SolSec 5.6 - 5- 63 Problems and Solutions Section 5.6 (5.67...

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5- 63 Problems and Solutions Section 5.6 (5.67 through 5.73) 5.67 Compare the resonant amplitude at steady state (assume a driving frequency of 100 Hz) of a piece of nitrite rubber at 50 ° F versus the value at 75 ° F. Use the values for η from Table 1. Solution: From Eq. (5.63), X F kj m = + () 0 2 1 ηω At resonance m k = ω so X F F = = 00 11 ηη The magnitude is X F k = 1 0 η At 50 ° , η = 0.5 and at 75 ° , η = 0.28, so k F X k F X 0 75 0 50 57 . 3 2 = = o o
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5- 64 5.68 Using Eq. (5.67), calculate the new modulus of a 0.5 × 0.01 × 1, piece of pinned-pinned aluminum covered with a 1-cm-thick piece of nitrite rubber at 75 ° F driven at 100 Hz. Solution: From Table 1.2, E 1 = 7.1 × 10 N/m 2 for aluminum. From Table 5.2, E 2 72 2 758 10 . N/m for nitrate rubber, Also, II e E E h H H == () () = × × = 1 3 2 2 1 7 10 4 2 2 1 1 3 0 05 1 0 01667 2 758 10 71 10 3 885 10 001 1 .. . . . . . m 4 From Eq. (5.67), E EI I eh h ej =+ + + + 11 22 2 2 2 10 13 1 1 7 136 10 . N/m 2 5.69 Calculate Problem 5.68 again at 50 ° F. What percent effect does this change in temperature have on the modulus of the layered material?
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SolSec 5.6 - 5- 63 Problems and Solutions Section 5.6 (5.67...

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