300 solved problems in geotechnical engineering

ii
300 Solved Problems
Soil / Rock Mechanics
and
Foundations Engineering
These notes are provided to you by Professor Prieto-Portar, and in exchange, he will
be grateful for your comments on improvements.
All problems are graded according to difficulty as follows:
* Easy; defines general principles; typical of the PE examination;
** Slightly more difficult; typical of Master’s level problems;
*** Professional level (“real-life”) problems.
by
Luis A. Prieto-Portar PhD, PE
Professor of Civil and Environmental Engineering
Florida International University, Miami, Florida
Former Professor, United States Military Academy (West Point)
Telephone 305-348-2825 / Fax 305-348-2802
Website: http://web.eng.fiu.edu/prieto/
Email: prietol@fiu.edu
© Copyright by L. Prieto-Portar, October, 2009

i
Table of Contents
Table of Contents................................................................................................................. i
Chapter 1 Soil Exploration.................................................................................................. 1
Symbols for Soil Exploration .......................................................................................... 1
*Exploration–01. Find the required number of borings and their depth. ....................... 2
*Exploration–02. The sample’s disturbance due to the boring diameter. ...................... 3
*Exploration–03. Correcting the SPT for depth and sampling method. ......................... 4
*Exploration–04. Three methods used for SPT depth corrections.................................. 6
*Exploration–05. SPT corrections under a mat foundation. ........................................... 7
*Exploration–06. The Shear Vane Test determines the in-situ cohesion........................ 9
*Exploration–07. Reading a soil boring log................................................................. 10
*Exploration–08: Using a boring log to predict soil engineering parameters............... 11
**Exploration–09. Find the shear strength of a soil from the CPT Report.................. 14
Chapter 2 Phase Relations of Soil..................................................................................... 16
Symbols for Phase Relations of soils ............................................................................ 16
Basic Concepts and Formulas for the Phases of Soils................................................... 17
*Phases of soils-01: Convert from metric units to SI and US units. ............................. 21
*Phases of soils–02: Compaction checked via the voids ratio.................................... 22
*Phases of soils–03: Value of the moisture when fully saturated................................. 23
*Phases of soils–04: Finding the wrong data. ............................................................... 24
*Phases of soils–05: Increasing the saturation of a soil. ............................................... 25
*Phases of soils–06: Find γd, n, S and Ww. .................................................................. 26
*Phases of soils–07: Use the block diagram to find the degree of saturation. .............. 27
*Phases of soils–08: Same as Prob-07 but setting the total volume V=1 m3
................ 28
*Phases of soils–09: Same as Problem #5 with a block diagram.................................. 29
*Phases of soils–10: Block diagram for a saturated soil. ............................................. 30
*Phases of soils–11: Find the weight of water needed for saturation. .......................... 31
*Phases of soils–12: Identify the wrong piece of data. ................................................. 32
*Phases of soils–13: The apparent cheapest soil is not!................................................ 33
*Phases of soils–14: Number of truck loads. ................................................................ 34
*Phases of soils–15: How many truck loads are needed for a project?......................... 35
*Phases of soils–16: Choose the cheapest fill supplier. ................................................ 36

ii
*Phases of soils–17: Use a matrix to the find the missing data..................................... 38
**Phases of soils–18: Find the voids ratio of“muck” (a highly organic soil)............... 40
Chapter 3 Classification of Soils and Rocks..................................................................... 41
Symbols for Classification of soils................................................................................ 41
*Classify–01: Percentage of each of the four grain sizes (G, S, M & C)...................... 42
*Classify–02: Coefficients of uniformity and curvature of granular soils.................... 43
*Classify-03: Classify two soils using the USCS.......................................................... 44
*Classify-04: Manufacturing a “new” soil.................................................................... 45
Classify – 05.................................................................................................................. 47
Classify – 06.................................................................................................................. 48
Classify – 07.................................................................................................................. 49
Classify – 08.................................................................................................................. 50
Classify – 09.................................................................................................................. 51
Classify – 10.................................................................................................................. 52
Classify – 11.................................................................................................................. 53
Chapter 4 Compaction and Soil Improvement.................................................................. 54
Symbols for Compaction............................................................................................... 54
*Compaction–01: Find the optimum moisture content (OMC). ................................... 55
*Compaction–02: Find maximum dry unit weight in SI units...................................... 57
*Compaction-03: What is the saturation S at the OMC? .............................................. 59
*Compaction-04: Number of truck loads required........................................................ 61
*Compaction-05: What is the saturation S at the OMC? .............................................. 62
*Compaction-06: Definition of the relative compaction (RC)...................................... 63
*Compaction-07: The relative compaction (RC) of a soil. ........................................... 64
*Compaction-08: Converting volumes from borrow pits and truck loads.................... 65
**Compaction-09: Ranges of water and fill required for a road................................... 66
**Compaction-10: Find the family of saturation curves for compaction...................... 68
**Compaction-11: Water needed to reach maximum density in the field. ................... 71
**Compaction-12: Fill volumes and truck load requirements for a levee. ................... 73
**Compaction-13: Multiple choice compaction problem of a levee. ........................... 75
Chapter 5 Permeability of Soils........................................................................................ 78
Symbols for Permeability.............................................................................................. 78
*Permeability–01: Types of permeability tests and common units............................... 79

iii
*Permeability-02: Use of Hazen’s formula to estimate the k of an aquifer. ................. 80
*Permeability-03: Flow in a sand layer from a canal to a river. ................................... 81
*Permeability-04: Find the equivalent horizontal permeability of two layers. ............. 82
*Permeability-05: Equivalent vertical and horizontal permeabilities. .......................... 83
*Permeability-06: Ratio of horizontal to vertical permeabilities. ................................. 84
*Permeability–07: Do not confuse a horizontal with a vertical permeability. .............. 85
*Permeability-08: Permeability as a function of the voids ratio e. ............................... 86
*Permeability–09: Uplift pressures from vertical flows................................................ 87
*Permeability-10: Capillary rise in tubes of differing diameters. ................................. 88
*Permeability-11: Rise of the water table due to capillarity saturation. ....................... 90
*Permeability-12: Find the capillary rise hc in a silt stratum using Hazen. .................. 91
*Permeability-13: Back-hoe trench test to estimate the field permeability................... 92
**Permeability-14: Seepage loss from an impounding pond........................................ 93
Chapter 6 Seepage and Flow-nets..................................................................................... 97
Symbols for Seepage and Flow-nets ............................................................................. 97
*Flownets-01: Correcting flawed flow-nets.................................................................. 98
*Flow-nets-02: A flow-net beneath a dam with a partial cutoff wall............................ 99
*Flow-nets-03: The velocity of the flow at any point under a dam. ........................... 100
*Flow-nets-04: Flow through an earth levee............................................................... 101
*Flow-nets-05: Finding the total, static and dynamic heads in a dam. ....................... 102
**Flow nets-06: Hydraulic gradient profile within an earth levee.............................. 103
**Flow-net-07: Flow into a cofferdam and pump size................................................ 105
*Flow-nets-08: Drainage of deep excavations for buildings....................................... 108
*Flow-nets-09: Dewatering a construction site........................................................... 110
*Flow-net-10: Dewatering in layered strata................................................................ 111
**Flownets-11: Flow through the clay core of an earth dam...................................... 113
Chapter 7 Effective Stresses and Pore Water Pressure................................................... 117
Symbols for Effective Stresses and Pore Water Pressure............................................ 117
*Effective Stress–01: The concept of buoyancy. ........................................................ 118
*Effective Stress–02: The concept of effective stress................................................. 119
*Effective Stress–03: The concept of effective stress with multiple strata................. 120
Effective Stress-03B.................................................................................................... 121
Chapter 8 Dams and Levees ........................................................................................... 122

iv
Symbols for Dams and Levees.................................................................................... 122
*Dams-01: Find the uplift pressure under a small concrete levee.............................. 123
*Dams-02: Determine the uplift forces acting upon a concrete dam. ......................... 124
Chapter 9 Stresses in Soil Masses................................................................................... 127
Symbols for Stresses in Soil Masses ........................................................................... 127
*Mohr-01: Simple transformation from principal to general stress state.................... 129
*Mohr – 02: Find the principal stresses and their orientation..................................... 130
*Mohr – 03: Find the principal stresses and their orientation..................................... 131
*Mohr – 04: ................................................................................................................. 132
*Mohr – 05: Normal and shear stress at a chosen plane. ............................................ 133
**Mohr – 07: Back figure the failure angle ................................................................ 134
*Mohr – 08: find the Principle pressure using Mohr .................................................. 135
*Mohr – 09: Relation between θ and φ. ...................................................................... 136
*Mohr – 10: ................................................................................................................. 137
*Mohr–11: ................................................................................................................... 138
*Mohr – 12: ................................................................................................................. 139
*Mohr – 13: Data from Mohr-Coulomb failure envelope........................................... 140
**Mohr – 14: ............................................................................................................... 141
*Mohr – 15: Derive the general formula for horizontal stress. ................................... 142
*Newmark–01: Stress beneath a tank at different depths............................................ 143
*Newmark-02: The stress below the center of the edge of a footing.......................... 144
*Newmark-03: Stress at a point distant from the loaded footing................................ 145
*Newmark-04: Stresses coming from complex shaped foundations........................... 146
*Newmark-05: Stress beneath a circular oil tank....................................................... 147
**Newmark-06: Use Newmark with a settlement problem. ....................................... 148
*Stress–01: Stress increase at a point from several surface point loads...................... 150
*Stress-02: Find the stress under a rectangular footing............................................... 151
*Stress-03: The effect of the WT on the stress below a rectangular footing............... 152
*Stress–04: Finding the stress outside the footing area............................................... 153
*Stress-05: Stress below a footing at different points................................................ 154
*Stress-06: Stress increase from a surcharge load of limited width............................ 155
*Stress-07: Finding a stress increase from a surface load of limited width. ............... 156
**Stress-08: Stress increase as a function of depth..................................................... 157

v
Chapter 10 Elastic Settlements ....................................................................................... 158
Symbols for Elastic Settlements.................................................................................. 158
*Elastic Settlement-01: Settlement (rutting) of a truck tire......................................... 159
*Elastic Settlement-02: Schmertmann method used for granular soils....................... 160
*Elastic Settlement-03: Schmertmann method used for a deeper footings................. 161
*Elastic Settlement-04: The 2:1 method to calculate settlement................................. 163
*Elastic Settlement-05: Differential settlement between two columns....................... 165
*Elastic Settlement-06: Compare the settlements predicted by the Boussinesq,
Westergaard, and the 2:1 methods............................................................................... 166
*Elastic Settlement-07: Schmertmann versus the strain methods............................... 169
*Elastic Settlement-08: The Schmertmann method in multiple strata. ....................... 170
**Elastic Settlement-09: Settlement of a mat foundation. .......................................... 172
Chapter 11 Plastic Settlements........................................................................................ 174
Symbols for Plastic Settlements .................................................................................. 174
*Plastic Settlement–01: Porewater pressure in a compressible soil............................ 175
*Plastic Settlement-02: Total settlement of a single layer. ......................................... 177
*Plastic Settlement-03: Boussinesq to reduce the stress with depth. .......................... 178
*Plastic Settlement -04: Surface loads with different units......................................... 180
*Plastic Settlement-05: Pre-consolidation pressure pc and index Cc........................... 181
*Plastic Settlement-06: Final voids ratio after consolidation...................................... 183
*Plastic Settlement-07: Settlement due to a lowered WT. .......................................... 184
*Plastic Settlement-08: The over-consolidation ratio (OCR)...................................... 185
**Plastic Settlement-09: Coefficient of consolidation Cv........................................... 186
*Plastic Settlement -10: Secondary rate of consolidation. .......................................... 188
*Plastic Settlement-11: Using the Time factor Tv....................................................... 189
*Plastic Settlement-12: The time rate of consolidation............................................... 190
*Plastic Settlement-13: Time of consolidation t.......................................................... 191
*Plastic Settlement-14: Laboratory versus field time rates of settlement. .................. 192
*Plastic Settlement-15: Different degrees of consolidation. ....................................... 193
**Plastic Settlement-16: Excavate to reduce the settlement. ...................................... 194
**Plastic Settlement-17: Lead time required for consolidation of surcharge. ............ 196
**Plastic Settlement-18: Settlement of a canal levee.................................................. 198
**Plastic Settlement-19: Differential settlements under a levee................................. 200
***Plastic Settlement-20: Estimate of the coefficient of consolidation cv.................. 202

vi
**Plastic Settlement-21: The apparent optimum moisture content............................. 204
**Plastic Settlement-22: The differential settlement between two buildings. ............ 205
**Plastic Settlement-23: Settlement of a bridge pier. ................................................. 210
Chapter 12 Shear Strength of Soils................................................................................. 212
Symbols for Shear Strength of Soils............................................................................ 212
*Shear strength–01: Maximum shear on the failure plane.......................................... 213
*Shear strength–02: Why is the maximum shear not the failure shear? ..................... 214
*Shear strength–03: Find the maximum principal stress σ1........................................ 215
*Shear strength–04: Find the effective principal stress............................................... 216
*Shear strength–05: Using the p-q diagram. ............................................................... 217
**Shear strength–06: Consolidated-drained triaxial test............................................. 218
**Shear strength–07: Triaxial un-drained tests........................................................... 220
**Shear strength-08: Consolidated and drained triaxial test....................................... 222
***Shear strength-09: Plots of the progressive failure in a shear-box........................ 224
**Shear strength-10: Shear strength along a potential failure plane........................... 227
***Shear strength-11: Use of the Mohr-Coulomb failure envelope. .......................... 228
***Shear strength-11b: Use of the Mohr-Coulomb failure envelope. ........................ 230
**Shear strength-12: Triaxial un-drained tests............................................................ 232
**Shear strength-12b: Triaxial un-drained tests.......................................................... 233
**Shear strength-13: Determine the principal stresses of a sample............................ 234
**Shear strength-13b: Determine the principal stresses of a sample.......................... 237
**Shear strength-14: Formula to find the maximum principal stress. ........................ 240
Chapter 13 Slope Stability .............................................................................................. 242
Symbols for Slope Stability......................................................................................... 242
*Slope-01: Factor of Safety of a straight line slope failure......................................... 243
*Slope-02: Same as Slope-01 but with a raising WT.................................................. 244
*Slope-03: Is a river embankment safe with a large crane?........................................ 245
*Slope-04: Simple method of slices to find the FS. .................................................... 246
**Slope-05: Method of slices to find the factor of safety of a slope with a WT......... 247
**Slope-06: Swedish slip circle solution of a slope stability...................................... 249
Chapter 14 Statistical Analysis of Soils.......................................................................... 252
Symbols for the Statistical Analysis of Soils............................................................... 252
Chapter 15 Lateral Pressures from Soils......................................................................... 253

vii
Symbols for Lateral Pressures from Soils ................................................................... 253
*Lateral-01: A simple wall subjected to an active pressure condition........................ 257
*Lateral–02: Compare the Rankine and Coulomb lateral coefficients....................... 258
*Lateral-03: Passive pressures using the Rankine theory. .......................................... 259
*Lateral-04: The “at-rest” pressure upon an unyielding wall...................................... 260
*Lateral-05: The contribution of cohesion to reduce the force on the wall. ............... 261
**Lateral-06: The effect of a rising WT upon a wall’s stability. ................................ 262
*Lateral-07: The effects of soil-wall friction upon the lateral pressure. ..................... 264
*Lateral-08: What happens when the lower stratum is stronger? ............................... 265
*Lateral-09: Strata with different parameters.............................................................. 266
*Lateral-10: The effects of a clay stratum at the surface. .......................................... 268
**Lateral-11: Anchoring to help support a wall.......................................................... 270
**Lateral-12: The effect of five strata have upon a wall............................................ 272
**Lateral-13: The stability of a reinforced concrete wall. ......................................... 274
***Lateral-14: Derive a formula that provides K and σH as a function of σv. ............ 277
**Lateral-15: The magnitude and location of a seismic load upon a retaining wall... 280
**Lateral-16: Seismic loading upon a retaining wall.................................................. 282
Chapter 16 Braced Cuts for Excavations........................................................................ 283
Symbols for Braced Cuts for Excavations................................................................... 283
*Braced-cuts-01: Forces and moments in the struts of a shored trench...................... 284
**Braced cuts-02: A 5 m deep excavation with two struts for support....................... 289
*Braced cuts-03: Four-struts bracing a 12 m excavation in a soft clay...................... 293
Chapter 17 Bearing Capacity of Soils............................................................................. 296
Symbols for the Bearing Capacity of Soils ................................................................. 296
*Bearing–01: Terzaghi’s bearing capacity formula for a square footing.................... 299
*Bearing–02: Meyerhof’s bearing capacity formula for a square footing. ................. 300
*Bearing–03: Hansen’s bearing capacity formula for a square footing...................... 301
*Bearing–04: Same as #01 but requiring conversion from metric units..................... 302
*Bearing–05: General versus local bearing capacity failures. .................................... 303
*Bearing–06: Comparing the Hansen and Meyerhof bearing capacities. ................... 304
*Bearing–07: Increase a footing’s width if the WT is expected to rise. .................... 305
**Bearing–08: The effect of the WT upon the bearing capacity. ............................... 307
*Bearing–09: Finding the gross load capacity. .......................................................... 309

viii
**Bearing–10: The effect of an eccentric load upon bearing capacity. ...................... 311
**Bearing–11: The effect of an inclined load upon the bearing capacity................... 312
**Bearing-12: Interpretation of borings to estimate a bearing capacity. .................... 314
Chapter 18 Shallow Foundations.................................................................................... 316
Symbols for Shallow Foundations............................................................................... 316
*Footings–01: Analyze a simple square footing. ........................................................ 318
*Footings–02: Add a moment to the load on a footing............................................... 322
*Footings–03: Find the thickness T and the As of the previous problem.................... 324
*Footings–04: Find the dimensions B x L of a rectangular footing............................ 329
*Footings–05: Design the steel for the previous problem........................................... 331
*Footings–06: Design a continuous footing for a pre-cast warehouse wall............... 335
**Footings–07: Design the footings of a large billboard sign..................................... 340
Chapter 19 Combined Footings ...................................................................................... 344
Symbols for Combined Footings................................................................................. 344
Chapter 20 Mat Foundations........................................................................................... 345
Symbols for Mat Foundations ..................................................................................... 345
*Mat Foundations–01: Ultimate bearing capacity in a pure cohesive soil.................. 346
Chapter 21 Deep Foundations - Single Piles .................................................................. 347
Symbols for Single Piles of Deep Foundations........................................................... 347
*Single-Pile–01: Pile capacity in a cohesive soil........................................................ 348
Chapter 22 Deep Foundations - Pile Groups and Caps................................................... 349
Symbols for Pile Groups and Caps of Deep Foundations ........................................... 349
**Pile-caps–01: Design a pile cap for a 9-pile cluster. ............................................... 350
Chapter 23 Deep Foundations: Lateral Loads ................................................................ 353
Symbols for Lateral Loads on Deep Foundations ....................................................... 353
**Lateral loads on piles-01: Find the lateral load capacity of a steel pile................... 354
Chapter 24 Reinforced Concrete Retaining Walls and Bridge Abutments..................... 358
Symbols for Reinforced Concrete Retaining Walls .................................................... 358
**RC Retaining Walls–01: Design a RC wall for a sloped backfill. .......................... 359
Chapter 25 Steel Sheet Pile Retaining Walls.................................................................. 367
Symbols for Steel Sheet Pile Retaining Walls............................................................. 367
**Sheet-pile Wall-01: Free-Earth for cantilevered walls in granular soils. ................ 368
Chapter 26 MSE (Mechanically Stabilized Earth) Walls ............................................... 373

ix
Symbols for Mechanically Stabilized Earth Walls...................................................... 373
**MSE Walls-01: Design the length L of geotextiles for a 16 ft wall........................ 374

x
Conversion of Units
Base SI Units
Derived SI Units
Quantity Derived SI Unit Name Symbol
area square meter m²
volume cubic meter m³
density kilogram per cubic meter kgm/m³
force kilogram-meter per square second Newtons N
moment of force Newton-meter N-m
pressure Newton per square meter Pascal Pa
stress Newton per square meter Pascal Pa or N/m²
work, energy Newton-meter joule J
power joule per second watt W
Multiplication Factor Prefix SI Symbol
1 000 000 000 giga G
1 000 000 mega M
1 000 kilo k
0.001 milli m
0.000 001 micro µ
0.000 000 001 Nano n
Quantity Unit Symbol
length meter m
mass kilograms (mass) kgm
force Newton N
time second s

xi
Conversion of SI Units to English Units
Lengths
Multiply by To
From inches feet yards miles
mm 3.94 x 10-2
3.28 x 10-3
1.09 x 10-3
6.22 x 10-7
cm 3.94 x 10-1
3.28 x 10-2
1.09 x 10-2
6.22 x 10-6
m 3.94 x 101
3.28 1.09 6.22 x 10-4
km 3.94 x 104
3.28 x 103
1.09 x 103
6.22 x 10-1
1 μm = 1 x 10-6
m
1 Å = 1 x 10-10
m = 3.28 x 10-10
feet
Area
Multiply by To
From square inches square feet square yards square miles
mm² 1.55 x 10-3
1.08 x 10-5
1.20 x 10-6
3.86 x 10-13
cm² 1.55 x 10-1
1.08 x 10-3
1.20 x 10-4
3.86 x 10-11
m² 1.55 x 103
1.08 x 101
1.20 3.86 x 10-7
km² 1.55 x 109
1.08 x 107
1.20 x 106
3.86 x 10-1
1 acre = 43,450 ft2
= 4,047 m2
= 0.4047 hectares
Volume
Multiply by To
From cubic inches cubic feet cubic yards quarts gallons
cm3
6.10 x 10-2
3.53 x 10-5
1.31 x 10-6
1.06 x 10-3
2.64 x 10-4
liter 6.10 x 101
3.53 x 10-2
1.31 x 10-3
1.06 2.64 x 10-1
m³ 6.10 x 104
3.53 x 101
1.31 1.06 x 103
2.64 x 102

xii
Force
Multiply by To
From ounces pounds kips tons (short)
dynes 1.405 x 10-7
2.248 x 10-6
2.248 x 10-9
1.124 x 10-9
grams 3.527 x 10-2
2.205 x 10-3
2.205 x 10-6
1.102 x 10-6
kilograms 3.527 x 101
2.205 2.205 x 10-3
1.102 x 10-3
Newtons 3.597 2.248 x 10-1
2.248 x 10-4
1.124 x 10-4
kilo-Newtons 3.597 x 103
2.248 x 102
2.248 x 10-1
1.124 x 10-1
tons (metric) 3.527 x 104
2.205 x 103
2.205 1.102
Pressure (or stress) σ
Multiply by To
From lb/in² lb/ft² kips/ft² tons (short)/ft² feet of water atmosphere
gm/cm² 1.422 x 10-2
2.048 2.048 x 10-3
1.024 x 10-3
3.281 x 10-2
9.678 x 10-4
kg/cm² 1.422 x 101
2.048 x 103
2.048 1.024 3.281 x 101
9.678 x 10-1
kN / m² 1.450 x 10-1
2.090 x 101
2.088 x 10-2
1.044 x 10-2
3.346 x 10-1
9.869 x 10-3
ton (metric)/m² 1.422 2.048 x 102
2.048 x 10-1
1.024 x 10-1
3.281 9.678 x 10-2
Torque (or moment) T or M
Multiply by To
From lb-in lb-ft kips-ft
gm-cm 8.677 x 10-4
7.233 x 10-5
7.233 x 10-8
kg-m 8.677 7.233 7.233 x 10-3
kN-m 9.195 x 103
7.663 x 102
7.663 x 10-1

xiii
Velocity v
Multiply by To
From ft / s ft / min mi / h
cm / s 3.281 x 10-2
1.9685 2.236 x 10-2
km / min 5.467 x 101
3.281 x 103
3.728 x 101
km / h 9.116 x 10-1
5.467 x 101
6.214 x 10-1
1 mile = 1,610 meters = 5,282.152 feet
Unit weight γ
Multiply by To
From lb / in³ lb / ft³
gm / cm³ 3.613 x 10-2
6.248 x 101
kg / m³ 3.613 x 10-5
6.248 x 10-2
kN / m³ 3.685 x 10-3
6.368 x 101
tons (metric ) / m³ 3.613 x 10-2
6.428 x 101
Power P
1 W = 1 J/sec = 1.1622 cal/hr = 3.41 Btu/hr = 0.0013 hp
1 hp = 745.7 W = 0.7457 kW
1 kW = 1.34 hp

300 solved problems in geotechnical engineering

1
Chapter 1
Soil Exploration
Symbols for Soil Exploration
CB → STP correction factor for the boreholes diameter.
CR → STP correction factor for the rod length.
CS → STP correction factor for the sampler type used.
cu → Soil’s un-drained cohesion.
Df → Depth of the foundation’s invert.
Em → The efficiency of the STP hammer.
N → The “raw” value of the STP (as obtained in the field).
po → The original vertical stress at a point of interest in the soil mass.
S → The number of stories of a building.
SPT→ Stands for “Standard Penetration Test”.
N60 →Corrected STP assuming 60% efficiency in the field.
N70 →Corrected STP assuming 70% efficiency in the field.
m→ Correction factor for the shear vane test using the clay’s Plasticity Index PI.

2
*Exploration–01. Find the required number of borings and their depth.
(Revised: Sept. 08)
A four story reinforced concrete frame office building will be built on a site where the soils are
expected to be of average quality and uniformity. The building will have a 30 m x 40 m footprint
and is expected to be supported on spread footing foundations located about 1 m below the ground
surface. The site appears to be in its natural condition, with no evidence of previous grading.
Bedrock is 30-m below the ground surface. Determine the required number and depth of the
borings.
Solution:
A reinforced concrete building is heavier than a steel framed building of the same size. Hence, the
design engineer will want soil conditions that are at least average or better. From Table-1 below, one
boring will be needed for every 300 m2
of footprint area. Since the total footprint area is 30 m x 40 m
=1,200 m2
, use four borings.
Table-2 provides the minimum depth required for the borings, 5 S0.7
+ D = 5(4)0.7
+ 1 = 14 m. Most
design engineers want one boring to go to a slightly greater depth to check the next lower stratum’s
strength.
In summary, the exploration plan will be 4 borings to a depth of 14 m.
Table-1 - Spacing of the exploratory borings for buildings on shallow foundations.
Structural footprint Area for Each Boring
Subsurface Conditions
(m2
) (ft2
)
Poor quality and / or erratic 200 2,000
Average 300 3,000
High quality and uniform 600 6,000
Table-2 - Depths of exploratory borings for buildings on shallow foundations.
Minimum Depth of Borings
(S = number of stories and D = the anticipated
depth of the foundation)
Subsurface Conditions
(m) (ft)
Poor and / or erratic 6S0.7
+ D 20S0.7
+ D
Average 5S0.7
+ D 15S0.7
+ D
High quality and uniform 3S0.7
+ D 10S0.7
+ D

3
*Exploration–02. The sample’s disturbance due to the boring diameter.
(Revised: Sept. 08)
The most common soil and soft rock sampling tool in the US is the Standard Split Spoon.
Split spoon tubes split longitudinally into halves and permit taking a soil or soft rock sample. The
tube size is designated as an NX. The NX outside diameter is Do = 50.8 mm (2 inches) and its
inside diameter is Di = 34.9 mm (1-3/8 inches). This small size has the advantage of cheapness,
because it is relatively easy to drive into the ground. However, it has the disadvantage of
disturbing the natural texture of the soil. In soft rocks, such as young limestone, it will destroy the
rock to such a degree that it may be classified as a “sand”.
A better sampler is the Shelby (or thin-tube sampler). It has the same outside diameter of 2 inches
(although the trend it to use 3 inches).
Compare the degree of sample disturbance of a US standard split-spoon sampler, versus the two
Shelby thin-tube samplers (2” and 3” outside diameters) via their area ratio Ar (a measure of
sample disturbance).
Solution:
( ) ( )
( )
( ) ( )
( )
2 2
2 2
2
2
2 2
2 2
2
2
11
The area ratio for a 2"-standard split-spoon sampler is,
2.0 1.38
(%) (100) (100)
1.38
The area ratio for a 2"-Shelby-tube sampler is,
2.0 1.875
(%) (100)
0
(100)
%
13.
1.875
8%
o i
r
i
o i
r
i
D D
A
D
D D
A
D
−
−
= = =
−
−
= = =
( ) ( )
( )
2 2
2 2
2
2
The area ratio for a 3"-Shelby-tube sampler is,
3.0 2.875
(%) (100) 8.9%
(100)
2.875
o i
r
i
D D
A
D
−
−
= = =
Clearly, the 3” O-D Shelby-tube sampler is the best tool to use.

4
*Exploration–03. Correcting the SPT for depth and sampling method.
(Revision Sept-08)
A standard penetration test (SPT) has been conducted in a loose coarse sand stratum to a depth of
16 ft below the ground surface. The blow counts obtained in the field were as follows: 0 – 6 in = 4
blows; 6 -12 in = 6 blows; 12 -18 in = 8 blows. The tests were conducted using a US-style donut
hammer in a 6 inch diameter boring with a standard sampler and liner. The effective unit weight
of the loose sand stratum is about 93.8 pcf.
Determine the corrected SPT if the testing procedure is assumed to only be 60% efficient.
Solution:
The raw SPT value is N = 6 + 8 = 14 (that is, only the last two sets of 6” penetrations).
The US-style donut hammer efficiency is Em = 0.45, and the other parameters are obtained from the
Tables provided on the next page: CB = 1.05, CS = 1.00, CR = 0.85.
With these values, the SPT corrected to 60% efficiency can use Skempton’s relation,
( )( )( )( )( )
60
0.45 1.05 1.00 0.85 14
9
0.60 0.60
m B S R
E C C C N
N = = =
Notice that the SPT value is always given as a whole number.
That corrected SPT N60 is then corrected for depth. For example, using the Liao and Whitman method
(1986),
( )
( )( )
( )
( )( )
2 2
60
60
2,000 / 2,000 /
9
16 93.8
10
lb ft lb ft
N N
depth effective unit weight ft pcf
= = =
Other methods for corrections are discussed in Exploration-04.

5
SPT Hammer Efficiencies (adapted from Clayton, 1990).
Country Hammer Type Release Mechanism Hammer Efficiency
Argentina donut cathead 0.45
Brazil pin weight hand dropped 0.72
China automatic trip 0.60
donut hand dropped 0.55
donut cat-head 0.50
Colombia donut cat-head 0.50
Japan donut Tombi trigger 0.78 - 0.85
donut cat-head + sp. release 0.65 - 0.67
UK automatic trip 0.73
US safety 2-turns on cat-head 0.55 - 0.60
donut 2-turns on cat-head 0.45
Venezuela donut cat-head 0.43
Correction Factors for the Boring Diameter, Sampling Method and Boring Rod Length
(adapted from Skempton, 1986).
Correction Factor Equipment Variables Value
Borehole diameter factor CB 65 – 115 mm (2.5 – 4.5 in) 1.00
150 mm (6 in) 1.05
200 mm (8 in) 1.15
Sampling method factor CS Standard sampler 1.00
Sampler without liner
(not recommended)
1.20
Rod length factor, CR 3 – 4 m (10 – 13 ft) 0.75
4 – 6 m (13 – 20 ft) 0.85
6 – 10 (20 – 30 ft) 0.95
>10 m (>30 ft) 1.00

6
*Exploration–04. Three methods used for SPT depth corrections.
(Revision Sept.-08)
A raw value of N = 40 was obtained from an SPT at a depth of 20 feet in a sand stratum that has a
unit weight of 135 lb/ft3
. Correct it only for depth.
Solution:
Any of these three methods will provide acceptable answers. Notice how similar their results are from
each other:
1. Using the Bazaraa Method (1967):
'
2
0
'
2
0
3
2 2
0
'
2
4
1.5 /
1 2
4
1.5 /
3.25 0.5
(20 )(135 / )
2.70 / 1.5 /
1000 /
4 4(40)
3.25 0.5 3.25 0.5(2.70 /
35
)
corrected
o
corrected
o
corrected
o
N
N if p kips ft and
p
N
N if p kips ft
p
ft lb ft
but p kips ft kips ft
lb kip
N
therefore N
p kips ft
= ≤
+
= ≥
+
= = >
= = =
+ +
2. Using the Peck Method (1974):
' 2
10 0
0
2
10 0
0
3
2 2
0
10 2
20
0.77log is in tons /
1915
0.77log is in /
(20 )(135 / )
1.35 / 2.70 /
2000 /
20
0.77log 0.90 (40)(0.90)
1.35 /
3
corrected N N
N
N corrected
N N C where C if p ft
p
or C if p kN m
p
ft lb ft
but p tons ft kips ft
lb ton
C N
tons ft
= =
=
= = =
∴ = = ∴ = = 6
3. The Liao-Whitman Method (1986), as used in Exploration-03,
' 2 '
2
2 2
2
2
2
100 2,000
with in / with in
96.1 /
(1.35 / ) 129.7 /
1 /
100 /
40
129.7
35
/
corrected o o
o o
o
corrected
psf
N N p kN m or N p psf
p p
kN m
but p tons ft kN m
ton ft
kN m
N
kN m
= =
⎛ ⎞
= =
⎜ ⎟
⎝ ⎠
∴ = =

7
Water Table
Sand+ gravel
N =26
T = 3.5’
N =25
Soft clay N =24
N =30
Medium sand
N =31
Hard clay
+8 7’
+5.2’
invert of
*Exploration–05. SPT corrections under a mat foundation.
(Revision Sept.-08)
Correct the SPT values shown below for an energy ratio of 60% using a high-efficient US-type
donut hammer in a 2”-diameter boring. The invert (bottom) of the mat foundation is at elevation
+5.2 feet.
+20’
Ground Surface
+13.2’
+10.9’
+10.0’
+4.1’
+0.0’
-10.0’
-20.0’

8
Solution:
Skempton proposed in 1986 the following correction for the sampling methods to the raw SPT value,
assuming that only 60% of the energy of the hammer drives the sampler,
60
.
0
60
N
C
C
C
E
N R
S
B
m
=
where: N60 = SPT N-value corrected for field procedures assuming 60% efficiency
Em = 0.60 → efficiency for a high-efficiency US-style safety hammer
CB = 1.00 → borehole diameter correction
CS = 1.00 → sampler correction, = 0.75 (10’-13’)
CR = 0.85 (13’-20’)→ rod length correction, = 0.95 (20’-30’), = 1.0 (>30’)
N = SPT-value recorded in the field by the driller (known as the “raw” SPT).
The depth correction is,
( )
( )( )
2
1 60
60
2,000 /
lb ft
N N
depth effectiveunit weight
=
At depth of +5.2 feet:
( ) ( )
( )( )
2
60 60
(0.60)(1)(1)(0.75)(26) 2,000 /
20
0.60
20 3
8 127 62.4
9
lb ft
N and N
ft pcf
= = = =
−
At +4.1’ ( ) ( )
( )( )
2
60 60
(0.60)(1)(1)(0.75)(25) 2,000 /
19
0.60
19 3
9 127 62.4
5
lb ft
N and N
ft pcf
= = = =
−
At+2.0’ ( ) ( )
( )( )
2
60 60
(0.60)(1)(1)(0.75)(24) 2,000 /
18
0.60 11 125 62.
1
4
18 3
lb ft
N and N
ft pcf
= = = =
−
At -1.0’ ( ) ( )
( )( )
2
60 60
(0.60)(1)(1)(0.85)(30) 2,000 /
26
0.60 14 126 62.
9
4
26 3
lb ft
N and N
ft pcf
= = = =
−
At -5.0’ ( ) ( )
( )( )
2
60 60
(0.60)(1)(1)(0.85)(31) 2,000 /
26
0.60 18 126 62.
4
4
26 3
lb ft
N and N
ft pcf
= = = =
−
At -10’ ( ) ( )
( )( )
2
60 60
(0.60)(1)(1)(0.95)(30) 2,000 /
29
0.60 23 126 62.
4
4
29 3
lb ft
N and N
ft pcf
= = = =
−
At -21’ ( ) ( )
( )( )
2
60 60
43
(0.60)(1)(1)(1)(43) 2,000 /
43
0.60 33 130 62.
41
4
lb ft
N and N
ft pcf
= = = =
−
Notice that the depth correction does not affect the deeper layers.

9
*Exploration–06. The Shear Vane Test determines the in-situ cohesion.
(Revision Sept.-08)
A shear vane tester is used to determine an approximate value of the shear strength of clay. The
tester has a blade diameter d = 3.625 inches and a blade height h = 7.25 inches. In a field test, the
vane required a torque of 17.0 ft-lb to shear the clay sample, which has a plasticity index of 47%
(PI = LL – PL). Determine the un-drained cohesion cu corrected for its plasticity.
2 3 2 3
17.0
168
( /2) ( /6) (0.3021 ) (0.6042 ) (0.3021 )
2 6
u
T ft lb
c psf
d h d ft ft ft
π
π
⋅
= = =
⎡ ⎤ ⎡ ⎤
+
⎣ ⎦ +
⎢ ⎥
⎣ ⎦
The plasticity index helps correct the raw shear vane test value (Bjerrum, 1974) through the graph
shown above. For a plasticity index of 47% read a correction factor μ = 0.80. Therefore,
(0.80)(168 4
) 13
u corrected u
c c psf psf
μ
− = = =
47

10
*Exploration–07. Reading a soil boring log.
(Revision Sept.-08)
Read the boring log shown below and determine, (1) the location of the phreatic surface, (2) the
depth of the boring and (3) the number of samples taken.
Solution:
(1) The phreatic surface (the water table) was not encountered in this boring and is noted at the bottom
of the report;
(2) The boring was terminated at 21 feet in depth; and
(3) Five samples were taken. Only one sample (#2) was used for laboratory tests (dry density and
moisture content). Samples #1 and #3 were complete split-spoon samples. Samples #4 and #5 were
incomplete split-spoon samples.

11
*Exploration–08: Using a boring log to predict soil engineering parameters.
(Revision Sept.-08)
Using the boring log and the SPT versus Soil Engineering Parameters Table shown on the next two
pages, answer these four questions:
(1) Correct the values of the SPT of Sample S-4 to a 70% sampling efficiency with a standard
sampling method and a US-donut hammer at elevation – 17 feet;
(2) Correct the same sample S-4 for depth assuming the unit weight is γ = 126 pcf;
(3) What are your estimates for the angle of internal friction and unit weight γ?
(4) What is the elevation (above sea level) of the groundwater and the elevation of the bottom of
the boring?
Solution:
(1) The log shows a value of N = 15 (Sample S-4) at elevation -16.5’; at elevation -17’ it has dropped a
small amount to N = 14. Notice that the “Legend” portion denotes that the sampler was a 2” O.D. split
spoon. Therefore, the sampling correction is,
( )( )( )( )( )
70
0.45 1.0 1.0 0.85 14
0.70 0.70
8
B S R
EC C C N
N = = ≈
(2) Correct the same sample S-4 for depth.
( )
( )
( )
( )( )
70 70
2000 2000
8
126 17
8
≈
= =
psf psf
N N
h psf
γ
(3) What are your estimates for the angle of internal friction and unit weight γ?
The log identifies this level at -17’ as a “brown and grey fine to medium SAND”. Use the Table
provided on page 23 to obtain an estimate of some of the engineering parameters for granular soils. Read
the SPT for medium sands; then go to the Medium column and read the value of “N = 8” to obtain the
values:
φ = 32º and γwet = 17 kN/m2
.
(4) What are the elevations (above sea level) of the groundwater and of the bottom of the boring?
- The boring did not report finding a ground water table.
- The bottom of the boring was at -36.5’ from the surface, or 347.0’ – 36.5’ = +310.5’.

13
Correlation between SPT values and some Engineering Parameters of Granular Soils
Description Very loose Loose Medium Dense
Very
dense
Dr
Relative
density 0 0.15 0.35 0.65 0.85
SPT fine 1 - 2 3 - 6 7 - 15 16 - 30
(N'70 ) medium 2 - 3 4 - 7 8 - 20 21 - 40 > 40
coarse 3 - 6 5 - 9 10 - 25 26 - 45
φ fine 26 - 28 28 - 30 30 - 34 33 - 38
medium 27 - 28 30 - 32 32 - 36 36 - 42 < 50
coarse 28 - 30 30 - 34 33 - 40 40 - 50
γwet pcf 70 - 102 89 - 115 108 - 128 108 -140
128 -
147
kN/m3
11 - 16 14 - 18 17 - 20 17 - 22
20 -
23
Note #1: These values are based on tests conducted at depths of about 6 m;
Note #2: Typical values of relative densities are about 0.3 to 0.7; values of 0 or 1.0 do not exist in nature;
Note #3: The value of the angle of internal friction is based on Φ = 28º + 15ºDr;
Note #4: The typical value of an excavated soil ranges from 11 to 14 kN/m3
;
Correlation between SPT values and some Engineering Parameters of Cohesive Soils
SPT - N70 Compressive Strength qu
Description
0 - 2 < 25 kPa Very soft – squeezes between fingers
Very young NC clay
3 - 5 25 - 50 kPa Soft – easily deformed by fingers
Young NC clay
6 - 9 50 - 100 kPa Medium
10 - 16 100 - 200 kPa Stiff – Hard to deform w/fingers
Small OCR – aged clay
17 - 30 200 - 400 kPa Very Stiff – Very hard w/fingers
Increasing OCR – older clays
> 30 > 400 kPa Hard – Does not deform w/fingers
Higher OCR – cemented clays

14
**Exploration–09. Find the shear strength of a soil from the CPT Report.
(Revision: Sept.-08)
Classify a soil from the data provided by the Cone Penetration Test (CPT) shown below at a depth
of 11 m. The clay samples recovered from that depth had γ = 20 kN/m3
and PI = Ip = 20. Compare
your estimate of the shear strength versus the lab test value of 550 kPa.
Solution.
Reading the data, q s ~ 400 kPa and q c ~ 11 MPa which results in a fR ~ 3%.
From the next chart, the soil appears to be a silty clay.

15
3
At a depth of 11 m, the in-situ pressure for a NC clay is,
(20 / )(11 ) 220
From the versus graph, for = 20 yields an ~ 17.5.
The un-drained shear strength is,
o
o
k p p k
u
c o
u
k
p
p z kN m m kPa
N I I N
s
q p
s
N
γ
= = =
−
= =
11,000 220
616 550 (a 12% error).
17.5
kPa kPa
kPa versus lab kPa
−
= =

16
Chapter 2
Phase Relations of Soil
Symbols for Phase Relations of soils
e → Voids ratio.
GS → Specific gravity of the solids of a soil.
n → Porosity.
S → Degree of saturation.
V → Total volume (solids + water + air).
Va → Volume of air.
VV → Volume of voids (water + air).
VS → Volume of solids.
VW → Volume of water.
w → Water content (also known as the moisture content).
WS → Weight of solids.
WW → Weight of water.
g→ Unit weight of the soil.
gd → Dry unit weight of the soil.
gb → Buoyant unit weight of the soil (same as g’).
gSAT→ Unit weight of a saturated soil.
gW → Unit weight of water

17
Basic Concepts and Formulas for the Phases of Soils.
(A) Volumetric Relationships:
1. - Voids ratio e
V
S
V
e
V
= 2-1
ranges from 0 to infinity.
Typical values of sands are: very dense 0.4 to very loose 1.0
Typical values for clays are: firm 0.3 to very soft 1.5.
2. - Porosity n
( )
100%
V
V
n
V
= 2.2
ranges from 0% to 100%.
The porosity provides a measure of the permeability of a soil.
The interrelationship of the voids ratio and porosity are given by,

18
1 1
n e
e and n
n e
= =
− +
2-3
3. - Saturation S
100%
W
V
V
S x
V
= 2-4
ranges from 0% to 100%.
(B) Weight Relationships:
4. - Water content w
100%
W
S
W
w x
W
= 2-5
Values range from 0% to over 500%; also known as moisture content.
5. – Unit weight of a soil γ
S W
S W A
W W
W
V V V V
γ
+
= =
+ +
2-6
The unit weight may range from being dry to being saturated.
Some engineers use “bulk density ρ” to refer to the ratio of mass of the solids and water
contained in a unit volume (in Mg/m3
). Note that,
.
W m
g g which is the equivalent of F ma
V V
γ ρ
= = = = 2-6
6. - Dry unit weight γd
1
S
d
W
V w
γ
γ = =
+
2-7
The soil is perfectly dry (its moisture is zero).
7. - The unit weight of water γw
3
( )
62.4 1 / 1 / 9.81 /
W
w
W
w
W
where g F ma
V
pcf g ml kg liter kN m
γ γ ρ
γ
= = =
= = = =

19
Note that the above is for fresh water. Salt water is 64 pcf, etc.
8. - Saturated unit weight of a soil γsat
0
S W
SAT
S W
W W
V V
γ
+
=
+ +
2-8
9. - Buoyant unit weight of a soil γb
'
b SAT w
γ γ γ γ
= = − 2-9
10. - Specific gravity of the solids of a soil G
S
S
w
G
γ
γ
= 2-10
Typical Values for the Specific Gravity of Minerals in Soils and Rocks
Mineral Composition Absolute specific gravity Gs
Anhydrite CaSO4 2.90
Barites BaSO4 4.50
Calcite, chalk CaCO3 2.71
Feldspar KALSi3O8 2.60 to 2.70
Gypsum CaSO4 2H2O 2.30
Hematite Fe2O3 5.20
Kaolinite Al4Si4O10(OH)8 2.60
Magnetite Fe3O4 5.20
Lead Pb 11.34
Quartz (silica) SiO2 2.65
Peat Organic 1.0 or less
Diatomaceous earth Skeletons of plants 2.00

20
Other useful formulas dealing with phase relationships:
( )
1
:
(1 ) ( ) (1 )
(1 )(1 )
1 1 1
:
( ) 1
1 1
1
1
1
S
s
dry
S w S w S w
S w
S
S w
SAT w
SAT d w s w s
s
Se wG
e
Unit weight relationships
w G G Se w G
G n w
wG
e e
S
Saturated unit weights
G e e w
e w e
w
n n G n G
wG
γ
γ
γ γ γ
γ γ
γ
γ γ
γ γ γ γ
=
= −
+ + +
= = = = − +
+ + +
+ +
⎛ ⎞⎛ ⎞
= = ⎜ ⎟⎜ ⎟
+ +
⎝ ⎠⎝ ⎠
⎛ ⎞
+
⎡ ⎤
= + = − + = ⎜ ⎟
⎣ ⎦ +
⎝ ⎠
( )
'
:
1
1 1 (1 ) ( )
1
w
SAT w
S w w s w
d s w
s
d SAT w SAT w
Dry unit weights
G eS eG
G n
w e e w S wG
e
n
e
γ
γ γ γ
γ γ γ
γ
γ γ
γ γ γ γ γ
= +
= = − = = =
+ + + +
⎛ ⎞
= − = −⎜ ⎟
+
⎝ ⎠
.

21
*Phases of soils-01: Convert from metric units to SI and US units.
(Revision: Oct.-08)
A cohesive soil sample was taken from an SPT and returned to the laboratory in a glass jar. It was
found to weigh 140.5 grams. The sample was then placed in a container of V = 500 cm3
and 423
cm3
of water were added to fill the container. From these data, what was the unit weight of the soil
in kN/m3
and pcf?
Solution.
Notice that the 140.5 grams is a mass. Therefore, the ratio of mass to volume is a density ρ,
3 3
3
2
3 3 2 3
3
3 3
3
140.5
1.82
(500 423)
1 1 1
17.
0
1.82 9.806
10 sec 10 1
0.2248
1000 1
17.9
1 1 35.
9 ( )
114
3
f f
f f
f
f
g g
m
V cm cm
g kg m kN cm
g
cm g N m
lbs
kN N m
m k
kN
SI un
N N ft
its
m
ρ
γ ρ
γ
= = =
−
⎛ ⎞
⎛ ⎞ ⎛ ⎞
⎛ ⎞⎛ ⎞
= = =
⎜ ⎟
⎜ ⎟ ⎜ ⎟
⎜ ⎟⎜ ⎟
⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠
⎝ ⎠⎝ ⎠
⎛ ⎞⎛ ⎞
⎛ ⎞
⎛ ⎞
= =
⎜ ⎟⎜ ⎟
⎜ ⎟
⎜ ⎟
⎝ ⎠⎝ ⎠ ⎝ ⎠
⎝ ⎠
( )
pcf US units

22
*Phases of soils–02: Compaction checked via the voids ratio.
(Revision: Sept.- 08)
A contractor has compacted the base course for a new road and found that the mean value of the
test samples shows w = 14.6%, GS = 2.81, and γ = 18.2 kN/m3
. The specifications require that e ≤
0.80. Has the contractor complied with the specifications?
Solution:
( ) ( )
( )
3
3
1 1
1
1
2.81 9.81
0.74 0.
1 0.146
1 1.74
18.2
1.74 1 0
8
.
.
7
,
4
S W S W
G w G w
e
e
kN
m
e
kN
m
e Yes the contractor has com
e
plied
γ γ
γ
γ
+ +
= ∴ + =
+
⎛ ⎞
+
⎜ ⎟
⎝ ⎠
+ = =
= − =
∴ = <

23
*Phases of soils–03: Value of the moisture when fully saturated.
(Revision: Oct.-08)
(1) Show that at saturation the moisture (water) content is ( )
( )
W
sat
sat W
n
w
n
γ
γ γ
=
−
.
(2) Show that at saturation the moisture (water) content is 1 1
⎛ ⎞
= −
⎜ ⎟
⎝ ⎠
sat w
d S
w γ
γ γ
Solution:
( )
( )
( ) ( )
(1) In a fully saturated soil the relation, becomes simply
because 1
1
1
1 1
(1 )
= =
= = =
−
= − +
⎡ ⎤
⎣ ⎦
⎡ ⎤
= − + = − + = +
⎡ ⎤ ⎢ ⎥
⎣ ⎦ −
⎣ ⎦
− =
S s
S
sat sat
sat w S
sat
S
w sat sat
sat
w sat
Se wG e wG
e n
S or G
w w n
but n G n
n n
rearranging n G n n n n
w n w
n
or n there
w
γ γ
γ
γ
γ
γ
(2) Again, in a fully satura
1
ted soi
1
l,
1
= = = =
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
+ − +
∴ = = = = −
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠
=
⎝ ⎠
−
⎛ ⎞
= −
⎜
⎝ ⎠
⎝ ⎠
⎟
w
sat
sat w
sat w
d
V w V w S w V
sat
S S S S S S
w V V V S S V S S
sat w w w
S S S S S
S
fore
V V V V
e
w
G V V W W
V V V V V V V V
w
W W W W W
or
n
w
n
w
γ γ γ
γ
γ
γ
γ
γ
γ
γ
γ
γ
γ
γ

24
*Phases of soils–04: Finding the wrong data.
(Revision: Oct.-08)
A geotechnical laboratory reported these results of five samples taken from a single boring.
Determine which are not correctly reported, if any.
Sample #1: w = 30%, γd = 14.9 kN/m3
, γs = 27 kN/m3
; clay.
Sample #2: w = 20%, γd = 18 kN/m3
, γs = 27 kN/m3
; silt.
Sample #3: w = 10%, γd = 16 kN/m3
, γs = 26 kN/m3
; sand.
Sample #4: w = 22%, γd = 17.3 kN/m3
, γs = 28 kN/m3
; silt.
Sample #5: w = 22%, γd = 18 kN/m3
, γs = 27 kN/m3
; silt.
Solution:
1 1
1
The water content is in error if it is greater than the saturated moisture, that is,
V w V w S w V V V S S
sat w w
S S S S S S S S
V S S
sat w
S S
SA
d
T
w
S
V V V V V V V V
e
w
G V V W W W W
V V V
w
W W
w w
γ
γ γ
γ γ γ
γ γ
γ
γ
⎛ ⎞ ⎛ ⎞
+ −
= = = = = =
⎜ ⎟
⎛ ⎞
= −
⎜ ⎟
⎝ ⎠ ⎝ ⎠
⎛ ⎞
+
= ⎜ ⎟
−
⎜ ⎟
⎝ ⎠
∴ ≤
⎝ ⎠
( )
( )
( )
( )
3
3
3
3
1 1
1 1
1) 9.81 / 30% 30%
14.9 27
1 1
2) 9.81 / 18.5% 20%
18 27
1 1
3) 9.81 / 24% 10%
16 26
1 1
4) 9.81 / 22.1% 22%
17.3 28
5)
w
d S
SAT
SAT
SAT
SAT
w kN m w GOOD
w kN m v w
w kN m w GOOD
w kN m
WRONG
w GOOD
γ
γ γ
⎛ ⎞
= −
⎜ ⎟
⎝ ⎠
⎛ ⎞
= − = = =
⎜ ⎟
⎝ ⎠
⎛ ⎞
= − = < =
⎜ ⎟
⎝ ⎠
⎛ ⎞
= − = > =
⎜ ⎟
⎝ ⎠
⎛ ⎞
= − = > =
⎜ ⎟
⎝ ⎠
( )
3 1 1
9.81 / 18.5% 22%
18 27
SAT WRONG
w kN m w
⎛ ⎞
= − = < =
⎜ ⎟
⎝ ⎠

25
*Phases of soils–05: Increasing the saturation of a soil.
A soil sample has a unit weight of 105.7 pcf and a saturation of 50%. When its saturation is
increased to 75%, its unit weight raises to 112.7 pcf.
Determine the voids ratio e and the specific gravity Gs of this soil.
Solution:
( )
( )( )
1
62.4( 0.50 )
105.7 (1)
1
62.4( 0.75 )
112.7 (2)
1
Solving explicitely for in equation (1),
105.7 1
0.50
62.4
Replace in equation (2) with the above relation from (1),
11
W S
S
S
s
s
s
G Se
e
G e
pcf
e
G e
and pcf
e
G
e
G e
G
γ
γ
+
=
+
+
∴ =
+
+
=
+
+
= −
∴ ( )( ) ( )( ) ( )( )
2.7 1 105.7 1 62.4 0.
0.814 6
25
2. 7
S
e e
e
e
and G
+ =
=
+ +
=
∴

26
*Phases of soils–06: Find γd, n, S and Ww.
The moist unit weight of a soil is 16.5 kN/m3
. Given that the w = 15% and Gs = 2.70, find:
(1) Dry unit weight γd,
(2) The porosity n,
(3) The degree of saturation S, and
(4) The mass of water in kgm/m3
that must be added to reach full saturation.
Solution:
( )( )
( )
( )
( )( )
( )
( )
d
d
3
16.5
) = = =
(1 + w) (1 + 0.15)
) From the table of useful relationships,
2.70 9.81
1 1.85 0.85
1 14.3
0.85
100%
1 1 0.85
0.15 2.70
) Since 100
0
kN
14.3
m
4
8
6%
. 5
s w s w
d
s
s
a
b
G G
e e
e
e
n
e
wG
c Se wG S
e
γ
γ
γ γ
γ
γ
= ∴ + = = = ∴ =
+
= = =
+ +
= ∴ = = =
( )( )
( )
( )
S w
sat 3
3
2
2.70 + 0.85 9.81
(G + e) kN
) = = = 18.8
1+e 1+0.85
The water to be added can be found from the relation
18.8 - 16.5 / 1,000 9.81 - /
mass of water
9.8 1
48
/
%
1 -
d
m
g
kN m N kg m
g kg m s kN
γ
γ
γ ρ
γ
ρ
=
⎡ ⎤⎛ ⎞
∴ = = ⎢ ⎥⎜ ⎟
⎝ ⎠
⎣ ⎦
3
2
2,3
= 40 m
kg
m
s
N
⎛ ⎞
⎜ ⎟
⎝ ⎠

27
*Phases of soils–07: Use the block diagram to find the degree of saturation.
A soil has an “in-situ” (in-place) voids ratio 1.87, 60%, 2.75
o N S
e w and G
= = = . What are the
γmoist and S? (Note: All soils are really “moist” except when dry, that is when w = 0%).
Solution: Set VS = 1 m3
(Note: this problem could also be solved by setting V = 1.0 m3
).
( ) ( )( )( )
( ) ( )( )
3
3 3
1.87
1.87 1 1.87 2.87
1
The "natural" water content is 0.60 0.60
1 2.75 9.81 / 26.98
0.60 0.60 26.98
V
o S V
S
w
N w s
s
s
s s
s s s S w
w w
w s
V
e V V V m
V
W
w W W
W
W
V
G W V G m kN m kN
W W
γ
γ
γ γ
∴ = = = ∴ = + = + =
= = ∴ =
= = ∴ = = =
= =
3 3
16.19
26.98 16.19 4
15.0
3.17
43.17
2.87
16.19
9.
8
81
8.
1.87
2%
s w
moist
w
w w
V V
kN
kN
W W W kN
W kN
V m
W
V
S
V
m
V
γ
γ
=
= + = + =
∴ = = =
⎛ ⎞
⎜ ⎟
⎝ ⎠
∴ = = = =

28
*Phases of soils–08: Same as Prob-07 but setting the total volume V=1 m3
.
(Revision: Oct.-08)
A soil has an “in-situ” (in-place) voids ratio 1.87, 60%, 2.75
o N S
e w and G
= = = . What are the
γmoist and S? (Note: All soils are really “moist” except when dry, that is when w = 0%).
Solution: Set V = 1 m3
(instead of Vs = 1 m3
used in Phases-07).
( ) ( )( )( )
3
3 3
1.87 1 1.87 2.87 0.348 0.652
The "natural" water content is 0.60 0.60
0.348 2.75 9.81 / 9.39
V
o S V S S S S V
S
w
N w s
s
s
s s
s s s S w
w w
V
but e but V m V V V V V V and V
V
W
w W W
W
W
V
G W V G m kN m kN
γ
γ
γ γ
= = ∴ = = + = + = ∴ = =
= = ∴ =
= = ∴ = = =
( ) ( )( )
3 3
0.60 0.60 9.39 5.63
9.39 5.63 15.02
15.0
1
5.63
9.81
0.6
15
5
.0
88.
2
0%
w s
s w
moist
w
w w
V V
kN
W W kN
W W W kN
W kN
V m
W
V
S
V V
m
γ
γ
= = =
= + = + =
∴ = = =
⎛ ⎞
⎜ ⎟
⎝ ⎠
∴ = = = =

29
*Phases of soils–09: Same as Problem #5 with a block diagram.
A soil sample has a unit weight of 105.7 pcf and a water content of 50%. When its saturation is
increased to 75 %, its unit weight raises to 112.7 pcf. Determine the voids ratio e and the specific
gravity Gs of the soil. (NB: This is the same problem as Phase–06, but solved with a block
diagram).
Solution:
3
2 1
3
3
Set 1
112.7 105.7 7.0 25%
21.0 75%
112.7 20.8 91.9
20.8
0.333
62.4
1
0.111 0.111 0.333 0.444
3
1 1 0.444 0.556
0.444
0.55
S
w
w
w
a w v a w
s v
V
S
V ft
lbs are of water
lbs are of water
W lb
W lb
V ft
pcf
V V ft V V V
V V
V
e
V
γ γ
γ
=
− = − =
∴
∴ = − =
= = =
= = ∴ = + = + =
∴ = − = − =
= =
( )
6
91.9
0.556 (62
0.80
2.
.4
65
)
S S
S
w S w
W lb
and G
V
γ
γ γ
=
= = = =

30
*Phases of soils–10: Block diagram for a saturated soil.
A saturated soil sample has a unit weight of 122.5 pcf and Gs = 2.70. Find dry
γ , e, n, and w.
Solution:
( )
( )
( )
( )
3
1
1
122.5 2
122.5
1
Combining equations (1) and (2) yields 1
62.4 2.70
27.0
27.0 0.433
62.4
95.5
95.5 0.56
2.70 (62.4 )
⎛ ⎞
= + = +
⎜ ⎟
⎝ ⎠
= + =
−
⎛ ⎞
= +
⎜ ⎟
⎝ ⎠
∴ = ∴ = = =
∴ = ∴ = = =
w
S
S w w
w S
S w
w
w
w
w w
w
S
S S
S
W
V V V W
G
W W W lb
W
W
pcf
W lb
W lb V ft
pcf
W lb
W lb V
G pcf
γ
γ
γ
3
3
95.5
0.76
7
95.5
1
0.433
0.56
4
7
0.433
0.433
1
27
0.283
95
43.3%
28
5
3%
.
.
∴ = = =
∴ = = =
∴ =
= = =
∴ = =
= =
S
dry
V
S
V
w
S
ft
W lb
V ft
V
e
V
V
n
V
W
n
w
w
W
pcf
γ

31
*Phases of soils–11: Find the weight of water needed for saturation.
Determine the weight of water (in kN) that must be added to a cubic meter of soil to attain a 95 %
degree of saturation, if the dry unit weight is 17.5 kN/m3
, its moisture is 4%, the specific gravity of
solids is 2.65 and the soil is entirely made up of a clean quartz sand.
Solution:
( )
3 3
3
3
3 3
3
1 7 .5 1 8 .2
1 1 0 .0 4
1 8 .2 (1 .0 4 )
1 7 .5 , 0 .7 0
1 7 .5 1 7 .5
0 .6 7 3
2 .6 5 (9 .8 1 / )
0 .7 0
0 .0 7 0 .2 5 7
(9 .8 1 / )
0 .0 7 0 .2
d
S w S S S
S w
S
S
S S w
w
w a s w
w
V
S
k N k N
m w m
W W W W w W W
W k N a n d W k N
W k N k N
V m
G k N m
W k N
V m V V V V m
k N m
V
e
V
γ γ
γ γ
γ γ
γ
= = = ∴ =
+ +
= = + = + =
∴ = =
= = = =
= = = ∴ = − − =
+
= =
( )
( )
( )( )
5 7
0 .4 9
0 .6 7 3
(0 .0 4 ) 2 .6 5
1 0 0 2 1 .6 %
0 .4 9
W e re q u ire a 9 5 % , th e re fo re ,
0 .9 5 0 .4 9
0 .1 7
2 .6 5
(0 .1 7 )(1 7 .5 ) 2 .9 8
0 .7 0
S
S
w S
w
w G
T h e e x is tin g S
e
S
S e
w
G
W w W k N k N
a lr e a d y h a v e W k N
m u s t a d d w a te r
=
= = =
=
= = =
= = =
=
∴ 2 .2 8 k N
=
Answer: Add 2.28 kN of water per m3
.

32
*Phases of soils–12: Identify the wrong piece of data.
A project engineer receives a laboratory report with tests performed on marine marl calcareous
silt). The engineer suspects that one of the measurements is in error. Are the engineer’s suspicions
correct? If so, which one of these values is wrong, and what should be its correct value?
3
3
18.4
26.1
40%
1.12
95%
S
kN
Given unit weight of sample
m
kN
unit weight of solids
m
w water content
e voids ratio
S degree of saturation
γ
γ
= =
= =
= =
= =
= =
Solution:
( )( )
( ) ( )
3
Check the accuracy of 4 out of 5 of the variables using,
0.95 1.12 1.06
26.1
0.4 1.06 Therefore, these four are correct.
9.81
The only possibly incorrect value is . Assume that = 1 m
S
S
S
w
Se wG Se
wG w
V
γ
γ
γ
= ∴ = =
= = = ∴
( )
( )
( )
( )
3
3 3 3
3
3
3
3
.
1 1
1.12 0 1.12 2
0.472 , 0.528 0.95 0.502
0.026
26.1 0.472 12.3
0.40 12.3 4.9
12.3 4.9 17.2
The
a w S
V
a w S
S
S V w V
a
S S S
w S
V m V V V
V
but e V V V
V
V m V m but V V m
V m
kN
W V m kN
m
kN
W wW kN
m
W kN kN kN
γ
= = + +
= = ∴ = − − +
∴ = = = =
∴ =
⎛ ⎞
∴ = = =
⎜ ⎟
⎝ ⎠
⎛ ⎞
= = =
⎜ ⎟
⎝ ⎠
= + =
3
3 3
refore, the actual unit weight of the soi
17
l is,
17.2
.
1
2 18.4
k
kN
W
V
N kN
m m
m
γ
∴ = = ≠
=

33
*Phases of soils–13: The apparent cheapest soil is not!
You are a Project Engineer on a large earth dam project that has a volume of 5x106
yd3
of select
fill, compacted such that the final voids ratio in the dam is 0.80. Your boss, the Project Manager
delegates to you the important decision of buying the earth fill from one of three suppliers. Which
one of the three suppliers is the most economical, and how much will you save?
Supplier A Sells fill at $ 5.28/ yd3
with e = 0.90
Supplier B Sells fill at $ 3.91/ yd3
with e = 2.00
Supplier C Sells fill at $ 5.19/ yd3
with e = 1.60
Solution:
Without considering the voids ratio, it would appear that Supplier B is cheaper than Supplier A by $1.37
per yd3
.
Therefore: To put 1yd3
of solids in the dam you would need 1.8 yd3
of soil.
For 1yd3
of solids from A you would need 1.9 yd3
of fill.
For 1yd3
of solids from B you would need 3.0 yd3
of fill.
For 1yd3
of solids from C you would need 2.6 yd3
of fill.
The cost of the select fill from each supplier is (rounding off the numbers):
( )( )
( )( )
( )( )
6 3
3
6 3
3
6 3
3
1.9 5.28$
5 10 $ 27,900,000
1.8
3.0 3.91$
5 10 $ 32,600, 000
1.8
2.6 5.19$
5 10 $ 37,500, 000
1.8
A yd
yd
B yd
yd
C yd
yd
⎛ ⎞
= ≈
⎜ ⎟
⎝ ⎠
⎛ ⎞
= ≈
⎜ ⎟
⎝ ⎠
⎛ ⎞
= ≈
⎜ ⎟
⎝ ⎠
Therefore Supplier A is the cheapest by about $ 4.7 Million compared to Supplier B.

34
*Phases of soils–14: Number of truck loads.
Based on the previous problem (Phases–13), if the fill dumped into the truck has an e = 1.2, how
many truck loads will you need to fill the dam? Assume each truck carries 10 yd3
of soil.
Solution:
3 3
3 3
3 3
3
1 1.2 which means that there is 1 yd of solids per 1.2 yd of voids.
1
2.2 1 .
10
4.54 .
The requ
V V
S V
S
V V
Set V e V
V
yd of soil for each yd of solids
yd of soil for each x yd in a truck load
x yd of solids per truck trip
= = = = =
∴
∴ =
( )( )
( )
( )
6 3 3
6 3
3
6 3
3
ired volume of solids in the dam is,
5 10 1
2.8 10
1.8
Therefore, (rounding off)
2.
616,80
8 10
4.54 /
0
solids
x yd of soil yd of solids
V x yd of solids
yd of soil
x yd of solids
Number of Truck trips
yd of solids truck trip
= =
− = =
−

35
*Phases of soils–15: How many truck loads are needed for a project?
You have been hired as the Project Engineer for a development company in South Florida to build
610 housing units surrounding four lakes. Since the original ground is low, you will use the
limestone excavated from the lake to fill the land in order to build roads and housing pads. Your
estimated fill requirements are 700,000 m3
, with a dry density equivalent to a voids ratio e = 0.46.
The “in-situ” limestone extracted from the lakes has an e = 0.39, whereas the limestone dumped
into the trucks has an e = 0.71. How many truckloads will you need, if each truck carries 10 m3
?
Solution:
3 3
3 3 3
3 3
Assume: 1 = = = = 0.46 in the compacted fill
1
The required 700,000 m of fill have 1.46 m of voids per each 1 m of solids
Therefore, the 700,000 m of fill have 479,400 m of
V V
S V
S
V V
V m e V m
V
= ∴
3 3
3 3
3
solids
Each truck carries 1.71 m of fill per 1 m solids
In order for the trucks to carry 479,000 m of solids they must carry 820,000 m of fill
Since each truck carries 10 m of fill,
The number of t
∴
3
3
820,000
ruck-loads = = .
10
82,000 truck-loads
m
m

36
*Phases of soils–16: Choose the cheapest fill supplier.
(Revised: Sept.-08)
A large housing development requires the purchase and placement of the fill estimated to be
200,000 cubic yards of lime-rock compacted at 95% Standard Proctor with an OMC of 10%. Two
lime-rock suppliers offer to fill your order: Company A has a borrow material with an in-situ γ =
115 pcf, w = 25%, GS = 2.70; Standard Proctor yields a maximum γd = 112 pcf; at a cost of
$0.20/yd3
to excavate, and $ 0.30/yd3
to haul. Company B has a borrow material with an in-situ γ =
120 pcf, w = 20%, GS = 2.70; Standard Proctor yields a maximum γd = 115 pcf; a cost of $0.22/yd3
to excavate, and $ 0.38/yd3
to haul.
(1) What volume would you need from company A?
(2) What volume would you need from company B?
(3) Which would be the cheaper supplier?
Solution:
(1) The key idea: 1 yd3
of solids from the borrow pit supplies 1 yd3
of solids in the fill.
(2) Pit A: WS = 92 lb, WW = 23 lb VW = 0.369 ft3
, VS = 0.546 ft3
, Va = 0.085 ft3
3 3
0.454
0.83 1.83 1.0 .
0.546
V
S
V
e yd of soil contains yd of solids
V
= = = ∴
Pit B: WS = 100 lb, WW = 20 lb, VW = 0.321 ft3
, VS = 0.594 ft3
, Va = 0.08 ft3
3 3
0.401
0.68 1.68 1.0 .
0.594
V
S
V
V
= = = ∴
(3) Material needed for fill from company A:
( ) ( )( )
3 3
3
3
0.95 1 0.95 112 1 0.10 117 106.4 , 10.6
0.37
0.59 1.59 1.0
0.63
200,000
125,800
1.59
d S w
V
S
w pcf W lb W lb
V
V
yd of fill
Site A requires yd of solids
γ γ
= + = + = ∴ = =
= = = ∴
∴ =
Material needed for fill from company B:
( ) ( )( )
3 3
3
3
0.95 1 0.95 115 1 0.10 120 109.1 , 10.9
0.35
0.54 1.54 1.0
0.65
200,000
130,000
1.54
d S w
V
S
w pcf W lb W lb
V
V
yd of fill
Site B requires yd of solids
γ γ
= + = + = ∴ = =
= = = ∴
∴ =

37
(4) a) Cost of using Company A:
( )( )
3
3
$0.50
125,800 1.83 $115,100
A
Cost yd
yd
⎛ ⎞
= =
⎜ ⎟
⎝ ⎠
Cost of using Company B:
( )( )
3
3
$0.60
130,000 1.68 $131,100
B
Cost yd
yd
⎛ ⎞
= =
⎜ ⎟
⎝ ⎠
Using Company A will save about $ 16,000.

38
*Phases of soils–17: Use a matrix to the find the missing data.
A contractor obtains prices for 34,000 yd3
of compacted “borrow” material from three pits: Pit #3
is $11,000 cheaper than Pit #2 and $39,000 cheaper than Pit #1. The fill must be compacted down
to a voids ratio of 0.7. Pit #1 costs $ 6.00/yd3
and Pit #3 costs $ 5.50/yd3
. Pits #2 and #3 reported
their voids ratios as 0.88 and 0.95 respectively. Use a matrix to find,
a) The missing unit cost C2 for Pit #2;
b) The missing voids ratio e for Pit #1;
c) The missing volume of fill V required from each pit; and
d) The amount paid by the contractor for each pit.
Solution:
A summary of the data provided is herein shown in matrix form,
The volume of solids Vs contained in the total volume of fill V = 34,000 yd3
can be found from,

39
( )
( ) ( )( )
( )( )
3 3
3
3
3 3 3
3 3
3
3
2
34,000
0.7 0.7 1 34,000 20,000
1.7
At Pit #3, 1 1 20,000 1 0.95
The total cost of Pit #3 is 39,000 $ 5.
39,000
$ 214,5
50/
A
00
t Pit #2:
V S S S S S
S
S
V V V V V V yd V yd of solids
V
e V V e yd
V
TC yd yd
V
yd of soil
V
= + = + = + = ∴ = =
= + ∴ = + = + =
= =
( ) ( )( ) 3
3
2 2 2
2 3
2
2 3
2
3
2
1
1
1 1 20,000 1 0.88
But, the total cost of Pit #2 is $11,000 $ 214,500
$ 225,500
The unit cost of Pi
37,600
t #2
37,600
At P
$ 225,500
$ 6.
it #1:
$ 6.
00
0
/
S
S
yd o
e V V e yd
TC
f soil
TC
yd
TC
TC
C
V yd
TC
V
= + ∴ = + = + =
− = = ∴
=
=
= = =
( ) ( )( )
2
3 3
3
3
3 3
1 1 1 1
42,250
28,000 225,500 28,000
0/ $ 6.00/ $ 6.00/
But, 1 20,000 1 42,250 1.11
S
yd of soil
TC
yd yd yd
V V e yd e yd e
+ +
= = =
= + = + = =
∴

40
**Phases of soils–18: Find the voids ratio of“muck” (a highly organic soil).
You have been retained by a local municipality to prepare a study of their “muck” soils. Assume
that you know the dry unit weight of the material (solids) γsm and the dry unit weight of the
organic solids γso. What is the unit weight γs of the combined dry organic mineral soil whose
organic content is M0? (The organic content is the percentage by weight of the dry organic
constituent of the total dry weight of the sample for a given volume.) What is the voids ratio e of
this soil if it is known that its water content is w and its degree of saturation is S?
Solution:
Set Ws = 1 unit and γs = s
s so sm
W 1
=
V (V + V )
(a) Assume Mo = Wo for a unit weight of the dry soil
Therefore 1 - Mo = Wm
o
so
M
γ
= volume of organic Vso solids
o
sm
(1 - M )
γ
= volume of mineral Vsm solids
The total unit weight is the weight of a unit volume.
Therefore S
γ =
( ) ( )
sm
so
o sm so so
o
o
so sm
1
=
M - ) +
1 - M
M
+
γ
γ
γ γ γ
γ γ
⎡ ⎤
⎢ ⎥
⎛ ⎞ ⎣ ⎦
⎜ ⎟
⎝ ⎠
(b)
( )
w
w
v
s s s
w weight of solids
weight of water
volume of water
S
S
V S
e = = = =
V V V Vs
γ
γ
⎛ ⎞
⎛ ⎞
⎛ ⎞
⎜ ⎟
⎜ ⎟
⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
Therefore
( )
( ) ( )
1
e =
1 -
w sm so
o w o sm so so
o
so sm
w
S w
M S M
M
γ γ γ
γ γ γ γ
γ γ
⎛ ⎞
⎜ ⎟
⎝ ⎠ =
− +
⎛ ⎞ ⎡ ⎤
⎣ ⎦
+
⎜ ⎟
⎝ ⎠

41
Chapter 3
Classification of Soils and Rocks
Symbols for Classification of soils
Cc → Coefficient of gradation (also coefficient of curvature).
Cu → Coefficient of uniformity.
RC→ Relative compaction.
Dx → Diameter of the grains (at % finer by weight).
Dr → Relative density of a granular soil.
e→ Voids ratio.
emin → Minimum voids ratio.
emax → Maximum voids ratio.
IP → Index of plasticity (also referred to as PI).
K→ Constant of the yield value.
LL→ Liquid limit.
PL→ Plastic limit.
SL→ Shrinkage limit.
V→ Volume of the soil sample.
W→ Weight of the soil sample.
γd(min)→Dry unit weight in loosest condition (voids ratio emax).
γd → In-situ dry unit weight (voids ratio e).
γd(max)→ Dry unit in densest condition (voids ratio emin)

42
*Classify–01: Percentage of each of the four grain sizes (G, S, M & C).
Determine the percentage of gravels (G), sands (S), silts (M) and clays (C) of soils A, B and C
shown below.
Solution:
Notice that the separation between gravels (G) and sands (S) is the #4 sieve which corresponds to a
particle size of 4.75 mm. The separation between sands (S) and silts (M) is the #200 sieve which
corresponds to a particle size of 0.075 mm. Finally, the separation between silts (M) and clays (C) is the
0.002 mm (or 2 micro-meters = 2 μm). These divisions are shown above through color differentiation.
Each soil A, B and C is now separated into the percentage of each:
Soil A: 2% G; 98% S; 0% M; 0%C. This soil is a uniform or poorly-graded sand (SP).
Soil B: 1% G; 61% S; 31% M; 7%C. This soil is a well-graded silty sand (SM).
Soil C: 0% G; 31% S; 57% M; 12%C. This soil is a well-graded sandy silt (M).

43
*Classify–02: Coefficients of uniformity and curvature of granular soils.
Determine the uniformity coefficient Cu and the coefficient of gradation Cc for soil A.
Solution:
From the grain distribution curve, D60 = 1.4 mm, D30 = 0.95 mm and D10 = 0.50 mm, therefore the
coefficients are,
( )
( )( )
2
2
60 30
10 60 10
0.95
1.40
2.8 1.29
0.50 1.40 0.50
U C
D D
mm
C and C
D mm D D
= = = = = =
A uniform soil has a coefficient of uniformity Cu less than 4, whereas a well-graded soil has a
uniformity coefficient greater than 4 for gravels and greater than 6 for sands. Since soil A has a low
value of 2.8, and it is sand, this corresponds to a poorly-graded sand (SP). Steep curves are uniform
soils (low Cu) whereas diagonal curves are well-graded soils (high Cu).
Smooth curved soils have coefficients of curvature Cc between 1 and 3, whereas irregular curves have
higher or lower values. Soils that are missing a type of soil (a gap) are called gap-graded (Cc will be less
than 1 or greater than 3 for gap-graded soils).
Therefore, this soil is classified as poorly-graded sand (or SP).

44
*Classify-03: Classify two soils using the USCS.
Use the grain-size distribution curve shown below to classify soils A and B using the USCS. Soil
B’s Atterberg limits are LL = 49% and PL = 45%?
Solution:
Classify Soil A:
For soil A, the distribution is G = 2%, S = 98%, M = 0% and C = 0%.
60
10
1.40
2.8
0.50
U
D mm
C
D mm
= = = , therefore, soil A is a poorly graded sand (SP).
Classify Soil B:
For soil B, the distribution is G = 0%, S = 61%, M = 35% and C = 4%.
60
10
0.45
90
0.005
U
D mm
C
D mm
= = = , therefore, soil A is very well graded silty sand (SM).

45
*Classify-04: Manufacturing a “new” soil.
A site has an unsuitable in-situ soil A that does not compact properly. In lieu of removing that soil
A, you have decided to improve it by mixing it with a borrow pit soil B to produce an improved
new soil C that will compact better.
You desire a coefficient of uniformity Cu of about 100 for the new soil C. Determine the relative
percentages of these two uniform soils A and B so that they will result in better graded soil C. Plot
your results.
The plots of soils A and B are as shown below,
Soil A is composed of 2% G, and 98% S: (6% coarse sand, 85% medium sand and 7% fine sand).
It is obviously a poorly graded sand (SP).
Soil B is composed of approximately 33% S, 55% M and 12% C. It is a well-graded sandy silt.
Consider several solutions as shown below with A/B ratios of 30/70, 35/65, 40/60 and 50/50. The
best is the 50/50 solution via D10 = 0.006 mm,
60 60
60
10
100 0.6
0.006
U
D D
C D mm
D mm
= = = ∴ =
The best fit is a 50% of A plus 50% of B mix.

47
Classify – 05
A sample of soil weights 1.5 N. Its clay fraction weighs 0.34 N. If its liquid limit is 60% and its
plastic limit is 26%, classify the clay.
Solution:
W = 1.5 N
Wclay = 0.34 N (or 23% of W)
Ip = PI = LL – PL = 60% – 26% = 34 %
34%
1.5
% 23%
P
I
A
of clay fraction
= = ≈
The activity number 1.5 falls above the U-line in Skempton’s diagram (see Classify-03). Therefore, this
is a CH clay, and is probably a member of the Montmorillonite family.

48
Classify – 06
During a hydrometer analysis a soil with a Gs = 2.60 is immersed in a water suspension with a
temperature of 24°C. An R = 43 cm is obtained after 60 minutes of sedimentation. What is the
diameter D of the smallest-size particles that have settled during that time?
Solution:
Using the table below, for Gs = 2.60 and T= 24°C, K= 0.01321.
( )
16.29 0.164 16.29 [0.164(43)] 9.2
L R cm
∴ = − = − =
9.2
0.01321 0.00517
60 min
L cm
D K mm
t
= = = = 5.2 x 10-3
mm ( a silt)
Table of constant K versus Temperature T (°C)
Temparature
(°C) 2.45 2.50 2.55 2.60 2.65 2.70 2.75 2.80
16 0.01510 0.01505 0.01481 0.01457 0.01435 0.01414 0.01394 0.01374
17 0.01511 0.01486 0.01462 0.01439 0.01417 0.01396 0.01376 0.01356
18 0.01492 0.01467 0.01443 0.01421 0.01399 0.01378 0.01359 0.01339
19 0.01474 0.01449 0.01425 0.01403 0.01382 0.01361 0.01342 0.01323
20 0.01456 0.01431 0.01408 0.01386 0.01365 0.01344 0.01325 0.01307
21 0.01438 0.01414 0.01391 0.01369 0.01348 0.01328 0.01309 0.01291
22 0.01421 0.01397 0.01374 0.01353 0.01332 0.01312 0.01294 0.01276
23 0.01404 0.01391 0.01358 0.01337 0.01317 0.01297 0.01279 0.01261
24 0.01388 0.01365 0.01342 0.01321 0.01301 0.01282 0.01264 0.01246
25 0.01372 0.01349 0.01327 0.01306 0.01286 0.01267 0.01249 0.01232
26 0.01357 0.01334 0.01312 0.01291 0.01272 0.01253 0.01235 0.01218
27 0.01342 0.01319 0.01397 0.01277 0.01258 0.01239 0.01221 0.01204
28 0.01327 0.01304 0.01283 0.01264 0.01244 0.01225 0.01208 0.01191
29 0.01312 0.01290 0.01269 0.01249 0.01230 0.01212 0.01195 0.01178
30 0.01298 0.01276 0.01256 0.01236 0.01217 0.01199 0.01182 0.01169
Gs

49
Classify – 07
The fines fraction of a soil to be used for a highway fill was subjected to a hydrometer analysis by
placing 20 grams of dry fines in a 1 liter solution of water (dynamic viscosity 0.01 Poise at 20
degrees centigrade). The specific gravity of the solids was 2.65.
a) Estimate the maximum diameter D of the particles found at a depth of 5 cm after a
sedimentation time of 4 hours has elapsed, if the solution’s concentration has reduced to 2
grams/ liter at the level.
At that moment,
b) What percentage of the sample would have a diameter smaller than D?
c) What type of soil is this?
Solution:
a) Using Stoke’s relation:
2
18
s w
v d
γ γ
η
−
= or
18 ( )
( )
(min)
s w
L cm
d mm
t
η
γ γ
⎛ ⎞
= ⎜ ⎟
− ⎝ ⎠
where t = 4 hours = 14,400 sec, L = 5 cm, 2
2
10 . .,
dyne s
Poise i e
cm
η − ⋅
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
( )( )
3 3
2.65 9.81 / 26 /
s
s s s w
w
but G G dynes cm dynes cm
γ
γ γ
γ
= ∴ = = =
( )
2
2
3
sec
18 10 5
0.020
9.81 (2.65 1.00)(14, 400 sec)
dynes
x cm
cm
d mm
dynes
cm
− ⋅
⎛ ⎞
⎜ ⎟
⎝ ⎠
= =
⎛ ⎞
−
⎜ ⎟
⎝ ⎠
b) The unit weight γ of the solution after 4 hours is,
3 3
3
3
2 [1000 2 / 2.65] 1 /
1.001 /
1000
weight of soil in solution g cm g x g cm
g cm
volume of solution cm
γ
+ −
= = =
The portion of soil having a diameter smaller than D is,
( ) ( )
3
1000 2.65 1
1 1.001 1 0.08
20 2.65 1
8 % .
s w
s w
V cm x
Portion smaller
W g
The remaining soil is only of the original sample
γ γ
γ
γ γ
⎛ ⎞
= • − = − =
⎜ ⎟
− −
⎝ ⎠
∴
c) The diameter d = 0.020 mm corresponds to a silt.

50
Classify – 08
The formula for the relative compaction Dr is,
min
max
max
e
e
e
e
Dr
−
−
=
Derive an equivalent equation as a function of dry unit weights, such that
( )
( )
( )
( )
)
(
max
min
max
min
)
(
field
d
d
d
d
d
field
d
r
D
γ
γ
γ
γ
γ
γ
−
−
=
Solution:
γd (min) = dry unit weight in loosest condition (voids ratio emax)
γd = in-situ dry unit weight (voids ratio e)
γd (max) = dry unit in densest condition (voids ratio emin)
where
e
G
V
W W
S
S
d
+
=
=
1
γ
γ and Dr = 0 = loose, to 1= very dense
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎥
⎦
⎤
⎢
⎣
⎡
−
−
=
⎥
⎦
⎤
⎢
⎣
⎡
⎥
⎦
⎤
⎢
⎣
⎡
−
−
=
−
−
=
)
(
max
min
max
min
)
(
max
min
max
min
max
min
min
1
1
1
1
field
d
d
d
d
d
field
d
d
d
d
d
d
d
d
d
d
d
r
D
γ
γ
γ
γ
γ
γ
γ
γ
γ
γ
γ
γ
γ
γ
γ
γ
For example, what is the RC (relative density) of a sand in the field if it was tested to be at 98%
Standard Proctor, its maximum unit weight was 18.8 kN/m3
and its minimum unit weight was 14.0
kN/m3
?
( ) ( ) 3
( )
( . P r .)
9 8 % 1 8 .4 /
1 8 .8
d fie ld d fie ld
d fie ld
d S td o c t
R C k N m
γ γ
γ
γ
= = = ∴ =
( )( )
( )( )
%
94
4
.
18
0
.
14
8
.
18
8
.
18
0
.
14
4
.
18
max
min
max
min
=
−
−
=
⎥
⎦
⎤
⎢
⎣
⎡
⎥
⎦
⎤
⎢
⎣
⎡
−
−
=
d
d
d
d
d
d
r
D
γ
γ
γ
γ
γ
γ

51
Classify – 09
The data obtained from relative density tests is shown below. Calculate the range of relative
densities.
Limiting γ Average γ in kN/m3
γ max 18.07 17.52
γ min 14.77 15.56
γ field 16.97
Solution:
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
=
n
n
r
D
γ
γ
γ
γ
γ
γ max
min
max
min
( )( ) 71
.
0
97
.
16
07
.
18
77
.
14
07
.
18
77
.
14
97
.
16
1 max
min =
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
=
r
D
high
low
range γ
γ
( )( ) 60
.
0
97
.
16
07
.
18
56
.
15
07
.
18
56
.
15
97
.
16
2 max
min =
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
=
r
D
high
avg
range γ
γ
( )( ) 83
.
0
97
.
16
52
.
17
77
.
14
52
.
17
77
.
14
97
.
16
3 max
min =
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
=
r
D
avg
low
range γ
γ
( )( ) 74
.
0
97
.
16
52
.
17
56
.
15
52
.
17
56
.
15
97
.
16
4 max
min =
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
=
r
D
avg
avg
range γ
γ
60% 83%
D
r
∴ ≤ ≤

52
Classify – 10
South Florida has two types of sand, a quartzitic sand (γs2 = 165.5 pcf) and calcareous sand (γs1 =
146.3 pcf). At a particular site, their voids ratios were found to be:
for the quartzitic sand, еmax = 0.98 and emin = 0.53
for the calcareous sand, еmax = 0.89 and emin = 0.62
These voids ratios were measured by using a mold with a diameter of 4 inches and a height of 4.59
inches. The dry quartzitic sand weight was 3.254 lbs, and the dry calcareous sand was 2.868 lbs.
Find their relative densities and dry unit weights. Comment on these results.
Solution:
By definition
min
max
max
e
e
e
e
D r
−
−
=
For the calcareous sand, 70
.
0
3
.
146
12
868
.
2
12
)
3
.
146
868
.
2
(
4
)
59
.
4
(
)
4
(
4
3
3
2
1
1
1
1
2
1 =
×
×
−
=
−
=
π
γ
γ
π
S
S
P
P
h
d
e
For the quartzitic sand, 70
.
0
6
.
165
12
254
.
3
12
)
6
.
165
254
.
3
(
4
)
59
.
4
(
)
4
(
4
3
3
2
2
2
2
2
2
2 =
×
×
−
=
−
=
π
γ
γ
π
S
S
P
P
h
d
e
Notice that 2
1 e
e =
.
0
62
.
0
89
.
0
70
.
0
89
.
0
1
=
−
−
=
r
D and
.
0
53
.
0
98
.
0
70
.
0
98
.
0
2
=
−
−
=
r
D
The two types of sand have different relative densities because the calcareous sand grains are
more tightly packed than the quartzitic sand grains.
.
86
7
.
0
1
3
.
146
1 1
1
1
=
+
=
+
=
e
s
d
γ
γ pcf (but )
70
.
0
1
=
r
D
.
97
7
.
0
1
6
.
165
1 2
2
2
=
+
=
+
=
e
s
d
γ
γ pcf (but )
62
.
0
1
2
=
r
D
As a result, the dry unit weight is greater for the soil with the lower relative density.

53
Classify – 11
Prove that emin = 0.35.
3 3 3
3
0.1179 0.1179(2 ) 0.943
4
3
tetr
sphere
V a R R
V R
π
= = =
=
The volume of the sphere occupied by the tetrahedron is:
( )
60
0.167 16.7%
360
sphere tetr
V
∴ ≡ = =
( )
( )
35
.
0
3
/
4
167
.
0
3
/
4
167
.
0
943
.
0
3
3
3
=
−
=
−
=
=
∴
R
R
R
V
V
V
V
V
e
sphere
sphere
S
V
π
π
ALTERNATE METHOD:
( )
3
3
3
3
3
min
3
2 2 2
2
4
6 3
2
2 2
3
0.35
2
3
cube sphere
sphere
Volume of cube d d
d
Volume of sphere d
d d
V V
e
V d
π
π
π
π
= =
⎛ ⎞
= =
⎜ ⎟
⎝ ⎠
⎛ ⎞
−⎜ ⎟
− ⎝ ⎠
= = =
2R
2d
2
d

54
Chapter 4
Compaction and Soil Improvement
Symbols for Compaction
e → Voids ratio.
GS→ Specific gravity of the solids of a soil.
n → Porosity of the soil.
OMC→ Optimum moisture content.
S → Degree of saturation.
V → (
3 4 3
1 9.44*10
30
ft m
−
≡ Standard Proctor mold, ASTM D-698).
Va→ Volume of air.
VS→ Volume of solids.
VV → Volume of voids (water + air).
w→ Water content.
VS→ Volume of soil sample.
γ→ Unit weight of the soil.
γd → Dry unit weight.
γb → Buoyant unit weight of the soil.
γSAT →Saturated unit weight of the soil.
γS → Unit weight of the solid.
γW → Unit weight of water.
γd.field → Dry unit weight in the field.

55
*Compaction–01: Find the optimum moisture content (OMC).
(Revision: Aug-08)
A Standard Proctor test has yielded the values shown below. Determine:
(5) The maximum dry unit weight and its OMC; remember V = 1/30 ft3
.
(6) The moisture range for 93% of maximum dry unit weight.
No Weight of wet soil (lb) Moisture %
1 3.26 8.24
2 4.15 10.20
3 4.67 12.30
4 4.02 14.60
5 3.36 16.80
Solution:
Formulas used for the calculations: and
1
d
W
V w
γ
γ γ
= =
+
W (lb) w(%) γ (lb/ft3
) γd (lb/ft3
)
3.26 8.24 97.8 90.35
4.15 10.20 124.5 113.0
4.67 12.30 140.1 124.8
4.02 14.60 120.6 105.2
3.36 16.80 100.8 86.30

56
90
95
100
105
110
115
120
125
130
7 8 9 10 11 12 13 14 15 16 17 18
w(%)
γ
d
γdmax = 124.8 pcf
Maximum dry unit weight = 124.8 pcf
OMC = 12.3 %
γfield = (0.93)(124.8) = 116.1 pcf

57
*Compaction–02: Find maximum dry unit weight in SI units.
(Revision: Aug-08)
Using the table shown below:
(7) Estimate the maximum dry weight of a sample of road base material, tested under Standard
Proctor ASTM D-698 (all weights shown are in Newton).
(8) Note that the volume
3
3 4 3
3
1 1
9.44 10
30 35.32
m
V ft m
ft
−
⎛ ⎞
= = ×
⎜ ⎟
⎝ ⎠
(9) Find the OMC.
(10) What is the appropriate moisture range when attaining 95% of Standard Proctor?
Trial No. 1 2 3 4 5
W(Newton) 14.5 15.6 16.3 16.4 16.1
ω (%) 20 24 28 33 37
Solution:
Trial No. 1 2 3 4 5
( )
3
m
kN
V
W
=
γ 15.4 16.5 17.3 17.4 17.1
( )
3
1 m
kN
d
ω
γ
γ
+
= 12.8 13.3 13.5 13.1 12.5
γd max = 13.5 kN/m3
OMC = 28 %
OMC = 28%
γd-field = 0.95(13.5) = 12.8 kN/m3

58
12
12.4
12.8
13.2
13.6
18 20 22 24 26 28 30 32 34 36 38
w(%)
γ
d
(kN/m)

59
*Compaction-03: What is the saturation S at the OMC?
The results of a Standard Compaction test are shown in the table below:
ω (%) 6.2 8.1 9.8 11.5 12.3 13.2
γ (kN/m3
) 16.9 18.7 19.5 20.5 20.4 20.1
ω
γ
γ
+
=
1
d
15.9 17.3 17.8 18.4 18.2 17.8
(11) a) Determine the maximum dry unit weight and the OMC.
(12) b) What is the dry unit weight and moisture range at 95% RC (Relative Compaction)?
(13) c) Determine the degree of saturation at the maximum dry density if Gs = 2.70.
Solution:
a) γd max = 18.4 kN/m3
, OMC = 11.5%
b) γd at 95% = (0.95)(18.4) = 17.5 kN/m3
The moisture range w for 95% RC is from 8.75% to 13.75%.
OMC = 11.5 %
c)
13
14.3
15.6
16.9
18.2
19.5
5 6.1 7.2 8.3 9.4 10.5 11.6 12.7 13.8 14.9 16
w%
Dry
Unit
Weight

60
( )( )( )
( )
max
max
0.115 2.70 18.4
9.8
0.71
18.4
2.7
9.8
s d
w
d
s
w
wG
S
G
γ
γ
γ
γ
= = =
⎛ ⎞
− −⎜ ⎟
⎝ ⎠
Saturation S = 71%

61
*Compaction-04: Number of truck loads required.
The in-situ moisture content of a soil is 18% and its moist unit weight is 105 pcf. The specific
gravity of the soil solids is 2.75. This soil is to be excavated and transported to a construction site,
and then compacted to a minimum dry weight of 103.5 pcf at a moisture content of 20%.
a) How many cubic yards of excavated soil are needed to produce 10,000 yd3
of compacted fill?
b) How many truckloads are needed to transport the excavated soil, if each truck can carry 20
tons?
Solution:
( ) ( )
3
105
89 103.5
1 1 0.18
103.5
) Volume to be excavated 10,000
8
borrow site construction site
borrow site borrow site construction site construction site
d construction site
d borrow site
W W
V V
pcf
pcf versus pcf
w
pcf
a yd
γ γ
γ
γ γ
=
=
= = = =
+ +
=
( )
3
3
3
3 3
11,630
9
27 105
11,630
) Number of truck loads 824
20 2,000
yd
pcf
feet lb
yd
yd feet
b
ton lb
truc
t
k to
ruck loads
n
⎛ ⎞
=
⎜ ⎟
⎝ ⎠
⎛ ⎞⎛ ⎞
⎜ ⎟⎜ ⎟
⎝ ⎠
⎝ ⎠
= =
⎛ ⎞⎛ ⎞
⎜ ⎟⎜ ⎟
⎝
−
⎠⎝ ⎠

62
*Compaction-05: What is the saturation S at the OMC?
A Standard Proctor test was performed on a clayey gravel soil; the test results are shown below.
Find the degree of saturation at the optimum condition; assume that Gs = 2.60.
Test 1 2 3 4 5 6 7
w% 3.00 4.45 5.85 6.95 8.05 9.46 9.90
d
w
γ
γ 1.94 2.01 2.06 2.09 2.08 2.06 2.05
γd kN/m3
19.4 20.1 20.6 20.9 20.8 20.6 20.5
Use γw = 10 kN/m3
for simplicity.
Solution:
( )( )
( )( )
It is known that 1
therefore, at the OMC th
The degree of saturation at th
e saturation is,
20.9
0.0695 2.60 0.74
2.6
e OMC is
0 10 20.9
s s s d
s
d d
d OMC
OMC OMC s
s d
wG
Se wG S but e
e
S w G
γ γ γ
γ γ
γ
γ γ
−
⎛ ⎞ −
= ∴ = = − =
⎜ ⎟
⎝ ⎠
⎛ ⎞
⎛ ⎞
= = =
⎜ ⎟
⎜ ⎟ ⎜ ⎟
− −
⎝ ⎠ ⎝ ⎠
74%.
1.8
1.85
1.9
1.95
2
2.05
2.1
2.15
2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5
w%
Dry
Unit
Weight

63
*Compaction-06: Definition of the relative compaction (RC).
The relative compaction (RC) of a sandy road base in the field is 90%. The maximum and
minimum dry unit weights of the sand are γd(max) = 20.4 kN/m3
and γd(min) = 13.9 kN/m3
. Determine
the field values of:
a) The dry unit weight in the field;
b) Relative density (of compaction) Dr;
c) The moist unit weight γ when its moisture content is 15%.
Solution:
The relative compaction RC is the dry unit weight obtained in the field, as compared to the Standard
Proctor obtained in the laboratory.
( )( )
( )
( )
( ) ( )
( )
(max)
( ) (min) (max)
(max) (min) ( )
3
) The relative compaction is,
0.90 0.90 20.4
20.4
) The relative density is,
18.4 13.9 2
20.
18.4 /
4 13.9
d field d field
d field
d
r
d field d d
r
d d d field
a RC
RC
b D
m
D
kN
γ γ
γ
γ
γ γ γ
γ γ γ
= = = ∴ = =
− −
= × =
− −
( ) ( ) 3
0.4
0.768
18.4
) The moist unit weight is,
1
76.8
18.4 1 0.
%
21.2 /
15
d
c
m
w kN
γ
γ γ
⎛ ⎞
= =
⎜ ⎟
⎝ ⎠
= + = + =

64
*Compaction-07: The relative compaction (RC) of a soil.
(Revision: Aug-08)
A Standard Proctor compaction test performed on a sample of crushed limestone (Gs = 2.70)
obtained a maximum dry unit weight of 90 pcf at OMC. A field compacted sample showed a
moisture of 28% and a unit weight of 103.7 pcf.
Find the relative compaction (RC).
Find the degree of saturation S of the field soil sample.
Solution:
( )
( )
( )( )
( )
( )( )
( )
max
The Relative Compacti
103.7
81.0
1 1 0.28
81.0
) 0.90
90.0
) 1
1
2.70 62.4
1 1 1.08
81.0
0.28 2.70
0.70
1.
on = 9
08
0%
moist
d field
d field
d
s w s w
d
d
s w
d
s
s
pcf
w
pcf
a RC
pcf
G G
b e
e
G
e
Se wG
wG
S
e
γ
γ
γ
γ
γ γ
γ
γ
γ
γ
= = =
+ +
⎛ ⎞
= = =
⎜ ⎟
⎝ ⎠
= ∴ + =
+
⎛ ⎞
∴ = − = − =
⎜ ⎟
⎝ ⎠
=
∴ = = = 70%
Saturation S =

65
*Compaction-08: Converting volumes from borrow pits and truck loads.
(Revision: Oct.-08)
An embankment for a highway 30 m wide and 1.5 m thick is to be constructed from a sandy soil,
trucked in from a borrow pit. The water content of the sandy soil in the borrow pit is 15% and its
voids ratio is 0.69. Specifications require the embankment to be compacted to a dry unit weight of
18 kN/m3
. Determine, for 1 km length of embankment, the following:
a) The dry unit weight of sandy soil from the borrow pit required to construct the embankment,
assuming that GS = 2.70;
b) The number of 10 m3
truckloads of sandy soil required to construct the embankment;
c) The weight of water per truck load of sandy soil; and
d) The degree of saturation of the in-situ sandy soil.
Solution:
( )( ) 3
)
3
3
( )
(
2.7(9.8)
a) The borrow pit's dry unit weight
1 1 0.69
b) The volume of the finished embankment 30 1.5 1 = 45x10 m
Volume of borrow pit soil requir
7
ed
15.
S w
d
d reqd
d borrow pit
G
e
V m m km lo g
N
m
n
k
γ
γ
γ
γ
= = =
+ +
=
= ( ) ( )
3 3
3 3
3
3
3
18
45 10
15.7
18 45 10
Number of truck trips
15.7 10
c) W eight of dry soil in 1 truck-load 10 15.
5,160 truck-loa
7 157
W eight of water = (0.15)(1 ) 23
7
ds
5
d
d
V x m
x m
m
kN
W m kN
m
wW kN
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
⎛ ⎞
⎛ ⎞
= =
⎜ ⎟
⎜ ⎟
⎝ ⎠ ⎝ ⎠
⎛ ⎞
= =
⎜ ⎟
⎝ ⎠
= =
( )( )
0.15 2.70
d) Degree of satu
.6 per truck load
59%
ration 0.59
0.69
S
wG
S
e
kN
= = = =

66
9
.7
γd(max)
98% S. Proctor
1
4
.3
0.52
0.49
0.46
0.43
0.41
0.34
0.33
0.32
0.30
0.29
S
=
1
0
0
%
108
110
112
114
116
118
120
122
124
126
5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
w%
γ
d
V
o
id
R
a
tio
P
o
ro
s
ity
**Compaction-09: Ranges of water and fill required for a road.
(Revision: Octt.-08)
From the Standard Proctor compaction curve shown below:
Give two possible reasons that may cause a Proctor test to cross the ZAV curve?
What is the water content range (in gallons) needed to build a street 1,000 feet long of compacted
16” base at 98% Standard Proctor for two lanes, each 12 ft wide?

67
:
)
1
1
Thecrossingof theZAV(zeroair voids) curveis duetoanincorrect assumptionof ,
and/or amiscalculationof thewater content and .
)Th
s w wet
zav zav
s
s
wet
Solution
G
a Since and
wG w
S
G
w
b
γ γ
γ γ
γ
= =
+
⎛ ⎞
+⎜ ⎟
⎝ ⎠
( ) ( )( ) 3
max
esoil volume requiredfor theroadis,
1
16inthick 24 wide 1,000 long 32,000
12
Thepeakdrydensity 119.5 at anOMC=12%, andtherangeof thewater content
is from9.7%to14.
d
V
ft
V ft ft ft
in
pcf
γ
⎛ ⎞
= =
⎜ ⎟
⎝ ⎠
=
( ) ( )( )
( )( )
max
3 6
3%for 98%of StandardProctor or 98% 0.98 119.5 117.1
Thetotal weight of soil inthepavement is 32,000 117.1 3.75 10
Theweight of water at thelowend(9.7%) is,
0
d
w
w
w
w
pcf
W V ft pcf x lbs
W
W
V
γ
γ
γ
= =
= = =
= =
( )
( )
6
3
6
3
3.75 10
.097
1.097 7.45
40,000
62.4
Theweight of water at thehighend(14.3%) is,
3.75 10
0.143
1.143 7.45
56,000
62.4
Therefore,
w
w
w
w
x lbs
gal
gallons
pcf ft
W
x lbs
W gal
V gallons
pcf ft
γ
⎛ ⎞
⎜ ⎟
⎛ ⎞
⎝ ⎠
⎜ ⎟
⎝ ⎠
⎛ ⎞
⎜ ⎟
⎛ ⎞
⎝ ⎠
= = ⎜ ⎟
⎝ ⎠
( )( )
theaveragevolumeof water is 1/2 56,000 40,000 48,000
Therefore, thevolumeof wate 48,000 8,000 .
r required
gallons
gallons
+ =
= ±

68
**Compaction-10: Find the family of saturation curves for compaction.
(Revision: Oct.-08)
This problem expands Compaction-05: A Modified Proctor compaction test is performed on a
clayey gravel road base. The solids have a specific gravity of 2.65. The compaction data yielded the
following binomial values for γd/γw versus w %:
w(%) 3.00 4.45 5.85 6.95 8.05 9.46 9.90
γd / γw 1.94 2.01 2.06 2.09 2.08 2.06 2.05
(14) Find the γd max and the OMC from the compaction curve;
(15) Find the degree of saturation at the above conditions.
(16) Calculate the percentage of air for a given porosity n and the saturation S.
(17) Find the equation that describes the points of equal saturation.
(18) Determine the equation for S = 100%
(19) Discuss the characteristics of this last curve and equation.
Solution:
a)
18
18.5
19
19.5
20
20.5
21
21.5
22
2 3 4 5 6 7 8 9 10 11
Water Content w(%)
γ
d
(kN/m
3
)
γd max = 20.9 kN/m
3
OMC = 7.7%

69
1
n
1-n
a
nS
A
W
S
b) ( )( )
max
max
2.09
7.7% 2.65 76%
2.65 2.09
s d
s
w d
wG
Se wG S
e
γ
γ γ
⎛ ⎞ ⎛ ⎞
= ∴ = = =
⎜ ⎟ ⎜ ⎟
− −
⎝ ⎠
⎝ ⎠
c)
( )
1
1 1 1 1
(1 )
V W W W
W
V
a
V V V V
ns V
V V V
a nS n nS n
a n S V
= = = =
= − − − = − − +
= − =
d) Consider a family of curves of equal saturation,
volume of water volume of water
volume of air+volume of water 1 volume of soil grains
1
d
w
d
w s
d
s
d s
w s w
S
w
w S
S
S
S
w
S
γ
γ
γ
γ γ
γ γ
γ
γ
γ
γ
γ
= = =
−
⎛ ⎞
⎜ ⎟
⎛ ⎞
⎝ ⎠
= ∴ + − ∴
⎜ ⎟
⎛ ⎞ ⎝ ⎠
−⎜
=
+
⎟
⎝ ⎠
That is, the percentage of air a is equal to the porosity n times the factor (1 - S), where
S is the degree of saturation.

70
If w = 0 all curves pass thru γs
Asymptotes
γd = (1-a2)γs
Curves can be plotted for varying values of a and saturation S.
a = 0
a1
a2
a3
0
2
4
6
8
10
12
0 3 6 9 12 15 18 21 24 27
w
γ
d
S = 100%
S1
S2
0
0.2
0.4
0.6
0.8
1
1.2
0 10 20 30 40
w
γ
d
These are hyperbolas with the w-axis as an asymptote.

71
**Compaction-11: Water needed to reach maximum density in the field.
(Revision: Aug-08)
A Standard Proctor test yields the values listed below for a soil with Gs = 2.71. Find:
(20) The plot of the dry unit weight versus the water content;
(21) The maximum dry unit weight;
(22) The optimum moisture content;
(23) The dry unit weight at 90% of Standard Proctor;
(24) The moisture range for the 90% value;
(25) The volume of water (in gallons) that must be added to obtain 1 cubic yard of soil at the
maximum density, if the soil was originally at 10% water content.
Solution:
a) The plot of γd (dry unit weight) versus w (water content):
85
90
95
100
105
110
115
120
125
0 10 20 30
w
dry
unit
weight,
γd
moisture range
(b) From the plot, the maximum dry unit weight is γdmax = 107.5 pcf.
(c) From the plot, the optimum moisture content is OMC = 20%.
(d) The 90% value of the maximum dry unit weight γdmax = (0.9) (107.5) = 96.8 pcf
w (%) 10 13 16 18 20 22 25
γ (pcf) 98 106 119 125 129 128 123
γd (pcf) 89 94 102.6 105.9 107.5 104.9 98.4
OMC=20%
100% Proctor
90% Proctor

72
(e) The moisture range for the 90% value is approximately from 13% to 26%.
(f) Soil at 10% moisture,
( )( )
0.10 0.10
89 0.10 0.1 89 9 89 9 98
w
w s
s
s w s s w
W
w W W
W
but W lb and W W lb W W W lb
= = ∴ =
= = = = ∴ = + = + =
Soil at 20% moisture,
( )( )
0.20 0.20
107.5 0.20 0.2 107.5 21.5 129
w
w s
s
s w s s w
W
w W W
W
but W lb and W W lb W W W lb
= = ∴ =
= = = = ∴ = + =
Therefore, need to add the following water: (21.5 lb) – (9 lb) = 12.5 lb/ft3
∴
3
3 3
3 3 3
40 /
Answer: Add 40 gallons of water per cubic yard of compacted soi
12.5 7.48 27
6
l
.4 1
.
2
lb ft gallons ft
Added water
f
ga
t lb
llons yd
ft yd
= =
Ws= 107.5
Ww=21.5
V = 1 ft3
W = 129 lb
S
W
A
Ws=89
lb
Ww=9
V = 1 ft3
W = 98 lb
S
W
A

73
**Compaction-12: Fill volumes and truck load requirements for a levee.
(Revision: Aug-08)
Your company has won a contract to provide and compact the fill material for an earth levee, with
the dimensions shown below. The levee fill is a silty clay soil to be compacted to at least 95% of
maximum Standard Proctor of γd = 106 pcf at an OMC of 18%. Your borrow pit has a silty clay
with an in-situ moist density of 112.1 pcf at 18%, and a Gs = 2.68. When the soil is excavated and
loaded on to your trucks, the voids ratio of the material is e = 1.47. Your trucks can haul 15 cubic
yards of material per trip.
(26) Determine the volume of fill required for the levee;
(27) Determine the volume required from the borrow pit;
(28) Determine the required number of truckloads.
Solution:
( )( ) ( )( ) ( )( ) ( )
2
3
1 1 450
a) Volume of levee = 20' 40' 20' 20' 20' 60' 23,300
2 2 27 /
b) To find the volume required from the borrow pit consider that the weight of solids is the same
in both
ft long
ft cy
ft cy
⎛ ⎞
⎡ ⎤
⎛ ⎞ ⎛ ⎞
+ + =
⎜ ⎟
⎜ ⎟ ⎜ ⎟
⎢ ⎥
⎝ ⎠ ⎝ ⎠
⎣ ⎦ ⎝ ⎠
( )( )
( )
( ) ( )
( ) ( )
( ) ( )
( )
( )
,
But or or
112.1
where 95 0.95 106 100.7
1 1 0.18
100.7
23,300
95
s borrow s levee
s
d s d d borrow borrow d levee levee
d borrow d levee
d levee
borrow levee
d borrow
W W
W
W V V V
V
pcf and pcf
w
pcf
V V cy
pcf
γ γ γ γ
γ
γ γ
γ
γ
=
= = =
= = = = =
+ +
∴ = = 24,700cy
⎡ ⎤
=
⎢ ⎥
⎣ ⎦

74
( )( )
( ) ( )
( ) ( )
( )
( )
( ) ( )
Number of truck-loads required is based on,
2.68 62.4
100.7 =
1 1 1.47
s hauled s levee
d hauled hauled d levee levee
d levee
hauled levee
d hauled
S w
d levee d hauled
W W
V V
V V
G
but pcf and
e
γ γ
γ
γ
γ
γ γ
=
=
⎛ ⎞
∴ = ⎜ ⎟
⎜ ⎟
⎝ ⎠
= =
+ +
( )
( ) 3 3
( )
3
3
= 67.7
100.7
23,300 34,700
67.7
34,7
Number of truck-loads 2,3
00
=
15 /
14
d levee
hauled levee
d hauled
hauled
pcf
pcf
V V yd yd
pcf
V yd
truck capacity yd truck load
γ
γ
⎛ ⎞ ⎛ ⎞
= = =
⎜ ⎟ ⎜ ⎟
⎜ ⎟ ⎝ ⎠
⎝ ⎠
⎛ ⎞
⎛ ⎞
= =
⎜ ⎟
⎜ ⎟
−
⎝ ⎠ ⎝ ⎠

75
**Compaction-13: Multiple choice compaction problem of a levee.
(Revision: Aug-08)
A town’s new reservoir will impound its fresh water with a small earth dam, rectangularly shaped
in plan. The perimeter of the dam will be 2,200 ft long by 1,750 ft wide, and its cross-section is
shown below in figure B. The dam is to be built with a silty clay soil, with a specific gravity of 2.70,
available from three different local sources. Specifications call for a compacted soil at the dam
with a dry unit weight of 97 pcf at an OMC of 31 percent. Assume all voids are totally devoid of
any gas (or air). The borrow suppliers quoted the following:
Pit Price ($/yd3
) Gs S (%) w (%)
A 1.05 2.69 65 22
B 0.91 2.71 49 22
C 0.78 2.66 41 18
Questions:
1. What is the dam’s cross-sectional area?
a) 675 ft2
b) 2,100 ft2
c) 2,350 ft2
d) 2,700 ft2
e) 2,550 ft2

76
2. What is the approximate total volume V of soil required for the dam?
a) 2,550 ft3
b) 300,000 yd3
c) 900,000 yd3
d) 1.22 Myd3
e) 0.75 Myd3
3. What is the approximate volume of water impounded (stored) in the dam?
a) 6,300,000 ft3
b) 134 acre-ft
c) 7 Mft3
d) 154 acre-ft
e) 45 · 106
gallons
4. What is the unit weight γ of soil of the compacted earth dam?
a) 127 pcf
b) 97 pcf
c) 86 pcf
d) 98 pcf
e) 128 pcf
5. What is the designed voids ratio e of the compacted soil in the dam?
a) 1.1
b) 0.92
c) 0.84
d) 0.79
e) 1.2
6. What is the voids ratio e of the material in pit A?
a) 0.98
b) 0.91
c) 1.02
d) 1.1
e) 0.72
7. What is the voids ratio e of the material in pit B?
a) 0.93
b) 1.22
c) 0.81
d) 1.01
e) 1.00

77
8. What is the voids ratio e of the material in pit C?
a) 1.10
b) 1.12
c) 1.08
d) 1.05
e) 1.17
9. Assume that the voids ratio e of pit A is 0.98. What is the equivalent volume required of pit A to
place 1.2 million cubic yards of compacted soil in the dam?
a) 1.29 Myd3
b) 1.20 Myd3
c) 0.95 Myd3
d) 0.97 Myd3
e) 0.96 Myd3
10. Assume that the voids ratio e of pit B is 0.81, what is the equivalent volume required of pit B to
a) 1.00 Myd3
b) 1.02 Myd3
c) 1.18 Myd3
d) 1.05 Myd3
e) 1.07 Myd3
11. Assume that the voids ratio e of pit C is 1.10, what is the equivalent volume required of pit C to
a) 1.34 Myd3
b) 1.37 Myd3
c) 1.25 Myd3
d) 1.23 Myd3
e) 1.21 Myd3
12. Which pit offers the cheapest fill?
a) Pit A
b) Pit B
c) Pit C
d) Both A and C
e) Both B and C

78
Chapter 5
Permeability of Soils
Symbols for Permeability
A→ area of a seepage surface.
C→ Hazen’s coefficient.
d→ Diameter of a capillary tube
Dx → Diameter of a soil % finer (represents % finer by weight)
e →The voids ratio.
GS → Specific gravity of the solids of a soil.
h→ Thickness of the aquifer.
H→ Thickness of the soil layer.
hC → Height of the rising capillary
i→ Hydraulic gradient.
k→ coefficient of permeability in D’Arcy’s equation .
kH → Coefficient of horizontal permeability..
kV → Coefficient of vertical permeability
L→ Distance of the hydraulic head loss.
Ludgeon→ Standard unit for permeability ( 4
10
sec
mm
−
).
Nf→ The number of flow channels in Forheimer’s equation.
Neq→ The number of equipotentials drops in Forheimer’s equation.
po →In-situ vertical pressure at any depth .
q→ Flow rate (ft3
per second per foot of width).
Q→ Total seepage (total flow).
Sc → Seepage capacity.
TS → Surface tension (typically given as 0.073 N per meter).
u → pore water pressure.
umax → Maximum pore water pressure.
γS →Unit weight of solids.
γW →Unit weight of water .
σV→ Vertical effective stress.

79
*Permeability–01: Types of permeability tests and common units.
(Revision: Aug-08)
(a) When is it appropriate to use a constant-head permeability test versus a falling-head
permeability test?
(b) What are the “standard” units of permeability versus the “common” unit?
Solution:
(a) The constant head test is performed for granular soils (gravels G, and sands S), whereas the falling
head test is used for fine-grained (cohesive) soils (silts M, and clays C).
(b) In Europe, the “standard” unit is the Ludgeon (10- 4
mm/sec), and in the USA the standard unit is the
Meinzer, which is the rate of flow in gallons per day through an area of 1 square foot under a hydraulic
gradient of unity (1 foot/foot).
The “common” unit of permeability is cm / sec.

80
*Permeability-02: Use of Hazen’s formula to estimate the k of an aquifer.
(Revision: Aug-08)
A test boring was performed at an elevation 955 feet MSL, and it found the phreatic surface
(water table) 5 feet below the ground surface. An aquifer stratum was identified, and a sample of
its soil showed the grain size distribution below. Estimate the permeability using Hazen’s formula
with the coefficient C = 12. A piezometer (measures the location of the WT) was installed 2500 feet
downstream from the boring, and showed its phreatic surface at elevation 942 feet MSL. If the
thickness of the aquifer was a uniform 12 feet between both points, estimate the quantity of flow
per foot of width in gallons/hour ).
45
.
7
1
( 3
gallons
ft ≈
Solution:
Allen Hazen’s (1893) formula for the permeability ( )
2 2
10
( ) 12 (0.16 ) 0.31
mm
k C D mm
s
= = =
The hydraulic head drop is (955 ft – 5 ft) – (942 ft) = 8 feet
Applying D’Arcy formula,
( )( )
3
3600 1 1 7.45 8
0.31 12 1
sec 25.4 12 1 2,500
1.05
h mm s in ft gal ft
q k A ft ft
L hour m
gallons
per ft width
m in hou
ft t r
f
⎛ ⎞⎛ ⎞
Δ ⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞
= = =
⎜ ⎟⎜ ⎟
⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠
GRAIN SIZE DISTRIBUTION DIAGRAM
D10
D10 = 0.16 mm

81
*Permeability-03: Flow in a sand layer from a canal to a river.
(Revision: Aug-08)
A canal and a river run parallel an average of 250 feet apart. The elevation of the water surface in
the canal is at +1050 feet and in the river at +1021 feet. A stratum of sand intersects both the river
and the canal below their water levels. The sand is 6 feet thick, and is sandwiched between strata
of impervious clay. Compute the seepage loss q from the canal in
3
ft
day mile
−
if the permeability of
the sand is 2 x10-3
.
sec
ft
Solution:
D’Arcy’s formula for q yields,
( ) ( )
3
3
635
86,400 (1050 1021) 5,280
2 (10) 6.0
250
,000
h ft s ft ft
q k A ft
L s day f
ft
q
day
t mi
mi e
le
l
−
=
⎛ ⎞⎡ ⎤
Δ − ⎛ ⎞
= = ⎜ ⎟ ⎜ ⎟
⎢ ⎥
⎝ ⎠
⎝
−
⎠⎣ ⎦
ELEV. +1050’
250’
ELEV. +1021’
CLAY
SAND
CLAY
6 ’

82
*Permeability-04: Find the equivalent horizontal permeability of two layers.
(Revision: Aug-08)
The topmost layer is loose, clean sand, 1 meter thick. Its vertical permeability kV can be estimated
using Hazen’s formula with C = 1.5 (to over-estimate) and the sieve analysis shown here. Its kH is
known to be approximately 500% of the kV. Below the sand stratum is a marine marl, 3 meters
thick, with a kV = kH = 10-6
m/s. What is the combined kHcomb for the upper 4 m in cm/sec?
Sand 1 m
.
Marl 3 m
Solution:
Use Hazen’s formula to find the permeability, where k is in cm/s if C ranges from 0.8-1.5. D10 is in mm.
Use the grain-size distribution curve of Permeability-02.
( )( ) ( )( )
1 1
2 2 -6
10
6 6
1 1 2
2
4
2
1
5 5( ) (5)(1.5)(0.16 mm) = 1.92 / = 1,920(10 ) /
The formula for combining seve
4.8 10 /
ral horizontal layers is,
1,920 10 / 1 10 / 3
1 3
H V
Hcomb
k k CD cm s m s
x m s m m s m
k H k H
k
H
cm
H m
s
m
− −
−
= = =
+
+
= = =
+
×
+

83
*Permeability-05: Equivalent vertical and horizontal permeabilities.
(Revision: Aug-08)
The soil profile shown below is typical of Miami-Dade County. Estimate the equivalent
permeabilities kV(eq) and kH(eq) in cm/sec, and the ratio of kH(eq) / kV(eq)..
1 m Fine Sand k = 1x10-1
cm/sec Pamlico Formation
5 m Porous Limestone k = 2x10-3
cm/sec Miami Formation
3 m Fine Sand k = 1x10-3
cm/sec Fort Thompson Formation
12 m Upper sandy Limestone k = 2x10- 4
cm/sec Fort Thompson Formation
Solution:
( )
[ ]
( )
( )
3
1 2 4
1 3 3 4
1 2 3 4
3
3
1 3 3
( ) 1 1 2 2 3 3 4 4
(1 5 3 12)
1 5 3 12
10 / 2 10 / 10 / 2 10 /
2,100
0.32 10 /
(6,551 10 )
1 1
(1)(10 ) (5)(2 10 ) (3)(10 ) (12)(
21
V EQ
H EQ
H m
k
H m m m m
H H H
cm s cm s cm s cm s
k k k k
cm
cm s
x s
k H k H k H k H k
H m
− − − −
−
− − −
+ + +
= =
+ + +
+ + +
× ×
= = ×
= + + + = + × + +
)
3
3
(
3
4
4.8 10 /
15
0.32 10 /
The horizontal permeabi
2 10 )
4.8 10 /
Therefore
lityis 15 times larger than the vertical permeability.
H EQ
H
V
k cm s
k cm s
k cm s
−
−
−
−
⎡ ⎤
×
=
×
⎣ ⎦
×
×
=
=

84
*Permeability-06: Ratio of horizontal to vertical permeabilities.
(Revision: Aug-08)
Estimate the ratio of the horizontal to the vertical permeability of these four strata.
Solution:
The equivalent horizontal permeability of all four layers is:
( )( )
( )
( )( ) ( )( )( ) ( )( ) ( )( )( )
3 4 5 3
( ) 1 1 2 2 3 3 4 4
4
( )
/
1
( ) 3 10 3 2 10 3 10 3 2 10
12
8 10 /sec
H eq
H eq
ft cm s
k k H k H k H k H
H ft
k x cm
− − − −
−
⎡ ⎤
= + + + = + + +
⎣ ⎦
=
The equivalent vertical permeability of all four layers is:
1 2 3 4
4
3
1 2 4
3 4 5 3
12
0.37 10 / sec
3' 3' 3' 3'
10 2 10 10 2 10
Veq
V V V V
H ft
k x cm
H
H H H
x x
k k k k
−
− − − −
= = =
+ + +
+ + +
Therefore the ratio of the horizontal to the vertical permeability is:
4
4
8 1 0 /
0 .3 7 1
2
0
2
/
H eq
V eq
k x cm s
k x cm s
−
−
= =
H1 = 3’
k1 = 10-3
cm/sec
k2 = 2x10-4
cm/sec H2 = 3’
k3 = 10-5
cm/sec H3 = 3’
k4 = 2x10-3
cm/sec H4 = 3’

85
*Permeability–07: Do not confuse a horizontal with a vertical permeability.
(Revision: Aug-08)
The soil layers below have a cross section of 100 mm x 100 mm each. The permeability of each soil
is: kA =10-2
cm/sec.; kB =3 x 10-3
cm/sec; kC = 4.9 x 10-4
cm/sec. Find the rate of water supply in
cm3
/hr.
Solution:
This is a trick drawing: it “looks” like a horizontal flow, but in reality it is a vertical flow because the
flow has to cross through every layer; it can not “bypass” any layer. Therefore, every soil layer has the
same flow v = v1 = v2 = v3 and the total head Δh = Δh1 + Δh2 + Δh3.
( )
3
( )
3
1 2
2 3 4
1 2
3
3
450
1.2 10 /sec
150 150 150
1 10 / 3 10 / 4.9 10 /
300 3,600sec
(0.0012 / ) (10 )(10 )
450 1
291 /
V eq
eq eq
H mm
k cm
H mm mm mm
H H
x cm s x cm s x cm s
k k k
h mm
q k iA k A cm s cm cm hour
cm
H mm hr
−
− − −
= = =
+ +
+ +
⎛ ⎞ ⎛ ⎞
Δ
∴ = = = =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
Δh=300 mm
hA hB
A B C v
H1=150 mm H2=150 mm H3=150 mm

86
*Permeability-08: Permeability as a function of the voids ratio e.
(Revision: Aug-08)
The coefficient of permeability of fine sand is 0.012 cm/sec at a voids ratio of 0.57. Estimate the
increased permeability using the Kozeny-Carman formula of this same sand when its voids ratio
has increased to 0.72.
Solution:
Using the Kozeny-Carman formula,
( )
( )
( )
3
3 3
1
1 1
3 3
2
2 2
2
2
1
1
0.57
1
0.012 1 0.57 0.544
0.72
1 1 0.72
0.012
0.544
Notice that since = 0.012 cm/s the permeability has almost doubled. A 26% increase
of the voids ratio has effe
0.022
cted a
/
e
k
e
e
k e
e
k k
e
k
k
cm s
=
+
+ +
∴ = = = =
+ +
∴ = =
doubling the permeability.

87
*Permeability–09: Uplift pressures from vertical flows.
(Revision: Aug-08)
The soil below is a dense well-graded clayey sand with γd = 112 pcf and Gs = 2.63, a permeability k
= 240 mm/min at a voids ratio of e = 0.85; the cross-sectional area of the tank is 36 ft2
. Find (a) the
seepage rate q in ft3
/min., and (b) the direction of the flow.
Solution:
a) The hydraulic gradient i and the voids ratio e are:
2
4 (2.63)(62.4 )
0.667 1 1 0.465
6 112
s w
d
G
h ft pcf
i and e
H ft pcf
γ
γ
Δ
= = = = − = − =
The Casagrande formula relates the known permeability k0.85 at e = 0.85 to an unknown
permeability k at any voids ratio e,
2 2
0.85
1 1
1.4 1.4(0.465) 240 0.239 / min
min 25.4 12
in
mm ft
k e k ft
mm in
⎛ ⎞⎛ ⎞
⎛ ⎞
= = =
⎜ ⎟⎜ ⎟
⎜ ⎟
⎝ ⎠⎝ ⎠⎝ ⎠
Therefore, the seepage rate q is:
2
3
0.238 ( 0.667)(36 )
m
5.74
m n
i i
n
ft
q kiA ft
ft
⎛ ⎞
= = − =
⎜ ⎟
⎠
−
⎝
b) The flow direction is UP.
Clayey sand
H1=2’
H2=6’
Δh = 4 feet

88
*Permeability-10: Capillary rise in tubes of differing diameters.
Determine the different heights hc that water will raise in three different capillary tubes, with
diameters:
d1 = 0.00075 mm (corresponding to a fine clay sized particle),
d2 = 0.075 mm (corresponding to the smallest sand sized particle) and
d3 = 0.75 mm (corresponding to a medium sand sized particle).
Assume that the surface tension is 0.075 N/m with an angle α = 3o
.
Solution:
The surface tension of water Ts ranges from about 0.064 to 0.075 N/m (0.0044 to 0.0051 lb/ft). In this
problem we have chosen the largest value. Notice that the negative sign indicates that the water has risen
due to the capillary tension.

89
( )( ) ( )( )
( )
( )( )
( )( )
( )( )
( )( )
( )( )
( )( )
3
3
3
3
s
2
1 7
2
2 5
2
3 4
-4 T c o s 4 0 .0 7 5 / c o s 3
=
9 .8 1 /
4 7 .5 1 0 / c o s 3
= =
7 .5 1 0 9 .8 1 /
4 7 .5 1
- 4 1 m
- 0 .4 1 4 1 0
0 / c o s 3
= =
7 .5 1 0 9 .8 1 /
4 7 .5 1 0 / c o s 3
= =
7 .5 1 0 9 .8 1 /
- 0 .0 4 1 4 1
c
w
m m
N m
h
d d k N m
x N m
h
x m k N m
x N m
h
x m k N m
m
x N m
m
h
x m k N m
α
γ
−
−
−
−
−
−
− °
=
− °
− °
− °
=
= m m

90
*Permeability-11: Rise of the water table due to capillarity saturation.
How much does the capillary water rise above the water table in a very fine sand (d = 0.1 mm)
if the surface tension force is To = 0.064 N/m with an α = 3º?
Solution:
( )
( )( )
4 3
4 0.064 / cos3
4 cos
1
0.
0 9.
2
81 /
6
o
c
w
N m
T
h
d m m
m
kN
α
γ −
− °
−
= = = −
z
100% Saturation zone
Capillary saturation zone
Capillary fringe zone
Discontinuous moisture zone
Capillary flow
Vapor flow
hc

91
*Permeability-12: Find the capillary rise hc in a silt stratum using Hazen.
(Revision: Aug-08)
Another method of determining the capillary rise in a soil is to use Hazen’s capillary formula. The
3 m thick dense silt layer shown below is the top stratum of a construction site, has an effective
diameter of 0.01 mm. What is the approximate height of the capillary rise in that silt stratum?
What are the vertical effective stresses at depths of 3 m and 8 m below the surface? The “free
ground water” level is 8 meters below the ground surface, the γS =26.5 kN/m3
, and the soil between
the ground surface and the capillary level is partially saturated to 50%.
Solution:
( )( )
10
15.3
In essence, the entire silt
0.0306 0.0306
1. The Hazen empirical formula fo
stratum is saturated through capillarit
r capillary rise is
0.2 0.2 0.01
2. For full saturation, 1
y.
00%,
C
S
h
D mm
S
Se wG
m
= = =
=
= ∴
( )( )( )
( )( )
( )( )( )
( )( )
( ) ( )( )
3
3
1
3
1 0.40 9.81
0.148
26.5
26.5 1.148
1
21.8
1 1.40
For 50% saturation,
0.5 0.40 9.81
0.074
26.5
26.5 1.074
1
20.3
1 1.40
Therefore,
' 3 2 1
0.3 6
S S
SAT S
S
S
V
V
Se
Se
w
G
w kN
kPa
e m
Se
w
w kN
e m
h
ω
ω
γ
γ
γ γ
γ
γ
γ γ
σ γ
σ
= = = =
+
⎛ ⎞
∴ = = =
⎜ ⎟
+
⎝ ⎠
= = =
+
⎛ ⎞
∴ = = =
⎜ ⎟
+
⎝ ⎠
∴ = = =
∴( ) ( ) ( )( ) ( )( )
1 2
8
' 3 20.3 17
5 2 0
1.8
SAT
h h kPa
γ γ
= + = + =
h2 = 5 m
h1 = 3 m
Dense silt
Clay

92
*Permeability-13: Back-hoe trench test to estimate the field permeability.
A common method of determining a site’s drainage capabilities is the constant-head trench
percolation test shown below. The trench is dug by a backhoe to roughly the dimensions shown.
The testing crew uses a water truck to fill the trench with water above the WT, and then they
attempt to maintain the head constant for about 10 minutes. The amount of water that has flowed
out during the test is Q (in gallons/minute). The seepage capacity Sc = Q/CL’H, where C is a units
conversion factor, L’ is the trench semi-perimeter (length plus width, in ft) and H is the head. The
units of the seepage capacity are commonly given in cfs/ft/ft. Based on the reported geometric
conditions shown below, and that the crew used 1,540 gallons during a 10 minutes test, what is the
surface seepage capacity of that site?
Solution:
( ) ( )
( )( )
( )( )
3
1,540
154 / minute
10min
The perimeter of the trench is ' 2 6 1.5 15 4.4 1.8 2.6
The seepage capacity of the surface stratum is ,
154 / min min
' 15 2.6 7.45
c
c
gal
Q gallons
L ft ft ft and H ft ft ft
S
gallons ft
Q
S
C L H ft ft ga
= =
= + = = − =
= =
⋅ ⋅ ( )( )
3
60 se
8 0 / /
c
8. 1
x cfs
llo
t
s
ft f
n
−
=
0.0’
4.4’
8.0’
1.8’

93
**Permeability-14: Seepage loss from an impounding pond.
(Revision: Aug-08)
Borings were taken at the site of an intended impoundment pond and the in-situ voids ratios at
various depths are shown in the figure below. A constant-head permeability test was performed on
sample #1 (which was 6” high and 2” in diameter) subjected to a pressure head of 27”: after 5
seconds, 50 grams of water were collected through the sample.
(29) Determine the permeability of sample #1;
A trial test of a vibroflotation (densification) probe was taken to a depth of 5 feet and showed a
densified voids ratio e = 0.55. If this densification ratio is attained to the full depth of 25 feet, at
Impoundment pond area = 4.4 miles2
ELEV
0’
5’ e = 0.750 (Sample 1)
Δh
10’ e = 0.217 (Sample 2)
Geomembrane
15’ e = 0.048 (Sample 3)
L
20’ e = 0.015 (Sample 4)
25’ e = 0.006 (Sample 5)

94
what depth (to the nearest 5 feet) would you place the bottom of the pond in order to keep the total
seepage Q below 50,000 gal/min?
Assume seepage only through the bottom of the pond, and that the pond is kept filled.
Solution:
Step 1: Determine the permeability k of sample # 1.
From Darcy; Q= k i A, therefore 2
4
q q
k
iA i d
π
= =
Since 50 grams of water is equivalent to 50 cm3
(for γw = 1g/cm3
)
but ,
3 3
50
10
5sec sec
cm cm
q = = , and
27
4.5
6
h in
i
L in
Δ
= = =
( )
( )
2
2
2
4 10 (10 / )
4
0.122
sec
4.5 6 2.54
a
cm
mm cm
q mm
in
k
i d cm
in
in
π
π
⎛ ⎞
⎜ ⎟
⎝ ⎠
∴ = = =
⎛ ⎞
⎜ ⎟
⎝ ⎠
Step 2: Determine the new permeability of the sand due to vibroflotation densification.
Using Casagrande’s relation
2
0.85
1.4
k k e
=
Therefore
( )
( )
2
2
2
2
0.55
* 0.122 0.066 /sec
0.75
a
b a
b
e
k k mm
e
= = =
Step 3: The ratio of densified permeability to in-situ permeability
0.066
0.54
0.122
b
a
k
k
= = ∴ Vibroflotation has reduced the permeability by half.
Step 4: Find the densified permeability at each sample depth,

95
2
0.85
1.4
k k e
= therefore 0.85 2
1.4
k
k
e
= 2
(0.122 /sec)
0.155 /sec
1.4(0.75 )
mm
mm
= =
The corresponding densified permeability
2
0.85
1.4
k k e
= (x ratio)
Depth (feet) Original eo Original k (mm/sec) Densified k (mm/sec)
0. 0. 0. 0.
-5 0.750 0.122 0.066
-10 0.217 0.010 0.005
-15 0.048 0.005 2.76x10-4
-20 0.015 5.08x10-5
2.74x10-5
-25 0.006 1.03x10-5
0.56x10-5
Step 5: Estimate the required depth h of the pond.
Q = 50,000 gallons/min (1ft3
/ 7.48 gal) = 6,680 ft3
/min.
Consider this rate to be constant.
From Darcy’s equation Q kiA
= or
k h Q
ki
L A
Δ
= =
( )( ) ( )( )
3
3
4
2
2
1
6,680
min 3.281 10
2.79 10
sec
sec
4.4 5280 1min 60
min
mm
ft
ft
Q mm
A ft
mile
mile
−
−
⎛ ⎞
⎛ ⎞
⎜ ⎟⎜ ⎟
×
⎝ ⎠⎝ ⎠
= = ×

96
-30
-25
-20
-15
-10
-5
0
0 1 2 3 4 5 6
Consider the depth -15 ft.
4 4
15
2.76 10 4.14 10 /sec
10 sec
h mm
k x x mm
L
− −
Δ ⎛ ⎞
= =
⎜ ⎟
⎝ ⎠
or at a depth of: -20 ft.
5 4
20
0.56 10 0.224 10
5 sec sec
h m m m m
k x x
L
− −
Δ ⎛ ⎞
= =
⎜ ⎟
⎝ ⎠
Therefore, place the bottom of pond at -19 feet.
( )
sec
10 4 mm
A
Q −
×
Depth
ft
18.7 ft
2.76

97
Chapter 6
Seepage and Flow-nets
Symbols for Seepage and Flow-nets

98
*Flownets-01: Correcting flawed flow-nets.
(Revision: Aug-08)
Do you recognize something wrong with each of the following flow-nets?
a) b)
c)
Solution:
a) Incorrectly drawn mesh, because two equipotential lines intersect each other (equipotential lines
and flowlines must intersect orthogonally to each other).
b) Incorrectly drawn mesh, because two flow-lines intersect each other (same as above).
c) The well should be at the center of the net (a sink or a source point).
Equipotential Lines
Flow Lines
Well
filter

99
*Flow-nets-02: A flow-net beneath a dam with a partial cutoff wall.
(Revision: Aug-08)
The completed flow net for the dam shown below includes a steel sheet-pile cutoff wall located at
the head-water side of the dam in order to reduce the seepage loss. The dam is half a kilometer in
width (shore to shore) and the permeability of the silty sand stratum is 3.5 x 10-4
cm/s. Find, (a) the
total seepage loss under the dam in liters per year, and (b) would the dam be more stable if the
cutoff wall was placed under its tail-water side?
Solution:
(a) Notice that ∆h = 6.0 m, the number of flow channels Nf = 3 and the equipotentials Neq = 10.
Using Forcheimer’s equation,
4 6 2
3
3.5 10 (6.0 ) 6.3 10 /sec/
sec 100 10
f
eq
N cm m
q k h m m per mof dam width
N cm
− −
⎛ ⎞
⎛ ⎞ ⎛ ⎞
= Δ = × = ×
⎜ ⎟
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
⎝ ⎠
Since the dam is 500 meters wide (shore-to-shore) the total flow Q under the dam is,
3
6 3 6
3
10 sec
500 6.3 10 / 100
sec 31.5 10
1
liters
Q Lq m m
m year
millionliters
year
− ⎛ ⎞⎛ ⎞
⎡ ⎤
= = × × =
⎜ ⎟⎜ ⎟
⎣ ⎦
⎝ ⎠
⎝ ⎠
b) No. Placing the cutoff wall at the toe would allow higher uplift hydrostatic pressures to develop
beneath the dam, thereby decreasing the dam’s stability against sliding toward the right (down-stream).
ψ1
ψ2
ψ3
IMPERVIOUS STRATUM (CLAY OR ROCK)
15 m
Δh = 6.0 m
2.0 m
10.0 m
17.0 m

100
*Flow-nets-03: The velocity of the flow at any point under a dam.
(Revision: 12 Oct.-08)
Using the flow net shown below, (1) determine the seepage underneath the 1,000 foot wide
concrete dam, and (2) the velocity at point “a” in feet/hour, where the height of the net’s square is
19 feet. The soil has a GS = 2.67, D10 = 0.01 mm. Overestimate the flow by using Hazen’s coefficient
C = 15 to determine the permeability k.
Solution:
2 2 2
10
Find the permeability using Hazen's formula:
( ) 15(10 ) 0.0015
sec
Using Forheimer's equation with flow lines = 5 and equipotentials = 12,
1
0.0015
sec 25.4
−
= = =
⎛ ⎞
⎛ ⎞
= Δ = ⎜ ⎟
⎜ ⎟
⎝ ⎠⎝ ⎠
f eq
f
eq
k
mm
k C D mm
N N
N mm in
q k h
N mm
( )
( )
( )( )
3
3
2
3
1 3,600 sec 5
30' 5' 0.185
12 1 12
1,000 0.185
The velocity at " " has a flow in only that channel, or /5,
18
0
5
.185
5
19 1
⎡ ⎤⎛ ⎞ ⎛ ⎞
− =
⎜ ⎟ ⎜ ⎟
⎢ ⎥ −
⎝ ⎠
⎣ ⎦⎝ ⎠
⎛ ⎞
∴ = = =
⎜ ⎟
⎝ ⎠
⎛
⎜
= =
⎝
ft
ft ft
in hour hr ft of dam
ft
Q Lq ft
hr
a q q
ft
q hr
v
A ft high ft wide
hr
0.002
⎞
⎟
≈
⎜ ⎟
⎜ ⎟
⎜ ⎟
⎠
ft
hr
Δh = 30 feet
5 feet
“a” 19’

101
*Flow-nets-04: Flow through an earth levee.
(Revision: Aug-08)
In western Miami-Dade County, the Everglades are contained with levees. Levee #111 runs
North-South about 2 kilometers west of Krome Avenue and its cross section is show below.
Laboratory tests indicate that the permeability of the 80-year old levee is 0.30 m/day. What is the
volume of water lost through the levee along each kilometer in m3
/day?
Cross-section of levee looking north.
Solution:
Using Forheimer’s equation,
( ) ( )
3
3
1,00 2,0
0 0.3 23
10
70
f
eq
N m
Q Lq L k h m m
m
da
da y
N y
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= = Δ = =
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎜ ⎟ ⎝ ⎠
⎝ ⎠
⎝ ⎠
DRAINAGE
C B
A
D
E
WT
50 m 12 m 50 m
2.7 m
112 m
2 m
23 m 2:1
2:1

102
*Flow-nets-05: Finding the total, static and dynamic heads in a dam.
(Revision: Aug-08)
Find the seepage through the earth dam shown below in gallons/day if the sieve analysis shows the
D10 to be 0.17 mm, and the dam is 1,200 feet wide. What is the pressure head at the top of the
aquiclude and at mid-dam (point A)?
10
2 2 2
10
10
Number of flow channels 3
Number of equipotential drops 7
Using Hazen's formula 15 15(0.17 ) 0.43
sec
Note: 8 15 for mm; to overestimate flows, use =15.
f
eq
N
N
mm
k CD D mm
C D C
=
=
= = = =
≤ ≤
Solution:
( ) ( )
3
6
1 1 7.5 3 86,400sec
1,200 0.43 40
sec 25.4 12 1 7
18.8 10
1
f
eq
g
N mm inch ft gallons
Q Lq L k h ft ft
N mm inches f
allon
t d y
s
Q
d y
a
a
⎛ ⎞ ⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞
⎛ ⎞ ⎛ ⎞
= = Δ =
⎜ ⎟ ⎜ ⎟⎜
= ×
⎟⎜ ⎟ ⎜ ⎟
⎜ ⎟ ⎜ ⎟
⎜ ⎟ ⎝ ⎠ ⎝ ⎠
⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠
⎝ ⎠
( )
( )
4.4
At point "A" the is 40
7
2
The at "A" is approximately 40
3
Therefore,
dynamic pressure head 25.1
static head 26.7
th = static + dynam
e total head 5
ic = 25.1 26.7 = 8
1.
ft
ft
ft f
feet
feet
e
t fe t
⎛ ⎞
=
⎜ ⎟
⎝ ⎠
=
+
rock toe
40’
25’ Clay (an aquiclude)
0 1 2 3 4 5 6
7
A

103
**Flow nets-06: Hydraulic gradient profile within an earth levee.
(Revision: Aug-08)
The cross-section of an earth dam 5,000 feet wide is shown below. Determine (a) the seepage flow
through the dam, in ft3
/ minute, (b) the hydraulic gradient in square I, and (c) the pore pressures
along a trial failure surface along the line ED.

104
Solution:
(a) From graph D10 = 0.04 mm. Using Hazen’s relation, with C = 15 to overestimate the permeability of
the dam,
( ) ( )( )
( ) ( )
2 2
10
3
15 0.04 0.024 /sec
1 1 86,400sec 3
5,000 0.024 40
sec 2
454,000
5.4 12 9
f
eq
k C D mm mm
N mm inch ft
Q Lq L k h ft ft
N mm inches d
ft
Q
day
ay
= = =
⎛ ⎞ ⎛ ⎞⎛ ⎞⎛ ⎞
⎛ ⎞ ⎛ ⎞
= = Δ =
⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟
⎜ ⎟ ⎜ ⎟
⎜ ⎟ ⎝ ⎠ ⎝ ⎠
⎝ ⎠
⎝ ⎠⎝ ⎠
⎝
=
⎠
(b) The gradient in square I is,
Δh 40/9
lI 11.2
0.40
iI = = =
(c) The pore pressures along ED are approximately,
at
40
9
40
9
3 40
9
40
9
40
9
40
9
40
9
0
277 psf = 0.28 ksf
555 psf = 0.55 ksf
832 psf = 0.83 ksf
1109 psf = 1.11 ksf
1387 psf = 1.39 ksf
1664 psf = 1.66 ksf
62.4 =
1
2
4
5
6
D u = 40' - 8 ( )
62.4 =
u = 40' - 7 ( ) 62.4 =
62.4 =
u = 40' - 3 ( )
u = 40' - 6 ( )
u = 40' - 5 ( )
u = 40' - 4 ( ) 62.4 =
62.4 =
1.80 ksf
) 62.4 = 1803 psf =
40' - 2.5
u = (
E 0
pore pressure

105
**Flow-net-07: Flow into a cofferdam and pump size.
(Revision: Aug-08)
A cofferdam is to be built in the middle of a bay to place the foundations of a tall television tower.
A plan area of the cofferdam is 30 m long by 10 m wide. A sample taken from the bay bottom was
subjected to a hydrometer analysis: 20 grams of bay bottom dry fines were mixed with 1 liter of
water. The specific gravity of the solids was found to be 2.65. The dynamic viscosity of water is 10-
2
Poise (dynes-sec/cm2
) at 20o
C. After 1 hour of precipitation, the hydrometer dropped 16 cm. The
soil is uniform in size, with 80 % passing the # 200 sieve.
(30) What type of soil was the sample?
(31) Will a large 3 m3
per minute pump be adequate to maintain a 1 m draw down below the
bay bottom? Use FS > 2.
Pump
CL
7 m
1 m
15 m
Clay stratum

107
Solution:
a) Use Stoke’s formula to find diameter of the bay bottom particles:
( ) ( )( )
2 3
10 2
18 10 sec 16
18 18
.
1 2.65 1 9.81 3600sec
0.070
S W W S
x dynes x cm x cm
L
D d
t G t cm dynes x
d mm
η η
γ γ γ
−
⎛ ⎞
⎛ ⎞ −
⎜ ⎟
= = = ⎜ ⎟
⎜ ⎟⎜ ⎟
− − −
⎝ ⎠⎝ ⎠
=
Therefore the soil is silt (0.075 mm to 0.002 mm).
b) Hazen’s formula permits us to estimate the permeability k of the soil:
2 2
10 15(0.07 ) 0.074
sec sec
mm mm
k CD mm
= = =
Use Forheimer’s formula to estimate the total flow Q into the cofferdam:
( )
3
2 5
3
3 3
5
3
3
4
7.4 10 8 29.6 10
sec 8 10
60
( )( ) 80 29.6 10 . 1.42
min min
3 / min
1.42 / mi
.11 2 !
n
2
f
eq
N mm m m
q k h x m x
N m
OKAY Pump is adeq
m m s
m s m
Q perimeter q m x
m s
m
Pump FS
m
uate
− −
−
⎛ ⎞ ⎛ ⎞⎛ ⎞
= Δ = =
⎜ ⎟ ⎜ ⎟⎜ ⎟
−
⎝ ⎠ ⎝ ⎠⎝ ⎠
⎛ ⎞
= = =
⎜ ⎟
−
⎝ ⎠
>
= =

108
*Flow-nets-08: Drainage of deep excavations for buildings.
(Revision: Aug-08)
A new office building will require a two-level underground parking garage. The plan size of the
site is 100 x 80 meters. Some of the soil properties are shown below.
a) At what depth of the excavation will the limestone (shear strength = 0.1 MN/m2
) have a
punching shear failure? Suggest using a 1m x 1m plug as a model.
b) What size pump do you need (m3
/minute) with a factor of safety of 3?
Pump
Elevation +0 m
Sand 3
1 7
k N
m
γ =
Elevation –1.0 m
Anchor
Limestone
3
19
kN
m
γ =
Nf = 3 A Neq = 8 Elevation –7 m
0 8
F2 F1
F3
Elevation –9 m
1
2 7
3
4 5 6
Sandstone 3
21
kN
m
γ = Elevation –11 m

109
Solution:
( ) ( ) ( )
( )
y ta n
2
3
2
2
a) T h e u p lift fo rce at p o in t A is fo u n d b y,
F = 0 - 0
6 9 .8 1 1 5 9
0 .1 4 4 0 0
5 9
0 .1 4 7 7 1
4 0 0
T h is c o rre sp o n d
u p lift sh ea r resis ce
u p lift w
sh e a r
F F
k N
F u A h A m m k N
m
M N k N
F A x m x
m m
x m w ith F S x m
γ
τ
=
⎛ ⎞
= = Δ = =
⎜ ⎟
⎝ ⎠
= = =
∴ = = = ∴ =
∑
( )( )
( ) ( )
2
2
1 0
3
s to e le v a tio n - 6 m .
b ) D e te rm in e th e flo w q u a n tity w ith F o rh e im e r's fo rm u la ,
1 2 .5 0 .0 9 5 0 .1 1 3
sec
5
3 6 0
3 6 0 0 .1 1 3 5 1 0
se c 8
f
q
N
a
Q q L L k h
b N
m m
w h e re k C D m m
h m
m m m
Q m m
m m m
−
⎛ ⎞
⎛ ⎞
= = Δ
⎜ ⎟
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
= = =
Δ =
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
∴ = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝
3
3
3
4 .6 / m in 5 / m in
3 1 5 .
m in
m
F o r a F S u se a p u m p
m m
= ≈
⎟
⎠
=

110
*Flow-nets-09: Dewatering a construction site.
(Revision: Aug-08)
The figure below shows a dewatering plan to build the foundations of an office building below the
water table and without sheet-piling. The plan area of the excavation is 400 m long by 100 m wide.
The soil has a D10 of 0.02 mm. What size pump do you need (gpm) with a Factor of Safety = 2?
Solution:
( ) ( )( )
( ) ( )
2 2
10
3
3
3
Notice that 8 , 3, 4 and = 1, 000 .
The permeability 15 0.02 0.006 /
3 60
200 800 0.006 8 2.16
4 10 1min min
2.16
m
f eq
f
eq
h m N N L perimeter m
k C D mm mm s
N mm m s m
Q L q L k h m m
N s mm
m
Q
Δ = = = =
= = =
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= = Δ = + =
⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎝ ⎠
=
3
3
7.45
in 0
60
i
30 n
.
0
m
ft gal
m f
a lons
t
g l
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
=
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠
⎝ ⎠
⎝ ⎠
Therefore, for a factor of safety of 2 use at least a 1,200 gallons per minute pump or
two 600 gallons per minute pumps.
Dewatering wells
Marshy soils
Δh = 2m Excavated site CL
Δh = 2m
1 Δh = 2m
2
3 Δh = 2m Lowered WT
Neq = 4
Nf = 3

111
*Flow-net-10: Dewatering in layered strata.
(Revision: Aug-08)
The figure below shows the profile of a square excavation (in plan view) in a layered soil, where
the vertical permeability is 5 x 10-5
m/s and the horizontal permeability is roughly ten times higher
than the vertical. Estimate the dewatering capacity requirements, in m3
/hour, to prevent the
excavation from flooding. The value of Δh is to scale, but you may use 10 m.
CL
40 m 40 m
10m
10m 80 m
Phreatic surface condition is CL
fulfilled along the top flow-line
GWL Sheet-pile wall
Datum for h
(½ square) h=0
h=1(1/3)m
h=2(2/3)m
h=4m
h=5(1/3)m
h=6(2/3)m
(½ square)
h=8m
h=9(1/3)m
(½ square)
(½ square)
Assumed recharged boundary h = 10 4m
Scale

112
Solution:
( )( ) ( )
( )( )
( ) ( ) ( )
3
4 5 4
4
3
10 , 4, 8
5 10 / 5 10 / 1.6 10 /
T he perim eter of the cofferdam 4 80 320
4
1.6 10 10 320
8
3, 600
0.253 911
f eq
x y
f
eq
h m N N
k k k x m s x m s m s
p m m
N m
Q qp k h p m m
N s
m s
Q
s r
r
m
h
h
− − −
−
Δ = = =
= = =
= =
⎡ ⎤
⎛ ⎞ ⎡ ⎤
⎛ ⎞ ⎛ ⎞
= = Δ =
⎢ ⎥
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎢ ⎥
⎜ ⎟ ⎝ ⎠ ⎝ ⎠
⎣ ⎦
⎢ ⎥
⎝ ⎠
⎣ ⎦
⎛ ⎞ ⎛ ⎞
= ⎜ ⎟ ⎜ ⎟
⎝
=
⎝ ⎠
⎠

113
**Flownets-11: Flow through the clay core of an earth dam.
(Revision: Aug-08)
An earth dam on a pervious but strong earth foundation has the cross-section shown in the figure
below. The core of the dam is sealed from the jointed rock foundation with a thin layer of grout.
(32) State the function and properties of the core shell and drains;
(33) What is the function of the grout between the core and foundation? Under what conditions
is it most important?
(34) Calculate the seepage quantity per foot of length of the dam through the dam, through the
foundation, and the total seepage quantity.
(35) What grading requirements should be specified for the inclined filter A?
(36) What minimum permeability k is required in the horizontal drain B to prevent saturation
from rising into the random fill zone? Give the results of k in ft/day.
Solution:
(a) The core is used to retain water within the dam, that is, to resist seepage. The material should be
relatively impermeable (clay) and should not shrink or swell excessively.
The shell provides the structural strength to support and protect the core. The material must be
more permeable than the core material, strong and durable.
h
1
=5
ft
L = 150 ft
10 ft
inclined filter A
horizontal drain B
Foundation layer permeability k2 = 0.1 ft/day
core
shell
grout
H=100
ft
Reservoir surface
dam permeability k1= 0.001 ft/day
Note: The grain size of core = 100% passes 1”, 15% size = 1/8”, and 85% size = 0.001 in.

114
The drains are provided to reduce the pore water pressures in the foundation and in the
embankment to increase stability. The drains also remove seepage water to reduce soil erosion.
The drain material must be permeable enough to permit drainage with a low head loss and yet
fine enough to keep the adjacent soil in place.
(b) The primary function of the grout between the core and foundation is to form an impervious
layer which prevents seepage along the contact surface. This becomes most important when the
ratio k2/k1 becomes large.
(c) Calculate the seepage Q by using the flow net shown in the figure.
1) Through the dam: Q = k1(Δh/L)b in ft3
/day/ft where b is the normal distance between
streamlines. The flow net divides the core into 4 zones (#1 at the bottom, #4 at the top).
4 4
Q = k1 ∑ ij bj = k1 ∑ (Δh)j (b)j
j=1 j=1
In zone #1, the flow net is nearly rectangular, so (b)1 = 2; for zone # 4, (b )j = 1
l l
The average head loss hl across the core in each zone
(p) + Zu = constant = 100’ on the upstream face
γ
On down stream face of core (p) = 0 is assumed in the drain,
γ
so that ZL + hL = constant = 100’ on the downstream face
Using an average ZL for each zone by scaling,
Zone#1 ZL = 2’ hL = 98’ ΔhL = 98
6 6
Zone#2 ZL = 10’ hL = 90’ ΔhL = 90
5.5 5.5
Zone#3 ZL = 25’ hL = 75’ ΔhL = 75
4.5 4.5
Zone#4 ZL = 55’ hL = 45’ ΔhL = 45
3 3
4
Q1 = k1 ∑Δhj (b/L)j = 0.001 [98’(2) + 90’ + 75’ + 45’] = 0.081 ft3
/day/ft of dam
j=1
6 5.5 4.5 3
2) Through the foundation Q2 = NF kh

115
Nd
where NF = number of flow paths, Nd = number of equipotential drops and
h = total head dissipated.
Q2 = 3 (0.1)(100) = 3.75 ft3
/day/ft of dam
8
3) The total seepage Q is therefore,
Q = Q1 + Q2 = 0.081 + 3.75 = 3.83 ft3
/day /ft of dam
(d) The grading requirements for the inclined filter A,
(1) Free drainage, require D15 (filter) ≥ 4 D15 (soil)
D15 (filter) ≥ 4 (0.001 in)
≥ 0.004 in
(2) To prevent erosion of the core material requires D15 (filter) ≥ 4 D85 (soil)
≥ 4 (0.001 in)
≥ 0.004 in
So 85% of the filter material must be coarser than 0.01” to 0.2”. The filter grain size grading
curve should be parallel to or flatter than the core material grading curve. See the graph on the
next page for one possible grading curve, which gives
100% passes = 10 inch
15% passes = 2 inch
85% passes = 0.01 inch
(e) The drain B must carry the total seepage flow Q = 3.83 ft3
/day/ft of dam calculated above. The
Dupuit formula for two-dimensional flow on a horizontal impervious boundary is
Q = k (h1
2
-h2
2
)
2L
where Q = 3.83ft3
/day/ft, L = 150 ft, h1= 5 ft and h2 < 5 ft.
At what value of h2 will it minimize k? Clearly it is when h2 = 0, although this does seem unrealistic
since we are saying that the flow at the lower end of the drain has zero depth. Nevertheless, it gives us a
minimum value, which is:

116
kmin = 2LQ = 2(150)(3.83) = 46 ft/day
h1
2
52

117
Chapter 7
Effective Stresses and Pore Water Pressure
Symbols for Effective Stresses and Pore Water Pressure

118
*Effective Stress–01: The concept of buoyancy.
(Revision: Aug-08)
What force is required to hold an empty box that has a volume of 1 cubic foot, just below the
water surface?
Solution:
The volume of the displaced water is 1 ft3
.
Therefore, the force is the weight of 1 ft3
of water = 62.4 lbs / ft3
.
What is the force required to hold the same box 10 feet below the surface?

119
*Effective Stress–02: The concept of effective stress.
(Revision: Aug-08)
A sample was obtained from point A in the submerged clay layer shown below. It was determined
that it had a w = 54%, and a Gs = 2.78. What is the effective vertical stress at A?
Solution:
The effective stress σ’ at the point A consists solely of the depth of the soil (not of the water)
multiplied by the soil buoyant unit weight.
( )
' ' '
'
'
In order to find there are a number of derivations, such as this one,
where the voids ratio can be found through
1
and noticing that 1 becaus
soil b SAT W
S W
W S
h where
G e
e Se wG
e
S
σ γ γ γ γ γ
γ
γ
γ γ
= = = −
+
⎡ ⎤
= − =
⎢ ⎥
+
⎣ ⎦
= ( )
( ) ( )
( )
( )
'
'
e the soil is 100% saturated, 0.54 (2.78)
2.78 0.54 (2.78) (9.81)
' 9.81 15
1 1 0.54 (2.78)
105
S
S W
soil W soil
kP
e wG
G e
h h m
e
a
γ
σ γ γ
σ
= =
⎡ ⎤
+
⎧ ⎫ ⎡ ⎤
+
⎡ ⎤
⎪ ⎪ ⎣ ⎦
= = − = −
⎢ ⎥
⎨
=
⎬
⎢ ⎥
+ +
⎪ ⎪ ⎢ ⎥
⎣ ⎦
⎩ ⎭ ⎣ ⎦
water
A
Saturated clay
hw = 25 m
hs = 15 m

120
*Effective Stress–03: The concept of effective stress with multiple strata.
(Revision: Aug-08)
The City of Houston, Texas has been experiencing a rapid lowering of its phreatic surface (draw-
downs) during the past 49 years due to large volumes of water pumped out of the ground by
industrial users.
a) What was the effective vertical stress at a depth of 15 m in 1960?
b) What is the effective stress at the same depth in 2009?
c) What happens to the ground surface as a result of the draw-downs?
Solution:
a)
[ ] [ ] [ ]
( ) [ ] [ ]
'
'
'
' ' ' ' ' ' '
(20.4)(3) 18.8 9.81)(3 (14.9 9.81)(6) (12.6 9.81)(3)
128
V SAT W
SAN D SILT A
V
C L Y
V
h
kPa
h h h w here
σ γ γ γ γ γ γ
σ
σ
γ
= + + + = −
= + − + − + −
⎡
=
⎤
⎣ ⎦
b) ( ) [ ]
'
(20.4)(6) 16.5)(6 (12.6 9.81)(3 30
) 2
V kPa
σ = + + − =
⎡ ⎤
⎣ ⎦
This is an 80% increase in stress due solely to a dropping water table.
c) The ground surface has also been lowered, due to the decreasing thickness of the sand and the
silt strata due to their loss of the volume previously occupied by the water.

121
Effective Stress-03B
Revision
In the soil profile shown below, show a plot of the pore water pressure and the effective stress
along the right margin of the figure, with numerical values at each interface. Pay heed to the
capillarity in the upper clay. Assume S = 50%.
Solution:
Δh = (62.4) x (4) = -250
Δh = (62.4) x (6) = 562
Δh = (110.7) x (6) = 664
Δh = {[(110-62.4) x (4)] + 664} = 854
Δh = {[(117-62.4) x (9)] + 854} = 854
Δh = 664 – 250 = 414
Δh = 0 + 854 = 854
Δh = 562 + 1346 = 1908
Depth (ft) u + σ' σ
0 0 0 0
6 -250 664 414
10 0 854 854
19 562 1346 1908

122
Chapter 8
Dams and Levees
Symbols for Dams and Levees

123
*Dams-01: Find the uplift pressure under a small concrete levee.
Calculate the uplift force at the base of the weir, per foot of width. Points A and B are at the
corners of the concrete levee.
Solution:
The dynamic head drop per equipotential is,
30 5
( ) 1.8 /
14
Pressure head at A = 30 ft + 8 ft - h (1.5) = 35.3 ft
Pressure head at B = 30 ft + 8 ft - h (10) = 20.1 ft
The upli
A B
eq
H H ft ft
h ft drop
N drops
− −
Δ Δ = = =
Δ
Δ
( ) 3
16
fting force is,
35.3 20.1
98' 62.4 169,000
2 2
9
A B
w
F
p p lb lb
F L
ft ft
kip
ft
γ
+ +
⎛ ⎞ ⎛ ⎞
= = = =
⎜ ⎟
⎜ ⎟
⎝ ⎠
⎝ ⎠

124
*Dams-02: Determine the uplift forces acting upon a concrete dam.
(Revision: Aug-08)
The uplift (hydrostatic) force under the concrete gravity dam shown below varies as a straight line
from 67% of the headwater pressure at the heel, to 100% of the tail-water at the toe. Assume γ
Concrete = 145 pcf
a) Determine the Factor of Safety against overturning; and
b) Determine the FS against sliding, if the sand that underlay the dam has 37o
ϕ = .

125
Solution:
Step 1: Determine all forces on the dam
0
=
Σ V
F
( )( ) ( )( ) ft
kips
ft
ft
ft
ft
ft
k
dam
of
weight
V /
395
,
7
300
260
2
1
40
300
145
.
0 3
1 =
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
=
0
0 =
∑M
( )( ) ( ) ( )
( )
toe
of
left
ft
k
k
ft
k
ft
k
O
toe
from
x 198
5655
1740
260
3
2
5655
280
1740
,
1 =
+
⎟
⎠
⎞
⎜
⎝
⎛
+
=
( )( ) ft
kips
ft
ft
ft
k
tion
toe
upon
water
of
weight
vertical
V /
97
52
60
2
1
0624
.
0
sec
: 3
2 =
⎟
⎠
⎞
⎜
⎝
⎛
=
( ) toe
of
left
ft
ft
O
toe
from
x 3
.
17
52
3
1
,
2 =
=
( ) ft
kips
ft
ft
k
headwater
from
force
lateral
H /
534
,
2
285
0624
.
0
2
1
:
2
3
1 =
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
( ) .
95
285
3
1
,
1 toe
the
above
ft
ft
O
toe
from
y =
=
( ) ft
kips
ft
ft
k
tailwater
from
force
lateral
H /
112
60
0624
.
0
2
1
:
2
3
2 =
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
( ) .
20
60
3
1
,
2 toe
the
above
ft
ft
O
toe
from
y =
=
Step 2: The hydrostatic uplift at the base of the dam.
( )( )
3
1
: 300 2,534 /
2
LEFT RIGHT
V uplift force p p ft kips ft
= + =
( ) ( )( )
0.67 0.67 0.0624 285 11.9
LEFT w
where the pressure p h ft ksf
γ
= = =
( ) ( )( )
1.00 1.0 0.0624 60 3.7
RIGHT w
and the pressure p h ft ksf
γ
= = =
( ) ( )
3 2
1
11.9 3.7 300 2,340 /
2
k
V ft kips ft
ft
∴ = + =
( ) ( ) ( )
( ) ( )
toe
of
left
ft
ft
ft
k
ft
k
ft
ft
k
ft
ft
ft
k
ft
k
ft
ft
ft
k
O
toe
from
x 3
.
176
300
7
.
3
9
.
11
2
1
300
7
.
3
300
3
2
300
7
.
3
9
.
11
2
1
2
300
300
7
.
3
,
2
2
2
2
2
2
3 =
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
+
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=

126
Step 3: The factor of safety (FS) against overturning (taken about the toe),
1
1
3
3
2
2
2
2
1
1
y
H
x
V
y
H
x
V
x
V
moments
g
overturnin
moments
resisting
FS
+
+
+
=
=
( )( ) ( )( ) ( )( )
( )( ) ( )( )
7395 198 97 17.3 112 20
2534 95 2340 176.3
2.2 2
overturning
F GOOD
S >
+ +
= =
+
Step 4: The F.S. against sliding,
( )
2 1 2 3
1
tan
sliding
H V V V
resisting forces
FS
driving forces H
φ
+ + −
= =
( ) ( )
112 7395 97 2340 tan37
2534
1.58 2
sliding
FS NOT GOOD ENOUGH
+ + − °
= = <

127
Chapter 9
Stresses in Soil Masses
Symbols for Stresses in Soil Masses
θ→ The angle of the plane of interest angle with respect to the major principal stress (σ1).
σmax→ Maximum normal axial stress.
σmin→ Minimum normal axial stress.
σθ→ The normal stress on a plane with an angle θ with respect to the major principal stress plane (σ1).
τmax→ maximum shear stress.
τminx→ Minimum shear stress.
τθ→ The shear stress on aplane with an angle θ respect to the major principal stress plane (σ1).
ф → The angle of internal friction of the soil.
τn→ Shear stress.
σn→ Normal stress.
qu→ Ultimate shear strength of a soil.
Symbols for Boussinesq Stresses
B→ Width of the loaded selected region.
L→ Length of the loaded selected region.
m→ The ratio (B/Z).
n→ The ratio (L/Z).
N→ Normal load carried by a foundation.
Dp→ Increased stress on the soil from a surface loaded area.
p→ Stress of the loaded area.
z→ Depth of the soil at the point of interest.
Symbols for Newmark
IV → The influence value in the Newmark’ chart (for example, a chart divided into 100 areas, each is IV=0.01.
AB→ Scale to the depth of interest to determine the size of the surface structure graph the Newmark’s graph.
C→ consolidation.
M→ Nomber of squares (enclosed in the Newmark’ chart).
po → The effective stress at the point of interest.
q→ The load of the footing.
σx→ stress at an specific point (x).
z→ Depth of the stratum stat of the soil.

129
*Mohr-01: Simple transformation from principal to general stress state.
(Revision: Jan-09)
A soil particle is found to be subjected to a maximum stress of 14.6 kN/m2
, and a minimum stress
of – 4.18 kN/m2
. Find the σ and τ on the plane of θ = 50° with respect to the major principal
stresses, and also find τmax.
(a) The graphical solution,
(b) The calculated solution,
( )
( )
1 3 1 3
2
1 3
2
1 3
maximum 2
14.6 4.18 14.6 4.18
cos2 cos2 50 3.6
2 2 2 2
14.6 4.18
sin 2 sin 2 50 9.2
2 2
9.4
2
kN
m
kN
m
kN
m
θ
θ
σ σ σ σ
σ θ
σ σ
τ θ
σ σ
τ
+ − − +
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= + = + ° =
⎜ ⎟ ⎜ ⎟
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
⎝ ⎠ ⎝ ⎠
− +
⎛ ⎞ ⎛ ⎞
= = ° =
⎜ ⎟
⎜ ⎟
⎝ ⎠
⎝ ⎠
−
⎛ ⎞
= =
⎜ ⎟
⎝ ⎠

130
*Mohr – 02: Find the principal stresses and their orientation.
Equations for the principal stresses in the elastic half-space shown below for a uniformly loaded
strip footing are as follows: σ1 =
q/π(α + sin α) and σ3 = q/π(α - sin
α).
The direction of the major
principal stress bisects the angle α.
Calculate the vertical stress σy, the
horizontal stress σx, and τxy at
point A if x = 0.75B and y = 0.5B
using Mohr’s diagram.
Solution:
τ =
σ =
σ + σ
σ =
σ , τ
94.77
2θ =
Ο
β
1
2
2
σ =
σ + σ
1 2
σ , τ
85.23
β =
Ο
47.38
Ο
σ =
σ =
3
θ
σ =
α + δ = arc tan 2.5 = 68.20° 2θ + β = 180°
δ = arc tan 0.5 = 26.57° β = 85°
68.20° - 26.57° = 41.63°
if q = 432.5 kPa then σ1 = (σx + σy)/2 + R = 100 + 70.7 = 170.7 kPa
σ3 = (σx + σy)/2 - R = 100 - 70.7 = 29.3 kPa
Β /2
α
δ
Β /2

131
*Mohr – 03: Find the principal stresses and their orientation.
(Revision: Sept-09)
Given the general stresses at a point in a soil, determine the principal stresses and show them on a
properly oriented element.

132
*Mohr – 04:
(Revision: Sept-09)
A sample of clean sand was retrieved from 7 m below the surface. The sample had been under a
vertical load of 150 kN/m2
, a horizontal load of 250 kN/m2
, and a shear stress of 86.6 kN/m2
. If the
angle θ between the vertical stress and the principal stress is 60°, what is the angle of internal
friction φ of this sample?
Solution:

133
*Mohr – 05: Normal and shear stress at a chosen plane.
(Revision: Sept-09)
Using a Mohr circle, determine the normal and shear stresses on the plane AB.
Solution:
σy = 90 lb/ft2
(a) σ1 = σv + σx +√ σv - σx
2
+ τ xy
2
2 2
- 40 lb/ft2
A σ3 = 90 + 150 - √ 90-125 2
+ (-40) 2
2 2
τn
B σx
σn 125 lb/ft2
τxy - 40 lb/ft2
(b) σn = σ1 + σ3 + σ1 - σ3 cos 2θ
2 2
= 151 + 64 + 151- 64 cos60°
2 2
τn = σ1 - σ3 sin 2θ
2
= 151 – 64 sin60°
2
θ 30°
σ1 = 151 psf
σ3 = 64 psf
σn = 129 psf
τn = 38 psf

134
**Mohr – 07: Back figure the failure angle
(Revised: Sept-09)
From the stress triangle shown below, find (a) the maximum and minimum principle stresses, (b)
the angle alpha, as shown, (c) the angle theta, and (d) the value for the maximum shear stress.
A Graphical Solution:
(a) σ1 = 25 kN/m2
and σ3 = 0
(b) α in the figure is in the angle 2θ between
(σ, τ) and σ3.
…… α = 2θ = 126.9˚
(c) …… θ = 63.4˚
and τmax= 12.5 kN/m2
OR B Analytical Solution:
1 θ
σ
σ
σ
σ
σ 2
COS
2
2
1
1
1
1
1
+
+
+
=
θ
σ
θ
σ
σ
2
cos
1
40
2
COS
2
2
20 1
+
=
∴
+
= 1
1
1+cos2θ
2 2
SIN
2
n θ
σ
σ
τ 3
1 −
=
2
2
SIN
20
2
SIN
2
10 1
θ
ϑ
θ
σ
=
∴
= 1
∴ 1 - 2 sin 2θ + cos 2θ = 0
…… θ = 63.4˚
….. 0
m
KN
25
126.9
sin
20
3
0
=
=
= σ
σ ,
/ 2
1
…… τmax= 12.5 kN/m2
B
β C
A β
20 kN/m2
10 kN/m2
σ3 = 0
τ
σ3=0
τmax = 12.5
σ1=25
(σ, τ)
(20, -10)
2θ
2θ

135
*Mohr – 08: find the Principle pressure using Mohr
(Revised Sept-09)
The temporary excavation shown below is braced with a steel tube strut. Every morning, a
misguided foreman tightens the screw mechanism on the strut “just to be safe”. The stress on a
soil particle at point A, just behind the wall, has been measured with a pressure sensor installed by
the Engineer. It now measures 40 kN/m2
. If the potential failure planes in the soil behind the wall
sustain 60° angles with respect to the vertical wall, estimate the normal and shear stresses at that
point A along a potential failure plane.
Solution:
At point A:
σv = hγ = (1.25 m) (16 kN/m3
) = 20 kN/m2
∴ σv is the minor principal stress at A,
Since θ = 60° is with respect to the major principal stress (σ1) plane, then σv=σ 3
∴ σθ = (σ1+σ3)/2 + (σ1-σ3)/2 cos 2θ = (40+20)/2 + (40-20)/2 cos 120°
∴ σθ = 25 kN/m2
and τθ = (σ1-σ3)/2 sin 2θ = (40-20)/2 sin 120°
∴ τθ = 8.7 kN/m2

136
*Mohr – 09: Relation between θ and φ.
(Revised Sept-09)
For a clean sand, prove that θ = 45°+ φ/2 using Mohr’s circle.
Solution:
For sand c=0.
By inspection in ΔOAB
(180°-2θ)+90°+φ=180°
∴ 2θ=90°+φ
∴ θ = 45° + φ/2
A failure test on a clean sand (i.e. c=0) shows that σ1=11.5 ksf and σ3=3.2 ksf at failure. Find
the angle φ for this sand.

137
*Mohr – 10:
(Revised Sept-09)
Determine the normal and shear stresses on the plane AB.
001 = (300+125)/2 = 212.5 psf
0102 = (300-125)/2 = 87.5 psf
01B = 552
87.52 + = 103 psf
σ1 = 212.5 + 103 = 315.5 psf
σ3 = 212.5 – 103 = 109.5 psf
πC0102 = tan-1
(55/87.5) = 32°
σn = 212.5 – 103 cos (32+40) = 181 psf
τn = 103 sin (32+40) = 98 psf

138
*Mohr–11:
(Revised Sept-09)
(37) Derive the equation that transforms a general state of stress to the principal state of stress.
(Hint: Use Mohr’s circle for a graphical solution).
(38) Determine the value of the major principal stress.
(39) Determine the angle θ between the major principal stress and the state of stress shown in
the figure above.

139
*Mohr – 12:
(Revised Sept-09)
Determine the maximum and minimum principal stresses, and the normal and shear stresses on
plane AB.
Maximum and minimum principal stresses:
Normal and shear stresses:

140
*Mohr – 13: Data from Mohr-Coulomb failure envelope.
(Revised Sept-09)
A soil sample has been tested and when plotted developed the Mohr-Coulomb envelope of failure
shown below. Find (1) axial stresses at failure, (2) the normal and shear stresses on the failure
plane, (3) the angle of failure with respect to the principal axis, and (4) the soil tensile strength.
Solution:
σ1f = qu = 300 Pa
σf = σθ = 86 Pa
τf = τθ = 136 Pa
θ = 115°/2 = 57.5°
qu = -123 Pa
+ σ
+ τ
25 °
96 kN/m2
(1.) σ1f = qu
+ σ
(4.) qu
+ τ
σ1
2θ = 115°
(3.) θ
(2.) σf = σθ
τf = τθ

141
**Mohr – 14:
(Revised Sept-09)
A dry sample of sand was tested in a triaxial test. The angle of internal friction was found to be
36°. If the minor principal stress was 300 kPa, at what value of maximum principal stress will the
sample fail? The same test was then performed on a clay sample that had the same φ, and
cohesion of 12 kPa. What was the new maximum principal stress?
Solution:
a) Failure will occur when the Mohr circle becomes tangent to the failure envelopes.
3
1
2
1
3
2
1
R
sin
d
(1 - sin )
sin sin
(1 sin )
sin sin
(1 sin )
tan (45 )
(1 sin ) 2
36
(300 )tan (45 ) 1160
2
R
d R R R
R
d R R R
kPa kPa
φ
φ
σ
φ φ
φ
σ
φ φ
σ φ φ
σ φ
σ
=
= − = − =
+
= + = + =
+
∴ = = °+
−
°
∴ = °+ =
b)
2
1 3
2
1
(cot )
( )tan (45 )
2
(12 )(cot36 ) 17
36
(300 17)tan 1,2
(45 ) 17
2
00kPa
H c
H H
H kPa kPa
φ
φ
σ σ
σ
=
+ = + °+
= ° ≈
°
= + °+ − =
τ
Φ = 36°
Φ
σ
τ
σ3
0
Φ
σ
0
τ’
T
d
R
σ
τ
Φ
R
H
σ3
c
σ1
d

142
*Mohr – 15: Derive the general formula for horizontal stress.
(Revised Sept-09)
Derive the general formula of the horizontal stress as a function of the vertical stress, cohesion and
the angle of internal friction.
Solution:
θ = 45° +
2
θ
∴ 2θ = 90° + φ
2
ad 3
1 σ
σ −
=
af
ad
sin ≈
φ (1)
2
cot
c)
af
af
af 3
1 σ
σ
θ
+
+
=
+
= ( (2)
cot
(c)
of
cot
c
of
of
c
tan θ
θ
θ =
∴
=
∴
= (3)
According to the Mohr’s circle properties:
2
-
cot
(c)
af
2
-
oa
2
-
oa 3
1
3
1
3
1
3
σ
σ
φ
σ
σ
σ
σ
σ +
=
∴
=
∴
+
= (4)From (1) & (4):
( )
( )
2
-
2
cot
c
sin
2
cot
c
2
-
sin 3
1
3
1
3
1
3
1
σ
σ
σ
σ
φ
φ
σ
σ
φ
σ
σ
φ =
⎟
⎠
⎞
⎜
⎝
⎛ +
+
∴
+
+
=
( ) φ
σ
σ
σ
σ
φ
φ
σ
σ
φ
σ
σ
φ
φ sin
2
2
-
cot
sin
(c)
2
-
sin
2
cot
sin
c 3
1
3
1
3
1
3
1 +
−
=
∴
=
+
+
( )
( ) ( )
)
3
3 σ
φ
σ
σ
φ
σ
φ
φ (
sin
-
)
(
(sin
-
cot
sin
c
2 1
1 +
=
∴
( ) ( )
φ
σ
φ
σ
φ
φ
φ sin
1
-
)
sin
-
(1
sin
cot
sin
c
2 1 +
+
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∴ 3
( )
1
σ
φ
φ
σ
φ
φ
sin
1
sin
1
sin
1
cos
c
2 3 =
−
+
+
−
∴
Since ( )
φ
φ
φ
+
≈
−
0
45
tan
sin
1
cos
and ⎟
⎠
⎞
⎜
⎝
⎛
+
≈
−
+
2
45
tan
sin
1
sin
1 0
2 φ
φ
φ
2
45
tan
2c
2
45
tan 0
0
2
3
1 ⎟
⎠
⎞
⎜
⎝
⎛
+
+
⎟
⎠
⎞
⎜
⎝
⎛
+
=
φ
φ
σ
σ

143
*Newmark–01: Stress beneath a tank at different depths.
A construction site has a surface layer of Aeolic sand 2 m thick, underlain by a 10 m thick clay
stratum. The project involves placing a wastewater treatment tank, 10 m x 10 m in plan, with a
contact pressure po = 400 kN/m². Find the stress down the centerline of the tank at the top and the
bottom of the clay stratum using Newmark’s influence chart shown below.
Solution:
The increase in the vertical stress is found from Δp = po M (IV) where M is the number of
“squares” enclosed in the Newmark chart and (IV) is the influence value.
For Δp1, AB = 2 m Æ Δp1 = po M(IV) = (400 kN/m²) (190) (0.005) Æ Δp1 = 380 kN/m²
For Δp2, AB = 12 m Æ Δp2 = po M(IV) = (400 kN/m²) (42) (0.005) Æ Δp2 = 84 kN/m² (which
represents a 78% reduction in the vertical stress).
y
x
Influence Value (IV) = 1/200 = 0.005
A B
B = 10 m for point #1
B = 10 m for point # 2
S
S
SA
A
AN
N
ND
D
D
C
C
CL
L
LA
A
AY
Y
Y
2
2
2 m
m
m
1
1
10
0
0 m
m
m
1
2
B
B
B =
=
= 1
1
10
0
0 m
m
m
AB = 2 m
AB = 12 m

144
*Newmark-02: The stress below the center of the edge of a footing.
(Revision: Aug-08)
Find the stress at the point A shown below, at a depth of 3 m below the edge of the footing. The
plan of the square footing has been plotted on top of the Newmark graph to a scale of AB = 3m
and placed in such a way that point A falls directly over the center of the chart.
Solution:
The number of elements inside the outline of the plan is about M = 45. Hence,
( )( )
( )( ) 2
660
( ) 45 0.005 .5
3
1
3
6 /
kN
p q M IV
m
N m
m
k
⎡ ⎤
Δ = = =
⎢ ⎥
⎣ ⎦
The number of elements inside the outline of the plan is about M = 45. Hence,
( )( )
( )( ) 2
660
( ) 45 0.005 .5
3
1
3
6 /
kN
p q M IV
m
N m
m
k
⎡ ⎤
Δ = = =
⎢ ⎥
⎣ ⎦

145
*Newmark-03: Stress at a point distant from the loaded footing.
(Revision: Aug-08)
The footing shown below has a load q = 1.8 ksf. Find the stress at a depth of 5 feet below the
footing invert, at the point C.
Solution:
Set AB = 5 and draw the footing to that scale. The number of affected areas M =8, therefore
Δp = q M (IV) = (1,800 psf)(8)(0.005) = 72 psf
Depth Point = Z
Influence Value (IV) = 0.005

146
*Newmark-04: Stresses coming from complex shaped foundations.
(Revision: Aug-08)
A small but heavy utility building will be placed over a 2 m thick sand stratum. Below the sand is
a clay stratum 2 m thick. Find the stress at points A and B in the clay stratum directly below point
C at the surface.
Solution:
( ) ( )( )( )
( ) ( )( )( )
( )
2
2
2
Point A (top of clay stratum) 100 136 0.005
Point B (bottom of clay) 100 100
68
50
59
0.005
70 50
2 2
A
B
A B
average
kN
m
kN
m
q q M IV
q q M IV
q q
q
kN
m
= = =
= = =
+
+
= = =

147
*Newmark-05: Stress beneath a circular oil tank.
(Revision: Aug-08)
A circular oil storage tank will be built at the shore of Tampa Bay. It will be 20 m in diameter, and
15 m high. The tank sits upon a 2 m thick sand deposit that rests upon a clay stratum 16 m thick.
The water table is at practically at the surface. Find the stress increase from a fully loaded tank, at
mid-clay stratum, (a) directly under the center of the tank, and (b) at its outer edge, using the
Newmark influence chart shown below.
Solution:
The contact stress is qo = (0.95)(9.81 kN/m3
)(15m) = 2
140 /
kN m
At mid-clay depth along the centerline of the tank (depth = 10 m) ∴OQ = 10 m
( ) ( )( ) ( )( )
'
2
( ) 140 9.81 10 680 0.001 28.5
v o w
kN
m
q h M IV
σ γ
= − = − =
⎡ ⎤
⎣ ⎦
Set AB 10 m Influence Value (IV) = 0.001

148
**Newmark-06: Use Newmark with a settlement problem.
(Revision: Aug-08)
A small but heavily loaded utility building has dimensions of 20 m x 20 m. It applies a uniform
load on its mat foundation of 100 kN/m2
. Its mat foundation sits 1 m below the surface. The soil
profile consists of 3 m of a dry sand, with γ = 17.5 kN/m3
under laid by a 5 m thick clay layer with
a γ = 18.5 kN/m3
, a moisture content of 22%, Cc = 0.30 and a Gs = 2.70. The clay stratum is under
laid by another sand stratum, and the phreatic surface coincides with the top of the clay stratum.
(40) Using the Newmark method, what are the new stresses at the top and bottom of the clay
stratum due to the building’s loading?
(41) What is the expected differential settlement between the building’s center and one of its
corners, in mm?
(42) If a laboratory sample 4” thick of the field clay attained 50% consolidation in 5 hours,
what time will the clay layer in the field attain 60% consolidation?
Solution:
( ) ( )( )( )
2
2
(1) Set AB = 2 m and observe that the building's foot-print covers the entire graph.
100 /
Set AB = 7 m and the number of units = 175
100 175 0.005 87.5 /
The stress at point C has drop
E
F
kN m
M
qM IV kN m
σ
σ
∴ Δ =
∴ Δ = = =
( )( )
( )
( )
2 2
2
ped to 50% of the stress at E,
50 / 0.96 50 48 /
100 87.5
Average stress in the clay stratum beneath the center 93.8
2
50 48
Average stress in the clay stratum beneath the corner 49
2
C D
kN m and kN m
kN
m
k
σ σ
∴ Δ = ∴ Δ = =
+
= =
+
= = 2
N
m
q =100 kN/m2
SAND
3 m
2 m
γ = 18.5 kN/m3
w = 22 %
CLAY
1 m
5 m
C E
D F
• •
• •

149
( )( )
(2) The differential settlement between the center and a corner of the building is,
0.30 2.70
0.22 2.70
0.59
1
The in-situ stress at mid-clay stratum before the building is built is
c S
s
o S o
C and G
wG
Se wG e
S
= =
= ∴ = = =
( )( ) ( )( )
( )( )
( )
( ) ( )
( )
2
,
3 17.5 2.5 18.5 9.8 74.3
Therefore,
0.30 5 74.3 93.8
the settlement at the center log log 0.33
1 1 0.59 74.3
the settlement at the corner
o sand sand clay at mid clay
C o
o
kN
p h h m m
m
m
C H p
H m
e p
C
H
γ γ
σ
−
= + = + − =
⎛ ⎞
+
⎛ ⎞
+Δ
Δ = = =
⎜ ⎟
⎜ ⎟ ⎜ ⎟
+ +
⎝ ⎠ ⎝ ⎠
Δ =
( )( )
( )
( ) ( )
( )
( )
2
0.30 5 74.3 49
log log 0.21
1 1 0.59 74.3
the differential settlement is 0.33 0.21
(3) The time required to attain 60% consolidation in th
0.12 1
e field i ,
2
s
0
C o
o
drained
v
m
H p
m
e p
H m m
T
c
m m
H
m
σ
σ ⎛ ⎞
+
⎛ ⎞
+Δ
= =
⎜ ⎟
⎜ ⎟ ⎜ ⎟
+ +
⎝ ⎠ ⎝ ⎠
Δ Δ = −
=
=
=
( )( ) ( )( )
( )
( )( )
( )( )
( )
2 2
2
2
0.26 2,500 0.18 2 25.4
5
Solving for the time of settlement in the field ,
0.26 2,500 5 1 1
1 24 365
0.18 2 2 4
2
5.
F
f
F
mm x mm
t t hours
t
mm hours day year
t
hours days
y
x mm
ears
= =
⎛ ⎞⎛ ⎞
= ≈
⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠

150
*Stress–01: Stress increase at a point from several surface point loads.
(Revision: Aug-08)
Point loads of 2000, 4000, and 6000 lbs act at points A, B and C respectively, as shown below.
Determine the increase in vertical stress at a depth of 10 feet below point D.
Solution.
Using the Boussinesq (1883) table on page 202 for vertical point loads, the vertical increase in stress
contributed by each at a depth z =10 feet is found by,
( )
1
5/ 2
2 2
2
3 1
2 / 1
z
P P
p I
z z
r z
π
⎧ ⎫
⎪ ⎪
Δ = =
⎨ ⎬
⎡ ⎤
⎪ ⎪
+
⎣ ⎦
⎩ ⎭
Increase in
the load at:
P
(lbs)
r (ft) z (ft) r/z I1
Δp
(psf)
Δp from A 2,000
(102
+52
) 1/2
=
11.18
10 1.12 0.0626 1.25
Δp from B 4,000
(102
+52
) 1/2
=
11.18
10 1.12 0.0626 2.50
Δp from C 6,000 5 10 0.50 0.2733 16.40
Total = 20.2 psf
Therefore, the vertical stress increase at D from the three loads A, B and C is 20.2 psf.
A 10 feet B
10 feet
C 5 feet D

151
*Stress-02: Find the stress under a rectangular footing.
(Revision: Aug-08)
Determine the vertical stress increase in a point at a depth of 6 m below the center of the invert of
a newly built spread footing, 3 m by 4 m in area, placed on the ground surface carrying a
columnar axial load of N = 2,000 kN.
Solution:
The Boussinesq solution for a rectangular loaded area only admits finding stresses below a corner
of the loaded area. Therefore, the footing must be cut so that the load is at a “corner” (shown as
the quarter of the area), where the reduced footing dimensions for the shaded area are B1 = 1.5 m
and L1 = 2.0 m.
( )( )
( )( )
1 1
4
2
4
4
1.5 2.0
0.25 0.33
z 6.0 z 6.0
Use the table and extrapolate and find 0.0345
2,000
(4 ) (4 ) 23
4 0.0345
3 4
/
= = = = = =
=
⎛ ⎞
⎛ ⎞
Δ = = = =
⎜ ⎟
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
o
z
B L
m m
m and n
m m
I
N kN
q q k
I
BL m
N
m
m
I
L = 4 m
B = 3 m
Depth = 6 m
N = 2,000 kN

152
WT-2
Depth = 4 m
WT-1
L = 5 m
B = 3 m
WT
N = 4,500 kN
*Stress-03: The effect of the WT on the stress below a rectangular footing.
(Revision: Aug-08)
Find the effective stress increase in the soil at a depth of 4 m below the footing, and then find the
increase in the stress due to a drop of the WT from originally 1 m below the footing to 5 m below
the footing.
Solution:
( )( )( )
2 2
4
2
4
4,500 1.5 2.5
300 0.375 0.625
15 4 4
0.076
) The total stress increase from the footing is,
(4 ) 300 4 0.076 91.2 /
and the effective stress when the is 1 m below t
o
N kN kN B m L m
q m and n
A m m z m z m
I
a
p q I kN m
WT
= = = = = = = = =
∴ =
Δ = = =
( ) ( )( )
' 2
he footing is,
91.2 3 9.8
) When the WT drops from -1 m to -5 m below the footing, the effective stress is
identical to the total stress. Therefore the effective stress increase s
/
i ,
61.8
o o
p p u m m
b
kN
Δ = Δ − = − =
2
91.2 / which is a 48% increase in stress.
o kN m
p
Δ =

153
*Stress–04: Finding the stress outside the footing area.
(Revision: Aug-08)
Find the vertical stress increase p
Δ below the point A at a depth z = 4 m.
Solution:
The stress increase, p
Δ , can be written as : 1 2
p p p
Δ = Δ − Δ
where 1
p
Δ = stress increase due to the loaded area shown in (b).
2
p
Δ = stress increase due to the loaded area shown in (c).
( )( )
2
1 4
2
2
For the loaded area shown in (b):
2 4
0.5 1.0
4 4
(150)(0.12) 18 /
Similarly, for the loaded area show in (c):
1 2
0.25 0.5
4 4
150 0.048 7.2 /
Therefor
B L
m and n
Z Z
p qI kN m
B L
m and n
Z Z
p kN m
= = = = = =
Δ = = =
= = = = = =
Δ = =
1 2
2
e, 18 7.2 10.8 /
p p p kN m
Δ = Δ − Δ = − =

154
*Stress-05: Stress below a footing at different points.
A clay sanitary pipe is located at a point C below the footing shown below. Determine the increase
in the vertical stress Δp at the depth of the pipe, which is z = 5 feet below the footing invert, and 3
feet away from its edge. The footing has a uniformly distributed load q = 1,800 psf.
Solution:
For the expanded 5’ x 13’ area,
4
5 13
1 2.6 therefore, = 0.200
5 5
B L
m and n I
Z Z
= = = = = =
For virtual 3’ x 5’ area
4
3 5
0.6 1 therefore, = 0.136
5 5
B L
m and n I
Z Z
= = = = = =
The increase in stress at point C from the footing is therefore,
'
4 4 2
( - ) 1,800 (0.200- 11
0.136) 5
lb
p q I I f
ft
ps
⎛ ⎞
Δ = = =
⎜ ⎟
⎝ ⎠
SECTION
.
A C
B
10 ft 3 ft
.
.
2 ft
4 ft
5 ft
PLAN VIEW
.
q = 1,800
5 ft
10 ft 3 ft
.
C
.
B A

155
*Stress-06: Stress increase from a surcharge load of limited width.
(Revision: Aug-08)
Calculate the stress increase at the point A due to the new road embankment.
Solution:
The contribution from the central portion of the fill is 2
p
Δ , whereas the contribution from the left
and right hand slopes are 1
p
Δ and 3
p
Δ respectively. Using Boussinesq,
2
'
15
)
'
15
(
2
B
x
2
p
1
1
1 =
=
Δ ⇒ 2
'
15
)
'
15
(
2
B
z
2
1
=
=
1
0.25
p
q
Δ
=
1
3
(0.25)(15') 120 450
lb
p psf
ft
⎛ ⎞
∴Δ = =
⎜ ⎟
⎝ ⎠
-------------------------------------------------------
1
'
25
)
'
5
.
12
(
2
B
x
2
p
2
2
2
−
=
−
=
Δ ⇒ 2
.
1
'
25
)
'
15
(
2
B
z
2
2
=
=
2
0.47
p
q
Δ
=
2
3
(0.47)(15') 120 846
lb
p psf
ft
⎛ ⎞
∴Δ = =
⎜ ⎟
⎝ ⎠
------------------------------------------------------------
3
.
5
'
15
)
'
40
(
2
B
x
2
p
3
3
3 =
=
Δ ⇒ 2
'
15
)
'
15
(
2
B
z
2
3
=
=
3
0.02
p
q
Δ
=
3
3
(0.02)(15') 120 306
lb
p psf
ft
⎛ ⎞
∴Δ = =
⎜ ⎟
⎝ ⎠
1 2 3 450 846 1,332
36
p p p psf
p
Δ = Δ + Δ + Δ = + + =
1
1
15 ft
Z
15 ft
2
p
Δ
1
p
Δ 3
p
Δ
1
1
25m

156
*Stress-07: Finding a stress increase from a surface load of limited width.
(Revision: Aug-08)
Determine the average stress increase below the center of the loaded area, between z = 3 m and z =
5 m.
Solution:
The stress increase between the required depths (below the corner of each rectangular area)
can be given as:
2 4 2 1 4 1 4 2 4 1
2 1
2 1
( ) ( ) ( ) ( )
( / )
( )( ) ( )( ) (5)( ) (3)( )
100
5 3
H H H H
avg H H
H I H I I I
p q
H H
⎡ ⎤
− −
⎡ ⎤
Δ = =
⎢ ⎥ ⎢ ⎥
− −
⎣ ⎦
⎣ ⎦
For I4(H2): m′ = B / H2 = 1.5 / 5 = 0.3
n′ = L / H2 = 1.5 / 5 = 0.3
For m′ = n′ = 0.3, I4(H2) = 0.038
For I4(H1): m′ = B / H1 = 1.5 / 3 = 0.5
n′ = L / H1 = 1.5 / 3 = 0.5
For m′ = n′ = 0.5, I4(H1) = 0.086
Therefore:
2 1
av(H /H )
(5)(0.038) (3)(0.086)
Δp 100 3
5 3
−
= × =
−
2
.4 kN/m
The stress increase between z = 3 m and z = 5 m below the center of the load area is equal to:
2 1
2
( / )
4 (4)(3.4) 13.6 /
avg H H
p kN m
Δ = =
5 m
PLAN VIEW
L
A
3 m
1.5 m 1.5 m
q = 100 kN/m
2
SECTION
z
3 m
3 m
A
B
.
.
.

157
**Stress-08: Stress increase as a function of depth.
(Revision: Aug-08)
The vertical stress σv in a soil at any depth below the surface can be estimated as a function of the
soil unit weight γ by the equation,
( )
100
0 0
(95 0.0007 )
Z
v v v
dz dz
σ γ σ σ
= = +
∫ ∫
If a particular stratum has a function γ = 95 + 0.0007 σv , where γ is in pcf and σv is in psf, find the
vertical stress at a depth of 100 feet below the surface.
Solution:
At Z = 100 ( )
135,800 1.0725 1
v
σ = −
9,840
v psf
σ
∴ =
( )
( )
( )
100
0 0
100
0
0
100
0.0007
0
Rearranging, and integrating by parts,
95 0.0007
1
ln 95 0.0007
0.0007
135,800 9,840
1
v
V
v
V
V
z
v
d
dz
z
e psf
σ
σ
σ
σ
σ
σ
=
+
+ =
= − =
∫ ∫

158
Chapter 10
Elastic Settlements
Symbols for Elastic Settlements
N → Raw value of the STP (obtained in the field).
qo → Contact pressure.
C1 → Embedment coefficient.
C2→ Creep correction factor.
ES → Soil elastic modulus.
Eeq→ Equivalent modulus.
Δ(Δe) → Differential settlement between adjacent foundation.
ΔHi→ Elastic settlement.
I→ Influence factor; essentially equivalent to the strain ε in the soil.
IZ→ Simplifying influence factor.
ε → Strain at mid stratum.
v → Poisson’s ratio.
MT→ Transverse moment.

159
*Elastic Settlement-01: Settlement (rutting) of a truck tire.
(Revision: Jan-08)
You are required to move a 60-ton truck-mounted crane onto your construction site. The front
wheels carry 20% of the load on tires inflated with 80 psi air pressure. Calculate the possible
rutting depth to your temporary jobsite road built from in-situ compacted medium sand. A
surface SPT test shows an N = 12 and the tire’s bearing area is roughly square. Use the
Schmertmann method to estimate the rutting.
Solution:
Each front tire has a square bearing area of BxB such that:
2 2
2
o
tire's load (0.5)(20%)(120,000 )
150 12.2
tire pressure q (80 / )
lb
B in B inches
lb in
= = = ∴ =
A rough estimate of the soil’s elastic modulus is Es = 14N = 14(12) = 168 ksf.
Since the sand is compacted, it is a dense sand, and the influence factor Iz is equivalent to the strain ε.
The strain reaches a maximum value of 0.6. Therefore, the average value of the strain is about 0.3
throughout its depth to 2B = 2(12.2 inches) = 24.4 inches. Since the crane loads are on the surface and
only for a few days, it is permissible to assume that there is no creep and therefore C1 = C2 = 1.
Therefore, for the single layer of soil, the rutting is,
2
1 2 2 2 2
0.30 144
(1)(1)(80
0
)( )(24.4 )
168 /
.5 .
1,000
o
s
lb in k
C C q dz i
inches of ruttin
n
E in k ft ft lb
g
ε
⎛ ⎞ ⎛ ⎞⎛ ⎞
Δ = = ×
⎜ ⎟ ⎜ ⎟⎜ ⎟
⎝
∴Δ
⎠
≈
⎝ ⎠
⎝ ⎠

160
*Elastic Settlement-02: Schmertmann method used for granular soils.
(Revision: Aug-08)
Estimate the settlement of a square footing placed on a fine, medium dense sand, embedded 4 ft
below the ground surface, for long-term use. Use the Schmertmann method.
Assume 14
s
E
N
≅ where Es is in ksf; used for fine medium sands.
Solution:
Layer ∆z E s I z = ε
(in) (ksf)
1 42 140 0.30 0.090
2 60 210 0.46 0.130
3 66 168 0.16 0.061
∑= 0.281
s
z
E
z
I Δ
( )
2 2
1 2
1
200
The contact pressure on the soil is, 4 (0.120 ) 3.60
49
The coefficients for the Schmertmann method are :
0.48
Depth factor 1.0 0.5 1.0 0.5 0.93
3.60
Cree
o f
f
o
Q k
q D ft kcf ksf
B ft
C and C
D
C
q
γ
γ
= − = − =
⎛ ⎞ ⎛ ⎞
= − = − =
⎜ ⎟ ⎜ ⎟
⎝ ⎠
⎝ ⎠
2
2
1 2
0
p factor 1.35 .
The Schmertmann formula for the elasti
1.
c settlement is,
(0.93)(1.35)(3. 2
60)(0.28 ) 7
1
B
o
C for a five year period
C C q z
E
inches
ε
=
Δ
∴Δ = Δ = =
∑

161
*Elastic Settlement-03: Schmertmann method used for a deeper footings.
(Revision: Aug-08)
Determine the elastic settlement of a deep spread footing after five years of the 3 ft. x 3 ft. footing
when it is placed on a uniform clean sand with γ = 110 pcf.
Solution:
( )
0 2
2
64
The contact pressure on the soil is, 7.11
3
Q kips
q ksf
B ft
= = =
The SPT value indicates that the soil is a loose sand. The modulus E for loose sand can be calculated
using the following formula:
( )
10 15
S
E N ksf
≈ +

162
The following table summarizes the data and calculations:
Layer
number
Layer’s
thickness
ΔZ (feet)
SPT
(average)
N
Soil’s
elastic
modulus
E (ksf)
Average
strain at
mid
stratum ε
εZΔZ/E
(ft3
/kip)
1 1.5 6 210 0.35 0.0025
2 4.5 7 220 0.30 0.0061
Σ = 0.0086
The correction factors are as follows:
1. Depth factor,
( )( )( )
( )
1
0
0.5 0.5 0.110 4
1 1 0.97
7.11 0.110 4
f
f
D
C
q D
γ
γ
⎛ ⎞ ⎛ ⎞
= − = − =
⎜ ⎟ ⎜ ⎟
⎜ ⎟
⎜ ⎟
− − ×
⎝ ⎠
⎝ ⎠
Since C1 > 0.5, this is FINE.
2. Creep factor, ( )
2
5
1 0.2log 1 0.2 log 1.34
0.1 0.1
yr
t
C
⎛ ⎞ ⎛ ⎞
= + = + =
⎜ ⎟ ⎜ ⎟
⎝ ⎠
⎝ ⎠
The total elastic settlement is,
( ) ( )( ) ( )( ) ( )
2
1 2
0
0.97 1.34 7.11 0.110 4 0.0086 0.07 0.84
B
o f
s
ft
CC q D dz i
E
n
ε
γ
Δ= − = − =
⎡ ⎤ =
⎣ ⎦
∑

163
*Elastic Settlement-04: The 2:1 method to calculate settlement.
(Revision: Aug-08)
Use the 2:1 method to find the average stress increase (Δq) due to the applied load Qu in the 5-foot
sand stratum directly beneath the footing. If ES = 400 ksf and v = 0.3, what is the expected
immediate settlement ΔHi?
Solution:
The settlement ΔH of an elastic media (the 5 foot thick sand stratum in this case) can be found from the
theory of elasticity as,
2
1
o w
s
v
H q B I
E
−
Δ =
For square and flexible footings the influence factor is about IW = 0.95. The 2:1 method essentially
assumes that the stress reduces vertically by a vertical slope of 2 units vertically to 1 unit horizontally.
The stress increase can be found by integrating the above equation,
( )
2
1
2
1 H
u
H
Q
q dz
H B Z
Δ =
+
∫
where H1 = 0 feet (the footing’s invert) to H2 = 5 fee (bottom of the sand stratum).
Qu=120 kips

164
( ) ( )
5
0
2 2
2
1 1 120 120
1.82
5 6 11
1 120 1 0.3 12
6 0.95 0.5
6
2
3 400
o w
s
Q kip kip
q ksf
H B z ft ft ft
but
v kip in
H q B I ft
E ft ksf ft
inches
⎛ ⎞
⎡ ⎤
Δ = = − =
⎜ ⎟
⎢ ⎥
+
⎣ ⎦ ⎝ ⎠
⎛ ⎞
⎛ ⎞ ⎛ ⎞
− −
Δ = = =
⎜ ⎟
⎜ ⎟ ⎜ ⎟
⎝ ⎠
⎝ ⎠ ⎝ ⎠

165
*Elastic Settlement-05: Differential settlement between two columns.
(Revision: Aug-08)
The allowable bearing capacity of a 30-ft thick, medium dense sand stratum (with φ = 36o
and γ =
112 pcf) is 3 ksf. Column A has a design load of 430 kips and column B has a design load of 190
kips. Select footing sizes and determine the differential settlement Δ(ΔH) between them. Is this
Δ(ΔH) acceptable for columns spaced 30 ft apart?
Solution:
Footing size for column A:
430
12
3
A
A
all
Q kips
B feet
q ksf
= = =
Footing size for column B:
190
8
3
B
B
all
Q kips
B feet
q ksf
= = =
A quick estimate of the ratio of settlement to the proportionality is 5
.
1
8
12
=
=
Δ
Δ
ft
ft
H
H
B
A
Therefore if the settlement at column B is ΔHB = 1 in., then the settlement at column A will be ΔHA =
1.5 in. Then,
( ) 1.5 1.0 0.5
H in in inches
Δ Δ = − =
and the rotation between the two columns is
( )
( )
0.5
( )
0.0014
12
30
in
H
L in
ft
ft
θ
Δ Δ
= = =
⎛ ⎞
⎜ ⎟
⎝ ⎠
Both of these values [Δ(ΔH) and θ] are acceptable, since Δ(ΔH) should be < 1" and θ < 0.0033.

166
*Elastic Settlement-06: Compare the settlements predicted by the Boussinesq,
Westergaard, and the 2:1 methods.
(Revision: Aug-08)
Compute the average stress Δq at mid-clay stratum, for the values shown below, using: (a)
Boussinesq's method, (b) Westergaard's method, and (c) the 2:1 method.
Also, determine the size of a square spread footing, in order to limit total settlement ΔH = ΔHi +
ΔHc + ΔHS to only 1.5 inches.
Estimate the initial settlement i s
= 0.05
H Z
Δ where Zs is the thickness of the granular
stratum beneath the footing in feet, to give ΔHi in inches.
Solution:
Assume an initial value of B = 10 feet.
The contact pressure qo of the footing is:
2
240
2.4
100
o
B
Q kips
q ksf
A ft
= = =
(a) Stress at mid-clay stratum using Boussinesq’s method (use the charts on page 205) for a square
footing:
( )( )
0.52 0.52 2.4 1.25
o
q
q ksf ksf
q
= ∴ = =
SAND
4’
5’
5’
CLAY
Q = 240 kips
ROCK
γS= 110 pcf
γ = 120 pcf
eo = 1.11
Cc = 0.42
WT

167
(b) Stress at mid-clay stratum using Westergaard’s method for a square footing:
( )( )
0.33 0.33 2.4 0.79
o
q
q ksf ksf
q
= ∴ = =
(c) Stress at mid-clay stratum using the 2:1 method for a square footing:
The depth (Z) from the footing invert to mid-clay is 7.5 feet:
( )
2
240
0.78
10 7.5
B Z
Q kips
q ksf
A ft ft
+
= = =
+
Note that the Boussinesq method provides the highest predicted stress. Since this would predict faster
consolidation rates, it is the least conservative method. Therefore, for this problem, use the 2:1 method's
stress of 0.78 ksf. The instantaneous settlement (ΔHi):
0.05 0.05 5 0.11
i s
H Z ft in
Δ = = =
The in-situ effective stress qo' at mid-clay layer, before placing the footing is:
'
3 3 3
0.11 9 0.120 0.0624 2.5 1.13
o s s c c
kip kip kip
q h h ft ft ksf
ft ft ft
γ γ
⎡ ⎤
⎛ ⎞ ⎛ ⎞
= + = × + − × =
⎢ ⎥
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
⎣ ⎦
Using q = 0.78 ksf from the 2:1 method, the total settlement ΔH is equal to the immediate ΔHi, plus the
consolidation ΔHc, and the secondary settlement ΔHs, but limited to no more than 1.50 inches.
1.50
i C s
H H H H inches
Δ = Δ + Δ + Δ =
B = 10 ft.
B+z
z = 7.5 ft.
qo

168
But ΔHi = 0.11 in, and ΔHs is negligible for this problem. Therefore the maximum permissible
consolidation settlement ΔHc is limited to:
( )
'
1.50 1.50 0.11 1.39
12
0.42 5
1.13
log 1.39 log 1.39
1 ' 1 1.11 1.
0. 9
1
3
3
c i
c o
c
o o
H in H in in in
or
in
ft
ft
C H q q ksf q
H in in
e q ksf
The q ks
refo e f
r
Δ = − Δ = − =
⎛ ⎞
×
⎜ ⎟ ⎛ ⎞
+ Δ + Δ
⎝ ⎠
Δ = = = =
⎜ ⎟
+ + ⎝
=
⎠
Δ
Using the 2:1 method:
240
26.6
19
0.34
B Z
Q Q kips
q B Z ft
A q ksf
Therefore B feet
+
=
Δ = + = = =
Δ
Since the initial B = 10 feet, the new value of 19 feet should be used to re-iterate towards a better
solution that converges.

169
*Elastic Settlement-07: Schmertmann versus the strain methods.
(Revision: Aug-08)
Compute the immediate settlement ΔHi using the Schmertmann formula using an average Δq
value (qv1 = 233.3 kPa, qv2 = 163.3 kPa, qv3 = 77.0 kPa, qv4 =44.0 kPa and qv5 = 28.0). Es at point A is
20,400 kPa, 5
.
0
=
B
Df
and C1 = C2 = 1. Compare the results with an alternate method
using w
s
o I
E
v
B
q
H ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
=
Δ
2
1
, where v = 0.3 and Iw = 0.95.
Solution:
1
2 3 1
The average stress from multiple layers is solved via this formula,
1.5 233.3 28.0
163.3 77.0 44.0
2 6 2
104
The elastic settlement v
n
v n
v
q q
H m kPa kPa
q q q q kPa kPa kPa
H m
q kPa
−
⎡ + ⎤
Δ ⎡ + ⎤
⎛ ⎞ ⎛ ⎞
Δ = + + + + = + + +
⎜ ⎟
⎜ ⎟
⎢ ⎥ ⎢ ⎥
⎝ ⎠
⎝ ⎠ ⎣ ⎦
⎣ ⎦
Δ =
K
( )( )( )( )
( )( )
1 2
2 2
1
ia Schmertmann is,
1000 104
(0.6 ) 1 1 0.6 3
20,400
The alternative method from the theory of elasticity would yield,
1 1 0.3
233
9.2
.3 3
20,400
s
v w
s
q mm kPa
H CC B m
E m kPa
v
H q B I kPa m
E kPa
mm
⎛ ⎞
Δ ⎛ ⎞
Δ = = =
⎜ ⎟
⎜ ⎟
⎝ ⎠⎝ ⎠
⎛ ⎞ ⎛ ⎞
− −
Δ = =
⎜ ⎟ ⎜
⎝ ⎠
⎝ ⎠
( )
1000
0. 1
95 0
mm
m
mm
⎛ ⎞
=
⎟ ⎜ ⎟
⎝ ⎠
SAND
1.5 m
5 m
1.5 m
ROCK
Q = 2,100 kN
3 m
A
B
B = 3m x 3m

170
*Elastic Settlement-08: The Schmertmann method in multiple strata.
(Revision: Aug-08)
Determine the elastic settlement using the Schmertmann method of the 10'x 10' footing as shown
below. Estimate the elastic modulus using ES = 10(N + 15), where ES is in ksf and N is the
corrected SPT value.
Solution:
The data from these strata are placed into a table below.
Layer No.
Layer
Thickness
ΔZ, (feet)
Soil Modulus
ES, (ksf)
IZ = ε
(average strain)
Z Z
I
(ft/kp)
E
Δ
1 5.0 360 0.35 0.00486
2 5.0 260 0.50 0.00962
3 2.5 300 0.35 0.00292
4 2.5 560 0.25 0.00112
5 5.0 190 0.10 0.00263
Σ = 2B = 20 ft Σ= 0.02115

171
( )( ) ( )( ) ( )
1
2
2
1 2
0
The Schmertmann coefficients are,
75
1.35
0.88 1.35 2.5 0.1 5 0.021 0.04
1 2 . 5
5 0 0
f
f
B
f
i
0.5(0.1)(5)
D
Thedepthcoefficient C 1- [0.5( )]= 1- = 0.
q - 1.5 -(0.1)(5)
D
Thetimecoe
f
ffici
t i
ent C
H = C C (q - ) dz
D n
E
γ
γ
ε
γ
⎡ ⎤
= ⎢
=
⎥
⎣ ⎦
=
⎡ ⎤
Δ = − =
⎣ ⎦
∑

172
**Elastic Settlement-09: Settlement of a mat foundation.
(Revision: Aug-08)
A mat foundation located 8 feet below grade supports a ten story building upon an area of 50 ft by
150 ft, and carries a uniform load of 6 ksf. For the soil profile conditions shown below, determine
the total settlement at the center and a corner of the foundation. The structure is of reinforced
concrete with column spacing at 25 ft. Is the calculated differential settlement acceptable?
Solution:
a) Using the Schmertmann method,
Layer Δz (in) Es (ksi) Iz =ε (Iz/Es)Δz (in3
/kip)
1 300 3.47 0.35 30.26
2 300 3.47 0.50 43.23
3 300 8.33 0.30 10.80
Σ = 84.29
( )
( )( )
( )( )
( ) ( )( ) ( )( ) ( )
1
2
2
1 2
0
0.100 8
1.0 0.5 1.0 0.5 0.92
6 0.100 8
( ) 1.0
0.92 1.0 6 0.100 8 84 3.78
.29
f
f
B
o f
s
D
Thedepth factor C
q D
Thetime factor creep C
C i
C n
q D dz
E
γ
γ
ε
γ
⎡ ⎤
= − = − =
⎢ ⎥
−
− ⎣ ⎦
=
Δ = − = − =
⎡ ⎤
⎣ ⎦
∑
b) Consolidation settlement Method A (take each layer at a time),
Set eo1 = 1.00, and from e = 125/γ, eo2 = 0.96, eo3 = 0.89.
For clay 1: qo = (0.100)(50) + (0.110 - 0.0624)25 + (0.125 - 0.0624)12.5 = 5.95 ksf
At the mid clay stratum is at 87.5 ft below the surface,
(ie. 87.5/50=1.75 B = 3.5 B/2) Δq = 0.12q = 0.12(6) = 0.72 ksf
At corner Δq = 0.09q = 0.09(6) = 0.54
For clay 2: qo = 5.95 + ( 0.125 + 0.0624)12.5 + (0.130 - 0.0624)12.5 = 7.58 ksf
1.48
log log
1
c o
1 10 10
o o
+ q
q 0.20(25x12) 5.95+0.72
C H
= = in
1+ 1+1 5.95
q
e
Δ
=
Δ
1.13
log
1 10
0.20(25x12) 5.95+0.54
= in
2 5.95
=
Δ

173
midlayer at 112 ft = 4.5 B/2, Δq . 0.07q = 0.42
Δq . 0.05q = 0.30
For clay 3: qo = 7.58 + (0.130 + 0.0624)12.5 + (0.140 - 0.624)12.5 = 8.65 ksf
at midlayer (137.5' / 5.5 B/2) Δq = 0.04q = 0.24 ksf
at corner Δq = 0.03q = 0.18 ksf
Method B: the equivalent layer equation
The total settlement on the centerline is,
Δ6 = 3.78 + 1.48 + 1.25 + 0.28 = 6.78 inches
and along the foundation edge,
Δedge = 3.78 + 1.13 + 0.96 + 0.21 = 6.08 inches
The differential settlement is Δ(Δ) = 0.70 inches
The allowable for reinforced concrete buildings Δ(Δ)< 0.003(span) = 0.003(25x12 inches)
= 0.90 inches
Therefore, the design is acceptable.
1.25
log
2 10
0.35(25x12) 7.58+0.42
= in at the centerline
1+0.96 7.58
=
Δ
0.96
log
2 10
0.35(25x12) 7.58+0.30
= in
1+0.96 7.58
=
Δ
0.28
log
3 10
0.15(25x12) 8.65+0.24
= in
1+0.89 8.65
=
Δ
0.21
log
3 10
0.15(25x12) 8.65+0.18
= in
1+0.89 8.65
=
Δ
log o
c
o o
+ q
q
e
= where e= C
H 1+ q
e
Δ
Δ Δ
Δ

174
Chapter 11
Plastic Settlements
Symbols for Plastic Settlements
e → Voids ratio.
ES → Soil elastic modulus.
CC→ Compression index.
OCR → Over-consolidation ratio (ratio of in-situ stress divided by the overburden stress).
H→ Depth of zone influence.
DHc→ Plastic settlement (also called primary consolidation).
DHtotal→ Total settlement of a structure.
DHs→ Second consolidation settlement.
Dp → Increasing pressure on the surface.
u → Pore water pressure.
Pe→ Pre-consolidation pressure of a specimen.
γSAT → Saturated unit weight of the soil.
γW → Unit weight of water.
U→ Degree of consolidation.

175
*Plastic Settlement–01: Porewater pressure in a compressible soil.
(Revision: Oct.-08)
a. How high will the water rise in the piezometer immediately after the application of the
surface load of 3 ksf?
b. What is the degree of consolidation from the 3 ksf at point A, when h =15 ft.?
c. Find h when the degree of consolidation at A is 60%.
Solution:
a) Assume a uniform increase of the initial excess pore water pressure throughout the 10-foot
thickness of the clay layer:
2
0
3,000
The pore water pressure is 3 48.
,000 /
62
1
.4
w
w
p
u p h lb ft h feet
γ
γ
Δ
= Δ = = ∴ = = =
h
Sand
Clay
Rock
Δ p = 3 ksf
Ground water table
20 feet
15 feet
4 feet
10 feet
A

176
b) The degree of consolidation at A is UA (%) when h = 15 feet:
0
(15 )(62.4 )
% 1 100 1 6
100
(48.1 )(62.4
9%
)
A
A
u ft pcf
U
u ft pcf
⎛ ⎞ ⎛ ⎞
= − = − =
⎜ ⎟ ⎜ ⎟
⎝ ⎠
⎝ ⎠
c) When UA = 60 %, what is the value of h?
( )( )
( )
( )
0
0.6 1 1
3,000
1 0.6 3,000 1,200
1,200
.2
6 4
9
.
1
2
A A
A
A
A
w
u u
U
u psf
u psf psf
psf
u
fee
h
pc
t
f
γ
⎛ ⎞ ⎛ ⎞
= = − = −
⎜ ⎟ ⎜ ⎟
⎝ ⎠
⎝ ⎠
∴ = − =
= = =

177
*Plastic Settlement-02: Total settlement of a single layer.
(Revision: Aug-08)
A new building is planned upon the site shown below. Assume that the clay solids have a specific
gravity of 2.67. Find the primary consolidation settlement if the clay is normally consolidated.
Solution:
( )( )
0.009( 10) 0.009(50 10) 0.36
1 0.34 2.67 0.91
,
( ) ( )
− − −
= − = − =
= = ∴ = = =
−
= + − + −
c
s s
o d sand dry sand sat w sat sand sat w mid
Skempton formula for the C LL
Se wG but S e wG
The stress of the clay at its mid stratum before the building was built is
p H H H
γ γ γ γ γ
( )( )
10 10
(0.11 )(8 ) (0.115 0.0624) (7 ) (0.120 0.0624) (8.5 )
1.74
,
17 12 (0.36) 1.74 1
log log
1 1 0.91 1.74
7.6
−
= + − + −
=
+ Δ +
Δ = = =
+ +
clay
o
c o
o o
kcf ft kcf ft kcf ft
p ksf
The consolidation settlement is
H
i
C p p
H
e p
nches
Δp = 1 ksf
Water Table
Sand
γ = 110 pcf,
γsat = 115 pcf
Clay
w = 34%, LL = 50%, γsat = 120 pcf
8 feet
7 feet
17 feet

178
*Plastic Settlement-03: Boussinesq to reduce the stress with depth.
(Revision: Aug-08)
Calculate the settlement of the 10-foot thick clay layer shown below that will result from the
column’s load carried by a 5-foot square footing. The clay is normally consolidated. Apply
Boussinesq's formula to find the reduction of the vertical stress with depth.
Solution:
0.009( 10) 0.009(40 10) 0.27
,
( ) ( )
(0.10 )(10 ) (0.120 0.0624)
− − − −
= − = − =
−
= + − + −
= + −
c
o d sand dry sand sat w sat sand sat w mid clay
o
Skempton formulaC LL
The stress of theclay at its mid stratumbeforethebuilding wasbuilt is
p H H H
p kcf ft
γ γ γ γ γ
(5 ) (0.110 0.0624) (5 )
1.53
+ −
=
o
kcf ft kcf ft
p ksf
The avg
p
Δ below the center of the footing between 1 15
z feet
= to 2 25
z feet
= is given as,
2 1
2
2 4( ) 1 4( )
4( )
2 1 2 2
4 , '
5
2.5
2
H H
avg H
H I H I B L
p q where I f m n
H H H H
B L feet
−
⎡ ⎤ ⎡ ⎤
′
Δ = = = =
⎢ ⎥ ⎢ ⎥
− ⎣ ⎦
⎣ ⎦
= = =
200 kips
Ground Water Table
Sand
γdry = 100 lb/ft3
120 lb/ft3
Clay
γsat =110 lb/ft3
, e = 1.0, LL = 40
10 feet
5 feet
10 feet
5’
5’ x 5’ footing

179
( )( )
2 1
2 1
1 1
4
4( ) 4( )
2 4( ) 1 4( )
2 1
2.5 2.5
0.167 0.167
15 15
From Boussinesq's Table for
0.05 0.075
The average pressure at mid-clay layer is thus given by,
200
4 4
5 5
H H
H H
avg
B L
m and n
H H
I
I and I
H I H I kips
p q
H H
′ ′
= = = = = =
= =
−
⎡ ⎤
Δ = =
⎢ ⎥ ′ ′
−
⎣ ⎦
( )( )
10 10
(25)(0.05) (15)(0.075)
0.4 400
25 15
,
10 12 (0.27) 1.53 0.40
log log
1 1 1 1.5
1.6
3
o avg
c
o o
ksf psf
The consolidation settlement is
p
i
p
H
nches
C
H
e p
⎛ ⎞ −
⎡ ⎤
= =
⎜ ⎟⎢ ⎥
⎜ ⎟ ′ ′
−
⎣ ⎦
⎝ ⎠
+ Δ +
Δ = = =
+ +

180
*Plastic Settlement -04: Surface loads with different units.
(Revision: Aug-08)
Find the total settlement under a building that applies the load shown below.
Solution:
Notice that the data provided does not include a unit weight for the clay stratum. Therefore, this value
must be determined through the other information provided.
( ) [ ]( )
( )
3
3
3
27.8 (1) 0.40(2.78) 1.11
(2.78) (1.11) 9.81
Therefore, the clay unit weight is 18.1
1 1 1.11
The effective unit weight for the clay is,
18.1 9.81 8.3
Als
= = = = = =
+
+
= = =
+ +
′ = − = − =
s w s s
s o w
sat
o
sat w
G kN m Se e wG
G e
kN m
e
kN m
γ γ
γ
γ
γ γ γ
3 3 3 2
o, the Skempton relation is, 0.009( 10) 0.009(45 10) 0.32
The stress at the mid-clay stratum,
(4.6 )(17.6 ) (6.0 )(10.4 ) (7.6 2)(8.3 ) 175 /
The consolidation settlement is,
1
= − = − =
′ = + + =
Δ =
+
c L
c
C w
m kN m m kN m m kN m kN m
HC
H
e
σ
10 10
(7.6 )(0.32) 175 120
log log
1 1.11 175
0.26 260
′+ Δ +
= =
′ +
=
o o
m
m mm
σ σ
σ
Ground Water Table
Sand
γ = 17.6 kN/m3
, γ’ = 10.4 kN/m3
4.6 m
6.0 m
7.6 m
q = 1.2 daN/cm2
Normally consolidated clay
Gs =2.78, w = 40%, PI = 15%, PL = 30%

181
*Plastic Settlement-05: Pre-consolidation pressure pc and index Cc.
(Revision: Aug-08)
The results of a laboratory consolidation test on a clay sample are given below:
Pressure, p
(kN/m2
) Void ratio, e
23.94 1.112
47.88 1.105
95.76 1.080
191.52 0.985
383.04 0.850
766.08 0.731
(43) a) Draw an e-log p plot.
(44) b) Determine the pre-consolidation pressure, pc .
(45) c) Find the compression index, Cc.
Solution:

182
b) Determine the pre-consolidation pressure, pc. From the e-log p plot,
2 2
2 2 1 1
2
500 0.8 300 117.5
0 /
.9 c
p kN m and e p kN m a p kN m
nd e
= = = = ∴ =
c) Find the compression index, Cc. From the slope of the graph,
1 2
2
1
0.
0.9 0.8
500
log
log
30
451
0
c
e e
C
p
p
− −
= = =
⎛ ⎞ ⎛ ⎞
⎜ ⎟
⎜ ⎟ ⎝ ⎠
⎝ ⎠

183
*Plastic Settlement-06: Final voids ratio after consolidation.
(Revision: Aug-08)
The clay stratum shown in the profile below has a total vertical stress of 200 kN/m2
at its mid-
height with a voids ratio of 0.98. When the vertical stress increases to 500 kN/m2
the voids ratio
decreases to 0.81. Find (a) the effective overburden pressure at mid-height of the compressible
clay layer, and (b) the voids ratio of the clay if the total pressure at its mid-height is 1000 kN/m2
.
Solution:
[ ] [ ] [ ]
3 3 3
1 2
2 1
2
) ' '
(132 )(8 ) (132 62.4) (20 ) (125.4 62.4) (11 )
0.98 0.81
) 0.427
log( ) log(500 200)
0.98
0.42
3.14
7
log
o dry sand gravel saturated sand gravel mid clay stratum
o
c
o
p ks
a p h h h
p lb ft ft lb ft ft lb ft ft
e e
b C
p
f
p
e
γ γ γ
− − −
= + +
= + − + −
− −
= = =
−
=
=
2
(1000 200)
0.68
e
∴ =
Elev. 760 ft.
Ground Water Table
Sand and Gravel
Unit Weight = 132 lb/ft3
Elev. 752 ft.
Elev. 732 ft.
Elev. 710 ft.
Sand and Gravel
Unit Weight = 132 lb/ft3
Elev. 721 ft.
Clay
Unit Weight = 125.4 lb/ft3

184
*Plastic Settlement-07: Settlement due to a lowered WT.
(Revision: Aug-08)
Find the settlement due to lowering of the phreatic surface from elevation 349.5’ to 344.0’ using
the boring report shown below.
Solution:
( )( ) ( )( )
( )( ) ( )( )
2 1
1
1
1
2
0.110 7 0.110 6
0.110 0.0624 3 0.110 0.0624 3
1.71
due to the lowering
1.853 1.16 0.693
log
1
0
0.030(10) 3 1.16
log 0.032
1 0.96 3
0.034
o
o
o
o
p pcf ft pcf ft
pcf ft pcf ft
p ksf
q
p p WT
H
q p q
C H p q
c
H
e p
H ft
H
= + +
− + −
=
Δ
Δ = + Δ
Δ = Δ − Δ = − =
+ Δ
Δ =
+
+
Δ = =
+
Δ =
(10) 3 0.693
log 0.157
1 0.96 3
versus 0.083ft the additional settlement.
The rising may reduce settlement.
ft
WT
+ =
+

185
*Plastic Settlement-08: The over-consolidation ratio (OCR).
(Revision: Sept-08)
Oedometer (consolidation) tests of several samples from the clay stratum yields the consolidation
curve shown below. Given that Gs = 2.65, find (a) the value of po, (b) The value of pc and (c) the
OCR of the clay.
Solution:
3
3
) In order to find the in-situ stress before the footing was built, we need to find the unit weight
of the sand stratum,
(9.1 / )(2.65)
14.6 /
1 1.65
The stress is found at mid-clay
= = =
−
o
b s
d
s
o
a p
G kN m
kN m
G
p
γ
γ
[ ] [ ] [ ]
( )
3 3 3
stratum,
( ) ( ) ( )
(2 )(14.6 / ) (4 )(9.1 / ) 79.4
(1.5 )(9.2 / )
′ ′
= + +
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
= + + =
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
o d dry sand sand clay
o
p h h h
p m kN m m kN m m kN m kPa
γ γ γ
1200 kN
WT
2 m
4 m
3 m
γb = 9.1 kN/m3
1.5m
4 m x 4 m
γb = 9.0 kN/m3
γb = 9.2 kN/m3
Clay
Sand
Sand and Gravel

186
**Plastic Settlement-09: Coefficient of consolidation Cv.
(Revision: Aug-08)
An odometer test was performed on a peat soil sample from an FDOT project in the Homestead
area. The results are shown below. The initial sample thickness is 20 mm, with two-way drainage
through porous stones, simulating field conditions. The vertical stress increment is 10 kPa.
Estimate of the coefficient of consolidation cv as,
t
H
C
ion
consolidat
of
t
Coefficien dr
v
4
3
)
(
2
=
Time (minutes) 0 0.32 0.64 1.28 2.40 4.80 9.60 16.0
Settlement
(mm.)
0 0.16 0.23 0.33 0.45 0.65 0.86 0.96
( )
2
1
(min)
=
time 0 0.57 0.8 1.13 1.55 2.19 3.10 4.0
(1) Plot a graph of settlement against the square root of time.
(2) Determine the value of the coefficient of consolidation cv (in m2/s).
(3) Estimate what could be a good estimate of the elastic modulus of this soil Eo (kPa)
(4) What sort of permeability k (in m/s) could you estimate for this soil (from Eo)?
Solution:
(1)
Time vs. Settlement
0
0.2
0.4
0.6
0.8
1
1.2
0 1 2 3 4 5
Time (min)^(1/2)
S
ettlem
en
t
(m
m
.)
3.40

187
(2)
1/2
2 2 2
7
2
2
3.40min ( ) 11.56min
3 3(10 ) (1min)(1 )
4 4(11.56min)(60sec)(1000
1.08 10
sec
)
x x
v
x
vt and time t
d mm m
c
t mm
m
−
= =
= = ×
=
(3)
3
3
2
1
( ) 0.05 5%
20
20
10 10
200 10
0.0
0
5
v v
v
o
v
mm
Voids ratio e e
mm
N
E Pa
e
a
m
kP
σ
− = = ∴ =
×
= = = × =
(4)
( )( )
( )
9
2
7
3
2
5.3
2.30 ' 1
1
1.08 10 9.81
200
10
sec
c z z
v v o
o w c z w w
v w
o
C k e k k
e c c E
e C e
m kN
c s m
k
E kN
m
m
σ σ
γ γ γ
γ
−
−
⎡ ⎤ ⎡ ⎤ ⎛ ⎞ ⎛ ⎞
⎛ ⎞
+
≈ ∴ = ∴ = =
⎜ ⎟ ⎜ ⎟
⎜ ⎟
⎢ ⎥ ⎢ ⎥
+ ⎝ ⎠
⎣ ⎦ ⎣ ⎦ ⎝ ⎠
×
⎝ ⎠
×
= = =

188
*Plastic Settlement -10: Secondary rate of consolidation.
(Revision: Aug-08)
An oedometer (consolidation) test is performed on a normally consolidated clay stratum that is 8.5
feet thick, and it found that the clay’s initial voids ratio was eo = 0.8 and its primary compression
index is Cc = 0.28. The in-situ stress at mid-clay layer is po = 2,650 psf, and the building exerts a
pressure through its mat foundation of 970 psf. The secondary compression index Cα = 0.02.
The time of completion of the primary settlement is approximately 18 months. What is the total
consolidation of the 8.5 foot clay stratum 5 years after the primary consolidation?
Solution:
( )( )( )
( )
'
'
2
1
The primary consolidation is,
0.28 8.5 12 /
' 2,650 970
log log
1 1 0.8 2,650
The secondary consolidation is,
log
1
We must find first
2
,
.15
Δ
⎛ ⎞
+ Δ +
⎛ ⎞
Δ = = =
⎜ ⎟ ⎜ ⎟
+ + ⎝ ⎠
⎝ ⎠
Δ
⎛ ⎞
Δ = ⎜ ⎟
+ ⎝ ⎠
p
c o
p
o o
s
s
p
p
H
ft in ft
C H p p
H
e p
H
C H t
H
inche
e
s
e t
α
( )
( )( )( )
( )
'
'
2
1
by finding the change in the voids ratio during primary consolidation,
' 2,650 970
log 0.8 0.28 log 0.76
2,650
0.02 8.5 0.18 12 / 5
log log
1 1 0.76 1.5
⎛ ⎞
+ Δ +
⎛ ⎞
= − Δ = − = − =
⎜ ⎟ ⎜ ⎟
⎝ ⎠
⎝ ⎠
−
⎛ ⎞ ⎛
∴ Δ = =
⎜ ⎟
+ +
⎝ ⎠
o
p o o c
o
s
p
p p
e e e e C
p
in ft
C H t
H
e t
α
The total consolidation settlement is thus = 2.15 0.5
0.59
2.7
9 4
⎞
=
⎜ ⎟
⎝ ⎠
Δ + Δ = + =
p s
H H
inches
inches

189
*Plastic Settlement-11: Using the Time factor Tv.
(Revision: Aug-08)
A 3-m thick, doubly-drained saturated stratum of clay is under a surcharge loading that
underwent 90% primary consolidation in 75 days. Find the coefficient of consolidation cv of this
clay in cm2
/sec.
Solution:
The clay layer has two-way drainage, and Tv = 0.848 for 90% consolidation.
2 2
(0.848)(150 )
(75 24 60 60)
= = =
× × ×
v dr
v
T H cm
c
t days
0.00294 cm2
/sec
0
10
20
30
40
50
60
70
80
90
100
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2
Timefactor,Tv
Percent
consolidation

190
*Plastic Settlement-12: The time rate of consolidation.
(Revision: Aug-08)
An oedometer (consolidation) test is performed on a 4” thick specimen, drained on top and
bottom. It was observed that 45 percent consolidation (Tv = 0.15) was attained in 78 hours.
Determine the time required to attain 70 percent consolidation (Tv = 0.40) in a job site where the
clay stratum is shown below.
Solution:
The coefficient of consolidation
t
H
T
c dr
v
v
2
= is the same for the lab and field samples.
( )
( )
( )
2 2 2
2 2
2
2
0.15 (2 ) (0.40)(12.5 12 / )
78
1 1 0.40 (12.5 12 / )
78
24 365 0.15 (2 )
⎛ ⎞ ⎛ ⎞
= = =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
×
∴ =
⎛ ⎞
⎛ ⎞ ×
⎛ ⎞ ⎛ ⎞
∴ = ⎜ ⎟
⎜ ⎟
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
⎝ ⎠ ⎝ ⎠
∴
v dr v dr v dr
v
laboratory field
field
fiel
field
T H T H T H
c
t t t
in ft in ft
hours t
day year ft in ft
t hours
hours days in
t 134 70%
=
d years to attain consolidation
WT
Building’s load
q = 4 ksf
20’
25’
SAND
CLAY
SAND

191
*Plastic Settlement-13: Time of consolidation t.
(Revision: Oct.-08)
Using the information derived from Problem 11, how long will it take a 30-mm thick undisturbed
clay sample obtained from the field to undergo 90% consolidation in the laboratory?
Solution:
The Time Factor Tv is the same 90% in the field as in the laboratory, therefore,
( )
90( ) 90
90 2
2 2
( )
2
90( ) 2
(75 24 60 60)
3,000 30
( ) ( )
648 seconds =10minute
2 2
(75 24 60 60) 15
(1,500 )
s
v field v v
dr field
field laboratory
lab
c t c days c t
T
mm mm
H
days mm
t
mm
⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥
× × ×
= = =
⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦
× × ×
∴ = =

192
*Plastic Settlement-14: Laboratory versus field time rates of settlement.
(Revision: Aug-08)
Laboratory tests on a 25mm thick clay specimen drained at both the top and bottom show that
50% consolidation takes place in 8.5 minutes.
(1) How long will it take for a similar clay layer in the field, 3.2 m thick, but drained at the top
only, to undergo 50% consolidation?
(2) Find the time required for the clay layer in the field as described in part (a) above, to reach a
65% consolidation.
Solution:
(1)
( )
2 2
( ) ( )
2
( ) 2
(3.2 ) (8.5min)
557,000min
25
2 10
8
0
3
0
7
field
lab
dr lab dr field
field
t
t
H H
m
t
mm
m
d s
m
ay
=
∴ = = =
⎛ ⎞
⎜ ⎟
×
⎝ ⎠
(2)
2
2
5
50
2 2
65
( 65) 5 2
25
(0.197)
2 1000
0.36 10
(8.5min)
(0.34)(3.2 )
961,400min
0.36 10 / min
668
v dr
v
dr
field
v
T H
d
c
t
T H m
t ays
c m
−
−
⎛ ⎞
⎜ ⎟
×
⎝ ⎠
= = = ×
∴ = = = =
×

193
*Plastic Settlement-15: Different degrees of consolidation.
(Revision: Aug-08)
A clay layer 20 feet thick sitting on top of granite bedrock, experiences a primary consolidation of
8.9 inches. Find:
(a) The degree of consolidation when the settlement reaches 2 inches.
(b) The time to reach 50% settlement if cv is 0.002 cm2/sec.
(c) The time for 50% consolidation if the clay stratum is doubly-drained?
Solution:
( )( )
( )
( )( )
( )
2
2
50
2
2
2
2
2
( ) % 100
8.9
(0.197) 20 30.48 /
( ) 0.197
1
0.002 /sec
360sec 24
(0.197) 10 30.48 /
( )
1
0.002 /se
22.5%
42
c
360sec 24
4
v
dr
v
v
dr
v
a U
ft cm ft
T
b T and t H
hr days
c
cm
h
ft cm ft
T
c t H
hr days
c
s
cm
h
day
⎛ ⎞
= =
⎜ ⎟
⎝ ⎠
⎡ ⎤
⎣ ⎦
= = = =
⎛ ⎞⎛ ⎞
⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
⎡ ⎤
⎣ ⎦
= =
⎛ ⎞⎛
⎜ ⎟
⎝ ⎠
106 days
=
⎞
⎜ ⎟
⎝ ⎠
0
10
20
30
40
50
60
70
80
90
100
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2
Time factor , Tv
Percent
consolidation
0
10
20
30
40
50
60
70
80
90
100
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2
Time factor , Tv
Percent
consolidation

194
**Plastic Settlement-16: Excavate to reduce the settlement.
(Revision: Oct.-08)
An oedometer test on a 1” thick, doubly-drained sample from the clay stratum (shown below)
attained 50% consolidation in 6.5 minutes. Find:
(a) The total differential settlement of the fully loaded tank.
(b) The time required for 75% consolidation in the field.
(c) The depth of excavation for minimal settlement.
Solution:
a) The surface load from the oil tank (neglecting the weight of the tank) is Δp:
(40 )(60 ) 2.4
oil
p h ft pcf ksf
γ
Δ = = =
Using the Boussinesq pressure diagram (next two pages) for '
75
=
B , provides the stress levels at
any point in the soil mass, thus,
The stress at point A ( )
0.91 0.90 2.4
p
⇒ Δ = = 2.2 ksf
point B ( )=
=
Δ
⇒ 4
.
2
43
.
0
43
.
0 p 1.0 ksf
The in-situ effective stress at point A was po, before the tank was built:
( )( ) ( )( )
10' 0.120 0.062 28.5' 0.110 0.062 1.95
o
p ksf
= − + − =
The settlements at point A (below the center of the tank), and point B (at the edge of the tank) are,

195
( )
( )
( )
( )
0.4 57'
' 1.95 2.2
log log 3.34
1 ' 1 1.27 1.95
0.4 57'
' 1.95 1.0
log log 1.85
1 ' 1 1.27 1.95
3.34 1.85 1.49
c o
A
o o
c o
B
o o
C H p p
H ft
e p
C H p p
H ft
e p
The differential settlement between A and B is ft ft f
⎛ ⎞
+ Δ +
⎛ ⎞
Δ = = =
⎜ ⎟ ⎜ ⎟
+ + ⎝ ⎠
⎝ ⎠
⎛ ⎞
+ Δ +
⎛ ⎞
Δ = = =
⎜ ⎟ ⎜ ⎟
+ + ⎝ ⎠
⎝ ⎠
∴ = − = 18in s
t che
=
b) Since the time required for consolidation is,
Tv for 50% = 0.197, and
Tv for 75% = 0.480, using the relationship between field and lab conditions, through
the coefficient of consolidation,
( )
( )( )
2
2
2 2
75 7
75
5
0.2 0.5
0.480 28.5
1
1
2
6.5min
3.8 75% .
field field lab field
v
lab
t years for consolidati
ft
in
o
Tv H Tv H ft
in
c
n
t t t
⎡ ⎤
⎛ ⎞
⎜ ⎟
⎢ ⎥
⎝ ⎠
⎣ ⎦
=
= ∴ = =
∴
c) The settlement can be reduced by excavating weight of soil equal to the weight of the
structure (note: total, not effective, since the water is also removed). Assume need to
excavate “x” feet from the clay layer.
( )( ) ( )
10' 0.120 0.110 2.4
total x ksf
σ = + =
Solving for x: ≅
=
∴ ft
x 9
.
10 11 feet of clay
Add the 11 feet of the clay to the 10 feet of the sand on top of the clay for a
11 10 21
Total excavation depth ft e
ft fe t
∴ = + =

196
**Plastic Settlement-17: Lead time required for consolidation of surcharge.
(Revision: Aug-08)
A common method used to accelerate the consolidation of a clay stratum is a sand surcharge, as
shown below. The surcharge load will force the clay to attain a large part of its settlement before
the structure with the same load is built. This method minimizes the settlement of the structure.
An office building is planned to be built on the site shown below. The total weight of the building
is 136,000 kips, spread over a square foundation 200 ft by 200 ft. Field tests showed that the clay
stratum has a liquid limit of 28 percent, an initial void ratio of 0.95, a γ = 130 pcf and a
consolidation coefficient of 10-3
in2
/second. The sand stratum has a CC = 0.01, a γ = 125 pcf and an
initial void ratio of 0.70. The sand surcharge has a γ = 115 pcf.
(d) Determine the total settlement at mid-clay under the center of the surcharge.
(e) The time required to attain 60% consolidation of the clay stratum (i.e. TV = 0.30). This is the
lead time required to place the surcharge before construction.
(f) The SPT in the sand stratum is N = 15.
Solution:
(a) The weight of the new building is estimated at Q = 136,000 kips. The surcharge will have to weigh
the same, spread over an area = 200’ x 200’ = 40,000 ft2
, using a fill with a unit weight of γ = 115
pcf. The unit pressure of the surcharge γsurcharge is,
2
3
136,000
3.4
40,000
, 115 ( ) 3.4 30
sur
sur sur
Q kips
q ksf
A ft
lb
but q h h ksf h feet high
ft
γ
= = =
= = = ∴ =
Both the sand and clay strata contribute to the settlement. However, the settlement from the clay
stratum is a consolidation settlement, such that,
WT
h ft
Sand Surcharge
(Area = 200’ x 200’)
15 ft
30 ft Sand
Clay
Impermeable rock
20 ft

197
0
Using Skempton's formula 0.009( -10) 0.009(28-10) 0.162
Also 0.95 20 240 .
The initial stress at mid-clay stratum is,
(0.125 )(15 ) (0.125 0.0624)(15) (0.130
c
o clay stratum
b b
s s c
C LL
e and H ft in
p h h h kcf ft
γ γ γ
= = =
= = =
= + = = + − +
( ) ( )
0
10 10
0 0
0.0624)(10) 3.5
The increased load from the surcharge is,
(0.115 )(30 ) 3.45
The settlement created by the surcharge is,
(240 )(0.162) 3.5 3.45
log log
1 1 0.95 3
c
c
ksf
p
p h kcf ft ksf
HC p p in
H
e p
γ
− =
Δ
Δ = = =
⎛ ⎞
+ Δ +
Δ = =
⎜ ⎟
+ +
⎝ ⎠
( )( ) 2
1 2
1 2
5.94
.5
The settlement from the sand stratum is an elastic settlement,
30 360 14 14 15 210 / ; 1 1
0.30
(1)(1)(3.4 )( )(360
210
sand S
o
inches
H ft in E N k ft depth factor C and creep factor C
C C q dz ksf
E ksf
ε
⎛ ⎞
=
⎜ ⎟
⎝ ⎠
= = ≈ = = = =
⎛ ⎞
Δ = =
⎜ ⎟
⎝ ⎠
( )( )
( )
2
2
3 2
) 1.75
The total settlement 5.94 1.75 7.69
b) The lead time required for the surcharge to accomplish its task is,
(0.30) 20 12 /1 1min 1 1
60sec 60min
10 /sec
dr
in in
in in in
t
T H ft in ft hr
v
t
c in
v
−
=
= + =
⎡ ⎤ ⎛ ⎞⎛ ⎞
⎣ ⎦
= = ⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠ 24
The total settlement that has taken place at 200 days (60% consolidati
200 60
on) is,
(60%) 1.7
%
5
3 (0.6)(5.94 ) .3
S C
day
hr
Total Time H H in
days for consolidat
in
ion
inches
⎛ ⎞
=
⎜ ⎟
⎝ ⎠
= Δ + Δ = + =

198
**Plastic Settlement-18: Settlement of a canal levee.
(Revision: Aug-08)
A uniform surcharge of sand 20 feet in height will be placed over the marl stratum as shown
below, in order to preconsolidate that layer for a future building. The in-situ voids ratio of the
marl is 0.59, and its index of compression can be found from a relation proposed by Sowers as Cc =
0.75 (eo – 0.30). Find the total settlement of the surcharge at its point A.
The coefficient of consolidation cv for the marl can be found from the relation,
( )
w
v
o
v
a
e
k
c
γ
+
=
1
ft2
/day
where the permeability s
cms
k /
10
0
.
1 5
−
×
= , and lb
ft
av /
10
9
.
2 2
4
−
×
= .
Find the time required (in days) for the marl to attain 50% consolidation.
0
10
20
30
40
50
60
70
80
90
100
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2
Time factor , Tv
Percent
consolidation

199
Solution.
( ) ( )
( )
( )
o
3
3
0.75 0.30 0.75 0.59 0.30 0.218
At midpoint of the marl, the in-situ stress is,
120 20 2,400
The increase in stress due to the surcharge is,
125 20
c o
o
c e
lb
h ft psf
ft
lb
h ft
ft
σ
σ γ
σ
σ γ
= − = − =
⎛ ⎞
= = =
⎜ ⎟
⎝ ⎠
Δ
⎛ ⎞
Δ = Δ = ⎜ ⎟
⎝ ⎠
( )( )( )
( )
10 10
2,500
Therefore, the consolidation settlement of the marl is,
40 12 / 0.218 2.4 2.5
log log 20.4
1 1 0.59 2.4
The time required for 50% of consolidation to take pl
c o
o o
psf
ft in ft
HC
inches
e
σ σ
σ
=
⎛ ⎞
+ Δ +
⎛ ⎞
∴Δ = = =
⎜ ⎟ ⎜ ⎟
+ + ⎝ ⎠
⎝ ⎠
( )( )
( )( )
5 3
2
4 2 3
ace is found through the coefficient of
consolidation ,
1.0 10 /sec 1 0.59
1 1 1 86.4 10 sec
2.5 /
2.54 12
2.9 10 / 62.4 /
The time factor for 5
v
o
v
v w
c
cm
e in ft
c k ft day
a cm in day
ft lb lb ft
γ
−
−
× +
⎛ ⎞ ⎛ ⎞
+ ×
⎛ ⎞⎛ ⎞
= = =
⎜ ⎟ ⎜ ⎟
⎜ ⎟⎜ ⎟
× ⎝ ⎠⎝ ⎠⎝ ⎠
⎝ ⎠
( )( )
2
2
2
0% consolidation is 0.2, therefore the time required is,
0.2 40
=
2.
2
5 /
1 8
dr
v
v
v
T t
T H ft
t
c f
a
t d y
y
a
d s
=
= =

200
**Plastic Settlement-19: Differential settlements under a levee.
(Revision: Aug-08)
An oedometer (consolidation) test was performed on a clay sample 3 cm high, drained on both
sides, and taken from mid-stratum shown below. Seventy percent consolidation was attained in
6.67 minutes. Find:
(b) The time required to attain 70% consolidation of the clay stratum.
(c) The magnitude of that settlement in that time.
Solution.
2
2 2 2
2
2
2
(a) Since the soil is the same clay in the laboratory and the field, and both are 70% consolidation,
6.6
field field
v dr v dr v dr
v
lab lab
laboratory field
lab F
field
L
t H
T H T H T H
c
t t t t H
t H
or t
H
⎡ ⎤ ⎡ ⎤
= = = ∴ =
⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦
= = ( )
( )
2
6
70% 10
700
7min 1.45 10 min
1.5
(b) The amount of settlement that takes place at 70% consolidation is,
' '
0.70 log
1 '
The in-situ stress at mid-clay stratum
2.76
c o
o o
cm
cm
C H p p
H
e p
years
⎛ ⎞
= × =
⎜ ⎟
⎝ ⎠
⎧ ⎫
⎛ ⎞
+ Δ
⎪ ⎪
Δ = ⎨ ⎬
⎜ ⎟
+
⎪ ⎪
⎝ ⎠
⎩ ⎭
( )( ) ( ) ( )
( )( )
( )
3 3 2
2
10 10
before the surcharge was applied was,
' 18 / 2 20 9.81 / 3.5 71.7 /
and ' 72 /
0.7 0.20 700
0.7 ' '
14.8 cms
71.7 72
70% log log =
1 ' 1 1 71.7
o i i
i
c o
o o
p h kN m m kN m m kN m
p kN m
cm
C H p p
H
e p
γ
∴ = = + − =
Δ =
⎛ ⎞
+ Δ +
⎛ ⎞
∴ Δ = =
⎜ ⎟ ⎜ ⎟
+ + ⎝ ⎠
⎝ ⎠
∑

201
Vertical stresses induced by a uniform load on a circular area

202
***Plastic Settlement-20: Estimate of the coefficient of consolidation cv.
(Revised Oct-09)
An oedometer test was performed on a peat soil sample from an FDOT project in the Homestead
area. The results are shown below. The initial sample thickness was 20 mm with two way
drainage through porous stones, simulating field conditions. The vertical stress increment is 10
kPa. Use an estimate of the coefficient of consolidation v
c as,
t
H
c dr
v
4
3 2
=
Given Data
Time(min) 0.00 0.32 0.64 1.28 2.40 4.80 9.60 16.00
Settlement (mm) 0.00 0.16 0.23 0.33 0.45 0.65 0.86 0.96
Time^1/2(min)^1/2 0.00 0.57 0.80 1.13 1.55 2.19 3.10 4.00
(a) Plot a graph of settlement against the square root of time.
(b) Determine the value of v
c (
s
m2
).
(c) Estimate what could be a good estimate of the elastic modulus of this soil, o
E (in kPa).
(d) What sort of permeability k (
m
s
) could you estimate for this soil (from o
E )?
Solution:
(a)
Settlement vs. Time^1/2
y = 0.2503x + 0.0376
0.00
0.20
0.40
0.60
0.80
1.00
1.20
0.00 1.00 2.00 3.00 4.00 5.00
Time^1/2(min)^1/2
Settlement
Delta
H
(mm)
0.96,3.69

203
(b) 2
/
1
min
69
.
3
=
x
t → x
t =13.62 min
sec
10
*
178
.
9
)
1000
sec)(
60
min)(
62
.
13
(
4
)
1
min)(
1
(
min)
10
(
3
4
3 2
8
2
2
2
2
m
mm
m
t
d
c
x
v
−
=
=
=
(c) The strain at the end of the loading is:
05
.
0
)
20
(
)
1
(
=
=
Δ
=
mm
mm
L
v
ε or 5%
kPa
m
N
E
v
v
o 200
05
.
0
10
*
10
2
3
=
=
=
ε
σ
Vertical Modulus
From Coduto (pages 390 and 391)
o
c
v
e
C
+
≈
1
ε but Coduto (page 424)
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ +
=
c
w
z
v
C
e
k
c 0
1
30
.
2
γ
σ
or
w
o
w
z
z
v
k
E
k
c
γ
γ
ε
σ
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
*note: (2.30 is for ln; use 1.0 for log)
s
m
m
kN
m
kN
m
E
c
k
o
w
v 9
2
3
2
8
10
*
50
.
4
200
81
.
9
sec
10
*
178
.
9
−
−
=
⎟
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
=
γ

204
**Plastic Settlement-21: The apparent optimum moisture content.
(Revised Oct-09)
Find the Cc and the “apparent” OCR for a superficial clay stratum that is 4m thick. The water is
at the surface. A sample of the clay has a water content of 27.9%, and a specific gravity of 2.65. An
oedometer test showed the following results:
Solution:
Cc = [Ca-Cb]/[log(O’z b)-log(O’z a)] = [0.675-0.600]/[log(930)-log(230)] = 0.123
Se = wGs (Saturation means S = 1)
e = wGs = (0.279)(2.65) = 0.73935 = 0.739
γ = [(Gs + e) γw]/[1 + e] = [(2.65 + 0.739)(9.81)]/[1 + 0.739] = 33.25/1.739 = 19.12
OCR = (98)/(19.12) = 5.13
Voids Ratio Vertical Effective Stress
0.736 15
0.733 28
0.73 60
0.675 230
0.638 480
0.6 930

205
**Plastic Settlement-22: The differential settlement between two buildings.
(Revised Oct-09)
Two tall buildings sit next to each other in the downtown area of Boston. They are separated by a
narrow 5 foot alley and are both 50 stories high (550 feet tall). They also have similar foundations,
which consist of a simple mat foundation. Each mat is a thick reinforced concrete slab, 5 feet thick
and 100 feet by 100 feet in plan view. The total load (dead + live + wind) of each building is
150,000 kips.
The mats are sitting upon a thick prepared stratum of carefully improved soil 40 feet thick, that
has an allowable bearing capacity of 17 ksf. Below the compacted fill stratum lays a medium to
highly plastic clay stratum 38 feet thick. Below the clay stratum is a thick layer of permeable sand.
The water table coincides with the interface between the improved soil and the clay. The dry unit
weight of improved soil is 110 pcf, whereas the in-situ unit weight of the clay is 121 pcf.
The clay has a specific gravity of 2.68. Also a moisture content of 32%, a PI of 52% and a PL of
12% before building “A” was built in 1975. The clay has a consolidation coefficient of 10-4
in2
/sec.
The second building “B” was finished by early 1995. Assume that each building’s mat rotates as a
rigid plate.
How much do you predict will building ‘A” drift towards building “B” by early 1996, in inches?
(Note: The drift of a building is its horizontal movement at the edge of the roof level due to lateral
loads such as wind or earthquakes, or the differential settlement due to unequal pressures.)
Given: P = 150,000 kips
qa = 17 ksf;
For clay: Gs = 2.68,
w = 32%,
P.I = 52%,
P.L = 12%,
γ =121 pcf,
cv = 10-4
in2
/sec
For the improved soil: γd =110 pcf

206
PLAN VIEW
A
BUILDING A
50 Stories = 550 ft
Completed 1975
150,000 kips
A
D C
PLAN VIEW
B
BUILDING B
50 Stories = 550 ft
Completed 1975
150,000 kips
B
F E
Fill
Clay
Sand
Diagram 1

207
Solution:
Initial Condition: Geostatic Stresses
σ’zo = ∑Hγ – u = (40)(121) – (40)(62.4) = 2,344 lb/ ft2
Induced Stresses:
To solve the induced stresses we will use the Boussinesq’s Method:
( )
( )
( )
[ ]
( )
( )
( )
( ) L
B
z
L
B
z
z
2
L
B
BLz
2
z
L
B
z
2
L
B
L
B
z
L
B
z
4
z
L
B
2BLz
q
f
f
f
f
f
f
2
2
f
2
f
f
f
induced
z 2
2
2
2
2
2
2
1
2
2
2
1
2
2
2
2
2
2
2
2
2
2
1
2
2
2
+
+
+
+
+
−
−
Π
+
+
+
+
+
×
+
+
+
Π
+
+
= −
/
/
sin
σ
( )( )
lbs
15,000
100
100
10
150,000
A
p
q
3
=
×
=
=
zf = 58 ft (at mid-clay)
Replacing the values of q, B, L and zf into the Boussinesq’s equation yields:
(σz )induced = 8008.65 lb/ ft2
, 1059.79 lb/ ft2
Exists differential stresses at the corners of the foundation.
Stresses due to fill:
(σz )fill = H fill γ fill = (110) (38) = 4,180 lb/ ft2
• The total final stresses at corner “C”
(σ’zf )C = σ’zo + (σz) induced + (σz) fill
= 2344 lb/ ft2
+ 8008.65 lb/ ft2
+ 4180 lb/ ft2
= 14,532.65 lb/ ft2
≈ 14,533 lb/ ft2
• The total final stresses at corner “D”
(σ’zf )D = σ’zo + (σz) induced + (σz) fill
= 2344 lb/ ft2
+ 1059.79 lb/ ft2
+ 4180 lb/ ft2
= 7,583.79 lb/ ft2
≈ 7,584 lb/ ft2
(Note: Both buildings A and B will experience the same final stress due to fill, loadings and geostatic
factors).
Time Factor
Determining the time factor for building A:
Tv = _4Cvt_
H2
dr
For 21 years: t = 662,256,000 sec
( )( ) 165.56
(40)
sec
0
662,256,00
in
10
4
Tv 2
-4
A =
=
sec
/
2
Degree of Consolidation
U = [1-10 –(0.085+Tv)/0.933
] x 100%
The degree of consolidation at TvA = 165.56 is:

208
U = [1-10 –(0.085+166)/0.933
] x 100% =100%
By early 1996, 21 yrs after the load was applied, the soils are completely consolidated.
Therefore,
( )ult
c
c
U
δ
δ
=
( )ult
c
c δ
δ =
Assuming normally consolidated soils
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
= ∑
0
f
z
z
0
c
c log
H
e
1
C
σ
σ
δ (Eqn 1-1)
1 + eo σ’zo
Cc = 0.009 (LL – 10)
LL = PL + PL = 52 + 12 = 64 %
Therefore, Cc = 0.009 (64 – 10) = 0.486
1
G
e
d
w
s
0 −
=
γ
γ
( )( ) 0.82
1
-
91.67
62.4
2.68
e
91.67
0.32
1
121
w
1
0
d =
=
∴
=
+
=
+
=
γ
γ
Recall that uniform stress does not exist beneath the corners of the mat foundation. Therefore, we will
have differential settlement.
At corner C:
H= 40 ft (σ’zf )C = 14,533 lb/ ft2
σ’zo = 2,344 lb/ ft2
Replacing values of H, (σ’zf )C , σ’zo, Cc and eo into (Eqn 1-1):
( ) in
101.55
ft
8.46
2,344
14,533
log
ft
40
0.82
1
0.486
c =
=
⎟
⎠
⎞
⎜
⎝
⎛
+
=
δ
At corner D:
H= 40 ft (σ’zf )D = 7584 lb/ ft2
σ’zo = 2,344 lb/ ft2

209
DRIFT=x
A
A
'
D C
D'
C'
?
?
550 ft
5.45 ft
100 ft
8.46 ft
3.01 ft
Diagram2
Replacing values of H, (σ’zf )D , σ’zo, Cc and eo into (Eqn 1-1):
( ) in
65.35
ft
5,45
2,344
7,584
log
ft
40
0.82
1
0.486
c =
=
⎟
⎠
⎞
⎜
⎝
⎛
+
=
δ
The Δδc = 8.46 -5.45 = 3.01 ft
Δ = tan -1
(3.01/ 100) =1.72º
Drift = x = 550 tan 1.72 = 16.52 ft = 200 inches.

210
**Plastic Settlement-23: Settlement of a bridge pier.
(Revision: Aug-08)
Estimate the average settlement from primary consolidation of the clay stratum under the center
of the bridge pier.
Solution:
Stress at mid-clay stratum:
3 2
3 2
3 2
3 1 9 .6 2 5 8 .9
7 9 .8 0 6 8 .6
2 9 .6 0 1 9 .2
k N k N
m x
m m
k N k N
m x
m m
k N k N
m x
m m
=
=
=
for a total 2
1 4 6 .7
o
kN
m
ρ =
Determine eo : From chart pc ~ 150 < po normally consolidated, eo = 0.81.
Determine :
or p
σ
Δ Δ

211
( )
4
3
4 2
4 5
0.286 0.500 0.071
10 10
28 10
4 4 0.071 99.5
8 10
m m
m and n I
m m
x kN
Q kN
I
A m x m m
σ
= = = = ∴ =
⎛ ⎞
Δ = = =
⎜ ⎟
⎝ ⎠
Determine the settlement ΔΗ :
10 10
0.31 146.7 99.5
log 4 log
1 1 0.81
0
1 .7
5
4
.1
6
c o
o o
C p p
m
e p
m
⎛ ⎞ ⎛ ⎞
+ Δ +
⎛ ⎞ ⎛ ⎞
ΔΗ = Η = =
⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
+ +
⎝ ⎠ ⎝ ⎠
⎝ ⎠ ⎝ ⎠
Q

212
Chapter 12
Shear Strength of Soils
Symbols for Shear Strength of Soils
c→ The cohesion of a soil particle.
cu→
p→
p’→
q→
q’→
u → Pore water pressure.
uc→
ud→
σ→ The normal axial stress.
σ1→ stress
σ1’→ stress
σ3’→ stress
σ3→ Confirming pressure
σd→
τ→ The shear stress.
τf→ The normal shear stress at a failure.
ф→ The angle of internal friction of the soil.
φ→ The angle of inclination of the plane of failure caused by the failure shear stress.
σ’→ Effective stress.

213
*Shear strength–01: Maximum shear on the failure plane.
(Revision: Oct-08)
A consolidated un-drained triaxial test was performed on a specimen of saturated clay with a
chamber pressure 3 2
2.0
kg
cm
σ = . At failure,
1 3 2 2
2.8 , 1.8 57
kg kg
u and the failure plane angle
cm cm
σ σ θ
− = = = °.
Calculate (1) the normal stress σ and (2) shear stress τ on the failure surface and (3) the
maximum shear stress on the specimen.
Solution:
1 2
3 2
2.8 2.0 4.8
2.0
kg
cm
kg
cm
σ
σ
= + =
=
Shear stress 1 3
2
4.8 2
sin 2 sin114 1.27
2 2
kg
cm
σ σ
τ θ
− −
⎛ ⎞
= = ° =
⎜ ⎟
⎝ ⎠
Normal stress 1 3 1 3
2
4.8 2 4.8 2
cos2 cos114 2.83
2 2 2 2
kg
cm
σ σ σ σ
σ θ
+ − + −
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= + = + ° =
⎜ ⎟ ⎜ ⎟
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
⎝ ⎠ ⎝ ⎠
Maximum shear 1 3
2
4.8 2
sin 2 1.4
2 2
MAX
kg
cm
σ σ
τ θ
− −
⎛ ⎞
= = =
⎜ ⎟
⎝ ⎠
at 45
θ = °

214
*Shear strength–02: Why is the maximum shear not the failure shear?
(Revised Oct-09)
Using the results of the previous Problem 01, and 2
24 , ' 0.80
kg
c
cm
φ = ° = , show why the sample
failed at 57 grades instead of the plane of maximum shear stress.
Solution:
On failure plane
( ) 2
57 2
' 2.83 1.8 1.03
' 'tan 0.8 (1.03tan 24 ) 1.27
kg
u
cm
kg
S c
cm
σ σ
σ φ
°
= − = − =
= + = + ° =
Compare that to τ = 1.27 kg/cm from the previous problem, and note that they are equal, and so for
both, 57 57 failed.
S τ
° °
= .
Now at the plane of maximum shear stress 45
θ = °
2
2
45 2
4.8 2 4.8 2
cos90 3.4
2 2
' 3.4 1.8 1.6
' 'tan 0.8 (1.60tan 24 ) 1.51
kg
cm
kg
cm
kg
s c
cm
σ
σ
σ φ
°
+ −
⎛ ⎞ ⎛ ⎞
= + ° =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
= − =
= + = + ° =
The shear strength at 45º is much larger than at 57º, therefore failure does not occur.

215
*Shear strength–03: Find the maximum principal stress σ1.
(Revised Oct-09)
Continuing with the data from the two previous problems, the same soil specimen is now loaded
slowly to failure in a drained test, that is u = 0, with 3 2
2.0
kg
cm
σ = . What will be the major principal
stress at failure?
Solution:
a) Analytically, in a drained test u = 0; at failure 3 3 2
' 2
kg
cm
σ σ
= = , on the failure plane 57
θ = ° .
( ) ( )
( ) ( )
1 1
1
1
1
1 2 2 2
' 2 ' 2
' 'tan 0.80 cos114 tan 24 1.426 0.132 '
2 2
' 2
sin114 0.457 ' 0.914
2
' 7.31 ; 3.6 2.38
s c
and
At failu
kg kg k
re s
g
and
cm cm cm
θ θ
σ σ τ
σ σ
σ φ σ
σ
τ σ
τ
⎡ + − ⎤
⎛ ⎞ ⎛ ⎞
= + = + + ° ° = +
⎜ ⎟ ⎜ ⎟
⎢ ⎥
⎝ ⎠ ⎝ ⎠
⎣ ⎦
−
⎛ ⎞
= ° = −
⎜ ⎟
⎝ ⎠
= = =
∴ =
b) Graphically,

216
*Shear strength–04: Find the effective principal stress.
(Revised Oct-09)
A drained triaxial test on a normally consolidated clay showed that the failure plane makes an
angle of 58˚ with the horizontal. If the sample was tested with a chamber confining pressure of
103.5 kN/m2
, what was the major principal stress at failure?
Solution:
45 58 45 26
2 2
φ φ
θ φ
= ° + ∴ ° = ° + ∴ = °
Using the equation that relates the major principal stress σ1 to the minor principal stress σ3, and with c =
0 (the value of cohesion for a normally consolidated clay),
( )
2 2
1
2
1 3
26
' 'tan 45 103.5 tan 45 26
2 2
5
kN
m
φ
σ σ
⎡ ⎤
⎛ ⎞ ⎛ ⎞
= °+ = °+ =
⎜ ⎟
⎜ ⎟ ⎢ ⎥
⎝ ⎠
⎝ ⎠ ⎣ ⎦

217
*Shear strength–05: Using the p-q diagram.
(Revised Oct-09)
Triaxial tests performed on samples from our Miami Pamlico formation aeolian sand, showed the
peak stresses listed below. Plot these values on a p-q diagram to find the value of the internal angle
of friction.
σ1 = 76 psi for σ3 = 15 psi p = 45.5, q = 30.5 psi (1)
Solution:
Remember that p = (σ1 + σ3)/2 and q = (σ1 – σ3)/2
From the p-q diagram:
4
4
2 4 2 .5
ta n 0 .6 6 8
3 6 2 .5
sin ta n 0 .6 6 8 4 2
q
p
α
φ α φ
≈ = =
= =
= ∴ °
q (psi)
300
4
3
2
1
200
100
α = 34°
100 200 300 400
p(psi)

218
**Shear strength–06: Consolidated-drained triaxial test.
(Revised Oct-09)
A consolidated-drained triaxial test was conducted on a normally consolidated clay. The results
are as follows: ( )
3 2 2
' 276 276
d f
kN kN
m m
σ σ
= Δ =
Determine:
(a) The angle of friction φ;
(b) The angle θ that the failure plane makes with the major principal plane, and
(c) The normal stress σ’ and shear stress τf on the failure plane.
Solution:
For normally consolidated soil the failure envelope equation is:
' ta n
f
τ σ φ
= because c = 0
For the triaxial test, the effective major and minor principal stresses at failure are as follows:
( )
1 1 3 2
' 276 276 552
d f
kN
m
σ σ σ σ
= = + Δ = + = and 3 3 2
' 276
kN
m
σ σ
= =
Part A.
The Mohr circle and the failure envelope are shown in the figure below, from which:
1 3
1 3
1 3 1 3
' '
' ' 552 276
2
sin 0.333
' ' ' ' 55
19.45
2 276
2
AB
O A
σ σ
σ σ
φ φ
σ σ σ σ
−
− −
= = = = = ∴
+ + +
= °

219
Part B.
19.45
45 45 54.7
2 2
φ
θ
°
= ° + = ° + = °
Part C.
1 3 1 3
( )
' ' ' '
' cos 2
2 2
on the failure plane
σ σ σ σ
σ θ
+ −
= +
and
1 3
' '
sin 2
2
f
σ σ
τ θ
−
=
Substituting the values of 1 3
2 2
' 552 , ' 276
kN kN
m m
σ σ
= = and 54.7
θ = ° in the preceding
equations,
( )
( )
2
2
552 276 552 276
' cos 2 54.7
2 2
552 276
sin 2 54
36
.
8
130
7
2
f
kN
m
N
m
and
k
σ
τ
+ −
= + =
−
= =

220
**Shear strength–07: Triaxial un-drained tests.
(Revised Oct-09)
Triaxial un-drained tests were performed on clay samples taken from the stratum shown below.
The tests were taken with pore water pressure measurements, and 2
' 20 , 24
kN
c and
m
φ
= = °
(a) Find the clay shear strength at its mid-stratum, and
(b) Find the effective and total stresses at the same level acting on a vertical face of a soil element.
Solution:
(a) For the gravel:
( )
( )
3
1 3 3 3 2
1 1 1 3 2
1 2 3 2
16 0.3 10 19
4 16 9 19 3.5 17.6 297
' (9 3.5) 297 10 12.5 173
' ' tan ' 20 173 tan 2 6
4 96.
sat d w
w
kN
n x
m
kN kN kN kN
For m m
m m m m
kN kN
For u m
m m
kN kN
S c
m
m
kN
m
γ γ γ
σ
σ σ σ γ
σ φ
= + = + =
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
∴ = + = =
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞
∴ = − = − + = − =
⎜ ⎟
⎝ ⎠
∴ = + = + ° =

221
(b) Since the clay is saturated,
( )
( ) 2
3 1 1 2
3 3 2 3
' ' 0.5 ' 0.5 173 86
' 86 13 3.5 10 251
o
kN
K
m
kN kN
and u m m
k
m
m m
N
σ σ σ
σ σ
= = = =
= + = + + =

222
**Shear strength-08: Consolidated and drained triaxial test.
(Revised Oct-09)
Two similar clay soil samples were pre-consolidated in triaxial equipment with a chamber
pressure of 600 kN/m2
. Consolidated-drained triaxial tests were conducted on these two
specimens. The following are the results of the tests:
Specimen 1: Specimen 2:
( )
3 2
2
100
410.6
d f
kN
m
kN
m
σ
σ
=
Δ = ( )
3 2
2
50
384.37
d f
kN
m
kN
m
σ
σ
=
Δ =
Find the values of the cohesion c and the angle of internal friction φ.
Solution:
For Specimen 1, the principal stresses at failure are,
( )
3 3 1 1 3
2 2
' 100 ' 100 410.6 510.6
d f
kN kN
and
m m
σ σ σ σ σ σ
= = = = + Δ = + =
Similarly, the principal stresses at failure for specimen 2 are
( )
3 3 1 1 3
2 2
' 50 ' 50 384.4 434.4
d f
kN kN
and
m m
σ σ σ σ σ σ
= = = = + Δ = + =

223
These two samples are over-consolidated. Using the relationship given by equation
2 1 1
1 3
' 'tan 45 2 tan 45
2 2
c
φ φ
σ σ
⎛ ⎞ ⎛ ⎞
= °+ + °+
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
Thus, for specimen 1
( ) ( ) 2 1 1
510.6 100 tan 45 2 tan 45
2 2
c
φ φ
⎛ ⎞ ⎛ ⎞
= ° + + ° +
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
and for specimen 2
( ) ( ) 2 1 1
434.4 50 tan 45 2 tan 45
2 2
c
φ φ
⎛ ⎞ ⎛ ⎞
= °+ + °+
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
Subtracting both equations ( ) ( ) 1
2 1
76.2 50 ta 45 1
n
2
2
φ
φ
⎛ ⎞
= ° + ∴
⎟
⎠
=
⎜
⎝
°
Substituting 1
φ into the equation,
( ) ( ) 2
2
12 12
510.6 100 tan 45 2 tan 45
2 2
510.6 15 14
2.5 2.47 5
kN
c
m
c
c
⎡ ⎤ ⎡ ⎤
⎛ ⎞ ⎛ ⎞
= °+ + °+
⎜ ⎟ ⎜ ⎟
⎢ ⎥ ⎢ ⎥
⎝ ⎠ ⎝ ⎠
⎣ ⎦ ⎣ ⎦
=
= + ∴

224
***Shear strength-09: Plots of the progressive failure in a shear-box.
(Revised Oct-09)
A soil test is performed in the shear-box shown below. The test data lists the stresses and
displacements. Assign positive normal stresses to compression and positive shear stresses are
counter-clockwise. Plot the Mohr circles of stress for each stage.
DISPLACEMENTS(mm)
Stage
x y
σ’ XX
(kpa)
σ’YY
(kpa)
τxy
(kpa)
τyx
(kpa)
a 0 0 30 70 0 0
b 0.30 -0.50 71 70 31.0 -31.3
c (peak) 2.50 -0.60 145 70 43.3 =49.0
d 3.00 -0.82 - - - -
e 10.00 1.50 90.6 70 24.5 -32.0
Solution:
For small displacements, the x and y planes remain perpendicular. Use a compass to locate by trial and
error the center of the Mohr circle. The center of the circle must lie on the σ’ axis, and it must be
equidistant from the two stress points( ) ( )
' , ' ,
xx xy yy yx
and
σ τ σ τ .
TABLE OF VALUES
STAGE
1 3
' '
'
2
S
σ σ
+
=
kPa
1 3
' '
2
σ σ
τ
−
=
kPa
'
S
τ Change in angle between
x + y plan (in degrees)
a 50 20 0.40 0˚
b 70.5 31.2 0.44 0˚
c (peak) 103 60 0.58 5.25˚
e 70 32 0.46 21˚

227
**Shear strength-10: Shear strength along a potential failure plane.
(Revision Oct-09)
An engineer is evaluating the stability of the slope in the figure below, and considers that the
potential for a shear failure occurs along the shear surface shown. The soil has an angle
°
= 30
'
φ and no cohesive strength. Compute the shear strength at point A along this surface when
the groundwater table is at level B, then compute the new shear strength if it rose to level C. The
unit weight of the soil is 120 lb/ft3
above the WT and 123 lb/ft3
below.
Solution:
When the groundwater table is at B:
2
2
3
3
2
3
4332
1248
)
20
(
123
)
26
(
120
'
1248
)
20
(
4
.
62
ft
lb
ft
lb
ft
ft
lb
ft
ft
lb
u
H
ft
lb
ft
ft
lb
z
u
z
w
z
=
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
−
=
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
=
∑γ
σ
γ
The potential shear surface is horizontal, so z
'
' σ
σ =
2 2
' tan ' 4332 tan 3 25 1
0 0
lb
s
f t
t
lb
f
σ φ
⎛ ⎞
= = ° =
⎜ ⎟
⎝ ⎠
When the groundwater table is at C:
2
2
3
3
2
3
3619
1997
)
32
(
123
)
14
(
120
'
1997
)
32
(
4
.
62
ft
lb
ft
lb
ft
ft
lb
ft
ft
lb
u
H
ft
lb
ft
ft
lb
z
u
z
w
z
=
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
−
=
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
=
∑γ
σ
γ
2 2
' tan ' 3619 tan 3 20 9
0 8
lb
s
f t
t
lb
f
σ φ
⎛ ⎞
= = ° =
⎜ ⎟
⎝ ⎠
Potential shearing surface
C▼
B▼
● A

228
***Shear strength-11: Use of the Mohr-Coulomb failure envelope.
(Revised Oct-09)
Samples have been obtained from both soil strata shown in the figure below. A series of shear
strength tests were then performed on both samples and plotted in diagrams below. The c’ and φ’
values obtained from these diagrams are shown in the figure below. Using this data, compute the
shear strength on the horizontal and vertical planes at points A, B, and C.
Solution:
Point A - Horizontal plane:
kPa
kPa
kPa
c
s
kPa
m
m
kN
m
m
kN
m
m
kN
u
H
z
6
.
41
28
tan
)
5
.
59
(
10
'
tan
'
'
8
.
59
'
)
1
.
1
)(
8
.
9
(
)
1
.
1
)(
5
.
17
(
)
0
.
3
)(
0
.
17
( 3
3
3
'
=
°
+
=
+
=
=
−
+
=
−
= ∑
φ
σ
σ
γ
σ
Point A - Vertical plane:
kPa
kPa
kPa
c
s
kPa
kPa
K z
z
1
.
27
28
tan
)
1
.
32
(
10
'
tan
'
'
1
.
32
)
5
.
59
)(
54
.
0
(
'
'
=
°
+
=
+
=
=
=
=
φ
σ
σ
σ

229
Using similar computations:
Point B vertical plane s = 57.2 kPa
Point B horizontal plane s = 35.5 kPa
Point C vertical plane s = 68.1 kPa
Point C horizontal plane s = 54.4 kPa
Commentary
At each point the shear strength on a vertical plane is less than that on a horizontal plane because
K < 1. In addition, the shear strength at point B is greater than that at point A, because the
effective strength is greater. The strength at point C is even higher than at point B because it is in
a new strata with different c’, φ’, and K values. Thus, the strength would increase gradually with
depth within each stratum, but change suddenly at the boundary between the two strata.
Draw the shear strength envelope for the ML stratum and then plot the upper half of the Mohr
circle for point A on this diagram. Assume the principal stresses act vertically and horizontally.
Failure envelope and Mohr’s circle

230
***Shear strength-11b: Use of the Mohr-Coulomb failure envelope.
(Revised Oct-09)
Samples have been obtained from both soil strata shown in the figure below. A series of shear
strength tests were then performed on both samples and plotted in diagrams below. The c’ and φ’
values obtained from these diagrams are shown in the figure below. Using this data, compute the
shear strength on the horizontal and vertical planes at points A, B, and C.
Solution:
Point A - Horizontal plane:
kPa
kPa
kPa
c
s
kPa
m
m
kN
m
m
kN
m
m
kN
u
H
z
6
.
41
28
tan
)
5
.
59
(
10
'
tan
'
'
8
.
59
'
)
1
.
1
)(
8
.
9
(
)
1
.
1
)(
5
.
17
(
)
0
.
3
)(
0
.
17
( 3
3
3
'
=
°
+
=
+
=
=
−
+
=
−
= ∑
φ
σ
σ
γ
σ
Point A - Vertical plane:
kPa
kPa
kPa
c
s
kPa
kPa
K z
z
1
.
27
28
tan
)
1
.
32
(
10
'
tan
'
'
1
.
32
)
5
.
59
)(
54
.
0
(
'
'
=
°
+
=
+
=
=
=
=
φ
σ
σ
σ

231
Using similar computations:
Point B vertical plane s = 57.2 kPa
Point B horizontal plane s = 35.5 kPa
Point C vertical plane s = 68.1 kPa
Point C horizontal plane s = 54.4 kPa
Commentary
At each point the shear strength on a vertical plane is less than that on a horizontal plane because
K < 1. In addition, the shear strength at point B is greater than that at point A, because the effective
strength is greater. The strength at point C is even higher than at point B because it is in a new strata
with different c’, φ’, and K values. Thus, the strength would increase gradually with depth within each
stratum, but change suddenly at the boundary between the two strata.
Draw the shear strength envelope for the ML stratum and then plot the upper half of the Mohr circle for
point A on this diagram. Assume the principal stresses act vertically and horizontally.
Failure envelope and Mohr’s circle

232
sandy gravel
n= 0.30
γ =16 kN
/m
3
clay γ =17.6 kN
/m
3
Impermeable rock stratum
H=4
H= 13m
H= 7m
σ1
σ3
**Shear strength-12: Triaxial un-drained tests.
(Revised Oct-09)
The test were taken with pure water pressure measurements and yield a c` = 20 kN
/m
3
, and φ=
24º
.Find (1) the clay shear strength at mid-stratum, and (2) the effective and total stresses at that
same level acting on a vertical face of a small element.
Solution:
For the gravel: γsat = γsat + nγw = [16 + (0.3)10] = 19 kN
/m
3
For the clay: σ1 = (4m)(16 kN
/m
3
)+ (9m)(19 kN
/m
3
)+ (3.5m)(17.6 kN
/m
3
)= 297 kN
/m
3
σ1
’
= σ1 − u = σ1 − γw (9 + 3.5) = 297 - 10 kN
/m
3
[12.5m] = 172 kN
/m
3
S = c’
+ σ1
’
tan φ = (20 KN
/m
3
+ 172 kN
/m
3
tan 24ο
) = 96.6 NM
/m
3
Since the clay is saturated, σ3
’
= k σ1
’
= 0.5(172) = 86 kN
/m
3
σ3 = σ3
’
+ u = 86 kN
/m
3
+ (86m + 86m) 10 kN
/m
3
= 251 kN
/m
3
τ
σ`
φ = 24ο
.
C` = 20 KN
/m
3
S= 96.6 kN
/m
3

233
**Shear strength-12b: Triaxial un-drained tests.
(Revised Oct-09)
The test were taken with pure water pressure measurements and yield a c` = 20 kN
/m
3
, and φ=
24º
.Find (1) the clay shear strength at mid-stratum, and (2) the effective and total stresses at that
same level acting on a vertical face of a small element.
Solution:
For the gravel: γsat = γsat + nγw = [16 + (0.3)10] = 19 kN
/m
3
For the clay: σ1 = (4m)(16 kN
/m
3
)+ (9m)(19 kN
/m
3
)+ (3.5m)(17.6 kN
/m
3
)= 297 kN
/m
3
σ1
’
= σ1 − u = σ1 − γw (9 + 3.5) = 297 - 10 kN
/m
3
[12.5m] = 172 kN
/m
3
S = c’
+ σ1
’
tan φ = (20 KN
/m
3
+ 172 kN
/m
3
tan 24ο
) = 96.6 NM
/m
3
Since the clay is saturated, σ3
’
= k σ1
’
= 0.5(172) = 86 kN
/m
3
σ3 = σ3
’
+ u = 86 kN
/m
3
+ (86m + 86m) 10 kN
/m
3
= 251 kN
/m
3
τ
σ`
φ = 24ο
.
C` = 20 KN
/m
3
S= 96.6 kN
/m
3
H=
H=
H=4
Clay γ =17.6 kN
/m
3
σ1
σ3
σ3
Sandy
gravel
n= 0.30
γ =16 kN
/m
3

234
**Shear strength-13: Determine the principal stresses of a sample.
(Revised Oct-09)
A clay layer, 20 feet thick is covered by a 40 foot sandy gravel stratum with a porosity of 30%, and
a dry unit weight of 103 pcf. Tests on the un-drained samples of the clay gave c = 2.9 psi, γSAT =
112 psf and φ' = 24o
. Find:
(a) the soil shear strength s = c + σ'tanφ' at the clay's midlevel (point A), and
(b) the effective and total stress acting on the vertical face of a soil element at the clay midlevel
(point A).
Solution:
(a) In order to find s, it is required to know the γsat of the sand.
n 0.30 0.30
e= = = = 0.429
1- n 1-0.30 0.70
and d
S
W
(1+e) 103(1+0.429)
= = = 2.36
G
62.4
γ
γ
S W
sat
( +e) (2.36 +0.429)62.4
G
= = = 122pcf
1+e 1.429
γ
γ
*Assume that the clay was normally consolidated to find σ' at midlevel in the clay (point A),
that is c = 0.
d=13'
H1=40'
H2=20' • A
γd = 103 pcf
n = 30%
c = 2.9 psi
γsat = 112 pcf
φ' = 24o

235
( )( ) ( )( ) ( )( )
( )( )
0.103 13 0.122 0.0624 27 0.122 0.0624 10 3.4
tan
tan
The pore water pressure is,
10 27 0.0624 2
A S S C
A
2
o
2 2
2 3
= + ’ + ’ ksf
h h h
’
and
s = c+ ’
’
lb kip 144 k
in
s = 2.9 x x +3.44 ( )= 0.42ksf +1.53ksfs = 1.95ksf
24
lb ft ft
in 10
u
u
σ γ γ γ
σ φ
⎡ ⎤ ⎡ ⎤
= + − + − =
⎣ ⎦ ⎣ ⎦
= + =
'
A
.3
Therefore, the toatl stre
5
ss is,
3 .
. 2.3 7
4
A k
s
u f
k f
s
σ σ
= + = + =
(b) To find the stress on the vertical face of the soil element at A, we find θ through a graphical solution
as follows,
o
o o o o
24
= + = + = _2 =
45 45 57 114
2 2
φ
θ θ
tan tan
1 o
c 0.42
= = = 0.939ksf
x
’ 24
φ
tan tan
2 o
s 1.95
= = = 4.38ksf
x
’ 24
φ
tan tan
o o
3 = s( )= 1.95( )= 0.868ksf
x 24 24

236
cos cos o
s 1.95
R = = = 2.13ksf
’ 24
φ
σΝ3 = x2 + x3 - x1 - R = 4.38 + 0.868 - 0.939 - 2.13= 2.18 ksf
σΝ1 = 2.18 + 2R = 2.18 + 2(2.13) = 6.44 ksf
σ3 = σΝ3 + u = 2.18 + 2.31 = 4.5 ksf

237
**Shear strength-13b: Determine the principal stresses of a sample.
(Revised Oct-09)
A clay layer, 20 feet thick is covered by a 40 foot sandy gravel stratum with a porosity of 30%, and
a dry unit weight of 103 pcf. Tests on the un-drained samples of the clay gave c = 2.9 psi, γSAT =
112 psf and φ' = 24o
. Find (1) the soil shear strength s = c + σ'tanφ' at the clay's midlevel (point A),
and (2) the effective and total stress acting on the vertical face of a soil element at the clay midlevel
(point A).
Solution:
In order to find s, it is required to know the γsat of the sand.
n 0.30 0.30
e= = = = 0.429
1- n 1-0.30 0.70
and d
S
W
(1+e) 103(1+0.429)
= = = 2.36
G
62.4
γ
γ
S W
sat
( +e) (2.36 +0.429)62.4
G
= = = 122pcf
1+e 1.429
γ
γ
* Assume that the clay was normally consolidated to find σ' at midlevel in the clay (point A), that is c =
0.
Clay
Rock
Sand
H= 20’
H=40’
D=13’
γd= 103 pct
n= 30%
γsat= 103 pct
c = 2.9 psi
φ' = 24o
• A

238
( )( ) ( )( ) ( )( )
( )( )
0.103 13 0.122 0.0624 27 0.122 0.0624 10 3.4
tan
tan
The pore water pressure is,
10 27 0.0624 2
A S S C
A
2
o
2 2
2 3
= + ’ + ’ ksf
h h h
’
and
s = c+ ’
’
lb kip 144 k
in
s = 2.9 x x +3.44 ( )= 0.42ksf +1.53ksfs = 1.95ksf
24
lb ft ft
in 10
u
u
σ γ γ γ
σ φ
⎡ ⎤ ⎡ ⎤
= + − + − =
⎣ ⎦ ⎣ ⎦
= + =
'
A
.3
Therefore, the toatl stre
5
ss is,
3 .
. 2.3 7
4
A k
s
u f
k f
s
σ σ
= + = + =
To find the stress on the vertical face of the soil element at A, we find θ through a graphical
solution as follows,
o
o o o o
24
= + = + = _2 =
45 45 57 114
2 2
φ
θ θ
tan tan
1 o
c 0.42
= = = 0.939ksf
x
’ 24
φ
tan tan
2 o
s 1.95
= = = 4.38ksf
x
’ 24
φ
tan tan
o o
3 = s( )= 1.95( )= 0.868ksf
x 24 24

239
cos cos o
s 1.95
R = = = 2.13ksf
’ 24
φ
σΝ3 = x2 + x3 - x1 - R = 4.38 + 0.868 - 0.939 - 2.13= 2.18 ksf
σΝ1 = 2.18 + 2R = 2.18 + 2(2.13) = 6.44 ksf
σ3 = σΝ3 + u = 2.18 + 2.31 = 4.5 ksf

240
**Shear strength-14: Formula to find the maximum principal stress.
(Revised Oct-09)
Derive the general formula that gives the value of the major principal stress σ1 as a function of the
minor principal stress σ3, the cohesion c and the angle of internal friction φ.
Solution:
From the figure,
(1)
fa = fo + oa = (c) cot φ +
2
3
1 σ
σ +
tan φ =
fo
c →
c
fo
= cot φ → fo = (c)(cot φ)
Using the properties of the Mohr circle,
oa = σ3 +
2
)
( 3
1 σ
σ −
→ oa =
2
)
( 3
1 σ
σ +
→ fa = (c) cot φ +
2
)
( 3
1 σ
σ +
(2)
Introducing (1) into (2): sin φ =
2
)
(
cot
)
(
2
)
(
3
1
3
1
σ
σ
φ
σ
σ
+
+
−
c
→ (sin φ)[(c) cot φ +
2
3
1 σ
σ +
] =
2
3
1 σ
σ −
sin φ =
fa
ad
2θ
c
ф
d
h
f
o a
τ3
σ3
τ1
σ1
θ = 45° +
2
φ
→ 2 θ = 90 + φ
ad =
2
3
1 σ
σ −
2
φ
45°

241
→[(c) sin φ cot φ] + [
2
3
1 σ
σ +
sin φ] =
2
3
1 σ
σ −
→ (c) sin φ cot φ = [
2
3
1 σ
σ −
] - [
2
3
1 σ
σ +
sin φ]
→ (2) [(c) sin φ cot φ] = [σ1 – (sin φ)( σ1)] – [σ3 + (sin φ)( σ3)]
→ (2) [(c) sin φ
φ
φ
sin
cos
] = σ1 [1 – (sin φ)] – σ3[1 + (sin φ)]
)
sin
1
(
)
sin
1
(
)
(
)
sin
1
(
cos
)
(
2
3
φ
φ
σ
φ
φ
−
+
+
−
c
= σ1
Since
φ
φ
sin
1
(
cos
−
≈ tan (45° +
2
φ
) and
)
sin
1
(
)
sin
1
(
φ
φ
−
+
≈ tan2
(45° +
2
φ
)
2
1 3 tan 45 2 tan 45
2 2
c
ϕ ϕ
σ σ
⎛ ⎞ ⎛ ⎞
= ° + + ° +
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠

242
Chapter 13
Slope Stability
Symbols for Slope Stability

243
*Slope-01: Factor of Safety of a straight line slope failure.
(Revision: Oct.-08)
A slope cut to 1.5H:1V will be made in a shale rock stratum that has bedding planes that have an
apparent dip of 16˚ (see the figure below). If the acceptable factor of safety against failure is at
least 2 along the lower-most bedding plane, is this slope stable? Use a unit weight of 20.1 kN/m3
,
and bedding strength parameters of c = 22 kPa and φ = 30˚.
Solution:
( )( ) 3
The traingule of rock above the potential slip plane has a weight per unit width,
1
85.0 11.3 20.1 9, 650
2
The length of the slip plane is,
85.0
88.4
cos16
Therefore,
Resisting For
W
kN kN
W m m
m m
L
m
L m
FS
⎛ ⎞
= =
⎜ ⎟
⎝ ⎠
= =
°
=
( )
( )
( )
2
cos tan
ces
Driving Forces sin
22 88.4 9, 650 cos16 tan 30
9, 650 sin 6
2
1
2.7
cL W
W
O
kN kN
m
m m
FS
kN
m
K
α φ
α
+ ⎡ ⎤
⎣ ⎦
= =
⎡ ⎤
⎛ ⎞ ⎛ ⎞
+ ° °
⎜ ⎟ ⎜ ⎟
⎢ ⎥
⎝ ⎠ ⎝ ⎠
⎣ ⎦
= =
⎛ ⎞
°
⎜
>
⎟
⎝ ⎠

244
*Slope-02: Same as Slope-01 but with a raising WT.
(Revision: Oct.-08)
In the previous problem the slope appeared to be stable with a factor of safety = 2.7. What
happens to that factor of safety if the water table rises to the level shown below? Use a unit weight
of 20.1 kN/m3
, and bedding strength parameters are reduced by the effective parameters of c’ = 15
kPa and φ’ = 20˚.
Solution:
T h e w e ig h t o f th e ro c k tria n g le p e r u n it w id th is s till 9 , 6 5 0
T h e le n g th o f th e s lip p la n e is s till 8 8 .4 .
T h e p o re w a te r p re s s u re is b a s e d o n a n e s tim a te o f its v a lu e a lo n g th e le n g th ,
a t w a te r
k N
W
m
L m
L
( )
( )
( )
( )
3
2
d e p th a b o v e th e p la n e th a t ra n g e fro m 0 to 3 .2 m ; c o n s e rv a tiv e ly ,
9 .8 1 3 .2 3 1 .4
' c o s ta n
R e s is tin g F o rc e s
D riv in g F o rc e s s in
1 5 8 8 .4 9 , 6 5 0 c o s 1
w
w w
z
k N
u z m k P a
m
c L W u
F S
W
k N k N
m
m m
F S
γ
α φ
α
⎛ ⎞
= = =
⎜ ⎟
⎝ ⎠
+ −
⎡ ⎤
⎣ ⎦
= = =
⎛ ⎞ ⎛ ⎞
+
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
=
( )
6 3 1 .4 ta n 2 0
9 , 6 5 0 s in 1 6
T h e c o m p u te d fa c to r o f s a fe ty o f 1 .7 6 is le s s th a n th e m in im u m a c c e p ta b le
v a lu e o f 2 , th e re fo re th is d e s ig n is N O T a c c e p ta b le .
1 .7 6 2
s in
k P a
k N
m
N G
N o tic e th a t a r i g W
⎡ ⎤
°
<
− °
⎢ ⎥
⎣ ⎦ =
⎛ ⎞
°
⎜ ⎟
⎝ ⎠
.
T
d e c r e a s e s th e s ta b ility o f th e s lo p e

245
*Slope-03: Is a river embankment safe with a large crane?
(Revision: Oct.-08)
Determine if the work site shown below is safe, provided you consider the minimum acceptable
factor of safety for the man-made waterfront slope shown below to be 2. Assume the arc radius is
80 feet; the circular lengths are AB = 22 feet and BC = 102 feet. The total weight of the soil per
unit width are Wsoil = 205 kips and Wcrane =70 kips. The site is located in a seismic zone with a
seismic coefficient of 0.15.
Solution:
Mr = R[S1(AB)+S2(BC)] = R[(C1’+σ1’tanφ1)AB + (C2’+σ2’tanφ2)BC]
= 80’[(0.2+0.125(8’)(tan40)22’ + (1.8+(0.130 – 0.064)(21)(tan15)(102)] = 80[23k + 221k] = 19,500 k-ft
g
)
(d
W
Vb
b
W
-
)
(d
W
-
Wb
M 2
ae
WV
1
wH
1
0 +
+
= 3
2 )
(
= 205(40)–(1/2)(0.064)(15)² - (0.064)[30(15)+(1/2)(40)(15)]15+70(55)+205(0.15)(50)
Mo = 8,200 - 7.2 –720 +3850 +1540 = 12,900 k-ft
Therefore: FS = Mr/Mo = 19,500/12,900 = 1.51 ÎNot Good!
Removing Crane Î Mo =9,050 k-ft
Therefore: FS = Mr/Mo = 19,500/9050 = 2.15 Î GOOD!
Clayey sand
c = 0.2 ksf
φ
Sandy clay
c = 1.8 ksf
φ
20’ 15’ 40’ 15’ 15’
15’
45’
A
B
C
SEA
Wsoil
Wcrane

246
*Slope-04: Simple method of slices to find the FS.
(Revision: Oct.-08)
The stability of a slope was analyzed by the method of slices. One of the trial curved surfaces
through the soil mass yielded the shearing and normal components of each slice as listed below.
The curved length of the trial curved surface is 40 feet, the soil parameters are c = 225 lb/ft2
and φ
= 15º. Determine the factor of safety along this trial surface.
Solution:
Slice
Number
Shearing Component
(W sin α) (lb/ft)
Normal Component
(W cos α) (lb/ft)
1 -38 306
2 -74 1410
3 124 2380
4 429 3050
5 934 3480
6 1570 3540
7 2000 3210
8 2040 2190
9 766 600
∑ = 7,751 lb/ft ∑ = 20,166 lb/ft
( cos ) tan (225 )(40
1.86
) 20,166
sin 7,75
2
1
cL W n psf ft
N
plf
FS
l
G
W p f
α φ
α
+ +
= = <
=
∑
∑

247
**Slope-05: Method of slices to find the factor of safety of a slope with a WT.
(Revision: Oct.-08)
A 30 ft tall, 1.5H:1V slope is to be built as shown below. The soil is homogeneous, with 2
' 400
lb
c
ft
=
and ' 29
φ = °. The unit weight is 119 pcf above the groundwater table, and 123 pcf below. Using
the ordinary method of slices, compute the factor of safety along the trial circle.
Solution:
Weights:
1
2
3
4
10.3
10.8 119 6, 620
2
10.3 12.5 5.2
9.4 119 9.4 123 15, 800
2 2
12.5 14.6 5.2 10.0
12.1 119 12.1 123 30, 800
2 2
5.0 12.9 8.0
2.9 17.0 7.1 17.8 1620
2 2
W lb
b ft
W lb
b ft
W lb
b ft
W lb
b ft
⎛ ⎞
= =
⎜ ⎟
⎝ ⎠
+
⎛ ⎞ ⎛ ⎞
= + =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
+ +
⎛ ⎞ ⎛ ⎞
= + =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
+
⎛ ⎞ ⎛ ⎞
= + =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
5
6
7
16.8 12.8 10.7 7.3
9.3 119 9.3 123 39, 900
2 2
12.8 9.9 7.3
7.6 119 7.6 123 26, 700
2 2
9.9
4.0 119 2, 400
2
W lb
b ft
W lb
b ft
W lb
b ft
+ +
⎛ ⎞ ⎛ ⎞
= + =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
+
⎛ ⎞ ⎛ ⎞
= + =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
⎛ ⎞
= =
⎜ ⎟
⎝ ⎠

248
Average pore water pressure at base of each slice:
1
2 2
0
5 .2
6 2 .4 1 6 0
2
u
lb
u
ft
=
⎛ ⎞
= =
⎜ ⎟
⎝ ⎠
3 2
4 2
5 2
6 2
7
5 .2 1 0 .0
6 2 .4 4 7 0
2
1 0 .0 1 0 .7
6 2 .4 6 5 0
2
1 0 .7 7 .3
6 2 .4 5 6 0
2
7 .3
6 2 .4 2 3 0
2
0
lb
u
ft
lb
u
ft
lb
u
ft
lb
u
ft
u
+
⎛ ⎞
= =
⎜ ⎟
⎝ ⎠
+
⎛ ⎞
= =
⎜ ⎟
⎝ ⎠
+
⎛ ⎞
= =
⎜ ⎟
⎝ ⎠
⎛ ⎞
= =
⎜ ⎟
⎝ ⎠
=
Slice ( )
W
lb
b ( )
Deg
α
2
'
lb
c
ft
⎛ ⎞
⎜ ⎟
⎝ ⎠ ( )
Deg
φ
2
lb
u
ft
⎛ ⎞
⎜ ⎟
⎝ ⎠
( )
l ft
( )
'
cos tan '
c l
W
ul
b
α φ
+
⎛ ⎞
−
⎜ ⎟
⎝ ⎠
( )
W
lb
b
1 6620 -18 400 29 0 11.4 8,000 -2,000
2 15,800 -7 400 29 160 9.5 11,700 -1,900
3 30,800 8 400 29 470 12.2 18,600 4,300
4 39,900 24 400 29 650 13.9 20,800 16,200
5 26,700 38 400 29 560 11.8 12,700 16,400
6 13,700 53 400 29 230 12.6 8,000 10,900
7 24,00 67 400 29 0 10.2 4,600 2,200
84,400
Σ = 46,100
Σ =
Therefore, the factor of safety is,
( )
' cos tan '
84,400
46
1
,100
.83 2
W
c l ul
b
FS
W
b
NG
α φ
⎛ ⎞
+ −
⎜ ⎟
⎝ ⎠
= = =
⎛ ⎞
⎜
<
⎟
⎝ ⎠
Note how slices # 1 and 2 have a negative α because they are inclined backwards.

249
**Slope-06: Swedish slip circle solution of a slope stability.
(Revision: Oct.-08)
Using the Swedish slip circle method, compute the factor of safety along the trial circle shown in
the figure below.
Solution:
Divide the slide mass into vertical slices as shown. One of the slice borders should be directly below the
center of the circle (in this case, the border between slices 2 and 3). For convenience of computations,
also draw a slice border wherever the slip surface intersects a new soil stratum and whenever the ground
surface has a break in slope. Then, compute the weight and moment arm for each slide using simplified
computations as follows:

250
Solution:
Weights
:
1
2
3
4
5
2.0
4.6 17.8 80
2
2.0 9.8
7.0 17.8 130
2
9.8 12.9
2.9 17.8 590
2
5.0 12.9 8.0
2.9 17.0 7.1 17.8 1620
2 2
5.0 10.3 8.0
7.2 17.0 7.2 17.8 1450
2 2
6
W kN
b m
W kN
b m
W kN
b m
W kN
b m
W kN
b m
W
b
⎛ ⎞
= =
⎜ ⎟
⎝ ⎠
+
⎛ ⎞
= =
⎜ ⎟
⎝ ⎠
+
⎛ ⎞
= =
⎜ ⎟
⎝ ⎠
+
⎛ ⎞ ⎛ ⎞
= + =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
+
⎛ ⎞ ⎛ ⎞
= + =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
=
7
10.3 9.8
0.8 17.0 140
2
9.8
5.1 17.0 420
2
kN
m
W kN
b m
+
⎛ ⎞
=
⎜ ⎟
⎝ ⎠
⎛ ⎞
= =
⎜ ⎟
⎝ ⎠
Moment arms:
1
2
3
4
5
6
7
4.6
7.0 8.5
3
7.0
3.5
2
2.9
1.5
2
7.1
2.9 6.5
2
7.1
2.9 7.1 10.9
2
0.8
2.9 7.1 7.2 17.6
2
5.1
2.9 7.1 7.2 0.8 19.7
3
d m
d m
d m
d m
d m
d m
d m
= − − = −
−
= = −
= =
= + =
= + + =
= + + + =
= + + + + =

251
Slice ( )
u
S kPa ( )
Deg
θ u
S θ W kN
b m
⎛ ⎞
⎜ ⎟
⎝ ⎠
( )
d m W
d
b
⎛ ⎞
⎜ ⎟
⎝ ⎠
1 80 -8.5 -690
2 130 -3.5 -450
3 80 76 6080 390 1.5 890
4 1620 6.5 10,530
5 1450 10.9 15,800
6 140 17.6 2,460
7
40 30 1200
420 19.7 8,280
7280
Σ = 36,830
Σ =
( )
2
2
23.6 7, 280
180 180 36,8
2
0
1.9
3
2
u
S
R
FS
W
Not Goo
d
d
b
θ π
π
= = =
⎛ ⎞
⎜ ⎟
⎝ ⎠
<
∑
∑

252
Chapter 14
Statistical Analysis of Soils
Symbols for the Statistical Analysis of Soils

253
Chapter 15
Lateral Pressures from Soils
Symbols for Lateral Pressures from Soils
Dx → Diameter of the grains distributed (represent % finer by weight).
e →The voids ratio.
H→ Maximum depth of excavation or thickness of a soil layer.
hsoil→ depth of the soil.
icritical→ Critical hydraulic gradient.
kH → Horizontal permeability.
kV → Vertical permeability.
u → pore water pressure.
σ’→ Effective stress.
σV’→ Vertical Effective stress.
g’→ Bouyant unit weight of a soil.
γSAT →Saturated unit weight of a soil
γW →Unit weight of water.
w→ water content.

254
Formulas and Figures for Lateral Stresses.
Figure for symbols used in the Coulomb earth pressures.
Coulomb’s lateral pressure coefficients Ka and Kp.
2
2
2
2
2
2
cos ( - θ)
sin( )sin( )
cos cos( ) 1
cos( )cos( )
cos ( θ)
sin( )sin( )
cos cos( ) 1
cos( )cos( )
a
p
K
K
φ
δ φ φ α
θ δ θ
δ θ θ α
φ
φ δ φ α
θ δ θ
δ θ α θ
=
⎡ ⎤
+ −
+ +
⎢ ⎥
+ −
⎣ ⎦
+
=
⎡ ⎤
− +
− −
⎢ ⎥
− −
⎣ ⎦

255
Ka for the case of θ = 0º and α = 0º.
Ka for the case where δ = 2/3 φ.

256
Kp for θ = 0º and α = 0º.
Failure modes for flexible walls (sheet-piling).

257
*Lateral-01: A simple wall subjected to an active pressure condition.
Consider a small 10-foot tall and 3 feet thick concrete retaining wall. The backfill behind the wall
will be from local sandy gravel with a dry unit weight of 115 pcf and an angle of internal friction
of 30 degrees. The wall will not have to retain water.
Estimate, (a) the lateral force on the wall from the backfill in an active pressure condition, (b) its
stability against overturning, and (c) its stability against sliding (use a Factor of Safety ≥ 2).
Solution:
( )( )( )
2 2 30
(a) The Rankine earth pressure coefficient is, tan 45 tan 45 0.33
2 2
The lateral pressure at the bottom of the wall is 0.115 10 0.33 0.38
The force against the
°
⎛ ⎞ ⎛ ⎞
= ° − = ° − =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
= = =
a
a a
active K
p hK kcf ft ksf
φ
γ
( ) ( )( )( )
( )
1
wall is 0.5 0.38 10
2
(b) The stability of the wall against overturning is found by taking moments about the point "O"
at the toe of the wa
9
ll,
1.
= = =
=
a a
F p h
resisting moment
Fa
kips per foot of wal
ctor of Saf FS
l
ety
o
( )( )( )( )( )
( )( )
( )
( )( )( )( )( )
3' 10' 1' 0.150 1.5
1.9 10/3
(c) The stability of the wall against sliding towards the left is found by,
1.
3' 10'
07 2
1' 0.150 tan30
1
= <
=
°
= =
kcf ft
verturning moment kips ft
kcf
resisting force
Factor of Safety FS
driving force
NG
( )
1.
.9
37 2
= <
kips
NG

258
*Lateral–02: Compare the Rankine and Coulomb lateral coefficients.
(Revision: Sept-2008)
(a) Compare the Rankine and Coulomb lateral earth pressure coefficients for a wall that retains a
granular backfill soil with φ = 35°, δ = 12°, θ = 0º and α = 20°. (Note: δ is the angle of friction
between the soil and the backside of the wall; α is the angle of the slope for the backfill behind
the wall and θ is the back of the wall’s angle with respect to the vertical).
(b) What is the passive earth force on the wall at failure if the wall is 10 m high, γ = 18.1 kN/m3
and c = 9 kN/m2
?
Solution:
(a) Rankine’s active and passive earth pressure coefficients,
2 2
2 2
35
tan (45 - ) tan (45 - )
2 2
35 1
tan (45 ) tan (45 ) Note
0.27
that
2
1
3.690
2
°
= ° = ° =
°
= °+ = °+ = =
a
p a
P
K
K K
K
φ
φ
Coulomb’s active and passive earth pressure coefficients,
2 2
2 2
2 2
2
2
cos ( - θ) cos (35 - 0)
sin( )sin( ) sin(12 35)sin(35 20)
cos cos( ) 1 cos 0 cos(12 0) 1
cos( )cos( ) co
0.3
s(12 0)cos(0 20)
cos ( θ)
sin( )sin
23
( )
cos cos( ) 1
cos(
a
p
K
K
φ
δ φ φ α
θ δ θ
δ θ θ α
φ
φ δ φ α
θ δ θ
= = =
⎡ ⎤ ⎡ ⎤
+ − + −
+ + + +
⎢ ⎥ ⎢ ⎥
+ − + −
⎣ ⎦ ⎣ ⎦
+
=
− +
− −
2
2 2
2
cos (35 0)
sin(35 12)sin(35
3
20)
cos 0 cos(12 0) 1
)cos( ) cos(12 0)cos(20 0)
.517
δ θ α θ
+
= =
⎡ ⎤ ⎡ ⎤
− +
− −
⎢ ⎥ ⎢ ⎥
− − − −
⎣ ⎦ ⎣ ⎦
When α = 0º, θ = 0º and δ = 0º the Coulomb formula becomes identical to Rankine’s.
(b) Therefore, the Rankine coefficient is 3.690 versus 3.517 for Coulomb’s. Using these values, the total
passive force Fp on the wall per unit length is,
( )( )( ) ( ) ( )( )
( )( )( ) ( ) ( )( )
2
2
2
2
2
2
' 0.5 2 0.5 18.1 10 3.690 2 9 10 3. 3,685 /
3,520
690
' 0.5 2 0.5 18.1 10 3.517 2 9 1 /
0 3.517
p p p
p p p
Rankine s F h K ch K
Coulomb s F h K ch K
kN m
kN m
γ
γ
= + = + =
= + = + =

259
*Lateral-03: Passive pressures using the Rankine theory.
(Revision: Sept-08)
Using the Rankine method, find the magnitude and location of the passive pressure force Fp with
respect to the heel of the wall (point B), exerted upon a temporary retaining wall by a large
jacking system (which is not shown in the figure).
Solution:

260
*Lateral-04: The “at-rest” pressure upon an unyielding wall.
(Revision: Sept-08)
Find the lateral “at-rest” force o
F on the wall and its location with respect to the top of the wall.
Given: Sand #1 has a unit weight of 105 pcf, c = 0 psf and φ = 30º; Sand #2 has a unit weight of 122
pcf, c = 0 psf and φ = 30º.
Solution:
( )( )( )
From Jaky's empirical relation, 1 sin ' 1 sin30 0.50
at = 0 feet ' 0 , because there is no surcharge loading upon the surface of Sand #1.
at = 10 feet ' ' 0.5 0.105 10 0.525
o
h o v
K
z ksf
z K kcf ft
φ
σ
σ σ
= − = − ° =
=
= = =
( ) ( )( ) ( )
( )( )
( )( ) ( )( ) ( )( ) ( )( )
1 2 3 4
at = 20 feet ' 0.5 0.105 10 0.122 0.0624 10 0.823
0.0624 10 0.624
1 1 1
0.525 10 0.525 10 0.302 10 0.624 10
2 2 2
2.63 5.25 12.
1 3. 2 5
.49 1
h
w w
o i
i
o
ksf
z ksf
h pcf ft ks
k
f
F f
i
F F F F
F
σ
σ γ
= + − =
⎡ ⎤
⎣ ⎦
= = =
= = + + + = + + +
= + + + =
∑
( )( ) ( )( ) ( )( ) ( )( )
2.63 6.67 5.25 15 1.49 16.67 3.12 16.67 173
1
.1
12.5 12.5
3 .
.8
p
ft
z ft from the top of
kip ft
z
kip kip
the wall
+ + +
=
−
= =

261
*Lateral-05: The contribution of cohesion to reduce the force on the wall.
(Revision: Sept-08)
A 21 foot high retaining wall supports a purely cohesive soil (φ = 0°) with a cohesion of 630 psf and
a unit weight of 113 pcf. Find:
(a) The Rankine active earth pressure on the wall.
(b) Estimate the depth of separation of the clay from the wall, and (c) find the lateral force upon
the wall whilst considering the clay separation.
Solution:
( )( )( ) ( )
2 2 2
3
a) The coefficient of active earth pressure is,
0
tan 45 tan 45 tan 45 1
2 2
The net active earth pressure on the wall is,
2
0.113 21 1 2 0.630 1 2.37 1.26 1.11
°
⎛ ⎞ ⎛ ⎞
= °− = °− = ° =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
= = −
= − = − =
a
a
a a a
K
p
p hK c K
kcf ft ksf ksf
φ
σ γ
( )
( )( )
2
b) The crack stops where the pressure is zero, = 0,
- 2 2
2 2 0.630
2
11.2
0.113 1
c) The total (Rankine) active earth force upn the wall is,
1
2
2
= ∴ =
∴ = = = =
= −
a
a a a a a
a
crack
a a
a
a a
p
p hK c K hK c K
c K ksf
c
h feet
K kcf
K
F
F H K
γ γ
γ γ
γ
( )
( )( ) ( ) ( )( )( )
( )
( )
2
2
2
2
but there is no contact on the wall where the tension crack exists, therefore
1 2 1 2
2 2
2 2
2 0.63
1
0.113 21 1 2 5.48
0.63 21 1
2 0.113
/
⎛ ⎞
= − − = − +
⎜ ⎟
⎜ ⎟
⎝ ⎠
= − + =
a
a a a a
a
a
cH K
c c
F HK c K H H cH K
K
ksf
F kcf ft ksf ft
kcf
k ft of wall
γ γ
γ
γ
21 ft
a
zK
γ - 2 a
c K

262
**Lateral-06: The effect of a rising WT upon a wall’s stability.
(Revision: Sept-08)
A 4 m wall retains a dry sand backfill with a unit weight of 18.3 kN/m3
, an angle of internal
friction of 36˚ and a porosity of 31%. The backfill is fully drained through weep holes.
1) What is the magnitude of the backfill force on a 1 m wide slice of wall if it is not allowed to
deflect?
2) What is the magnitude of the backfill force on the same 1 m wide slice, if the wall does deflect
enough to develop a Rankine active earth pressure condition?
3) What is the new force on the wall, and its location from its heel, if the wall’s weep holes are
clogged and the water table now rises to within 1 m of the ground surface behind the wall?
Solution:
( )
0
2
2
3
1) No deflection of the wall means the soil is "at rest" and 1 - sin 1 - sin 36 0.41
½ ½ 18.3 4 (0.41)
2) When th
60
o d o
K
kN
The forc kN per meter of wall
e F h K m
m
φ
γ
= = ° =
⎛ ⎞
= = =
⎜ ⎟
⎝ ⎠
( )
2 2
2
2
3
e wall deflects to the left sufficiently to develop an active pressure condition,
36
tan 45 tan 45 0.26
2 2
½ ½ 18.3 4 (0.26)
3) The buoya
38
nt
a
a d a
K
kN
The force F h K m kN per meter
m
of wall
φ
γ
°
⎛ ⎞ ⎛ ⎞
= ° − = ° − =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
⎛ ⎞
= = =
⎜ ⎟
⎝ ⎠
( ) ( ) 3
a
3
weight ' of the flooded sand is,
' 18.3 (0.31) 9.81 9.81 11.5
The stress at point "a" is =0, and at "b" which is 1 meter below the surface,
(18.3 )(1 )(0.26) 4.8
sat w d w w
b d a
kN
n
m
kN
hK m
m
γ
γ γ γ γ γ γ
σ
σ γ
= − = − − = + − =
= = = 2 2
1
/ ½(4.8 / )(1 ) 2.4 /
kN m F kN m m kN m
∴ = =

263
2 2
2
3
2 2
3
3
(18.3 )(1 )(0.26) 4.8 / (4.8 / )(3 ) 14.4 /
' (11.5 )(3 )(0.26) 9.0 / ½(9.0 / )(3 ) 13.5 /
The water pressure and force,
bc d a
c a
w w
kN
hK m kN m F kN m m kN m
m
kN
hK m kN m F kN m m kN m
m
h
σ γ
σ γ
σ γ
= = = ∴ = =
= = = ∴ = =
= 2 2
4
3
4
1
(9.81 )(3 ) 29.4 / ½(29.4 / )(3 ) 44.1 /
Therefore 74.4 /
The location of the resulta
i
kN
m kN m F kN m m kN
N
m
R k
m
F m
=
= = ∴ = =
= =
∑
( )( ) ( )( ) ( )( ) ( )( )
( )
1 1 2 2 3 3 4 4
1.17 from the bottom of the wal
nt is ,
2.4 3.33 14.4 1.5 13.5 1 44.1 1
74.4
The percent increase in load upon the wall due to flooding is,
74.4 3
l.
96%
8
38
.
y m
increa
y
m m m
Fd F d F d F d
y
R
kN kN
F
kN
se
+ + +
+ + +
= =
−
Δ = =
=

264
*Lateral-07: The effects of soil-wall friction upon the lateral pressure.
(Revision: Sept-08)
A 7.0 m high retaining wall has a horizontal backfill of dry sand with a unit weight of 17.2 kN/m3
and an angle of internal friction φ = 32˚. The wall is cast-in-place concrete, with a friction angle δ
= 20˚. Ignoring the effect of the passive pressure upon the toe of the footing, find the magnitude of
the active earth force upon a length of wall equal to 3.5 m assuming Rankine conditions.
Solution:
2 2
2 3
The force applied to the wall first requires the coefficient of active earth pressure,
32
tan (45 ) tan (45 ) 0.307
2 2
The horizontal force per unit width of wall is,
½ ½(17.2 / )(7 )
a
H
H a
K
F
F h K kN m m
φ
γ
°
= °− = ° − =
= = 2
(0.307) 129.5 /
The is related to the total force R on the wall as a function of the angle of wall friction ,
129.5 /
cos 138 /
cos cos20
We are asked what is the total force ever
H
H
H
kN m
F
F kN m
F R R kN m
δ
δ
δ
=
= ∴ = = =
°
( )
y 3.5 m,
3.5 138 ( 4
3. 8
5 ) 2
Total Active Force every kN
m m
= =
FV
R
δ
FH

265
*Lateral-08: What happens when the lower stratum is stronger?
(Revision: Sept-08)
Calculate the active force Fa and its location ŷ with respect to the heel of the 6 m wall (point A), for
the worst case (clogged weep holes).
2 2
li
.
T h e w o r s t a c t i v e p r e s s u r e lo a d o c c u r s w h e n t h e w a t e r t a b le r a i s e s t o t h e t o p o f t h e w a ll.
3 0
t a n ( 4 5 ) t a n ( 4 5 ) 0 .3 3 3
2 2
a s a n d
a
S o l u t i o n
K
K
ϕ °
= ° − = ° − =
( )
2 2
m e s to n e
1 1 1
9 0
t a n ( 4 5 ) t a n ( 4 5 ) 0 t h e li m e s t o n e d o e s n o t lo a d t h e w a ll.
2 2
T h e = 9 0 i s r e a lly a c o m b i n a t i o n o f s h e a r a n d c o h e s i o n ( " c e m e n t a t i o n " ) .
' ( - ) (1 8 .5 - 9 .8 ) 3 ( 0 .3 3 ) 8 .7
a S A T w a
p h K h K k
ϕ
φ
γ γ γ
°
= ° − = ° − = ∴
°
= = = = 2
2
2
1 1 1
2 2
/
( 9 .8 ) ( 6 ) 5 8 .8 /
½ ( 0 .5 ) (8 .7 ) ( 3 ) 1 3 .1 /
½ ( 0 .5 ) ( 5 8 .8 ) ( 6 ) 1 7 6 .4 /
w
t o ta l
N m
p H m k N m
F p h m k N m
F p H m k N m
F
γ
= = =
= = =
= = =
( )( ) ( )( )
( )
1 1 2 2
= 1 8 9 .5 /
4 1 3 .1 2 1 7 6 .4
T h e lo c a t i o n 2 .1 f r o m A .
1 8 9 .5
to t a l
k N m
m m
y F y F
y m
F
+
+
= = =
H = 6 m
0
Medium dense sand
Weep holes
P li
worst load case
γsat = 18.5 kN/m3
φ = 30°
γ = 21.2 kN/m3
φ = 90°
3 m
3 m
A
WT
1m
+
ŷ
F2
F1
H
h1
h2
p1
p2
Ftotal
A

266
*Lateral-09: Strata with different parameters.
(Revised Oct-09)
Draw the pressure diagram on the wall in an active pressure condition, and find the resultant Ftotal
on the wall and its location with respect to the top of the wall.
Solution:
Step 1
Ka1 = tan2 (45°- 30°/2) = 0.333
Ka 2 = tan2 (45°- 40°/2) = 0.217
Step 2
The stress on the wall at point a is: pa = q Ka 1 = (2.5) (0.333) = 0.83 ksf
The stress at b (within the top stratum) is: pb’+ = (q + γ’h) Ka 1
= [2.5 + (0.115 - 0.0624) (10’)] [0.333] = 1.01 ksf
The stress at b (within bottom stratum) is: pb’ - = (q + γ’h) Ka 2
= [2.5 + (0.115 – 0.0624) (10’)] [0.217] = 0.66 ksf
The stress at point c is: pc’ = [q + (γ’h)1 + (γ’h)2] Ka 2
= [2.5 + (0.115 – 0.0624) (10’) + (0.125 – 0.0624)(10’)] [0.217] = 0.79 ksf
The pressure of the water upon the wall is: pw = γwh = (0.0624) (20’) = 1.25 ksf
Step 3
H = 20’
w.t.
q = 2.5 ksf
10’
c = 0
γ = 125 pcf
ф = 40°
+
0.83
0.83
1
2
4
0.66
3
5
1.25
a
b
c
0.13
c = 0
γ = 115 pcf
ф = 30°
10’
0.66 0.18

267
The forces from each area:
F1 = (10’) (0.83) = 8.30 kips/ft
F2 = ½ (10’)(0.18) = 0.90 kips/ft
F3 = (10’) (0.66) = 6.60 kips/ft
F4 = ½ (10’)(0.13) = 0.65 kips/ft
F5 = ½ (1.25) (20’) = 12.5 kips/ft
Ftotal = 29.0 kips/ft
Step 4
The location of forces ŷ is at:
ksf
0.66
29
12.5
3
40
0.65
3
50
6.6
15
0.9
3
20
8.3
5
y =
⋅
+
⋅
+
⋅
+
⋅
+
⋅
=
ˆ
The stress at point c is: ŷ = 11.2 feet from top of wall

268
*Lateral-10: The effects of a clay stratum at the surface.
The sheet pile wall shown below is flexible enough to permit the retained soil to develop an active
earth pressure condition. Calculate the magnitude of the resultant Ftotal of the active force above
the point “A” upon the wall. Assume Rankine conditions.
Solution:
Notice that the vertical pressure diagram will always increase in magnitude, but the horizontal pressures
are governed by the Ka coefficient, which may increase or decrease the pressures on the wall.
Lateral load from the surcharge
σc+ = Ka1 q = (0.70)(0.84 kcf) = 0.59 ksf
σc- = -2c 1
a
K = -2(0.5) 70
.
0 = -0.84 ksf
∴∑ σc = 0.59 - 0.84 = -0.25 ksf
σb+ = Ka 1 γ h – 2c 1
a
K + q Ka
= (0.7) (0.11) (20’) – (2) (0.50) 70
.
0
= 1.29 ksf
10’
20’
Surcharge q = 0.84 ksf
no water present
Sandy clay
c = 500 psf
φ = 10 °
Dense sand
c = 0
φ = 40°
γ = 130 pcf
3.25’
16.75’
-0.84
-0.25
4
2
1
3
+0.48
+1.29
+0 48 +0 77
A a
0
b +
b -
c +

269
σb- = Ka 2 γ h – 0 = (0.22) (0.11) (20’) = 0.48 ksf
σa = 0.48 + Ka 2 γ h = 0.48 + (0.22)(0.13)(10’) = 0.48 + 0.29 = 0.77 ksf
Ka 1 = tan2
(45° - φ /2) = tan2
40° = 0.70 F1 = ½ (-0.25)(3.25’) = - 0.41 k/ft (tension).
Ka 2 = tan2
(45° - 40° / 2) = tan2
25° = 0.22 F2 = ½ (1.29)(16.75’) = +10.80 k/ft
F3 = (0.48)(10’) = + 4.80 k/ft
F4 = ½ (0.29)(10’) = + 1.45 k/ft
Ftotal = +16.6 kip/ft

270
**Lateral-11: Anchoring to help support a wall.
The wall shown below will be used to retain the sides of an excavation for the foundations of a
large building. The engineer has decided to use earth anchors in lieu of braces or rakers to
stabilize the wall.
(1) What is the minimum distance x from the anchor to behind the wall?
(2) What is your recommended factor of safety for the anchor? What is an economical load for
the anchor?
x
Solution:
(1) The anchor must be beyond the passive slip plane, or (x) tan 30º = 19’ or x = 33 feet.
(2) Ka = tan2
(45º - φ/2) = 0.33 and Kp = tan2
(45º + φ/2) = 3.0
The active force upon the wall per unit width Fa is:
Fa = ½γH2
Ka -2cH a
K = ½(0.105)(24’)2(0.33) - 2(0.15)(24) 33
.
0 = 5.84 kip/ft with the force
located at ŷ = ⅓(19’) = 6.33’ above point O (note that the tensile portion does not load the wall).
The potential passive failure force (from the anchor) on the wall Fp is:
Fp = ½γH2
Kp + 2cH p
K = ½ (0.105)(24)2
(3) + 2(0.15)(24) 3 = 103 kip/ft
24’
5’
φ = 30°
c = 150 psf
O
Grouted anchor A

271
The factor of safety should be the same for an active failure as a passive failure. Therefore, a simple
equation could be written as, ( )
( )
p
a
F
F FS
FS
= or (FS)2
=
103
17.6 4.2
5.84
p
a
F kips
FS
F kips
= = ∴ =
Note that this corresponds to a load in the anchor of (5.84)(4.2) = 24.5 kips/ft (which is the same as
using the passive force = (103)/(4.2) = 24.5 kips/ft). The horizontal spacing of the anchors is not
influenced by this analysis, and depends on cost factors. A common spacing would be 10 feet, which
means A = 245 kips.

272
**Lateral-12: The effect of five strata have upon a wall.
(Revision Oct-09)
Plot the pressure diagram and find the resultant force F and its location under an active pressure
condition.
At h=0’ p1 = q K1a = (2) (0.307) = 0.614 ksf
at h = -6’ Δp2 = γ1h K1a = (0.110)(6) (0.307) = 0.203 ksf
at h = -8’ Δp3 = (γ2 - γw)h K2a = (0.125 - 0.0624)(2)(0.333) = 0.417 ksf
at h = -(8+dh)’ = [q + (γ1) 6’ + (γ2 - γw) 2’] K3a – 2c √(K3a from p = γh Ka - 2c√Ka
= [2 + (0.11)6’ + (0.125 – 0.0624)2’](0.704) – 2(0.6)(0.84) = 0.95 ksf
at h= -17’ Δp4 = (γ3 - γw)h K3a = (0.126-0.0624)(9)(0.704) = 0.403 ksf ∴0.95+0.403 = 1.35 ksf
at h = -(17 + dh)’ = [2 + 0.66 + 0.125 + (0.0626) (9)](1) –2(0.8)(1) = 1.76 ksf
at h = -25’ Δp5 = (γ4 - γw)h K4a = (0.120 - 0.0624)(8)(1) = 0.46 ksf ∴ 1.76 + 0.46 = 2.22 ksf
at h = -(25 + dh)’ = [2 + 0.66 + 0.125 + 0.572 + 8(0.120 – 0.0624)](0.49) – 2(0.4)(0.7) = 1.13 ksf
at h = -30’ Δp6 = (γ5-γw)h K5a = (0.120-0.0624)(5) (0.49) = 0.141 ksf ∴1.31+0.14 = 1.45 ksf

273
F1 = (0.614)(6) = 3.68 kips The resultant R is, R = ∑ Fi = 57.1 kips
F2 = 0.5(0.203)(6) = 0.61 kips
F3 = (0.817)(2) = 1.63 kips
F4 = 0.5(0.042)(2) = 0.04 The location of R is…….∑M0 = 0 (about 0)
F5 = (0.95)(9) = 8.55 kips 57.09(y) = (3.68)(27) + (0.61)(26) = (1.63)(23)
F6 = 0.5(0.40(9) = 1.80 kip
F7 = (1.758)(8) = 14.1 kips
F8 = 0.5(0.461)(8) = 1.84 kips
F9 = (1.31)(5) = 6.55 kips
F10 = 0.5(0.141)(5) = 0.35 kips
F11 = 0.51(1.50)(24) = 18.0 kips
57.1 kips
∴ y = 611 / 57.1 = 10.7 feet above “0”

274
**Lateral-13: The stability of a reinforced concrete wall.
(Revised Oct-09)
Calculate the Factor of Safety against, (a) overturning, (b) sliding, and (c) bearing capacity
failures.
γ1 16.8
kN
m
3
=
γconc 23.6
kN
m
3
= φ1 32
o
= c1 0
=
γ2 17.6
kN
m
3
= φ2 28
o
= c2 30
kN
m
2
=
1
2
3
4
1.5 m 0.6 m 3.5 m
0.75
m
H = 8
m
0.4
0.62m
0.96m
α = 10ο
0
y φ c
H = 9.58m

275
=
−
+
−
−
=
'
cos
cos
cos
'
cos
cos
cos
cos
2
2
2
2
φ
β
β
φ
β
β
α
Ka
2 2
2 2
cos10 cos 10 cos 32
cos10
cos10 cos 10 cos 32
o o o
o
o o o
− −
=
+ −
0.322
Fa = (1/2) H2
γ1 Ka = (1/2)(9.58 m)2
(16.8 kN/m3
)(0.322) = 248 kN/m
Fv = Fa sin10° = (248 kN/m)(0.174) = 43.1 kN/m
Fh = Fa cos10° = (248 kN/m)(0.985) = 244 kN/m
a) The factor of safety against overturning is found by taking moments about point “O”.
The resisting moment against overturning is MR,
MR = 23.6 kN/m3
[(0.4m)(8m)(1.90m) + (1/2)(0.2m)(8m)(1.63m) + (0.96m)(5.6m)(2.8m)] (1m)
+ 16.8 kN/m3
[(3.5m)(8m)(3.85m) + (1/2)(0.617m)(3.5m)(4.43m)] (1m)
+ 43.1 kN/m (5.6m)(1m) = 2661 kN-m
and the overturning moment is MO = Fh (1/3) H’ = 244 kN/m (9.58m)(1/3) = 777 kN-m
FSO = MR / MO = 3 .42
b) The factor of safety (FSS) against sliding failure,
K1 = K2 = 2/3
Kp = tan2
( 45° + 28°/2 ) = 2.77
Fp = (1/2) γ2 H2
Kp + 2 c2 H Kp
= (0.5)(2.77)(17.6 kN/m3
)(1.75 m)2
+ (2)(30kN/m2
) 77
.
2 (1.75 m) = 249 kN/m
the driving force = Fh = 244 kN/m
the resisting force = FR = ΣV tan(2/3)(28) + (5.6)(2/3)(30) = 355 kN/m
FSS = Fh / FR = 1.46
c) the factor of safety (FSBC) against a bearing capacity failure,

276
2
B Mr Mo
e
V
Σ − Σ
= − =
Σ
=
−
−
=
749
777
2661
8
.
2
e 0.31 m
qtoe = =
⎥
⎦
⎤
⎢
⎣
⎡
+
=
⎥
⎦
⎤
⎢
⎣
⎡
+
Σ
6
.
5
)
31
.
0
(
6
1
6
.
5
749
6
1
B
e
B
V
178 kN/m2
B’ = B – 2e = 5.6m – 2 (0.31m) = 4.98 m
qu = (1/2) γ2 B’ Nγ Fγd
s
Fγi
d
+ C2 Nc Fcd
s
Fci
d
+ q Nq Fqd
s
Fqi
d
q = γ2 D = (17.6)(1.75) = 30.8 kN/m2
using φ2 = 28° → Nc = 25.8
Nq = 14.7
Nγ = 16.7
Fcd = 14
.
1
98
.
4
75
.
1
4
.
0
1
'
4
.
0
1 =
+
=
+
B
Df
=
⎟
⎠
⎞
⎜
⎝
⎛
Σ
= −
V
Ph
1
tan
ψ tan-1
(244 / 749) = 18.04°
Fqi = Fci = ⎟
⎠
⎞
⎜
⎝
⎛
−
90
1
ψ 2
= 0.96
Fγi =
2
1 ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
φ
ψ
=
2
28
04
.
18
1 ⎟
⎠
⎞
⎜
⎝
⎛
− = 0.58
Fγd = 1
Fqd = ( ) ⎟
⎠
⎞
⎜
⎝
⎛
−
+
B
Df
2
sin
1
tan
2
1 φ
φ = 1 + 2 tan28°(1-sin28°)2
(1.75/5.6)) =
= 1.08
qu = (1/2)(17.6)(4.98)(16.7)(1)(0.58) + (30)(25.8)(1.14)(0.96) + (30.8)(14.7)(1.08)(0.96) =
= 1740 kN/m2
FSBC = qu / qtoe = 1740 / 178.00 = 9.78

277
***Lateral-14: Derive a formula that provides K and σH as a function of σv.
(Revised Oct-09)
Using the Mohr-Coulomb failure criterion combined with Rankine’s theory, find the coefficient of
active earth pressure Ka as an investigation of the stress conditions in soil at a state of plastic
equilibrium (in other words, when the soil mass is on the verge of failure).
Solution:
The definition of an “active pressure” condition is when σh decreases until it touches point D on the
Mohr-Coulomb failure envelope.
Find σa:
From the figure,
sin φ =
OC
AO
CD
AC
CD
+
=
CD = radius of failure circle = ⎟
⎠
⎞
⎜
⎝
⎛ −
=
⎟
⎠
⎞
⎜
⎝
⎛ −
2
2
3
1 a
v σ
σ
σ
σ
AO = c cot φ
OC = ⎟
⎠
⎞
⎜
⎝
⎛ −
=
⎟
⎠
⎞
⎜
⎝
⎛ +
2
2
3
1 a
v σ
σ
σ
σ
Substituting values into the equation for CD, AO and OC gives:

278
sin φ =
⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛ +
+
⎟
⎠
⎞
⎜
⎝
⎛ −
2
cot
2
a
v
a
v
c
σ
σ
φ
σ
σ
Rearrange the equation to make σa the subject:
This gives:
2c.cos φ + ⎟
⎠
⎞
⎜
⎝
⎛ +
2
a
v σ
σ
sin φ =
2
a
v σ
σ −
Τhis then gives:
σa (sin φ + 1) = σv (1 – sin φ) − 2c.cos φ
∴σa = ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
−
φ
φ
φ
φ
σ
sin
1
cos
2
sin
1
sin
1
c
v
Solution of the trigonometric expressions:
2α + (90° + φ) = 180°
α =
2
90 φ
+
°
= 45° –
2
φ
φ
φ
cos
sin
1+
= tan (α + φ)
= tan (45° − φ
φ
+
2
)
= ⎟
⎠
⎞
⎜
⎝
⎛
+
°
2
45
tan
φ
=
⎟
⎠
⎞
⎜
⎝
⎛
+
°
⎟
⎠
⎞
⎜
⎝
⎛
+
°
2
45
cos
2
45
sin
φ
φ
⎟
⎠
⎞
⎜
⎝
⎛
+
°
⎟
⎠
⎞
⎜
⎝
⎛
+
°
=
+
∴
2
45
sin
2
45
cos
sin
1
cos
φ
φ
φ
φ
But for any complementary angles β and (90° - β), cos β = sin (90° - β).
Thus, cos (45°
2
φ
 sin (45° -
2
φ
and sin (45°
2
φ
 cos (45°
2
φ


279
⎟
⎠
⎞
⎜
⎝
⎛
−
°
⎟
⎠
⎞
⎜
⎝
⎛
−
°
=
+
∴
2
45
cos
2
45
sin
sin
1
cos
φ
φ
φ
φ
= ⎟
⎠
⎞
⎜
⎝
⎛
−
°
2
45
tan
φ
( ) ( )
⎟
⎠
⎞
⎜
⎝
⎛
−
°
=
+
=
+
−
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
+
×
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
−
=
+
−
∴
2
45
tan
sin
1
cos
sin
1
sin
1
sin
1
sin
1
sin
1
sin
1
sin
1
sin
1 2
2
2
2
2
2
φ
φ
φ
φ
φ
φ
φ
φ
φ
φ
φ
σa = σv tan2
(45° -
2
φ
) - 2c tan (45° -
2
φ
)
∴ σa = (γz) Ka - 2c ( Ka ) ½
; where Ka = tan2
(45° -
2
φ
)
Using this equation, the slip planes can be described by the grid of lines shown below:
Rankine’s expression
gives the effective
Δ

280
**Lateral-15: The magnitude and location of a seismic load upon a retaining wall.
(Revision: Sept-08)
The reinforced concrete retaining wall shown below will be subjected to a horizontal seismic load
of 0.2 g without a vertical component. Determine,
(a) The magnitude of the active earth force Pa on the wall;
(b) The magnitude of the earthquake active earth force Pae on the wall;
(c) The location of the resultant of both forces.
Solution.
( )
( )
( ) ( )
( ) ( )
( )
( ) ( )
2
2
2
2
2
Calculate the coefficient of active earth pressure (Coulomb) using,
36 , 0 , 90 2/3 24 ,
sin '
sin ' sin '
sin sin 1
sin sin
sin 90 36
sin 36 2
sin 90 sin 90 24 1
a
a
a
K
and
K
K
φ α β δ φ
β φ
φ δ φ α
β β δ
β δ α β
= ° = ° = ° = = °
+
=
⎡ ⎤
+ −
− +
⎢ ⎥
− +
⎢ ⎥
⎣ ⎦
°+ °
=
°+
° °− ° +
( ) ( )
( ) ( )
( )
( )
( ) ( )
( )
( )
( )( )
( )
2
2
2 2
4 sin 36 0
sin 90 24 sin 0 90
sin 126 0.654
0.2346
sin 60 sin 36 0.866 0.588
sin 66 1 0.914 1
sin 66 0.914
a
K
⎡ ⎤
° °− °
⎢ ⎥
°− ° °+ °
⎢ ⎥
⎣ ⎦
°
= = =
⎡ ⎤ ⎡ ⎤
° °
° + +
⎢ ⎥ ⎢ ⎥
°
⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦
H = 5 m
0
ΔPae increase due to the earthquake load located at 0.6H
Pae the earthquake load
Pa lateral load from the soil located at 0.33H
Dense sand γ = 18 kN/m3
φ = 36°
1 m

281
( )( )( ) ( )
( )
2
2 3
1 1
The active earth force is,
1
0.5 18 / 5 0.2346
2
Calculate the earthquake coefficient of active earth pressure (Coulomb) ,
0.2,
53 /
0, 2/3 2/3 36 24 ' tan tan
1
a
a a
ae
h
h V
V
k
P
P H K kN m m
K
k
k
N m
k and
k
γ
δ φ θ − −
= = =
⎡ ⎤
= = = = ° = ° = =
⎢ ⎥
−
⎣ ⎦
( )
( )
( )
( ) ( )
( ) ( )
( )
( ) ( )
( ) ( )
( ) ( )
2
2
2
2
2
2
0.2 11.3
sin '
sin sin '
cos 'sin sin ' 1
sin ' sin
sin 36 90 11.3
0.372
sin 36 24 sin 36 0
sin 90 sin 90 24 1
sin 90 24 sin 0 90
The Mononobe-Okabe eart
ae
ae
K
K
φ β θ
φ δ φ θ α
θ β β θ δ
β δ θ α β
= °
+ −
∴ =
⎡ ⎤
+ − −
− − +
⎢ ⎥
− − +
⎢ ⎥
⎣ ⎦
°+ °− °
= =
⎡ ⎤
°+ ° °− °
° °− ° +
⎢ ⎥
°− ° °+ °
⎢ ⎥
⎣ ⎦
( )( )( ) ( )
2
2 3
hquake active earth force is,
1
(1 ) 0.5 18 / 5 (1 0) 0.372
2
The earthquake force is 83.7 53 30.7 /
The location of the resultant earthquake force is, found by l
3.7
o
8 /
ae
ae V ae
ae ae a
P
P H k K kN m m
P P P kN m
z
kN m
γ
= − = − =
Δ = − = − =
( )
( )( ) ( ) ( ) ( )( )( ) ( )( )( )
( )
2 2
2
cating the force at a height
0.6H above the base of the wall; the active earth force is obviously 0.33 above the base.
0.6 5 30.7 / 1/3 5 53 /
0.6 1/3
8 .
2.
3 7
1
/
ae
a
ae a
ae
P
P H
m kN m m kN m
H P H P
z
P kN m
m
+
Δ +
= = =

282
**Lateral-16: Seismic loading upon a retaining wall.
(Revision: Aug-08)
The reinforced concrete retaining wall shown below will be designed to a horizontal seismic
loading of 0.2 g. Assume no vertical seismic component (kv=0). Determine,
(a) The weight of the wall Ww under static conditions;
(b) The weight of the wall under seismic conditions, for zero lateral displacement;
(c) The weight of the wall under seismic conditions, for a lateral displacement = 1.5 inches.
Solution.
H = 5 m
0
Dense sand γ = 16 kN/m3
φ = 36°
δ = 2/3 φ = 24º
1m

283
Chapter 16
Braced Cuts for Excavations
Symbols for Braced Cuts for Excavations

284
*Braced-cuts-01: Forces and moments in the struts of a shored trench.
(Revision: Sept-08)
You have been asked by a contractor to design the internal supports (struts) of a temporary utility
trench, as shown below. In order to design the steel horizontal strut shown, you must first find the
force and moment on one of them, if they are spaced every 4 m horizontally.
Two triaxial laboratory tests were performed on samples of the clayey sand. The first sample had
a confining pressure of 0 kN/m2
, and the sample reached failure with a deviator stress of 90
kN/m2
. (N.B.: the deviator stress is the additional vertical stress required to reach failure, i.e. s-1 to
s-3). The second sample had its confining stress increased to 30 kN/m2
. The deviator stress needed
to attain failure was 160 kN/m2
.
Further laboratory tests show that this clayey sand had an in-situ voids ratio of 0.46 at a moisture
of 34% (assume Gs = 2.65). Show all your calculations.

285
Effective Stress Mohr’s Circle for failure Angle
From the Mohr’s Circle, we can get that φ2 = 32 o
Gs = 2.65 ; γW = 9810 N/m2 Î ( )( ) ( )( )
46
.
0
1
9810
65
.
2
1
s
+
=
+
=
e
G W
s γ
γ Î γS = γ2 = 17.8 kN/m2
=
1
A
K )
2
25
45
(
tan2
o
o
− =
0.406
2
A
K )
2
32
45
(
tan 2
o
o
−
= =
0.307
τ (kN/m2
)
σ (kN/m2
)
)
2
45
(
tan2
φ
−
= o
A
K

286
Pa = (q) (KA1) = (90kN/m2
) (.406) Î 36.54 kN/m2
Pb’+ = [KA1 (q + γ1h1)] = [(.406) (15kN/m2 x 3m)] Î 54.81 kN/m2
Pb’- = [KA2 (q + (γ2-γW) h] = [(.307) (90 + (17.8-9.81) (3)] Î 34.99 kN/m2
Pc = [(q + γ1h1 + (γ2-γW) h2] KA2 = .307 [90 + (15)(3) + (17.8-9.81)(2)]Î 46.35 kN/m2
PW = γW hW = (9.81)(2) Î 19.62 kN/m2
Location of the Forces (with respect to the top datum):
F1: 3m (1/2) = 1.5m
F2: 3m (2/3) = 2.0m
F3: 3m + 2m (1/2) = 4.0m
F4: 3m + 2m (2/3) = 4.33m
F5: 3m + 2m (2/3) = 4.33m
Magnitude of the Forces:
F1 = (Pa)( h1) = (36.54 kN/m2
)(3m) = 109.6 kN/m
F2 = (Pb+- Pa)( h1/2) = (54.81-36.54)(3/2) = 27.4 kN/m
1
3
2
4 5
2 m
3 m
F1
F3
F4 F5
F2
36.54
34.99
18.27
11.36 19.62

287
F3 = (Pb-)( h2) = (34.99 kN/m2
) (2m) = 69.98 kN/m
F4 = (Pc - Pb-)( h2/2) = (46.35-34.99)(2/2) = 11.36 kN/m
F5 = (PW ) (hW/2) = (19.62)(2/2) = 19.62 kN/m
=
+
+
+
+
=
∑ 5
4
3
2
1 F
F
F
F
F
F 237.96 kN/m
Ftot = ( )
∑F (space b/t struts) = (237.96kN/m)(4m) Î 951.84 kN
Located at
( ) ( ) ( ) ( ) ( ) m
y f 66
.
2
96
.
237
33
.
4
62
.
19
33
.
4
36
.
11
4
98
.
69
2
4
.
27
5
.
1
6
.
109
=
+
+
+
+
=
0
=
∑ c
M Where C is located at the bottom of the trench along with RA
RB is located at the end of the strut.
Î RB (3m) - 951.84 kN (2.34m) = 0
Î RB = 742.44 kN
ÎRA = 209.40 kN

288
Shear Diagram
Moment Diagram
742.44 kN
-209.40 kN
490.0kN-m
0 kN
0 kN-m
0 kN-m
2.34 m
0.66 m

289
**Braced cuts-02: A 5 m deep excavation with two struts for support.
(Revision: Sept-08)
Design a braced excavation for a large sanitary sewer force-main, which is a reinforced concrete
pipe with a diameter of 3 m. The trench should be 5 m deep and 5 m wide. The phreatic surface is
below bottom of excavation. The SPT for the silty clay is Navg = 20, and γ = 17 kN/m³
. Assume φ = 0.
Solution:
Use Stroud’s relation to estimate the un-drained cohesion of the soil (the previous problem provided the
shear strength):
cu = KN = (3.5 kN/m²) (20) = 70 kN/m².
Therefore,
( )( )
( )
4
4
17 5
In this problem, 1.21 4 this is a stiff clay
70
u
u
u
H
if theclayis soft tomedium
c
H
if theclayis stiff
c
H
c
γ
γ
γ
>
≤
= = < ∴
Also, since γH/ cu < 6, the sheet-piling should extend at least 1.5 m below bottom.

290
Step 1. Establish the lateral earth pressure distribution.
( )( ) 2
Using Peck's (1967) apparent pressure envelope, we must choose the larger of,
4
(1) 1
(2) 0.3 0.3 17 5 25.5 /
u
a
a
c
p H
H
p H kN m
γ
γ
γ
⎡ ⎤
⎛ ⎞
= −
⎢ ⎥
⎜ ⎟
⎝ ⎠
⎣ ⎦
= = =
The location of the top strut should be less then the depth of the tensile crack zc. Since φ = 0,
Ka → Ka = 1. therefore σ3 = σa = (γ)(zc)Ka - 2c Ka
therefore zc = 2c/γ = 2(70 kN/m²)/ 17 kN/m³ = 8.2 m >> 0.6 m OK
Step 2: Determine the lateral loads at strut locations and excavation bottom.
Isolation the left portion between the surface and strut #2.
∑ MF’2 = 0 = F1(1.16m)-(0.5)(1.25m)(26)[0.51+1.25/3]-(0.51)(26)[0.51/2] = 0
therefore, F1 = 15.9 kN/m
∑ Fy = 0 = -15.9 + 1/2 (1.25)(26)+(0.51)(26)- F’2 = 0 therefore, F’2 = 13.6 kN/m
Isolating the right portion between strut #2 and the trench bottom, by symmetry
F2
2
= F1
2 = 13.6 kN/m ∑ Fy = 0 = - F2
2
+ (3.75-0.51)(26)- F3
therefore, F3 = 70.6 kN/m

291
Step 3: Find the maximum moment Mmax in the sheet-piling.
Finding moments at A, B, & C (that is, the areas under the shear diagram):
MA = ½(0.60)(12.48)(0.60/3) = 0.75 kN-m/m
MB = ½(1.25)(26)(1.25/3)-15.9(0.65) = 3.56 kN-m/m
MC = (2.71)(26)(2.71/2) = 96 kN-m/m
Obviously, Mmax = 96 kN-m/m
Step 4: Select the steel-piling .
Assume fy = 50 ksi = 345 MN/m², therefore σallow = 50%fy = 172 MN/m²
The required section modulus S
S = MMax/ σall = 96 kN-m/ 172,000 kN/m² = 0.00056m³ = 56 m³/ m-105
Choose a PDA-27 section, which provides 57.46 m³/ m-105
.
Step 5: Select the horizontal waler at each strut level.
At strut level #1 the load F1 is 16 kN/m. Select the horizontal spacings to be 4 m. (May use 3 m to
reduce steel size, but increases the difficulty of placing the concrete pipes).
Mmax = F1s²/8 = (16)(4)²/8 = 32 kN-m (where s is the spacing)
therefore, Swale at 1 = Mmax/σallow = 32 kN-m/ 172,000 kN/m² = 18.6 m³/ m-105
At strut level #2 the load is 27.2 kN/m; the spacing s is = 4 m.
Mmax= F2s²/8 = (27.2)(4)²/8 = 54.4 kN-m
Therefore, Swale at 2 = Mmax/σallow = 54.4 kN-m/ 172,000 kN/m² = 31.6 m³/ m-105
Notes: 1. The bottom of the trench has the highest lateral load, with 70.6 kN per every meter. Propose to cast a concrete “mud” slab at the
bottom of the trench. Design the thickness of the slab (diaphragm).
2. Wales are commonly channels or WF beams. Design the steel pipe wales and the struts, calculated in Step 6 below.
Step 6: Select the struts.
Level # 1 strut = F1s = (16 kN/m)(4m) = 64 kN
Level # 2 strut = 2 F2s = (27.2 kN/m)(4m) = 109 kN (Design the steel for the struts).
Step 7: Check for possible heave of the excavation bottom.

292
Braced cuts in clay may become unstable if the bottom heaves upward and fails a section of wall.
FSagainst heaving = [cNc(0.84 + 0.16 B/L)]/ γH = (70)(6.4)(0.84)/(17)(5) = 4.4 > 2 O.K.
Step 8: Expected lateral yielding of the sheet-piling and ground settlement behind the wall.
Expect δh from 5 to 10 cms.
δγ from 1 to 5 cms.

293
*Braced cuts-03: Four-struts bracing a 12 m excavation in a soft clay.
(Revision: Sept-08)
A four-strut braced sheet pile installation is designed for an open cut in a clay stratum, as shown
below. The struts are spaced longitudinally (in plan view) at 4.0 m center to center. Assume that
the sheet piles are pinned or hinged at strut levels B and C.
Find: 1. The lateral earth pressure diagram for the braced sheet pile system.
2. The loads on struts A, B, C, and D.
Solution:
From Terzaghi and Peck (1967), a clay is soft, medium or stiff,
4
4 1
4 0.2 0.4
u
a
u
a
u
c
H
if theclayis soft tomedium then H
c H
H
if theclayis stiff then H to H
c
γ
σ γ
γ
γ
σ γ γ
⎛ ⎞
> = −
⎜ ⎟
⎝ ⎠
≤ =

294
( )( )
( )
2
2
3
2
96 /
Determine the cohesion from Mohr's circle 48 /
2 2
17.3 / 12
4.33 4
48 /
Peck (1969) provided a criterion for soft to medium clays,
4
1 (17
u
u
u
u
a
q kN m
c kN m
kN m m
H
this is a soft tomediumclay
c kN m
c
p H
H
γ
γ
γ
= = =
∴ = = > ∴
⎛ ⎞
= − =
⎜ ⎟
⎝ ⎠
2
3 2
3
(4)(48 / )
.3 / )(12 ) 1 15.48 /
(17.3 / )(12 )
kN m
kN m m kN m
kN m m
⎡ ⎤
− =
⎢ ⎥
⎣ ⎦
The lateral earth pressure diagram for the braced sheet pile system in soft clays is,
2. In the free body diagram, part (a), 0
B
M =
∑
( )( )( )( ) ( )( )( ) ( )( )
2 2
3.0 1.5
1 15.48 / 3.0 4.0 1.5 1.5 15.48 / 4.0 3.0 0
2 3 2
A
m m
kN m m m m m kN m m F m
+ − =
⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
100.6
A
F kN
∴ =
From 0
H =
∑
( )( )( )( )
2
1
1 1.5 4.5 15.48 / 4.0 100.6 85.2
2
B
F m m kN m m kN kN
= + − =
In the free body diagram, part (b)
( )( )( )( )
2
2 1
1 3.0 15.48 / 4.0 92
2
B C
F F m kN m m kN
= = =
In the free body diagram, part (c), 0
C
M =
∑
( )( ) ( )( )( )
2 4.5
3.0 4.5 15.48 / 4.0 0
2
D
m
F m m kN m m
⎛ ⎞
− =
⎜ ⎟
⎝ ⎠
209.0
D
F kN
∴ =

295
From 0
H =
∑
( )( )( )
2
2 4.5 15.48 / 4.0 0
C D
F F m kN m m
+ − =
( )( )( )
2
2 4.5 15.48 / 4.0 209.0 69.6
C
F m kN m m kN kN
= − =
Therefore,
85.2 92.9
100.6
178.1
162.5
209
92.9 69.
.
6
0
A
B
C
D
F kN
F kN
F kN
F
kN k
k
N
N
k
k
N N
=
=
= + =
= + =

296
Chapter 17
Bearing Capacity of Soils
Symbols for the Bearing Capacity of Soils

297
Bearing Capacity Factors for General Shear
Terzaghi Meyerhof Hansen
Angle φ
(Degrees)
Angle φ
(Radians) Kpγ Nc Nq Nγ Nc Nq Nγ Nc Nq Nγ
0 0.0000 10.18 5.70 1.00 0.00 5.10 1.00 0.00 5.10 1.00 0.00
1 0.0175 10.61 6.00 1.10 0.08 5.38 1.09 0.00 5.38 1.09 0.00
2 0.0349 11.07 6.30 1.22 0.18 5.63 1.20 0.01 5.63 1.20 0.01
3 0.0524 11.56 6.62 1.35 0.28 5.90 1.31 0.02 5.90 1.31 0.02
4 0.0698 12.07 6.97 1.49 0.39 6.19 1.43 0.04 6.19 1.43 0.05
5 0.0873 12.61 7.34 1.64 0.51 6.49 1.57 0.07 6.49 1.57 0.07
6 0.1047 13.19 7.73 1.81 0.65 6.81 1.72 0.11 6.81 1.72 0.11
7 0.1222 13.80 8.15 2.00 0.80 7.16 1.88 0.15 7.16 1.88 0.16
8 0.1396 14.44 8.60 2.21 0.96 7.53 2.06 0.21 7.53 2.06 0.22
9 0.1571 15.13 9.09 2.44 1.15 7.92 2.25 0.28 7.92 2.25 0.30
10 0.1745 15.87 9.60 2.69 1.35 8.34 2.47 0.37 8.34 2.47 0.39
11 0.1920 16.65 10.16 2.98 1.58 8.80 2.71 0.47 8.80 2.71 0.50
12 0.2094 17.49 10.76 3.29 1.84 9.28 2.97 0.60 9.28 2.97 0.63
13 0.2269 18.38 11.41 3.63 2.12 9.81 3.26 0.74 9.81 3.26 0.78
14 0.2443 19.33 12.11 4.02 2.44 10.37 3.59 0.92 10.37 3.59 0.97
15 0.2618 20.36 12.86 4.45 2.79 10.98 3.94 1.13 10.98 3.94 1.18
16 0.2793 21.46 13.68 4.92 3.19 11.63 4.34 1.37 11.63 4.34 1.43
17 0.2967 22.65 14.56 5.45 3.63 12.34 4.77 1.66 12.34 4.77 1.73
18 0.3142 23.92 15.52 6.04 4.13 13.10 5.26 2.00 13.10 5.26 2.08
19 0.3316 25.30 16.56 6.70 4.70 13.93 5.80 2.40 13.93 5.80 2.48
20 0.3491 26.80 17.69 7.44 5.34 14.83 6.40 2.87 14.83 6.40 2.95
21 0.3665 28.42 18.92 8.26 6.07 15.81 7.07 3.42 15.81 7.07 3.50
22 0.3840 30.18 20.27 9.19 6.89 16.88 7.82 4.07 16.88 7.82 4.13
23 0.4014 32.10 21.75 10.23 7.83 18.05 8.66 4.82 18.05 8.66 4.88
24 0.4189 34.19 23.36 11.40 8.90 19.32 9.60 5.72 19.32 9.60 5.75
25 0.4363 36.49 25.13 12.72 10.12 20.72 10.66 6.77 20.72 10.66 6.76
26 0.4538 39.01 27.09 14.21 11.53 22.25 11.85 8.00 22.25 11.85 7.94
27 0.4712 41.78 29.24 15.90 13.15 23.94 13.20 9.46 23.94 13.20 9.32
28 0.4887 44.85 31.61 17.81 15.03 25.80 14.72 11.19 25.80 14.72 10.94
29 0.5061 48.26 34.24 19.98 17.21 27.86 16.44 13.24 27.86 16.44 12.84
30 0.5236 52.05 37.16 22.46 19.75 30.14 18.40 15.67 30.14 18.40 15.07
31 0.5411 56.29 40.41 25.28 22.71 32.67 20.63 18.56 32.67 20.63 17.69
32 0.5585 61.04 44.04 28.52 26.20 35.49 23.18 22.02 35.49 23.18 20.79
33 0.5760 66.40 48.09 32.23 30.33 38.64 26.09 26.17 38.64 26.09 24.44
34 0.5934 72.48 52.64 36.50 35.23 42.16 29.44 31.15 42.16 29.44 28.77
35 0.6109 79.40 57.75 41.44 41.08 46.12 33.30 37.15 46.12 33.30 33.92
36 0.6283 87.33 63.53 47.16 48.11 50.59 37.75 44.43 50.59 37.75 40.05
37 0.6458 96.49 70.07 53.80 56.62 55.63 42.92 53.27 55.63 42.92 47.38
38 0.6632 107.13 77.50 61.55 67.00 61.35 48.93 64.07 61.35 48.93 56.17
39 0.6807 119.59 85.97 70.61 79.77 67.87 55.96 77.33 67.87 55.96 66.76
40 0.6981 134.31 95.66 81.27 95.61 75.31 64.20 93.69 75.31 64.20 79.54
41 0.7156 151.89 106.81 93.85 115.47 83.86 73.90 113.99 83.86 73.90 95.05
42 0.7330 173.09 119.67 108.75 140.65 93.71 85.37 139.32 93.71 85.37 113.96
43 0.7505 198.99 134.58 126.50 173.00 105.11 99.01 171.14 105.11 99.01 137.10
44 0.7679 231.10 151.95 147.74 215.16 118.37 115.31 211.41 118.37 115.31 165.58
45 0.7854 271.57 172.29 173.29 271.07 133.87 134.87 262.74 133.87 134.87 200.81
46 0.8029 323.57 196.22 204.19 346.66 152.10 158.50 328.73 152.10 158.50 244.65
47 0.8203 391.94 224.55 241.80 451.28 173.64 187.21 414.33 173.64 187.21 299.52
48 0.8378 484.34 258.29 287.85 600.15 199.26 222.30 526.45 199.26 222.30 368.67
49 0.8552 613.53 298.72 344.64 819.31 229.92 265.50 674.92 229.92 265.50 456.40
50 0.8727 801.95 347.51 415.15 1155.97 266.88 319.06 873.86 266.88 319.06 568.57

298
The bearing capacity of a soil is its ability to carry loads without failing in shear. There are four major
methods to predict failure. The fist method was developed by Karl Terzaghi in 1943. Field tests in
Canada by Meyerhof (1963) lead to modification factors. Finally, Brinch Hansen in Denmark (1970)
and Vesic in the USA modified these factor to a greater refinement.
These bearing capacity factors are based on these three authors:
Terzaghi (1943):
( )
( )
( )
'
'
2
0.75 /2 tan
2
, 1.3 0.4
0.5
, ,
cos 45 / 2
1 cot
tan
2
ult c q
ult c q
f
q
c q
For square footings q c N qN BN
For continuous or wall footings q c N qN BN
where q D and the factors are
a
N where a e
a
N N
K
N
γ
γ
π φ φ
γ
γ
γ
γ
ϕ
φ
φ
−
= + +
= + +
=
= =
° −
= −
= 2
1
cos
pγ
φ
⎛ ⎞
−
⎜ ⎟
⎝ ⎠
Meyerhof (1963):
( )
( )
( ) ( )
tan 2
, 0.4
, 0.4
,
tan 45 / 2
1 cot
1 tan 1.4
ult c sc dc q sq dq s d
ult c ic dc q iq dq i d
q
c q
q
For vertical loads q cN F F qN F F BN F F
and for inclined loads q cN F F qN F F BN F F
and the factors are
N e
N N
N N
γ γ γ
γ γ γ
π φ
γ
γ
γ
φ
φ
φ
= + +
= + +
= °−
= −
= −
Brinch Hansen (1970):
( )
( )
( )
tan 2
, 0.4
,
tan 45 / 2
1 cot
1.5 1 tan
ult c sc dc ic q sq dq iq s d i
q
c q
q
The general equation q cN F F F qN F F F BN F F F
and the factors are
N e
N N
N N
γ γ γ γ
π φ
γ
γ
φ
φ
φ
= + +
= °−
= −
= −

299
*Bearing–01: Terzaghi’s bearing capacity formula for a square footing.
(Revision: Sept-08)
The square footing shown below must be designed to carry a 294 kN load. Use Terzaghi’s bearing
capacity formula to determine B of the square footing with a Factor of Safety =3.
Solution:
'
Terzaghi's formula for the ultimate bearing capacity of a square footing is,
1.3 0.4
The allowable bearing capacity with the factor of safety of 3 is,
3
ult
ult c q f
all
ult
all
q
q c N qN BN where q D
q
q
q
γ
γ γ
= + + =
= ( )
( )
( ) ( )
'
2 2
'
2
2
1 294
1.3 0.4
3
294 1
1.3 0.4
3
For =35º, =57.8, =41.4, and =41.1.
Substituting these values into Terzaghi's equation, we get
294 1
0 18.15)(1 (41.4) (0.4) 1
3
c q all
c q
c q
W kN
c N qN BN and q
B B
or c N qN BN
B
N N N
B
γ
γ
γ
γ
γ
φ
= + + = =
= + +
= + + ( )
2
3 2
8.15) (41.1
294
250.5 99.5
2.52 2.96 0.90
0
B
B
B
B B
B m
⎡ ⎤
⎣ ⎦
= +
+ − = =
∴
W
W = 294 kN
γ = 18.15 kN/m3
φ = 35°
c = 0
B
Df = 1 m

300
*Bearing–02: Meyerhof’s bearing capacity formula for a square footing.
(Revision: Sept-08)
The square footing shown below must be designed to carry a 294 kN load. Use Meyerhof’s bearing
capacity formula to determine B with a factor of safety =3.
Solution:
'
Meyerhof's formula for the ultimate bearing capacity of a square footing is,
0.4
Since the load is vertical, all three inclination factors
ult
ult c sc dc ic q sq dq iq s d i f
i
q
q c N F F F qN F F F BN F F F where q D
F
γ γ γ γ
γ γ
= + + =
( ) ( )
2 2
= = =1.
1
1 tan 1 tan35 1.70 1 0.4 1 0.4(1) 0.6
1
1
1 2tan 1 sin 1 2 tan35 (1 sin35 ) 1.25 1
The allowable bearing capacity with the f
c iq i
sq s
f
dq d
all
F F
B B
F and F
L L
D
F and F
B B
q
γ
γ
γ
φ
φ φ
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= + = + ° = = − = − =
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞
= + − = + ° − ° ≈ =
⎜ ⎟ ⎜ ⎟
⎝ ⎠
⎝ ⎠
( )
( )
'
2 2
'
2
actor of safety of 3 is,
1 294
0.4
3 3
294 1
0.4
3
For = 35º, = 46.12, = 33.30, and = 37.15.
Substituting these val
ult
all c sc dc q sq dq s d all
c sc dc q sq dq s d
c q
q W kN
q c N F F qN F F BN F F and q
B B
or c N F F qN F F BN F F
B
N N N
γ γ γ
γ γ γ
γ
γ
γ
φ
= = + + = =
= + +
( ) ( ) ( )( ) ( )( )( )
2
3
2
ues into Meyerhof''s equation, we get
294 1
0 18.15)(1 (33.3) 1.7 1.25 (0.4) 18.15) (37.15 0.6 1
3
294
428.1 53.9 0.65
4 7.94 5.45 0
B
B
B or B B m
B
B
⎡ ⎤
= + +
⎣ ⎦
= + + =
− = ∴
W
W = 294 kN
γ = 18.15 kN/m3
φ = 35°
c = 0
B
Df = 1 m

301
*Bearing–03: Hansen’s bearing capacity formula for a square footing.
(Revision: Sept-08)
The square footing shown below must be designed to carry a 294 kN load. Use Brinch Hansen’s
bearing capacity formula to determine B with a factor of safety =3.
Solution:
'
Hansen's formula for the ultimate bearing capacity of a square footing is,
0.4
Since the load is vertical, all three inclination factors =
ult
ult c sc dc ic q sq dq iq s d i f
ic
q
q c N F F F qN F F F BN F F F where q D
F
γ γ γ γ
γ γ
= + + =
( ) ( )
2 2
= =1.
1
1 tan 1 tan35 1.7 1 0.4 1 0.4(1) 0.6
1
1
1 2tan 1 sin 1 2 tan35 (1 sin35 ) 1.255 1
The allowable bearing capacity with the fac
iq i
sq s
f
dq d
all
F F
B B
F and F
L L
D
F and F
B B
q
γ
γ
γ
φ
φ φ
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= + = + ° = = − = − =
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞
= + − = + ° − ° ≈ =
⎜ ⎟ ⎜ ⎟
⎝ ⎠
⎝ ⎠
( )
( )
'
2 2
'
2
tor of safety of 3 is,
1 294
0.4
3 3
294 1
0.4
3
For = 35º, = 46.12, = 33.30, and = 33.92.
Substituting these value
ult
all c sc dc q sq dq s d all
c sc dc q sq dq s d
c q
q W kN
q c N F F qN F F BN F F and q
B B
or c N F F qN F F BN F F
B
N N N
γ γ γ
γ γ γ
γ
γ
γ
φ
= = + + = =
= + +
( ) ( ) ( )( ) ( )( )( )
2
3
2
s into Hansen's equation, we get
294 1
0 18.15)(1 (33.3) 1.7 1.255 (0.4) 18.15) (33.92 0.6 1
3
294
429.8 49.25 8.73 5.9 0.7
7 0
0
B
B
B or B B B
B
m
=
⎡ ⎤
= + +
⎣ ⎦
= + + − = ∴
W
W = 294 kN
γ = 18.15 kN/m3
φ = 35°
c = 0
B
Df = 1 m

302
*Bearing–04: Same as #01 but requiring conversion from metric units.
(Revision: Sept-08)
The square footing shown below must be designed to a load of 30,000 kgm. Using a factor of safety
of 3 and using Terzaghi’s method, determine the size B of the square footing.
Solution:
( )
( )
( )
3
3 2
3
The soil density 1,850 / converts to a unit weight via ,
1,850 9.81
18.15 / and the load to be supported by the footing is,
1,000 /
30,000 9.81
m
m
kgm m g like F ma
kg m
m s
g kN m
N kN
m
kg
W ma
ρ γ ρ
γ ρ
= = =
⎛ ⎞⎛ ⎞
⎜ ⎟
⎜ ⎟
⎝ ⎠
⎝ ⎠
= = =
= =
( )
( )
2
'
'
2 2
'
2
294
1,000 /
Terzaghi's ultimate bearing capacity of a square footing is given by,
1.3 0.4
1 294
1.3 0.4
3 3
294 1
1.3 0.4
3
ult c q
ult
all c q all
c q
s
kN
N kN
q c N qN BN
q P
q c N qN BN and q
B B
or c N qN BN
B
γ
γ
γ
γ
γ
⎛ ⎞
⎜ ⎟
⎝ ⎠ =
= + +
∴ = = + + = =
= + +
( )
( ) ( ) ( ) 3 2
2
For = 35º, = 57.8, = 41.4, and = 41.1,
294 1
0 18.15)(1 (41.4) (0.4) 18.15) (41.1 2.52 2.96 0
3
0.90
c q
N N N
B
B
B
m
B
B
γ
γ
φ
⎡ ⎤
= + + ∴ + −
=
=
⎣ ⎦
W
m = 30,000 kgm
ρ = 1,850 kg/m3
φ = 35°
c = 0
B
Df = 1 m

303
*Bearing–05: General versus local bearing capacity failures.
(Revision: Sept-08)
Using Terzaghi’s method, distinguish between the value of the local shear failure versus the
general shear failure.
Solution:
'
Terzahi's general bearing capacity failure of a square footing is,
1.3 0.4
28 31.6, 17.8, 15.0 (0.115)(2) 0.23
,
1.3(0.30)(31.6) (0.23)(17.8) 0.4(0.11
ult C q
c q f
ult
q c N qN BN
For N N N and q D ksf
Therefore
q
γ
γ
γ
φ γ
= + +
= ° = = = = = =
= + +
' ' ' ' '
5)(2.5)(15.0) 18.1
To find the value of the bearing capacity of a local shear failure, the cohesion and angle
of internal friction are reduced by two-thirds,
1.3 ' 0.4
ult local c q
ksf
q c N qN BN where c
γ
γ
−
=
= + +
( )
' ' ' '
'
2 2
(0.30) 0.2
3 3
2 2
( ) (28 ) 18.7 16.2, 6.5 4.52
3 3
1.3 (0.2)(16.2) (0.23)(6.5) (0.4)(0.115)(2.5)(4.52) 6.2
1 .1
8
ult general failure ult local fai
c q
ult local
c ksf
and which give N N
q ksf v
and N
q k
s
s
ersu
f
q
γ
ϕ ϕ
−
− −
= = =
= = ° = ° = = =
∴ +
=
= + =
6.2 ( )
lure ksf Almost a three to one
= − −

304
*Bearing–06: Comparing the Hansen and Meyerhof bearing capacities.
(Revision: Sept-08)
Compare the results of the Hansen and the Meyerhof bearing capacity formulas to the results of a
field test that took a rectangular footing to failure when the load reached 1,863 kN. Given B = 0.5
m, L = 2.0 m, c = 0, φtriaxial = 42° and γ’ = 9.31 kN/m3
(the WT is at the surface).
Solution:
( )( )
1,863
1,863 was the field measured failure load.
0.5 2.0
(1) The Hansen formula predicts an ultimate bearing capacity of,
0 0.5
' 1.5
ult
ult
ult q qs qd s d
ps tr
P kN
q kPa
BL m m
q qN F F BN F F
Lee s adjustment formula is
γ γ γ
γ
φ φ
= = =
= + +
= ( )
2 2
17 1.5 42 17 46
46 , 158.5 244.65
0.5
1 tan 1 tan 46 1.26
2
0.5
1 0.4 1 0.4 0.9
2
0.5
1 2 tan (1 sin ) 1 2 tan 46 (1 sin 46 ) 1.16
0.5
iaxial
q
qs
s
f
qd
For N and N
B
F
L
B
F
L
D
F
B
γ
γ
ϕ
ϕ
ϕ ϕ
− ° = ° − ° = °
= ° = =
⎛ ⎞ ⎛ ⎞
= + = + ° =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞
= − = − =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞
= + − = + ° − ° =
⎜ ⎟ ⎜ ⎟
⎝ ⎠
⎝ ⎠
1.0
0 (9.31)(0.5)(159)(1.27)(1.16) (0.5)(9.31)(0.5)(245)(0.9)(1.0)
1, 485 1,863 ( 20%)
(2) The Meyerhof formula with = 46º, = 158.5 and = 328.7
d
ult
ult
q
F
q
q kPa versus kPa measured Hansen underestimates by
N N
γ
γ
φ
=
∴ = + +
∴ =
( )( ) ( )
( )
3,
0 0.5
0 (9.31) 0.5 158.5 (1.27) 1.16 (0.5)(9.31)(0.5)(328.73)(0.9)(1.0)
1,782 1,863 4 % .
ult q qs qd s d
ult
ult
q qN F F BN F F
q
q kPa versus kPa Meyerhof underestimates by
γ γ γ
γ
= + +
∴ = + +
∴ =
Pult = 1,863 kN
▼ WT
B = 0.5 m
Df = 0.5 m

305
*Bearing–07: Increase a footing’s width if the WT is expected to rise.
(Revision: Sept-08)
Use Meyerhof’s bearing capacity formula (with a factor of safety = 3) to select a footing’s width B
if, (a) the water table is as shown below, and (b) if the water table rises to the ground surface?
The soil has a unit weight of 112 pcf, a moisture of 10%, φ = 25º, a cohesion cu = 240 psf and a
specific gravity of solids of Gs = 2.68.
Solution:
( )
( )
SAT
3 3 3
(a) Find to determine ',
112
101.8
1 1.10
101.8
1 0.61 1 0.61 0.39
2.68 62.4
101.8 (0.39) 62.4 126.2
dry
s
dry s
N s w s w
s v s
v
sat dry w dry w sat
W
pcf and V
w G G
set V ft V ft V V V ft
V
but n pcf
V
γ γ
γ
γ
γ
γ γ
γ γ γ γ γ γ
= = = = =
+
= ∴ = = ∴ = − = − =
⎛ ⎞
= + = + ∴ = + =
⎜ ⎟
⎝ ⎠
2
' 126.2 62.4 63.8
Try = 5.7 feet with Meyerhof's equation,
' ( ) ( ) 0.5 ( )
, 1
10 tan 45
2
sat w
ult c cs cd ci q qs qd qi s d i
ci qi i
p
and pcf
B
q c N F F F qN F F F BN F F F
where the load inclination factors F F and F
For K
γ γ γ γ
γ
γ γ γ
γ
ϕ
ϕ
= − = − =
= + +
=
⎛ ⎞
≥ ° = ° +
⎜ ⎟
⎝ ⎠
2 25
tan 45 2.46,
2
5.7
1 (0.2) 1 (0.2) (2.46) 1.49
5.7
4
1 (0.2) 1 (0.2) 2.46 1.22
5.7
4
1 (0.1) 1 (0.1) 2.46 1.11
5.7
1 (0
cs p
f
cd p
f
qd d p
qs s
therefore
B
F K
L
D
F K
B
D
F F K
B
F F
γ
γ
°
⎛ ⎞
= ° + =
⎜ ⎟
⎝ ⎠
⎛ ⎞ ⎛ ⎞
= + = + =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞
= + = + =
⎜ ⎟ ⎜ ⎟
⎝ ⎠
⎝ ⎠
⎛ ⎞ ⎛ ⎞
= = + = + =
⎜ ⎟ ⎜ ⎟
⎝ ⎠
⎝ ⎠
= = +
5.7
.1) 1 (0.1)( )(2.46) 1.25
5.7
p
B
K
L
⎛ ⎞
= + =
⎜ ⎟
⎝ ⎠

306
25
20.7, 10.7, 6.77
' ( ) ( ) 0.5 ( )
(0.24)(20.7)(1.49)(1.22)(1) (0.112)(4)(10.7)(1.25)(1.11)(1) (0.5)(0.112
c q
ult c cs cd ci q qs qd qi s d i
ult
The Meyerhof bearing capacity factors for are
N N and N
q c N F F F qN F F F BN F F F
q
γ
γ γ γ γ
φ
γ
= °
= = =
= + +
= + +
2 2
)(5.7)(6.67)(1.25)(1.11)(1)
18.6
18.6 200
6.2 32.25 5.7
3 6.2
(b) When the water table rises to the ground sur
5
face
.7 .
,
ult
ult
all
all
Therefore the choic
q ksf
q Q
q ksf therefore B ft
e of B ft was a good cho
B ft
i
q
ce
FS
=
=
= = = = = = ∴ =
( )( )
need a larger footing; try = 7.0 feet.
7
1 0.2 1 0.2 (2.46) 1.49
7
1.49
4
1 0.1 1 0.1 2.46 1.09
7
1.25
(0.24)(20.7) 1.49 1.18 (0
cd p
cs
qd d p
qs s
ult
B
B
F K
L
F same as above
D
F F K
B
F F same as above
q
γ
γ
⎛ ⎞ ⎛ ⎞
= + = + =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
=
⎛ ⎞ ⎛ ⎞
= = + = + =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
= =
= + ( )( )( )( ) ( )( )( )
2 2
.062) 4 10.7 1.25 1.09 (0.5)(0.062)(7) 6.67 1.09 1.25
16.62
16.62 200
5.54 36.1 6.01
3 5.54
Iterate once more, and fin = 7.5 f t
d ee .
ult
ult
all
all
q ksf
q Q
q ksf and B ft B ft
FS q
B
+
=
= = = = = = ∴ =

307
**Bearing–08: The effect of the WT upon the bearing capacity.
(Revision: Sept-08)
Using the Hansen method, what are the ultimate and allowable bearing capacities for the footing
shown below if you require a factor of safety of at least 2?
Solution:
Always use the effective unit weight of water in the bearing capacity formulas. The average effective
weight e
γ of the soil can also be given by the formula:
( ) ( )
2
2 2
3
3
'
2
35
(0.5) tan 45 (0.5)(2.5) tan 45 2.40
2 2
0.85
1
18.10
16.5
1 1 0.10
w
e w wet w
w
dry
wet
dry s
s
d
H d H d
H H
where H B m
and d depth to the WT below the footing invert m
Set the total volume V m
kN
and V
w m G
γ
γ γ
ϕ
γ
γ
γ
= − + −
°
⎛ ⎞ ⎛ ⎞
= ° + = ° + =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
= =
=
= = = =
+ + ( )
( )
( )
( )
( ) ( )
( )
3
3
3
2
2 2 3
16.5
0.63
(2.68) 9.8
1.0 1 0.63 0.37 16.5 (0.37) 9.8 20.1
0.85 18.10 20.1 9.8
(2)(2.40 0.85) 2.40 0.85 12.6
2.4 2.4
wet
v s sat dry wet
e
m
kN
V V m and n
m
kN
m
γ
γ γ γ
γ
= =
= − = − = = + = + =
⎡ ⎤ ⎡ ⎤
−
∴ = − + − =
⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦
Using Hansen’s method with φ = 35º, the bearing capacity factors are Nq = 33.3 and 33.92.
Nγ =

308
( ) ( )
2
2
2.5
1 tan 1 tan 35 1.70
2.5
1
1 2 tan 1 sin 1 2 tan 35 1 sin 35 1.10
2.5
2.5
1 0.4 1 0.4( ) 0.6
2.5
1.0
Therefore, the ultimate and allow able bearing capacities are,
qs
f
qd
s
d
ult
B
F
L
D
F
B
B
F
L
F
q
γ
γ
φ
φ φ
⎛ ⎞
= + = + ° =
⎜ ⎟
⎝ ⎠
⎛ ⎞ ⎛ ⎞
= + − = + − =
⎜ ⎟ ⎜ ⎟
⎝ ⎠
⎝ ⎠
= − = − =
=
o o
( )( )( )( ) ( )( )( )( )( )
0 ( ) 0.5 ( )
(18.1) 1.0 33 1.70 1.10 (0.5) 12.6 2.5 34 0.6 1
1, 497
2
1, 497
749
q qs q d e s d
ult
ult
all
qN F F BN F F
q
q kPa
kPa
q
γ γ γ
γ
= + +
= +
=
= =

309
*Bearing–09: Finding the gross load capacity.
(Revision: Sept-08)
Use the Hansen formula to determine the gross normal load N on the column shown below using a
factor of safety of 3.
Solution:
The Hansen formula for a footing is,
0.5
The inclination factors , , and are all equal to 1 because the load is vertical.
For = 32 , = 35.49, = 23.18
ult c cs cd q qs qd y ys yd
ci qi i
c q
q cN F F qN F F BN F F
F F F
N N
γ
γ
φ
= + +
°
( )
( )
( )
2
f
and = 20.79 and / = 1
1 1 23.20/ 35.50 1.65
1 tan 1 0.62 1.62
1 0.4 1 0.4 0.60
1 2tan 1 sin 1 (2)(0.62)(0.22)(1) 1.273 for D / 1
1
1
tan
q
cs
c
qs
ys
f
qd
yd
qd
cd qd
q
N B L
N
F
N
F
B
F
L
D
F B
B
F
F
F F
N
γ
φ
φ φ
⎛ ⎞
= + = + =
⎜ ⎟
⎝ ⎠
= + = + =
⎛ ⎞
= − = − =
⎜ ⎟
⎝ ⎠
⎛ ⎞
= + − = + = ≤
⎜ ⎟
⎝ ⎠
=
−
= −
1 1.273
1.273 1.292
23.20 0.62
φ
⎡ ⎤ ⎡ − ⎤
⎛ ⎞
= − =
⎢ ⎥ ⎜ ⎟
⎢ ⎥
×
⎝ ⎠
⎢ ⎥ ⎣ ⎦
⎣ ⎦
γ =18.1 kN/m3
φ =32°
γ t = 21 07 kN/m3
1.22 m
0.61 m
0.61 m
N

310
( )( ) ( )( )
( )( )( )( )( )( )
3 2
2
2
The is located above the footing, therefore,
0.61 18.1 / 0.61 21.07 9.81 17.9 /
(17.9)(1.62)(1.273)(23.20) 0.5 0.6 21.07 9.81 1.22 20.8 1 981 /
Therefore,
981 /
3
3 3
ult
ult
all
WT
q m kN m m kN m
q kN m
q kN m
q
= + − =
∴ = + − =
⎛ ⎞
⎛ ⎞
= = =
⎜ ⎟
⎜ ⎟
⎝ ⎠ ⎝ ⎠
2
2 2 2
27 /
Hence, the total gross load is,
(327 48
/ )(1 2 7
.2 )
all
kN m
N
N q B kN m N
m k
= = =

311
**Bearing–10: The effect of an eccentric load upon bearing capacity.
(Revision: Sept-08)
A rectangular footing measures 5 feet by 2.5 feet. Determine the gross ultimate load Qult applied
eccentrically upon the footing, and the ultimate bearing capacity of the soil qult, given that γ = 115
pcf, c = 0 andφ = 30°.
Solution:
( ) ( )
( )
The effective width footing width '= - 2 = 2.5 - 2 0.2 = 2.1 ft
and the effective length '= - 2 = 5 - 2(0.4) = 4.2 ft.
Meyerhof's ultimate bearing capacity formula with = 0 is,
0 0.5
x
y
ult q qs qd
B B e
L L e
c
q qN F F B Nγ
γ ′
= + +
( )
( )( )
( ) ( )( )
'
'
2
30 , 18.4 15.67
2.1
1 tan 1 0.58 1.29
4.2
2
1 2 tan 30 1 sin 30 1.275
2.1
2.1
1 0.4 1 0.4 0.8
4.2
1
2)(0.115 (18.4) 1.29 1.275 (0.5)(0.11
ys d
q
qs
qd
ys
yd
ult
F F
For N and N
B
F
L
F
B
F
L
F
q
γ
γ
φ
φ
= = =
⎛ ⎞ ⎛ ⎞
= + = + =
⎜ ⎟ ⎜ ⎟
⎝ ⎠
⎝ ⎠
⎛ ⎞
= + ° − ° =
⎜ ⎟
⎝ ⎠
′
⎛ ⎞ ⎛ ⎞
= − = − =
⎜ ⎟ ⎜ ⎟
′
⎝ ⎠ ⎝ ⎠
=
= +
o
( )( )
( ) ( )( )
5)(2.1)(15.67) 0.8 1 8.47
, 8.47 2.1)(4.2 74.73
ult ult
ksf
Henc kips
e Q q B L
=
′ ′
= = =

312
**Bearing–11: The effect of an inclined load upon the bearing capacity.
(Revision: Sept-08)
A square 8’ x 8’ footing is loaded with an axial load of 400 kips and Mx = 200 ft-kips, My = 120 ft-
kips. Un-drained triaxial tests (the soil is not saturated) gave φ = 33º and c = 200 psf. The footing
depth Df = 6.0 feet, the soil unit weight is 115 pcf, and the WT was not found.
Use the Hansen equation with the Meyerhof reduction factors and a FS = 3 to find the
Solution:
Eccentricities
120
0.3
400
y
x
M ft k
e feet
Q
−
= = = and
200
0.5
400
x
y
M ft k
e feet
Q
−
= = =
2 8' 1' 7
r y
B B e feet
∴ = − = − = and 2 8' 0.6' 7.4
r x
L L e feet
= − = − = (ie. Lr > Br)
Adjusting theφ from triaxial ( )
tr
φ to a plane-strain value ( )
ps
φ via Lee’s formulation,
( )
1.1 1.1 32.7 36
ps tr
φ φ
≅ = ° = °
tan36 2 36
tan 45 37.8
2
q
N eπ ° °
⎛ ⎞
= °+ =
⎜ ⎟
⎝ ⎠
( ) ( )
1 cot 36.8 cot36 50.6
c q
N N φ
= − = ° =
( ) ( ) ( )
( ) ( )
1 tan 1.4 36.8 tan50.4 44.4
1.5 1 tan 1.5 36.8 tan36 40.1
q
q
N N
N N
γ
γ
φ
φ
= − = ° =
= − = ° =
( )
7
1 0.2 1 0.2 3.85 1.73
7.4
r
c p
r
B
S K
L
⎛ ⎞
∴ = + = + =
⎜ ⎟
⎝ ⎠
( ) 6
1 0.2 1 0.2 3.85 1.34
7
c p
r
D
and d K
B
⎛ ⎞ ⎛ ⎞
= + = + =
⎜ ⎟ ⎜ ⎟
⎝ ⎠
⎝ ⎠
Since 10
φ > °, 1.0 1.0
q q
S S and d d
γ γ
= ≅ = = .
Hansen’s 0.5
ult c c c c c c q q q q q q q
q BN S d i g b cN S d i g b q N S d i g b
γ γ γ γ γ γ
γ
= + +
Vertical axial load = 400 kips
Mx = 200 ft-kips
My = 120 ft-kips

313
Also i = g = b = 1.0 for this problem, since 0
α = = i (inclination factor f / load Q with t vertical) 4
η =
g (ground factor with t inclined ground on side of footing)
b (base factor with t inclined ground under the footing)
( )( )( )( ) ( )( )( )( ) ( )( )( )( )
0.5 0.115 7 40.1 1 0.200 50.6 1.73 1.34 0.115 6 37.8 1
16.1 23.5 26.1 65.7
ult
ult
q
q ksf
= + + =
= + + =
65.7
21.9
3
ult
all
q
q ksf
FS
= = =
1 1
2 2
e
1 1
2 2
e
0.3
R 1 1 0.81
8
0.5
R 1 1 0.75
8
x
x
y
y
e
B
e
B
⎛ ⎞ ⎛ ⎞
= − = − =
⎜ ⎟
⎜ ⎟
⎝ ⎠
⎝ ⎠
⎛ ⎞ ⎛ ⎞
= − = − =
⎜ ⎟ ⎜ ⎟
⎝ ⎠
⎝ ⎠
( )( )( ) ( )( )( )
2
e e
R R 21.9 8 8 0.81 0.75 851
all all x y
Q q B x kips
= = =
2
851
6
13 3
4
.
all
all
q s
Q
k
B
f
⎛ ⎞
= = =
⎜ ⎟
⎝ ⎠
(The contact load
400
13 6.1
851
o
q ksf
⎛ ⎞
= =
⎜ ⎟
⎝ ⎠
)

314
**Bearing-12: Interpretation of borings to estimate a bearing capacity.
(Revision: Sept-08)
Use the boring logs show below to recommend an allowable soil pressure qall for the footings
located in the vicinity of elevation 284, boring No. 2?
The building is a four-story (five on the low side) office building with column loads around 160
kips. State your reasons.
Brown silty clay
Topsoil Fine brown silty sand - small gravel
Fine brown silty sand - trace of coarse sand
Fine to medium brown silty
sand -some small to medium
gravel
295
290
285
280
270
280
275
74
69
32
39
22
25
8
Hard
Boring No.2 Boring No.3 Boring No.4
Boring No.5
Elevation
288.0
Elevation
290.6
Elevation
292.8
Elevation
295.0
Elevation
296.6
Got
firmer
Got
Firmer
Cohesive
Sandy
Dark
brown
Got Firmer
Got Firmer
Got
Firmer
Got
Firmer
6 in.
boulder
Hard
Got
Firmer
Got
Firmer
Hard
34
38
36
27
10
6
34
38
47
16
7
4
69
51
46
71
14
5
67
62
71
29
25
13
7
1. All elevations are in accordance with plot furnished by architect.
2. Borings were made using standard procedures with 2-in. -OD split spoon.
4.No water encountered in any of the borings.
3. Figures to the right of each boring log indicate the numbeer of blows required
to drive the 2-in.-OD split spoon 12 in. using a 140 lb weight falling 30 in.
Notes:
Solution:
It is presumed that all the building’s footings will be placed at roughly elevation 284 or thereabouts.
This is fine for the building area covered by borings # 3, 4 and 5 because they have good SPT values.
Meyerhof has proposed formulas for the allowable bearing capacity adjusted so that the settlement is
limited to 1-inch. These formulas are:
( ) for B 4 ft
4
all D
N
q K
= ≤

315
( )
2
1
for B> 4ft
6
all D
N B
q K
B
+
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
where 1 0.33 1.33
f
D
D
K
B
⎛ ⎞
= + ≤
⎜ ⎟
⎝ ⎠
For the silty sand use N=
47 51 71
56.33 56
3
+ +
= (#3, 4, and 5)
Let’s assume B=4.5 ft and Df =0
qall =
56
6
( )
2
4.5 1
1 13.9
45
ksf
+
⎛ ⎞
=
⎜ ⎟
⎝ ⎠
This suggests that a B = 4.5 feet is excessive since
q0 =
2
B
Q
=
160
7.9 13.9 ksf
20.25
all
kips
q
sf
= =
Assume B < 4 ft, say B~ 3.5 ft , and use formula
qall= ( )
4
D
N
K kd =1+ 0.33Df /B
56
4
⎟
⎠
⎞
⎜
⎝
⎛
+
B
Df
33
.
0
1 and Df = 0 qall = 14 ksf
∴ qo =
2
B
Q
= 2
160
(3.5)
kips
= 13.06 ≅ 13 ksf 14 ksf OK
≤
For footings in area of borings # 1 and #2, they will be deeper by 1-story (ie. for 5-story building). That
places the shallow foundation at elevation 274 ft. This area will have bearing in the same strata. N= 32
and using B= 3.50’ and Df = 4.5’
qall=
4
N
kd kd =1+0.33Df /B 1.33
≤ Kd =
32
4
0.33 4.5
1
3.50
x
⎛ ⎞
+
⎜ ⎟
⎝ ⎠
1.33
=
10.64 13 ksf NOT GOOD
all
q ksf
= <
Let’s use B= 3.90 feet 10.64 ksf
all
q = 0 2
10.51 ksf 10.64 ksf
Q
q
B
⎛ ⎞
= = ≤
⎜ ⎟
⎝ ⎠
Use B = 3.90 feet.

316
Chapter 18
Shallow Foundations
Symbols for Shallow Foundations

317
Properties of Reinforcing Steel (British and SI units).

318
*Footings–01: Analyze a simple square footing.
(Revision: Sept-08)
Design a square reinforced concrete footing for a column 15”x15” with 4 # 8 rebars, and,
'
100 3,000
120 60,000
4 10 2.5
L c
L y
allowable ult
D kips f psi
L kips f psi
q ksf from q ksf and FS
= =
= =
= = =
Solution:
1) Footing size for service loads:
220
7.42
4
a
Q kips
B feet
q ksf
= = = therefore use B = 7.5 feet.
2) Check ultimate parameters: that is the actual soil pressure qo under Qult,
( ) ( )
1.2 1.6 1.2 100 1.6 120 120 192 312
ULT L L
Q D L kips
= + = + = + =
( )
2
2
312
5.5
7.5
U
O
Q
q ksf
B ft
∴ = = = < 10 ksf for ult
q …GOOD
3) Compute the allowable concrete shear strength ( )
c
v allowable
4 ' 0.75
all C
v f where for shear and torsion
φ ϕ
= = ,
( )
4 0.75 3,000 164 23.7
C all
v v psi psi ksf
= = = ≈
4) Find d, the effective depth, (in this case two-way shear governs) in feet.
15”
B = 7.5 feet
15” + d

319
( )
2 2 2
2
2 2
2
0
4 4 4
5.5 5.5 15 15 5.5
23.7 23.7 7.5 0
4 4 12 12 4
1.25 3.0 0
o o o
c c
q q q
d v d v w B w
d d
d d
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
+ + + − − =
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎡ ⎤
⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞
+ + + − − =
⎢ ⎥
⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎢ ⎥
⎣ ⎦
+ − =
Which yield to two solutions for d =
1.22
2.47
ft
ft
+
−
and using the modified formula equation 2a:
When the column has a rectangular area bxc, the formula is,
( )
( ) ( )
2
2 2
2
7.5 5.5
15 15
4 2 4 2 0
12 12 23.7
1.25 3.26 0 0.39
o
c
BLq
d b c d d d
v
d d which yields d ft
⎡ ⎤
⎛ ⎞ ⎛ ⎞
+ + − = + + − =
⎢ ⎥
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎢ ⎥
⎝ ⎠ ⎣ ⎦
+ − = = +
Use the largest d = 1.22 feet = 14.6 inches; round-out to d = 15 inches. It is not necessary to check for
wide-beam shear on a square footing.
5) Compute the area of steel AS for flexure.
Unit strip of the cantilever arm =
15
7.5
12
3.13
2 2
B w
L L ft
⎛ ⎞
⎛ ⎞
−⎜ ⎟
⎜ ⎟
−
⎛ ⎞ ⎝ ⎠
⎝ ⎠
= ∴ = =
⎜ ⎟
⎝ ⎠
L
1 ft
w
15”
B = 7.5 ft
0
q

320
The cantilever moment:
( ) ( )
2
2
2
5.5 3.13 12
359
2 2
o
U
k
ft in
ft
q L
M in kips
⎡ ⎤
⎛ ⎞
⎢ ⎥
⎜ ⎟
⎛ ⎞ ⎝ ⎠
⎣ ⎦
= = = −
⎜ ⎟
⎝ ⎠
( )
( )( )
0.9
2
60
1.96
0.85 ' 0.85 3 12
U S y
S y S
S
C
d a
M A f where for tension
A f A
a A
f b in
φ φ
−
⎛ ⎞
= =
⎜ ⎟
⎝ ⎠
= = =
Substituting a into the U
M equation above,
( )
( )
( )( )
2
2
2
0.9
2 0.9 2
359
15 1.96
0.9 60 2
15 0.98 5.99 0
15.3 6.11 0
0.41
U
U S y S
y
S
S
S S
s s
S
M
d a d a
M A f A
f
A
in kip
A in
ksi
A A
A A
A in per foot of footing
− −
⎛ ⎞ ⎛ ⎞
= ∴ =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
− ⎛ ⎞
= −
⎜ ⎟
⎝ ⎠
− − =
− + =
=
The total steel required across the footing is S
A = 7.5 ft (0.41 in2
/ft) = 3.08 in2
(Check ACI 10.5.1) for
minimum steel and ACI 7.12 for temperature and shrinkage,
( )
'
min
min
0.41
0.0023 0.0018
12 15
3
0.0018
200
0.0018
S
c
y
y
A
bd
f
bh or
f
f
ρ
ρ
ρ
= = = >
= =
= =
Therefore, S
A = 7.5 (0.0023)(12)(15) = 3.1 in2
or 0.59 in2
per foot of footing
For B = 90” (7.5’) use 6 # 7 bars ( S
A = 3.18 in2
) @ 12 inches on center
or 5 # 8 bars ( S
A = 3.95 in2
) @ 12 inches on center
Check for Development length d
L (ACI-318-08.12.3), and the embedment length of the dowels.

322
*Footings–02: Add a moment to the load on a footing.
(Revision: Sept-08)
The footing shown below is a square footing with the dimensions and loads as shown.
a) Compute the load’s eccentricity e.
b) Check the bearing pressure at ultimate load.
c) Calculate the wide beam shear.
d) Determine the flexural moment for a strip of footing 1 foot wide.
( )
4 1.5
all
q ksf with a FS
= = , ' 3
C
f ksi
= , 60
y
f ksi
= . P = DL + LL
Solution:
a) Compute the load eccentricity:
20
0.121
165
M
e ft
P
= = =
( )
max min
6 0.121'
6 165
1 1
7 ' 7 ' 7 '
3.72 3.02
P e
q
A B x
q ksf and q ksf
⎛ ⎞
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= ± = ± =
⎜ ⎟
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= =
b) Check bearing pressure at ultimate load: 1.2(100) 1.6(65) 224
u
Q kips
= + =
max min
224 224
3.72 5.05 3.02 4.1
165 165
u u
all all
Q Q
q q ksf and q q ksf
P P
⎛ ⎞ ⎛ ⎞
⎛ ⎞ ⎛ ⎞
= = = = = =
⎜ ⎟ ⎜ ⎟
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
⎝ ⎠ ⎝ ⎠
Then consider all
q = 5.1 ksf < all
q (FS) = 4 (1.5) = 6 ksf, GOOD.
P
M
P = 165 kips
M = 20 ft-kips
DL = 100 kips
LL = 65 kips d 3’-d
15”
T
7 feet

323
c) Calculate the allowable one-way (wide beam) shear; assume a d = 1 ft;
5.1 4.1
4.6
2
q
+
= =
7 1
( ) 4.6( 1 ) 9.2
2 2 2 2
U
B c
V q T kips
= − − = − − = Equation (1)
'
'
2 110
(9.2)(1000)
9.3 12
(0.75)(110)(12)
2
c
u
c
The allowable shear f psi
V
The required d inches in assumed
f b
φ
ϕ
= =
= = = ≈
Equation (2)
d) Determine the flexural moment for a strip 1 foot wide:
3 2
0 0
5.9 0.157
25.9
2
x
x x ft kip
M V dx dx
ft
⎛ ⎞
− −
= = =
⎜ ⎟
⎝ ⎠
∫ ∫
V
U
3’ -
d
4.1 ksf
5.1 ksf

324
*Footings–03: Find the thickness T and the As of the previous problem.
(Review:
Find, 1) The soil pressure under the footing for the given loads,
2) The footing thickness T, and
3) The flexural steel reinforcement As.
Given: 60
y
f ksi
= , ' 3
C
f ksi
= , 4
a
q ksf
= , 65
L
D kips
= , 100
L
L kips
= , 20
M ft kips
= −
L L
N D L
= +
The actual soil pressure o
q (versus the allowable soil bearing capacity):
( )( )
( )
( )( )
3
3
4
4
max min
7' 7'
200
12 12
20 . 3.5
165
7' 7' 200
3.7 3.0 4
o
o
o o
N Mc bd
q where I ft
A I
k ft ft
k
q
x ft
q ksf and q ksf ksf allowable GOOD
= ± = = =
= ± =
= = <
M
N
T
7’
7’
c=3.5
x
My
y

325
b) The ultimate load on the soil:
( ) ( )
( )
( )( )
( )( )
4
max min
1.2 1.6 1.2 65 1.6 100 238
1.6 1.6 20 . 32
32 . 3.5
238
7 ' 7 ' 200
5.9 4.7
u L L
u LL
u u
u
u u
Q D L kips kips kips
M M ft kips ft kips
ft kips ft
Q M c kips
q
A I ft
q ksf and q ksf
= + = + =
= = = ⋅
= ± = ± =
= =
My
N
x
3.0 ksf
3.7 ksf
My
Q
x
4.7 ksf
5.9 ksf

326
a) Determine the thickness of the footing T.
Check wide-beam shear: Assume d =12”, (controls for rectangular footing).
From 0
y
f =
∑ , the ultimate shear is
82,600
u
V lbs
= , but 2 '
u c
V f bd
φ
= ,
( )( ) ( )
82, 600
10.6
2 ' 0.85 2 3000 84
u
c
V lbs
d in
f b in
φ
= = = < 12 in assumed.
Check for punching shear (controls for square footings, such as this one).
d
T
3”
3” minimum cover
7’ My
Q
d
12”
2’
d
w =12”
Vu
Critical
Section
c =
12”
b0 /4
d/2
7’
Critical Section
ACI 11.11.1.2

327
4 '
u c o
V f b d
φ
= , where bo is the perimeter of the critical section; 4 2
2
o
d
b c
⎛ ⎞
⎛ ⎞
= + ⎜ ⎟
⎜ ⎟
⎝ ⎠
⎝ ⎠
( )
( )
( )( )( )( )( )
5.9 7 7 2 2 1000
14.8
4 ' 0.85 4 3000 4 24
u
c
ksf ft x ft ft x ft
V
d in
f b in
φ
−
∴ = = = assumed;
must increase d.
Recalculate by trying d = 14.0 in.
( )
( )
( )( )( )( )( )
5.9 7 7 2.17 2.17 1000
13.5
4 ' 0.85 4 3000 4 26
u
c
ksf ft x ft ft x ft
V
d in
f b in
φ
−
∴ = = =
d = 13.5 in < 14.0 in, therefore is Good!!
The footing thickness T = 14 in + 1 bar diameter (1”) + 3 in (cover) = 18 in.
Footing dimensions: 7.5 feet x 7.5 feet x 18 in.
Finding the flexural reinforcement As:
( )( )( )
2
2
5.9 7 3
186 .
2 2
s
u
ksf ft ft
q Bl
M kip ft
= = =
( )
0.85 '
s y
c
A f
a where b B
f b
= = then
( )( )( )
0.85 3000 7
0.85 '
0.3
60,000
c
s
y
ft a
f Ba
A a
f
= = =
The ultimate moment Mu is given by:
2
u s y
a
M A f d
φ
⎛ ⎞
= −
⎜ ⎟
⎝ ⎠
with Mu = 186 kip-ft , and 0.9
φ = .
( )
2
2
2 2
186 . 0.9 0.3 60 144 1.16
2
0.07 0.3 0.021
s
kip in a
kip ft a ft ft
in ft
a ft and A a
⎛ ⎞
⎛ ⎞ ⎛ ⎞
∴ = − =
⎜ ⎟
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
⎝ ⎠
= ∴ = =
Percentage of steel
( )( )
2
0.021
0.00257
7 1.167
s s
A A ft
p
bd Bd ft ft
= = = = > min 0.0018
p =

328
2 2
min 0.021 3.024
s
A ft in
= = Therefore, use 6 # 7 bars at 12 inches.
e) Check development length, Ld (ACI 12.2.2) 0.04 0.0004
'
y
d b b y
c
f
L A d f
f
⎛ ⎞
= >
⎜ ⎟
⎜ ⎟
⎝ ⎠
( )
( )( )
0.44 60, 000
0.04 0.0004 0.75 60, 000
3000
19.3 18
d b
d
L A
L in in
⎛ ⎞
= >
⎜ ⎟
⎝ ⎠
= >
( ) ( )
provided
12
Actual 7 6 3 75
2 1
d
c in
L B cover ft in in in
ft
⎛ ⎞
= − − = − − =
⎜ ⎟
⎝ ⎠
f) Check bearing strength on concrete (ACI 318-02 10.15)
Bearing strength = 2
0.85 'c g
g
A
f A
A
φ
⎛ ⎞
⎜ ⎟
⎜ ⎟
⎝ ⎠
where 2
2
g
A
A
⎛ ⎞
≤
⎜ ⎟
⎜ ⎟
⎝ ⎠
Ag is the area of the column and A2 is the area of the footing.
Then ( ) ( )( )
2
2
2
3
0.85 ' 0.85 0.70 144 2 514
c g
g
A kips
f A in kips
A in
φ
⎛ ⎞ ⎛ ⎞
= =
⎜ ⎟ ⎜ ⎟
⎜ ⎟ ⎝ ⎠
⎝ ⎠
But, the factored column load U = 261 kips < 514 kips. Good !
g) Dowels to column:
Since bearing strength is adequate, a minimum area of dowels should be provided across the interference
of the column and footing (ACI 318-02 15.8.2.1)
Minimum area of steel = 0.005 (area of column) = 0.005 (144 in2
) = 0.72 in2
.
Use 4 # 4 bars ( )
2 2
0.20 4 0.80
S
A in in
= =
h) Final design

329
*Footings–04: Find the dimensions B x L of a rectangular footing.
(Revision: Sept-08)
Find the footing dimensions B x L to carry a moment induced by winds of 800 kN-m.
Solution:
Select a test value for B x L. Set B x L = B2
and check the increase in soil pressure due to wind
( )
2 2
2
(800 800 )
8 2.82
200
800
0.5 6 0.5 3
1600
a
kN kN
N
B m B m
kN
q
m
M
e m L m
N
+
= = = ∴ =
⎛ ⎞
⎜ ⎟
⎝ ⎠
= = = ∴ ≥ =
If L = 3 m, try for max 2 avg
q q
≅ iterate by trying footing: 2.5 m x 4 m = B x L.
( )
2
max
1600
160
10
6 0.5
1600
1 280
6 10 4
1
avg
kN
N
q kPa
A m
N
q kPa
e
BL
L
= = =
⎡ ⎤
= = + =
⎢ ⎥
⎛ ⎞ ⎣ ⎦
+
⎜ ⎟
⎝ ⎠
Note that max
q exceeds avg
q by 33 %, therefore increase the area.
Try using 2.75 m x 4.5 m dimensions;

330
( )
( )
2
max
1600
130
2.75 4.5
6 0.5
130 1 217
6 4.5
1
avg
kN
N
q kPa
A m
N
q kPa
e
BL
L
= = =
⎡ ⎤
= = + =
⎢ ⎥
⎛ ⎞ ⎣ ⎦
+
⎜ ⎟
⎝ ⎠
Iterate again, with B = 3.0 m and L = 5.0 m
( )
( )
max
1600
107
3 5
6 0.5
1600
1 171
6 15 5
1
avg
kN
N
q kPa
A m m
N
q kPa GOOD
e
BL
L
= = =
⎡ ⎤
= = + =
⎢ ⎥
⎛ ⎞ ⎣ ⎦
+
⎜ ⎟
⎝ ⎠
The footing dimensions are B = 3 m by L = 5 m.

331
*Footings–05: Design the steel for the previous problem.
(Revision: Sept-08)
Design the previous footing using ' 21 415
c y
f MPa and f MPa
= = .
Solution.
1) Check the ultimate pressures:
( ) ( ) ( ) ( )
1.2 1.6 1.2 800 1.6 800 2240
u L L
N D L kN
= + = + =
800
0.5
1600
M
e m
N
= = =
( )
( )
max
min
6 0.5
2240
1 239
6 15 5
1
6 0.5
2240
1 60
6 15 5
1
u
N
q kPa
e
A
L
N
q kPa
e
A
L
⎡ ⎤
⎛ ⎞
= = + =
⎢ ⎥
⎜ ⎟
⎛ ⎞ ⎝ ⎠⎣ ⎦
+
⎜ ⎟
⎝ ⎠
⎡ ⎤
⎛ ⎞
= = − =
⎢ ⎥
⎜ ⎟
⎛ ⎞ ⎝ ⎠⎣ ⎦
−
⎜ ⎟
⎝ ⎠
q max = 239 kPa and q min = 60 kPa

332
2
2240
149 200
15
avg all
kN
q kPa q kPa
m
= = < = Good
2a) Calculate footing depth T based on punching shear, for a ' 21
c
f MPa
= , and
1.29
c
V MPa
=
.
Using the simplified equation for a square footing:
( )
( )
2
2
15 149
4 2 0.5 0.5 0
2240
0.5 0.25 0 0.50
x
T m m T
T T T m
+ + − =
+ − = ∴ ≅
2b) Calculate footing depth T based on wide beam shear:
The shear is calculated from the outer edge of footing ( x = 0 ) inwards towards a distance d
from the column ( x = 2.225 – d ) :
( )
( )
2.25
2
0 0 0
2
2
40.2
266 40.2 266
2
598 266 20.1 2.25
1290
40.8 24.7 0 0.60
2 2
T
x x
c
dv qdx
x
V qdx x dx x
V T T
V
V T T T m
T T
−
=
⎧ ⎫
⎛ ⎞
⎪ ⎪
= = − = −
⎨ ⎬
⎜ ⎟
⎪ ⎪
⎝ ⎠
⎩ ⎭
= − − − =
= = ∴ + − = =
∫ ∫
We will use the highest of (2a) or (2b), therefore T = 0.60 m.

333
3) a) Find AS ( Longitudinal ):
( )( )
415
23.3
0.85 ' 0.85 21 1
s y s
s
c
A f A
a A
f b
= = =
2.25
2.25 2.25 2 2 3
0 0 0
40.2 266 40.2
266 597 .
2 2 6
u
x x x
M Vdx x dx kN m
⎛ ⎞ ⎡ ⎤
= = − = − =
⎜ ⎟ ⎢ ⎥
⎝ ⎠ ⎣ ⎦
∫ ∫
( )
2 2
2
2 2
23.3 597
0.6 0.0515 1.37 10 0
2 0.9 415
28.2
u
u s y s
y
s s s
s
M
a a
M A f d A d
f
A A A x
cm
A
m
φ
φ
−
⎛ ⎞ ⎛ ⎞
= − ∴ − =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
⎛ ⎞
− = = − + =
⎜ ⎟
⎝ ⎠
=
Check
( )
0.00282
0.0047
1 0.6
s
A
p
bd
= = > 0.002 GOOD
3) b) Find AS (transverse): Use the high average.
( ) ( )
( ) ( )
max
2
2
2
149 266 20.37
3 0.5
0.5 215
2
2
168 .
2 2 2
ave
u
q q q kPa
B
q
wl
M kN m
= + = + =
⎛ ⎞
−
⎛ ⎞
−
⎡ ⎤ ⎜ ⎟
⎜ ⎟
⎢ ⎥ ⎜ ⎟
⎝ ⎠
⎣ ⎦ ⎝ ⎠
= = = =
and
2
2
0.0515 0.000386 0 7.61
2
u
s s s s
y
M
a cm
A d A A A
f m
φ
⎛ ⎞
− = ∴ − + = ∴ =
⎜ ⎟
⎝ ⎠
Check for
( )
4 2
7.61 10
0.00126 0.002
1 0.6
s
A x m
bT
ρ
−
= = = <
( )( ) ( )( )
2
2
Use minimum 0.002 0.002 1 0.6 0.0012 12
s
cm
m
A b T
m m
∴ = = = =
4) Check footing:
Longitudinal steel: 28.2
2
cm
m
(3 m) = 84.6 cm2
2
3
2
1
8.46 10
10, 000
m
x m
cm
−
=
Therefore use 17 # 25 mm bars 17.6
at cms o c
−

335
*Footings–06: Design a continuous footing for a pre-cast warehouse wall.
(Revision: Sept-08)
Design a continuous footing for the warehouse wall with the loads shown below:
Solution:
1) Assume an initial footing thickness T = 12”:
From ACI 7.7.2, the minimum cover is 3” from the steel to the footing invert. Assume # 4 bars.
Therefore, d = 12” - 3” - 0.5” / 2 = 8.75”
Roof
Precast concrete wall
Overhead Crane
12 in
4 ft
Floor Slab
12 in
T=12 in
B

336
Find the ultimate soil pressure qult.
The actual soil pressure from the structure is,
qo = qall – qconcrete – qsoil-above footing
qo = 2 ksf – (1 ft )( 0.15 k/ ft3
) – ( 3 ft )( 0.110 k/ ft3
) = 2 – 0.15 – 0.33 = 1.52 ksf
Estimate B = Q/ qo = 4.2 k/ 1.52 ksf = 2.76 ft. per unit length. Therefore, assume B = 3 ft.
Ultimate load, U = 1.2 DL +1.6 LL = 1.2 ( 3 ) + 1.6 ( 1.2 ) = 4.0 + 2.0 = 6.0 kips
Therefore, the soil pressure at ultimate loads qu is:
qu = U / ( B )( 1 ) = 6.0 kips/ ( 3 ft )( 1 ft ) = 2.0 ksf < qult = 4 ksf Æ GOOD
2) Check the shear strength of the footing:
The critical section for shear section occurs at a distance d from the face of the wall.
(ACI 15.5 & ACI 11.11.1 ).
( ) ( )
2.1
12 1
o
u o
q ksf
V d q
=
= −
b =12
in
T
B
a
d
b =12”
in
12” 12”
B = 3 ft
d
12 in - d
Vu
qo = 2.1 ksf
T

337
The ultimate shear Vu = (12” – 8.75”) ( 1 ft / 12 in ) x 2.1 k/ ft2
= 0.57 kips / ft of wall
The concrete shear strength must be: bd
f'
2
V
V c
c
u φ
φ =
≤
ft
/
kips
0.57
wall
of
ft
/
kips
9.8
8.75in
12in
3000
2
0.85
Vu >>
=
⋅
⋅
⋅
⋅
=
Since Vu << ØVc , we can reduce the thickness of the footing T from 1.0 ft to say 0.85 ft (~10”).
Therefore, d = 10” – 3” – 0.5 / 2 = 6.5” > 6” = dmin from ACI 15.4
Rechecking, GOOD
wall
of
ft
/
kips
7.3
6.5in
12in
3000
2
0.85
Vu →
=
⋅
⋅
⋅
⋅
=
Therefore, the total thickness T = d + bar diameter + 3” = 10” + 1” + 3” = 14”
3) Design the flexural reinforcement, As ( ACI 15.4):
Mu = qo ℓ2
/ 2 where ℓ = 12”
Mu = (2.1 ksf x 1 ft2
)/ 2 = 1.05 kip-ft / ft of wall
But, a = Asfy / [ 0.85 f’c ( b ) ] = [ (60 ksi ) As ) ] / [ 0.85 ( 3 ksi )( 12 in ) ] = 1.96 As (inches)
and, Mn = Asfy ( d – a/2) = As ( 60 k / in2
) ( d – a / 2 ) = 60 As ( 6.5 in – 9.8 As )
But, Mu = Ø Mn = 0.9 Mn
Therefore, 1.05 kips-in / in = 0.9 (60 k / in2
) As ( 6.5 in – 9.8 As )
53 As
2
– 351As – 1.05 = 0
Two possible answers: As ( 1 ) = 6.6 in2
per ft. of wall
As ( 2 ) = 0.003 in2
per ft. of wall
The percentages of steel with As ( 1 ) & As ( 2 ) ( ACI 7.12.2.1 )

338
ρ1 = As(1) / bd = 6.6 in2
/ (12 in )( 6.5 in) = 0.085
ρ2 = As(2) / bd = 0.003 in2
/ (12 in )( 6.5 in) = 0.0004 < 0.0018 minimum steel
The maximum steel percentages allowed ρmax = 0.75 ρb, where
ρb = ( 0.85 f’c / fy ) β ( 87,000 / ( 87,000 + fy ) )
= ( 0.85 ( 3 ) / 60 ) 0.85 ( 87,000 / ( 87,000 + 60,000 ) = 0.021
therefore,
ρmax = 0.75 ρb = 0.75 ( 0.021 ) = 0.016
Note that ρ1 = 0.085 > ρmax = 0.016 therefore, use ρmin = 0.0018
Therefore,
As = ρmin b d = ( 0.0018 )( 12 in )( 6.5 in )
= 0.14 in2
per ft. of wall Æ use 1 # 4 every ft. of wall ( As = 0.20 in2
)
4) Check the development length, Ld ( ACI 12.2 ):
8.8"
3000
60,000
0.20
0.04
f
d
0.0004
than
less
not
(But
f'
f
A
0.04
L y
b
c
y
b
d =
⋅
⋅
=
= )
Ld = 8.8” or 12” Æ Clearly, 12” controls.
Presently we have 12” – 3” cover = 9” < 12”. Therefore, we are missing 3” on each side.
Increase the footing width B by 6” to B = 3.5 ft.
(Note that increasing B, reduces qo, and the design could be further optimized.)

339
Therefore,
Use minimum steel in longitudinal direction, to offset shrinkage and temperature effects ( ACI
7.12 ):
As = ( 0.0018 )( b )( d ) = 0.0018 ( 42 in )( 6.5 in ) = 0.49 in2
Provide 3 # 4 bars at 12” o.c. ( As = 0.60 in2
)
B = 3.5 ft
T = 13”
As = 1 # 4 @ 12” along the wall

340
24" STEEL COLUMN 40 ' HIGH
WITH 1" THICK WALLS
20'
20'
D
32'
P = 10 k
B
X
Y
Z
**Footings–07: Design the footings of a large billboard sign.
(Revised Oct-09)
Design a spread footing for the billboard sign shown below using FBC-2004 and ASCE 7- 02.
Ignore the torsion and the wind load on the column, and the water table.
Given: γ = 150 pcf c = 150 pcf φ = 20° V = 146 mph
solution:
STEP #1: Find the wind load as per ASCE 7-02, assuming an Exposure C, Category I.
qz = 0.00256 Kz (IV)2
Kz = 0.98
I = 1.05
V = 146 mph
Gh = 1.26
Cf = 1.2
The sign shape factor is
M 32
= = 1.6
N 20
Therefore:
(34 psf) (1.26) (1.2) (32ft x 20ft)
= = 32.4 kips
1000
F
F = qz Gh Cf Af ∴ qz = 52 psf

341
.
Step #2: Calculate loads on footing
Weight of steel column = γs L A = 0.49
Mx = 10 kips x 15’ = 150 k-ft
My = 32.4 k x 30’ = 972 k-ft
Mz = 32.4 k x 15’ = 486 k-ft
Total (normal) load N = 10 k + 5 k = 15 kips
Step #3: Calculate the footing’s bearing capacity using Hansen’s formula.
c (cohesion) = 0.150 ksf
q = γDf = (embedment pressure) = (0.130 ksf)(3 ft) = 0.39 ksf
B = (footing width – initial assumptions) = 5 ft
L = (footing length – initial assumptions) = 15 ft
Nq (factor for embedment at φ = 20°) = eπ tan φ
tan2
(45+φ/2) = 6.40
Nc (factor for cohesion at φ = 20°) = (Nq – 1)cot φ = 14.83
Nγ (factor for width at φ = 20°) = 1.5 (Nq – 1)tan φ = 2.95
Fsq = (shape factor for embedment) = 1.0 + (B/L) sin φ = 1.11
Fsc = (shape factor for cohesion) = 1.0 + (Nq / Nc) (B/L) = 1.14
Fsγ = (shape factor for width) = 1.0 – 0.4 (B/L) = 0.867
Fdq = (depth factor for embedment) = 1 + 2 tan φ(1 – sin φ)2
(Df/B) = 1.19
Fdc = (depth factor for cohesion) = 1.0 + 0.4 (Df / B) = 1.24
Fdγ = (depth factor for width) = 1.0
Fic = (inclination factor) =
1 - H
0.5 -
(Af Ca)
⎛ ⎞
⎜ ⎟
⎝ ⎠
where ca = (0.6 to 1.0) c
Fiq = [ 1 – (0.5 H) / (V +Af Ca cot φ)]d
where 2 ≤ d ≤ 5
Fiγ = [1 – (0.7 H) / (V +Af Ca cot φ)]α
qult = c’ Nc Fsc Fdc Fic +⎯q Nq Fsq Fdq Fiq + 0.5 γ B’ Nγ Fsγ Fdγ Fiγ

342
Step #4: Assume:
B = 10’
L = 50’
D = 3’
B/L = 0.2
FS = 3.0
SC = 1.0 + (0.431 x 0.2) = 1.09
DC = 1.0 + (0.4)(3/10) = 1.12
Q = (130)x3 = 390
SQ = 1.0 + 0.2 sin 20 = 1.07
SJ = 1.0 – 0.4(0.2) = 0.92
DQ = 1 + (0.315)(3/10) = 1.09
( )
y x
max,min 2 2
- 6P e (3x10x50x0.150)
- 6P e
= + + where P = 15 kips + = 240 kips
B L BL FTG WT
P
q
BL
⎛ ⎞ ⎛ ⎞
⎜ ⎟
⎜ ⎟
⎝ ⎠
⎝ ⎠
y
x
x
y
M 972
e = = = 4.05'
P 240
M 150
e = = = 0.625'
P 240
max,min
240kips - 6(240)(0.625) - 6(240)(4.05)
= + + = 0.89 < 2.5 GOOD
500 102 x 50 10 x 502
q ksf ksf
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
Step #5: Check out (long direction),
MOT = 32.4 Kips (30+3) = 1069 kip-ft
MR = 5 Kips x 25’ + 225 Kips x 25’ = 5750 kip-ft
F.S. = 5750 / 1069 = 5.4 >> 1.5 GOOD
Check Sliding RS = Σ V tan f + CB = 240 tan f + 150 (10) = 1587 kips >> 32.4 kips
#3
LOAD COMBINATION = 0.75 (1.2D + 1.6L + 1.7W)
FACTORED LOADS: Pu = 1.05 x 240 Kips = 252 kips
Mu-x = 1.05 x 150 kip-ft = 158 kip-ft
Mu-y = 1.275 x 972 kip-ft = 1239 kip-ft
∴ MAX,MIN 2 2
336 6 x 336 x 0.625 6 x 336 x 4.9
Q = = 0.989 ksf, 0.019 ksf
10 x 50 10 x 50 10 x 50
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
± ±
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
Check beam shear: D = 36” – 4” = 32”
VU = 0.019x(50 – 21.33) + 0.15(50-21.33)(0.989 – 0.019) = 14.4 kip per feet
Punching shear will not govern by observation.

343
#3 Design for flexure in long direction
f’C = 3000 PSI and fY = 60000 PSI
A = AS x FY / 0.85 F’C B AS = MU x 12 / F FY (D – A/2)
A = 1.64 in. AS = 0.83 in2
AS 0.83
R = = = 0.0022 > 0.0018 OK
BD 12 x 32
⎛ ⎞
⎜ ⎟
⎝ ⎠
USE # 7 @ 8” O/C, As = 0.90 in2
OK
Therefore; ( )( )
1.64
132 -
2
= 0.9 60 = 126 kip-ft > 116 kip-ft GOOD
12
U
M
⎛ ⎞
⎜ ⎟
⎝ ⎠
Use a footing 10 feet x 50 feet x 3 feet thick with # 7 bars @ 8” on-center top and bottom, each
way.
NOTE: In lieu of such a large and expensive footing, a short drilled shaft would be an
efficient and inexpensive foundation. That alternative will be covered in the drilled shaft
section later in this course.

344
Chapter 19
Combined Footings
Symbols for Combined Footings

345
Chapter 20
Mat Foundations
Symbols for Mat Foundations

346
*Mat Foundations–01: Ultimate bearing capacity in a pure cohesive soil.
(Revision: Sept-08)
Determine the ultimate bearing capacity of a mat foundation measuring 45 feet long by 30 feet
wide placed 6.5 feet below the surface and resting upon a saturated clay stratum with cu = 1,950
lb/ft2
and φ = 0º.
Solution:
Mat foundations in purely cohesive soils have the following ultimate bearing capacity:
( )
( )
( )
( )
ult(net)
ult(net)
0.4
0.195
= 5.14 (1+ )(1+ )
0.195 30 0.40 6.5
= 5.14(1.95 ) [1+ ] [1 + ] =
45
12
30
f
u
D
B
q c
L B
ft ft
q ksf
ft ft
ksf

347
Chapter 21
Deep Foundations - Single Piles
Symbols for Single Piles of Deep Foundations

348
*Single-Pile–01: Pile capacity in a cohesive soil.
(Revision: Oct.-08)
A concrete pile 20 m long with a cross section of 381 mm x 381 mm is fully embedded in a
saturated clay stratum. The clay has γsat = 18.5 kN/m3
, φ=0º and cu = 70 kN/m2
. The water table lies
below the tip of the pile. Determine the allowable capacity of the pile for a FS = 3 using the α-
method.
Solution:
( ) ( ) ( )
point
The ultimate capacity of the pile is given by the simple formula,
( ) ( )
Notice that the value of the cohesion is reduced by the " " fac
ult
ult shaft p p u p u c u
Q
Q Q Q A q c perimeter L A c N c perimeter L
α
α
= + = + = +
( ) ( )
2 2 2
tor found in the graph below,
0.38 (70 / )(10.97) 0.75 (70 / )4(0.38 )(20 ) 1,890
The allowable capacity i
6
s,
1,89
3
30
0
3
ult
ult
all
Q m kN m kN m m m kN
Q kN
Q kN
∴ = + =
= = =

349
Chapter 22
Deep Foundations - Pile Groups and Caps
Symbols for Pile Groups and Caps of Deep Foundations

350
**Pile-caps–01: Design a pile cap for a 9-pile cluster.
(Revision: Oct-08)
Design a pile cap footing to support an 18” square column subjected to a live load reaction of 180
kips and a dead load reaction of 160 kips at service loads. The testing laboratory recommends an
ultimate pile load of 70 kips per pile, and a service pile load of 42 kips per pile. The vertical steel in
the column consists of 12 No.7 bars. Use ƒ‫׳‬c = 3000 psi, ƒy = 40,000 psi, and 12” diameter piles.
Solution.
Since the footing weight will be about 3 kips/pile, the net service load per pile is 42.0-3.0 = 39.0 kips/
pile. The number of piles required in N=W/P = 340/39 = 8.7, or 9 piles. Use a pile pattern as shown in
Fig. 1. The net ultimate load is used to design the footing; thus Wu = (1.4) (160) + (1.7) (180) = 530 kips,
and the load per pile is Pu = 530/9 = 58.9, say 59.0 kips/pile, which is less than the maximum ultimate
load, 70 kips/ pile. Punching shear around a single pile often governs the footing depth determination,
except in cases in which the loads are small. In this case, it will be shown that beam shear governs.
Referring to Fig. 2, we calculate the punching shear stress. After several trials, assume d = 19.5″.
The shear perimeter is bo = π(12 + d) = 99.0″. The permissible shear force around the pile will be,
Vc = 4√f‫׳‬c bod = 4√3000 (99) (19.5) / 1000 = 423 kips
Since the actual shear force is the nominal pile reaction, Pn = Pu/φ = 59.0/0.85 = 69.4 kips < 423 kips,
the pile will not punch through the pile cap (footing).
Figure 1 Figure 2

351
Figure 3 Figure 4
Perimeter shear (punching shear) must now be checked around the column in a similar manner. In this
case, all of the nominal pile reactions outside of the critical section plus any partial reactions outside of
the critical section will contribute to punching shear for the column. Refer to Figure 3. Assuming No. 6
bars will be used, clearance above the pile butts will be 3″ and embedment of the piles will be 6″. The
total dept required will be 28.75″. For practical reasons use 29″; this furnishes an effective depth d =
19.625″. Thus c = a + b = 18.0 + 19.625 = 37.625″ and bo = 4(37.625) = 150.0″. Hence, Vou = 472 kips
on 8 piles outside of the critical section as shown on Fig. 3. The permissible punching shear force (βc =
18/18 < 2) is given by (6.12) as
Vc = 4 √3000 (150) (19.625) / 1000 = 644.9 kips
The force to be resisted is Vn = Vou/φ = 472/0.85 = 555.3 kips; therefore the pile cap (footing) is
satisfactory for punching shear. Beam shear must now be checked. Refer to Fig. 4. Three piles exist
beyond the critical section, so Vu = (3) (59.0) = 177.0 kips. Since b = B= 8′-6″ = 102″, the permissible
beam shear (one-way shear) force on the critical section is
Vc = 2√ƒ′c bd = 2 √3000 (102) (19.625) / 1000 = 219.3 kips
The force to be resisted is the nominal shear force, Vn = Vu / φ = 177/ 0.85 = 208.2 kips. Hence the
footing is satisfactory for beam shear. The bending moment about the face of the column must now be
investigated. Refer to Fig. 4
Mu = (177) (27/12) = 398.3 ft-kips
φ Ru = Mu / bd2
= 398.3 x 12,000 / (102)(19.625)2
= 121.67 psi

352
Table 5.2 for ƒ ′c = 3000 psi and ƒy = 40,000 psi, discloses the fact that the steel ration required is less
than the minimum steel ration, ρ min = 200 / ƒy = 0.005. Further, if the steel ration required is increased
by 1/3, it will still be less than ρ min. It would appear that 4/3 times the required steel ration would satisfy
the 1983 ACI Code. However, the Code does not permit un-reinforced (plain concrete) pile caps. Since
any section having less than minimum reinforcement is usually considered to be un-reinforced, the
minimum are of steel will be provided. Thus,
As = (200/ƒy) bd = (200/ 40,000) (102) (19.925) = 10.0in.2
Use seventeen No. 7 bars (As = 10.2 in.2
).
The 1983 ACI Code is not explicit concerning minimum steel for footings. Hence, some structural
engineers use 0.002bh for minimum steel area if ƒy 40,000 psi and 0.0018bh if ƒy = 60,000 psi. This
corresponds to temperature and shrinkage reinforcement requirements. The assumed footing weight
must finally be checked. The total weight is
WF = (8.5) (8.5) (29) (12.5) / 1000 = 26.2 kips
And the weight per pile is 26.2/ 9 = 2.91 kips / pile. The assumed weight of 3.0 kips / pile is most
satisfactory. The final details are shown below

353
Chapter 23
Deep Foundations: Lateral Loads
Symbols for Lateral Loads on Deep Foundations

354
**Lateral loads on piles-01: Find the lateral load capacity of a steel pile.
(Revision: Oct-08)
Determine the lateral load capacity Qg of a steel H-pile (HP 250 x 0.834) fully embedded to a depth
of 25 m in very dense submerged sand. The top end of the pile is allowed to deflect laterally 8 mm.
For simplicity assume that there is no moment applied to the top of the pile (that is, Mg = 0).
Solution:
The subgrade modulus ks is a description of the reaction of the soil mass to vertical loads. The
modulus of horizontal subgrade reaction nh is a function of ks at any depth z,
z h
k n z
=
nh modulus of horizontal subgrade reaction
Type of soil lb/in3
kN/m3
Dry or moist sand
- loose 6.5 to 8.0 1,800 to 2,200
- medium 20 to 25 5,500 to 7,000
- dense 55 to 65 15,000 to 18,000
Submerged sand
- loose 3.5 to 5.0 1,000 to 1,400
- medium 12 to 18 3,500 to 4,500
- dense 32 to 45 9,000 to 12,000

355
From this table and the soil conditions noted above, choose nh = 12 MN/m3
for the modulus. Now
choose the parameters for the steel H-pile,
For future reference, this is the same table in British units,
For the steel HP 250 x 0.834 pile, the moment of inertia about the strong axis is Ip = 123 x 10-6
m4
,
its modulus of elasticity is Ep = 207 x 106
kN/m2
, the steel’s yield strength is Fy = 248 MN/m2
and
the pile depth d1 = 0.254 m.

356
The characteristic length T of a pile-soil system is given by,
( )( )
( )
6 6
5
5
207 10 123 10
1 16
12 000
p p
h
x x
E I
T . m
n ,
−
= = =
Therefore, the ratio L / T = 25 m / 1.16 m = 21.6 > 5, so this is a long pile.
The formula for the pile’s top end lateral deflection Δ at any depth z is given by,
3 2
0
g g
z z g
p p p p
Q T M T
Δ A B but M
E I E I
= + =
In this problem we are given this value of Δ = 8 mm, and we want to find the allowable lateral load
Qg, at a depth z = 0, where the coefficient Az is taken from a table of coefficients kz = nh z for long
piles.

357
The magnitude of the lateral load Qg limited by the displacement condition only is,
( )( )( )
( )( )
6 2 6 4
3
3
207 10 123 10
2 435 1 1
5
6
4
p p
g
z
0.008 m x kN / m x m
Δ(E I )
Q
T . .
kN
A
−
∴ = = =
Since the value of the allowable lateral load Qg found above is based on the limiting displacement
conditions only, and ignores that the pile has a moment capacity, that moment capacity at any
depth z is found through,
z m g
M A Q T
=
The table above shows that the maximum value of Am at any depth is 0.772. The maximum
allowable moment that the pile can carry is,
( )
( )
( )
( )
( )( )
6 4
2
1
123 10
248 240
2 0 254 2
240
268
0 772 1 16
p
max y
max
g
m
x m
I
M F MN / m kN m
d / . m /
kN m
M
Q kN
A T . . m
−
= = = −
−
∴ = = =
This last value of Qg emanating from the moment capacity is much larger than the value of Qg = 54
kN found for the deflection criterion.
Therefore use, Qg = 54 kN.

358
Chapter 24
Reinforced Concrete Retaining Walls and
Bridge Abutments
Symbols for Reinforced Concrete Retaining Walls

359
1'
2'
15'
16'-6"
1'-6"
18.33'
1.83'
Pv
Ph
3
1
1
2
3
4
5
6
6.11'
3' 1'-6" 5'-6"
10'
**RC Retaining Walls–01: Design a RC wall for a sloped backfill.
(Revision: Oct-08)
Design a reinforced concrete wall with a backfill γ = 125 pcf, an allowable soil bearing capacity of
qall = 3 ksf, and a friction at the base of φ = 30º. Design the wall and check for it’s stability under
working loads. (Note: All loads, shears and moments are per linear ft. of retaining wall).
Solutio
n:
2
Step 1: Find
The active la
tan 4
The presure
a
b a
K
p HK
γ
⎛
= ⎜
⎝
=
The forces o
½(12)(18
½(39)(1
Step 2: Stabi
v
h
P
P
=
=
=
Moment
About A
Area Area Force Arm Moment

360
(kip) (ft) (kip-ft)
1 ½ x 5.5x 1.83 = 5.03 x 0.125 = 0.63 1.83 1.2
and2 5.5 x 15.0 = 82.5 x 0.125 = 10.31 2.75 28.4
3 1.0 x 15.0 = 15.0 x 0.150 = 2.25 6.00 13.5
4 ½ x 0.5 x 15.0 = 3.75 x 0.150 = 0.56 6.67 3.7
5 10.0 x 1.5 = 15.0 x 0.150 = 2.25 5.00 11.3
6 3.0 x 2.0 = 6.0 x 0.125 = 0.75 8.50 6.4
Pv 2.00
Ph ΣH = 6.60 6.11 40.4
ΣV=18.75 ΣM=104.9
Location of Resultant
From point A, 104.9 = 5.6 ft
18.75
then e = 5.6 – 10 = 0.6 ft o.k. < B
2 6
Soil Pressure at Toe of Base
qmax = 18.75 (1 + 6 x 0.6 ) = 1.875 (1 + 0.36) = 2.55 ksf OK < 3 ksf
10 10
Check F against Sliding
Shear available along base = 18.75 kips x 0.58 = 10.9 kips
Passive force at toe
Use S = 2/3 (30º) = 20º , Pp = ( cos δ ) = 5.8 kips
½ γ H²
Pp = 5.8 (0.125) (3.5)² = 4.7 kips
2 (0.940)
Min. F = 10.9 = 1.7 kips , Max F = 10.9 + 4.7 = 15.6 = 2.4 kips OK without Key
6.6 6.6 6.6

361
Step 2: Design parameters.
Load Factors
Stem – Use 1.7 Ph
Base (toe and heel) – distribute ΣV uniformly over front B/3
Concrete and Steel Data
Capacity reduction factors: 0.90(flexure); 0.85(shear)
F’c = 3,000psi x 0.85 = 2,550 psi (for stress block)
Vc = 2 √3,000 = 110 psi
fy = 40,000 psi ; ρmin = 0.005 ; ρmax = 0.0278 ; ρshrinkage = 0.002
ld = 0.04 Ab (40,000) = 29.2 Ab (bottom bars) x 1.4 = 40.9 (top bars)
√3,000
Step 3: Design the stem of the wall.
Vertical Reinforcement
Ph = 1.7 ½ 39 (15)² = 7.46
M = 7.46 x 5 x 12 = 448 kip-in
Use 6” batter on front, then t = 12 + 6 = 18”
Use d = 18 – 4 = 14”
Assume arm = d – a/2 = 13”
T = 448/13 = 34.5 kip
As = 34.5 / (40 x 0.90) = 0.96 in²/ft
At bottom of wall Use #6 @ 5” ctrs
As = 1.06 in²/ft
ρ=1.06/(14 x 12) = 0.0063 >0.005 and <0.0278 OK
Pv
Ph
14"
5'
15'
12"

362
2"
8'-10"
6'
143
kip-in
286
kip-in
248
kip-in
448
kip-in
496
kip-in
133
kip-in
17
kip-in
0
5'
10'
15'
0
#6 @ 10"ctrs
#6 @ 5"ctrs
Ld = 40.9 (0.44) = 1' - 6"
top of base
14" plus std. hook
Check compressive stress block:
C = T = 34.5 / (2.55 x 12 x 0.90) = αmin = 1,25 in OK < 2.0
Check Shear:
7.46 / (14 x 12 x 0.85) = 52psi OK < 110 psi
Moment computations for steel cutoff:
At bottom : M = 448 kips
At 10’ level : M = 448 (10/15)³ = 133 kip-in
At 5’ level : M = 448 (5/15)³ = 17 kip-in
Resisting moment of steel :
#6 @ 5” ctrs. ; As = 1.06 in/ft
At bottom : Mr = 1.06 (40) (0.90) (13) = 496 kip-in
At top : Mr = 106 (40) (0.90) (7.5) = 286 kip-in

363
Compute level of cutoff for #6 @ 10” ctrs. :
As = 0.53 in²/ft ; ρ = 0.53 / (±11 x 12) = 0.004 > 0.002 < 0.005
Therefore, cutoff and develop bars above level where:
M = 0.53 (40) (±11) (0.90) = 158 kip-in
1.33
M = 158kip-ft @ ± 10’-6” level
Developmental length = 1’-6”
Level of Cutoff = 9’-0”
Step 4: Design the toe of the base of the wall.
Distribute ΣV over front B/3. Assume t = 18” , d =14”
ΣV = 18.75 kips = qmax = 5.63 ksf
(10 / 3)
wt. of soil over toe = 2.0 (0.125) = 0.25 ksf
wt of concrete base = 1.5 (0.150) = 0.23 ksf
Net toe pressure for design = qmax - wt. of soil over toe - wt of concrete base
= 5.63ksf - 0.25ksf - 0.23 ksf = 5.15 ksf
V = 5.15 x (3.0) = 15.45 kips
M = 15.45 x (3/2) x 12 =278 kip-in
Assume arm = 13 in
T = 278/13 = 21.4 kips = 0.59 in²/ft
(40 x 0.9)
Check ρ = 0.59 = 0.0035 > 002, shrinkage OK
5.15 kip
1'-6"
3'
d =14"
front face of stem

364
14 x 12
Ρ< 0.005, so increase As by 1/3 , then As = 0.59 x 1.33 = 0.80 in²/ft
use #6 @ 6” ctrs. As = 0.88 in²/ft
Compressive stress block and shear OK by inspection after stem computations
Development length : Extend full base width, therefore ld OK

365
Step 5: Design the heel of the base of the wall.
At Stem:
Weight of soil above heel at back face of stem = 15.0 (0.125) = 1.88 ksf
Weight of concrete base = 1.5 (0.150) = 0.23 ksf
Net pressure for design = 1.88 + 0.23 = 2.11 ksf
At Back:
Weight of soil above hell at back = 16.83 (0.125) = 2.10 ksf
Weight of concrete base = 0.23 ksf
Net pressure for design = 2.33 ksf
V1 = ½ (2.11) (5.5) = 5.80 kips
V2 = ½ (2.33) (5.5) = 6.41 kips
Total V = 5.80 + 6.41 = 12.21 kips
M1= 5.8 (1/3) (5.5) (12) = 128 kip-in
M2 = 6.41 (2/3) (5.5) (12)=282 kip-in
Total M = 128 + 282 = 410 kip-in
Assume arm = 13”
T= 410/13 = 31.5 kip
As = 31.5kip = 0.88 in²/ft
(40 x 0.9)
Use #6 @ 6” ctrs. As = 0.88 in²/ft ρ > 0.002 and >0.005 and < 0.0278 OK
Compressive stress block and shear OK
Development length: Extend full base width, therefore ld OK
Horizontal shrinkage steel in stem:
1'-6"
5'-6"
d =14"
2.33 ksf
2.11 ksf

366
Required: 0.002 (±15) (12) = 0.36 in²/ft of height
Use #4 @ 9” ctrs. front As = 0.27 in²/ft
Use #4 @ 18” ctrs back As = 0.13 in²/ft
Total As = 0.27 + 0.13 = 0.40 in²/ft
Horizontal shrinkage steel in base: Use #4 @ 12 ctrs. top and bottom As = 0.40 in²/ft
Step 6: Finished sketch of the wall.
3
1
12"
16'-6"
2'
1'-6"
3' 1'-6" 5'-6"
Vert.
bars,
Vert.
bars,
back
face
back
face
#6
@
5"
#6
@
10"
Provide 3" clear to
all bars except 11
2 "
clear to bars in
front face of stem
#4 @ 2'-0"
6" drains
@ 10'-0"
#6 @ 6" 2"x6" key
#6 @ 6"
#4 @ 12"
#4@ 9" front face
#4@18" back face

367
Chapter 25
Steel Sheet Pile Retaining Walls
Symbols for Steel Sheet Pile Retaining Walls

368
**Sheet-pile Wall-01: Free-Earth for cantilevered walls in granular soils.
(Updated: 9 April 2008)
Select an appropriate steel sheet-pile section and its total length L to retain a medium sand backfill for
the conditions shown below without an anchoring system.
Solution:
Step 1. Determine the pressures on the cantilevered wall.
115
65
35
0 27 3 70
=
=
= °
= =
'
a p
pcf
pcf
therefore
K . and K .
γ
γ
φ

369
Step 2. Determine the depth of embedment D.
The procedure will follow the following steps:
a) From statics, find the dimension z shown in the figure of previous page,
b) Assume a value for D,
c) Calculate z,
d) Use z to check if ΣM = 0; adjust D for convergence.
( )( )( )
( ) ( )( )
( ) ( )
( )
( ) ( )( )( )
1
2 1
1
115 14 0 27 435
435 65 0 27 435 17 6
65 6 29 435
409 435
65 6 29 115 14 6 56
409 10 560
= = =
= +
= + = +
= − − = −
= −
= − +
= +
= +
A a
A A a
E p a A
J p a p
p HK . psf
p p DK
. D . D
p D K K p D .
D
p D K K HK
D . .
D ,
γ
γ
γ
γ γ
1
2
10 5
435
620
14 860
3 860
3 860 435 10 5 14 435
=
=
=
=
=
− −
A
A
J
E
Try D . feet,
p psf ,
p psf ,
p , psf
p , psf
, .

370
Step 3. Determine the maximum moment Mmax (point of zero shear).
( )
( ) ( ) ( ) ( )
( ) ( ) ( )
( )
( )
( ) ( ) ( )
1 1 2 1 2
1 1 2 2
1 1
2 2
1 1
1 0
0
1
0
2 2 2 2
2 0
1
2 3 2 6
Σ =
+ + − =
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
+ + + + − + =
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
− −
=
+
Σ =
⎛ ⎞ ⎛ ⎞
⎛ ⎞
+ + + + − +
⎜ ⎟ ⎜ ⎟
⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
H
A A A E J E A
E A
E J
A A E J E
F
Area BAA Area AA A F Area ECJ Area EA A
D z D
Hp p p p p p p
Solving for z,
p p D Hp
z
p p
M about any point, say F ,
H D z
H p D p p p p
( ) ( )
2 2
2 2 1 0
6 6
⎛ ⎞ ⎛ ⎞
+ − =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
A A A
D D
p p p
( ) ( )( )
( )( )
( )( )
( )
1
1 1
2 1
2
1 2
435
1 06
65 6 29
1 1
435 14 3 040
2 2
1 1
435 1 0 218
2 2
1
2
= = =
−
= = =
= = =
− = +
A
'
p a
A
A
'
p a
p
y . feet
.
K K
P p H , lb
P p y . lb
also,
K K x P P
γ
γ
( )
( )
( )
( )( )
( )( )
( ) ( )
( )
( )
1 2
2
2
3
1 1 2 2 3 3
1
2
3
1 1 2 2 3 3
2 2 3 040 218
16 4 0
65 6 29
1
4 3 280
2
3
2
3
3
2 1 0
14 4 0
3 040 1 0 4 0 218 4 0 3 280
3 3 3
29
+ +
∴ = = = ∴ =
−
= − =
= + −
= + +
= +
=
⎛ ⎞
⎛ ⎞ ⎛ ⎞
= + − = + + + + −
⎜ ⎟
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
⎝ ⎠
=
'
p a
'
p a
max
max
max
P P ,
x x . feet
.
K K
P K K , lb
M Pl P l P l
where,
H
l y x
y
l x
x
l
. .
M Pl P l P l , . . . ,
M
γ
γ
300 1 030 4 360 26 000 312
+ − = − = −
, , , , ft lb in kips

372
Step 5. Determine the sheet-pile length L.
The total sheet-pile length L = H + D = 14.0 + 13.5 = 27.5 feet
( )( )
( )
3
3
45 29
26 000 12
10 8
29 000
116 10 7
− − =
= = =
∴ − =
s
max
s
If instead , an Ex Ten steel sec tion is used , f ksi
,
M
The required sec tion mod ulus S . in
f ,
Use a MP that provides a S . in at a lower cos t.
This is shown plotted with the red lines.

373
Chapter 26
MSE (Mechanically Stabilized Earth) Walls
Symbols for Mechanically Stabilized Earth Walls

374
**MSE Walls-01: Design the length L of geotextiles for a 16 ft wall.
(Revision: Oct-08)
Determine the length L of a geotextile-reinforcing for the 16 foot high temporary retaining wall
shown below. Also determine the required vertical spacing of the reinforcing layers Sv and the
required lap length ll. The geotextile chosen has an allowable strength of σG of 80 lb/inch, and the
available granular backfill has a unit weight of γ = 110 lb/ft3
and the angle of internal friction is φ
= 36°.
Comment: These geotextile walls are usually used for temporary civil works, such as detour
roadways, temporary abutments or excavation walls, etc. If the wall must become permanent,
then the face is stabilized and the plastic geotextile is protected from ultra-violet light through a
layer of shotcrete.
Solution:
The active pressure on the wall is,
( ) ( )
2 2
45 2 45 36 2 0 26
a a v a a
K K z where K tan / tan ( / ) .
σ σ γ ϕ
= = = °− = °− ° =
16 feet

375
Step 1: Find the vertical spacing SV of each layer of geotextile, where the Factor of Safety (FS) is
generally chosen between 1.3 to 1.5 for temporary walls. Permanent walls should use at least FS ≥
2. For this problem choose FS = 1.5.
( )
( )
( )
( )( )( )( )
( )
( )
( )( )( )( )
80 12
2 8 34
110 8 0 26 1 5
80 12
1 9 22
110 12 0 26 1 5
G G
V
a a
G
V
a
G
V
a
S
FS zK FS
At a depth of z = 8 feet from the top,
x lb / ft
S . feet inches
zK FS . .
At a depth of z = 12 feet from the top,
x lb / ft
S . feet inches
zK FS . .
At a
σ σ
σ γ
σ
γ
σ
γ
= =
= = = ≈
= = = ≈
( )
( )
( )( )( )( )
80 12
1 4 17
110 16 0 26 1 5
G
V
a
depth of z = 16 feet from the top,
x lb / ft
S . feet inches
zK FS . .
σ
γ
= = = ≈
Notice how the spacing becomes denser the deeper we go below the surface; choose SV = 20 inches
from z = 0 to z = 8 feet; below z = 8 feet use SV = 16 inches throughout, as shown in the figure.

376
Step 2: Find the length of each layer of geotextile L, which is composed of two parts, lR which is the
length of the geotextile within the Rankine failure zone, and le which is the effective length of the
geotextile beyond the failure zone (see the first figure, page 408). Again use FS = 1.5, and the angle
φF is the soil-to-geotextile angle of friction, which is usually assumed to be 2/3 φ of the soil. Other
values can be used, and a few are shown in this table,
Therefore, the length L of the geotextile layer is,
( )
( )
( ) ( )
( )
( )( )
( )
( )( )
0 26 1 5
45 2 2 45 36 2 2 0 445
0 51 0 438
V a V
R e
F
V
H z S K FS H z S . .
L l l
tan / tan tan / .
L . H z . S
φ ϕ
− −
= + = + = +
°− °− °
= − +
From this equation, we can now prepare a table with the required lengths.
z SV (0.51)(H-z) (0.438SV) L
inches feet feet feet feet feet
16 1.33 1.67 7.48 0.731 8.21
56 4.67 1.67 5.78 0.731 6.51
76 6.34 1.67 4.93 0.731 5.66
96 8.00 1.67 4.08 0.731 4.81
112 9.34 1.33 3.40 0.582 3.99
144 12.00 1.33 2.04 0.582 2.66
176 14.67 1.33 0.68 0.582 1.26
Based on this table, use L = 8.5 feet for z ≤ 8 feet and L = 4.0 feet for z > 8 feet.
Step 3: Find lap length ll for the geotextile, but never smaller than 3 feet,

377
( ) ( )( )
( )( )
( )
2
3
0 26 1 5
0 219
4 4 36
20
0 219 0 219 0
3
365 3
12
l
l
V a V
V
v
V
l
F
S FS S . .
l . S
tan tan
For a depth of z=16 inches,
inches
l . S . . feet feet
in / ft
There l f
fore, u eet
se
σ
σ ϕ
= = =
⎡ ⎤
°
⎣
=
⎦
⎛ ⎞
= = = <
⎜ ⎟
⎝ ⎠
Comment:
These MSE problems commonly use Rankine’s active pressure coefficient. However, the actual
value of K must depend on the degree of restraint of the type of reinforcement, as shown in this
figure:

300 solved problems in geotechnical engineering

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300 solved problems in geotechnical engineering