To calculate the reinforcement ratio p, assume a 7 in. slab with an effective depth of 6.0 in., with No. 6 (No. 19) bars spaced 5.5 in. center to center. Imperial Metric bar Linear Mass Density Nominal diameter Nominal area bar size size (soft) ko (in) (mm) (in) (mm') 21 No. 6 0.167 0 250 = % 6.35 0.249 0.05 32 N3 No. 10 0.376 0.560 0.375 =% 9.53 0.11 71 84 No. 13 0.668 0.994 0.500 4 12.7 0.20 129 #5 No. 16 1.043 1.552 0.625 = % 15.9 0.31 200 N6 No 19 1.502 2.235 0.750 = % 19.1 0.44, 284 #7 No 22 2.044 3.042 0.875= 1%222 0.6087 #8 No. 25 2.670 3.973 1.000 = % 25.4 0.79 510 #9 No 29 3.400 5.060 1.128 28.7 1.00 645 #10 No 32 4.303 6.404 1.270 32.3 1.27 819
To calculate the reinforcement ratio p, assume a 7 in. slab with an effective depth of 6.0 in., with No. 6 (No. 19) bars spaced 5.5 in. center to center. Imperial Metric bar Linear Mass Density Nominal diameter Nominal area bar size size (soft) ko (in) (mm) (in) (mm') 21 No. 6 0.167 0 250 = % 6.35 0.249 0.05 32 N3 No. 10 0.376 0.560 0.375 =% 9.53 0.11 71 84 No. 13 0.668 0.994 0.500 4 12.7 0.20 129 #5 No. 16 1.043 1.552 0.625 = % 15.9 0.31 200 N6 No 19 1.502 2.235 0.750 = % 19.1 0.44, 284 #7 No 22 2.044 3.042 0.875= 1%222 0.6087 #8 No. 25 2.670 3.973 1.000 = % 25.4 0.79 510 #9 No 29 3.400 5.060 1.128 28.7 1.00 645 #10 No 32 4.303 6.404 1.270 32.3 1.27 819
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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