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Water - Cement Ratio

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.

naomi-ramirez
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Water - Cement Ratio

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  1. Water - Cement Ratio

  2. 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.

  3. 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

  4. 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

  5. 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

  6. 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)

  7. 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

  8. 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.

  9. QUESTIONS

  10. % 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.

  11. 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

  12. 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%

  13. 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!!

  14. Pass Out Day 1 Mix Design Problems

  15. 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)

  16. 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%

  17. 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

  18. PROBLEM SOLUTION 5. SSD sand weight ? 1720 / 1.062 = 1620 pounds SSD sand

  19. PROBLEM SOLUTION • Gallons of Water from Wet Sand • 1720 – 1620 = 100 pounds / 8.33 = 12.0 gallons

  20. HOMEWORK PROBLEM

  21. QUESTIONS

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