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: How much cement is needed?
W/C = 0.500, and Water = 250 pounds
How much cement is needed?
250 / = 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:
= 567 pounds
Convert Metered Water Into Pounds:
27.5 X 8.33 = 229 pounds
Add free water = 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
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 = = (2.80 X 62.4)
0.533 X 100 = 55%
% Voids = = 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:
, = 27,245
Divide by unit weight:
27, = 7.1 cu. yd.
( X 27)

16. PROBLEM SOLUTION 2. Water / Cement = Ratio 1866 / 4060 = 0.460
3. (A) % Solid
= 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
= 5.4% 5.4 / 100 = 0.054
0.054 X 1102 = 59.5 pounds
= 1162 pounds (batch weight)
Dry Sand:
0.5 / 100 = 0.005
0.005 X 1102 = 5.5 pounds
= 1096 pounds

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