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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.

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Presentation on theme: "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."— Presentation transcript:

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2 Water - Cement Ratio

3 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.

4 W/C Ratio Cont. §Example: W/C = 0.500, and Water = 250 pounds How much cement is needed? 250 / = 500 pounds of ``````````````````````cement

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

6 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 l Determine the design w/c ratio and the batched w/c ratio

7 SOLUTION: Design W/C Add Cement And Fly Ash: = 567 pounds Convert Design Water Into Pounds: 33.5 X 8.33 = 279 pounds Plug Into Formula W/C = Ratio: 279 / 567 = (carry answer to three places after decimal)

8 Batched W/C Ratio Add Cement And Fly Ash: = 567 pounds Convert Metered Water Into Pounds: 27.5 X 8.33 = 229 pounds Add free water = 279 Lbs 279 / 567 = 0.492

9 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.

10 QUESTIONS

11 % 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.

12 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

13 Example: Dry Rodded Unit Weight:96.6 pcf Specific Gravity Of Agg.:2.80 % Solids = 96.6 = (2.80 X 62.4) X 100 = 55% % Voids = = 45%

14 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!!

15 Pass Out Day 1 Mix Design Problems

16 PROBLEM SOLUTION 1.Water: = 224 gals 224 X 8.33 = 1866 pounds §Add all material: , = 27,245 § Divide by unit weight: 27,245 = 7.1 cu. yd. ( X 27)

17 PROBLEM SOLUTION 2. Water / Cement = Ratio 1866 / 4060 = (A) % Solid 88.6 = X 100 = 51% (2.79 X 62.4) (B) % Void = 100 – 51 = 49%

18 PROBLEM SOLUTION 4. Wet Sand: = 5.4%5.4 / 100 = X 1102 = 59.5 pounds = 1162 pounds (batch weight) Dry Sand: 0.5 / 100 = X 1102 = 5.5 pounds = 1096 pounds

19 PROBLEM SOLUTION 5. SSD sand weight ? 1720 / = 1620 pounds SSD sand

20 PROBLEM SOLUTION 6.Gallons of Water from Wet Sand l 1720 – 1620 = 100 pounds / 8.33 = 12.0 gallons

21 HOMEWORK PROBLEM

22 QUESTIONS


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