Page 437 - Thomson, William Tyrrell-Theory of Vibration with Applications-Taylor _ Francis (2010)

424 Random Vibrations Chap. 13

Thus, by squaring and substituting into Eq. (13.2-4), we find the mean square value
of y is
-lliot
y^ = lim + 2HH* + H*e ) dt

^0
== = F^,\H{cd)V (13.3-6)
In the preceding equation, the first and last terms become zero because of
T ^ 00 in the denominator, whereas the middle term is independent of T.
Equation (13.3-6) indicates that the mean square value of the response is equal to
the mean square excitation multiplied by the square of the absolute values of the
frequency response function. For excitations expressed in terms of Fourier series
with many frequencies, the response is the sum of terms similar to Eq. (13.3-6).
Example 13,3-1
A single-DOF system with natural frequency = yjk/m and damping ^ = 0.20 is
excited by the force
F{t) = F cos + F cos (x)^t + F cos \(x)^t
= ^ Fcosmoj^t
m - 1 / 2 , 1 , 3 / 2
Determine the mean square response and compare the output spectrum with
that of the input.
Solution: The response of the system is simply the sum of the response of the single-DOF
system to each of the harmonic components of the exciting force.
x (i) = X! |//(w&))|Fcos(wü)„i - </>„)
1/2, 1,3/2
where
l/k 1.29
|w ( K ,) l = k
/ 9/ I 6 + (0.20)

\H(co„)\ = lA 2.50
k
•\/4{0.20Ÿ
l/k_________ 0.72
__________
|W ( K )I =
/ 25/ I 6 + 9(0.20/ ^
4>x/2 ^ ^ 0.083t7

(/>! = tan ^00 = O.5 O77
- 1 -12^'
/
<>3/2 = tan = — 0 .1 4 2 7 T