Equations Flashcards
Whats is Newtons Second Law
sum F = ma
Rotational equivalent of Newtons Second Law
sum MG= IG * alpha
general equation for sinusoidal motion
x = a sin(wt + phi)
where a is amplitude, w is frequency, t is time and phi phase shift
alternative form of sinusoidal motion
x = Acos(wt) + Bsin(wt)
where A = a sin(phi) and B = a cos (phi)
Velocity equation and how is it found
x. = -Awsin(wt) + Bwcos(wt)
by taking the time derivative
Acceleration equation
x.. = - Aw^2 * cos (wt) - Bw^2 * sin(wt) = -w^2 x
eulers equation of motion
x = a*e^jwt
x. = jwae^jwt = jw x
x. . = -w^2e^jwt = -w^2 x
undamped natural frequency
wn = sqrt (k/m)
Equation of free vibrations
x(t) = Acos(wnt) + Bsin(wnt)
natural frequency is
the preferred vibration frequency of an undamped system
equation of motion of damped system
mx.. + cx. + kx = 0
characteristic equation of motion for damped system
ms^2 + cs + k =0
critical damping equation
cc = 2 sqrt (km) = 2m wn
damping ration =
c/cc
characteristic equation of motion for undamped system
mx.. + kx = 0
Overdamped system equation
see book
damped natural frequency
wc = wn sqrt ( 1 - zeta^2)
underdamped system repsonse
see book
log decrement =
ln (xn-1/xn)
alternative form of underdamped free vibration equation
see book
log decrement simplified form (time period)
zeta * wn * taud
log decrement simplified form
2 PI() Zeta/ sqrt(1 - zeta^2)
logDec for several cycles
logdec = 1 /n * ln (xo/xn)
low damping zeta from logdec
zeta = ln(x0/xn) / wn * tdrop
where tdrop is the time taen to fall from x0 to xn
Force vibration equation X/F =
1/ (k -w^2 m + jwc)
what is the magnification factor equal for an undamped forced vibration system
1/ (1 - r^2)
Dynamic magnification =
X / Xstat where X stat = F/k
normalised frequency r =
frequency ratio
w / wn
equation of motion of forced damped system
mx.. + cx. + kx = f(t)
Force vibration equation using system terms X/Xtat =
1/ (1 - r^2 + j *2 * zeta * r)
Magnitude of damped frequency response
Xdyn / X stat = 1/sqrt((1-r^2)^2 + (2zetar)^2)
phase angle =
tan(phi) = 2zetar/(1-r^2)
When does the maximum response occur for damped forced vibrations
r = sqrt ( 1 - 2*zeta^2) when zeta is less than 1/root 2
What is the maximum response equal to at zeta less than 1/root 2
Xk/F = 1 / 2*zeta * sqrt ( 1 - zeta^2)
Q =
mag @max X dyn / X stat = 1 /2*zeta * sqrt (1-zeta^2)
Q at low damping
Q = 1/2*zeta
Force transmission equation =>
FT/F = (k + jwc)/(k - w^2m + jwc)
Force transmission equation in system properties =>
FT/F = (1 + j2zetar)/(1-r^2 + j2zetar)
equation of motion for base motion
mx.. + c(x.-y.) + k(x-y) = 0
Z/Y =
w^2 m / k - w^2 m + jwc
r^2 / 1 - r^2 + j2zeta*r
where Z = X - Y
Displacement response of rotating unbalance
X = Me/m * (r^2) / (1 - r^2 + j2zeta*r)
equation of motion of system excited by rotating unbalance
mx.. + cx. + kx = Mew^2 *exp (jwt)
where e is eccentricity of the unbalance
and M the mass of the eccentricity
Force due to spring
F = kx
Work done/strain energy stored in a spring
W or U = 1/2 k x^2
When mass elements are rigidly connected together equivalent mass for translational motion is
sum of the masses
meq = m1 + m2 + m3
Rigid connection for inertia,
rigidly connected together for rotation about a point
J0 = (JG1 + m1r1^2) + (JG2 + m1r2^2) + (JG3 + m1*r3^2)
Equivalent Rotation Inertia Jeq =
Jeq = mR^2 + J0
Equivalent translational mass meq =
m + J0/R^2
Equivalent kinetic energy equation =
T = sum 1/2 mx.^2 + sum 1/2 Jtheta.^2
Energy dissipated in damper
delta W = pi() * c * w * X^2
Method for equivalent damper
use delta W = pi() * c * w * X^2, and related X^2 for each damper using similar triangles
simplest representation in hysteretic damping
DETLAW = pi() h X^2
an equivelent viscous unit for hysteretic damping
ceq = h/w = eta * keq / w
spring and hysteretic damper under the general harmonic excitation equation of motion
f = h/w *x. + kx Fe^jwt = h/w * jwX*e^jwt + kX*e^jwt
For hysteretic damper model what does F/X =
jh + k = k(1 + j*eta)
What is the complex stiffness of hysteretic damping
k(1 + j*eta)
SDOF system with hysteric damped equation
mx.. + ceq *x. + keq * x = 0
Reaction force of coulomb damping
f = -sign(x.) mu N where the sign function takes the sign of the velocity ie if velocity is -ve f = mu N
what is the drop in amplitude of coulomb damping per cycle
4 * mu * N / k
What is the equation of the line coulomb damping will follow
xline = x0 - (4 * mu * N / k) t/T
energy dissipated per cycle by coulomb damping
DELTAW = 4 * mu * N * X
Equivalent damping for coulomb ceq =
4 * mu * N / pi() * w * X
Relative Velocity
vB/A = w * rAB
Vector dot product or scalar product =
a dot b = mag a * mag b * cos theta
Component of force F in direction r
F dot r / mag of r
vector cross product
mag a cross b = mag a * mag b * sin theta
equation for tangential acc
(aB/A)t = alpha x rB/A
equation for normal acc
(aB/A)n = w x ( w x rB/A)
va =
vb + va/b = vb + w * ra/b
aa =
ab + aa/b = ab + alpha x ra/b + w x (w x ra/b)
Rotating frame of reference velocity equation
va = vb + w x r + vrel
Rotating frame of reference acceleration equation
aa = ab + w. x r + w x (w x r) + 2 w x vrel + arel
Equation for resultant moment
Sum Mg = Ig alpha
where Mg is the resultant moment Ig the mass moment of inertia alpha the angular acceleration
Linear momentum =
L = mv
L. =
d/dt (mv) = ma = sum F
symbol of angular momentum =
H0 = r x mv
Rate of change of angular momentum =
H0. = r x mv. = sum of moments
sum of moments relationship with sum of forces
Sum of M0 = r x sum of Forces
sum of moment about an arbitrary point
sum MP = sum rate of change of angular momentum + rhoG (dist from arbitrary point to centre of mass) x m ag
Sum of moments about an arbitrary point in a planar system
sum MP = IG alpha + rhoG x maG
what is the acceleration of single out of balance mass m located at radius r from axis of rotation
under constant velocity rotating
a = -w^2 r
Force applied to rotor with single out of balance mass m located at radius r from axis of rotation
F = m w^2 r
needs to be equal and opposite
moment vector of out of balance masses in multiple planes what is this system
M = d x F = d x mw^2 r = dk x mw^2 ri = dmw^2r j
dynamically unbalanced
How do you obtain bearing loads from tabular method
mag F = mrw^2 angle will = theta
Moment needed do gyroscopic motion
M = I * omega x w
where I is mass moment of inertia
spin axis rotates in the same direction
as the gyroscopic moment
