Design 1m3 concrete

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/m
3
)
 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/m
3
γA= 1620 kg/m
3

 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 1m
3
of concrete

Step 1:
Select slump: 25-50 mm (with HRWR)
Required compressive strength:
/
=
/
t
:
0.9
=70 UIF
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 m
3
concrete is: =E
Ln
L= 0.68× 1620 = 1101.6 ( )

2. Step 4: Estimate the mixing water and air content

Water content is adjusted (l/m
3
)
- Void content of sand (fineness aggregate) (V -%)
n=)1−
E
2
1
2×10
0
4×100%=)1−
1650
2.59×10
0
4×100%=36.3%
- Adjusted water content (A -litter)
A=(n−35)×4.7=(36.3−35)×4.7=6.11 ị
Based on slump of 25 to 50 mm and max. size of 12.7 mm, require water for 1 m
3
concrete is 174 l
(Table 6.18)
Total water for 1 m
3
of concrete is H=174 + 6.11 = 180.11 ị
Step 5:
W/(C+P) to determine (C+P) content
From table 6.20, the
ả
d8x
ratio for concrete with High range water reducer is 0.316



H
M+I
=0.316→M+I=
H
0.316
=
180.11
0.316
=570
Step 6: Determine fine aggregate (sand):
- Volume of cement :n
d=
(d8 x)
f=×
=
Bk9

=0.181 ρ
0

- Volume of coarse aggregate: n
=
( )
f ×
=
.w

=0.399 ρ
0

- Volume of Water : n
ả=
(ả)
fγ
=
.69.

=0.18011 ρ
0

- Volume of air content : n
= 2%=0.02 ρ
0
(Table 6.18)
Then Volume of sand:
n
2= 1ρ
0
–(n
d+ n
+ n
ả+ n
)=1−(0.181+0.399+0.18011+0.02)=0.21989
 Dry weight of sand: z = n
2× 1
2×1000651.95=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
z= 569.52×1.064 = 606

3. Total water need for 1 m
3
of concrete is:
H= 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.