SolSec 5.9

# SolSec 5.9 - 5 77 Problems Section 5.9(5.86 through 5.88...

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5- 77 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 r G G . . = + + 1 1 2 2 2 2 η η ω = 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 1 0 21 1 2 0 21 2 2 2 2 2 . ( . ) ( . ) = + + r G G This is solved numerically in the following Mathcad session: From the plot, any value of r greater then about 2.5 will do the trick. Choosing r = 2.5 yields ω ω n k m k = = = = = 3 5 35 3 5 10 100 3 300 . . ( ) N/m

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5- 78 5.87 A machine part is driven at 40 Hz at room temperature. The machine has a mass
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