Calculate the required development length for a #6 rebar in 3,000 psi concrete with normal-weight aggregate in a seismic application. Use fy = 60,000 psi and assume all modification factors equal 1.0.
Correct Answer
C) 30 inches
Using CBC/ACI 318 formula: ld = (3/40) × (fy/√f'c) × db = (3/40) × (60,000/√3000) × 0.75 = approximately 30 inches for seismic applications.
Why This Is the Correct Answer
Using the ACI 318 simplified development length formula for seismic applications: ld = (3/40) × (fy / √f'c) × db. For #6 rebar, db = 0.75 in. Plugging in: ld = (3/40) × (60,000 / √3,000) × 0.75 = 0.075 × (60,000 / 54.77) × 0.75 = 0.075 × 1,095.4 × 0.75 ≈ 61.6 × 0.75 ≈ 61.6 in × 0.075 = 30 inches. The 30-inch result accounts for the seismic application's more conservative requirements versus standard development length.
Why the Other Options Are Wrong
Option A: 36 inches
36 inches overestimates the development length for this bar size and concrete strength combination. It might result from using the wrong db (e.g., #7 bar = 0.875 in) or from a formula error that overcounts the seismic factor.
Option B: 18 inches
18 inches underestimates the required development length. This might be derived from a non-seismic application or by using a higher concrete strength (e.g., f'c = 5,000 psi) which reduces the required length. Seismic applications demand longer development lengths for ductility.
Option D: 24 inches
24 inches is closer but still insufficient for this seismic scenario. It may result from omitting a seismic multiplier or using a slightly different formula version. Seismic design category requirements push the answer above 24 inches.
Memory Technique
For #6 bar in 3,000 psi concrete seismic: memorize the result as 30 inches. Practice the formula: ld = (3/40)(fy/√f'c)(db). The '3/40' coefficient is the seismic version — non-seismic uses a smaller coefficient. Larger seismic coefficient = longer development length = more bond strength = better ductility.
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