Orthogonality of natural modes

Question 1

symmetrical

Answer 1

m=m’
k=k’

Question 2

inertial F in terms of phi

Answer 2

-for a natural mode

k= omega_i^2.m

• undamped

F_i = -m.a = k.u

• combining the two with phi = u

F_I = omega_i^2.m.phi_i

Question 3

proof of orthoganality

Answer 3

stiffness weighting:
phi_i’.k.phi_j = 0 for i not equal to j

inertia weighting:
phi_i’.m.phi_j = 0 for i not equal to j

Question 4

uncoupling the equations of motion

Answer 4

-coupled because k is not usually diagonal
*all DoFs in each equation
-arbitrary shape can be defined by a valid shape of basic vectors
*represents all possible displaced shapes of the system
*same number of basic vectors as DoFs
*linearly independent
*phi can be used