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Response Spectrum Concept Questions and Answers

Answer: C

Because the peaks are combined statistically the signs are lost, so quantities that must be combined consistently, such as axial force and moment in a column, need special treatment. Time history analysis retains the signs.

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32.

A structure of period 0.4 s sits at a site where the design spectrum gives Sa/g = 2.5 on the plateau. For a zone factor of 0.24, importance factor 1.0 and response reduction factor 5, the design horizontal seismic coefficient is

Answer: D

Correct answer: 0.06

Substituting into the Indian code expression: \[ A_h = \frac{Z}{2}\cdot\frac{I}{R}\cdot\frac{S_a}{g} = \frac{0.24}{2}\times\frac{1.0}{5}\times 2.5 \] \[ = 0.12 \times 0.2 \times 2.5 = 0.06. \] The factor of two in the denominator converts the zone factor, which is the maximum considered earthquake level, to the design basis earthquake, and omitting it doubles the answer.

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33.

A building of seismic weight 10 000 kN has a design horizontal seismic coefficient of 0.05. Its design base shear is

Answer: A

Correct answer: 500 kN

\[ V_B = A_h W = 0.05 \times 10\,000 = 500\ \text{kN.} \] The base shear is simply the seismic coefficient times the seismic weight, and it is then distributed up the height of the building in proportion to \( W_i h_i^2 \) in the static method. Note that the seismic weight is not the full gravity load - only part of the imposed load is included.

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34.

A system has a period of 0.5 s and reads a spectral acceleration of 0.4 g. Its spectral displacement is closest to

Answer: B

Spectral displacement is the spectral acceleration divided by the square of the circular frequency. With a period of 0.5 s the circular frequency is 12.57 rad/s, so 3.924 divided by 158 gives about 0.0249 m, or 24.9 mm.

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35.

Doubling the mass of a single-degree-of-freedom system multiplies its natural period by

Answer: C

Correct answer: 1.41

\[ T \propto \sqrt{m} \quad\Rightarrow\quad \frac{T_2}{T_1} = \sqrt{2} = 1.41. \] Doubling the mass therefore lengthens the period by about 41 per cent, not by a factor of two. It also increases the seismic weight, so the force demand may rise even though the longer period moves the structure to a lower point on the spectrum.

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36.

Two buildings of period 0.3 s and 3.0 s stand at the same site. On a typical design spectrum the one attracting the larger spectral acceleration is

Answer: D

Correct answer: the 0.3 s building

A period of 0.3 s falls on or near the plateau of a typical design spectrum, where the ordinate is at its maximum. At 3.0 s the ordinate has fallen well down the descending branch, so the flexible building attracts a much smaller acceleration - though a much larger displacement, which must be provided for at joints and separations.

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Answer: A

Stiffening shortens the period, which for many buildings moves them up the spectrum towards the plateau and increases the acceleration demand, even though drift falls. This is one of the ways earthquake loading differs from gravity or wind.

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38.

Match the period range with the controlling ground motion parameter.

1. Very short period   2. Intermediate period   3. Very long period

P. Peak ground displacement   Q. Peak ground acceleration   R. Peak ground velocity

Select the correct matching.

Answer: B

A very stiff system follows the ground and its acceleration approaches the ground's; a very flexible one stays still and its relative displacement approaches the ground displacement; velocity governs the middle. That is the structure of every spectrum.

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Answer: C

Correct answer: Higher damping shortens the natural period appreciably

This is the incorrect statement. The damped period is \( T_d = T/\sqrt{1-\zeta^2} \), and at the damping ratios found in buildings, five to ten per cent, the change is a fraction of one per cent - entirely negligible. What damping does affect strongly is the amplitude of the response and therefore the spectral ordinates.

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Answer: D

Correct answer: false, since it gives only peak values

A response spectrum is a plot of the maximum response of each oscillator, and the maximum is all it retains - the sign and the instant at which it occurred are discarded in the computation. Knowing how the structure moves at each instant requires a time-history analysis, which is precisely the extra information that method provides.

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