Three groups of students from the Geotechnical Engineering class collected soil-aggregate samples for laboratory testing from a recycled aggregate processing plant in Palm Beach County, Florida. Three samples, denoted by Soil A , Soil B , and Soil C , were collected from three locations of the aggregate stockpile, and sieve analyses were conducted (see Figure 2.35). (a) (b) Figure 2.35 (a) Soil-aggregate stockpile; (b) sieve analysis ( Courtesy of Khaled Sobhan, Florida Atlantic University, Boca Raton, Florida ) a. Determine the coefficient of uniformity and the coefficient of gradation for Soils A , B , and C . b. Which one is coarser: Soil A or Soil C ? Justify your answer. c. Although the soils are obtained from the same stockpile, why are the curves so different? ( Hint : Comment on particle segregation and representative field sampling.) d. Determine the percentages of gravel, sand and fines according to Unified Soil Classification System.

Question

Want to see more full solutions like this?

Answer 1

Textbook Question

Chapter 2, Problem 2.1CTP

Three groups of students from the Geotechnical Engineering class collected soil-aggregate samples for laboratory testing from a recycled aggregate processing plant in Palm Beach County, Florida. Three samples, denoted by Soil A, Soil B, and Soil C, were collected from three locations of the aggregate stockpile, and sieve analyses were conducted (see Figure 2.35).

Chapter 2, Problem 2.1CTP, Three groups of students from the Geotechnical Engineering class collected soil-aggregate samples , example 1

(a)

Chapter 2, Problem 2.1CTP, Three groups of students from the Geotechnical Engineering class collected soil-aggregate samples , example 2

(b)

Figure 2.35 (a) Soil-aggregate stockpile; (b) sieve analysis (Courtesy of Khaled Sobhan, Florida Atlantic University, Boca Raton, Florida)

a. Determine the coefficient of uniformity and the coefficient of gradation for Soils A, B, and C.

b. Which one is coarser: Soil A or Soil C? Justify your answer.

c. Although the soils are obtained from the same stockpile, why are the curves so different? (Hint: Comment on particle segregation and representative field sampling.)

d. Determine the percentages of gravel, sand and fines according to Unified Soil Classification System.

(a)

Expert Solution

To determine

Calculate the coefficient of uniformity $(Cu)$ and coefficient of gradation $(Cc)$ for soils A, B, and C.

Answer to Problem 2.1CTP

The uniformity coefficient of soil A is $26.42_$ .

The coefficient of gradation of soil A is $3.64_$ .

The uniformity coefficient of soil B is $35_$ .

The coefficient of gradation of soil B is $3.35_$ .

The uniformity coefficient of soil C is $46.67_$ .

The coefficient of gradation of soil C is $1.15_$ .

Explanation of Solution

Sketch the grain size distribution curve for soils A, B, and C as shown in Figure 1.

Principles of Geotechnical Engineering (MindTap Course List), Chapter 2, Problem 2.1CTP

Refer to Figure 1.

For soil A:

The diameter of the particle corresponding to $60%$ finer $(D60)$ is $14 mm$ .

The diameter of the particle corresponding to $30%$ finer $(D30)$ is $5.2 mm$ .

The diameter of the particle corresponding to $10%$ finer $(D10)$ is $0.53 mm$ .

For soil B:

The diameter of the particle corresponding to $60%$ finer $(D60)$ is $8.1 mm$ .

The diameter of the particle corresponding to $30%$ finer $(D30)$ is $2.5 mm$ .

The diameter of the particle corresponding to $10%$ finer $(D10)$ is $0.23 mm$ .

For soil C:

The diameter of the particle corresponding to $60%$ finer $(D60)$ is $7 mm$ .

The diameter of the particle corresponding to $30%$ finer $(D30)$ is $1.1 mm$ .

The diameter of the particle corresponding to $10%$ finer $(D10)$ is $0.15 mm$ .

Calculate the uniformity coefficient $(Cu)$ using the relation.

$Cu=D60D10$ (1)

For soil A:

Substitute $14 mm$ for $D60$ and $0.53 mm$ for $D10$ in Equation (1).

$Cu=140.53=26.42$

Hence, the uniformity coefficient for soil A is $26.42_$ .

For soil B:

Substitute $8.1 mm$ for $D60$ and $0.23 mm$ for $D10$ in Equation (1).

$Cu=8.10.23=35$

Hence, the uniformity coefficient for soil B is $35_$ .

For soil C:

Substitute $7 mm$ for $D60$ and $0.15 mm$ for $D10$ in Equation (1).

$Cu=70.15=46.67$

Hence, the uniformity coefficient for soil C is $46.67_$ .

Calculate the coefficient of gradation $(Cc)$ using the relation.

$Cc=D302D60×D10$ (2)

For soil A:

Substitute $14 mm$ for $D60$ , $5.2 mm$ for $D30$ , and $0.53 mm$ for $D10$ in Equation (2).

$Cc=5.2214×0.53=27.045.512=3.64$

Hence, the coefficient of gradation for soil A is $3.64_$ .

For soil B:

Substitute $8.1 mm$ for $D60$ , $2.5 mm$ for $D30$ , and $0.23 mm$ for $D10$ in Equation (2).

$Cc=2.528.1×0.23=6.251.863=3.35$

Hence, the coefficient of gradation for soil B is $3.35_$ .

For soil C:

Substitute $7 mm$ for $D60$ , $1.1 mm$ for $D30$ , and $0.15 mm$ for $D10$ in Equation (2).

$Cc=1.127×0.15=1.211.05=1.15$

Therefore, the coefficient of gradation for soil C is $1.15_$ .

(b)

Expert Solution

To determine

State which of the soil is coarser from soil A and C.

Answer to Problem 2.1CTP

Soil A is coarser than soil C.

Explanation of Solution

Refer to part (a).

The uniformity coefficient of soil A is $26.42$ .

The uniformity coefficient of soil C is $46.67$ .

The percent of soil finer than $1 mm$ for soil A is $20%$ .

The percent of soil finer than $1 mm$ for soil C is $47%$ .

Hence, a higher percentage of soil C is finer than soil A.

Hence, soil A is coarser than soil C.

(c)

Expert Solution

To determine

Explain the reason for curve different of soil A, B and C if it is obtained from same stockpile.

Explanation of Solution

The particle-size distribution curve shows the range of particle sizes present in a soil and the type of distribution of various-size particles.

Refer to Figure 1.

Particle separation of coarser and finer particles may take place in aggregate stockpiles. This makes representative sampling difficult.

Therefore, the particle-size distribution curve is different for soils A, B, and C.

(d)

Expert Solution

To determine

Calculate the percentages of gravel, sand, and fines according to the Unified Soil Classification System.

Answer to Problem 2.1CTP

The percentage of gravel for soil A is $71%_$ .

The percentage of sand for soil A is $28%_$ .

The percentage of fines for soil A is $1%_$ .

The percentage of gravel for soil B is $55%_$ .

The percentage of sand for soil B is $43%_$ .

The percentage of fines for soil B is $2%_$ .

The percentage of gravel for soil C is $47%_$ .

The percentage of sand for soil C is $50%_$ .

The percentage of fines for soil C is $3%_$ .

Explanation of Solution

Refer to Figure 1.

For soil A.

The percent passing through $4.75 mm$ sieve is $29%$ .

The percent passing through $0.075 mm$ sieve is $1%$ .

Calculate the percentage of gravel as shown below.

$Percent of gravel=100−percent passing through 4.75 mm sieve=100−29=71%$

Hence, the percentage of gravel is $71%_$ .

Calculate the percentage of sand as shown below.

$Percent of sand=(percent passing through 4.75 mm sieve)−(percent passing through 0.075 mm sieve)=29−1=28%$

Hence, the percentage of sand is $28%_$ .

Calculate the percentage of fines as shown below.

$Percent of fines=percent passing through 0.075 mm sieve−0=1−0=1%$

Hence, the percentage of fines is $1%_$ .

Refer to Figure 1.

For soil B.

The percent passing through $4.75 mm$ sieve is $45%$ .

The percent passing through $0.075 mm$ sieve is $2%$ .

Calculate the percentage of gravel as shown below.

$Percent of gravel=100−percent passing through 4.75 mm sieve=100−45=55%$

Hence, the percentage of gravel is $55%_$ .

Calculate the percentage of sand as shown below.

$Percent of sand=(percent passing through 4.75 mm sieve)−(percent passing through 0.075 mm sieve)=45−2=43%$

Hence, the percentage of sand is $43%_$ .

Calculate the percentage of fines as shown below.

$Percent of fines=percent passing through 0.075 mm sieve−0=2−0=2%$

Hence, the percentage of fines is $2%_$ .

Refer to Figure 1.

For soil C.

The percent passing through $4.75 mm$ sieve is $53%$ .

The percent passing through $0.075 mm$ sieve is $3%$ .

Calculate the percentage of gravel as shown below.

$Percent of gravel=100−percent passing through 4.75 mm sieve=100−53=47%$

Hence, the percentage of gravel is $47%_$ .

Calculate the percentage of sand as shown below.

$Percent of sand=(percent passing through 4.75 mm sieve)−(percent passing through 0.075 mm sieve)=53−3=50%$

Hence, the percentage of sand is $50%_$ .

Calculate the percentage of fines as shown below.

$Percent of fines=percent passing through 0.075 mm sieve−0=3−0=3%$

Hence, the percentage of fines is $3%_$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

*13-12. The control linkage for a machine consists of two L2 steel rods BE and FG, each with a diameter of 1 in. If a device at G causes the end G to freeze up and become pin connected, determine the maximum horizontal force P that could be applied to the handle without causing either of the two rods to buckle. The members are pin connected at A, B, D, E, and F. P 12 in. C G F B 2 in. E 4 in. 4 in. A D + -15 in.- + -20 in.-

13-31. The steel bar AB has a rectangular cross section. If it is pin connected at its ends, determine the maximum allowable intensity w of the distributed load that can be applied to BC without causing bar AB to buckle. Use a factor of safety with respect to buckling of 1.5. Est = 200 GPa, σy 360 MPa. = B W 5 m 3 m -20 mm 30 mm x x A 20 mm y C

Problem 1: A man-made 30 ft tall, 1.5:1 slope is to be build as shown in the figure. The soil is homogeneous with shear strength parameters c = 400 psf and φ = 290 . The moist unit weight of the soil is 119 pcf above the groundwater table and the saturated unit weight is 123 pcf below. Using the ordinary method of slices, compute the Factor of Safety (FS) along the trial failure surface shown. (Hint: Please note the unit weight is changing within the same slice.) Note 1: Use the same number of slices and dimensions as provided. Document ALL the calculations of weight (W) for each slice. Note 2: Document your solutions by following the same approach illustrated in the class, including a summary table showing all the variables and calculations involved in assessing FS.

Answer 2

Textbook Question

Chapter 2, Problem 2.1CTP

Three groups of students from the Geotechnical Engineering class collected soil-aggregate samples for laboratory testing from a recycled aggregate processing plant in Palm Beach County, Florida. Three samples, denoted by Soil A, Soil B, and Soil C, were collected from three locations of the aggregate stockpile, and sieve analyses were conducted (see Figure 2.35).

Chapter 2, Problem 2.1CTP, Three groups of students from the Geotechnical Engineering class collected soil-aggregate samples , example 1

(a)

Chapter 2, Problem 2.1CTP, Three groups of students from the Geotechnical Engineering class collected soil-aggregate samples , example 2

(b)

Figure 2.35 (a) Soil-aggregate stockpile; (b) sieve analysis (Courtesy of Khaled Sobhan, Florida Atlantic University, Boca Raton, Florida)

a. Determine the coefficient of uniformity and the coefficient of gradation for Soils A, B, and C.

b. Which one is coarser: Soil A or Soil C? Justify your answer.

c. Although the soils are obtained from the same stockpile, why are the curves so different? (Hint: Comment on particle segregation and representative field sampling.)

d. Determine the percentages of gravel, sand and fines according to Unified Soil Classification System.

(a)

Expert Solution

To determine

Calculate the coefficient of uniformity $(Cu)$ and coefficient of gradation $(Cc)$ for soils A, B, and C.

Answer to Problem 2.1CTP

The uniformity coefficient of soil A is $26.42_$ .

The coefficient of gradation of soil A is $3.64_$ .

The uniformity coefficient of soil B is $35_$ .

The coefficient of gradation of soil B is $3.35_$ .

The uniformity coefficient of soil C is $46.67_$ .

The coefficient of gradation of soil C is $1.15_$ .

Explanation of Solution

Sketch the grain size distribution curve for soils A, B, and C as shown in Figure 1.

Principles of Geotechnical Engineering (MindTap Course List), Chapter 2, Problem 2.1CTP

Refer to Figure 1.

For soil A:

The diameter of the particle corresponding to $60%$ finer $(D60)$ is $14 mm$ .

The diameter of the particle corresponding to $30%$ finer $(D30)$ is $5.2 mm$ .

The diameter of the particle corresponding to $10%$ finer $(D10)$ is $0.53 mm$ .

For soil B:

The diameter of the particle corresponding to $60%$ finer $(D60)$ is $8.1 mm$ .

The diameter of the particle corresponding to $30%$ finer $(D30)$ is $2.5 mm$ .

The diameter of the particle corresponding to $10%$ finer $(D10)$ is $0.23 mm$ .

For soil C:

The diameter of the particle corresponding to $60%$ finer $(D60)$ is $7 mm$ .

The diameter of the particle corresponding to $30%$ finer $(D30)$ is $1.1 mm$ .

The diameter of the particle corresponding to $10%$ finer $(D10)$ is $0.15 mm$ .

Calculate the uniformity coefficient $(Cu)$ using the relation.

$Cu=D60D10$ (1)

For soil A:

Substitute $14 mm$ for $D60$ and $0.53 mm$ for $D10$ in Equation (1).

$Cu=140.53=26.42$

Hence, the uniformity coefficient for soil A is $26.42_$ .

For soil B:

Substitute $8.1 mm$ for $D60$ and $0.23 mm$ for $D10$ in Equation (1).

$Cu=8.10.23=35$

Hence, the uniformity coefficient for soil B is $35_$ .

For soil C:

Substitute $7 mm$ for $D60$ and $0.15 mm$ for $D10$ in Equation (1).

$Cu=70.15=46.67$

Hence, the uniformity coefficient for soil C is $46.67_$ .

Calculate the coefficient of gradation $(Cc)$ using the relation.

$Cc=D302D60×D10$ (2)

For soil A:

Substitute $14 mm$ for $D60$ , $5.2 mm$ for $D30$ , and $0.53 mm$ for $D10$ in Equation (2).

$Cc=5.2214×0.53=27.045.512=3.64$

Hence, the coefficient of gradation for soil A is $3.64_$ .

For soil B:

Substitute $8.1 mm$ for $D60$ , $2.5 mm$ for $D30$ , and $0.23 mm$ for $D10$ in Equation (2).

$Cc=2.528.1×0.23=6.251.863=3.35$

Hence, the coefficient of gradation for soil B is $3.35_$ .

For soil C:

Substitute $7 mm$ for $D60$ , $1.1 mm$ for $D30$ , and $0.15 mm$ for $D10$ in Equation (2).

$Cc=1.127×0.15=1.211.05=1.15$

Therefore, the coefficient of gradation for soil C is $1.15_$ .

(b)

Expert Solution

To determine

State which of the soil is coarser from soil A and C.

Answer to Problem 2.1CTP

Soil A is coarser than soil C.

Explanation of Solution

Refer to part (a).

The uniformity coefficient of soil A is $26.42$ .

The uniformity coefficient of soil C is $46.67$ .

The percent of soil finer than $1 mm$ for soil A is $20%$ .

The percent of soil finer than $1 mm$ for soil C is $47%$ .

Hence, a higher percentage of soil C is finer than soil A.

Hence, soil A is coarser than soil C.

(c)

Expert Solution

To determine

Explain the reason for curve different of soil A, B and C if it is obtained from same stockpile.

Explanation of Solution

The particle-size distribution curve shows the range of particle sizes present in a soil and the type of distribution of various-size particles.

Refer to Figure 1.

Particle separation of coarser and finer particles may take place in aggregate stockpiles. This makes representative sampling difficult.

Therefore, the particle-size distribution curve is different for soils A, B, and C.

(d)

Expert Solution

To determine

Calculate the percentages of gravel, sand, and fines according to the Unified Soil Classification System.

Answer to Problem 2.1CTP

The percentage of gravel for soil A is $71%_$ .

The percentage of sand for soil A is $28%_$ .

The percentage of fines for soil A is $1%_$ .

The percentage of gravel for soil B is $55%_$ .

The percentage of sand for soil B is $43%_$ .

The percentage of fines for soil B is $2%_$ .

The percentage of gravel for soil C is $47%_$ .

The percentage of sand for soil C is $50%_$ .

The percentage of fines for soil C is $3%_$ .

Explanation of Solution

Refer to Figure 1.

For soil A.

The percent passing through $4.75 mm$ sieve is $29%$ .

The percent passing through $0.075 mm$ sieve is $1%$ .

Calculate the percentage of gravel as shown below.

$Percent of gravel=100−percent passing through 4.75 mm sieve=100−29=71%$

Hence, the percentage of gravel is $71%_$ .

Calculate the percentage of sand as shown below.

$Percent of sand=(percent passing through 4.75 mm sieve)−(percent passing through 0.075 mm sieve)=29−1=28%$

Hence, the percentage of sand is $28%_$ .

Calculate the percentage of fines as shown below.

$Percent of fines=percent passing through 0.075 mm sieve−0=1−0=1%$

Hence, the percentage of fines is $1%_$ .

Refer to Figure 1.

For soil B.

The percent passing through $4.75 mm$ sieve is $45%$ .

The percent passing through $0.075 mm$ sieve is $2%$ .

Calculate the percentage of gravel as shown below.

$Percent of gravel=100−percent passing through 4.75 mm sieve=100−45=55%$

Hence, the percentage of gravel is $55%_$ .

Calculate the percentage of sand as shown below.

$Percent of sand=(percent passing through 4.75 mm sieve)−(percent passing through 0.075 mm sieve)=45−2=43%$

Hence, the percentage of sand is $43%_$ .

Calculate the percentage of fines as shown below.

$Percent of fines=percent passing through 0.075 mm sieve−0=2−0=2%$

Hence, the percentage of fines is $2%_$ .

Refer to Figure 1.

For soil C.

The percent passing through $4.75 mm$ sieve is $53%$ .

The percent passing through $0.075 mm$ sieve is $3%$ .

Calculate the percentage of gravel as shown below.

$Percent of gravel=100−percent passing through 4.75 mm sieve=100−53=47%$

Hence, the percentage of gravel is $47%_$ .

Calculate the percentage of sand as shown below.

$Percent of sand=(percent passing through 4.75 mm sieve)−(percent passing through 0.075 mm sieve)=53−3=50%$

Hence, the percentage of sand is $50%_$ .

Calculate the percentage of fines as shown below.

$Percent of fines=percent passing through 0.075 mm sieve−0=3−0=3%$

Hence, the percentage of fines is $3%_$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Textbook Question

Chapter 2, Problem 2.1CTP

Three groups of students from the Geotechnical Engineering class collected soil-aggregate samples for laboratory testing from a recycled aggregate processing plant in Palm Beach County, Florida. Three samples, denoted by Soil A, Soil B, and Soil C, were collected from three locations of the aggregate stockpile, and sieve analyses were conducted (see Figure 2.35).

Chapter 2, Problem 2.1CTP, Three groups of students from the Geotechnical Engineering class collected soil-aggregate samples , example 1

(a)

Chapter 2, Problem 2.1CTP, Three groups of students from the Geotechnical Engineering class collected soil-aggregate samples , example 2

(b)

Figure 2.35 (a) Soil-aggregate stockpile; (b) sieve analysis (Courtesy of Khaled Sobhan, Florida Atlantic University, Boca Raton, Florida)

a. Determine the coefficient of uniformity and the coefficient of gradation for Soils A, B, and C.

b. Which one is coarser: Soil A or Soil C? Justify your answer.

c. Although the soils are obtained from the same stockpile, why are the curves so different? (Hint: Comment on particle segregation and representative field sampling.)

d. Determine the percentages of gravel, sand and fines according to Unified Soil Classification System.

(a)

Expert Solution

To determine

Calculate the coefficient of uniformity $(Cu)$ and coefficient of gradation $(Cc)$ for soils A, B, and C.

Answer to Problem 2.1CTP

The uniformity coefficient of soil A is $26.42_$ .

The coefficient of gradation of soil A is $3.64_$ .

The uniformity coefficient of soil B is $35_$ .

The coefficient of gradation of soil B is $3.35_$ .

The uniformity coefficient of soil C is $46.67_$ .

The coefficient of gradation of soil C is $1.15_$ .

Explanation of Solution

Sketch the grain size distribution curve for soils A, B, and C as shown in Figure 1.

Principles of Geotechnical Engineering (MindTap Course List), Chapter 2, Problem 2.1CTP

Refer to Figure 1.

For soil A:

The diameter of the particle corresponding to $60%$ finer $(D60)$ is $14 mm$ .

The diameter of the particle corresponding to $30%$ finer $(D30)$ is $5.2 mm$ .

The diameter of the particle corresponding to $10%$ finer $(D10)$ is $0.53 mm$ .

For soil B:

The diameter of the particle corresponding to $60%$ finer $(D60)$ is $8.1 mm$ .

The diameter of the particle corresponding to $30%$ finer $(D30)$ is $2.5 mm$ .

The diameter of the particle corresponding to $10%$ finer $(D10)$ is $0.23 mm$ .

For soil C:

The diameter of the particle corresponding to $60%$ finer $(D60)$ is $7 mm$ .

The diameter of the particle corresponding to $30%$ finer $(D30)$ is $1.1 mm$ .

The diameter of the particle corresponding to $10%$ finer $(D10)$ is $0.15 mm$ .

Calculate the uniformity coefficient $(Cu)$ using the relation.

$Cu=D60D10$ (1)

For soil A:

Substitute $14 mm$ for $D60$ and $0.53 mm$ for $D10$ in Equation (1).

$Cu=140.53=26.42$

Hence, the uniformity coefficient for soil A is $26.42_$ .

For soil B:

Substitute $8.1 mm$ for $D60$ and $0.23 mm$ for $D10$ in Equation (1).

$Cu=8.10.23=35$

Hence, the uniformity coefficient for soil B is $35_$ .

For soil C:

Substitute $7 mm$ for $D60$ and $0.15 mm$ for $D10$ in Equation (1).

$Cu=70.15=46.67$

Hence, the uniformity coefficient for soil C is $46.67_$ .

Calculate the coefficient of gradation $(Cc)$ using the relation.

$Cc=D302D60×D10$ (2)

For soil A:

Substitute $14 mm$ for $D60$ , $5.2 mm$ for $D30$ , and $0.53 mm$ for $D10$ in Equation (2).

$Cc=5.2214×0.53=27.045.512=3.64$

Hence, the coefficient of gradation for soil A is $3.64_$ .

For soil B:

Substitute $8.1 mm$ for $D60$ , $2.5 mm$ for $D30$ , and $0.23 mm$ for $D10$ in Equation (2).

$Cc=2.528.1×0.23=6.251.863=3.35$

Hence, the coefficient of gradation for soil B is $3.35_$ .

For soil C:

Substitute $7 mm$ for $D60$ , $1.1 mm$ for $D30$ , and $0.15 mm$ for $D10$ in Equation (2).

$Cc=1.127×0.15=1.211.05=1.15$

Therefore, the coefficient of gradation for soil C is $1.15_$ .

(b)

Expert Solution

To determine

State which of the soil is coarser from soil A and C.

Answer to Problem 2.1CTP

Soil A is coarser than soil C.

Explanation of Solution

Refer to part (a).

The uniformity coefficient of soil A is $26.42$ .

The uniformity coefficient of soil C is $46.67$ .

The percent of soil finer than $1 mm$ for soil A is $20%$ .

The percent of soil finer than $1 mm$ for soil C is $47%$ .

Hence, a higher percentage of soil C is finer than soil A.

Hence, soil A is coarser than soil C.

(c)

Expert Solution

To determine

Explain the reason for curve different of soil A, B and C if it is obtained from same stockpile.

Explanation of Solution

The particle-size distribution curve shows the range of particle sizes present in a soil and the type of distribution of various-size particles.

Refer to Figure 1.

Particle separation of coarser and finer particles may take place in aggregate stockpiles. This makes representative sampling difficult.

Therefore, the particle-size distribution curve is different for soils A, B, and C.

(d)

Expert Solution

To determine

Calculate the percentages of gravel, sand, and fines according to the Unified Soil Classification System.

Answer to Problem 2.1CTP

The percentage of gravel for soil A is $71%_$ .

The percentage of sand for soil A is $28%_$ .

The percentage of fines for soil A is $1%_$ .

The percentage of gravel for soil B is $55%_$ .

The percentage of sand for soil B is $43%_$ .

The percentage of fines for soil B is $2%_$ .

The percentage of gravel for soil C is $47%_$ .

The percentage of sand for soil C is $50%_$ .

The percentage of fines for soil C is $3%_$ .

Explanation of Solution

Refer to Figure 1.

For soil A.

The percent passing through $4.75 mm$ sieve is $29%$ .

The percent passing through $0.075 mm$ sieve is $1%$ .

Calculate the percentage of gravel as shown below.

$Percent of gravel=100−percent passing through 4.75 mm sieve=100−29=71%$

Hence, the percentage of gravel is $71%_$ .

Calculate the percentage of sand as shown below.

$Percent of sand=(percent passing through 4.75 mm sieve)−(percent passing through 0.075 mm sieve)=29−1=28%$

Hence, the percentage of sand is $28%_$ .

Calculate the percentage of fines as shown below.

$Percent of fines=percent passing through 0.075 mm sieve−0=1−0=1%$

Hence, the percentage of fines is $1%_$ .

Refer to Figure 1.

For soil B.

The percent passing through $4.75 mm$ sieve is $45%$ .

The percent passing through $0.075 mm$ sieve is $2%$ .

Calculate the percentage of gravel as shown below.

$Percent of gravel=100−percent passing through 4.75 mm sieve=100−45=55%$

Hence, the percentage of gravel is $55%_$ .

Calculate the percentage of sand as shown below.

$Percent of sand=(percent passing through 4.75 mm sieve)−(percent passing through 0.075 mm sieve)=45−2=43%$

Hence, the percentage of sand is $43%_$ .

Calculate the percentage of fines as shown below.

$Percent of fines=percent passing through 0.075 mm sieve−0=2−0=2%$

Hence, the percentage of fines is $2%_$ .

Refer to Figure 1.

For soil C.

The percent passing through $4.75 mm$ sieve is $53%$ .

The percent passing through $0.075 mm$ sieve is $3%$ .

Calculate the percentage of gravel as shown below.

$Percent of gravel=100−percent passing through 4.75 mm sieve=100−53=47%$

Hence, the percentage of gravel is $47%_$ .

Calculate the percentage of sand as shown below.

$Percent of sand=(percent passing through 4.75 mm sieve)−(percent passing through 0.075 mm sieve)=53−3=50%$

Hence, the percentage of sand is $50%_$ .

Calculate the percentage of fines as shown below.

$Percent of fines=percent passing through 0.075 mm sieve−0=3−0=3%$

Hence, the percentage of fines is $3%_$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Textbook Question

Chapter 2, Problem 2.1CTP

Three groups of students from the Geotechnical Engineering class collected soil-aggregate samples for laboratory testing from a recycled aggregate processing plant in Palm Beach County, Florida. Three samples, denoted by Soil A, Soil B, and Soil C, were collected from three locations of the aggregate stockpile, and sieve analyses were conducted (see Figure 2.35).

Chapter 2, Problem 2.1CTP, Three groups of students from the Geotechnical Engineering class collected soil-aggregate samples , example 1

(a)

Chapter 2, Problem 2.1CTP, Three groups of students from the Geotechnical Engineering class collected soil-aggregate samples , example 2

(b)

Figure 2.35 (a) Soil-aggregate stockpile; (b) sieve analysis (Courtesy of Khaled Sobhan, Florida Atlantic University, Boca Raton, Florida)

a. Determine the coefficient of uniformity and the coefficient of gradation for Soils A, B, and C.

b. Which one is coarser: Soil A or Soil C? Justify your answer.

c. Although the soils are obtained from the same stockpile, why are the curves so different? (Hint: Comment on particle segregation and representative field sampling.)

d. Determine the percentages of gravel, sand and fines according to Unified Soil Classification System.

(a)

Expert Solution

To determine

Calculate the coefficient of uniformity $(Cu)$ and coefficient of gradation $(Cc)$ for soils A, B, and C.

Answer to Problem 2.1CTP

The uniformity coefficient of soil A is $26.42_$ .

The coefficient of gradation of soil A is $3.64_$ .

The uniformity coefficient of soil B is $35_$ .

The coefficient of gradation of soil B is $3.35_$ .

The uniformity coefficient of soil C is $46.67_$ .

The coefficient of gradation of soil C is $1.15_$ .

Explanation of Solution

Sketch the grain size distribution curve for soils A, B, and C as shown in Figure 1.

Principles of Geotechnical Engineering (MindTap Course List), Chapter 2, Problem 2.1CTP

Refer to Figure 1.

For soil A:

The diameter of the particle corresponding to $60%$ finer $(D60)$ is $14 mm$ .

The diameter of the particle corresponding to $30%$ finer $(D30)$ is $5.2 mm$ .

The diameter of the particle corresponding to $10%$ finer $(D10)$ is $0.53 mm$ .

For soil B:

The diameter of the particle corresponding to $60%$ finer $(D60)$ is $8.1 mm$ .

The diameter of the particle corresponding to $30%$ finer $(D30)$ is $2.5 mm$ .

The diameter of the particle corresponding to $10%$ finer $(D10)$ is $0.23 mm$ .

For soil C:

The diameter of the particle corresponding to $60%$ finer $(D60)$ is $7 mm$ .

The diameter of the particle corresponding to $30%$ finer $(D30)$ is $1.1 mm$ .

The diameter of the particle corresponding to $10%$ finer $(D10)$ is $0.15 mm$ .

Calculate the uniformity coefficient $(Cu)$ using the relation.

$Cu=D60D10$ (1)

For soil A:

Substitute $14 mm$ for $D60$ and $0.53 mm$ for $D10$ in Equation (1).

$Cu=140.53=26.42$

Hence, the uniformity coefficient for soil A is $26.42_$ .

For soil B:

Substitute $8.1 mm$ for $D60$ and $0.23 mm$ for $D10$ in Equation (1).

$Cu=8.10.23=35$

Hence, the uniformity coefficient for soil B is $35_$ .

For soil C:

Substitute $7 mm$ for $D60$ and $0.15 mm$ for $D10$ in Equation (1).

$Cu=70.15=46.67$

Hence, the uniformity coefficient for soil C is $46.67_$ .

Calculate the coefficient of gradation $(Cc)$ using the relation.

$Cc=D302D60×D10$ (2)

For soil A:

Substitute $14 mm$ for $D60$ , $5.2 mm$ for $D30$ , and $0.53 mm$ for $D10$ in Equation (2).

$Cc=5.2214×0.53=27.045.512=3.64$

Hence, the coefficient of gradation for soil A is $3.64_$ .

For soil B:

Substitute $8.1 mm$ for $D60$ , $2.5 mm$ for $D30$ , and $0.23 mm$ for $D10$ in Equation (2).

$Cc=2.528.1×0.23=6.251.863=3.35$

Hence, the coefficient of gradation for soil B is $3.35_$ .

For soil C:

Substitute $7 mm$ for $D60$ , $1.1 mm$ for $D30$ , and $0.15 mm$ for $D10$ in Equation (2).

$Cc=1.127×0.15=1.211.05=1.15$

Therefore, the coefficient of gradation for soil C is $1.15_$ .

(b)

Expert Solution

To determine

State which of the soil is coarser from soil A and C.

Answer to Problem 2.1CTP

Soil A is coarser than soil C.

Explanation of Solution

Refer to part (a).

The uniformity coefficient of soil A is $26.42$ .

The uniformity coefficient of soil C is $46.67$ .

The percent of soil finer than $1 mm$ for soil A is $20%$ .

The percent of soil finer than $1 mm$ for soil C is $47%$ .

Hence, a higher percentage of soil C is finer than soil A.

Hence, soil A is coarser than soil C.

(c)

Expert Solution

To determine

Explain the reason for curve different of soil A, B and C if it is obtained from same stockpile.

Explanation of Solution

The particle-size distribution curve shows the range of particle sizes present in a soil and the type of distribution of various-size particles.

Refer to Figure 1.

Particle separation of coarser and finer particles may take place in aggregate stockpiles. This makes representative sampling difficult.

Therefore, the particle-size distribution curve is different for soils A, B, and C.

(d)

Expert Solution

To determine

Calculate the percentages of gravel, sand, and fines according to the Unified Soil Classification System.

Answer to Problem 2.1CTP

The percentage of gravel for soil A is $71%_$ .

The percentage of sand for soil A is $28%_$ .

The percentage of fines for soil A is $1%_$ .

The percentage of gravel for soil B is $55%_$ .

The percentage of sand for soil B is $43%_$ .

The percentage of fines for soil B is $2%_$ .

The percentage of gravel for soil C is $47%_$ .

The percentage of sand for soil C is $50%_$ .

The percentage of fines for soil C is $3%_$ .

Explanation of Solution

Refer to Figure 1.

For soil A.

The percent passing through $4.75 mm$ sieve is $29%$ .

The percent passing through $0.075 mm$ sieve is $1%$ .

Calculate the percentage of gravel as shown below.

$Percent of gravel=100−percent passing through 4.75 mm sieve=100−29=71%$

Hence, the percentage of gravel is $71%_$ .

Calculate the percentage of sand as shown below.

$Percent of sand=(percent passing through 4.75 mm sieve)−(percent passing through 0.075 mm sieve)=29−1=28%$

Hence, the percentage of sand is $28%_$ .

Calculate the percentage of fines as shown below.

$Percent of fines=percent passing through 0.075 mm sieve−0=1−0=1%$

Hence, the percentage of fines is $1%_$ .

Refer to Figure 1.

For soil B.

The percent passing through $4.75 mm$ sieve is $45%$ .

The percent passing through $0.075 mm$ sieve is $2%$ .

Calculate the percentage of gravel as shown below.

$Percent of gravel=100−percent passing through 4.75 mm sieve=100−45=55%$

Hence, the percentage of gravel is $55%_$ .

Calculate the percentage of sand as shown below.

$Percent of sand=(percent passing through 4.75 mm sieve)−(percent passing through 0.075 mm sieve)=45−2=43%$

Hence, the percentage of sand is $43%_$ .

Calculate the percentage of fines as shown below.

$Percent of fines=percent passing through 0.075 mm sieve−0=2−0=2%$

Hence, the percentage of fines is $2%_$ .

Refer to Figure 1.

For soil C.

The percent passing through $4.75 mm$ sieve is $53%$ .

The percent passing through $0.075 mm$ sieve is $3%$ .

Calculate the percentage of gravel as shown below.

$Percent of gravel=100−percent passing through 4.75 mm sieve=100−53=47%$

Hence, the percentage of gravel is $47%_$ .

Calculate the percentage of sand as shown below.

$Percent of sand=(percent passing through 4.75 mm sieve)−(percent passing through 0.075 mm sieve)=53−3=50%$

Hence, the percentage of sand is $50%_$ .

Calculate the percentage of fines as shown below.

$Percent of fines=percent passing through 0.075 mm sieve−0=3−0=3%$

Hence, the percentage of fines is $3%_$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

Calculate the coefficient of uniformity $(Cu)$ and coefficient of gradation $(Cc)$ for soils A, B, and C.

Answer to Problem 2.1CTP

The uniformity coefficient of soil A is $26.42_$ .

The coefficient of gradation of soil A is $3.64_$ .

The uniformity coefficient of soil B is $35_$ .

The coefficient of gradation of soil B is $3.35_$ .

The uniformity coefficient of soil C is $46.67_$ .

The coefficient of gradation of soil C is $1.15_$ .

Explanation of Solution

Sketch the grain size distribution curve for soils A, B, and C as shown in Figure 1.

Principles of Geotechnical Engineering (MindTap Course List), Chapter 2, Problem 2.1CTP

Refer to Figure 1.

For soil A:

The diameter of the particle corresponding to $60%$ finer $(D60)$ is $14 mm$ .

The diameter of the particle corresponding to $30%$ finer $(D30)$ is $5.2 mm$ .

The diameter of the particle corresponding to $10%$ finer $(D10)$ is $0.53 mm$ .

For soil B:

The diameter of the particle corresponding to $60%$ finer $(D60)$ is $8.1 mm$ .

The diameter of the particle corresponding to $30%$ finer $(D30)$ is $2.5 mm$ .

The diameter of the particle corresponding to $10%$ finer $(D10)$ is $0.23 mm$ .

For soil C:

The diameter of the particle corresponding to $60%$ finer $(D60)$ is $7 mm$ .

The diameter of the particle corresponding to $30%$ finer $(D30)$ is $1.1 mm$ .

The diameter of the particle corresponding to $10%$ finer $(D10)$ is $0.15 mm$ .

Calculate the uniformity coefficient $(Cu)$ using the relation.

$Cu=D60D10$ (1)

For soil A:

Substitute $14 mm$ for $D60$ and $0.53 mm$ for $D10$ in Equation (1).

$Cu=140.53=26.42$

Hence, the uniformity coefficient for soil A is $26.42_$ .

For soil B:

Substitute $8.1 mm$ for $D60$ and $0.23 mm$ for $D10$ in Equation (1).

$Cu=8.10.23=35$

Hence, the uniformity coefficient for soil B is $35_$ .

For soil C:

Substitute $7 mm$ for $D60$ and $0.15 mm$ for $D10$ in Equation (1).

$Cu=70.15=46.67$

Hence, the uniformity coefficient for soil C is $46.67_$ .

Calculate the coefficient of gradation $(Cc)$ using the relation.

$Cc=D302D60×D10$ (2)

For soil A:

Substitute $14 mm$ for $D60$ , $5.2 mm$ for $D30$ , and $0.53 mm$ for $D10$ in Equation (2).

$Cc=5.2214×0.53=27.045.512=3.64$

Hence, the coefficient of gradation for soil A is $3.64_$ .

For soil B:

Substitute $8.1 mm$ for $D60$ , $2.5 mm$ for $D30$ , and $0.23 mm$ for $D10$ in Equation (2).

$Cc=2.528.1×0.23=6.251.863=3.35$

Hence, the coefficient of gradation for soil B is $3.35_$ .

For soil C:

Substitute $7 mm$ for $D60$ , $1.1 mm$ for $D30$ , and $0.15 mm$ for $D10$ in Equation (2).

$Cc=1.127×0.15=1.211.05=1.15$

Therefore, the coefficient of gradation for soil C is $1.15_$ .

(b)

Expert Solution

To determine

State which of the soil is coarser from soil A and C.

Answer to Problem 2.1CTP

Soil A is coarser than soil C.

Explanation of Solution

Refer to part (a).

The uniformity coefficient of soil A is $26.42$ .

The uniformity coefficient of soil C is $46.67$ .

The percent of soil finer than $1 mm$ for soil A is $20%$ .

The percent of soil finer than $1 mm$ for soil C is $47%$ .

Hence, a higher percentage of soil C is finer than soil A.

Hence, soil A is coarser than soil C.

(c)

Expert Solution

To determine

Explain the reason for curve different of soil A, B and C if it is obtained from same stockpile.

Explanation of Solution

The particle-size distribution curve shows the range of particle sizes present in a soil and the type of distribution of various-size particles.

Refer to Figure 1.

Particle separation of coarser and finer particles may take place in aggregate stockpiles. This makes representative sampling difficult.

Therefore, the particle-size distribution curve is different for soils A, B, and C.

(d)

Expert Solution

To determine

Calculate the percentages of gravel, sand, and fines according to the Unified Soil Classification System.

Answer to Problem 2.1CTP

The percentage of gravel for soil A is $71%_$ .

The percentage of sand for soil A is $28%_$ .

The percentage of fines for soil A is $1%_$ .

The percentage of gravel for soil B is $55%_$ .

The percentage of sand for soil B is $43%_$ .

The percentage of fines for soil B is $2%_$ .

The percentage of gravel for soil C is $47%_$ .

The percentage of sand for soil C is $50%_$ .

The percentage of fines for soil C is $3%_$ .

Explanation of Solution

Refer to Figure 1.

For soil A.

The percent passing through $4.75 mm$ sieve is $29%$ .

The percent passing through $0.075 mm$ sieve is $1%$ .

Calculate the percentage of gravel as shown below.

$Percent of gravel=100−percent passing through 4.75 mm sieve=100−29=71%$

Hence, the percentage of gravel is $71%_$ .

Calculate the percentage of sand as shown below.

$Percent of sand=(percent passing through 4.75 mm sieve)−(percent passing through 0.075 mm sieve)=29−1=28%$

Hence, the percentage of sand is $28%_$ .

Calculate the percentage of fines as shown below.

$Percent of fines=percent passing through 0.075 mm sieve−0=1−0=1%$

Hence, the percentage of fines is $1%_$ .

Refer to Figure 1.

For soil B.

The percent passing through $4.75 mm$ sieve is $45%$ .

The percent passing through $0.075 mm$ sieve is $2%$ .

Calculate the percentage of gravel as shown below.

$Percent of gravel=100−percent passing through 4.75 mm sieve=100−45=55%$

Hence, the percentage of gravel is $55%_$ .

Calculate the percentage of sand as shown below.

$Percent of sand=(percent passing through 4.75 mm sieve)−(percent passing through 0.075 mm sieve)=45−2=43%$

Hence, the percentage of sand is $43%_$ .

Calculate the percentage of fines as shown below.

$Percent of fines=percent passing through 0.075 mm sieve−0=2−0=2%$

Hence, the percentage of fines is $2%_$ .

Refer to Figure 1.

For soil C.

The percent passing through $4.75 mm$ sieve is $53%$ .

The percent passing through $0.075 mm$ sieve is $3%$ .

Calculate the percentage of gravel as shown below.

$Percent of gravel=100−percent passing through 4.75 mm sieve=100−53=47%$

Hence, the percentage of gravel is $47%_$ .

Calculate the percentage of sand as shown below.

$Percent of sand=(percent passing through 4.75 mm sieve)−(percent passing through 0.075 mm sieve)=53−3=50%$

Hence, the percentage of sand is $50%_$ .

Calculate the percentage of fines as shown below.

$Percent of fines=percent passing through 0.075 mm sieve−0=3−0=3%$

Hence, the percentage of fines is $3%_$ .

Answer 8

(a)

Expert Solution

To determine

Calculate the coefficient of uniformity $(Cu)$ and coefficient of gradation $(Cc)$ for soils A, B, and C.

Answer to Problem 2.1CTP

The uniformity coefficient of soil A is $26.42_$ .

The coefficient of gradation of soil A is $3.64_$ .

The uniformity coefficient of soil B is $35_$ .

The coefficient of gradation of soil B is $3.35_$ .

The uniformity coefficient of soil C is $46.67_$ .

The coefficient of gradation of soil C is $1.15_$ .

Explanation of Solution

Sketch the grain size distribution curve for soils A, B, and C as shown in Figure 1.

Principles of Geotechnical Engineering (MindTap Course List), Chapter 2, Problem 2.1CTP

Refer to Figure 1.

For soil A:

The diameter of the particle corresponding to $60%$ finer $(D60)$ is $14 mm$ .

The diameter of the particle corresponding to $30%$ finer $(D30)$ is $5.2 mm$ .

The diameter of the particle corresponding to $10%$ finer $(D10)$ is $0.53 mm$ .

For soil B:

The diameter of the particle corresponding to $60%$ finer $(D60)$ is $8.1 mm$ .

The diameter of the particle corresponding to $30%$ finer $(D30)$ is $2.5 mm$ .

The diameter of the particle corresponding to $10%$ finer $(D10)$ is $0.23 mm$ .

For soil C:

The diameter of the particle corresponding to $60%$ finer $(D60)$ is $7 mm$ .

The diameter of the particle corresponding to $30%$ finer $(D30)$ is $1.1 mm$ .

The diameter of the particle corresponding to $10%$ finer $(D10)$ is $0.15 mm$ .

Calculate the uniformity coefficient $(Cu)$ using the relation.

$Cu=D60D10$ (1)

For soil A:

Substitute $14 mm$ for $D60$ and $0.53 mm$ for $D10$ in Equation (1).

$Cu=140.53=26.42$

Hence, the uniformity coefficient for soil A is $26.42_$ .

For soil B:

Substitute $8.1 mm$ for $D60$ and $0.23 mm$ for $D10$ in Equation (1).

$Cu=8.10.23=35$

Hence, the uniformity coefficient for soil B is $35_$ .

For soil C:

Substitute $7 mm$ for $D60$ and $0.15 mm$ for $D10$ in Equation (1).

$Cu=70.15=46.67$

Hence, the uniformity coefficient for soil C is $46.67_$ .

Calculate the coefficient of gradation $(Cc)$ using the relation.

$Cc=D302D60×D10$ (2)

For soil A:

Substitute $14 mm$ for $D60$ , $5.2 mm$ for $D30$ , and $0.53 mm$ for $D10$ in Equation (2).

$Cc=5.2214×0.53=27.045.512=3.64$

Hence, the coefficient of gradation for soil A is $3.64_$ .

For soil B:

Substitute $8.1 mm$ for $D60$ , $2.5 mm$ for $D30$ , and $0.23 mm$ for $D10$ in Equation (2).

$Cc=2.528.1×0.23=6.251.863=3.35$

Hence, the coefficient of gradation for soil B is $3.35_$ .

For soil C:

Substitute $7 mm$ for $D60$ , $1.1 mm$ for $D30$ , and $0.15 mm$ for $D10$ in Equation (2).

$Cc=1.127×0.15=1.211.05=1.15$

Therefore, the coefficient of gradation for soil C is $1.15_$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

Calculate the coefficient of uniformity $(Cu)$ and coefficient of gradation $(Cc)$ for soils A, B, and C.

Answer 13

Answer to Problem 2.1CTP

The uniformity coefficient of soil A is $26.42_$ .

The coefficient of gradation of soil A is $3.64_$ .

The uniformity coefficient of soil B is $35_$ .

The coefficient of gradation of soil B is $3.35_$ .

The uniformity coefficient of soil C is $46.67_$ .

The coefficient of gradation of soil C is $1.15_$ .

Answer 14

Explanation of Solution

Sketch the grain size distribution curve for soils A, B, and C as shown in Figure 1.

Principles of Geotechnical Engineering (MindTap Course List), Chapter 2, Problem 2.1CTP

Refer to Figure 1.

For soil A:

The diameter of the particle corresponding to $60%$ finer $(D60)$ is $14 mm$ .

The diameter of the particle corresponding to $30%$ finer $(D30)$ is $5.2 mm$ .

The diameter of the particle corresponding to $10%$ finer $(D10)$ is $0.53 mm$ .

For soil B:

The diameter of the particle corresponding to $60%$ finer $(D60)$ is $8.1 mm$ .

The diameter of the particle corresponding to $30%$ finer $(D30)$ is $2.5 mm$ .

The diameter of the particle corresponding to $10%$ finer $(D10)$ is $0.23 mm$ .

For soil C:

The diameter of the particle corresponding to $60%$ finer $(D60)$ is $7 mm$ .

The diameter of the particle corresponding to $30%$ finer $(D30)$ is $1.1 mm$ .

The diameter of the particle corresponding to $10%$ finer $(D10)$ is $0.15 mm$ .

Calculate the uniformity coefficient $(Cu)$ using the relation.

$Cu=D60D10$ (1)

For soil A:

Substitute $14 mm$ for $D60$ and $0.53 mm$ for $D10$ in Equation (1).

$Cu=140.53=26.42$

Hence, the uniformity coefficient for soil A is $26.42_$ .

For soil B:

Substitute $8.1 mm$ for $D60$ and $0.23 mm$ for $D10$ in Equation (1).

$Cu=8.10.23=35$

Hence, the uniformity coefficient for soil B is $35_$ .

For soil C:

Substitute $7 mm$ for $D60$ and $0.15 mm$ for $D10$ in Equation (1).

$Cu=70.15=46.67$

Hence, the uniformity coefficient for soil C is $46.67_$ .

Calculate the coefficient of gradation $(Cc)$ using the relation.

$Cc=D302D60×D10$ (2)

For soil A:

Substitute $14 mm$ for $D60$ , $5.2 mm$ for $D30$ , and $0.53 mm$ for $D10$ in Equation (2).

$Cc=5.2214×0.53=27.045.512=3.64$

Hence, the coefficient of gradation for soil A is $3.64_$ .

For soil B:

Substitute $8.1 mm$ for $D60$ , $2.5 mm$ for $D30$ , and $0.23 mm$ for $D10$ in Equation (2).

$Cc=2.528.1×0.23=6.251.863=3.35$

Hence, the coefficient of gradation for soil B is $3.35_$ .

For soil C:

Substitute $7 mm$ for $D60$ , $1.1 mm$ for $D30$ , and $0.15 mm$ for $D10$ in Equation (2).

$Cc=1.127×0.15=1.211.05=1.15$

Therefore, the coefficient of gradation for soil C is $1.15_$ .

Answer 15

(b)

Expert Solution

To determine

State which of the soil is coarser from soil A and C.

Answer to Problem 2.1CTP

Soil A is coarser than soil C.

Explanation of Solution

Refer to part (a).

The uniformity coefficient of soil A is $26.42$ .

The uniformity coefficient of soil C is $46.67$ .

The percent of soil finer than $1 mm$ for soil A is $20%$ .

The percent of soil finer than $1 mm$ for soil C is $47%$ .

Hence, a higher percentage of soil C is finer than soil A.

Hence, soil A is coarser than soil C.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

State which of the soil is coarser from soil A and C.

Answer 20

Answer to Problem 2.1CTP

Soil A is coarser than soil C.

Answer 21

Explanation of Solution

Refer to part (a).

The uniformity coefficient of soil A is $26.42$ .

The uniformity coefficient of soil C is $46.67$ .

The percent of soil finer than $1 mm$ for soil A is $20%$ .

The percent of soil finer than $1 mm$ for soil C is $47%$ .

Hence, a higher percentage of soil C is finer than soil A.

Hence, soil A is coarser than soil C.

Answer 22

(c)

Expert Solution

To determine

Explain the reason for curve different of soil A, B and C if it is obtained from same stockpile.

Explanation of Solution

The particle-size distribution curve shows the range of particle sizes present in a soil and the type of distribution of various-size particles.

Refer to Figure 1.

Particle separation of coarser and finer particles may take place in aggregate stockpiles. This makes representative sampling difficult.

Therefore, the particle-size distribution curve is different for soils A, B, and C.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

Explain the reason for curve different of soil A, B and C if it is obtained from same stockpile.

Answer 27

Explanation of Solution

The particle-size distribution curve shows the range of particle sizes present in a soil and the type of distribution of various-size particles.

Refer to Figure 1.

Particle separation of coarser and finer particles may take place in aggregate stockpiles. This makes representative sampling difficult.

Therefore, the particle-size distribution curve is different for soils A, B, and C.

Answer 28

(d)

Expert Solution

To determine

Calculate the percentages of gravel, sand, and fines according to the Unified Soil Classification System.

Answer to Problem 2.1CTP

The percentage of gravel for soil A is $71%_$ .

The percentage of sand for soil A is $28%_$ .

The percentage of fines for soil A is $1%_$ .

The percentage of gravel for soil B is $55%_$ .

The percentage of sand for soil B is $43%_$ .

The percentage of fines for soil B is $2%_$ .

The percentage of gravel for soil C is $47%_$ .

The percentage of sand for soil C is $50%_$ .

The percentage of fines for soil C is $3%_$ .

Explanation of Solution

Refer to Figure 1.

For soil A.

The percent passing through $4.75 mm$ sieve is $29%$ .

The percent passing through $0.075 mm$ sieve is $1%$ .

Calculate the percentage of gravel as shown below.

$Percent of gravel=100−percent passing through 4.75 mm sieve=100−29=71%$

Hence, the percentage of gravel is $71%_$ .

Calculate the percentage of sand as shown below.

$Percent of sand=(percent passing through 4.75 mm sieve)−(percent passing through 0.075 mm sieve)=29−1=28%$

Hence, the percentage of sand is $28%_$ .

Calculate the percentage of fines as shown below.

$Percent of fines=percent passing through 0.075 mm sieve−0=1−0=1%$

Hence, the percentage of fines is $1%_$ .

Refer to Figure 1.

For soil B.

The percent passing through $4.75 mm$ sieve is $45%$ .

The percent passing through $0.075 mm$ sieve is $2%$ .

Calculate the percentage of gravel as shown below.

$Percent of gravel=100−percent passing through 4.75 mm sieve=100−45=55%$

Hence, the percentage of gravel is $55%_$ .

Calculate the percentage of sand as shown below.

$Percent of sand=(percent passing through 4.75 mm sieve)−(percent passing through 0.075 mm sieve)=45−2=43%$

Hence, the percentage of sand is $43%_$ .

Calculate the percentage of fines as shown below.

$Percent of fines=percent passing through 0.075 mm sieve−0=2−0=2%$

Hence, the percentage of fines is $2%_$ .

Refer to Figure 1.

For soil C.

The percent passing through $4.75 mm$ sieve is $53%$ .

The percent passing through $0.075 mm$ sieve is $3%$ .

Calculate the percentage of gravel as shown below.

$Percent of gravel=100−percent passing through 4.75 mm sieve=100−53=47%$

Hence, the percentage of gravel is $47%_$ .

Calculate the percentage of sand as shown below.

$Percent of sand=(percent passing through 4.75 mm sieve)−(percent passing through 0.075 mm sieve)=53−3=50%$

Hence, the percentage of sand is $50%_$ .

Calculate the percentage of fines as shown below.

$Percent of fines=percent passing through 0.075 mm sieve−0=3−0=3%$

Hence, the percentage of fines is $3%_$ .

Answer 29

(d)

Expert Solution

Answer 30

(d)

Expert Solution

Answer 31

Expert Solution

Answer 32

To determine

Calculate the percentages of gravel, sand, and fines according to the Unified Soil Classification System.

Answer 33

Answer to Problem 2.1CTP

The percentage of gravel for soil A is $71%_$ .

The percentage of sand for soil A is $28%_$ .

The percentage of fines for soil A is $1%_$ .

The percentage of gravel for soil B is $55%_$ .

The percentage of sand for soil B is $43%_$ .

The percentage of fines for soil B is $2%_$ .

The percentage of gravel for soil C is $47%_$ .

The percentage of sand for soil C is $50%_$ .

The percentage of fines for soil C is $3%_$ .

Answer 34

Explanation of Solution

Refer to Figure 1.

For soil A.

The percent passing through $4.75 mm$ sieve is $29%$ .

The percent passing through $0.075 mm$ sieve is $1%$ .

Calculate the percentage of gravel as shown below.

$Percent of gravel=100−percent passing through 4.75 mm sieve=100−29=71%$

Hence, the percentage of gravel is $71%_$ .

Calculate the percentage of sand as shown below.

$Percent of sand=(percent passing through 4.75 mm sieve)−(percent passing through 0.075 mm sieve)=29−1=28%$

Hence, the percentage of sand is $28%_$ .

Calculate the percentage of fines as shown below.

$Percent of fines=percent passing through 0.075 mm sieve−0=1−0=1%$

Hence, the percentage of fines is $1%_$ .

Refer to Figure 1.

For soil B.

The percent passing through $4.75 mm$ sieve is $45%$ .

The percent passing through $0.075 mm$ sieve is $2%$ .

Calculate the percentage of gravel as shown below.

$Percent of gravel=100−percent passing through 4.75 mm sieve=100−45=55%$

Hence, the percentage of gravel is $55%_$ .

Calculate the percentage of sand as shown below.

$Percent of sand=(percent passing through 4.75 mm sieve)−(percent passing through 0.075 mm sieve)=45−2=43%$

Hence, the percentage of sand is $43%_$ .

Calculate the percentage of fines as shown below.

$Percent of fines=percent passing through 0.075 mm sieve−0=2−0=2%$

Hence, the percentage of fines is $2%_$ .

Refer to Figure 1.

For soil C.

The percent passing through $4.75 mm$ sieve is $53%$ .

The percent passing through $0.075 mm$ sieve is $3%$ .

Calculate the percentage of gravel as shown below.

$Percent of gravel=100−percent passing through 4.75 mm sieve=100−53=47%$

Hence, the percentage of gravel is $47%_$ .

Calculate the percentage of sand as shown below.

$Percent of sand=(percent passing through 4.75 mm sieve)−(percent passing through 0.075 mm sieve)=53−3=50%$

Hence, the percentage of sand is $50%_$ .

Calculate the percentage of fines as shown below.

$Percent of fines=percent passing through 0.075 mm sieve−0=3−0=3%$

Hence, the percentage of fines is $3%_$ .

Answer 35

Ch. 2 - For a gravel with D60 = 0.48 mm, D30 = 0.25 mm,...Ch. 2 - Prob. 2.2P Ch. 2 - Prob. 2.3P Ch. 2 - The following are the results of a sieve analysis....Ch. 2 - Repeat Problem 2.4 with the following data. 2.4...Ch. 2 - Repeat Problem 2.4 with the following data. 2.4...Ch. 2 - Repeat Problem 2.4 with the following data. 2.4...Ch. 2 - The following are the results of a sieve and...Ch. 2 - Repeat Problem 2.8 using the following data. 2.8...Ch. 2 - Repeat Problem 2.8 using the following data. 2.8...

Videos

Answer to Problem 2.1CTP

Explanation of Solution

Answer to Problem 2.1CTP

Explanation of Solution

Explanation of Solution

Answer to Problem 2.1CTP

Explanation of Solution

Want to see more full solutions like this?

Chapter 2 Solutions