Dynamic Analysis of Structures Questions and Answers

Answer: A

Correct answer: 3.7 per cent

Halving over three cycles gives the decrement per cycle:

\[ \delta = \frac{1}{3}\ln 2 = \frac{0.6931}{3} = 0.2310 \]

\[ \xi \approx \frac{\delta}{2\pi} = \frac{0.2310}{6.283} = 0.037 = 3.7\% \]

Averaging over several cycles is standard practice, since it reduces the effect of noise on any single peak. The distractor 16.7 per cent takes the halving as a simple fraction spread over three cycles, ignoring both the logarithm and the \( 2\pi \).

Answer: B

Correct answer: the shorter natural period

Period varies inversely with the square root of stiffness:

\[ T = 2\pi\sqrt{\frac{m}{k}} \qquad \frac{T_2}{T_1} = \frac{1}{\sqrt{2}} = 0.707 \]

so doubling the stiffness shortens the period by about 29 per cent. The square root matters: stiffening a structure is an inefficient way of changing its period, and on a typical design spectrum a shorter period often means a higher spectral acceleration, so the stiffer building may attract more force.

Answer: C

Correct answer: the horizontal response usually governs the design

Structures are designed to carry gravity, so they already have substantial vertical stiffness and strength, and their vertical periods are very short. Lateral resistance is the scarce commodity, and it is horizontal shaking that produces the demands buildings fail under.

The simplification is not universal: vertical motion matters for long-span floors and cantilevers, and for structures near the fault. Columns certainly have axial stiffness and vertical ground motion certainly exists - the neglect is a judgement about what governs.

Answer: D

Correct answer: Adding mass at the upper floors

Period rises with the square root of mass, so adding mass lengthens it - and adding it high in the building is most effective, since the upper floors have the largest modal displacement.

The other three all add stiffness or reduce height, which shortens the period. Note that the change is rarely worth making deliberately: extra mass attracts extra seismic force in proportion, so the lengthened period is bought at a price the spectrum may not repay.

Answer: A

Correct answer: Adding stiff infill walls throughout

Period falls with increasing stiffness, and infill walls are very stiff in their own plane - often stiffer than the frame surrounding them, whether or not the designer intended them to participate.

The other three all add mass, which lengthens the period. Note the danger the third option hints at: infill removed at ground level only creates a soft storey, concentrating the whole drift demand in one level, which is among the most common causes of collapse.

Answer: B

Correct answer: dies away under damping, leaving the steady-state part

The response to a harmonic force has two parts: a transient at the damped natural frequency, set by the initial conditions, and a steady state at the forcing frequency.

Damping removes the transient within a few cycles, after which only the steady state remains - which is why steady-state amplitude is what design generally considers. The transient still matters for short-duration loading, where the motion may never reach a steady state at all.

Answer: C

Correct answer: smaller than the static response to the same peak force

The magnification factor is the ratio of dynamic to static response, so a value below one means the dynamic response is the smaller.

It is entirely possible and it is what happens at high frequency ratios: when the forcing is much faster than the structure can respond, the mass cannot follow and the amplitude falls away. That is precisely the regime vibration isolation works in, above a frequency ratio of \( \sqrt{2} \), where the transmitted force is less than the applied one.

Answer: D

Correct answer: 25 Hz

Convert revolutions per minute to cycles per second:

\[ \frac{1500}{60} = 25\ \mathrm{Hz} \]

An unbalanced rotor produces a force at exactly this frequency, and its harmonics at multiples of it. The figure is the starting point for any isolation calculation, since the mounting must be designed relative to it - and it is also the frequency the structure must avoid having a natural frequency near.

Answer: A

Correct answer: 5 Hz or lower

Isolation requires a frequency ratio above \( \sqrt{2} \), and effective isolation a ratio of about three or more. At 25 Hz forcing:

\[ \frac{25}{5} = 5 \]

which gives a comfortable margin. A mounting at 25 Hz would sit exactly at resonance and amplify the force badly, and a stiffer mounting still would be worse. The consequence is that isolators must be soft, with a large static deflection - the practical constraint on how low the mounting frequency can be taken.

Answer: B

A heavier block has a lower natural frequency and a smaller response to the same unbalanced force, which is why such blocks are massive. It also raises the frequency ratio, moving the system further into the isolation region.