Question

Chapter 12, Problem 3 We want to produce 100 ft of concrete with a ratio of 1:1.9:3.8 (by weight) of cement, sand, and gravel. What are the required amounts of the components if 5.5 gal of water per sack of cement is to be used? Assume the free moisture contents of the sand and gravel are 3 and 0 percent, respe gravities of the cement, sand, and gravel are 3.15, 2.65, and 2.65, respectively. (One sack of cement weighs 94 lb and 1 ft water = 7.48 gal.) Give answers for the cement in sacks, the sand and gravel in pounds, and the water in gallons.

Answer 1

Hi,

Solution as follows:

Each component volume can be calculated by absolute volume method.

w Absolute volume V=GX 62.43

Where,

G = Specific Gravity of the material.

V = Volume in cubic feet

W = Weight in lbs

Assume the weight of cement as W.

STEP 1: Absolute volume of cement:

Given Data:

· Specific gravity of cement = 3.15

WW Absolute volume of cement V=GX 62.43 = 3,15 x 62.43 = U.

STEP 2: Absolute volume of Water:

Water to cement ratio = 5.5 gallon per sack

1 ft3 = 7.48 gallon, 1 sack = 94 lbs, weight of water = 62.43 lbs/ft3

5.5 62.43 Water cement ratio = 7.4894=

Weight of water = 0.488 W

0.488 x W Absolute volume of water V = 1 x 62.43 = 0.007817W

STEP 3: Absolute volume of Sand:

Given Data:

· Sand to cement ratio = 1.9

· Specific gravity of Sand = 2.65

Weight of Sand = 1.9 W

1.9 x W Absolute volume of Sand V = 765 x 62.43 = 0.0

Step 4: Absolute volume of Gravel:

Given Data:

· Gravel to cement ratio = 3.8

· Specific gravity of Gravel = 2.65

Weight of Gravel = 3.8 W

3.8 x W Absolute volume of Gravel V = 26 coa = 0.022969W

STEP 5: Required Weight of cement:

The sum of above volumes will be the volume of materials required.

Hence,

W(0.005085 +0.007817 +0.011485 + 0.022969) = 100

:: 0.047356W = 100

W 2111.67 = 2111.67 lbs = -= 22.5 sack 94

Step 6: Adjustment for free water in Sand:

Weight of Sand = 1.9 W = 1.9 x 2111.67 = 4012.17 lbs

Free Moisture content of Sand = 3.00%

Weight of free water in Sand = 4012.17 x 3% = 120.37 lbs

Total weight of Sand = 4012.17 + 120.37 = 4132.54 lbs

Step 7: Weight of gravel:

Weight of Gravel = 3.8 W = 3.8 x 2111.67 = 8024.35 lbs

STEP 8: Final weight of water required:

Weight of water = 0.488W = 0.488 x 2111.67 = 1030.49 lbs

Less Free water from Sand = 120.37 lbs

Net weight of water required =1030.49 – 120.37 = 910.13 lbs

1 ft3 = 7.48 gallon, Unit weight of water = 62.43 lbs/ft3

Volume of water in gallons =- 910.13 X 7.48 = 109 gallons 62.43

Final Results	Required Qty
Cement	22.5 sack
Water	109 gallons
Gravel	8024.35 lbs
Sand	4132.54 lbs

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