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.

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Water - Cement Ratio

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Water cement ratio

Water - Cement Ratio


Water cement ratio1

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

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 cont1

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

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

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

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

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.


Questions

QUESTIONS


Solids and voids

% 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

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

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

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

Pass Out Day 1 Mix Design Problems


Problem solution

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 solution1

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 solution2

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 solution3

PROBLEM SOLUTION

5. SSD sand weight ?

1720 / 1.062 = 1620 pounds SSD sand


Problem solution4

PROBLEM SOLUTION

  • Gallons of Water from Wet Sand

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


Homework problem

HOMEWORK PROBLEM


Questions1

QUESTIONS


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