Modal Analysis of MDOF Systems

Calculation of the natural frequencies and the corresponding natural modes of vibration are important in developing a general method of dynamic analysis called the Modal Analysis. This method decomposes the dynamic system into different SDOF systems after solving the eigenvalue problem for natural frequencies and natural modes and considers the individual modes separately to obtain the total solution.

The Modal Analysis uses a very important characteristic of the modal vectors, i.e., the orthogonality conditions. The derivation of the orthogonality conditions is avoided here, but they are available in any

standard text on Structural Dynamics. If th ni and nj are the i and j natural frequencies of an undamped

system and th

and j are the i and j modes of vibration, then if j i, the mass and stiffness matrices satisfy the following orthogonality conditions

K i =0 ………………..(10.1)

M i =0 ………………..(10.2)

where the superscripts T indicate the transpose of the matrices. If j = i, the ratio of the products T

and th

M i give the square of the i natural frequency of the system; i.e.,

T ni =( i K i )/( i M i ) ………………..(10.3)

Choosing the displacement vector as the summation of a number (equal to the DOF) of variable separable vectors [using Eq. (9.3)]

u(t) = q i (t) i

2 so that the governing equations of motion M d 2 u/dt + K u = f(t) …………………….….(8.7)

2 can be written as, (M d 2 q

i /dt i +Kq i i ) = f(t)

2 2 Pre-multiplying (10.4) by T

j (M d q i /dt i +Kq i i )= j f(t)

Using the orthogonality equations T (

M i )d q i /dt +( i K i )q i = i f(t) ……...………….….(10.6)

where T

i M i is called the ‘modal mass’ M i , i K i the ‘modal stiffness’ K i and i f(t) the ‘modal load’

f th

i for the i mode of the system. Eq. (10.6) is an uncoupled differential equation that can be solved to get q i (t) as a function of time.

Since i is already known by solving the eigenvalue problem, q i (t) can be inserted in Eq. (9.3) and summing up similar components gives u(t). Therefore, the main advantage of the orthogonality conditions is to uncouple the equations of motion so that they can be solved as separate SDOF systems.

For a damped system, the damping matrix C can also be formed to satisfy orthogonality condition; i.e.,

C i =0 ………………..(10.7)

This can be possible if the matrix C is proportional to the mass matrix M or the stiffness matrix K, or more rationally a combination of the two; i.e.,

C=a 0 M+a 1 K

Thus formulated, the equation of motion for the i th mode can be written as

M i )d q i /dt +( i C i ) dq i /dt +( i K i )q i = i f(t)

The ratio ( T

C i )/( i M i )=2 i ni

The modal damping ratio, T

i =( i C i )/( i M i )/(2 ni ) ..……………(10.10)

Modal Analysis is helpful in illustrating the basic features of MDOF system, and is preferred in analytical works. However for practical purposes, it has many drawbacks. The solution of the eigenvalue problem can be cumbersome for large dynamic systems, and the method cannot be applied for nonlinear systems.

Example 10.1 For the 2-storied building system described in Example 9.1, calculate the dynamic displacement vector if

step loads of 25 kips are applied at both stories when the system is at rest; i.e., f 1 (t) = f 2 (t) = 25 kips.

Solution The mass and stiffness matrices of the system are given by

while the damping matrix C = 0

From Example 9.1, the natural frequencies of the system are found to be n1 = 3.09 rad/sec, and n2 = 8.09 rad/sec, while the modal vectors are given by

The modal masses are, M 2

1 = 1 M 1 = 3.618 k-sec /ft, M 2 = 2 M 2 = 1.382 k-sec /ft

The modal stiffnesses are, K T

1 = 1 K 1 = 34.55 k/ft, K 2 = 2 K 2 = 90.45 k/ft

The modal loads are, f T

1 (t) = 1 f = 65.45 k, f 2 (t) = 2 f = 9.55 k

The uncoupled modal equations of motion are

2 3.618 d 2 q

1 /dt + 34.55 q 1 = 65.45

2 1.382 d 2 q

2 /dt + 90.45 q 2 = 9.55

The solution of these equations starting ‘at rest’ is q 1 (t) = (65.45/34.55) [1 – cos (3.09t)] = 1.894 [1– cos (3.09t)] and q 2 (t) = (9.55/90.45) [1 – cos (8.09t)] = 0.1056 [1– cos (8.09t)]

u(t) = q i (t) i =q 1 (t) 1 +q 2 (t) 2

u(t) = 1.894 [1 – cos (3.09t)] + 0.1056 [1– cos (8.09t)]

u 1 (t) = 1.894 [1 – cos (3.09t)] + 0.1056 [1– cos (8.09t)]

2 u (t) = 3.065 [1 – cos (3.09t)] – 0.065 [1– cos (8.09t)]

The displacements are plotted with time in Fig. 10.1 and 10.2. Fig. 10.1 shows the contribution of the two modes to the total displacements, which are shown in Fig. 10.2. The figures indicate that u 2 is larger in

this case, and by far the bigger contributions to the displacements come from the 1 st mode of vibration.

u1, Mode1 u 1 u2, Mode1 u 2 u u1, Mode2 1 u u2, Mode2 2

Time (sec)

Fig. 12.1: Contribution of various Modes

Time (sec) Fig. 12.2: Total Responses

Example 10.2 Calculate the modal damping ratios for the 2-storied system described in Example 10.1 if dampers

equivalent to ones for a 5% damped SDOF system are included in each story; i.e., c 1 =c 2 = 0.5 k-sec/ft.

Solution

c 1 +c 2 –c 2 1.0 –0.5

–c 2 2 c –0.5 0.5

The modal dampings are, C T

1 = 1 C 1 = 0.691 k-sec/ft, C 2 = 2 C 2 = 1.809 k-sec/ft

2 In Example 10.1, the modal masses were calculated to be M 2

1 = 3.618 k-sec /ft, M 2 = 1.382 k-sec /ft, and in Example 9.1, the natural frequencies of the system were found to be

n1 = 3.09 rad/sec, and n2 = 8.09 rad/sec

Using Eq. (10.10), the modal damping ratios are

1 =C 1 /(2M 1 n1 ) = 0.691/(2 3.618 3.09) = 0.0309

2 =C 2 /(2M 2 n2 ) = 1.809/(2 1.382 8.09) = 0.0809

1 is lower while 2 is greater than 0.05. Particularly the damping ratio of the second mode is much higher, which helps to suppress it even further.

Modal Analysis of MDOF Systems