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# Class Notes for Aerospace at Ryerson University

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##### AER 403 Lecture Notes - Lecture 20: Optical Table, Hovercraft, Aircraft Engine Premium

AER 403 W20 Motion transmission isolation In motion transmission isolation, there is no internal vibration source. The excitation is the external support motion. Hence we need to look at the support motion vibration....

Aerospace
AER 403
Fengfeng Xi
##### AER 403 Lecture Notes - Lecture 7: Gear Train, Involute, Tamiya Corporation Premium

Mechanisms & Vibrations Gears Course outline Speed relation Gear geometry Gear train Text: Chapter 7 24, | 2 Course outline Speed relation Gear geometry Gear train Text: Chapter 5 24, | 3 ...

Aerospace
AER 403
Fengfeng Xi
##### AER 403 Lecture Notes - Lecture 21: Vibration Isolation, Accelerometer, Hertz Premium

AER 403 W21 Example 10.2 vibration isolation An engine of 100 kg is mounted on a spring of k=700 kN/m. The unbalance induced vibration is 350 N at 3000 rpm, and =0.2. Required to determine: 1) Force induced vibratio...

Aerospace
AER 403
Fengfeng Xi
##### AER 403 Lecture Notes - Lecture 13: Damping Ratio, Interval Ratio, Quasi Premium

AER 403 W13 Particular solution Now we have X sin(t-+) = Fsint Obviously, X = F = F/[(k-m 2)2 + (c)2]1/2 t-+ = t = = tan-1c/(k-m 2) 8, | 10 Relation with free vibration Re-arranging X = (F/k)/[(1-2m/k...

Aerospace
AER 403
Fengfeng Xi
##### AER 403 Lecture Notes - Lecture 18: Vibration Isolation, Interval Ratio, Inverse Trigonometric Functions Premium

Mechanisms & Vibrations Forced Vibrations Course outline Support motion Vibration isolation Vibration sensors Text: Chapter 3 22, | 2 Course outline Support motion Vibration isolation Vibration ...

Aerospace
AER 403
Fengfeng Xi
##### AER 403 Lecture Notes - Lecture 14: Interval Ratio, Centrifugal Force Premium

AER 403 W14 Example 9.1 (Cont.) Free vibration x X e nt sin( t ) 1d1 X1 = (C21 +C22)1/2; 1 = tan-1(C2/C1) d=(1- 2)1/2n= (1-0.12)1/2 20 = 19.89 where C1 = (v(0)+nx(0))/d; C2=x(0) v(0) = F/m = F2n/k = 202...

Aerospace
AER 403
Fengfeng Xi
##### AER 403 Lecture Notes - Lecture 17: Inverse Trigonometric Functions Premium

AER 403 W17 Determination of balance masses Summing all the moments about the second balance plane leads to Mx/2 = MR1ox z1o + MR2ox z2o + MR1x z1 =0 My/2 = MR1oy z1o + MR2oy z2o + MR1y z1 =0 Then MR1x = -(MR1ox z1...

Aerospace
AER 403
Fengfeng Xi
##### AER 403 Lecture 15: AER 403 W15 Premium

AER 403 W15 Course outline Forced harmonic vibration Rotating unbalance Rotor balancing Static balancing Dynamic balancing Vibration assignment 2 Text: Chapter 3 8, | 29 Rotor balancing The unbalan...

Aerospace
AER 403
Fengfeng Xi
##### AER 403 Lecture 12: AER 403 W12 copy Premium

AER 403 W12 Mechanisms & Vibrations Forced Vibrations Course outline Forced harmonic vibration Rotating unbalance Rotor balancing Static balancing Dynamic balancing Vibration assignment 2 Text: Chapter...

Aerospace
AER 403
Fengfeng Xi
##### AER 403 Lecture Notes - Lecture 16: Trial Balance, Asteroid Family Premium

AER 403 W16 Dynamic balancing theory The equation for imbalance induced vibration is md2x/dt2 + c dx/dt + kx = moeo2sint With imbalance Amplitude: MXo/moeo = (/n)2/ /[(1-(/n)2)2 + (2(/n))2]1/2 Phase: o=tan-1(2(/n))...

Aerospace
AER 403
Fengfeng Xi
##### AER 403 Lecture Notes - Lecture 19: Sine Wave, Inverse Trigonometric Functions, Vibration Isolation Premium

AER 403 W19 Superposition method X2=(F/k)/[(1-(/n)2)2+(2(/n))2]1/2cos(t-) =Y(c/k)/[(1- (/n)2)2+(2(/n))2]1/2cos(t-) =Y(2(/n))/[(1- (/n)2)2+(2(/n))2]1/2cos(t-) =Arctan[2(/n)/(1- (/n)2)] Total solution X=x1 + x2 =Y/[(...

Aerospace
AER 403
Fengfeng Xi
##### AER 403 Lecture Notes - Lecture 10: Viscosity, Damping Ratio, Bes Premium

Effective mass A spring-mass system has a concentrated mass. A continuous system like a spring or a beam can be modeled as a spring- mass system using the concept of effective mass (lumped mass). The effective ma...

Aerospace
AER 403
Fengfeng Xi
##### AER 403 Lecture Notes - Lecture 11: Damping Ratio, Vibration, Inverse Trigonometric Functions Premium

Damping ratio Notethatc=cC c/2m = (cC/2m) = n n = (k/m)1/2 since cC = 2(km)1/2 = 2m(k/m)1/2 = 2mn Now s1,2 = -c/2m [(c/2m)2 k/m]1/2= [-(2-1)1/2]n Solution is rewritten as x e nt (Ae( Decay term (no o...

Aerospace
AER 403
Fengfeng Xi
##### AER 403 Lecture Notes - Lecture 9: Inverse Trigonometric Functions Premium

Fourier transform 26, | 8 Course outline Oscillatory motion Free vibration (natural frequency) Effective mass Damped free vibration Vibration measurement Vibration lab 1 and assignment 1 Text: Chapters 1...

Aerospace
AER 403
Fengfeng Xi
##### AER 403 Lecture Notes - Lecture 12: Damping Ratio, Accelerometer Premium

Logarithmic decrement ln(x/x)ln[ent/en(td)]lnend 12 nd Since 2/(12)1/2 dn then [2 /(12)1/2 ]2 /(12)1/2 nn If is obtained from measurement, can be determined as 2 21/2 /[(2) ] For <<1 2 / 2 Not...

Aerospace
AER 403
Fengfeng Xi
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