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## Mix Design Concrete School

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### Concrete School Weight - Volume ... 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 ... – PowerPoint PPT presentation

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Title: Mix Design Concrete School

1
Mix DesignConcrete School
• Weight - Volume Relationships

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)

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

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.

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.

6
Terms
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

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)

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

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

10
Method ASpecific 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)

11
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12
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

13
Vb
Va
Wa
14
Method B
• The volume of the material is equal to the volume
of the water displaced
• Formula Wa / (Vb - Va)

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.

16
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

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

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 34.5 gals 4.61 Cu.Ft.
• 7.5

19
Determining Absolute vol pg 58 ex.
20
Absolute volume
• Remember the weight of one cubic foot of material
melted is determined by multiplying the
materials specific gravity by 62.4.
• In the case of cement the melted weight will be
196.56 (3.15 x 62.4).

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.

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

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
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

25
Field Unit Weight
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

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

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.

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 matl in mix 150.16 lbs/cu.ft
• 25.38 cu.ft. in mix w/o air
(theoretical
• unit weight)

29
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

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.

31
Nominal 10 cu.yd. batch
pg
32
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

33
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.

34
Weight Volume Problem No.1
35
Problem 1
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 / 27.01 145.06 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
37
Weight Volume Problem No. 2
38
Problem 2
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 / 27.04 146.71 pcf
Field UW 147.95 pcf
Theor UW 27.04-1.6225.42 3967/25.42 156.06 pcf
Air by UW 156.06-147.95 x100 156.06 5.2
40
Weight Volume Problem No. 3
41
Problem 3
42
• 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
43
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
44
Weight Volume Problem No. 4
45
Problem 4
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 288 Lbs. Water / 588 Lbs.cem.
.490
• Calculated Unit Wt 4060 / 27.05
150.09 pcf
• Field Unit Weight 149.20 pcf

Problem 4
47
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
48
Weight Volume Problem No. 5
49
Problem 5
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 275 Lbs water
/ 677 Lbs. Cem. .406 Calculated Unit Wt
3928 / 27.01 145.43 pcf Field Unit Wt
147.10 pcf
Problem 5
51
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
52
Weight Volume problem No. 6
53
Problem 6
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 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
55
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
56
Weight Volume Problem 7
57
Problem 7
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 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
59
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
60
Weight Volume Problem No. 8
61
Problem 8
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 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
63
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
64
TERMS I SHOULD KNOW
• Absolute Volume
• Cement Factor
• Consistency
• Setting Time
• Specific Gravity
• Yield
• STOP AND LET ME COPY THOSE DOWN!

65
QUESTIONS?
66
Homework Problem
67
Weight Vol Homework
68
Wt vol Homework Ans
69
• 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

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

71
Quiz
72
Weight Vol Quiz
73
74
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
75
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

76
Bonus Questions
77
• 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

78
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