Geotechnical Engineering

Question 1

What is a D10 value and how is it obtained?

Answer 1

D10, AKA effective grain size, is the particle diameter (in mm) for which only 10% of the particles are finer.
- 10% of the soil mass passes through this sieve size

It is obtained through a sieve analysis

Question 2

Why is D10 important?

Answer 2

• Permeability: It is used in Hazens equation to estimate the hydraulic conductivity (permeability) of sandy soils.
• Filter Design: When designing filters (drainage layers), D helps determine the appropriate gradation of materials to prevent soil migration.
• Soil Uniformity: Uniformity Coefficient, which gives insight into the range of particle sizes. Cu = D60/D10.

Question 3

What soils are defined as coarse grained soils?

Answer 3

Gravel and Sand.

It is classified by its particle size which can be obtained by determining the D10 in a sieve analysis.

Question 4

What soils are defined as fine-grained soils?

Answer 4

Inorganic silts and clays.

It is classified by its particle size which can be obtained by a hydrometer test. The hydrometer test is based on Stoke’s Law, whihc relates the speed of a particle falling out of suspension to its diameter and soil density.

Question 5

Hazens Uniformity Coefficient is Cu = D60/D10. What does the Cu value tell you

Answer 5

A low Cu indicates a uniform soil. Uniform soils has most of its particles around the same size.

A high Cu indicates a Well-Graded soil. Well graded soils have a wide range of particle sizes, which generally improves compaction and stability.

Sands:
- Cu < 4 -> Uniform sand
- Cu >= 6 -> Well-Graded sand

Gravels
- Cu < 4 Uniform Gravel
- Cu> 4 Well-graded Gravekl

Question 6

What makes clays different than silts and coarse grained soils.

Answer 6

Clays display the characteristic of plasticity. Clays exhibit a high plasticity index. Clays also have a smaller particle size.

Plasticity is how a soil interacts with water and how it can be molded or deformed without cracking or changing volume significantly. Silts have low to no plasticity and sands are non-plastic. Sands do not change consistency with moisture.

Question 7

There are two soil classification systems. Name them and the differences between the two.

Answer 7

AASHTO (American Association of State Highway Transportation Officials) is based on sieve analysis, plasticity index, and liquid limit. It rates the materials based off their suitability for roadway subgrades. The best soils are classified as A-1 and the worst (highly organics) A-8.

USCS (Unified Soil Classification System) is also based on grain size distribution, liquid limit, and plasticity index of the soil. This is where you get the symbols: GW, SP, SM, ML, CL, etc.
- The symbols each contain two letters. The first represents the particle size, and the second is a decriptive modifier. Gravels (G), Sands (S), Silts (M), Clays (C). Descriptors: Well-Graded (W), CLay (C), Poorly Graded (P), Silt (M). For USCS use the chart in the attached images to determine if it is “above the A line) which requires PI and LL

Question 8

What is a loam

What is an alluvium

Answer 8

A loam is a type of soil that is considered ideal for plant growth befcause is has a balanced mixture of the three main soil types: sand, silt, clay. A typical loam soil contains roughly 40% sand, 40% silt, and 20% clay. It has good drainage, good nutrient holding capacity, and is workable.

An alluvium is a term used in geology to describe loose, unconsolidated sediments that have been deposited by running water, typically in riverbeds, floodplains or deltas. Alluviums have a low bearing capacity, are subject to liquefaction, and are highly erodible.

Question 9

What is a non-invasive geophysical method of analyzing a subsurface. It uses high frequency electromagnetic waves to detect and map subsurface structures and features.

Answer 9

Ground Penetrating Radar (GPR)

It can locate utilities, voids, structural features like rebar and post tension cables in concrete.

Question 10

Name 3 In-Situ Tests that are direct/invasive.

Answer 10

Cone Penetration Test (CPT) - Effective for soft clays, sands, and layered soils. Can also be used to estimate undrained shear strength, friction angle, and modulus. Disadvantage: No sample is retrieved.

Standard Penetration Test (SPT) - Test best suited for sands. Estimates clay shear strength are crude and should not be used in design. Determines relative density, consistency, and bearing capacity of soils.

Vane Shear Test - Measures the undrained shear strength of saturated fine grained soils, particularly soft clays. Not suitable for sandy or gravelly soils.

Borehole Shear Test

Question 11

Describe the procedure for SPT

Answer 11

Split-Spoon sampler is driven into the soil at the bottom of a borehole using a 140 lb hammer dropped from 30”.

The number of hammer blows required to drive the sampler 18 inches is record as the N value or blow count. The first 6 inches it is driven down is not counted towards the blow count as there is often loose or disturbed material at the bottom of the borehole.

N < 4 -> Very loose/soft soil
N = 4-10 -> Loose/Medium Soft Soil
N = 10-30 -> Medium Dense/Firm Soil
N > 30 -> Dense/Stiff Soil

• Make sure to convert to N60 to account for efficienct of the hammer
• Disturbs the soil during sampling
• Less reliable in soft clays or gravel
• Most commonly used in-situ soil test
• CPT provides reliable and continuous data.
• ASTM D1586

Question 12

What is the density of water in lb/ft^3 and kg/m^3

Answer 12

pw = 62.4 lb/ft^3 = 1000 kg/m^3

Question 13

What is the water content at which a soil changes from a liquid state to a plastic state.

Answer 13

Liquid Limit

“Plastic State” is when a soil becomes moldable.

It is determined using the Casagrande Cup Method.

Question 14

What is the water content at which a soil changes from a plastic state to a semi-solid state.

Answer 14

Plastic Limit

“Semi-solid State” is when a soil becomes crumbly.

It is determined by rolling a small soil sample into a 3 mm diameter thread.The plastic limit is the moisture content at which a soil begins to crumble when rolled.

Question 15

Aeolian soils are deposited by ___

Alluvial soils are deposited by ___

Answer 15

Aeoliab -> Wind

Alluvial -> Flowing water

Question 16

Give the different unit weights of water

Does 1 cm^3 of water have a mass of 1 gram?

Answer 16

yw = 9.81 kN/m^3 = 1 g/cm^3 = 62.4 lb/ft^3

Yes, 1 cm^3 of water = mass of 1 gram of water
- Mass = Density x Volume = 1 g/cm^3 x 1 cm^3 = 1g

Question 17

What describes the flow of water through a permeable aquifer or soil

Answer 17

Darcy’s Law

q = kia

q = flow rate
k = hydraulic conductivity
i = hydraulic gradient
A = cross sectional area of aquifer

Question 18

Describe how soils are typically compacted in the field

Answer 18

Soils are compacted to increase stability, strength, enhance resistance to erosion, decreases permeability, and decrease compressibility.

The soil is placed in lifts of a few inches to a few feet thick and then mechanically compacting the lifts. The thickness of the lifts depends on the soil type and compaction equipment.

Equipment: Smooth-wheel roller, rubber-tired roller, grid roller, sheepsfoot roller.

Question 19

What is the basic laboratory test used to determine the maximum dry density of compacted soils

Answer 19

Proctor Test

Question 20

What is the difference between allowable stress design (ASD) and load and resistance factor design (LRFD)

Answer 20

ASD: Uses FS to reduce capacity

LRFD:

Question 21

What is it called when clayey soils are used at lower loadings and/or higher moisture contents than existed prior to excavation

Answer 21

Swell

A small percentage may need to be added calculated borrow-soil volumes to account for swell.

Soils with large swell potential (expansive/fat clays) are probably not suitable for foundation fill.

Question 22

What is the formula for effective stress and total stress

Answer 22

Effective Stress = σ’ = σ − u

Total Stress = σ = ρgz
z=depth
g=gravity
p=density

Above the water table, total and effective stresses are equal and pore water pressure is 0. Below the water table, the pore water exerts a buoyant force.

u = pore water pressure = γw⋅h
- Unit weight of water = 62.4 lb/ft or 9.81 kN/m
- h = height of water above point

Question 23

What is the units for a pascal

Answer 23

1 Pa = 1 N/m^2

Question 24

What is the assumed cohesion of sands and gravels

Answer 24

0

Gravels and sands are considered cohesionless soils. Silty sands (SM) are also near 0. The strength in these soils comes from internal friction.

Question 25

What is shear strength?

Answer 25

Shear strength is a soils ability to resist sliding failure. τ=c+σ′tanϕ - τ = shear strength - c = cohesion - σ′ = effective normal stress - ϕ = angle of internal friction

Question 26

When would you use a hollow stem auger?

Answer 26

Hollow stem augers are used for shallow drilling when drilling through unconsolidated or soft soils where hole stability is a concern. The auger acts as a temporary casing, preventing bore hole collapse. No drilling fluid is needed. You would use this method to install monitoring wells or piezometers. Not suitable for hard rock or dense materials. Not ideal for deep borings compared to rotary methods (depths greater than 100-150').

Question 27

When would you use a solid stem auger instead of a hollow stem?

Answer 27

Use a solid stem auger when it is known that the subsurface consists of cohesive soils that wont collapse (clays, silts). With a solid stem, you cant take in-situ samples. So it is better used when quick shallow hole is needed with no need for sampling. Hollow stem augers allows for a sampler to be inserted through the hollow center.

Question 28

When is direct rotary drilling utilized.

Answer 28

Direct rotary drilling is used when deep borings are required in soil or rock. It can drill through mixed layers without switching methods. Can penetrate dense rock with the right bit. Samples can be taken with the direct rotary drilling method. Drilling has to be paused to take the sample. More depth capacity compared to augers. Drilling fluid or mud supports the hole and prevents it from collapsing in. The drilling fluid also cools off and lubricates the bit. It also carries the cuttings to the surface. Sometimes air is used instead of fluid -> AKA air rotary. Air rotary is used for hard rock formations, when the subsurface materials are cohesive enough to where fluid isn't needed to support the borehole, in areas where there are fractured rock and the drilling fluid could be lost, cold regions where drilling fluid could freeze. Air should not be used where sand/sandy silt subsurface conditions are known, below the water table, if samples are needed

Question 29

What does drilling fluid consist of? Name some different types

Answer 29

Water, Bentonite, and Additives tailored to the site Types of drilling fluids: Water based mud, Bentonite slurry, polymer based fluid,

Question 30

What soil sampling is disturbed and what is undisturbed? What soil properties can be determined from disturbed sampling?

Answer 30

Disturbed: Auger, split spoon (SPT), borrow pits Undisturbed (In-Situ): Shelby tube sampler, Coring (undisturbed rock) Disturbed soils samples can determine: Soil classification, gradation, and moisture content.

Question 31

When is a borrow pit used?

Answer 31

A borrow pit used when local material isn't suitable and natural material is excavated from one location and used as construction fill at another site. It typically is within a few miles from the site. From specs: Proposed borrow material should consist of clay, clayey sand, or sandy clay. PI between 8 and 30, LL < 50. Non-expansive.

Question 32

When classifying a soil, what are the two lab tests that are required.

Answer 32

Sieve Analysis - Required for USCS and AASHTO Atterburg Limits - Liquid Limit and Plastic Limit. Required for USCS and AASHTO. Moisture Content is typically collected as well although it is not necessaary to classify a soil as it gives context to Atterberg limits. - Moisture Content (%) = Weight of water/Dry weight of soil x 100

Question 33

Shelby Tube verse Split Spoon

Answer 33

Shelby Tubes are designed to collect undisturbed samples. Split Spoons are designed to collect disturbed samples typically via SPT. Since it is an undisturbed sample in shelby tube, strength, consolidation and permeability can be measured. Split spoon samples are used for classification and index testing/atterberg limits. Shelby Tube is a thin walled, typically 3 inches in diameter.

Question 34

What is the purpose of a direct shear test How does a direct shear test work

Answer 34

Measure shear strength - Max shear stress a sample of soil can resist before failure. Also measures cohesion and angle of internal friction. How it Works: 1. Sample is placed inside split shear box 2. Normal vertical load is applied 3. Apply horizontal shear force at a constant displacement rate. - Shear force and horizontal displacement are measured 4. Test continues until the sample fails on the horizontal plane Direct Shear test is good for coarse grained soils - Provides direct measurement of shear vs. displacement.

Question 35

When would you use a triaxial test instead of a direct shear test

Answer 35

- When testing a fine grained, cohesive sample - When you need to measure pore water pressure. - Triaxial gives you a full stress-strain curve

Question 36

What are the tests that can be performed to determine soil strength

Answer 36

Direct Shear Test Triaxial Stress Test Vane Shear Test Unconfined Compression Test

Question 37

Hydraulic Conductivity is also referred to as

Answer 37

Permeability

Question 38

Effective Stress does not include __

Answer 38

Pore Water Pressure Effective Stress is a long-term condition when the water pressure has had time to dissipate from the system. Eff Stress = Total Stress - Pore Pressure

Question 39

What are the 3 main types of settlement?

Answer 39

Immediate (elastic) Settlement - Happens immediately after loading. Occurs in granular soils and dry clays Primary Consolidation - Occurs in saturated cohesive soils (clay). Results from the expulsion of pore water over time. Time dependent and is governed by Terzhagis Theory. Secondary Consolidation - Occurs after primary consolidation is complete. More significant in highly plastic or organic clays.

Question 40

What is the gradual reduction in volume of a saturated soil due to water being squeezed out from its pores under sustained loading.

Answer 40

Consolidation.

Question 41

How is pore pressure determined in a piezometer How is pore pressure determined in at a specific soil depth.

Answer 41

u = γwater * h Pore pressure =Unit weight of water times the height of the water in the standpipe Pore pressure at specific depth below the surface = Unit weight of water times the depth of the soil below the water table. At or above the water table u = 0

Question 42

What is bentonite?

Answer 42

Bentonite is a swelling clay. When hydrated, it absorbs large amounts of water and expands. It has a very low permeability and a high plasticity. Treating a soil with bentonite -> main purpose is to reduce permeability. Improves seal and containment.

Question 43

What are geotextiles and what are their purpose?

Answer 43

Geotextiles are fabrics used to retain soil. They are used to prevent the mixing of different soil layers (mixing reduces the support strength). They provide a filtering function, allowing free passage of water while restraining soil movement.

Question 44

What is a sudden drop in shear strength that occurs in saturated cohesionless soil particles such as sand?

Answer 44

Liquefaction - The lower shear strength results in a decrease in bearing capacity. When the shear strength drops to zero, liquefactions occurs.

Question 45

How is total stress calculated? How is total stress calculated when below the water table?

Answer 45

σ=γ⋅z Total stress = Unit Weight of the soil x depth below ground surface. Below the ground surface: σ′ = σ − u - Effective Stress = Total Stress - Pore Pressure Above water table -> u = 0, σ′ = σ Always calculate total stress as the starting point.

Question 46

What is the term for the water content at which further drying out of the soil does not decrease the volume of the soil.

Answer 46

Shrinkage Limit - It is the third atterburg limit.

Question 47

When loaded, how does the effective stress of clay and sands change over time

Answer 47

Sands: Effective Stress increases rapidly initially, due to excess pore pressure draining immediately. As time increases, effective stress stays constant. Clays: Effective Stress increases very slowly over time. Low permeability, so drainage is very slow. As the clay consolidates, pore pressure dissipates, effective stress gradually increases. Effective stress is the stress carried by the soil particles themself.

Question 48

For clays, when loaded, does pore pressure increase or decrease over time.

Answer 48

Clays are undrained, meaning water cant escape fast enough when loaded. So, Initially, pore pressure increases. Over time, as the water slowly drains out, pore pressure gradually decreases. Over time -> effective stress increases, pore pressure decreases, settlement occurs.

Question 49

Atterberg limit values are a form of __

Answer 49

Moisture content

Question 50

Describe the ralationship between the unit weight of water and the density of water.

Answer 50

Density of water = ρwater = 1000 kg/m^3 Unit weight of water = γwater = 62.4 lb/ft^3 = 9.81 KN/m^3 Direct Relationship: γ=ρ⋅g Density is the MASS per unit volume. Unit Weight is the WEIGHT per unit volume.

Question 51

What is the maximum and minimum principal stress

Answer 51

σ₁ = the maximum principal stress σ₃ = the minimum principal stress In the principal stress formula, there is a plus/minus. You do the plus for σ₁ and the minus for σ₃

Question 52

What does a pocket penetrometer measure

Answer 52

Field test for estimating undrained shear strength of cohesive soils.

Question 53

What does it mean if a soil is undrained What does it mean if a soil is unconsolidated

Answer 53

An undrained soil means no water is allowed to flow in or out of the soil. It's a short term condition where the pore water pressure carries part of the load instead of being dissipated. An unconsolidated soil means the soil has not experienced any time dependent compression under loading.

Question 54

For a triaxial stress test, give the differences in what each measures: Unconsolidated Undrained (UU) Consolidated Drained (CD) Consolidated Undrained (CU)

Answer 54

There are two phases for the triaxial shear test: Consolidation phase and shearing phase. There is a valve within the test cell that regulated drainage. When the valve is open, the test is considered drained resulting in a pore pressure of 0. When the valve is closed, the test is undrained resulting in excess pore pressure. UU: No consolidation, No drainage. Calculates Total Stress. The drainage valve stays closed throughout the whole test but since it is not consolidated, pore pressure is not needed. CD: Since the sample is drained during the whole test, including both shearing and consolidation, water flows freely and no excess pore pressure is built up. Pore pressure is zero and allows for the measurement of effective stress. CU: Drainage valve is open during consolidation but then closed during shearing. This allows for the calculation of both total stress and shear stress.

Question 55

What is the purpose of lime stabilization with soil. What type of soils is it primarily used on?

Answer 55

Improves soil strength and stiffness Reduces Plasticity Dries wet soils Reduces swell potential Lime stabilization is commonly used clayey soils

Question 56

What is a surcharge loading

Answer 56

The application of an additional vertical load on the ground surface to simulate or account for future loads or to accelerate soil consolidation. The surcharge is any load applied to the ground surface that is not part of the permanent structure being built. It can be either temporary or permanent.

Question 57

In terms of a soil's phase properties, when a soil is saturated what is the relation between Vv and Vw

Answer 57

In the case of a saturated soil, all voids are filled with water and Vv = Vw

Question 58

Rank Silt, Gravel, Sand, and Clays in terms of largest particle size to smallest

Answer 58

Largest - Smallest Gravel, Sand, Silt, Clay