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Dynamic Analysis of Structures Questions and Answers

Answer: C

Correct answer: \( 2\pi\xi \)

The exact relation is

\[ \delta = \frac{2\pi\xi}{\sqrt{1-\xi^2}} \]

and for light damping the square root is close to one, leaving \( \delta \approx 2\pi\xi \).

The approximation is excellent at practical levels: at 5 per cent damping the error is about a tenth of one per cent. Inverting it gives the working formula \( \xi \approx \delta/2\pi \), which is how a decay record is turned into a damping ratio.

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12.

Successive peaks in a free vibration record measure 12 mm and 9 mm. The logarithmic decrement is nearest to

Answer: D

Taking \( \delta = \ln(12/9) = \ln 1.333 = 0.288 \). Dividing the two amplitudes without taking the logarithm gives 1.333, which is the amplitude ratio itself and not the decrement, and it would imply damping several times larger than the record actually shows.

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13.

From a logarithmic decrement of 0.288, the damping ratio is nearest to

Answer: A

Correct answer: 4.6 per cent

Invert the light-damping approximation:

\[ \xi \approx \frac{\delta}{2\pi} = \frac{0.288}{6.283} = 0.0458 = 4.6\% \]

The distractor 28.8 per cent takes the decrement itself for the ratio, forgetting the factor of \( 2\pi \). A result near 5 per cent is entirely typical of a concrete structure, which is a useful check - a decrement of 0.288 means successive peaks in the ratio \( e^{0.288} = 1.33 \), a visible but gentle decay.

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Answer: B

Correct answer: the width of the resonance peak at \( 1/\sqrt{2} \) of its height

The method uses the shape of the frequency response curve. The frequencies either side of resonance at which the amplitude falls to \( 1/\sqrt{2} \) of the peak bound a band whose width gives the damping:

\[ \xi = \frac{f_2 - f_1}{2f_n} \]

Light damping gives a sharp peak and a narrow band, heavy damping a broad one. It requires a forced vibration test with a swept frequency, where the logarithmic decrement needs only a decay record - the two methods suit different experimental means.

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15.

A forced vibration test gives half-power frequencies of 4.9 Hz and 5.1 Hz about a resonant frequency of 5.0 Hz. The damping ratio is

Answer: C

Applying \( \xi = (f_2 - f_1)/2f_n = (5.1 - 4.9)/(2 \times 5.0) = 0.2/10 = 0.02 \), so the structure is damped at two per cent of critical. A narrow resonance peak indicates light damping and a broad one heavy damping, which is why the width is the quantity measured.

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Answer: D

Correct answer: \( 1/2\xi \)

At resonance the exact magnification is \( 1/(2\xi\sqrt{1-\xi^2}) \), which for light damping reduces to \( 1/2\xi \).

The relation shows how completely damping governs the resonant response: at 5 per cent the amplification is 10, at 2 per cent it is 25, and as \( \xi \to 0 \) it grows without bound. Away from resonance damping matters far less, which is why an accurate damping estimate is needed only where the forcing is close to a natural frequency.

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17.

A harmonic force is applied at the natural frequency of a structure damped at 2 per cent of critical. The steady-state amplitude is about

Answer: A

Correct answer: 25 times the static deflection

Applying the resonant magnification:

\[ \frac{1}{2\xi} = \frac{1}{2 \times 0.02} = 25 \]

so the steady-state amplitude is 25 times the deflection the same force would produce statically. The figure explains why resonance is treated as a design condition in its own right: a modest force sustained at the natural frequency produces displacements far beyond anything the static analysis suggests, and only damping limits them.

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Answer: B

Damping bounds the amplitude at roughly \( 1/2\xi \) times the static value, whereas an undamped system would grow without limit. Real structures always have some damping, so the response is large rather than infinite.

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19.

Vibration isolation of a machine from its supporting floor becomes effective when the ratio of forcing frequency to natural frequency exceeds

Answer: C

Correct answer: \( \sqrt{2} \)

The transmissibility curve crosses unity at a frequency ratio of \( \sqrt{2} \); below that the mounting amplifies the transmitted force, and above it the force is reduced.

So an isolator is effective only when the machine runs well above the mounting's natural frequency, which means soft mounts for low-speed machinery. It also means the machine passes through resonance during run-up and run-down, where the transmitted force peaks - the reason such passages are made quickly.

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20.

Transmissibility equal to unity means the force transmitted to the support is

Answer: D

Unity marks the crossover at a frequency ratio of \( \sqrt{2} \), where the mounting neither amplifies nor attenuates. Isolation is only achieved beyond that point. Below that ratio a soft mounting amplifies the very force it was meant to reduce.

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