation of motion of satellite under the assumption<br />
that its mass is much smaller than the mass of the large body (say earth) it is rotating around. Hence we can<br />
use ν = GM and the equation of motion reduces to<br />
¨⃗r + µ r 3⃗r = 0<br />
In the above equation, the vector ⃗r is the relative vector from the center of the earth to the center of the satellite.<br />
The reason the center of earth is used as the origin of the inertial frame of reference is due to the assumption<br />
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<br />
satellite. Hence the median center of mass between the earth and the satellite is taken to be the center of earth.<br />
This is an approximation, but a very good approximation.<br />
The first step is to dot product the above equation with ˙⃗r giving<br />
˙⃗r · µ<br />
r 3 ⃗r = µ ṙ<br />
r 2<br />
˙⃗r · ¨⃗r + ˙⃗r · µ<br />
r3⃗r = 0 (1)<br />
( )<br />
ṙ2<br />
2<br />
and we also see that<br />
And there is the main trick. We look ahead and see that ˙⃗r · ¨⃗r = ṙ¨r but ṙ¨r = d dt<br />
but µ ṙ = d ( −µ<br />
)<br />
r 2 dt r Hence equation 1 above can be written as<br />
Hence<br />
(<br />
d v<br />
2<br />
dt 2 − r )<br />
= 0<br />
µ<br />
E = v2<br />
2 − r µ<br />
Where E is a constant, which is the total energy of the satellite.<br />
18.7 Earth satellite Transfer orbits<br />
18.7.1 Hohmann transfer<br />
This diagram shows the Hohmann transfer<br />
72
73
18.7.2 Bi-Elliptic transfer orbit<br />
74
18.7.3 semi-tangential elliptical transfer<br />
18.8 Rocket engines, Hohmann transfer, plane change at equator<br />
two cases: Hohmann transfer, 2 burns, or semi-tangential. All burns at equator.<br />
75
76
18.8.1 Rocket equation with plane change not at equator<br />
77
18.9 Spherical coordinates<br />
78
18.10 interplanetary transfer orbits<br />
18.10.1 interplanetary hohmann transfer orbit, case one<br />
The following are the steps to accomplish the above. The first stage is getting into the Hohmann orbit from<br />
planet 1, then reaching the sphere of influence of the second planet. Then we either do a fly-by or do a parking<br />
orbit around the second planet. These steps below show how to reach the second planet and do a parking orbit<br />
around it.<br />
79
The input is the following.<br />
1. µ 1 planet one standard gravitational parameter<br />
2. µ 2 planet two standard gravitational parameter<br />
3. µ sun standard gravitational parameter for the sun 1.327 × 10 8 km<br />
4. r 1 planet one radius<br />
5. r 2 planet two radius<br />
6. alt 1 original satellite altitude above planet one. For example, for LEO use 300 km<br />
7. alt 2 satellite altitude above second planet. (since goal is to send satellite for circular orbit around second<br />
planet)<br />
8. R 1 mean distance of center of first planet from the sun. For earth use AU = 1.495978 × 10 8 km<br />
9. R 2 mean distance of center of second planet from the sun. For Mars use 1.524 AU<br />
10. SOI 1 sphere of influence for first planet. For earth use 9.24 × 10 8 km<br />
11. SOI 2 sphere of influence for second planet.<br />
Given the above input, there are the steps to achieve the above maneuver<br />
1. Find the burn out distance of the satellite r bo = r 1 + alt 1<br />
2. Find satellite speed around planet earth (relative to planet) V sat =<br />
√<br />
µ1<br />
r bo<br />
3. Find Hohmann ellipse a = R 1+R 2<br />
2<br />
4. Find speed of satellite at perigee relative to sun V perigee =<br />
√<br />
µ sun<br />
(<br />
2<br />
R 1<br />
− 1 a<br />
)<br />
5. Find speed of earth (first planet) relative to sun V 1 =<br />
√<br />
µsun<br />
R 1<br />
6. Find escape velocity from first planet V ∞,out = V perigee − V 1<br />
7. Find burn out speed at first planet by solving the energy equation V 2<br />
bo<br />
2 − µ 1<br />
r bo<br />
8. Find ∆V 1 needed at planet one ∆V 1 = V bo − V sat<br />
9. Find e the eccentricity of the escape hyperbola e =<br />
√<br />
1 + V 2 ∞V 2<br />
bo r2 bo<br />
µ 2 1<br />
10. Find the angle with the path of planet one velocity vector η = arccos ( − 1 e<br />
11. Find the dusk-line angle θ = 180 0 − η<br />
)<br />
= V 2 ∞,out<br />
2<br />
− µ 1<br />
SOI 1<br />
for V bo<br />
The above completes the first stage, now the satellite is in the Hohmann transfer orbit. Assuming it reached<br />
the orbit of the second planet ahead of it as shown in the diagram above. Now we start the second stage to land<br />
the satellite on a parking orbit around the second planet at altitude alt 2 above the surface of the second planet.<br />
These are the steps needed.<br />
1. Find the apogee speed of the satellite V apogree =<br />
√<br />
µ sun<br />
(<br />
2<br />
R 2<br />
− 1 a<br />
)<br />
80
2. Find speed of second planet V 2 =<br />
√<br />
µsun<br />
R 2<br />
3. Find V ∞ entering the second planet sphere of influence V ∞,in = V 2 − V apogree<br />
4. Find burn in radius where the satellite will be closest to the second planet. r bo = r 1 + alt 2<br />
5. Find burn out speed at second planet by solving the energy equation V 2<br />
bo<br />
2 − µ 2<br />
r bo<br />
6. Find impact parameter b on entry to second planet SOI b = r boV bo<br />
V ∞,in<br />
7. Find the required satellite speed around the second planet V sat =<br />
√<br />
µ2<br />
r bo<br />
8. Find ∆V 2 needed at planet two ∆V 2 = V sat − V bo<br />
9. Find e the eccentricity of the approaching hyperbola on second planet e =<br />
10. Find the angle with the path of planet two velocity vector η = arccos ( − 1 e<br />
11. Find the dusk-line angle for second planet θ = 2η − 180 0<br />
18.11 rendezvous orbits<br />
18.11.1 Two satellite, walking rendezvous using Hohmann transfer<br />
)<br />
√<br />
= V 2 ∞,in<br />
2<br />
− µ 2<br />
SOI 2<br />
for V bo<br />
1 + V 2 ∞ V 2<br />
bo r2 bo<br />
µ 2 2<br />
81
Algorithm 1 Hohmann Walking Rendezvous Orbit, case 1<br />
1: function hohmann walking rendezvous(θ 0 , r, altitude, µ)<br />
2: θ 0 := θ 0<br />
π<br />
180<br />
⊲ convert from degrees to radian<br />
3: r a := r + altitude<br />
4: T := 2π√<br />
r 3 aµ<br />
⊲ period of circular orbit<br />
5: n := 1<br />
6: done:=false<br />
7: while not(done) do<br />
(<br />
8: TOF := N − θ 0<br />
9: a := solve<br />
10: r p := 2a − r a<br />
11: if rp < r then<br />
12: N = N + 1<br />
13: else<br />
14: done:=true<br />
15: end if<br />
16: end while<br />
2π<br />
)<br />
T<br />
(<br />
T OF = 2π<br />
√<br />
17: V befor := µ<br />
h<br />
√<br />
18: V after := µ ( 2<br />
h − 1 )<br />
a<br />
19: ∆V := 2(V after − V before )<br />
20: return (TOF, ∆V )<br />
21: end function<br />
√<br />
An example implementation is below<br />
)<br />
a 3<br />
µ<br />
for a<br />
1 hohmannRendezvousSameOrbit[\[Theta]00_, r_, alt_, mu_] :=<br />
2 Module[{\[Theta]0 = \[Theta]00*Pi/180, n = 1, delT, v1, v2, period, a,<br />
3 rp, ra, done = False, vBefore, vAfter},<br />
4 ra = r + alt;<br />
5 period = 2 Pi Sqrt[ra^3/mu];<br />
6 While[Not[done],<br />
7 delT = (n - \[Theta]0 /(2 Pi)) period ;<br />
8 a = First@Select[a /. NSolve[delT == 2 Pi Sqrt[a^3/mu], a], Element[#, Reals] &];<br />
9 rp = 2 a - ra;<br />
10 If[rp < r,(*we hit the earth, try again*)<br />
11 n = n + 1,<br />
12 done = True<br />
13 ]<br />
14 ];<br />
15 vBefore = Sqrt[mu/h];<br />
16 vAfter = Sqrt[mu (2/h - 1/a)];<br />
17 {delT, 2 (vAfter - vBefore)}<br />
18 ]<br />
And calling the above<br />
82
1 mu = 324859;<br />
2 alt = 1475.776;<br />
3 r = 6052;<br />
4 \[Theta]0 = 3.80562; (*degree*)<br />
5 hohmannRendezvousSameOrbit[\[Theta]0, r, alt, mu]<br />
6 {7123.89, -0.0467913}<br />
18.11.2 Two satellite, separate orbits, rendezvous using Hohmann transfer, coplaner<br />
83
Algorithm 2 Hohmann rendezvous algorithm, case 1<br />
1: function hohmann rendezvous 1(θ 0 , r a , r b , µ)<br />
π<br />
2: θ 0 := θ 0 180<br />
⊲ convert from degrees to radian<br />
3: a := ra+r b<br />
2<br />
⊲ Hohmann orbit semi-major axes<br />
√<br />
4: TOF := π<br />
5: θ H := π<br />
6: ω a :=<br />
a 3<br />
µ<br />
⊲ time of flight on Hohmann orbit<br />
( ( ) ) 3/2<br />
1 − ra+r b<br />
2r b<br />
√ µ<br />
r 3 a<br />
⊲ required phase angle before starting Hohmann transfer<br />
⊲ angular speed of lower rad/sec<br />
7: ω b := √ µ<br />
⊲ angular speed of higher satellite rad/sec<br />
rb<br />
3<br />
8: if θ 0 ≤ θ H then ⊲ adjust initial angle if needed<br />
9: θ 0 := θ 0 + 2π<br />
10: end if<br />
11: wait time := θ 0−θ H<br />
ω a−ω b<br />
⊲ how long to wait before starting Hohmann transfer<br />
12: wait time := wait time + TOF ⊲ now ready to go, add Hohmann transfer time<br />
13: return wait time<br />
14: end function<br />
An example implementation is below<br />
1 hohmann_rendezvous_1:= proc({<br />
2 theta::numeric:=0,<br />
3 r1::numeric:=0,<br />
4 r2::numeric:=0,<br />
5 mu::numeric:=3.986*10^5})<br />
6 local theta0,thetaH,TOF,a,omega1,omega2,wait_time;<br />
7 theta0 := evalf(theta*Pi/180);<br />
8 a := (r1+r2)/2;<br />
9 TOF := Pi*(sqrt(a^3/mu));<br />
10 omega1 := sqrt(mu/r1^3);<br />
11 omega2 := sqrt(mu/r2^3);<br />
12 thetaH := evalf(Pi*(1-((r1+r2)/(2*r2))^(3/2)));<br />
13 if theta0
18.11.3 Two satellite, separate orbits, rendezvous using bi-elliptic transfer, coplaner<br />
In this transfer, the lower (fast satellite) does not have to wait for phase lock as in the case with Hohmann<br />
transfer. The transfer can starts immediately. There is a free parameter N that one select depending on fuel<br />
cost requiments or any limitiation on the first transfer orbit semi-major axes distance required. One can start<br />
with N = 0 and adjust as needed.<br />
85
Algorithm 3 Hohmann rendezvous algorithm, case 2<br />
1: function hohmann rendezvous 2(θ 0 , r 1 , r 2 , N, µ)<br />
π<br />
2: θ 0 := θ 0 180 (<br />
⊲ convert from degrees to radian<br />
( ) ) 3/2<br />
3: θ H := π 1 − r1 +r 2<br />
2r 2<br />
⊲ Find Hohmann ideal phase angle before transfer<br />
4: if θ 0 = θ H and N = 0 then ⊲ adjust for special case<br />
5: a := r 2+r 1<br />
2 (√<br />
6: TOF := π<br />
)<br />
a 3<br />
µ<br />
7: else<br />
8: ω 2 := √ µ<br />
r2<br />
3<br />
9: t 2 := (2π−θ 0)+2πN<br />
ω 2<br />
10: a 1 := rt+r 1<br />
2<br />
11: a 2 := rt+r 2<br />
2<br />
12: TOF := π<br />
(√<br />
a 3<br />
1<br />
µ<br />
+ a3 2<br />
µ<br />
)<br />
⊲ angular speed of slower satellite in rad/sec<br />
⊲ find time of light of the slower satellite<br />
⊲ time of flight for the fast satellite<br />
13: r t := solve (t 2 = T OF ) for r t ⊲ Solve numerically for r t<br />
14: end if<br />
15: return TOF<br />
16: end function<br />
An example implementation is below<br />
1 hohmann_rendezvous_2:= proc({<br />
2 theta::numeric:=0,<br />
3 r1::numeric:=0,<br />
4 r2::numeric:=0,<br />
5 N::nonnegint:=0,<br />
6 mu::numeric:=3.986*10^5})<br />
7 local theta0,thetaH,TOF;<br />
8 theta0 := theta*Pi/180;<br />
9 thetaH := Pi*(1-((r1+r2)/(2*r2))^(3/2));<br />
10 if theta0 = thetaH and N = 0 then<br />
11 proc()<br />
12 local a:=(r1+r2)/2;<br />
13 TOF:= Pi*(sqrt(a^3/mu));<br />
14 end proc()<br />
15 else<br />
16 proc()<br />
17 local t2,a1,a2,rt,omega2;<br />
18 omega2 := sqrt(mu/r2^3);<br />
19 t2 := ((2*Pi-theta0)+2*Pi*N)/omega2;<br />
20 a1 := (rt+r1)/2;<br />
21 a2 := (rt+r2)/2;<br />
22 TOF := Pi*(sqrt(a1^3/mu)+sqrt(a2^3/mu));<br />
23 rt := op(select(is, [solve(t2=TOF,rt)], real));<br />
24 end proc()<br />
25 fi;<br />
26 eval(TOF);<br />
27 end proc:<br />
And calling the above for two different cases gives<br />
86
1 TOF:=hohmann_rendezvous_2(theta=0,r1=6678,r2=6878,N=0):<br />
2 evalf(TOF/(60*60)); #in hrs<br />
3 1.576892101<br />
4 TOF:=hohmann_rendezvous_2(theta=160,r1=6678,r2=6878,N=1):<br />
5 evalf(TOF/(60*60)); #in hrs<br />
6 2.452943266<br />
87
18.12 Semi-tangential transfers, elliptical, parabolic and hyperbolic<br />
88
89
18.13 Lagrange points<br />
90
18.14 Orbit changing by low contiuous thrust<br />
18.15 References<br />
1. Orbital mechanics for Engineering students, second edition, by Howard D. Curtis<br />
2. Orbital Mechanics, Vladimir Chobotov, second edition, AIAA<br />
3. Fundamentals of Astrodynamics, Bates, Muller and White. Dover1971<br />
91
19 Dynamic of flights<br />
19.1 Wing geometry<br />
length of mean<br />
aerodynamic chord<br />
projected on the<br />
root chord<br />
Leading edge<br />
sweep angle<br />
root chord<br />
centroid of area of<br />
one half of wing<br />
located on mean<br />
aerodynamic chord<br />
local chord<br />
wing tip chord<br />
wing1.vsdx<br />
Nasser M. Abbasi<br />
020914<br />
half wing span<br />
92
C r below is the core chord of the wing.<br />
A.C.<br />
tail<br />
control-fixed<br />
static margin<br />
apex of<br />
wing<br />
wing<br />
A.C.<br />
mac<br />
c.g<br />
positive for<br />
static<br />
stability<br />
wing_4_generic.vsdx<br />
Nasser M. Abbasi<br />
020914<br />
This is a diagram to use to generate equations of longitudinal equilibrium.<br />
This distance is called the stick-fixed static margin k m = (h n − h) ¯c Must be positive for static stability<br />
93
tail mean<br />
aerodynamic<br />
chord<br />
The whole Airplane<br />
neutral point (aft c.g.<br />
for static stability)<br />
Wing mean<br />
aerodynamic chord<br />
km<br />
forces.vsd<br />
updated 2/3/<br />
14<br />
Nasser M.<br />
Abbasi<br />
19.2 Summary of main equations<br />
This table contain some definitions and equations that can be useful.<br />
# equation meaning/use<br />
94
1<br />
C L = ∂C L<br />
∂α α<br />
= C Lα α<br />
= aα<br />
C L is lift coefficient. α is angle<br />
of attack. a is slope ∂C L<br />
∂α<br />
which<br />
is the same as C Lα<br />
2 C Lw = C Lwα α wing lift coefficient<br />
3 C D = C Dmin + kC 2 L<br />
drag coefficient<br />
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<br />
due to wing only about the<br />
C.G. of the airplane assuming<br />
small α w . This is simplified<br />
more by assuming C Dw α w ≪<br />
C Lw and (C L α w − C Dw ) ≪ 1<br />
5 C mw = C macw + C Lw (h − h nw ) simplified wing Pitching moment<br />
C mwb = C macwb + C Lwb (h − h nw )<br />
6<br />
= C macwb + ∂C L wb<br />
∂α wb<br />
α wb (h − h nw )<br />
= C macwb + a wb α wb (h − h nw )<br />
simplified pitching moment coefficient<br />
due to wing and body<br />
about the C.G. of the airplane.<br />
α wb is the angle of attack<br />
7 C Lt = Lt<br />
1<br />
2 ρV 2 S t<br />
C Lt is the lift coefficient generated<br />
by tail. S t is the tail area.<br />
V is airplane air speed<br />
8 L = L wb + L t total lift of airplane. L wb is lift<br />
due to body and wing and L t<br />
is lift due to tail<br />
9 C L = C Lwb + St<br />
S C L t<br />
coefficient of total lift of airplane.<br />
C Lwb is coefficient of lift<br />
due to wing and body. C Lt is<br />
lift coefficient due to tail. S is<br />
the total wing area. S t is tail<br />
area<br />
10<br />
1<br />
M t = −l t L t = −l t C Lt 2 ρV 2 S t pitching moment due to tail<br />
about C.G. of airplane<br />
11 C mt = Mt = − lt¯c<br />
S t<br />
1<br />
2 ρV 2 S t¯c S C L t<br />
= −V H C Lt pitching moment coefficient<br />
due to tail. V H = lt¯c<br />
S t<br />
S<br />
is called<br />
tail volume<br />
12<br />
V H = lt¯c<br />
S t<br />
S<br />
¯V H = ¯l t¯c<br />
S t<br />
S<br />
introducing ¯V H bar tail volume<br />
which is V H but uses ¯l t instead<br />
of l t . Important note. V H<br />
depends on location of C.G.,<br />
but ¯V H does not. ¯lt = l t +<br />
(h − h nwb ) ¯c<br />
95
13 C mt = − ¯V H C Lt + C Lt<br />
S t<br />
S (h − h n wb<br />
) pitching moment coefficient<br />
due to tail expressed using ¯V H .<br />
This is the one to use.<br />
14 C mp pitching moment coefficient<br />
due to propulsion about airplane<br />
C.G.<br />
15 C m = C mwb + C mt + C mp total airplane pitching moment<br />
coefficient about airplane C.G.<br />
16<br />
C m = C mwb + C mt + C mp<br />
[<br />
]<br />
= C macwb + C Lwb (h − h nw )<br />
C L<br />
+ [ − ¯V H C Lt + C Lt<br />
S t<br />
S (h − h n wb<br />
) ] + C mp<br />
{ ( }} ) {<br />
simplified total Pitching moment<br />
coefficient about airplane<br />
S t<br />
= C macwb + C Lwb + C Lt (h − h nw ) −<br />
S<br />
¯V H C Lt + C mp<br />
C.G.<br />
= C macwb + C L (h − h nw ) − ¯V H C Lt + C mp<br />
17<br />
∂C m<br />
∂α<br />
= ∂Cmac wb<br />
∂α<br />
C mα = ∂Cmac wb<br />
∂α<br />
18 h n = h nwb − 1 ∂C L<br />
∂α<br />
+ ∂C L<br />
∂α<br />
(h − h n w<br />
) − ¯V H<br />
∂C Lt<br />
∂α<br />
+ C Lα (h − h nw ) − ¯V H<br />
∂C Lt<br />
∂α<br />
( ∂Cmacwb<br />
∂α<br />
− ¯V H<br />
∂C Lt<br />
∂α<br />
)<br />
+<br />
∂Cmp<br />
∂α<br />
+ ∂Cmp<br />
∂α<br />
+<br />
∂Cmp<br />
∂α<br />
derivative of total pitching moment<br />
coefficient C m w.r.t airplane<br />
angle of attack α<br />
location of airplane neutral<br />
point of airplane found by setting<br />
C mα = 0 in the above<br />
equation<br />
19<br />
∂C m<br />
∂α<br />
= ∂C L<br />
∂α (h − h n)<br />
C mα = C Lα (h − h n )<br />
rewrite of C mα in terms of h n .<br />
Derived using the above two<br />
equations.<br />
20 k n = h n − h static margin. Must be Positive<br />
for static stability<br />
19.2.1 Writing the equations in linear form<br />
The following equations are derived from the above set of equation using what is called the linear form. The main<br />
point is to bring into the equations the expression for C Lt written in term of α wb . This is done by expressing the<br />
∂C Lwb<br />
tail angle of attack α t in terms of α wb via the downwash angle and the i t angle.<br />
∂α wb<br />
in the above equations are<br />
replaced by a wb and ∂C L t<br />
∂α t<br />
is replaced by a t . This replacement says that it is a linear relation between C L and<br />
the corresponding angle of attack. The main of this rewrite is to obtain an expression for C m in terms of α wb<br />
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<br />
from now on.<br />
# equation meaning/use<br />
96
1<br />
2<br />
3<br />
C Lwb = ∂C L wb<br />
∂α wb<br />
α wb<br />
= a wb α wb<br />
C Lt = a t α t<br />
C mp = C m0 p + ∂C mp<br />
∂α α<br />
α t = α wb − i t − ɛ<br />
ɛ = ɛ 0 + ∂ɛ<br />
∂α α wb<br />
C Lt = a t α t<br />
[ (<br />
= a t α wb 1 − ∂ɛ ) ]<br />
− i t − ɛ 0<br />
∂α<br />
a wb is constant, represents<br />
∂C L wb<br />
∂α wb<br />
and C m0p<br />
is propulsion pitching<br />
moment coeff. at zero<br />
angle of attack α<br />
main relation that associates<br />
α wb with α t . α wb<br />
is the wing-body angle<br />
of attack, ɛ is downwash<br />
angle at tail, and<br />
i t is tail angle with horizontal<br />
reference (see diagram)<br />
Lift due to tail expressed<br />
using α wb and<br />
ɛ (notice that α t do not<br />
show explicitly)<br />
(<br />
4 a = a wb<br />
[1 + at S t<br />
a wb S 1 −<br />
∂ɛ<br />
∂α) ] a defined for use with<br />
overall lift coefficient<br />
5<br />
C L =<br />
a wb α { }}<br />
wb<br />
{<br />
C Lwb<br />
+ St<br />
S C L t<br />
= a wb α wb + St<br />
S a t<br />
= aα<br />
= (C L ) αwb =0 + aα wb<br />
[<br />
αwb<br />
(<br />
1 −<br />
∂ɛ<br />
∂α<br />
)<br />
− it − ɛ 0<br />
]<br />
overall airplane lift using<br />
linear relations<br />
6 α = α wb − at S t<br />
a S (i t + ɛ 0 )<br />
overall angle of attack<br />
α as function of the<br />
wing and body angle of<br />
attack α wb and tail angles<br />
cl_apha.vsdx<br />
Nasser M. Abbasi<br />
021 214<br />
7<br />
C m = C m0 + ∂Cm<br />
∂α α = C m0 + C mα α<br />
C m = ¯C m0 + ∂Cm<br />
∂α α wb = ¯C m0 + C mα α wb<br />
overall airplane pitch<br />
moment. Two versions<br />
one uses α wb and one<br />
uses α<br />
97
8<br />
9<br />
( )<br />
C mα = a (h − h nwb ) − a t ¯VH 1 −<br />
∂ɛ ∂C<br />
∂α + mp<br />
∂α<br />
C mα = a wb (h − h nwb ) − a t V H<br />
(<br />
1 −<br />
∂ɛ<br />
∂α<br />
)<br />
+<br />
∂C mp<br />
∂α<br />
C m0 = C macwb + C mop + a t ¯VH (ɛ 0 + i t ) [ 1 − at S t<br />
a S<br />
¯C m0 = C macwb + ¯C mop + a t V H (ɛ 0 + i t )<br />
( )]<br />
1 −<br />
∂ɛ<br />
∂α<br />
∂C m<br />
∂α<br />
Two versions of<br />
one for α wb and one one<br />
uses α<br />
C m0 is total pitching<br />
moment coef. at zero<br />
lift (does not depend on<br />
C.G. location) but ¯C m0<br />
is total pitching moment<br />
coef. at α wb = 0<br />
(not at zero lift). This<br />
depends on location of<br />
C.G.<br />
10<br />
¯Cm0p = C m0p + (α − α wb ) ∂Cmp<br />
∂α<br />
h n = h nwb + at ¯V<br />
( )<br />
a H 1 −<br />
∂ɛ<br />
∂α −<br />
1 ∂C mp<br />
a ∂α<br />
11<br />
= h nwb +<br />
a t (<br />
a wb<br />
[1+ a t S t<br />
a wb S<br />
1− ∂ɛ<br />
∂α<br />
)] ¯VH<br />
(<br />
1 −<br />
∂ɛ<br />
∂α<br />
)<br />
−<br />
1<br />
a wb<br />
[1+ a t S t<br />
a wb S<br />
(<br />
)] ∂Cmp<br />
1− ∂ɛ ∂α<br />
∂α<br />
Used to determine h n<br />
19.3 definitions<br />
1. Remember that for symmetric airfoil, when the chord is parallel to velocity vector, then the angle of attack<br />
is zero, and also the left coefficient is zero. But this is only for symmetric airfoil. For the common campbell<br />
airfoil shape, when the chord is parallel to the velocity vector, which means the angle of attack is zero,<br />
there will still be lift (small lift, but it is there). What this means, is that the chord line has to tilt down<br />
more to get zero lift. This extra tilting down makes the angle of attack negative. If we now draw a line<br />
from the right edge of the airfoil parallel to the velocity vector, this line is called the zero lift line (ZLL)<br />
see diagram below.<br />
Just remember, that angle of attack (which is always the angle between the chord and the velocity vector,<br />
the book below calls it the geometrical angle of attack) is negative for zero lift. This is when the airfoil is<br />
not symmetric. For symmetric airfoil, ZLL and the chord line are the same. This angle is small, −3 0 or so.<br />
Depending on shape. See Foundations of Aerodynamics, 5th ed, by Chow and Kuethe, here is the diagram.<br />
98
2. stall from http://en.wikipedia.org/wiki/Stall_(flight)<br />
In fluid dynamics, a stall is a reduction in the lift coefficient<br />
generated by a foil as angle of attack increases.[1] This occurs when<br />
the critical angle of attack of the foil is exceeded. The critical<br />
angle of attack is typically about 15 degrees, but it<br />
may vary significantly depending on the fluid, foil, and Reynolds number.<br />
3. Aerodynamics in road vehicle wiki page<br />
4. some demos relating to airplane control http://demonstrations.wolfram.com/ControllingAirplaneFlight/<br />
http://demonstrations.wolfram.com/ThePhysicsOfFlight/<br />
5. http://www.americanflyers.net/aviationlibrary/pilots_handbook/chapter_3.htm<br />
6. Lectures Helicopter Aerodynamics and <strong>Dynamics</strong> by Prof. C. Venkatesan, Department of Aerospace<br />
Engineering, IIT Kanpur http://www.youtube.com/watch?v=DKWj2WzYXtQ&list=PLAE677E56C97A7C7D<br />
7. http://avstop.com/ac/apgeneral/terminology.html has easy to understand definitions airplane geometry.<br />
”The MAC is the mean average chord of the wing”<br />
8. http://www.tdmsoftware.com/afd/afd.html airfoil design software<br />
99
19.4 images and plots collected<br />
These are diagrams and images collected from different places. References is given next to each image.<br />
Perpendicular to c.g. velocity vector V<br />
Body x-axes<br />
(chord of<br />
airplane<br />
Zero left line<br />
for airplane<br />
Current<br />
velocity<br />
vector of<br />
airplane c.g.<br />
Parallel to V<br />
vector originate<br />
from A.C. and not<br />
C.G.<br />
Angle of attack of thrust<br />
Absolute angle of attack<br />
Weight of<br />
airplane.<br />
Vertically down<br />
By Nasser M. Abbasi<br />
my_drawing.vsd<br />
1/24/14<br />
100
http://en.wikipedia.org/wiki/Flight_dynamics_%28aircraft%29<br />
Dihedral angle<br />
Reference: http://en.wikipedia.org/wiki/Dihedral_%28aircraft%29<br />
101
definitions<br />
From Performance, Stability, <strong>Dynamics</strong>, and Control of Airplanes<br />
By Baudu N. Pamadi<br />
I do not see viscosity here?<br />
http://en.wikipedia.org/wiki/Drag_coefficient<br />
This below from http://www.grc.nasa.gov/WWW/k-12/UEET/StudentSite/dynamicsofflight.html<br />
102
http://www.grc.nasa.gov/WWW/k-12/UEET/StudentSite/dynamicsofflight.html<br />
http://www.grc.nasa.gov/WWW/k-12/airplane/alr.html<br />
103
http://www.grc.nasa.gov/WWW/k-12/airplane/alr.html<br />
http://www.grc.nasa.gov/WWW/k-12/airplane/alr.html<br />
From http://en.wikipedia.org/wiki/Lift_coefficient and http://en.wikipedia.org/wiki/File:Aeroforces<br />
104
http://en.wikipedia.org/wiki/Lift_coefficient<br />
http://en.wikipedia.org/wiki/File:Aeroforces.svg<br />
svg<br />
from http://adg.stanford.edu/aa241/drag/sweepncdc.html<br />
http://en.wikipedia.org/wiki/Lift_%28force%29<br />
105
http://adg.stanford.edu/aa241/drag/sweepncdc.html<br />
Images from http://adamone.rchomepage.com/cg_calc.htm and Flight dynamics principles by Cook, 1997.<br />
106
http://adamone.rchomepage.com/cg_calc.htm<br />
From http://chrusion.com/BJ7/SuperCalc7.html<br />
107
From http://www.willingtons.com/aircraft_center_of_gravity_calcu.html<br />
108
From http://www.solar-city.net/2010/06/airplane-control-surfaces.html nice diagram that shows<br />
clearly how the elevator causes the pitching motion (nose up/down). From same page, it says ”The purpose of<br />
109
the flaps is to generate more lift at slower airspeed, which enables the airplane to fly at a greatly reduced speed<br />
with a lower risk of stalling.”<br />
Images from flight dynamics principles, by Cook, 1997.<br />
110
Page 184, flight dynamics principles, by Cook<br />
Page 41, flight dynamics principles, by Cook<br />
Images from Performance, stability, dynamics and control of Airplanes. By Pamadi, AIAA press. Page 169.<br />
and http://www.americanflyers.net/aviationlibrary/pilots_handbook/chapter_3.htm<br />
111
From: Performance, stability, dynamics and control of<br />
Airplanes. By Pamadi, AIAA press. Page 169<br />
http://www.americanflyers.net/aviationlibrary/pilots_handbook/chapter_3.htm<br />
Image from http://www.americanflyers.net/aviationlibrary/pilots_handbook/chapter_3.htm<br />
112
Image from http://www.americanflyers.net/aviationlibrary/pilots_handbook/chapter_3.htm<br />
113
The profile drag of a streamlined object held<br />
in a fixed position relative to the airflow increases<br />
approximately as the square of the velocity;<br />
thus, doubling the airspeed increases the drag four times,<br />
and tripling the airspeed increases the drag nine times.<br />
The amount of induced drag varies inversely<br />
as the square of the airspeed.<br />
From the foregoing discussion, it can be noted that parasite drag increases<br />
as the square of the airspeed, and induced drag varies inversely as<br />
the square of the airspeed.<br />
lift varies directly with the wing area<br />
a wing with a planform area of 200 square feet lifts twice as<br />
much at the same angle of attack as a wing with an area of 100<br />
square feet.<br />
http://www.americanflyers.net/aviationlibrary/pilots_handbook/chapter_3.htm<br />
Image from FAA pilot handbook and http://www.youtube.com/watch?v=8uT55aei1NI<br />
114
From FAA pilot’s handbook<br />
I do not understand the above<br />
From FAA pilot’s handbook<br />
http://www.youtube.com/watch?v=8uT55aei1NI<br />
Image http://www.youtube.com/watch?v=8uT55aei1NI and http://www.youtube.com/user/DAMSQAZ?<br />
feature=watch<br />
115
http://www.youtube.com/watch?v=8uT55aei1NI<br />
http://www.youtube.com/user/DAMSQAZ?feature=watch<br />
A=area of wing<br />
D=density of air<br />
V=wind speed relative to<br />
wing<br />
116
zero-lift angle<br />
The angle of attack at which an airfoil does not produce any lift. Its value is generally less than zero unless<br />
the airfoil is symmetrical.<br />
This from Foundations of aerodynamics, 5 th ed. By Chow and Kuethe<br />
http://www.answers.com/topic/zero-lift-angle<br />
These are from text :foundation of aerodynamics by Kuethe and Chow<br />
1. the center of pressure is at ¼ chord for all values of lift coeff.<br />
2. center of pressure (c.p.) of a force is defined as the point about which the moment vanishes.<br />
3. The geometric angle of attack is defined as the angle between the flight path and the chord line of the airfoil.<br />
When the geometric angle of attack is zero, the lift coeff. Is zero.<br />
4. The point about which the moment coeff. Is independent of the angle of attack is called the aerodynamic<br />
center of the section. (a.c.)<br />
5. a.c. is at the ¼ chord line point.<br />
6. The value of the angle of attack that makes the lift coeff. Zero is called the angle of zero lift (Z.L.L.)<br />
There are from Foundations<br />
of aerodynamics, 5 th ed. By<br />
Chow and Kuethe<br />
19.5 Some strange shaped airplanes<br />
Image http://edition.cnn.com/2014/01/16/travel/inside-airbus-beluga/index.html?hpt=ibu_c2<br />
117
Image from http://edition.cnn.com/2014/01/16/travel/inside-airbus-beluga/index.html?hpt=ibu_<br />
c2<br />
Image from http://www.nasa.gov/centers/dryden/Features/super_guppy.html<br />
118
Image from http://www.aerospaceweb.org/question/aerodynamics/q0130.shtml ”Boeing Pelican ground<br />
effect vehicle”<br />
19.6 links<br />
1. https://3dwarehouse.sketchup.com/search.html?redirect=1&tags=airplane<br />
19.7 references<br />
1. Etkin and Reid, <strong>Dynamics</strong> of flight, 3rd edition.<br />
2. Cook, Flight <strong>Dynamics</strong> principles, third edition.<br />
3. Lecture notes, EMA 523 flight dynamics and control, University of Wisconsin, Madison by Professor<br />
Riccardo Bonazza<br />
4. Kuethe and Chow, Foundations of Aerodynamics, 4th edition<br />
119