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SolSec3_5 - 3 54 Problems and Solutions Section 3.5(3.39...

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3- 54 Problems and Solutions Section 3.5 (3.39 through 3.42) 3.39 Calculate the mean-square response of a system to an input force of constant PSD, S 0 , and frequency response function H ! ( ) = 10 3 + 2 j ! ( ) Solution: Given: S ff = S 0 and H ! ( ) = 10 3 + 2 j ! The mean square of the response can be found from Eqs (3.66) and (3.68): x 2 = E x 2 ! " # \$ = %& & H ( ( ) 2 S ff ( ( ) d ( x 2 = S 0 %& & 10 3 + 2 j ( 2 d ( Using Eq. (3.67) yields x 2 = 50 ! S 0 3

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3- 55 3.40 Consider the base excitation problem of Section 2.4 as applied to an automobile model of Example 2.4.1 and illustrated in Figure 2.16. In this problem let the road have a random stationary cross section producing a PSD of S 0 . Calculate the PSD of the response and the mean-square value of the response. Solution: Given: S ff = S 0 From example 2.4.1: m = 1007 kg, c = 2000 kg/s, k = 40,000 N/m ! = c 2 km = 2000 2 40000 i 1007 = 0.157 (underdamped) So, H ! ( ) = 1 k " m ! 2 + jc ! = 1 4 # 10 4 " 1007 ! 2 + 2000 j ! H ! ( ) 2 = 1 4 # 10 4 " 1007 ! 2 ( ) 2 + 2000 ( ) 2 j ! 2 H ! ( ) 2 = 1 1.01 # 10 6 ! 4 " 4.06 # 10 7 ! 2 + 1.6 # 10 9 The PSD is found from equation (3.62): S xx ! ( ) = H ! ( ) 2 S ff ! ( ) S xx ! ( ) = 1 1.01 " 10 6 ! 4 # 8.46 " 10 7 ! 2 + 1.6 " 10 9 The mean square value is found from equation (3.68): x 2 = E x 2 ! " # \$ = %& & H ( ( ) 2 S ff ( ( ) d ( x 2 = S 0 %& & 1 4 ) 10 4 % 1007 ( 2 + 2000 j ( 2 d ( Using equation (3.70) yields
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SolSec3_5 - 3 54 Problems and Solutions Section 3.5(3.39...

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