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Solution to Mix Design Problem: Determining Concrete Mix Proportions, Exercises of Civil Engineering

Aligarh Muslim UniversityCivil Engineering

A step-by-step solution to a mix design problem, determining the required compressive strength, water-cement ratio, mixing water requirement, air content, cement content, coarse and fine aggregate content, and adjustments for aggregate moisture. The solution includes calculations based on tables and formulas.

Typology: Exercises

2012/2013

Uploaded on 05/07/2013

angarika
angarika 🇮🇳

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SOLUTION TO MIX DESIGN PROBLEM
DESIGN PROCEDURES:
(1) Determine Required Average Compressive Strength
Since the standard deviation of the compressive strength is not known,
f’cr = f’c + 1200 (Table 9-11)
= 4800 + 1200 = 6000 psi (10 pt.)
(2) Determine Water Cement Ratio
w/c for strength (Table 9-3) : .41
w/c for exposure (Tables 9-1, 9-2): N.A.
Use w/c = 0.41 (10 pt.)
(3) Determine Mixing Water Requirement & Air Content
Mixing water (Table 9-5): 340 lb / yd3 (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: 340/0.41 = 829 lb/yd3 (10 pt.)
(5) Determine Coarse Aggregate Content
Dry-rodded volume (Table 9-4) 0.66 yd3/yd3
(F.M. of sand = 2.4, Nom. Max. Size = 3/4 inch)
Dry wt. of C.A.per yd3 of concrete:
0.66 X 27 X 88 = 1568 lb. (10 pt.)
(6) Determine Fine Aggregate Content
Absolute Volume of ingredients (per yd3)
Water 340/62.4 = 5.449 ft3
Cement 829/(3.15X62.4) = 4.218 ft3
C.A. 1568/(2.35X62.4) = 10.693 ft3
Air 0.02 X 27 = 0.540 ft3
Total = 20.900 ft3 (10 pt.)
Absolute Volume of Sand = 27 – 20.900 = 6.100 ft3
Dry Wt. of sand = 6.100 X 2.61 X 62.4 = 993 lb. (10 pt.)
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SOLUTION TO MIX DESIGN PROBLEM

DESIGN PROCEDURES:

(1) Determine Required Average Compressive Strength

Since the standard deviation of the compressive strength is not known, f’cr = f’c + 1200 (Table 9-11) = 4800 + 1200 = 6000 psi (10 pt.)

(2) Determine Water Cement Ratio w/c for strength (Table 9-3) :.

w/c for exposure (Tables 9-1, 9-2): N.A. Use w/c = 0.41 (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: 340/0.41 = 829 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 88 = 1568 lb. (10 pt.)

(6) Determine Fine Aggregate Content Absolute Volume of ingredients (per yd^3 ) Water 340/62.4 = 5.449 ft^3 Cement 829/(3.15X62.4) = 4.218 ft^3 C.A. 1568/(2.35X62.4) = 10.693 ft^3 Air 0.02 X 27 = 0.540 ft^3 Total = 20.900 ft 3 (10 pt.) Absolute Volume of Sand = 27 – 20.900 = 6.100 ft^3 Dry Wt. of sand = 6.100 X 2.61 X 62.4 = 993 lb. (10 pt.)

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(7) Adjustments for Agg. Moisture

Wt. of agg. In natural moisture condition: Coarse Agg.: 1568 X 1.025 = 1607 lb Fine Agg.: 993 X 1.002 = 995 lb (10 pt.)

Additional water needed by agg.:

Coarse Agg.: 1568 X (0.049 - 0.025) = 37.6 lb Fine Agg.: 9935 X (0.004 - 0.002) = 2.0 lb 40 lb Adjusted mixing water: 340 + 40 = 380 lb (10 pt.)

Summary of Mix Ingredients for 1 yd^3 of concrete Water 380 lb Cement 829 lb Coarse Agg. 1607 lb Fine Agg. 995 lb 3811 lb (5 pt.)

Calculated Unit Wt.: 3811/27 = 141.1 pcf (5 pt.)

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