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Unformatted text preview: 5- 72Problems and Solutions Section 5.6 (5.67 through 5.73) 5.67Compare the resonant amplitude at steady state (assume a driving frequency of 100 Hz) of a piece of nitrite rubber at 50F versus the value at 75F. Use the values for from Table 5.2. Solution: From equation (5.63), X=Fk1+!j( )"m#2At resonance !=kmso X=Fk1!"j( )!1=Fk"jThe magnitude is X=1!Fk"#$%&At 50, = 0.5 and at 75, = 0.28, so X50!=2FkX75!=3.57Fk5- 735.68Using Equation (5.67), calculate the new modulus of a 0.05 0.01 1, piece of pinned-pinned aluminum covered with a 1-cm-thick piece of nitrite rubber at 75F driven at 100 Hz. Solution: From Table 1.2, E1= 7.1 1010N/m2for aluminum. From Table 5.2, E2=2.758!107N/m2for nitrate rubber, Also, I=I1=130.05( )1( )3=0.01667 m4e2=E2E1=2.758!1077.1!1010=3.885!10"4h2=H2H1=0.010.01=1From Equation (5.67), E=E1I1I1+e2h22+3 1+h2( )2e2h21+e2j2!"#$%&=7.1361010N/m25.69Calculate Problem 5.68 again at 50F. What percent effect does this change in temperature have on the modulus of the layered material? temperature have on the modulus of the layered material?...
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This note was uploaded on 02/15/2010 for the course ME 3222 taught by Professor Tang during the Spring '10 term at UConn.
- Spring '10