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Exam (elaborations)

Solution Manual for Soil Mechanics and Foundations (3rd Edition) by Muniram Budhu

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This complete solution manual provides step-by-step answers to selected problems from Soil Mechanics and Foundations (3rd Edition) by Muniram Budhu. It covers fundamental and applied topics such as soil properties, effective stress, permeability, consolidation, shear strength, slope stability, lateral earth pressure, retaining structures, and foundation design — aligned with modern geotechnical practice. Essential for students in civil and geotechnical engineering, this manual reinforces theoretical understanding with detailed calculations and practical design applications, supporting coursework, lab assignments, and exam preparation. soil mechanics solution manual, budhu 3rd edition answers, geotechnical engineering problems solved, permeability and consolidation solutions, effective stress soil examples, shear strength soil mechanics, slope stability problem solving, lateral earth pressure answers, retaining wall design problems, shallow and deep foundations, soil testing and classification, civil engineering geotech solutions, bearing capacity examples, foundation engineering textbook answers, budhu soil mechanics solutions

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Chapters 2 to 16 Covered




SOLUTIONS

,Exercise 2.1 (a) What are the three layers that make up the earth’s internal structure?
(b) What is the composition of each of the layers?

Solution 2.1

a) The core, the mantle and the crust
b) Studies from elastic waves generated by earthquakes showed that the earth has a
core of heavy metals mostly iron. The mantle consists of two parts; upper mantle
which is solid rock and lower mantle which is molten rock. Above the upper mantle
is the crust. The materials that compose the earth’s crust are sediments and rock.




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,Exercise 2.2 (a) Describe the differences among the three key groups of rocks.
(b) Explain why sedimentary rocks are of particular importance to
geotechnical engineers.
(c) Are rock masses homogeneous and continuous? Explain.

Solution 2.2

a) There are three main groups of rocks;
i. Igneous rocks. They are formed from magma emitted from
volcanoes that have cooled and solidified.
ii. Sedimentary rocks. They are formed from sediments animals and
plant materials that are deposited in the water or on land on the
earth’s surface and then subjected to pressures and heat.
iii. Metamorphic rocks. They are formed deep within the earth’s
crust from the transformation of igneous, sedimentary, and
even existing metamorphic rocks into denser rocks.
b) They cover 75% of the earth’s surface area with an average thickness of
0.8 km.
c) Rock masses are seldom homogeneous and continuous.




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,Exercise 2.3 (a) How are soils formed?
(b) What are the agents responsible for weathering of rocks?


Solution 2.3

a) Soils are formed from the physical and chemical weathering of rocks.
a. Physical weathering involves reduction of size without any change in the
original composition of the parent rock.
b. Chemical weathering causes both reductions in size and chemical alteration of
the original parent rock
b) Main agents responsible for physical weathering are; exfoliation, unloading,
erosion, freezing and thawing. Main agents responsible for chemical weathering
are; hydration, carbonation, and oxidation




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,Exercise 2.4 (a) What is a mineral?
(b) Describe the differences among the three main soil minerals.
(c) Why are silicates the most common minerals?


Solution 2.4

a) Minerals are crystalline materials and make up the solids constituent of a soil.
b) Clays are composed of three main types of mineral—kaolinite, illite, and
montmorillonite. The clay minerals consist of silica and alumina sheets that are
combined to form layers. The bonds between layers play a very important role in
the mechanical behavior of clays. The bond between the layers in montmorillonite
is very weak compared with kaolinite and illite. Water can easily enter between the
layers in montmorillonite, causing swelling.
c) Silicates are by far the most common minerals in the earth’s crust and mantle.




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,Exercise 2.5 Why does Montmorillonite undergo large volume change in contact with water?

Solution 2.5

Montmorillonite layers are held together by weak Van der waals forces and exchangeable
ions. Water can easily enter the bond and separate the layers in montmorillonite, causing
swelling.




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,Exercise 2.8 (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 2.8

a) During deposition, the mineral particles are arranged into structural
frameworks that is called soil fabric
b) The spaces between the mineral particles are called voids.
c) Voids occupy a large proportion of the soil volume. As an example; the
amount of settlement depends on how much we compress the volume of
voids. The rate at which the settlement occurs depends on the
interconnectivity of the voids.
d) A flocculated structure, formed under a freshwater environment, results
when many particles tend to orient perpendicular to each other. A dispersed
structure is the result when a majority of the particles orient parallel to each
other.




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,Exercise 2.9 (a) What are the six categories of soil types identified in the ASTM
classification system?
(b) For which soil type is surface forces important? Why?
(c) ) What is adsorbed water?
(d) Can you remove the adsorbed water by oven drying at 105o C? Explain.

Solution 2.9

a) Clay, Silt, Sand, Gravel, Cobbles and Boulders
b) Surface forces on fine-grained soils are important. Because they have
larger surface areas than coarse-grained soils
c) A thin layer of water, called absorbed water, is bonded to the mineral
surfaces of soils.
d) No. Drying of most soils, with the exception of gypsum, using an oven for
which the standard temperature is 105 ± 5° C, cannot remove the adsorbed
water.




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,Exercise 2.10 A particle size analysis on a soil sample yields the following data.

Sieve no. 3/8” 4 10 20 60 200 Pan

Sieve size (mm) 9.53 4.75 2.0 0.84 0.25 0.074 —
Mass retained 0 310 580 380 260 680 210
(grams)
(a) Plot the particle size distribution curve.
(b) Determine the amount of coarse-grained and fine-grained soils in the soil.

Solution 2.10

A B C D E F
Sieve no. Opening Mass retained % retained % retained % Finer
(mm) (grams) (%) (%) (%)
100 x 100 -
Mr Mr/Mt column D column E
3/8" 9.93 0 0.0 0.0 100.0
4 4.75 310 12.8 12.8 87.2
10 2 580 24.0 36.8 63.2
20 0.85 380 15.7 52.5 47.5
60 0.25 260 10.7 63.2 36.8
200 0.075 680 28.1 91.3 8.7
Pan 210 8.7
SUM 2420 100.0
Mt = 2420


b) Fine-grained soils = 8.7% and coarse-grained soils = 91.3%




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