SolSec 6.7 - 6- 52 Problems and Solutions Section 6.7 (6.53...

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6- 52 Problems and Solutions Section 6.7 (6.53 through 6.63) 6.53 Calculate the response of Example 6.7.1 for l = 1 m, E = 2.6 × 10 10 N/m 2 and ρ = 8.5 × 10 3 kg/m 3 . Plot the response using the first three modes at x = l /2, l /4, and 3 l /4. How many modes are needed to represent accurately the response at the point x = l /2? Solution: wxt l et x n n t nn n n (,) . ( ) cos sin . =− + = 002 1 22 1 001 2 1 σ ωσ ω Where π ρ ωω n dn n n l E = = = () . 21 2 0 9999 For l = 1 m E 26 10 85 10 10 3 . . N/m kg/m 2 3 Response using first three modes at x ll l = 24 3 4 ,, plotted below. Three modes accurately represents the response at x l = 2 . The error between a three and higher mode approximation is less than 0.2%.
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6- 54 6.54 Repeat Example 6.7.1 for a modal damping ratio of ζ n = 0.01. Solution: Using ζ n = 0.01 and the frequency given in the example ωω ζ ω ρ σ dn nn n n n nE E =− = = = 1 0 995 21 2 2 ., l The time response is then Tt Ae t t dn n n ( ) sin( ) . =+ 01 ωφ and the total solution is: wxt Ae t n x n t dn n n n ( , ) sin( )sin () . = 1 2 π l The initial conditions are: wx x t (,) . 00 0 1 == l m and Therefore: 001 . sin sin x Ax n l = φσ Multiply by sin σ m x and integrate over the length of the bar to get 1 2 123 1 2 . sin , , ,. .. + m m mm Am l l φ From the velocity initial condition A x tn n n dn n n n ( , ) . sin cos sin 0 1 1 + [] = Again, multiply by sin σ m x and integrate over the length of the bar to get A mn n dn n ( . sin cos ) −+ = 2 0 l Since A m is not zero this yields: tan sin cos . ... n n n n n =⇒ = = ° 1 9 9499 1 4706 84 3 3 rad Substitution into the equation from the displacement initial condition yields: A m m m nm m = ++ 1 1 0 0201 1 22 1 1 . sin . ll σφ The solution is then e t n x m m t dn n n n . ( ) sin( )sin . + + = 1 2 1 1
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6- 55 6.55 Repeat Problem 6.53 for the case of Problem 6.54. Does it take more or fewer modes to accurately represent the response at l /2? Solution: Use the result given in 6.54 and l E = 1 26 10 85 10 10 3 m N/m kg/m 2 3 . . ρ The response is plotted below at x ll l = 42 3 4 ,, . An accurate representation of the response is obtained with three modes. The error between a three mode and a higher mode representation is always less than 0.2%. The results here are from Mathcad:
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6- 57 6.56
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This note was uploaded on 02/11/2010 for the course MECHANICAL ms316 taught by Professor Abduljaba during the Spring '09 term at Kalamazoo.

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SolSec 6.7 - 6- 52 Problems and Solutions Section 6.7 (6.53...

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