Water - Cement Ratio

Water/Cement RatioThe 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 poundsHow much cement is needed?250 / 0.500 = 500 pounds of ``````````````````````cement

W/C Ratio Cont.Example:W/C = 0.500, and Cement = 600 poundsHow much water is needed?0.500 X 600 = 300 pounds of water300 pounds / 8.33 = 36.0 gallons

Water/Cementitious ProblemCement Used in Mix 436 poundsFly Ash in Mix 131 poundsMaximum Water 36.0 gallonsTotal Water 33.5 gallonsMetered Water 27.5 gallonsFree Water in aggregates 50 poundsDetermine the design w/c ratio and the batched w/c ratio

SOLUTION: Design W/CAdd Cement And Fly Ash:436 + 131 = 567 poundsConvert Design Water Into Pounds:33.5 X 8.33 = 279 poundsPlug 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 poundsConvert Metered Water Into Pounds:27.5 X 8.33 = 229 poundsAdd free water 229 + 50 = 279 Lbs279 / 567 = 0.492

W/C Ratio with IceDetermine 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

% Solids And VoidsIn 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 pcfSpecific 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 AggregateDurabilityHydrationPhSaturated Surface DrySet RetarderUnit WeightWater / Cement RatioTHAT IS ENOUGH FOR A MONDAY!!

Pass Out Day 1 Mix Design Problems

PROBLEM SOLUTION1.Water: 209 + 15 = 224 gals224 X 8.33 = 1866 poundsAdd all material:4060+7733 +13,586 +1866 = 27,245 Divide by unit weight: 27,245 = 7.1 cu. yd.(142.10 X 27)

PROBLEM SOLUTION2. Water / Cement = Ratio1866 / 4060 = 0.460

3. (A) % Solid 88.6 = 0.508 X 100 = 51%(2.79 X 62.4) (B) % Void = 100 51 = 49%

PROBLEM SOLUTION4. Wet Sand:5.9 - 0.5 = 5.4%5.4 / 100 = 0.0540.054 X 1102 = 59.5 pounds1102 + 60 = 1162 pounds (batch weight)Dry Sand:0.5 / 100 = 0.0050.005 X 1102 = 5.5 pounds 1102 - 6.0 = 1096 pounds

PROBLEM SOLUTION5. SSD sand weight ?

1720 / 1.062 = 1620 pounds SSD sand

PROBLEM SOLUTIONGallons of Water from Wet Sand

1720 1620 = 100 pounds / 8.33 = 12.0 gallons

HOMEWORK PROBLEM

QUESTIONS

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