1. Cement: TYPE I, Specific Gravity=3.15 Coarse Aggregate: (BSG)SSD= 2.70
Example 1:
The concrete is required for an exterior column to be located above ground level in an area where substantial freezing-thawing 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

2. 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 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 This does not exceed the limits based on durability in TABLE ACI-3b.

3. 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 dry-rodded 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

4. 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/ = m3
Cement: 423/(1000x3.15) = m3
Coarse Aggregate (SSD): 1018/(1000x2.70) = m3
Air: = 0.06 m3
TOTAL: = m3
Therefore, the fine aggregate must occupy a volume of
= m3
The SSD weight of fine agg. = x 2.65 x 1000 = 700 kg/m3

5. 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( ) = 1033 kg/m3
Fine aggregate (wet) = 700( ) = 729 kg/m3
Surface moisture contributed by the coarse aggregate and fine aggregate are ( ) = 1.5% and ( ) = 4.2%, respectively.
Therefore, to maintain the water-cement ratio constant, the water to be added to the mix becomes
(0.015) - 700(0.042) = 120 kg/m3 or
165 - ( ) - ( ) = 120 kg/m3

6. 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.

7. 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) = kg
F.A.(wet)= 14.58kg

8. 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) = kg
F.A.(wet) = kg
TOTAL = kg
The slump and unit weight measured were 15 mm and kg/m3, respectively.
Increased water necessitates certain adjustments:
Yield = 46.40/2295 = m3
Total mixing water = = kg (0.305 kg coming from CA and kg from FA)
So, water in 1 m3 becomes
3.593/ =178 kg/m3.

9. The slump measured was 15 mm
The slump measured was 15 mm. However, the required slump was 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/ = 1023 kg
Coarse Aggregate (SSD) = 1023/ = 1008 kg
Fine aggregate (wet) = 14.58/ = 722 kg
Fine aggregate (SSD) = 722/ = 693 kg

10. Therefore, the new weights for 1 m3 of the mix become;
Water =136 kg
Cement = 462 kg
Coarse Aggregate (wet) = 1023 kg
Fine Aggregate (wet) = 722 kg

11. 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: mm.
Dmax: 25 mm.
Specific gravity of cement: 3.15
Grading of both aggregates is satisfactory.
The properties of aggregates:
CA FA
Bulk specific gravity (SSD)
Absorption (%)
Total moisture (%)
Dry-rodded unit weight, kg/m3 -
Fineness modulus

12. 1. Slump is given 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 m (Table ACI-4)
Dry weight of CA = 0.69x1600 = 1104 kg/m3
SSD weight of CA = 1104x ( ) = 1110 kg/m3

13. 7. FA Content
Vw = 195/ (1x1000) = m3
Vc = 320/ (3.15x1000) = m3
VCA = 1110/ (2.68x1000) = m3
Vair = m = m3
VFA = = m3
Weight of FA, SSD = (0.274) (2.62) (1000)= 718 kg/m3

14. 8. Moisture correction
Water kg/m3
Cement kg/m3
Coarse aggregate (SSD) 1110 kg/m3
Fine aggregate (SSD) 718 kg/m3
= 2343 kg/m3
FA (wet) = 718 (1+0.05)/ (1+0.01) = 746 kg
=28 kg
CA (wet) = 1110 (1.02)/ (1.005) = 1127 kg
=17 kg
W= =150 kg
Water kg/m3
Coarse aggregate (wet) 1127 kg/m3
Fine aggregate (wet) 746 kg/m3

15. 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)
Absorption (%)
Total moisture (%)
Dry-rodded unit weight, kg/m3 -
Fineness modulus

16. 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

17. VW = 205/ (1x1000) = m3
VC = 456/ (3.0x1000) = m3
VCA = 1097/ (2.7x1000) = m3
Vair = m3
= m3
VFA = = m3
Weight of FA, SSD = (0.222) (2.55) (1000) = 566 kg/m3

18. Water kg/m3
Cement kg/m3
Coarse aggregate (SSD) 1097 kg/m3
Fine aggregate (SSD) 566 kg/m3
= 2324 kg/m3
FA (wet) = 566 (1.04)/ (1.025) = kg
=8.3 kg
CA (air-dry) = 1097 (1.02)/ (1.03) = kg
=10.7 kg
W= =207.4 kg
Water kg/m3
Coarse aggregate (air-dry) kg/m3
Fine aggregate (wet) kg/m3

19. 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 x0.03=6.22 kg
Cement 456x0.03= kg
Coarse aggregate (field) kg
Fine aggregate (field) kg
= kg

20. To increase slump (from 10. 5 to 16 cm) 5. 5 cm roughly 5
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= =6.55 kg
FA (SSD) = (1.025)/(1.04)= kg
=0.25 kg
CA (SSD) = (1.03)/(1.02)= kg
=0.32 kg
W= =6.48 kg
W/C=0.45  C=14.4 kg
=0.72 kg cement should be added.

21. Assume that the slump and unit weight of the 2
Assume that the slump and unit weight of the 2. trial batch were measured as 16 cm and 2305 kg/m3, respectively. Thus,
=70.77 kg
Yield=70.77/2305= 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

22. 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)
Absorption (%)
Total moisture (%)
Dry-rodded unit weight, kg/m3 -
Fineness modulus

23. 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) = kg/m3

24. VW = m3
VC = 333.3/(3.15x1000) = m3
VCA = /(2.7x1000) = m3
Vair = m3
= m3
VFA = = m3
Weight of FA, SSD = (0.274) (2.55) (1000)= kg/m3

25. Water kg/m3
Cement kg/m3
Coarse aggregate (SSD) kg/m3
Fine aggregate (SSD) kg/m3
= kg/m3
Moisture correction
FA (wet) = (1.052)/ (1.017)= kg
=24.0 kg
CA (air-dry) = (1.005)/(1.01)= kg
=5.5 kg
W= =131.5 kg
Water kg/m3
Coarse aggregate (air-dry) kg/m3
Fine aggregate (wet) kg/m3

26. 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.

27. Water=350x0.5=175 kg
VW = m3
VC = 350/(3.15x1000) = m3
Vair = m3
= m3
Vagg=VFA+VCA=0.704 m3 Eq. I
VFA(2.4)/[VFA(2.4)+VCA(2.6)]=0.45
VCA=1.128VFA Eq. II

28. 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 kg/m3
Cement kg/m3
Coarse aggregate (SSD) 962 kg/m3
Fine aggregate (SSD) 792 kg/m3
= 2279 kg/m3

29. 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 CA
If the temperature of mixing water is 60C and the temperature of solid ingredients is 3C, determine the temperature of fresh concrete.
where W: weight, T: temperature, a: aggregate, c: cement, m: mixing water, f: free water (surface moisture)

30. FA is wet and contains some free water (surface moisture)
FA (wet) = 792 (1.05)/1.02=815 kg
Surface moisture= =23 kg
CA is air-dry.
CA (air-dry) =962(1.01)/1.02=953 kg
=9 kg water is needed
W= =161 kg mixing water
Total aggregate= =1745 kg