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

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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%_.

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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%_.

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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%_.

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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%_.

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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...

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