Dynamic Responses - Mechanics of Structures: Variational and Computational Methods - page 624

10.4.3 Base Excitation

Suppose a prescribed displacement U is applied t o the base of the frame of Fig. 10.2. This
displacement is such that the base translates as U ( t ) in the X and Z directions. The nodal

displacements V

would then be

V =

c ] T [ U + U Xb + U + U Zb +

b U

+ U Xc + U + U Zc +

a T U = V + (10.92)

where a T [110110] T

c ] T

= and V = [ U Xb U Zb

b U Xc U Zc

are the displacements

relative to the moving base.
Suppose there are no applied forces. Then, the governing equilibrium equations would
be of the form M V +

=

0 , since Newton's second law applies to absolute accelerations
of the masses. With Eq. (10.92) this leads to

KV

Ma T U M V

+ KV =− (10.93)

Thus, the equations of motion including base input have the same general form as Eq. (10.68)
except the forcing function takes the special form

Ma T U P =− (10.94)

The physical responses are still given by Eq. (10.81). It follows that Eq. (10.84) would now
appear as

Ma T U
M i

φ i

2

q i + 2 ζ ω q i ˙ + ω

i q i

=− (10.95)

i i

The quantity

Ma T
M i

φ i MPF i = (10.96)

is often referred to as the modal participation factor for the i th mode. To be more precise,
for the case at hand, this is a translational modal participation factor. This factor depends
on the mode shapes, the mass distribution, and the direction of input. As indicated in
Eq. (10.95), the magnitude of the excitation force, and, hence, the modal response, is directly
proportional to the modal participation factor associated with the mod e. Thus, q i can be
calculated as the product of MPF i and the solution of Eq. (10.95) due to

U only.

10.5 Direct Integration of the Equations of Motion

The equations of motion

M V C V + + KV = P (10.97)

can be solved using integration directly without employing modal superposition, which
applies for linear responses only. Such direct integration methods use step-by-step numerical
integration.