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.
 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 pounds
How much cement is needed?
250 / 0.500 = 500 pounds of
`````````````````````` cement
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
Water/Cementitious Problem
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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
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)
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
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.
QUESTIONS
% 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.
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:
Specific Gravity Of Agg.:
% Solids =
% Voids =
96.6 pcf
2.80
96.6
= 0.553
(2.80 X 62.4)
0.533 X 100 = 55%
100 - 55 = 45%
Terms I Should Know….
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Abrasion Resistance of an Aggregate
Durability
Hydration
Ph
Saturated Surface Dry
Set Retarder
Unit Weight
Water / Cement Ratio
THAT IS ENOUGH FOR A MONDAY!!
Pass Out Day 1 Mix Design
Problems
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)
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%
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
PROBLEM SOLUTION
5. SSD sand weight ?
1720 / 1.062 = 1620 pounds SSD sand
PROBLEM SOLUTION
6. Gallons of Water from Wet Sand
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1720 – 1620 = 100 pounds / 8.33 = 12.0
gallons
HOMEWORK PROBLEM
QUESTIONS