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Ch 4 HW Solutions

# Ch 4 HW Solutions - Chapter 4 Single Degree-of-Freedom...

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Unformatted text preview: Chapter 4 Single Degree-of-Freedom Vibration: General Loading and Advanced Topics Problems for Section 4.1 &Arbitrary Loading: Laplace Transform 1. Solve the following equations of motion using the Laplace transform approach: (a) & y + 2 _ y + 3 y = 5cos3 t; y (0) = 3 ; _ y (0) = 4 (b) 4& y + 5 _ y + 5 y = 4 u ( t ) ; y (0) = 1 ; _ y (0) = 1 , where u ( t ) is the unit step function, (c) 3& y + 3 _ y + 6 y = 3 e & t + 2cos3 t; y (0) = 2 ; _ y (0) = 4 (d) & y + _ y + y = F ( t ) ; y (0) = 0 ; _ y (0) = 0 , where F ( t ) is given by a square wave function with maximum amplitude 1 and period 1, (e) & y + 2 _ y + 3 y = cos3 t + cos5 t; y (0) = 0 ; _ y (0) = 0 . Solution: The Laplace transform of the general equation m & y + c _ y + ky = F ( t ) is m & s 2 Y ( s ) & sy (0) & _ y (0) ¡ + c ( sY ( s ) & y (0)) + kY ( s ) = F ( s ) ; where y (0) and _ y (0) are the initial conditions, and where Y ( s ) and F ( s ) are the Laplace transforms of y ( t ) and F ( t ) ; respectively. Then, Y ( s ) = F ( s ) + m ( sy (0) + _ y (0)) + cy (0) ms 2 + cs + k : The inverse Laplace transform of Y ( s ) gives y ( t ) : Let us now do the individual problems. (a) & y + 2 _ y + 3 y = 5 cos 3 t , y (0) = 3 , y (0) = 4 In this case, m = 1 ; c = 2 ; k = 3 ; and F ( s ) = 5 s= ( s 2 + 9) where F ( s ) is obtained from the Laplace transform table (Table A.2). The Laplace transform of the response, Y ( s ) ; is given by Y ( s ) = ( 5 s s 2 +9 ) + 3 s + 10 s 2 + 2 s + 3 = 3 s 3 + 10 s 2 + 32 s + 90 ( s 2 + 9)( s 2 + 2 s + 3) : The inverse transform can be found using a symbolic program. Alternatively, to perform the inversion by hand, we need to rewrite Y ( s ) using the method of partial fraction expansion. Y ( s ) can be rewritten as 3 s 3 + 10 s 2 + 32 s + 90 ( s 2 + 9)( s 2 + 2 s + 3) ¡ As + B ( s 2 + 9) + Cs + D ( s 2 + 2 s + 3) ; 1 2 CHAPTER 4 SDOF VIBRATION: GENERAL LOADING where the assumed numerators depend on highest order of the numerator of the original fraction. Equating, we &nd that 3 s 3 + 10 s 2 + 32 s + 90 = ( As + B ) & s 2 + 2 s + 3 ¡ + ( Cs + D ) & s 2 + 9 ¡ = ( A + C ) s 3 + ( B + D + 2 A ) s 2 + (3 A + 2 B + 9 C ) s + 3 B + 9 D: Matching coe¢ cients, we obtain A = & 5 = 12 ; B = 5 = 4 ; C = 41 = 12 ; D = 115 = 12 : Then, Y ( s ) = & 5 s= 12 + 5 = 4 ( s 2 + 9) + 41 s + 115 12( s 2 + 2 s + 3) : To transform Y ( s ) to y ( t ) ; &rst we must rewrite Y ( s ) in the following form, Y ( s ) = & 5 12 s ( s 2 + 9) + 5 12 3 ( s 2 + 9) + 41 12 s + 1 ¢ ( s + 1) 2 + & p 2 ¡ 2 £ + 74 12 1 p 2 p 2 ¢ ( s + 1) 2 + & p 2 ¡ 2 £ : The inverse transform is then given by y ( t ) = L & 1 [ Y ( s )] = & 5 12 cos3 t + 5 12 sin3 t + 41 12 e & t cos p 2 t + 37 p 2 12 e & t sin p 2 t = & : 417cos3 t + 0 : 417sin3 t + exp( & t )(3 : 42cos1 : 41 t + 4 : 36sin1 : 41 t ) : The response plot is shown at the end....
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Ch 4 HW Solutions - Chapter 4 Single Degree-of-Freedom...

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