• Author / Uploaded
• NABIL HUSSAIN

Citation preview

UE15ME352: Mechanical Vibrations PES University Department of Mechanical Engineering UE15ME352: Mechanical Vibrations Question Bank: Partial Solutions

Unit II Chapter 9: Vibration Control 1. An automobile moving on a rough road, in the form of a sinusoidal surface, is modeled as a spring-mass system, as shown in Figure 1. The sinusoidal surface has a wave length of 5 m and an amplitude of Y = 1mm. If the mass of the automobile, including the passengers, is

1,500 kg and the stiffness of the suspension system (k) is 400 kN/m,

determine the range of speed (v) of the automobile in which the passengers perceive the vibration. Suggest possible methods of improving the design for a more comfortable ride of the passengers.

Fig. 1 Solutions:

λ = 5   = 1 = 1 × 10   = 

 400 × 10 = = 16.3299 /  1,500

 = 2 = ∴=

2

=

2 × 1,000 ×  = 0.3491  / 3,600 × 5

 0.3491  = = 0.02138  16.3299

λ

# 1 = ;  |1 −  & |

Dr. C V Chandrashekara, Professor and Lead: Design Domain, Dept. of Mechanical Engineering, PESU, B’lore

UE15ME352: Mechanical Vibrations  1 × 10 ∴#= = |1 −  & | |1 − 0.02138 & |

# () (* ∞, ,ℎ) )*.(* /( * 012 () (* 3)* ∴ |1 − 0.02138 & | = 0 ∴ 4 = 56. 7787 9:/;


&

1 + 2ζ 2.5 =  2ζ &

&

b. Second part:

90% of isolation means, QJ = 0.1

@ operating speed, QJ = 0.1 ,  = 600 / = 62.832 / ζ = 0.2182

1 + 2ζ & QJ =  1 −  & & + 2ζ

&

1 + 2 × 0.2182 ×  & 0.1 =  1 −  & & + 2 × 0.2182 ×   C − 20.8595 & − 99 = 0 ∴  = 4.9844

&

Dr. C V Chandrashekara, Professor and Lead: Design Domain, Dept. of Mechanical Engineering, PESU, B’lore

UE15ME352: Mechanical Vibrations ∴  =

 62.832 = = 12.6057 /  4.9844

∴  = & = 81.5494 × 12.6057& = 12,958.49 @/  G = 2 ζ = 2 × 81.5494 × 12.6057 × 0.2182 G = 448.61@ − /

5. A dishwashing machine weighing 80kg operates at 300 rpm. Find the minimum static deflection of an isolator that provides 60 percent isolation. Assume that the damping in the isolator is negligible. Solutions: For an undamped system, transmissibility ratio is given by,

QJ =

1 |1 −  & |

and it can be approximated as, (Ref: SS Rao book)

1 + Q] 1 + 0.4 = = 3.5 Q] 0.4 ∴  = 1.8708 2@ 2 × 300 = = = 31.42 / 60 60  31.42 ∴  = = = 16.7949 /  1.8708 P 9.81 WXY = & = = 0.03477  = 34.77   16.7949& & ≈

6. A washing machine of mass 50 kg operates at 1,200 rpm. Find the maximum stiffness of an isolator that provides 75 percent isolation. Assume that the damping ratio of the isolator is 7 percent. 2@ 2 × 1,200 = = 125.664 / 60 60

Solutions: =

Damping ratio, 7%, ξ= 0.07 Transmissibility ratio,

Dr. C V Chandrashekara, Professor and Lead: Design Domain, Dept. of Mechanical Engineering, PESU, B’lore

UE15ME352: Mechanical Vibrations QJ =

E1 + 2ξ

E1 −  &

&

&

+ 2ξ

&

75% isolation means, QJ = 0.25

E1 + 2 × 0.07 × 

0.25 =

E1 −  &

∴  = 2.773  =

&

&

+ 2 × 0.07 × 

&

 125.664 = = 55.1803 /  2.773

∴  = & = 50 × 55.1803& ∴  = 1,52,243 @/

7. It is found that an exhaust fan, of mass 80 kg and operating speed 1,000 rpm, produces a repeating force of 10,000 N on its rigid base. If the maximum force transmitted to the base is to be limited to 2,000 N using an undamped isolator, determine (a) the maximum permissible stiffness of the isolator that serves the purpose; (b) the steady-state amplitude of the exhaust fan with the isolator that has the maximum permissible stiffness; and (c) the maximum amplitude of the exhaust fan with isolation during start-up. 2@ 2 × 1,000 = = 104.72 / 60 60

Solutions: =

2,000 N out of 10,000N force is transmitted means, Transmission ratio of the isolator (Tf), Q] = But, & ≈

2,000 = 0.2 10,000

1 + Q] 1 + 0.2 = Q] 0.2

∴  = 2.4494 ∴  =

 104.72 = = 42.75 /  2.4494

Dr. C V Chandrashekara, Professor and Lead: Design Domain, Dept. of Mechanical Engineering, PESU, B’lore

UE15ME352: Mechanical Vibrations ∴  = & = 80 × 42.75& ∴  = 1,46,205 @/

10,000 _` ^=^ ^ #=^ &  −  1,46,205 − 80 × 104.72& # = 0.01368 

_` #=^ ^=a  −  &

_`

&  b1 − & c 

a

For which,  =  and # → ∞

8. An engine is mounted on a rigid foundation through four springs. During operation, the engine produces an excitation force at a frequency of 3,000 rpm. If the weight of the engine causes the springs to deflect by 10 mm, determine the reduction in the force transmitted to the foundation 9. A printed circuit board of mass 1 kg is supported to the base through an undamped isolator. During shipping, the base is subjected to a harmonic disturbance (motion) of amplitude 2 mm and frequency 2 Hz. Design the isolator so that the displacement transmitted to the printed circuit board is to be no more than 5 percent of the base motion. Solutions:

 = 2 = 2 × 2 = 4 / # 1 =^ ^ 1 −  &  5% =

5 1 =^ ^ 1 −  & 100

∴  = 4.5826 ∴  =

 4 = = 2.7422 /  4.5826

∴  = & = 1 × 2.7422& = 7.5196 @/

Dr. C V Chandrashekara, Professor and Lead: Design Domain, Dept. of Mechanical Engineering, PESU, B’lore

UE15ME352: Mechanical Vibrations

Unit III Chapter 5: Two Degree-Of-Freedom System Problem No.1: Answer at different stages: Step 4: matrix form:  + & − T & # −& 0 e T f g Th = i j & # 0 & +  − &  −& & Step 6: Frequency Equation:

T & C − kT + & & + & +  T l& + kT + & & +  − && l = 0 Step 7: Natural frequencies:

∴ T& , && =

kmn omp qp omp omr qn l±Lkmn omp qp omp omr qn lp Cqn qp tmn omp mp omr mpp u &qn qp

,ℎ)), T  & are known as the natural frequencies of the system Step 8: Mode shapes of the system: First Mode: #T T & e f =  T + & − T T& #& Second Mode: #T & & e f = T + & − T && #& Problem No.2: Answer at different stages: Step 4: Matrix form:  + & − T & −& # 0 e T f g Th = i j & #& 0 & − &  −& Step 6: Frequency Equation:

T & C − kT + & & + & T l& + kT + & & − && l = 0 Step 7: Natural frequencies: ∴ T& , && =

−kT + & & + & T l ± EkT + & & + & T l& − 4T & kT + & & − && l 2T &

First Mode: #T T & e f =  T + & − T T& #& Second Mode: #T & & e f = T + & − T && #&

Problem No.3: Similar to Problem No. 2 and substitute the value of mass and stiffness appropriately Problem No.4: Similar to Problem No. 1 Problem No.5: Similar to Problem No. 2 and substitute the value of mass and stiffness appropriately Dr. C V Chandrashekara, Professor and Lead: Design Domain, Dept. of Mechanical Engineering, PESU, B’lore

UE15ME352: Mechanical Vibrations

Problem No.6, 7, 8, 9 and 10: Try yourself Problem No.11: Answer at different stages: Matrix form: 3P − B& −P # 0 e f g Th = i j & #& 0 P − B −P Frequency Equation: B & C − 4& BP & + 2P & Natural Frequencies: P T& , && = v2 ∓ √2x B P P ∴ T = 0.7654L  & = 1.8478L B B Problem No.12: Answer at different stages: Matrix form: B & + PB − B & & −B & # 0 e f g Th = i j & & & & 0 B + 2PB − 4B  #& −B Frequency Equation: 4B & C − 5B & + 6PB & + 3PB + 2P& = 0 Problem No.13: Answer at different stages: Matrix form: 3P  P b + c − & − # 0  B y B z g Th = i j # P P  0 & − b + c − & B B  Frequency Equation: 4P 2 & 2P& 4P   & C  − b + c +{ & + + |=0 B  B B  & Problem No.14, 15, 16, 17, 18, 19, 20 and 21: Try yourself Problem No.22: Solved Example 5.5 page No. 485 SS Rao Problem No.23, 24, 25 and 26: Try yourself Problem No.27: Solved Example 5.6 page No. 491 SS Rao

Dr. C V Chandrashekara, Professor and Lead: Design Domain, Dept. of Mechanical Engineering, PESU, B’lore

UE15ME352: Mechanical Vibrations

Unit IV Chapter 6: Multi-degree of freedom systems 9. The dynamic matrix [D] of a system is given by, 1 3 2 1 k}l = ~3 5 2.5 3 3 5 5.5 Evaluate the fundamental frequency and mode shape of the characteristic equation, €#‚ = k}l#‚ Using matrix iteration method

Solutions: Let, x1 = 1, x2 = 2 and x3 = 3 1st Iteration: 끃‚& = k}lƒ‚T ƒT ƒT 1 3 2 1 ƒ ƒ λ „ & … = ~3 5 2.5 „ & … ƒ & 3 3 5 5.5 ƒ T ƒT 1 3 2 1 1 1 10 10 1 ƒ λ „ & … = ~3 5 2.5 „2… = „20.5… = „2.05… 3 3 ƒ & 3 3 5 5.5 3 29.5 2.95 nd 2 Iteration: 끃‚ = k}lƒ‚& ƒT 1 1 3 2 1 1 10.05 10.05 1 ƒ λ „ & … = ~3 5 2.5 „2.05… = „20.63… = „2.05… 3 3 ƒ 3 3 5 5.5 2.95 2.93 29.48 3rd Iteration: 끃‚C = k}lƒ‚ ƒT 1 1 3 2 1 1 10.03 10.03 1 ƒ λ „ & … = ~3 5 2.5 „2.05… = „20.58… = „2.05… 3 3 ƒ C 3 3 5 5.5 2.93 29.37 2.93 λT =

10.03 1 = &  ∴ _†)(B _)‡†)Gˆ, T = 0.5469 /)G 3 T 1

D/B.(†) 0(.* = „2.05…

Mode Shape:

2.93

Practice all Class Work Problems

Dr. C V Chandrashekara, Professor and Lead: Design Domain, Dept. of Mechanical Engineering, PESU, B’lore