Question 1 of 5: Aggregate Requirements and Tension-Test Properties

Question 1: Aggregate Requirements and Tension-Test Properties (20 marks)

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.

Part (a) I — Desirable characteristics of aggregate for Portland cement concrete (5 marks)

Aggregate occupies 60 to 80 per cent of the volume of a normal concrete, so its properties govern the volume stability, the strength ceiling and much of the durability of the finished member. The requirements below are the ones CSA A23.1/A23.2 and ASTM C33 actually specify, and each is desirable for a reason that can be traced to a mechanism in the hardened concrete.

1. Adequate strength, hardness and stiffness. The particles must be stronger and stiffer than the hydrated cement paste that surrounds them, because in a normal-strength concrete the failure path runs through the paste and the interfacial transition zone rather than through the stone. A weak or friable aggregate — a soft shale, a chert, a poorly cemented sandstone — caps the achievable compressive strength no matter how rich the mix, and its low modulus increases the creep and elastic shortening of columns and prestressed members.

2. Soundness and freeze–thaw durability. In the Canadian climate the aggregate must survive repeated freezing of the water held in its pores. Soundness is measured by the magnesium or sodium sulfate test (CSA A23.2-9A, ASTM C88) or by unconfined freeze–thaw (CSA A23.2-24A); the desirable aggregate has low absorption, a small proportion of critically sized fine pores and a low loss in these tests, so that it does not cause D-cracking or popouts at the surface.

3. A well-graded particle size distribution with a suitable nominal maximum size. Continuous grading from the maximum size down to the fines minimises the void content of the aggregate skeleton, and the paste only has to fill those voids plus provide a lubricating film. Less void means less paste, which means less water, less cement, less heat of hydration, less drying shrinkage and lower cost. The nominal maximum size is limited by clear cover, bar spacing and member thickness (CSA A23.1 limits it to one fifth of the narrowest dimension, three quarters of the clear spacing between bars, and one third of a slab thickness).

4. Favourable particle shape and surface texture. Equidimensional, cubical particles pack better and demand less water than flat or elongated ones, which also bridge across reinforcement and break under compaction. A moderately rough texture improves the mechanical keying between paste and stone and therefore the flexural strength, while a completely smooth, glassy surface gives a weak interfacial transition zone. Flat and elongated particles are normally limited to about 15 per cent by mass.

5. Cleanliness and chemical stability. The aggregate must be free of clay coatings, silt, organic matter, soft particles, chlorides and sulfates, all of which interfere with the paste–aggregate bond, increase water demand or attack the reinforcement. Chemical stability also means the aggregate must not be alkali-reactive: reactive siliceous or carbonate rocks combined with a high-alkali cement produce expansive alkali–aggregate reaction, and CSA A23.2-27A prequalification or the use of supplementary cementing materials is required where such aggregates cannot be avoided.

Part (a) II — Desirable characteristics of aggregate for asphalt concrete (5 marks)

In hot-mix asphalt the binder film is thin and viscoelastic, so the aggregate skeleton, not the binder, carries the traffic load through particle-to-particle contact. The emphasis therefore shifts from paste compatibility to internal friction, degradation resistance and the binder–aggregate interface.

1. Toughness and abrasion resistance. The aggregate must survive the roller during construction and millions of load repetitions afterwards without degrading into fines, because breakdown fills the voids, stiffens the gradation and destroys the air-void structure. The Los Angeles abrasion test (ASTM C131) and Micro-Deval (ASTM D6928) are the acceptance tests; a maximum LA loss of about 35 to 45 per cent is typical for surface courses.

2. Angularity, with crushed faces, and a rough surface texture. Shear resistance in a mix comes from aggregate interlock, and interlock comes from angular, rough particles. Specifications require a minimum percentage of fractured faces on the coarse fraction and a minimum uncompacted void content on the fine fraction. Rounded natural gravel and rounded natural sand give mixes that tender under the roller and rut under traffic.

3. Durability and resistance to stripping. Besides soundness against freeze–thaw, the aggregate must have an affinity for asphalt rather than for water. Siliceous, hydrophilic aggregates lose adhesion when moisture reaches the interface, and the mix ravels and potholes; the tensile strength ratio test (AASHTO T 283) screens for it, and hydrated lime or a liquid anti-stripping agent is added where the aggregate fails.

4. A gradation that plots within the design band, with controlled dust. The blend must have the right nominal maximum size for the lift thickness and must produce enough voids in the mineral aggregate to hold the design binder film without flushing. The material passing the 0.075 mm sieve is separately controlled, and the dust-to-effective-binder ratio is normally kept between 0.6 and 1.2 — too much dust stiffens and embrittles the mastic, too little leaves the mix tender.

5. Cleanliness, low absorption and few flat or elongated particles. Clay coatings prevent the binder from wetting the stone. A highly absorptive aggregate soaks up binder that was intended to coat the particles, so the effective binder content and the film thickness fall and the mix becomes brittle and permeable. Flat and elongated particles (limited by ASTM D4791, commonly to 10 per cent at a 5:1 ratio) fracture under compaction and reduce the stability of the skeleton.

Part (b) — Tension-test properties of metals A and B (10 marks)

Given. The tension-test stress–strain curves of two metals, plotted to fracture on axes of stress in ksi (0 to 150) against strain in in/in (0 to 0.14). Metal A is the solid curve and metal B the dashed curve. The curve was read off the examination figure at the values tabulated below.

Find. For each metal: the proportional limit, the yield stress at a 0.002 in/in offset, the ultimate strength, the modulus of resilience, the toughness, and which of the two is the more ductile.

[Figure not reproduced: Figure 1.1 — The examination stress–strain curves redrawn to scale from the paper, with the two 0.002 in/in offset lines and their intersections with the curves marked. Metal A (solid) fractures at 0.079 in/in; metal B (dashed) carries on to 0.117 in/in. See the official exam paper.]

Approach. Read the elastic slope from the straight portion of each curve, use it to construct the 0.002 offset line and locate the offset yield point, take the peak of each curve as the ultimate strength, and obtain resilience and toughness as the areas under the elastic branch and under the whole curve respectively.

1. Establish the elastic modulus of each metal from the straight branch. The modulus is the slope of the initial linear part of the curve, $$E=\frac{\Delta\sigma}{\Delta\varepsilon}$$ For metal A the straight branch runs from the origin to about 50 ksi at 0.0025 in/in, and for metal B to about 45 ksi at 0.0045 in/in, so $$E_A=\frac{50}{0.0025}=20\,000\ \text{ksi}\qquad E_B=\frac{45}{0.0045}=10\,000\ \text{ksi}$$ These are the values used to construct the offset lines. They are consistent with metal A being a steel and metal B a lighter alloy of about one-half the stiffness.
2. Read the proportional limit (item I). The proportional limit is the stress at which the plot first departs from that straight line — the last point at which Hooke's law still describes the material. Reading the departure point on each curve gives $$\boxed{\sigma_{pl,A}\approx 50\ \text{ksi at }\varepsilon=0.0025,\qquad \sigma_{pl,B}\approx 45\ \text{ksi at }\varepsilon=0.0045}$$ Metal B reaches almost the same proportional limit as metal A but needs nearly twice the strain to do so, because its modulus is half as large.
3. Construct the 0.002 offset and read the yield stress (item II). A line of slope E is drawn from the strain axis at 0.002 in/in; the yield stress is the ordinate where that line cuts the curve: $$\sigma=E\,(\varepsilon-0.002)$$ For metal A the line rises at 20 000 ksi and meets the curve at a strain of about 0.0051, and for metal B the 10 000 ksi line meets the curve at about 0.0070, giving $$\boxed{\sigma_{y,A}\approx 61.5\ \text{ksi},\qquad \sigma_{y,B}\approx 50.2\ \text{ksi}}$$ Both offset yield stresses sit above the corresponding proportional limits, as they must, because the offset construction deliberately allows 0.2 per cent permanent strain.
4. Read the ultimate strength (item III). The ultimate (tensile) strength is the highest ordinate the curve reaches. Metal A climbs steadily to the end of its record and metal B flattens out near the end of its own, so $$\boxed{\sigma_{u,A}\approx 130\ \text{ksi},\qquad \sigma_{u,B}\approx 74\ \text{ksi}}$$ Neither curve shows a falling branch before fracture, so on this plot the ultimate strength and the fracture strength coincide; a real engineering curve would show a short necking drop for a ductile metal, and its omission here is a drafting simplification.
5. Compute the modulus of resilience (item IV). Resilience is the elastic strain energy stored per unit volume up to the proportional limit — the triangular area under the straight branch: $$U_r=\tfrac{1}{2}\,\sigma_{pl}\,\varepsilon_{pl}=\frac{\sigma_{pl}^{2}}{2E}$$ Substituting each pair of values, $$U_{r,A}=\tfrac{1}{2}(50)(0.0025)=0.0625\ \text{ksi}\qquad U_{r,B}=\tfrac{1}{2}(45)(0.0045)=0.1013\ \text{ksi}$$ Because 1 ksi of area under a stress–strain curve is 1000 in·lb per cubic inch, $$\boxed{U_{r,A}\approx 63\ \tfrac{\text{in}\cdot\text{lb}}{\text{in}^{3}},\qquad U_{r,B}\approx 101\ \tfrac{\text{in}\cdot\text{lb}}{\text{in}^{3}}}$$ Metal B is the more resilient of the two even though it is the weaker, because resilience rewards a high proportional limit combined with a low modulus, and B has half the stiffness of A.
6. Compute the toughness (item V). Toughness is the total energy absorbed per unit volume up to fracture, that is the whole area under the curve. Dividing each curve into strips and applying the trapezoidal rule, $$U_T=\int_{0}^{\varepsilon_f}\sigma\,d\varepsilon\approx\sum \tfrac{1}{2}\left(\sigma_i+\sigma_{i+1}\right)\left(\varepsilon_{i+1}-\varepsilon_i\right)$$ which gives 8.06 ksi for metal A over the range 0 to 0.079 and 7.52 ksi for metal B over 0 to 0.117, that is $$\boxed{U_{T,A}\approx 8\,100\ \tfrac{\text{in}\cdot\text{lb}}{\text{in}^{3}},\qquad U_{T,B}\approx 7\,500\ \tfrac{\text{in}\cdot\text{lb}}{\text{in}^{3}}}$$ A quick sanity check confirms the order of magnitude: the mean ordinate of curve A is roughly 102 ksi over a strain of 0.079, and 102 times 0.079 is about 8.1 ksi. The two toughnesses are close because A trades a shorter strain record for much higher stresses while B trades lower stresses for a longer one.
7. Compare the ductility (item VI). Ductility is the amount of permanent deformation a material sustains before fracture, and on a stress–strain plot it is read directly as the strain at fracture: $$\varepsilon_{f,A}=0.079,\qquad \varepsilon_{f,B}=0.117,\qquad \frac{\varepsilon_{f,B}}{\varepsilon_{f,A}}=1.48$$ $$\boxed{\text{Metal B is the more ductile, by about 48 per cent in elongation at fracture}}$$ Metal B stretches roughly half as far again as metal A before it breaks, and it does so on a long, nearly flat plastic plateau, which is the visual signature of ductile behaviour. Note that the greater ductility does not make B the tougher material here: A's far higher stress level more than compensates for its shorter strain record.