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 cu.ft
1 x 62.4 pcf

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.
Weight Melted Abs vol
Cement 3.15
588 196.56
2.99
#57 2.65
1808 165.36
10.93
Sand 2.63
1129 164.11
6.88
Water 1.00
286
Air
Total
6%
3811 lbs
62.4
4.58
.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 = 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
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
18,080 lbs
12,760 lbs
6006 lbs
5 lbs
Water through plant 230gals x 8.33 = 1916
Water through truck 10gals x 8.33 = 83 Lbs
Total Weight 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. 3.15 564 196.56 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
C/F
W/C Ratio
Cal UW
Field UW
564/94 = 6/0 bags
27/4.50 = 6.0
288 / 564
3918 / 27.01
27.01/ 6.0 = 4.50 cf
6.0 x 94 = 564 Lbs
.511
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 Sp.Gr.
Weight Melted
Cem
#67
Sand
3.15
2.88
2.62
Water 1.00
Air 6%
714
1940
1015
298
196.56
179.712
163.488
62.4
.06 x 27
3967 27.04
Abs
3.63
10.80
6.21
4.78
1.62
Problem 2

Yield
C/F
W/C Ratio
Cal UW
714 / 94 = 7.60
27/3.56 = 7.6
298 / 714 =
3967 / 27.04 =
27.04 / 7.60 =
7.6 x 94 =
3.56 cf
714
.417
146.71 pcf
Field UW 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 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
Air
2.85 1,853 177.84 10.42
Sand 2.61 1187 162.864 7.29
Water 1.00 300 62.4 4.81
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

Weight Vol Homework
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

Wt vol Homework Ans
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

•
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

Weight Vol 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 Answer
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
Air 6% .06 x 27
4.60
1.62
Totals 3968 (287 water) 27.00

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