Chapter 3 - Free Vibration with Viscous Damping Flashcards
What are Dampers?
Devices which give a force proportional to velocity. Often modelled as dashpots.
What force do dampers create?
When two bodies approach each other with a velocity x(dot), a damper will seek to push them apart with a force cx(dot).
Can you draw a mass, spring, damper system and its fbd? What is the resulting e.o.m?
YES OR NO. mx(double dot) + cx(dot) + kx = 0.
what is the damping ratio equation?
ζ = c/2mωn
Dividing through the e.o.m by m gives?
x(double dot) + 2ζωnx(dot) + ωn^(2)x = 0
Assuming a solution looks like x=Ae^(p.t), what is the resulting general solution for the e.o.m?
do the sums….. x = Ae^[-ζωn + ωnsqrt(ζ^(2)-1)]t + Be^[-ζωn - ωnsqrt(ζ^(2)-1)]t
What does the nature of the damped behaviour depend on?
The value of ζ (damping ratio).
What happens to the solution when ζ <1.
This is called subcritical damping. The term in the square root goes complex so the solution has to be re-written to account for this:
x = Ae^[-ζωn + ωn.i.sqrt(1-ζ^(2))]t + Be^[-ζωn - ωn.i.sqrt(1-ζ^(2))]t
When and why do we introduce the idea of damped natural frequency?
We need to consider that, unlike coulomb damping, viscous damping will affect the frequency of the system. So we introduce damped natural frequency: ωd = ωnsqrt(1-ζ^(2)).
Inserting damped natural frequency into our solution gives?
x = Ae^[-ζωn + i.ωd]t + Be^[-ζωn - i.ωd]t
=> x= e^(-ζ.ωn.t)[Ae^(i.ωd.t) + Be^(-i.ωd.t)]
Using euler’s equation on this solution gives?
x=e^(-ζ.ωn.t)[Pcos(ωd.t) + Qsin(ωd.t)]
Using the Harmonic addition theorem on the term in square brackets gives?
x=e^(-ζ.ωn.t)[Rsin(ωd.t+ phi)]
What is the damped system a sum of?
An oscillatory component and an exponential-decay component. DRAW DIAGRAMS.
What happens to the solution if damping ratio is = 1?
Critical damping. x= Ae^(-ωnt) + Be^(-ωnt) = Ce^(-ωnt).
Can you draw and describe the response?
DRAW. No oscillatory term. Exponential decay. The mass will come to a halt just as it reaches the neutral position. This is the quickest way to ‘stop’ the motion without overshoot, but it never really stops.
