3501as1-f10-solution

3501as1-f10-solution - (10 pt.) (6) Determine Fine...

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SOLUTION TO MIX DESIGN PROBLEM Fall 2010 DESIGN PROCEDURES: (1) Determine Required Average Compressive Strength Since the standard deviation of the compressive strength is not known, f = cr = f = cr + 1200 (Table 9-11) = 4700 + 1200 = 5900 psi (10 pt.) (2) Determine Water Cement Ratio w/c for strength (Table 9-3) : .417 (by interpolation) (w/c = .41 at 6000 psi; w/c = .48 at 5000 psi) w/c for exposure (Tables 9-1, 9-2): N.A. Use w/c = 0.42 (10 pt.) (3) Determine Mixing Water Requirement & Air Content Mixing water (Table 9-5): 340 lb / yd 3 (3-4 in. slump, non-air-entra., nom. max. agg. size: 3/4 in.) (5 pt.) Air content (Table 9-5): 2.0% (5 pt.) (4) Calculate Required Cement Content Cement based on w/c: 340/0.42 = 810 lb/yd 3 (10 pt.) (5) Determine Coarse Aggregate Content 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 89 = 1586 lb.
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Unformatted text preview: (10 pt.) (6) Determine Fine Aggregate Content Absolute Volume of ingredients (per yd 3 ) Water 340/62.4 = 5.449 ft 3 Cement 810/(3.15X62.4) = 4.121 ft 3 C.A. 1586/(2.34X62.4) = 10.862 ft 3 Air 0.02 X 27 = 0.540 ft 3 Total = 20.972 ft 3 (10 pt.) Absolute Volume of Sand = 27 20.972 = 6.028 ft 3 Dry Wt. of sand = 6.028 X 2.60 X 62.4 = 978 lb. (10 pt.) (7) Adjustments for Agg. Moisture Wt. of agg. In natural moisture condition: Coarse Agg.: 1586 X 1.02 = 1618 lb Fine Agg.: 978 X 1.002 = 980 lb (10 pt.) Additional water needed by agg.: Coarse Agg.: 1586 X (0.045 - 0.02) = 40 lb Fine Agg.: 978 X (0.004 - 0.002) = 2 lb 42 lb Adjusted mixing water: 340 + 42 = 382 lb (10 pt.) Summary of Mix Ingredients for 1 yd 3 of concrete Water 382 lb Cement 810 lb Coarse Agg. 1618 lb Fine Agg. 980 lb 3790 lb (5 pt.) Calculated Unit Wt.: 3790/27 = 140.4 pcf (5 pt.) ________________________________________...
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3501as1-f10-solution - (10 pt.) (6) Determine Fine...

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