Calculate the seismic design force for a single-story building with a weight of 150,000 pounds in Honolulu (Seismic Design Category D, SDS = 0.75).
Correct Answer
A) 28,125 pounds
Seismic base shear V = Cs × W, where Cs = SDS/(R/I). For typical construction, Cs ≈ 0.1875, so V = 0.1875 × 150,000 = 28,125 pounds.
Why This Is the Correct Answer
Option A (28,125 pounds) is correct. The seismic base shear formula is V = Cs × W, where Cs is the seismic response coefficient and W is the total seismic weight. For this problem, using a typical response modification factor R = 4 and importance factor I = 1.0 for ordinary construction, Cs = SDS / (R/I) = 0.75 / (4/1) = 0.1875. Therefore V = 0.1875 × 150,000 = 28,125 pounds.
Why the Other Options Are Wrong
Option B: 56,250 pounds
56,250 pounds results from using Cs = 0.375, which corresponds to SDS / (R/I) where R = 2. An R value of 2 is unusually low for a typical structure and would apply to systems with very limited ductility. Using too low an R value overestimates the seismic force significantly.
Option C: 18,750 pounds
18,750 pounds results from using Cs = 0.125, corresponding to SDS / (R/I) = 0.75 / 6 = 0.125. An R value of 6 applies to more ductile systems (e.g., special moment frames), which would be over-credit for a standard structure, underestimating the required seismic force.
Option D: 37,500 pounds
37,500 pounds results from using Cs = 0.25, equivalent to SDS / (R/I) = 0.75 / 3. While some structural systems have R = 3, this is not the standard assumption for typical single-story construction. This answer represents a mid-range error in R value selection.
Memory Technique
Seismic base shear: V = Cs × W. Find Cs = SDS ÷ (R ÷ I). The standard assumption for typical construction is R = 4, I = 1. So Cs = 0.75 ÷ 4 = 0.1875. Then V = 0.1875 × 150,000 = 28,125 lbs. Think of R as a 'reduction credit' for ductility — higher R = more ductile = lower required force.
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