Vibration of undamped MDoF

Question 1

natural modes of vibration

Answer 1

-the system is able to vibrate freely in this mode with the ratio of displacements of ant two masses being unchanged in time
*complete motion can be defined by superposition of the motion of individual modes
-same as the no of DoFs
-each has a characteristic shape and natural f

Question 2

synchronous solution

Answer 2

• floors vibrate at the same rate

u(t) = phi(Asin(wt) + Bcos(wt))
*used to get to the eigenvalue problem

Question 3

stiffness eigen value problem

Answer 3

|k - omega^2m|phi = 0

det|k - omega^2m| = 0
-leads to a linear dependency in that no value of beta makes the equations of the top one independent
-only phi1/phi2 can be found not their acc. values
* phi1 always set to 1
*phi_ij = disp. at position i (row) due to mode j (column)

beta = w^2m/k –> eigenvalue
phi –> eigenvector
Ti = spi/w_i

Question 4

general solution

Answer 4

u(t) = phi[s(t).A + c(t).B]
-A/B are constants found using the initial conditions
-s/c have off diagonals zero with diag. as sin or cos(w_i.t)

Question 5

flexibility eigenvalue problem

Answer 5

-used for beams and where shear distortion needs to be included, not possible with the stiffness approach
-mode shapes found the same way as in the stiffness approach

[ 1/(w^2) . I - FM]phi = 0
-F and M are always symmetric but their product is not which limits this methods use in some eigenvalue comp. programs

lambda = 1/w^2
D = EI/ML^3

Derivation:
-at a fixed time, equivalent loads can be treated as statics that cause displacements

u(t) = FP(t)

-equation of motion can be rewritten as

F[P - Ma] = u
*P is zero for free vibrations
*the bracket gives the net equivalent static F

-equate these two equations and apply the synchronous solution
-divide through by omega^2

Question 6

fundamental mode

Answer 6

has the smallest omega and the largest lambda