Example 1:
The concrete is required for an exterior column to be located
above ground level in an area where substantial freezingthawing may occur. It is required to have an average 28-day
compressive strength of 35MPa. For the conditions of
placement, the slump should be between 25 and 50 mm, and
the maximum aggregate size should not be larger than 20mm.
The material properties are as follows:
Cement: TYPE I, Specific Gravity=3.15
Coarse Aggregate: (BSG)SSD= 2.70
Absorption Capacity=1.0%
Total Moisture=2.5%
Dry-rodded Unit Weight=1600 kg/m3
Fine Aggregate: (BSG)SSD=2.65
Absorption Capacity=1.3%
Total Moisture=5.5%
Fineness Modulus=2.70
The sieve analyses of both coarse and fine aggregates fall
within the specified limits. The mix design will be carried
out using the sequence of steps outlined:
Step 1. Choice of Slump: The slump is given as 25-50 mm.
Step 2. Maximum Aggregate Size: Dmax is given as 20 mm.
Step 3. Estimation of Mixing Water and Air Content: Since
the concrete will be exposed to freezing and thawing, it
must be air entrained. From TABLE ACI-2, the air content
recommended for extreme exposure is 6%; the water
requirement is 165 kg/m3.
Step 4. Water-Cement Ratio: From TABLE ACI-3a, the
estimate for the required W/C ratio to give a 28-day
compressive strength of 35 MPa is 0.39. This does not
exceed the limits based on durability in TABLE ACI-3b.
Step 5. Calculation of Cement Content: The required
cement content based on the results of Step 3 and 4, is
165/0.39=423kg/m3.
Step 6. Estimation of Coarse Aggregate Content:
Interpolating in TABLE ACI-4 for the fineness
modulus of fine aggregate of 2.70, the volume of dryrodded coarse aggregate per unit volume of concrete is
0.63.
Therefore, the oven-dry weight of the coarse aggregate is
0.63xl600=1008kg/m3
The SSD weight is 1008xl.01=1018kg/m3
Step 7. Estimation of Fine Aggregate Content: Knowing the
weights and specific gravities of the water, cement, and coarse
aggregate, and knowing the air volume, the volumes per m3
occupied by different ingredients can be determined as follows:
Water: 165/1000
Cement: 423/(1000x3.15)
Coarse Aggregate (SSD): 1018/(1000x2.70)
Air:
TOTAL:
= 0.165 m3
= 0.134 m3
= 0.377 m3
= 0.06 m3
= 0.736 m3
Therefore, the fine aggregate must occupy a volume of
1 - 0.736 = 0.264 m3
The SSD weight of fine agg. = 0.264 x 2.65 x 1000 = 700 kg/m3
Step 8. Adjustment for Moisture in the Aggregate: Since the
aggregates will neither be SSD or OD in the field, it is
necessary to adjust the aggregate weights for the amount of
water combined in the aggregate. Note that dry aggregates will
absorb water and wet aggregates will give extra water. For the
given moisture contents, the adjusted aggregate weights
become
Coarse aggregate (wet) = 1018(1.025 - 0.01) = 1033 kg/m3
Fine aggregate (wet) = 700(1.055-0.013)
= 729 kg/m3
Surface moisture contributed by the coarse aggregate and fine
aggregate are (2.5 - 1.0) = 1.5% and (5.5 -1.3) = 4.2%,
respectively.
Therefore, to maintain the water-cement ratio constant, the water
to be added to the mix becomes
165 - 1018(0.015) - 700(0.042) = 120 kg/m3
or
165 - (1033 - 1018) - (729 - 700) = 120 kg/m3
Thus, the estimated batch weights per m3 of concrete are:
WATER (to be added)
= 120 kg
CEMENT
= 423 kg
COARSE AGGREGATE (wet)
= 1033 kg
FINE AGGREGATE (wet)
= 729 kg
TOTAL
= 2305 kg/m3.
Step 9. Trial Batch: A trial mix (of much less volume) has
to be prepared using the proportions calculated.
If any of the desired concrete properties are not achieved in
this trial batch, these properties must be adjusted. If very
large adjustments appear to be indicated, it is probably
best to redesign the mix.
Let us assume that, a laboratory trial batch of 0.02 m3
volume was decided to be produced for this case.
So the weights for the trial batch are
Water = 2.4 kg
Cement = 8.46 kg
C.A.(wet) = 20.66 kg
F.A.(wet)= 14.58kg
However, it was observed that, it becomes necessary to
increase the water to 2.7 kg in order to obtain a workable
mix. Thus, the new weights of the ingredients in the trial
batch with sufficient workability become
Water = 2.7 kg
Cement = 8.46 kg
C.A.(wet) = 20.66 kg
F.A.(wet) = 14.58 kg
TOTAL = 46.40 kg
The slump and unit weight measured were 15 mm and
2295 kg/m3, respectively.
Increased water necessitates certain adjustments:
Yield = 46.40/2295 = 0.0202 m3
Total mixing water = 2.70 + 0.305 + 0.588 = 3.593 kg (0.305
kg coming from CA and 0.588 kg from FA)
So, water in 1 m3 becomes
3.593/0.0202 =178 kg/m3.
The slump measured was 15 mm. However, the required
slump was 25 - 50 mm. In order to reach the required
value, 2 kg per m3 more water should be added to the mix.
Thus, total water content of the mix becomes 180 kg/m3.
Since the water content of the mix is increased, the cement
content should also be adjusted to maintain a constant W/C
ratio. Otherwise the strength will be lower than the
required. Thus,
Cement = 180/0.39 = 462 kg/m3.
Since the workability of the concrete was sufficient, there is
no need for other adjustments and the weights of the
aggregates are as follows:
Coarse Aggregate (wet) = 20.66/0.0202 = 1023 kg
Coarse Aggregate (SSD) = 1023/1.015 = 1008 kg
Fine aggregate (wet)
= 14.58/0.0202 = 722 kg
Fine aggregate (SSD)
= 722/1.042
= 693 kg
Therefore, the new weights for 1 m3 of the mix
become;
Water
Cement
Coarse Aggregate (wet)
Fine Aggregate (wet)
=136 kg
= 462 kg
= 1023 kg
= 722 kg
EXAMPLE 2:
A concrete mix design is required. The concrete will not be
subjected to freezing and thawing or to sulfate attack.
The average 28-day compressive strength: 25 MPa.
The required slump: 75-100 mm.
Dmax: 25 mm.
Specific gravity of cement: 3.15
Grading of both aggregates is satisfactory.
The properties of aggregates:
Bulk specific gravity (SSD)
Absorption (%)
Total moisture (%)
Dry-rodded unit weight,
Fineness modulus
CA
2.68
0.5
2.0
1600 kg/m3
-
FA
2.62
1.0
5.0
2.6
1. Slump is given 75-100 mm
2. Dmax: 25 mm given
3. Water and air contents
W: 195 kg/m3 (Table ACI-2)
Air Content: 1.5% (Table ACI-2)
4. W/C ratio
w/c = 0.61 (Table ACI-3a)
5. Cement Content
C= 195/0.61 = 320 kg/m3
6. CA Content
Dry rodded volume of CA = 0.69 m3
(Table ACI-4)
Dry weight of CA = 0.69x1600 = 1104 kg/m3
SSD weight of CA = 1104x (1+0.005) = 1110 kg/m3
7. FA Content
Vw = 195/ (1x1000) =
Vc = 320/ (3.15x1000) =
VCA = 1110/ (2.68x1000) =
Vair =
0.195 m3
0.102 m3
0.414 m3
0.015 m3
= 0.726 m3
VFA = 1-0.726 = 0.274 m3
Weight of FA, SSD = (0.274) (2.62) (1000)=
718 kg/m3
8. Moisture correction
Water
Cement
Coarse aggregate (SSD)
Fine aggregate (SSD)
195 kg/m3
320 kg/m3
1110 kg/m3
718 kg/m3
= 2343 kg/m3
FA (wet) = 718 (1+0.05)/ (1+0.01) = 746 kg
746-718=28 kg
CA (wet) = 1110 (1.02)/ (1.005) = 1127 kg
1127-1110=17 kg
W=195-28-17=150 kg
Water
150 kg/m3
Cement
320 kg/m3
Coarse aggregate (wet)
1127 kg/m3
Fine aggregate (wet)
746 kg/m3
= 2343 kg/m3
EXAMPLE 3:
Design a concrete mix for a reinforced wall which will be
exposed to sulfate attack with no risk of frost action. The
required 28-day compressive strength is 23 MPa; from the
experience standard deviation is expected to be 4.69 MPa.
The required slump is 160 mm. Compute the field weights
of the ingredients for 1 m3 of the concrete mixture,
assuming that:
Dmax: 25 mm.
Specific gravity of cement: 3.0.
CA
FA
Bulk specific gravity (SSD)
2.7
2.55
Absorption (%)
3.0
2.5
Total moisture (%)
2.0
4.0
Dry-rodded unit weight,
1500 kg/m3 Fineness modulus
2.4
No frost action; non-air entrained concrete
Water= 205 kg, air= 1.5%
fcm=fck+1.48.S=30 MPa
From Table ACI-3a W/C=0.54
From Table ACI-3b W/C=0.45
Select lowest value of W/C ratio!
Cement=205/0.45=456 kg
Dry rodded VCA=0.71 m3
CA (dry) = 0.71(1500) =1065 kg
CA (SSD) =1065(1.03) =1097 kg
VW = 205/ (1x1000) =
VC = 456/ (3.0x1000) =
VCA = 1097/ (2.7x1000) =
Vair =
0.205 m3
0.152 m3
0.406 m3
0.015 m3
= 0.778 m3
VFA = 1-0.778 = 0.222 m3
Weight of FA, SSD = (0.222) (2.55) (1000) =
566 kg/m3
205 kg/m3
456 kg/m3
1097 kg/m3
566 kg/m3
= 2324 kg/m3
FA (wet) = 566 (1.04)/ (1.025) = 574.3 kg
574.3-566=8.3 kg
CA (air-dry) = 1097 (1.02)/ (1.03) = 1086.3 kg
1097-1086.3=10.7 kg
W=205-8.3+10.7=207.4 kg
Water
207.4 kg/m3
Cement
456 kg/m3
Coarse aggregate (air-dry) 1086.3 kg/m3
Fine aggregate (wet)
574.3 kg/m3
= 2324 kg/m3
Water
Cement
Coarse aggregate (SSD)
Fine aggregate (SSD)
Assume that a 0.03 m3 trial batch of the concrete is
prepared to show a slump of 10.5 cm. Make
necessary adjustments in the weight of ingredients
so that W/C ratio and FA/CA ratio of the mix
remain constant.
Assume that for increasing slump by 1 cm, about 2
kg/m3 water is required.
1. Trial batch weights:
Water
207.4x0.03=6.22 kg
Cement
456x0.03= 13.68 kg
Coarse aggregate (field)
32.59 kg
Fine aggregate (field)
17.23 kg
= 69.72 kg
To increase slump (from 10.5 to 16 cm) 5.5 cm roughly
5.5x2=11 kg/m3 water should be added. For the 0.03 m3
trial batch:
11x0.03=0.33 kg
2. Trial batch weights:
W=6.22+0.33=6.55 kg
FA (SSD) = 17.23 (1.025)/(1.04)= 16.98 kg
17.23-16.98=0.25 kg
CA (SSD) = 32.59 (1.03)/(1.02)= 32.91 kg
32.91-32.59=0.32 kg
W=6.55-0.32+0.25=6.48 kg
W/C=0.45  C=14.4 kg
14.4-13.68=0.72 kg cement should be added.
Assume that the slump and unit weight of the
2. trial batch were measured as 16 cm and
2305 kg/m3, respectively. Thus,
14.4+6.55+32.59+17.23=70.77 kg
Yield=70.77/2305=0.0307 m3
14.4/0.0307=469 kg
6.55/0.0307=213 kg
32.59/0.0307=1062 kg
17.23/0.0307=561 kg
= 2305 kg/m3
EXAMPLE 4:
Design a concrete mixture to have 3-5 cm slump and a
characteristic 28-day compressive strength of fck=20 MPa.
Concrete will be exposed to severe freezing-thawing and to
seawater. Material properties are as follows:
Dmax: 37.5mm
Type V cement, specific gravity: 3.15.
CA
FA
Bulk specific gravity (SSD)
2.7
2.55
Absorption (%)
1.0
1.7
Total moisture (%)
0.5
5.2
Dry-rodded unit weight,
1500 kg/m3 Fineness modulus
2.5
Standard deviation is not known;
fcm=fck+f=20+6=26 MPa
Concrete should be air entrained.
28-D=26 MPaW/C=0.5
From freezing-thawing point of view W/C=0.5
From seawater action point of view W/C=0.45
Water=150 kg/m3
Air=5.5%
Cement= 150/0.45=333.3 kg
Dry rodded volume of CA = 0.74 m3
Dry weight of CA = 0.74x1500 = 1110 kg/m3
SSD weight of CA = 1110x (1.01) = 1121.1 kg/m3
VW =
0.150 m3
VC = 333.3/(3.15x1000) = 0.106 m3
VCA = 1121.1/(2.7x1000) = 0.415 m3
Vair =
0.055 m3
= 0.726 m3
VFA = 1-0.726 = 0.274 m3
Weight of FA, SSD = (0.274) (2.55) (1000)=
698.7 kg/m3
Water
Cement
Coarse aggregate (SSD)
Fine aggregate (SSD)
150 kg/m3
333.3 kg/m3
1121.1 kg/m3
698.7 kg/m3
= 2303.1 kg/m3
Moisture correction
FA (wet) = 698.7 (1.052)/ (1.017)= 722.7 kg
722.7-698.7=24.0 kg
CA (air-dry) = 1121.1 (1.005)/(1.01)= 1115.6 kg
1121.1-1115.6=5.5 kg
W=150+5.5-24.0=131.5 kg
Water
131.5 kg/m3
Cement
333.3 kg/m3
Coarse aggregate (air-dry) 1115.6 kg/m3
Fine aggregate (wet)
722.7 kg/m3
= 2303.1 kg/m3
EXAMPLE 5:
Job specifications dictate the following values for a
concrete mix:
Cement content: 350 kg/m3
W/C=0.5
FA (SSD)/Total agg. (SSD)=0.45 (by weight)
Air content: 1% (assumed)
Materials properties are as follows:
Specific gravity of cement: 3.15
SSD bulk specific gravity of FA and CA: 2.4 and 2.6,
respectively.
Water=350x0.5=175 kg
VW =
VC = 350/(3.15x1000) =
Vair =
Vagg=VFA+VCA=0.704 m3
0.175 m3
0.111 m3
0.01 m3
= 0.296 m3
Eq. I
VFA(2.4)/[VFA(2.4)+VCA(2.6)]=0.45
VCA=1.128VFA
Eq. II
Solving I and II simultaneously;
VFA=0.33 m3
VCA=0.37 m3
FASSD=0.33(2.4)(1000)=792 kg
CASSD=0.37(2.6)(1000)=962 kg
Water
Cement
Coarse aggregate (SSD)
Fine aggregate (SSD)
175 kg/m3
350 kg/m3
962 kg/m3
792 kg/m3
= 2279 kg/m3
EXAMPLE 6:
A concrete mixture consists of
C=350 kg/m3
W=175 kg/m3
FA (SSD) = 792 kg/m3
CA (SSD) = 962 kg/m3
Aggregate properties are as follows:
Abs. Cap. (%)
Total moisture (%)
FA
2
5
CA
2
1
If the temperature of mixing water is 60C and the
temperature of solid ingredients is 3C, determine the
temperature of fresh concrete.
T
0.22(TaWa  TcWc )  TmWm  T f W f
0.22(Wa  Wc )  Wm W f
where W: weight, T: temperature, a: aggregate, c: cement,
mixing water, f: free water (surface moisture)
m:
FA is wet and contains some free water (surface
moisture)
FA (wet) = 792 (1.05)/1.02=815 kg
Surface moisture=815-792=23 kg
CA is air-dry.
CA (air-dry) =962(1.01)/1.02=953 kg
962-953=9 kg water is needed
W=175+9-23=161 kg mixing water
Total aggregate=792+953=1745 kg
0.22(350  1745)3  23x3  161x60
T
 17.2 C
0.22(350  1745)  23  161