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CHAPTER 3

ACI CONCRETE MIX DESIGN AND

ALLOWABLE FIELD ADJUSTMENTS

ACI Concrete Mix Design

In 1963 the Virginia Department of Transportation realized that a de! nite need existed to adopt a standard method of concrete design. Contractors, Producers, and the Department itself had used many varying methods of concrete designs.

The Department has adopted the ACI absolute volume method of design, and requires that this method be used in the design of all normal weight concrete mixes.

Example: Absolute Volume = Weight (Sp. Gr.) X 62.4 lb/ft3

For purposes of establishing concrete proportions and calculating yields, we will not concern ourselves with bulk yield or bulk volumes of aggregate, cement, etc., but with the Absolute Volume of these materials. This means the volume of material is solid and without voids.

For example: 94 lbs. of cement in a bulk state occupies approximately 1 cubic foot of volume; however, the Absolute Volume of 94 lbs. of cement is only approximately 0.48 cubic feet. (This means the cement is consolidated without voids.) It is the latter volume which 94 lbs. of cement actually occupies in a batch of concrete.

As a further explanation, we will calculate the absolute volume occupied by cement in a cubic yard of concrete which contains 588 lbs. of cement.

EXAMPLE: The weight of cement is: 588 lbs. Cement has a speci! c gravity of 3.15 (this means cement is 3.15 times heavier than an equal volume of water)

A solid cubic foot of cement then weighs 3.15 x 62.4 lb/ft3 of water = 196.56 lb/ft3

So the Absolute Volume, or the space occupied by 588 lbs. of cement, will be:

588 lb. = 2.99 ft3

196.56 lb/ft3

Before calculations can be started for a concrete design, there are certain items that must be known or available to the person doing the design work. Those items are as follows:

1. Class of concrete to be designed.

2. Fine aggregate: a. Speci! c gravity b. Fineness modulus

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3. Coarse aggregate: a. Maximum size aggregate b. Speci! c Gravity c. Unit Weight (dry rodded unit weight)

4. From VDOT Speci! cations: a. Cement factor (minimum cement content) b. W.C. ratio (maximum W/C ratio) c. Air content (mean air content) d. Nominal Maximum size aggregate% '

Other information ( ) * ACI Table A1.5.3.6 (volume of coarse aggregate per unit of volume of concrete) (Page 3-6)

b.Type of cement and alkali content c. TL-27 (Concrete Mix Design Form)(Page 3-7) d. ACI Mix design work sheet may be used (Page 3-3) e. Source of all materials going into the mix

(For example purposes, the Fineness Modulus and Speci! c Gravity for all design problems in this study guide are taken from the Aggregate Data Sheets on Pages 3-32 and 3-33. For an updated list, please see the Aggregate Quality List, which is published annually in the Materials Division Manual of Instruction.)

HELPFUL CONVERSION FACTORS

One cubic foot of water = 7.5 gallons = 62.4 lbs.One bag of cement = 94 lbs. (42.6 kg)Speci! c gravity of cement is 3.15Speci! c gravity x 62.4 = Absolute Volume1.308 cubic yards = One cubic meter1 gallon of water = 8.33 lbs. = 3.78 liters1 gallon per yard = 5 liters per meterOne cubic yard = 27 cubic feetOne bag of cement = one cubic foot (loose volume)One bag of cement = 0.48 cubic feet (absolute volume)

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TABLE A1.5.3.6 VOLUME OF COARSE AGGREGATE PER UNIT OF VOLUME OF CONCRETE (SI)

Nominal Maximum size of

Aggregate

Volume of dry-rodded coarse aggregate *per unit volume of concrete for different

! neness moduli** of ! ne aggregate

Metric English 2.40 2.50 2.60 2.70 2.80 2.90 3.00

9.5 mm 3/8 0.50 0.49 0.48 0.47 0.46 0.45 0.44

12.5 mm ½ 0.59 0.58 0.57 0.56 0.55 0.54 0.53

19.0 mm ¾ 0.66 0.65 0.64 0.63 0.62 0.61 0.60

25.0 mm 1 0.71 0.70 0.69 0.68 0.67 0.66 0.65

37.5 mm 1 ½ 0.75 0.74 0.73 0.72 0.71 0.70 0.69

50 mm 2 0.78 0.77 0.76 0.75 0.74 0.73 0.72

75 mm 3 0.82 0.81 0.80 0.79 0.78 0.77 0.76

150 mm 6 0.87 0.86 0.85 0.84 0.83 0.82 0.81

* Volumes are based on aggregates in dry-rodded condition as described in ASTM C 29

** See ASTM Method 136 for calculation of ! neness modulus

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Mix designs must incorporate a mineral admixture. The amount of mineral admixture varies according to the alkali content of the cement and the type of mineral admixture selected for the concrete mix. Blended cements such as IP or IS already have mineral admixtures in them and don�t require any additional admixtures. Road and Bridge Speci! cation Section 217.02(a) lists the percent replacement guidelines of cement with mineral admixtures based on the alkali content of the cement.

Section 217.02(a) Cementitious materials shall be a blend of mineral admixtures and Portland cement or a blended cement. In overlay concretes, expansive hydraulic cement is permitted in lieu of Portland cement. Portland cement (Types I, II, III) blended cements (Type IP, Type IS) or expansive cement (Type K) shall comply with Section 214. Flyash, ground granulated iron blast-furnace slag, silica fume or metakaolin shall conform to Section 215. As a portion of the cementitious material, Table 1 lists the minimum percents of speci! c pozzolans required by mass of the cementitious material depending on the alkali content of the cement. Any other mineral admixture or any other amount or combination of mineral admixtures may be used if approved by the Engineer. As a portion of the cementitious material, the " y ash content shall not exceed 30 percent for Class F, the ground granulated blast-furnace slag content shall not exceed 50 percent and the silica fume content shall not exceed 10 percent unless approved by the Engineer. Class C Flyash or other pozzolans may be used provided the contractor demonstrates that the percent usage of Class C Flyash or other pozzolans have a maximum expansion of 0.15% according to ASTM C227 at 56 days using borosilicate glass as aggregate. Blended cements require no further pozzolan additions to meet minimum pozzolan content to compensate for the alkali-silica reaction.

Up to 7 percent silica fume may be added to all combinations of cementitious materials to reduce early permeability without approval by the Engineer. Other silica fume additions must be approved by the Engineer.

Table 1 � Minimum percent pozzolan required by mass of cementitious material and are based upon the alkali content of the cement.

Total Alkalies of Cement less than or equal to 0.75%

Total Alkalies of Cement greater than 0.75% and less

than or equal to 1%

Class F Flyash 20% 25%

GGBF Slag 40% 50%

Silica Fume 7% 10%

Metakaolin 7% 10%

! "# $ % & ' % ( $

EXAMPLE - Using the ACI worksheet provided, let us design a VDOT Class A-3 general use mix using Type IS cement. This mix will be for 1 yd3. No pozzolan will be added as Type IS is a blended cement.

Conditions:Minimum Cement Content: 588 lbs.(Table II-17)Maximum Water-cement ratio: 0.49 lb water per lb cement (Table II-17)Nominal maximum size aggregate: 1 in (Table II-17)Air Content: 6% ± 2% (Table II-17)Slump: 1 � 5 in (Table II-17)Fineness Modulus of Sand: 2.70 (! ne aggr. data sheet)Sp. Gr. of Fine Aggregate: 2.66 (! ne aggr. data sheet)Sp. Gr. of Coarse Aggregate: 2.61 (coarse aggr. data sheet)Dry-rodded Unit Weight of C.A.: 104 lb/ft3 (lab results)Sp. Gr. of Type IS Cement: 3.05 (from supplier)

There are ! ve materials going into this mix:

l. Cement2. Water3. Air4. Coarse Aggregate5. Fine Aggregate

Solve for the Absolute Volume of each of the ! ve materials in the mix. The combined volume must total one cubic yard (27.00 ft3).

Quantities for three of the ! ve materials are given by the speci! cation.

These are: 1. Cement 2. Water

3. Air

This means then that you only have to solve for the quantities of two ingredients:

(1) Coarse aggregate, and (2) Fine Aggregate

Let us now solve the absolute volume of each material, remembering the total must be 27.00 cubic feet for all material.

1. Cement: 588 lb (by Spec.)

588 lb (lb per 1 yd3)3.05 (sp.gr. cement) X 62.4 (weight of 1 ft3 of water) =

588 = 3.09 ft3 = Absolute volume

3.05 X 62.4

2. Water: By speci! cations, the maximum water is 0.49 lb water per lb cement. If the cement content for a yd3 is 588 lb, the maximum design water will be:

588 x 0.49 = 288 lbs.

! " # $ # " % & " ' #

To ! nd absolute volume:

288 = 4.62 ft3 = Absolute volume 1.00 X 62.4 4.62 ft3 x 7.5 (gallons of water in 1 ft3) = 34.6 gallons

3. Air: The target air content is 6%. To ! nd the absolute volume of air:

0.06 x 27 ft3 = 1.62 ft3 (Absolute Volume)

As air will not weigh anything, it will not have a speci! c gravity, so we have solved for the 6% volume displaced by the air in a cubic yard. As we said previously, three of the ! ve materials are given, namely:

1. Cement2. Water, and3. Air

Now, we must solve for coarse aggregate:

4. Coarse Aggregate: From Table A1.5.3.6 of ACI, the percent of the total mix that should be coarse aggregate is found by taking the 1 inch nominal maximum size aggregate designated by the speci! cation, and locating it on the left side of the table. The F.M. of the sand is found across the top of Table A1.5.3.6. We said that the F.M. of the sand was 2.70. Across the top of the columns, we ! nd 2.60 and 2.80. If the F.M. happens to be 2.70, we must interpolate the value of 2.70. In this case, the value will be half way between 0.69 (F.M. = 2.60) and 0.67 (F.M. = 2.80), which for 1 inch would be 0.68. So, 68% of the cubic yard mix will be coarse aggregate in a dry-rodded condition.

To convert this to volume:

0.68 (dry-rodded condition of the coarse aggregate)x 27.0 ft3 (volume) 18.36 ft3 = dry-rodded volume

To determine the weight of coarse aggregate going into the mix:

18.36 ft3 (dry rodded unit weight)x 104 lb/ft3 (dry-rodded volume)1909 lbs. = design weight of coarse aggregate

To determine the absolute volume:

1909 lbs. = 11.72 ft3 = Absolute volume2.61 X 62.4 lb/ft3

! " "# $ " % & " ' $

5. Fine Aggregate: To solve for the amount of ! ne aggregate, work the problem in reverse as compared to the other materials. First, total the absolute volume of the other four materials.

Design Absolute VolumeCement 588 lbs. 3.09 ft3

Water 288 lbs. 4.62 ft3

Air 6% 1.62 ft3

Coarse Aggregate 1909 lbs. 11.72 ft3

TOTAL 21.05 ft3

If four materials total 21.05 ft3, it is proper to assume that the ! ne aggregate will ! ll the remaining volume of a cubic yard (27.00 ft3).

To ! nd the volume the ! ne aggregate will occupy:

27.00 ft3 � 21.05 ft3 =5.95 ft3

Now , multiply the volume of ! ne aggregate times speci! c gravity of ! ne aggregate times 62.4 (unit weight of water in lb/ft3):

5.95 ft3 x 2.66 x 62.4 = 988 lbs.

This is the design weight of the ! ne aggregate.

The class A3 general use concrete mix design as shown on the TL-27 will be as follows:

Cement 588 lbs. Water 288 lbs. Air 6 % Coarse aggregate 1909 lbs. Fine Aggregate 988 lbs.

! " # # $ " % & " ' $

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! " # $ " % & " ' $

GRANULATED IRON BLAST FURNACE SLAG

Speci! cation:

Shall conform to ASTM C989, Grade 100 or 120Replaces up to 50% of the cement

Reaction:

Reactive within itself and also reacts with the free lime in the cement

Advantages:

Cheaper than cementUtilization of a waste productReduces heat of hydration (less cement)Improves sulfate resistanceReduces alkali-silica reactionGives higher strength at later ages

Disadvantages:

Another mixtureScaling and drying shrinkage may be increasedEarly strengths retarded, particularly in cool weather

Note: When forms are stripped, the concrete will be discolored with greens, blues, and blacks, like ink blots, but will bleach fairly rapidly.

! " # $ % " & ' " ( %

ACI Concrete Mix Design UtilizingGround Granulated Iron Blast-Furnace Slag

When Ground Granulated Blast-Furnace Slag is utilized as an additive in concrete, it must conform to the requirements of ASTM C 989, Grade 100 or 120.

Ground Granulated Blast-Furnace Slag shall not exceed 50 percent of the total design cement weight speci! ed in Table II-17. The method of design is very similar to that used in the previous mix design in this section.

EXAMPLE: Using the ACI worksheet provided, let us design a VDOT A3 general use mix with slag.

Conditions:Minimum Portland Cement Content: 588 lbs.(Table II-17)Alkali Content of Cement: 0.82 (from supplier)Maximum Water Cement Ratio: 0.49 lb water per lb Cement.(Table II-17)Nominal maximum size aggregate: 1 in (Table II-17)Air Content: 6% ± 2% (Table II-17)Slump: 1 � 5 in (Table II-17)Sp. Gr. of Slag: 2.90 (Speci! c gravity of slag will vary; therefore, the most current gravity should be obtained from the District Materials Section)Fineness Modulus of Sand: 2.70 (! ne aggr. data sheet)Sp. Gr. of Fine Aggregate: 2.66 (! ne aggr. data sheet)Sp. Gr. of Coarse Aggregate: 2.61 (coarse aggr. data sheet)Dry rodded unit weight of C.A.: 104 lb/ft3 (Lab Results)Sp. Gr. of Cement: 3.15 (from supplier)

Let us now solve for the absolute volume of each material, remembering the total must be 1 yd3 (27.00 ft3) for all materials.

The minimum cement content is 588 lbs.; however, a portion of the cement will be replaced with slag. We need to refer to Section 217.02(a) Table 1 of the Speci! cations. The alkali content of the cement is 0.82%; therefore, 50% of the cement will be replaced by slag.

1. Cement: 294 lbs. (this is equal to 50% of 588 lbs., which is the design weight for cement on Class A3 General Use mixes.)

294 lbs. 3.15 (sp.gr. of port. cem.) X 62.4 (weight of 1 ft3 of water)


294 = 1.50 ft3 (absolute volume)3.15 X 62.4

2. Slag: 294 lbs.(this is equal to 50% of 588 lbs., which is design weight for cement on Class A3 General use mixes.)


294 lbs. = 1.62 ft3

2.90(sp.gr. for this slag) X 62.4 (weight of 1 ft3 of water)

! " #$ % " # & " ' %

3. Water: By speci! cations, the maximum water is 0.49 lb water per lb. cementitious material. For this mix, you would consider the cementitious weight to be: 294 lbs. Cement + 294 lbs. Slag = 588 lbs.

588 x 0.49 = 288 lbs.


288 = 4.62 ft3 = Absolute Volume1.00 x 62.4

4.62 ft3 x 7.5 (gallons of water in 1 ft3) = 34.6 gallons


0.06 x 27 ft3 = 1.62 ft3 = Absolute Volume

As air will not weigh anything, it will not have a speci! c gravity, so we have solved for the 6% volume displaced by the air in a cubic yard.

5. Coarse Aggregate: A factor of 0.68 is obtained from Table A1.5.3.6 of ACI by using the Fineness Modulus of the sand and the nominal maximum size of the coarse aggregate in the method described on Page 3-10. To convert this to volume, we say 0.68 x 27 ft3 = 18.36 ft3 (this volume is dry rodded). With the dry rodded unit weight given as 104 lb/ft3, determine the weight of coarse aggregate going into the mix by the following:

18.36 ft3 x 104 lb/ft3 = 1909 lbs. (design wt. of C.A.)

The speci! c gravity of the coarse aggregate was 2.61. The absolute volume is determined by the following:

1909 lbs. = 11.72 ft3 = Absolute Volume2.61 X 62.4 (wt. of ft3 of water)

6. Fine Aggregate: To solve for the amount of ! ne aggregate, work the problem in reverse as compared to the other materials. First, total the absolute volume of the other ! ve materials:


Slag 294 lbs. 1.62 ft3


Air 6 % 1.62 ft3

Coarse Aggr. 1909 lbs. 11.72 ft3 TOTAL 21.08 ft3

! " # $ % " & ' " ( %

If the ! ve materials total 21.08 ft3, it is proper to assume that the ! ne aggregate will ! ll the remaining volume of a cubic yard. To ! nd the volume the ! ne aggregate will occupy, solve the following:

27.00 ft3 � 21.08 ft3 = 5.92 ft3

Now, multiply the volume of ! ne aggregate times the speci! c gravity of ! ne aggregate times 62.4 (unit weight of water):

5.92 ft3 x 2.66 x 62.4 = 983 lbs. (design weight of ! ne aggregate)

The Class A3 general use concrete mix utilizing slag will be shown on the TL-27 as follows:

Cement - 294 lbs.Slag - 294 lbs.Water - 288 lbs.Air - 6 %Coarse Aggregate - 1909 lbs.Fine Aggregate - 983 lbs.

! " #$ % " & ' " ( %

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` K \ P [ K \ P 3 B C J [ O J T K E D RK J \ @ _ T K J P = : = + * [ K O J @ X E D RK L J @ @ E D R` K O J @ X E D R; J + J ? J \ K E D R 6 K J T T , S J . I J 6 T K J P [ \ X - 3 B C J, N . . 8 , = 8 7 ] N + 9 = 5 = 5 8 , a ? A = : * 8 I + 9 ) 8) 8 4 8 9 = K \ P 3 B C J< + = 8 I K X X 3 B C J Z 1 - P J T H 2 3 C J+ 5 I T A) J + . . I J O \ @ \ 3 B C J ` b c c c c c c c c c d e b c c c c c c c c c c d; J + . . I J \ X - 3 B C J ` b c c c c c c c c c d e b c c c c c c c c c c d+ 7 7 * J 4 + = * , J C 3 2 H [ K \ P 3 B C J[

! " # $ % " & ' " ( %

FLY ASH

Production:

Waste product of coal ! red electrical utility

Speci! cations:

Must meet ASTM C618 Class F with maximum loss on ignition of 6%Replaces up to 30% of the cement

Reaction:

Reacts with the free lime (calcium hydroxide) given off by cement during hydration, and water to produce cementitious material.

Advantages:

Utilization of a waste productReduces energy for production of cementReduces heat of hydrationImproves workabilityImproves sulfate resistanceReduces alkali-silica aggregate reactionCosts less than cement

Disadvantages:

Another admixtureScaling and drying shrinkage may be increasedSlower strength gainAir content dif! cult to control 1. Finer than cement 2. Unburned carbon

! " #$ % " & ' " ( %

ACI Concrete Mix Design Utilizing Fly AshWhen " y ash is utilized as an additive in concrete, it must conform to the

requirements of ASTM C618 Class F or Class C, except that the Loss on Ignition shall be limited to a maximum of 6%. Class F, " y ash shall replace 20 to 30% by weight of the design cement depending on the alkali content of the cement used. The minimum total cementitious materials are speci! ed in Table II-17. The method of design is very similar to that used in the previous mix designs in this chapter.

EXAMPLE: Using the ACI worksheet provided, let us design a VDOT Class A3 General mix using " y ash.

Conditions:Minimum Cement Content: 588 lbs. (Table II-17)Alkali Content of Cement: 0.65% (from supplier)Maximum Water-Cement ratio: 0.49 lbs. water per lb. cement (Table II-17)Nominal Maximum Size Aggregate: 1 in (Table II-17)Air Content : 6% ± 2% (Table II-17)Slump: 1 - 5� (Table II-17)Sp. Gr. of " y ash: 2.25 (Speci! c gravity of " y ash will vary. The most current gravity from the manufacturer/supplier should be used).Fineness Modulus of Sand: 2.70 (! ne aggr. data sheet)Sp. Gr. of Fine Aggregate: 2.66 (! ne aggr. data sheet)Sp. Gr. of Coarse Aggregate: 2.61 (coarse aggr. data sheet)Dry-rodded Unit Weight of C.A.: 104 lb/ft3 (lab results)Sp.Gr. of Cement: 3.15 (from supplier)

Let us now solve the absolute volume of each material, remembering the total must be 27.00 ft3 for all materials.

The minimum cement content is 588 lbs.; however, a portion of the cement will be replaced with " y ash. We need to refer to Section 217.02(a) Table 1 of the Speci! cations. The alkali content of the cement is 0.65%; therefore, 20% of the cement will be replaced by " y ash.

1. Cement: 470 lbs. (this is equal to 80% of 588 lbs., which is the design wt. of cement of a Class A3 General Mix).

470 lbs. (this is design weight for 1 yd3) 3.15 (sp.gr. of cement) X 62.4(weight of 1 ft3 of water)

470 lbs. = 2.39 ft3 Absolute volume 3.15 X 62.4

2. Fly ash: 118 lbs. (this is equal to 20% of 588 lbs., which is design wt. for cement of a Class A3 General use mix).


118 lbs. = 0.84 ft3

2.25(sp.gr.for this " y ash)X 62.4(weight of 1 ft3 of water)

! " # " # $ % & $ ' #

3. Water: By speci! cation, the maximum water is 0.49 lbs. water per lb. cementitious material. For this mix, you would consider the cementitious weight to be 588 lbs. Cement + Fly Ash (470 lbs. + 118 lbs.).

588 x 0.49 = 288 lbs.


288 = 4.62 ft3 = Absolute volume1.00 x 62.4

4.62 ft3 x 7.5 (gallons of water in 1 ft3) = 34.6 gallons


0.06 x 27.00 ft3 = 1.62 ft3 = Absolute Volume

As air will not weigh anything, it will not have a speci! c gravity, so we have solved for the 6% volume displaced by the air a cubic yard.

5. Coarse Aggregate: A factor of 0.68 is obtained from Table A1.5.3.6 of ACI by using the Fineness Modulus of the sand and the nominal maximum size of the coarse aggregate as described on Page 3-10. To convert this to volume, 0.68 x 27.00 ft3 = 18.36 ft3 (this volume is dry rodded). With the dry rodded unit weight given as 104 lb/ft3, determine the weight of coarse aggregate going into the mix by the following:

18.36 ft3 x 104 lb/ft3 = 1909 lbs. (this is the design mass of C.A.)

The speci! c gravity of the coarse aggregate was 2.61. The absolute volume is determined by the following:

1909 = 11.72 ft3 = Absolute Volume2.61 X 62.4 (wt. of ft3 of water)

6. Fine Aggregate: To solve for the amount of ! ne aggregate, we will work the problem in reverse as compared to the other materials. First, we must total the absolute volume of the other ! ve materials:


Fly Ash 118 lbs. 0.84 ft3


Air 6 % 1.62 ft3

Coarse Aggr. 1909 lbs. 11.72 ft3 TOTAL 21.19 ft3

! " #" $ # % & # ' $

If the ! ve materials total 21.19 ft3, it is proper to assume that the ! ne aggregate will ! ll the remaining volume of a cubic yard. To ! nd the volume the ! ne aggregate will occupy, solve the following:

27.00 ft3 � 21.19 ft3 = 5.81 ft3

Now, multiply the volume of ! ne aggregate times the speci! c gravity of ! ne aggregate times 62.4 (unit weight of water):

5.81 ft3 x 2.66 x 62.4 = 964 lbs. (design mass of ! ne aggregate)

The Class A3 general use concrete mix utilizing " y ash will be shown on the TL-27 as follows:

Cement - 470 lbs.Fly Ash - 118 lbs.Water - 288 lbs.Air - 6 %Coarse Aggregate - 1909 lbs.Fine Aggregate - 964 lbs.

! " " " # $ % & $ ' #

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Allowable Field AdjustmentsIf the quantities calculated by ACI absolute volume method do not give the

required workability and consistency in the ! eld, the mix can be adjusted by an allowable interchange of coarse aggregate and ! ne aggregate. The interchange of coarse aggregate and ! ne aggregate may vary up to 5 percent (by weight), but neither may be changed more than 5 percent. When an interchange of aggregate is needed, the ! ne aggregate, normally being of less weight than coarse aggregate, is increased or decreased 5 percent and then the coarse aggregate is changed by an equal volume so the design will be 27 ft3.

For example, when the ! rst load of concrete using the Class A3 general use design (as shown below) arrived on the project, the slump was 2 inches. The contractor desired a higher slump. In order to accomplish this, the surface areas of the aggregate must be decreased as much as is allowable, which will make the mix as coarse as possible and remain within the speci! cation requirements.

Class A3 General Mix Design:

IP Cement - 588 lbs. Sp. gr. F.A. = 2.67 Water - 288 lbs. Sp. gr. C.A. = 2.61 Air - 6 % F.M. of F.A. = 2.70 #57 - 1965 lbs. Unit wt. of C.A. = 107 lbs/ft3

F.A. - 930 lbs. Sp. gr. Cement = 3.03

The F.A., being of less weight than the C.A., will be decreased by 5% as follows:

F.A. = 930 x 0.05 = 47 lbs.

Original Wt. of F.A. = 930 lbs.Less 5% = - 47 lbs.New Wt. of F.A. 883 lbs.

The C.A. must then be increased the same volume that the F.A. is decreased so the design will remain 27 ft3. This is accomplished as follows:

47 lbs. F.A./[2.67(F.A. Sp. Gr.) X 62.4] = 0.28 ft3

0.28 X [2.61 (C.A. Sp. Gr.) X 62.4] = 46 lbs. of C.A. to be added

Original wt. of C.A. #57 1965 lbs.Plus wt. C.A. to be increased + 46 lbs.Net wt. of C.A. 2011 lbs.

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The adjusted design quantities are:

Cement 588 lbs. Air 6 % Water 288 lbs. #57 2011 lbs. F.A. 883 lbs.

After these adjustments are made, the design should be checked to make sure it yields 27.00 ft3

Absolute Volume (ft3)

IP Cement - 588 lbs. 588 = 3.11 3.03 X 62.4

Air - 6% 0.06 X 27 = 1.62

Water � 288 lbs. 288 = 4.62 1.00 X 62.4

C. A. - #57 = 2011 lbs. 2011 = 12.35 2.61 X 62.4

F. A. - 883 lbs. 883 = 5.30 2.67 X 62.4

TOTAL 27.00

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Allowable Field Adjustment Example Problem No. 1

The following Class A3 General mix design produced a harsh mix. The contractor desires to reduce the harshness. Show the aggregate adjustments that may be made within the speci! cation requirements.

IP Cement - 588 lbs. F. A. Sp. Gr. - 2.64 Air - 6.0% F. A. F. M. - 3.00 Water - 288 lbs. C. A. Sp. Gr. - 2.62 F. A. - 998 lbs. C. A. Unit Wt.- 108 lb/ft3

C. A. No. 57 - 1895 lbs. Cement Sp. Gr. - 3.03

SOLUTION:

998 X 0.05 = 50 lbs. (This weight must be added to the weight of sand)

998 + 50 = 1048 lbs. (Total weight of sand)

50 = 0.30 ft3 0.30 X 2.62 X 62.4 = 49 lbs.2.64 X 62.4 (The weight of C.A. must be decreased by this amount)

1895 - 49 = 1846 lbs. (Total weight of C.A. due to increase in the sand)

ADJUSTED MIX DESIGN

Cement - 588 lbs.Air - 6.0%Water - 288 lbs.F. A. - 1048 lbs. C. A. - No. 57 - 1846 lbs.


The following Class A3 General mix design produced a slump of 3 inches. The contractor desires a 4� slump. Show the aggregate adjustments that may be made within the speci! cation requirements.

Cement - 470 lbs. F. A. Sp. Gr. - 2.62 Air - 6% F. A. F. M. - 2.70 Water - 288 lbs. C. A. Sp. Gr. - 3.04 F. A. - 1189 lbs. C. A. Unit Wt.- 106 lb/ft3

C. A. No. 57 - 1946 lbs. Cement Alkali Content - 0.65% Flyash - 118 lbs. Flyash Sp. Gr. - 2.25

SOLUTION:

1189 X 0.05 = 59 lbs. (This weight must be subtracted from the weight of sand)

1189 � 59 = 1130 lbs. (Total weight of sand)

59 = 0.36 ft3 0.36 X 3.04 X 62.4 = 68 lbs. 2.62 X 62.4 (This weight must be added to C.A. weight)

1946 + 68 = 2014 lbs. (Total weight of C. A.)

ADJUSTED MIX DESIGN

Cement - 470 lbs. Air - 6% Water - 288 lbs. F. A. - 1130 lbs. C. A. - No. 57 - 2014 lbs. Flyash - 118 lbs.

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The following Class A4 General mix design produced a harsh mix. The contractor desires to reduce the harshness. Show the aggregate adjustments that may be made within the speci! cation requirements.

Cement - 381 lbs. F. A. Sp. Gr. - 2.60 Air - 6.5% F. A. F. M. - 2.93 Water - 286 lbs. C. A. Sp. Gr. - 2.67 F. A. - 1063 lbs. C. A. Unit Wt. - 96.5 lbs/ft3

C. A. No. 57 - 1798 lbs. Cement Alkali Content - 0.70%Slag - 254 lbs.

SOLUTION:

1063 X 0.05 = 53 lbs. (This weight must be added to the weight of sand)

1063 + 53 = 1116 lbs. (Total weight of sand)

53 = 0.33 ft3 0.33 X 2.67 X 62.4 = 55.0 lbs.2.60 X 62.4 (The weight of C. A. must be decreased by this amount)

1798 - 55 = 1743 lbs. (Total weight of C. A. due to increase in the sand)

ADJUSTED MIX DESIGN

Cement - 381 lbs. Air - 6.5% Water - 286 lbs. F. A. - 1116 lbs. C. A. - No. 57 - 1743 lbs. Slag - 254 lbs.

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VIRGINIA DEPARTMENT OF TRANSPORTATION

RESULTS OF QUALITY TEST ON

COMMERCIALLY PRODUCED

COARSE AGGREGATES

(THE VALUES THAT FOLLOW ARE TO BE USED FOR TRAINING PURPOSES ONLY. FOR ACTUAL VALUES, REFER TO THE LATEST PUBLISHED LIST.)

L. A. ABRASION LOSSPRODUCER AND LOCATION SP. GR. ABS.(%) A B C Sadler Materials 2.62 0.9 37.6 38.0 ----Richmond, VA

Lone Star Industries 2.63 0.8 39.2 38.0 34.0Dock St., Richmond, VA

Fredericksburg Sand & GravelFredericksburg, VA 2.64 0.3 37.5 35.5 ----

Virginia Traprock, Inc. 3.04 0.4 ---- 19.1 26.7Leesburg, VA

West Sand and Gravel 2.60 1.0 ---- 39.3 37.1Richmond, VA

Bull Run Stone Co. 2.67 1.5 ---- 14.3 ----Manassas, VA

Dillon, E & Co. 2.83 0.4 23.1 ---- ----Swords Creek, VA

Lone Jack 2.81 0.3 ---- 18.9 21.8Glasgow, A

Shenandoah Asphalt 2.59 1.0 ---- 30.3 39.3Vesuvius, VA

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VIRGINIA DEPARTMENT OF TRANSPORTATION

FINE AGGREGATE DATA

(THE VALUES THAT FOLLOW ARE TO BE USED FOR TRAINING PURPOSES ONLY. FOR ACTUAL VALUES, REFER TO THE LATEST PUBLISHED LIST.)

SOUNDNESSPRODUCER AND LOCATION SP. GR. ABS.(%) F.M. LOSS Chickahominy, Inc. 2.62 1.0 2.7 7.9Charles City, VA

Sadler Materials 2.64 0.5 3.0 7.0Richmond, VA

Fredericksburg Sand & Gravel 2.62 0.8 2.7 11.9Fredericksburg, VA

Lone Star Industries 2.64 0.6 2.8 5.8Dock St., Richmond, VA

West Sand and Gravel 2.64 0.9 2.7 6.7Richmond, VA

Aylett Sand & Gravel 2.62 0.2 2.9 4.6Aylett, VA

Dillon E. & Co. 2.80 1.1 3.1 13.8Swords Creek, VA

Stuart M. Perry, Inc. 2.67 1.0 2.8 9.1Winchester, VA

Wilson Quarries 2.83 0.5 2.8 6.8Horsepasture, VA

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Chapter 3Study Problems) * + , - . / 0 1 - 2 3 4 5 6 7 8 0 9 : 6 ; <= > ? @ A B C D E C ? ? F G A > B > H C E I @ J K ? @ B A C L M N > O O P > I > B Q R H S I @ C B Q D > I > B Q D S !" @ Q # C B C E $ C E @ P S B Q > B Q S R % ! & ' ! ( E M F ? # ) * N > P @ + P H C , @ Q M S R - ! . / ) @ ? C , C @ E C 0 E >@ R B > P > ? ? C H M R H S I 1 S B > J 2 > ? S K H P > ? ) O B P ! D S C H ? > C A A H > A C Q > " @ E E P S I > R H S I 3 S B >* Q C H O B 4 K ? Q H @ > ? ) 2 @ P # I S B 4 ) 5 F ! = H M H S 4 4 > 4 K B @ Q " > @ A # Q S R P S C H ? > C A A H > A C Q >@ ? & % . E 0 6 R Q 7 ! ( @ B > C A A H > A C Q > " @ E E P S I > R H S I 8 > ? Q * C B 4 C B 4 H C , > E ) 2 @ P # I S B 4 )5 F ! 9 ? > C A A H > A C Q > 4 C Q C ? # > > Q ? R S K B 4 S B N C A > . : . - C B 4 . : . . !D 3 F * * 1 O ; = < * O =1 > = O ( O < = 8 O L ?( O = < F 2 < F L < D > F 2 * < F 2 < F L <( ! 1 ! = 2 @ 2 > = = < = 9 = O L 8 L !* A ! 2 ! * A ! 2 != > 1 O = F 3 1 F ; ! * O B < D ! F ! L F C 3 < F & ! / ! . ! D ( F D L > 2> L ? < 2 = F L F = < < = < = ( > 2 * A < D O F 3 = < * O = *E 9 F = L O L @ > ( D > F 2 * < F 2 < F L <L F C 3 < F & ! / ! . ! D ; - F R Q 7 ; 9 = O L 8 L ! G E 0 ? !H I J K L M N O P K L M Q O JA > 2 L 3 F = = D < 1 < = L E 0 ? ! G R Q 7. ! & / J D - ! G8 F L < 2 E 0 ? ! G R Q 7& ! % % J D - ! GF O 2 R J - F G R Q 7& % %D ! F 2 ! 3 0 ? ! G R Q 7* A ! 2 ! ; D - ! GF = = O L O > = F 3 1 F L < 2 O F 3 * G R Q 7G R Q 7L > L F 3 G R Q 7- F ! % % R Q 7: R Q 7( ! F ! R Q 7 ; * A ! 2 ! ; D - ! G G E 0 ? !* 9 < * L < = E 9 F = L O L O < * S / R L > 3 < 2 F = D <D < 1 < = L E 0 ? !8 F L < 2 E 0 ? ! S H A C E ? !F O 2 RD ! F 2 ! 3 0 ? ! : T U V T U( ! F 2 ! 3 0 ? ! : T U V T UF = = 3 ! 1 F L 3 * ! GG

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