Dynamic Analysis of Structures Questions and Answers

Answer: C

Rotational degrees of freedom at joints often carry negligible mass, and condensing them out shrinks the eigenvalue problem without material loss of accuracy. The stiffness contribution of those degrees of freedom is retained in the condensed matrix.

Answer: D

Two translations and a rotation describe the whole floor if it cannot deform in plan, which is a good approximation for most reinforced concrete slabs. Long narrow or heavily perforated floors may require a flexible diaphragm model instead.

Answer: A

The idealisation gives one lateral degree of freedom per floor and a tridiagonal stiffness matrix, which makes hand analysis tractable. It suits framed buildings with stiff floors and stops being reasonable for slender shear-wall structures.

Answer: B

Two lateral degrees of freedom give two frequencies and two mode shapes: the first with both floors moving together and the second with them moving in opposition. The count follows the degrees of freedom, not the number of members.

Answer: C

Correct answer: in opposite directions

Higher modes are characterised by sign changes in the mode shape, and the second mode of a two-storey building has exactly one: the floors move out of phase, one to the left as the other goes right.

The fundamental mode has both moving together, with the upper floor further. The number of sign changes rises with the mode number, which is the general pattern - and it is why higher modes contribute little net base shear, the opposing floor forces partly cancelling.

Answer: D

Correct answer: an arbitrary forcing function, by superposing impulse responses

Duhamel's integral treats the load as a succession of infinitesimal impulses, computes the free vibration each would produce, and sums them:

\[ u(t) = \frac{1}{m\omega_d}\int_0^t p(\tau)e^{-\xi\omega_n(t-\tau)}\sin\omega_d(t-\tau)\,d\tau \]

Because it is built on superposition it requires linearity, but it places no restriction on the shape of the load. Harmonic, step, blast and earthquake loading are all covered by the same expression.

Answer: A

Correct answer: behaves inelastically during the excitation

The integral superposes the responses to a sequence of impulses, and superposition holds only for a linear system.

Once yielding begins the stiffness depends on the displacement history, the response to two loads is no longer the sum of the responses to each, and the whole construction fails. Step-by-step numerical integration is used instead, updating the stiffness at each step. Multiple degrees of freedom and any damping ratio are handled comfortably; nonlinearity alone defeats it.

Answer: B

Correct answer: handles inelastic behaviour and irregular records naturally

Step-by-step integration advances the solution one small interval at a time, so the stiffness can be updated whenever a member yields - something superposition cannot accommodate.

An earthquake record is also a long irregular series of digitised values, which the marching scheme consumes directly. It is not exact for any time step, it needs the damping like any other method, and it produces the whole history from which the peak is then read.

Answer: C

Correct answer: average acceleration method

With \( \beta = 1/4 \) and \( \gamma = 1/2 \), Newmark's scheme assumes the acceleration is constant over the step at the average of its end values - hence the name, and the alternative label "constant average acceleration".

The choice is important because it is unconditionally stable, so the time step is chosen for accuracy alone. Setting \( \beta = 1/6 \) gives the linear acceleration method, which is more accurate but only conditionally stable.

Answer: D

Correct answer: \( \gamma \ge 1/2 \) and \( \beta \ge \gamma/2 \)

These conditions make the scheme unconditionally stable, so the solution does not diverge however large the time step. The average acceleration method, with \( \gamma = 1/2 \) and \( \beta = 1/4 \), satisfies them exactly.

Stability is not accuracy: a stable scheme with too large a step gives a bounded but wrong answer, which is why a step of about a tenth of the shortest period of interest is still recommended. Elastic behaviour is not required, and that is precisely why the method suits inelastic analysis.