Design 1m3 concrete

1. / Department of Civil Engineering Subject: CONSTRUCTION OF MATERIALS
International University (IU) Date: 21/10/2014
Name: Nguyễn Thị Hồng Hải
Student ID: CECEIU12018
Design mix proportion of concrete
I. Sand properties
 Fineness modulus FM = 2.90.
 Bulk specific gravity (over dry) dry ρS = 2.59.
 Absorption based on dry weight Abs = 1.1%.
 Dry unit weight γS = 1650 (kg/m3
)
 Moisture content in sand = 6.4%. M = 6.4%
II. Aggregate properties
 Maximum size: 12.7 mm
 Bulk specific gravity (over dry) dry ρA = 2.76
 Absorption based on dry weight Abs = 0.70%
 DRUW = 1620 kg/m3
γA= 1620 kg/m3
 Moisture content = 0.5%. M = 0.5%
III. Cement:
 Bulk specific gravity (over dry) dry ρA = 3.15
Assumption:
- The slum from 25 to 50 mm (with HRWR)
- Create 1m3
of concrete
Step 1:
Select slump: 25-50 mm (with HRWR)
Required compressive strength:
=
0.9
= 70
Step 2: Select maximum size of aggregate
From table 6.16,
Maximum size of aggregate = 12.7 mm
Step 3: Select optimum coarse content
From table 6.17, the fractional volume of oven-dry rodded coarse aggregate is 0.68  Dry weight of coarse
aggregate for 1 m3
concrete is: = = 0.68 × 1620 = 1101.6 ( )

2. Step 4: Estimate the mixing water and air content
Water content is adjusted (l/m3
)
- Void content of sand (fineness aggregate) (V -%)
= 1 −
× 10
× 100% = 1 −
1650
2.59 × 10
× 100% = 36.3%
- Adjusted water content (A -litter)
= ( − 35) × 4.7 = (36.3 − 35) × 4.7 = 6.11
Based on slump of 25 to 50 mm and max. size of 12.7 mm, require water for 1 m3
concrete is 174 l
(Table 6.18)
Total water for 1 m3
of concrete is = 174 + 6.11 = 180.11
Step 5:
W/(C+P) to determine (C+P) content
From table 6.20, the ratio for concrete with High range water reducer is 0.316
+
= 0.316 → + =
0.316
=
180.11
0.316
= 570
Step 6: Determine fine aggregate (sand):
- Volume of cement : =
( )
×
= = 0.181
- Volume of coarse aggregate: =
( )
×
=
.
= 0.399
- Volume of Water : =
( )
=
.
= 0.18011
- Volume of air content : = 2% = 0.02 (Table 6.18)
Then Volume of sand:
= 1 – ( + + + ) = 1 − (0.181 + 0.399 + 0.18011 + 0.02) = 0.21989
 Dry weight of sand: = × × 1000651.95 = 0.21989 × 2590 = 569.52
Because moisture content is 0.5% and absorption content of 0.7%, therefore normal weight of
coarse aggregate is = 1101.6 × 1.005 = 1107.1
Moisture content is 6.4% and absorption content of 1.1%, therefore normal weight of sand is
= 569.52 × 1.064 = 606

3. Total water need for 1 m3
of concrete is:
= 180.11 + (0.7 − 0.5) ×
1101.6
100
− (6.4 − 1.1) ×
569.52
100
= 152.13
Dry condition (kg) Wet condition (kg)
Water 180.11 152.13
Cement and admixture 570 570
Coarse Aggregate 1101.6 1107.1
Sand 569.52 606
Step 7: Determine P.
In laboratory, P is determine 10, 15, 20, 25, 30 or 35 % of cement.
Some trial batches are done to determine fine mix proportion of compound for concrete.
Step 8: Controlling slump by adding HRWR
0.23kg to 0.45kg / 45 kg cement weight.
Adding larger 0.45kg/45kg cement weight can cause to segregation
0.23 kg HRWR/ 45kg cement can increase 30 mm in slump.
0.45 kg HRWR/45 kg cement can increase 254 mm in slump.