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Unformatted text preview: 6Problems and Solutions Section 6.7 (6.53 through 6.63)6.53Calculate the response of Example 6.7.1 for l= 1 m, E= 2.6 1010N/m2and = 8.5 103kg/m3. Plot the response using the first three modes at x= l/2, l/4, and 3l/4. How many modes are needed to represent accurately the response at the point x= l/2?Solution:w(x,t)=0.0222(129+10.012=1Where=(2 1292== 0.9999Forl= 1 mE= 2.6 1010/2= 8.5 103/3Response using first three modes at x=2,4,34plotted below.Three modes accurately represents the response atx=2. The error between a three and higher mode approximation is less than 0.2%.5265366.54Repeat Example 6.7.1 for a modal damping ratio of n= 0.01.Solution:Using n= 0.01 and the frequency given in the example=12= 0.995,=2 12=The time response is then Tn(t)=0.1(+29and the total solution is:w(x,t)=0.1(+29=1(2 1292 The initial conditions are:w(x,0)= 0.01(,029= 0Therefore:0.01xl=Multiply by sinmxand integrate over the length of the bar to get0.01(129+12=2= 1,2,3,...From the velocity initial conditionwt(x,0)= 0 =0.1+=1Again, multiply by sinmxand integrate over the length of the bar to getAm(0.1+292= 0Since Amis not zero this yields:tan==130.1= 9.9499 = 1.4706 = 84.3Substitution into the equation from the displacement initial condition yields:Am=0.0122(129+11=0.020122(129+1The solution is thenw(x,t)=0.0122(129+10.1(+29=1(2 1292 5466.55Repeat 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 andl= 1 = 2.6 1010/2= 8.5 103/3The response is plotted below atx=4,2,34. 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:5565666.56Calculate the form of modal damping ratios for the clamped string of equation (6.151) and the clamped membrane of equation (6.152).(6....
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 Winter '09
 Hill

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