mix design concrete school weight - volume relationships
TRANSCRIPT
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.
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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.
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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
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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)
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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)
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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
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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.
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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
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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.
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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
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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.
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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.
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%
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%
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
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
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
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
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
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!
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
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
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