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Problems Section 5.9 (5.86 through 5.88) 5.86: ω = 35 rad/s ! " r # # % & ' * ! η are: η = 0.21

This document discusses recalculating a vibration isolator design example using a different equation, selecting an isolator for a machine part driven at 40 Hz, and comparing the transmissibility ratio equations from the text and a window. 1) It recalculates a previous example design using a new dynamic shear modulus equation and selects an isolator with a natural frequency below 20 Hz. 2) For a 100 kg machine part driven at 40 Hz, it determines that 75 isolator mounts are needed, each supporting less than 3 lbs, and selects possible isolators from a table. 3) It shows that the two transmissibility ratio equations only differ in that one uses dynamic shear modulus terms where the other uses static

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29 views3 pages

Problems Section 5.9 (5.86 through 5.88) 5.86: ω = 35 rad/s ! " r # # % & ' * ! η are: η = 0.21

This document discusses recalculating a vibration isolator design example using a different equation, selecting an isolator for a machine part driven at 40 Hz, and comparing the transmissibility ratio equations from the text and a window. 1) It recalculates a previous example design using a new dynamic shear modulus equation and selects an isolator with a natural frequency below 20 Hz. 2) For a 100 kg machine part driven at 40 Hz, it determines that 75 isolator mounts are needed, each supporting less than 3 lbs, and selects possible isolators from a table. 3) It shows that the two transmissibility ratio equations only differ in that one uses dynamic shear modulus terms where the other uses static

Uploaded by

sonti11
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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5- 89 Problems Section 5.9 (5.86 through 5.88) 5.86 Reconsider Example 5.2.

1, which describes the design of a vibration isolator to protect an electronic module. Recalculate the solution to this example using equation (5.92). Solution: If data sheets are not available use G! =G/2. One of many possible designs is given. From the example we have T.R. = 0.5, m = 3 kg and ! = 35 rad/s = 5.57 Hz. From equation (5.92): 1 + !2 T. R. = = 0.5 2 % '1 " r2 G# ( * + !2 & G$ # ) From Table 5.2 for 75F and frequency of 10 Hz (the closest value listed), the value of E and " are: E = 2.068 x 107 N/m2 and " = 0.21 9 2 Thus G = E/3 = 6.89 x 10 N/m using the approximation suggested after equation (5.86). They dynamic shear modulus is estimated from plots such as Figure 5.38 to be G! =G/2. Thus equation 5.92 becomes 1 + (0.21) 2 2 0.5 = 2 # & G " %1 ! r 2 ( + (0.21)2 G " $ 2' This is solved numerically in the following Mathcad session:

5- 90 From the plot, any value of r greater then about 2.5 will do the trick. Choosing r ! 35 k = 2.5 yields ! n = = " = 10 " k = 100(3) = 300 N/m 3.5 3.5 m

5- 91

5.87

A machine part is driven at 40 Hz at room temperature. The machine has a mass of 100 kg. Use Figure 5.42 to determine an appropriate isolator so that the transmissibility is less than 1. Solution: Given f = 40 Hz, m = 100 kg or about 220 lbs. and T.R. <1. The maximum static load per mount is 3 lbs. Therefore the system would require a minimum of 73 mounts. Assume then that 75 mounts are used. Thus

220# = 2.9# per mount 75


For the isolator, fn <0.5 f = 0.5(40) = 20 Hz. Therefore the fn of the isolator must be less then 20 Hz. Referring to the performance characteristics of the table in Figure 5.42 yields 4 possible isolator choices: AM 001-2,3,17,18

5.88

Make a comparison between the transmissibility ratio of Window 5.1 and that of equation (5.92). Solution: Comparing equation (5.92) with Window 5.1 yields: Window 5.1: Equation (5.92):
T. R. = 1 + (2!r) 2 (1 " r2 ) 2 + (2!r ) 2

% '1 " r2 G# ( * + !2 & G$ #) Comparing the two equations yields G# ! = 2"r and %1 G$ #
2

T. R. =

1 + !2

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