# SolSec 5_9 - 5 89 Problems Section 5.9(5.86 through 5.88...

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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 ω = G ’/2. One of many possible designs is given. From the example we have T.R. = 0.5, m = 3 kg and ω = 35 rad/s = 5.57 Hz. From equation (5.92): T . R . = 1 + ! 2 1 " r 2 # G # G \$ % & ( ) * 2 + 2 = 0.5 From Table 5.2 for 75°F and frequency of 10 Hz (the closest value listed), the value of E and η are: E = 2.068 x 10 7 N/m 2 and η = 0.21 Thus G’ = E /3 = 6.89 x 10 9 N/m 2 using the approximation suggested after equation (5.86). They dynamic shear modulus is estimated from plots such as Figure 5.38 to be G ω = G ’/2. Thus equation 5.92 becomes 0.5 2 = 1 + (0.21) 2 1 ! r 2 " G # \$ % & ( 2 + (0.21) 2 This is solved numerically in the following Mathcad session:

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5- 90 From the plot, any value of r greater then about 2.5 will do the trick. Choosing r = 2.5 yields ! n = 3.5 = 35 3.5 " k m = 10 " k = 100(3) = 300 N/m
5- 91 5.87 A machine part is driven at 40 Hz at room temperature. The machine has a mass
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SolSec 5_9 - 5 89 Problems Section 5.9(5.86 through 5.88...

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