300 Solved Problems in 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 how to improve them. All problems are graded according to difficulty as follows: *
Easy; define general principles; typical of the PE examination questions;
** Slightly more difficult; typical of Master’s level problems; *** Professional level design problems; typical of real-life problems.
by Luis A. Prieto-Portar PhD, PE Professor of Civil and Environmental Engineering Florida International University
Telephone 305-348-2825 / Fax 305-348-2802 Website:
http://web.eng.fiu.edu/prieto / E-mail:
[email protected] © Copyright by L. Prieto-Portar, March 2010
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Table of Contents Table of Contents ............................................................................................................... iii Chapter 1 Soil Exploration.................................................................................................. 1 Symbols for Soil Exploration .......................................................................................... 1 *Exploration–01. Find the required number of borings and their depth. ....................... 2 *Exploration–02. Propose the number of samples from boring data. ............................. 3 * Exploration-03. What is a pressure cell? ...................................................................... 4 * Exploration-04. In-situ instrumentation used in the field. ............................................ 5 *Exploration–05. Which sampling method disturbs the soil the least? ........................... 6 *Exploration–06. Correcting the SPT for sampling and depth. ...................................... 7 *Exploration–07. Three methods used for SPT depth corrections. ................................. 8 *Exploration–08. Applying SPT corrections to a mat foundation. ................................. 9 *Exploration–09: The RQD versus fractured percentage of a rock sample. ................. 11 *Exploration–10. The Shear Vane Test determines the in-situ cohesion. ..................... 12 *Exploration–11. Using the SPT to estimate engineering parameters. ......................... 13 *Exploration–12. Reading a soil boring log. ................................................................ 14 *Exploration –13. Read a soil boring. ........................................................................... 15 *Exploration–14: Using a boring log to estimate soil engineering parameters. ............ 16 *Exploration–15. Find the shear strength of a soil from the CPT. ............................... 18 Chapter 2 Phase Relations of Soil..................................................................................... 21 Symbols for Phase Relations of soils ............................................................................ 21 Basic Concepts and Formulas for the Phases of Soils. .................................................. 22 *Phases of soils-01: Convert from metric units to SI and US units. ............................. 26 *Phases of soils–02: Compaction checked via the voids ratio e. ................................ 27 *Phases of soils–03: Finding the moisture of a soil. ..................................................... 28 *Phases of soils–04: Finding the wrong data. ............................................................... 29 *Phases of soils–05: Dry unit weight and degree of saturation. .................................... 30 *Phases of soils–06: Increasing the saturation of a soil. ............................................... 31 **Phases of soils – 07: Comparing two samples from a boring log.............................. 32 *Phases of soils–08: How much water is added to a truck to offset losses? ................ 34 *Phases of soils–09: Find γd, n, S and Ww. .................................................................. 35 *Phases of soils–10: Continued previous Phases-09 problem, adding water. ............. 36 iii
*Phases of soils–11: Use the block diagram to find the degree of saturation. .............. 37 *Phases of soils–12: Same as Phases-11 but setting the total volume V=1 m3. ............ 38 *Phases of soils–13: Same as Phases-06 but using a block diagram............................. 39 *Phases of soils–14: Block diagram for a saturated soil. .............................................. 40 *Phases of soils–15: Find the weight of water needed for saturation. .......................... 41 *Phases of soils–16: Identify the wrong piece of data. ................................................. 42 *Phases of soils–17: The apparent cheapest soil is not! ................................................ 43 *Phases of soils–18: Number of truck loads for Phases-17........................................... 44 *Phases of soils–19: How many truck loads are needed for a project? ......................... 45 *Phases of soils–20: Choose the cheapest fill supplier. ................................................ 46 *Phases of soils–21: Use a matrix to the find the missing data. .................................... 48 **Phases of soils–22: Find the voids ratio of “muck” (a highly organic soil). ............. 49 Chapter 3 Classification of Soils and Rocks ..................................................................... 50 Symbols for Classification of soils ................................................................................ 50 *Classify–01: Percentage of each of the four grain sizes (G, S, M & C). ..................... 51 *Classify–02: Coefficients of uniformity and curvature of granular soils. ................... 52 *Classify-03: Classify two soils using the USCS.......................................................... 53 *Classify-04: Manufacturing a “new” soil. ................................................................... 54 Classify – 05 .................................................................................................................. 56 Classify – 06 .................................................................................................................. 57 Classify – 07 .................................................................................................................. 58 Classify – 08 .................................................................................................................. 59 Classify – 09 .................................................................................................................. 60 Classify – 10 .................................................................................................................. 61 Classify – 11 .................................................................................................................. 62 Chapter 4 Compaction and Soil Improvement.................................................................. 63 Symbols for Compaction ............................................................................................... 63 *Compaction–01: Find the optimum moisture content (OMC). ................................... 64 *Compaction–02: Find maximum dry unit weight in SI units. ..................................... 65 *Compaction-03: What is the saturation S at the OMC? .............................................. 66 *Compaction-04: Number of truck loads required. ....................................................... 67 *Compaction-05: What is the saturation S at the OMC? .............................................. 68 *Compaction-06: Definition of the relative compaction (RC). ..................................... 69 iv
*Compaction-07: The relative compaction (RC) of a soil. ........................................... 70 *Compaction-08: Converting volumes from borrow pits and truck loads. ................... 71 **Compaction-09: Ranges of water and fill required for a road. .................................. 72 **Compaction-10: Find the family of saturation curves for compaction. ..................... 74 **Compaction-11: Water needed to reach maximum density in the field. ................... 77 **Compaction-12: Fill volumes and truck load requirements for a levee. ................... 79 **Compaction-13: Multiple choice compaction problem of a levee. ........................... 81 Chapter 5 Permeability of Soils ........................................................................................ 84 Symbols for Permeability .............................................................................................. 84 *Permeability–01: Types of permeability tests and common units. .............................. 85 *Permeability-02: Use of Hazen’s formula to estimate the k of an aquifer. ................. 86 *Permeability-03: Flow in a sand layer from a canal to a river. ................................... 87 *Permeability-04: Find the equivalent horizontal permeability of two layers. ............. 88 *Permeability-05: Equivalent vertical and horizontal permeabilities. .......................... 89 *Permeability-06: Ratio of horizontal to vertical permeabilities. ................................. 90 *Permeability–07: Do not confuse a horizontal with a vertical permeability. .............. 91 *Permeability-08: Permeability as a function of the voids ratio e. ............................... 92 *Permeability–09: Uplift pressures from vertical flows................................................ 93 *Permeability-10: Capillary rise in tubes of differing diameters. ................................. 94 *Permeability-11: Rise of the water table due to capillarity saturation. ....................... 96 *Permeability-12: Find the capillary rise hc in a silt stratum using Hazen. .................. 97 *Permeability-13: Back-hoe trench test to estimate the field permeability................... 98 **Permeability-14: Seepage loss from an impounding pond. ....................................... 99 Chapter 6 Seepage and Flow-nets ................................................................................... 103 Symbols for Seepage and Flow-nets ........................................................................... 103 *Flownets-01: Correcting flawed flow-nets. ............................................................... 104 *Flow-nets-02: A flow-net beneath a dam with a partial cutoff wall. ......................... 105 *Flow-nets-03: The velocity of the flow at any point under a dam. ........................... 106 *Flow-nets-04: Flow through an earth levee. .............................................................. 107 *Flow-nets-05: Finding the total, static and dynamic heads in a dam. ....................... 108 **Flow nets-06: Hydraulic gradient profile within an earth levee. ............................. 109 **Flow-net-07: Flow into a cofferdam and pump size................................................ 111 *Flow-nets-08: Drainage of deep excavations for buildings. ...................................... 113 v
Anchor............................................................................................................................. 113 Limestone ........................................................................................................................ 113 *Flow-nets-09: Dewatering a construction site. .......................................................... 115 *Flow-net-10: Dewatering in layered strata. ............................................................... 116 **Flownets-11: Flow through the clay core of an earth dam. ..................................... 118 Chapter 7 Effective Stresses and Pore Water Pressure ................................................... 122 Symbols for Effective Stresses and Pore Water Pressure............................................ 122 *Effective Stress–01: The concept of buoyancy. ........................................................ 123 *Effective Stress–02: The concept of effective stress. ................................................ 124 *Effective Stress–03: The concept of effective stress with multiple strata. ................ 125 Effective Stress-03B .................................................................................................... 126 Chapter 8 Dams and Levees ........................................................................................... 127 Symbols for Dams and Levees .................................................................................... 127 *Dams-01: Find the uplift pressure under a small concrete levee. ............................. 128 *Dams-02: Determine the uplift forces acting upon a concrete dam. ......................... 129 Chapter 9 Stresses in Soil Masses ................................................................................... 132 Symbols for Stresses in Soil Masses ........................................................................... 132 *Stress-01: Reading the Boussinesq charts. ................................................................ 141 *Stress-02: 2:1 Method to find the Stress below a Rectangular Footing. ................... 142 *Stress-03: Summation of vertical and line loads using Boussinesq. ........................ 143 *Stress-04: The influence of point loads on a culvert. ................................................ 144 *Stress–05: Stress increase at a point from several surface point loads. ..................... 146 *Stress-06: Find the stress under a rectangular footing. .............................................. 147 *Stress-07: The effect of the WT on the stress below a rectangular footing. .............. 148 *Stress–08: Finding the stress outside the footing area. .............................................. 149 *Stress-09: Stress below a footing at different points. ............................................... 150 *Stress-10: Stress increase from a surcharge load of limited width. ........................... 151 *Stress-11: Finding a stress increase from a surface load of limited width. ............... 152 **Stress-12: Stress increase as a function of depth. .................................................... 153 *Newmark–01: Stress beneath a tank at different depths............................................ 154 *Newmark-02: The stress below the center of the edge of a footing. ......................... 155 *Newmark-03: Stress at a point distant from the loaded footing. ............................... 156 *Newmark-04: Stresses coming from complex shaped foundations........................... 157 vi
*Newmark-05: Stress beneath a circular oil tank. ...................................................... 158 **Newmark-06: Use Newmark with a settlement problem. ....................................... 159 *Mohr-01: Find the angle of internal friction φ of sand. ............................................. 161 *Mohr-02: Simple transformation from principal to general stress state. ................... 162 *Mohr–03: Find the principal stresses and their orientation. ...................................... 163 *Mohr–04: Find the principal stresses and their orientation. ...................................... 164 *Mohr-05: Find the angle of internal friction. ............................................................ 165 *Mohr – 06: Normal and shear stress at a chosen plane. ............................................ 166 *Mohr–07: Find the maximum and minimum stresses on a given plane. ................... 167 **Mohr – 08: Back figure the failure angle ................................................................ 168 *Mohr – 09: find the Principle pressure using Mohr................................................... 169 *Mohr –10: Relation between θ and φ. ....................................................................... 170 *Mohr – 11: Normal and shear stresses on any plane. ................................................ 171 *Mohr–12: Derive the transformation from general to principal stresses. .................. 172 *Mohr – 13: Tensile stress from the Mohr-Coulomb failure envelope. ...................... 173 **Mohr–14: Maximum principal stress from triaxial test. .......................................... 174 *Mohr – 15: Derive the general formula for horizontal stress. ................................... 175 Chapter 10 Elastic Settlements ....................................................................................... 176 Symbols for Elastic Settlements .................................................................................. 176 *Elastic Settlement-01: Settlement (rutting) of a truck tire. ........................................ 177 *Elastic Settlement-02: Schmertmann method used for granular soils. ...................... 178 *Elastic Settlement-03: Schmertmann method used for a deeper footings. ................ 179 *Elastic Settlement-04: The 2:1 method to calculate settlement. ................................ 181 *Elastic Settlement-05: Differential settlement between two columns. ...................... 183 *Elastic Settlement-06: Compare Boussinesq, Westergaard, and the 2:1 methods. ... 184 *Elastic Settlement-07: Schmertmann versus the strain methods. .............................. 187 *Elastic Settlement-08: The Schmertmann method in multiple strata. ....................... 188 **Elastic Settlement-09: Settlement of a mat foundation. .......................................... 190 Chapter 11 Plastic Settlements........................................................................................ 192 Symbols for Plastic Settlements .................................................................................. 192 *Plastic Settlement–01: Porewater pressure in a compressible soil. ........................... 193 *Plastic Settlement-02: Total settlement of a single layer. ......................................... 195 *Plastic Settlement-03: Boussinesq to reduce the stress with depth. .......................... 196 vii
*Plastic Settlement -04: Surface loads with different units. ........................................ 198 *Plastic Settlement-05: Pre-consolidation pressure pc and index Cc. .......................... 199 *Plastic Settlement-06: Final voids ratio after consolidation. ..................................... 201 *Plastic Settlement-07: Settlement due to a lowered WT. .......................................... 202 *Plastic Settlement-08: The over-consolidation ratio (OCR)...................................... 203 **Plastic Settlement-09: Coefficient of consolidation Cv. .......................................... 204 *Plastic Settlement -10: Secondary rate of consolidation. .......................................... 206 *Plastic Settlement-11: Using the Time factor Tv. ...................................................... 207 *Plastic Settlement-12: The time rate of consolidation. .............................................. 208 *Plastic Settlement-13: Time of consolidation t.......................................................... 209 *Plastic Settlement-14: Laboratory versus field time rates of settlement. .................. 210 *Plastic Settlement-15: Different degrees of consolidation. ....................................... 211 **Plastic Settlement-16: Excavate to reduce the settlement. ...................................... 212 **Plastic Settlement-17: Lead time required for consolidation of surcharge. ............ 214 **Plastic Settlement-18: Settlement of a canal levee. ................................................. 216 **Plastic Settlement-19: Differential settlements under a levee. ................................ 218 ***Plastic Settlement-20: Estimate of the coefficient of consolidation cv. ................. 219 **Plastic Settlement-21: The apparent optimum moisture content. ............................ 221 **Plastic Settlement-22: The differential settlement between two buildings. ............ 222 **Plastic Settlement-23: Settlement of a bridge pier. ................................................. 227 Chapter 12 Shear Strength of Soils ................................................................................. 229 Symbols for Shear Strength of Soils............................................................................ 229 *Shear strength–01: Maximum shear on the failure plane. ......................................... 230 *Shear strength–02: Why is the maximum shear not the failure shear? ..................... 231 *Shear strength–03: Find the maximum principal stress σ1. ....................................... 232 *Shear strength–04: Find the effective principal stress. .............................................. 233 *Shear strength–05: Using the p-q diagram. ............................................................... 234 **Shear strength–06: Consolidated-drained triaxial test. ............................................ 235 **Shear strength–07: Triaxial un-drained tests. .......................................................... 237 **Shear strength-08: Consolidated and drained triaxial test. ...................................... 239 ***Shear strength-09: Plots of the progressive failure in a shear-box. ....................... 241 **Shear strength-10: Shear strength along a potential failure plane. .......................... 244 ***Shear strength-11: Use of the Mohr-Coulomb failure envelope. .......................... 245 viii
***Shear strength-11b: Use of the Mohr-Coulomb failure envelope. ........................ 247 **Shear strength-12: Triaxial un-drained tests. ........................................................... 249 **Shear strength-12b: Triaxial un-drained tests. ......................................................... 250 **Shear strength-13: Determine the principal stresses of a sample. ........................... 251 **Shear strength-13b: Determine the principal stresses of a sample. ......................... 254 **Shear strength-14: Formula to find the maximum principal stress. ........................ 257 Chapter 13 Slope Stability .............................................................................................. 259 Symbols for Slope Stability ......................................................................................... 259 *Slope-01: Factor of Safety of a straight line slope failure. ........................................ 260 *Slope-02: Same as Slope-01 but with a raising WT. ................................................. 261 *Slope-03: Is a river embankment safe with a large crane? ........................................ 262 *Slope-04: Simple method of slices to find the FS. .................................................... 263 **Slope-05: Method of slices to find the factor of safety of a slope with a WT. ........ 264 **Slope-06: Swedish slip circle solution of a slope stability. ..................................... 266 Chapter 14 Statistical Analysis of Soils .......................................................................... 269 Symbols for the Statistical Analysis of Soils............................................................... 269 Chapter 15 Lateral Pressures from Soils ......................................................................... 270 Symbols for Lateral Pressures from Soils ................................................................... 270 *Lateral-01: A simple wall subjected to an active pressure condition. ....................... 274 *Lateral–02: Compare the Rankine and Coulomb lateral coefficients....................... 275 *Lateral-03: Passive pressures using the Rankine theory. .......................................... 276 *Lateral-04: The “at-rest” pressure upon an unyielding wall...................................... 277 *Lateral-05: The contribution of cohesion to reduce the force on the wall. ............... 278 **Lateral-06: The effect of a rising WT upon a wall’s stability. ................................ 279 *Lateral-07: The effects of soil-wall friction upon the lateral pressure. ..................... 281 *Lateral-08: What happens when the lower stratum is stronger? ............................... 282 *Lateral-09: Strata with different parameters. ............................................................. 283 *Lateral-10: The effects of a clay stratum at the surface. .......................................... 285 **Lateral-11: Anchoring to help support a wall. ......................................................... 287 **Lateral-12: The effect of five strata have upon a wall. ........................................... 289 **Lateral-13: The stability of a reinforced concrete wall. ......................................... 291 ***Lateral-14: Derive a formula that provides K and σH as a function of σv. ............ 294 **Lateral-15: The magnitude and location of a seismic load upon a retaining wall. .. 297 ix
**Lateral-16: Seismic loading upon a retaining wall. ................................................. 299 Chapter 16 Braced Cuts for Excavations ........................................................................ 300 Symbols for Braced Cuts for Excavations ................................................................... 300 *Braced-cuts-01: Forces and moments in the struts of a shored trench. ..................... 301 **Braced cuts-02: A 5 m deep excavation with two struts for support. ...................... 306 *Braced cuts-03: Four-struts bracing a 12 m excavation in a soft clay. ..................... 310 Chapter 17 Bearing Capacity of Soils ............................................................................. 313 Symbols for the Bearing Capacity of Soils ................................................................. 313 *Bearing–01: Terzaghi’s bearing capacity formula for a square footing. ................... 316 *Bearing–02: Meyerhof’s bearing capacity formula for a square footing. ................. 317 *Bearing–03: Hansen’s bearing capacity formula for a square footing. ..................... 318 *Bearing–04: Same as #01 but requiring conversion from metric units. .................... 319 *Bearing–05: General versus local bearing capacity failures. .................................... 320 *Bearing–06: Comparing the Hansen and Meyerhof bearing capacities. ................... 321 *Bearing–07: Increase a footing’s width if the WT is expected to rise. .................... 322 **Bearing–08: The effect of the WT upon the bearing capacity. ............................... 324 *Bearing–09: Finding the gross load capacity. .......................................................... 326 **Bearing–10: The effect of an eccentric load upon bearing capacity. ...................... 328 **Bearing–11: The effect of an inclined load upon the bearing capacity. .................. 329 **Bearing-12: Interpretation of borings to estimate a bearing capacity. .................... 331 Chapter 18 Shallow Foundations .................................................................................... 333 Symbols for Shallow Foundations ............................................................................... 333 *Footings–01: Analyze a simple square footing. ........................................................ 335 *Footings–02: Add a moment to the load on a footing. .............................................. 339 *Footings–03: Find the thickness T and the As of the previous problem. ................... 341 *Footings–04: Find the dimensions B x L of a rectangular footing. ........................... 346 *Footings–05: Design the steel for the previous problem. .......................................... 348 *Footings–06: Design a continuous footing for a pre-cast warehouse wall. .............. 352 **Footings–07: Design the footings of a large billboard sign..................................... 357 Chapter 19 Combined Footings ...................................................................................... 361 Symbols for Combined Footings ................................................................................. 361 **Combined Footing-01: Rectangular footing with an eccentric column. ................. 362 **Combined Footing-02: Combined footing on the property-line. ............................. 369 x
**Combined Footing-03: Design a combined footing. ............................................... 373 **Combined Footing-04: Optimization via a trapezoidal footing............................... 380 **Combined Footing-05: Combined footing. ............................................................. 386 **Combined Footing–06: Strap footing (or grade beam) between two columns........ 388 Chapter 20 Mat Foundations ........................................................................................... 395 Symbols for Mat Foundations ..................................................................................... 395 *Mat Foundations–01: Ultimate bearing capacity in a pure cohesive soil. ................. 396 *Mat Foundations–02: Ultimate bearing capacity in a granular soil........................... 397 *Mat Foundations–03: Find the depth Df for a fully compensated mat. ..................... 398 *Mat Foundations–04: The consolidation settlement of a mat foundation. ................ 399 *Mat Foundations–05: Settlement of a rigid mat. ....................................................... 400 *Mat Foundations–06: Mat foundation carries a large transfer girder. ....................... 401 **Mat Foundations–07: Design a small mat for an office building. ........................... 402 Chapter 21 Deep Foundations - Single Piles .................................................................. 408 Symbols for Single Piles of Deep Foundations ........................................................... 408 *Single-Pile–01: Pile capacity in a cohesive soil. ....................................................... 409 *Single-Pile–02: Pile capacity in granular soils. ......................................................... 410 *Single-Pile–02b: The friction of a concrete pile in granular soils. ........................... 411 *Single-Pile–03: The capacity of a driven concrete pile. ............................................ 412 *Single-Pile–04: The driving criterion for a concrete pile. ......................................... 413 *Single-Pile–05: The Meyerhof and Vesic methods for pile capacity. ....................... 414 *Single-Pile–06: Tension or pull-out capacity in granular soils. ................................ 416 *Single-Pile–07: Tension (uplift) capacity of a pile in granular soils. ....................... 417 ***Single-Pile–08: The settlement of a drilled shaft in cohesive soils. ..................... 418 ***Single-Pile–09: The performance of steel pipe piles. ............................................ 422 Chapter 22 Deep Foundations - Pile Groups and Caps................................................... 426 Symbols for Pile Groups and Caps of Deep Foundations ........................................... 426 **Pile-caps–01: Design a pile cap for a 9-pile cluster. ............................................... 427 **Pile cap-02: Pile group efficiency and lateral load capacity. .................................. 429 ***Pile caps–03: Settlement of the combined pile-cap and pile cluster. .................... 431 Chapter 23 Deep Foundations: Lateral Loads ................................................................ 435 Symbols for Lateral Loads on Deep Foundations ....................................................... 435 **Lateral loads on piles-01: Find the lateral load capacity of a steel pile. .................. 436 xi
**Lateral loads on piles-02: Find the lateral load capacity of a steel pile. (US Units) 441 **Lateral loads on piles-03: The Broms method for pile lateral loads. ...................... 445 ***Lateral loads on piles-03: Lateral load capacity of a mooring bollard. ................. 449 Chapter 24 Reinforced Concrete Retaining Walls and Bridge Abutments..................... 453 Symbols for Reinforced Concrete Retaining Walls .................................................... 453 **RC Retaining Walls–01: Design a RC wall for a sloped backfill. .......................... 454 **RC Retaining Walls–02: Design a RC bridge abutment wall. ............................... 462 **RC Retaining Walls–02: Design a RC bridge abutment wall. ................................ 464 Chapter 25 Steel Sheet Pile Retaining Walls .................................................................. 468 Symbols for Steel Sheet Pile Retaining Walls............................................................. 468 **Sheet-pile Wall-01: Free-Earth for cantilevered walls in granular soils. ................ 469 **Sheet-pile Wall-02: Free-Earth for cantilevered walls in cohesive soils................. 474 ** Sheet-pile Wall–03: Free-Earth Method f/anchored walls in granular soils. ......... 477 ** Sheet-pile Wall–04: Free-Earth Method anchored walls in cohesive soils. ........... 483 Chapter 26 MSE (Mechanically Stabilized Earth) Walls ............................................... 485 Symbols for Mechanically Stabilized Earth Walls ...................................................... 485 **MSE Walls-01: Design the length L of geotextiles for a 16 ft wall. ....................... 486
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Conversion of Units 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
Multiplication Factor
Base SI Units Quantity
Unit
Symbol
length
meter
m
mass
kilograms (mass)
kgm
force
Newton
N
time
second
s
Derived SI Units Quantity
Derived SI Unit
area
square meter
m²
volume
cubic meter
m³
density
kilogram per cubic meter
kgm/m³
force
Name
kilogram-meter per square second Newtons
Symbol
N
moment of force
Newton-meter
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
xiii
N-m
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
xiv
Conversion of SI Units to English Units 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
xv
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
Conversion of SI Units to English Units
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 100
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
xvi
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.
1
*Exploration–01. Find the required number of borings and their depth. (Revised: Feb-109)
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 100 ft x 120 ft footprint and is expected to be supported on spread footing foundations located about 3 ft below the ground surface. The site appears to be in its natural condition, with no evidence of previous grading. Bedrock is 100-ft 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. Table-1 below suggests one boring for every 3,000 ft2 of building footprint area. Since the total footprint area is 100 ft x 120 ft =12,000 ft2, use four borings. From Table-2 the minimum depth required for the borings, 15S0.7 + D = 15(4)0.7 + 3 = 43 ft. Most engineers want one of the borings to go to a slightly greater depth to check the next lower stratum’s strength.
Table-1 Spacing exploratory borings for medium to heavy buildings. 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
700
7,000
Table-2 Depths of exploratory borings for buildings on shallow foundations. Minimum Depth of Borings Subsurface Conditions
(S = number of stories and D = the anticipated depth of the foundation) (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
2
*Exploration–02. Propose the number of samples from boring data. (Revised: Feb-10)
Use the three borings shown below taken at parking lot to suggest the minimum number of samples required to design the foundations of a light two-story office building. Suggest the approximate depth of these samples. Solution: The legend identifies four distinct layers. Also notice that the water table (WT) is around elevations -7.0’ to -7.5’, assuming that the ground elevation is at +0.0 feet. Casting spread footings will be cheaper above the WT. Notice that the top Layer #4 (limestone) seems to have adequate bearing capacities for lightly loaded spread footings since its SPTs range from 15 to 43. If three borings are taken, then three (3) samples are the minimum required, at about elevation – 4 feet, because it will permit casting spread footings 3-feet thick without constructing them below the WT. Prudence suggests taking samples of Layers #2 and #3 too, below the limestone but these extra samples are above the minimum required.
3
* Exploration-03. What is a pressure cell? (Revision: Feb-10)
a) b) c) d) e)
What device is used to measure the total pressure within a soil mass? (a) Diaphragm and hydraulic earth pressure cells. (b) Flat plate dilatometers. (c) The Iowa borehole shear test. (d) Pressuremeter.
Solution: The correct answer is (a). a) Earth pressure cells are devices that are permanently embedded in the soil mass to attempt to measure the total stress at that point. A pressure cell may be installed in a fill or an embankment dam to determine (1) the distribution, (2) the magnitude, and (3) the direction of the stress. Other structures that commonly use pressure cells are retaining walls, culverts, piles, slurry walls and beneath shallow foundations. There are two types: (1) the diaphragm cells are a stiff circular membrane that deflects with an external soil pressure. Inside the cell is a vibrating wire transducer that detects the deformation via a change in the vibration rate of the steel string. (2) The hydraulic cell is filled with a fluid under high pressure that detects a minor change in the outside pressure. b) flat plate dilatometer is a device inserted into the ground by a truck mounted rig, to measure the soil’s modulus of elasticity and lateral stress at the point of testing. It is then pushed down another 6” to 8” for another measurement. The device expands its cell through a gas pumped from the surface, and is also known as the Marchetti dilatometer. c) The Iowa borehole shear test is a device inserted into a 3”-diameter borehole to a depth slightly below the point of interest. Then its serrated cylinder halves are expanded in the hole with gas from the surface. The serrated cylinder is pulled and its load-displacements are measured. These results yield the in-situ shear strength, and indirectly φ and c. d) A pressuremeter (PMT) is a device similar to the Iowa borehole shear test, and it too measures the undrained in-situ shear strength su of a soil and its static shear modulus G’ at a specific location in a borehole. It is also known in the literature as the Ménard test.
4
* Exploration-04. In-situ instrumentation used in the field. (Revision: Feb-10)
What instruments would you use to measure the deformation of a soil in the field? (a) Settlement plate and piezometer. (b) Penetrometer and CPT. (c) Inclinometer and probe extensometer. (d) Load cells and strain gages. Solution: The correct answer is (c). A definition of each term is as follows: a) A settlement plate is a thick steel plate placed close to the surface in order to load it and measure the load-settlement of the soil at that point. It is used to predict the behavior of a shallow footing at that point by obtaining its soil modulus (stress-strain relationship). b) A piezometer (also known as a pore pressure cell) is a device that is inserted into the ground and sealed within the ground to measure the groundwater pressure at a specific point only. It is different from an observation well because the latter has no subsurface seals and it creates a vertical connection between strata. c) A penetrometer is a solid steel tip and rod that is either pushed into the ground at a constant rate, or driven by blows (like an SPT) in order to (indirectly) measure the shear strength of the soil. There are many types, such as the Dutch cone, the electric friction, electric-piezo/friction and the seismic cones. The CPT simply stands for cone penetration test, and is widely used in lieu of the SPT for soft clays and silts and in fine sands. d) An inclinometer is a device that monitors the deformation of a soil orthogonally to the axis of the inclinometer’s pipe. A probe inside the pipe contains a gravity-sensing transducer designed to measure the inclination with respect to the vertical. e) A probe extensometer is a devise for monitoring the deformation or changing distance between two or more points within a soil mass along a pipe. Inside the pipe is a probe that measures the deformation either mechanically or electrically. f) A load cell, also known as a dynamometer, is a device placed into a structure to measure the amount of load that passes through the cell. They are used in piles, drilled shafts, and earth/rock anchors. g) A strain gage is used when a load cell cannot be interposed in the structure due to geometrical reasons, and measures the load indirectly via the strain, provided the stress-strain relationship is known.
5
*Exploration–05. Which sampling method disturbs the soil the least? (Revised: Feb-10)
Which soil sampling method disturbs the soil the least? a) b) c) d)
The Standard Penetration Test (SPT) with a spilt spoon sampler. A Standard Penetrometer Test. A Cone Penetrometer Test (CPT). A Shelby tube.
Solution: The answer is (d) A Shelby tube (also known as a “thin tube sampler”). Answer (a) is wrong. The most common soil and soft rock sampling tool is the Standard Split Spoon, that splits longitudinally into halves and permit taking a soil or soft rock sample. The tube’s 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 being 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 the rock is often classified as “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 the 3 inch sampler). Compare the degree of sample disturbance of a Standard split-spoon sampler versus the two popular Shelby thin-tube samplers (2” and 3” outside diameters) via their area ratio Ar which is a measure of sample’s disturbance.
The area ratio for a 2"-Standard split-spoon sampler is, D2 − D2 ( 2.0 ) − (1.38) (100) = 110% Ar (%) = o 2 i (100) = 2 Di (1.38) 2
2
The area ratio for a 2"-Shelby-tube sampler is, D2 − D2 ( 2.0 ) − (1.875) (100) = 13.8% Ar (%) = o 2 i (100) = 2 Di (1.875) 2
2
The area ratio for a 3"-Shelby-tube sampler is, Do2 − Di2 ( 3.0 ) − ( 2.875) (100) = 8.9% Ar (%) = (100) = 2 Di2 ( 2.875) 2
2
Clearly, the 3” O-D Shelby-tube sampler is the best tool to use. Answers (b) and (c) are also wrong because penetrometers do not obtain soil samples.
6
*Exploration–06. Correcting the SPT for sampling and depth. (Revision Sept-09)
A standard penetration test (SPT) has been conducted in a loose coarse sand stratum to a depth of 16 feet 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 Table 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, ' = N 60
E m C B C S C R N ( 0.45 )(1.05 )(1.00 )( 0.85 )(14 ) = =9 0.60 0.60
Notice that the SPT value is always given as a whole number. That corrected SPT N’60 is then corrected for depth. For example, using the Liao and Whitman method (1986), ' N corrected − 60 = N 60
2, 000 lb / ft 2 = (9 ) ( depth )( effective unit weight )
2, 000 lb / ft 2 (16 ft )( 93.8 pcf
)
= 10
SPT Hammer Efficiencies (adapted from Clayton, 1990) US Safety Hammers
2-turns on cat-head
0.60
US Donut Hammers
2-turns on cat-head
0.45
Correction Factors for the Boring Diameter, Sampling Method and Boring Rod Length 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
Standard sampler
1.00
Sampling method factor CS
Sampler without liner (not recommended) Rod length factor, CR
7
1.20
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
*Exploration–07. Three methods used for SPT depth corrections. (Revision: Feb-10)
An N’60 = 40 was corrected for sampling errors at a depth of 20 feet in a sand stratum that has a unit weight of 135 lb/ft3. Correct it 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):
N corrected
4N ' = 1 + 2 po
N corrected = but
p0 =
if p0 ≤ 1.5 kips / ft 2
4N ' 3.25 + 0.5 po
and
if p0 ≥ 1.5 kips / ft 2
(20 ft )(135 lb / ft 3 ) = 2.70 kips / ft 2 > 1.5kips / ft 2 1000 lb / kip
therefore N corrected =
4N ' 4(40) = = 35 3.25 + 0.5 po 3.25 + 0.5(2.70 kips / ft 2 )
2. Using the Peck Method (1974):
N corrected = N 'CN
where CN = 0.77 log10
20 if p0 is in tons / ft 2 p0
or C N = 0.77 log10
1915 if p0 is in kN / m 2 p0
(20 ft )(135 lb / ft 3 ) = 1.35 tons / ft 2 = 2.70 kips / ft 2 2000 lb / ton 20 ∴ CN = 0.77 log10 = 0.90 ∴ N corrected = (40)(0.90) = 36 1.35 tons / ft 2 but
p0 =
3. The Liao-Whitman Method (1986), as used in Exploration-05, N corrected = N ' but
100 with po in kN / m 2 po
or
= N'
2, 000 psf with po in psf po
96.1 kN / m 2 2 po = (1.35 tons / ft 2 ) = 129.7 kN / m 2 1 ton / ft
∴ N corrected = 40
100 kN / m 2 = 35 129.7 kN / m 2
8
*Exploration–08. Applying SPT corrections to a mat foundation. (Revision: Feb-10)
Correct the SPT values shown below for a US-type safety hammer in a 2”-diameter boring.
9
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, N 60 =
Em C B C S C R N 0.60
See the sampling corrections Table on page 15. The depth correction is, 2, 000 lb / ft 2 ( depth )( effective unit weight )
( N 1 )60 = N 60 At N 60 =
depth
of
(0.6)(1)(1)(0.75)(26) = 20 and ( N )60 = ( 20 ) 0.60
+5.2 2, 000 lb / ft = 39 (8 ft )(127 − 62.4 pcf )
At +4.1’ N 60 =
(0.6)(1)(1)(0.75)(25) = 19 and ( N )60 = (19 ) 0.60
2, 000 lb / ft 2 = 35 ( 9 ft )(127 − 62.4 pcf )
At+2.0’ N 60 =
(0.6)(1)(1)(0.75)(24) = 18 and ( N )60 = (18 ) 0.60
2, 000 lb / ft 2 = 31 (11 ft )(125 − 62.4 pcf )
At -1.0’
N 60 =
(0.6)(1)(1)(0.85)(30) = 26 and ( N )60 = ( 26 ) 0.60
2, 000 lb / ft 2 = 39 (14 ft )(126 − 62.4 pcf )
(0.6)(1)(1)(0.85)(31) = 26 and ( N )60 = ( 26 ) 0.60
2, 000 lb / ft 2 = 34 (18 ft )(126 − 62.4 pcf )
(0.6)(1)(1)(0.95)(30) At -10’ N 60 = = 29 and ( N )60 = ( 29 ) 0.60
2, 000 lb / ft 2 = 34 ( 23 ft )(126 − 62.4 pcf )
At -5.0’ N 60 =
At -21’ N 60 =
feet:
2
(0.6)(1)(1)(1)(43) = 43 and ( N )60 = ( 43) 0.60
2, 000 lb / ft 2 = 41 ( 33 ft )(130 − 62.4 pcf )
Notice that the effect of the depth correction decreases with depth.
10
*Exploration–09: The RQD versus fractured percentage of a rock sample. (Revision: Feb-10)
What is the rock description of a sample that has a 40% fractured percentage? Solution: The SPTs of rock samples usually provide values of N > 100. The exception is in soft or partially decomposed sandstone and limestone (grainstones). In these types of rock, it is necessary to use the “rock quality designation” (RQD) as a measure of the quality of the rock. The rock quality is based on the percentage of the sample recovered (what field crews call the “fractured percentage”). The RQD is defined as,
RQD =
∑ ( length of recovered pieces equal or larger than 4" long ) Theoretical length of the rock core
A RQD = 1 would indicate it was an excellent quality intact rock mass, or in essence, equivalent to a 100%. The table below shows all the RQD quality descriptions.
In-Situ Rock Quality Designation (based on Deere, 1963) RQD
Description
0 – 0.25
very poor
0.15
0.25 – 0.50
poor
0.20
0.50 – 0.75
fair
0.25
0.75 – 0.90
good
0.3 – 0.7
0.90 – 1.00
excellent
0.7 – 1.0
Unconf comp. strength
Based on this table, a RQD = 0.4 is considered a poor quality rock.
11
qu-field / qlab
*Exploration–10. 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.
47
cu =
T 17.0 ft ⋅ lb = = 168 psf 3 2 (0.3021 ft ) (0.6042 ft ) (0.3021 ft )3 π (d h / 2) + (d / 6) + π 2 6 2
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, cu − corrected = µ cu = (0.80)(168 psf ) = 134 psf
12
*Exploration–11. Using the SPT to estimate engineering parameters. (Revision: Feb-10)
Estimate the angle of internal friction and the unit weight of a medium dense sand located at a depth of 20 feet below the surface, where the SPT N’60 = 9? Solution.
The corrected SPT N 60' = 9 is based on 60% efficiency. However, the Table provided is based on 70% efficiency. Therefore, convert via the following relation, N 70' = N 60'
( 0.60 ) = 9 ( 0.60 ) = 8 ( ) ( 0.70 ) ( 0.70 )
(remember that the SPT values are whole integers)
Correlation between SPT values and some Granular Soils Engineering Parameters Description
Very loose
Loose
Medium
Dense
Very dense
0.85
Dr
Relative density
0
0.15
0.35
0.65
SPT
Very loose - loose
1-2
3-6
7 - 15
16 - 30
N'70
Medium
2-3
4-7
8 - 20
21 - 40
Dense –Very Dense
3-6
5-9
10 - 25
26 - 45
Very loose - loose
26 - 28
28 - 30
30 - 34
33 - 38
Medium
27 - 28
30 - 32
32 - 36
36 - 42
Dense –Very Dense
28 - 30
30 - 34
33 - 40
40 - 50
pcf
70 - 102
89 - 115
108 - 128
108 -140
128 -147
kN/m3
11 - 16
14 - 18
17 - 20
17 - 22
20 - 23
φ
γwet
> 40
< 50
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.
Notice that an SPT N’70 = 8 for a medium dense sand permits estimating the angle of internal friction φ = 32º and a moist unit weight of 108 pcf.
13
*Exploration–12. 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.
14
*Exploration –13. Read a soil boring. (Revision: Aug.-09)
Read the boring shown below, and determine: (1) at what depth is the phreatic surface? (2) At what depth was the boring terminated? (3) What type of sampler was used? (4) What is the lowest SPT in the limestone?
(1) The depth of the phreatic surface (WT) is at -12 feet; (1) The depth of the phreatic surface (WT) is at -12 feet; (2) The boring was terminated at a depth of 15 feet below the surface; (3) Split spoon sampler; (4) N = 20.
15
*Exploration–14: Using a boring log to estimate soil engineering parameters. (Revision Aug.-09)
Using the boring log and the SPT versus Soil Engineering Parameters Table to 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,
N 70 =
EC B C S C R N ( 0.45 )(1.0 )(1.0 )( 0.85 )(14 ) = ≈8 0.70 0.70
(2) Correct the same sample S-4 for depth.
( N 70 ) = N 70
2000 psf = (8 ) (γ h )
2000 psf ≈8 (126 )(17 ) psf
(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 in Exploration-10 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’. 16
17
*Exploration–15. Find the shear strength of a soil from the CPT. (Revision: Aug.-09)
Classify the soil at depth 11 m below the surface using the cone penetration test (CPT) data shown below. The clay samples separately recovered at that depth had γ = 20 kN/m3 and Ip= 20. Estimate the shear strength su, from the graphs, and compare to laboratory tests on the samples that showed shear strength of 550 kPa.
Solution. The CPT is a superior field test to determine soil engineering parameters because it provides continuous readings, that discriminate between point resistance versus shaft friction. Its major disadvantage is that samples are not obtained. From the data, q s = 400 kPa and q c = 11 MPa which gives a fR ~ 3%
18
From the next chart, the soil appears to be silty clay.
19
At a depth of 11 m, po = γ z = (20 kN / m3 )(11 m) = 220 kPa and from the N k versus I p graph, for I p = 20 yields an N k ~ 17.5. The un-drained shear strength su is, su =
qc − po 11, 000 kPa − 220 kPa = = 616 kPa Nk 17.5
Compare with the laboratory tested value of 550 kPa (a 12% error).
20
Chapter 2 Phase Relations of Soil Symbols for Phase Relations of soils e→ GS →
Voids ratio. 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
21
Basic Concepts and Formulas for the Phases of Soils. (A) Volumetric Relationships:
1. - Voids ratio e e=
VV VS
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
n=
VV (100% ) 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, 22
e=
n 1− n
and
n=
e 1+ e
2-3
3. - Saturation S S=
VW x100% VV
2-4
ranges from 0% to 100%.
(B) Weight Relationships: 4. - Water content w w=
WW x100% WS
2-5
Values range from 0% to over 500%; also known as moisture content.
5. – Unit weight of a soil
γ γ=
WS + WW W = V VS + VW + VA
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,
γ=
W m = ρg = g V V
which is the equivalent of F = ma.
2-6
6. - Dry unit weight γd
γd =
WS γ = V 1+ w
The soil is perfectly dry (its moisture is zero). 7. - The unit weight of water
γw =
WW VW
γw where γ = ρ g ( F = ma )
γ w = 62.4 pcf = 1 g / ml = 1 kg / liter = 9.81 kN / m3 23
2-7
Note that the above is for fresh water. Salt water is 64 pcf, etc.
8. - Saturated unit weight of a soil γsat WS + WW VS + VW + 0
2-8
γ b = γ ' = γ SAT − γ w
2-9
γ SAT =
9. - Buoyant unit weight of a soil γb
10. - Specific gravity of the solids of a soil G
GS =
γS γw
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
Pb
11.34
SiO2
2.65
Organic
1.0 or less
Skeletons of plants
2.00
Lead Quartz (silica) Peat Diatomaceous earth
24
Other useful formulas dealing with phase relationships: Se = wGS e=
γs −1 γ dry
Unit weight relationships : (1 + w)GS γ w (GS + Se)γ w (1 + w)GS γ w = = = GS γ w (1 − n)(1 + w) wGS 1+ e 1+ e 1+ S Saturated unit weights :
γ=
γ SAT =
(GS + e)γ w e 1 + w = γ w 1+ e w 1 + e 1+ w
γ SAT = γ d + nγ w = (1 − n ) Gs + n γ w = Gsγ w 1 + wGs γ SAT = γ '+ γ w Dry unit weights :
γd =
γ 1+ w
= Gsγ w (1 − n ) =
GS γ w eS γ w eGsγ w = = 1 + e (1 + e) w ( S + wGs )
e
γ d = γ SAT − nγ w = γ SAT − γ w 1+ e .
25
*Phases of soils-01: Convert from metric units to SI and US units. (Revision: Feb-10)
A cohesive soil sample was taken from an SPT and returned to the laboratory in a glass jar. It had a mass of 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 identified as a mass. The ratio of mass to volume is a density ρ,
ρ=
m 140.5 g g = = 1.82 3 V (500 − 423) cm cm 3
g γ = ρ g = 1.82 3 cm
1 kg 3 10 g
m 1 kN 9.806 sec 2 10 3 N
3
2 kN 10 cm = 17.9 3 ( SI units ) m 1 m 3
kN 1000 N 0.2248 lbs f 1 m γ = 17.9 3 = 114 pcf (U S units ) m 1 kN 1N 3.28 f 08 t
26
*Phases of soils–02: Compaction checked via the voids ratio e. (Revision: Feb-10)
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 = 115.8 lb/ft3. The specifications require that e ≤ 0.80. Has the contractor complied with the specifications? (a) Yes, the contractor has satisfied with more than required. (b) Yes, barely satisfied. (c) No, the contractor has not attained the required compaction. (d) Not enough data. Solution:
γ =
GS γ W (1 + w ) G γ (1 + w ) ∴ 1+ e = S W γ 1+ e
lb 2.81 62.4 f3 (1 + 0.146 ) ft 1+ e = = 1.74 lb f 115.8 3 ft e = 1.74 − 1 = 0.74 ∴ e = 0.74 < 0.80
27
Yes, the contractor has complied; answer (a).
*Phases of soils–03: Finding the moisture of a soil. (Revision: Feb-10)
A container has a sample of moist soil with a mass of 23.42 g. The soil is dried in an oven with its original container. After drying, it is reweighed, and its mass is 19.81 g. If the container has a mass of 1.73 g, what is the soil’s moisture content? a) 18% b) 20% c) 30% d) 85% Solution: mmoist = 23.42 g − 1.73 g = 21.69 g mdry = 19.81g − 1.73 g = 18.08 g ∴ mwater w=
= 3.61g
Ww mw 3.61g = = = 19.96% ≈ 20% which is (b) Ws mS 18.08 g
28
*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:
wsat =
V V + VS − VS e VV γ w VV γ w VS γ wVV = = = = γw V = γw V GS VS γ S VS 1 WS WS WS WS
V + VS VS 1 1 − wsat = γ w V = γw − WS WS γd γS The water content is in error if it is greater than the saturated moisture, that is, 1 1 ∴ w ≤ wSAT = γ w − γd γS 1 1 − = 30% = w = 30% GOOD 1) wSAT = ( 9.81 kN / m3 ) 14.9 27 1 1 2) wSAT = ( 9.81 kN / m3 ) − = 18.5% v < w = 20% WRONG 18 27 1 1 3) wSAT = ( 9.81 kN / m3 ) − = 24% > w = 10% GOOD 16 26 1 1 4) wSAT = ( 9.81 kN / m3 ) − = 22.1% > w = 22% GOOD 17.3 28 1 1 5) wSAT = ( 9.81 kN / m3 ) − = 18.5% < w = 22% WRONG 18 27
29
*Phases of soils–05: Dry unit weight and degree of saturation. (Revision: Aug.-09)
During compaction, the engineer needs to know the relationship between the saturation of the soil and the ensuing dry density. As a rule, the maximum dry density is in the range of 70% to 85% saturation. Express the theoretical dry unit weight as a function of the saturation S. Solution:
The two basic equations of phase relationships are, Gγ γd = S W (1) and 1+ e Se = wGS ( 2) Solving for e in equation (2), and substituting that value into (1), yields SGS γ W Gγ γd = S W or γ d = wGS S + wGS 1+ S
30
*Phases of soils–06: Increasing the saturation of a soil. (Revision: Aug.-09)
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:
γ =
γ W ( GS + Se ) 1+ e
62.4(GS + 0.50e) 1+ e 62.4(GS + 0.75e) and 112.7 pcf = 1+ e
∴
105.7 pcf =
(1) (2)
Solving exp licitely for Gs in equation (1), Gs =
(105.7 )(1 + e ) − 0.50e 62.4
Replace Gs in equation (2) with the above relation from (1), ∴
(112.7 )(1 + e ) = (105.7 )(1 + e ) + ( 62.4 )( 0.25e )
∴ e = 0.814 and
31
GS = 2.67
**Phases of soils – 07: Comparing two samples from a boring log. (Revision Feb-10)
Two clay samples were taken from a boring to a laboratory and yielded the following: Weight of the sample
Sample #1
Sample #2
0.48 N
0.68 N
-5
3
Volume
3 x 10 m
4.3 x 10-5 m3
Dry weight (at 105ºC)
0.30 N
0.40 N
Find: a) Their unit weights γ and moisture contents w; b) Their voids ratios e if the unit weight of solids in the lab was 27 kN/m3; c) Their degrees of saturation S; d) Their relative variation of volumes ∆V / V, if before each sample was taken to the lab their unit weight of solids in the field was actually 27.5 kN/m3.
Solution.
a) Their unit weights and moisture contents: Ww W − WS ( 0.48 − 0.3) N W 0.48 N 3 γ1 = 1 = = 16 kN / m and w = = = (100% ) = 60% 1 V1 3 (10−5 ) m3 WS WS 0.3 N
γ2 =
W W − WS ( 0.68 − 0.4 ) N W2 0.68 N = = 15.8 kN / m3 and w2 = w = = (100% ) = 70% −5 3 V2 4.3 (10 ) m WS WS 0.4 N
b) Their voids ratios if the unit weight of solids in the lab was 27 kN/m3: 32
γd =
WS1
γd =
WS2
1
2
V1
=
γS 0.3 N 27 kN / m3 3 3 10, 000 / 10 / 1 = N m = kN m ∴ e = − = − 1 = 1.7 1 γd 10kN / m3 3 (10 −5 ) m3
=
γS 0.4 N 27 kN / m3 3 3 9,30 0 N / m 9.3 kN / m e 1 = = ∴ = − = − 1 = 1.9 2 γd 9.3kN / m3 4.3 (10−5 ) m3
V2
c) Their degrees of saturation: w G ( 0.60 )( γ S / γ w ) ( 0.60 )( 27 / 9.81) S1 = 1 s = = = 97% e1 (1.7 ) (1.7 ) S2 =
w2Gs ( 0.70 )( γ S / γ w ) ( 0.70 )( 27 / 9.81) = = = 100% e2 (1.9 ) (1.9 )
d) Relative variation of the volume: γ s −lab γ s − field
27 kN / m3 W 9.81kN / m3 ) (1 + 0.6 ) = 15.41kN / m3 = ( γ w )(1 + w ) = 3 ( V1− field 27.5kN / m
γ s −lab γ s − field
27kN / m3 W γ + = 1 w 9.81kN / m3 ) (1 + 0.7 ) = 16.37 kN / m3 = ( )( ) w 3 ( 27.5 kN / m V 2 − field
γ 1− field = γ d (1 + w) = ∴ V1− field =
W
γ 1− field
−5 0.48 N ∆V1 V1− field − V1 ( 3.334 − 3) (10 ) −5 3 = = 3.114 (10 ) m and = = (100% ) = 3.8% 15.41kN / m3 V1 V1 3 (10−5 )
γ 2− field = γ d (1 + w) = ∴ V2− field =
W
γ 2− field
∆V2 V2− field − V2 ( 4.334 − 4.3) (10 0.68 N = = 4.153 (10−5 ) m3 and = = 3 16.37kN / m V2 V2 4.3 (10−5 )
−5
)
(100% ) = 3.4%
Notes: The relative variation of the volume of each sample before taking to the lab if its unit weight of solids in the field was 27.5 kN/m3. (γsolid)field = (γs)f = 27.5 kN/m3 [(γs)f] / γs = 27.5/27 = 0.981818181=> therefore a 1.82% loss of water The 1.82% loss is from water loss, either through evaporation or leakage. Since the water content w is 60% and 70% for samples #1 and #2 respectively, there is no net effect on the volume of the samples taken. The dry weight of solid is independent of the amount of water since by definition it is the weight of the solid without any water and the volume of solid is therefore unaffected. The volume of water however does change, but since the water content or moisture content is 60% and 70% which falls within 0% < w < 500%, the relative total volume of the field sample should be the same as the total volume of the lab sample – noting of course the weight differences due to the 1.82% loss of water. The unit weights of the samples would need to be adjusted because of the total weight variation – in this case the lab values of 0.48 N and the 0.68 N is only a percentage of the field values (slightly less than 100%.) A simple correction of this data would yield closer to true results.
33
*Phases of soils–08: How much water is added to a truck to offset losses? (Revision: Feb-10)
The weight of an empty truck is 3,000 lbs. It is filled with 1,000 lbs of soil. The original moisture of the soil was 12%, but during transportation the moisture dropped to 11.5%. 1) How much water (in gallons) did the soil lose during transportation? 2) How much water (in gallons) must be added to the truck to arrive at the site with a 15% moisture? Solution:
For a moisture content w = 12%, W W w = W = 0.12 but W = 1, 000lbs = WS + WW = W + WW = 9.33WW WS 0.12 W 1, 000lbs 107.2lb = 107 lb and WS = W = = 893 lb 9.33 0.12 0.12 For the lowered moisture content of w = 11.5%, W W w = W = 0.115 but W = 1, 000lbs = WS + WW = W + WW = 9.696WW WS 0.115
∴WW =
∴WW =
1, 000lbs = 103 lb 9.696
1 ft 3 7.48 gallons ∴ the truck lost 4 lbs of water = (4lb ) = 0.48 gallons 1 ft 3 62.4lb To arrive at the site with a moisture content of w = 15%, W w = W = 0.15 ∴WW = 0.15WS = ( 0.15 )( 893lb ) = 134lb WS The truck load needs to add 134 -103 = 31 lbs of water ≡ 3.71gallons
34
*Phases of soils–09: Find γd, n, S and Ww. (Revision: Feb-10)
The moist unit weight of a soil is 105 lb/ft3. Given that the w = 15% and Gs = 2.70, find: a) Dry unit weight γd, b) The porosity n, c) The degree of saturation S, and Solution:
γ
105 lb = 91.3 3 (1 + w) (1 + 0.15) ft b) From the table of useful relationships, a) γ d =
γd = n=
=
Gs γ w Gγ ( 2.70 )( 62.4 ) = 1.85 ∴ e = 0.85 ∴ 1+ e = s w = γd 1+ e ( 91.3)
e 0.85 = (100% ) = 46% 1 + e 1 + 0.85
c ) Since Se = wGs ∴ S =
35
wGs ( 0.15 )( 2.70 ) = (100 ) = 48% e ( 0.85 )
*Phases of soils–10: Continued previous Phases-09 problem, adding water. (Revision: Feb-10)
Using the previous problem Phases-09, how much mass of water in lbm/ft3 must be added to reach full saturation. Solution:
The mass of water that must be added can be found from the relation,
γ = ρg
or
ρ=
γ
which is the density of water (mass per unit volume) g However, this γ is the saturated unit weight, or
( G + e ) γ w = ( 2.70 + 0.85 ) ( 62.4lb / γ sat = S
ft 3 )
= 119.7lb / ft 3
1 + 0.85 3 lbm γ (119.7 - 62.4 ) lb / ft ∴ ρ ( mass of water ) = = = 1.78 2 g ft 3 32.2 ft / s 1+ e
36
*Phases of soils–11: Use the block diagram to find the degree of saturation. (Revision: Feb-10)
A soil has an “in-situ” (in-place) voids ratio eo = 1.87, wN = 60%, and GS = 2.75 .
What are the
γmoist and S? All soils are really “moist” except when dry, that is when w = 0%.
Set VS = 1 m3 (Note: this problem could also be solved by setting V = 1.0 m3).
Solution:
∴ eo =
VV 1.87 = = 1.87 ∴ V = VS + VV = 1 + 1.87 = 2.87 m 3 VS 1
The "natural" water content is
Gs =
∴
∴
37
wN =
Ww = 0.60 Ws
∴ Ww = 0.60Ws
Ws Vs
γs = ∴ Ws = Vs ( GS γ w ) = (1 m 3 ) ( 2.75 ) ( 9.81 kN / m 3 ) = 26.98 kN γw γw Ww = 0.60 (Ws ) = ( 0.60 )( 26.98 )
= 16.18 kN
W = Ws + Ww = 26.98 + 16.19
= 43.16 kN
W 43.16 kN kN = = 15.0 3 3 V 2.87 m m Ww 16.19 γ V 9.81 S= w = w = = 88.1% VV VV 1.87
γ moist =
*Phases of soils–12: Same as Phases-11 but setting the total volume V=1 m3. (Revision: Feb-10)
A soil has an “in-situ” (in-place) voids ratio eo = 1.87, wN = 60%, and GS = 2.75 . What are the
γmoist and S?
Set V = 1 m3 (instead of Vs = 1 m3 used in Phases-11).
Solution: but eo =
VV = 1.87 ∴ but V = 1 m 3 = VS + VV = VS + 1.87VS = 2.87VS VS
∴ VS = 0.348 and VV = 0.652 The "natural" water content is
Gs =
∴
Ww = 0.60 ∴ Ww = 0.60Ws Ws
Ws Vs
γs = ∴ Ws = Vs ( GS γ w ) = ( 0.348 m 3 ) ( 2.75 ) ( 9.81 kN / m 3 ) = 9.39 kN γw γw
γ moist =
S=
Ww = 0.60 (Ws ) = ( 0.60 )( 9.39 )
= 5.63 kN
W = Ws + Ww = 9.39 + 5.63
= 15.02 kN
W 15.0 kN kN = = 15.0 3 3 V 1m m Ww
∴
wN =
Vw γ w = VV VV
5.63 9.8 = = 88.1% 0.652
38
*Phases of soils–13: Same as Phases-06 but using a block diagram. (Revision: Feb-10)
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.
Solution:
S et V = 1 ft 3 γ 2 − γ 1 = 112.7 − 105.7 = 7.0 lbs are 25% of w ater ∴ ∴
21.0 lbs are 75% of w ater
W S = 112.7 − 20.8 Vw =
Ww
γw
=
=
91.9 lb
20.8 lb = 0.333 ft 3 62.4 pcf
1 V w = 0.111 ft 3 ∴ V v = V a + V w = 0.111 + 0.333 = 0.444 3 V s = 1 − V v = 1 − 0.444 = 0.556
Va = ∴
e= and
39
GS =
VV 0.444 = = 0.80 VS 0.55 6
γS WS 91.9 lb = = = 2. 65 γ w V S γ w ( 0.556 ) (62 .4 )
*Phases of soils–14: Block diagram for a saturated soil. (Revision: Feb-10)
A saturated soil sample has a unit weight of 122.5 pcf and Gs = 2.70. Find γ dry , e, n, and w.
Solution: Set the total volume V = 1 ft 3 and rearranging terms, 1 WS + Ww γ w GS W = W S + W w = 122.5 lb
(1 )
V = VS + Vw =
(2 )
Combining equations (1) and (2) yields
1=
1 122.5 − W w + Ww ( 62.4 pcf ) 2.70
∴ W w = 27.0 lb ∴ V w =
Ww
∴ W S = 95.5 lb ∴ V S =
WS 95.5 lb = = 0.567 ft 3 G S γ w ( 2.70 ) (62.4 pcf )
γw
=
27.0 lb = 0.433 ft 3 62.4 pcf
WS 95.5 lb = = 95.5 pcf V 1 ft 3
∴
γ dry =
∴
e=
VV 0.433 = = 0.764 VS 0.567
∴
n=
VV 0.433 = = 0.433 V 1
n = 43.3%
∴
w=
Ww 27 = = 0.2 83 WS 9 5 .5
w = 28 . 3 %
40
*Phases of soils–15: Find the weight of water needed for saturation. (Revision: Sept.-08)
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:
kN γ γ kN = = ∴ γ = 1 8 .2 3 1+ w 1 + 0 .0 4 m m3 W = 1 8 .2 = W S + W w = W S + w W S = (1 .0 4 ) W S
γ
d
= 1 7 .5
∴ W S = 1 7 .5 k N , VS = Vw =
WS
γ
γ
=
1 7 .5 k N = G Sγ w
=
0 .7 0 k N = 0 .0 7 m ( 9 .8 1 k N / m 3 )
S
Ww
W w = 0 .7 0 k N
and
w
1 7 .5 k N = 0 .6 7 3 m ( 2 .6 5 ) ( 9 .8 1 k N / m 3 ) 3
∴
V a = V − V s − V w = 0 .2 5 7 m
VV 0 . 0 7 + 0 .2 5 7 = 0 .4 9 = VS 0 .6 7 3
e =
( 0 .0 4 ) ( 2 . 6 5 ) wGS = (1 0 0 ) = 2 1 . 6 % e 0 .4 9 W e re q u ire a S = 9 5 % , th e re fo re ,
T h e e x is tin g S =
w =
Se = GS
( 0 . 9 5 ) ( 0 .4 9 ) 2 .6 5
= 0 .1 7
W w = w W S = ( 0 . 1 7 ) (1 7 . 5 k N )
= 2 .9 8 k N
a lr e a d y h a v e W w
= 0 .7 0 k N
∴
= 2 .2 8 k N
m u s t a d d w a te r 3
Answer: Add 2.28 kN of water per m .
41
3
3
*Phases of soils–16: Identify the wrong piece of data. (Revision: Feb-10)
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? G iven
kN m3 kN γ S = unit w eight of solids = 26.1 m3 w = w ater content = 40%
γ = unit w eight of sam ple = 18.4
e = voids ratio = 1.12 S = degree of saturation = 95%
Solution:
Check the accuracy of 4 out of 5 of the variables using, Se = wGS ∴ Se = ( 0.95 )(1.12 ) = 1.06 wGS = ( w )
γS 26.1 = ( 0.4 ) = 1.06 ∴ Therefore, these four are correct. γw 9.81
The only possibly incorrect value is γ . Assume that V = 1 m3 .
(1)
V = 1 m3 = Va + Vw + VS but e =
VV = 1.12 VS
∴
0 = − Va − Vw + 1.12VS ( 2 )
∴ VS = 0.472 m3 , VV = 0.528 m3 but Vw = 0.95VV = 0.502 m3 ∴ Va = 0.026 m3 kN ∴WS = γ SVS = 26.1 3 ( 0.472 m3 ) = 12.3 kN m kN = 4.9 kN Ww = wWS = ( 0.40 ) 12.3 3 m W = 12.3 kN + 4.9 kN = 17.2 kN Therefore, the actual unit weight of the soil is, ∴γ =
W 17.2 kN kN kN = = 17.2 3 ≠ 18.4 3 3 V 1m m m
42
*Phases of soils–17: The apparent cheapest soil is not! (Revision: Feb-10)
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 $1.37 per yd3. But,
Therefore: To put 1yd3 of solids in the dam you would need 1.8 yd3 of soil. For 1yd3 of solids from A you would need 1.9 yd3 of fill. For 1yd3 of solids from B you would need 3.0 yd3 of fill. For 1yd3 of solids from C you would need 2.6 yd3 of fill. The cost of the select fill from each supplier is (rounding off the numbers): A=
5.28$ 1.9 ( 5 ) (10 6 yd 3 ) 3 ≈ $ 27, 900, 000 1.8 yd
B=
3.91$ 3.0 ( 5 ) (10 6 yd 3 ) 3 ≈ $ 32, 600, 000 1.8 yd
C =
5.19$ 2.6 ( 5 ) (10 6 yd 3 ) 3 ≈ $ 37, 500, 000 1.8 yd
Therefore Supplier A is the cheapest by about $ 4.7 Million compared to Supplier B.
43
*Phases of soils–18: Number of truck loads for Phases-17. (Revision: Feb-10)
Based on the previous problem data (Phases–17), 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 20 yd3 of soil.
Solution:
Set VS = 1 e =
VV VV = = VV = 1.2 which means that there is 1 yd 3 of solids per 1.2 yd 3 of voids. VS 1
∴ 2.2 yd 3 of soil for each 1 yd 3 of solids. 10 yd 3 of soil for each x yd 3 in a truck load ∴
x = 4.54 yd 3 of solids per truck trip.
The required volume of solids in the dam is, Vsolids
( 5x10 =
6
yd 3of soil )(1 yd 3of solids ) 3
1.8 yd of soil
= 2.8 x106 yd 3 of solids
Therefore, (rounding off) Number of Truck − trips =
( 2.8x10
( 4.54
6
yd 3of solids )
yd 3of solids / truck − trip )
= 616,800
44
*Phases of soils–19: How many truck loads are needed for a project? (Revision: Feb-10)
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:
Assume: VS = 1 m3 ∴ e =
VV VV = = VV = 0.46 m3 in the compacted fill VS 1
The required 700,000 m3 of fill have 1.46 m 3 of voids per each 1 m3 of solids Therefore, the 700,000 m3 of fill have 479,400 m3 of solids Each truck carries 1.71 m3 of fill per 1 m3 solids In order for the trucks to carry 479,000 m3 of solids they must carry 820,000 m3 of fill Since each truck carries 10 m3 of fill, 820, 000 m3 ∴ The number of truck-loads = = 82,000 truck-loads. 10 m3
A=
5.28$ 1.9 ( 5 ) (10 6 yd 3 ) 3 ≈ $ 27, 900, 000 1.8 yd
B=
3.91$ 3.0 ( 5 ) (10 6 yd 3 ) 3 ≈ $ 32, 600, 000 1.8 yd
C =
5.19$ 2.6 ( 5 ) (10 6 yd 3 ) 3 ≈ $ 37, 500, 000 1.8 yd
Therefore Supplier A is the cheapest by about $ 4.7 Million compared to Supplier B.
45
*Phases of soils–20: Choose the cheapest fill supplier. (Revised: Feb-10)
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 e=
VW = 0.369 ft3, VS = 0.546 ft3, Va = 0.085 ft3
VV 0.454 = = 0.83 ∴ 1.83 yd 3of soil contains 1.0 yd 3of solids. VS 0.546 Pit B: WS = 100 lb, WW = 20 lb, VW = 0.321 ft3, VS = 0.594 ft3, Va = 0.08 ft3
e=
VV 0.401 = = 0.68 ∴ 1.68 yd 3of soil contains 1.0 yd 3of solids. VS 0.594
(3) Material needed for fill from company A:
γ = 0.95γ d (1 + w ) = 0.95 (112 )(1 + 0.10 ) = 117 pcf ∴ WS = 106.4 lb, Ww = 10.6 lb e=
VV 0.37 = = 0.59 ∴ 1.59 yd 3of soil contains 1.0 yd 3 of solids VS 0.63
∴ Site A requires
200, 000 yd 3 of fill = 125,800 yd 3 of solids 1.59
Material needed for fill from company B:
γ = 0.95γ d (1 + w ) = 0.95 (115 )(1 + 0.10 ) = 120 pcf ∴ WS = 109.1 lb, Ww = 10.9 lb e=
VV 0.35 = = 0.54 ∴ 1.54 yd 3of soil contains 1.0 yd 3 of solids VS 0.65
∴ Site B requires
200, 000 yd 3 of fill = 130, 000 yd 3 of solids 1.54 46
(4) a) Cost of using Company A:
$0.50 Cost A = (125,800 yd 3 ) (1.83) = $ 115,100 3 yd Cost of using Company B:
$0.60 Cost B = (130, 000 yd 3 ) (1.68 ) = $ 131,100 3 yd
Using Company A will save about $ 16,000.
47
*Phases of soils–21: Use a matrix to the find the missing data. (Revision: Sept.-08)
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 above converted into matrix form, with missing data in red.
The volume of solids Vs contained in the total volume of fill V = 34,000 yd3 can be found from, V = VV + VS = 0.7VS + VS = VS ( 0.7 + 1) = 34, 000 yd 3 ∴ VS =
At Pit #3,
34, 000 = 20, 000 yd 3 of solids 1.7
V3 = 1 + e3 ∴ V3 = VS (1 + e3 ) = ( 20, 000 yd 3 ) (1 + 0.95 ) = 39, 000 yd 3 of soil VS
The total cost of Pit #3 is TC3 = ( 39, 000 yd 3 )( $ 5.50 / yd 3 ) = $ 214,500
At Pit #2:
V2 = 1 + e2 ∴V2 = VS (1 + e2 ) = ( 20, 000 yd 3 ) (1 + 0.88 ) = 37, 600 yd 3 of soil VS
But, the total cost of Pit #2 is TC2 − $ 11, 000 = TC3 = $ 214,500 The unit cost of Pit #2 C2 =
At Pit #1: V1 =
∴ TC2 = $ 225, 500
TC2 $ 225,500 = = $ 6.00 / yd 3 3 V2 37, 600 yd
TC1 TC + 28, 000 225, 500 + 28, 000 = 2 = = 42, 250 yd 3 of soil 3 $ 6.00 / yd $ 6.00 / yd 3 $ 6.00 / yd 3
But, V1 = VS (1 + e1 ) = ( 20, 000 yd 3 ) (1 + e1 ) = 42, 250 yd 3
∴
e1 = 1.11
48
**Phases of soils–22: Find the voids ratio of “muck” (a highly organic soil). (Revision: Feb-10)
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: γs =
Set Ws = 1 unit and
Ws 1 = Vs (Vso + Vsm )
(a) Assume Mo = Wo for a unit weight of the dry soil Therefore 1 - Mo = Wm Mo
γ so
= volume of organic Vso solids
(1 - M o )
γ sm
= volume of mineral Vsm solids
The total unit weight is the weight of a unit volume.
Therefore γ S =
Mo γ so
γ sm 1 = γ so M o ( γ sm - γ so ) + γ so ) 1 - Mo ) ( + γ sm
weight of water w ( weight of solids ) volume of water γ wS γ wS Vv S (b) e = = = = Vs Vs Vs Vs
Therefore e =
49
w wγ sm γ so γ w S (1) = Mo (1 - M o ) γ w S M o (γ sm − γ so ) + γ so + γ sm γ so
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)
50
*Classify–01: Percentage of each of the four grain sizes (G, S, M & C). (Revision: Sept.-08)
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).
51
*Classify–02: Coefficients of uniformity and curvature of granular soils. (Revision: Sept.-08)
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,
D 1.40 mm CU = 60 = = 2.8 and D10 0.50 mm
D302 ( 0.95) = 1.29 CC = = D60 D10 (1.40 )( 0.50 ) 2
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). 52
*Classify-03: Classify two soils using the USCS. (Revision: Sept.-08)
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%.
CU =
D60 1.40 mm = = 2.8 , therefore, soil A is a poorly graded sand (SP). D10 0.50 mm
Classify Soil B: For soil B, the distribution is G = 0%, S = 61%, M = 35% and C = 4%.
CU =
53
D60 0.45 mm = = 90 , therefore, soil A is very well graded silty sand (SM). D10 0.005 mm
*Classify-04: Manufacturing a “new” soil. (Revision: Sept.-08)
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,
CU =
D60 D60 = 100 = ∴ D60 = 0.6mm 0.006mm D10
The best fit is a 50% of A plus 50% of B mix.
54
55
Classify – 05 (Revision: Sept.-09)
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 %
A =
IP 34% = ≈ 1.5 % of clay fraction 23%
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.
56
Classify – 06 (Revision: Sept.-09)
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.
∴
L = 16.29 − ( 0.164 R ) = 16.29 − [0.164(43)] = 9.2 cm
D=
K
L = 0.01321 t
9.2 cm = 0.00517 mm = 5.2 x 10-3 mm ( a silt) 60 min
Table of constant K versus Temperature T (°C) Temparature (°C) 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
57
Gs 2.45 0.01510 0.01511 0.01492 0.01474 0.01456 0.01438 0.01421 0.01404 0.01388 0.01372 0.01357 0.01342 0.01327 0.01312 0.01298
2.50 0.01505 0.01486 0.01467 0.01449 0.01431 0.01414 0.01397 0.01391 0.01365 0.01349 0.01334 0.01319 0.01304 0.01290 0.01276
2.55 0.01481 0.01462 0.01443 0.01425 0.01408 0.01391 0.01374 0.01358 0.01342 0.01327 0.01312 0.01397 0.01283 0.01269 0.01256
2.60 0.01457 0.01439 0.01421 0.01403 0.01386 0.01369 0.01353 0.01337 0.01321 0.01306 0.01291 0.01277 0.01264 0.01249 0.01236
2.65 0.01435 0.01417 0.01399 0.01382 0.01365 0.01348 0.01332 0.01317 0.01301 0.01286 0.01272 0.01258 0.01244 0.01230 0.01217
2.70 0.01414 0.01396 0.01378 0.01361 0.01344 0.01328 0.01312 0.01297 0.01282 0.01267 0.01253 0.01239 0.01225 0.01212 0.01199
2.75 0.01394 0.01376 0.01359 0.01342 0.01325 0.01309 0.01294 0.01279 0.01264 0.01249 0.01235 0.01221 0.01208 0.01195 0.01182
2.80 0.01374 0.01356 0.01339 0.01323 0.01307 0.01291 0.01276 0.01261 0.01246 0.01232 0.01218 0.01204 0.01191 0.01178 0.01169
Classify – 07 (Revision: Sept.-09)
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: v =
where
γs −γw 2 d 18η
t = 4 hours = 14,400 sec,
but G s =
γs γw
d =
or
d (mm) =
18η L(cm) γ s − γ w t (min)
dyne ⋅ s L = 5 cm, η = 10−2 Poise i.e., cm 2
∴ γ s = G sγ w =
( 2.65 ) ( 9.81
dynes / cm 3
)
= 26
dynes / cm 3
dynes ⋅ sec 18 x10 −2 ( 5 cm ) cm 2 = 0.020 mm dynes 9.81 (2.65 − 1.00)(14, 400 sec) 3 cm
b) The unit weight γ of the solution after 4 hours is,
γ=
weight of soil in solution 2 g + [1000cm3 − 2 g / 2.65] x 1 g / cm3 = = 1.001 g / cm3 volume of solution 1000 cm3
The portion of soil having a diameter smaller than D is,
Portion smaller =
γ γ V • s w W γs −γw
(γ
− 1) =
1000 cm 3 2.65 x 1 (1.001 − 1) = 0.08 20 g 2.65 − 1
∴ The remaining soil is only 8 % of the original sample.
c) The diameter d = 0.020 mm corresponds to a silt. 58
Classify – 08 (Revision: Sept.-09)
Dr =
The formula for the relative compaction Dr is,
emax − e emax − emin
Derive an equivalent equation as a function of dry unit weights, such that Dr =
(γ
d ( field )
(γ d max
− γ d min ) (γ d max ) − γ d min ) (γ d ( field ) )
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
γd =
WS G S γ W = V 1+ e
1 Dr =
γ d min 1
γ d min
−
−
and Dr = 0 = loose, to 1= very dense
1
γd 1
γ d max
γ − γ d min γ d max γ d ( field ) − γ d min γ d max = d = γ d max − γ d min γ d γ d max − γ d min γ d ( field )
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?
RC = 98% =
γ γ
d ( fie ld )
d ( S td . P r o c t .)
=
γ
d ( fie ld )
1 8 .8
∴ γ
d ( fie ld )
γ − γ d min γ d max (18.4 − 14.0 )(18.8) Dr = d = 94% = γ d max − γ d min γ d (18.8 − 14.0 )(18.4 )
59
= 1 8 .4 k N / m
3
Classify – 09 (Revision: Sept.-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 16.97
γ field Solution:
γ −γ γ Dr = n min max γ max − γ min γ n
16.97 − 14.77 18.07 range 1 (low γ min )(high γ max ) Dr = = 0.71 18.07 − 14.77 16.97 16.97 − 15.56 18.07 range 2 (avg γ min )(high γ max ) Dr = = 0.60 18 . 07 − 15 . 56 16 . 97 16.97 − 14.77 17.52 range 3 (low γ min )(avg γ max ) Dr = = 0.83 − 17 . 52 14 . 77 16 . 97 16.97 − 15.56 17.52 range 4 (avg γ min )(avg γ max ) Dr = = 0.74 17.52 −15.56 16.97
∴
60%
≤
D r
≤
83%
60
Classify – 10 (Revision: Sept.-09)
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
Dr =
e max − e e max − e min πd 2 h 4
For the calcareous sand, e1 =
−
π (4) 2 (4.59)
P1
γ S1
P1
2.868 ) × 12 3 4 146.3 = 0.70 2.868 × 12 3 146.3
=
γ S1 πd 2 h 4
For the quartzitic sand, e2 =
−
π (4) 2 (4.59)
P2
γ S2
P2
=
γ S2
−(
3.254 ) × 12 3 4 165.6 = 0.70 3.254 × 12 3 165.6 −(
Notice that e1 = e2 For the calcareous sand, Dr1 =
For the quartzitic sand, Dr2 =
0.89 − 0.70 = 0.70 and 0.89 − 0.62
0.98 − 0.70 = 0.62 0.98 − 0.53
The two types of sand have different relative densities because the calcareous sand grains are more tightly packed than the quartzitic sand grains. For the calcareous sand, γ d1 =
For the quartzitic sand, γ d 2 =
γs
1
1 + e1
γs
2
1 + e2
=
146.3 = 86.1 pcf (but Dr1 = 0.70) 1 + 0 .7
=
165.6 = 97.4 pcf (but Dr21 = 0.62) 1 + 0 .7
As a result, the dry unit weight is greater for the soil with the lower relative density. 61
Classify – 11 (Revision: Sept.-09)
Prove that emin = 0.35.
Vtetr = 0.1179a 3 = 0.1179(2 R )3 = 0.943R 3 4 Vsphere = π R 3 3 The volume of the sphere occupied by the tetrahedron is:
∴ 2R
∴
60 = 0.167 = 16.7% 360 3 V V− V 3 −V
Vsphere ( tetr ) ≡ e=
VV = VS
sphere
V sphere
=
(
)
0.943R − 0.167 4 / 3 π R 3 = 0.35 0.167 4 / 3πR 3
(
)
ALTERNATE METHOD:
d 2
(
Volume of cube = d 2
2d
)
3
= d32 2
π d3 2 3 Volume of sphere = 4 = πd 6 3 2 2d 2 − π d 3 Vcube − Vsphere 3 = 0.35 emin = = 2 3 Vsphere πd 3 e = 0.35 OK !!!
62
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. 3 −4 3 V → ( 1 30 ft ≡ 9.44*10 m Standard Proctor mold, ASTM D-698).
Va→
Volume of air.
VS→
Volume of solids.
VV → Volume of voids (water + air). VW → Volume of water. 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.
63
*Compaction–01: Find the optimum moisture content (OMC). (Revision: Aug-08)
A Standard Proctor test has yielded the values shown below. Determine: a) The maximum dry unit weight and its OMC; remember V = 1/30 ft3. b) 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: γ = W (lb)
γ (lb/ft3)
w(%)
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
130
W γ and γ d = V 1+ w
γdmax = 124.8 pcf
125
120
Maximum dry unit weight = 124.8 pcf 115
γd
OMC = 12.3 %
110
105
γfield = (0.93)(124.8) = 116.1 pcf
100
95
90 7
8
9
10
11
12
13
14
15
16
17
18
w(%) 64
*Compaction–02: Find maximum dry unit weight in SI units. (Revision: Aug-08)
Using the table shown below: a) 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). 1 3 1 m3 −4 3 b) Note that the volume V = ft = 9.44 × 10 m 30 35.32 ft 3 c) Find the OMC. d) 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.
γ = γd =
(
W kN m3 V
γ 1+ ω
)
(kN m ) 3
1
2
3
4
5
15.4
16.5
17.3
17.4
17.1
12.8
13.3
13.5
13.1
12.5
γd max = 13.5 kN/m
γd max = 13.5 kN/m3
3
OMC = 28 %
OMC = 28%
13.6
γd-field = 0.95(13.5) = 12.8 kN/m3
γ d (kN/m)
13.2
12.8
12.4
12 18
20
22
24
26
28
w(%)
65
30
32
34
36
38
*Compaction-03: What is the saturation S at the OMC? (Revision: Sept.-08)
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
15.9
17.3
17.8
18.4
18.2
17.8
γd =
γ 1+ ω
e) a) Determine the maximum dry unit weight and the OMC. f) b) What is the dry unit weight and moisture range at 95% RC (Relative Compaction)? g) 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%.
19.5
Dry Unit Weight
18.2
16.9
15.6
14.3
13 5
6.1
7.2
8.3
9.4
10.5
11.6
12.7
13.8
14.9
16
w%
γd max = 13.5 kN/m3 OMC = 11.5 % c) wGsγ d max
S=
γw γ Gs − d max γw
( 0.115 )( 2.70 )(18.4 ) ( 9.8) = = 0.71 18.4 2.7 − 9.8
Saturation S = 71%
66
*Compaction-04: Number of truck loads required. (Revision: Sept.-08)
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:
Wborrow site = Wconstruction site Vborrow siteγ borrow site = Vconstruction siteγ construction site
γ
105 pcf = 89 pcf versus γ d ( construction site ) = 103.5 pcf 1 + w 1 + 0.18 103.5 pcf 3 a ) Volume to be excavated = 10, 000 yd 3 = 11, 630 yd 89 pcf
γ d (borrow site) =
=
27 feet 3 105 lb (11, 630 yd ) yd 3 feet 3 = 824 truck − loads b) Number of truck loads = 20 ton 2, 000 lb truck ton 3
67
*Compaction-05: What is the saturation S at the OMC? (Revision: Sept.-08)
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
1.94
2.01
2.06
2.09
2.08
2.06
2.05
19.4
20.1
20.6
20.9
20.8
20.6
20.5
γd
γw
γd kN/m3
Use γw = 10 kN/m3 for simplicity.
Solution:
γ γ −γ wGs but e = s − 1 = s d γd e γd therefore, at the OMC the saturation is,
It is known that Se = wGs ∴ S =
γ 20.9 SOMC = wOMC Gs d −OMC = ( 0.0695 )( 2.60 ) = 0.74 2.6 0 10 − 20.9 ( )( ) γs −γd The degree of saturation at the OMC is 74%. 2.15
2.1
Dry Unit Weight
2.05
2
1.95
1.9
1.85
1.8 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%
68
*Compaction-06: Definition of the relative compaction (RC). (Revision: Sept.-08)
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.
a ) The relative compaction RC is, RC = 0.90 =
γ d ( field ) γ d ( field ) = ∴ γ d ( field ) = ( 0.90 )( 20.4 ) = 18.4 kN / m3 γ d (max) 20.4
b) The relative density Dr is, Dr =
γ d ( field ) − γ d (min) γ d (max) (18.4 − 13.9 ) 20.4 × = = 0.768 = 76.8% γ d (max) − γ d (min) γ d ( field ) ( 20.4 − 13.9 ) 18.4
c) The moist unit weight γ is,
γ = γ d (1 + w ) = 18.4 (1 + 0.15 ) = 21.2 kN / m3
69
*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:
γ moist
103.7 = 81.0 pcf 1 + w 1 + 0.28 γ d ( field ) 81.0 pcf a ) RC = = = 0.90 γ d max 90.0 pcf
γ d ( field ) =
=
The Relative Compaction = 90% Gγ Gγ b) γ d = s w ∴ 1 + e = s w γd 1+ e G γ ( 2.70 )( 62.4 ) ∴ e = s w − 1 = − 1 = 1.08 ( 81.0 ) γd Se = wGs ∴
S=
wGs ( 0.28 )( 2.70 ) = = 0.70 Saturation S = 70% e (1.08)
70
*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:
GSγ w 2.7 (9.8) kN = = 15. 7 3 1+ e 1 + 0.69 m b) T he volum e of the finished em bankm ent V = 30 m (1.5 m )(1km lo n g ) = 45x10 3 m 3
a) T he borrow pit's dry unit w eight γ d =
V olum e of borrow pit soil requir ed =
γ d ( reqd ) γ d ( borrow
pit )
(V ) =
18 3 3 ( 45 x10 ) m 15.7
3 3 45 x10 m = 5,160 truck-loa ds 3 10 m kN c) W eight of dry soil in 1 truck-load W d = 10 m 3 15. 7 3 = 157 kN m
N um ber of truck trips
18 = 15.7
W eight of w ater = w W d = (0.15)(1 5 7 kN ) = 23 .6 kN per truck load d) D egree of satu ration S =
71
wGS ( 0.15 )( 2.70 ) = 0.59 = 59% = e 0.69
**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? 126
124
122
S
γd(max)
120
= 10
0.29
0.43
0.30
0%
0.41
0.46 114
P o ro s ity
116
V o id R a tio
γd
9 .7
98% S. Proctor
1 4 .3
118
0.32
0.49
0.33
0.52
0.34
112
110
108 5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
w%
72
Solution : Gsγ w γ and γ zav = wet wG 1+ w 1+ s S The crossing of the ZAV (zero air voids) curve is due to an incorrect assumption of Gs , a) Since γ zav =
and/or a miscalculation of the water content w and γ wet . b) The soil volume V required for the road is, 1 ft 3 V = (16inthick) ( 24 ft wide)(1,000 ft long) = 32,000 ft 12 in The peak dry density γ d max = 119.5 pcf at an OMC=12%, and the range of the water content
is from 9.7% to 14.3% for 98% of Standard Proctor or ( 98%) γ d max = ( 0.98)(119.5) = 117.1pcf The total weight of soil in the pavement is W = Vγ = ( 32,000 ft3 ) (117.1pcf ) = 3.75x106 lbs The weight of water Ww at the low end (9.7%) is, 3.75x106 lbs ( 0.097) 1.097 7.45gal Ww Vw = = ≃ 40,000gallons 3 γw 62.4 pcf ft The weight of water Ww at the high end (14.3%) is, 3.75x106 lbs ( 0.143) 1.143 7.45gal Ww Vw = = ≃ 56,000gallons 3 γw 62.4 pcf ft Therefore, the average volume of water is (1/2)( 56,000 + 40,000) = 48,000gallons Therefore, the volume of water required = 48,000 ± 8,000 gallons.
73
**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
h) i) j) k) l) m)
Find the γd max and the OMC from the compaction curve; Find the degree of saturation at the above conditions. Calculate the percentage of air for a given porosity n and the saturation S. Find the equation that describes the points of equal saturation. Determine the equation for S = 100% Discuss the characteristics of this last curve and equation.
Solution: a) 22
21.5
γd max = 20.9 kN/m
21
OMC = 7.7%
20.5 3
γd (kN/m )
3
20
19.5
19
18.5
18 2
3
4
5
6
7
8
9
10
11
Water Content w(%)
74
b)
Se = wGs ∴ S =
wGs γ d max 2.09 = 7.7% 2.65 ( )( ) = 76% e γ w −γ d max 2.65 − 2.09
c)
A
a n
W
nS
1
1-n
S ns =
VV VW VW VW = = = VW V VV V 1
a = 1 − nS − (1 − n ) = 1 − nS − 1 + n a = n(1 − S ) = Va 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.
d) Consider a family of curves of equal saturation,
S=
volume of water volume of water = = volume of air+volume of water 1 − volume of soil grains
γ w d γ w S γ Sγ s S = w ∴γ d + − S ∴ d = γ w wγ s + Sγ w γ γw γs 1− d γs
75
Curves can be plotted for varying values of a and saturation S.
12
10
8
γd
γd = (1-a2)γs 6
Asymptotes
a=0
4 a1 a2 a3
2
0 0
3
6
9
12
15
18
21
24
27
w
1.2
If w = 0 all curves pass thru γs
1
γd
0.8
0.6
0.4 S = 100% S1 0.2
S2
0 0
10
20
30
40
w
These are hyperbolas with the w-axis as an asymptote. 76
**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: n) o) p) q) r) s)
The plot of the dry unit weight versus the water content; The maximum dry unit weight; The optimum moisture content; The dry unit weight at 90% of Standard Proctor; The moisture range for the 90% value; 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. 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
Solution: a) The plot of γd (dry unit weight) versus w (water content): 125 120
dry unit weight, γd
115 110
100% Proctor
105
90% Proctor
100 95
OMC=20%
90 85 0
10
20
w
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 (e) The moisture range for the 90% value is approximately from 13% to 26%. 77
30
(f) Soil at 10% moisture,
A W
V = 1 ft3
W = 98 lb
S
w = 0.10 =
Ww=9
Ws=89 lb
Ww ∴ Ww = 0.10Ws Ws
but Ws = 89 lb and Ww = 0.10Ws = ( 0.1)( 89 ) = 9 lb ∴ W = Ws + Ww = 89 + 9 = 98 lb Soil at 20% moisture,
A V = 1 ft3
W S
Ww=21.5 W = 129 lb Ws= 107.5
w = 0.20 =
Ww ∴ Ww = 0.20Ws Ws
but Ws = 107.5 lb and Ww = 0.20Ws = ( 0.2 )(107.5 ) = 21.5 lb ∴W = Ws + Ww = 129 lb Therefore, need to add the following water: (21.5 lb) – (9 lb) = 12.5 lb/ft3
∴
Added water =
12.5lb ft 3 7.48 gallons 27 ft 3 = 40 gallons / yd 3 ft 3 62.4 lb ft 3 1 yd 3
Answer: Add 40 gallons of water per cubic yard of compacted soil.
78
**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. a) Determine the volume of fill required for the levee; b) Determine the volume required from the borrow pit; c) Determine the required number of truckloads.
Solution: 450 ft long 1 1 a) Volume of levee = ( 20')( 40') + ( 20')( 20') + ( 20')( 60') ( ft 2 ) = 23,300 cy 3 2 2 27 ft / cy b) To find the volume required from the borrow pit consider that the weight of solids is the same in both, But γ d =
Ws (borrow) = Ws ( levee ) Ws V
or Ws = γ dV
112.1 = 95 pcf and 1 + w 1 + 0.18 γ 100.7 pcf = Vlevee d (levee) = ( 23,300 cy ) γ d (borrow) 95 pcf
where γ d (borrow) = ∴ Vborrow
79
γ
or γ d (borrow)Vborrow = γ d (levee)Vlevee
=
γ d (levee ) = ( 0.95)(106 ) = 100.7 pcf = 24,700cy
Number of truck-loads required is based on, Ws ( hauled ) = Ws ( levee )
γ d ( hauled )Vhauled = γ d ( levee )Vlevee γ d ( levee ) ∴ Vhauled = Vlevee γ d ( hauled ) but γ d ( levee ) = 100.7 pcf
and γ d ( hauled ) =
GS γ w ( 2.68 )( 62.4 ) = 67.7 pcf = 1 + e 1 + 1.47
γ d ( levee ) 100.7 pcf Vhauled = Vlevee = ( 23,300 yd 3 ) = 34, 700 yd 3 γ 67.7 pcf d ( hauled ) Vhauled 34, 700 yd 3 Number of truck-loads = = = 2,314 3 truck capacity 15 yd / truck − load
80
**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 81
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
82
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 place 1.2 million cubic yards of compacted soil in the dam? 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 place 1.2 million cubic yards of compacted soil in the dam? 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
83
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 ( 10 −4 mm
sec
).
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.
84
*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.
85
*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 (1 ft 3 ≈ 7.45 gallons ). Solution: Allen Hazen’s (1893) formula for the permeability k = C ( D10 ) 2 = (12 ) (0.16mm)2 = 0.31
mm s
The hydraulic head drop is (955 ft – 5 ft) – (942 ft) = 8 feet Applying D’Arcy formula, ∆h mm 3600s 1in 1ft 7.45gal 8 ft gallons q = k A= 0.31 per ft width (12 ft)(1ft) =1.05 3 hour L sec hour 25.4mm12in 1ft 2,500 ft
D10
D10 = 0.16 mm GRAIN SIZE DISTRIBUTION DIAGRAM
86
*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 ft 3 if the permeability of of impervious clay. Compute the seepage loss q from the canal in day − mile ft the sand is 2 x10-3 . sec 250’ ELEV. +1050’
ELEV. +1021’
CLAY 6’
SAND CLAY
Solution: D’Arcy’s formula for q yields,
ft 86, 400 s (1050 − 1021) ft ∆h 5, 280 ft 6.0 A = ( 2 ) (10) −3 ft ( ) L s day 250 ft mile ft 3 q = 635, 000 day − mile
q=k
87
*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
1m .
Marl
3m
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.
k H1 = 5kV1 = 5(CD102 ) = (5)(1.5)(0.16 mm) 2 = 1.92 cm / s = 1,920(10-6 ) m / s The formula for combining several horizontal layers is, k Hcomb
−6 −6 k1 H1 + k2 H 2 (1,920 x10 m / s ) (1 m ) + (10 m / s ) ( 3 m ) = = = 4.8 ×10−4 cm / s H1 + H 2 1m + 3 m
88
*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).. 1m
Fine Sand
k = 1x10-1 cm/sec
Pamlico Formation
5m
Porous Limestone
k = 2x10-3 cm/sec
Miami Formation
3m
Fine Sand
k = 1x10-3 cm/sec
Fort Thompson Formation
12 m Upper sandy Limestone
k = 2x10- 4 cm/sec
Fort Thompson Formation
Solution:
kV (EQ) =
=
H H1 H2 H3 H4 + + + k1 k2 k3 k4
( 2,100 cm) 3
=
(1+ 5 + 3 +12) m 1m 5m 3m 12 m + + + 10−1cm / s 2×10−3 cm / s 10−3 cm / s 2×10−4 cm / s
= 0.32×10−3 cm / s
(6,551x10 )s 1 1 (1)(10−1) + (5)(2×10−3 ) + (3)(10−3 ) + (12)(2×10−4 ) kH ( EQ) = [ H1k1 + H2k2 + H3k3 + H4k4 ] = H ( 21 m) kH ( EQ) = 4.8×10−3 cm / s Therefore
kH 4.8×10−3 cm / s = = 15 kV 0.32×10−3 cm / s
The horizontal permeability is 15 times larger than the vertical permeability.
89
*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.
H1 = 3’
-3
k1 = 10 cm/sec k2 = 2x10-4 cm/sec
H2 = 3’
k3 = 10-5 cm/sec
H3 = 3’
k4 = 2x10-3 cm/sec
H4 = 3’
Solution: The equivalent horizontal permeability of all four layers is:
kH(eq) =
( ft )( cm/ s) 3 10−3 + 3 2 10−4 + 3 10−5 + 3 2 10−3 1 (k1H1 + k2H2 + k3H3 + k4H4 ) = ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) H (12 ft )
kH(eq) = 8x10−4 cm/sec The equivalent vertical permeability of all four layers is:
kVeq =
H H1 H 2 H 3 H 4 + + + kV1 kV2 kV3 kV4
=
12 ft 3' 3' 3' 3' + + −5 + −3 −4 10 2 x10 10 2 x10 − 3
= 0.37 x10 − 4 cm / sec
Therefore the ratio of the horizontal to the vertical permeability is:
k H eq k V eq
8 x1 0 − 4 c m / s = = 22 0 .3 7 x 1 0 − 4 c m / s
90
*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.
∆h=300 mm hA
A
H1=150 mm
hB
B
C
H2=150 mm
H3=150 mm
v
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 ∆h = ∆h1 + ∆h2 + ∆h3.
kV ( eq ) =
H 450mm = = 1.2 (10 −3 ) cm / sec 150mm 150mm 150mm H1 H 2 H 3 + + + + −2 −3 k1 k2 k3 1x10 cm / s 3 x10 cm / s 4.9 x10−4 cm / s
∴ q = keq iA = keq
91
300 mm 3, 600sec ∆h 3 A = (0.0012 cm / s ) (10 cm )(10 cm ) = 291 cm / hour H 450 mm 1 hr
*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,
e3 k= (1 + e ) 3 e13 ( 0.57 ) k 1 + e1 0.012 ∴ 1 = = = 1 + 0.573 = 0.544 3 e2 k2 k2 ( 0.72 ) 1 + e2 1 + 0.72 0.012 ∴ k2 = = 0.022 cm / s 0.544 Notice that since k1 = 0.012 cm/s the permeability has almost doubled. A 26% increase
of the voids ratio has effected a doubling the permeability.
92
*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.
∆h = 4 feet
H1=2’
H2=6’
Clayey sand
Solution: a) The hydraulic gradient i and the voids ratio e are: i=
∆h 4 ft = = 0.667 and H 2 6 ft
e=
Gsγ w
γd
−1 =
(2.63)(62.4 pcf ) − 1 = 0.465 112 pcf
The Casagrande formula relates the known permeability k0.85 at e = 0.85 to an unknown permeability k at any voids ratio e,
mm 1 in 1 ft k = 1.4e 2 k0.85 = 1.4(0.465) 2 240 = 0.239 ft / min min 25.4 mm 12 in Therefore, the seepage rate q is:
ft ft 3 2 q = kiA = 0.238 (−0.667)(36 ft ) = −5.74 min min b) The flow direction is UP.
93
*Permeability-10: Capillary rise in tubes of differing diameters. (Revision: Sept.-2008)
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.
94
hc =
h1 =
-4 ( T s ) ( c o s α dγ w
)
=
− 4 ( 0 .0 7 5 N / m ) ( c o s 3 ° ) d ( 9 .8 1 k N / m 3 )
− 4 ( 7 .5 x 1 0 − 2 N / m
) (c o s 3 ° ) = - 4 1
m
( 7 .5 x 1 0 m ) ( 9 .8 1 k N / m ) − 4 ( 7 .5 x 1 0 N / m ) ( c o s 3 ° ) = - 0 .4 1 m = ( 7 .5 x 1 0 m ) ( 9 .8 1 k N / m ) − 4 ( 7 .5 x 1 0 N / m ) ( c o s 3 ° ) = - 0 .0 4 1 m ( 7 .5 x 1 0 m ) ( 9 .8 1 k N / m ) −7
3
−2
h2 =
−5
3
410m m
−2
h3 =
95
−4
3
= 41 m m
*Permeability-11: Rise of the water table due to capillarity saturation. (Revision: Sept.-08)
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:
hc =
−4To cos α −4 ( 0.064 N / m ) cos 3° = = −0.26 m −4 3 dγ w 1 0 m 9. 81 kN / m ( )( )
Discontinuous moisture zone Vapor flow
Capillary fringe zone
hc
Capillary saturation zone
Capillary flow
100% Saturation zone z
96
*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%. Dense silt
h1 = 3 m
Clay
h2 = 5 m
Solution:
1.
The Hazen empirical formula for capillary rise is hC =
0.0306 0.0306 = = 15.3 m 0.2 D10 ( 0.2 )( 0.01 mm )
In essence, the entire silt stratum is saturated through capillarity. 2. For full saturation, S = 100%, Se = wGS ∴ w =
Se Seγ ω (1)( 0.40 )( 9.81) = = = 0.148 GS 26.5 γS
kN 1 + w ( 26.5 )(1.148 ) ∴ γ SAT = γ S = 21.8 3 = m 1.40 1+ e For 50% saturation, w=
Seγ ω
γS
=
( 0.5 )( 0.40 )( 9.81) = 0.074 26.5
kN 1 + w ( 26.5 )(1.074 ) ∴γ = γ S = 20.3 3 = 1.40 m 1+ e Therefore, ∴ (σ V ')3 = h1γ = ( 3 )( 20.3 ) = 61 kPa ∴ (σ V ')8 = h1γ + ( h2 ) γ SAT = ( 3 )( 20.3 ) + ( 5 )( 21.8 ) = 170 kPa
97
*Permeability-13: Back-hoe trench test to estimate the field permeability. (Revision: Sept.-08)
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?
0.0’ 1.8’ 4.4’
8.0’
Solution:
1,540 gal = 154 gallons / minute 10 min The perimeter of the trench is L ' = 2 ( 6 ft + 1.5 ft ) = 15 ft and H = ( 4.4 ft − 1.8 ft ) = 2.6 ft
Q =
The seepage capacity of the surface stratum is Sc ,
154 gallons / min ( ft 3 ) ( min ) Q Sc = = = 8.8 x10−3 cfs / ft / ft C ⋅ L '⋅ H (15 ft )( 2.6 ft ) ( 7.45 gallons )( 60 sec )
98
**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. Impoundment pond area = 4.4 miles2 ELEV 0’
5’
e = 0.750 (Sample 1)
∆h 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)
d) 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 99
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
k=
q 4q = iA iπ d 2
Since 50 grams of water is equivalent to 50 cm3 (for γw = 1g/cm3)
but , q =
∴
50cm3 cm3 = 10 , 5sec sec
and
i=
∆h 27in = = 4.5 L 6in
( 4 ) 10
cm (10mm / cm) 4q mm in ka = = = 0.122 2 2 iπ d sec cm 2 4.5π ( 6in ) 2.54 in
Step 2: Determine the new permeability of the sand due to vibroflotation densification.
Using Casagrande’s relation
k = 1.4k0.85 e 2
( 0.55 ) = 0.066 mm / sec ea2 kb = ka * 2 = 0.122 2 eb ( 0.75 ) 2
Therefore
Step 3: The ratio of densified permeability to in-situ permeability
kb 0.066 = = 0.54 ka 0.122
∴
Vibroflotation has reduced the permeability by half.
Step 4: Find the densified permeability at each sample depth, 100
k = 1 .4 k 0 .8 5 e 2 therefore k0.85 =
k (0.122mm / sec) = 0.155 mm / sec 2 = 1.4e 1.4(0.752 )
The corresponding densified permeability
k = 1.4 k 0.85 e 2
(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
-20
0.015
5.08x10
-25
0.006
1.03x10
-4
-5
2.74x10
-5
-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
ki =
k ∆h Q = L A
1mm ft 3 6, 680 − 3 min 3.281× 10 ft Q = = 2.79 × 10−4 mm 2 sec A ( 4.4mile2 ) 5280 ft mile (1min ) 60 sec min
(
101
)
(
)
Consider the depth -15 ft.
k
∆h 15 −4 mm −4 = 2.76 x10 = 4.14 x10 mm / sec L 10 sec or at a depth of: -20 ft.
k
∆h 20 −5 m m −4 m m = 0 .5 6 x1 0 = 0 .2 2 4 x1 0 L 5 se c se c Therefore, place the bottom of pond at -19 feet.
Q A
(× 10
−4
mm
sec
)
0 0
1
2
3
4
5
6
-5
-10
Depth ft -15
18.7 ft -20
2.76 -25
-30
102
Chapter 6 Seepage and Flow-nets Symbols for Seepage and Flow-nets
103
*Flownets-01: Correcting flawed flow-nets. (Revision: Aug-08)
Do you recognize something wrong with each of the following flow-nets? a)
b)
filter
c)
Equipotential Lines Flow Lines Well
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).
104
*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?
15 m
∆h = 6.0 m
2.0 m
ψ1 10.0 m
17.0 m
ψ3
ψ2 IMPERVIOUS STRATUM (CLAY OR ROCK)
Solution: (a) Notice that ∆h = 6.0 m, the number of flow channels Nf = 3 and the equipotentials Neq = 10.
Forcheimer’s
Using
q = k ∆h
Nf N eq
equation,
cm m 3 −6 2 = 3.5 × 10−4 (6.0 m) = 6.3 ×10 m / sec/ per m of dam width sec 100 cm 10
Since the dam is 500 meters wide (shore-to-shore) the total flow Q under the dam is,
103 liters sec million liters 6 Q = Lq = 500 m 6.3 × 10−6 m3 / sec 31.5 × 10 = 100 3 year year 1 m 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).
105
*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.
∆h = 30 feet feet
5 feet
“a” 19’
Solution:
Find the permeability k using Hazen's formula: mm k = C ( D10 ) 2 = 15(10 −2 mm) 2 = 0.0015 sec Using Forheimer's equation with flow lines N f = 5 and equipotentials N eq = 12, q = k ∆h
Nf N eq
mm 1 in 1 ft 3, 600 sec ft 3 5 = 0.0015 ( 30 '− 5') = 0.185 sec 25.4 mm 12 in 1 hour hr − ft of dam 12
ft 2 ft 3 ∴ Q = Lq = (1, 000 ft ) 0.185 = 185 hr hr The velocity at "a" has a flow q in only that channel, or q/5, 0.185 ft 3 q ft 5 hr v= = ≈ 0.002 A (19 ft high )(1 ft wide ) hr 106
*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?
50 m
12 m
50 m
WT 2m
C
23 m
B
2:1
2:1 D
A
DRAINAGE
E 112 m
2.7 m
Cross-section of levee looking north. Solution: Using Forheimer’s equation,
N Q = Lq = L k ∆h f N eq
107
m m3 3 = (1, 000 m ) 0.3 ( 23 m ) = 2, 070 da y day 10
*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)? Number of flow channels N f = 3 Number of equipotential drops N eq = 7 Using Hazen's formula k = CD 102 = 15 D102 = 15(0.17 mm ) 2 = 0.43
mm sec
Note: 8 ≤ C ≤ 15 for D10 mm; to overestimate flows, use C =15.
40’
rock toe 7 0
25’
1
2
A3
4
5
6
Clay (an aquiclude)
Solution:
N mm 1inch 1 ft 7.5 gallons 3 86,400sec Q = Lq = L k∆h f = (1,200 ft ) 0.43 40 ft ) ( 3 Neq sec 25.4 mm 12inches 1 ft 7 1day gallons Q = 18.8×106 day
4.4 At point "A" the dynamic pressure head is ( 40 ft ) = 25.1 feet 7 2 The static head at "A" is approximately ( 40 ft ) = 26.7 feet 3 Therefore, the total head = static + dynamic = 25.1 ft + 26.7 ft = 51.8 feet
108
**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.
109
Solution: (a) From graph D10 = 0.04 mm. Using Hazen’s relation, with C = 15 to overestimate the permeability of the dam,
k = C ( D10 ) = (15 )( 0.04 mm ) = 0.024 mm / sec 2
2
Nf Q = Lq = L k ∆h N eq ft 3 Q = 454, 000 day
mm 1inch 1 ft 86, 400 sec 3 = ( 5, 000 ft ) 0.024 ( 40 ft ) day sec 25.4 mm 12 inches 9
(b) The gradient in square I is,
iI =
∆h 40/9 = = 0.40 lI 11.2
(c) The pore pressures along ED are approximately, at
pore pressure
E
0
0
1
u=
40' - 2.5
(
40 9
)
62.4 =
1803 psf =
1.80 ksf
2
u=
40' - 3
(
40 9
)
62.4 =
1664 psf =
1.66 ksf
u=
40' - 4
(
40 9
)
62.4 =
1387 psf =
1.39 ksf
4
u=
40' - 5
(
40 9
)
62.4 =
1109 psf =
1.11 ksf
5
u=
40' - 6
(
40 9
)
62.4 =
832 psf =
0.83 ksf
6
u=
40' - 7
(
40 9
)
62.4 =
555 psf =
0.55 ksf
D
u=
40' - 8
(
40 9
)
62.4 =
277 psf =
0.28 ksf
3
110
**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 102 Poise (dynes-sec/cm2) at 20oC. After 1 hour of precipitation, the hydrometer dropped 16 cm. The soil is uniform in size, with 80 % passing the # 200 sieve. e) What type of soil was the sample? f) 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
7m
1m
15 m
Clay stratum
111
Solution: a) Use Stoke’s formula to find diameter of the bay bottom particles:
D10 = d =
18η L 18η . = γ S − γ W t γ W ( GS − 1) t
18 x10−2 dynes − sec x cm3 x 16cm cm 2 ( 2.65 − 1) ( 9.81dynes x 3600sec )
d = 0.070 mm 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:
k = CD102
mm mm = 15(0.07 mm ) 2 = 0.074 sec sec
Use Forheimer’s formula to estimate the total flow Q into the cofferdam:
q = k ∆h
Nf N eq
3 4 m − 2 mm −5 m = 7.4 x10 ( 8m ) 3 = 29.6 x10 m−s sec 8 10 m m
3 60 s m3 −5 m Q = ( perimeter )( q ) = 80 m 29.6 x10 . = 1.42 m − s min min 3 m 3 / min = 2.11 > 2 OKAY Pump is adeq uate ! Pump FS = 1.42 m 3 / min
112
*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 γ = 1 7
kN m3
Elevation –1.0 m
Anchor
Limestone γ = 19
kN m3
A
Nf = 3
Neq = 8 8
0
F2
Elevation –7 m
F1
F3
Elevation –9 m
1 2
7 3 4
113 Sandstone
5
6 γ = 21
kN m3
Elevation –11 m
Solution:
a ) T h e u p li f t f o r c e a t p o i n t A i s f o u n d b y ,
∑
Fy = 0
F u p l if t - F s h e a r
F u p l if t = u A =
(∆ hγ w ) A
F s h e a r = τ A = 0 .1
MN m2
r e s is ta n c e
= 0
kN (1 m m3 kN (4 x ) m 2 = 4 0 0 x m =
( 6 m ) 9 .8 1
59 = 0 .1 4 7 m w i t h F S = 7 ∴ 400 T h i s c o r r e s p o n d s t o e le v a t i o n - 6 m .
∴
x =
2
)=
59 kN
x =1m
b ) D e t e r m i n e t h e f lo w q u a n t i t y w i t h F o r h e i m e r 's f o r m u la , a N Q = qL = Lk b N w here
f q
∆ h
k = C D 120 = (1 2 .5 ) ( 0 .0 9 5 m m
)
2
= 0 .1 1 3
mm sec
∆h = 5m mm 3 m 60 −3 3 3 (5 m ) 10 = 4 .6 m / m i n ≈ 5 m / m i n s e c 8 m m m 3 m F o r a F S = 3 u se a 15 pum p. m in
∴
Q =
( 3 6 0 m ) 0 .1 1 3
114
*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? Dewatering wells
Marshy soils
∆h = 2m
Excavated site
CL
∆h = 2m ∆h = 2m
1 2 3
∆h = 2m
Lowered WT
Neq = 4 Nf = 3
Solution: N otice that ∆ h = 8 m , N f = 3, N eq = 4 and L = perim eter = 1, 000 m . T he perm eability k = C ( D10 ) = (15 )( 0.02 m m ) = 0.006 m m / s 2
N Q = L q = L k ∆h f N eq
2
mm m3 3 m 60 s = 200 + 800 m 0.006 8 m = 2.16 ) ) 3 ( ( s m in 4 10 m m 1 m in 3
m3 ft 7.45 gal g a l lons Q = 2.16 = 60 0 3 m in 0 . 30 m ft m in 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. 115
*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 ∆h is to scale, but you may use 10 m. CL 40 m 10m 10m
40 m 80 m
Phreatic surface condition is fulfilled along the top flow-line GWL
CL
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
116
Solution:
∆h = 10m , N k =
kxk y =
f
= 4 , N eq = 8
( 5 x1 0
−4
m / s )( 5 x1 0 − 5 m / s ) = 1 .6 (1 0 − 4 m / s )
T h e p erim eter o f th e co fferd am p = ( 4 )( 8 0 m ) = 3 2 0 m Nf Q = qp = k ∆h N eq
−4 m 4 p = (1 .6 ) 1 0 (1 0 m ) ( 3 2 0 m ) s 8
m 3 3, 6 0 0 s m3 Q = 0 .2 5 3 = 911 hr s hr
117
**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. g) State the function and properties of the core shell and drains; h) What is the function of the grout between the core and foundation? Under what conditions is it most important? i) Calculate the seepage quantity per foot of length of the dam through the dam, through the foundation, and the total seepage quantity. j) What grading requirements should be specified for the inclined filter A? k) 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. Reservoir surface
H=100 ft
inclined filter A
10 ft
dam permeability k1= 0.001 ft/day
shell core
L = 150 ft
horizontal drain B
Note: The grain size of core = 100% passes 1”, 15% size = 1/8”, and 85% size = 0.001 in. Foundation layer permeability k2 = 0.1 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.
118
h1=5 ft
grout
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(∆h/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 ∑ (∆h)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’
∆hL = 98 6
Zone#2 ZL = 10’ hL = 90’
∆hL = 90 5.5
Zone#3 ZL = 25’ hL = 75’
5.5
∆hL = 75 4.5
Zone#4 ZL = 55’ hL = 45’
6
4.5
∆hL = 45 3
3
4
Q1 = k1 ∑∆hj (b/L)j = 0.001 [98’(2) + 90’ + 75’ + 45’] = 0.081 ft3/day/ft of dam j=1
119
6
5.5
4.5
3
2) Through the foundation Q2 = NF kh 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 (h12-h22) 2L where Q = 3.83ft3/day/ft, L = 150 ft, h1= 5 ft and h2 < 5 ft.
120
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: kmin = 2LQ = 2(150)(3.83) = 46 ft/day h1 2
121
52
Chapter 7 Effective Stresses and Pore Water Pressure Symbols for Effective Stresses and Pore Water Pressure
122
*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?
123
*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?
hw = 25 m water Saturated clay
hs = 15 m 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.
σ ' = γ ' hsoil
where γ ' = γ b = γ SAT − γ W
In order to find γ ' there are a number of derivations, such as this one, ( GS + e ) γ W − γ W where the voids ratio e can be found through Se = wGS 1 + e and noticing that S = 1 because the soil is 100% saturated, e = wGS = ( 0.54 ) (2.78)
γ' =
2.78 + ( 0.54 ) (2.78) (9.81) ( GS + e ) γ W − h = − 9.81 γ (15 m ) W soil + + 1 e 1 0.54 (2.78) ( )
σ ' = γ ' hsoil = σ ' = 105 kPa
124
*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 (drawdowns) 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)
σ V' = [γ h + γ ' h ' ]S A N D + [γ ' h ' ]S IL T + [γ ' h ' ]C L A Y
w h ere
γ ' = γ SAT − γ W
σ V' = ( 2 0 .4 )(3) + (1 8 .8 − 9 .8 1)(3 ) + [ (1 4 .9 − 9 .8 1)(6 ) ] + [ (1 2 .6 − 9 .8 1)(3) ] σ V' = 1 2 8 kP a b)
σ V' = (20.4)(6) + (16.5)(6 ) + [ (12.6 − 9.81)(3) ] = 230 kPa
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.
125
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: ∆h = (62.4) x (4) = -250 ∆h = (62.4) x (6) = 562 ∆h = (110.7) x (6) = 664 ∆h = {[(110-62.4) x (4)] + 664} = 854 ∆h = {[(117-62.4) x (9)] + 854} = 854 ∆h = 664 – 250 = 414 ∆h = 0 + 854 = 854 ∆h = 562 + 1346 = 1908
u + σ'
Depth (ft)
σ
0
0
0
0
6
-250
664
414
10
0
854
854
19
562
1346
1908 126
Chapter 8 Dams and Levees Symbols for Dams and Levees
127
*Dams-01: Find the uplift pressure under a small concrete levee. (Revision: Sept.-08)
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, ∆ (∆h) =
HA − HB N eq
=
30 ft − 5 ft = 1.8 ft / drop 14 drops
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 uplifting force F is, lb lb kip p + pB 35.3 + 20.1 F = L A γ w = 98 ' ( 62.4 ) 3 = 169, 000 = 16 9 2 2 ft ft ft 128
*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 o b) Determine the FS against sliding, if the sand that underlay the dam has ϕ = 3 7 .
129
Solution: Step 1: Determine all forces on the dam ΣFV = 0
k 1 V1 = weight of dam = 0.145 3 (300 ft )(40 ft ) + (260 ft )(300 ft ) = 7,395 kips / ft ft 2 ∑M0 = 0 x1 , from toe O =
(1740 k )(280 ft ) + (5655 k ) 2 (260 ft ) 3 (1740 k + 5655 k )
= 198 ft left of toe
V2 : vertical weight of water upon toe sec tion = 0.0624
k 1 (60 ft )(52 ft ) = 97 kips / ft 3 ft 2
1 (52 ft ) = 17.3 ft left of toe 3 1 k 2 H 1 : lateral force from headwater = 0.0624 3 (285 ft ) = 2,534 kips / ft ft 2 1 y1 , from toe O = (285 ft ) = 95 ft above the toe. 3 1 k 2 H 2 : lateral force from tailwater = 0.0624 3 (60 ft ) = 112 kips / ft 2 ft 1 y 2 , from toe O = (60 ft ) = 20 ft above the toe. 3 x 2 , from toe O =
Step 2: The hydrostatic uplift at the base of the dam. V3 : uplift force =
1 ( pLEFT + pRIGHT )( 300 ft ) = 2, 534 kips / ft 2
where the pressure pLEFT = ( 0.67 ) γ w h = 0.67 ( 0.0624 )( 285 ft ) = 11.9ksf and the pressure pRIGHT = (1.00 ) γ w h = 1.0 ( 0.0624 )( 60 ft ) = 3.7 ksf ∴ V3 =
1 k (11.9 + 3.7 ) 2 ( 300 ft ) = 2,340 kips / ft 2 ft
k k k 300 ft 1 2 3.7 2 (300 ft ) + 11.9 2 − 3.7 2 (300 ft ) (300 ft ) ft ft ft 2 2 3 x3 , from toe O = = 176.3 ft left of toe k 1 k k 3.7 2 (300 ft ) + 11.9 2 − 3.7 2 (300 ft ) 2 ft ft ft
130
Step 3: The factor of safety (FS) against overturning (taken about the toe),
FS =
V x + V2 x 2 + H 2 y 2 resisting moments = 1 1 overturning moments V3 x3 + H 1 y1
FSoverturning =
( 7395 )(198 ) + ( 97 )(17.3) + (112 )( 20 ) = 2.2 > 2 ( 2534 )( 95 ) + ( 2340 )(176.3)
GOOD
Step 4: The F.S. against sliding, FS sliding =
FS sliding =
131
resisting forces H 2 + (V1 + V2 − V3 ) tan φ = driving forces H1
(112 ) + ( 7395 + 97 − 2340 ) tan 37° = 1.58 < 2 NOT GOOD ENOUGH 2534
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. γ→ Unit weight 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. GS→ Specific gravity of the solids of a soil. 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. γ→ Unit weight of the soil. 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. qu→ Ultimate shear strength of a soil. σx→ stress at an specific point (x). z→ Depth of the stratum stat of the soil.
132
133
134
Boussinesq Isobars for a Square Footing (left) versus a Strip Footing (right).
135
Boussinesq Influence factor I1 (the influence from a point load on a surface upon a point inside a semi-infinite body).
P ∆pz = 2 z
1 3 P = 2 I1 5/ 2 2 2π ( r / z ) + 1 z
136
Vertical Stress caused by a Line Load.
Vertical Stress caused by a Strip Load.
137
Boussinesq Influence factor I4 (the influence of a rectangular loaded area upon a point A beneath a corner of the rectangular area).
138
Boussinesq Influence factor I4 (alternative).
139
Vertical Stresses induced by a uniform load upon a circular area.
140
*Stress-01: Reading the Boussinesq charts. (Revision: Feb-10)
Case-1: A single column with a load of 1,000 kips is supported by a square footing 10 feet by 10 feet in plan view. Case 2: A 10 feet wide continuous (wall) footing that carries 10 kips per linear feet. Find the stress increase in a soil mass at the following depths: (a) 5 feet and 10 feet below the center of the footing, (b) 5 feet and 10 feet below the edge of the footing, and (c) 5 feet beyond the edge of the footing, both 5 and 10 feet below the ground surface: Solution. Use the charts on pages 202 through 204 to read the answers. Case 1: The stress at the square footing’s invert is the column load divided by the footing area.
σ=
Q 1, 000 kips = = 10 ksf B2 100 ft 2
This is 100% of the stress at the footing invert. It decreases as the stress spreads out in all directions below the footing. a) The stress at 5 feet below the center of the footing is
σ = ( 0.6 )(10 ksf ) = 6.0 ksf
b) The stress at 10 feet below the center of the footing is
σ = ( 0.32 )(10 ksf ) = 3.2 ksf
σ = ( 0.4 )(10 ksf ) = 4.0 ksf c) The stress at 5 feet below the edge of the footing is The stress at 10 σ = ( 0.25 )(10 ksf ) = 2.5 ksf feet below the edge of the footing is d) The stress at 5 feet beyond the edge of the footing, and 5 feet below the surface is σ = ( 0.07 )(10 ksf ) = 0.7 ksf The stress at 5 feet beyond the edge of the footing, and 10 feet below σ = ( 0.1)(10 ksf ) = 1.0 ksf the surface is Case 2: The stress at the footing’s invert is simply 1 ksf. Therefore,
σ = ( 0.82 )(1 ksf ) = 0.82 ksf a) The stress at 5 feet below the center of the footing is The stress at 10 σ = ( 0.54 )(1 ksf ) = 0.54 ksf feet below the center of the footing is σ = ( 0.4 )(1 ksf ) = 0.4 ksf b) The stress at 5 feet below the edge of the footing is The stress at 10 feet σ = ( 0.4 )(1 ksf ) = 0.4 ksf below the edge of the footing is c) The stress at 5 feet beyond the edge of the footing, and 5 feet below the surface is σ = ( 0.12 )(1 ksf ) = 0.12 ksf d) The stress at 5 feet beyond the edge of the footing, and 10 feet below the surface is σ = ( 0.18 )(1 ksf ) = 0.18 ksf
141
*Stress-02: 2:1 Method to find the Stress below a Rectangular Footing. (Revision: Feb-10)
A rectangular footing carries a single column and measures B = 10 feet wide by L = 14 feet long. The total load of the footing and column results in a uniform load of 10,500 pounds per square feet at the footing’s invert. Determine the stress increase in the soil mass from this footing at a depth of z = 25 feet below the footing’s invert. Solution. The simple 2:1 method assumes that the stress is reduced from 100% at the footing’s invert to decreasing values at increasing depths due to the spreading out of the stress over an increasing area, based on the 2 vertical: 1 horizontal geometry. The stress at z= 25 feet is given by,
∆σ z =
(10 ,500 psf )(10 ft )(14 ft ) ( B + z )( L + z ) (10 ft + 25 ft )(14 ft + 25 ft ) P
=
= 1, 080 psf versus 10 , 500 psf at the footing' s invert .
142
*Stress-03: Summation of vertical and line loads using Boussinesq. (Revision: Feb-10)
A rail gantry crane has been installed close to an existing water main (point A). A pressure cell placed at point A measured a vertical pressure increase of 35 psf due to the two line loads and the point load P = 50 kips. What is the magnitude of the line load q2 (lb/ft)? Line load = q2
P = 50 k
Line load = 750 lb/ft
?s z A
Solution. The pressure increase at “A” due to the concentrated load P1 is found by Boussinesq, 3P z 3 3 ( 50, 000 lb ) ∆σ 1 = = 2π L5 2π
(
( 3 ft )
3
16 + 3 2
2
)
= 0.56 psf
5
The pressure increase at “A” due to the line load q1 is also found by Boussinesq,
∆σ 2 =
2q1 z 3
π ( x2 + z 2 )
2
=
2 ( 750 lb / ft )( 3 ft )
3
2 2 π (12 ft ) + ( 3 ft )
2
= 0.55 psf
Similarly, the pressure increase at “A” due to the line load q2 is,
∆σ 3 =
2q2 z 3
π ( x2 + z 2 )
2
=
2q2 ( 3 ft )
3
2 2 π ( 4 ft ) + ( 3 ft )
2
= 0.0275q2 psf
Use the Law of Superposition to add these three pressures; they must equal 35 psf, ∆σ 1 + ∆σ 2 + ∆σ 3 = 0.56 + 0.55 + 0.0275q2 = 35 psf ∴ q2 = 1, 232 psf
143
*Stress-04: The influence of point loads on a culvert. (Revision: Feb-10)
The truck parked over the drainage culvert box has the rearmost axle directly over the center of the box. The next axle is 48 inches to the front of the rearmost axle. Each axle has four (4) tires and weighs 18,000 lbs. Calculate the stress contributed by each tires of the truck.
Solution. The vertical stress increase from each tire can be found using the Boussinesq point load formula and/or its influence factor I1 (table on page 205). The formula is shown below, with the data from Tire #2 as an example,
∆σ V-truck
P 3 = 2 z 2π
4, 500lb 3 1 = 5/ 2 2 r 2 ( 2 ft ) 2π + 1 z
1 = 95.0 psf 5/ 2 2 ft 2 + 1 2 ft
144
All eight (8) tires will contribute an increase in the vertical stress, as shown in the following table, Tire #
x (ft)
y (ft)
r (ft)
z (ft)
r/z
P (lbs)
1
1
0
1
2
0.50
4500
307.5
2
2
0
2
2
1.00
4500
95.0
3
8
0
8
2
4.00
4500
0.451
4
9
0
9
2
4.50
4500
0.258
5
1
4
4.123
2
2.06
4500
8.506
6
2
4
4.472
2
2.24
4500
6.092
7
8
4
8.944
2
4.47
4500
0.266
8
9
4
9.849
2
4.92
4500
0.168
Total stress
The total vertical stress increase from the truck tires is 418 psf.
145
418 psf
*Stress–05: Stress increase at a point from several surface point loads. (Revision: Feb-10)
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.
A
10 feet
B
10 feet
C
5 feet
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,
P ∆pz = 2 z
1 3 P = 2 I1 5/ 2 2 2 π r / z ) +1 z (
Increase in the load at:
P (lbs)
r (ft)
z (ft)
r/z
I1
∆p (psf)
∆p from A
2,000
(102+52) 1/2 = 11.18
10
1.12
0.0626
1.25
∆p from B
4,000
(102+52) 1/2 = 11.18
10
1.12
0.0626
2.50
∆p 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. 146
*Stress-06: Find the stress under a rectangular footing. (Revision: Feb-10)
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. N = 2,000 kN
B=3m
L=4m
Depth = 6 m
B1 1.5m L 2.0m = = 0.25 and n = 1 = = 0.33 z 6.0m z 6.0m Use the table and extrapolate and find I 4 = 0.0345
m=
2, 000 kN N ∆qz = qo (4 I 4 ) = (4 I ) = 4 )( 0.0345 ) = 23 kN / m 2 ( 4 ( 3 m )( 4 m ) BL
147
*Stress-07: The effect of the WT on the stress below a rectangular footing. (Revision: Feb-10)
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. N = 4,500 kN
B=3m
WT L=5m
WT-1
Depth = 4 m
WT-2
Solution:
q=
N 4,500 kN kN = = 300 A 15 m 2 m2
m=
B 1.5 m L 2.5 m = = 0.375 and n = = = 0.625 z 4m z 4m
∴ I 4 = 0.076 a) The total stress increase from the footing is, ∆po = q(4 I 4 ) = ( 300 )( 4 )( 0.076 ) = 91.2 kN / m 2 and the effective stress when the WT is 1 m below the footing is, ∆po' = ∆po − u = ( 91.2 ) − ( 3 m )( 9.8 ) = 61.8 kN / m 2 b) 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 is, ∆po = 91.2 kN / m 2 which is a 48% increase in stress.
148
*Stress–08: Finding the stress outside the footing area. (Revision: Feb-10)
Find the vertical stress increase ∆p below the point A at a depth z = 4 m. Solution: The stress increase, ∆p , can be written as :
∆p = ∆p1 − ∆p2
where ∆p1 = stress increase due to the loaded area shown in (b). ∆p2 = stress increase due to the loaded area shown in (c).
For the loaded area shown in (b): B 2 L 4 m = = = 0.5 and n = = = 1.0 Z 4 Z 4 ∆p1 = qI 4 = (150)(0.12) = 18 kN / m 2 Similarly, for the loaded area show in (c): B 1 L 2 m = = = 0.25 and n = = = 0.5 Z 4 Z 4 ∆p2 = (150 )( 0.048) = 7.2 kN / m 2 Therefore, ∆p = ∆p1 − ∆p2 = 18 − 7.2 = 10.8 kN / m 2
149
*Stress-09: Stress below a footing at different points. (Revision: Sept.-08)
A clay sanitary pipe is located at a point C below the footing shown below. Determine the increase in the vertical stress ∆p 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. q = 1,800 psf
. .
. . .
B
4 ft
5 ft 2 ft
A
.
B
A
10 ft
C
3 ft
10 ft
PLAN VIEW
5 ft
C
3 ft
SECTION VIEW
Solution: For the expanded 5’ x 13’ area, m=
B 5 = = 1 and Z 5
n=
For virtual 3’ x 5’ area
L 13 = = 2.6 therefore, I 4 = 0.200 Z 5 m=
B 3 = = 0.6 and Z 5
n=
L 5 = = 1 therefore, I 4 = 0.136 Z 5
The increase in stress at point C from the footing is therefore, lb ∆p = q ( I 4 - I 4' ) = 1,800 2 (0.200 - 0.136) = 115 psf ft
150
*Stress-10: 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. ∆p2 25m
15 ft
1
1
∆p 1
1
1
∆p 3
Z 15 ft
The contribution from the central portion of the fill is ∆p2 whereas the contribution from the left and right hand slopes are ∆p1 and ∆p 3 respectively. Using Boussinesq,
Solution: The contribution from the central portion of the fill is ∆p2 , whereas the contribution from the left and right hand slopes are ∆p1 and ∆p 3 respectively. Using Boussinesq, ∆p1 ⇒
2x1 2(15') 2z 2(15') = =2 ∴ = =2 B1 15' B1 15'
∆p1 lb = 0.25 ∴∆p1 = (0.25)(15') 120 3 = 450 psf q ft
------∆p2 ⇒
2x2 2(−12.5') 2z 2(15') = = −1 ∴ = = 1.2 B2 25' B2 25'
∆p 2 lb = 0.47 ∴∆p 2 = (0.47)(15') 120 3 = 846 psf q ft
-------∆p3 ⇒
2x3 2(40') 2z 2(15') ∆p 3 lb = = 5.3 ∴ = =2 = 0.02 ∴∆p 3 = (0.02)(15') 120 3 = 306 psf q ft B3 15' B3 15'
∆p = ∆p1 + ∆p 2 + ∆p 3 = 450 + 846 + 36 = 1, 332 psf
151
*Stress-11: Finding a stress increase from a surface load of limited width. (Revision: Feb-10)
Determine the average stress increase below the center of the loaded area, between z = 3 m and z = 5 m. q = 100 kN/m2
3m 5m
1.5 m
. .
z 1.5 m
B
3m
L
.
A,
A
SECTION VIEW
3m
PLAN VIEW
Solution: The stress increase between the required depths (below the corner of each rectangular area) can be given as: ( H )( I ( H )) − ( H1 )( I 4 ( H1 )) (5)( I 4 ( H 2 )) − (3)( I 4 ( H1 )) ∆pavg ( H 2 / H1 ) = q 2 4 2 = 100 H 2 − H1 5−3 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:
∆ p av(H 2 /H 1) = 100 ×
(5)(0.038) − (3)(0.086) = 3 .4 kN /m 2 5−3
The stress increase between z = 3 m and z = 5 m below the center of the load area is equal to:
4∆pavg ( H 2 / H1 ) = (4)(3.4) = 13.6 kN / m 2
152
**Stress-12: Stress increase as a function of depth. (Revision: Feb-10)
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,
σv =
Z
∫ γ (σ ) dz v
100
=
0
∫ (95 + 0.0007σ
v
) dz
0
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:
Rearranging, and integrating by parts, σv
∫ 0
dσ v = ( 95 + 0.0007σ V )
100
∫ dz 0
1 σ ln ( 95 + 0.0007σ V )0 V = z100 0 0.0007
σ v = 135,800 ( e0.0007 z − 1)
100
At Z = 100
∴
= 9,840 psf
σ v = 135,800 (1.0725 − 1)
σ v = 9,840 psf (Revision: Sept.-08)
153
0
*Newmark–01: Stress beneath a tank at different depths. (Revision: Feb-10)
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: BB == 1100 m m B = 10 m for point #1
SSA AN ND D
22 m m 1
C CLLA AY Y
1100 m m
B = 10 m for point # 2
2 x
AB = 2 m
A
B
AB = 12 m
Influence Value (IV) = 1/200 = 0.005
y
The increase in the vertical stress is found from ∆p = po M (IV) where M is the number of “squares” enclosed in the Newmark chart and (IV) is the influence value. For ∆p1, AB = 2 m ∆p1 = po M(IV) = (400 kN/m²) (190) (0.005) ∆p1 = 380 kN/m² For ∆p2, AB = 12 m ∆p2 = po M(IV) = (400 kN/m²) (42) (0.005) ∆p2 =
84 kN/m²
(This decrease in stress represents a 78% reduction in the vertical stress).
154
*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,
6 6 0 kN 2 ∆ p = q M ( IV ) = ( 4 5 )( 0 .0 0 5 ) = 1 6 .5 k N / m ( 3 m )( 3 m ) The number of elements inside the outline of the plan is about M = 45. Hence,
6 6 0 kN 2 ∆ p = q M ( IV ) = ( 4 5 )( 0 .0 0 5 ) = 1 6 .5 k N / m ( 3 m )( 3 m )
155
*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. Influence Value (IV) = 0.005
Depth Point = Z
Solution:
Set AB = 5 and draw the footing to that scale. The number of affected areas M =8, therefore ∆p = q M (IV) = (1,800 psf)(8)(0.005) = 72 psf
156
*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:
kN m2 kN Point B (bottom of clay) q B = q M ( IV ) = (100 )(100 )( 0.005 ) = 50 2 m q + qB ( 70 + 50 ) kN qaverage = A = = 59 2 2 2 m Point A (top of clay stratum) q A = q M ( IV ) = (100 )(136 )( 0.005 ) = 68
157
*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. Set AB 10 m
Influence Value (IV) = 0.001
Solution: The contact stress is qo = (0.95)(9.81 kN/m3)(15m) = 140 kN / m 2 At mid-clay depth along the centerline of the tank (depth = 10 m) ∴OQ = 10 m
σ v' = ( qo − γ w h ) M ( IV ) = 140 − ( 9.81)(10 ) ( 680 )( 0.001) = 28.5
kN m2
158
**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. l) Using the Newmark method, what are the new stresses at the top and bottom of the clay stratum due to the building’s loading? m) What is the expected differential settlement between the building’s center and one of its corners, in mm? n) 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? q =100 kN/m2 3m
1m 2m
SAND C
•
E
•
γ = 18.5 kN/m3
5m
•D
CLAY
w = 22 %
•
F
Solution:
(1) Set AB = 2 m and observe that the building's foot-print covers the entire graph. ∴ ∆ σ E = 100 kN / m 2 Set AB = 7 m and the number of units M = 175 ∴ ∆ σ F = qM ( IV ) = (100 )(175 )( 0.005 ) = 87.5 kN / m 2 The stress at point C has dropped to 50% of the stress at E, ∴ ∆ σ C = 50 kN / m 2
and ∴ ∆ σ D = ( 0.96 )( 50 ) = 48 kN / m 2
Average stress in the clay stratum beneath the center = Average stress in the clay stratum beneath the corner =
159
(100 + 87.5 ) 2 ( 50 + 48 ) 2
= 93.8
= 49
kN m2
kN m2
(2) The differential settlement between the center and a corner of the building is, Cc = 0.30 and GS = 2.70 wGs ( 0.22)( 2.70) = = 0.59 S 1 The in-situ stress at mid-clay stratum before the building is built is, Seo = wGS ∴ eo =
po = γ sand hsand + γ clay hat mid −clay = ( 3 m)(17.5) + ( 2.5 m)(18.5 − 9.8) = 74.3
kN m2
Therefore, the settlement at the center ∆H =
p + ∆σ ( 0.30)( 5m) ( 74.3) + ( 93.8) CC H log o log = 0.33 m = 1+ e 74.3 ( ) po 1 + ( 0.59)
the settlement at the corner ∆H =
p + ∆σ ( 0.30)( 5m) ( 74.3) + ( 49) CC H log o log = 0.21m = 1+ e 74.3 ( ) po 1+ ( 0.59)
the differential settlement is ∆ ( ∆H ) = 0.33 m − 0.21 m = 0.12 m = 120 mm (3) The time required to attain 60% consolidation in the field is,
( 0.26)( 2,500 mm) = ( 0.18)( 2x25.4 mm) T H2 cv = σ drained = t tF ( 5 hours ) 2
2
Solving for the time of settlement in the field t f ,
( 0.26)( 2,500 mm) ( 5 hours ) 1day 1 year ≈ 2 years tF = 2 1 ( 0.18)( 2x25.4 mm) 24 hours 365 days 2
160
*Mohr-01: Find the angle of internal friction φ of sand. (Revision: Jan-09)
The horizontal stress on a soil particle at a depth of 130 ft is ⅓ of its vertical stress. If the average unit weight of the intervening soil mass above is 116.4 pcf, and the shear stress on the soil particle’s failure plane is 30 psi, what is the soil’s average angle of friction?
Solution: The vertical stress is the major principal stress σ1 and the horizontal stress is the minor principal stress σ3. The shear stress is related to these two principal stresses via the angle θ, which is the angle to the failure plane with respect to the principal stress σ1. These are related by the equation,
τ=
σ1 − σ 3 2
sin 2θ
From this relation, the angle of internal friction is found from θ,
ϕ = 2θ − 90° Therefore,
σ 1 = γh = (116.4lb / ft 3 )(130 ft )
1 ft 2 = 105 psi 2 144in
σ 3 = σ 1 = (0.33)(105 psi ) = 35 psi 1 3
From τ =
σ1 − σ 2 2
sin 2θ
we have 30 psi =
∴ 2θ = 120 ∴ ϕ = 2θ − 90 = (120 ) − 90 = 30°
161
105 psi − 35 psi sin 2θ 2
*Mohr-02: Simple transformation from principal to general stress state. (Revision: Feb-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,
σ θ = 36kN / m 2 τ = 9 . 2 kN / m 2
σ 3 = −4.18kN / m 2
σ 1 = 146 kN / m 2
(b) The calculated solution,
kN σ1 + σ 3 σ1 − σ 3 14.6 − 4.18 14.6 + 4.18 + cos 2 ( 50° ) = 3.6 2 + cos 2θ = 2 2 m 2 2 kN σ −σ 14.6 + 4.18 τ θ = 1 3 sin 2θ = sin 2 ( 50° ) = 9.2 2 2 m 2
σθ =
kN σ1 − σ 3 = 9.4 2 m 2
τ maximum =
162
*Mohr–03: Find the principal stresses and their orientation. (Revision: Sept.-08)
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 σ = q/ π ( α sin 3 Β /2 Β /2 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.
δ α
Solution:
σ = σ1 + σ2
σ ,τ
2
β = 85.23
σ =
σ =
θ β 2θ = 94.77
σ =
Ο
47.38
σ =
Ο
σ1 + σ3
σ =
2
σ ,τ τ = α + δ = arc tan 2.5 = 68.20° δ = arc tan 0.5 = 26.57°
2θ + β = 180°
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
163
Ο
*Mohr–04: 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.
164
*Mohr-05: Find the angle of internal friction. (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:
165
*Mohr – 06: 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: - 40 lb/ft
2
σy = 90 lb/ft2
θ=30° τn σx = 125 lb/ft2 σn τxy = - 40 lb/ft2
(a) σ1 =
(b) σ y +σ x 2
σ y −σx 2 + + τ xy 2 2
2
90 + 125 90 − 125 2 + σ1 = + (−40) 2 2 σ 1 = 151 psf
σ3 =
σ y +σ x 2
σ y −σ x 2 − + τ xy 2 2
σn =
σ1 + σ 3
σ1 − σ 3
cos 2θ 2 2 151 + 64 151 − 64 + cos 60 = 129 psf σn = 2 2 σ −σ τ n = 1 3 sin 2θ 2 151 − 64 sin 60 = 38 psf τn = 2 +
90 + 125 90 − 125 2 − σ3 = + (−40) 2 2 σ 3 = 64 psf 2
166
*Mohr–07: Find the maximum and minimum stresses on a given plane. (Revision: Feb-10)
Determine the normal and shear stresses on plane AB.
Solution: 2 2 σ1 σ X + σ Y σ 1 = 922 psf 400 + 750 σ Y − σ X 400 − 750 2 2 τ = ± + = ± + − = ( 300) XY 2 σ3 2 2 2 σ 3 = 228 psf
Knowing the principal stresses permits to calculate the stresses on the normal plane,
σn =
σ1 + σ 3
+
σ1 − σ 3
cos 2θ =
922 + 228 922 − 228 + cos 2 ( 45° ) = 575 psf 2 2
2 2 σ −σ 922 − 228 τ n = 1 3 sin 2θ = sin 2 ( 45° ) = 347 psf 2 2
167
**Mohr – 08: 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. B
σ3 = 0
20 kN/m2 10 kN/m2 A
A
β
β
Graphical Solution:
C
OR
B Analytical Solution:
τ τmax = 12.5 σ1 =
1
20 = 2θ
σ1 +σ1 σ1 2
σ1=25
σ3=0
2 +
σ1 2
+
σ1 +σ1 2
COS 2θ
COS 2θ ∴ σ1 =
40 1 + cos 2θ
1+cos2θ
τn =
2
σ1 −σ3 2
2θ
SIN 2θ
2
10 =
(σ, τ)
σ1 2
SIN 2θ
∴ ϑ1 =
20 SIN 2θ
(20, -10)
σ1 = 25 kN/m2
(b)
α in the figure is in the angle 2θ between (σ, τ) and σ3.
……
θ = 63.4˚
…… α = 2θ = 126.9˚
…..
σ1 =
……
τmax= 12.5 kN/m2
(c)
and
σ3 = 0
∴ 1 - 2 sin 2θ + cos 2θ = 0
(a)
20 = 25 KN / m2 , sin 126.90
σ3 =0
…… θ = 63.4˚ and
τmax= 12.5 kN/m2
168
*Mohr – 09: 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
169
*Mohr –10: 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 ∆OAB (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.
170
*Mohr – 11: Normal and shear stresses on any plane. (Revised Sept-09)
Determine the normal and shear stresses on the plane AB.
Solution. 001 = (300+125)/2 = 212.5 psf 0102 = (300-125)/2 = 87.5 psf 01B = 87.52 + 552 = 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
171
*Mohr–12: Derive the transformation from general to principal stresses. (Revised Sept-09)
o) 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). p) Determine the value of the major principal stress. q) Determine the angle θ between the major principal stress and the state of stress shown in the figure above.
172
*Mohr – 13: Tensile stress from the 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. +τ
(2.) σf = σθ
+τ
τf = τθ
(3.) θ 25 ° 2θ = 115° 96 kN/m2
+σ
σ1
(4.) qu
Solution: σ1f = qu = 300 Pa σf = σθ = 86 Pa τf = τθ = 136 Pa θ = 115°/2 = 57.5° qu = -123 Pa
173
+σ
(1.) σ1f = qu
**Mohr–14: Maximum principal stress from triaxial test. (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: τ
τ Φ = 36°
T R 0
Φ σ3
σ
Φ 0
σ d
τ’
a)
Failure will occur when the Mohr circle becomes tangent to the failure envelopes.
R = sin φ d R (1 - sin φ ) −R= R sin φ sin φ R (1+ sin φ ) σ1 = d + R = +R=R sin φ sin φ σ (1+ sin φ ) φ ∴ 1= = tan2 (45° + ) σ3 (1 − sin φ ) 2
σ3 = d − R =
∴σ1 = (300 kPa) tan2 (45° +
36° ) = 1160 kPa 2
b)
H = c(cot φ )
φ σ1 + H = (σ 3 + H ) tan2 (45° + ) 2 H = (12 kPa)(cot 36°) ≈ 17 kPa
σ1 = (300 +17) tan2 (45° +
36° ) −17 = 1,200kPa 2 174
*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
θ = 45° +
ad =
∴ 2θ = 90° + φ
2
σ1 −σ3 2
sinφ ≈
ad af
(1)
af = af + af = (c) cotθ +
tan θ =
c of
σ1 +σ3
(2)
2
of = cot θ ∴ of = (c) cot θ c
∴
(3)
According to the Mohr’s circle properties: oa = σ 3 +
σ1 -σ3 2
∴ oa =
σ1 -σ3 sin φ =
2
(c )cot φ + σ 1 + σ 3
σ1 -σ3 2
∴ af = (c) cot φ +
σ1 -σ3
(4)From (1) & (4):
2
σ +σ3 σ1 -σ3 ∴ sinφ (c )cot φ + 1 = 2 2
2
(c ) sin φ cot φ + σ 1 + σ 3 sin φ = σ 1 - σ 3 ∴ 2
2
2
∴ (c) sin φ cot φ =
((c ) sin φ cot φ ) = σ 1 - (sin φ ( σ 1 ) - (σ 3 + sin φ ( σ 3 ) )
cot φ ∴ 2 (c ) sin φ = σ 1 (1 - sin φ ) - σ 3 + (1 + sin φ ) sin φ
Since
cos φ ≈ tan 45 0 + φ 1 − sin φ
(
)
and
φ φ σ 1 = σ 3 tan 2 45 0 + + 2c tan 45 0 +
175
2
2
σ1 -σ3 2
−
σ1 +σ3
∴ 2
2
sin φ
(c ) cos φ + σ
1 − sin φ
1 + sin φ φ ≈ tan 2 45 0 + 1 − sin φ 2
3
1 + sin φ = σ1 1 − sin φ
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. ∆(∆e) → Differential settlement between adjacent foundation. ∆Hi→ Elastic settlement. I→
Influence factor; essentially equivalent to the strain ε in the soil.
IZ→
Simplifying influence factor.
ε→
Strain at mid stratum.
γ→ Unit weight of the soil. v→
Poisson’s ratio.
MT→ Transverse moment.
176
*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:
B2 =
tire's load (0.5)(20%)(120, 000 lb) = = 150 in 2 ∴ B = 12.2 inches 2 tire pressure q o (80 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,
ε 144 in 2 lb 0.30 k ∆ = C1C2 qo dz = (1)(1)(80 2 )( )(24.4 in) × 2 2 in 168 k / ft ft 1, 000 lb Es ∴∆ ≈ 0.5 inches of rutting.
177
*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
Es ≅ 14 where Es is in ksf; used for fine medium sands. N
Solution:
Layer ∆z
1 2 3
(in) 42 60 66
Es (ksf) 140 210 168
Iz=ε
I z ∆z Es
0.30 0.46 0.16 ∑=
0.090 0.130 0.061 0.281
Q 200 k − Df γ = − ( 4 ft ) (0.120 kcf ) = 3.60 ksf 2 B 49 ft 2 The coefficients for the Schmertmann method are C1 and C2 : The contact pressure on the soil is, qo =
D γ 0.48 Depth factor C1 = 1.0 − 0.5 f = 1.0 − 0.5 = 0.93 3.60 qo Creep factor C2 = 1.35 for a five year period . The Schmertmann formula for the elastic settlement ∆ is, 2B
ε
∴∆ = C1C2 qo ∑ ∆z = (0.93)(1.35)(3.60)(0.281) = 1.27 inches 0 E 178
*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:
The contact pressure on the soil is, q0 =
Q 64 kips = = 7.11 ksf B 2 ( 3 ft )2
The SPT value indicates that the soil is a loose sand. The modulus E for loose sand can be calculated using the following formula:
ES ≈ 10 ( N + 15 ) ksf 179
The following table summarizes the data and calculations:
Average strain at mid
εZ∆Z/E
Layer’s thickness ∆Z (feet)
SPT (average) N
Soil’s elastic modulus E (ksf)
stratum ε
(ft /kip)
1
1.5
6
210
0.35
0.0025
2
4.5
7
220
0.30
0.0061
Layer number
3
Σ = 0.0086 The correction factors are as follows: 0.5γ D f 1. Depth factor, C1 = 1 − q −γ D f 0
( 0.5 )( 0.110 )( 4 ) = 1 − = 0.97 7.11 − ( 0.110 × 4 )
Since C1 > 0.5, this is FINE.
t yr 5 2. Creep factor, C2 = 1 + 0.2log = 1 + ( 0.2 ) log = 1.34 0.1 0.1 The total elastic settlement is,
ε
∆= CC 7.11−( 0.110)( 4)( 0.0086) = 0.07 ft = 0.84in 1 2 ( qo −γ Df ) ∑ dz = ( 0.97)(1.34) 0 Es 2B
180
*Elastic Settlement-04: The 2:1 method to calculate settlement. (Revision: Aug-08)
Use the 2:1 method to find the average stress increase (∆q) 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 ∆Hi?
Qu=120 kips
Solution: The settlement ∆H of an elastic media (the 5 foot thick sand stratum in this case) can be found from the theory of elasticity as,
1 − v2 ∆H = qo B Iw Es 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,
∆q =
1 H
H2
Qu
∫ ( B + Z ) dz H1
2
where H1 = 0 feet (the footing’s invert) to H2 = 5 fee (bottom of the sand stratum). 181
5
1 Q 1 120kip 120kip ∆q = = − = 1.82ksf H B + z 0 5 ft 6 ft 11 ft but 1 − 0.32 120 kip 12 in 1 − v2 ∆H = qo B Iw = 6 ft 0.95 ( ) ( ) = 0.52 inches 2 3 6 400 Es ft ksf ft
182
*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 ∆(∆H) between them. Is this ∆(∆H) acceptable for columns spaced 30 ft apart?
Solution: Footing size for column A:
BA =
QA = qall
430 kips = 12 feet 3 ksf
Footing size for column B:
BB =
QB 190 kips = = 8 feet qall 3 ksf
A quick estimate of the ratio of settlement to the proportionality is
∆H A 12 ft = = 1.5 ∆H B 8 ft
Therefore if the settlement at column B is ∆HB = 1 in., then the settlement at column A will be ∆HA = 1.5 in. Then,
∆ (∆H ) = 1.5 in − 1.0 in = 0.5 inches and the rotation between the two columns is
θ=
∆ ( ∆H ) = L
( 0.5
in )
12in ( 30 ft ) ft
= 0.0014
Both of these values [∆(∆H) and θ] are acceptable, since ∆(∆H) should be < 1" and θ < 0.0033.
183
*Elastic Settlement-06: Compare the settlements predicted by the Boussinesq, Westergaard, and the 2:1 methods. (Revision: Aug-08)
Compute the average stress ∆q 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 ∆H = ∆Hi + ∆Hc + ∆HS to only 1.5 inches. Estimate the initial settlement ∆ H i = 0.05 Z s where Zs is the thickness of the granular stratum beneath the footing in feet, to give ∆Hi in inches.
Solution: Assume an initial value of B = 10 feet.
Q = 240 kips
γS= 110 pcf
4’
5’
SAND
5’
CLAY
WT γ = 120 pcf eo = 1.11 Cc = 0.42
ROCK
The contact pressure qo of the footing is:
qo =
Q 240 kips = = 2.4 ksf AB 100 ft 2
(a) Stress at mid-clay stratum using Boussinesq’s method (use the charts on page 205) for a square footing: q = 0.52 ∴ q = ( 0.52 )( 2.4ksf ) = 1.25 ksf qo 184
(b) Stress at mid-clay stratum using Westergaard’s method for a square footing: q = 0.33 ∴ q = ( 0.33)( 2.4 ksf ) = 0.79 ksf qo (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:
B = 10 ft.
z = 7.5 ft.
qo
B+z
q=
Q 240 kips = = 0.78 ksf AB + Z (10 ft + 7.5 ft ) 2
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 (∆Hi): ∆H i = 0.05 Z s = 0.05 5 ft = 0.11 in
The in-situ effective stress qo' at mid-clay layer, before placing the footing is:
kip kip kip qo ' = γ s hs + γ c hc = 0.11 × 9 ft + 0.120 3 − 0.0624 3 × 2.5 ft = 1.13 ksf 3 ft ft ft Using q = 0.78 ksf from the 2:1 method, the total settlement ∆H is equal to the immediate ∆Hi, plus the consolidation ∆Hc, and the secondary settlement ∆Hs, but limited to no more than 1.50 inches. ∆H = ∆H i + ∆H C + ∆H s = 1.50 inches 185
But ∆Hi = 0.11 in, and ∆Hs is negligible for this problem. Therefore the maximum permissible consolidation settlement ∆Hc is limited to: ∆H c = 1.50 in − ∆H i = 1.50 in − 0.11 in = 1.39 in or
∆H c =
Cc H q + ∆q log o = 1.39 in = 1 + eo qo ' '
12 in 1.13 ksf + ∆q ft log = 1.39 in 1 + 1.11 1.13 ksf
( 0.42 ) 5 ft ×
Therefore ∆q = 0.39 ksf
Using the 2:1 method:
∆q =
Q AB + Z
B+Z =
Q 240 kips = = 26.6 ft ∆q 0.34 ksf
Therefore B = 19 feet Since the initial B = 10 feet, the new value of 19 feet should be used to re-iterate towards a better solution that converges.
186
*Elastic Settlement-07: Schmertmann versus the strain methods. (Revision: Aug-08)
Compute the immediate settlement ∆Hi using the Schmertmann formula using an average ∆q 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 Df 20,400 kPa, = 0.5 and C1 = C2 = 1. Compare the results with an alternate method B 1− v2 I w , where v = 0.3 and Iw = 0.95. using ∆H = qo B E s
Q = 2,100 kN B = 3m x 3m 1.5 m 1.5 m
A
3m B
SAND
5m ROCK
Solution:
The average stress from multiple layers is solved via this formula, ∆qv =
1.5m 233.3kPa + 28.0kPa ∆H q1 + qn 163.3 77.0 44.0 + q + q + … + q = + kPa + kPa + kPa 2 3 n 1 − 6m H 2 2
∆qv = 104 kPa The elastic settlement via Schmertmann is, ∆H = C1C2 (0.6B)
∆q 1000 mm 104 kPa = (1)(1)( 0.6)( 3 m) = 9.2 mm Es m 20, 400 kPa
The alternative method from the theory of elasticity would yield, 1 − v2 1 − 0.32 1000 mm ∆H = qv1B I w = ( 233.3 kPa )( 3 m) ( 0.95) = 10 mm m 20, 400 kPa Es
187
*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 Layer No.
Thickness
1 2 3 4 5
∆Z, (feet) 5.0 5.0 2.5 2.5 5.0 Σ = 2B = 20 ft
Soil Modulus
IZ = ε
ES, (ksf)
(average strain)
I Z ∆Z (ft/kp) E
360 260 300 560 190
0.35 0.50 0.35 0.25 0.10 Σ=
0.00486 0.00962 0.00292 0.00112 0.00263 0.02115
188
The Schmertmann coefficients are, The depth coefficient C1 =1- [0.5(
γDf 0.5(0.1)(5) )]= 1- = 0.75 q -γ D f 1.5 - (0.1)(5)
The time coefficient C2 = 1.35 2B
ε
∆H i = C1C2 (q - γ D f )∑ dz = ( 0.88 )(1.35 ) 2.5 − ( 0.1)( 5 ) ( 0.02115 ) = 0.042 ft = 0.05 in 0 E
189
**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 ∆z (in) Es (ksi) Iz =ε (Iz/Es)∆z (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
( 0.100 )( 8 ) = 1.0 − 0.5 = 0.92 6 0.100 8 − q − D γ ( )( ) ( f ) The time factor (creep ) C2 = 1.0 The depth factor C1 = 1.0 − 0.5
2B
ε
0
Es
∆ = C1C2 ( qo − γ D f ) ∑
γ Df
dz = ( 0.92 )(1.0 ) 6 − ( 0.100 )( 8 ) ( 84.29 ) = 3.78 in
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) ∆q = 0.12q = 0.12(6) = 0.72 ksf
∆1 =
q + ∆q 0.20(25x12) 5.95 +0.72 CcH 1 = = 1.48 in log 10 o log 10 1+ e o 1+1 5.95 qo
At corner ∆q = 0.09q = 0.09(6) = 0.54 ∆1 =
0.20(25x12) 5.95+0.54 = 1.13 in log 10 2 5.95
For clay 2: qo = 5.95 + ( 0.125 + 0.0624)12.5 + (0.130 - 0.0624)12.5 = 7.58 ksf 190
midlayer at 112 ft = 4.5 B/2, ∆q . 0.07q = 0.42 ∆2 =
0.35(25x12) 7.58 +0.42 = 1.25 in at the centerline log 10 1+0.96 7.58
∆q . 0.05q = 0.30 ∆2 =
0.35(25x12) 7.58 +0.30 = 0.96in log 10 1+0.96 7.58
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) ∆q = 0.04q = 0.24 ksf ∆3 =
0.15(25x12) 8.65+0.24 = 0.28in log 10 1+0.89 8.65
at corner ∆q = 0.03q = 0.18 ksf ∆3 =
0.15(25x12) 8.65+0.18 = 0.21in log 10 1+0.89 8.65
Method B: the equivalent layer equation
∆ ∆e = H 1+ e o
where ∆e = C c log
q o + ∆q qo
The total settlement on the centerline is, ∆6 = 3.78 + 1.48 + 1.25 + 0.28 = 6.78 inches
and along the foundation edge, ∆edge = 3.78 + 1.13 + 0.96 + 0.21 = 6.08 inches
The differential settlement is ∆(∆) = 0.70 inches The allowable for reinforced concrete buildings ∆(∆)< 0.003(span) = 0.003(25x12 inches) = 0.90 inches
Therefore, the design is acceptable.
191
Chapter 11 Plastic Settlements Symbols for Plastic Settlements e→
Voids ratio.
ES →
Soil elastic modulus.
CC→
Compression index.
GS→
Specific gravity of the solids of a soil.
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.
192
*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%.
∆ p = 3 ksf h
Ground water table
20 feet 15 feet
Sand
4 feet 10 feet
Clay
A
Rock
Solution: a) Assume a uniform increase of the initial excess pore water pressure throughout the 10-foot thickness of the clay layer:
The pore water pressure is u0 = ∆p = γ w h = 3,000 lb / ft 2 ∴ h =
193
∆p
γw
=
3,000 = 48.1 feet 62.4
b) The degree of consolidation at A is UA (%) when h = 15 feet:
u (15 ft )(62.4 pcf ) U A % = 1 − A 100 = 1 − 100 = 69% (48.1 )(62.4 ) u ft pcf 0
c) When UA = 60 %, what is the value of h?
u uA U A = 0.6 = 1 − A = 1 − u0 3, 000 psf ∴ u A = (1 − 0.6 )( 3, 000 psf ) = 1, 200 psf h=
uA
γw
=
(1, 200 psf ) = 19.2 feet ( 62.4 pcf )
194
*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.
∆p = 1 ksf Sand 8 feet
γ = 110 pcf, Water Table
7 feet
γsat = 115 pcf
Clay 17 feet
w = 34%, LL = 50%, γsat = 120 pcf
Solution:
Skempton formula for the C c = 0.009( LL − 10) = 0.009(50 − 10) = 0.36 Se = wG s
but S = 1 ∴ e = wG s = ( 0.34 )( 2.67 ) = 0.91
The stress of the clay at its mid − stratum before the building was built is , p o = γ d − sand H dry − sand + (γ sat − γ w ) H sat − sand + (γ sat − γ w ) H mid − clay = (0.11kcf )(8 ft ) + (0.115 − 0.0624) kcf (7 ft ) + (0.120 − 0.0624) kcf (8.5 ft ) p o = 1.74 ksf The consolidation settlement is , ∆H =
195
HCc p + ∆ p (17 )(12 ) (0.36) 1.74 + 1 log 10 o = log 10 = 7.6 i nches 1 + eo po 1 + 0.91 1.74
*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.
200 kips
5’ 10 feet
5’ x 5’ footing Sand Ground Water Table
γdry = 100 lb/ft3 γsat = 120 lb/ft3
5 feet
Clay 10 feet
γsat =110 lb/ft3, e = 1.0, LL = 40
Solution:
Skempton formula Cc = 0.009(LL −10) = 0.009(40 −10) = 0.27 The stress of the clay at its mid − stratumbefore the building was built is, po = γ d −sand Hdry−sand + (γ sat − γ w )Hsat −sand + (γ sat − γ w )Hmid −clay po = (0.10kcf )(10 ft) + (0.120 − 0.0624)kcf (5 ft) + (0.110 − 0.0624)kcf (5 ft ) po = 1.53 ksf The ∆pavg below the center of the footing between z1 = 15 feet to z2 = 25 feet is given as,
H 2 I 4( H 2 ) − H1 I 4( H1 ) B L ∆pavg = 4q ,n' = where I 4( H 2 ) = f m′ = H 2 − H1 H2 H2 5 B = L = = 2.5 feet 2 196
m′ =
B 2.5 = = 0.167 and H1 15
n′ =
L 2.5 = = 0.167 H1 15
From Boussinesq's Table for I 4 I 4( H 2 ) = 0.05 and
I 4( H1 ) = 0.075
The average pressure at mid-clay layer is thus given by, 200 kips (25)(0.05) − (15)(0.075) H 2 I 4( H 2 ) − H1 I 4( H1 ) ∆pavg = 4q = 4 ′ ′ = 0.4 ksf = 400 psf ′ ′ H − H 5 5 − 25 15 ( )( ) 2 1 The consolidation settlement is, ∆H =
197
p + ∆pavg (10 )(12 ) (0.27) HCc 1.53 + 0.40 = = 1.6 inches log10 o log10 po 1 + eo 1+1 1.53
*Plastic Settlement -04: Surface loads with different units. (Revision: Aug-08)
Find the total settlement under a building that applies the load shown below.
q = 1.2 daN/cm2
4.6 m Ground Water Table Sand 6.0 m
γ = 17.6 kN/m3, γ’ = 10.4 kN/m3 Normally consolidated clay
7.6 m
Gs =2.78, w = 40%, PI = 15%, PL = 30%
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.
γ s = γ wGs = 27.8 kN m3
Se = (1)e = wGs = 0.40(2.78) = 1.11
Therefore, the clay unit weight is γ sat =
( Gs + eo ) γ w = [(2.78) + (1.11)] ( 9.81) = 18.1 kN 1 + eo 1 + (1.11)
m3
The effective unit weight for the clay is,
γ ′ = γ sat − γ w = 18.1 − 9.81 = 8.3 kN m3 Also, the Skempton relation is, Cc = 0.009( wL − 10) = 0.009(45 − 10) = 0.32 The stress at the mid-clay stratum,
σ ′ = (4.6m)(17.6 kN m3 ) + (6.0m)(10.4 kN m3 ) + (7.6m 2)(8.3 kN m3 ) = 175kN / m 2 The consolidation settlement is, HCc σ ′ + ∆σ (7.6m)(0.32) 175 + 120 ∆H = log10 = log10 = 0.26 m = 260 mm σ o′ 1 + eo 1 + 1.11 175 198
*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
r) a) Draw an e-log p plot. s) b) Determine the pre-consolidation pressure, pc . t) c) Find the compression index, Cc.
Solution:
b) Determine the pre-consolidation pressure, pc. From the e-log p plot, 199
p2 = 500 kN m 2 and e2 = 0.8
p1 = 300 kN m 2 and e1 = 0.9 ∴ pc = 117.5 kN / m 2
c) Find the compression index, Cc. From the slope of the graph,
Cc =
e1 − e2 0.9 − 0.8 = = 0.451 p2 500 log log 300 p1
200
*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 midheight 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.
Elev. 760 ft. Sand and Gravel Unit Weight = 132 lb/ft3
Elev. 752 ft. Ground Water Table Sand and Gravel Unit Weight = 132 lb/ft3
Elev. 732 ft.
Clay Unit Weight = 125.4 lb/ft
3
Elev. 721 ft. Elev. 710 ft.
Solution:
a ) po = [γ h ]dry sand − gravel + [γ ' h ]saturated sand − gravel + [γ ' h ]mid −clay stratum po = (132 lb ft 3 )(8 ft ) + (132 − 62.4) lb ft 3 (20 ft ) + (125.4 − 62.4) lb ft 3 (11 ft ) po = 3.14 ksf b) Cc = 0.427 =
201
e1 − e2 0.98 − 0.81 = = 0.427 log( p2 p1 ) log(500 200)
0.98 − e2 ∴ e2 = 0.68 log(1000 200)
*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: po = ( 0.110 pcf )( 7 ft ) + ( 0.110 pcf )( 6 ft ) +
( 0.110 − 0.0624 pcf )( 3 ft ) + ( 0.110 − 0.0624 pcf )( 3 ft ) po = 1.71ksf ∆q + ∆p due to the lowering WT H ∆q2 = ∆p − ∆q1 = 1.853 − 1.16 = 0.693 ∆p =
C H c log po + ∆q1 1+ e po 0 0.030(10) ∆H1 = log 3 + 1.16 = 0.032 ft 1 + 0.96 3 0.034(10) ∆H 2 = log 3 + 0.693 = 0.157 ft 1 + 0.96 3 versus 0.083ft the additional settlement.
∆H1 =
The rising WT may reduce settlement.
202
*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: 1200 kN
1.5m 2m 4m
3m
WT
4mx4m
Sand
γb = 9.1 kN/m3
Clay
γb = 9.2 kN/m3
Sand and Gravel
γb = 9.0 kN/m3
a) In order to find the in-situ stress po before the footing was built, we need to find the unit weight of the sand stratum,
γ bGs
(9.1kN / m3 )(2.65) γd = = = 14.6 kN / m3 Gs −1 1.65 The stress po is found at mid-clay stratum, po = [ h(γ d )]dry(sand ) + [ h(γ ′)]sand + [ h(γ ′)]clay
po = (2m)(14.6kN / m3 ) + (4m)(9.1kN / m3 ) + (1.5m)(9.2kN / m3 ) = 79.4 kPa
203
**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,
3H dr Coefficient of consolidat ion(C v ) = 4t
2
Time (minutes)
0
0.32
0.64
1.28
2.40
4.80
9.60
16.0
Settlement
0
0.16
0.23
0.33
0.45
0.65
0.86
0.96
0
0.57
0.8
1.13
1.55
2.19
3.10
4.0
(mm.) time = (min)
a) b) c) d)
( 12 )
Plot a graph of settlement against the square root of time. Determine the value of the coefficient of consolidation cv (in m2/s). Estimate what could be a good estimate of the elastic modulus of this soil Eo (kPa) What sort of permeability k (in m/s) could you estimate for this soil (from Eo)?
Solution: (1)
Time vs. Settlement Time (min)^(1/2) 0
1
2
3
4
5
0 S ettlem en t (m m .)
0.2 0.4 0.6 0.8
3.40
1 1.2
(2)
204
vt x = 3.40 min1/ 2
and
time(t x ) = 11.56 min
2 3d 2 3(10mm) 2 (1min)(1m) 2 −7 m cv = = = 1.08 ×10 4t x 4(11.56 min)(60sec)(1000mm) 2 sec
(3)
Voids − ratio (ev ) =
1 mm = 0.05 ∴ ev = 5% 20 mm
σv
10 × 103 N Eo = = = 200 × 103 Pa = 200 kPa 2 ev 0.05 m (4)
e≈
2.30σ z ' k 1 + e σ z k k Cc ∴ cv = ∴ cv = = Eo 1 + eo γw Cc ez γ w γw
(
)(
)
2 1.08 × 10−7 m 9.81 kN 3 cvγ w s m = 5.3 × 10−9 m k= = Eo sec 200 kN 2 m
205
(
)
*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:
The primary consolidation ∆H p is, po' + ∆p ' ( 0.28 )( 8.5 ft )(12in / ft ) Cc H 2, 650 + 970 ∆H p = log log = = 2.15 inches ' 1 + eo p 1 + 0.8 2, 650 ( ) o The secondary consolidation ∆H s is, t Cα H log 2 1 + ep t1 We must find e p first, by finding the change in the voids ratio during primary consolidation,
∆H s =
po' + ∆p ' 2, 650 + 970 e p = eo − ∆e = eo − Cc log = 0.8 − ( 0.28 ) log = 0.76 ' p 2, 650 o t ( 0.02 )( 8.5 − 0.18 )(12in / ft ) C H 5 ∴ ∆H s = α log 2 = log = 0.59 inches 1 + ep t 1 + 0.76 1.5 ( ) 1 The total consolidation settlement is thus =∆H p + ∆H s = 2.15 + 0.59 = 2.74 inches
206
*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. Tv H dr2 (0.848)(150 cm) 2 = = 0.00294 cm2/sec t (75 days × 24 × 60 × 60)
Percent consolidation
cv =
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 Time factor , Tv
207
1
1.1 1.2
*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.
Building’s load
q = 4 ksf
20’
WT SAND CLAY
25’
SAND
Solution: The coefficient of consolidation cv =
cv = ∴
TvH dr2 is the same for the lab and field samples. t
Tv H dr 2 Tv H dr 2 Tv H dr 2 = = t t laboratory t field
( 0.15) (2 in) ( 78 hours )
2
=
(0.40)(12.5 ft × 12 in / ft ) 2 t field
2 1day 1 year 0.40 (12.5 ft × 12in / ft ) ∴ t field = ( 78 hours ) (2in) 2 24hours 365days 0.15 ∴ t field = 134 years to attain 70% consolidation
208
*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,
T90 =
cv t90( field ) H dr2 ( field )
∴ t90( lab ) =
209
c (75days × 24 × 60 × 60) cvt90 = v = 3, 000mm 2 30mm 2 ( ( ) ) 2 field 2 laboratory
(75days × 24 × 60 × 60) (15mm ) (1,500mm) 2
2
= 648 seconds =10 minutes
*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. a) 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? b) Find the time required for the clay layer in the field as described in part (a) above, to reach a 65% consolidation.
Solution: (1)
tlab H dr2 ( lab )
=
∴ t( field ) =
t( field ) H dr2 ( field ) (3.2m) 2 (8.5 min) 25mm 2 ×1000mm
2
= 557, 000 min = 387 days
(2) 2
cv =
Tv H dr2 t50
∴ t( field 65)
25 (0.197) 2 ×1000 = 0.36 × 10−5 = (8.5 min)
T65 H dr2 (0.34)(3.2m) 2 = = = 961, 400 min = 668 days cv 0.36 × 10−5 m 2 / min
210
*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: (d) The degree of consolidation when the settlement reaches 2 inches. (e) The time to reach 50% settlement if cv is 0.002 cm2/sec. (f) The time for 50% consolidation if the clay stratum is doubly-drained?
Solution: 2 (a ) U % = 100 = 22.5% 8.9 (b) T50 = 0.197 and
t=
Tv cv
(0.197) ( 20 ft )( 30.48cm / ft ) = = 424 days 1hr days 2 0.002 / sec cm ( ) 360 sec 24h 2
H
2 dr
(0.197) (10 ft )( 30.48cm / ft ) = 106 days 1hr days 2 0.002 cm / se c ( ) 360 sec 24h 2
Percent consolidation
T (c) t = v H dr2 = cv
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
0.8
0.9
1
1.1
1.2
Percent consolidation
Time factor , Tv
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
Time factor , Tv
211
**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 ∆p: ∆p = hγ oil = (40 ft )(60 pcf ) = 2.4 ksf Using the Boussinesq pressure diagram (next two pages) for B = 75' , provides the stress levels at any point in the soil mass, thus, The stress at point A ⇒ 0.91∆p = 0.90 ( 2.4 ) = 2.2 ksf point B ⇒ 0.43∆p = 0.43(2.4) = 1.0 ksf The in-situ effective stress at point A was po, before the tank was built:
po = (10 ')( 0.120 − 0.062 ) + ( 28.5')( 0.110 − 0.062 ) = 1.95 ksf The settlements at point A (below the center of the tank), and point B (at the edge of the tank) are,
212
p '+ ∆p 0.4 ( 57 ' ) Cc H 1.95 + 2.2 log o log = = 3.34 ft 1 + eo 1.95 po ' (1 + 1.27 ) p '+ ∆p 0.4 ( 57 ' ) CH 1.95 + 1.0 ∆H B = c log o log = = 1.85 ft 1 + eo p ' 1 + 1.27 1.95 ( ) o ∆H A =
∴ The differential settlement between A and B is = 3.34 ft − 1.85 ft = 1.49 ft = 18inches 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
Tv field H 2 field t75
=
Tvlab H 2 field tlab
ft ( 0.2 ) 0.5in 2 0.480 )( 28.5 ft ) ( 12in ∴ cv = = 6.5 min t 75
∴ t75 = 13.8 years for 75% consolidation. 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.
σ total = (10 ')( 0.120 ) + x ( 0.110 ) = 2.4 ksf Solving for x:
∴ x = 10.9 ft ≅ 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
∴ Total excavation depth = 11 ft + 10 ft = 21 feet
213
**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.
h ft
Sand Surcharge (Area = 200’ x 200’)
WT
15 ft 30 ft
20 ft
Sand
Clay Impermeable rock
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, Q 136, 000kips = = 3.4 ksf A 40, 000 ft 2 lb but , qsur = γ sur h = 115 3 (h) = 3.4 ksf ∴ h = 30 feet high ft Both the sand and clay strata contribute to the settlement. However, the settlement from the clay stratum is a consolidation settlement, such that, qsur =
214
Using Skempton's formula Cc = 0.009( LL -10) = 0.009(28 -10) = 0.162 Also eo = 0.95 and
H clay stratum = 20 ft = 240 in.
The initial stress at mid-clay stratum is, p0 = γ s h + γ sb h = γ cb h = (0.125kcf )(15 ft ) + (0.125 − 0.0624)(15) + (0.130 − 0.0624)(10) = 3.5 ksf The increased load from the surcharge ∆p is, ∆p = γ h = (0.115kcf )(30 ft ) = 3.45 ksf The settlement created by the surcharge is, ∆H c =
p + ∆p (240in)(0.162) HCc 3.5 + 3.45 log10 0 log10 = = 5.94 inches (1 + e0 ) (1 + 0.95) 3.5 p0
The settlement from the sand stratum is an elastic settlement, H sand = 30 ft = 360in ES ≈ 14 N = (14 )(15 ) = 210k / ft 2 ; depth factor C1 = 1 and
creep factor C2 = 1
0.30 ε ∆ = C1C2 qo dz = (1)(1)(3.4ksf )( )(360 in) = 1.75in 210 ksf E The total settlement = 5.94in + 1.75in = 7.69in b) The lead time t required for the surcharge to accomplish its task is, 2 T H dr2 (0.30) 20 ft 12 in /1 ft ( )( ) 1min 1hr 1day t= v = = 200 days for 60% consolidation −3 2 c 60sec 60 min 24hr 10 in / sec ) ( v The total settlement that has taken place at 200 days (60% consolidation) is, Total Time = ∆H S + (60%) ∆H C = 1.73in + (0.6)(5.94in) = 5.3 inches
215
**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,
cv =
k (1 + eo ) 2 ft /day av γ w
where the permeability k = 1.0 × 10 −5 cms / s , and a v = 2.9 × 10 −4 ft 2 / lb .
Percent consolidation
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
216
Solution.
cc = 0.75 ( eo − 0.30 ) = 0.75 ( 0.59 − 0.30 ) = 0.218 At midpoint of the marl, the in-situ stress σ o is,
σ o = γ h = 120
lb ( 20 ft ) = 2, 400 psf ft 3
The increase in stress due to the surcharge ∆σ is,
lb ∆σ = γ∆h = 125 3 ( 20 ft ) = 2,500 psf ft Therefore, the consolidation settlement of the marl is, σ + ∆σ ( 40 ft )(12in / ft )( 0.218 ) HCc 2.4 + 2.5 log10 o log10 = = 20.4inches 1 + eo (1 + 0.59 ) 2.4 σo The time required for 50% of consolidation to take place is found through the coefficient of consolidation cv ,
∴∆ =
1.0 ×10−5 cm / sec ) (1 + 0.59 ) 1in 1 ft 86.4 ×103 sec ( 1 + eo 2 cv = k = = 2.5 ft / day −4 2 3 day avγ w ( 2.9 ×10 ft / lb )( 62.4lb / ft ) 2.54cm 12in The time factor for 50% consolidation is Tv = 0.2, therefore the time t required is,
t=
217
Tv H dr2 cv
( 0.2 )( 40 ft ) =
2
2.5 ft 2 / day
= 128 days
**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.
(a) Since the soil is the same clay in the laboratory and the field, and both are 70% consolidation, t field H 2field Tv H dr 2 Tv H dr 2 Tv H dr 2 cv = = = ∴ = 2 t tlab H lab t laboratory t field 2
or t field
t H 2 700cm 6 = lab 2F = ( 6.67 min ) = 1.45 × 10 min = 2.76 years HL 1.5cm
(b) The amount of settlement that takes place at 70% consolidation is, C H p '+ ∆p ' ∆H 70% = ( 0.70 ) c log10 o po ' 1 + eo The in-situ stress at mid-clay stratum before the surcharge was applied was, ∴ po ' = ∑ γ i hi = (18kN / m3 ) ( 2m ) + ( 20 − 9.81) kN / m3 ( 3.5m ) = 71.7kN / m 2 i
and ∆p ' = 72 kN / m 2 ∴ 70% ∆H =
p '+ ∆p ' 0.7 ( 0.20 )( 700cm ) 0.7Cc H 71.7 + 72 log10 o log10 = = 14.8 cms 1 + eo (1 + 1) 71.7 po '
218
***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 cv as,
3H dr2 cv = 4t 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. m2 ). s (c) Estimate what could be a good estimate of the elastic modulus of this soil, E o (in kPa). m (d) What sort of permeability k ( ) could you estimate for this soil (from E o )? s
(b) Determine the value of cv (
Solution: e) Settlement vs. Time^1/2 Time^1/2(min)^1/2
Settlement Delta H (mm)
0.00 0.00
2.00
3.00
4.00
0.20 0.40 y = 0.2503x + 0.0376 0.60 0.80 1.00 1.20
219
1.00
0.96,3.69
5.00
f)
t x = 3.69 min 1 / 2 → t x =13.62 min
2 3d 2 3(10 min) 2 (1 min)(1m) 2 −8 m cv = = = 9.178 * 10 4t x sec 4(13.62 min)(60 sec)(1000mm) 2 g) The strain at the end of the loading is: ∆ (1mm) εv = = = 0.05 or 5% L (20mm)
Eo =
σ v 10 * 10 3 N = = 200kPa Vertical Modulus εv 0.05m 2
From Coduto (pages 390 and 391) ε v ≈
Cc but Coduto (page 424) 1 + eo
2.30σ z k 1 + e0 σ k or cv = z γ w Cc εz γw *note: (2.30 is for ln; use 1.0 for log)
k = E o γw
cv =
k=
cv γ w Eo
m 2 kN 9.178 *10 −8 9 . 81 3 sec m m = = 4.50 *10 −9 kN s 200 2 m
220
**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:
Voids Ratio
Vertical Effective Stress
0.736
15
0.733
28
0.73
60
0.675
230
0.638
480
0.6
930
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
221
**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
222
Diagram 1 PLAN VIEW A
PLAN VIEW B
BUILDING A
BUILDING B
50 Stories = 550 ft
50 Stories = 550 ft
Completed 1975
Completed 1975
150,000 kips
150,000 kips
A
B
D
C
F
Fill Clay Sand
223
E
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:
(σ z )induced = q=
[(
(
q 2BLz f B 2 + L2 + z 2f
( (
)
)
1/ 2
)] × B
4 Π z 2f B 2 + L2 + z 2f + B 2 L2
)
2
+ L2 + 2 z 2f
B2 + L2 + z 2f
−1
+ Π − sin −
(
2 BLz f B2 + L2 + 2 z 2f
(
)
)
1/ 2
z 2f B 2 + L2 + z 2f + B 2 L2
p 150,000 × 10 3 = = 15,000 lbs A (100 )(100 )
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_ H2dr For 21 years: t = 662,256,000 sec Tv A =
(
)
4 10 -4 in 2 / sec (662,256,000 sec ) = 165.56 (40)2
Degree of Consolidation U = [1-10 –(0.085+Tv)/0.933] x 100% The degree of consolidation at TvA = 165.56 is: 224
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,
U=
δc (δ c )ult
δ c = (δ c )ult Assuming normally consolidated soils
δc = ∑
σz Cc H log f σz 1 + e0 0 1 + eo
(Eqn 1-1)
σ’zo
Cc = 0.009 (LL – 10) LL = PL + PL = 52 + 12 = 64 % Therefore, Cc = 0.009 (64 – 10) = 0.486 e0 =
γd =
Gsγ w
γd γ
1+ w
−1
=
121 (2.68 )(62.4 ) - 1 = 0.82 = 91.67 ∴ e0 = 1 + 0.32 91.67
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):
δc =
0.486 (40 ft ) log 14,533 = 8.46 ft = 101.55 in 1 + 0.82 2,344
At corner D: H= 40 ft 225
(σ’zf )D = 7584 lb/ ft2
σ’zo = 2,344 lb/ ft2
Replacing values of H, (σ’zf )D , σ’zo, Cc and eo into (Eqn 1-1):
δc =
0.486 (40 ft ) log 7,584 = 5,45 ft = 65.35 in 1 + 0.82 2,344
The ∆δc = 8.46 -5.45 = 3.01 ft
A
Diagram 2
DRIFT = x
550 ft
A '
?
D
C
5.45 ft 8.46 ft
D' ?
3.01 ft C'
100 ft
∆ = tan -1 (3.01/ 100) =1.72º Drift = x = 550 tan 1.72 = 16.52 ft = 200 inches.
226
**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: kN kN = 5 8 .9 3 m m2 kN kN 7 m x 9 .8 0 = 6 8 .6 3 m m2 kN kN 2 m x 9 .6 0 = 1 9 .2 3 m m2
3 m x 1 9 .6 2
for a total
ρ o = 1 4 6 .7
kN m2
Determine eo : From chart pc ~ 150 < po normally consolidated, eo = 0.81. Determine ∆σ or ∆p :
227
5m = 0.5 00 ∴ I 4 = 0.07 1 10m 28 x10 3 kN Q kN ∆σ = 4 I4 = 4 ( 0.07 1 ) = 9 9.5 2 A m 8 m x 10 m
m =
4m = 0.28 6 10 m
a nd
n=
Q
Determine the settlement ∆Η :
C p + ∆p 0.31 146.7 + 99.5 ∆Η = Η c log10 o = 4m log10 = 0.15 m 1 + 1 + 0.81 1 4 6 .7 e p o o
228
Chapter 12 Shear Strength of Soils Symbols for Shear Strength of Soils c→ The cohesion of a soil particle. cu→ p→ p’→ q→ q’→ qu→ Ultimate shear strength of a soil. u → Pore water pressure. u c→ 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.
229
*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 kg chamber pressure σ 3 = 2.0 2 . At failure, cm kg kg σ 1 − σ 3 = 2.8 2 , u = 1.8 2 and the failure plane angle θ = 57° . cm cm 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:
σ 1 = 2.8 + 2.0 = 4.8 σ 3 = 2.0
kg cm2
kg cm 2 σ1 − σ 3
Shear stress
τ=
Normal stress
σ =
Maximum shear
2
kg 4.8 − 2 sin 2θ = sin114° = 1.27 cm 2 2
kg σ1 + σ 3 σ1 − σ 3 4.8 + 2 4.8 − 2 + cos114° = 2.83 + cos 2θ = cm 2 2 2 2 2
τ MAX =
σ1 − σ 3 2
kg 4.8 − 2 sin 2θ = = 1.4 cm 2 2
at
θ = 45°
230
*Shear strength–02: Why is the maximum shear not the failure shear? (Revised Oct-09)
Using the results of the previous Problem 01, and φ = 24°, c ' = 0.80
kg , show why the sample cm 2
failed at 57 grades instead of the plane of maximum shear stress.
Solution: On failure plane
σ ' = (σ − u ) = 2.83 − 1.8 = 1.03
kg cm 2
S57° = c '+ σ ' tan φ = 0.8 + (1.03 tan 24°) = 1.27
kg cm 2
Compare that to τ = 1.27 kg/cm from the previous problem, and note that they are equal, and so for both, S 57 ° = τ 57 ° failed. . Now at the plane of maximum shear stress θ = 4 5 ° kg 4.8 + 2 4.8 − 2 + cos 90° = 3.4 2 cm 2 2 kg σ ' = 3.4 − 1.8 = 1.6 2 cm
σ =
s45° = c '+ σ ' tan φ = 0.8 + (1.60 tan 24°) = 1.51
kg cm 2
The shear strength at 45º is much larger than at 57º, therefore failure does not occur.
231
*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 kg slowly to failure in a drained test, that is u = 0, with σ 3 = 2.0 2 . What will be the major principal cm stress at failure?
Solution: a) Analytically, in a drained test u = 0; at failure σ '3 = σ 3 = 2
kg , on the failure plane θ = 57° . cm 2
σ ' + 2 σ '1 − 2 s = c '+ σ ' tan φ = 0.80 + 1 + cos114° tan 24° = (1.426 ) + ( 0.132 ) σ '1 2 2 σ ' −2 and τ = 1 sin114° = ( 0.457 ) σ '1 − ( 0.914 ) 2 kg kg kg At failure s = τ ∴ σ '1 = 7.31 2 ; σ θ = 3.6 2 and τ θ = 2.38 2 cm cm cm b) Graphically,
232
*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:
θ = 45° +
φ
∴
2
φ
58° = 45° + ∴ 2
φ = 26°
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),
σ 1 ' = σ 3 ' tan 2 45° +
233
φ1
2 26 kN = (103.5 ) tan 45° + = 265 2 2 2 m
*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)
σ1 = 148 psi
for σ3 =
30 psi
p = 89.0,
q = 59.0 psi
(2)
σ1 = 312 psi
for σ3 =
60 psi
p = 186.0,
q = 126.0 psi
(3)
σ1 = 605 psi
for σ3 = 120 psi
p = 362.5,
q = 242.5 psi
(4)
Solution: Remember that p = (σ1 + σ3)/2 and q = (σ1 – σ3)/2
q (psi) 300
4
α = 34°
200
3 100
2 1
p(psi) 100
200
300
400
From the p-q diagram:
ta n α ≈
q4 2 4 2 .5 = = 0 .6 6 8 p4 3 6 2 .5
s i n φ = t a n α = 0 .6 6 8
∴
φ = 42°
234
**Shear strength–06: Consolidated-drained triaxial test. (Revised Oct-09)
A consolidated-drained triaxial test was conducted on a normally consolidated clay. The results kN kN σ '3 = 276 2 are as follows: ( ∆ σ d ) f = 276 2 m m 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:
τ
f
= σ ' ta n φ
because c = 0
For the triaxial test, the effective major and minor principal stresses at failure are as follows:
σ '1 = σ 1 = σ 3 + ( ∆ σ d ) f = 276 + 276 = 552
kN m2
and
σ '3 = σ 3 = 276
kN m2
Part A. The Mohr circle and the failure envelope are shown in the figure below, from which:
σ '1 − σ '3 σ '1 − σ '3 AB 552 − 276 2 sin φ = = = = = 0 .3 3 3 σ '1 + σ '3 σ '1 + σ '3 5 5 2 + 2 7 6 OA 2
235
∴
φ = 1 9 .4 5 °
Part B.
θ = 45° +
φ 2
= 45° +
1 9 .4 5 ° = 5 4 .7 ° 2
Part C.
σ'
( on the failure plane )
and τ
f
=
=
σ '1 + σ '3
σ '1 − σ '3 2
2
+
σ '1 − σ '3 2
cos 2θ
sin 2θ
Substituting the values of σ '1 = 552
kN , m2
σ '3 = 276
kN m2
and θ = 5 4 .7 ° in the preceding
equations,
σ' = and
552 + 276 552 − 276 kN + cos 2 ( 54.7 ) = 36 8 2 m 2 2 kN 552 − 276 τf = sin 2 ( 54 .7 ) = 130 2 2 m
236
**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. kN The tests were taken with pore water pressure measurements, and c ' = 20 2 , and φ = 24 ° m
(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:
kN m3 kN kN kN kN ∴ For σ 1 = 4m 16 3 + 9m 19 3 = 3.5 17.6 3 = 297 2 m m m m kN kN ∴ For σ '1 = σ 1 − u = σ 1 − γ w (9 + 3.5) = 297 − 10 3 (12.5m ) = 173 2 m m kN kN kN ∴ S = c '+ σ '1 tan φ ' = 20 2 + 173 3 tan 24° = 96.6 2 m m m
γ sat = γ d + nγ w = (16 + 0.3x10 ) = 19
237
b) Since the clay is saturated,
σ '3 = K oσ '1 = 0.5σ '1 = 0.5 (173) = 86 and σ 3 = σ '3 + u = 86
kN m2
kN kN kN + (13m + 3.5m )10 3 = 251 2 2 m m m
238
**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:
σ 3 = 100
( ∆σ d ) f
Specimen 2:
kN m2
= 410.6
σ 3 = 50
kN m2
(∆σ d ) f
kN m2
= 384.37
kN m2
Find the values of the cohesion c and the angle of internal friction φ.
Solution: For Specimen 1, the principal stresses at failure are,
σ 3 ' = σ 3 = 100
kN m2
and σ 1 ' = σ 1 = σ 3 + ( ∆σ d ) f = 100 + 410.6 = 510.6
Similarly, the principal stresses at failure for specimen 2 are
σ 3 ' = σ 3 = 50 239
kN m2
and σ 1 ' = σ 1 = σ 3 + ( ∆σ d ) f = 50 + 384.4 = 434.4
kN m2
kN m2
These two samples are over-consolidated. Using the relationship given by equation
σ 1 ' = σ 3 ' tan 2 45° +
φ1
φ1 + 2c tan 45° + 2 2
Thus, for specimen 1
( 510.6 ) = (100 ) tan 2 45° +
φ1
φ1 + 2c tan 45° + 2 2
and for specimen 2
( 434.4 ) = ( 50 ) tan 2 45° +
φ1
φ1 + 2c tan 45° + 2 2
( 76.2 ) = ( 50 ) ta n 2 45 ° +
Subtracting both equations
φ1
∴ φ1 = 1 2 ° 2
Substituting φ 1 into the equation,
12 12 + 2c tan 45° + 2 2
( 510.6 ) = (100 ) tan 2 45° +
510.6 = 152.5 + 2.47c ∴ c = 145
kN m2
240
***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
σ’ XX
σ’YY
τxy
τyx
x
y
(kpa)
(kpa)
(kpa)
(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 (σ ' xx , τ xy ) and
(σ '
yy
, τ yx ) .
TABLE OF VALUES STAGE
S'=
σ 1 '+ σ 3 ' 2
τ=
σ 1 '− σ 3 ' 2
τ
Change in angle between
S'
x + y plan (in degrees)
kPa
kPa
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˚
241
242
243
**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 φ ' = 30° 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.
Potential shearing surface C▼ B▼
●A
Solution: lb lb u = γ z z w = 62.4 3 (20 ft ) = 1248 2 ft ft When the groundwater table is at B: lb lb lb lb σ ' z = ∑ γH − u =120 3 (26 ft ) + 123 3 (20 ft ) − 1248 2 = 4332 2 ft ft ft ft
The potential shear surface is horizontal, so σ ' = σ ' z lb lb s = σ ' tan φ ' = 4332 2 tan 30° = 2501 2 f t f t
When the groundwater table is at C: lb lb u = γ z z w = 62.4 3 (32 ft ) = 1997 2 ft ft lb lb lb lb σ ' z = ∑ γH − u =120 3 (14 ft ) + 123 3 (32 ft ) − 1997 2 = 3619 2 ft ft ft ft
lb lb s = σ ' tan φ ' = 3619 2 tan 30° = 2089 2 ft ft
244
***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:
σ ' = ∑ γH − u = (17.0
kN kN kN )(3.0m) + (17.5 3 )(1.1m) − (9.8 3 )(1.1m) 3 m m m
σ ' z = 59.8 kPa s = c' + σ ' tan φ ' = 10kPa + (59.5kPa ) tan 28° = 41.6 kPa Point A - Vertical plane:
σ ' z = Kσ ' z = (0.54)(59.5 kPa) = 32.1 kPa s = c' + σ ' tan φ ' = 10 kPa + (32.1 kPa ) tan 28° = 27.1 kPa
245
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
246
***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:
σ ' = ∑ γH − u = (17.0
kN kN kN )( 3 . 0 m ) + ( 17 . 5 )( 1 . 1 m ) − ( 9 . 8 )(1.1m) m3 m3 m3
σ ' z = 59.8 kPa s = c' + σ ' tan φ ' = 10kPa + (59.5kPa ) tan 28° = 41.6 kPa Point A - Vertical plane:
σ ' z = Kσ ' z = (0.54)(59.5 kPa) = 32.1 kPa s = c' + σ ' tan φ ' = 10 kPa + (32.1 kPa ) tan 28° = 27.1 kPa 247
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
248
**Shear strength-12: Triaxial un-drained tests. (Revised Oct-09)
Triaxial un-drained tests were performed on clay samples taken from the stratum shown below. The test were taken with pure water pressure measurements and yield a c` = 20 kN/m3, 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. H=4
sandy gravel n= 0.30
H= 13m
γ =16 kN/m3
σ1 clay
σ3
H= 7m
γ =17.6 kN/m3
Impermeable rock stratum
Solution: For the gravel: γsat = γsat + nγw = [16 + (0.3)10] = 19 kN/m3 For the clay:
σ1 = (4m)(16 kN/m3 )+ (9m)(19 kN/m3 )+ (3.5m)(17.6 kN/m3 )= 297 kN/m3
σ1’ = σ1 − u = σ1 − γw (9 + 3.5) = 297 - 10 kN/m3 [12.5m] = 172 kN/m3
S = c’ + σ1’ tan φ = (20 KN/m3 + 172 kN/m3 tan 24ο) = 96.6 NM/m3 Since the clay is saturated,
σ3’ = k σ1’ = 0.5(172) = 86 kN/m3
σ3 = σ3’ + u = 86 kN/m3 + (86m + 86m) 10 kN/m3 = 251 kN/m3 τ φ = 24ο.
S= 96.6 kN/m3 C` = 20
KN
/m
3
σ`
249
**Shear strength-12b: Triaxial un-drained tests. (Revised Oct-09)
Triaxial un-drained tests were performed on clay samples taken from the stratum shown below. The test were taken with pure water pressure measurements and yield a c` = 20 kN/m3, 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. H=4
Sandy gravel
H=
n= 0.30
σ1 σ3 σ3
H= 7m
γ =16 kN/m3 Clay
γ =17.6 kN/m3
Solution: For the gravel: γsat = γsat + nγw = [16 + (0.3)10] = 19 kN/m3 For the clay:
σ1 = (4m)(16 kN/m3 )+ (9m)(19 kN/m3 )+ (3.5m)(17.6 kN/m3 )= 297 kN/m3
σ1’ = σ1 − u = σ1 − γw (9 + 3.5) = 297 - 10 kN/m3 [12.5m] = 172 kN/m3
S = c’ + σ1’ tan φ = (20 KN/m3 + 172 kN/m3 tan 24ο) = 96.6 NM/m3 Since the clay is saturated,
σ3’ = k σ1’ = 0.5(172) = 86 kN/m3
σ3 = σ3’ + u = 86 kN/m3 + (86m + 86m) 10 kN/m3 = 251 kN/m3 τ φ = 24ο.
S= 96.6 kN/m3 C` = 20 KN/m3
σ`
250
**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).
d=13'
H1=40'
H2=20'
•A
γd = 103 pcf n = 30% c = 2.9 psi γsat = 112 pcf φ' = 24o
Solution: a) In order to find s, it is required to know the γsat of the sand. e=
n 0.30 0.30 γ (1+ e) 103(1+0.429) = = = 0.429 and G S = d = = 2.36 1- n 1 - 0.30 0.70 62.4 γW
γ sat =
( G S + e)γ W (2.36 +0.429)62.4 = = 122pcf 1+ e 1.429
*Assume that the clay was normally consolidated to find σ' at midlevel in the clay (point A), that is c = 0.
251
σ’ A = γ h S + γ’ h S + γ’ h C = ( 0.103)(13) + ( 0.122 − 0.0624 )( 27 ) + ( 0.122 − 0.0624 )(10 ) = 3.4ksf and s = c + σ’ A tan φ’ kip 144in 2 k x + 3.44 2 ( tan 24 o )= 0.42ksf +1.53ksfs = 1.95ksf 2 3 ft ft in 10 lb The pore water pressure u is, s = 2.9
lb
x 2
u = (10 + 27 )( 0.0624 ) = 2.3ksf Therefore, the toatl stress is,
σ A = σ A' + u = 3.4 + 2.3 = 5.7ksf 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 φ θ = 45 o + = 45 o + 24 = 57 o_2θ = 114 o
2
2
x1 =
c 0.42 = = 0.939ksf tan φ’ tan 24 o
x2 =
s 1.95 = = 4.38ksf tan φ’ tan 24 o
o o x 3 = s( tan 24 )= 1.95( tan 24 )= 0.868ksf
252
R=
s 1.95 = = 2.13ksf cos φ’ cos 24 o
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
253
**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
D=13’
H=40’
Sand
γd= 103 pct n= 30%
•
γsat= 103 pct
A
H= 20’
Clay
c = 2.9 psi
φ' = 24o Rock
(point A).
Solution: In order to find s, it is required to know the γsat of the sand. e=
n 0.30 0.30 γ (1+ e) 103(1+0.429) = = = 0.429 and G S = d = = 2.36 1- n 1 - 0.30 0.70 62.4 γW
γ sat =
( G S + e)γ W (2.36 +0.429)62.4 = = 122pcf 1+ e 1.429
* Assume that the clay was normally consolidated to find σ' at midlevel in the clay (point A), that is c = 0.
254
σ’ A = γ h S + γ’ h S + γ’ h C = ( 0.103)(13) + ( 0.122 − 0.0624 )( 27 ) + ( 0.122 − 0.0624 )(10 ) = 3.4ksf and s = c + σ’ A tan φ’ kip 144in 2 k x + 3.44 2 ( tan 24 o )= 0.42ksf +1.53ksfs = 1.95ksf 2 3 ft ft in 10 lb The pore water pressure u is, s = 2.9
lb
x 2
u = (10 + 27 )( 0.0624 ) = 2.3ksf Therefore, the toatl stress is,
σ A = σ A' + u = 3.4 + 2.3 = 5.7ksf To find the stress on the vertical face of the soil element at A, we find θ through a graphical solution as follows,
o φ θ = 45 o + = 45 o + 24 = 57 o_2θ = 114 o
2
x1 =
c 0.42 = = 0.939ksf tan φ’ tan 24 o
x2 =
s 1.95 = = 4.38ksf tan φ’ tan 24 o
o o x 3 = s( tan 24 )= 1.95( tan 24 )= 0.868ksf
255
2
R=
s 1.95 = = 2.13ksf cos φ’ cos 24 o
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
256
**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: h
θ = 45° +
d
ad =
φ 2
→ 2 θ = 90 + φ
σ1 − σ 3 2
φ
ф
2 2θ
f c τ3
o
τ1
a
σ3
σ1
45°
From the figure, sin φ =
ad fa
(1)
fa = fo + oa = (c) cot φ +
tan φ = c fo
→
σ1 + σ 3
fo = cot φ c
2
→
fo = (c)(cot φ)
Using the properties of the Mohr circle,
oa = σ3 +
(σ 1 − σ 3 ) 2
→
oa =
(σ 1 + σ 3 ) 2
→
fa = (c) cot φ +
(σ 1 − σ 3 ) 2 Introducing (1) into (2): sin φ = (σ + σ 3 ) (c) cot φ + 1 2 257
→
(σ 1 + σ 3 ) 2
(sin φ)[(c) cot φ +
(2)
σ1 + σ 3 2
]
=
σ1 − σ 3 2
→[(c) sin φ cot φ] + [ σ
1
→ (2) [(c) sin φ cot φ]
= [σ1 – (sin φ)( σ 1)] – [σ3 + (sin φ)( σ 3)]
→ (2) [(c) sin φ
cos φ sin φ
+σ3 σ −σ3 sin φ] = 1 2 2
]=
→
(c) sin φ cot φ =
[σ
1
−σ3 2
] - [σ
1
+σ3 sin φ] 2
σ1 [1 – (sin φ)] – σ3[1 + (sin φ)]
2(c) cos φ (1 + sin φ ) + (σ 3 ) = σ1 (1 − sin φ ) (1 − sin φ )
Since
cos φ φ ≈ tan (45° + ) 2 (1 − sin φ
σ 1 = σ 3 tan 2 45° +
and
φ (1 + sin φ ) ≈ tan2 (45° + ) (1 − sin φ ) 2
ϕ
ϕ + 2 c tan 45 ° + 2 2
258
Chapter 13 Slope Stability Symbols for Slope Stability
259
*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:
T he traingule of rock above the potential slip plane has a weight W per unit width, kN kN 1 ( 85.0 m )(11.3 m ) 20.1 3 = 9, 650 2 m m T he length L of the slip plane is,
W =
85.0 m = 88.4 m cos16 ° T herefore, L=
FS =
Resisting For ces cL + (W ) cos α tan φ = = Driving Forces (W ) sin α
kN kN 22 2 ( 88.4 m ) + 9, 650 cos16 ° tan 30 ° m m FS = = 2.7 > 2 kN 9, 650 sin 1 6 ° m
OK
260
*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 W o f th e r o c k tr ia n g le p e r u n it w id th is s till 9 , 6 5 0
kN m
T h e l e n g t h L o f t h e s li p p la n e i s s t i l l 8 8 . 4 m . T h e p o r e w a t e r p r e s s u r e i s b a s e d o n a n e s t i m a t e o f i t s v a lu e a l o n g t h e l e n g t h L , a t w a t e r d e p t h z w a b o v e t h e p l a n e t h a t r a n g e f r o m 0 t o 3 . 2 m ; c o n s e r v a t i v e ly , kN z w = 9 .8 1 (3 . 2 m ) = 3 1 . 4 k P a m3 c ' L + (W ) c o s α − u t a n φ R e s is tin g F o rc e s FS = = = D riv in g F o rc e s (W ) s i n α
u = γ
w
kN 9 , 6 5 0 m c o s 1 6 ° − ( 3 1 .4 k P a ) t a n 2 0 ° FS = = 1 .7 6 < 2 kN ° 9 , 6 5 0 s i n 1 6 m T h e c o m p u t e d f a c t o r o f s a f e t y o f 1 .7 6 i s l e s s t h a n t h e m i n i m u m a c c e p t a b l e v a lu e o f 2 , t h e r e f o r e t h i s d e s i g n i s N O T a c c e p t a b le . N o t i c e t h a t a r i s i n 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. kN 15 (8 8 . 4 m m2
261
)+
NG
*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.
20’
15’
40’
15’
15’
A Wcrane
45’
Clayey sand SEA c = 0.2 ksf
B
φ = 40° C 15’
Wsoil Sandy clay c = 1.8 ksf
φ = 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 W (d ) M 0 = Wb1 - WwH (d 1 ) - WWV (b2 ) + Vb3 + ae 2 g = 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
Not Good!
Removing Crane Mo =9,050 k-ft Therefore: FS = Mr/Mo = 19,500/9050 = 2.15
GOOD! 262
*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:
FS =
263
Slice
Shearing Component
Normal Component
Number
(W sin α) (lb/ft)
(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
cL + ∑ (W cos α ) tan nφ
∑W sin α
=
(225 psf )(40 ft ) + 20,166 plf = 1.86 < 2 NG 7, 751 plf
**Slope-05: Method of slices to find the factor of safety of a slope with a WT. (Revision: Oct.-08)
lb ft 2 and φ ' = 29° . 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.
A 30 ft tall, 1.5H:1V slope is to be built as shown below. The soil is homogeneous, with c ' = 400
Solution: Weights: W1 lb 1 0 .3 = 1 0 .8 119 = 6, 620 b 2 ft W2 1 0 .3 + 1 2 .5 = 9 .4 b 2 W3 b W4 b
lb 5 .2 1 1 9 + 9 .4 1 2 3 = 1 5, 8 0 0 ft 2 lb 1 2 .5 + 1 4 .6 5 .2 + 1 0 .0 = 1 2 .1 1 1 9 + 1 2 .1 1 2 3 = 3 0, 8 0 0 2 2 ft lb 5 .0 1 2 .9 + 8 .0 = 2 .9 1 7 .0 + 7 .1 1 7 .8 = 1 6 2 0 2 ft 2
W5 lb 1 6 .8 + 12 .8 10 .7 + 7.3 = 9.3 11 9 + 9.3 1 23 = 3 9, 90 0 b 2 2 ft W6 lb 12 .8 + 9 .9 7.3 = 7 .6 1 19 + 7.6 1 2 3 = 2 6, 70 0 b 2 ft 2 W7 9 .9 = 4 .0 b 2
lb 1 19 = 2, 4 00 ft
264
Average pore water pressure at base of each slice: u1 = 0 lb 5 .2 u2 = 6 2 .4 = 1 6 0 ft 2 2 lb 5 .2 + 1 0 .0 u3 = 6 2 .4 = 4 7 0 2 ft 2 lb 1 0 .0 + 1 0 .7 u4 = 6 2 .4 = 6 5 0 2 f t2 lb 1 0 .7 + 7 .3 u5 = 6 2 .4 = 5 6 0 2 ft 2 lb 7 .3 u6 = 6 2 .4 = 2 3 0 2 f t2 u7 = 0
α
φ
c 'l +
( Deg )
lb c ' 2 ft
( Deg )
lb u 2 ft
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
Σ = 84, 400
Σ = 46,100
Slice
W ( lb) b
1
l
( ft )
W ( cosα − ul ) tan φ ' b
Therefore, the factor of safety is, W c ' l + ( cos α − ul ) tan φ ' 84, 400 b FS = = = 1.83 < 2 NG 46,100 W b
Note how slices # 1 and 2 have a negative α because they are inclined backwards.
265
W ( lb ) b
**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:
266
Solution: Weights
W1 2 .0 = 4 .6 b 2
kN 1 7 .8 = 8 0 m W2 kN 2 .0 + 9 .8 = 7 .0 1 7 .8 = 1 3 0 b 2 m W3 kN 9 .8 + 1 2 .9 = 2 .9 1 7 .8 = 5 9 0 b m 2 :
W4 5 .0 1 2 .9 + 8 .0 = 2 .9 1 7 .0 + 7 .1 1 7 .8 = 1 6 2 0 b 2 2 W5 5 .0 + 1 0 .3 8 .0 = 7 .2 1 7 .0 + 7 .2 1 7 .8 = 1 4 5 0 b 2 2 W 6 kN 1 0 .3 + 9 .8 = 0 .8 1 7 .0 = 1 4 0 b 2 m W7 9 .8 = 5 .1 b 2
kN 1 7 .0 = 4 2 0 m
Moment arms:
d 1 = − 7 .0 − d2 = d3 = d4 = d5 = d6 = d7 =
267
4 .6 = − 8 .5 m 3
− 7 .0 = − 3 .5 m 2 2 .9 = 1 .5 m 2 7 .1 2 .9 + = 6 .5 m 2 7 .1 2 .9 + 7 .1 + = 1 0 .9 m 2 0 .8 = 1 7 .6 m 2 .9 + 7 .1 + 7 .2 + 2 5 .1 2 .9 + 7 .1 + 7 .2 + 0 .8 + = 1 9 .7 m 3
kN m kN m
Slice
Su ( kPa )
θ ( Deg )
Su θ
W b
kN m
W d b
1
80
-8.5
-690
2
130
-3.5
-450
390
1.5
890
4
1620
6.5
10,530
5
1450
10.9
15,800
140
17.6
2,460
420
19.7
8,280
3
80
76
6080
6 40
7
30
1200
Σ = 36,830
Σ = 7280
FS =
d ( m)
π R2 180
∑Sθ W ∑ b d u
=
π ( 23.6 ) 180
2
7, 280 = 1.9 2 < 2 36, 8 3 0
Not Goo d
268
Chapter 14 Statistical Analysis of Soils Symbols for the Statistical Analysis of Soils
269
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. GS→ Specific gravity of the solids of a soil. 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. VW → Volume of water. w→ water content.
270
Formulas and Figures for Lateral Stresses.
Figure for symbols used in the Coulomb earth pressures. Coulomb’s lateral pressure coefficients Ka and Kp.
Ka =
Kp =
271
cos2 (φ - θ) sin(δ + φ )sin(φ − α ) cos2θ cos(δ + θ ) 1 + cos( + )cos( − ) δ θ θ α cos2 (φ + θ)
2
sin(φ − δ )sin(φ + α ) cos2θ cos(δ − θ ) 1 − cos(δ − θ )cos(α − θ )
2
Ka for the case of θ = 0º and α = 0º.
Ka for the case where δ = 2/3 φ.
272
Kp for θ = 0º and α = 0º.
Failure modes for flexible walls (sheet-piling).
273
*Lateral-01: A simple wall subjected to an active pressure condition. (Revision: Sept.-08)
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:
φ 30° (a) The Rankine active earth pressure coefficient is, K a = tan 2 45° − = tan 2 45° − = 0.33 2 2 The lateral pressure at the bottom of the wall is pa = γ hK a = ( 0.115 kcf )(10 ft )( 0.33) = 0.38 ksf The force against the wall is Fa =
1 ( pa ) h = ( 0.5 )( 0.38)(10 ) = 1.9 kips per foot of wall 2
(b) The stability of the wall against overturning is found by taking moments about the point "O" at the toe of the wall, Factor of Safety ( FS ) =
( 3')(10 ')(1')( 0.150 kcf )(1.5 ft ) = 1.07 < 2 resisting moment = overturning moment (1.9 kips )(10 / 3 ft )
NG
(c) The stability of the wall against sliding towards the left is found by, Factor of Safety ( FS ) =
resisting force ( 3')(10 ')(1' )( 0.150 kcf )( tan 30° ) = = 1.37 < 2 driving force (1.9 kips )
NG
274
*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,
φ 35° ) = 0.271 K a = tan 2 (45° - ) = tan 2 (45° 2 2 φ 35° ) = 3.690 Note that K p = tan 2 (45° + ) = tan 2 (45° + 2 2
Ka =
1 KP
Coulomb’s active and passive earth pressure coefficients, Ka =
Kp =
cos2 (φ - θ) sin(δ + φ )sin(φ − α ) cos2θ cos(δ + θ ) 1 + cos(δ + θ )cos(θ − α ) cos2 (φ + θ)
2
=
cos2 (35 - 0) sin(12 + 35)sin(35 − 20) cos2 0 cos(12 + 0) 1 + cos(12 + 0)cos(0 − 20) cos2 (35 + 0)
2
= 0.323
= = 3.517 2 2 sin( − )sin + sin(35 12)sin(35 ( ) − + 20) φ δ φ α 2 cos2θ cos(δ − θ ) 1 − cos 0 cos(12 − 0) 1 − cos( δ − θ )cos( α − θ ) cos(12 − 0)cos(20 − 0) 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,
Rankine ' s Fp = 0.5γ h2 K p + 2ch K p = ( 0.5)(18.1)(10) ( 3.690) + 2 ( 9)(10) 3.690 = 3,685 kN / m2 2
Coulomb ' s Fp = 0.5γ h2 K p + 2ch K p = ( 0.5)(18.1)(10) ( 3.517) + 2 ( 9)(10) 3.517 = 3,520 kN / m2 2
275
*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:
276
*Lateral-04: The “at-rest” pressure upon an unyielding wall. (Revision: Sept-08)
Find the lateral “at-rest” force F o 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, K o = 1 − sin φ ' = 1 − sin 30° = 0.50
σ ' = 0 ksf , because there is no surcharge loading upon the surface of Sand #1. at z = 10 feet σ 'h = K oσ 'v = ( 0.5 )( 0.105kcf )(10 ft ) = 0.525 ksf at z = 0 feet
at z = 20 feet σ 'h = ( 0.5 ) ( 0.105 )(10 ) + ( 0.122 − 0.0624 )10 = 0.823 ksf
σ w = γ w h = ( 0.0624 pcf )(10 ft ) = 0.624 ksf 1 1 1 ( 0.525)(10 ) + ( 0.525 )(10 ) + ( 0.302 )(10 ) + ( 0.624 )(10 ) 2 2 2 i kip Fo = 2.63 + 5.25 + 1.49 + 3.12 = 12.5 ft
Fo = ∑ f i = F 1 + F 2 + F 3 + F 4 =
z=
( 2.63)( 6.67 ) + ( 5.25 )(15 ) + (1.49 )(16.67 ) + ( 3.12 )(16.67 ) = 173.1 kip − ft 12.5 kip
z = 13.8 ft from the top of the wall.
277
12.5 kip
*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.
21 ft
γ zK a
-
2c K a
Solution:
a) The coefficient of active earth pressure is,
φ 0° K a = tan 2 45° − = tan 2 45° − = tan 2 45° = 1 2 2 The net active earth pressure pa on the wall is, pa = σ 3 = γ hK a − 2c K a = ( 0.113 kcf )( 21 ft )(1) − 2 ( 0.630ksf ) 1 = 2.37 − 1.26 = 1.11 ksf b) The crack stops where the pressure is zero, pa = 0, pa = γ hK a - 2c K a ∴ γ hK a = 2c K a ∴ hcrack =
2c K a 2 ( 0.630ksf ) 2c = = = 11.2 feet γ Ka γ K a ( 0.113kcf )(1)
c) The total (Rankine) active earth force upn the wall Fa is, 1 Fa = γ H 2 K a − 2cH K a 2 but there is no contact on the wall where the tension crack exists, therefore Fa =
(
1 γ HK a − 2c K a 2
) H − γ 2cK
a
1 2c 2 2 = γ H − 2cH K a + γ 2
2 ( 0.63ksf ) 1 2 Fa = ( 0.113kcf )( 21 ft ) (1) − 2 ( 0.63ksf )( 21 ft )(1) + = 5.48 k / ft of wall 2 ( 0.113kcf ) 2
278
**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: 1) No deflection of the wall means the soil is "at rest" and K 0 = 1 - sin φ = 1 - sin 36 ° = 0.41 kN 2 The force Fo = ½ γ d h 2 K o = ½ 18.3 3 ( 4 m ) (0.41) = 60 kN per meter of wall m 2) When th e wall deflects to the left sufficiently to develop an active pressure condition,
φ 36 ° K a = tan 2 45 ° − = tan 2 45 ° − = 0.26 2 2 kN 2 The force Fa = ½ γ d h 2 K a = ½ 18.3 3 ( 4 m ) (0.26) = 38 kN per meter of wall m 3) The buoya nt weight γ ' of the flooded sand is, kN γ ' = γ sat − γ w = γ d − nγ w − γ w = 18.3 + (0.31) ( 9.81) − ( 9.81) = 11.5 3 m The stress at point "a" is σ a =0, and at "b" which is 1 meter below the surface, σ b = γ d hK a = (18.3
279
kN )(1m )(0.26) = 4.8 kN / m 2 m3
∴ F1 = ½(4.8 kN / m 2 )(1m ) = 2.4 kN / m
kN )(1m)(0.26) = 4.8 kN / m 2 ∴ F2 = (4.8kN / m 2 )(3m) = 14.4 kN / m 3 m kN σ c = γ ' hK a = (11.5 3 )(3m)(0.26) = 9.0 kN / m 2 ∴ F3 = ½(9.0kN / m 2 )(3m) = 13.5 kN / m m The water pressure and force,
σ bc = γ d hK a = (18.3
σ w = γ w h = (9.81
kN )(3m) = 29.4 kN / m 2 m3
∴ F4 = ½(29.4kN / m 2 )(3m) = 44.1 kN / m R =
Therefore
4
∑F
= 74.4 kN / m
i =1
The location of the resultant is y, F1d1 + F2 d 2 + F3 d 3 + F4 d 4 ( 2.4 )( 3.33m ) + (14.4 )(1.5m ) + (13.5 )(1m ) + ( 44.1)(1) = R 74.4 y = 1.17 m from the bottom of the wall. y=
The percent increase in load upon the wall due to flooding is, ∆F =
( 74.4kN − 38kN ) = 38kN
96% increase.
280
*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: The force applied to the wall first requires the coefficient of active earth pressure,
φ 32° K a = tan 2 (45° − ) = tan 2 (45° − ) = 0.307 2 2 The horizontal force FH per unit width of wall is, FH = ½γ h 2 K a = ½(17.2kN / m3 )(7 m) 2 (0.307) = 129.5 kN / m The FH is related to the total force R on the wall as a function of the angle of wall friction δ , FH 129.5kN / m = = 138 kN / m cos δ cos 20° We are asked what is the total force every 3.5 m, FH = R cos δ ∴ R =
Total Active Force every 3.5 m = (138 ) (3.5m) = 482kN
FH δ
FV
281
R
*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).
1m WT
worst load case γsat = 18.5 kN/m3
Medium dense sand
3m
φ = 30°
H=6m
γ = 21.2 kN/m3
Weep holes
3m
φ = 90°
Porous limestone A 0
S o lu tio n . 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 ta b le r a i s e s t o th e to p o f t h e w a ll. ϕ 30° K a sa nd = ta n 2 ( 4 5 ° − ) = ta n 2 ( 4 5 ° − ) = 0 .3 3 3 2 2 ϕ 90° K a lim e s to n e = t a n 2 ( 4 5 ° − ) = ta n 2 ( 4 5 ° − ) = 0 ∴ t h e li m e s to n e d o e s n o t lo a d th e w a ll. 2 2 T h e φ = 9 0 ° i s r e a lly a c o m b in a ti o n o f s h e a r a n d c o h e s io n ( " c e m e n t a ti o n " ) . p 1 = γ ' h1 K p2 = γ
w
a
= (γ
SAT
-γ
w
) h1 K
a
= (1 8 . 5 - 9 . 8 ) ( 3 ) ( 0 . 3 3 ) =
8 .7 k N / m
=
5 8 .8 k N / m
H = (9 .8 )( 6 m )
F1 = ½ p 1 h1 = ( 0 . 5 ) ( 8 . 7 ) ( 3 m )
=
2
1 3 .1 k N / m
F 2 = ½ p 2 H = ( 0 .5 )(5 8 .8 )( 6 m ) F to ta l
2
= 1 7 6 .4 k N / m = 1 8 9 .5 k N / m
T h e lo c a tio n y =
h1
y1 F1 + y 2 F 2 = F to ta l
+ F1
H
(4 m
) (1 3 . 1 ) + ( 2 m (1 8 9 . 5 )
) (1 7 6 . 4
)
= 2 .1 m fro m A .
Ftotal
p1 h2
F2 A
ŷ
p2 282
*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. q = 2.5 ksf
w.t.
a
0.83
c=0
γ = 115 pcf
H = 20’
10’
1
ф = 30° b c=0
γ = 125 pcf ф = 40°
2
0.83 0.66
10’
3
0.18
+ 5
4
c
1.25 0.66
0.13
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: The stress at b (within the top stratum) is:
pa = q Ka 1 = (2.5) (0.333) = 0.83 ksf 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] The pressure of the water upon the wall is: Step 3 283
= 0.79 ksf
pw = γwh = (0.0624) (20’) = 1.25 ksf
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:
yˆ =
5 ⋅ 8.3 + 20 ⋅ 0.9 + 15 ⋅ 6.6 + 50 ⋅ 0.65 + 40 ⋅ 12.5 3 3 3 29
The stress at point c is:
= 0.66 ksf
ŷ = 11.2 feet from top of wall
284
*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.
Surcharge q = 0.84 ksf
-0.25 -0.84
0
c+ 3.25’
1
no water present 20’
Sandy clay
16.75’
2
c = 500 psf
φ 10’
= 10 °
b+
Dense sand
+0.48
b-
+1.29
3
c=0
A
φ = 40°
a
γ = 130 pcf
+0.48
Lateral load from the surcharge
σc+ = Ka1 q = (0.70)(0.84 kcf)
= 0.59 ksf
σc- = -2c K a1 = -2(0.5) 0.70 = -0.84 ksf ∴ ∑ σc = 0.59 - 0.84
= -0.25 ksf
σb+ = Ka 1 γ h – 2c K a 1 + q Ka = (0.7) (0.11) (20’) – (2) (0.50) 0.70 = 1.29 ksf
285
4
+0.77
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
286
**Lateral-11: Anchoring to help support a wall. (Revision: Sept.-08)
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. a) What is the minimum distance x from the anchor to behind the wall? b) What is your recommended factor of safety for the anchor? What is an economical load for the anchor? x
5’
Grouted anchor A
24’
φ = 30° c = 150 psf O
. 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 = ½γH2Ka -2cH K a = ½(0.105)(24’)2(0.33) - 2(0.15)(24) 0.33 = 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 = ½γH2Kp + 2cH K p = ½ (0.105)(24)2(3) + 2(0.15)(24) 3 = 103 kip/ft
287
The factor of safety should be the same for an active failure as a passive failure. Therefore, a simple Fp Fp 103 kips equation could be written as, Fa ( FS ) = or (FS)2 = = = 17.6 ∴ FS = 4.2 ( FS ) Fa 5.84 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.
288
**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’
∆p2 = γ1h K1a = (0.110)(6) (0.307) = 0.203 ksf
at h = -8’
∆p3 = (γ2 - γw)h K2a = (0.125 - 0.0624)(2)(0.333) = 0.417 ksf = [q + (γ1) 6’ + (γ2 - γw) 2’] K3a – 2c √(K3a
at h = -(8+dh)’
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’
∆p4 = (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)’ at h = -25’
∆p5 = (γ4 - γw)h K4a = (0.120 - 0.0624)(8)(1) = 0.46 ksf ∴ 1.76 + 0.46 = 2.22 ksf
at h = -(25 + dh)’ at h = -30’
289
= [2 + 0.66 + 0.125 + (0.0626) (9)](1) –2(0.8)(1) = 1.76 ksf
= [2 + 0.66 + 0.125 + 0.572 + 8(0.120 – 0.0624)](0.49) – 2(0.4)(0.7) = 1.13 ksf
∆p6 = (γ5-γw)h K5a = (0.120-0.0624)(5) (0.49) = 0.141 ksf ∴1.31+0.14 = 1.45 ksf
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
∴ y = 611 / 57.1 = 10.7 feet above “0”
57.1 kips
290
**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.
0.4 m
α = 10ο 0.62m
4
H = 8 m
1 H = 9.58m
2 1.5 m 0.6 m
3
3.5 m
0.75 m
0.96m
0
γ1 =
16.8
kN m
γ2 =
γ conc =
3
17.6
23.6
kN m
kN m
291
yφc
3
3
φ1 =
32
φ2 =
28
o
o
c1 =
0
c2 =
30
kN m
2
Ka = cos α
cos β − cos 2 β − cos 2 φ ' cos β + cos 2 β − cos 2 φ '
= cos10o
cos10o − cos 2 10o − cos 2 32o cos10o + cos 2 10o − cos 2 32o
= 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) 2.77 (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,
292
e=
B ΣMr − ΣMo 2661 − 777 − = e = 2 .8 − = 0.31 m 2 ΣV 749
qtoe =
ΣV B
6e 749 6(0.31) 2 1 + B = 5.6 1 + 5.6 = 178 kN/m
B’ = B – 2e = 5.6m – 2 (0.31m) = 4.98 m qu = (1/2) γ2 B’ Nγ Fγds Fγid + C2 Nc Fcds Fcid + q Nq Fqds Fqid q = γ2 D = (17.6)(1.75) = 30.8 kN/m2 using φ2 = 28° → Nc = 25.8 Nq = 14.7 Nγ = 16.7 Fcd = 1 + 0.4
Df 1.75 = 1 + 0 .4 = 1.14 B' 4.98
Ph -1 = tan (244 / 749) = 18.04° ΣV
ψ = tan −1
ψ Fqi = Fci = 1 − 2 = 0.96 90 ψ Fγi = 1 − φ
2
2
18.04 = 1 − = 0.58 28
Fγd = 1 2 Df 2 Fqd = 1 + 2 tan φ (1 − sin φ ) = 1 + 2 tan28°(1-sin28°) (1.75/5.6)) = B
= 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
293
***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 =
CD CD = AC AO + OC
σ −σ3 σv −σa CD = radius of failure circle = 1 = 2 2
AO = c cot φ σ +σ3 σv −σ a OC = 1 = 2 2
Substituting values into the equation for CD, AO and OC gives:
294
σv −σa 2 sin φ = σv +σ a c cot φ + 2
Rearrange the equation to make σa the subject: This gives:
σ −σ a σ +σa 2c.cos φ + v sin φ = v 2 2 Τhis then gives: σa (sin φ + 1) = σv (1 – sin φ) − 2c.cos φ 1 − sin φ cos φ ∴σa = σ v − 2c 1 + sin φ 1 + sin φ
Solution of the trigonometric expressions: 2α + (90° + φ) = 180° α=
90° + φ φ = 45° – 2 2
1 + sin φ = tan (α + φ) cos φ = tan (45° −
φ 2
+φ )
φ = tan 45° + 2 φ sin 45° + 2 = φ cos 45° + 2 φ cos 45° + cos φ 2 ∴ = φ 1 + sin φ sin 45° + 2 But for any complementary angles β and (90° - β), cos β = sin (90° - β). Thus, cos (45°
295
φ 2
sin (45° -
φ 2
and sin (45°
φ 2
cos (45°
φ 2
φ sin 45° − cos φ φ 2 ∴ = = tan 45° − φ 1 + sin φ 2 cos 45° − 2
∴
1 − sin φ 1 − sin φ 1 + sin φ 12 − sin 2 φ cos 2 φ φ × = = = = tan 2 45° − 2 2 1 + sin φ 1 + sin φ 1 + sin φ (1 + sin φ ) 2 (1 + sin φ )
σa = σv tan2 (45° -
φ 2
) - 2c tan (45° -
φ 2
)
Rankine’s expression gives the effective pressure at failure!!!
∴ σa = (γz) Ka - 2c ( Ka ) ½ ; where Ka = tan2 (45° -
φ 2
)
Using this equation, the slip planes can be described by the grid of lines shown below:
∆
296
**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.
1m
∆Pae increase due to the earthquake load located at 0.6H Pae the earthquake load H=5m
Pa
lateral load from the soil located at 0.33H
Dense sand γ = 18 kN/m3 φ = 36°
0
Solution.
Calculate the coefficient of active earth pressure (Coulomb) K a using,
φ = 36°, α = 0°, β = 90° and δ = 2 / 3φ = 24°, Ka =
sin 2 ( β + φ ') sin (φ '+ δ ) sin (φ '− α ) sin β sin ( β − δ ) 1 + sin ( β − δ ) sin (α + β ) sin 2 ( 90° + 36° )
2
2
Ka =
2
sin ( 36° + 24° ) sin ( 36° − 0° ) sin ( 90° ) sin ( 90° − 24° ) 1 + sin ( 90° − 24° ) sin ( 0° + 90° ) sin 2 (126° ) 0.654 Ka = = = 0.2346 2 2 sin ( 60° ) sin ( 36° ) ( 0.866 )( 0.588) sin ( 66° ) 1 + ( 0.914 ) 1 + sin ( 66° ) ( 0.914 ) 2
297
The active earth force Pa is, 1 2 Pa = γ H 2 K a = ( 0.5 ) (18kN / m3 ) ( 5m ) ( 0.2346 ) = 53 kN / m 2 Calculate the earthquake coefficient of active earth pressure (Coulomb) K ae , k kh = 0.2, kV = 0, δ = 2 / 3φ = 2 / 3 ( 36° ) = 24° and θ ' = tan −1 h = tan −1 ( 0.2 ) = 11.3° 1 − kV sin 2 (φ + β − θ ') ∴ K ae = 2 φ δ φ θ α sin + sin − ' − ( ) ( ) cos θ 'sin 2 β sin ( β − θ '− δ ) 1 + sin ( β − δ − θ ') sin (α + β ) sin 2 ( 36° + 90° − 11.3° )
K ae =
sin ( 36° + 24° ) sin ( 36° − 0° ) sin 2 ( 90° ) sin ( 90° − 24° ) 1 + sin ( 90° − 24° ) sin ( 0° + 90° ) The Mononobe-Okabe earthquake active earth force Pae is,
2
= 0.372
1 2 Pae = γ H 2 (1 − kV ) K ae = ( 0.5) (18kN / m3 ) ( 5m ) (1 − 0) ( 0.372 ) = 83.7 kN / m 2 The earthquake force is ∆Pae = Pae − Pa = 83.7 − 53 = 30.7 kN / m The location of the resultant earthquake force z is, found by locating the force Pae at a height
0.6H above the base of the wall; the active earth force Pa is obviously ( 0.33) H above the base. 2 2 0.6 H )( ∆Pae ) + (1/ 3) H ( Pa ) ( 0.6 )( 5m ) ( 30.7kN / m ) + (1/ 3)( 5m ) ( 53kN / m ) ( z= = = 2.1 m
Pae
(83.7kN / m ) 2
298
**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.
1m
H=5m
Dense sand γ = 16 kN/m3
φ = 36° δ = 2/3 φ = 24º
0
Solution.
299
Chapter 16 Braced Cuts for Excavations Symbols for Braced Cuts for Excavations
300
*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.
301
Effective Stress Mohr’s Circle for failure Angle
τ (kN/m2)
σ (kN/m2)
From the Mohr’s Circle, we can get that φ2 = 32 o
Gs = 2.65 ; γW = 9810 N/m2
φ
KA = tan2 (45o − ) 2
γs =
(G s )(γ W ) (2.65)(9810) = 1+ e
1 + 0.46
γS = γ2 = 17.8 kN/m2 o
K A 1 = tan 2 (45o − 25 ) = 0.406 2 o K A 2 = tan 2 ( 45 o − 32 ) = 0.307 2
302
Pa = (q) (KA1) = (90kN/m2) (.406) 36.54 kN/m2
Pb’+ = [KA1 (q + γ1h1)] = [(.406) (15kN/m2 x 3m)] 54.81 kN/m2 Pb’- = [KA2 (q + (γ2-γW) h] = [(.307) (90 + (17.8-9.81) (3)] 34.99 kN/m2 Pc = [(q + γ1h1 + (γ2-γW) h2] KA2 = .307 [90 + (15)(3) + (17.8-9.81)(2)] 46.35 kN/m2 PW = γW h W = (9.81)(2) 19.62 kN/m2 36.54
1
2m
F1 F2 2 18.27
3m
3
F3 F4
34.99
4 11.36
F5 5 19.62
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 303
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 ) (h W/2) = (19.62)(2/2) = 19.62 kN/m
∑F = F
1
Ftot =
(∑ F )(space b/t struts) =
Located at y f =
∑M
c
+ F2 + F3 + F4 + F5 = 237.96 kN/m
(237.96kN/m)(4m) 951.84 kN
109.6(1.5) + 27.4(2 ) + 69.98(4 ) + 11.36(4.33) + 19.62(4.33) = 2.66m 237.96
= 0 Where C is located at the bottom of the trench along with RA RB is located at the end of the strut.
RB (3m) - 951.84 kN (2.34m) = 0 RB = 742.44 kN
RA = 209.40 kN 304
Shear Diagram 742.44 kN
0 kN -209.40 kN Moment Diagram 490.0kN-m
0 kN-m
0 kN-m 0.66 m
305
2.34 m
**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, if if
γH cu
γH cu
> 4 the clay is soft to medium ≤ 4 the clay is stiff
In this problem,
γH cu
=
(17 )( 5 ) = 1.21 < 4 ∴ this is a stiff clay ( 70 )
Also, since γH/ cu < 6, the sheet-piling should extend at least 1.5 m below bottom. 306
Step 1. Establish the lateral earth pressure distribution. Using Peck's (1967) apparent pressure envelope, we must choose the larger of, 4c (1) pa = γ H 1 − u γ H (2) pa = 0.3γ H = 0.3 (17 )( 5 ) = 25.5 kN / m 2 The location of the top strut should be less then the depth of the tensile crack zc. Since φ = 0, Ka →
Ka = 1. therefore therefore
σ3 = σa = (γ)(zc)Ka - 2c Ka
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 F22 = F12 = 13.6 kN/m
∑ Fy = 0
therefore, F3 = 70.6 kN/m
307
= - F22 + (3.75-0.51)(26)- F3
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. reduce steel size, but increases the difficulty of placing the concrete pipes).
(May use 3 m to
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.
308
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.
309
*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,
if if
γH
4c > 4 the clay is soft to medium then σ a = γ H 1 − u cu γH γH σ a = 0.2γ H to 0.4γ H ≤ 4 the clay is stiff then cu
310
Determine the cohesion from Mohr's circle cu = ∴
γH cu
(17.3 kN / m ) (12 m ) = 4.33 > 4 = ( 48 kN / m )
qu 96 kN / m 2 = = 48 kN / m 2 2 2
3
2
∴ this is a soft to medium clay
Peck (1969) provided a criterion for soft to medium clays, 4c pa = γ H 1 − u γH
(4)(48 kN / m2 ) 3 2 = .3 kN / m )(12 m ) 1 − (17 = 15.48 kN / m 3 (17.3 kN / m )(12 m)
The lateral earth pressure diagram for the braced sheet pile system in soft clays is,
2. In the free body diagram, part (a),
∑M
B
=0
( 1 2 ) (15.48 kN / m ) ( 3.0 m )( 4.0 m ) 1.5 m 3.03 m + (1.5 m ) (15.48 kN / m ) ( 4.0 m ) 1.52 m − ( F 2
2
∴ FA = 100.6 kN From
∑H =0
( 2 ) (1.5 m + 4.5 m ) (15.48 kN / m ) ( 4.0 m ) − 100.6 kN = 85.2 kN
FB1 = 1
2
In the free body diagram, part (b)
( 2 ) (3.0 m ) (15.48 kN / m ) ( 4.0 m ) = 92 kN
FB 2 = FC1 = 1
In the free body diagram, part (c),
2
∑M
C
=0
( FD )( 3.0 m ) − ( 4.5 m ) (15.48 kN / m 2 ) ( 4.0 m )
4.5 m =0 2
∴ FD = 209.0 kN 311
A
)( 3.0 m ) = 0
From
∑H =0
FC 2 + FD − ( 4.5 m ) (15.48 kN / m2 ) ( 4.0 m ) = 0 FC 2 = ( 4.5 m ) (15.48 kN / m 2 ) ( 4.0 m ) − 209.0 kN = 69.6 kN
Therefore,
FA = 100.6 kN FB = 85.2 kN + 92.9 kN = 178.1 kN FC = 92.9 kN + 69.6 kN = 162.5 kN FD = 209.0 kN
312
Chapter 17 Bearing Capacity of Soils Symbols for the Bearing Capacity of Soils
313
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 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37
0.0000 0.0175 0.0349 0.0524 0.0698 0.0873 0.1047 0.1222 0.1396 0.1571 0.1745 0.1920 0.2094 0.2269 0.2443 0.2618 0.2793 0.2967 0.3142 0.3316 0.3491 0.3665 0.3840 0.4014 0.4189 0.4363 0.4538 0.4712 0.4887 0.5061 0.5236 0.5411 0.5585 0.5760 0.5934 0.6109 0.6283 0.6458
10.18 10.61 11.07 11.56 12.07 12.61 13.19 13.80 14.44 15.13 15.87 16.65 17.49 18.38 19.33 20.36 21.46 22.65 23.92 25.30 26.80 28.42 30.18 32.10 34.19 36.49 39.01 41.78 44.85 48.26 52.05 56.29 61.04 66.40 72.48 79.40 87.33 96.49
5.70 6.00 6.30 6.62 6.97 7.34 7.73 8.15 8.60 9.09 9.60 10.16 10.76 11.41 12.11 12.86 13.68 14.56 15.52 16.56 17.69 18.92 20.27 21.75 23.36 25.13 27.09 29.24 31.61 34.24 37.16 40.41 44.04 48.09 52.64 57.75 63.53 70.07
1.00 1.10 1.22 1.35 1.49 1.64 1.81 2.00 2.21 2.44 2.69 2.98 3.29 3.63 4.02 4.45 4.92 5.45 6.04 6.70 7.44 8.26 9.19 10.23 11.40 12.72 14.21 15.90 17.81 19.98 22.46 25.28 28.52 32.23 36.50 41.44 47.16 53.80
0.00 0.08 0.18 0.28 0.39 0.51 0.65 0.80 0.96 1.15 1.35 1.58 1.84 2.12 2.44 2.79 3.19 3.63 4.13 4.70 5.34 6.07 6.89 7.83 8.90 10.12 11.53 13.15 15.03 17.21 19.75 22.71 26.20 30.33 35.23 41.08 48.11 56.62
5.10 5.38 5.63 5.90 6.19 6.49 6.81 7.16 7.53 7.92 8.34 8.80 9.28 9.81 10.37 10.98 11.63 12.34 13.10 13.93 14.83 15.81 16.88 18.05 19.32 20.72 22.25 23.94 25.80 27.86 30.14 32.67 35.49 38.64 42.16 46.12 50.59 55.63
1.00 1.09 1.20 1.31 1.43 1.57 1.72 1.88 2.06 2.25 2.47 2.71 2.97 3.26 3.59 3.94 4.34 4.77 5.26 5.80 6.40 7.07 7.82 8.66 9.60 10.66 11.85 13.20 14.72 16.44 18.40 20.63 23.18 26.09 29.44 33.30 37.75 42.92
0.00 0.00 0.01 0.02 0.04 0.07 0.11 0.15 0.21 0.28 0.37 0.47 0.60 0.74 0.92 1.13 1.37 1.66 2.00 2.40 2.87 3.42 4.07 4.82 5.72 6.77 8.00 9.46 11.19 13.24 15.67 18.56 22.02 26.17 31.15 37.15 44.43 53.27
5.10 5.38 5.63 5.90 6.19 6.49 6.81 7.16 7.53 7.92 8.34 8.80 9.28 9.81 10.37 10.98 11.63 12.34 13.10 13.93 14.83 15.81 16.88 18.05 19.32 20.72 22.25 23.94 25.80 27.86 30.14 32.67 35.49 38.64 42.16 46.12 50.59 55.63
1.00 1.09 1.20 1.31 1.43 1.57 1.72 1.88 2.06 2.25 2.47 2.71 2.97 3.26 3.59 3.94 4.34 4.77 5.26 5.80 6.40 7.07 7.82 8.66 9.60 10.66 11.85 13.20 14.72 16.44 18.40 20.63 23.18 26.09 29.44 33.30 37.75 42.92
0.00 0.00 0.01 0.02 0.05 0.07 0.11 0.16 0.22 0.30 0.39 0.50 0.63 0.78 0.97 1.18 1.43 1.73 2.08 2.48 2.95 3.50 4.13 4.88 5.75 6.76 7.94 9.32 10.94 12.84 15.07 17.69 20.79 24.44 28.77 33.92 40.05 47.38
38 39 40 41 42 43 44 45 46 47 48 49 50
0.6632 0.6807 0.6981 0.7156 0.7330 0.7505 0.7679 0.7854 0.8029 0.8203 0.8378 0.8552 0.8727
107.13 119.59 134.31 151.89 173.09 198.99 231.10 271.57 323.57 391.94 484.34 613.53 801.95
77.50 85.97 95.66 106.81 119.67 134.58 151.95 172.29 196.22 224.55 258.29 298.72 347.51
61.55 70.61 81.27 93.85 108.75 126.50 147.74 173.29 204.19 241.80 287.85 344.64 415.15
67.00 79.77 95.61 115.47 140.65 173.00 215.16 271.07 346.66 451.28 600.15 819.31 1155.97
61.35 67.87 75.31 83.86 93.71 105.11 118.37 133.87 152.10 173.64 199.26 229.92 266.88
48.93 55.96 64.20 73.90 85.37 99.01 115.31 134.87 158.50 187.21 222.30 265.50 319.06
64.07 77.33 93.69 113.99 139.32 171.14 211.41 262.74 328.73 414.33 526.45 674.92 873.86
61.35 67.87 75.31 83.86 93.71 105.11 118.37 133.87 152.10 173.64 199.26 229.92 266.88
48.93 55.96 64.20 73.90 85.37 99.01 115.31 134.87 158.50 187.21 222.30 265.50 319.06
56.17 66.76 79.54 95.05 113.96 137.10 165.58 200.81 244.65 299.52 368.67 456.40 568.57
314
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):
For square footings,
qult = 1.3c ' N c + qN q + 0.4γ BNγ
For continuous or wall footings
qult = c ' N c + qN q + 0.5γ BNγ
where, q = γ D f Nq =
and the factors are,
a2 a cos 2 ( 45° − ϕ / 2 )
where a = e(
0.75π −φ / 2 ) tan φ
N c = ( N q − 1) cot φ Nγ =
tan φ K pγ − 1 2 2 cos φ
Meyerhof (1963):
For vertical loads,
qult = cN c Fsc Fdc + qN q Fsq Fdq + 0.4γ BNγ Fsγ Fd γ
and for inclined loads, qult = cN c Fic Fdc + qN q Fiq Fdq + 0.4γ BNγ Fiγ Fdγ and the factors are, N q = eπ tan φ tan 2 ( 45° − φ / 2 ) N c = ( N q − 1) cot φ
Nγ = ( N q − 1) tan (1.4φ ) Brinch Hansen (1970):
The general equation,
qult = cN c Fsc Fdc Fic + qN q Fsq Fdq Fiq + 0.4γ BNγ Fsγ Fd γ Fiγ
and the factors are, N q = eπ tan φ tan 2 ( 45° − φ / 2 ) N c = ( N q − 1) cot φ
Nγ = 1.5 ( N q − 1) tan φ
315
*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.
W = 294 kN Df = 1 m
γ = 18.15 kN/m3
φ = 35°
W
c =0
B Solution: Terzaghi's formula for the ultimate bearing capacity qult of a square footing is, qult = 1.3c ' N c + qN q + 0.4γ BNγ
where q = D f γ
The allowable bearing capacity qall with the factor of safety of 3 is,
(
)
(
)
qult 1 = 1.3c ' N c + qN q + 0.4γ BNγ and 3 3 294 1 = 1.3c ' N c + qN q + 0.4γ BNγ or B2 3 For φ =35º, N c =57.8, N q =41.4, and Nγ =41.1. qall =
qall =
W 294 kN = B2 B2
Substituting these values into Terzaghi's equation, we get 294 1 = ( 0 ) + (18.15)(1) (41.4) + (0.4) (18.15) B (41.1) B2 3 294 = 250.5 + 99.5 B B2 B 3 + 2.52 B 2 − 2.96 = 0 ∴ B = 0.90 m
316
*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.
W = 294 kN Df = 1 m
γ = 18.15 kN/m3
φ = 35°
W
c =0
B Solution:
Meyerhof's formula for the ultimate bearing capacity qult of a square footing is, qult = c ' N c Fsc Fdc Fic + qN q Fsq Fdq Fiq + 0.4γ BN γ Fsγ Fd γ Fiγ
where q = D f γ
Since the load is vertical, all three inclination factors Fic =Fiq =Fiγ =1. B 1 B Fsq = 1 + tan φ = 1 + tan 35° = 1.70 and Fsγ = 1 − 0.4 = 1 − 0.4(1) = 0.6 L 1 L 2 Df 2 1 Fdq = 1 + 2 tan φ (1 − sin φ ) = 1 + 2 ( tan 35° ) (1 − sin 35°) ≈ 1.25 and Fd γ = 1 B B The allowable bearing capacity qall with the factor of safety of 3 is,
(
)
(
)
qult 1 ' = c N c Fsc Fdc + qN q Fsq Fdq + 0.4γ BN γ Fsγ Fd γ 3 3 294 1 ' or = c N c Fsc Fdc + qN q Fsq Fdq + 0.4γ BN γ Fsγ Fd γ B2 3 For φ = 35º, N c = 46.12, N q = 33.30, and N γ = 37.15. qall =
and
qall =
W 294 kN = B2 B2
Substituting these values into Meyerhof''s equation, we get 294 1 = ( 0 ) + (18.15)(1) (33.3) (1.7 )(1.25 ) + (0.4) (18.15) B (37.15 )( 0.6 )(1) B2 3 294 = 428.1 + 53.94 B or B 3 + 7.94 B − 5.45 = 0 ∴ B = 0.65 m 2 B
317
*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.
W = 294 kN Df = 1 m
γ = 18.15 kN/m3
φ = 35°
W
c =0
B Solution:
Hansen's formula for the ultimate bearing capacity qult of a square footing is, qult = c ' N c Fsc Fdc Fic + qN q Fsq Fdq Fiq + 0.4γ BN γ Fsγ Fd γ Fiγ
where q = D f γ
Since the load is vertical, all three inclination factors Fic =Fiq =Fiγ =1. B 1 B Fsq = 1 + tan φ = 1 + tan 35° = 1.7 and Fsγ = 1 − 0.4 = 1 − 0.4(1) = 0.6 L 1 L 2 Df 2 1 Fdq = 1 + 2 tan φ (1 − sin φ ) = 1 + 2 ( tan 35° ) (1 − sin 35°) ≈ 1.255 and Fdγ = 1 B B The allowable bearing capacity qall with the factor of safety of 3 is,
(
)
(
)
qult 1 ' = c N c Fsc Fdc + qN q Fsq Fdq + 0.4γ BN γ Fsγ Fdγ 3 3 294 1 ' or = c N c Fsc Fdc + qN q Fsq Fdq + 0.4γ BN γ Fsγ Fd γ B2 3 For φ = 35º, N c = 46.12, N q = 33.30, and N γ = 33.92. qall =
and
qall =
W 294 kN = B2 B2
Substituting these values into Hansen's equation, we get 294 1 = ( 0 ) + (18.15)(1) (33.3) (1.7 )(1.255 ) + (0.4) (18.15) B (33.92 )( 0.6 )(1) B2 3 294 = 429.8 + 49.25 B or B 3 + 8.73B − 5.97 = 0 ∴ B = 0.70 m 2 B
318
*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.
m = 30,000 kgm Df = 1 m
W
ρ = 1,850 kg/m3 φ = 35° c =0
B Solution:
The soil density ρ = 1,850 kgm / m3 converts to a unit weight via γ = ρ g ( like F = ma ) , kg m m 1,850 m3 9.81 s 2 = 18.15 kN / m3 and the load to be supported by the footing is, γ = ρg = (1, 000 N / kN ) W = ma =
( 30, 000 kg m ) 9.81 (1, 000 N / kN )
m s 2
= 294 kN
Terzaghi's ultimate bearing capacity of a square footing is given by, qult = 1.3c ' N c + qN q + 0.4γ BN γ
(
)
qult 1 P 294 = 1.3c ' N c + qN q + 0.4γ BN γ and qall = 2 = 2 3 3 B B 294 1 or = 1.3c ' N c + qN q + 0.4γ BN γ 2 B 3 For φ = 35º, N c = 57.8, N q = 41.4, and N γ = 41.1,
∴ qall =
(
)
294 1 = ( 0 ) + (18.15)(1) (41.4) + (0.4) (18.15) B (41.1) ∴ B 3 + 2.52 B 2 − 2.96 = 0 2 B 3 B = 0.90 m
319
*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, qult = 1.3c ' NC + qN q + 0.4γ BNγ For φ = 28° N c = 31.6, N q = 17.8, Nγ = 15.0 and q = γ D f = (0.115)(2) = 0.23 ksf Therefore, qult = 1.3(0.30)(31.6) + (0.23)(17.8) + 0.4(0.115)(2.5)(15.0) = 18.1 ksf 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, ' ' ' ' qult −local = 1.3c ' N c + qN q + 0.4γ BN γ
2 2 where c' = c = (0.30) = 0.2 ksf 3 3
2 2 and ϕ ' = (ϕ ) = (28°) = 18.7° which give N c' = 16.2, N q' = 6.5 and Nγ' = 4.52 3 3 ' ∴ qult −local = (1.3) (0.2)(16.2) + (0.23)(6.5) + (0.4)(0.115)(2.5)(4.52) = 6.2 ksf qult − general failure = 18.1 ksf
versus qult −local failure = 6.2 ksf
( Almost a three − to − one)
320
*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).
Pult = 1,863 kN ▼ WT
Df = 0.5 m B = 0.5 m
Solution: qult =
Pult 1, 863 kN = = 1, 863 kPa BL ( 0.5 m )( 2.0 m )
was the field measured failure load.
(1) The Hansen formula predicts an ultimate bearing capacity of, qult = 0 + qN q Fqs Fqd + 0.5γ BN γ Fγ s Fγ d Lee ' s adjustment formula is φ ps = 1.5φtriaxial − 17 ° = 1.5 ( 42 ° ) − 17 ° = 46 ° For ϕ = 46 °, N q = 158.5 and N γ = 244.65 B 0.5 Fqs = 1 + tan ϕ = 1 + tan 46 ° = 1.26 L 2 B 0.5 Fγ s = 1 − 0.4 = 1 − 0.4 = 0.9 L 2 D 0.5 Fqd = 1 + 2 tan ϕ (1 − sin ϕ ) 2 f = 1 + 2 tan 46 °(1 − sin 46 °) 2 = 1.16 0.5 B Fγ d = 1.0 ∴ qult = 0 + (9.31)(0.5)(159)(1.27)(1.16) + (0.5)(9.31)(0.5)(245)(0.9)(1.0) ∴ qult = 1, 485 kPa versus 1, 863 kPa measured ( Hansen underestimates by 20%) (2) The Meyerhof formula with φ = 46º, N q = 158.5 and N γ = 328.7 3, qult = 0 + qN q Fqs Fqd + 0.5γ BN γ Fγ s Fγ d ∴ qult = 0 + (9.31) ( 0.5 )(158.5 ) (1.27) (1.16 ) + (0.5)(9.31)(0.5)(328.73)(0.9)(1.0) ∴ qult = 1, 782 kPa versus 1, 863 kPa ( Meyerhof underestimates by 4 % ) .
321
*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: (a) Find γ SAT to determ ine γ ',
γ dry =
γ 1 + wN
=
γ dry Ws 112 = 101.8 pcf and V s = = 1.10 G sγ w G sγ w
set V = 1 ft 3 ∴ V s =
101.8 = 0.61 ft 3 ∴ V v = V − V s = 1 − 0.61 = 0.39 ft 3 2.68 ( 62.4 )
V but γ sat = γ dry + nγ w = γ dry + v γ w ∴ γ sat = 101.8 + (0.39) ( 62.4 ) = 126.2 pcf V and γ ' = γ sat − γ w = 126.2 − 62.4 = 63.8 pcf T ry B = 5.7 feet with M eyerhof's equation, q ult = c ' N c ( Fcs Fcd Fci ) + qN q ( Fqs Fqd Fqi ) + 0.5γ BN γ ( Fγ s Fγ d Fγ i ) w here the load inclination factors Fci , Fqi and Fγ i = 1 25 ° ϕ K p = tan 2 45 ° + = tan 2 45 ° + = 2.46, therefore 2 2 B 5.7 Fcs = 1 + (0.2) K p = 1 + (0.2) (2.46) = 1.49 L 5.7 For ϕ ≥ 10 °
Df Fcd = 1 + (0.2) B
4 K p = 1 + (0.2) 2.46 = 1.22 5.7
Df 4 Fqd = Fγ d = 1 + (0.1) K p = 1 + (0.1) 2.46 = 1.11 5.7 B 5.7 B Fqs = Fγ s = 1 + (0 .1) K p = 1 + (0.1)( )(2.46) = 1.25 5.7 L 322
The Meyerhof bearing capacity factors for φ = 25° are N c = 20.7, N q = 10.7, and N γ = 6.77 qult = c ' N c ( Fcs Fcd Fci ) + qN q ( Fqs Fqd Fqi ) + 0.5γ BN γ ( Fγ s Fγ d Fγ i ) qult = (0.24)(20.7)(1.49)(1.22)(1) + (0.112)(4)(10.7)(1.25)(1.11)(1) + (0.5)(0.112)(5.7)(6.67)(1.25)(1.11)(1) qult = 18.6 ksf qall =
qult 18.6 Q 200 = = 6.2 ksf therefore B 2 = = = 32.25 ft 2 ∴ B = 5.7 ft FS 3 qall 6.2
Therefore the choice of B = 5.7 ft was a good choice. (b) When the water table rises to the ground surface, need a larger footing; try B = 7.0 feet. B 7 Fcd = 1 + 0.2 K p = 1 + 0.2 (2.46) = 1.49 L 7 Fcs = 1.49 same as above D 4 Fqd = Fγ d = 1 + 0.1 K p = 1 + 0.1 2.46 = 1.09 B 7 Fqs = Fγ s = 1.25 same as above qult = (0.24)(20.7) (1.49 )(1.18 ) + (0.062) ( 4 )(10.7 )(1.25 )(1.09 ) + (0.5)(0.062)(7) ( 6.67 )(1.09 )(1.25 ) qult = 16.62 ksf qall =
qult 16.62 Q 200 = = 5.54 ksf and B 2 = = = 36.1 ft 2 ∴ B = 6.01 ft FS 3 qall 5.54
Iterate once more, and find B = 7.5 feet.
323
**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 γ e of the soil can also be given by the formula: dw γ' 2 γ wet + 2 ( H − d w ) 2 H H 35 ° ϕ where H = (0.5) B tan 45 ° + = (0.5)(2.5) tan 45 ° + = 2.40 m 2 2 and d w = depth to the WT below the footing invert = 0.85 m
γ e = (2H − dw )
Set the total volume V = 1 m 3
γ wet
γ dry 16.5 = = 0.63 m 3 G s γ wet (2.68) ( 9.8 )
18.10 kN = 16.5 3 1 + 0.10 m
and
Vs =
Vv = 1.0 − V s = 1 − 0.63 = 0.37 m 3
and
γ sat = γ dry + nγ wet = 16.5 + (0.37) ( 9.8 ) = 20.1
γ dry =
1+ w
=
kN m3
0.85 (18.10 ) 20.1 − 9.8 kN 2 ∴ γ e = ( (2)(2.40 − 0.85) ) + ( 2.40 − 0.85 ) = 12.6 3 2 2 m ( 2.4 ) ( 2.4 ) Using Hansen’s method with φ = 35º, the bearing capacity factors are Nq = 33.3 and Nγ = 33.92.
324
tan 3 5 ° = 1 .7 0 2 D f 2 1 = 1 + 2 tan φ (1 − sin φ ) = 1 + 2 tan 3 5 (1 − sin 3 5 ) = 1 .1 0 2 .5 B B 2 .5 = 1 − 0 .4 = 1 − 0 .4 ( ) = 0 .6 L 2 .5 = 1 .0
Fq s = 1 + Fq d Fγ s Fγ d
B 2 .5 tan φ = 1 + L 2 .5
T h erefo re, th e u ltim ate an d allo w ab le b earin g cap acities are, q u lt = 0 + q N q ( F q s F q d ) + 0 .5 γ e B N γ ( Fγ s Fγ d ) q u lt = (1 8 .1) (1 .0 )( 3 3 )(1 .7 0 )(1 .1 0 ) + (0 .5) (1 2 .6 )( 2 .5 )( 3 4 )( 0 .6 )(1 ) q u lt = 1, 4 9 7 kP a q a ll =
325
1, 4 9 7 = 7 4 9 kP a 2
*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.
N γ =18.1 kN/m3 φ =32°
0.61 m
c=0
0.61 m
γsat = 21.07 kN/m3 1.22 m
Solution: The Hansen formula for a footing is, qult = cN c Fcs Fcd + qN q Fqs Fqd + 0.5γ BN y Fys Fyd The inclination factors Fci , Fqi , and Fγ i are all equal to 1 because the load is vertical. For φ = 32° , N c = 35.49, N q = 23.18 and N γ = 20.79 and B / L = 1 Nq Fcs = 1 + = 1 + ( 23.20 / 35.50 ) = 1.65 Nc Fqs = 1 + tan φ = 1 + 0.62 = 1.62 B Fys = 1 − 0.4 = 1 − 0.4 = 0.60 L 2 Df Fqd = 1 + 2 tan φ (1 − sin φ ) = 1 + (2)(0.62)(0.22)(1) = 1.273 for D f / B ≤ 1 B Fyd = 1 (1 − Fqd ) 1 − 1.273 Fcd = Fqd − = 1.273 − = 1.292 N q tan φ 23.20 × 0.62
326
The WT is located above the footing, therefore, q = ( 0.61m ) (18.1 kN / m3 ) + ( 0.61m )( 21.07 − 9.81) = 17.9 kN / m 2 ∴ qult = (17.9)(1.62)(1.273)(23.20) + ( 0.5 )( 0.6 )( 21.07 − 9.81)(1.22 )( 20.8 )(1) = 981 kN / m 2 Therefore, 2 q 981kN / m 2 qall = ult = = 327 kN / m 3 3 Hence, the total gross load N is,
N = qall B 2 = (327 kN / m 2 )(1.22m) 2 = 487 kN
327
**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 and φ = 30°.
Solution: The effective width footing width B ' = B - 2 e x = ( 2.5 ) - 2 ( 0.2 ) = 2.1 ft and the effective length L ' = L - 2 e y = ( 5 ) - 2(0.4) = 4.2 ft.
M eyerhof's ultimate bearing capacity formula with c = 0 is, q ult = 0 + qN q Fqs Fqd + 0.5γ B ′N γ F ys Fγ d For φ = 30 , N q = 18.4 and N γ = 15.67 B' 2.1 Fqs = 1 + ' tan φ = 1 + ( 0.58 ) = 1.29 4.2 L 2 2 Fqd = 1 + 2 ( tan 30 ° )(1 − sin 30 ° ) = 1.275 2.1 B′ 2.1 F ys = 1 − 0.4 = 1 − 0.4 = 0.8 L′ 4.2 F yd = 1 q ult = ( 2)(0.115 ) (18.4) (1.29 )(1.275 ) + (0.5)(0.11 5)(2.1)(15.67) ( 0.8 )(1 ) = 8.47 ksf Henc e ,
Q ult = q ult ( B ′L ′ ) = ( 8.47 )( 2.1)(4.2 ) = 74.73 kips
328
**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 ftkips. 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
Vertical axial load = 400 kips Mx = 200 ft-kips My = 120 ft-kips
Solution: Eccentricities
ex =
My Q
=
120 ft − k = 0.3 feet 400
∴ Br = B − 2ey = 8 '− 1' = 7 feet
and
and
ey =
M x 200 ft − k = = 0.5 feet Q 400
Lr = L − 2ex = 8 '− 0.6 ' = 7.4 feet
(ie. Lr > Br)
Adjusting the φ from triaxial (φtr ) to a plane-strain value (φ ps ) via Lee’s formulation,
φ ps ≅ 1.1φtr = 1.1( 32.7°) = 36° 36° N q = eπ tan 36° tan 2 45° + = 37.8 2
Nc =
(N
q
− 1) cot φ = ( 36.8 ) cot 36° = 50.6
Nγ = ( N q − 1) tan (1.4φ ) = ( 36.8 ) tan 50.4° = 44.4 Nγ = 1.5 ( N q − 1) tan φ = 1.5 ( 36.8 ) tan 36° = 40.1
∴
and
B Sc = 1 + 0.2 K p r Lr
7 = 1.73 = 1 + 0.2 ( 3.85 ) 7.4
D d c = 1 + 0.2 K p Br
Since φ > 10° , S q = Sγ ≅ 1.0 Hansen’s 329
= 1 + 0.2 and
(
)
6 3.85 = 1.34 7
d q = dγ = 1.0 .
qult = 0.5γ BN γ Sγ d γ iγ g γ bγ + cN c S c d c ic g c bc + qq N q S q d q iq g q bq
Also i = g = b = 1.0 for this problem, since α = 0 = i (inclination factor f / load Q with t vertical) η = 4 g (ground factor with t inclined ground on side of footing) b (base factor with t inclined ground under the footing) qult = 0.5 ( 0.115 )( 7 )( 40.1)(1) + ( 0.200 )( 50.6 )(1.73 )(1.34 ) + ( 0.115 )( 6 )( 37.8 )(1) = qult = 16.1 + 23.5 + 26.1 = 65.7 ksf
q all =
qult 65.7 = = 21.9 ksf FS 3 1
1
1
1
R ex
e 2 0.3 2 = 1− x = 1− = 0.81 8 B
Rey
e 2 0.5 2 =1− y = 1− = 0.75 8 B
Qall = qall ( B 2 ) ( R e x ) ( R e y ) = 21.9 ( 8 x8 )( 0.81)( 0.75 ) = 851 kips qall =
Qall 851 = = 13.3 ksf B 2 64
400 (The contact load qo = 13 = 6.1 ksf ) 851
330
**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.
Topsoil
Fine brown silty sand - small gravel
Brown silty clay
Fine brown silty sand - trace of coarse sand
Fine to medium brown silty sand -some small to medium gravel
Boring No.5 Boring No.2
Boring No.3
Elevation 295.0
295 Elevation 290.6
Elevation 292.8
5 4
290
Elevation 288.0
Sandy
8
Got 10 Firmer
285 Got Firmer 25 Got Firmer 22
280
Cohesive 275
Got firmer
Got 27 Firmer 36 6 in. boulder 38 34
7 13
14
7 brown 16
25 71 46
29 Hard 71
51 47 Hard 69 38 34
62 67
39 32
Hard 270
Got Firmer Got Firmer
Dark 6
Got Firmer
Elevation 296.6
Boring No.4
69 74
Notes:
1. All elevations are in accordance with plot furnished by architect. 2. Borings were made using standard procedures with 2-in. -OD split spoon. 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. 4.No water encountered in any of the borings.
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: qall =
331
N ( K D ) for B ≤ 4 ft 4
N B +1 ( K D ) for B> 4ft 6 B 2
qall =
Df ≤ 1.33 B
where K D = 1 + 0.33
For the silty sand use N=
47 + 51 + 71 = 56.33 ≃ 56 (#3, 4, and 5) 3
Let’s assume B=4.5 ft and Df =0 56 4.5 + 1 = (1) = 13.9 ksf 6 45 2
qall
q0
=
Q B2
=
This suggests that a B = 4.5 feet is excessive since
160kips = 7.9 ≪ qall = 13.9 ksf 20.25sf
Assume B < 4 ft, say B~ 3.5 ft , and use formula qall=
N ( KD ) 4
∴ qo
=
kd =1+ 0.33Df /B
56 0.33Df 1 + and Df = 0 qall = 14 ksf B 4
160 kips Q = = 13.06 ≅ 13 ksf ≤ 14 ksf OK 2 (3.5)2 B
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=
N kd 4
kd =1+0.33Df /B ≤ 1.33
qall = 10.64ksf < 13 ksf Let’s use B= 3.90 feet
Kd =
32 4
0.33 x 4.5 1 + = 1.33 3.50
NOT GOOD Q qall = 10.64 ksf q0 = 2 = 10.51 ksf ≤ 10.64 ksf B
Use B = 3.90 feet.
332
Chapter 18 Shallow Foundations Symbols for Shallow Foundations
333
Properties of Reinforcing Steel (British and SI units).
334
*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,
D L = 100 kips
f c' = 3, 000 psi
LL = 120 kips
f y = 60, 000 psi
q allowable = 4 ksf from qult = 10 ksf and FS = 2.5 Solution: 1) Footing size for service loads: Q 220 kips B= = = 7.42 feet therefore use B = 7.5 feet. 4 ksf qa
2) Check ultimate parameters: that is the actual soil pressure qo under Qult, QULT = 1.2 DL + 1.6 LL = 1.2 (100 ) + 1.6 (120 ) = 120 + 192 = 312 kips
∴ qO =
QU 312 = = 5.5 ksf < 10 ksf for qult …GOOD 2 2 B ( 7.5 ft )
3) Compute the allowable concrete shear strength vc ( allowable ) vall = 4φ f 'C
where ϕ = 0.75 for shear and torsion ,
vC = vall = 4 ( 0.75 ) 3, 000 psi = 164 psi ≈ 23.7 ksf 4) Find d, the effective depth, (in this case two-way shear governs) in feet.
15” B = 7.5 feet
15” + d
335
q d 2 vc + o 4
qo + d vc + 4
2 2 qo w − (B − w ) = 0 4 2 5.5 5.5 15 2 15 5.5 2 d 23.7 + + d 23.7 + − 7.5 − =0 4 4 12 12 4 d 2 + 1.25d − 3.0 = 0 Which yield to two solutions for d =
+1.22 ft −2.47 ft
and using the modified formula equation 2a:
When the column has a rectangular area bxc, the formula is,
( 7.5 ) 2 ( 5.5 ) BLqo 15 15 2 4d + 2 ( b + c ) d − =0 = 4d + 2 + d − 12 12 23.7 vc 2
d 2 + 1.25d − 3.26 = 0 which yields d = +0.39 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.
B−w Unit strip of the cantilever arm = =L 2
∴
15 7.5 − 12 L= = 3.13 ft 2
B = 7.5 ft
15” 1 ft
w
L
q0 336
The cantilever moment:
MU
k 2 5.5 3.13 12 ft in ( ) ( ) 2 ft q o L2 = 359 in − kips = = 2 2
d −a M U = φ AS f y where φ = 0.9 for tension 2 AS f y AS ( 60 ) = = 1.96 AS a= 0.85 f 'C b 0.85 ( 3 )(12 in ) Substituting a into the M U equation above,
d −a M U = ( 0.9 ) AS f y 2
∴
MU d −a = AS ( 0.9 ) f y 2
A 359 in − kip = AS 15 in − 1.96 S 2 ( 0.9 )( 60 ksi ) 15 AS − 0.98 AS 2 − 5.99 = 0 As2 − 15.3 As + 6.11 = 0 AS = 0.41 in 2 per foot of footing The total steel required across the footing is AS = 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,
ρ=
AS 0.41 = = 0.0023 > 0.0018 bd 12 (15 )
ρ min =
3 f c' = 0.0018bh or fy
ρ min =
200 = 0.0018 fy
Therefore, AS = 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 ( AS = 3.18 in2 ) @ 12 inches on center or 5 # 8 bars ( AS = 3.95 in2 ) @ 12 inches on center Check for Development length Ld (ACI-318-08.12.3), and the embedment length of the dowels.
337
338
*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.
M P = 165 kips
P
M = 20 ft-kips DL = 100 kips d
LL = 65 kips
3’-d
15”
T
7 feet
d) Determine the flexural moment for a strip of footing 1 foot wide.
qall = 4 ksf ( with a FS = 1.5) ,
f 'C = 3 ksi ,
f y = 60 ksi .
P = DL + LL
Solution: a) Compute the load eccentricity: e =
M 20 = = 0.121 ft P 165
6 ( 0.121' ) 6 e 165 P q = 1 ± = = 1 ± B 7 ' x7 ' 7' A q max = 3.72 ksf and q min = 3.02 ksf b) Check bearing pressure at ultimate load: Q qall = qmax u P
Qu = 1.2(100) + 1.6(65) = 224 kips
224 = 5.05 ksf = 3.72 165
and
Q qall = qmin u P
224 = 4.1 ksf = 3.02 165
Then consider qall = 5.1 ksf < qall (FS) = 4 (1.5) = 6 ksf, GOOD.
339
c) Calculate the allowable one-way (wide beam) shear; assume a d = 1 ft; q = VU = q (
B c 7 1 − T − ) = 4.6( − 1 − ) = 9.2 kips 2 2 2 2
5.1 + 4.1 = 4.6 2
Equation (1)
The allowable shear = 2φ f c' = 110 psi The required d =
Vu 2ϕ f c' b
=
(9.2)(1000) = 9.3 inches ≈ 12 in assumed (0.75)(110)(12)
Equation (2)
V
3’ - d 4.1 ksf 5.1 ksf
d) Determine the flexural moment for a strip 1 foot wide: x
M =
∫ Vdx = 0
5 .9 x − 0 .1 5 7 x 2 ∫0 2 3
ft − k ip d x = 2 5 .9 ft
340
*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. M N
T
Given: f y = 60 ksi , f 'C = 3 ksi , qa = 4 ksf , DL = 65 kips , LL = 100 kips , M = 20 ft − kips N = DL + LL The actual soil pressure qo (versus the allowable soil bearing capacity): y 7’ c=3.5 My
7’
N Mc qo = ± A I qo = qo
341
165k ( 7 ' x7 ' )
max
x
bd 3 ( 7 ' )( 7 ' ) where I= = = 200 ft 4 12 12 ( 20k . ft )( 3.5 ft ) = ± 200 ft 4
= 3.7 ksf
3
and
qo
min
= 3.0 ksf < 4 ksf allowable
GOOD
My N
x
3.0 ksf
3.7 ksf
b) The ultimate load on the soil: Qu = 1.2 D L + 1.6 LL = 1.2 ( 65 kips ) + 1.6 (100 kips ) = 238 kips M u = 1.6 M LL = 1.6 ( 20 ft .kips ) = 32 ft ⋅ kips
qu = qu
Qu M u c 238 kips ( 32 ft .kips )( 3.5 ft ) ± = ± = A I 200 ft 4 ( 7 ' )( 7 ' )
max
= 5.9 ksf
and
qu
min
= 4.7 ksf
My Q
x
4.7 ksf
5.9 ksf
342
a) Determine the thickness of the footing T.
d T 3” 3” minimum cover Check wide-beam shear: Assume d =12”, (controls for rectangular footing). Q
7’
My
w =12” 12” Critical Section
d
2’
Vu d
∑f
From Vu = 82, 600 lbs , but d =
Vu
φ2
f 'c b
=
= 0 , the ultimate shear is
Vu = φ 2 f 'c bd ,
82, 600 lbs = 10.6 in < 12 in assumed. ( 0.85 )( 2 ) 3000 ( 84 in )
Check for punching shear (controls for square footings, such as this one).
d/2 c = 12” Critical Section ACI 11.11.1.2
b0 /4 7’ 343
y
d Vu = φ 4 f 'c bo d , where bo is the perimeter of the critical section; bo = 4 c + 2 2 ∴
d =
Vu
φ 4 f 'c b
=
( 5.9
ksf ( 7 ft x 7 ft − 2 ft x 2 ft )1000 )
( 0.85 )( 4 ) (
3000
) ( 4 )( 24
in )
= 14.8 in assumed;
must increase d. Recalculate by trying d = 14.0 in. ∴
d =
Vu
φ 4 f 'c b
=
( 5.9
ksf ( 7 ft x 7 ft − 2.17 ft x 2.17 ft )1000 )
( 0.85 )( 4 ) (
3000
) ( 4 )( 26 in )
= 13.5 in
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:
q Bl 2 ( 5.9 ksf Mu = s = 2 a=
As f y 0.85 f 'c b
( where
)( 7 ft )( 3 ft ) 2
b = B)
then
2
= 186 kip . ft As =
0.85 f 'c Ba 0.85 ( 3000 )( 7 ft )( a ) = = 0.3a fy 60, 000
a The ultimate moment Mu is given by: M u = φ As f y d − with Mu = 186 kip-ft , and φ = 0.9 . 2 kip in 2 a ∴ 186 kip . ft = 0.9 ( 0.3 a ft ) 60 144 1.16 − ft = 2 2 in ft 2 a = 0.07 ft and ∴ As = 0.3 a = 0.021 2
Percentage of steel
p=
As A 0.021 ft 2 = s = = 0.00257 > pmin = 0.0018 bd Bd ( 7 ft )(1.167 ft )
344
As
min
= 0.021 ft 2 = 3.024 in 2
Therefore, use 6 # 7 bars at 12 inches.
e) Check development length, Ld (ACI 12.2.2)
0.44 ( 60, 000 ) Ld = 0.04 Ab 3000 Ld = 19.3 in > 18 in Actual Ld
( provided )
= B−
>
fy Ld = 0.04 Ab f' c
>
0.0004d b f y
0.0004 ( 0.75 )( 60, 000 )
c 12in − ( cover ) = 7 ft − 6 in − 3 in = 75 in 2 1 ft
f) Check bearing strength on concrete (ACI 318-02 10.15) A Bearing strength = 0.85φ f 'c A g 2 where A g Ag is the area of the column Then
and
A2 ≤ 2 Ag
A2 is the area of the footing.
A 3 kips 2 0.85φ f 'c Ag 2 = 0.85 ( 0.70 ) (144 in ) ( 2 ) = 514 kips 2 A in g
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 h) Final design
345
AS = 0.20 in 2 ( 4 ) = 0.80 in 2
*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
N (800 kN + 800 kN ) = = 8 m2 kN qa 200 2 m M 800 e= = = 0.5 m ∴ N 1600 B2 =
∴
B = 2.82 m
L ≥ 6 ( 0.5 ) = 3 m
If L = 3 m, try for qmax ≅ 2qavg iterate by trying footing: 2.5 m x 4 m = B x L. q avg = q max =
N 1600 kN = = 160 kPa A 10 m 2 N 6e BL 1 + L
=
1600 10
6 ( 0.5 ) 1 + = 280 kPa 4
Note that qmax exceeds qavg by 33 %, therefore increase the area. Try using 2.75 m x 4.5 m dimensions;
346
q avg =
N 1600 kN = = 130 kPa A 2.75 ( 4.5 ) m 2
q max =
6 ( 0.5 ) = 130 1 + = 217 kPa 6e 4.5 BL 1 + L N
Iterate again, with B = 3.0 m and L = 5.0 m q a vg = q m ax =
N 1 6 0 0 kN = = 1 0 7 kP a A 3 m (5 m ) N 6e BL 1 + L
=
1600 15
6 ( 0 .5 ) 1 + = 1 7 1 kP a 5
The footing dimensions are B = 3 m by L = 5 m.
347
GOOD
*Footings–05: Design the steel for the previous problem. (Revision: Sept-08)
Design the previous footing using f ' c = 21 MPa and f y = 415 MPa .
Solution. 1) Check the ultimate pressures:
N u = 1.2 ( DL ) + 1.6 ( LL ) = 1.2 ( 800 ) + 1.6 ( 800 ) = 2240 kN
e=
M 800 = = 0.5 m N 1600 2240 6 ( 0.5 ) = = 239 kPa 1 + 5 6e 15 A 1 + L N 2240 6 ( 0.5 ) = = = 60 kPa 1 − 5 6e 15 A 1 − L
qmax =
qmin
Nu
q max = 239 kPa and q min = 60 kPa 348
qavg =
2240 kN = 149 kPa 15 m 2
<
qall = 200 kPa Good
2a) Calculate footing depth T based on punching shear, for a Vc = 1.29 MPa
f 'c = 21 MPa , and
.
Using the simplified equation for a square footing: 4T 2 + 2 ( 0.5 m + 0.5 m ) T − T 2 + 0.5T − 0.25 = 0
∴
(15
x 149 ) =0 2240 T ≅ 0.50 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 ) :
dv = qdx 40.2 x 2 V = ∫ qdx = ∫ ( 266 − 40.2 x ) dx = 266 x − 2 0 0 x
x
2.25 − T
0
V = 598 − 266T − 20.1 ( 2.25 − T ) = 2
V =
Vc 1290 = 2T 2T
∴
T 2 + 40.8T − 24.7 = 0
T = 0.60 m
We will use the highest of (2a) or (2b), therefore T = 0.60 m.
349
a=
3) a) Find AS ( Longitudinal ):
Mu =
2.25
2.25
0
0
∫ Vdx = ∫
As f y 0.85 f 'c b
40.2 x 2 266 x − 2
a M u = φ As f y d − 2
∴
=
415 As = 23.3 As 0.85 ( 21)(1) 2.25
266 x 2 40.2 x 3 dx = 2 − 6 0
= 597 kN .m
a Mu As d − = 2 φ fy
23.3 597 As 0.6 − = As 2 − 0.0515 As + 1.37 x10 − 2 = 0 = 2 0.9 415 ( ) As = 28.2 Check
cm 2 m
p=
As 0.00282 = 0.0047 > 0.002 GOOD 1 ( 0.6 ) bd
3) b) Find AS (transverse): Use the high average.
q=
( q ave + q max ) = (149 + 266 ) = 20.37
kPa
2 2 ( 3 − 0.5 ) ( B − 0.5 ) 215 q 2 2 wl 2 = = 168 kN .m Mu = = 2 2 2
a M and As d − = u 2 φ fy
Check for
∴
ρ=
∴
As 2 − 0.0515 As + 0.000386 = 0 ∴
As = 7.61
cm 2 m
As 7.61 x 10 −4 m 2 = = 0.00126 < 0.002 1 ( 0.6 ) bT
Use minimum
As = 0.002 ( b )(T ) = 0.002 (1)( 0.6 ) = 0.0012
m2 cm 2 = 12 m m
4) Check footing: Longitudinal steel:
28.2
cm 2 1m 2 (3 m) = 84.6 cm2 = 8 .4 6 x 1 0 − 3 m 2 m 1 0, 0 0 0 c m
Therefore use 17 # 25 mm bars at 17.6 cms o − c
350
Sketch:
351
*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:
Roof
Precast concrete wall Overhead Crane
12 in Floor Slab 4 ft
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. Therefore, d = 12” - 3” - 0.5” / 2 = 8.75”
Assume # 4 bars.
12 in T=12 in
B
352
b =12 in a T
d
B 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 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 ).
qo = 2.1 ksf
B = 3 ft
Vu = (12 − d ) qo (1) 12”
b =12” in
12”
12 in - d
d Vu
T
qo = 2.1 ksf 353
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: Vu ≤ φ Vc = φ 2 f' c bd
Vu = 0.85 ⋅ 2 ⋅ 3000 ⋅ 12in ⋅ 8.75in = 9.8 kips / ft of wall >> 0.57 kips / ft 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,
Vu = 0.85 ⋅ 2 ⋅ 3000 ⋅ 12in ⋅ 6.5in = 7.3 kips / ft of wall → GOOD
Therefore, the total thickness T = d + bar diameter + 3” = 10” + 1” + 3” = 14” 3) Design the flexural reinforcement, As ( ACI 15.4): M u = qo ℓ 2 / 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 As2 – 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 ) 354
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 use 1 # 4 every ft. of wall ( As = 0.20 in2 ) 4) Check the development length, Ld ( ACI 12.2 ):
Ld = 0.04 Ab
fy f' c
(But not less than 0.0004 d b f y ) = 0.04 ⋅ 0.20 ⋅
60,000 = 8.8" 3000
Ld = 8.8” or 12” 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.) 355
Therefore,
B = 3.5 ft T = 13” As = 1 # 4 @ 12” along the wall
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 )
356
**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 32'
20'
P = 10 k 24" STEEL COLUMN 40 ' HIGH WITH 1" THICK WALLS
20' Z Y X
D B
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
F = qz Gh Cf Af
V = 146 mph Gh = 1.26 Cf = 1.2 The sign shape factor is Therefore:
357
F=
M 32 = = 1.6 N 20
(34 psf) (1.26) (1.2) (32 ft x 20 ft) = 32.4 kips 1000
∴ qz = 52 psf
.
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) = 0.5 -
1-H (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
358
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 qmax,min = ex =
- 6P e ( (3x10x50x0.150) ) = 240 kips P - 6P e x + 2 y + where P = 15 kips + 2 BL FTG WT BL BL
My
P Mx ey = P
972 = 4 .0 5 ' 2 40 150 = = 0.6 25 ' 24 0
=
240kips - 6(240)(0.625) - 6(240)(4.05) qmax,min = + + = 0.89 ksf < 2.5 ksf GOOD 500 102 x 50 10 x 502 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
336 6 x 336 x 0.625 6 x 336 x 4.9 ∴ Q MAX,MIN = ± ± = 0.989 ksf, 0.019 ksf 2 10 2 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. 359
#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
R=
AS 0.83 = = 0.0022 > 0.0018 OK BD 12 x 32
USE # 7 @ 8” O/C,
Therefore;
As = 0.90 in2 OK
1.64 132 2 M U = ( 0.9 )( 60 ) = 126 kip-ft > 116 kip-ft GOOD 12
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.
360
Chapter 19 Combined Footings Symbols for Combined Footings
361
**Combined Footing-01: Rectangular footing with an eccentric column. (Revision: Sept-08)
Find a suitable rectangular combined footing for the conditions shown below. Given
f 'c = 3 ksi
f y = 60 ksi
qa = 2.5 ksf (qult = 10 ksf) and use ACI 318-05.
Column #1
Column #2
12 in x 12 in
16 in x 16 in = 1.33 ft x 1.33 ft
4 # 7 bars
6 # 8 bars
DL = 80 kips
DL = 120 kips
LL = 60 kips
LL = 110 kips
140 kips
230 kips
Edge of column #1 is at the Property Line (PL); spacing between columns is 20 ft on-center.
Solution: STEP 1: Convert column loads to ultimate; then, convert qall to qult.
Ratio:
P1 = 140
P1u = 1.2 ( 80 ) + 1.6 ( 60 )
P1 = 230
P2 u = 1.2 (120 ) + 1.6 (110 ) = 320 kips
∑ = 370
= 512 kips
U ltimate U 512 = = = 1.38 Service S 370
qult =
= 192 kips
U 512 = 1.38 ( 2.5 ) = 3.45 ksf ( q all ) = S 370 362
STEP 2:
∑M
o
Determine footing dimensions L and B
=0
R x = P1 x1 + P2 x2 where the moments are taken about the left corner. 512 x = 192 ( 0.5 ) + 320 ( 20.5 ) x = 12.98 ft ,
say : 13 ft
Thus , L = 2 x = 26 ft
The rounding to 26 ft will prevent complete closure of the moment diagram, but negligibly. B=
( P1u + P2 u ) = Lq u
512 = 5.70 ft 26 ( 3.45 )
Use 5 ft 9 in. For construction; B = 5.75 feet.
q ' = 3.45 ( 5.70 ) = 19.66
kip ft
STEP 3: Draw shear (V) and moment (M) diagrams. The column loads are treated as concentrated loads acting at the centers of the columns. The shear and moments diagrams are shown below.
363
STEP 4: Determine the footing depth T based on one-way (wide-beam) and two-way (punching) shear, as shown in figure below. Note that the punching analysis reflects on a three-sided section for column 1 and four-sided section for column 2. We shall first determine d via a wide-beam analysis, and then check for punching shear. From the shear diagram, the maximum shear is near column 2 at a distance d from the face of column 2.
V = 204.4 − 19.66 ( d ) , Also, from vc =
V , Bd
Bdvc = 204.44 − 19.66 ( d )
ACI 318-05 φ = 0.75 In our case, Hence,
vc = 2φ
f 'c = 2 ( 0.75 ) 3000 = 82.16 psi = 11.8 ksf ,
( 5.73 )(13.4 ) d = 204.40 − 19.66 d . Solving d
= 2.11 ft = 25.30 in
Using d = 25.30 in we check diagonal tension at column 1. V=column load less upward soil pressure = 192 − Aqu = 219 − A ( 3.45 )
d A = 12 + (12 + d ) = (12 + 12.6 )(12 + 25.3 ) = 917.5 in 2 = 6.37 ft 2 2 d Hence, V= 192 - (6.37)(3.45) =167.5 kips, Perimeter = 2 12 + + (12 + d ) = 86.50 in = 7.2 ft 2
vc =
167.5 = 11.02 ksi ( 7.2 )( 2.11)
At column 2
( 41.3) A= 144
<
4φ
2
= 11.84 ft 2 and
f 'c = 23.7 ksf OK
( 41.3 ) Perimeter = 4 = 13.76 ft 12
V = 320 − 11.84 ( 3.45 ) = 279.20 kips vc =
305.76 = 9.61 ksi (13.76 )( 2.11)
<
4φ f 'c
=
26.8 ksf
OK
364
STEP 5: Determine the longitudinal reinforcement steel, As:
For f y = 60 ksi , f c' = 3 ksi , M u = 842 k − ft , B = 5.75 ft = 69 in ; d = 25.3 in R=
M u (12, 000) (842)(12, 000) = = 0.0763 ' 2 f c bd (3000)(69)(25.3) 2
From this value w ≈ 0.08
ρ =
wf c' = 0.0039 > 0.0033 < ρ max fy
GOOD
As = ρ Bd = (0.0039)(69)(25.3) = 6.81 in 2 Use 11 # 8 bars (As = 8.63 in2) at 6 in o.c. across top of footing (at approximately 4 in from each Side). Provided that 1/3 of the bars extend the full length of the footing, the bars could be cut off as dictated by the moment requirements (moment diagram). However, the saving is not worth the effort (engineering, fabrication, placing, etc.). Thus, typically, all bars will run the full length of the footing. Based on
200 , As fy
min
= ( 0.0033 )( 25.3 )(12 ) = 1.00 in 2 . For the positive moment,
+As = 1.07 x 5.25 = 5.75 in2. Use 8 # 8 bars; As = 6.28 in2 > 5.75 in2; this is larger by the required positive moment. Place these bars at 8.5 in. c.c. and say 4 in. from bottom. Also, based on the moment diagram, running 1/3 of As (three bars) the full length of the footing satisfies both code and moment requirements. The other five bars could be cut off at, say, half length and placed on the right half (column 2 side) of the footing.
STEP 6: Determine As in the short direction. Reference is made to the figure below. The widths s1 = 12 + 0.75(25.3) = 31.00 in. = 2.58 ft and s2 = 16 + 1.5(25.3) = 54.0 in. = 4.5 ft.
s1
B
s2
1
2 l1
2.58 ft 365
l2
15.77 ft
4.50 ft
3.15 ft
Column #1
P 192 = = 12.94 ksf B s1 ( 5.75 )( 2.58 )
q'= l1 =
( 5.68 − 1 ) 2.37 ft 2
2 12 M 1 = (12.94 )( 2.37 ) = 436.0 k − in 2
Placing the transverse steel above the positive longitudinal steel, d = (25.92 – 1) = 25.92 in
As ( 25.92 − 0.98 As ) =
Thus,
461.92 ( 0.9 )( 60 )
Solving
0.334 in 2 0.895 in 2 = ft s1
width.
As = 2.75 in 2 (for 0.0033)
Based on pmin
s1
s2
1
B
As =
2 l1
l2
2.68 ft
4.70 ft
15.47 ft
3.15 ft
See the figure above to determine transverse steel reinforcement for the footing. As = 3.00 in 2
Thus, use 5 # 7 bars, at 8 inches c.c. ld =
0.04 ( 0.60 )( 60, 000 ) 3000
= 26.3 in
ld
available
>
2.75 in 2
= 2.34 ( 2 ) = 28 in
For column 2: Transverse + As . s2 = 4.70 ft ;
q' =
355 = 13.30 psf ( 4.70 )( 5.68 )
l2 =
( 5.75 − 1.33) = 2.21 ft
M2
(13.30 )( 2.21) (12 ) = 389.7 =
2
2
2
k − in
Using d = 25.92 in (transverse bars on top of longitudinal)
366
As ( 25.92 − 0.92 As ) =
375.71 = 4.82 in 2 Based on pmin 0.9 ( 60 )
Use nine # 7 bars at 6.5 in. c.c.
As =
4.82 in 2 width. s2
As = 5.40 in 2
Development length is same as above. Based on pmin
As in 15.47 ft − sec tion = 15.87 in 2 , Thus use 27 # 7 bars at 6.75 in. c.c.
STEP 7: Evaluate dowel steel. Column 1 Allowable
f c = 0.85 ( 0.65 ) f 'c = 1.65 ksi (ACI 10.15) P = 12 x 12 x 1.65 = 238 > 192 kips (bearing capacity of the column)
Thus, dowel size, should be one size smaller than column bars; use 4-#6 As = 1.76 in 2 Column 2: In our case, f c' = ( 0.85)( 0.65)(16 )(16 )
3000 = 424 ksi > 320 ksi GOOD 1000
Use four # 6 bars; As = 1.76 in 2 > (16 )(16 )( 0.005) = 1.28 in 2
STEP 8: Make a drawing showing design details.
367
GOOD
An alternative design is of course, a strap footing,
Typical configuration of a strap footing.
368
**Combined Footing-02: Combined footing on the property-line. (Revision: Sept-08)
Design the combined rectangular footing shown below using the assumptions given.
f 'c = 3.5 ksi; f y = 50 ksi; soil qa = 5 ksf , SF=3, qultimate = 15 ksf ;
Solution: STEP 1: Convert column loads to ultimate; convert qa to qultimate . P1 = 255k
P1 ult = 1.4 (80k ) + 1.7 (175k ) = 410k
P2 = 330k
P2 ult = 1.4 (130k ) + 1.7 (200k ) = 522k
∑ = 585
= 932
369
k
STEP 2: Determine the footing dimensions L and B: R x = P1 x1 + P2 x2 or
932k x = 410k (0.75') + 522 k (18'+ 0.75')
x = 10.8 feet
11 feet and L = 2 x = 2 (11 ft ) = 22 feet
say
Note: Rounding the 10.8’ to 11’ will prevent the closure of the moment diagram, but thi is acceptable with hand calculations. A computer program is better.
B=
P1ult + P2ult Lqo
410k + 522 k = = 5.31 feet B = 5'− 4" in construction (22 ')(7.97 ksf )
Finally q '(soil reaction) = qo B = 7.97
k ft 2
x 5.31 ft = 42.3 k ft along L.
STEP 3: Draw the shear V and moment M diagrams.
370
STEP 4: Determine the footing depth D. Shear at a distance d from the face of column 2, V = 341k − (42.3k ft ) (d ) But also the shear stress Vc in concrete is Vc =
V ∴ BdVc = 341k − (42.3d ) k Bd
But Vc = 2∅ f 'c = 2 (0.85) 3500 = 100.6 psi = 14.5 ksf (for wide beam shear) ∴ (5.31 ft ) d (14.5ksf ) = 341k − 42.3d ∴ d = 2.86 ft = 34.3 inches Using d = 34.3” check diagonal tension at column # 1
V = Column load less upward soil pressure V = 410 k − Aqult = 410k − A(7.97) d
but A = (18"+ ) (18"+ d ) = (18"+ 17.1") (18"+ 34.3") 2
A = 1836 in 2 = 12.75 ft 2 Hence V = 410 k − (12.75 ft 2 ) (7.97 ksf ) = 308.4k
d Perimeter = 2 (18"+ ) + (18"+ d ) = 2 (35.1") (52.3") 2 = 122.5" = 10.2 ft ∴
Vc =
for diagonal V 308.4k = = 10.6 ksf < 4φ f 'c tension shear ( perim) d (10.2 ') (2.86 ')
4 (0.85) 3500 201 psi 29 ksf
OK
Checking around column # 2 Area A = (24"+ d ) 2 = 3400in 2 = 23.6 ft 2 Perimeter = 4 (24"+ d ) = 4 (58.3") = 233.2" = 19.4 ft Shear V = 522 - (23.6 ft ) (7.97 ksf ) = 334 k k
∴
371
Vc =
2
V 334k = = 6.0 ksf < 4∅ f 'c ( perim) d (19.4 ') (2.86 ') 29 ksf
OK
STEP 5: Determine the reinforcing steel As (longitudinal).
a M As (d − ) = but 2 ∅f y B ∴
a=
As f y 0.85 f 'c
=
As (50 ksi ) 1 = 1.40 As 0.85 (3.5 ksi ) 12"
As (34.3"- 0.70 As ) = [(1646 k - ft ) (12 in ft )] [(0.9) (50) (5.31')]
As = 2.5 in 2 ft
width
for balanced steel As 2.5in 2 200 ρ= = = 0.006 > db (34.3") (12") fy 200 50000 > 0.004 >
OK
also < ρ max = 0.016 AS = 2.5 in 2 x B
in 2 (5.31 ft ) = 13.3 in 2 of steel ft Use 11 #10 bars (1.27 in 2 each x 11 = 14 in 2 ) at 6” spacing center-to-center. total = 2.5
372
**Combined Footing-03: Design a combined footing. (Revised Nov-09)
Design the combined footing shown below using (ACI 318-02), given f’c = 3 ksi and fy = 60 ksi, DL = 140 k
DL = 180 k
LL = 90 k
LL = 110 k
Solution: Compute the footing dimensions:
P1 = 140k + 90k = 230k P2 = 180k + 110k = 290k The location of the resultant is,
X =
373
P2 L c − c ( 2 9 0 ) (1 6 ) = = 8 .9 2 fe e t P1 + P2 290 + 230
The footing length L is, L = 2(
L c1 13 + x) = 2( + 8 .9 2 ) = 1 8 . 9 3 ' 2 12
The footing area is,
A=
∴ u s e L = 1 9 fe e t
P1 + P2 230 k + 290 k = = 148.6 ft 2 qall 3.5 ksf
A 148.6 ft 2 The footing width is, B = = = 7.82 feet; use 8 feet 19 ft L
Therefore footing size is 8 feet x 19 feet. 2) Convert the allowable soil pressure, qall to an equivalent ultimate load value, qu : Under column #1, P1 = 1.2 DL + 1.6 LL = 1.2(140 K ) + 1.6(90 K ) = 312 K Under column #2, P2 = 1.2 DL + 1.6 LL = 1.2(180 K ) + 1.6(110 K ) = 392 K Overall factor = ( P1 + P2 ) ∑ N
=
(312 K + 392 K ) 520 K = 1.35
That is, P = 1.35( DL + LL) ∴ qult = (3.5 ksf )(1.35) = 4.725 ksf 3) Find the shears and moments along the footing, and draw their diagram.
∑P
u
= Pu1 + Pu 2 = 312 kips + 392 kips = 704 kips
The load per foot of the beam (w) =
Check:
Vu 704 kips kips = = 37.05 L ft 19 ft
qult ( B ) = (4.725)(8 ft ) = 37.8 k / ft > 37.05 k / ft
GOOD
w = 37.05 k/ft
1.0833 ft x
374
Note: These two values would be exact if we corrected qult to 5.28 ksf. Integrating from x = 0 to x = 1.083 ft q = 37.8 −
V=∫
1.083
0
312 k kip = −250.3 1.083 ft ft
qdx = 250.3 x + c0 , where c0 = 0 since V = 0 at X = 0
∴ V = ( −250.3)(1.083) = −271 kips a very minor difference 250.3 X 2 + c1 , where c1 = 0, since M = 0 at X = 0 2 250.3 2 ∴M = (1.083) = −146.7 kip-ft 2 M = ∫ Vdx =
Integrating from 1.083' < Χ ≤ 15.9 ' q=
37.05 kips ft
V = ∫ qdx = 37.05 − 312 M = ∫ Vdx = 37.05( x 2 2) − 312( x − 0.54)
The maximum moment, and its location is found by setting the shear equal to zero in terms: V = 0 ∴ V = 37.5 x − 312 = 0 ∴ x = 8.42 feet ∴ M max =(37.05)[
(8.42) 2 ] − (312)(8.42 − 0.542) = −1145 kip − ft 2
Similarly, integrate from 15.9 ft < x ≤ 17.17 ft , and from 17.17 ft < x ≤ 18.93 ft
375
The resulting shear and moment diagrams are shown below.
376
Find the depth d for wide beam shear, at the location of the largest shear (that is, at the interior of column #2, V max = 277.5 kips). The allowable concrete stress VC is given by ACI 318-02- 114.1, vC = 2∅ f c ' = (2)(0.75) 3000 = 82.16 psi = 11.83 ksf where φ = 0.75 Vbeam = Vmax − wd = 277.5 − 37.05d Vconcrete = BdvC = 8d (11.83 ksf ) = 94.64d
Equating both: Vbeam =Vconcrete ∴ 277.5=94.64d + 37.05d ∴ d = 2.10 feet = 25.3 inches; therefore h = d + 3 + bar diameter = 25.3 + 3 + 1 = 29.3 use 30 inches
Check the punching shear: vc = 4∅ f c' = (4)(0.75) 3000 = 164.3 psi = 23.66 ksf At column #1, the perimeter p in shear is,
2.1 ) + (1.08 + 2.1) = 7.5 feet 2 = pdVc = 7.5(2.1)(23.66) = 373 kips provided > 349 kip needed = P1 GOOD
perimeter p = 2(1.08 + Vshear capacity
Similarly, at column #2, the perimeter in shear is,
perimeter p = 4(1.25 + 2.1) = 13.4 feet Vshear capacity concrete
377
= (13.4)(2.1)(23.66) = 666 kips provided > 392 kips needed = Pu 2
GOOD
5) Compute the area As of steel required. Longitudinal steel will be required both at the top of the footing between columns (positive moment steel), and at the bottom in the cantilevered part. Also required is transverse steel perpendicular to longitudinal axis, plus shrinkage and temperature steel. Negative moment (flexural steel): M u (max) = −1082 kip − ft for a footing width of B = 8 ft = 96 inches and d = 25.3 inches R=
M u (12, 000) (1082)(12,000) = = 0.07 which gives w ≃ 0.008 ' 2 f c bd (3000)(96)(25.3)2
wf c' (0.008)(3) ρ= = = 0.0004 < ρ min (0.0033) therefore use ρ min = 0.0033 fy (60) As = ρ min bd = (0.0033)(96)(25.3) = 8.01 in 2 Use 11#8 bars at 6" o − c (this provides 11x 0.79 = 8.69 in 2 > 8.01 in 2 required ) For temperature steel As = 0.002 Bh = 0.002(96)(30) = 5.76 in 2 These top bars must be hooked (see ACI 12.1, 2) at column #1’s end; remember to use 3”cover. Positive steel (at column #2)
MU =
83.9 kip − ft = 10.5 kip -ft ft 8 ft
which requires As = 0.09 in 2 per foot
This area of steel is obviously too low, therefore use same as above 11#8, etc. Design for shrinkage steel requirements and spacing of bars (ACI 7.4.3). Transverse steel (short direction)
For column #1 (a=13” wide):
q u lt =
Pu 1 + Pu 2 312 + 392 = = 4 .6 3 k s f BL (8 )(1 9 )
25.3 d a + = 13 + = 25.65 in = 2.14 ft 2 2
d for steel = 25.3 + 0.56 –1.69 (for #9 bars)
∴ d = 24.2 inches
The length of cantilever Lc (to calculate bending moment) is, Lc =
(8
ft − 1.083 ft ) = 3.46 ft 2
and ∴ M =
qL2 4.63 (3.46) 2 = = 27.7 ft -kips 2 2
The required area of steel As ,
378
R=
M (12, 000) (27.7)(12, 000) = = 0.0018 which gives a w ≃ 0.003 ' 2 f c bd (3000)(96)(25.3) 2
ρ=
wf c' (0.003)(3) = = 0.00015 << 0.002 of ρ min fy (60)
therefore use As min
As min = ρ min bd = (0.002)(96)(25.3) = 4.85 in 2 Use # 6 bars at 12 inches on center both top and bottom. 7) Sketch of the completed design:
379
**Combined Footing-04: Optimization via a trapezoidal footing. (Revised Nov-09)
Design a trapezoid-shaped footing, with columns at ends of footing as shown below, where f’c = 21 MPa, fy = 414 MPa, qall = 120 kPa and L = 7 m.
7m
a
b
1
2
Column #1 (0.3 m x 0.3 m)
Column #2 (0.4 m x 0.4 m)
4 # 25-mm bars
6 # 25-mm bars
DL = 355 kN
DL = 535 kN
LL = 265 kN
LL = 490 kN
Solution: Step 1. Convert P1 and P2 to ultimate loads P1u and P2u, and qall to qult:
∑ Ratio =
P1 = 620
P1u=1.4(355) + 1.7(625) = 947
P2 = 1025
P2u=1.4(535) + 1.7(490) = 1582
= 1645
= 2529 kN
ultimate.loads 2529 = = 1.54 actual.loads 1645
qult = qall x 1.54 = 184.5 kPa Step 2. Determine a and b (the two footing widths):
∑M
O
=0
x=
[(947 )(0.15) + (1582)(6.8)] = 4.31 m
A=
(a + b )L = (a + b )7 = 3.5(a + b )
2529
For the trapezoid,
Also, A =
2
2
∑ P = 2529 = 13.7 m qu
184.5
380
Thus,
13.71 = 3.5(a + b)
From Simplifying,
or (a + b) = 3.92 m
L (a + 2b ) x = 3 (a + b )
or
7 3 (a + 2b ) 4.31 = (a + b )
0.8471a – 0.1529b = 0
Simultaneous solution of Equations 1 and 2 yields, b = 3.32 m;
a = 0.60 m
Check via substituting in the expression for A or x ; for A, 3.5(3.32 + 0.6) = 13.71;
OK
Step 3. Draw the shear V and moment M diagrams on the next sheet.
381
(1)
(2)
Step 4. Determine d. First, we’ll evaluate wide-beam shear at the narrow end, and then
check for diagonal tension at column 2. From the shear equation, V = 121(x − 0.15) + 35.8(x − 0.15) − 930 2
At d from the inner face of column 1 , x = 0.3 + d. Hence, V = 121(d + 0.15) + 35.8(d + 0.15) − 930 2
V = 35.8d 2 + 131.74d − 911 (3)
Or
vc = 660 KPa =
Also,
V bd
b at a distance d from inner face of column 1 is b = 0 .6 +
(d + 0.3)(3.32 − 0.6) = 0.3885d + 0.716 7
660 =
Thus,
V V = (0.3885d + 0.716)d 0.3885d 2 + 0.716d
(
V = 256.4d 2 + 472.56d
or
Equating equations #3 and #4,
)
(4)
(
)
256.4d 2 + 472.56d = 35.8d 2 + 131.74d − 911
292.2d 2 + 604.3d − 911 = 0 which yields, d = 1.01 m
or
Check vc ; at x = 1.01 + 0.3 = 1.31 m b = 0.392 + 0.716 = 1.108 V = 121(1.16 ) + 35.8(1.16 ) − 930 = −741.5 2
vc =
741.5 = 662 which is close to 660 kPa (1.108)(1.01)
OK
By inspection, required d at the large end is less than 1.01 m. Step 5. Determine the flexural steel area based on moments at several points,
a M Use As d − = 2 0 .9 f y b a=
As f y 0.85 f ' c b
=
400 As = 22.4 As 0.85(21)(1)
at x = 1.5 m, we have 3 22.4 As 1115 As1.01 − = 2.65 × 10 − = 2 (0.9 )(400 )(1000(1.18))
Or
As − 0.09018 As + 0.000234 = 0 2
382
7m
3.32
0.6
2.5 3.5 3.32 Possible Arrangement of As Solving As = 0.0036m 2 = 36
cm 2 width m
Checking percent steel,
ρ=
As 0.0036 = = 0.00356 > ρ min ; OK < ρmax bd (1)(1.01)
Thus total As at x = 1.5m is ( As )(b ) = (36)(1.18) = 42 cm2/m Similarly for points at x = 2.5 m; 3.84 m; 5m. The results are summarized below As cm2 x (m)
Width b
M
Per 1 m
For Width
min.*
No. of
(m)
Kn-m
Width
b
1.5
1.18
1115
36
42
42
8
2.5
1.57
1695
31
48.5
48.5
10
3.84
2.09
2005 max
27.3
57
57
12
5
2.54
1725
19
48.5
51
1
25-mm Bars
* As based on ρmin is probably debatable, particularly near the wide end, since b varies so greatly. Thus using 12-25-mm bars are deemed as satisfactory (no cutoffs) for the 4 m on “right” side. Step 6. Ast in the transverse direction. The effective “transverse beam width “ s2 can be
approximated as before.
s 2 = 0.4 + 0.75(d ) = 1.16m 383
3.095 L2 = − 0.4 = 1.147 2 qL2 P M = = A 2 strip
s1
(1.147 )2 , etc. 2
s2
7m
2.87 m
3.32 m
As’ ≤ As, based on ρmin
By inspection , Thus, Asmin = (5-25-mm bars)
Asmin = (0.002)(1.01) = 20 cm2 / m Practically, its advisable to space these bars at about 0.25 m even spacing for the entire 7 m length, with varying length, for approximately 3 m near the wide end, to about 0.4 m near the narrow end. These bars would be tied to the +As at about 0.2 m from the bottom of excavation. Other configurations of placement are, of course, possible. Step 7 Four dowels, 25-mm bars are recommended for each column (see Example 7.1) Step 8. Detail the drawing. The following figure shows a suitable layout. Because of the irregular shape, some practical improvisation is frequently a practical necessity.
384
4 25-mm bars
4 25-mm bars
12 25-mm bars placed as shown
●
● ●
●
●
25-mm bars at 0.25 m c.c. (varied lengths) ●
●
●
●
●
●
●
●
●
●
● ●
●
●
●
Fig. 7.11 Illustration of final design details for footing in Example 7.2
385
**Combined Footing-05: Combined footing. (Revised Nov-09)
Design a combined footing for the following conditions:
f ’c = 3,750 psi, fy = 20,000 psi and qall = 8.5 ksf. Dead and Live Loads DL = 1050k
+
DL = 1210k
=
1375k
LL = 325
+
LL = 615
=
1825k
Total load = 1375k +
TL= 1825k
=
3200k
Solution:
= 1825 * 24’ = 13.5’ + 2.5 =
Locate centroid: x
3200
16.0’ Length: 2 * 16.0’ = 32.0’ Clearance: 32.0’ – 26.5’ = 5.5’ 3200 32 x 8.5 = 11.8 ≈ 12.0’ Width = Combined Load:
Length x qallow
Required depth: d =
6018 0.298x12
= 42”
(3” added for cover + 1” margin for error) d = 46” V=
1175000 = 175 psi 12(12) 7 * (54) 8
Longitudinal Steel: Bottom steel + As =
Top steel
6018 = 90.85 in2 (1.44)(54)
ΣO =
1175000 = 104.26 in2. 280( 7 ) * 54 8
- As =
1010 = 12.3 in2 1.44(57) 386
Combined Footing Transverse steel:
M = 5’ x 32’ x 8.33 ksf x 2.5 = 3330 kip-ft. As =
3330 = 41.6 in2 1.44(55)
x
1825 = 23.8 in2 3200 1375 3200
ΣO =
(5)(32)(8330) = 100 in 280( 7 ) * 55 8
Use 15- #11 bars in left column at 5.5” c/c. 12- #11 bars in right column at 10” c/c. Web reinforcing:
Vc = 0.075(12)(12) ( 78 ) 55 V’ = V – Vc = 1175 – 515 = 660 k 6 sets of #6 stirrups
Av = 5.28 in2
5.28(16)( 7 ) * 55 Avfujd 8 Spacing = s = = = 6 in. V' 660
387
For 12” spacing:
V’ = 330 + 515 = 845 kips
For 24” spacing:
V’ = 165 + 515 = 680 kips
=
515
k
**Combined Footing–06: Strap footing (or grade beam) between two columns. (Revision: Sept-08)
Design a strap-footing given f’c = 3.5 ksi, fy = 60 ksi, and qult = 3.85 ksf. Column #1
Column #2
12 in x 12 in with 4 # 7 bars
16 in x 16 in = 1.33 ft x 1.33 ft with 6 # 8 bars
DL = 80 kips
DL = 120 kips
LL = 60 kips
LL = 110 kips
Solution: P1 = 1.2 DL + 1.6 LL = 1.2( 80 kips) + 1.6( 60 kips) = 192 kips. P2 = 1.2 DL + 1.6 LL = 1.2(120 kips) + 1.6(110 kips) = 320 kips. The left edge of column #1 is placed at the property line (P.L.).
P1 = 140 kips and P1u = 192 kips P2 = 230 kips and P2u = 320 kips
∑ →= 370 kips
→= 512 kips
Ratio = 512 / 370 = 1.38
and qult = 3.85 ksf
Try e = 3 feet. From Σ M2 =0 , we have (192)(25) – R1(22) = 0, From
Σ M1 = 0, 3220(22) – (192)(3) – R2(22) = 0,
From
Σ Fy = 0,
R1 = 218.2 kips R2 = 293.8 kips
218.2 + 293.8 = 512 kips. 388
Step 2: L1 = 2(0.5 + e) = 2(0.5 + 3) = 7 feet B1 = R1 / (L1)(qult) = 218.2 / (7)(3.85) = 8 feet L1 = 0.875 B1 The ratio appears reasonable. If a specific clearance between footings is required, then adjustment of L1 and L2 would be in order. If we select to make footing # 2 a square, B=L=
R2 293.8 = = 8.73 feet qu 3.85
Step 3: Draw the shear and moment diagrams.
389
Round out to 9 feet.
Step 4: Design the strap using the figure above and V = 23.6 kips and M = 493 k-ft. While a deeper strap is more efficient than a wider but shallower one, a too narrow strap may pose some practical or functional concerns (for example, limited space for reinforcement, lateral or torsional stiffness, etc). The recommended range for strap footing widths b is, L' 2 fe e t ≤ b ≥ fe e t 10
In addition, the effective depth d, should be 50 to 100 % larger than theoretically needed, in order to reduce the number of reinforcing bars, provide spacing and minimize the effects of an uneven excavation and perhaps even eliminate the need for stirrups. Let L’ = 14 feet, the distance between footings. In this problem L’/10 = 1.4 ft; hence use b = 2 ft. V = vc bd
where
vc = 2φ f 'c = (2)(0.75) 3500 = 88.7 psi = 12.8 ksf
Thus, from 23.60 = (12.7 ksf)(2 ft) d,
d = 0.92 feet. From experience use d = 24 in.
With M u = 493 kpi − ft from the moment diagram, R=
M u (12, 000) (493)(12, 000) = = 0.122 which gives w = 0.149 ' 2 f c bd (3500)(24)(24) 2
ρ=
wf c' (0.149)(3.5) = = 0.0087 therefore fy (60)
As = ρ bd = (0.0087)(24)(24) = 5.0 in 2
Use 7 # 8 bars ( which provide As = 5.5 in 2 ) at 3.25 inches o − c, extended to the footings. Step 5. Design of the footings using the figure below. In footing #1, for b = 8 feet and Determine the section thickness d for the strap. With vc = 12.7 ksf, Solving,
(8 ft)(12.70 ksf)d = 161.2 – 30.8 d (see the shear diagram).
d = 1.22 feet = 14.6 inches.
390
Check diagonal tension with V = 192 kips
1+0.75d =23 in 3.83 ft 40 in
8 ft
#1
#2
12 in + d = 28
12 in + d/2=20
7.0
14.0
9.0
d A= 12+ (12+d ) = (12 + 7.3)(12+14.6 ) = 513.4 in 2 = 3.56 ft 2 2
where
Perimeter = 2 (12 + 7.3) + (12 + 14.6) = 65 in = 5.43 feet Hence, V = 192 - 18.73 = 173.3 kips and vc =
V 173.3 = = 26.14 > 4φ f 'c =(4)(0.75) 3500 = 25.55 ksf pd (5.43)(1.22)
Not Good
Therefore, increase d to 16 inches. Therefore, A = 2 (20)(28) = 1120 in2 = 7.78 ft2 Perimeter = 2 (12 + 8) + (12 + 16) = 68 in = 5.67 feet Hence, V = 192 – (7.78)(3.85) = 184 kips and
vc =
V 184 = = 24.4 < 25.55 ksf GOOD pd (5.43)(1.33)
The transverse steel. In the transverse direction, a disproportionate percentage of the column load P1 is carried by the left side of the footing #1. Hence, concentrate more of the transverse steel in that segment of the footing (as done for similar cases of rectangular and trapezoidal footings). On the other hand, footing #2 resembles a spread footing condition, and therefore will be treated that way. Thus, for footing #1, assume the shaded width s, as 12 in. + 0.75d = 2 feet, as shown in the figure.
391
9 ft
q=
192 = 10.66 ksf the critical section for flexure at the face of the column. (2)(8)
10.66(3.5) 2 = 65.3 k-ft 2 (65.3)(12, 000) M (12, 000) R= = = 0.0091 which gives w = 0.011 ' 2 (3500)(8 x12)(16)2 f c bd
M=
ρ=
wf c' (0.011)(3.5) = = 0.00064 < 0.0033 = ρ min fy (60)
therefore use ρ min
As = ρ min bd = (0.0033)(96)(16) = 5.0 in 2 Since the reinforcing from the strap is the same, add 3 rebars (#8s) at each side of the footing as shown in the final sketch of the footings with the strap. Footing #2. When comparing column loads, the d = 1 ft appears low for footing #2 when d = 16 inches was needed for footing 1. Check diagonal tension using d = 16 in. similar to footing # 1. A = (16 + 16)2 = 1024 in2 = 7.11 ft2 Perimeter = 4 (16 +16) = 128 inches = 10.67 feet
V = 320 – 7.11(3.85) = 292.6 kips vc =
V (297.6 kip ) = = 20.6 ksf < 4φ f c'=(4)(0.75) 3500=25.55 ksf pd (10.67 ft )(1.33 ft )
Good
Use d = 16 inches = 1.33 feet for both footings. As = ρ min bd = (0.0033)(108)(16) = 5.7 in 2 The maximum moment is 502 k-ft and occurs in footing #1 versus the moment 493 k-ft that occurs at footing-strap juncture. Therefore extend the seven # 8 bars into footing #1 and add six more #8s (three on each side) to carry the slightly larger moment. In the transverse direction,
392
vc =
V 292.6 = = 20.61 ksf < 4φ f c'=(4)(0.75) 3500=25.55 kips Good pd (10.67)(1.33)
qult ( L' )2 (3.85 ksf)(3.83 ft)2 = = 28.3 k-ft 2 2 M (12, 000) (28.3)(12,000) R= = = 0.0035 which gives w = 0.004 ' 2 (3500)(8 x12)(16)2 f c bd
M=
ρ=
wf c' (0.004)(3.5) = = 0.0002 < 0.0033 = ρ min fy (60)
therefore use ρ min
As = ρ min bd = (0.0033)(108)(16) = 5.7 in 2 In the longitudinal (axial direction), M = 502 k-ft,
R=
M (12, 000) (502)(12,000) = = 0.062 which gives w = 0.071 ' 2 f c bd (3500)(8 x12)(16)2
wf c' (0.071)(3.5) ρ= = = 0.0042 > 0.0033 = ρ min fy (60) As = ρ bd = (0.0042)(108)(16) = 7.25 in 2 Use 13 # 8 bars (As = 10.20 in2) and extend the 7 # 8 bars from the strap the full length (7 feet) and add 3 # 8 bars at approximately 16 inches c.c. on each side of strap bars.
393
394
Chapter 20 Mat Foundations Symbols for Mat Foundations
395
*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:
0.4 D f 0.195 B )(1+ ) L B 0.195 ( 30 ft ) 0.40 ( 6.5 ft ) = 5.14(1.95 ksf ) [1+ ] [1 + ] = 12 ksf ( 45 ft ) ( 30 ft )
qult(net) = 5.14 cu (1+ qult(net)
396
*Mat Foundations–02: Ultimate bearing capacity in a granular soil. (Revision: Sept-08)
What will be the net allowable bearing capacity of a mat foundation 15 m long by 10 m wide, embedded 2 m into a dry sand stratum with a corrected SPT to 55% efficiency N55 = 10? It is desired that the allowable settlement is ∆Hall = 30 mm.
Solution: The allowable bearing capacity of a mat foundation in granular soils was proposed by Meyerhof (with a Factor of Safety of 3) to be based on the SPT corrected to a 55% efficiency as,
qall = 12.5 N 55 [1 +
0.33D f B
][
0.33 ( 2 ft ) ∆H all 30 mm ] = 12.5 N55 [1 + ][ ] = 151 kN / m 2 25.4 mm 25.4 mm (10 ft )
An alternate formula is, ∆H all 30 mm qall = 15.9 N 55 [ ] = 15.9 N 55 [ ] = 188 kN / m2 25.4 mm 25.4 mm Since qall = 151 kN / m 2 is the smaller of the two, choose this one for the answer.
397
*Mat Foundations–03: Find the depth Df for a fully compensated mat. (Revision: Sept-08)
The mat shown above is 30 m wide by 40 m long. The live and dead load on the mat is 200 MN. Find the depth Df for a fully compensated foundation placed upon a soft clay with a unit weight γ = 18.75 kN/m3.
Df
Q
Solution: The net soil pressure q under the mat is the load from the building Q over the entire mat, minus the weight of the soil excavated D f γ , Q − Df γ A When the mat is fully compensated, the weight of the soil W excavated is equal to the weight of the newly imposed building Q, in other words q = 0 and therefore,
q=
Df =
Q [200x103 kN ] = =9m Aγ [(30 m)(40 m)(18.75 kN / m3 )]
398
*Mat Foundations–04: The consolidation settlement of a mat foundation. (Revision: Sept-08)
The mat foundation shown below is 30 m wide by 40 m long. The total dead plus live load on the mat is 200 MN. Estimate the consolidation settlement at the center of the foundation; Cc and eo of the normally consolidated clay are 0.28 and 0.9 respectively.
Solution: The net load per unit area q is, q=
( 200, 000kips ) − 2m 15.7 kN / m3 = 135 kN / m 2 Q − Df γ = ( )( ) A ( 30m )( 40m )
The pressure at mid-clay (depth of 18 m below the mat) is found via Boussinesq as, z 18m z 18m L 40m m= = = 0.6 n = = = 0.45 and = = 1.33 ∴∆pm = 0.66q B 30m B 40m B 30m ∆pm = 0.66q = ( 0.66 ) (135 kN / m 2 ) = 89 kN / m 2 The in-situ stress at mid-clay layer before the mat foundation is built is, po = ( 4m ) (15.7 kN / m3 ) + (13m )(19.1 − 9.81) ( kN / m3 ) + ( 3m )(18.6 − 9.81) ( kN / m3 ) = 210kN / m 2 The consolidation (plastic) settlement is, ∆H =
399
( 210 ) + ( 89 ) p + ∆pm ( 0.28 )( 6, 000mm ) Cc H log10 o log = = 136 mm 10 1 + eo po 1 + ( 0.9 ) ( 210 )
*Mat Foundations–05: Settlement of a rigid mat. (Revision: Sept-08)
A building is to be supported by a rigid reinforced concrete mat foundation, whose dimensions are 20 m wide by 50 m long. The load on the mat is to be uniformly distributed with a magnitude of 65 kPa. The mat rests on a deep deposit of saturated clay with an approximate un-drained Young’s modulus Eu = 40 MPa and a Poisson ratio ν = 0.4. Estimate the immediate settlement at the center and corner of the mat.
Solution: Since the mat foundation is stiff, use the rigid factor Cs is found from the L x B ratio, L 50 = = 2.5, which by interpolation in the chart provides a Cs = 1.20. B 20 One of the possible equations for immediate settlement ∆Hi is this one, ∆Hi = C s q B (
1 - µ2 1 - (0.4)2 2 ) = (1.20)(65 kN/m )(20 m) [ ] = 0.033 m Eu 40x103 kN / m 2
Since the mat is assumed to be rigid, the surface settlement at both the center and at the corners of the mat, are the same, which is 33 mm.
400
*Mat Foundations–06: Mat foundation carries a large transfer girder. (Revision: Sept-08)
A transfer girder carries a load of 9,000 kips. It is supported by two square reinforced concrete columns, 30 in x 30 in. The columns are to be supported by a mat foundation upon a soil that has an allowable bearing capacity qa = 4 ksf. Using a load factor of 1.5, f’c = 4 ksi, concrete shear vc = 31 ksi and fy = 60 ksi, design the width and length B by B of the mat and its thickness T.
Solution: The square mat has dimensions BxB, B2qa = Q therefore B =
9000 = 47.4 feet 2
The effective depth d of the mat can be found using the approximate formula, 4d2 + 2(b + c)d =
BLq where b = c = 30 in = 2.5 feet, and the column load = 2,000 kips vc
BLq = column factored load = 1.5(2,000) = 3,000 kips Therefore, 4d2 + 2(2.5 + 2.5)d =
3000 or 31
d2 + 2.5d = 24.193 or d = 3.8 feet
The thickness T = d + 0.4 (for cover and reinforcing) = 4.25 feet = 50 inches
401
**Mat Foundations–07: Design a small mat for an office building. (Revision: Sept-08)
A small office building with the column loads shown below is founded 3 m deep into a sand stratum with a unit weight of 18 kN/m3. The foundation is the mat shown below. All the columns are 0.5 m x 0.5 m. The concrete strength is f ’c = 20.7 MN/m2 and the steel yield strength is fy = 413.7 MN/m2. Determine their reinforcement requirements in the y-direction only.
402
Solution: Step 1: Find the soil pressures, the location of the soil reaction’s resultant and the eccentricities in the x and y directions.
The service load = (400 × 2) + (500 × 2) + 450 + 350 + (1500 × 4) + (1200 × 2) = 11, 000 kN The moments of inertia of the mat in the x and y-directions are,
xy 3 16.5 × 21.53 IX = = = 13, 670 m 4 12 12
x 3 y 16.53 x 21.5 IY = = = 8, 048 m 4 12 12 To find the eccentricity in x and y directions, take moments about the axes. For the eccentricity about the y’-axis, take
∑M
y'
=0
(11,000)x’ = (8 m)(500+1500+1500+500)+(16 m)(450+1200+1200+350) x’ = 7.56 m, which translated to the mat’s centroid gives ex = For the eccentricity about the x’-axis, take
∑M
x'
16.5 − 7.56 = 0.69 m 2
=0
(11,000)y’ = (7 m)(1500+1500+1200) + (14 m)(1500+1500+1200) + (21)(400+500+450) y’ = 10.60 m, translated to the mat’s centroid gives ey =
21.5 − 10.60 = −0.15 m 2
Step 2. Find the soil reaction pressures. Let us factor the applied loads: 1.7(Service Loads) = 1.7(11,000 kN) = 18,700 kN The two eccentricities ex and ey create moments about the centroid. The soil reaction is no longer uniform, and varies linearly between the columns. These moments are: Mx = R ey = (18,700 kN)(0.15 m) = 2,805 kN-m My = R ex = (18,700 kN)(0.69 m) = 12,903 kN-m The soil reaction pressure at any point under the mat is found from the relation: q= 403
R M yx Mx y 18, 700 12, 903( x) 2,805 ( y ) ± ± = ± ± A Iy Ix 16.5 × 21.5 8, 048 13, 665
Therefore,
q = 52.7 ± 1.6 x ± 0.21 y
kN / m 2
Step 3. Using the equation for q, prepare a table of its value at points A through J.
POINT
R/A
x
1.6x
y
0.21y
q
(kN/m )
(m)
(m)
(m)
(m)
(kN/m2)
A
52.7
-8
-12.8
10.5
2.21
42.11
B
52.7
0
0
10.5
2.21
54.91
C
52.7
8
12.8
10.5
2.21
67.71
D
52.7
8
-8.16
-10.5
-2.21
63.29
E
52.7
0
0
-10.5
-2.21
50.50
F
52.7
-8
-12.8
-10.5
-2.21
37.7
G
52.7
-4
-6.4
10.5
2.21
48.51
H
52.7
-4
-6.4
-10.5
-2.21
44.10
I
52.7
4
6.4
10.5
2.21
61.31
J
52.7
4
6.4
-10.5
-2.21
56.9
2
Step 4. Determine the effective depth d and the thickness T of the mat. a) Check a critical edge column (for example, one of the 1.5 MN at the left edge): U = factored column load = 1.2(1.0) + 1.6(0.5) = 2 MN or 1.7(1.5) =2.55 MN b0 = critical perimeter = 2(0.5 m + d/2) + (0.5 + d) = (1 + d) + 0.5 + d = 1.5 + 2d Using φVC ≥ Vu (from ACI 318-05) and fc’ = 20.7 MN/m2 (3 ksi),
φ (0.34) f ' ( b d ) = (0.85)(0.34) 20.7 × (1.5 + 2d )(d ) ≥ U = 2 MN o c d2 + 0.75 d – 0.76 ≥ 0
d=
−0.75 ± (0.75)2 − 4(1)(−0.76) − b ± b 2 − 4ac = = 0.57 m 2a 2(1)
b) Check the largest corner column (the 0.45 MN at top right corner): d = 0.36 m (This does not control). c) Check the most critical internal column (the 1.5 MN): 404
b0 = 4(0.5 + d) = 2 + 4d 0.85(0.34)( 20.7 )(2 + 4d )(d ) = 2 MN 5.26d 2 + 2.63d − 2 = 0
d = 0.415m ∴ use d = 23 inches or 585 mm and T = 23+3+1 = 27 in or 686 cm
Step 5. Find the average soil reaction for each strip: Strip AGHF (W = 4.25 m)
q1 =
q A + qG 42.11 + 48.5 kN = = 45.31 2 2 2 m
q2 =
qH + qF 44.1 + 37.7 kN = = 41 2 2 2 m
Strip GIJH (W = 8 m)
q1 =
48.51 + 54.91 + 61.31 kN = 54.91 2 3 m
q2 =
56.9 + 50.5 + 44.1 kN = 50.5 2 3 m
Strip ICDJ (W = 4.25 m)
q1 =
61.31 + 67.71 kN = 64.51 2 2 m
q2 =
63.29 + 56.9 kN = 60.1 2 2 m
Soil reaction AGHF = ½ (45.31+41)(4.25)(21.5) = 3943 kN Soil reaction GIJH = ½ (54.91+50.5)(8)(21.5) = 9065 kN Soil reaction ICDH = ½ (64.51+60.1)(4.25)(21.5) = 5693 kN
∑ Fy = 3943 + 9065 + 56903 = 18,700 kN Strip GIJH. 405
Q1 = 1.7(500) = 850 kN Q2 = 1.7(1500) = 2550 kN Q3 = 1.7(1500) = 2550 kN Q4 = 1.7(500) = 850 kN
54.91x8 m = 43.93 kN/m 433.4 kN/m
50.50x8 m = 404 kN/m 415.76 kN/m 427.52 kN/m 421.64 kN/m 409.9 kN/m
Step 6. Find the maximum positive moments for each span at midpoints a, b & c.
∑ Ma = 0
∑ M A = (850 × 3.5 m) − ∴ M A = −302.36
∑M
b
=0
( 439.3 + 433.4 ) (3.52 ) + M 2
2
A
=0
kN-m/m
(
)
2 (439.3 + 415.76) 10.5 + Mb = 0 ∑ M b = (850 ×10.5m) + ( 2550 × 3.5m ) − 2 2
∴ M b = 5718 kN-m/m
Step 7. Calculate maximum negative moment at d, column B, see page 449:
( )
2 (439.3 + 427.52) 7 + Md = 0 ∑ M d = (850 × 7m) − 2 2
∴ M d = 4668.5 kN-m/m
Step 8. Design the strip for flexure: 406
d = 23 inches = 685 mm , f `c = 3 ksi (20.7 MN / m 2 ), f y = 60 ksi (413.7 MN / m 2 ) a=
As f y
AS =
0.85 f `c b
Mu φ ⋅ f y (d − a 2)
say Mu = 5718 kN-m/ 8 m = 715 kN-m/m = 161 k-ft/ft say a = 3.3 in , As required = 1.68 in2/ft Try #9 @ 6” o.c.
As = (1)(12/6) = 2 in2 > 1.68 in2
∴ Use #9 @ 6” bottom, As required = 1.68 in2
ρmin = 200/60000 = 0.0033
As-min= 0.0033(23)(12) = 0.91 in2 < 1.68 in2
As-min Bottom → 1.68 in2 Top
→ 1.3 in2
Negative moment:
Mu = 4668.5 / 8 m = 584 KN-m/m = 131.6 k-ft/ft Say a = 2.54 in ,
As required = 1.3 in2/ft
Try #9 @ 9” o.c.
As = (1)(12/9) = 1.33 in2 > 1.30 in2
Good
∴ Use #9 @ 9” top. Use top and bottom reinforcing throughout the mat in the y-direction. Step 9. Sketch the mat’s cross-sections and reinforcement.
407
Good
Chapter 21 Deep Foundations - Single Piles Symbols for Single Piles of Deep Foundations
408
*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 αmethod.
Solution: The ultimate capacity of the pile Qult is given by the simple formula,
Qult = Qpoint + Qshaft = Ap q p + ( cu ) ( perimeter ) L = Ap ( cu ) N c + α ( cu ) ( perimeter ) L Notice that the value of the cohesion is reduced by the "α " factor found in the graph below,
∴ Qult = ( 0.38m ) (70 kN / m 2 )(10.97) + ( 0.75 ) (70 kN / m 2 )4(0.38m)(20m) = 1,890 kN 2
The allowable capacity is,
Qall =
409
Qult 1,890kN = = 630 kN 3 3
*Single-Pile–02: Pile capacity in granular soils. (Revision: Oct.-8)
Find the ultimate load of a slightly tapered precast pre-stressed concrete pile, given Pile length
L = 50 ft
Width of pile
b = 16 in
Bearing capacity factor
Nq* = 55
Unit weight of sand
γ = 110 lb/ft3
Effective sand’s friction angle
φ = 30°
Effective earth coefficient
K = 1.3
Soil-pile friction angle
δ = 0.8φ
Solution:
The ultimate load Qult is found from the formula, Qult = Q p + Qshaft = Ap q N q * + 16 Qult = b 2 ( D f h )( N q *) = ( ft )2 (0.110kcf ) ( 50 ft ) (55) = 538 kips 12 b) Ultimate frictional resistance Unit skin friction at depth z =0, fs(0) = 0 ksf Unit skin friction at depth z = L’ Where
L’ = 15D = 15(16 in)/12 = 20 ft
fs(L’) = Kσ’o tanδ = (1.3)(2.2 ksf)(tan24°) = 1.27 ksf Where
σ’o = γL’ = (110 lb/ft3)(20 ft)/1000 = 2.2 ksf
Frictional resistance from z = 0 to z = L’ = 20 ft
Qs(0-L’) = pL’fs(avg) = (5.33 ft)(20 ft)(0.635 ksf) = 67.7 kips Where
p = 4b = 4(16 in)/12 = 5.33 ft
fs(avg) = (fs(L’) - fs(0))/2 = (1.27 ksf – 0 ksf)/2 = 0.635 ksf Frictional resistance from z = L’ = 20 ft to z = L = 50 ft
Qs(L’-L) = p(L-L’)fs(L’) = (5.33 ft)(50 ft -20 ft)(1.27 ksf) = 203 kips Total frictional resistance
Qs = Qs(0-L’) + Qs(L’-L) = 67.7 kips + 203 kips = 271 kips The total pile capacity = 56 + 68 + 203 = 327 kips.
410
*Single-Pile–02b: The friction of a concrete pile in granular soils. (Revised Nov-09)
Find the length of a 14” square pile required to carry a 50 ton load embedded in a firm sand (N = 25). Given: Design load Side length of the square pile
D = 1.167 ft
Atmospheric pressure
Pa = 2 ksi
Avg. corrected SPT at 10D above and 4D below pile point
(N1)60 = 25
Average standard penetration resistance
( N 1)60 = 25
Factor of safety
FS = 3
Solution: Ultimate load Qult = Qdesign FS = (100)(3) = 300 kips Length of the pile is found from Meyerhof’s method,
Qult = 4Pa(N1)60Ap + 0.02Pa( N 1)60As 300 = 4(2)(25)(1.36) + 0.02(2)(25)(4.67)L = 272 + 4.67L
L = (300 – 272) / 4.67 = 6 feet where Ap = D2 = (1.167)2 = 1.36 ft2 and As = 4DL = (4)(1.167)L = 4.67L
411
Qdesign = 100 kips
*Single-Pile–03: The capacity of a driven concrete pile. (Revision: Sept.-08)
Find the ultimate and the design (allowable) capacities of a prestressed precast concrete pile (PSPC) driven with a 72 blows/feet criterion. The piles are 30 feet long and have 14” square cross sections. The hammer available is a LinkBelt 520 with a ram weight of 5,070 lbs and a drop height of 5.2 feet. The hammer efficiency is 0.9, and the contractor intends to use scrap plywood in place of cushion blocks (hence, n = 0.4).
L = 30 ft
Given: Concrete pile length Width of pile
b = 14 inches
Weight of the ram
Wram = 5,070 lbs
Height of the fall of the ram
h = 5.2 feet
Hammer efficiency
e = 0.9
Penetration of pile per hammer blow
s = 0.17 inches / blow
Coefficient of restitution
n = 0.4
Elastic compression of the pile cap
sc = 0.37 in
Elastic compression of the pile itself
sp = 0.18 in
Elastic compression of the soil
ss = 0.10 in
Factor of safety
FS = 3
Solution:
The ultimate load capacity Qult of a driven pile is found through Hilley's formula, Qult =
(W + s ) (W
12Wram h eeff
s + 0.5 ( sc + s p
ram
s
+ n 2W pile )
ram
+ W pile )
The weight of the pile W pile is, 2
14in W pile = b Lγ concrete = ( 30 ft )( 0.150kcf ) = 6.13kips 12in / ft ( 5.07 k ) + ( 0.4 )2 ( 6.13k ) 12 ( 5.07 kips )( 5.2 ft ) ( 0.9 ) = 311 kips ∴ Qult = ( 0.17in ) + 0.5 ( 0.37in ) + ( 0.18in ) + ( 0.10in ) ( 5.07k ) + ( 6.13k ) 2
The service load is commonly a factor of 2 to 3 from the ultimate pile capacity: ∴ Qservice =
Qult 311 kips = = 104 kip (52tons) 3 3
412
*Single-Pile–04: The driving criterion for a concrete pile. (Revised Nov-09)
A highway bridge will be founded on 14” x 14” square pre-stressed pre-cast (PSPC) concrete piles at mid-stream. Establish the driving criterion for a 50 ton design capacity for these end bearing piles using MKT S-14 pile driving hammer with a plywood cushion. Given: Design load
Qdesign = Qall = 100 kips
Concrete pile length
L = 50 ft
Width of pile
b = 14 in
Factor of safety
FS = 3
Weight of the ram
Wr = 14 kips
Hammer energy
HE = 37.5 ft-kips
Height of fall of the ram
h = HE / Wram = 37.5 /14 = 2.67 ft
Hammer efficiency
e = 0.9
Coefficient of restitution for plywood
n = 0.4
Elastic compression of the pile cap
sc = 0.37 in
Elastic compression of the pile itself
sp = 0.006L = 0.006(50) = 0.30 in
Elastic compression of the soil
ss = 0.10 in
Solution: Ultimate load: Qult = Qall FS = (100)(3) = 300 kips Weight of the pile Wp = b2Lwc = (14 in/12 in/ft)2(50 ft)(150 lb/ft3) = 10.21 kips Hiley’s dynamic pile driving formula is,
Qult =
W + n 2W p 12 Wr he ×X r s + 1/2(S c + S p + S s ) Wr + W p
300 =
12 (14 )(2.67 )(0.9 ) 14 + 0.4 2 (10.21) ×X s + 1/2(0.37 + 0.30 + 0.10 ) 14 + 10.21
300(s + 0.385) = (403.7)(0.646) s = 0.484 in / blow ⇒ 0.484 / 12 = 0.0403 ft / blow The pile driving criterion is, 1 / s = 1 / 0.0403 = 24.79 ⇒ 25 blows / foot
413
*Single-Pile–05: The Meyerhof and Vesic methods for pile capacity. (Revised Nov-09)
For the pile described below, find: a) the ultimate point load using the Meyerhof method, b) the ultimate point load using the Vesic method, c) an approximate ultimate point load method, d) the ultimate shaft friction resistance, and e) the total design load. Given: Pile length in sand
L = 25 m
Diameter of pile
D = 0.305 m
Atmospheric pressure
Pa = 100 kN/m2
Bearing capacity factor
Nq* = 150
Bearing capacity factor
Nσ* = 71
Reduced rigidity of the soil
Irr = 50
Dry unit weight of sand from 0 to 5 m
γdry = 16 kN/m3
Effective soil friction angle from 0 to 5 m
φ = 32°
Cohesion of soil supporting pile tip at 5 m
c’ = 0
Water table
WT = 5 m
Unit weight of water
γw = 9.81 kN/m3
Unit weight of saturated sand from 5 to 10 m
γsat = 18.4 kN/m3
Effective soil friction angle from 5 to 10 m
φ = 32°
Cohesion of soil supporting pile tip at 10 m
c’ = 0
Unit weight of saturated sand from 10 to 25 m
γsat = 19.9 kN/m3
Effective soil friction angle from 10 to 25 m
φ = 37°
Cohesion of soil supporting pile tip at 25 m
c’ = 0
Effective earth coefficient
K = 1.4
Soil-pile friction angle
δ = 0.7φ
Factor of safety
FS = 4
Solution: a) Ultimate point load using Meyerhof’s method
Qp1 = Apqp = (0.073 m2)(5,652 kN/m2) = 413 kN Where
Ap = πD2/4 = 3.14(0.305 m)2/4 = 0.073 m2 N q∗ = 150 for φ =37 and pA = 100 kN/m 2 (atmospheric pressure)
qp = 0.5paNq*tanφ’ = 0.5(100)(150)tan37° = 5,652 kN/m2
414
b) Ultimate point load using Vesic’s method
Qp2 = Apqp = (0.073 m2)(11,644 kN/m2) = 850 kN Ap = πD2/4 = 3.14(0.305 m)2/4 = 0.073 m2 Nσ∗ = 71 for I R = 50
Where
qp = σo’Nσ* = 164(71) 11,644 kN/m2 and
σo’ = [(1+2Ko)/3]q’ = {[1+2(0.398)]/3}(274) = 164 kN/m2 Ko = 1 - sinφ = 1 - sin37° = 0.398
and
q’ = (5-0) γdry + (10-5)(γsat-γw) + (25-10)(γsat-γw) = (5)(16) + (5)(18.4-9.81) + (15)(19.9-9.81) = 274 kN/m2 c) Approximate ultimate point load: (Average between the Meyerhof and Vesic methods)
Qp = (Qp1 + Qp2)/2 = (413 + 850)/2 = 631 kN d) Ultimate frictional resistance Unit skin friction at depth z =0, fs(0) = 0 kN/m2 Unit skin friction at depth z = L’ Where
L’ = 15D = 15(0.305) = 4.575 m Use
L’ = 5 m
fs(L’) = Kσ’o tanδ = (1.4)(80 kN/m2)(tan22.4°) = 46.16 kN/m2 Where
σ’o = γdryL’ = (16 kN/m3)(5 m) = 80 kN/m2 δ = 0.7φ = 0.7(32°) = 22.4°
Frictional resistance from z = 0 to z = L’ = 5 m
Qs(0-L’) = pL’fs(avg) = (0.958 m)(5 m)(23.08 kN/m2) = 110.55 kN Where
p = πD = 3.14(0.305 m) = 0.958 m
fs(avg) = (fs(L’) - fs(0))/2 = (46.16 – 0)/2 = 23.08 kN/m2 Frictional resistance from z = L’ = 5 m to z = 10 m
Qs(L’-L) = p(10-L’)fs(L’) = (0.958 m)(10 m-5 m)(46.16 kN/m2) = 221.1 kN Frictional resistance from z = 10 m to z = L = 25 m
Qs(L’-L) = p(L-10)fs(L’) = (0.958 m)(25 m-10 m)(46.16 kN/m2) = 663.32 kN Total frictional resistance
Qs = Qs(0-L’) + Qs(L’-10) + Qs(10-L) = 110.55 kN + 221.1 kN + 663.32 kN
Qs = 995 kN e) Total design load
Qall = (Qp + Qs)/FS = (631 kN + 995 kN)/4 = 406 kN
415
*Single-Pile–06: Tension or pull-out capacity in granular soils. (Revised Nov-09)
Find the allowable pullout capacity for a 10.4 m long by 406 mm square concrete pile embedded in a sand stratum. Given: Pile length in sand
L = 10.4 m
Side of square pile
D = 0.406 m
Unit weight of sand
γ = 15.8 kN/m3
Effective soil friction angle
φ = 30°
Effective earth coefficient
K = 1.4
Soil-pile friction angle
δ = 0.7φ
Factor of safety
FS = 4
Solution: Unit skin friction at depth z =0, fs(0) = 0 kN/m2 Unit skin friction at depth z = L’ where L’ = 8D = 8(0.406) = 3.25 m
fs(L’) = Kσ’o tanδ = (1.4)(51.35 kN/m2)(tan 21°) = 27.59 kN/m2 Where
σ’o = γL’ = (15.8 kN/m3)(3.25 m) = 51.35 kN/m2 δ = 0.7φ = 0.7(30°) = 21°
Frictional resistance from z = 0 to z = L’ = 3.25 m
Qs(0-L’) = pL’fs(avg) = (1.624 m)(3.25 m)(13.8 kN/m2) = 72.84 kN Where
p = 4D = 4(0.406 m) = 1.624 m
fs(avg) = (fs(L’) - fs(0))/2 = (27.59 – 0)/2 = 13.8 kN/m2 Frictional resistance from z = L’ = 3.25 m to z = 10.4 m
Qs(L’-L) = p(10.4-L’)fs(L’) = (1.624 m)(10.4 m-3.25 m)(27.59 kN/m2) = 320.36 kN Total pullout capacity Qs = Qs(0-L’) + Qs(L’-L) = 72.84 kN + 320.36 kN = 393.2 kN Allowable pullout capacity Qall = Qs/FS = 393.2 kN / 4 = 98.3 kN
416
*Single-Pile–07: Tension (uplift) capacity of a pile in granular soils. (Revised Nov-09)
A square concrete pile (406 mm x 406 mm) is driven into a sand stratum that has a γ = 15.8 kN/m3, a relative density of 70% and a φ = 30° 30°. If the pile has reached refusal at a depth of 10.4 feet long, estimate its allowable pullout capacity T using a FS = 4.
Solution:
L The criterion that = 8, means that Lcr = 8 (0.406 m) = 3.25 m D cr Since L > Lcr the ultimate pullout capacity Tult is provided by the formula, 1 Tult = (perimeter)(γ)(Lcr) 2 (Kw ) tanδ + (perimeter)(γ)(Lcr)(Kw) tanδ (L - Lcr) 2 δ Dr = = 0.70, in other words δ = 0.70(φ) = 0.70(30°) = 21° and assume Kw =1.4 φ Therefore, 1 Tult = (0.406 m x 4)(15.8 kN/m 3 )(3.25 m)2 (1.4) tan21˚ 2 + (0.406 m x 4)(15.8 kN/m 3 )(3.25 m)(1.4) tan 21˚ (10.4 m – 3.25 m) = 73 + 320 = 393 kN The allowable pullout capacity is Tall =
417
Tult 393 = = 98 kN FS 4
***Single-Pile–08: The settlement of a drilled shaft in cohesive soils. (Revised Nov-09)
To illustrate the calculation of the settlement of a single pile, consider the case of a large bored pile in clay. The pile has been tested by Whitaker and Cooke (1966) and is denoted as Pile F in the graphs shown here. The pile details are as follows:
L = 39.9 ft (12.2 m) [shaft length = 36.5 ft (11.1 m)] D = 2.0 ft (0.61 m) db = 4.0 ft (1.2m) Ep = 3.0 X 106 lb/sq in. (19.3 X 104 tons/sq ft, 20.67 x 106 kN/m2) The soil details are as follows: Along shaft, c u= 1.2 tons/sq ft (129 kN/m2) At base,
c u = 1.4 tons/sq ft (150 kN/m2)
Average Es along shaft = 10,500 lb/sq in (675 tons/sq ft, 72, 400 kN/m2)
vs = 0.5 (assumed undrained conditions) Solution: From the above data,
L/d = 19.5, db/d = 2.0
K = 3 X 106/1.05 X 104 = 285
Determination of I and β From Fig. 21.1, Io = 0.085 (for L/d = 19.5 and db/d = 2.0) From Fig. 21.2, RK = 1.35 (for K = 285) From Fig. 21.3, Rv = 1.0 Treating the pile as a floating pile in a deep soil layer, Rh = 1.0 PI From ρ = , I = 0.085 X 1.35 X 1.0 X 1.0 = 0.115 Es d From Fig. 21.4, βo = 0.215 where I = IoRKRhRv ρ = settlement of pile head P = applied axial load Io = settlement-influence factor for incomp pile in semi-infinite mass, for vs = 0.5. RK = correction factor for pile compressibility From Fig. 21.5, CK = 0.76 From Fig. 21.6, Cv = 1.0 β = 0.215 X 0.76 X 1 = 0.164
418
Fig. 21.2-Compressibility correction factor for settlement, RK. Fig. 21.1 - Settlement-influence factor Io.
Fig. 21.3-Poisson’s ratio correction factor for settlement, Rv.
419
Fig. 21.4 - Proportion of base load, βo.
Fig. 21.5 - Compressibility correction factor for base load, CK.
Fig. 21.6-Poisson’s ratio correction factor for settlement, Rv. 420
Determination of Ultimate Shaft and Base Resistance Assuming ca/cu = 0.33,
Psu = π X 2.0 X 36.5 X 1.2 X 0.33 = 92 tons (920 kN) Pbu = 9.0 X 1.4 X 12.6 = 158 tons (1580 kN)
Determination of Overall Load-Settlement Curve The total load at ultimate shaft yield is PYl =
P SU 92 = = 110 tons (1100 kN) 1 − β 0.836
I The settlement at ultimate yield shaft is ρ Y 1= E sd
0.115 * 110 * 12 (PYl ) = = 0.112 in. 2.0 * 675
The settlement at ultimate failure of pile [Pu = 92 + 158 = 250 tons (2500 kN)] is
ρu = Psu + Pbu
ρu = 1.088 in. (27.6 mm)
I Pbu P β L 0.115 158 + Pbu − su = * * (1 − β ) Ap E p 2.0 * 675 0.165 β
ρu = E sd
0.165 39.5 * 12 12 + 158 − 92 * = 1.088 * 0.835 3.14 * 19.3 * 10
The load-settlement curve is thus drawn as two straight lines, the first joining the origin to the point P = 110 tons, ρ = 0.112 in., and the second joining the latter point to the point P = 250 tons, ρ = 1.088 in. This computed curve is compared with the measured curve in Fig. 21.7
Fig. 21.7 – Comparisons between observed and predicted load-settlement curves. Tests of Whitaker and Cooke (1966) 421
***Single-Pile–09: The performance of steel pipe piles. (Revised Nov-09)
Piles are being considered for a power plant, specifically a closed-ended steel pipe pile, 24.4 m long and 0.32 m in diameter, driven into a fine sand grading to coarse and having an average standard penetration value, N, of about 16. The water table was 3.4 m below the surface. On the basis of the available data, the following values were adopted: Bulk unit weight above water table = 17.3 kN/m3. Submerged unit weight below water table = 7.8 kN/m3. Angle of internal friction angle prior to installation:
φ′1 = 25° (0 – 2.4 m) φ′1 = 32° (2.4 – 18.3 m) φ′1 = 30° (18.3 –20.8 m) φ′1 = 33° (> 20.8 m) Solution: (Reference to Poulos and Davis). First, find the ultimate skin resistance (given by φ =
3 φ′1 +10), as follows: 4
φ = 28.75° (0 – 2.4 m) φ = 34° (2.4m – 18.4 m) φ = 32.5° (18.3 – 20.8 m) φ = 34.75° (>20.8 m) The values of Ks tan φ′a are 1.00 (0 –2.4 m), 1.30 (2.4 – 18.3 m), 1.18 (18.3 – 20.8 m), 1.35 (20.8). If it is assumed that the critical depth is less than 2.4 m below the surface, then for φ = 28. 75°, zc/d = 5.0; that is, zc = 5.0 x 0.32 = 1.56 m. Thus, the assumption is justified. At the critical depth, the effective overburden stress is :
σ′vc = 1.56 x 17.3 = 26.99 kN/m2 Because the pile has uniform diameter, Fw = 1. For the ultimate base resistance, the value of φ given by φ ′ + 40 φ= is 36.5°. The value of Nq is 98. 2 L
′ Nq −W The formula for the ultimate load capacity of a single pile Pu = ∫ Fw Cσ ν′ K S tan φ a′ + Abσ vb 0
422
0 + 26.99 × 1.56 + 26.99 × (2.4 − 1.56 ) 2 0.32 2 Pu = π × 0.32× 1.00 + 26.99 × (18.3 − 2.4 ) × 1.30 + 26.99 + 26.99 × π × × 98 4 × (20.8 − 18.3) × 1.18 + 26.99 × (24.4 − 20.8) × 1.35 Pu = 816 + 213 = 1029 kN (115.6 tons)
This compares well with the measured value of 1,112 kN (125 tons) from a load test. The second type of pile (B) considered was a Raymond Standard Pile, 10 m long, with a head diameter of 0.55 m and a tip diameter of 0.20 m. The pile taper w= 1°, so Fw = 3.35 (0-2.4 m), and Fω = 4.1(2.4-18.3 m). Assuming again that the critical depth is above 2.4 m, zc / d = 5.0 as before, and taking an average value of d of 0.51, zc = 2.55 m, that is, greater than 2.4 m. However, the difference is negligible and hence zc will be taken as 2.55 m. At this level,
σ ν′c = 2.55 × 17.3 = 44.12 kN/m2 At 2.4 m,
σ ν′ = 2.4 × 17.3 = 41.52 kN/m2 Since the pile tip is founded in the second stratum, φ from
φ ′ + 40
is 36° and the corresponding value of 2 Nq is 88. Substituting into Pu and using, for simplicity, the mean diameter of the pile in the upper 2.4 m and the lower 7.6 m,
L
′ Nq −W Pu = ∫ Fw Cσ ν′ K S tan φ a′ + Abσ vb 0
0 + 41.52 (41.52 + 44.12 ) Pu = × 0.15 + 44.12 × (10 − 2.55) × π × 0.51 × 3.35 × 1.00 × 2.4 + 2 2 2 0.20 × π × 0.33 × 4.1 × 1.30 + 44.12 × π × × 88 4
Pu = 2243 kN (252.2 tons)
The measured ultimate load for this pile was 2,400 kN (270 tons).
423
Fig 21.8 – Values of zc/d and Ks tan φ’a for piles in sand.
424
Figure 21.9 - Pile taper factor Fω (after Nordlund, 1963)
Figure 21.10 - Relationship between Nq and Ø (after Berezantzev et al., 1961)
425
Chapter 22 Deep Foundations - Pile Groups and Caps Symbols for Pile Groups and Caps of Deep Foundations
426
**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. 22.1. 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. 22.2, 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).
Fig 22.2 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 Fig. 22.3. Assuming Fig. 22.1 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. 22.3. 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
427
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. 22.4. 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
Fig. 22.4 Fig. 22.3 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. 22.4 Mu = (177) (27/12) = 398.3 ft-kips
φ Ru = Mu / bd2 = 398.3 x 12,000 / (102)(19.625)2 = 121.67 psi When ƒ ′c = 3000 psi and ƒy = 40,000 psi, the steel ration required is less than the minimum steel ration, ρmin = 200 / ƒy = 0.005. Further, if the steel ration required is increased by 1/3, it will still be less than ρmin. 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 in fig. 22.5.
Fig. 22.5 428
**Pile cap-02: Pile group efficiency and lateral load capacity. (Revised Nov-09)
A pile group is constructed in a sand stratum (plan shown below) with piles of φ =18 in and a center-to-center spacing of 36 inches. Find the pile group efficiency η using the Converse-Labarre formula; Piles are acting in friction only; select a hammer to drive them; Determine the allowable load-bearing capacity of the pile group (for a FS = 4); Determine the allowable horizontal load on the pile cap, and the collective bending moment capacity.
Solution: a) The Converse-Labarre group efficiency η is,
d (n − 1)m + ( m − 1)n where ξ = arctan ( in degrees ) 90mn s d 18" ∴ ξ = arctan = arctan = 26.6° s 36" (3 − 1) ( 3) + (3 − 1) ( 3) η = 1 − ( 26.6 ) = 1 − ( 26.6 )( 0.015 ) = 0.61 or 61% 90 ( 3)( 3)
η = 1− ξ
b) If pre-cast pre-stressed concrete piles are used, assume a pile/hammer ratio,
Wp Wr
429
=
2 1 1 π k 2 ∴ Wr = W p = [ (1.5 ft ) ( 20 ft ) 1.5 3 = 2.9 kips 1 2 2 4 ft
Choose a hammer from a list of available hammers.
c) QW
Vulcan INI00 McKiernan-Terry DE30 LinkBelt 312 Foster K13 Berminghammer B225 Mitsubishi M14 = QP + QS = QS (for friction piles only) 20
QW = Q S = π D (C u + κγ tan φ ∫ Hdh ) use κ = 1 0
Q w = π D (1)(0.125) tan φ but η =
H2 20 2 = π D (0.125) tan 38 ° = 92 kips 2 2
Q u group Q group = 0 . 61 = u Q u SINGLE × NO . PILES 92 kips × 9
∴ Q u group = 505 kips but Q all =
Q u group 505 = FS 4
∴ Q all = 126 kips d) For granular soils
H u = 1 .15γ soil L2 DK P = 1 .15 ( 0 .125 )( 20 ) 2 (1 .5 ) tan 2 ( 45 ° +
38 ° ) =362.5 kips 2
H all =90 kips Mu =
2 2 H u L = (362 .5 )( 20 ) = 4833 k-ft 3 3
M all = 1,210 k-ft
430
***Pile caps–03: Settlement of the combined pile-cap and pile cluster. (Revised Nov-09)
A square group of 25 piles extends between depths of 1 m and 13 m in a deposit of stiff clay 25 m thick over-laying rock. The piles are 0.6 m in diameter and are spaced at 2 m on centers in the group. The un-drained shear strength of the clay at pile base level is 170 kN/m2 and the average value of un-drained strength over the depth of the piles is 105 kN/m2. The adhesion coefficient α is 0.45, Eu is 65 MN/m2 and mv is 0.07 m2/MN. The pore pressure coefficient A is 0.24. If the total load on the pile group is 12,000 kN, determine the load factor and the total settlement.
Solution: 1. At the base level, cu = 170 kN/m2. Therefore,
qf = 9 (cu) = 9 · 170 = 1530 kN/m2
2. Over the depth of the piles, cu = 105 kN/m2. Therefore,
fs = α (cu) = 0.45 · 105 = 47 kN/m2
3. For a single pile the ultimate load is given by Qf = (Ab) qf + (As) fs = [(π/4) x 0.62 x 1530] + (π x 0.6 x12 x 47) = 432 + 1063 = 1495 kN 4. The ultimate load on the group assuming single pile failure and a group efficiency of 100% , → QGROUP = 25 x 1495 = 37,375 kN
431
5. The width of the group is 8.6 m. Therefore the ultimate load on the group assuming a block failure, and taking full un-drained strength on the perimeter:
Qult = (8.62 x 1530) + (4 x 8.6 x 12 x 105) = 113 159 + 43 344 = 156,500 kN Hence the load factor is 37,375 / 12,000 = 3.1. Even if the re-molded strength were taken on the perimeter there would be no likelihood of block failure. However, settlement is likely to be the limiting criterion. Referring to the figure below, the equivalent raft is located 8 m (2/3 x 12 m) below the top of the piles. The width of the equivalent raft is 12.6 m. The load on the equivalent raft (12 000 kN) is spread at a slope of 1:2 to the underlying clay
Equivalent raft concept The pressure on the equivalent raft is: q = 12 000 / 12.62 = 76 kN/m2 The immediate settlement is determined using the graphs for the coefficients for vertical displacements, shown below. Now,
H/B = 16/12.6 = 1.3 D/B = 9/12.6 = 0.7 L/B = 1
432
Coefficients for vertical displacement. (Reproduced from N. Janbu, L. Bjerrum and B. Kjaernsli (1956) Norwegian Geotechnical Institute Publication No. 16, by permission.)
433
Therefore µ1 = 0.50 and µ0 = 0.80; thus ∆i
= (µ0
µ1 ) (qB/Eµ) = 0.80 x 0.50 x [(76 x 12.6)/65] = 6 mm
To calculate the consolidation settlement, the clay below the equivalent raft will be divided into four sublayers each of thickness H = 4m. The pressure increment (∆σ) at the center of each sublayer is equal to the load of 12,000 kN divided by the spread area and is shown in the Table below. The settlement coefficient is obtained from the last graph that is shown below. The diameter of a circle having the same area as the equivalent raft is 14.2 m. Thus H/B = 16/14.2 = 1.1. Then from the graph with the settlement coefficients for A = 0.24 and H/B = 1.1, the value of µ is 0.52 and the consolidation settlement is:
∆c = µsoed = 0.52 x 36.9 = 19 mm Layer
Z (m)
Area (m2)
∆σ (kN/m2)
mv∆σH
1
2
14. 62
56.3
15.8
2
6
18. 62
34.7
9.7
3
10
22. 62
23.5
6.6
4
14
26. 62
17
4.8
(mm)
Soed =
The total settlement is:
36.9
∆ = ∆i + ∆c = 6 + 19 = 25 mm
Settlement coefficient. Reproduced from R.F Scott (1963). 434
Chapter 23 Deep Foundations: Lateral Loads Symbols for Lateral Loads on Deep Foundations
435
**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,
k z = nh z nh modulus of horizontal subgrade reaction lb/in3
kN/m3
- 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
- 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
Type of soil Dry or moist sand
Submerged sand
436
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,
437
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.
The characteristic length T of a pile-soil system is given by,
T=
5
EpI p nh
( 207 x10 )(123 x10 ) = 1.16 m −6
6
=
5
(12 , 000 )
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 ∆ at any depth z is given by,
∆ = Az
Q gT 3 EpIp
+ Bz
M gT 2 EpIp
but M g = 0
In this problem we are given this value of ∆ = 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.
438
The magnitude of the lateral load Qg limited by the displacement condition only is,
∴ Qg =
∆(E p I p ) AzT 3
( 0.008 m ) ( 207 x106 kN / m 2 )(123 x10−6 m 4 ) = = 54 kN 3 ( 2.435 )(1.16 )
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,
M z = Am Q gT 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,
M max = Fy ∴ Qg =
439
Ip d1 / 2
(
= 248 MN / m
2
(123 x10
) ) ( 0.254 m / 2 ) = 240 kN − m −6
( 240 kN − m ) = 268 kN M max = AmT ( 0.772 )(1.16 m )
m4
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
440
**Lateral loads on piles-02: Find the lateral load capacity of a steel pile. (US Units) (Revised Nov-09)
Determine the lateral load capacity Qg of a steel H-pile (HP 10 x 57) fully embedded to a depth of 82 feet in very dense submerged sand. The top end of the pile is allowed to deflect laterally .31 inches. For simplicity assume that there is no moment applied to the top of the pile (that is, Mg = 0). Figure 1:
Solution: Step 1: The sub-grade modulus (ks) is a description of the reaction of the soil mass to vertical loads. The modulus of horizontal sub-grade reaction nh is a function of ks at any depth z, kz = nh z From Table 1 for the soil conditions noted, choose nh = 45 lb/in3 for the modulus.
Table 1 - nh = modulus of horizontal sub-grade reaction 441
Now choose the parameters for the steel H-pile,
Table 2: H-pile parameters (British Units)
For the steel HP 10 x 57 pile: the moment of inertia about the strong axis is Ip = 294 in4, (SI: 123 x 10-6 m4) its modulus of elasticity is Ep = 3.01 x 107 lbf/in2, (SI: 207 x 106 kN/m2) the steel’s yield strength is Fy= 3.60 x 104 lbf/in2, (SI: Fy = 248 MN/m) and the pile depth d1 = 10 in. (SI: 0.254m) The characteristic length T of a pile-soil system is given by,
T =5
Ep I p np
=5
(3.01e7lbf / in 2 )(294in 4 ) (45lb / in3 )
= 5 198680414in5 = 45.67in = 3.81 ft Therefore, the ratio
L / T = 82 ft / 3.81 ft = 21.5 > 5, so this is a long pile. The formula for the pile’s top end lateral deflection _ at any depth z is given by,
442
In this problem we are given this value of ∆ = 0.31 inches, 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.
The magnitude of the lateral load Qg limited by the displacement condition only is,
∴ Qg =
∆( E p I p ) AzT 3
(0.31in)(3.01e7lbf / in 2 )(294in 4 ) = = 12139lbf (4090)(3.81)3
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,
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, Working backwards from Qg = 268 kN = 60248 lbf then Am = 0.768
443
SI Units:
British Units:
M max = Fy
Ip d1 / 2
= (3.60e 4lbf / in 2 )
(294in 4 ) (10in) / 2
= 2,116,800lbf − in = 176,400lbf − ft ∴ Qg =
M max (176,400lbf − ft ) = = 60,000lbf AmT (.772)(3.81 ft )
This last value of Qg emanating from the moment capacity is much larger than the value of Qg = 12139 lbf found for the deflection criterion. Therefore use, Qg = 12139 lbsf
444
**Lateral loads on piles-03: The Broms method for pile lateral loads. (Revised Nov-09)
For the previous problem, use the Broms method to determine the lateral load capacity Qg of a flexible 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 free and allowed to only deflect 8 mm. The unit weight of the soil is γ = 18 kN/m3 and the angle of internal friction is φ = 35°.
Solution: The moment at steel’s yield failure is,
M y = S( Fy ) =
) 248 x10 F ) ( ( /2 ( 0.254 m ) / 2
Ip d1
y
(123 x10 =
−6
m4
3
)
kN / m 2 = 240 kN − m
Therefore, the unitless ratio My D γ Kp 4
=
( 240 kN − m )
( 0.250 m )
4
35° 18 kN / m tan 45° − 2
(
3
)
= 926
2
Broms’ graphs are used to determine the ultimate lateral resistance for long piles in sand shown in (a), or in clays shown in (b). We use (a):
445
Using Broms’ graph (a) for granular soils, the ratio of 926 and a ratio of e / D = 0 yields an ultimate lateral resistance ratio of approximately 140, therefore,
35° 3 Qultimate = 140 K p D 3γ = 140 tan 2 45° − ( 0.250 m ) 18 kN / m 3 = 581 kN 2 and the deflection η of the head is given by,
(
η= 5
nh = EpI p
12 MN / m 3
5
( 207 x10 )(123 x10 ) 6
−6
)
= 0.86 m −1
η L = ( 0.86 m −1 ) ( 25 m ) = 21.5 and using
e = 0 which is a free-headed pile. L
Using Broms’ graph for estimating the deflection of the pile head in sands (a) and in clays (b), we find for sands a value of 0.15, Broms’ graphical solutions for estimating the deflection of the pile head in granular soils is shown below, and for cohesive soils below that one,
446
447
Using figure (a) for the granular soils, the ratio yields,
(
∆ EpIp
) (n ) 3/ 5
2/5
≈ 0.15 by interpolation
h
Qg L
∴ Qg =
(
∆ EpI p
) (n ) 3/ 5
2/ 5
h
0.15 L
( 0.008 m ) ( 207 x106 )(123 x10−6 ) (12 , 000 ) Qg = 0.15 ( 25 m ) 3/ 5
2/ 5
= 40 kN
This value 40 kN is much smaller than the 581 kN found above, therefore,
The lateral capacity of the pile Qg = 40 k.
448
***Lateral loads on piles-03: Lateral load capacity of a mooring bollard. (Revised Nov-09)
Design a steel sheet-pile cellular dolphin for a mooring bollard that will be subjected to an equivalent horizontal load of 60 kips. The live load upon the dolphin will be 250 psf, and assume that there are negligible wave forces because it is located in a shielded area. The mooring geometry limits the diameter of the dolphin to 20 feet or less. Choose a FS = 2.5. The site conditions are shown below.
Solution: Step 1: Determine the Required Depth of Embedment D of the steel sheet-pile cell. For φ = 30° Ka = 0.33 and Kp = 3.0 and Kp – Ka = 2.67
449
(
)
Apply the factor of safety to the K p − K a factor,
(K
− Ka
p
)
=
(K
p
− Ka
)
actual
=
( 3.0 − 0.33) = ( 2.67 ) = 1.07
FS 2.5 2.5 Using Tschebotarioff's analysis, the forces E1 and E 2 are, reduced
E1 = area ( abc ) 2d E 2 = area ( cfg ) 2d
where d is the cell diameter = 20 feet In granualar soils, the maximum pressure pmax is
(
)
pmax = K p − K a γ D = (1.07 ) γ D
The embedment depth D is usually estimated through trial and error. The depth (ac) is estimated by having the force E1 be a bit larger than the mooring force P. Now locate the point (f) so that the force E2 will satisfy equilibrium, or
∑F
H
=0
P + E2 = E1
Taking moments about the center of the circular base yields,
∑M
a
P h = ( E1 + E2 )( 0.318) d tan (δ ) + E2e2 − E1e1
=0
and with δ = 20° ,
P h=E2 ( 0.12d + e2 ) − E1 ( e1 − 0.12d )
First try D = 20 feet. This yields ac = 16 feet and cf = 4 feet Therefore, 1 (16' )(16' )( 0.065 kcf )( 40' ) = 333 kips 2 1 E2 = ( 4' )( 2.67' )( 0.065 kcf )( 20 )( 40' ) = 278 kips 2 Check equilibrium, ∑ FH = 0 P + E2 = E1 ∴ 60 kips + 278 kips ≈ 333 kips GOOD
E1 =
∑M
a
= 0 P h=E2 ( 0.12d + e2 ) − E1 ( e1 − 0.12d )
(16' ) = 10.7 feet and e2 = 16' + 23 ( 4' ) = 18.7 feet ( 60 kips )( 42 ft ) = ( 278 kips ) ( 0.12 )( 20' ) + 18.7' − ( 333 kips ) 10.7' − ( 0.12 )( 20' ) 2 , 520 kips − ft < ( 5 , 866 − 2 , 764 ) kips − ft = 3,102 kips − ft
where e1 = ∴ ∴
2 3
∴ the D = 20-foot embedment is conservative (could be reduced).
450
Step 2: Check the cell’s Stability against the Over-turning Moment.
The forces F1 and F2 are ,
F1 = E1 tan δ = E1 tan ( 23 ϕ ) = ( 333 kips )( 0.36 ) = 121 kips
F2 = E2 tan δ = E 2 tan ( 23 ϕ ) = ( 278 kips )( 0.36 ) = 101 kips
and the distance f = ( 0.318 ) d = ( 0.318 )( 20' ) = 6.36 feet Find the total weight of the cell above a plane throuogh point A above,
π D2
π ( 20' )
2
( 2' )( 0.150 kcf ) + ( 4' )( 0.106 kcf ) + ( 55' )( 0.065 kcf ) = 1, 350 kips h= 4 4 Note that W ≡ R in the figure.
W =
The total moment about point A is,
∑M ∑M
A
= ( 62' )( 60 k ) − ( 9.3' )( 333 k ) − ( 6.36' )(121 k ) − ( 6.36' )(101 k ) + (1.33' )( 278 k )
A
= −400 kip − ft
Therefore, the reactive force R has an eccentricity e of, M −400 kip − ft M A = Re ∴ e = A = = 0.25 ft R 1, 350 kip This eccentricity is negligible, and clearly the resultant falls in the middle third. The cell is stable with respect to the over-turning moment.
451
Step 3: Check the cell’s Stability against Horizontal Sliding. It is clear from the figure that the cell will tilt before it will slide. Therefore, this analysis is unnecessary.
Step 4: Check the Bearing Capacity of the soil beneath the cell. The ultimate bearing capacity of a soil under a cellular dolphin is given by the formula shown below. The bearing capacity factors are functions of the angle of internal friction φ = 30°, Nq = 22.5 and Nγ = 20.
The ultimate bearing capacity of the soil beneath the cell is, 1 1 qultimate = qN q + ( 0.6 ) γ dN γ = γ D f N q + ( 0.6 ) γ dN γ 2 2 1 qultimate = ( 0.065 )( 20' )( 22.5 ) + ( 0.6 )( 0.065 )( 20' )( 20 ) = 37 ksf 2 The total load pressure ptotal from the cell is,
ptotal = ( 2' )( 0.150 pcf ) + ( 4' )( 0.106 pcf ) + ( 55' )( 0.065 pcf ) + ( 0.250 psf ) = 4.6 ksf
The Factor of Safety is FS BearingCapacity =
qultimate 37 ksf = = 8 GOOD 4.6 ksf ptotal
Step 5: Check the steel Sheet-piling inter-lock Tension. The maximum internal pressure at ground level is,
Use an "at-rest" lateral pressure rather than an "active" pressure, Ko = 1− tanϕ = 1− tan( 30°) = 0.50 Therefore, the pressure pa is given by, pa = Ko ∑( γ z ) = ( 0.50) ( 2' )(150 pcf ) + ( 4' )( 0.106 pcf ) + ( 35' )( 65 pcf ) + 250 = 1,625 psf The tension is, 1 1 T = pa D = (1,625 psf )( 20' ) = 16, 250 lb / feet = 1,350 lb / inch 2 2 Use a PSA-23 steel sheet-pile. The wall requires 46 pieces. The actual diameter is 19 feet 6.25 inches.
452
Chapter 24 Reinforced Concrete Retaining Walls and Bridge Abutments Symbols for Reinforced Concrete Retaining Walls
453
**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). 3 1'
1
2
15'
18.33'
3
16'-6"
1.83'
1
Pv Ph
6.11'
6
1'-6"
2'
4
5 3'
1'-6"
5'-6" 10'
Solution:
Step 1: Find the lateral pressures and forces. The active lateral earth pressure K a is,
φ 30° 2 K a = tan 2 45° − = tan 2 45° − = tan ( 30° ) = 0.33 2 2 The presure at the base (invert) of the wall footing is, pb = γ HK a = ( 0.120kcf )(18.33 ft )( 0.33) = 0.733ksf The forces on the wall (per unit length) are, Pv = ½(12)(18.33)² = 2.0 kips Ph = ½(39)(18.33)² = 6.6 kips Step 2: Stability analysis for sliding and overturning.
454
Moment About A Area
Area
Force (kip)
Arm (ft)
Moment (kip-ft)
1
½ x 5.5x 1.83
= 5.03 x 0.125 =
0.63
1.83
1.2
2
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 Pv
3.0 x 2.0
= 6.0 x 0.125 =
0.75 2.00
8.50
6.4
6.11
40.4
ΣH = 6.60
Ph
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 =
Pp =
Min F =
455
cos δ = 5.8 kips 1/2 γ H 2
(5.8 )(0.125 )(3.5 )2 = 4.7 kips (2 )(0.940 )
10.9 10.9 + 4.7 = 1.7 kips , Max F = = 2.4 kips 66 66
OK without Key
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 12"
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” 15'
T = 448/13 = 34.5 kip As = 34.5 / (40 x 0.90) = 0.96 in²/ft
Pv Ph
At bottom of wall Use #6 @ 5” ctrs
5'
As = 1.06 in²/ft ρ=1.06/(14 x 12) = 0.0063 >0.005 and <0.0278 OK
14"
456
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
#6 @ 10"ctrs 8'-10"
17 kip-in
0
2"
143 kip-in
286 kip-in
At top : Mr = 106 (40) (0.90) (7.5) = 286 kip-in
133 kip-in
5'
Ld = 40.9 (0.44) = 1' - 6"
10'
6'
#6 @ 5"ctrs
top of base
457
496 kip-in
448 kip-in
0
248 kip-in
15'
14" plus std. hook
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
1'-6"
3'
front face of stem
d =14"
V = 5.15 x (3.0) = 15.45 kips 5.15 kip
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 458
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
459
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
2.33 ksf 2.11 ksf
1'-6"
V1 = ½ (2.11) (5.5) = 5.80 kips d =14"
5'-6"
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: 460
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" #4@ 9" front face #4@18" back face Vert. bars, back face #6 @ 10"
Provide 3" clear to all bars except 112 " clear to bars in front face of stem
16'-6"
#4 @ 2'-0"
3'
1'-6"
5'-6"
#6 @ 6"
2"x6" key
#6 @ 6"
1'-6"
2'
Vert. bars, back face #6 @ 5"
6" drains @ 10'-0"
461
#4 @ 12"
**RC Retaining Walls–02: Design a RC bridge abutment wall. (Revised Nov-09)
Design a bridge abutment for a backfill of γ = 110 lb/ft3, an equivalent surcharge of 3 feet, an allowable soil pressure of 3.5 kips/ft2, and an allowable shear between soil and base of 45% of the vertical shear.
L v =5.26 k
4
L h =0.24 k
3.88'
12" 12"
5
6
12'-0"
3
12" 5'-6"
7
2 3'-6"
12"
1'-6"
4'-0"
2'-6"
1
6'-0"
3'-0"
9"
8 A 7'-0" Note:
All loads and moments are per lineal foot of abutment.
Solution: Phs = 30(3.0)(12) = 1.08 kips Ph = 1/2(30)(122) = 2.16 kips
462
Stability Computations: Moments about A
Area 1 2 3 4 5 6 7 8
2.5 x 10.5 x 0.110 1.0 x 5.5 x 0.110 1/2 x 1.0 x 1.0 x 0.110 0.75 x 3.0 x 0.150 2.0 x 1.0 x 0.150 1/2 x 1.0 x 1.0 x 0.150 1.0 x 6.5 x 0.150 7.0 x 1.5 x 0.150
= = = = = = = =
Lv
Force 2.89 0.61 0.06 0.34 0.30 0.08 0.97 1.58
Arm 1.25 3.00 2.83 2.88 3.50 3.17 4.00 3.50
Moment 3.61 1.83 0.15 0.98 1.05 0.25 3.88 5.53
5.26
3.88
20.40
Lh
0.24
9.00
2.16
Phs
1.08
6.00
6.48
Ph
2.16
4.00
8.64
SH = 3.48 kips
SV = 12.09 kips
S MA= 54.96 ft-kips
Location of Resultant From point A,
54.96 7 .0 B = 4.55’ then e = 4.55 = 1.05’ < 12.09 2 6
Soil Pressure at Base: At toe qmax =
12.09 6 × 1.05 1 + 7 7
qmax = 1.73 (1 + 0.9) = 3.3 k/ft2 < 3.5
ok
At heel qmin = 1.73 (1 - 0.9) = 0.2 k/ft2 Sliding: Shear available along base = 12.09 k x 0.45 = 5.43 kips Factor of safety =
463
5.43 = 1.6 > 1.5 3.48
ok
ok
**RC Retaining Walls–02: Design a RC bridge abutment wall. (Revised Nov-09)
Design a bridge abutment for a backfill of _ = 110 lb/ft3, an equivalent surcharge of ___ feet, an allowable soil pressure of 3.5 kips/ft2, and an allowable shear between soil and base of 45% of the vertical shear.
All loads and moments are per lineal foot of abutment.
Solution: Phs = 30(3.0)(12) = 1.08 kips Ph = 1/2(30)(122) = 2.16 kips
464
Stability Computations: Moments about A Area
Arm
Moment
1
2.5x10.5x0.11
=
2.89
1.25
3.61
2
1x5.5x0.11
=
0.61
3
1.83
3
1/2x1x1x0.11
=
0.06
2.83
0.15
4
0.75x3x0.15
=
0.34
2.88
0.98
5
2x1x0.15
=
0.3
3.5
1.05
6
.5x1x1x0.15
=
0.08
3.17
0.25
7
1x6.5x.15
=
0.97
4
3.88
8
7x1.5x0.15
=
1.58
3.5
5.53
Lv
5.26
3.88
20.4
Lh
0.24
9
2.16
Phs
1.08
6
6.48
Ph
2.16
4
8.64
H
465
Force
3.48kips
V
12.09kips
MA
54.96kips
Location of Resultant
From point A,
54.96 7.0 B = 4.55 ft then e = 4.55 = 1.05 ft < 12.09 2 6
OK
Soil Pressure at Base:
Sliding:
Design the stem:
466
Vertical Reinforcement Ph + Phs = 8.64 +6.48 = 15.12 M = 15.12 x 12/3 x 12 = 726 kip – in Use 6” batter on front, then t = 9 + 6 = 15 “ Assume arm 15 – 4 = 11” T = 726/11 = 66 kip As = 66/ 40 x 0.9 = 1.83 in^2/ft At bottom of wall Use #5 @ 2in =1.86in^2/ft or #9 at 6” c.c. = 2.00in^2/ft p = 2.00/ 14 x 9 = 0.015 >.005 < 0.0278 OK Check compressive stress block C = T = 66kip / 2.55 x 12 x 0.9 = 2.39 >2 NOT GOOD Check Shear 15.12 / 14x9 x .75 = 160psi > 110 psi. Need Shear Design for stirrups Resisting Moment Top : 2 x 40 x .9 x 9 = 648 kip –in Bottom : 2 x 40 x.9 x 14 = 1008 kip – in Horizontal shrinkage in stem Required = 0.002 x 14 x 9 = 0.336in^2/ft Use #4@9” c.c front As = 0.27in^2/ft Use #4@18” c.c back As = 0.13in^2/ft Total As = 0.27 + 0.13 = 0.40
467
Chapter 25 Steel Sheet Pile Retaining Walls Symbols for Steel Sheet Pile Retaining Walls
468
**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.
γ = 115 pcf γ ' = 65 pcf φ = 35° therefore K a = 0.27 and K p = 3.70
Solution: Step 1. Determine the pressures on the cantilevered wall.
Step 2. Determine the depth of embedment D. The procedure will follow the following steps: a) b) c) d) 469
From statics, find the dimension z shown in the figure of previous page, Assume a value for D, Calculate z, Use z to check if ΣM = 0; adjust D for convergence.
(1)
ΣFH = 0
Area ( BAA1 ) + Area ( AA1 A2 F ) + Area ( ECJ ) − Area ( EA1 A2 ) = 0
1 D z D Hp A1 + ( p A1 + p A 2 ) + ( pE + pJ ) − ( pE + p A 2 ) = 0 2 2 2 2 Solving for z , z=
( 2)
( pE − pA1 ) D − Hp1 p E + pJ ΣM = 0 about any po int, say F ,
D2 z2 D2 D2 1 H ( H ) pA1 D + + ( pA1 ) + ( pE + pJ ) − ( pE + pA2 ) + ( p A2 − pA1 ) = 0 2 3 2 6 6 6
p A1 = γ HK a = (115 )(14 )( 0.27 ) = 435 psf p A 2 = p A1 + γ DK a
= ( 435 ) + ( 65 )( 0.27 ) D = 435 + 17.6 D
(
)
pE = γ D K p − K a − p A1 = 65 D ( 6.29 ) − 435
= 409 D − 435
(
)
pJ = γ D K p − K a + γ HK p
= 65 D ( 6.29 ) + (115 )(14 )( 6.56 ) = 409 D + 10 , 560
Try D = 10.5 feet , p A1 = 435 psf , p A 2 = 620 psf , pJ = 14 , 860 psf pE = 3 , 860 psf therefore, z =
( 3,860 − 435)(10.5) − (14 )( 435) = 1.60 14 , 860 + 3 , 860
feet
(10.5 ) (10.5 ) 1 ∑ M F = 2 (14 )( 435)(10.5 + 4.67 ) + ( 435) 2 + ( 620 − 435) 6 2
+ ( 3 , 860 + 14 , 860 )
(1.60 )
2
(10.5)
2
2
− ( 3 , 860 + 620 ) = −730 ft − lb 6 6 This value is close to zero, therefore adopt D = 10.5 feet .
Increase D by 30% as a factor of safety ,∴ D = 13.5 feet .
470
Step 3. Determine the maximum moment Mmax (point of zero shear).
y=
p A1
γ ( K p − Ka ) '
=
435 = 1.06 feet ( 65)( 6.29 )
1 1 p A1 H = ( 435 )(14 ) = 3 , 040 lb 2 2 1 1 P2 = p A1 y = ( 435 )(1.0 ) = 218 lb 2 2 also, 1 ' γ K p − K a x 2 = P1 + P2 2 P1 =
(
∴ x2 =
2 ( P1 + P2 )
γ ( K p − Ka ) '
(
=
2 ( 3 , 040 + 218 )
( 65)( 6.29 )
)
= 16 ∴ x = 4.0 feet
)
1 2 P3 = γ ' K p − K a ( 4 ) = 3 , 280 lb 2 M max = P1l1 + P2 l2 − P3 l3 where , H + y+ x 3 2y +x l2 = 3 x l3 = 3 l1 =
2 (1.0 ) 14 4 .0 M max = P1l1 + P2 l2 − P3l3 = ( 3 , 040 ) + 1.0 + 4.0 + ( 218 ) + 4.0 − ( 3 , 280 ) 3 3 3 M max = 29 , 300 + 1, 030 − 4 , 360 = 26 , 000 ft − lb = 312 in − kips
471
472
If instead , an Ex − Ten − 45 steel sec tion is used , f s = 29 ksi The required sec tion mod ulus S =
M max ( 26 , 000 )(12 ) = = 10.8 in3 fs ( 29 , 000 )
∴ Use a MP − 116 that provides a S = 10.7 in3 at a lower cos t . This is shown plotted with the red lines.
Step 5. Determine the sheet-pile length L. The total sheet-pile length L = H + D = 14.0 + 13.5 = 27.5 feet
473
**Sheet-pile Wall-02: Free-Earth for cantilevered walls in cohesive soils. (Updated: 9 April 2008)
Determine the required length L of steel sheet-piles for the site conditions shown below for an unanchored wall.
γ = 120 pcf γ ' = 60 pcf c = 750 psf
φ = 0°
Solution: Step 1. Determine all the pressures on the cantilevered wall.
Step 2. Apply a Factor of Safety = 1.5 to the cohesive strength of the soil. F S = 1 .5 =
750 c red u ced
∴
c red u ced =
750 = 50 0 p sf 1 .5
q u = 2 c = 2 (5 0 0 ) = 1 , 0 0 0 p s f T h e fin a l s tr e n g th ( lo n g − te r m s tr e n g th o f th e s o il ) b e c o m e s , c = 0
φ = 27° T h e c r itic a l h e ig h t H H
c
=
4c
γ
=
4 (5 0 0 )
(1 2 0 )
c
is ,
= 1 6 .7 fe e t > 1 4 fe e t
474
Step 3. Determine the wall pressures.
H o = po int of zero pressure =
2c
γ
=
2 ( 500 )
(120 )
= 8.3 feet
H − H o = 14.0 − 8.3 = 5.7 feet
1 ) γ H − 2c = (120 )(14.0 ) − 2 ( 500 ) = 680 psf
2 ) 4c − γ H = 4 ( 500 ) − (120 )(14.0 ) = 320 psf
3 ) 4c + γ H = 4 ( 500 ) + (120 )(14.0 ) = 3 , 680 psf Step 4. 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 slide #12, b) Assume a value for D, c) Calculate z, d) Use z to check if ΣM = 0; adjust D for convergence.
(1)
ΣFH = 0
Area ( O' A' A ) + Area ( CJE ) − Area ( BAFE ) = 0
8c ( z ) 1 − ( 4c − γ H ) D = 0 ( γ H − 2c )( H − H o ) + 2 2 Solving for z , z=
( 2)
2 D ( 4c − γ H ) − ( γ H − 2c )( H − H o ) 8c ΣM = 0 about any po int, say F ,
∑ MF =
H − Ho 1 ( γ H − 2c )( H − H o ) D + 2 3
D2 8cz 2 c H + − 4 − γ ( ) =0 6 2
The method of solution is as follows, 1) Assume a depth of penetration D, 2) Calculate the value of z (from ΣFH = 0), 3) Substitute z into the ΣMF = 0 equation and check to see if it is zero, 4) Adjust D and reiterate if necessary.
475
z=
2 D ( 4c − γ H ) − ( γ H − 2c )( H − H o ) 8c
=
2 ( 320 ) D − 5.7 ( 680 ) 8 ( 500 )
= 0.16 D − 0.97
and H − H o 8cz 2 1 D2 ∑ M F = 2 (γ H − 2c )( H − H o ) D + 3 + 6 − ( 4c − γ H ) 2 = 0 680 )( 5.7 )( D + 1.90 ) 8 ( 500 ) z 2 320 D 2 ( = + = =0 2 6 2 1938 ( D + 1.90 ) + 666.7 z 2 − 160 D 2 = 0 ∴ z2 =
160 D 2 − 1938 ( D + 1.90 ) 666.7
Try D = 14.0 feet ,
therefore , z = ( 0.16 )(14.0 ) − ( 0.97 ) = 1.27 feet
Check ∑ M F =0
(160 )(14.0 ) − (1,938)(14.0 + 1.90 ) = 0.82 = 2
z
2
666.7 z = 0.9 feet < 1.27 feet ∴ not satisfactory
Try D = 14.5 feet ,
therefore , z = ( 0.16 )(14.5 ) − ( 0.97 ) = 1.35 feet
Check ∑ M F =0
(160 )(14.5) − (1, 938)(14.5 + 1.90 ) = 2.79 = 2
z
2
666.7 z = 1.65 feet > 1.27 feet ∴ acceptable.
Step 5. Determine the required sheet-pile length L. The total sheet-pile length L is, L = H + D = 14 + 14.5 = 28.5 feet.
476
** Sheet-pile Wall–03: Free-Earth Method f/anchored walls in granular soils. (Updated: 9 April 2008).
Select an appropriate steel sheet-pile section and its total length L to retain a sand backfill for the conditions shown below without an anchoring system.
Sand backfill : γ = 110 pcf γ ' = 60 pcf φ = 34° ∴ K a = 0.28
In − situ medium sand : γ ' = 65 pcf
φ ' = 34.5° ∴ K 'a = 0.26 and K 'p = 3.85 Solution: Step 1. Determine the pressures on the wall.
pB = γ H1 K a = (110 )(10 )( 0.28 ) = 308 psf
pC 1 = pB + γ H w K a = ( 308 ) + ( 60 )( 26 )( 0.28 ) = 308 + 437 = 745 psf pC 2 = [γ H1 + γ H w ] K a' = (110 )(10 ) + ( 60 )( 26 ) ( 0.26 ) = 692 psf
(
)
pE = γ D1 K p − K a = ( 65 ) D1 ( 6.37 ) = 414 D1 Determine the pressure on the wall from the surcharge, a ) above point C : psurcharg e = 300 K a = ( 300 )( 0.28 ) = 84 psf b ) below point C : psurcharg e = 300 K a' = ( 300 )( 0.26 ) = 78 psf Determine the pressure on the wall due to the line load , K 4 .0 = = 0.111 H 36 pH = ( 0.55 ) Q = ( 0.55 )( 2 , 000 ) = 1,100 lbs
using Lecture #18 theory, m =
with its resultant at = ( 0.60 ) H = ( 0.60 )( 36 ) = 21.6 feet from the dredge line Locating the point y, 477
y=
pC 2
γ '(K − K ' p
' a
( 692 ) = 1.67 ) ( 65)( 6.37 ) =
feet
The resultant forces from the pressures are,
1 1 H 1 pB = (10 )( 308 ) = 1, 540 plf 2 2 P2 = H w pB = ( 26 )( 308 ) = 8 , 008 plf P1 =
1 1 H w ( pC 1 − pB ) = ( 26 )( 745 − 308 ) = 5 , 681 plf 2 2 1 1 P4 = ypC 2 = (1.67 )( 692 ) = 578 plf 2 2 Psurcharg e ( + C ) = ( H 1 + H w ) psurcharg e = ( 36 )( 84 ) = 3 , 024 plf P3 =
Psurcharg e ( − C ) = Dpsurcharg e = ( D1 + 1.67 )( 78 ) = 78 D1 + 130 plf P5 =
1 1 pDE D1 = ( 414 D1 ) D1 = 207 D12 2 2
Step 2. Determine the depth of penetration D.
Solve for D using
∑M
Force (lb)
ap
= 0, Arm (ft)
Moment (ft-lb)
1)
1,540
- 2.33
-3,588
2) 3)
8,008 5,681
+14.00 +18.33
+112,112 +104,133
4) Sur(+C)
578 3,024
+26.56 +9.00
+15,930 +27,216
D + 1.67 + 27.0 Sur(-C) 78D1 +130 + 1 39D12 +2,236D1 +3,619 2 Line-load 1,100 +5.40 +5,940
5)
2D 207D12 - 27+1.67+ 1 - 138D13 + 5 , 935 D12 3 ∑ ( +M ) = 39 D12 + 2 , 236 D1 + 268 , 950
but
∑ ( -M ) = 138 D + 5 , 935 D ∑ ( +M ) = ∑ ( − M )
(
Passive
3 1
2 1
)
+ 3 , 588
39 D12 + 2 , 236 D1 + 268 , 950 = 138 D13 + 5 , 935 D12 + 3, 588 D13 + 42.72 D12 − 16.20 D1 − 1, 923 = 0
∴ D1 = 6.50 feet
The total depth of penetration D = D1 + y = 6.50 + 1.67 = 8.2 feet. Apply a Factor of Safety of about 30% to the calculated depth D = 11.0 feet.
478
Step 3. Determine the tie rod’s tension T. T = Active forces – Passive forces = 1,540 + 8,008 + 5,681 + 578 + 3,024 + (78)(6.5) + 1,100 – (207)(6.5)2 = 11,870 lb Apply a Factor of Safety of about 30% to the tie rod tension, so T = 16,000 lb.
Step 4. Determine the maximum moment Mmax in the sheet-pile. Let the point of zero shear be at a distance “x” below the low water level.
1 (16.8) x 2 2 2 ∴ x + 37 x − 1,100 = 0 ∴ x = 20 feet 11,870 = 1, 540 + 1,100 + 308 x +
The moments are, (1 )
− (1, 540 )( 20 + 3.33) = − 35 , 900
(2)
( 308)( 20 ) -
(3)
(16.8)( 20 ) ( 20 ) = -
2
2
= − 61, 600
2
(4)
− 22 , 400 2 + ( 11,870 )( 20 + 1.0 ) = + 249 , 000
( 5 ) - ( 1,100 )( 20 + 10 − 14.4 ) =
− 17 ,150
Maximum moment
+112,000 ft-lb = 112 ft-kip = 1,340 in-kip
=
Step 5. Select the sheet-pile section. Selecting the USS Ex-Ten 50 steel provides an allowable stress = 32 ksi. The required section modulus S is,
S=
479
M max 1, 340 in − kips = = 42 in3 2 f all 32 kips / in
Initially, assume that there is no moment reduction due to the flexibility of the steel sheet-piling. Therefore, select a section MZ-38, that has a S = 46.8 in3 / foot of wall.
480
Step 6. Apply Rowe’s Theory of Moment Reduction. (1) Select the appropriate moment reduction curve (see next slide #12) corresponding to the relative density of the sand. For this case, the curve for medium compact and coarse grained sand is used. (H + D )4 for the sheet-pile sections being considered. Calculate the ratio P = EI Obtain the ratio,
M design M max imum
From the curves (next slide) calculate the design moment and the stress for the sheet pile that was selected on page 472.
( H + D) P= EI
481
4
( 36 + 11)(12 ) 3 , 373 = = 6 I ( 30 )(10 ) I 4
Pile Sections M Z-32
M Z-38 Section m odulus S (per foot) M om . of Inertia I (per foot) 3 , 373 P = I R atio = M de sign / M m ax . M design = R atio( M m ax . ) = Ratio (1,340) M Stress = design S
46.8 in
3
280.8 in 4
M Z-27
3
30.2 in 3
220.4 in 4
184.2 in 4
38.3 in
12.0
15.3
18.3
0.73
0.67
0.62
900 in-kip
833 in-kip
980 in-kip 980 in-kip 46 . 8 in 3 21.0 ksi
900 in-kip 38 . 3 in 3 23.5 ksi
833 in-kip 30 . 2 in 3 27.6 ksi
Based on Rowe’s Theory of Moment Reduction, the following sheet-pile selections may be used,
MZ-38 Regular carbon grade steel, fs = 21.0 ksi < 25 ksi MZ-32 Regular carbon grade steel, fs = 23.5 ksi < 25 ksi MZ-27 Ex-Ten 45 grade steel,
fs = 27.6 ksi < 29 ksi
Step 7. Establish the length of the steel sheet-pile L.
L = H1 + Hw + D = 10 feet + 26 feet + 11 feet = 47 feet.
482
** Sheet-pile Wall–04: Free-Earth Method anchored walls in cohesive soils. (Updated: 9 April 2008).
Find the anchor force T and the steel sheet-pile length L for these conditions,
Sand backfill
γ =115 pcf and γ ' =60 pcf φ =32°∴ K a = 0.33
In-situ soft clay
γ =120 pcf and γ '=65 pcf c=400 psf
φ =0°∴ K a = 1.0
In − situ medium soft clay
γ '=65 pcf c=850 psf
φ =0°∴ K a = 1.0 Solution: Step 1. Introduce a Factor of Safety for the cohesion. The shear strength of the cohesion is typically reduced through a Factor of Safety of between 1.5 to 2.0; let us use FS = 1.7,
cmeasured 850 psf = = 500 psf FS 1.7 Step 2. Determine the pressures on the wall. creduced =
p1 = γ HK a = (115 )( 6 )( 0.33) = 228 psf
pB −a = γ ( H − H B ) K a = (115 )( 6 )( 0.33) + ( 60 )( 2.5 )( 0.33) = 228 + 50 = 278 psf pB −b = γ ( H − H B ) − 2c = (115 )( 6 ) + ( 60 )( 2.5 ) − ( 2 )( 400 ) = 40 psf pC −c = pB −b + γ H B = 40 + ( 65 )(11.5 ) = 788 psf
pC −d = γ H − 2c = (115 )( 6 ) + ( 60 )( 2.5 ) + ( 65 )(11.5 ) − ( 2 )( 500 ) = 588 psf pC − f = 2c = 2 ( 500 ) = 1, 000 psf
pC − net = pC − f − pC −d = 1, 000 − 588 = 412 psf 483
Step 3. Find the resultant forces on the wall.
(115)( 6 ) ( 0.33) = 684 plf 1 P1 = γ H 2 K a = 2 2 1 1 P2 = ( p1 + pB − a )( H a − H B ) = ( 228 + 278 )( 2.5 ) = 633 plf 2 2 1 1 P3 = ( pB − b + pC −c )( H B ) = ( 40 + 788 )(11.5 ) = 4 , 760 plf 2 2 and the c.g. at 8.5 ft + 7.48 ft = 15.98 ft from point A. 2
P4 = 412 D Step 4. Find the depth of embedment D.
From statics, the following conditions must be satisfied,
∑ F = P + P + P − P −T = 0 ( 2 ) ∑ M ( about the anchor tie rod ) = 0 (1)
∴
H
∑M
ap
1
2
3
4
D = ( 684 )(1) − ( 633)(1 + 1.29 ) − ( 4 , 760 )(15.98 − 5.0 ) + 412 D + 15 = 0 2
Solving the two equations, D 2 + 30 D − 257.4 = 0
∴ D = 7.0 feet The required length L of the sheet-piling is, L= H + D = 20 + 7 = 27 feet Step 5. Find the tensile force in the anchor tie rod T.
Using the equation for
∑F
H
= 0,
T = 6 , 077 − ( 412 )( 7 ) ≈ 3 , 200 lb per foot If the anchors are spaced every 20 feet, each anchor carries, A = T s = ( 3,200 )( 20 ) = 64 , 000 lbs = 64 kips
484
Chapter 26 MSE (Mechanically Stabilized Earth) Walls Symbols for Mechanically Stabilized Earth Walls
485
**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.
16 feet
Solution: The active pressure on the wall is,
σ a = K aσ v = K a ( γ z ) where K a = tan2 ( 45° − ϕ / 2 ) = tan2 ( 45° − 36° / 2 ) = 0.26 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. SV =
σG σG = σ a FS ( γ zK a ) FS
At a depth of z = 8 feet from the top, SV =
σG
(γ zK a ) FS
=
( 80 x12 lb / ft ) = 2.8 feet ≈ 34 inches (110 )(8)( 0.26 )(1.5 )
At a depth of z = 12 feet from the top, SV =
σG
(γ zK a ) FS
=
( 80 x12 lb / ft ) = 1.9 (110 )(12 )( 0.26 )(1.5)
feet ≈ 22 inches
At a depth of z = 16 feet from the top, SV =
σG
(γ zK a ) FS
=
( 80 x12 lb / ft ) = 1.4 (110 )(16 )( 0.26 )(1.5)
feet ≈ 17 inches
486
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.
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,
S ( 0.26 )(1.5 ) ( H − z ) + SV K a ( FS ) = (H − z) + V tan ( 45° − φ / 2 ) 2 tan ϕ F tan ( 45° − 36° / 2 ) 2 ( 0.445 ) L = ( 0.51)( H − z ) + 0.438 SV L = l R + le =
487
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, ll =
SV σ a ( FS )
4σ v tan ϕ F
=
SV ( 0.26 )(1.5 )
4 tan ( 23 ) ( 36° )
= 0.219 SV
For a depth of z=16 inches, 20 inches l l = 0.219 SV = ( 0.219 ) = 0.365 feet < 3 feet 12 in / ft Therefore, use ll = 3 feet
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:
488
