Homework 5 answers

Homework 5 answers - Total = 19.65 cf Fine aggregate volume...

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Homework 5
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Water Cement Ratio = 0.46
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Coefficient of Permeability = 5
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Weight of cement = weight of water / w/c Weight of cement = 295/.48 Weight of cement = 615 lbs Volume of course aggregate = .68 X 27 = 18.36 cf/cy OD weight is 18.36 X 100 = 1836 lbs Weight SSD = 1836 (1+ .5) = 2754 lbs Estimation of fine aggregate content Water = 295 / 62.4 = 4.73 cf Cement = 615 / (3.15 x 62.4) = 3.13 cf Coarse aggregate = 1836 / (2.68 x 62.4) = 10.98 cf
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Air = 3% x 27 = .81 cf
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Unformatted text preview: Total = 19.65 cf Fine aggregate volume = 27 – 19.65 = 7.35 cf SSD weight of fine agg. = 7.35 x 2.59 x 62.4 = 1188 lbs Adjustment for moisture: Coarse aggregate = 2754 x 1.00 = 2754 lb/cy Fine aggregate = 1188 x 1.03 = 1224 lb/cy Water required = 295 – 1188 x .03 = 259.36 lb/cy Batch Design Water ----------------260 lb/cy Cement--------------615 lb/cy Coarse Aggregate—2754 lb/cy Fine Aggregate-----1224 lb/cy...
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This note was uploaded on 06/22/2009 for the course CM 3400 taught by Professor Hassan during the Spring '09 term at LSU.

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Homework 5 answers - Total = 19.65 cf Fine aggregate volume...

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