The ingredient weights for making 1 yd (cyd) of concrete by assuming aggregates in SSD state are given below. The volume of air is 2% of the total concrete volume. Coarse Aggregates: 1,575 lb; Fine Aggregates: 1,100 lb; Cement: 700 lb; Water: 315 lb Given, the BSGssp of both coarse and fine aggregates is 2.4, specific gravity of cement is 3.15 and the density of water is 62.4 lb/ft?. 1 yd = 3 ft. Part 1: After performing a trial batch, it is determined that we need to increase the mix water content from 315 lb to 328.5 lb to correct the slump. For the same ingredients, recalculate all the ingredient weights in SSD state to achieve the concrete with the same compressive strength. Assume the same volume of air. If you state in your solution that the weight of a particular ingredient remains the same before and after the mix redesign, clearly state the reason behind your assumption for full credit. Note: You do NOT need any mix design table for solving this problem. Part 2: If the moisture content of coarse aggregates is -1% (note the minus sign) and the moisture content of fine aggregates is -2% (note the minus sign), determine the batch weights (also called stockpile or stock weights) of all ingredients for the redesigned concrete of Part 1. Part 3: Determine the expected unit weight and yield of the redesigned concrete of Part 1.

Explanation:

Ans) Given batch weight of each component :

Cement = 700 lb

Water = 315 lb

Coarse aggregate = 1575 lb

Fine aggregate = 1100 lb

Part 1) Amount of water = 328.5 lb

Amount of water is needed to be increased if the aggregates has absorption capacity, To maintain constant water cement ratio, the mixing water is increased because some of the water is absorbed by aggregates.

Amount of water absorbed = 328.5 lb - 315 lb = 13.5 lb

Total amount of aggregates = 1575 + 1100 = 2675 lb

=> % Absorption capacity = 13.5 x 100 / 2675 = 0.5 %

Hence, new amount of Coarse aggregate = (1 - 0.005) x 1575 lb = 1567.125 lb

New amount of fine aggregate = (1 - 0.005) x 1100 = 1094.5 lb

Since, water cement ratio is maintained constant , amount of cement remains unchanged

=> Volume of water = 328.5 / 62.4 = 5.26 ft3

=> Volume of cement = 700 / (3.15 x 62.4) = 3.56 ft3

=> Volume of coarse aggregate = 1567.125 / (2.4 x 62.4) = 10.46 ft3

=> Volume of fine aggregate = 1100 / (2.4 x 62.4) = 7.34 ft3

Volume of air = 2% = 0.02 x 27 = 0.54 ft3

Total concrete volume = 5.26 + 3.56 + 10.46 + 7.34 + 0.54 \approx 27 ft3 = 1 yd3

Hence, calculated amount of each component is correct

Part 2) We know, minus sign indicated that the aggregate will absorb some moisture from concrete, hence mixing water amount needed to be corrected .

=> Amount of water absorbed by coarse aggregate = 0.01 x 1567.125 lb = 15.67 lb

=> Amount of water absorbed by fine aggregate = 0.02 x 1094.50 lb = 21.89 lb

Total amount of water absorbed = 15.67 + 21.89 = 37.56 lb

To maintain same water cement ratio, amount of mixing water is needed to be increased

=> Corrected amount of mixing water = 328.5 lb + 37.56 lb = 366 lb

=> Corrected amount of coarse aggregate = (1 - 0.01) x 1567.125 = 1551.45 lb

=> Corrected amount of fine aggregate = (1 - 0.02) x 1094.5 = 1072.6 lb

Part 3) We know,

Unit weight = Sum of weight of each material / Total volume

=> Sum of weight = 366 + 700 + 1551.45 + 1072.6 = 3690.05 lb

Total volume = 1 yd3 or 27 ft3

=> Expected Unit Weight = 3690.05 lb / 27 ft3 = 136.67 lb/ft3

Also, Concrete Yield = Weight of all components / Unit weight of concrete

=> Yield = 3690.05 / 136.67 = 27 ft3 or 1 yd3

the clapeyron equation (also called the clausius-clapeyron equation) relates the slope of a reaction line on a phase diagram to fundamental thermodynamic properties.

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