Mix Design Concrete School Weight - Volume Relationships 1

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) 2

Conversion Factors One bag of cement equals 0.48 cu ft (absolute volume) Four bags of cement equals one barrel. 3

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. 4

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. 5

Terms Remember One cu ft of water weighs 62.4 pounds Pg 53 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 6

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) 7

Terms Absolute Volume of Water: = 62.4 lbs = 1 cu.ft 1 x 62.4 pcf Pg 54 Absolute Volume of Water: = 62.4 lbs = 1 cu.ft 1 x 62.4 pcf 8

HOW TO CALCULATE THE SPECIFIC GRAVITY OF A MATERIAL Weight - Volume HOW TO CALCULATE THE SPECIFIC GRAVITY OF A MATERIAL 9

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) 10

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Method B The weight of the material in air is Wa 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 12

Vb Va Wa 13

Method B The volume of the material is equal to the volume of the water displaced Formula: Wa / (Vb - Va) 14

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. 15

Absolute Volume of an Aggregate Weight Volume Pg 57 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 16

Weight Volume Problem Find the absolute volume of 288 lbs of water 288 1 x 62.4 = 4.62 cu.ft. 17

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 18

Determining Absolute vol pg 58 ex. 19

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). 20

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. 21

Absolute Volume Yield 588 = 6.26 bags cem. 27.00 cuft = 4.31cuft Pg 59 Yield 588 = 6.26 bags cem. 27.00 cuft = 4.31cuft 94 6.26 bags cem 22

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

Calculated Unit weight 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 24

Field Unit Weight (Wt. Concrete + Wt Bucket) - Wt Bucket Pg 59 (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 25

Field Unit Weight 94.8 - 23.2 = 140.39 lbs/cuft .51 (Actual Unit Weight) 26

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. 27

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) 28

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 29

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. 30

Nominal 10 cu.yd. batch pg 31

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 32

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. 33

Weight Volume Problem No.1 34

Problem 1 35

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% 36

Weight Volume Problem No. 2 37

Problem 2 38

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% 39

Weight Volume Problem No. 3 40

Problem 3 41

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 42

% 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 43

Weight Volume Problem No. 4 44

Problem 4 45

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 46

% 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 47

Weight Volume Problem No. 5 48

Problem 5 49

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 50

% 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 51

Weight Volume problem No. 6 52

Problem 6 53

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 54

% 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 55

Weight Volume Problem 7 56

Problem 7 57

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 58

% 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 59

Weight Volume Problem No. 8 60

Problem 8 61

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 62

% 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 63

TERMS I SHOULD KNOW Absolute Volume Cement Factor Consistency Setting Time Specific Gravity Yield STOP AND LET ME COPY THOSE DOWN! 64

QUESTIONS? 65

Homework Problem 66

Weight Vol Homework 67

Wt vol Homework Ans 68

Calculated Unit Wt 8,209 Lbs / 54.78 cu.ft. = 149.85 pcf 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 69

% Air by Unit Wt No Air therefore Theoretical = Calculated 149.85 - 147.95 x 100 = 1.3% 149.85 70

Quiz 71

Weight Vol Quiz 72

Weight vol Quiz Answer 73

Calculated Unit Wt 3968 / 27 = 146.96 pcf 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 74

% Air by Unit Wt 27.00 - 1.62 = 25.38 cu.ft. 3968 / 25.38 = 156.34 pcf 75

Bonus Questions 76

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 77

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