Dynamic Analysis of Structures Questions and Answers

Answer: C

Correct answer: 1-Q, 2-P, 3-R

The three expressions build on one another. Natural circular frequency is \( \sqrt{k/m} \), depending only on stiffness and mass.

Critical damping is \( 2\sqrt{km} \) - note the product rather than the ratio, and the factor of two. Damped frequency is \( \omega_n\sqrt{1-\xi^2} \), the undamped value reduced by an amount that is negligible at practical damping. Recognising which combination of \( k \) and \( m \) each uses settles the first two at once.

Answer: D

The matrices come first, the eigenvalue problem next, then the independent modal equations, and finally the superposition back to physical coordinates. The last step is an algebraic sum in time, unlike the statistical combination used with a spectrum.

Answer: A

Correct answer: A is false but R is true

The reason is correct and it is exactly what defeats the assertion. Modal superposition rests on decomposing the response into independent modal contributions and adding them, which requires linearity.

A yielding structure has a stiffness that depends on its displacement history, so the modes are not constant and the contributions cannot be superposed. Nonlinear time-history analysis or pushover analysis must be used instead. A true reason refuting rather than supporting the assertion - each must be judged separately.

Answer: B

Stability and accuracy are different properties: a large step gives a bounded but badly wrong answer. The reason states the meaning of unconditional stability correctly, which is precisely why it does not license any step size.

Answer: C

Correct answer: False, because the damped and undamped frequencies differ negligibly at practical damping levels

The relation is \( \omega_d = \omega_n\sqrt{1-\xi^2} \), and at 5 per cent damping the factor is 0.99875 - a difference of about an eighth of one per cent in the period.

So damping does lower the frequency, but by an amount no measurement would resolve, which is why codes use the undamped period throughout. The second option overstates the case in the other direction: damping has a large effect on the amplitude of response, particularly near resonance, and only a negligible one on the period.

Answer: D

Correct answer: False, since refining the model changes the distribution of response rather than guaranteeing a larger total

A refined model redistributes the response among more modes and represents the structure's behaviour more faithfully; it does not add force. The total mass is unchanged, and it is mass that attracts seismic demand.

Including higher modes may raise or lower the base shear, since their contributions partly cancel - higher modes have sign changes in their shapes. Codes set a minimum base shear precisely because a refined analysis can give a lower value than the simplified method.

Answer: A

Correct answer: the peak responses of different modes occur at the same instant

Each mode has its own period, so its peak arrives at its own moment. Adding the modal peaks arithmetically would assume a coincidence that does not occur and would badly overstate the response.

That is why modal combination rules exist: square root of the sum of squares for well-separated periods, and complete quadratic combination where periods are close enough for the modes to interact. The other three statements are all true of modal analysis.

Answer: B

Correct answer: The ground motion record for the site

The eigenvalue problem \( (K - \omega^2 M)\phi = 0 \) contains only the mass and stiffness matrices, and the boundary conditions determine how those matrices are assembled.

Periods and mode shapes are therefore properties of the structure alone, independent of any loading - which is why they are computed once and reused for every load case. The ground motion enters afterwards, when the modal responses are evaluated against a spectrum or a record.

Answer: C

Since \( \omega_n = \sqrt{k/m} \), the frequency follows the square root of stiffness, so it rises by about forty-one per cent. The period falls by the same factor. Halving the mass would have exactly the same effect on the frequency as doubling the stiffness.

Answer: D

Correct answer: 0.71 s

Period varies with the square root of mass:

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

\[ T_2 = 0.5 \times 1.414 = 0.71\ \mathrm{s} \]

The distractor 1.00 s doubles the period, treating the relation as linear. The square root also explains why adding mass is an inefficient way of lengthening a period: doubling the mass buys only 41 per cent more period, and attracts more seismic force in proportion to the mass added.