For that same four-storey building, the force at the top storey is nearest to
Answer: A
Taking sixteen of the thirty parts gives \( 800 \times 16/30 = 426.7 \) kN. Slightly more than half the base shear is applied at the topmost level, which is why the overturning moment on the lower storeys is so much larger than a uniform distribution would suggest.
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An elastic analysis gives a base shear of 2,400 kN. Designing with a response reduction factor of 4 gives a design base shear of
Answer: B
The factor divides the elastic demand, so \( 2{,}400/4 = 600 \) kN. The remaining three quarters of the demand is expected to be absorbed by inelastic deformation, which is legitimate only if the detailing to sustain it has been provided.
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A structure is designed with R = 3 rather than R = 5, everything else unchanged. Its design base shear becomes
Answer: C
Since the factor appears in the denominator, reducing it from five to three multiplies the force by five thirds, that is by about sixty-seven per cent. The lower factor goes with less demanding detailing, so the extra strength is what pays for the reduced ductility.
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A hospital and an ordinary office building are identical in every respect and stand in the same zone. If the office is designed for 500 kN, the hospital is designed for
Answer: D
The importance factor of 1.5 multiplies the design force directly, so \( 500 \times 1.5 = 750 \) kN. The intention is that the hospital remains usable after the earthquake rather than merely avoiding collapse, which is a more demanding objective.
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A frame 24 m tall of reinforced concrete without infill is analysed. Its code period is nearest to
Answer: A
Applying \( T = 0.075 \times 24^{0.75} = 0.075 \times 10.84 = 0.81 \) s. A useful check is that doubling the height multiplies the period by \( 2^{0.75} \), about 1.68, rather than by two, which keeps tall buildings from appearing slower than they are.
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A storey drift of 18 mm is computed in a storey 3.0 m high. The code drift limit is
Answer: B
The limit is 0.004 times 3,000 mm, that is 12 mm, so the computed drift is half again as large as permitted and the frame must be stiffened. Drift rather than strength frequently governs the design of tall flexible frames.
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To bring a frame within the drift limit, the most direct measure is to
Answer: C
Larger columns, added shear walls or diagonal bracing all reduce drift by stiffening the storey in which the movement is occurring. Adding mass would increase the demand instead, and altering the importance or reduction factors changes the paperwork rather than the behaviour of the frame.
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A building has a static eccentricity of 0.9 m and a plan dimension of 20 m perpendicular to the direction of shaking. The design eccentricity from the amplified expression is
Answer: D
Applying \( 1.5 \times 0.9 + 0.05 \times 20 = 1.35 + 1.0 = 2.35 \) m. The alternative expression gives \( 0.9 - 1.0 = -0.1 \) m, and both cases have to be checked because either of them can govern a particular element on one side of the plan or the other.
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Which one of the following does NOT appear in the expression for the design horizontal seismic coefficient?
Answer: A
The expression is \( A_h = (Z/2)(I/R)(S_a/g) \), and the material grade enters nowhere in it. Grade affects the capacity the structure can be given, not the demand the code places upon it, which is why a richer mix is no substitute for a properly proportioned lateral system.
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Two buildings differ only in that one is founded on rock and the other on soft soil. The difference in their design forces arises through
Answer: B
Soil type selects which spectrum curve applies, and soft soil generally gives a larger ordinate over the range of periods most buildings occupy. The zone factor is a property of the region and does not change with the ground conditions at a particular plot.
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