Free vibrations and impulse loads

Question 1

free vibrations

Answer 1

occur when the structure is disturbed from static equilibrium and oscillates without any external excitation

Question 2

Undamped free vibration solutions

Answer 2

Amplitude and Phase Angle:
-ansantz: u(t) = Acos(omega_n.t - phi)

fn = omega_n/2pi = revolutions per unit time
Tn = 1/fn = natural period

-unknowns found by considering static equilibrium disturbed by initial conditions u0/v0:

A = sqrt(u0^2 + (v0/omega_n)^2)
tan(phi) = v0/(u0.omega_n)

Trigonometric Functions:
-ansantz: u(t) = A1cos(omega_n.t ) +A2sin(omega_n.t )

A1 = u0
A2 = v0/omega_n

Exponential Functions:
-ansantz: u(t) = exp(lambda.t)
-solution:
*u(t) = sum over i of Ci.exp(lambda_i.t)
*u(t) = (C1 +C2)cos(omega_n.t) + i(C1-C2)sin(omega_n.t)
–> this corresponds to the trigonometric solution

Question 3

Damped free vibrations

Answer 3

-damping causes oscillations to subside
-effects are lumped into one viscous damper

Exponential Function:
Cr = 2sqrt(KM) = critical damping
squiggle = c/Cr
- <1 is underdamped: omega_d = omega_n.sqrt(1 - squiggle6”)
- = 1 is critically damped
- >1 is overdamped

u(t) = epx(-squiggle.omega_n.t).(A1cos(omega_d.t) + A2sin(omega_d.t))
A1 = u0
A2 = (v0 + squiggle.omega_n.u0)/omega_d

Amplitude and Phase Angle:

u(t) = A.exp(-squiggle.omega_n.t).cos(omega_d.t - phi)

A = sqrt(u0^2 +[ (v0 + squiggle.omega_n.u0)/omega_d]^2)
-from A, the oscillations have a decreasing amplitude with the envelop defined by the exp. term

tan(phi) = (v0 + squiggle.omega_n.u0)/(omega_d.u0)

Td = Tn/sqrt(1 - squiggle^2)
-damped period is longer than the undamped period

Question 4

logarithmic decrement

Answer 4

-used to determine the damping ratio
-uses the amplitude and phase angel solution

u0/un = exp(squiggle.omega_n.N.Td)
delta = ln(u0/un) then solve for squiggle
*for squiggle<0.2 assume sqrt(1-squigle^2) = 1

Question 5

impulse loads

Answer 5

-numerical methods used to remove assumptions of periodicity and stationary structural characteristics
-impulse is applied for a short amount of time relative to the structures natural f

F hat = impulse magnitude = integral over the duration of F with respect to t [Ns]
-the same magnitude can be achieved with a large F and small t or large t and small f:

F hat = F.delta t

-F hat = m.delta v from a discretised form of F=ma
*can be reduced to an equivalent initial velocity
*response defined in terms of the Trigonometric functions

Question 6

arbitrary loading

Answer 6

-F(t) is split into impulse loads acting at t=tau for d.tau then integrated
-du(t)=… comes from F hat in terms of newtons second law
-convolution integral

u(t) = integral from 0 to t of p(tau)h(t-tau) d.tau
-p(tau) is the loading
-undamped:

u(t) = integral from 0 to t of F(tau)/(m.omega_n). sin(omega_n(t - tau)) d.tau

-damped:

u(t) = 1/(m.omega_n.sqrt(1-squiggle^2)) times the integral from 0 to t of F(tau).exp(-squiggle. omega_n. (t-tau)) . sin(sqrt(1-squiggle^2).omega_n(t - tau)) d.tau

-useful rule:
integral of exp(alpha.y).sin(beta y) dy =
exp(alpha.y)/(alpha^2 + beta^2) . (alpha.sin(beta.y) - beta.cos(beta.y))