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ch3sol-fourier-class

ch3sol-fourier-class - 61 Problems for Section 3.6 Periodic...

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61 Problems for Section 3.6 °Periodic but Not Harmonic Excitation 31. A machine is loaded by a periodic °sawtooth±shaped force, as depicted in Figure 3.50. The load is assumed to have existed for a very long time. Model this force using: (a) a one-term Fourier series, (b) a three-term Fourier series, and (c) a ²ve-term Fourier series. In each case solve for the response as a function of time. System properties are: m = 1 kg, k = 9 N/cm, ° = 0 : 15 , T = 1 s, and A = 1 cm. Figure 3.50: Sawtooth loading. Solution: First of all, we realize that this problem can be solved exactly by solving the problem in discrete time intervals such that m ° x 1 + c _ x 1 + kx 1 = A T t for 0 ° t ° T m ° x 2 + c _ x 2 + kx 2 = A T t ± A for T ° t ° 2 T . . . The displacements and velocities are continuous at t = nT for n = 1 ; 2 ; ² ² ² : Here, we use the Fourier expansion for the periodic forcing. The sawtooth function is given by At=T; 0 ° t ° T: We ²rst have to evaluate the Fourier coe¢ cients as follows: a 0 = 2 T Z T 0 At T dt = A a p = 2 T Z T 0 At T cos p! T tdt = 2 T 2 A cos p! T T + p! T (sin p! T T ) T ± 1 p 2 ! 2 T b p = 2 T Z T 0 At T sin p! T tdt = ± 2 T 2 ( ± sin p! T T + p! T (cos p! T T ) T ) A p 2 ! 2 T : Note that the integration by part was used to evaluate a p and b p : For example, Z T 0 t |{z} u cos p! T tdt | {z } dv = t |{z} u sin p! T t p! T | {z } v ° ° ° ° ° ° ° ° T 0 ± Z T 0 1 dt |{z} du sin p! T t p! T | {z } v :

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62 CHAPTER 3 SDOF VIBRATION: WITH DAMPING Since ! T = 2 ±=T; and p is an integer, cos p! T T = cos 2 ±p = 1 ; and sin p! T T = 0 : Therefore, a p = 0 and b p = ± A=p±: The Fourier series representation of the force is then F ( t ) = A 2 ± 1 X p =1 A sin p 2 ± T t: The one-, three-, and ²ve-term expansions are respectively, F 1 ( t ) = A 2 ± A ± sin !
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