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1.1 Describe the processes responsible for the formation of soils from rock.
Solution 1.1
The main processes are physical and chemical weathering of rocks. Physical weathering involves
reduction of size without any change in the original composition of the parent rock by processes such as
exfoliation, unloading, erosion, freezing, and thawing. Chemical weathering causes both reductions in
size and chemical alteration of the original parent rock. The main agents responsible for chemical
weathering are hydration, carbonation, and oxidation. Often, chemical and physical weathering take
place in concert.
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1.2 (a) What is a mineral?
(b) Describe the differences among the three main soil minerals.
(c) Which mineral group is most important for soils and why?
Solution 1.2
(a) Minerals are crystalline materials and make up the solids constituent of a soil. Minerals are
classified according to chemical composition and structure.
(b) Kaolinite has a structure that consists of one silica sheet and one alumina sheet bonded
together into a layer about 0.72 nm thick and stacked repeatedly. The layers are held together
by hydrogen bonds. Tightly stacked layers result from numerous hydrogen bonds. Illite consists
of repeated layers of one alumina sheet sandwiched by two silicate sheets. The layers, each of
thickness 0.96 nm, are held together by potassium ions. Montmorillonite (smectite) has a
structure similar to illite, but the layers are held together by weak van der Waals forces. Sodium
smectite can absorb enough water to cause the particles to fully separate. Calcium smectites do
not usually absorb enough water to cause particle separation because of their divalent cations.
Montmorillonite is often called a swelling or expansive clay.
(c) Silicates, because this group of minerals is abundant on earth.
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1.3 Which of the three main clay minerals undergo large volume change in contact with water and why?
Solution 1.3
Montmorillonite. The layers in montmorillonite are bonded by van der Waals forces, which are very
weak. Water can infiltrate the bond pushing the mineral layers apart.
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1.4 (a) What is soil fabric?
(b) What is the name for the spaces between mineral particles?
(c) Why are the spaces between mineral particles important to geo-engineers?
(d) Explain the differences between a flocculated and a dispersed structure?
Solution 1.4
(a) Soil fabric refers to the structural arrangement of soil particles.
(b) Voids
(c) They contain liquids and gases, and control the settlement and strength of soils. If the voids
space decreases the soil settles. The pressure of the water and gases in the voids affects the
amount of stresses on the soil particles.
(d) A flocculated structure, formed in a freshwater environment, results when many particles tend
to orient perpendicular to one another. A dispersed structure occurs when a majority of the
particles orient parallel to one another.
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1.5 Describe the differences among alluvial, collovial, glacial and lateritic soils.
Solution 1.5
Alluvial soils are fine sediments that have been eroded from rock and transported by water, and have
settled on river and stream beds.
Collovial soils (collovium) are soils found at the base of mountains that have been eroded by the
combination of water and gravity.
Glacial soils are mixed soils consisting of rock debris, sand, silt, clays, and boulders.
Lateritic soils are residual soils that are cemented with iron oxides and are found in tropical regions.
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1.6 (a) What are the two types of surface forces in clayey soils.
(b) What is adsorbed water?
(c) Can you remove the adsorbed water by oven drying at 105o
C? Explain.
Solution 1.6
(a) The surface forces on clay particles are London–van der Waals forces ( attracting forces) and
repelling forces from the diffuse double layer
(b) Adsorbed water is water that is bonded to the mineral surfaces.
(c) No, because it is bonded to the mineral surfaces.
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1.7 What is the shape of the hole in a standard sieve?
Solution 1.7
Square.
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1.8 What tests would you specify to determine the grain size of a sand that contains fine-grained soils?
Solution 1.8
A sieve analysis test followed by a hydrometer test on the particles less than 0.075 mm (passing the No.
200 sieve).
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1.9 In a sieve analysis test, the amount of soil retained on all sieves 0.425 mm and above is 100 grams.
The total mass used in the test is 500 grams,
(a) Determine the percentage of the soil greater than 0.425 mm ( #40 sieve).
(b) Determine the percentage finer than 0.425 mm.
Solution 1.9
(a) the percentage of the soil greater than 0.425 = 100 x 100/500 = 20%
(b) the percentage finer = 100 - %retained = 100 -20 = 80%
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1.10 The data from a particle size analysis on a sample of a dry soil at a depth of 0.5 m near a
mountain range are given in the table below.
Sieve opening (mm) 9.53 4.75 2.0 0.84 0.425 0.15 0.074 Pan
Mass retained (grams) 0 31 38 58 126 120 68 58
(a) What is the total mass of the soil retained on all sieves including the pan?
(b) If the total mass used at the start of the test is 500 grams, what is the percentage loss? Explain why
this loss occurred in the test.
(c) Plot the particle size distribution curve.
(d) What are the percentages of coarse-grained and fine-grained soils in the sample.
Solution 1.10
Initial mass 500 grams
Sieve no. 3/8" 4 10 20 40 100 200 Pan
Opening
(mm)
9.93 4.75 2 0.85 0.425 0.15 0.075
Mass retained
(grams)
0 31 38 58 126 120 68 58
A B C D E F
Sieve no.
Opening
(mm)
Mass retained
(grams)
% retained
(%)
S% retained
(%)
% Finer
(%)
Mr 100 x Mr/Mt S column D 100 - column E
3/8" 9.93 0 0.0 0.0 100.0
4 4.75 31 6.2 6.2 93.8
10 2 38 7.6 13.8 86.2
20 0.85 58 11.6 25.5 74.5
40 0.425 126 25.3 50.7 49.3
100 0.15 120 24.0 74.7 25.3
200 0.075 68 13.6 88.4 11.6
Pan 58 11.6
SUM 499 100.0
(a) Mt = 499 grams
(b) Loss = 500 - 499 = 1 grams
% loss = 1 /500 x 100 = 0.2
This loss usually occurs from soil sticking in the holes of the sieves.
(c) See figure below for particle size distribution plot
(d) 88.4
11.6
% coarse-grained soil =
% fine-grained soil =
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1.11 The effective depth measured in a hydrometer test after 8 minutes is 1 cm. (a) Determine the
average particle size if K is 0.01341, and (b) Identify the soil type (e.g. silt or clay) corresponding to
the average particle size.
Solution 1.11
1
0.01341 0.0047 mm
8D
z
D K
t
  
Using the USCS classification, this soil is a silt (0.075 mm to 0.002 mm).
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1.12 The following results were obtained from sieve analyses of two soils.
Mass (grams)
Sieve opening
(mm)
Soil A Soil B
4.75 0 0
2.00 20.2 48.1
0.85 25.7 219.5
0.425 60.4 67.3
0.15 98.1 137.2
0.075 127.2 22.1
168.2 5.6
Hydrometer tests on these soils gave the following results.
% Finer
Particle size
(mm) Soil A Soil B
0.05 22.6 1
0.01 13.8 0.9
0.005 12.2 0.8
0.002 5 0.5
(a) Plot the gradation curve for each soil on the same graph.
(b) How much coarse-grained and fine-grained soils are in each soil?
(c) What are the percentages of clay and silt in each soil according to USCS?
(d) Determine D10 for each soil.
(e) Determine the uniformity coefficient and the coefficient of concavity for each soil.
(f) Describe the gradation curve (e.g. well graded) for each soil?
Solution 1.12
(a to e)
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Soil A
A B C D E F
Sieve no. Opening
(mm)
Weight
retained
(gm)
% retained
(%)
S% retained
(%)
% Finer
(%)
Gravel (%) 0.0
Mr 100 x Mr/Mt S column D 100 - column E Sand (%) 66.3
3/8" 9.93 0 0.0 0.0 100.0 Silt + Clay (%) 33.7
4 4.75 0 0.0 0.0 100.0 D60 0.16
10 2 20.2 4.0 4.0 96.0 D10 0.0038
20 0.85 25.7 5.1 9.2 90.8 D30 0.007
40 0.425 60.4 12.1 21.3 78.7 Cu 42.1
100 0.15 98.1 19.6 40.9 59.1 Cc 0.081
200 0.075 127.2 25.5 66.3 33.7
Pan 168.2 33.7
SUM 499.800 100.0
Soil B
A B C D E F
Sieve no. Opening
(mm)
Weight
retained
(gm)
% retained
(%)
S% retained
(%)
% Finer
(%)
Gravel (%) 0.0
Mr 100 x Mr/Mt S column D 100 - column E Sand (%) 98.9
3/8" 9.93 0 0.0 0.0 100.0 Silt + Clay (%) 1.1
4 4.75 0 0.0 0.0 100.0 D60 1.1
10 2 48.1 9.6 9.6 90.4 D10 0.18
20 0.85 219.5 43.9 53.5 46.5 D30 0.38
40 0.425 67.3 13.5 67.0 33.0 Cu 6.1
100 0.15 137.2 27.5 94.5 5.5 Cc 0.73
200 0.075 22.1 4.4 98.9 1.1
Pan 5.6 1.1
SUM 499.800 100.0
Particle size (mm) Soil A Soil B
0.075 33.7 1.1
0.05 22.6 1
0.01 13.8 0.9
0.005 12.2 0.8
0.002 5 0.5
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Soil A: Clay: (clay + silt + 33.7%); ; 5 % <0.002 mm; Clay = 5%, 28.7% silt
Soil B: Clay: (clay + silt = 1.1%); 0.5 % <0.002 mm; Clay = 0.5%, Silt = 0.6% silt
f) Both soils can be described as well-graded based on Cu (≥ 4 (gravel), ≥ 6 (sand)-> well graded) and Cc
(Range 1 to 3 -> well graded).
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1.13 Why do geo-engineers plot particle distribution curves on a semi-log scale with particle size on
the abscissa (logarithmic scale) versus % finer on the ordinate (arithmetic scale)? Is there any
theoretical justification for this? Would the shape of the grain size graph be different if arithmetic
rather than semi-log scale is used?
Solution 1.13
The semi-log plot is used for convenience because the particle size distribution in natural soils can be
very large. There is no theoretical justification. Yes, the shape of the grain size graph will be different
because the log scale tends to squeeze the particle size distribution laterally.
https://www.book4me.xyz/solution-manual-soil-mechanics-fundamentals-budhu/
1.14 If a soil consists of sand and fines, would drying the soil and then sieving it through a standard
stack of sieves give accurate results on the fines content? Justify your answer.
Solution 1.14
No. The fines will cluster together or adhere to the sand.
https://www.book4me.xyz/solution-manual-soil-mechanics-fundamentals-budhu/
1.15 If you have to select a soil for a roadway that requires good drainage qualities, what type of soil
would you select and why?
Solution 1.15
A coarse-grained soil such as sand or gravel because coarse-grained soils generally have higher bearing
(load) capacity, lower settlement, and better drainage than fine-grained soils.
https://www.book4me.xyz/solution-manual-soil-mechanics-fundamentals-budhu/
1.16 A house foundation consists of a concrete slab casted on a clay soil. The homeowner planted
vegetation near one side of the foundation and watered it regularly, sometimes excessively. She
noticed that this side of the foundation curled upwards, the concrete slab cracked and several
cracks appeared on the wall. What do you think is likely the predominant mineral in the clay soil?
Justify your answer.
Solution 1.16
Montmorillonite. This clay mineral expands in contact with water. The expansion will push the edges of
the slab upwards causing the concrete slab to crack.
https://www.book4me.xyz/solution-manual-soil-mechanics-fundamentals-budhu/

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Solution Manual for Soil Mechanics Fundamentals – Muni Budhu

  • 1. 1.1 Describe the processes responsible for the formation of soils from rock. Solution 1.1 The main processes are physical and chemical weathering of rocks. Physical weathering involves reduction of size without any change in the original composition of the parent rock by processes such as exfoliation, unloading, erosion, freezing, and thawing. Chemical weathering causes both reductions in size and chemical alteration of the original parent rock. The main agents responsible for chemical weathering are hydration, carbonation, and oxidation. Often, chemical and physical weathering take place in concert. https://www.book4me.xyz/solution-manual-soil-mechanics-fundamentals-budhu/ Access full Solution Manual click on link
  • 2. 1.2 (a) What is a mineral? (b) Describe the differences among the three main soil minerals. (c) Which mineral group is most important for soils and why? Solution 1.2 (a) Minerals are crystalline materials and make up the solids constituent of a soil. Minerals are classified according to chemical composition and structure. (b) Kaolinite has a structure that consists of one silica sheet and one alumina sheet bonded together into a layer about 0.72 nm thick and stacked repeatedly. The layers are held together by hydrogen bonds. Tightly stacked layers result from numerous hydrogen bonds. Illite consists of repeated layers of one alumina sheet sandwiched by two silicate sheets. The layers, each of thickness 0.96 nm, are held together by potassium ions. Montmorillonite (smectite) has a structure similar to illite, but the layers are held together by weak van der Waals forces. Sodium smectite can absorb enough water to cause the particles to fully separate. Calcium smectites do not usually absorb enough water to cause particle separation because of their divalent cations. Montmorillonite is often called a swelling or expansive clay. (c) Silicates, because this group of minerals is abundant on earth. https://www.book4me.xyz/solution-manual-soil-mechanics-fundamentals-budhu/
  • 3. 1.3 Which of the three main clay minerals undergo large volume change in contact with water and why? Solution 1.3 Montmorillonite. The layers in montmorillonite are bonded by van der Waals forces, which are very weak. Water can infiltrate the bond pushing the mineral layers apart. https://www.book4me.xyz/solution-manual-soil-mechanics-fundamentals-budhu/
  • 4. 1.4 (a) What is soil fabric? (b) What is the name for the spaces between mineral particles? (c) Why are the spaces between mineral particles important to geo-engineers? (d) Explain the differences between a flocculated and a dispersed structure? Solution 1.4 (a) Soil fabric refers to the structural arrangement of soil particles. (b) Voids (c) They contain liquids and gases, and control the settlement and strength of soils. If the voids space decreases the soil settles. The pressure of the water and gases in the voids affects the amount of stresses on the soil particles. (d) A flocculated structure, formed in a freshwater environment, results when many particles tend to orient perpendicular to one another. A dispersed structure occurs when a majority of the particles orient parallel to one another. https://www.book4me.xyz/solution-manual-soil-mechanics-fundamentals-budhu/
  • 5. 1.5 Describe the differences among alluvial, collovial, glacial and lateritic soils. Solution 1.5 Alluvial soils are fine sediments that have been eroded from rock and transported by water, and have settled on river and stream beds. Collovial soils (collovium) are soils found at the base of mountains that have been eroded by the combination of water and gravity. Glacial soils are mixed soils consisting of rock debris, sand, silt, clays, and boulders. Lateritic soils are residual soils that are cemented with iron oxides and are found in tropical regions. https://www.book4me.xyz/solution-manual-soil-mechanics-fundamentals-budhu/
  • 6. 1.6 (a) What are the two types of surface forces in clayey soils. (b) What is adsorbed water? (c) Can you remove the adsorbed water by oven drying at 105o C? Explain. Solution 1.6 (a) The surface forces on clay particles are London–van der Waals forces ( attracting forces) and repelling forces from the diffuse double layer (b) Adsorbed water is water that is bonded to the mineral surfaces. (c) No, because it is bonded to the mineral surfaces. https://www.book4me.xyz/solution-manual-soil-mechanics-fundamentals-budhu/
  • 7. 1.7 What is the shape of the hole in a standard sieve? Solution 1.7 Square. https://www.book4me.xyz/solution-manual-soil-mechanics-fundamentals-budhu/
  • 8. 1.8 What tests would you specify to determine the grain size of a sand that contains fine-grained soils? Solution 1.8 A sieve analysis test followed by a hydrometer test on the particles less than 0.075 mm (passing the No. 200 sieve). https://www.book4me.xyz/solution-manual-soil-mechanics-fundamentals-budhu/
  • 9. 1.9 In a sieve analysis test, the amount of soil retained on all sieves 0.425 mm and above is 100 grams. The total mass used in the test is 500 grams, (a) Determine the percentage of the soil greater than 0.425 mm ( #40 sieve). (b) Determine the percentage finer than 0.425 mm. Solution 1.9 (a) the percentage of the soil greater than 0.425 = 100 x 100/500 = 20% (b) the percentage finer = 100 - %retained = 100 -20 = 80% https://www.book4me.xyz/solution-manual-soil-mechanics-fundamentals-budhu/
  • 10. 1.10 The data from a particle size analysis on a sample of a dry soil at a depth of 0.5 m near a mountain range are given in the table below. Sieve opening (mm) 9.53 4.75 2.0 0.84 0.425 0.15 0.074 Pan Mass retained (grams) 0 31 38 58 126 120 68 58 (a) What is the total mass of the soil retained on all sieves including the pan? (b) If the total mass used at the start of the test is 500 grams, what is the percentage loss? Explain why this loss occurred in the test. (c) Plot the particle size distribution curve. (d) What are the percentages of coarse-grained and fine-grained soils in the sample. Solution 1.10 Initial mass 500 grams Sieve no. 3/8" 4 10 20 40 100 200 Pan Opening (mm) 9.93 4.75 2 0.85 0.425 0.15 0.075 Mass retained (grams) 0 31 38 58 126 120 68 58 A B C D E F Sieve no. Opening (mm) Mass retained (grams) % retained (%) S% retained (%) % Finer (%) Mr 100 x Mr/Mt S column D 100 - column E 3/8" 9.93 0 0.0 0.0 100.0 4 4.75 31 6.2 6.2 93.8 10 2 38 7.6 13.8 86.2 20 0.85 58 11.6 25.5 74.5 40 0.425 126 25.3 50.7 49.3 100 0.15 120 24.0 74.7 25.3 200 0.075 68 13.6 88.4 11.6 Pan 58 11.6 SUM 499 100.0 (a) Mt = 499 grams (b) Loss = 500 - 499 = 1 grams % loss = 1 /500 x 100 = 0.2 This loss usually occurs from soil sticking in the holes of the sieves. (c) See figure below for particle size distribution plot (d) 88.4 11.6 % coarse-grained soil = % fine-grained soil = https://www.book4me.xyz/solution-manual-soil-mechanics-fundamentals-budhu/
  • 12. 1.11 The effective depth measured in a hydrometer test after 8 minutes is 1 cm. (a) Determine the average particle size if K is 0.01341, and (b) Identify the soil type (e.g. silt or clay) corresponding to the average particle size. Solution 1.11 1 0.01341 0.0047 mm 8D z D K t    Using the USCS classification, this soil is a silt (0.075 mm to 0.002 mm). https://www.book4me.xyz/solution-manual-soil-mechanics-fundamentals-budhu/
  • 13. 1.12 The following results were obtained from sieve analyses of two soils. Mass (grams) Sieve opening (mm) Soil A Soil B 4.75 0 0 2.00 20.2 48.1 0.85 25.7 219.5 0.425 60.4 67.3 0.15 98.1 137.2 0.075 127.2 22.1 168.2 5.6 Hydrometer tests on these soils gave the following results. % Finer Particle size (mm) Soil A Soil B 0.05 22.6 1 0.01 13.8 0.9 0.005 12.2 0.8 0.002 5 0.5 (a) Plot the gradation curve for each soil on the same graph. (b) How much coarse-grained and fine-grained soils are in each soil? (c) What are the percentages of clay and silt in each soil according to USCS? (d) Determine D10 for each soil. (e) Determine the uniformity coefficient and the coefficient of concavity for each soil. (f) Describe the gradation curve (e.g. well graded) for each soil? Solution 1.12 (a to e) https://www.book4me.xyz/solution-manual-soil-mechanics-fundamentals-budhu/
  • 14. Soil A A B C D E F Sieve no. Opening (mm) Weight retained (gm) % retained (%) S% retained (%) % Finer (%) Gravel (%) 0.0 Mr 100 x Mr/Mt S column D 100 - column E Sand (%) 66.3 3/8" 9.93 0 0.0 0.0 100.0 Silt + Clay (%) 33.7 4 4.75 0 0.0 0.0 100.0 D60 0.16 10 2 20.2 4.0 4.0 96.0 D10 0.0038 20 0.85 25.7 5.1 9.2 90.8 D30 0.007 40 0.425 60.4 12.1 21.3 78.7 Cu 42.1 100 0.15 98.1 19.6 40.9 59.1 Cc 0.081 200 0.075 127.2 25.5 66.3 33.7 Pan 168.2 33.7 SUM 499.800 100.0 Soil B A B C D E F Sieve no. Opening (mm) Weight retained (gm) % retained (%) S% retained (%) % Finer (%) Gravel (%) 0.0 Mr 100 x Mr/Mt S column D 100 - column E Sand (%) 98.9 3/8" 9.93 0 0.0 0.0 100.0 Silt + Clay (%) 1.1 4 4.75 0 0.0 0.0 100.0 D60 1.1 10 2 48.1 9.6 9.6 90.4 D10 0.18 20 0.85 219.5 43.9 53.5 46.5 D30 0.38 40 0.425 67.3 13.5 67.0 33.0 Cu 6.1 100 0.15 137.2 27.5 94.5 5.5 Cc 0.73 200 0.075 22.1 4.4 98.9 1.1 Pan 5.6 1.1 SUM 499.800 100.0 Particle size (mm) Soil A Soil B 0.075 33.7 1.1 0.05 22.6 1 0.01 13.8 0.9 0.005 12.2 0.8 0.002 5 0.5 https://www.book4me.xyz/solution-manual-soil-mechanics-fundamentals-budhu/
  • 15. Soil A: Clay: (clay + silt + 33.7%); ; 5 % <0.002 mm; Clay = 5%, 28.7% silt Soil B: Clay: (clay + silt = 1.1%); 0.5 % <0.002 mm; Clay = 0.5%, Silt = 0.6% silt f) Both soils can be described as well-graded based on Cu (≥ 4 (gravel), ≥ 6 (sand)-> well graded) and Cc (Range 1 to 3 -> well graded). https://www.book4me.xyz/solution-manual-soil-mechanics-fundamentals-budhu/
  • 16. 1.13 Why do geo-engineers plot particle distribution curves on a semi-log scale with particle size on the abscissa (logarithmic scale) versus % finer on the ordinate (arithmetic scale)? Is there any theoretical justification for this? Would the shape of the grain size graph be different if arithmetic rather than semi-log scale is used? Solution 1.13 The semi-log plot is used for convenience because the particle size distribution in natural soils can be very large. There is no theoretical justification. Yes, the shape of the grain size graph will be different because the log scale tends to squeeze the particle size distribution laterally. https://www.book4me.xyz/solution-manual-soil-mechanics-fundamentals-budhu/
  • 17. 1.14 If a soil consists of sand and fines, would drying the soil and then sieving it through a standard stack of sieves give accurate results on the fines content? Justify your answer. Solution 1.14 No. The fines will cluster together or adhere to the sand. https://www.book4me.xyz/solution-manual-soil-mechanics-fundamentals-budhu/
  • 18. 1.15 If you have to select a soil for a roadway that requires good drainage qualities, what type of soil would you select and why? Solution 1.15 A coarse-grained soil such as sand or gravel because coarse-grained soils generally have higher bearing (load) capacity, lower settlement, and better drainage than fine-grained soils. https://www.book4me.xyz/solution-manual-soil-mechanics-fundamentals-budhu/
  • 19. 1.16 A house foundation consists of a concrete slab casted on a clay soil. The homeowner planted vegetation near one side of the foundation and watered it regularly, sometimes excessively. She noticed that this side of the foundation curled upwards, the concrete slab cracked and several cracks appeared on the wall. What do you think is likely the predominant mineral in the clay soil? Justify your answer. Solution 1.16 Montmorillonite. This clay mineral expands in contact with water. The expansion will push the edges of the slab upwards causing the concrete slab to crack. https://www.book4me.xyz/solution-manual-soil-mechanics-fundamentals-budhu/