HW2soln

HW2soln - CE 573: Structural Dynamics HW#2 Solution Problem...

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CE 573: Structural Dynamics HW#2 Solution Problem #1 Given : A viscously damped structure is set into free vibration with an initial velocity. The resulting damped oscillations are shown above. Required : (a) Determine the logarithmic decrement, the damping factor ξ , and the undamped natural period of vibration for this structure. (b) Suppose that the displacement measurements have an accuracy of in. (You may assume that the time measurements have negligible error.) Determine the range of possible values for the damping factor. Discuss the influence that the measurement errors have on 0.02 ± . Solution : (a) I should point out that, since all of the results will depend upon the accuracy with which you read the displacement values off of the graph, it is expected that there will be some variations in your answers. As long as your results are reasonably close, they will be accepted. Let’s start by estimating the logarithmic decrement δ . I chose to work with the positive peaks, starting with peak #1 (at seconds): 0.08 t 1 0.60 u inches. Next, I looked at peak #4 (at seconds): 0.60 t 4 0.18 u inches. Thus, the value of can be evaluated as: 11 0 . 6 0 ln ln 0.40 30 . 1 8 i ik u ku + ⎛⎞ === ⎜⎟ ⎜⎟ ⎝⎠ Answer Based on this, the damping factor can be estimated according to the formulas we derived in class: () 22 2 2 0.40 0.064 4 0.40 4 δπ π == = + + Answer
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Finally, we can estimate T as follows. The length of time between peak #4 and peak #1 is approximately . There are three cycles of damped oscillation during this time, so the damped natural period can be estimated as 14 0.60 0.08 seconds 0.52 seconds t ∆≈ − = 0.173 3 d t T ≈= seconds. But is related to the natural period of vibration T via d T () ( ) 2 2 0.173 seconds 1 0.064 0.172 seconds. 1 d T TT ξ =⇒ = − = Answer (b) Since the range in relates directly to the range in δ , I will discuss the range in first. Notice that, since depends on a ratio of peak amplitudes, the largest possible value of comes when the amplitude ratio is as large as possible, while the smallest possible value of occurs when the amplitude ratio is as small as possible. The measurement errors give us a way to estimate these largest and smallest ratios. (I will use the same peaks as in part (a) to estimate the extreme ratios.) For the largest amplitude ratio, we want to assume that the measured peak #1 is underestimated while the measured peak #4 is overestimated. This gives us: 11 44 max max max 22 max 0.60 0.02 0.62 in. 0.18 0.02 0.16 in. 10 . 6 2 ln 0.4515 0.072. 30 . 1 6 4 real measured real measured uu u u δξ δπ =+ = + = =− = = ⎛⎞ ⇒≈ = ⇒= = ⎜⎟ ⎝⎠ + For the smallest amplitude ratio, we want to assume that the measured peak #1 is overestimated while the measured peak #4 is underestimated. This gives us: min min min min 0.60 0.02 0.58 in. 0.18 0.02 0.20 in. . 5 8 ln 0.3549 0.056. . 2 0 4 real measured real measured u u = =−= = + = = = + Thus, based on the measurements used above and the measurement error, we can estimate the range of damping ratios as 0.056 0.072 ≤≤ Answer . Notice that, as a percentage of our assumed “correct” damping ratio 0.064 = , the range of damping ratios goes from 12.5% below to 10.8% above the
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This note was uploaded on 05/09/2008 for the course CE 573 taught by Professor Whalen during the Fall '05 term at Purdue.

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HW2soln - CE 573: Structural Dynamics HW#2 Solution Problem...

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