Mix Design Concrete 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

Pg 53  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.

Terms

Pg 53  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

Terms

Pg 54  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)

Terms

Absolute Volume of Water: Pg 54 = 62.4 lbs 1 x 62.4 pcf = 1 cu.ft

Weight - Volume HOW TO CALCULATE THE SPECIFIC GRAVITY OF A MATERIAL

Method A Specific Gravity

Pg 55  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)

Method B

Pg 56  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

Wa Vb Va

Method B

 The volume of the material is equal to the volume of the water displaced  Formula: Wa / (Vb - Va)

Weight- Volume

Pg 57  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

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 7.5

= 4.61 Cu.Ft.

Determining Absolute vol pg 58 ex.

Material Sp.Gr.

Cement 3.15

#57 Sand Water Air Total 2.65

2.63

1.00

6% Weight Melted 588 196.56

1808 1129 286 165.36

164.11

62.4

Abs vol 2.99

10.93

6.88

4.58

3811 lbs .06 x27 1.62

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

Pg 58 

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 59

Absolute Volume

 Yield 588 = 6.26 bags cem. 27.00 cuft = 4.31cuft

94 6.26 bags cem

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 27.00 cu ft = 141.15 lbs/cuft

Pg 59

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

Field Unit Weight

 94.8 - 23.2 .51

= 140.39 lbs/cuft (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

Pg 61  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.

Nominal 10 cu.yd. batch

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

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 #67’s 2.82 Sand 2.62 564 1966 1100 196.56 2.87 175.968 11.17 163.488 6.73 Water 1.00 288 62.4 4.62 Air 6% .06 x27 1.62 3918 27.01 Problem 1

Yield

C/F W/C Ratio

564/94 = 6/0 bags

27/4.50 = 6.0

288 / 564 3918 / 27.01

27.01/ 6.0 =

6.0 x 94 =

4.50 cf

564 Lbs .511

Cal UW Field UW 145.06 pcf 147.95 pcf Theor. UW 27.01-1.62 = 25.39

% Air by UW 154.31 – 147.95 x 100 154.31

3918/25.39 = 154.31 pcf 4.1%

Weight Volume Problem No. 2

Mat’l Cem #67 Sand Water Air Sp.Gr.

3.15

2.88

2.62

1.00

6% Weight 714 1940 1015 298 Melted 196.56

179.712

163.488

62.4

.06 x 27 Abs 3.63

10.80

6.21

4.78

1.62

3967 27.04

Problem 2

Yield

C/F

714 / 94 = 7.60

27/3.56 = 7.6

27.04 / 7.60 = 3.56 cf

7.6 x 94 = 714 .417

W/C Ratio Cal UW Field UW 298 / 714 = 3967 / 27.04 = 146.71 pcf 147.95 pcf Theor UW 27.04-1.62=25.42

% Air by UW 156.06-147.95 x100 156.06

3967/25.42= 156.06 pcf 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 Water 1.00 Air 6% 1197 300 162.24 7.38 62.4 4.81 .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 = •

W/C Ratio

544 300 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

#67

2.88

Sand Water

1.00

Air

2.62 6% 588 1956 1228 288 196.56 3.00 189.696 10.31 163.488 7.51 62.4 4.62 .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 = •

W/C Ratio

588 Lbs 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

#67

2.79

Sand Water

1.00

Air

2.60 6% 677 1,895 174.096 10.88 1081 275 196.56 3.44 162.24 6.66 62.4 4.41 .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

Cem

Sp.Gr. Weight 3.15 564 Melted 196.56 Abs 2.87

#67

2.85 1,853 177.84 10.42

Sand Water

2.61 1.00 1187 300 162.864 62.4 7.29 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 Sand

2.71 2.63 1782 1200 169.10 10.54 164.112 7.31

Water

1.00

Air

6% 267 62.4 4.28 .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 =

W/C Ratio

639 Lbs 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

153.19

x 100 = 5.6 % Air 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

Sand Water

1.00

Air

2.64 6% 1,948 1100 287 173.472 11.23 164.736 6.68 62.4 4.60 .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 =

W/C Ratio

564 Lbs 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

153.62

x 100 = 6.1% 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

Weight Vol Homework Mat’l Sp.Gr. Weight Melted Abs Cem 3.15

Stone 2.70

1135 4013 Sand 2.60

Water 1.00

Air none 2420 77gals Totals

Wt vol Homework Ans Mat’l Sp.Gr. Weight Melted Abs Cem 3.15

1135 196.56

5.77

Stone 2.70

Sand 2.60

Water 1.00

Air none 4013 2420 168.48

162.24

77gals 7.5

23.82

14.92

10.27

Totals 8209

(641 water)

54.78

•

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

149.85

x 100 = 1.3%

Quiz

Weight Vol Quiz Mat’l Sp.Gr. Weight Melted

Cem

3.15

Stone

2.74

Sand Water

1.00

Air

2.83

6% 715 1883 1083 34.5gal

Totals Abs

Weight vol Quiz Answer Mat’l Sp.Gr. Weight Melted Abs

Cem

3.15

Stone Sand Air

2.74

2.83

Water

1.00

6% 715 1883 1083 196.56

164.74

176.59

34.5gal 7.5

.06 x 27 3.64

11.01

6.13

4.60

1.62

Totals 3968

(287 water)

27.00

Yield

715 94 = 7.6 bags 27.00

7.6 bags = 3.55 cu.ft. / bag

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

156.34

= 5.6% Air

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