Response Spectrum Concept Questions and Answers

Answer: A

With \( \omega = \sqrt{k/m} = \sqrt{16{,}000{,}000/40{,}000} = 20 \) rad/s, the period is \( T = 2\pi/\omega = 0.314 \) s. Note that the stiffness must be converted from kN/m to N/m before the root is taken, which is where most slips occur.

Answer: B

Correct answer: twice the original value

\[ T \propto \sqrt{m} \quad\Rightarrow\quad \frac{T_2}{T_1} = \sqrt{4} = 2. \] Quadrupling the mass doubles the period. Note that the quadrupled mass also quadruples the seismic weight, so unless the longer period moves the structure far enough down the spectrum, the base shear rises substantially.

Answer: C

Base shear is the spectral acceleration coefficient times the seismic weight, so \( 0.6 \times 2{,}000 = 1{,}200 \) kN. Because the acceleration is already expressed as a multiple of \( g \), no further division by \( g \) is needed - doing so is a common error.

Answer: D

The response reduction factor divides the elastic demand, giving \( 1{,}600/4 = 400 \) kN, on the understanding that ductile detailing will let the structure absorb the rest inelastically. Multiplying by the factor instead inverts the intention of the clause.

Answer: A

The chain \( S_v = \omega S_d \) and \( S_a = \omega S_v = \omega^2 S_d \) links the three, which is precisely what allows all of them to be plotted on a single tripartite logarithmic sheet. They are not independent readings.

Answer: B

With \( \omega = 2\pi/0.4 = 15.7 \) rad/s, \( S_a = \omega^2 S_d = 246.7 \times 0.020 = 4.93 \) m/s², or about half of \( g \). Forgetting to convert millimetres to metres shifts the answer by three orders of magnitude.

Answer: C

A very flexible structure barely moves while the ground moves beneath it, so the relative displacement approaches the ground displacement itself. At the other extreme a rigid structure moves with the ground and its acceleration approaches the peak ground acceleration.

Answer: D

Correct answer: the peak ground acceleration

A structure of very short period is effectively rigid: it moves with the ground rather than deforming relative to it, so its absolute acceleration equals the ground’s. The spectrum therefore starts at the peak ground acceleration at zero period, rises to the plateau as the structure becomes flexible enough to amplify the motion, and falls away beyond it.

Answer: A

The three regions of a spectrum are acceleration-sensitive, velocity-sensitive and displacement-sensitive in that order as the period grows. Recognising which region a structure sits in tells you which ground-motion measure controls its demand.

Answer: B

The matrices come first, the eigenvalue problem then yields periods and mode shapes, the spectrum is read at each period, and only then are the modal contributions combined. Reading the spectrum before the periods are known is impossible.