# Water - Cement Ratio - PowerPoint PPT Presentation

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Water - Cement Ratio. Water/Cement Ratio. The number of pounds of water per pound of cement. A low ratio means higher strengths, a high ratio means lower strengths.

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Water - Cement Ratio

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## Water - Cement Ratio

### Water/Cement Ratio

• The number of pounds of water per pound of cement.

• A low ratio means higher strengths, a high ratio means lower strengths.

• For NCDOT, the ratio depends on the class of concrete, whether an air agent is used or not, and the shape of the stone - rounded or angular.

### W/C Ratio Cont.

• Example:

W/C = 0.500, and Water = 250 pounds

How much cement is needed?

250 / 0.500 = 500 pounds of ``````````````````````cement

### W/C Ratio Cont.

• Example:

W/C = 0.500, and Cement = 600 pounds

How much water is needed?

0.500 X 600 = 300 pounds of water

300 pounds / 8.33 = 36.0 gallons

### Water/Cementitious Problem

• Cement Used in Mix – 436 pounds

• Fly Ash in Mix – 131 pounds

• Maximum Water – 36.0 gallons

• Total Water – 33.5 gallons

• Metered Water – 27.5 gallons

• Free Water in aggregates – 50 pounds

• Determine the design w/c ratio and the batched w/c ratio

### SOLUTION: Design W/C

Add Cement And Fly Ash:

436 + 131 = 567 pounds

Convert Design Water Into Pounds:

33.5 X 8.33 = 279 pounds

Plug Into Formula W/C = Ratio:

279 / 567 = 0.492

(carry answer to three places after decimal)

### Batched W/C Ratio

Add Cement And Fly Ash:

436 + 131 = 567 pounds

Convert Metered Water Into Pounds:

27.5 X 8.33 = 229 pounds

Add free water 229 + 50 = 279 Lbs

279 / 567 = 0.492

### W/C Ratio with Ice

• Determine the W/C ratio if 68 pounds of ice is used to lower the temperature of the concrete.

• The W/C ratio remains the same because the quantity of total water does not change.

### % Solids And Voids

• In determining mix designs, you must use an aggregate dry rodded unit weight.

• This weight is determined at the lab.

• In the procedure for determining this weight, only the coarse aggregate is used.

• Therefore, there is a % of solids and a % of voids in the container.

### Formula : % Solids & Voids

% Solids:

Dry Rodded Unit Weight

(Spec. Gravity) X (62.4)

The Answer Is Then Multiplied Times 100

To get % Voids:

Subtract % Solids from 100

### Example:

Dry Rodded Unit Weight:96.6 pcf

Specific Gravity Of Agg.:2.80

% Solids = 96.6 = 0.553 (2.80 X 62.4)

0.533 X 100 = 55%

% Voids = 100 - 55 = 45%

### Terms I Should Know….

• Abrasion Resistance of an Aggregate

• Durability

• Hydration

• Ph

• Saturated Surface Dry

• Set Retarder

• Unit Weight

• Water / Cement Ratio

• THAT IS ENOUGH FOR A MONDAY!!

## Pass Out Day 1 Mix Design Problems

### PROBLEM SOLUTION

1.Water: 209 + 15 = 224 gals

224 X 8.33 = 1866 pounds

• Add all material:

4060+7733 +13,586 +1866 = 27,245

• Divide by unit weight:

27,245 = 7.1 cu. yd.

(142.10 X 27)

### PROBLEM SOLUTION

2. Water / Cement = Ratio

1866 / 4060 = 0.460

3. (A) % Solid

88.6 = 0.508 X 100 = 51%

(2.79 X 62.4)

(B) % Void = 100 – 51 = 49%

### PROBLEM SOLUTION

4. Wet Sand:

5.9 - 0.5 = 5.4%5.4 / 100 = 0.054

0.054 X 1102 = 59.5 pounds

1102 + 60 = 1162 pounds (batch weight)

Dry Sand:

0.5 / 100 = 0.005

0.005 X 1102 = 5.5 pounds

1102 - 6.0 = 1096 pounds

### PROBLEM SOLUTION

5. SSD sand weight ?

1720 / 1.062 = 1620 pounds SSD sand

### PROBLEM SOLUTION

• Gallons of Water from Wet Sand

• 1720 – 1620 = 100 pounds / 8.33 = 12.0 gallons