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Standard Deviation - Civil Engineering - Assignment, Exercises of Civil Engineering

Aligarh Muslim UniversityCivil Engineering

The assignment of civil engineering contain these points:Standard Deviation, Compressive Strength, Required Average, Water Cement Ratio, Interpolation, Mixing Water Requirement, Air Content, Cement Content, Coarse Aggregate, Fine Aggregate Content

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 known,
fcr = f’c + 1.34 S = 4200 + 1.34 (350) = 4669 psi
fcr = f’c + 2.33 S - 500 = 4200 + 2.33 (350) -500 = 4516 psi
Use fcr = 4700 psi (10 pt.)
(2) Determine Water Cement Ratio
w/c for strength (Table 9-3) : .43 (by interpolation)
(w/c = .57 at 4000 psi; w/c = .48 at 5000 psi)
w/c for exposure (Tables 9-1, 9-2): N.A.
Use w/c = 0.51 (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: 340/0.51 = 667 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 89 = 1586 lb. (10 pt.)
(6) Determine Fine Aggregate Content
Absolute Volume of ingredients (per yd3)
Water 340/62.4 = 5.449 ft3
Cement 667/(3.15X62.4) = 3.393 ft3
C.A. 1586/(2.36X62.4) = 10.770 ft3
Air 0.02 X 27 = 0.540 ft3
Total = 20.152 ft3 (10 pt.)
Absolute Volume of Sand = 27 – 20.152 = 6.848 ft3
Dry Wt. of sand = 6.848 X 2.61 X 62.4 = 1115 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 known, fcr = f’c + 1.34 S = 4200 + 1.34 (350) = 4669 psi fcr = f’c + 2.33 S - 500 = 4200 + 2.33 (350) -500 = 4516 psi Use fcr = 4700 psi (10 pt.)

(2) Determine Water Cement Ratio w/c for strength (Table 9-3) : .43 (by interpolation) (w/c = .57 at 4000 psi; w/c = .48 at 5000 psi) w/c for exposure (Tables 9-1, 9-2): N.A. Use w/c = 0.51 (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: 340/0.51 = 667 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. (10 pt.)

(6) Determine Fine Aggregate Content Absolute Volume of ingredients (per yd^3 ) Water 340/62.4 = 5.449 ft^3 Cement 667/(3.15X62.4) = 3.393 ft^3 C.A. 1586/(2.36X62.4) = 10.770 ft^3 Air 0.02 X 27 = 0.540 ft^3 Total = 20.152 ft^3 (10 pt.) Absolute Volume of Sand = 27 – 20.152 = 6.848 ft^3 Dry Wt. of sand = 6.848 X 2.61 X 62.4 = 1115 lb. (10 pt.)

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(7) Adjustments for Agg. Moisture

Wt. of agg. In natural moisture condition: Coarse Agg.: 1586 X 1.022 = 1621 lb Fine Agg.: 1115 X 1.002 = 1117 lb (10 pt.)

Additional water needed by agg.:

Coarse Agg.: 1586 X (0.041 - 0.022) = 30.1 lb Fine Agg.: 1115 X (0.003 - 0.002) = 1.1 lb 31 lb Adjusted mixing water: 340 + 31 = 371 lb (10 pt.)

Summary of Mix Ingredients for 1 yd^3 of concrete Water 371 lb Cement 667 lb Coarse Agg. 1621 lb Fine Agg. 1117 lb 3776 lb (5 pt.)

Calculated Unit Wt.: 3776/27 = 139.9 pcf (5 pt.)

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