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Next: Response to Sinusoidal Ground Up: Solutions to the Differential Previous: Free Response: Single Story

Free Response: Multiple Stories

Returning to our three story building problem, we will assume that the solution position vector will be of the form . With the trial function , the velocities are , and the accelerations are . Substituting this trial solution into the differential equations of motion and rearranging terms, we obtain the matrix equation
(28)

This expression will be satisfied if mode shapes = 0 (the trivial solution), or if the matrix is not invertible, i.e., it's determinant is equal to zero.

Because our building's stiffness-mass-damper system has more than one mass, the solution of the equation (28) requires solving det = 0 for lambda. Substituting the mass, damping, and stiffness matrices into this determinant equation, we get a sixth degree polynomial. Unfortunately, unlike the single degree of freedom system, there is no way to solve the determinant equation for lambda in terms of mass, damping, and stiffness. Luckily, though, if numerical values exist for mass, damping, and stiffness, they can be used to solve the equation instead.

When dealing with the physical model, we do have numerical values. Because the building is under-damped, be have three complex-conjugate pairs of values for lambda.

It is interesting to note that in our model, because the damping matrix is proportional to the mass matrix (), the real parts of the three roots three values of lambda are all equal to (derived from equation 25). Additionally, the damping ratios are , , and .

Furthermore, since the damping matrix is of the form (ie., damping is stiffness and mass proportional), it can be shown that the natural frequencies of a multi-mass system are independent of damping. In Figure 6 below, the three natural frequencies, first natural frequency, second natural frequency, and third natural frequency are plotted with respect to the square root of the ratio of the story shear stiffness, stiffness, to the floor mass mass, . The natural frequencies for this structure are approximately , , and .

Figure 6: The three natural frequencies of the three story shear-building model as a function of the interstory-stiffness to floor-mass ratio, stiffness to mass ratio.
graph of natural frequencies as a function of the stiffness-mass ratio

You can use the frequency calculator at right to input different values of mass, stiffness, and damping and find out what the model building's natural frequencies would be in that case. You can see that changing the damping doesn't affect the natural frequencies, just the damped natural frequencies. Also, note the relationships graphed in Figure 6 can be determined from this calculator. What other concepts from this section can be illustrated with this calculator?
Frequency Calculator
Stiffness Constant Mass of each floor Damping

First Floor Second Floor Third Floor
Natural Frequencies:
Damped Natural Frequencies:
Damping Ratios:


With knowledge of natural frequencies and damping ratios, we may now numerically determine the constants . This may be done by substituting a natural frequency and a damping ratio into equation (27) and then substituting lambda into equation (28) and solving for an amplitude. For every natural frequency a natural frequency and damping ratio a damping ratio, i=1,2,3, the vector imaginary part of amplitude is zero. The fact that the vectors amplitudes are real in this problem is another desirable property of proportionally damped systems. With the simplification that the vectors amplitudes are real, the solutions become

position vector =  
velocity =  


Table 1: Modal amplitudes for the three-story building model.
Story first modal amplitude second modal amplitude third modal amplitude
1 0.328 0.737 0.591
2 0.591 0.328 -0.737
3 0.737 -0.591 0.328

The vectors amplitudes vector are the mode-shape vectors. Every mode-shape vector has an associated natural frequency and damping ratio. If we were to apply an initial set of displacements that was proportional to one of the mode-shape vectors, then the free response of the structure would occur at only that mode's natural frequency. In general, however, the initial conditions will not be proportional to a particular mode shape, but may always be represented by a linear combination of mode-shapes, for example, .

In general, we will assume that the free displacement response of our three-story building is made up of the sum of three distinct damped oscillations, each with a different frequency frequency and damping damping, . This trial solution can be written as , where the three-by-three modal matrix modal matrix is a column-wise combination of the vectors first mode shape vector, second mode shape vector and third mode shape vector. If the initial condition inital position vector is only in a single mode, then the free response will remain entirely in that mode.


Quick Quiz: A three-story building of the type shown in Figure 3 has k=8x10^7N/m and m=2x10^5kg. Which of the following is a natural frequency?

36 Hz
20 Hz
13 Hz
6 Hz


next up previous
Next: Response to Sinusoidal Ground Up: Solutions to the Differential Previous: Free Response: One Story
Henri P Gavin
2002-03-30