2 Water Cement
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Transcript of 2 Water Cement
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Water - Cement RatioWater - Cement Ratio
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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.
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W/C Ratio Cont.
Example:W/C = 0.500, and Water = 250 poundsHow much cement is needed?
250 / 0.500 = 500 pounds of `````````````````````` cement
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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
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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 pounds
Determine the design w/c ratio and the batched w/c ratio
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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)
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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 Lbs279 / 567 = 0.492
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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.
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QUESTIONS
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% 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.
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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
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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%
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Terms I Should Know….Abrasion Resistance of an AggregateDurabilityHydrationPhSaturated Surface DrySet RetarderUnit WeightWater / Cement RatioTHAT IS ENOUGH FOR A MONDAY!!
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Pass Out Day 1 Mix Design Problems
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PROBLEM SOLUTION1.Water: 209 + 15 = 224 gals
224 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)
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PROBLEM SOLUTION2. 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%
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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
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PROBLEM SOLUTION5. SSD sand weight ?
1720 / 1.062 = 1620 pounds SSD sand
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PROBLEM SOLUTION
6. Gallons of Water from Wet Sand
1720 – 1620 = 100 pounds / 8.33 = 12.0 gallons
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HOMEWORK PROBLEM
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QUESTIONS