Question 5 of 7

Question 5 20 marks — (a) 8, (b) 12

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A wet aggregate sample of known wet and oven-dry weight with a stated absorption, and a sieve analysis of a second sample:

Find. For part (a), the free (surface) water expressed as a percentage; for part (b), the percent passing each sieve, plotted on a 0.45 power gradation chart together with the maximum-density line.

Approach. Part (a) rests on the four aggregate moisture states — oven dry, air dry, saturated surface dry (SSD) and wet — because absorption is defined at the SSD condition on the oven-dry weight, so the absorbed water can be separated from the total water and what remains is free water. Part (b) is a routine sieve computation followed by a plot on the 0.45 power abscissa, on which the Fuller maximum-density gradation becomes a straight line through the origin.

Part (a) — free water in the wet sample

1. Find the total water carried by the sample. Everything lost on oven drying is water, whether it sat inside the aggregate pores or on the particle surfaces: $$W_{total} = W_{wet} - W_{OD} = 300.0 - 280.0 = 20.0\text{ N}$$
2. Find the absorbed water from the definition of absorption. Absorption is the water held in the permeable pores when the aggregate is saturated surface dry, expressed as a percentage of the oven-dry weight: $$W_{abs} = \frac{A_b}{100}\,W_{OD} = 0.020(280.0) = 5.6\text{ N}$$ so the SSD weight of this sample is $280.0 + 5.6 = 285.6$ N. Because the sample weighs 300.0 N, it is wetter than SSD, which confirms that free water is present rather than the aggregate being partially dry.
3. Separate the free (surface) water. Free water is the total water less the water the pores absorb: $$W_{free} = W_{total} - W_{abs} = 20.0 - 5.6 = 14.4\text{ N}$$
4. Express the free water as a percentage on the conventional basis. Aggregate moisture contents are, by convention and by ASTM C566, quoted on the oven-dry weight: $$\text{Free moisture} = \frac{W_{free}}{W_{OD}} \times 100 = \frac{14.4}{280.0} \times 100 = \boxed{5.14\%}$$ For comparison, the total moisture content on the same basis is $20.0/280.0 = 7.14\%$, and the two differ by exactly the 2.0% absorption, which is the arithmetic check on the whole calculation.
5. Give the alternative reading of the question. The question asks for the free water “in the original wet sample”. If that phrase is taken literally as a fraction of the 300.0 N wet weight rather than of the oven-dry weight, $$\frac{14.4}{300.0} \times 100 = 4.80\%$$ Both numbers describe the same 14.4 N of surface water. The 5.14% figure on the oven-dry basis is the one that batching corrections use, so it is the value carried into the results table, with 4.80% quoted alongside it.
6. State why the number matters. In a concrete or asphalt plant the free water is the quantity that must be deducted from the batch water (for concrete) or driven off in the dryer (for asphalt). At 5.14% free moisture, a 1000 kg batch of this aggregate carries about 51 kg of surface water; ignoring it would raise the water-cement ratio enough to cost several megapascals of 28-day strength.

Part (b) — sieve analysis and the 0.45 power chart

1. Total the masses retained. Summing the eleven sieves and the pan, $$W_{total} = 0 + 400 + 900 + 500 + 1000 + 900 + 900 + 700 + 600 + 600 + 200 + 50 = 6750\text{ g}$$ The pan mass must be included; leaving it out inflates every percent passing.
2. Accumulate the retained masses and convert to percent passing. For each sieve the cumulative mass retained on it and on all coarser sieves is subtracted from the total: $$P_i = \frac{W_{total} - \sum_{j \le i} W_{ret,j}}{W_{total}} \times 100$$ Applied to the 4.75 mm sieve, for which the cumulative retained mass is $0 + 400 + 900 + 500 + 1000 = 2800$ g, $$P_{4.75} = \frac{6750 - 2800}{6750} \times 100 = 58.52\%$$ The full set is tabulated in the results below.
3. Identify the maximum and nominal maximum aggregate sizes. All the material passes the 25 mm sieve, so the maximum size is 25 mm. The first sieve to retain more than 10% of the total is the 12.5 mm sieve, which retains $900/6750 = 13.3\%$; one size larger gives a nominal maximum size of 19 mm. These two sizes set where the maximum-density line is drawn and which specification band the mixture would be judged against.
4. Construct the 0.45 power chart. The abscissa is the sieve opening raised to the 0.45 power, on which the Fuller-Thompson maximum-density gradation $P = 100(d/D)^{0.45}$ plots as a straight line from the origin to 100% at the maximum size $D$. With $D = 25$ mm the line passes through, for example, $100(4.75/25)^{0.45} = 47.4\%$ at the 4.75 mm sieve and $100(0.30/25)^{0.45} = 13.7\%$ at the 0.30 mm sieve.
5. Plot the gradation and read what it says about the mixture. Plotted against that line (see the figure), the measured gradation rides consistently above the maximum-density line from 19 mm down to about 0.30 mm — 58.5% against 47.4% at the 4.75 mm sieve, 45.2% against 34.6% at 2.36 mm — which marks a fine-graded mixture carrying more intermediate material than the densest packing would use. Below 0.30 mm the curve then plunges through the line, reaching only 3.70% at 0.15 mm and 0.74% at 0.075 mm against maximum-density values of 10.0% and 7.3%.
6. Draw the engineering conclusion. That deficiency at the fine end is the significant finding. An asphalt concrete normally requires roughly 4% to 8% passing the 0.075 mm sieve to develop the mastic that stiffens the binder and fills the remaining voids; at 0.74% this aggregate has almost no dust. As graded, the mixture would be prone to high air voids, a tender mix during compaction, and poor moisture resistance. The practical remedy is to add mineral filler or a manufactured fine, or to blend in a screening with a substantial minus-0.150 mm fraction, and then to re-run the analysis.