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t3501-f08-002-mix-design-solution

# t3501-f08-002-mix-design-solution - 3(4 pt Air 0.02 X 27 =...

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1 Solutions to Problem of Test No. 2, Fall 2008 ( 1) Determine Mixing Water Requirement & Air Content Mixing water (Table 9-5) : 315 lb / yd 3 (1-2 in. slump, non-air-entra., nom. max. agg. size: 3/4 in.) (5 pt.) Air content (Table 9-5) : 2.0% (4 pt.) (2) Calculate Required Cement Content (5 pt.) Cement based on w/c : 3150/0.42 = 750 lb/yd 3 (3) Determine Coarse Aggregate Content (5 pt.) Dry-rodded volume (Table 9-4) 0.66 yd 3 /yd 3 (F.M. of sand = 2.4, Nom. Max. Size = 3/4 inch) Dry wt. of C.A.per yd 3 of concrete: 0.66 X 27 X 98 = 1746 lb. (4) Determine Fine Aggregate Content Absolute Volume of ingredients (per yd 3 ) Water 315/62.4 = 5.048 ft 3 Cement 750/(3.15X62.4) = 3.816 ft 3 C.A. 1746/(2.37X62.4) = 11.806 ft
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Unformatted text preview: 3 (4 pt.) Air 0.02 X 27 = 0.540 ft 3 Total = 21.210 ft 3 Absolute Volume of Sand = 27 – 21.21 = 5.79 ft 3 Dry Wt. of sand = 5.79 X 2.60 X 62.4 = 939 lb. (5 pt.) (5) Adjustments for Agg. Moisture Wt. of agg. In natural moisture condition: (4 pt.) Coarse Agg.: 1746 X 1.035 = 1807 lb Fine Agg.: 939 X 1.012 = 950 lb Additional water needed by agg.: Coarse Agg.: 1746 X (0.04 - 0.035) = 8.7 lb Fine Agg.: 939 X (0.003 - 0.012) = -8.5 lb 0.2 lb Adjusted mixing water: 315 +0.2 = 315 lb (4 pt.) Summary of Mix Ingredients for 1 yd 3 of concrete Water 315 lb Cement 750 lb Coarse Agg. 1807 lb (2 pt.) Fine Agg. 950 lb 3822 lb Calculated Unit Wt.: 3822/27 = 141.6 pcf (2 pt.)...
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