Dynamic Analysis of Structures Questions and Answers

Answer: A

Correct answer: its number of degrees of freedom

The eigenvalue problem \( (K - \omega^2 M)\phi = 0 \) is of the order of the model, so an \( n \)-degree system has \( n \) natural frequencies and \( n \) mode shapes.

The count is a property of the model, not of the building: the same structure idealised with one lumped mass per floor and with a full finite element mesh yields quite different numbers. A real structure is continuous and has infinitely many; discretisation is what makes the count finite.

Answer: B

Correct answer: the determinant of \( K - \omega^2 M \) vanish

The equation is homogeneous, so a non-zero \( \phi \) exists only if the matrix is singular:

\[ \det(K - \omega^2 M) = 0 \]

Expanding the determinant gives the characteristic polynomial in \( \omega^2 \), whose \( n \) roots are the natural frequencies. Substituting each root back gives the corresponding mode shape, determined only up to a scale factor - which is why mode shapes are normalised rather than absolute.

Answer: C

Orthogonality with respect to \( M \) and \( K \) is what uncouples the equations of motion and makes modal analysis possible. Orthogonality with respect to damping holds only for particular forms of the damping matrix.

Answer: D

Transforming to modal coordinates leaves a set of independent single-degree equations, one per mode, each solved exactly as in the simple case. Damping must be of a form that diagonalises too, which is why classical damping is assumed.

Answer: A

Writing \( C = a_0M + a_1K \) guarantees that the mode shapes diagonalise the damping matrix as well, keeping the equations uncoupled. The two constants are chosen to give the desired damping ratio at two selected frequencies.

Answer: B

The mass-proportional part dominates at low frequency and the stiffness-proportional part at high, so the curve dips between the two anchor frequencies. Choosing those anchors sensibly is part of setting up a time-history analysis.

Answer: C

Lumping the mass at nodes gives a diagonal matrix that is cheap to invert, while the consistent formulation uses the element shape functions and is fully populated. The lumped form is very widely used in earthquake analysis.

Answer: D

Correct answer: how strongly base motion excites that mode

The participation factor measures the coupling between the ground motion and a particular mode - how much of the base excitation that mode responds to.

In a regular building the fundamental mode has much the largest factor, which is why it dominates the response and why codes allow simplified methods based on it. The related effective modal mass says what fraction of the total mass a mode mobilises, and codes require enough modes to account for 90 per cent of it.

Answer: A

Correct answer: the fundamental mode

The first or fundamental mode has the lowest natural frequency and therefore the longest period, and in a regular building it is the simple sway shape with all floors moving the same way.

It usually carries the greatest participation factor and effective mass, so it dominates the seismic response. Higher modes have shorter periods, more complex shapes with sign changes up the height, and smaller participation - though they matter for tall or irregular buildings.

Answer: B

Correct answer: an upper bound on the true fundamental frequency

Rayleigh's method computes a frequency from an assumed deflected shape, and the constraint imposed by assuming a shape stiffens the system. Any assumed shape other than the true mode therefore gives a frequency above the true fundamental.

The bound is one-sided and it is useful: the estimate is conservative in the sense of erring high, and the error is second order in the shape error, so even a crude assumption - the static deflection under self weight is the usual choice - gives a good result.