Determine the amount of rebar needed for the CMU fence in Problem 9. Allow for two inches of cover. Add 10 percent for lap and waste to the horizontal bars.
What will be an ideal response?
The vertical bars are calculated as follows:
Number of spaces = 250? × 12 in per ft / 32? = 94 spaces
Add 1 to get the number of vertical bars – Use 95 bars
The vertical bars are 6?6? long (6?8? – 2?).
Total length of vertical bars (lf) = 95 bars × 6.5? per bar = 618 lf
Total weight of vertical bars (pounds) = 618 lf × 1.043 pounds per lf = 645 pounds
Four horizontal bars are needed; two at 4? and two at 6?8?.
Total length of horizontal bars (lf) = 250? × 4 bars = 1,000 lf
Total weight of horizontal bars (pounds) = 1,000 lf × 0.668 pounds per lf = 668 pounds
Add for waste and lap.
Total weight of horizontal bars (pounds) = 668 pounds × 1.10 = 735 pounds
Total weight of rebar (pounds) = 645 pounds + 735 pounds = 1,380 pounds
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