With scalar coefficients lecture notes 74 061606 1225

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Unformatted text preview: = 0 −2500 2500 Initial Conditions: 1.0 { x(0)} = 0 0 20.0 { x(0)} = 0 0 Forcing Function: 0 { f (t )} = −75 sin(30t ) 0 Complementary Solution Results: λ1 = −0.1848 + 6.0760 j λ2 = −1.6417 + 18.0458 j λ3 = −3.6795 + 26.8767 j 1.0000 {ψ 1} = 1.5435 1.8763 1.0000 {ψ 2 } = 0.5722 −0.9933 1.0000 {ψ 3} = −0.7863 0.3105 Example Solution - Particular Part: s 2 [ M ] + s [C ] + [ K ] { X p ( s )} = { F ( s )} Formulating the positive frequency portion of the solution at 30 rad s ( s = j 30 ): 0 −4000 + 1500 j −3000 − 900 j 0 −3000 − 900 j −7100 + 1650 j −2500 − 750 j X (30 j ) = 37.5 j } { p −2500 − 750 j −8300 + 750 j 0 0 −0.0041 + 0.0011 j { X p (30 j )} = 0.0036 − 0.0045 j −0.0016 + 0.0009 j Lecture Notes -78- 06/16/06 12:25 PM Mechanical Vibrations I Therefore: −0.0041 + 0.0011 j −0.0041 − 0.0011 j j 30t { x p (t )} = 0.0036 − 0.0045 j e + 0.0036 + 0.0045 j e− j 30t −0.0016 + 0.0009 j −0.0016 − 0.0009 j Or: −0.0082 −0.0022 { x p (t )} = 0.0072 cos(30t ) + 0.0090 sin(30t ) −0.0031 −0.0018 Example Solution - Complementary Part: While the complementary solution of the matrix differential equation for linear vibration problems always involves complex conjugate pair solutions, it is easier, when finding the numerical solution using MATLAB, to ignore this and determine the solution without this assumption. Once you find the solution, a check for complex conjugate pairs serves as a verification that no other mistakes have occurred. Therefore: { x(t )} = { xc (t )} + { x p (t )} { x(t )} = ∑ α r {ψ r } e r =1 6 λr t −0.0041 + 0.0011 j −0.0041 − 0.0011 j j 30t + 0.0036 − 0.0045 j e + 0.0036 + 0.0045 j e − j 30t −0.0016 + 0.0009 j −0.0016 − 0.0009 j Rewriting in matrix notation: Lecture Notes -79- 06/16/06 12:25 PM Mechanical Vibrations I { x(t )} = eλ t {ψ }1 1 eλ2t {ψ }2 eλ3t {ψ }3 eλ4t {ψ }4 eλ5t {ψ }5 α1 α 2 α eλ6t {ψ }6 3 α 4 α 5 α 6 −0.0041 + 0.0011 j −0.0041 − 0.0011 j j 30t + 0.0036 − 0.0045 j e + 0.0036 + 0.0045 j e − j 30t −0.0016 + 0.0009 j −0.0016 − 0.0009 j Taking the time derivative to get velocity: { x(t )} = λ1eλ t {ψ }1 λ2eλ t {ψ }2 λ3eλ t {ψ }3 1 2 3 −0.0041 + 0.0011 j + j 30 0.0036 − 0.0045 j e j 30t −0.0016 + 0.0009 j Evaluating displacement and velocity at t=0: α1 α 2 α λ4 eλ4t {ψ }4 λ5eλ5t {ψ }5 λ6 eλ6t {ψ }6 3 α 4 α 5 α 6 −0.0041 − 0.0011 j − j 30 0.0036 + 0.0045 j e − j 30t −0.0016 − 0.0009 j 1.0 0 = {ψ }1 0 {ψ }2 {ψ }3 20.0 0 = λ1 {ψ }1 λ2 {ψ }2 0 λ3 {ψ }3 α1 α 2 −0.0082 α {ψ }4 {ψ }5 {ψ }6 3 + 0.0071 α 4 −0.0031 α 5 α 6 α1 α 2 −0.0639 α λ4 {ψ }4 λ5 {ψ }5 λ6 {ψ }6 3 + 0.2722 α 4 −0.0542 α 5 α 6 Lecture Notes -80- 06/16/06 12:25 PM Mechanical Vibrations I α1 1.0082 α −0.0071 2 0.0031 α 3 = [ A] 20.0639 α 4 −0.2722 α 5 0.0542 α 6 Where: 1.0000 −0.7863 0.3105 [ A] = −3.6795 − 26.8767 j 2.8932 + 21.1335 j −1.1425 − 8...
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