## Preview Extract

This Document Contains Chapters 17 to 18 Chapter 17 Soil Engineering for Highway Design 17-1 Determine the void ratio of a soil if its bulk density is 125 lb/ft3 and it has a moisture content of 24 percent. The specific gravity of the soil particles is 2.75. Also, determine its dry density and degree of saturation. Solve for γd, dry density, using Equation 17.11. γd = γ / (1 + w) γd = 125 / (1 + 0.24) γd = 100.81 lb/ft3 Solve for void ratio, e, using Equation 17.13. This equation can be modified to use dry density when the degree of saturation is zero (no moisture is present) as follows: γd = (Gs γw ) / 1 + e 1 + e = Gs γw / γd e = ( Gs γw / γd ) – 1 e = ((2.75)(62.4) / 100.81) – 1 e = 0.702 Solve for the degree of saturation, S. S = w Gs / e S = (0.24)(2.75) / 0.702 S = 0.940 17-2 A soil has a bulk density of 139 lb/ft3 and a dry density of 122 lb/ft3, and the specific gravity of the soil particles is 2.70. Determine: (a) moisture content, (b) degree of saturation, (c) void ratio, and (d) porosity. Solve for moisture content, w, using Equation 17.11. γd = γ / (1 + w) 1 + w = γ / γd w = (γ / γd ) – 1 w = (139 / 122) – 1 = 0.139 = 13.9 % Solve for void ratio, e, using Equation 17.13. This equation can be modified to use dry density when the degree of saturation is zero (no moisture is present) as follows: γd = (Gs γw ) / 1 + e 1 + e = Gs γw / γd e = ( Gs γw / γd ) – 1 e = ((2.7)(62.4) / 122) – 1 e = 0.381 Solve for the degree of saturation, S. S = w Gs / e S = (0.139)(2.7) / 0.381 S = 0.985 = 98.5% Solve for porosity, n, using Equation 17.4. n = e / (1 + e) n = (0.381)/(1 + 0.381) n = 0.276 17-3 The weight of a sample of saturated soil before drying is 2.8 lb and after drying is 2.1 lb. If the specific gravity of the soil particles is 2.6, determine: (a) moisture content, (b) void ratio, (c) porosity, (d) bulk density, and (e) dry density. (a) Solve for moisture content, w, using Equation 17.6. Calculate the weight of the water in the soil sample. W = Ww + Ws Ww = 2.8 – 2.1 Ww = 0.7 lbs = (Ww / Ws) (100) w = (0.7 / 2.1) (100) w = 33.33 % (b) Solve for void ratio, e, using Equation 17.3. First, the volume of water and the volume of solid must be determined. γw = Ww / Vw Vw = Ww / γw Vw = 0.7 / 62.4 Vw = 0.0112 ft3 Gs γw = Ws / Vs Vs = Ws / (Gs γw ) Vs = 2.1 / (2.6)(62.4) Vs = 0.0129 ft3 e = Vv / Vs = (Vw + Va) / Vs Va = 0 (since the sample is saturated, no air voids are assumed to be present). e = 0.0112 / 0.0129 e = 0.867 (c) Solve for porosity, n, using Equation 17.4. n = e / (1 + e) n = (0.867)/(1 + 0.867) n = 0.464 Calculate the bulk density, γ, using Equation 17.9 γ = W / V γ = 2.8 / (0.0112 + 0.0129) γ = 2.8 / 0.0241 γ = 116.2 lb/ft3 Solve for the dry density, γd , using Equation 17.11 γd = Ws / V γd = 2.1 / (0.0112 + 0.0129) γd = 87.1 lb/ft3 17-4 A moist soil has a moisture content of 12.5 percent, weighs 43.2 lb, and occupies a volume of 0.35 ft3. The specific gravity of the soil particles is 2.6. Find: (a) bulk density, (b) dry density, (c) void ratio, (d) porosity, (e) degree of saturation, and (f) volume occupied by water (ft3). Solve for the bulk density, using Equation 17.9. γ = W / V γ = 43.2 lb / 0.35 ft3 γ = 123.4 lb/ft3 Solve for the dry density, using Equation 17.11. γd = γ / (1 + w) γd = 123.4 / (1 + 0.125) γd = 109.7 lb/ft3 Solve for void ratio, e, using Equation 17.13. = [((1 + w) (Gs)(γw)) / γ] – 1 e = [((1 + 0.125) (2.6) (62.4)) / 123.4] – 1 e = 0.479 Solve for porosity, n, using Equation 17.4. n = e / (1 + e) n = 0.479 / (1 + 0.479) n = 0.324 Compute the degree of saturation, S. S = Vw / Vv = (w)(Gs) / e S = (0.125)(2.6) / 0.479 S = 0.678 = 67.8% Determine the volume occupied by water, using Equation 17.7. w = Ww / Ws Ww = (w/(1 + w)) W Ww = (0.125/1.125)(43.2) = 4.80 lb Vw = Ww / γw = (4.80 lb)/(62.4 lb/ft3) Vw = 0.0769 ft3 17-5 The moist weight of 0.15 ft3 of soil is 18.6 lb. If the moisture content is 17 percent and the specific gravity of soil solids is 2.62, find the following: (a) bulk density, (b) dry density, (c) void ratio, (d) porosity, (e) degree of saturation, and (f) the volume occupied by water (ft3). Solve for the bulk density, using Equation 17.9 γ = W / V γ = 18.6 lb / 0.15 ft3 γ = 124 lb/ft3 Solve for the dry density, using Equation 17.11. γd = γ / (1 + w) γd = 124 / (1 + 0.17) γd = 106 lb/ft3 Solve for void ratio, e, using Equation 17.13. = (((1 + w) (Gs)(γw)) / γ) – 1 e = (((1 + 0.17) (2.62) (62.4)) / 124) – 1 e = 0.543 (d) Solve for porosity, n, using Equation 17.4. n = e / (1 + e) n = 0.543 / (1 + 0.543) n = 0.352 (e) Compute the degree of saturation, S. S = Vw / Vv = (w)(Gs) / e S = (0.17)(2.62) / 0.543 S = 0.821 = 82.1% (f) Determine the volume occupied by water, using Equation 17.7. w = Ww / Ws Ww = (w/(1 + w)) W Ww = (0.17/1.17)(18.6) = 2.70 lb Vw = Ww / γw = (2.70 lb)/(62.4 lb/ft3) Vw = 0.043 ft3 17-6 The moist weight of 0.19 ft3 of soil is 21.6 lb. If the moisture content is 18 percent and the specific gravity of soil solids is 2.55, find the following: (a) bulk density, (b) dry density, (c) void ratio, (d) porosity, (e) degree of saturation, and (f) the volume occupied by water (ft3). Solve for the bulk density, using Equation 17.9 γ = W / V γ = 21.6 lb / 0.19 ft3 γ = 114 lb/ft3 Solve for the dry density, using Equation 17.11. γd = γ / (1 + w) γd = 113.68 / (1 + 0.18) γd = 96 lb/ft3 Solve for void ratio, e, using Equation 17.13. e = [((1 + w) (Gs)(γw)) / γ] – 1 e = (((1 + 0.18) (2.55) (62.4)) / 113.68) – 1 e = 0.652 (d) Solve for porosity, n, using Equation 17.4. n = e / (1 + e) n = 0.652 / (1 + 0.652) n = 0.395 (e) Compute the degree of saturation, S. S = Vw / Vv = (w)(Gs) / e S = (0.18)(2.55) / 0.652 S = 0.704 = 70.4% (f) Determine the volume occupied by water, using Equation 17.7. w = Ww / Ws Ww = (w/(1 + w)) W Ww = (0.18/1.18)(21.6) = 3.295 lb Vw = Ww / γw = (3.295 lb)/(62.4 lb/ft3) Vw = 0.053 ft3 17-7 A liquid limit test conducted in the laboratory on a sample of soil gave the following results listed below. Determine the liquid limit of this soil from a plot of the flow curve. Number of Blows (N) Moisture Content (%) 20 45.0 28 43.6 30 43.2 35 42.8 40 42.0 The liquid limit is defined as the moisture content after 25 blows of the standardized testing equipment. From the following figure it can be seen that the liquid limit is approximately 44%. 17-8 A plastic limit test for a soil showed that moisture content was 19.2%. Data from a liquid limit test were as follows: Number of Blows (N) Moisture Content (%) 14 42.0 19 40.8 27 39.1 (a) Draw the flow curve and obtain the liquid limit. (b) What is the plasticity index of the soil? The liquid limit is defined as the moisture content after 25 blows of the standardized testing equipment. From the following figure it can be seen that the liquid limit is approximately 39.5%. Determine the plasticity index using Equation 17.14. PI = LL – PL PI = 39.5 – 19.2 PI = 20.3 17-9 A plastic limit test for a soil showed that moisture content was 18.0%. Data from a liquid limit test were as follows: Number of Blows (N) Moisture Content (%) 16 39.8 21 39.2 26 38.4 30 36.9 Draw the flow curve and obtain the liquid limit. (b) What is the plasticity index of the soil? (a) The liquid limit is defined as the moisture content after 25 blows of the standardized testing equipment. From the following figure it can be seen that the liquid limit is approximately 38.5%. Determine the plasticity index using Equation 17.14. PI = LL – PL PI = 38.5 – 18.0 PI = 20.5 17-10 The following results were obtained by a mechanical analysis. Classify the soil using the AASHTO classification system and give the group index. Sieve Analysis, % Finer than No. 10: 98%, than No. 40: 81%, than No. 200: 38%; liquid limit = 42; plastic limit = 23. Use Table 17.1 to classify this soil. Since more than 35% of the material is passing No. 200 sieve, the liquid limit is greater than 41, the plasticity index is greater than 11, the soil is Group A-7. Since the plasticity index (19) is greater than the liquid limit minus 30, the subgroup is A-7-6. Determine the Group Index using Equation 17.18. GI = (F – 35)[0.2 + 0.005(LL – 40)] + 0.01(F – 15)(PI – 10) GI = (38 – 35)[0.2 + 0.005(42 – 40)] + 0.01(38 – 15)(19 – 10) GI = 2.7; use GI = 3 The soil can be classified as A-7-6(3). 17-11 The following results were obtained by a mechanical analysis. Classify the soil using the AASHTO classification system and give the group index. Sieve Analysis, % Finer than No. 10: 84%, than No. 40: 58%, than No. 200: 8%; soil is not plastic. Use Table 17.1 to classify this soil. Since less than 35% of the material is passing No. 200 sieve, there is no liquid limit, and the soil is not plastic, the soil is Group A-3, and there is no associated group index. 17-12 The following results were obtained by a mechanical analysis. Classify the soil using the AASHTO classification system and give the group index. Sieve Analysis, % Finer than No. 10: 99%, than No. 40: 85%, than No. 200: 71%; liquid limit = 55; plastic limit = 21. Use Table 17.1 to classify this soil. Since more than 35% of the material is passing No. 200 sieve, the liquid limit is greater than 41, the plasticity index is greater than 11, the soil is Group A-7. Since the plasticity index (34) is greater than the liquid limit minus 30, the subgroup is A-7-6. Determine the Group Index using Equation 17.18. GI = (F – 35)[0.2 + 0.005(LL – 40)] + 0.01(F – 15)(PI – 10) GI = (71 – 35)[0.2 + 0.005(55 – 40)] + 0.01(71 – 15)(34 – 10) GI = 23.34 Use GI = 23 The soil can be classified as A-7-6(23). 17-13 The following results were obtained by a mechanical sieve analysis. Classify the soil using the Unified Soil Classification System (USCS). Sieve Analysis, % Passing by Weight than No. 4: 30%, than No. 40: 40%, than No. 200: 30%; liquid limit = 33; plastic limit = 12. Use Table 17.3. Since more than half of the material is larger than No. 200 sieve, the material is Coarse-grained sand. Since more than half of the soil is greater than No. 4 sieve, the soil is either gravels with fines or clean gravels. Since this soil is above the "A" line on the Atterberg plot (with LL=33 and PI=21) and the PI is greater than 7, the group symbol is GC. Therefore, the soil can be classified as GC – Clayey Gravels. 17-14 The following results were obtained by a mechanical sieve analysis. Classify the soil using the Unified Soil Classification System (USCS). Sieve Analysis, % Passing by Weight than No. 4: 4%, than No. 40: 44%, than No. 200: 52%; liquid limit = 29; plastic limit = 11. Use Table 17.3. Since the less than half of the material is larger than No. 200 sieve and the Liquid Limit is less than 50, the soil is in the silts and clays division. Since this soil is above the "A" line on the Atterberg plot (with LL=29 and PI=11), and the PI is greater than 7, the group symbol is CL. The soil can be classified as CL – Sandy Clays. 17-15 The following results were obtained by a mechanical sieve analysis. Classify the soil using the Unified Soil Classification System (USCS). Sieve Analysis, % Passing by Weight than No. 4: 11%, than No. 40: 24%, than No. 200: 65%; liquid limit = 44; plastic limit = 23. Use Table 17.3. Since the less than half of the material is larger than No. 200 sieve and the Liquid Limit is less than 50, the soil is in the silts and clays division. Since this soil is above the "A" line on the Atterberg plot (with LL=29 and PI=11), and the PI is greater than 7, the group symbol is CL. The soil can be classified as CL – Sandy Clays. 17-16 The following results were obtained by a mechanical sieve analysis. Classify the soil using the Unified Soil Classification System (USCS). Sieve Analysis, % Passing by Weight than No. 4: 24%, than No. 40: 72%, than No. 200: 4%; liquid limit = 26; plastic limit = 8. Use Table 17.3. Since more than half of the material is larger than No. 200 sieve, the material is Coarse-grained sand. Since more than half of the soil is greater than No. 4 sieve, the soil is either gravels with fines or clean gravels. Since the soil is above the “A” line on the Atterberg plot (with LL = 26 and PI = 18) and the PI is greater than 7, the group symbol is GC. Therefore, the soil can be classified as GC – Clayey Gravels.

Following are the results of a sieve analysis: Sieve Percent No. Finer 4 100 10 92 20 82 40 67 60 58 80 38 100 22 200 6 Pan – Plot the grain-size distribution curve for this sample. Determine D10, D30, and D60. Calculate the uniformity coefficient, Cu. Calculate the coefficient of gradation, Cc. (a)Grain-size distribution are shown as follows, From the preceding figure, D10 = 0.100 mm D30 = 0.160 mm D60 = 0.279 mm The coefficient of uniformity can be calculated using Equation 17.19: Cu = (D60 / D10) Cu = 0.279 / 0.100 Cu = 2.79 The coefficient of curvature can be calculated using Equation 17.20: Cc = D302 / (D60)(D10) Cc = (0.160)2 / (0.279)(0.100) Cc = 0.92 17-18 Following are the results of a sieve analysis: Sieve Percent No. Finer 4 100 10 90 20 80 40 70 60 40 80 29 100 19 200 10 Pan -- Plot the grain-size distribution curve for this sample. Determine D10, D30, and D60. Calculate the uniformity coefficient, Cu. Calculate the coefficient of gradation, Cc. The grain-size distribution curve is shown as follows, From the preceding figure, D10 = 0.075 mm D30 = 0.180 mm D60 = 0.347 mm The coefficient of uniformity can be calculated using Equation 17.19: Cu = (D60 / D10) Cu = 0.347 / 0.075 Cu = 4.63 The coefficient of curvature can be calculated using Equation 17.20: Cc = D302 / (D60)(D10) Cc = (0.180)2 / (0.347)(0.075) Cc = 1.24 Following are the results of a sieve analysis: Sieve Percent No. Finer 4 99 10 87 20 77 40 65 60 32 80 23 100 14 200 7 Pan – Plot the grain-size distribution curve for this sample. Determine D10, D30, and D60. Calculate the uniformity coefficient, Cu. Calculate the coefficient of gradation, Cc. The grain-size distribution curve is shown as follows, From the preceding figure, D10 = 0.113 mm D30 = 0.237 mm D60 = 0.396 mm The coefficient of uniformity can be calculated using Equation 17.19: Cu = (D60 / D10) Cu = 0.396 / 0.113 Cu = 3.50 The coefficient of curvature can be calculated using Equation 17.20: Cc = D302 / (D60)(D10) Cc = (0.237)2 / (0.396)(0.113) Cc = 1.26 17-20 The results of a compaction test on samples of soil that are to be used for an embankment on a highway project are listed below. Determine the maximum dry unit weight of compaction and the optimum moisture content. Sample No. Moisture Content Bulk Density(lb/ft3) 4.8 135.1 7.5 145.0 7.8 146.8 8.9 146.4 9.7 145.3 For each sample, calculate the dry density and then plot the results to estimate maximum dry unit weight and optimum moisture content, as shown below. Dry density can be calculated using Equation 17.11. γ d = γ / (1 + w)

Sample w γ γd

1 4.8 135.1 128.9

2 7.5 145.0 134.9

3 7.8 146.8 136.2

4 8.9 146.4 134.4

5 9.7 145.3 132.5

From the above graph, the maximum dry unit weight of the soil is approximately 136.2 lb/ft3, and the optimum moisture content is approximately 8.0%. The results of a compaction test on samples of soil that are to be used for an embankment on a highway project are listed below. Determine the maximum dry unit weight of compaction and the optimum moisture content. Sample No. Moisture Content Bulk Density (lb/ft3) 15 114.7 16 118.6 19 120.4 22 120.1 25 118.3 26 117.1 For each sample, calculate the dry density and then plot the results to estimate maximum dry unit weight and optimum moisture content, as shown below. Dry density can be calculated using Equation 17.11. γ d = γ / (1 + w)

Sample w γ γd

1 15 114.7 99.7

2 16 118.6 102.2

3 19 120.4 101.2

4 22 120.1 98.4

5 25 118.3 94.6

6 26 117.1 92.9

From the above graph, the maximum dry unit weight of the soil is approximately 102.2 lb/ft3, and the optimum moisture content is approximately 16%. 17-22 The results obtained from a seismic study along a section of the centerline of a highway are shown below. Estimate the depths of the different strata of soil and suggest the type of soil in each stratum. Distance of Impulse to Geophone Time for Wave Arrival (ft) (10–3 sec) 25 20 50 40 75 60 100 68 125 74 150 82 175 84 200 86 225 88 250 90 From the data provided, plot the following graph: Determine the soil type in each stratum. u1 = 75 / (60 × 10–3) u1 = 1,250 ft / sec u1 = 381 m / sec The soil in this stratum is probably sand or loess. u2 = (150 – 75) / ((82 – 60) × 10–3) u2 = 3,409 ft / sec u2 = 1,039 m / sec The soil in this stratum is probably loam or clay.

u3 = (250 – 150) / ((90 – 82) × 10–3) u3 = 12,500 ft / sec u3 = 3,810 m / sec The soil in stratum is probably sandstone, limestone, or slate and shale. Next, determine the depths for each layer of soil. Sin ∝ = u1 / u2 Sin ∝ = 1,250 / 3,409 Sin ∝ = 0.367 ∝ = 21.52º, cos ∝ = 0.93 OP = 35 × 10–3 sec (from graph above) Use Equation 17.21 to determine depth, H1 = ((35 × 10–3 sec) × 1,250) / (2 × 0.93) H1 = 23.5 ft Therefore, the depth for the first stratum is 23.5 ft. Next, determine the depth for the second layer of soil. Sin β = u2 / u3 Sin β = 3,409 / 12,500 Sin β = 0.273 β = 15.83º, Cos β = 0.962 PT = 35 × 10–3 sec (from graph above) Use Equation 17.22 to determine depth, H2 = ((35 × 10–3 sec) × 3,409) / (2 × 0.962) H2 = 62.0 feet Therefore, the depth for the second stratum is 62.0 feet. Chapter 18 Bituminous Materials 18-1 Briefly describe the process of distillation by which asphalt cement is produced from crude petroleum. Also describe in detail how to obtain asphalt binders that can be used to coat highly siliceous aggregates. Asphalt cement is obtained from the distillation of crude petroleum by either fractional or destructive distillation. Fractional distillation processes involve the separation of the different materials in the crude petroleum without significant changes in the chemical composition of each material. The different volatile materials are removed at higher and higher temperatures until the petroleum asphalt is left as residue. Either steam or vacuum is used to increase the temperature. Destructive distillation, or cracking, involves the application of high temperatures and pressure and results in chemical changes. Cracking processes are used when larger amounts of the light fractions (e.g. motor oil) are needed. The asphaltic material obtained from cracking is not widely used in paving because it is more susceptible to weather changes. Asphalt emulsions are produced by breaking asphalt cement into minute particles and dispersing them in water with an emulsifier. The particles are of like electrical charge and therefore do not coalesce. Aggregates containing a high percentage of siliceous materials are electronegative and therefore cationic emulsions are more effective. 18-2 Describe both the factors that influence the durability of asphalt materials and the effect of each factor has on the materials. The durability of an asphaltic material is dependent on its ability to resist weathering. Some of the factors that influence weathering are oxidation, volatilization, temperature, exposed surface area, and age hardening. Oxidation is a chemical reaction that causes gradual and permanent hardening and eventually considerable loss of the plastic characteristics of the asphalt material. Volatilization is the evaporation of the lighter hydrocarbons from the asphaltic materials which causes loss of the plastic characteristics of the material. As temperatures increase, so do the rates of oxidation and volatilization. The rate of hardening is directly proportional to the ratio of the surface area to the volume. Age hardening occurs when an asphaltic material is heated and allowed to cool. A gel-like structure is formed that hardens over time, significantly within the first few hours after cooling and decreasing to an almost negligible amount in a year. 18-3 Discuss the factors to consider in producing a rapid-curing cutback asphaltic material. The rate of curing indicates the time that should elapse before a cutback will attain a consistency thick enough for the binder to perform satisfactorily. The rate of curing is affected by both inherent and external factors. The important inherent factors are the volatility of the solvent, the quantity of solvent in the cutback, and the consistency of the base material. In a rapid-curing material, gasoline or naphtha is used as the solvent because each evaporates quickly and therefore the material cures faster than with other solvents. External factors include temperature, ratio of surface area to volume, and wind velocity across the exposed surface. The higher any of these factors is, the higher the rate of curing. 18-4 Results obtained from laboratory tests on a sample of RC-250 asphalt cement are given. Determine whether the properties of this material meet the AASHTO specifications for this type of material; if not, note the differences.

Property Specification Test Results Meet (Y/N)

Kinematic viscosity 250-500 centistokes 260 cs Y

Flash Point Min 27°C (80°F) 75°F N

Distillate % to 680°F To 437°F To 500°F To 600°F Min 35 Min 60 Min 80 35 54 75 Y N N

Residue to 680°F Min 65 64 N

Ductility at 77°F Min 100 cm 95 cm N

Absolute viscosity at 140ºF 600-2400 poises 750 poises Y

Solubility % Min 99.0 95 N

While the sample does meet some of the specifications, overall, it does not meet specifications. 18-5 Given the specifications for an asphaltic concrete mixture and the results of a sieve analysis, determine the proportion of different aggregates to obtain the required gradation. Coarse aggregates: 60% Fine aggregates: 35% Filler: 5%

Passing sieve Retained on sieve % by Weight

Coarse Fine Filler

¾ in. ½ in. 4 — —

½ in. ⅜ in. 36 — —

⅜ in. No. 4 40 — —

No. 4 No. 10 15 6 —

No. 10 No. 40 5 32 —

No. 40 No. 80 — 33 5

No. 80 No. 200 — 29 40

No. 200 — — — 55

Total 100 100 100

The following table shows the sieve analysis data and the computations required to obtain the required gradation.

Passing sieve Retained on sieve % by Weight

Coarse Fine Filler Total

¾ in. ½ in. 0.6(4) = 2.4 — — 2.4

½ in. ⅜ in. 0.6(36) = 21.6 — — 21.6

⅜ in. No. 4 0.6(40) = 24.0 — — 24.0

No. 4 No. 10 0.6(15) = 9.0 0.35(6) = 2.1 — 11.1

No. 10 No. 40 0.6(5) = 3.0 0.35(32) = 11.2 — 14.2

No. 40 No. 80 — 0.35(33) = 11.55 0.05(5) = 0.25 11.8

No. 80 No. 200 — 0.35(29) = 10.15 0.05(40) = 2.0 12.15

No. 200 — — — 0.05(55) = 2.75 2.75

Total 60.0 35.0 5.0 100.0

18-6 Given the particle size distributions of two aggregates and the required limits of particle size distribution for the mix, determine a suitable ratio for blending the two aggregates to obtain an acceptable combined aggregate.

Sieve Size Aggregate A Aggregate B Required Mix (%)

3/4 in. 100 98 96-100

3/8 in. 80 76 65-80

No. 4 50 45 40-55

No. 10 43 33 35-40

No. 40 20 30 15-35

No. 200 4 8 5-8

Observation of the results of the sieve analysis indicate that aggregate B appears to more closely conform to the required mix for most sizes; therefore, the initial mix design should contain more than 50% aggregate B. The initial trial will be 40% aggregate A and 60% aggregate B. The following table shows the sieve analysis data and the computations required to obtain the required gradation.

Sieve Size Aggregate A Aggregate B Mix (%) Required (%) OK?

3/4 in. (0.4)(100) (0.6)(98) 98.8 96-100 Y

3/8 in. (0.4)(80) (0.6)(76) 77.6 65-80 Y

No. 4 (0.4)(50) (0.6)(45) 47.0 40-55 Y

No. 10 (0.4)(43) (0.6)(33) 37.0 35-40 Y

No. 40 (0.4)(20) (0.6)(30) 26.0 15-35 Y

No. 200 (0.4)(4) (0.6)(8) 6.4 5-8 Y

The proposed design (40% aggregate A and 60% aggregate B) is acceptable. Note that other combinations are also possible. 18-7 Given the particle size distributions of two aggregates and the required limits of particle size distribution for the mix, determine a suitable ratio for blending the two aggregates to obtain an acceptable combined aggregate.

Sieve Size Aggregate A Aggregate B Required Mix (%)

3/4 in. 100 95 98-100

3/8 in. 80 74 75-85

No. 4 56 43 50-60

No. 10 42 32 37-47

No. 40 24 29 25-35

No. 200 5 12 7-10

Observation of the results of the sieve analysis indicate that aggregate A appears to more closely conform to the required mix for most sizes; therefore, the initial mix design should contain more than 50% aggregate A. The initial trial will be 70% aggregate A and 30% aggregate B. The following table shows the sieve analysis data and the computations required to obtain the required gradation.

Sieve Size Aggregate A Aggregate B Mix (%) Required (%) OK?

3/4 in. (0.7)(100) (0.3)(95) 98.5 98-100 Y

3/8 in. (0.7)(80) (0.3)(74) 78.2 75-85 Y

No. 4 (0.7)(56) (0.3)(43) 52.1 50-60 Y

No. 10 (0.7)(42) (0.3)(32) 39.0 37-47 Y

No. 40 (0.7)(24) (0.3)(29) 25.5 25-35 Y

No. 200 (0.7)(5) (0.3)(12) 7.1 7-10 Y

The proposed design (70% aggregate A and 30% aggregate B) is acceptable. Note that other combinations are also possible. 18-8 Given four different types of aggregates to be used to produce a blended aggregate for use in the manufacture of asphaltic concrete, determine the bulk specific gravity of the aggregate mix.

Material Percent by Weight Bulk Specific Gravity

A 38 2.62

B 42 2.58

C 12 2.63

D 8 2.54

The bulk specific gravity of the proposed mix is simply a weighted average of the bulk specific gravities of the aggregates in the blend. Gsb = (0.38)(2.62) + (0.42)(2.58) + (0.12)(2.63) + (0.08)(2.54) G = 2.60 18-9 If the specific gravity of the asphalt cement used in a sample of asphalt concrete mix is 0.95, the maximum specific gravity of the mix is 2.58, and the mix contains 6.5 percent by weight of asphalt cement, determine the effective specific gravity of the mixture. Use Equation 18.7. 100−Pb Gse = (100/Gmm) −(P Gb / b) Pb = 6.5; Gmm = 2.58; Gb = 0.95 Gse = Gse= 2.93 18-10 The table below lists data used in obtaining a mix design for an asphalt paving mixture. If the maximum specific gravity of the mixture is 2.41 and the bulk specific gravity is 2.35, determine: the bulk specific gravity of aggregates in the mix the asphalt absorbed the effective asphalt content of the paving mixture (d) the percent voids in the mineral aggregate VMA.

Material Specific Gravity Mix Proportion

Asphalt cement 1.06 6.5

Coarse aggregate 2.55 51.5

Fine aggregate 2.66 34.8

Mineral filler 2.64 7.2

Use Equation 18.5 to determine bulk specific gravity. P P Pca + fa + mf Gsb = P P Gbca Gbfa Gbmf Gsb = (51.5 + 34.8 + 7.2) / [(51.5/2.55) + (34.8/2.66) + (7.2/2.64)] Gsb = 2.598 Use Equation 18.9 to determine asphalt absorption. Pba =100 G Gse − sb Gb G Gse sb First, Gea must be calculated using Equation 18.7, Gse = 100 − Pb (100/Gmm) − (Pb /Gb) Gse = (100 – 6.5) / [(100/2.41) – (6.5/1.06)] Gse = 2.644 Pba =100 Gse −Gsb Gb G Gse sb Pba = 100[(2.644 – 2.598) / (2.644)(2.598)]1.06 Pba = 0.71% Use Equation 18.10 to determine effective asphalt content Pbe = −Pb P Pba s 100 Pbe = 6.5 – (0.71/100)(100-6.5) Pbe = 5.84 Use Equation 18.11 to determine VMA VMA =100− GmbPs Gsb VMA = 100 – (2.35)(93.5)/2.598 VMA = 15.43% 18-11 For hot-mix, hot-laid asphaltic concrete mixtures, if the asphaltic content is specified as 5 to 7 percent, how is the optimum percentage determined? To determine the optimum asphalt content, each potential mix would be evaluated. Mixes created with asphalt contents of 5%, 5.5%, 6%, 6.5%, and 7% would be developed. For these mixes, test results would be used to plot curves for each of the following five variables as function of the range of asphalt content values used in the test mixes: Marshall stability, flow, voids in mineral aggregates, voids filled with asphalt, and percent air voids in the compacted mixture. Curves, such as those shown in Figure 18.13, could then be developed by plotting these values are also against the corresponding asphalt percentages. The five plots are now used to determine the optimum asphalt percentage. A suitable starting point would be to identify the asphalt content that would result in 4% air voids. While this value of asphalt content may represent that optimal asphalt content value when examining the other four curves, if would result in a mix that meets the criteria in Table 18.7, then this design would be acceptable. If all of the criteria in Table 18.7 are not met using the asphalt content value associated with 4% air voids, then adjustments to asphalt content can be made in attempt to meet these criteria. 18-12 The aggregate mix used for the design of an asphalt paving concrete consists of 42% coarse aggregates, 51% fine aggregates, and 7% mineral fillers. If the respective bulk specific gravities of these materials are 2.60, 2.71, and 2.69, and the effective specific gravity of the aggregates is 2.82, determine the optimum asphalt content as a percentage of the total mix using results obtained using the Marshall method as shown in the following table. The specific gravity of the asphalt cement is 1.02. First, compute the bulk specific gravity and associated bulk density of each possible mix.

Asphalt % Gbcm γ

5.5 1325.3/(1325.3 – 785.6) = 2.46 153.5

6.0 1330.1/(1330.1 – 793.3) = 2.48 154.8

6.5 1336.2/(1336.2 – 800.8) = 2.50 156.0

7.0 1342.0/(1342.0 – 804.5) = 2.50 156.0

7.5 1347.5/(1347.5 – 805.1) = 2.48 154.8

Next, compute VMA for each possible mix using Equation 18.11. VMA =100− GmbPs Gsb Gbcm and Pta are known for each possible mix; however, Gsb must be computed, using Equation 18.5, in order to find VMA. P P Pca + fa + mf Gsb = P P Gbca Gbfa Gbmf

For 5.5% asphalt content:

Gmb = 2.46 Pca = 0.42(94.5) = 39.7

Pta = 94.5 Pfa = 0.51(94.5) = 48.2 Pmf = 0.07(94.5) = 6.6

Gsb = (39.7 + 48.2 + 6.6) / [(39.7/2.60) + (48.2/2.71) + (6.6/2.69)] Gsb = 2.66 VMA = 100 – (2.46)(94.5) / 2.66 VMA = 12.61% For 6.0% asphalt content: Gmb = 2.48 Pca = 0.42(94) = 39.5 Pta = 94 Pfa = 0.51(94) = 47.9 Pmf = 0.07(94) = 6.6 Gsb = (39.5 + 47.9 + 6.6) / [(39.5/2.60) + (47.9/2.71) + (6.6/2.69)] Gsb = 2.66 VMA = 100 – (2.48)(94) / 2.66 VMA = 12.36% For 6.5% asphalt content: Gmb = 2.50 Pca = 0.42(93.5) = 39.3 Pta = 93.5 Pfa = 0.51(93.5) = 47.7 Pmf = 0.07(93.5) = 6.5 Gsb = (39.3 + 47.7 + 6.5) / [(39.3/2.60) + (47.7/2.71) + (6.5/2.69)] Gsb = 2.66 VMA = 100 – (2.50)(93.5) / 2.66 VMA = 12.12% For 7.0% asphalt content: Gmb = 2.50 Pca = 0.42(93) = 39.1 Pta = 93.0 Pfa = 0.51(93) = 47.4 Pmf = 0.07(93) = 6.5 Gsb = (39.1 + 47.4 + 6.5) / [(39.1/2.60) + (47.4/2.71) + (6.5/2.69)] Gsb = 2.66 VMA = 100 – (2.5)(93) / 2.66 VMA = 12.59% For 7.5% asphalt content: Gmb = 2.48 Pca = .42(92.5) = 38.9 Pta = 92.5 Pfa = .51(92.5) = 47.2 Pmf = .07(92.5) = 6.5 Gsb = (38.9 + 47.2 + 6.5) / [(38.9/2.60) + (47.2/2.71) + (6.5/2.69)] Gsb = 2.66 VMA = 100 – (2.48)(92.5) / 2.66 VMA = 13.76% Now compute % air voids for each mix using Equation 18.12, Va =100Gmm −Gmb Gmm Gmm must be computed using Equation 18.8, Gmm = 100 (P Gs / se) + (P Gb / b) For 5.5% asphalt content: Gmp = 100 / [(94.5/2.82) + (5.5/1.02)] Gmp = 2.57 Va = 100 (2.57 – 2.46)/2.57 Va = 4.28% For 6.0% asphalt content: Gmp = 100 / [(94.0/2.82) + (6.0/1.02)] Gmp = 2.55 Va = 100 (2.55 – 2.48)/2.55 Va = 2.75% For 6.5% asphalt content: Gmp = 100 / [(93.5/2.82) + (6.5/1.02)] Gmp = 2.53 Va = 100 (2.53 – 2.5)/2.53 Va = 1.19% For 7.0% asphalt content: Gmp = 100 / [(93.0/2.82) + (7.0/1.02)] Gmp = 2.51 Va = 100 (2.51 – 2.50)/2.51 Va = 0.40% For 7.5% asphalt content: Gmp = 100 / [(92.5/2.82) + (7.5/1.02)] Gmp = 2.49 Va = 100 (2.49 – 2.48)/2.49 Va = 0.40% Now compute VFA for each mix using Equation 18.13, VFA=100VMA V − a VMA For 5.5% asphalt content: VFA = 100 / [(12.61 – 4.28) /12.61] VFA = 66.1% For 6.0% asphalt content: VFA = 100 / [(12.36 – 2.75) /12.36] VFA = 77.8% For 6.5% asphalt content: VFA = 100 / [(12.12 – 1.19) /12.12] VFA = 90.2% For 7.0% asphalt content: VFA = 100 / [(12.59 – 0.40) /12.59] VFA = 96.8% For 7.5% asphalt content: VFA = 100 / [(13.76 – 0.40) /13.76] VFA = 97.1% From these calculations, plot VFA vs. % asphalt, VMA vs. asphalt, Va vs. asphalt. Also, using the given data, plot stability vs. % asphalt, and flow vs. % asphalt. At 4% voids, AC is approximately 5.6% Checking this against the required values in Table 18.7: At 5.6% AC Stability ~ 1810 Flow ~ 13 VMA ~ = 12.5% Assuming medium traffic, this mix is acceptable. 18-13 Determine the asphalt absorption of the optimum mix found in Problem 18-12. Asphalt absorption is calculated using Equation 18.9 Pba =100 Gse −Gsb Gb G Gse sb Pba = 100 [(2.82 – 2.66) / (2.82)(2.66)]1.02 Pba = 2.18%

18-14 The latitude at the location where a high-speed rural road is to be constructed is 35°. The expected ESAL is 32 × 106. The seven-day average high temperature is 53°C, and the low air temperature is -18°C. If the standard deviations for the high and low temperatures are ±2°C and ±1˚C respectively, and the depth of the pavement is 155 mm, determine an appropriate asphalt binder for a reliability of 98%. Use Equation 18.13 to determine the high pavement temperature at a depth of 20 mm, T20mm = (Tair – 0.00618 Lat2 + 0.2289Lat + 42.2)(0.9545)) – 17.78 Note: for Tair, with a reliability of 98%, use 53 + (2)(2) = 57 T20mm = (57 – 0.00618(35)2 + 0.2289(35) + 42.2)(0.9545)) – 17.78 T20mm = 77.32°C From Table 18.12, preliminarily select PG 58-22 as the binder. With ESAL > 10 × 106, increase temperature grade of binder by one. Therefore, select PG 64-22 as the appropriate binder. 18-15 An urban expressway is being designed for a congested area in Washington, D.C. It is expected that most of the time traffic will be moving at a slow rate. If the anticipated ESAL is 8 × 106, determine an appropriate asphalt binder for this project. Assume design reliability is 98%. From Table 18.13, preliminarily select PG 58-16 as the binder. With slow traffic as a design criterion, increase temperature grade of binder by one. With ESAL < 10 × 106, no adjustment is required. Therefore, select PG 64-16 as the appropriate binder. 18-16 The table below shows properties of three trial aggregate blends that are to be evaluated so as to determine their suitability for use in a Superpave mix. If the nominal maximum sieve of each aggregate blend is 19 mm, determine the initial trial asphalt content for each of the blends.

Property Trial Blend 1 Trial Blend 2 Trial Blend 3

Gsb 2.698 2.696 2.711

Gse 2.765 2.766 2.764

Assume: Pb = 0.05 Ps = 0.95 Gb = 1.02 Va = 0.04 Since Gsb and Gse are given, the trial percentage of asphalt binder can be found using Equations 18.17, 18.18, and 18.19. For trial blend 1: Use Equation 18.17, Vba = Ps (1−Va ) [ 1 − 1 ] = 0.95(1− 0.04) [ 1 − 1 ]= 0.0209 ( Pb + Ps ) Gsb Gse (0.05 + 0.95 ) 2.698 2.765 Gb Gse 1.02 2.765 Use Equation 18.18, Vbe = 0.176 – 0.0675 log Sn = 0.176 – 0.067 log (19) = 0.0903 Use Equation 18.20, Ws = Ps (1−Va ) = 0.95(1− 0.04) = 2.323 Pb + Ps (0.05 + 0.95 ) Gb Gse 1.02 2.765 Use Equation 18.19, Pbi =100 G Vb ( be +Vba ) = 100 1.02(0.090 + 0.021) = 0.0466 (G Vb ( be +Vba )) +Ws 1.02(0.090 + 0.021) + 2.323 For trial blend 2: Use Equation 18.17, Vba = Ps (1−Va ) [ 1 − 1 ] = 0.95(1− 0.04) [ 1 − 1 ]= 0.0218 Pb + Ps ) Gsb Gse (0.05 + 0.95 ) 2.696 2.766 ( Gb Gse 1.02 2.766 Use Equation 18.18, Vbe = 0.176 – 0.0675 log Sn = 0.176 – 0.067 log (19) = 0.0903 Use Equation 18.20, Ws = Ps (1−Va ) = 0.95(1− 0.04) = 2.324 Pb + Ps (0.05 + 0.95 ) Gb Gse 1.02 2.766 Use Equation 18.19, Pbi =100 G Vb ( be +Vba ) = 100 1.02(0.090 + 0.022) = 0.0469 (G Vb ( be +Vba )) +Ws 1.02(0.090 + 0.022) + 2.324 For trial blend 3: Use Equation 18.17, Vba = Ps (1−Va ) [ 1 − 1 ] = 0.95(1− 0.04) [ 1 − 1 ]= 0.0164 Pb + Ps ) Gsb Gse (0.05 + 0.95 ) 2.711 2.764 ( Gb Gse 1.02 2.764 Use Equation 18.18, Vbe = 0.176 – 0.0675 log Sn = 0.176 – 0.067 log (19) = 0.0903 Use Equation 18.20, Ws = Ps (1−Va ) = 0.95(1− 0.04) = 2.322 Pb + Ps (0.05 + 0.95 ) Gb Gse 1.02 2.764 Use Equation 18.19, Pbi =100 G Vb ( be +Vba ) = 100 1.02(0.090 + 0.016) = 0.0634 (G Vb ( be +Vba )) +Ws 1.02(0.090 + 0.016) + 2.322 Solution Manual for Traffic and Highway Engineering Nicholas J. Garber, Lester A. Hoel 9781133605157