SECTION III. ABSOLUTE-VOLUME METHOD

BASIC GUIDELINES

3-15. Concrete mixtures can be proportioned using absolute volumes. This method was detailed in ACI 211.1-81. For this procedure, select the W/C ratio, slump, air content, and the maximum size of the aggregate. Then estimate the water requirement using Table 3-3. Addition information is required before making calculations such as the specific gravities of the FA and CA, the dry-rodded unit weight (DRUW) of the CA, and the fineness modulus of the FA. If the maximum size of the aggregate and the fineness modulus of the FA are known, estimate the volume of dry-rodded CA per cubic yard. From Table 3-6, calculate the quantities per cubic yard of water, cement, CA, and air, and the dry-rodded unit weight of the CA. Finally, subtract the sum of the absolute volumes of these materials in cubic feet from 27 cubic feet per cubic yard to give the specific volume of the FA.

Table 3-6. Volume of CA per cubic yard of concrete

Maximum
Size of
Aggregate,
in Inches
Fineness Modulus of FA
2.40 2.60 2.80 3.00
CA, in Cubic Feet Per Cubic Yard
3/8 13.5 13.0 12.4 11.9
1/2 15.9 15.4 14.8 14.3
3/4 17.8 17.3 16.7 16.2
1 19.2 18.6 18.1 17.6
1 1/2 20.2 19.7 19.2 18.6
2 21.1 20.5 20.0 19.4
3 22.1 21.6 21.1 20.5
NOTE: Volumes are based on aggregates in a dry-rodded condition, as described in Method of Test for Unit Weight of Aggregate (ASTM C29). These volumes are selected from empirical relationships to produce concrete with a degree of workability suitable for usual reinforced construction. For less workable concrete, such as that required for concrete pavement construction, the volume may be increased about 10 percent. When placement is to be by pump, volume should be decreased about 10 percent.

EXAMPLE USING THE ABSOLUTE-VOLUME METHOD

3-16. For a retaining wall, determine the mix proportions using the following specifications and conditions:

• Required 28-day compressive strength of 3,000 psi.
• Maximum size aggregate is 3/4 inch.
• Exposure condition will be moderate freeze-thaw exposure (exposed to air).
• Fineness modulus of the FA is 2.70.
• Specific gravity of portland cement is 3.15.
• Specific gravity of the FA (SGFA) is 2.66.
• Specific gravity of the CA (SGCA) is 2.61.
• Dry-rodded unit weight (DRUW) of the CA is 104 lb/cu ft.
• DRUW of the FA is 103 lb/cu ft.
• Slump is 3 inches.
• Cement type is IA.

Step 1. Estimate the air content (refer to Table 3-3) using the following information:

• Aggregate size = 3/4 inch
• Air-entrained concrete
• Moderate exposure
• Air content = 5 percent

Step 2. Estimate the mixing-water content refer to Table 3-3) using the following information:

• Slump = 3 inches
• Aggregate size = 3/4 inch
• Air-entrained concrete
• Mixing water = 305 lb/cu yd

Step 3. Determine the W/C ratio. Check the W/C ratios for strength, durability, and watertightness using the following information:

• Strength is determined by:
• Mix design strength = 3,000 psi
• Specified compressive strength.
• Design strength.
• W/C ratio (refer to Table 3-2) = 0.46
• Durability (refer to Table 3-1) is determined by:
• Frost-resistant concrete.
• All other structures - W/C = 0.50
• Watertightness (refer to Table 3-1) is determined by:
• Freshwater exposure - W/C = 0.50
• Selection of the lowest W/C ratio - W/C = 0.46

Step 4. Calculate cement content (C) using the formula below:

Step 5. Calculate the CA content (refer to Table 3-6) by referring to the following information:

• CA size = 3/4 inch
• Fineness modulus of FA = 2.7 (given)
• Determine volume of CA by interpolation by referring to the following information:

• Weight of CA.
G = Volume of CA x unit weight of CA
G = 17 x 104 lb/cu ft
G = 1,768 lb/cu yd

Where G = the weight of the CA in lbs / cu yd

Step 6. Calculate the FA content by the absolute method using the following formula:

Summary

• Cement = 3.37 cu ft/cu yd
• Water = 4.89 cu ft/cu yd
• CA = 10.86 cu ft/cu yd
• Air = 1.35 cu ft/cu yd
• Partial volume (PV) = 20.47 cu ft/cu yd
• (PV is the sun of water, cement, CA and air)
• Absolute volume of FA = 27 cu ft/cu yd - PV
• 27 cu ft - 20.47 cu ft/cu yd = 6.53 cu ft/cu yd
• Weight of FA = absolute vol of FA x SGFA x 62.4 lbs / cu ft =
• 6.53 cu ft / cu yd x 2.66 x 62.4 lbs / cu ft = 1,083.9 lbs / cu yd

Step 7. Determine the first trial batch. Determine the proportions for the first trial batch (for 1 cubic yard) by converting the absolute volumes to dry volumes.

If needed, mix more trial batches to obtain the desired slump and air content while keeping the W/C ratio constant.

VARIATION IN MIXTURES

3-17. The mixture proportions will vary somewhat depending on the method used. The variation is due to the empirical nature of the methods and does not necessarily imply that one method is better than another. Since the methods begin differently and use different procedures, the final proportions will vary slightly. This is to be expected and further points out the necessity of trial mixtures in determining the final mixture proportions. For variations in a mixture, such as for concrete used in slabs or other flatwork, there are minimum cement requirements, depending on the maximum size of the aggregates (see Table 3-7)

Table 3-7. Concrete Used in flatwork

MaximumSize of
Aggregate,in Inches
Cement, in Pounds
Per Cubic Yard
1 1/2 470
1 520
3/4 540
1/2 590
3/8 610
NOTES:
1. 1 inch = 25 mm
2. 100 lb/cu yd = 60 kg/m3

3-18. The initial mix design assumes that the aggregates are SSD; that is, neither the FAs nor the CAs have any free water on the surface that would be available as mixing water. This is a laboratory condition and seldom occurs in the field. The actual amount of water on the sand and gravel can only be determined from the material at the mixing site. Furthermore, the moisture content of the aggregates will change over a short period of time; therefore, their condition must be monitored and appropriate adjustments made as required. A good field test for estimating the free surface moisture (FSM) on FA is contained in Appendix C. CAs are free-draining and rarely hold more than 2 percent (by weight) FSM even after heavy rains. FA have a tendency to bulk (expand in volume) when wetted and when the mass is disturbed. This factor becomes very important if the concrete is being batched at a mixer by volume (the initial mix design must then be adjusted). The procedure for adjusting the mixing water and sand bulking due to FSM is as follows:

Step 1. Determine the approximate FSM of the FA by the squeeze test (Appendix C).

Step 2. Estimate the FSM of the CA by observation. Usually, 2 percent FSM is the maximum amount that gravel will hold without actually dripping.

Step 3. Multiply the percentages of FSM on the aggregates by their respective weights per cubic yards. This will yield the weight of the FSM on the aggregates.

Step 4. Divide the total weight of the FSM by 8.33 pounds per gallon to determine the number of gallons of water. Subtract those gallons from the water requirements in the original mix design.

• If you are batching your concrete mix by weight, you need to account for the weight contributed by the FSM by increasing the total weights of the aggregates per cubic yard by the weights of the FSM.
• If you are batching your concrete by volume, you must increase the volume of the FA by the bulking factor (BF) determined from Figure 3-3. The formula for volume increase is as follows:

Vwet = Vdry x (1 + BF)

Where:

V = volume
BF = bulking factor (CAs do not bulk, therefore no adjustment is necessary.)

Figure 3-3. Bulking factor curves

SAMPLE PROBLEM FOR MOISTURE ADJUSTMENT ON AGGREGATES

3-19. Using the final mix proportions as determined in paragraph 3-15, adjust the design mix to account for 6 percent FSM on the FA (fineness modulus equals 2.7) and 2 percent FSM on the CA. The original mix design was:

• Cement = 7.05 sacks (Type IA)
• Water = 36.6 gal
• CA = 17.0 cu ft or 1,768 lb/cu yd
• FA = 10.5 cu ft or 1,083 lb/cu yd
• Air content = 5%
Step 1. Determine the amount of water (in gallons) on the aggregate.
• CA = 1,768 lb/cu yd x 0.02 = 35.38 lb/cu yd of water
• FA = 1,083 lb/cu yd x 0.06 = 64.98 lb/cu yd of water
• Total weight of water = 100.36 lb/cu yd
• Convert to gallons:

Step 2. Reduce the original amount of mixing water by the amount contributed by the aggregates as determined in step 1. Therefore, 36.6 gallons minus 12 gallons equals 24.6 gallons of water that must be added to the mix.
Step 3. Adjust the weights of the aggregates by the amount contributed by the water.
CA = 1,769 lb/cu yd + 35.38 lb/cu yd = 1,803.3 lb/cu yd
FA = 1,083 lb/cu yd + 64.98 lb/cu yd = 1,148 lb/cu yd
Step 4. Adjust the volume of the FA to reflect the BF. From Figure 3-3, the fineness modulus equals 2.70. The FA is considered a medium sand. Select the appropriate moisture content across the bottom of the figure, read up to the appropriate sand curve and then read the correct BF on the left edge. For this example, FSM equals 6 percent; read up the figure until it intersects the medium sand line, then read left to the BF of 0.28.

The increase in FA volume is then:

Vwet = Vdry x (1 + BF) = 10.5 cu ft x (1 + .28) = 13.44 cu ft

Step 5. The adjusted mix design to account for the actual field conditions is now:
• Cement = 7.05 sacks (Type IA)
• Water = 24.6 gal
• CA = 17.0 cu ft or 1,803.3 lb/cu yd
• FA = 13.44 cu ft or 1,148 lb/cu yd
• Air content = 5%

It is important to check the moisture content of the aggregates and make appropriate adjustments as conditions change, such as after rains, after periods of dryness, and after the arrival of new materials. This quality-control step assures that the desired concrete is produced throughout the construction phase.

MATERIALS ESTIMATION

3-20. After designing the mix, use the following steps to estimate the total amounts of materials needed for the job:

Step 1. Determine the total volume (in cubic yards) of concrete to be placed.
Step 2. Determine the extra amount that will be added for waste. If the total volume is 200 cubic yards or less, add l0 percent. If the total volume is greater than 200 cubic yards, add 5 percent.
Step 3. Determine the total amounts of cement, FA, and CA, by multiplying the amounts of these components needed for 1 cubic yard by the adjusted total volume. Then convert the answer to tons and round up to the next whole number.
Step 4. Determine the required amount of water needed for the job. Water is required on concrete projects not only for mixing but for wetting the forms, tools, and curing the concrete.

A planning factor of 8 gallons of water for each sack of cement is usually sufficient; however, not all of this water will be used for the concrete.

SAMPLE PROBLEM FOR MATERIALS NEEDED FOR MIX DESIGN

3-21. Using the mix design, determine the total amount of materials needed to construct the retaining wall shown in Figure 3-4. The 1 cubic yard mix design is recapped below.

Figure 3-4. Retaining wall

• Cement = 7.05 sacks (Type IA)
• Water = 36.6 gal
• CA = 17.0 cu ft or 1,768 lb/cu yd
• FA = 10.5 cu ft or 1,083 lb/cu yd
• Air content = 5%
Step 1. Determine the total volume of concrete required. An easy way to do this is to break the project up into simple geometric shapes. Divide the retaining wall into two sections, the wall portion and the footing. A close examination of Figure 3-4 shows the wall cross-section is a trapezoid that is 14 feet, 9 inches high, 8 inches wide on one end, and 1 foot wide on the other.

Step 2. Calculate the waste factor (since the volume is less than 200 cubic yards, the waste factor is 10 percent).
• Total volume + waste = 63.7 cu yd x 1.10 = 70.1 cu yd
Step 3. Determine the amounts of cement and aggregates needed.
• Cement 7.05 sacks/cu yd x 70.1 cu yd = 494.2 sacks = 495 sacks

NOTE: Round values up to the next whole number since you cannot order partial sacks.

• CA 1,768 lb/cu yd x 70.1 cu yd = 123,936.8 lb or 62 tons
• FA 1,083 lb/cu yd x 70.1 cu yd = 75,918.3 lb or 38 tons
Step 4. Determine the amount of water required for cleanup and mixing.

Water required = 495 sacks x 8 gallons per sack = 3,960 gallons

SUMMARY

3-22. Summary of the amounts of materials to be ordered for the project are as follows:

• Cement = 495 sacks
• Water = 3,960 gallons
• CA = 62 tons
• FA = 38 tons