Mix DesignConcrete School

Weight - Volume Relationships

Conversion Factors

One Cubic foot of water = 7.5 gallons One Cubic foot of water = 62.4 lbs One Gallon of water = 8.33 Lbs One Cubic yard = 27 cu ft One bag of cement = 94 lbs One bag of cement equals one cu ft (loose

vol)

Conversion Factors

One bag of cement equals 0.48 cu ft (absolute volume)

Four bags of cement equals one barrel.

Basic Mathematical Terms Related to Volume Unit Weight - The weight of one cubic foot

of material. For concrete, the weight in pounds of one cubic foot of plastic concrete

Dry Rodded Unit Weight - The weight in pounds of one cubic foot of stone compacted in a container by rodding.

Pg 53

Terms

Cement Yield - The volume of concrete in cubic feet produced from one bag of cement.

Absolute volume - The volume of material in a voidless state.

Specific gravity - The ratio of the weight of a given volume of material to the weight of an equal volume of water.

Pg 53

Terms

Remember One cu ft of water weighs 62.4 pounds One gallon of water weighs 8.33 pounds

62.4 lbs/cu.ft = 7.5 gals per cu.ft 8.33 lbs/gal

Pg 53

Terms

If we know the weight and specific gravity of a material, the absolute volume can be calculated:

Absolute Volume = Weight of Material

(Sp.Gr.) x (62.4 pcf)

Pg 54

Terms

Absolute Volume of Water:

= 62.4 lbs = 1 cu.ft

1 x 62.4 pcf

Pg 54

Weight - Volume

HOW TO CALCULATE THE SPECIFIC GRAVITY OF A

MATERIAL

Method ASpecific Gravity The weight of the material in air is Wa The weight of the material in water is Ww The specific gravity equals the weight of

the material in air divided by the difference of the weight in air and the weight in water.

Formula: Wa / (Wa - Ww)

Pg 55

Method B

The weight of the material in air is Wa Pour it into a calibrated flask The original volume of water in the flask

was Va The final volume of water of water and

material is Vb

Pg 56

Vb

Va

Wa

Method B

The volume of the material is equal to the volume of the water displaced

Formula: Wa / (Vb - Va)

Weight- Volume

The specific gravity of any material multiplied by 62.4 lbs is the Unit Weight of that material. It is the weight of one cubic foot of solid material if it were melted.

Pg 57

Weight Volume

Absolute Volume of an Aggregate

= Weight of aggregate Weight of one cubic foot of aggregate

melted Weight of aggregate = Absolute Volume

Sp.Gr. of agg. X 62.4

Pg 57

Weight Volume Problem

Find the absolute volume of 288 lbs of water

288

1 x 62.4

= 4.62 cu.ft.

Absolute volume

When water is given in gallons rather than pounds the absolute volume is calculated by dividing the gallons by 7.5 (gallons water in one cubic foot).

Gallons of Water

Abs vol = 34.5 gals = 4.61 Cu.Ft.

7.5

Determining Absolute vol pg 58 ex.

Material Sp.Gr. Weight Melted Abs volCement 3.15 588 196.56 2.99#57 2.65 1808 165.36 10.93Sand 2.63 1129 164.11 6.88Water 1.00 286 62.4 4.58Air 6% .06 x27 1.62Total 3811 lbs 27.00cf

Absolute volume

Remember the weight of one cubic foot of material melted is determined by multiplying the material’s specific gravity by 62.4.

In the case of cement the melted weight will be 196.56 (3.15 x 62.4).

Definitions

Yield - The volume of concrete (cubic feet) produced from one bag of cement.

C/F - The number of pounds of cement per cubic yard.

W/C - The pounds of water per pound of cement in a concrete mix.

Unit Weight - Pounds per cu.ft. of concrete.

Pg 58

Absolute Volume

Yield

588 = 6.26 bags cem. 27.00 cuft = 4.31cuft

94 6.26 bags cem

Pg 59

Cement Factor

C/F = 27.00 cu ft = 6.26 bags

4.31 cu ft / bag of cement

6.26 bags x 94 Lbs = 588 Lbs of cem per cy

Absolute Volume

W/C

34.3 gal x 8.33 = 286 lbs water = .486

588 lbs cem. Calculated Unit weight

3811 lbs material = 141.15 lbs/cuft

27.00 cu ft

Field Unit Weight

(Wt. Concrete + Wt Bucket) - Wt Bucket

Volume of Bucket

= Unit Wt of Fresh Concrete Example:

Weight of Unit Wt Bucket - 23.2 lbs

Volume of Bucket - .51 cuft

Weight of concrete & bucket - 94.8

Pg 59

Field Unit Weight

94.8 - 23.2 = 140.39 lbs/cuft

.51 (Actual Unit Weight)

Computing % Air by Unit Wt

Formula for Computing % Air

Theoretical Unit Wt - Actual Unit Wt x 100

Theoretical Unit Wt

Theoretical Unit Weight is the air free unit weight.

Theoretical Unit Wt

Example 27.00 cu.ft. in concrete mix for 1 cu.yd. 1.62 cu.ft. in concrete with 6% air. 25.38 cu.ft. in the mix without air. 3811 lbs mat’l in mix = 150.16 lbs/cu.ft

25.38 cu.ft. in mix w/o air (theoretical

unit weight)

Percent Air

T = Theoretical Unit Weight A = Actual Field Unit Weight T - A x 100 = % air

T 150.16 - 140.39 x 100 = 6.5% Air

150.16

Checking Yield of a Mix

The yield of a batch of concrete is the total volume occupied by fresh concrete.

Yield in cu.ft. is determined by dividing the total weight in pounds of all ingredients going into the batch by the unit weight in lbs/cu.ft. of the fresh concrete.

To convert to cu.yds. Divide by 27 cu.ft.

Pg 61

Nominal 10 cu.yd. batch

Wet stone 18,080 lbs Wet Sand 12,760 lbs Cement 6006 lbs Admixture 5 lbs Water through plant 230gals x 8.33 = 1916 Water through truck 10gals x 8.33 = 83 Lbs Total Weight 38,850 lbs

pg

Checking yield

Air content by pressure meter = 5.2% Unit weight of fresh concrete = 140.50 pcf

yield in cu.ft. = 38,850 lbs = 276.50 cu.ft.

140.50 lb/cu.ft. yield in cu yd = 276.50 / 27.00 = 10.24

cuyd

Checking Yield

Yield per nominal cu.yd. 276.50 cu.ft./batch = 27.65 cu.ft./cu.yd

10 cu. Yds This batch over yield by .65 cu.ft./cu.yd.

Measured air content is not used in the calculations.

Weight Volume Problem No.1

Mat’ls Sp.Gr. Weight Melted Abs.

Cem. 3.15 564 196.56 2.87

#67’s 2.82 1966 175.968 11.17

Sand 2.62 1100 163.488 6.73

Water 1.00 288 62.4 4.62

Air 6% .06 x27 1.62

3918 27.01

Problem 1

Yield 564/94 = 6/0 bags 27.01/ 6.0 = 4.50 cf

C/F 27/4.50 = 6.0 6.0 x 94 = 564 Lbs

W/C Ratio 288 / 564 .511

Cal UW 3918 / 27.01 145.06 pcf

Field UW 147.95 pcf

Theor. UW 27.01-1.62 = 25.39 3918/25.39 = 154.31 pcf

% Air by UW 154.31 – 147.95 x 100 154.31

4.1%

Weight Volume Problem No. 2

Mat’l Sp.Gr. Weight Melted Abs

Cem 3.15 714 196.56 3.63

#67 2.88 1940 179.712 10.80

Sand 2.62 1015 163.488 6.21

Water 1.00 298 62.4 4.78

Air 6% .06 x 27 1.62

3967 27.04

Problem 2

Yield 714 / 94 = 7.60 27.04 / 7.60 = 3.56 cf

C/F 27/3.56 = 7.6 7.6 x 94 = 714

W/C Ratio 298 / 714 = .417

Cal UW 3967 / 27.04 = 146.71 pcf

Field UW 147.95 pcf

Theor UW 27.04-1.62=25.42 3967/25.42= 156.06 pcf

% Air by UW 156.06-147.95 x100 156.06

5.2%

Weight Volume Problem No. 3

Mat’l Sp.Gr. Weight Melted Abs

Cem 3.15 545 196.56 2.77

#67 2.85 1853 177.84 10.42

Sand 2.60 1197 162.24 7.38

Water 1.00 300 62.4 4.81

Air 6% .06 x 27 1.62

Totals 3895 27.00

Problem 3

•Yield 545 = 5.8 bags cem. 27.00cu.ft = 4.66 cu.ft.

94 5.8 bags

C/F 27/ 4.66 = 5.79 5.79 x 94 = 544

•W/C Ratio 300 Lbs water / 545 Lbs.cem = .550

•Calculated Unit Wt. 3895 Lbs / 27.00 cu.ft. = 144.26 pcf

•Field Unit Wt 144.35 pcf

Problem 3

% Air by Unit Wt

27.00 cu.ft. - 1.62 cu.ft. = 25.38 cu.ft.

3895 Lbs / 25.38 cu.ft = 153.47 pcf

153.47 - 144.35 x 100 = 5.9% air

153.47

Problem 3

Weight Volume Problem No. 4

Mat’l Sp.Gr. Weight Melted Abs

Cem 3.15 588 196.56 3.00

#67 2.88 1956 189.696 10.31

Sand 2.62 1228 163.488 7.51

Water 1.00 288 62.4 4.62

Air 6% .06 x 27 1.62

Totals 4060 27.05

Problem 4

•Yield 588 = 6.26 bags cem 27.05cu.ft = 4.32 cu.ft.

94 6.26 bags cem

C/F 27/ 4.32 = 6.25 x 94 = 588 Lbs

•W/C Ratio 288 Lbs. Water / 588 Lbs.cem. = .490

•Calculated Unit Wt 4060 / 27.05 = 150.09 pcf

•Field Unit Weight 149.20 pcf

Problem 4

% Air by Unit Wt

27.05 cu.ft. - 1.62 cu.ft. = 25.43 cf.

4060 Lbs / 25.43 cu.ft. = 159.65 pcf

159.65- 149.20 x 100 = 6.5% air

159.65

Problem 4

Weight Volume Problem No. 5

Mat’l Sp.Gr. Weight Melted Abs

Cem 3.15 677 196.56 3.44

#67 2.79 1,895 174.096 10.88

Sand 2.60 1081 162.24 6.66

Water 1.00 275 62.4 4.41

Air 6% .06 x 27 1.62

Totals 3928 27.01

Problem 5

Yield 677 = 7.2 bags cem. 27.01 Cu.ft. = 3.75 cu.ft.

94 7.2 bags

C/F 27/3.75 = 7.2 x 94 = 677 Lbs

W/C Ratio 275 Lbs water / 677 Lbs. Cem. = .406

Calculated Unit Wt 3928 / 27.01 = 145.43 pcf

Field Unit Wt 147.10 pcf

Problem 5

% Air by Unit Wt

27.01 cu.ft. - 1.62 cu.ft = 25.39 cf.

3928 Lbs. / 25.39 cu.ft. = 154.71 pcf

154.71- 147.10 x 100 = 4.9% air

154.71

Problem 5

Weight Volume problem No. 6

Mat’l Sp.Gr. Weight Melted Abs

Cem 3.15 564 196.56 2.87

#67 2.85 1,853 177.84 10.42

Sand 2.61 1187 162.864 7.29

Water 1.00 300 62.4 4.81

Air 6% .06 x 27 1.62

Totals 3904 27.01

Problem 6

Yield 564 = 6.0 bags cem. 27.01 cu.ft. = 4.50 cu.ft.

94 6.0 bags

CF 27/4.50 = 6 x 94 = 564 Lbs

W/C Ratio 300 Lbs. Water / 564 Lbs cem. = .532

Calculated Unit Wt 3,904 Lbs. / 27.00 cu.ft. = 144.54 pcf

Field Unit Wt 141.50 pcf

Problem 6

% Air by Unit Wt

27.01 cu.ft. - 1.62 cu.ft. = 25.39 cu.ft.

3,904 Lbs. / 25.39 cu.ft. = 153.76 pcf

153.76 - 141.50 x 100 = 8.0 % Air

153.76

Problem 6

Weight Volume Problem 7

Mat’l Sp.Gr. Weight Melted Abs

Cem 3.15 639 196.56 3.25

#67 2.71 1782 169.10 10.54

Sand 2.63 1200 164.112 7.31

Water 1.00 267 62.4 4.28

Air 6% .06 x27 1.62

Totals 3,888 27.00

Problem 7

Yield 639 = 6.80 bags cem 27.00 cu.ft = 3.97 cu.ft.

94 6.80 bags

C/F 27/3.97 = 6.80 x 94 = 639 Lbs

W/C Ratio 267 Lbs water / 639 Lbs cem. = .418

Calculated Unit Wt. 3,888 Lbs. / 27.00 cu.ft. = 144.00 pcf

Field Unit Wt 144.62 pcf

Problem 7

% Air by Unit Wt

27.00 cu.ft. - 1.62 cu.ft. = 25.38 cu.ft.

3,888 Lbs. / 25.38 Lbs = 153.19 pcf.

153.19 - 144.62 x 100 = 5.6 % Air

153.19

Problem 7

Weight Volume Problem No. 8

Mat’l Sp.Gr. Weight Melted Abs

Cem 3.15 564 196.56 2.87

#67 2.78 1,948 173.472 11.23

Sand 2.64 1100 164.736 6.68

Water 1.00 287 62.4 4.60

Air 6% .06 x27 1.62

Totals 3,899 27.00

Problem 8

Yield 564 = 6.0 bags cem. 27.00 cu.ft = 4.50 cu.ft.

94 6.0 bags

C/F 27 / 4.50 = 6.0 x 94 = 564 Lbs

W/C Ratio 287 Lbs water / 564 lbs cem. = .509

Calculated Unit Wt. 3,899 Lbs / 27.00 cu.ft. = 144.41 pcf

Field Unit Wt. 144.20 pcf

Problem 8

% Air by Unit Wt

27.00 cu.ft. - 1.62 cu.ft. = 25.38 cu.ft.

3,899 Lbs / 25.38 cu.ft. = 153.62 pcf

153.62 - 144.20 x 100 = 6.1%

153.62

Problem 8

TERMS I SHOULD KNOW

Absolute Volume Cement Factor Consistency Setting Time Specific Gravity Yield STOP AND LET ME COPY THOSE DOWN!

Homework Problem

Mat’l Sp.Gr. Weight Melted Abs

Cem 3.15 1135

Stone 2.70 4013

Sand 2.60 2420

Water 1.00 77gals

Air none

Totals

Weight Vol Homework

Mat’l Sp.Gr. Weight Melted Abs

Cem 3.15 1135 196.56 5.77

Stone 2.70 4013 168.48 23.82

Sand 2.60 2420 162.24 14.92

Water 1.00 77gals 7.5 10.27

Air none

Totals 8209(641 water) 54.78

Wt vol Homework Ans

•Yield 1135 = 12.07 bags 54.78 = 4.54 cu.ft./ bag

94 12.07 bags

•CF 27/4.54 = 5.95 x 94 = 559 Lbs

•W/C 641 Lbs water / 1135 Lbs cem = .565

•Calculated Unit Wt 8,209 Lbs / 54.78 cu.ft. = 149.85 pcf

•Field Unit Wt 147.95 pcf

% Air by Unit Wt

No Air therefore Theoretical = Calculated

149.85 - 147.95 x 100 = 1.3%

149.85

Quiz

Mat’l Sp.Gr. Weight Melted Abs

Cem 3.15 715

Stone 2.74 1883

Sand 2.83 1083

Water 1.00 34.5gal

Air 6%

Totals

Weight Vol Quiz

Mat’l Sp.Gr. Weight Melted Abs

Cem 3.15 715 196.56 3.64

Stone 2.74 1883 164.74 11.01

Sand 2.83 1083 176.59 6.13

Water 1.00 34.5gal 7.5 4.60

Air 6% .06 x 27 1.62

Totals 3968(287 water) 27.00

Weight vol Quiz Answer

Yield 715 = 7.6 bags 27.00 = 3.55 cu.ft. / bag

94 7.6 bags

CF 27/3.55 = 7.61 X 94 = 715 pounds

W/C 287 / 715 = .401

Calculated Unit Wt 3968 / 27 = 146.96 pcf

Field Unit Weight 147.52 pcf

% Air by Unit Wt

27.00 - 1.62 = 25.38 cu.ft.

3968 / 25.38 = 156.34 pcf

156.34 - 147.52 = 5.6% Air

156.34

Bonus Questions

How many cubic feet are in one gallon? 7.5 gal / cu.ft. 1 / 7.5 = .133 cf / gal How many cubic feet are in 94 Lbs cement? 94 / (3.15 x 62.4) = .478 = .48 How many cubic feet of air are in one cubic

yard of concrete with one percent air? 1/ 100 x 27 = .27 cu.ft. How many gallons are in one cubic foot? 7.5 gallons / cu.ft. How many Lbs of water are in one cu.ft.? 62.4