1. Mix Design Concrete School
Weight - Volume Relationships 1

2. 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

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

4. 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

5. 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

6. 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

7. 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

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

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

10. 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

11. 11

12. 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

13. Vb Va Wa 13

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

15. 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

16. 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

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

18. 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 = gals = Cu.Ft.
7.5 18

19. Determining Absolute vol pg 58 ex.19

20. 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 (3.15 x 62.4). 20

21. 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

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

23. 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

24. Calculated Unit weight
Absolute Volume
W/C
34.3 gal x 8.33 = 286 lbs water =
588 lbs cem.
Calculated Unit weight
3811 lbs material = lbs/cuft
27.00 cu ft 24

25. 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 lbs
Volume of Bucket cuft
Weight of concrete & bucket 25

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

27. 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

28. 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 = lbs/cu.ft
25.38 cu.ft. in mix w/o air (theoretical
unit weight) 28

29. Percent Air T = Theoretical Unit Weight A = Actual Field Unit Weight
T - A x 100 = % air T
x 100 = 6.5% Air
150.16 29

30. 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

31. Nominal 10 cu.yd. batch pg 31

32. Checking yield Air content by pressure meter = 5.2%
Unit weight of fresh concrete = pcf
yield in cu.ft. = 38,850 lbs = cu.ft.
lb/cu.ft.
yield in cu yd = / = cuyd 32

33. Checking Yield Yield per nominal cu.yd.
cu.ft./batch = 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

34. Weight Volume Problem No.134

35. Problem 1 35

36. 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
pcf
Field UW
pcf
Theor. UW
= 25.39
3918/25.39 =
pcf
% Air by UW
– x 100
154.31
4.1% 36

37. Weight Volume Problem No. 237

38. Problem 2 38

39. 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 / =
pcf
Field UW
pcf
Theor UW
=25.42
3967/25.42=
pcf
% Air by UW
x100
156.06
5.2% 39

40. Weight Volume Problem No. 340

41. Problem 3 41

42. Yield 545 = 5.8 bags cem. 27.00cu.ft = 4.66 cu.ft. 94 5.8 bags
C/F 27/ 4.66 = x 94 = 544
W/C Ratio Lbs water / 545 Lbs.cem = .550
Calculated Unit Wt Lbs / cu.ft. = pcf
Field Unit Wt pcf
Problem 3 42

43. % Air by Unit Wt 27.00 cu.ft. - 1.62 cu.ft. = 25.38 cu.ft.
3895 Lbs / cu.ft = pcf
x = % air
153.47
Problem 3 43

44. Weight Volume Problem No. 444

45. Problem 4 45

46. 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 Lbs. Water / 588 Lbs.cem. =
Calculated Unit Wt / = pcf
Field Unit Weight pcf
Problem 4 46

47. % Air by Unit Wt 27.05 cu.ft. - 1.62 cu.ft. = 25.43 cf.
4060 Lbs / cu.ft. = pcf
x = % air
159.65
Problem 4 47

48. Weight Volume Problem No. 548

49. Problem 5 49

50. 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 Lbs water / 677 Lbs. Cem. = .406
Calculated Unit Wt / = pcf
Field Unit Wt pcf
Problem 5 50

51. % Air by Unit Wt 27.01 cu.ft. - 1.62 cu.ft = 25.39 cf.
3928 Lbs. / cu.ft. = pcf
x 100 = % air
154.71
Problem 5 51

52. Weight Volume problem No. 652

53. Problem 6 53

54. 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 Lbs. Water / 564 Lbs cem. =
Calculated Unit Wt ,904 Lbs. / cu.ft. = pcf
Field Unit Wt pcf
Problem 6 54

55. % Air by Unit Wt 27.01 cu.ft. - 1.62 cu.ft. = 25.39 cu.ft.
3,904 Lbs. / cu.ft. = pcf
x 100 = % Air
153.76
Problem 6 55

56. Weight Volume Problem 7 56

57. Problem 7 57

58. 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 Lbs water / 639 Lbs cem. =
Calculated Unit Wt. 3,888 Lbs. / cu.ft. = pcf
Field Unit Wt pcf
Problem 7 58

59. % Air by Unit Wt 27.00 cu.ft. - 1.62 cu.ft. = 25.38 cu.ft.
3,888 Lbs. / Lbs = pcf.
x = % Air
153.19
Problem 7 59

60. Weight Volume Problem No. 860

61. Problem 8 61

62. 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 Lbs water / 564 lbs cem. =
Calculated Unit Wt. 3,899 Lbs / cu.ft. = pcf
Field Unit Wt pcf
Problem 8 62

63. % Air by Unit Wt 27.00 cu.ft. - 1.62 cu.ft. = 25.38 cu.ft.
3,899 Lbs / cu.ft. = pcf
x = %
153.62
Problem 8 63

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

65. QUESTIONS? 65

66. Homework Problem 66

67. Weight Vol Homework 67

68. Wt vol Homework Ans 68

69. Calculated Unit Wt 8,209 Lbs / 54.78 cu.ft. = 149.85 pcf
Yield = bags = cu.ft./ bag
bags
CF 27/4.54 = 5.95 x 94 = 559 Lbs
W/C Lbs water / Lbs cem =
Calculated Unit Wt 8,209 Lbs / cu.ft. = pcf
Field Unit Wt pcf 69

70. % Air by Unit Wt No Air therefore Theoretical = Calculated
x = %
149.85 70

71. Quiz 71

72. Weight Vol Quiz 72

73. Weight vol Quiz Answer 73

74. Calculated Unit Wt 3968 / 27 = 146.96 pcf
Yield = bags = cu.ft. / bag
bags
CF 27/3.55 = 7.61 X 94 = 715 pounds
W/C 287 / =
Calculated Unit Wt / = pcf
Field Unit Weight pcf 74

75. % Air by Unit Wt 27.00 - 1.62 = 25.38 cu.ft. 3968 / 25.38 = 156.34 pcf75

76. Bonus Questions 76

77. How many cubic feet are in one gallon?
7.5 gal / cu.ft / 7.5 = .133 cf / gal
How many cubic feet are in 94 Lbs cement?
94 / (3.15 x 62.4) = = .48
How many cubic feet of air are in one cubic yard of concrete with one percent air?
1/ 100 x 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

78. 78