STATE OF TENNESSEE - EcoC²Secoccs.com/Geotechnical Engineering- Final with solutions...Reference: Publication No. FHWA NHI-06-088 SOILS AND FOUNDATIONS Reference Manual – Volume

Geotechnical Engineering

4/12/2017

FE REVIEW COURSE – SPRING 2017

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Geotechnical Knowledge

• 9-14 problems ▫ Geology ▫ Index properties and soil classifications ▫ Phase relations (air-water-solid) ▫ Laboratory and field tests ▫ Effective stress (buoyancy) ▫ Stability of retaining walls (e.g., active pressure/passive pressure) ▫ Shear strength ▫ Bearing capacity (cohesive and noncohesive) ▫ Foundation types (e.g., spread footings, deep foundations, wall footings,

mats) ▫ Consolidation and differential settlement ▫ Seepage/flow nets ▫ Slope stability (e.g., fills, embankments, cuts, dams) ▫ Soil stabilization (e.g., chemical additives, geosynthetics)

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Classification of Rocks

4


Characterizing Rock Mass Quality

5


Sieve Analysis

6


Sieve Sizes

7


Sieve Analysis

Tests: Mechanical and Hydrometer (< No. 200 sieve)

0.075 mm or #200 sieve

8


Sieve Analysis • Uniformity Coefficient

𝐶𝐶𝑢𝑢 = 𝐷𝐷60𝐷𝐷10

Smaller Cu, more uniform, steeper curve

• Coefficient of Curvature

𝐶𝐶𝑐𝑐 = 𝐷𝐷30 2

𝐷𝐷10 𝐷𝐷60

Cc between 1 and 3 is well graded so long as Cu > 4 (gravels) and Cu > 6 (sands)

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Example Problem

10


Example Problem

11


Atterberg Limit Test

Commonly Used as an Indicator of: • Potential Shrink/Swell Problems • Potential Constructability Problems (i.e. wet pumping soils, dry-

heave prone soils, etc.)

12


Soil Classification Systems

AASHTO and Unified are the two primary soil classification systems to be familiar with. Both are based on the following factors.

1. Sieve analysis. The sample’s particle size distribution.

2. Atterberg Limits

A. Liquid limit (LL). The water content corresponding to the transition between the plastic and liquid state. A liquid limit of 100 means that the moisture weighs as much as the dry soil.

B. Plastic limit (PL). The range of moisture content over which the soil is between a semisolid and plastic state.

C. Plasticity index (PI). The range of moisture content over which the soil is plastic.

CERC13

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AASHTO Classification System

7-13 CERC13

200 sieve = 0.075 mm

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Example Problem

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Example Problem

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Unified Soil Classification

7-16 CERC13

The USCS groups are designated by a group symbol and a corresponding group name. Group symbols contain two letters, the first representing the most significant particle size fraction, the second a descriptive modifier.

17


Unified Soil Classification

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Example Problem

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Example Problem

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Soil Phases

The components of soils can be expressed by weight or by volume. Note that ρsat = ρd (1 + wsat). For U.S. calculations, γ (unit weight) can be substituted for ρ.

7-20 CERC13

Void ratio (e) = VV / VS

Porosity (n) = VV / VT

e = n / (1 – n) Saturation (S) = VW / VV

Specific Gravity (SG) = Ɣs / Ɣw 𝜌𝜌𝑡𝑡 = 𝑀𝑀𝑡𝑡𝑉𝑉𝑡𝑡

=𝑀𝑀𝑠𝑠+𝑀𝑀𝑤𝑤𝑉𝑉𝑡𝑡

𝜌𝜌𝑠𝑠 = 𝑀𝑀𝑠𝑠𝑉𝑉𝑠𝑠

𝜌𝜌𝑑𝑑 = 𝜌𝜌1+𝑤𝑤

=𝑀𝑀𝑠𝑠𝑉𝑉𝑡𝑡

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Mass-Volume Relationships

7-21 CERC13

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Mass-Volume Relationships

CERC13 7-22

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Example Problem

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Example Problem

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Example Problem

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Example Problem

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Effective Stress • Effective Stress, σ ′ is in units of pressure and is given by this equation: σ ′ =σ−u • Total stress, σ , total amount of stress due to soil at depth you are considering: σ=ρ g z

ρ soil density g acceleration of gravity (e.g. 32.2 ft/s^2) z distance of surface, or beginning of the soil section, to the point you are considering

• Pore water pressure, u , upward pressure due to buoyant force of water in water table. It is

subtracted so that effective stress is less than total stress due to buoyant force counteracting it.

u=ρw g zw ρw water density, 62.4 lb/ft^3 zw depth from water table elevation

• Pore water pressure is ZERO above the water table and starts to have a value only BELOW the

water table line. • It is important to understand that the total stress and effective stress are EQUAL above the

water table and that the effective stress is less than the effective stress below the water table.

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Example Problem

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Example Problem

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Proctor Test

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Proctor Test-Soil Compaction • Compaction

▫ densification of soil by removal of air and rearrangement of soil particles through addition of mechanical energy.

• Energy exerted by compaction forces soil to fill available, and additional frictional forces between the soil particles improves mechanical properties the soil.

• Degree of compaction of a soil ▫ measured by its dry unit weight, γd. ▫ dry unit weight after compaction increases as moisture

content (ω) increases, ▫ after optimum moisture content (ωopt) percentage is

exceeded, any added water will result in a reduction in dry unit weight because pore water pressure (pressure of water in-between each soil particle) will be pushing soil particles apart.

32


Permeability Tests

Permeability Permeability (hydraulic conductivity) has the same units as velocity Coefficient of permeability is dependent on void ratio, grain-size distribution, pore-size distribution, roughness of mineral particles, fluid viscosity, and degree of saturation. Constant Head Test Difference of head between inlet and outlet remains constant Water collected in a flask is measured in quantity over a time period. More suitable for coarse-grained soils that have a higher coefficient of permeability. Falling Head Test Falling head test uses a similar procedure to the constant head test, but the head is not kept constant. Permeability is measured by decrease in head over a specified time. This test is more appropriate for fine-grained soils.

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Permeability or Flow of Water Through Porous Media

K (in in/sec) is the main parameter for flow through soils. There are separate lab tests and equations for sands and clays. • For sands, use the constant head permeameter test with • For clays, use the falling head permeameter test with V is the volume of water used in the test t is the time period over which the test was conducted

7-33 CERC13


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P145 FE Reference Handbook V9.4

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Seepage loss through the sandy silt below earth dam (K=3.66 cm/hr)

After B. Muhunthan Washington State University

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38

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Falling Head Permeability

After Dr. Jerry Vandevelde

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After Dr. Jerry Vandevelde

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Flow Net

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Flow Net Components

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a) What is the rate of seepage (per foot of length) below the dam?

b) What is the total head at Point A?

c) What is the uplift pressure at the bottom of the dam at point A?

A concrete dam is constructed on a fine sand layer, as shown on the figure below. The sand is underlain by impervious bedrock. The sand is 40 feet thick and has a coefficient of permeability of 10-4 ff/min and a saturated unit weight of 122 pounds per cubic foot. The bottom of the dame is 5 feet below the top of the sand layer. The tail water is at the ground surface, elevation 100, and the level of the head water is 120.

After ANUJ CHOUDHARI

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45

NOTE THAT THE DRAWING IN THE SOLUTION HAS AN EXTRA FLOW CHANNEL. (it was accidently)

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The flow net around a sheet pile wall driven into a sand soil is shown in the figure. The saturated unit weight of the sand soil is 17 kN/m3. If the average vertical length of the flow net cells at the exit side is 2.8 m, determine the factor of safety against seepage liquefaction.

A) 1.8 B) 1.2 C) 0.6 D) 2.6 After Dr. David Arellano University of Memphis

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48

P145 FE Reference Handbook V9.4


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Shear Strength Page 149 in FE Handbook

(psf)

(psf)

is the shear strength

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ɸ = 𝒕𝒕𝒕𝒕𝒕𝒕−𝟏𝟏 ( 𝜟𝜟 𝑺𝑺𝑺𝑺𝑺𝑺𝒕𝒕𝑺𝑺 𝑺𝑺𝒕𝒕𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝜟𝜟 𝑵𝑵𝑵𝑵𝑺𝑺𝑵𝑵𝒕𝒕𝑵𝑵 𝑺𝑺𝒕𝒕𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺

)

Reference: Publication No. FHWA NHI-06-088 SOILS AND FOUNDATIONS Reference Manual – Volume I

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Direct Shear Test • In the direct shear test, a shear load is applied to a box sample.

7-52 CERC13

53


Direct Shear Test Example Problem

Hint 1: cohesion = 0

1 1 2 2

Hint 2: solve for ɸ

54


Solution:

55


Triaxial Test

56


Triaxial Test

57


Triaxial Test Example Problem

effective minor stress, σ3 = 200 kPa

effective major stress,

Lets plot it!

σ1 = 200 kPa + 468 kPa = 668 kPa

σ3 = 200 kPa σ1 = 668 kPa

ɸ= ?

What is the cohesion value (y-intercept?

0 434 kPa

sin ɸ= 234/434

ɸ = 32.6

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Unconfined Compressive Strength Test

The unconfined compressive strength test is a triaxial test without the lateral load on the sample. It cannot be used on cohesionless soils.

7-58 CERC13

59


Unconfined Compressive Strength Example Problem

Hint: The overall volume doesn’t change. Solve for the diameter at failure

See Solutions Set

60


Solution:

61


Bearing Capacity Page 149 in FE Handbook

CLAYS (undrained) cohesion > 0 ɸ= 0

SANDS (drained) cohesion = 0 ɸ > = 0

C- ɸ soil cohesion = 500 psf ɸ = 20 deg

Nγ =0

Nq =1.0

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Bearing Capacity

63


Bearing Capacity Page 149 in FE Handbook

One thing to look out for is the ground water elevation

γ’ = γT - γwater

Example Problem: Find the Ultimate Bearing Capacity

64



65


cohesion = 500 psf

ɸ = 20 deg

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γ’ = ?

γ’ = γT - γwater

γ’ = 120 pcf – 62.4 pcf = 57.6 pcf

qult = (500 psf * 17.7) + (57.6 pcf * 5’ * 7.4) + (1/2 * 57.6 pcf * 6’ * 5.0)

qult = 11,845.2 psf

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Effective Footing Area –Eccentric Loads

Reference: Reference: AASHTO LRFD 2012 Bridge Design Specifications 6th Ed (US)

68


Shallow Foundations – Example Problem • A shallow rectangular foundation (B= 8 ft, L = 4 ft) transmits an eccentric

load of 95 kips to the soil. What is the Factor of Safety if the ultimate bearing capacity of the soil is 8530 psf? eB = 0.50 ft

eL= 0.25 ft A) 2.9 B) 1.5 C) 2.2 D) 2.5

FOS = Ultimate Bearing Capacity Applied Pressure

Find the Applied Pressure = Force / Effective Area

Effective Area = (B- 2eB ) * (L- 2eL ) = (8’- 2(0.5)) * (4’ - 2(0.25)) = 24.5 ft2

95,000 lbs

Applied Pressure = 95,000 lbs 24.5 ft2 = 3878 psf

FOS = Ultimate Bearing Capacity Applied Pressure

= 8530psf 3878 psf

= 2.2

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Consolidation and Settlement Page 147-148 in FE Handbook

Reference: Michael R. Lindeburg, FE Civil Review Manual, 3rd Edition, Professional Publications, Inc.,

70


Consolidation Tests

71


Consolidation

Reference: Das, Braja- Principles of Foundation Engineering, 6th Edition

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Page 148 in FE Handbook

73


Consolidation – Example Problem

Reference: Six-Minute Solutions for Civil PE Exam Geotechnical Depth Problems, 3rd Edition

74


Solution:

75


Consolidation Page 147 in FE Handbook

Calculate the consolidation settlement following the newly constructed embankment. The embankment causes a stress increase of 800 psf at the midpoint of the clay layer

BEDROCK (IMPERVIOUS)

Normally consolidated clay: Unit weight = 120 pcf

Gravel: Unit Weight = 130 pcf

CC=0.4 e0 =0.75

Embankment

76


Solution:

77


Pressure at Depth A concentrated vertical load of 6000 lbf is applied at the ground surface. What is the increase in vertical pressure 3.5 ft below the surface and 4 ft from the line of action of the force?

Reference: Civil Engineering Reference Manual for the PE Exam, 15th Edition, Lindeberg

3.5’

4’

78


Pressure from Applied Loads: Area of Influence

Reference: SOIL MECHANICS (version Fall 2008) Presented by: Jerry Vandevelde, P.E. Chief Engineer GEM Engineering, Inc. 1762 Watterson Trail Louisville, Kentucky

∆σz = 20,000 lbs (8’ + 7’)*(5’+7’)

∆σz = 111.1 psf

79


Boussinesq’s Method - Stresses Beneath Foundations

A 4’ x 4’ square footing applies a load of 15 kips at the surface. What is the increase in pressure at 8 feet below the center of the footing?

z= 8’ B= 4’ 𝜟𝜟σ = 𝟎𝟎.𝟏𝟏𝟏𝟏 𝒙𝒙 ( 𝟏𝟏𝟏𝟏,𝟎𝟎𝟎𝟎𝟎𝟎 𝑵𝑵𝒍𝒍

𝟒𝟒 𝒇𝒇𝒕𝒕 𝒙𝒙 𝟒𝟒 𝒇𝒇𝒕𝒕 )

𝜟𝜟σ = 𝟏𝟏𝟎𝟎𝟏𝟏.𝟏𝟏𝟏𝟏𝟏𝟏 𝒑𝒑𝑺𝑺𝒇𝒇

Refe: Civil Engineering Reference Manual for the PE Exam, 15th Edition, Lindeberg


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81


Types of Retaining Wall Structures

• There are many different types of walls, but two distinct types:

▫ Fill Retaining Walls ▫ Cut Retaining Walls

82


Fill Retaining Wall Structures

• For the Exam you will likely see a gravity, semi-gravity, a cantilever, and/or an mechanically stabilized earth (MSE) wall.

83


Cut Retaining Wall Structures

• Cut wall for the exam would mostly be a soldier pile or anchor wall.

84


Consideration of Retained Soil

• The two main classifications are cohesive and granular soils.

• Key Concepts:

▫ Cohesive Soils – Clay Type Soils, must account for cohesion in calculations

– C is usually given ,but may have to calculate the cohesion based on the undrained shear strength. In which C=Su=(Undrained Shear Strength (qu)/2)

– Slow Draining ▫ Granular Soils (noncohesive soils or cohensionless soils) – Sand and gravel

type soils, c=0 typically no cohesion or cohesion is negligible.

– Free Draining

• The nature of the backfilled or retained soil greatly affects the design of retaining walls.

85


Earth Pressure

• Force per unit area exerted by soil on the retaining wall.

• Primarily understood to mean “ Horizontal Earth Pressure” ▫ Active Earth Pressure ▫ Passive Earth Pressure ▫ At-Rest Earth Pressure

• Two Main Theories

▫ Rankine Earth Pressure

– Disregards friction between the soil and the wall. ▫ Coulomb

– Accounts for friction between the soil and the wall.

86


Rankine’s Earth Pressure

• Civil Engineering (Page 149) • Know how to use this

information given by the FE handbook and be able to determine the horizontal active and passive earth pressure.

• Note that this is for a smooth wall with cohesion-less soil C=0 and a level backfill.

87


Effective Stress in Retaining Walls

• Above the water table ▫ σv=ɣ(H1) ▫ Where:

– σ is vertical stress, psf (Pa) – ɣ is unit weight of the soil, pcf

(N/m^3) – H is the thickness of the layer, ft.

(m)

• Below the water table ▫ σ’v= σ- μ =(ɣ1(H1))+(ɣ2(H2))- (ɣw(H2)) Or Simplified : ▫ σ’v=ɣ1(H1)+ɣb(H2) ▫ Where:

Ɣb=ɣ-ɣw

Ɣw = 62.4 lbf/ft^3 (9.8 N/m^3)

• Civil Engineering (Page 149) • Must Account for Water Table!

88


Surcharge in Retaining Walls

• Any additional loading applied externally to the soil is known as surcharge.

• Types of Surcharges: ▫ Point ▫ Line ▫ Uniform (Referenced in NCESS

Handbook) • Unsurcharged pressure plus the

surcharge .

• Does not increase the pore water pressure.

• Civil Engineering (Page 149) • Must Account for Surcharge!

89


Retaining Wall - Example 1

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• Since the soil is being compressed, the passive earth pressure coefficient is required.

• The passive earth pressure coefficient is

▫ Kp=tan2(45˚+ Φ/2)

▫ Kp=tan2(45˚+ 28˚/2)= 2.77

– Answer: (D) 2.8

• If you got the answer to be (A) you calculated the active earth pressure, which would be correct if there was no force acting on the front of the wall.

91


Retaining Wall – Example 2

• The Cantilever Retaining Wall shown is under design.

• What is most nearly the active resultant per unit length of the wall? ▫ (A) 280 lbf/ft. ▫ (B) 1900 lbf/ft. ▫ (C) 4300 lbf/ft. ▫ (D) 8500 lbf/ft.

92


Retaining Wall – Example 2

93


Retaining Wall – Example 2 (Extended)

94



• After several years of service the drains of the retaining wall from example 2 plugged resulting in a water table 7 foot above the footing level as shown below. Calculate the resulting overturning moment, per unit length of wall, about point A.

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Retaining Wall – Example 3 (Solution)

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100


Slope Failure Resistance

101


Slope Failure Example

• An excavated slope in uniform soil is shown. The soil has a cohesion of 28.2 kPa and an internal friction angle of 20 degrees, The weight of the soil above the trial failure plane is 5000 kN/m.

• What is most nearly the factor of safety against sliding? ▫ (A) 1.6 ▫ (B) 2.4 ▫ (C) 3.0 ▫ (D) 3.4

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4/12/2017


STATE OF TENNESSEE - EcoC²Secoccs.com/Geotechnical Engineering- Final with solutions...Reference: Publication No. FHWA NHI-06-088 SOILS AND FOUNDATIONS Reference Manual – Volume

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