This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
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 50°F versus the value at 75°F. Use the values for ηfrom 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"#$%&’At 50°, η= 0.5 and at 75°, η= 0.28, so X50!=2FkX75!=3.57Fk5- 735.68Using Equation (5.67), calculate the new modulus of a 0.05 ×0.01 ×1, piece of pinned-pinned aluminum covered with a 1-cm-thick piece of nitrite rubber at 75°F driven at 100 Hz. Solution: From Table 1.2, E1= 7.1 ×1010N/m2for aluminum. From Table 5.2, E2=2.758!107N/m2for nitrate rubber, Also, I=I1=130.05( )1( )3=0.01667 m4e2=E2E1=2.758!1077.1!1010=3.885!10"4h2=H2H1=0.010.01=1From Equation (5.67), E=E1I1I1+e2h22+3 1+h2( )2e2h21+e2j2!"#$%&=7.136’1010N/m25.69Calculate Problem 5.68 again at 50°F. 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?...
View Full Document
- Spring '10
- Hertz, e2 h2