{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Section 5.6

Section 5.6 - 5 72Problems and Solutions Section 5.6(5.67...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the 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

{[ snackBarMessage ]}

Page1 / 8

Section 5.6 - 5 72Problems and Solutions Section 5.6(5.67...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon bookmark
Ask a homework question - tutors are online