300 Solved ProblemsSoil / Rock Mechanics and Foundations Engineering These notes are provided to you by Professor Prieto-Portar, and in exchange, he will be grateful for your comments on improvements. All problems are graded according to difficulty as follows: * Easy; defines general principles; typical of the PE examination; ** Slightly more difficult; typical of Master’s level problems; *** Professional level (“real-life”) problems. by Luis A. Prieto-Portar PhD, PE Professor of Civil and Environmental Engineering Florida International University, Miami, Florida Former Professor, United States Military Academy (West Point) Telephone 305-348-2825 / Fax 305-348-2802 Website: http://web.eng.fiu.edu/prieto/ Email:
[email protected] © Copyright by L. Prieto-Portar, October, 2009 ii Table of Contents Table of Contents................................................................................................................. i Chapter 1 Soil Exploration.................................................................................................. 1 Symbols for Soil Exploration .......................................................................................... 1 *Exploration–01. Find the required number of borings and their depth. ....................... 2 *Exploration–02. The sample’s disturbance due to the boring diameter. ...................... 3 *Exploration–03. Correcting the SPT for depth and sampling method. ......................... 4 *Exploration–04. Three methods used for SPT depth corrections.................................. 6 *Exploration–05. SPT corrections under a mat foundation. ........................................... 7 *Exploration–06. The Shear Vane Test determines the in-situ cohesion........................ 9 *Exploration–07. Reading a soil boring log................................................................. 10 *Exploration–08: Using a boring log to predict soil engineering parameters............... 11 **Exploration–09. Find the shear strength of a soil from the CPT Report. ................. 14 Chapter 2 Phase Relations of Soil..................................................................................... 16 Symbols for Phase Relations of soils ............................................................................ 16 Basic Concepts and Formulas for the Phases of Soils................................................... 17 *Phases of soils-01: Convert from metric units to SI and US units. ............................. 21 *Phases of soils–02: Compaction checked via the voids ratio. ................................... 22 *Phases of soils–03: Value of the moisture when fully saturated. ................................ 23 *Phases of soils–04: Finding the wrong data. ............................................................... 24 *Phases of soils–05: Increasing the saturation of a soil. ............................................... 25 *Phases of soils–06: Find γd, n, S and Ww. .................................................................. 26 *Phases of soils–07: Use the block diagram to find the degree of saturation. .............. 27 *Phases of soils–08: Same as Prob-07 but setting the total volume V=1 m3. ............... 28 *Phases of soils–09: Same as Problem #5 with a block diagram.................................. 29 *Phases of soils–10: Block diagram for a saturated soil. ............................................. 30 *Phases of soils–11: Find the weight of water needed for saturation. .......................... 31 *Phases of soils–12: Identify the wrong piece of data. ................................................. 32 *Phases of soils–13: The apparent cheapest soil is not!................................................ 33 *Phases of soils–14: Number of truck loads. ................................................................ 34 *Phases of soils–15: How many truck loads are needed for a project?......................... 35 *Phases of soils–16: Choose the cheapest fill supplier. ................................................ 36 i *Phases of soils–17: Use a matrix to the find the missing data..................................... 38 **Phases of soils–18: Find the voids ratio of“muck” (a highly organic soil). .............. 40 Chapter 3 Classification of Soils and Rocks..................................................................... 41 Symbols for Classification of soils................................................................................ 41 *Classify–01: Percentage of each of the four grain sizes (G, S, M & C)...................... 42 *Classify–02: Coefficients of uniformity and curvature of granular soils. ................... 43 *Classify-03: Classify two soils using the USCS.......................................................... 44 *Classify-04: Manufacturing a “new” soil. ................................................................... 45 Classify – 05 .................................................................................................................. 47 Classify – 06 .................................................................................................................. 48 Classify – 07 .................................................................................................................. 49 Classify – 08 .................................................................................................................. 50 Classify – 09 .................................................................................................................. 51 Classify – 10 .................................................................................................................. 52 Classify – 11 .................................................................................................................. 53 Chapter 4 Compaction and Soil Improvement.................................................................. 54 Symbols for Compaction ............................................................................................... 54 *Compaction–01: Find the optimum moisture content (OMC). ................................... 55 *Compaction–02: Find maximum dry unit weight in SI units. ..................................... 57 *Compaction-03: What is the saturation S at the OMC? .............................................. 59 *Compaction-04: Number of truck loads required........................................................ 61 *Compaction-05: What is the saturation S at the OMC? .............................................. 62 *Compaction-06: Definition of the relative compaction (RC)...................................... 63 *Compaction-07: The relative compaction (RC) of a soil. ........................................... 64 *Compaction-08: Converting volumes from borrow pits and truck loads. ................... 65 **Compaction-09: Ranges of water and fill required for a road................................... 66 **Compaction-10: Find the family of saturation curves for compaction...................... 68 **Compaction-11: Water needed to reach maximum density in the field. ................... 71 **Compaction-12: Fill volumes and truck load requirements for a levee. ................... 73 **Compaction-13: Multiple choice compaction problem of a levee. ........................... 75 Chapter 5 Permeability of Soils ........................................................................................ 78 Symbols for Permeability .............................................................................................. 78 *Permeability–01: Types of permeability tests and common units............................... 79 ii *Permeability-02: Use of Hazen’s formula to estimate the k of an aquifer. ................. 80 *Permeability-03: Flow in a sand layer from a canal to a river. ................................... 81 *Permeability-04: Find the equivalent horizontal permeability of two layers. ............. 82 *Permeability-05: Equivalent vertical and horizontal permeabilities. .......................... 83 *Permeability-06: Ratio of horizontal to vertical permeabilities. ................................. 84 *Permeability–07: Do not confuse a horizontal with a vertical permeability. .............. 85 *Permeability-08: Permeability as a function of the voids ratio e. ............................... 86 *Permeability–09: Uplift pressures from vertical flows................................................ 87 *Permeability-10: Capillary rise in tubes of differing diameters. ................................. 88 *Permeability-11: Rise of the water table due to capillarity saturation. ....................... 90 *Permeability-12: Find the capillary rise hc in a silt stratum using Hazen. .................. 91 *Permeability-13: Back-hoe trench test to estimate the field permeability................... 92 **Permeability-14: Seepage loss from an impounding pond........................................ 93 Chapter 6 Seepage and Flow-nets..................................................................................... 97 Symbols for Seepage and Flow-nets ............................................................................. 97 *Flownets-01: Correcting flawed flow-nets. ................................................................. 98 *Flow-nets-02: A flow-net beneath a dam with a partial cutoff wall............................ 99 *Flow-nets-03: The velocity of the flow at any point under a dam. ........................... 100 *Flow-nets-04: Flow through an earth levee............................................................... 101 *Flow-nets-05: Finding the total, static and dynamic heads in a dam. ....................... 102 **Flow nets-06: Hydraulic gradient profile within an earth levee.............................. 103 **Flow-net-07: Flow into a cofferdam and pump size................................................ 105 *Flow-nets-08: Drainage of deep excavations for buildings....................................... 108 *Flow-nets-09: Dewatering a construction site. .......................................................... 110 *Flow-net-10: Dewatering in layered strata. ............................................................... 111 **Flownets-11: Flow through the clay core of an earth dam. ..................................... 113 Chapter 7 Effective Stresses and Pore Water Pressure................................................... 117 Symbols for Effective Stresses and Pore Water Pressure............................................ 117 *Effective Stress–01: The concept of buoyancy. ........................................................ 118 *Effective Stress–02: The concept of effective stress. ................................................ 119 *Effective Stress–03: The concept of effective stress with multiple strata................. 120 Effective Stress-03B .................................................................................................... 121 Chapter 8 Dams and Levees ........................................................................................... 122 iii Symbols for Dams and Levees .................................................................................... 122 *Dams-01: Find the uplift pressure under a small concrete levee.............................. 123 *Dams-02: Determine the uplift forces acting upon a concrete dam. ......................... 124 Chapter 9 Stresses in Soil Masses................................................................................... 127 Symbols for Stresses in Soil Masses ........................................................................... 127 *Mohr-01: Simple transformation from principal to general stress state.................... 129 *Mohr – 02: Find the principal stresses and their orientation. .................................... 130 *Mohr – 03: Find the principal stresses and their orientation. .................................... 131 *Mohr – 04: ................................................................................................................. 132 *Mohr – 05: Normal and shear stress at a chosen plane. ............................................ 133 **Mohr – 07: Back figure the failure angle ................................................................ 134 *Mohr – 08: find the Principle pressure using Mohr .................................................. 135 *Mohr – 09: Relation between θ and φ. ...................................................................... 136 *Mohr – 10: ................................................................................................................. 137 *Mohr–11: ................................................................................................................... 138 *Mohr – 12: ................................................................................................................. 139 *Mohr – 13: Data from Mohr-Coulomb failure envelope........................................... 140 **Mohr – 14: ............................................................................................................... 141 *Mohr – 15: Derive the general formula for horizontal stress. ................................... 142 *Newmark–01: Stress beneath a tank at different depths............................................ 143 *Newmark-02: The stress below the center of the edge of a footing. ......................... 144 *Newmark-03: Stress at a point distant from the loaded footing. ............................... 145 *Newmark-04: Stresses coming from complex shaped foundations........................... 146 *Newmark-05: Stress beneath a circular oil tank....................................................... 147 **Newmark-06: Use Newmark with a settlement problem. ....................................... 148 *Stress–01: Stress increase at a point from several surface point loads...................... 150 *Stress-02: Find the stress under a rectangular footing............................................... 151 *Stress-03: The effect of the WT on the stress below a rectangular footing............... 152 *Stress–04: Finding the stress outside the footing area............................................... 153 *Stress-05: Stress below a footing at different points. ............................................... 154 *Stress-06: Stress increase from a surcharge load of limited width............................ 155 *Stress-07: Finding a stress increase from a surface load of limited width. ............... 156 **Stress-08: Stress increase as a function of depth..................................................... 157 iv Chapter 10 Elastic Settlements ....................................................................................... 158 Symbols for Elastic Settlements .................................................................................. 158 *Elastic Settlement-01: Settlement (rutting) of a truck tire......................................... 159 *Elastic Settlement-02: Schmertmann method used for granular soils....................... 160 *Elastic Settlement-03: Schmertmann method used for a deeper footings. ................ 161 *Elastic Settlement-04: The 2:1 method to calculate settlement................................. 163 *Elastic Settlement-05: Differential settlement between two columns....................... 165 *Elastic Settlement-06: Compare the settlements predicted by the Boussinesq, Westergaard, and the 2:1 methods............................................................................... 166 *Elastic Settlement-07: Schmertmann versus the strain methods. .............................. 169 *Elastic Settlement-08: The Schmertmann method in multiple strata. ....................... 170 **Elastic Settlement-09: Settlement of a mat foundation. .......................................... 172 Chapter 11 Plastic Settlements........................................................................................ 174 Symbols for Plastic Settlements .................................................................................. 174 *Plastic Settlement–01: Porewater pressure in a compressible soil. ........................... 175 *Plastic Settlement-02: Total settlement of a single layer. ......................................... 177 *Plastic Settlement-03: Boussinesq to reduce the stress with depth. .......................... 178 *Plastic Settlement -04: Surface loads with different units......................................... 180 *Plastic Settlement-05: Pre-consolidation pressure pc and index Cc........................... 181 *Plastic Settlement-06: Final voids ratio after consolidation...................................... 183 *Plastic Settlement-07: Settlement due to a lowered WT. .......................................... 184 *Plastic Settlement-08: The over-consolidation ratio (OCR)...................................... 185 **Plastic Settlement-09: Coefficient of consolidation Cv. .......................................... 186 *Plastic Settlement -10: Secondary rate of consolidation. .......................................... 188 *Plastic Settlement-11: Using the Time factor Tv. ...................................................... 189 *Plastic Settlement-12: The time rate of consolidation............................................... 190 *Plastic Settlement-13: Time of consolidation t.......................................................... 191 *Plastic Settlement-14: Laboratory versus field time rates of settlement. .................. 192 *Plastic Settlement-15: Different degrees of consolidation. ....................................... 193 **Plastic Settlement-16: Excavate to reduce the settlement. ...................................... 194 **Plastic Settlement-17: Lead time required for consolidation of surcharge. ............ 196 **Plastic Settlement-18: Settlement of a canal levee.................................................. 198 **Plastic Settlement-19: Differential settlements under a levee. ................................ 200 ***Plastic Settlement-20: Estimate of the coefficient of consolidation cv.................. 202 v **Plastic Settlement-21: The apparent optimum moisture content............................. 204 **Plastic Settlement-22: The differential settlement between two buildings. ............ 205 **Plastic Settlement-23: Settlement of a bridge pier. ................................................. 210 Chapter 12 Shear Strength of Soils................................................................................. 212 Symbols for Shear Strength of Soils............................................................................ 212 *Shear strength–01: Maximum shear on the failure plane. ......................................... 213 *Shear strength–02: Why is the maximum shear not the failure shear? ..................... 214 *Shear strength–03: Find the maximum principal stress σ1. ....................................... 215 *Shear strength–04: Find the effective principal stress............................................... 216 *Shear strength–05: Using the p-q diagram. ............................................................... 217 **Shear strength–06: Consolidated-drained triaxial test............................................. 218 **Shear strength–07: Triaxial un-drained tests. .......................................................... 220 **Shear strength-08: Consolidated and drained triaxial test. ...................................... 222 ***Shear strength-09: Plots of the progressive failure in a shear-box. ....................... 224 **Shear strength-10: Shear strength along a potential failure plane. .......................... 227 ***Shear strength-11: Use of the Mohr-Coulomb failure envelope. .......................... 228 ***Shear strength-11b: Use of the Mohr-Coulomb failure envelope. ........................ 230 **Shear strength-12: Triaxial un-drained tests............................................................ 232 **Shear strength-12b: Triaxial un-drained tests.......................................................... 233 **Shear strength-13: Determine the principal stresses of a sample. ........................... 234 **Shear strength-13b: Determine the principal stresses of a sample. ......................... 237 **Shear strength-14: Formula to find the maximum principal stress. ........................ 240 Chapter 13 Slope Stability .............................................................................................. 242 Symbols for Slope Stability......................................................................................... 242 *Slope-01: Factor of Safety of a straight line slope failure......................................... 243 *Slope-02: Same as Slope-01 but with a raising WT.................................................. 244 *Slope-03: Is a river embankment safe with a large crane? ........................................ 245 *Slope-04: Simple method of slices to find the FS. .................................................... 246 **Slope-05: Method of slices to find the factor of safety of a slope with a WT......... 247 **Slope-06: Swedish slip circle solution of a slope stability. ..................................... 249 Chapter 14 Statistical Analysis of Soils.......................................................................... 252 Symbols for the Statistical Analysis of Soils............................................................... 252 Chapter 15 Lateral Pressures from Soils......................................................................... 253 vi ............... ...................... 305 **Bearing–08: The effect of the WT upon the bearing capacity............................................ 264 *Lateral-08: What happens when the lower stratum is stronger? ............................ 258 *Lateral-03: Passive pressures using the Rankine theory.......... ............................ ........................... 296 *Bearing–01: Terzaghi’s bearing capacity formula for a square footing............... 307 *Bearing–09: Finding the gross load capacity. .............................. 296 Symbols for the Bearing Capacity of Soils ... 253 *Lateral-01: A simple wall subjected to an active pressure condition................... .............................. ............................................ 283 Symbols for Braced Cuts for Excavations............................. 280 **Lateral-16: Seismic loading upon a retaining wall.............. 274 ***Lateral-14: Derive a formula that provides K and σH as a function of σv......... 299 *Bearing–02: Meyerhof’s bearing capacity formula for a square footing..................... 300 *Bearing–03: Hansen’s bearing capacity formula for a square footing.................. Symbols for Lateral Pressures from Soils .................................... 260 *Lateral-05: The contribution of cohesion to reduce the force on the wall............................................ 261 **Lateral-06: The effect of a rising WT upon a wall’s stability. 309 vii ......................................... ....................... 272 **Lateral-13: The stability of a reinforced concrete wall.. 268 **Lateral-11: Anchoring to help support a wall................................... 293 Chapter 17 Bearing Capacity of Soils... 262 *Lateral-07: The effects of soil-wall friction upon the lateral pressure............................. 304 *Bearing–07: Increase a footing’s width if the WT is expected to rise.... .......... 302 *Bearing–05: General versus local bearing capacity failures............ ............................... 283 *Braced-cuts-01: Forces and moments in the struts of a shored trench............................................... .......... 277 **Lateral-15: The magnitude and location of a seismic load upon a retaining wall...................... 265 *Lateral-09: Strata with different parameters................ ................................... ... 259 *Lateral-04: The “at-rest” pressure upon an unyielding wall.......................... 284 **Braced cuts-02: A 5 m deep excavation with two struts for support........................................ ............ ................ ................................................................................................ 270 **Lateral-12: The effect of five strata have upon a wall. 289 *Braced cuts-03: Four-struts bracing a 12 m excavation in a soft clay...... ............................................. 301 *Bearing–04: Same as #01 but requiring conversion from metric units........ ............... 266 *Lateral-10: The effects of a clay stratum at the surface............................................... 282 Chapter 16 Braced Cuts for Excavations .................... 257 *Lateral–02: Compare the Rankine and Coulomb lateral coefficients.............................. 303 *Bearing–06: Comparing the Hansen and Meyerhof bearing capacities...................................... .................................... 329 *Footings–05: Design the steel for the previous problem....................... 316 *Footings–01: Analyze a simple square footing............................................... 316 Symbols for Shallow Foundations............................................................. .................................. ..................................... 322 *Footings–03: Find the thickness T and the As of the previous problem........Pile Groups and Caps..................................................... 373 viii .... 345 Symbols for Mat Foundations ...... 318 *Footings–02: Add a moment to the load on a footing................................................................... 367 Symbols for Steel Sheet Pile Retaining Walls........................................... ......................................................................................................................... 358 Symbols for Reinforced Concrete Retaining Walls .................... ....... 346 Chapter 21 Deep Foundations ........................... 348 Chapter 22 Deep Foundations ........................ .... ...................................... ...................... 368 Chapter 26 MSE (Mechanically Stabilized Earth) Walls ................................. 358 **RC Retaining Walls–01: Design a RC wall for a sloped backfill.................................................... 367 **Sheet-pile Wall-01: Free-Earth for cantilevered walls in granular soils...................................................................... 345 *Mat Foundations–01: Ultimate bearing capacity in a pure cohesive soil........................................................................... 340 Chapter 19 Combined Footings ........ **Bearing–10: The effect of an eccentric load upon bearing capacity............................................................................................................................................ 349 Symbols for Pile Groups and Caps of Deep Foundations ........ 349 **Pile-caps–01: Design a pile cap for a 9-pile cluster............ 312 **Bearing-12: Interpretation of borings to estimate a bearing capacity........................... 314 Chapter 18 Shallow Foundations .......................................................... ..... 359 Chapter 25 Steel Sheet Pile Retaining Walls...................Single Piles .................... 354 Chapter 24 Reinforced Concrete Retaining Walls and Bridge Abutments.......................................... 353 Symbols for Lateral Loads on Deep Foundations ................................................... 350 Chapter 23 Deep Foundations: Lateral Loads ................................................. 331 *Footings–06: Design a continuous footing for a pre-cast warehouse wall............................................. ....................................................................... 347 *Single-Pile–01: Pile capacity in a cohesive soil.......... 344 Chapter 20 Mat Foundations.. 324 *Footings–04: Find the dimensions B x L of a rectangular footing............ 353 **Lateral loads on piles-01: Find the lateral load capacity of a steel pile..................... 344 Symbols for Combined Footings.................... 311 **Bearing–11: The effect of an inclined load upon the bearing capacity............................................... 335 **Footings–07: Design the footings of a large billboard sign........ 347 Symbols for Single Piles of Deep Foundations .. ... 373 **MSE Walls-01: Design the length L of geotextiles for a 16 ft wall.................... ...................... 374 ix .................. Symbols for Mechanically Stabilized Earth Walls................ energy Newton-meter joule J power joule per second watt W x .001 milli m 0.000 000 001 Nano n 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 Name Symbol area square meter m² volume cubic meter m³ density kilogram per cubic meter kgm/m³ force kilogram-meter per square second Newtons N moment of force Newton-meter N-m pressure Newton per square meter Pascal Pa stress Newton per square meter Pascal Pa or N/m² work.000 001 micro µ 0.Conversion of Units Multiplication Factor Prefix SI Symbol 1 000 000 000 giga G 1 000 000 mega M 1 000 kilo k 0. 53 x 10-2 1.53 x 10-5 1.4047 hectares Volume Multiply by To From cubic inches cubic feet cubic yards quarts gallons cm3 6.64 x 102 xi .10 x 10-2 3.94 x 10-2 3.06 x 103 2.86 x 10-11 m² 1.22 x 10-4 km 3.86 x 10-7 km² 1.20 3.08 x 10-5 1.20 x 106 3.28 x 10-10 feet Area Multiply by To From square inches square feet square yards square miles mm² 1.94 x 101 3.08 x 107 1.55 x 10-3 1.047 m2 = 0.09 x 103 6.94 x 104 3.55 x 10-1 1.Conversion of SI Units to English Units Lengths Multiply by To From inches feet yards miles mm 3.22 x 10-1 1 μm = 1 x 10-6 m 1 Å = 1 x 10-10 m = 3.20 x 10-4 3.450 ft2 = 4.10 x 101 3.28 1.06 x 10-3 2.55 x 103 1.53 x 101 1.31 1.09 6.09 x 10-3 6.31 x 10-3 1.86 x 10-13 cm² 1.64 x 10-1 m³ 6.06 2.10 x 104 3.64 x 10-4 liter 6.31 x 10-6 1.86 x 10-1 1 acre = 43.09 x 10-2 6.28 x 10-2 1.28 x 103 1.08 x 101 1.22 x 10-7 cm 3.55 x 109 1.28 x 10-3 1.20 x 10-6 3.22 x 10-6 m 3.08 x 10-3 1.94 x 10-1 3. 233 x 10-8 kg-m 8.527 x 10-2 2.677 7.248 x 102 2.678 x 10-1 kN / m² 1.678 x 10-4 kg/cm² 1.090 x 101 2.205 1.248 x 10-1 2.248 x 10-9 1.102 x 10-3 Newtons 3.102 x 10-6 kilograms 3.663 x 10-1 xii .422 2.205 x 10-3 1.233 7.088 x 10-2 1.422 x 10-2 2.527 x 101 2.048 1.248 x 10-6 2.124 x 10-4 kilo-Newtons 3.281 x 101 9.124 x 10-9 grams 3.048 x 103 2.Conversion of SI Units to English Units Force Multiply by To From ounces pounds kips tons (short) dynes 1.281 9.044 x 10-2 3.281 x 10-2 9.195 x 103 7.678 x 10-2 Torque (or moment) T or M Multiply by To From lb-in lb-ft kips-ft gm-cm 8.048 x 10-1 1.663 x 102 7.024 x 10-3 3.248 x 10-1 1.205 x 10-3 2.124 x 10-1 tons (metric) 3.248 x 10-4 1.422 x 101 2.048 x 102 2.205 2.869 x 10-3 ton (metric)/m² 1.048 2.450 x 10-1 2.405 x 10-7 2.205 x 103 2.527 x 104 2.677 x 10-4 7.102 Pressure (or stress) σ Multiply by To From lb/in² lb/ft² kips/ft² tons (short)/ft² feet of water atmosphere gm/cm² 1.597 2.024 x 10-1 3.233 x 10-3 kN-m 9.205 x 10-6 1.597 x 103 2.024 3.048 x 10-3 1.233 x 10-5 7.346 x 10-1 9. 41 Btu/hr = 0.9685 2.116 x 10-1 5.467 x 101 6.236 x 10-2 km / min 5.7457 kW 1 kW = 1.214 x 10-1 1 mile = 1.152 feet Unit weight γ Multiply by To From lb / in³ lb / ft³ gm / cm³ 3.1622 cal/hr = 3.368 x 101 tons (metric ) / m³ 3.613 x 10-2 6.248 x 10-2 kN / m³ 3.281 x 10-2 1.428 x 101 Power P 1 W = 1 J/sec = 1.248 x 101 kg / m³ 3.0013 hp 1 hp = 745.613 x 10-2 6.282.613 x 10-5 6.281 x 103 3.34 hp xiii .610 meters = 5.7 W = 0.467 x 101 3.Conversion of SI Units to English Units Velocity v Multiply by To From ft / s ft / min mi / h cm / s 3.728 x 101 km / h 9.685 x 10-3 6. . 1 . Chapter 1 Soil Exploration Symbols for Soil Exploration CB → STP correction factor for the boreholes diameter. 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. SPT→ Stands for “Standard Penetration Test”. CS → STP correction factor for the sampler type used. N → The “raw” value of the STP (as obtained in the field). CR → STP correction factor for the rod length. Em → The efficiency of the STP hammer. Df → Depth of the foundation’s invert. cu → Soil’s un-drained cohesion. S → The number of stories of a building. po → The original vertical stress at a point of interest in the soil mass. In summary. Since the total footprint area is 30 m x 40 m =1.7 + D 20S0. the design engineer will want soil conditions that are at least average or better.000 Average 300 3. (Revised: Sept. Determine the required number and depth of the borings.7 + 1 = 14 m. Hence. 08) A four story reinforced concrete frame office building will be built on a site where the soils are expected to be of average quality and uniformity.Depths of exploratory borings for buildings on shallow foundations. The site appears to be in its natural condition.7 + D Average 5S0. 5 S0.7 + D High quality and uniform 3S0.7 + D 15S0. one boring will be needed for every 300 m2 of footprint area. with no evidence of previous grading. Solution: A reinforced concrete building is heavier than a steel framed building of the same size. use four borings.000 Table-2 .Spacing of the exploratory borings for buildings on shallow foundations.*Exploration–01.7 + D 10S0. Find the required number of borings and their depth.7 + D 2 . From Table-1 below. Table-2 provides the minimum depth required for the borings. the exploration plan will be 4 borings to a depth of 14 m.200 m2.7 + D = 5(4)0. Table-1 .000 High quality and uniform 600 6. The building will have a 30 m x 40 m footprint and is expected to be supported on spread footing foundations located about 1 m below the ground surface. Most design engineers want one boring to go to a slightly greater depth to check the next lower stratum’s strength. Structural footprint Area for Each Boring Subsurface Conditions (m2) (ft2) Poor quality and / or erratic 200 2. Minimum Depth of Borings (S = number of stories and D = the anticipated Subsurface Conditions depth of the foundation) (m) (ft) Poor and / or erratic 6S0. Bedrock is 30-m below the ground surface. ( 3.875) (100) = 13. the 3” O-D Shelby-tube sampler is the best tool to use. The sample’s disturbance due to the boring diameter. It has the same outside diameter of 2 inches (although the trend it to use 3 inches).0 ) − (1. 08) The most common soil and soft rock sampling tool in the US is the Standard Split Spoon. ( 2. However. (Revised: Sept.875) 2 Di2 The area ratio for a 3"-Shelby-tube sampler is. 3 . Solution: The area ratio for a 2"-standard split-spoon sampler is. Compare the degree of sample disturbance of a US standard split-spoon sampler.8% 2 2 Do2 − Di2 Ar (%) = (100) = (1.38) 2 Di2 The area ratio for a 2"-Shelby-tube sampler is.*Exploration–02.9 mm (1-3/8 inches). The NX outside diameter is Do = 50.875) (100) = 8.0 ) − (1. A better sampler is the Shelby (or thin-tube sampler). Split spoon tubes split longitudinally into halves and permit taking a soil or soft rock sample.875) 2 Di2 Clearly. In soft rocks. it has the disadvantage of disturbing the natural texture of the soil.38) (100) = 110% 2 2 Do2 − Di2 Ar (%) = (100) = (1. ( 2. This small size has the advantage of cheapness. versus the two Shelby thin-tube samplers (2” and 3” outside diameters) via their area ratio Ar (a measure of sample disturbance).0 ) − ( 2.8 mm (2 inches) and its inside diameter is Di = 34. because it is relatively easy to drive into the ground.9% 2 2 Do2 − Di2 Ar (%) = (100) = ( 2. such as young limestone. The tube size is designated as an NX. it will destroy the rock to such a degree that it may be classified as a “sand”. only the last two sets of 6” penetrations).85.8 pcf ) Other methods for corrections are discussed in Exploration-04. The effective unit weight of the loose sand stratum is about 93.60 0.00 )( 0. With these values. and the other parameters are obtained from the Tables provided on the next page: CB = 1. CS = 1. The US-style donut hammer efficiency is Em = 0.85 )(14 ) N 60 = = =9 0. That corrected SPT N60 is then corrected for depth. The tests were conducted using a US-style donut hammer in a 6 inch diameter boring with a standard sampler and liner. Determine the corrected SPT if the testing procedure is assumed to only be 60% efficient. the SPT corrected to 60% efficiency can use Skempton’s relation. 4 . CR = 0. Correcting the SPT for depth and sampling method.45. For example.*Exploration–03.8 pcf.00.05 )(1. 2.45 )(1.60 Notice that the SPT value is always given as a whole number. using the Liao and Whitman method (1986). 000 lb / ft 2 2. 6 -12 in = 6 blows. (Revision Sept-08) A standard penetration test (SPT) has been conducted in a loose coarse sand stratum to a depth of 16 ft below the ground surface. Solution: The raw SPT value is N = 6 + 8 = 14 (that is. The blow counts obtained in the field were as follows: 0 – 6 in = 4 blows.05. 000 lb / ft 2 ( N )60 = N 60 = (9 ) = 10 ( depth )( effective unit weight ) (16 ft )( 93. Em C B CS C R N ( 0. 12 -18 in = 8 blows. 1990).50 Japan donut Tombi trigger 0. 1986).00 5 . CR 3 – 4 m (10 – 13 ft) 0.72 China automatic trip 0.45 Brazil pin weight hand dropped 0.20 (not recommended) Rod length factor.85 6 – 10 (20 – 30 ft) 0. Correction Factor Equipment Variables Value Borehole diameter factor CB 65 – 115 mm (2.5 – 4.60 donut hand dropped 0.0.85 donut cat-head + sp.73 US safety 2-turns on cat-head 0. Sampling Method and Boring Rod Length (adapted from Skempton.78 .45 Venezuela donut cat-head 0.50 Colombia donut cat-head 0.00 150 mm (6 in) 1.95 >10 m (>30 ft) 1.0. Country Hammer Type Release Mechanism Hammer Efficiency Argentina donut cathead 0.43 Correction Factors for the Boring Diameter.0.5 in) 1.55 donut cat-head 0.15 Sampling method factor CS Standard sampler 1.75 4 – 6 m (13 – 20 ft) 0.60 donut 2-turns on cat-head 0.05 200 mm (8 in) 1.55 . SPT Hammer Efficiencies (adapted from Clayton.00 Sampler without liner 1.67 UK automatic trip 0. release 0.65 . The Liao-Whitman Method (1986).7 kN / m 2 6 .77 log10 if p0 is in tons / ft 2 p0 1915 or C N = 0.35 tons / ft 2 = 2.25 + 0. 000 psf N corrected = N ' with po in kN / m 2 or = N' with po in psf po po ⎛ 96.25 + 0. Using the Peck Method (1974): 20 N corrected = N 'C N where C N = 0.1 kN / m 2 ⎞ but po = (1.35 tons / ft 2 3.70 kips / ft 2 2000 lb / ton 20 ∴ C N = 0.35 tons / ft 2 ) ⎜ 2 ⎟ = 129.77 log10 if p0 is in kN / m 2 p0 (20 ft )(135 lb / ft 3 ) but p0 = = 1.25 + 0. Solution: Any of these three methods will provide acceptable answers.*Exploration–04. (Revision Sept. Three methods used for SPT depth corrections.7 kN / m 2 ⎝ 1 ton / ft ⎠ 100 kN / m 2 ∴ N corrected = 40 = 35 129.-08) A raw value of N = 40 was obtained from an SPT at a depth of 20 feet in a sand stratum that has a unit weight of 135 lb/ft3.5 kips / ft 2 3. as used in Exploration-03.77 log10 = 0. Using the Bazaraa Method (1967): 4N ' N corrected = if p0 ≤ 1.5(2.5kips / ft 2 1000 lb / kip 4N ' 4(40) therefore N corrected = = = 35 3.90 ∴ N corrected = (40)(0. Correct it only for depth.5 po 3. Notice how similar their results are from each other: 1.5 po (20 ft )(135 lb / ft 3 ) but p0 = = 2.70 kips / ft 2 > 1. 100 2.90) = 36 1.70 kips / ft 2 ) 2.5 kips / ft 2 and 1 + 2 po 4N ' N corrected = if p0 ≥ 1. 2 feet.1’ N =26 N =25 +0. +20’ Ground Surface +13.0’ Hard clay 7 .0’ -20.0’ Soft clay N =24 N =30 Medium sand N =31 -10.5’ +4.9’ Water Table +8 7’ +10.0’ +5.*Exploration–05.2’ invert of Sand+ gravel T = 3.-08) Correct the SPT values shown below for an energy ratio of 60% using a high-efficient US-type donut hammer in a 2”-diameter boring. The invert (bottom) of the mat foundation is at elevation +5. (Revision Sept.2’ +10. SPT corrections under a mat foundation. 4 pcf ) At -5.85 (13’-20’)→ rod length correction.60)(1)(1)(0.75 (10’-13’) CR = 0.60 ( 23 ft )(126 − 62.00 → sampler correction.4 pcf ) Notice that the depth correction does not affect the deeper layers.4 pcf ) At +4.85)(30) = 26 and ( N )60 = ( 26 ) 2. 000 lb / ft 2 = 35 0.75)(25) = 19 and ( N )60 = (19 ) 2. = 0.60)(1)(1)(0.4 pcf ) At+2.60 ( 8 ft )(127 − 62. N 60 = 0.60)(1)(1)(1)(43) = 43 and ( N )60 = ( 43) 2.95)(30) = 29 and ( N )60 = ( 29 ) 2. 8 . 000 lb / ft 2 = 41 0. 000 lb / ft 2 = 39 0.60)(1)(1)(0.2 feet: (0.0 (>30’) N = SPT-value recorded in the field by the driller (known as the “raw” SPT).00 → borehole diameter correction CS = 1.60 ( 9 ft )(127 − 62.4 pcf ) At -1.75)(26) 2. 000 lb / ft 2 = 34 0.60 ( 33 ft )(130 − 62.4 pcf ) At -10’ N 60 = (0.0’ N 60 = (0.60)(1)(1)(0.75)(24) = 18 and ( N )60 = (18 ) 2. 000 lb / ft 2 = 34 0.60 where: N60 = SPT N-value corrected for field procedures assuming 60% efficiency Em = 0.85)(31) = 26 and ( N )60 = ( 26 ) 2.95 (20’-30’). 000 lb / ft 2 N 60 = = 20 and ( N )60 = ( 20 ) = 39 0. 000 lb / ft 2 = 31 0.4 pcf ) At -21’ N 60 = (0.60 (14 ft )(126 − 62. 000 lb / ft 2 ( N1 )60 = N 60 ( depth )( effective unit weight ) At depth of +5.0’ N 60 = (0.60 → efficiency for a high-efficiency US-style safety hammer CB = 1.60)(1)(1)(0.0’ N 60 = (0. = 1.1’ N 60 = (0.Solution: Skempton proposed in 1986 the following correction for the sampling methods to the raw SPT value.60 (18 ft )(126 − 62.60 (11 ft )(125 − 62. Em C B C S C R N assuming that only 60% of the energy of the hammer drives the sampler. 2. The depth correction is. = 0.60)(1)(1)(0. 3021 ft) (0. which has a plasticity index of 47% (PI = LL – PL). cu−corrected = μcu = (0.6042 ft) (0. (Revision Sept. Determine the un-drained cohesion cu corrected for its plasticity. 1974) through the graph shown above. The Shear Vane Test determines the in-situ cohesion.25 inches.625 inches and a blade height h = 7. In a field test. For a plasticity index of 47% read a correction factor μ = 0.80)(168 psf ) = 134 psf 9 .*Exploration–06. 47 T 17.3021 ft)3 ⎤ 2 π⎢ + ⎥ ⎣ 2 6 ⎦ The plasticity index helps correct the raw shear vane test value (Bjerrum.0 ft ⋅ lb cu = = =168 psf π ⎡⎣(d h / 2) + (d /6)⎤⎦ 2 3 ⎡ (0.0 ft-lb to shear the clay sample. the vane required a torque of 17.-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. Therefore.80. and (3) Five samples were taken. (1) the location of the phreatic surface. Only one sample (#2) was used for laboratory tests (dry density and moisture content). (Revision Sept. Samples #4 and #5 were incomplete split-spoon samples.-08) Read the boring log shown below and determine. Reading a soil boring log.*Exploration–07. 10 . (2) the depth of the boring and (3) the number of samples taken. (2) The boring was terminated at 21 feet in depth. Solution: (1) The phreatic surface (the water table) was not encountered in this boring and is noted at the bottom of the report. Samples #1 and #3 were complete split-spoon samples. (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. . Read the SPT for medium sands. Use the Table provided on page 23 to obtain an estimate of some of the engineering parameters for granular soils.70 (2) Correct the same sample S-4 for depth.The bottom of the boring was at -36.0)( 0. 11 .5’.0’ – 36. at elevation -17’ it has dropped a small amount to N = 14.-08) Using the boring log and the SPT versus Soil Engineering Parameters Table shown on the next two pages. ECBCS CR N ( 0. Notice that the “Legend” portion denotes that the sampler was a 2” O.*Exploration–08: Using a boring log to predict soil engineering parameters. (4) What are the elevations (above sea level) of the groundwater and of the bottom of the boring? .5’ from the surface. 2000 psf 2000 psf ( N 70 ) = N 70 = (8 ) ≈8 (γ h ) (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”.5’.85)(14) N70 = = ≈8 0.45)(1. 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. split spoon.0)(1.70 0. then go to the Medium column and read the value of “N = 8” to obtain the values: φ = 32º and γwet = 17 kN/m2. (2) Correct the same sample S-4 for depth assuming the unit weight is γ = 126 pcf.D. the sampling correction is.The boring did not report finding a ground water table.5’ = +310. (Revision Sept. or 347. Therefore. 12 . N70 Compressive Strength qu Description 0-2 < 25 kPa Very soft – squeezes between fingers Very young NC clay 3-5 25 .30 30 .38 medium 27 .35 0.18 17 .28 30 .34 33 .30 (N'70 ) medium 2-3 4-7 8 . Correlation between SPT values and some Engineering Parameters of Granular Soils Very Description Very loose Loose Medium Dense dense Relative Dr density 0 0.42 < 50 coarse 28 . Note #3: The value of the angle of internal friction is based on Φ = 28º + 15ºDr.50 kPa Soft – easily deformed by fingers Young NC clay 6-9 50 .45 φ fine 26 .85 SPT fine 1-2 3-6 7 .28 28 . Note #4: The typical value of an excavated soil ranges from 11 to 14 kN/m3.40 40 .0 do not exist in nature.25 26 .30 30 . values of 0 or 1.102 89 .115 108 .32 32 .7.30 200 .34 33 . Correlation between SPT values and some Engineering Parameters of Cohesive Soils SPT .3 to 0.16 100 .40 > 40 coarse 3-6 5-9 10 .65 0.200 kPa Stiff – Hard to deform w/fingers Small OCR – aged clay 17 .128 108 -140 147 20 - kN/m3 11 .22 23 Note #1: These values are based on tests conducted at depths of about 6 m.15 16 .16 14 . Note #2: Typical values of relative densities are about 0.50 128 - γwet pcf 70 .20 17 .100 kPa Medium 10 .20 21 .15 0.400 kPa Very Stiff – Very hard w/fingers Increasing OCR – older clays > 30 > 400 kPa Hard – Does not deform w/fingers Higher OCR – cemented clays 13 .36 36 . 14 . From the next chart. q s ~ 400 kPa and q c ~ 11 MPa which results in a fR ~ 3%.-08) Classify a soil from the data provided by the Cone Penetration Test (CPT) shown below at a depth of 11 m. (Revision: Sept. the soil appears to be a silty clay. Solution.**Exploration–09. Reading the data. The clay samples recovered from that depth had γ = 20 kN/m3 and PI = Ip = 20. Compare your estimate of the shear strength versus the lab test value of 550 kPa. Find the shear strength of a soil from the CPT Report. 5. the in-situ pressure po for a NC clay is. po = γ z = (20 kN / m 3 )(11 m ) = 220 kPa From the N k versus I p graph.At a depth of 11 m. Nk 17.5 15 . for I p = 20 yields an N k ~ 17. qc − po 11. The un-drained shear strength su is. 000 kPa − 220 kPa su = = = 616 kPa versus lab = 550 kPa (a 12% error). w→ Water content (also known as the moisture content). VV → Volume of voids (water + air). VS → Volume of solids. V→ Total volume (solids + water + air). gb → Buoyant unit weight of the soil (same as g’). WS → Weight of solids. g→ Unit weight of the soil. Va → Volume of air. gW → Unit weight of water 16 . gSAT→ Unit weight of a saturated soil. gd → Dry unit weight of the soil. VW → Volume of water. GS → Specific gravity of the solids of a soil. n→ Porosity. Chapter 2 Phase Relations of Soil Symbols for Phase Relations of soils e→ Voids ratio. S→ Degree of saturation. WW → Weight of water. Typical values of sands are: very dense 0.5.Porosity n VV n= (100% ) 2. The interrelationship of the voids ratio and porosity are given by.Basic Concepts and Formulas for the Phases of Soils.0 Typical values for clays are: firm 0.3 to very soft 1.4 to very loose 1. 2.Voids ratio e VV e= 2-1 VS ranges from 0 to infinity. .2 V ranges from 0% to 100%. 17 . The porosity provides a measure of the permeability of a soil. . (A) Volumetric Relationships: 1. .Saturation S VW S= x100% 2-4 VV ranges from 0% to 100%. n e e= and n= 2-3 1− n 1+ e 3. (B) Weight Relationships: 4.Water content w WW w= x100% 2-5 WS Values range from 0% to over 500%. – Unit weight of a soil γ W WS + WW γ= = 2-6 V VS + VW + VA The unit weight may range from being dry to being saturated. .Dry unit weight γd WS γ γd = = 2-7 V 1+ w The soil is perfectly dry (its moisture is zero). 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). . W m γ= = ρg = g which is the equivalent of F = ma. Note that. .The unit weight of water γw WW γw = where γ = ρ g ( F = ma ) VW γ w = 62. 5. 7.4 pcf = 1 g / ml = 1 kg / liter = 9. also known as moisture content. 2-6 V V 6.81 kN / m3 18 . 71 Feldspar KALSi3O8 2.Buoyant unit weight of a soil γb γ b = γ ' = γ SAT − γ w 2-9 10. Salt water is 64 pcf.50 Calcite. etc.Specific gravity of the solids of a soil G γS GS = 2-10 γw Typical Values for the Specific Gravity of Minerals in Soils and Rocks Mineral Composition Absolute specific gravity Gs Anhydrite CaSO4 2. Note that the above is for fresh water.65 Peat Organic 1.60 Magnetite Fe3O4 5. 8.20 Lead Pb 11.0 or less Diatomaceous earth Skeletons of plants 2.30 Hematite Fe2O3 5. .60 to 2.70 Gypsum CaSO4 2H2O 2.20 Kaolinite Al4Si4O10(OH)8 2. chalk CaCO3 2.Saturated unit weight of a soil γsat WS + WW γ SAT = 2-8 VS + VW + 0 9. . .00 19 .34 Quartz (silica) SiO2 2.90 Barites BaSO4 4. Other useful formulas dealing with phase relationships: Se = wGS γs e= −1 γ dry Unit weight relationships : (1 + w)GS γ w (GS + Se)γ w (1 + w)GS γ w γ= = = = GS γ w (1 − n)(1 + w) 1+ e 1+ e wGS 1+ S Saturated unit weights : (GS + e)γ w ⎛ e ⎞⎛ 1 + w ⎞ γ SAT = = ⎜ ⎟⎜ ⎟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 : γ GS γ w eS γ w eGsγ w γd = = Gsγ w (1 − n ) = = = 1+ w 1 + e (1 + e) w ( S + wGs ) ⎛ e ⎞ γ d = γ SAT − nγ w = γ SAT − ⎜ ⎟γ w ⎝ 1+ e ⎠ . 20 . *Phases of soils-01: Convert from metric units to SI and US units. (Revision: Oct.-08) A cohesive soil sample was taken from an SPT and returned to the laboratory in a glass jar. It was found to weigh 140.5 grams. The sample was then placed in a container of V = 500 cm3 and 423 cm3 of water were added to fill the container. From these data, what was the unit weight of the soil in kN/m3 and pcf? Solution. Notice that the 140.5 grams is a mass. Therefore, the ratio of mass to volume is a density ρ, m 140.5 g f gf ρ= = = 1.82 V (500 − 423) cm 3 cm 3 3 ⎛ g ⎞ ⎛ 1 kg f ⎞ ⎛ m ⎞ ⎛ 1 kN ⎞ ⎛ 1 0 2 cm ⎞ kN γ = ρ g = ⎜ 1.82 f 3 ⎜ ⎟⎜ 3 ⎟ ⎜ 9.806 ⎟ ⎜ ⎟ ⎜ ⎟ = 17.9 3 ( SI un its ) ⎠ ⎝ 10 g f ⎟⎠ ⎝ 2 3 ⎝ cm sec ⎠ ⎝ 10 N ⎠ ⎝ 1 m ⎠ m ⎛ kN ⎞ ⎛ 1000 N ⎞ ⎛ 0.2248 lbs f ⎞ ⎛ 1 m 3 ⎞ γ = ⎜ 17.9 3 ⎟ ⎜ ⎟⎜ ⎟⎜ 3 ⎟ = 114 pcf (US units ) ⎝ m ⎠ ⎝ 1 kN ⎠ ⎝ 1N ⎠ ⎝ 35.3 ft ⎠ 21 *Phases of soils–02: Compaction checked via the voids ratio. (Revision: Sept.- 08) A contractor has compacted the base course for a new road and found that the mean value of the test samples shows w = 14.6%, GS = 2.81, and γ = 18.2 kN/m3. The specifications require that e ≤ 0.80. Has the contractor complied with the specifications? Solution: GS γ W (1 + w ) G γ (1 + w ) γ = ∴ 1+ e = S W 1+ e γ ⎛ kN ⎞ 2.81 ⎜ 9.81 3 ⎟ (1 + 0.146 ) ⎝ m ⎠ 1+ e = = 1.74 kN 18.2 3 m e = 1.74 − 1 = 0.74 ∴ e = 0.74 < 0.8 Yes , the contractor has complied . 22 *Phases of soils–03: Value of the moisture when fully saturated. (Revision: Oct.-08) (1) Show that at saturation the moisture (water) content is wsat = ( nγ W ) . ( γ sat − nγ W ) ⎛ ⎞ (2) Show that at saturation the moisture (water) content is wsat = γ w ⎜ 1 − 1 ⎟ ⎝ γd γS ⎠ Solution: (1) In a fully saturated soil the relation, Se = wGS becomes simply e = wGs e n because S = 1 or GS = = wsat wsat (1 − n ) but γ sat = γ w ⎡⎣(1 − n ) GS + n ⎤⎦ γ sat ⎡ n ⎤ n rearranging = ⎡⎣(1 − n ) GS + n ⎤⎦ = ⎢(1 − n ) + n⎥ = +n γw ⎣ wsat (1 − n) ⎦ wsat γ sat n nγ w or −n = therefore wsat = γw wsat γ sat − nγ w e VV γ w VV γ w VS γ wVV (2) Again, in a fully saturated soil, wsat = = = = GS VS γ S VS 1 WS WS γ wVV ⎛V ⎞ ⎛ VV + VS − VS ⎞ ⎛ VV + VS VS ⎞ ∴ wsat = = γw ⎜ V ⎟ = γw ⎜ ⎟ = γw ⎜ − ⎟ WS ⎝ WS ⎠ ⎝ WS ⎠ ⎝ WS WS ⎠ ⎛ 1 1⎞ or wsat = γ w ⎜ − ⎟ ⎝ γd γS ⎠ 23 *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: e VV γ w VV γ w VS γ wVV ⎛V ⎞ ⎛ V + VS − VS ⎞ wsat = = = = = γw ⎜ V ⎟ = γw ⎜ V ⎟ GS VS γ S VS 1 WS WS ⎝ WS ⎠ ⎝ WS ⎠ ⎛ V + VS VS ⎞ ⎛ 1 1 ⎞ wsat = γw ⎜ 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 ⎞ 1) wSAT = ( 9.81 kN / m3 ) ⎜ − ⎟ = 30% = w = 30% GOOD ⎝ 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 ⎠ 24 *Phases of soils–05: Increasing the saturation of a soil. (Revision: Sept.-08) 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 W ( GS + Se ) γ = 1+ e 62.4(GS + 0.50e) ∴ 105.7 pcf = (1) 1+ e 62.4(GS + 0.75e) and 112.7 pcf = (2) 1+ e 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 GS = 2.67 25 *Phases of soils–06: Find γd, n, S and Ww. (Revision: Sept.-08) The moist unit weight of a soil is 16.5 kN/m3. Given that the w = 15% and Gs = 2.70, find: (1) Dry unit weight γd, (2) The porosity n, (3) The degree of saturation S, and (4) The mass of water in kgm/m3 that must be added to reach full saturation. Solution: γ 16.5 kN a) γ d = = = 14.3 (1 + w) (1 + 0.15) m3 b) From the table of useful relationships, γd = Gs γ w Gγ ∴ 1+ e = s w = ( 2.70 )( 9.81) = 1.85 ∴ e = 0.85 1+ e γd (14.3) e 0.85 n= = (100% ) = 46% 1 + e 1 + 0.85 wGs ( 0.15 )( 2.70 ) c ) Since Se = wGs ∴ S = = (100 ) = 48% e ( 0 .8 5 ) d ) γ sat = (G S + e) γ w = ( 2.70 + 0.85 )( 9.81) = 18.8 kN 1+e 1+0.85 m3 The water to be added can be found from the relation γ = ρ g ⎡ (18.8 - 16.5 ) kN / m 3 ⎤ ⎛ 1, 000 N ⎞ ⎛ 9.81kg -m / s 2 ⎞ γ kg m ∴ ρ ( mass of water ) = = ⎢ 2 ⎥⎜ ⎟⎜ ⎟ = 2, 340 g ⎣ 9.81 kg - m / s ⎦ ⎝ 1 kN ⎠ ⎝ N ⎠ m3 26 *Phases of soils–07: Use the block diagram to find the degree of saturation. (Revision: Sept.-08) 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? (Note: All soils are really “moist” except when dry, that is when w = 0%). Solution: Set VS = 1 m3 (Note: this problem could also be solved by setting V = 1.0 m3). VV 1.87 ∴ eo = = = 1.87 ∴ V = VS + VV = 1 + 1.87 = 2.87 m 3 VS 1 Ww The "natural" water content is wN = = 0.60 ∴ Ww = 0.60Ws Ws Ws γs Vs Gs = = ∴ 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 ) = 16.19 kN W = Ws + Ww = 26.98 + 16.19 = 43.17 kN W 43.17 kN kN ∴ γ moist == 3 = 15.0 3 V 2.87 m m Ww ⎛ 16.19 ⎞ V γ ⎜ ⎟ S= w = w =⎝ 9.81 ⎠ ∴ = 88.2% VV VV 1.87 27 -08) A soil has an “in-situ” (in-place) voids ratio eo = 1. and GS = 2.60 (Ws ) = ( 0. Solution: Set V = 1 m3 (instead of Vs = 1 m3 used in Phases-07).39 kN γw γw Ww = 0. What are the γmoist and S? (Note: All soils are really “moist” except when dry. that is when w = 0%).39 ) = 5.60 ∴ Ww = 0.75 . VV but eo = = 1.81 ⎠ ∴ S= w = w =⎝ = 88.0% VV VV 0.*Phases of soils–08: Same as Prob-07 but setting the total volume V=1 m3.652 VS Ww The "natural" water content is wN = = 0.652 28 .0 3 V 1m m Ww ⎛ 5.348 m 3 ) ( 2.87 ∴ but V = 1 m 3 = VS + VV = VS + 1.0 kN kN ∴ γ moist == 3 = 15.81 kN / m 3 ) = 9.63 ⎞ V γ ⎜ ⎟ 9.60Ws Ws Ws γs Vs Gs = = ∴ Ws = Vs ( GS γ w ) = ( 0. (Revision: Oct.39 + 5. wN = 60%.87.87VS = 2.63 kN W = Ws + Ww = 9.60 )( 9.75 ) ( 9.87VS ∴ VS = 0.02 kN W 15.63 = 15.348 and VV = 0. Determine the voids ratio e and the specific gravity Gs of the soil. (NB: This is the same problem as Phase–06.556 VV 0.9 lb Ww 20.7 = 7. but solved with a block diagram). 65 γ w V S γ w ( 0.7 − 20.*Phases of soils–09: Same as Problem #5 with a block diagram.8 lb Vw = = = 0.444 = 0.0 lbs are 25% of w ater ∴ 21.111 ft 3 ∴ V v = V a + V w = 0. its unit weight raises to 112. Solution: Set V = 1 ft 3 γ 2 − γ 1 = 112.444 3 ∴ V s = 1 − V v = 1 − 0.111 + 0.7 pcf. (Revision: Sept.7 pcf and a water content of 50%.9 lb and GS = = = = 2.55 6 γS WS 91.8 = 91.4 pcf 1 Va = V w = 0.333 = 0. When its saturation is increased to 75 %.80 VS 0.444 e= = = 0.556 ) (62 .333 ft 3 γw 62.7 − 105.-08) A soil sample has a unit weight of 105.0 lbs are 75% of w ater ∴ W S = 112.4 ) 29 . 5 pcf and Gs = 2.70 ) (62. e.433 n = 43.4 pcf ) ⎝ 2.5 − W w ⎞ Combining equations (1) and (2) yields 1= ⎜ + Ww ⎟ ( 62. (Revision: Sept.4 pcf WS 95.56 7 ft 3 G S γ w ( 2.70.5 lb ∴ W S = 95. and w. 5 30 .5 pcf V 1 ft 3 VV 0.3% V 1 Ww 27 ∴ w= = = 0. 3% WS 95 .56 7 VV 0.4 pcf ) WS 95. n.5 lb ∴ γ dry = = = 95. Solution: 1 ⎛ WS ⎞ V = VS + Vw = ⎜ + W w ⎟ (1) γ w ⎝ GS ⎠ W = W S + W w = 122.*Phases of soils–10: Block diagram for a saturated soil. Find γ dry .283 w = 28 .0 lb ∴ V w = = = 0.-08) A saturated soil sample has a unit weight of 122.433 ∴ n= = = 0.76 4 VS 0.433 ft 3 γw 62.0 lb ∴ W w = 27.5 lb (2) 1 ⎛ 122.433 ∴ e= = = 0.5 lb ∴ V S = = = 0.70 ⎠ Ww 27. 2 = W S + W w = W S + w W S = (1 .4 9 VS 0 . 2 5 7 e = = = 0 .0 7 + 0 .0 4 ) ( 2 .9 8 k N a lr e a d y h a v e W w = 0 .4 9 W e re q u ire a S = 9 5 % .5 k N 1 7 .6 5 ) (9 .7 0 k N WS 1 7 .1 7 ) (1 7 .6 7 3 wGS ( 0 .5 k N .2 m 3 1+ w 1 + 0 .6 % e 0 .9 5 ) ( 0 . its moisture is 4%.5 k N VS = = = = 0 .4 9 ) = 0 .-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.2 8 k N Answer: Add 2.0 4 ) W S ∴ W S = 1 7 .5 kN/m3. (Revision: Sept. 31 .6 5 W w = w W S = ( 0 .6 5 ) T h e e x is tin g S = = (1 0 0 ) = 2 1 . the specific gravity of solids is 2. Solution: kN γ γ kN γ d = 1 7 .*Phases of soils–11: Find the weight of water needed for saturation.65 and the soil is entirely made up of a clean quartz sand. w = Se = ( 0 .0 7 m 3 ∴ V a = V − V s − V w = 0 .7 0 k N ∴ m u s t a d d w a te r = 2 .8 1 k N / m 3 ) Ww 0 . if the dry unit weight is 17. th e re fo re .8 1 k N / m 3 ) VV 0 .1 7 GS 2 .28 kN of water per m3.6 7 3 m 3 γ S G Sγ w (2 .2 5 7 m 3 γ w ( 9 .5 k N ) = 2 .5= = ∴ γ = 1 8 . and W w = 0 .7 0 k N Vw = = = 0 .0 4 m3 W = 1 8 . 528 m3 but Vw = 0.2 3 ≠ 18. Se = wGS ∴ Se = ( 0.-08) A project engineer receives a laboratory report with tests performed on marine marl calcareous silt).2 kN kN kN ∴γ = = 3 = 17.4 ) = 1.3 3 ⎟ = 4. (Revision: Sept.502 m3 ∴ Va = 0.95VV = 0. The engineer suspects that one of the measurements is in error. V = 1 m3 = Va + Vw + VS (1) VV but e = = 1. Are the engineer’s suspicions correct? If so.06 ∴ Therefore.81 The only possibly incorrect value is γ . which one of these values is wrong.1 wGS = ( w ) = ( 0.12 ∴ 0 = − Va − Vw + 1.*Phases of soils–12: Identify the wrong piece of data.4 m3 kN γ S = unit w eight of solids = 26. these four are correct.40 ) ⎜12.12 S = degree of saturation = 95% Solution: Check the accuracy of 4 out of 5 of the variables using.3 kN ⎝ m ⎠ ⎛ kN ⎞ Ww = wWS = ( 0.1 3 ⎟ ( 0.9 kN = 17.1 m3 w = w ater content = 40% e = voids ratio = 1.06 γS 26. W 17. γw 9.3 kN + 4. and what should be its correct value? kN G iven γ = unit w eight of sam ple = 18.12 ) = 1.95 )(1.472 m3 .2 kN Therefore.9 kN ⎝ m ⎠ W = 12. Assume that V = 1 m3 .12VS ( 2 ) VS ∴ VS = 0. the actual unit weight of the soil is.472 m3 ) = 12.026 m3 ⎛ kN ⎞ ∴WS = γ SVS = ⎜ 26.4 3 V 1m m m 32 . VV = 0. 19/ yd3 with e = 1.8 ⎝ yd ⎠ 2. 000 1.6 ⎛ 5. For 1yd3 of solids from C you would need 2.90 Supplier B Sells fill at $ 3. For 1yd3 of solids from B you would need 3.91/ yd3 with e = 2. 600.60 Solution: Without considering the voids ratio.9 ⎛ 5.6 yd3 of fill.-08) You are a Project Engineer on a large earth dam project that has a volume of 5x106 yd3 of select fill.*Phases of soils–13: The apparent cheapest soil is not! (Revision: Sept.80.7 Million compared to Supplier B.00 Supplier C Sells fill at $ 5. 900.0 yd3 of fill. 000 1.0 ⎛ 3. 000 1. Which one of the three suppliers is the most economical.19$ ⎞ C= ( 5 ) (10 6 yd 3 ) ⎜ 3 ⎟ ≈ $ 37. compacted such that the final voids ratio in the dam is 0.8 ⎝ yd ⎠ 3. Therefore: To put 1yd3 of solids in the dam you would need 1. the Project Manager delegates to you the important decision of buying the earth fill from one of three suppliers. Your boss.9 yd3 of fill.8 yd3 of soil.28$ ⎞ A= ( 5 ) (10 6 yd 3 ) ⎜ 3 ⎟ ≈ $ 27. 33 .28/ yd3 with e = 0. For 1yd3 of solids from A you would need 1.91$ ⎞ B= ( 5 ) (10 6 yd 3 ) ⎜ 3 ⎟ ≈ $ 32.8 ⎝ yd ⎠ Therefore Supplier A is the cheapest by about $ 4. and how much will you save? Supplier A Sells fill at $ 5. 500.37 per yd3. it would appear that Supplier B is cheaper than Supplier A by $1. The cost of the select fill from each supplier is (rounding off the numbers): 1. 54 yd 3of solids / truck − trip ) 34 .54 yd 3 of solids per truck trip.8 x106 yd 3 of solids 3 1.2 which means that there is 1 yd 3 of solids per 1. Vsolids = ( 5x10 6 yd 3of soil )(1 yd 3of solids ) = 2.2 yd 3 of soil for each 1 yd 3 of solids. (Revision: Sept. Solution: VV VV Set VS = 1 e = = = VV = 1.8 yd of soil Therefore. VS 1 ∴ 2.2 yd 3 of voids. 10 yd 3 of soil for each x yd 3 in a truck load ∴ x = 4.8x10 6 yd 3of solids ) = 616.800 ( 4.-08) Based on the previous problem (Phases–13). if the fill dumped into the truck has an e = 1.2.*Phases of soils–14: Number of truck loads. (rounding off) Number of Truck − trips = ( 2. how many truck loads will you need to fill the dam? Assume each truck carries 10 yd3 of soil. The required volume of solids in the dam is. 000 m3 of fill have 479.46. the 700. 10 m3 35 . whereas the limestone dumped into the trucks has an e = 0. if each truck carries 10 m3? Solution: VV VV Assume: VS = 1 m3 ∴ e = = = VV = 0.71.000 m 3 of fill Since each truck carries 10 m3 of fill.000 m3 of fill have 1.46 m3 of voids per each 1 m3 of solids Therefore. Since the original ground is low.46 m3 in the compacted fill VS 1 The required 700. The “in-situ” limestone extracted from the lakes has an e = 0. How many truckloads will you need. Your estimated fill requirements are 700. you will use the limestone excavated from the lake to fill the land in order to build roads and housing pads. with a dry density equivalent to a voids ratio e = 0.400 m3 of solids Each truck carries 1.*Phases of soils–15: How many truck loads are needed for a project? (Revision: Sept. 000 m3 ∴ The number of truck-loads = = 82. 820.39.000 m3 of solids they must carry 820.-08) You have been hired as the Project Engineer for a development company in South Florida to build 610 housing units surrounding four lakes.000 m3.71 m3 of fill per 1 m3 solids In order for the trucks to carry 479.000 truck-loads. Ww = 10. Va = 0.454 e= = = 0.9 lb VV 0.321 ft3.800 yd 3 of solids 1. w = 25%.401 e= = = 0. 000 yd 3 of fill ∴ Site B requires = 130.54 36 . (2) Pit A: WS = 92 lb. VS = 0.0 yd 3 of solids VS 0.10 ) = 120 pcf ∴ WS = 109. Standard Proctor yields a maximum γd = 112 pcf.000 cubic yards of lime-rock compacted at 95% Standard Proctor with an OMC of 10%. w = 20%.70.68 yd 3of soil contains 1.*Phases of soils–16: Choose the cheapest fill supplier. VW = 0.83 yd 3of soil contains 1. Va = 0. VS 0.0 yd 3 of solids VS 0.08 ft3 VV 0.70.369 ft3. VS 0. Standard Proctor yields a maximum γd = 115 pcf. 000 yd 3 of fill ∴ Site A requires = 125.30/yd3 to haul. WW = 20 lb. and $ 0.63 200.59 ∴ 1.085 ft3 VV 0.83 ∴ 1.37 e= = = 0.59 Material needed for fill from company B: γ = 0. (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.95 (115 )(1 + 0. VS = 0. Two lime-rock suppliers offer to fill your order: Company A has a borrow material with an in-situ γ = 115 pcf.546 ft3.0 yd 3of solids.59 yd 3of soil contains 1. Company B has a borrow material with an in-situ γ = 120 pcf. 000 yd 3 of solids 1.95γ d (1 + w ) = 0.546 Pit B: WS = 100 lb.594 (3) Material needed for fill from company A: γ = 0.10 ) = 117 pcf ∴ WS = 106.4 lb.68 ∴ 1. WW = 23 lb VW = 0.65 200. (Revised: Sept.35 e= = = 0.22/yd3 to excavate.38/yd3 to haul. at a cost of $0.20/yd3 to excavate.1 lb.95 (112 )(1 + 0.-08) A large housing development requires the purchase and placement of the fill estimated to be 200. a cost of $0. Ww = 10.95γ d (1 + w ) = 0.54 yd 3of soil contains 1.54 ∴ 1.594 ft3. GS = 2. GS = 2. and $ 0.0 yd 3of solids.6 lb VV 0. 50 ⎞ Cost A = (125.68 ) ⎜ 3 ⎟ = $ 131.(4) a) Cost of using Company A: ⎛ $0.60 ⎞ Cost B = (130.83) ⎜ 3 ⎟ = $ 115.100 ⎝ yd ⎠ Cost of using Company B: ⎛ $0.000. 37 . 000 yd 3 ) (1.100 ⎝ yd ⎠ Using Company A will save about $ 16.800 yd 3 ) (1. 000 cheaper than Pit #1. a) The missing unit cost C2 for Pit #2.000 yd3 can be found from. 38 .50/yd3.00/yd3 and Pit #3 costs $ 5. Pit #1 costs $ 6. (Revision: Sept. The volume of solids Vs contained in the total volume of fill V = 34. Solution: A summary of the data provided is herein shown in matrix form. Pits #2 and #3 reported their voids ratios as 0. The fill must be compacted down to a voids ratio of 0.7.000 yd3 of compacted “borrow” material from three pits: Pit #3 is $11.88 and 0.-08) A contractor obtains prices for 34. b) The missing voids ratio e for Pit #1. Use a matrix to find. and d) The amount paid by the contractor for each pit.*Phases of soils–17: Use a matrix to the find the missing data.000 cheaper than Pit #2 and $39. c) The missing volume of fill V required from each pit.95 respectively. 88 ) = 37.00 / yd 3 V2 37.7VS + VS = VS ( 0. the total cost of Pit #2 is TC2 − $ 11. 000 yd 3 of soil VS The total cost of Pit #3 is TC3 = ( 39. 000 yd 3 ) (1 + e1 ) = 42.500 V2 At Pit #2: = 1 + e2 ∴V2 = VS (1 + e2 ) = ( 20.500 ∴ TC2 = $ 225. V1 = VS (1 + e1 ) = ( 20. 34.500 + 28. 600 yd 3 of soil VS But. 000 225.50 / yd 3 ) = $ 214.7 + 1) = 34.00 / yd But. 000 V = VV + VS = 0.500 The unit cost of Pit #2 C2 = = 3 = $ 6. 000 yd 3 )( $ 5. = 1 + e3 ∴ V3 = VS (1 + e3 ) = ( 20. 000 yd 3 ) (1 + 0. 250 yd 3 ∴ e1 = 1.500 TC2 $ 225. 000 yd 3 ) (1 + 0.00 / yd $ 6. 000 yd 3 ∴ VS = = 20.95 ) = 39. 000 = TC3 = $ 214.11 39 . 600 yd TC1 TC + 28. 000 At Pit #1: V1 = 3 = 2 3 = 3 = 42. 250 yd 3 of soil $ 6.7 V3 At Pit #3.00 / yd $ 6. 000 yd 3 of solids 1. -08) You have been retained by a local municipality to prepare a study of their “muck” soils.γ so ) + γ so ) ⎦ ⎜ + ⎟ ⎝ γ so γ sm ⎠ ⎛ volume of water ⎞ ⎛ weight of water ⎞ ⎛ w ( weight of solids ) ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ Vv ⎝ S ⎠ ⎝ γ wS ⎠ ⎝ γ wS ⎠ (b) e = = = = Vs Vs Vs Vs ⎛ w ⎞ ⎜ ⎟ ⎝ γ w S (1) ⎠ = wγ sm γ so Therefore e = ⎛ Mo + (1 . 1 ⎡ γ sm ⎤ Therefore γ S = = γ so ⎢ ⎥ ⎛ Mo (1 .M o ) ⎞ γ w S ⎡⎣ M o (γ sm − γ so ) + γ so ⎤⎦ ⎜ ⎟ ⎝ γ so γ sm ⎠ 40 .Mo = Wm Mo = volume of organic Vso solids γ so (1 .Mo ) ⎞ ⎣ M o ( γ sm . Assume that you know the dry unit weight of the material (solids) γsm and the dry unit weight of the organic solids γso.**Phases of soils–18: Find the voids ratio of“muck” (a highly organic soil). 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.M o ) = volume of mineral Vsm solids γ sm The total unit weight is the weight of a unit volume. (Revision: Sept.) 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: Ws 1 Set Ws = 1 unit and γs = = Vs (Vso + Vsm ) (a) Assume Mo = Wo for a unit weight of the dry soil Therefore 1 . γd → In-situ dry unit weight (voids ratio e). γd(min)→Dry unit weight in loosest condition (voids ratio emax). PL→ Plastic limit. emin → Minimum voids ratio. Chapter 3 Classification of Soils and Rocks Symbols for Classification of soils Cc → Coefficient of gradation (also coefficient of curvature). SL→ Shrinkage limit. RC→ Relative compaction. emax → Maximum voids ratio. LL→ Liquid limit. Dx → Diameter of the grains (at % finer by weight). W→ Weight of the soil sample. V→ Volume of the soil sample. γd(max)→ Dry unit in densest condition (voids ratio emin) 41 . e→ Voids ratio. K→ Constant of the yield value. Dr → Relative density of a granular soil. IP → Index of plasticity (also referred to as PI). Cu → Coefficient of uniformity. This soil is a well-graded sandy silt (M).-08) Determine the percentage of gravels (G). This soil is a well-graded silty sand (SM). silts (M) and clays (C) of soils A. M & C). sands (S). 7%C.75 mm. 31% M.002 mm (or 2 micro-meters = 2 μm). This soil is a uniform or poorly-graded sand (SP). 12%C. Finally.*Classify–01: Percentage of each of the four grain sizes (G. (Revision: Sept. Solution: Notice that the separation between gravels (G) and sands (S) is the #4 sieve which corresponds to a particle size of 4. 42 .075 mm. 98% S. Soil C: 0% G. B and C shown below. 0% M. 61% S. 31% S. 0%C. Each soil A. B and C is now separated into the percentage of each: Soil A: 2% G. S. The separation between sands (S) and silts (M) is the #200 sieve which corresponds to a particle size of 0. 57% M. the separation between silts (M) and clays (C) is the 0. These divisions are shown above through color differentiation. Soil B: 1% G. 95 mm and D10 = 0. Solution: From the grain distribution curve.-08) Determine the uniformity coefficient Cu and the coefficient of gradation Cc for soil A. Since soil A has a low value of 2.50 mm. and it is sand. D30 = 0. 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).40 )( 0.4 mm.40 mm D302 CU = = = 2. (Revision: Sept. Smooth curved soils have coefficients of curvature Cc between 1 and 3. 43 .8 and CC = = D10 0. therefore the coefficients are. D60 = 1.*Classify–02: Coefficients of uniformity and curvature of granular soils. this corresponds to a poorly-graded sand (SP). whereas irregular curves have higher or lower values. ( 0. this soil is classified as poorly-graded sand (or SP). whereas a well-graded soil has a uniformity coefficient greater than 4 for gravels and greater than 6 for sands.50 ) A uniform soil has a coefficient of uniformity Cu less than 4.29 2 D60 1. Therefore.8.95) = 1.50 mm D60 D10 (1. Steep curves are uniform soils (low Cu) whereas diagonal curves are well-graded soils (high Cu). 40 mm CU = = = 2.8 . D60 0. D10 0. therefore. the distribution is G = 0%. M = 0% and C = 0%.45 mm CU = = = 90 . therefore. D10 0. S = 61%. (Revision: Sept.*Classify-03: Classify two soils using the USCS. soil A is very well graded silty sand (SM). the distribution is G = 2%. S = 98%. M = 35% and C = 4%. soil A is a poorly graded sand (SP).50 mm Classify Soil B: For soil B. Soil B’s Atterberg limits are LL = 49% and PL = 45%? Solution: Classify Soil A: For soil A. D60 1.005 mm 44 .-08) Use the grain-size distribution curve shown below to classify soils A and B using the USCS. Soil B is composed of approximately 33% S. 55% M and 12% C.*Classify-04: Manufacturing a “new” soil. It is obviously a poorly graded sand (SP). D60 D60 CU = = 100 = ∴ D60 = 0. 35/65. The plots of soils A and B are as shown below. and 98% S: (6% coarse sand.6mm D10 0. 40/60 and 50/50.006 mm. In lieu of removing that soil A. Determine the relative percentages of these two uniform soils A and B so that they will result in better graded soil C.-08) A site has an unsuitable in-situ soil A that does not compact properly.006mm The best fit is a 50% of A plus 50% of B mix. The best is the 50/50 solution via D10 = 0. 45 . 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. 85% medium sand and 7% fine sand). It is a well-graded sandy silt. Soil A is composed of 2% G. (Revision: Sept. Plot your results. Consider several solutions as shown below with A/B ratios of 30/70. 46 . this is a CH clay.-09) A sample of soil weights 1.5 falls above the U-line in Skempton’s diagram (see Classify-03). Its clay fraction weighs 0. classify the clay. Therefore. 47 .Classify – 05 (Revision: Sept. If its liquid limit is 60% and its plastic limit is 26%.5 N Wclay = 0.34 N (or 23% of W) Ip = PI = LL – PL = 60% – 26% = 34 % IP 34% A = = ≈ 1.5 N. Solution: W = 1. and is probably a member of the Montmorillonite family.5 % of clay fraction 23% The activity number 1.34 N. 01323 20 0.164 R ) = 16.01328 0.01439 0.01386 0.45 2. ∴ L = 16.01253 0.29 − [0.01397 0.01301 0.01365 0.01321.164(43)] = 9.01332 0.01264 0.01457 0.01272 0.01342 0.01505 0.01376 0.01339 19 0.01230 0.55 2.65 2.50 2.01372 0.01353 0.01235 0.01319 0.01312 0.01276 0.01342 0.01414 0.01325 0.01269 0. K= 0.01374 0.01321 = 0.01408 0.01474 0.01312 0.01357 0.01291 22 0.01365 0.01225 0.01218 27 0.01304 0.01414 0.01421 0.60 is immersed in a water suspension with a temperature of 24°C.01456 0.01244 0.01312 0.01421 0.01249 0.01321 0.01510 0.01256 0.01204 28 0.29 − ( 0.01492 0.01232 26 0.01236 0.80 16 0.01294 0.01397 0.01358 0.01267 0.01443 0.01391 0.01337 0. for Gs = 2.01307 21 0.01169 48 .01221 0.01435 0.01277 0.01282 0.01327 0.01239 0.01261 24 0.Classify – 06 (Revision: Sept.01334 0. What is the diameter D of the smallest-size particles that have settled during that time? Solution: Using the table below.01208 0.01396 0.01378 0.75 2.01394 0.01306 0.01286 0. An R = 43 cm is obtained after 60 minutes of sedimentation.01359 0.01246 25 0.01348 0.01342 0.2 x 10-3 mm ( a silt) t 60 min Table of constant K versus Temperature T (°C) Temparature Gs (°C) 2.60 and T= 24°C.01417 0.01195 0.01399 0.01481 0.01199 0.01382 0.01344 0.01279 0.01449 0.01283 0.01178 30 0.01182 0.01258 0.01431 0.01361 0.01212 0.01356 18 0.01217 0.01486 0.01403 0.01425 0.01264 0.01291 0.01369 0.01309 0.01297 0.01249 0.01388 0.60 2.01298 0.01404 0.01327 0.01391 0.00517 mm = 5.01511 0.01191 29 0.-09) During a hydrometer analysis a soil with a Gs = 2.2 cm D= K = 0.70 2.01438 0.01467 0.2 cm L 9.01317 0.01290 0.01374 17 0.01462 0.01276 23 0.01349 0. V γ γ 1000 cm 3 ⎛ 2.020 mm corresponds to a silt.e. 400 sec) ⎝ cm ⎠ b) The unit weight γ of the solution after 4 hours is.01 Poise at 20 degrees centigrade). η = 10−2 Poise ⎜ i. b) What percentage of the sample would have a diameter smaller than D? c) What type of soil is this? Solution: γs −γw 2 18η ⎛ L(cm) ⎞ a) Using Stoke’s relation: v = d or d (mm) = ⎜ ⎟ 18η γ s − γ w ⎝ t (min) ⎠ ⎛ dyne ⋅ s ⎞ where t = 4 hours = 14. ⎟ ⎝ cm 2 ⎠ γs but G s = ∴ γ s = G sγ w = ( 2. if the solution’s concentration has reduced to 2 grams/ liter at the level.65] x 1 g / cm3 γ= = = 1.65 − 1. weight of soil in solution 2 g + [1000cm3 − 2 g / 2.001 g / cm3 volume of solution 1000 cm3 The portion of soil having a diameter smaller than D is.81 dynes / cm 3 ) = 26 dynes / cm 3 γw ⎛ dynes ⋅ sec ⎞ 18 x10 − 2 ⎜ ⎟ ( 5 cm ) ⎝ cm 2 ⎠ d = = 0.400 sec.00)(14. At that moment.65. The specific gravity of the solids was 2.65 ) ( 9.65 x 1 ⎞ Portion smaller = • s w (γ − 1) = ⎜ ⎟ (1.65 − 1 ⎠ ∴ The remaining soil is only 8 % of the original sample.020 mm ⎛ dynes ⎞ 9..001 − 1) = 0. 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.81 ⎜ 3 ⎟ (2.Classify – 07 (Revision: Sept. L = 5 cm. 49 .-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. c) The diameter d = 0.08 W γs −γw 20 g ⎝ 2. -09) emax − e The formula for the relative compaction Dr is.8) Dr = ⎢ d ⎥⎢ ⎥= = 94% ⎣ γ d max − γ d min ⎦ ⎣ γ d ⎦ (18.Classify – 08 (Revision: Sept.4 − 14.4 k N / m 3 γ d ( S t d .0)(18.0)(18.8 − 14. Dr = emax − emin Derive an equivalent equation as a function of dry unit weights.) 1 8 . its maximum unit weight was 18. P r o c t . what is the RC (relative density) of a sand in the field if it was tested to be at 98% Standard Proctor. such that (γ d ( field ) − γ d min ) (γ d max ) Dr = (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 WS G S γ W γd = = and Dr = 0 = loose.8 kN/m3 and its minimum unit weight was 14.4) 50 . to 1= very dense V 1+ e 1 1 − γ d min γd ⎡ γ − γ d min ⎤ ⎡ γ d max ⎤ ⎡ γ d ( field ) − γ d min ⎤ ⎡ γ d max ⎤ Dr = =⎢ d ⎥⎢ ⎥=⎢ ⎥⎢ ⎥ 1 − 1 ⎣ γ d max − γ d min ⎦ ⎣ γ d ⎦ ⎣ γ d max − γ d min ⎦ ⎣⎢ γ d ( field ) ⎦⎥ γ d min γ d max For example.8 d ( fie ld ) ⎡ γ − γ d min ⎤ ⎡ γ d max ⎤ (18.0 kN/m3? γ γ RC = 98% = d ( fie ld ) = d ( fie ld ) ∴ γ = 1 8 . Calculate the range of relative densities.97 − 14.77 15.-09) The data obtained from relative density tests is shown below.56 γ field 16.56 ⎞⎛ 18.52 − 15.07 ⎞ range 1 (low γ min )(high γ max ) Dr = ⎜⎜ ⎟⎟⎜ ⎟ = 0.97 − 14.56 ⎞⎛ 17.52 γ min 14.77 ⎞⎛ 18.Classify – 09 (Revision: Sept.77 ⎠⎝ 16. Limiting γ Average γ in kN/m3 γ max 18.83 ⎝ 17 .71 ⎝ 18.07 ⎞ range 2 (avg γ min )(high γ max ) Dr = ⎜⎜ ⎟⎟⎜ ⎟ = 0.56 ⎠⎝ 16. 97 ⎠ ⎛ 16.07 17. 56 ⎠⎝ 16 .60 ⎝ 18 .97 ⎠ ∴ 60% ≤ D ≤ 83% r 51 .97 − 15. 52 − 14 .97 Solution: ⎛ γ −γ ⎞⎛ γ max ⎞ Dr = ⎜⎜ n min ⎟⎟⎜⎜ ⎟⎟ ⎝ γ max − γ min ⎠⎝ γ n ⎠ ⎛ 16.97 ⎠ ⎛ 16. 97 ⎠ ⎛ 16.07 − 14.77 ⎞⎛ 17.97 − 15.52 ⎞ range 4 (avg γ min )(avg γ max ) Dr = ⎜⎜ ⎟⎟⎜ ⎟ = 0. 77 ⎠⎝ 16 .52 ⎞ range 3 (low γ min )(avg γ max ) Dr = ⎜⎜ ⎟⎟⎜ ⎟ = 0. 07 − 15 .74 ⎝ 17. 70 P1 2.868 × 12 3 γ S1 146. еmax = 0.62 0.62 These voids ratios were measured by using a mold with a diameter of 4 inches and a height of 4.6 For the quartzitic sand.868 − −( ) × 12 3 4 γ S1 4 146. e2 = = = 0. and the dry calcareous sand was 2.7 As a result.53 The two types of sand have different relative densities because the calcareous sand grains are more tightly packed than the quartzitic sand grains.62) 1 + e2 1 + 0.70 For the calcareous sand.98 − 0. γs 146.254 lbs.3 πd 2 h P2 π (4) 2 (4.254 × 12 3 γ S2 165. At a particular site.4 pcf (but Dr21 = 0.89 − 0.7 γs 165.5 pcf) and calcareous sand (γs1 = 146. 52 .6 For the quartzitic sand.70 For the quartzitic sand.3 For the calcareous sand.254 − −( ) × 12 3 4 γ S2 4 165.89 − 0. Solution: e max − e By definition Dr = e max − e min πd 2 h P1 π (4) 2 (4.98 and emin = 0.98 − 0.70 and 0. their voids ratios were found to be: for the quartzitic sand. Find their relative densities and dry unit weights. the dry unit weight is greater for the soil with the lower relative density.868 lbs.-09) South Florida has two types of sand. γ d1 = 1 = = 86. a quartzitic sand (γs2 = 165. Dr1 = = 0.1 pcf (but Dr1 = 0.53 for the calcareous sand.59 inches.3 For the calcareous sand.Classify – 10 (Revision: Sept.70 P2 3. Comment on these results. e1 = = = 0. γ d 2 = 2 = = 97.70) 1 + e1 1 + 0.59) 3. еmax = 0.3 pcf).59) 2. The dry quartzitic sand weight was 3.89 and emin = 0.62 0. Dr2 = = 0.6 Notice that e1 = e2 0. 1179(2 R )3 = 0.35 VS V sphere ( 0.-09) Prove that emin = 0. Vtetr = 0.167 = 16.7% 360 2R ∴ e= VV = = ( V −V sphere 0.Classify – 11 (Revision: Sept.943R 3 4 Vsphere = π R 3 3 The volume of the sphere occupied by the tetrahedron is: 60 ∴ Vsphere ( tetr ) ≡ = 0.943 R 3 − 0.35 emin = = Vsphere 2 3 πd 3 2d 53 .167 4 / 3πR 3 ) ALTERNATE METHOD: d 2 ( ) 3 Volume of cube = d 2 = d32 2 ⎛ πd3 ⎞ 2 3 Volume of sphere = 4 ⎜ ⎟ = πd ⎝ 6 ⎠ 3 ⎛2 ⎞ 2d 2 − ⎜ π d 3 ⎟ Vcube − Vsphere ⎝3 ⎠ = 0.35.1179a 3 = 0.167 4 / 3 π R 3 ) = 0. Chapter 4 Compaction and Soil Improvement Symbols for Compaction e → Voids ratio. γS → Unit weight of the solid. γb → Buoyant unit weight of the soil. γd → Dry unit weight. γd. VS→ Volume of solids. VV → Volume of voids (water + air). GS→ Specific gravity of the solids of a soil. 54 .field → Dry unit weight in the field. ASTM D-698). w→ Water content.44*10 m Standard Proctor mold. γ→ Unit weight of the soil. V → ( 1 30 ft ≡ 9. VS→ Volume of soil sample. 3 −4 3 Va→ Volume of air. OMC→ Optimum moisture content. S → Degree of saturation. n → Porosity of the soil. VW → Volume of water. γSAT →Saturated unit weight of the soil. γW → Unit weight of water. 80 100.24 97.0 4.15 10.2 3. No Weight of wet soil (lb) Moisture % 1 3.36 16. (6) The moisture range for 93% of maximum dry unit weight.20 3 4.30 55 .20 124.67 12.8 90. remember V = 1/30 ft3.26 8.6 105.36 16.67 12. Determine: (5) The maximum dry unit weight and its OMC.24 2 4.35 4.8 4.80 Solution: W γ Formulas used for the calculations: γ = and γ d = V 1+ w W (lb) w(%) γ (lb/ft3) γd (lb/ft3) 3.*Compaction–01: Find the optimum moisture content (OMC).60 120.26 8.02 14.5 113. (Revision: Aug-08) A Standard Proctor test has yielded the values shown below.02 14.30 4 4.15 10.8 86.30 140.1 124.60 5 3. 93)(124.8 pcf 115 110 OMC = 12.3 % γd 105 100 γfield = (0.8) = 116.8 pcf 125 120 Maximum dry unit weight = 124.1 pcf 95 90 7 8 9 10 11 12 13 14 15 16 17 18 w(%) 56 . 130 γdmax = 124. 3 16.1 12.1 γd = γ (kN ) 3 12. tested under Standard Proctor ASTM D-698 (all weights shown are in Newton). (10) What is the appropriate moisture range when attaining 95% of Standard Proctor? Trial No.5 15.32 ft ⎠ (9) Find the OMC. 1 2 3 4 5 W(Newton) 14.3 13.95(13.*Compaction–02: Find maximum dry unit weight in SI units.8 13.8 kN/m3 57 .5) = 12.5 13.5 kN/m3 γd max = 13.4 16.5 17.5 kN/m 3 OMC = 28 % OMC = 28% γd-field = 0.5 1+ ω m γd max = 13.4 16.4 17.3 17.6 16. 1 2 3 4 5 γ = W kN V ( m3 ) 15. 1 3 ⎛ 1 m3 ⎞ (8) Note that the volume V = ft ⎜ 3 ⎟ = 9.1 ω (%) 20 24 28 33 37 Solution: Trial No.44 × 10−4 m3 30 ⎝ 35. (Revision: Aug-08) Using the table shown below: (7) Estimate the maximum dry weight of a sample of road base material. 13.2 γd (kN/m) 12.8 12.6 13.4 12 18 20 22 24 26 28 30 32 34 36 38 w(%) 58 . 8 14.8 11.1 9. Solution: a) γd max = 18. *Compaction-03: What is the saturation S at the OMC? (Revision: Sept.5 12.4 kN/m3.5 kN/m3 The moisture range w for 95% RC is from 8.4 18.3 17.-08) The results of a Standard Compaction test are shown in the table below: ω (%) 6.6 12.95)(18.9 15.5 kN/m3 OMC = 11.6 14.9 16 w% γd max = 13.4 20.75% to 13.4 10.5 20.5 20.3 9.2 γ (kN/m3) 16.5 11.5 18.3 13 5 6.2 17.9 18.2 8.4) = 17.7 13. (12) b) What is the dry unit weight and moisture range at 95% RC (Relative Compaction)? (13) c) Determine the degree of saturation at the maximum dry density if Gs = 2.70.8 γd = 1+ ω (11) a) Determine the maximum dry unit weight and the OMC.2 Dry Unit Weight 16.1 7.75%. OMC = 11.8 18.3 13.1 γ 15.7 19.5% b) γd at 95% = (0. 19.9 17.5 % c) 59 .2 8. 4 ⎞ Gs − d max 2.4 ) S= γw = ( 9.115)( 2.70 )(18.8) = 0.7 − ⎜ ⎟ γw ⎝ 9. wGsγ d max ( 0.71 Saturation S = 71% γ ⎛ 18.8 ⎠ 60 . 630 yd ) ⎜ yd 3 ⎟ ⎜ feet 3 ⎟ 3 b) Number of truck loads = ⎝ ⎠⎝ ⎠ = 824 truck − loads ⎛ 20 ton ⎞⎛ 2.000 yd3 of compacted fill? b) How many truckloads are needed to transport the excavated soil.-08) The in-situ moisture content of a soil is 18% and its moist unit weight is 105 pcf. a) How many cubic yards of excavated soil are needed to produce 10.5 pcf = 1 + w 1 + 0. 630 yd 3 ⎝ 89 pcf ⎠ ⎛ 27 feet 3 ⎞ ⎛ 105 lb ⎞ (11. The specific gravity of the soil solids is 2. 000 lb ⎞ ⎜ ⎟⎜ ⎟ ⎝ truck ⎠⎝ ton ⎠ 61 .*Compaction-04: Number of truck loads required. (Revision: Sept.75. if each truck can carry 20 tons? Solution: Wborrow site = Wconstruction site Vborrow siteγ borrow site = Vconstruction siteγ construction site γ 105 pcf γ d (borrow site) = = 89 pcf versus γ d ( construction site ) = 103.5 pcf ⎞ a ) Volume to be excavated = 10.18 ⎛ 103. 000 yd 3 ⎜ ⎟ = 11. This soil is to be excavated and transported to a construction site.5 pcf at a moisture content of 20%. and then compacted to a minimum dry weight of 103. 0695 )( 2.01 2.5 4 4.60.74 ⎝ γs −γd ⎠ ⎝ ( 2.85 1.90 γd γw 1.5 11 11.95 8.15 2.08 2.5 7 7.6 20.9 20.05 γd kN/m3 19.9 ⎞ SOMC = wOMC Gs ⎜ d −OMC ⎟ = ( 0.85 6.60 ) ⎜⎜ ⎟⎟ = 0.05 Dry Unit Weight 2 1.05 9.9 ⎠ The degree of saturation at the OMC is 74%.5 9 9. assume that Gs = 2. 2.5 5 5.6 20.1 2.95 1.4 20.06 2.8 2 2.5 3 3. Find the degree of saturation at the optimum condition.5 8 8.45 5. Solution: wGs ⎛γ ⎞ γ −γ It is known that Se = wGs ∴ S = but e = ⎜ s − 1⎟ = s d e ⎝ γd ⎠ γd therefore.5 w% 62 .6 0 )(10 ) − 20.00 4.1 20.5 10 10.-08) A Standard Proctor test was performed on a clayey gravel soil.5 Use γw = 10 kN/m3 for simplicity.*Compaction-05: What is the saturation S at the OMC? (Revision: Sept.09 2.9 1. the test results are shown below.06 2.46 9.8 20. Test 1 2 3 4 5 6 7 w% 3. at the OMC the saturation is.94 2. ⎛γ ⎞ ⎛ 20.5 6 6. 4 − 13.-08) The relative compaction (RC) of a sandy road base in the field is 90%.*Compaction-06: Definition of the relative compaction (RC). b) Relative density (of compaction) Dr.90 = = ∴ γ d ( field ) = ( 0.4 kN/m3 and γd(min) = 13.4 ) = 18.9 kN/m3.2 kN / m3 63 .15 ) = 21. The maximum and minimum dry unit weights of the sand are γd(max) = 20.9 ) ⎜⎝ 18.4 (1 + 0.768 = 76.4 ⎞ Dr = × = = 0.8% γ d (max) − γ d (min) γ d ( field ) ( 20.4 kN / m3 γ d (max) 20. Determine the field values of: a) The dry unit weight in the field. γ = γ d (1 + w ) = 18.4 − 13. (Revision: Sept. γ d ( field ) − γ d (min) γ d (max) (18. Solution: The relative compaction RC is the dry unit weight obtained in the field.90 )( 20. c) The moist unit weight γ when its moisture content is 15%. a) The relative compaction RC is.4 ⎟⎠ c) The moist unit weight γ is. γ d ( field ) γ d ( field ) RC = 0. as compared to the Standard Proctor obtained in the laboratory.9 ) ⎛ 20.4 b) The relative density Dr is. 4 ) ∴ e = ⎜ s w − 1⎟ = − 1 = 1. Find the degree of saturation S of the field soil sample.70 Saturation S = 70% e (1.0 pcf ⎞ a ) RC = = = 0.28 )( 2.08) 64 .7 pcf.90 γ d max ⎜⎝ 90.7 γ d ( field ) = = = 81.70) obtained a maximum dry unit weight of 90 pcf at OMC. (Revision: Aug-08) A Standard Proctor compaction test performed on a sample of crushed limestone (Gs = 2. A field compacted sample showed a moisture of 28% and a unit weight of 103. Solution: γ moist 103. Find the relative compaction (RC).0 ) Se = wGs wGs ( 0.*Compaction-07: The relative compaction (RC) of a soil.70 ) ∴ S= = = 0.0 pcf 1 + w 1 + 0.08 ⎝ γd ⎠ ( 81.28 γ d ( field ) ⎛ 81.70 )( 62.0 pcf ⎟⎠ The Relative Compaction = 90% Gsγ w Gγ b) γ d = ∴ 1+ e = s w 1+ e γd ⎛Gγ ⎞ ( 2. The water content of the sandy soil in the borrow pit is 15% and its voids ratio is 0. 7 3 ⎟ = 157 kN ⎝ m ⎠ W eight of w ater = w W d = (0. assuming that GS = 2. Specifications require the embankment to be compacted to a dry unit weight of 18 kN/m3. 7 3 1+ e 1 + 0.*Compaction-08: Converting volumes from borrow pits and truck loads.59 = 59% e 0. for 1 km length of embankment.-08) An embankment for a highway 30 m wide and 1. Solution: GSγ w 2.5 m thick is to be constructed from a sandy soil. c) The weight of water per truck load of sandy soil.70 ) = 0. and d) The degree of saturation of the in-situ sandy soil.69. b) The number of 10 m3 truckloads of sandy soil required to construct the embankment.8) kN a) T he borrow pit's dry unit w eight γ d = = = 15. trucked in from a borrow pit.15)(1 5 7 kN ) = 23 .7 ⎠ ⎛ kN ⎞ c) W eight of dry soil in 1 truck-load W d = 10 m 3 ⎜ 15. (Revision: Oct. Determine.69 m b) T he volum e of the finished em bankm ent V = 30 m (1.6 kN per truck load d) D egree of satu ration S = wGS = ( 0. the following: a) The dry unit weight of sandy soil from the borrow pit required to construct the embankment.70.7 (9.7 ⎠ ⎞ ⎛ 45 x10 m ⎞ 3 3 ⎛ 18 N um ber of truck trips =⎜ ⎟⎜ 3 ⎟ = 5.160 truck-loa ds ⎠ ⎝ 10 m ⎝ 15.69 65 .15 )( 2.5 m )(1km lo n g ) = 45x10 3 m 3 γ d ( reqd ) (V ) = ⎛⎜ 18 ⎞ V olum e of borrow pit soil requir ed = ⎟ ( 45 x10 ) m 3 3 γ d ( borrow pit ) ⎝ 15. 000 feet long of compacted 16” base at 98% Standard Proctor for two lanes.32 114 0.-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 1.43 0.33 112 0.3 9 .34 110 108 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 w% 66 .7 γd 116 P o ro s ity V o id R a tio 0. **Compaction-09: Ranges of water and fill required for a road. (Revision: Octt.30 1 4 .41 0.52 0. Proctor 0.46 0.49 0.29 10 0% 118 98% S. each 12 ft wide? 126 124 122 120 γd(max) S = 0. 000) = 48. ⎛ 3.98)(119.3%) is.143) ⎜ ⎟ Ww ⎝ 1.000 ft 3 ⎝ 12 in ⎠ The peak dry density γ d max =119.000 gallons.000 + 40.5) = 117. and the range of the water content is from 9.000 ft long) = 32. ⎛ 1 ft ⎞ V = (16inthick) ⎜ ⎟ ( 24 ft wide)(1.4 pcf 3 ⎝ ft ⎠ Therefore.143 ⎠ ⎛ 7. b) The soil volume V required for the road is.000gallons Therefore.000gallons γw 62. ⎛ 3.000gallons γw 62.75x106 lbs The weight of water Ww at the low end (9.7%) is.75x106 lbs ⎞ ( 0.7% to 14.3% for 98% of Standard Proctor or ( 98%) γ d max = ( 0. 67 .45gal ⎞ Vw = = ⎜ ⎟ 56.4 pcf ⎝ ft 3 ⎠ The weight of water Ww at the high end (14.1pcf ) = 3.000 ± 8.75x106 lbs ⎞ ( 0.45gal ⎞ Vw = = ⎜ ⎟ 40.097 ⎠ ⎛ 7.1pcf The total weight of soil in the pavement is W = Vγ = ( 32.5 pcf at an OMC=12%.097) ⎜ ⎟ Ww ⎝ 1. the average volume of water is (1/2)( 56. and/or a miscalculation of the water content w and γ wet .Solution : Gsγ w γ a) Since γ zav = and γ zav = wet 1+ ⎛⎜ s ⎞⎟ wG 1+ w ⎝ S⎠ The crossing of the ZAV (zero air voids) curve is due to an incorrect assumption of Gs .000 ft3 ) (117. the volume of water required = 48. 46 9.5 OMC = 7.45 5.65. (Revision: Oct.01 2.5 3 γd max = 20.08 2.-08) This problem expands Compaction-05: A Modified Proctor compaction test is performed on a clayey gravel road base.5 19 18. The compaction data yielded the following binomial values for γd/γw versus w %: w(%) 3. (16) Calculate the percentage of air for a given porosity n and the saturation S. (17) Find the equation that describes the points of equal saturation. (18) Determine the equation for S = 100% (19) Discuss the characteristics of this last curve and equation. Solution: a) 22 21.5 18 2 3 4 5 6 7 8 9 10 11 Water Content w(%) 68 .90 γd / γw 1. The solids have a specific gravity of 2.05 (14) Find the γd max and the OMC from the compaction curve.06 2. (15) Find the degree of saturation at the above conditions.94 2.05 9.7% γd (kN/m ) 3 20 19.06 2.00 4.**Compaction-10: Find the family of saturation curves for compaction.95 8.9 kN/m 21 20.09 2.85 6. wGs ⎛ γ d max ⎞ ⎛ 2. where S is the degree of saturation. volume of water volume of water S= = = 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 ⎞ 1− ⎜ d ⎟ ⎝γw γs ⎠ γ w wγ s + Sγ w ⎝ γs ⎠ 69 .09 ⎠ c) A a n nS 1 W 1-n S VV VW VW VW ns = = = = VW V VV V 1 a = 1 − nS − (1 − n ) = 1 − nS − 1 + n a = n(1 − S ) = Va That is.65) ⎜ ⎟ = 76% b) e ⎝ γ w −γ d max ⎠ ⎝ 2. the percentage of air a is equal to the porosity n times the factor (1 . d) Consider a family of curves of equal saturation.7%)( 2.65 − 2.09 ⎞ Se = wGs ∴ S = ⎜ ⎟ = ( 7.S). 6 γd 0. Curves can be plotted for varying values of a and saturation S.4 S = 100% S1 0. 12 10 8 γd = (1-a2)γs γd 6 4 a=0 Asymptotes a1 a2 2 a3 0 0 3 6 9 12 15 18 21 24 27 w 1.2 1 If w = 0 all curves pass thru γs 0.8 0.2 S2 0 0 10 20 30 40 w These are hyperbolas with the w-axis as an asymptote. 70 . 9 107. (22) The optimum moisture content.4 Solution: a) The plot of γd (dry unit weight) versus w (water content): 125 120 115 dry unit weight. w (%) 10 13 16 18 20 22 25 γ (pcf) 98 106 119 125 129 128 123 γd (pcf) 89 94 102. (25) The volume of water (in gallons) that must be added to obtain 1 cubic yard of soil at the maximum density. Find: (20) The plot of the dry unit weight versus the water content. the maximum dry unit weight is γdmax = 107. (d) The 90% value of the maximum dry unit weight γdmax = (0.9 98.71. (Revision: Aug-08) A Standard Proctor test yields the values listed below for a soil with Gs = 2.5) = 96. the optimum moisture content is OMC = 20%.5 104. (24) The moisture range for the 90% value. (c) From the plot.8 pcf 71 . (21) The maximum dry unit weight. if the soil was originally at 10% water content. γd 110 100% Proctor 105 100 90% Proctor 95 90 OMC=20% 85 0 10 20 w 30 moisture range (b) From the plot. (23) The dry unit weight at 90% of Standard Proctor.9) (107.6 105.5 pcf.**Compaction-11: Water needed to reach maximum density in the field. A Ww=21.10Ws Ws but Ws = 89 lb and Ww = 0.20Ws = ( 0. (f) Soil at 10% moisture. 72 .5 W V = 1 ft3 W = 129 lb S Ws= 107.20Ws Ws but Ws = 107.2 )(107. need to add the following water: (21.5 ) = 21.10Ws = ( 0.5 lb and Ww = 0.(e) The moisture range for the 90% value is approximately from 13% to 26%.5lb ft 3 7.48 gallons 27 ft 3 Added water = = 40 gallons / yd 3 ∴ 3 ft 62.5 lb ∴W = Ws + Ww = 129 lb Therefore.5 lb) – (9 lb) = 12.1)( 89 ) = 9 lb ∴ W = Ws + Ww = 89 + 9 = 98 lb Soil at 20% moisture.5 Ww w = 0. A W Ww=9 V = 1 ft3 W = 98 lb S Ws=89 lb Ww w = 0.10 = ∴ Ww = 0.20 = ∴ Ww = 0.5 lb/ft3 12.4 lb ft 3 1 yd 3 Answer: Add 40 gallons of water per cubic yard of compacted soil. 300 cy ) ⎢ ⎥ = 24.1 pcf at 18%. 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%. the voids ratio of the material is e = 1. Solution: ⎡⎛ 1 ⎞ ⎛1⎞ ⎤ ⎛ 450 ft long ⎞ a) Volume of levee = ⎢⎜ ⎟ ( 20')( 40') + ( 20')( 20') + ⎜ ⎟ ( 20')( 60') ⎥ ( ft 2 ) ⎜ 3 ⎟ = 23. with the dimensions shown below. When the soil is excavated and loaded on to your trucks.18 γ ⎡100.**Compaction-12: Fill volumes and truck load requirements for a levee. Your trucks can haul 15 cubic yards of material per trip.7 pcf 1 + w 1 + 0. (27) Determine the volume required from the borrow pit.68.1 where γ d (borrow) = = = 95 pcf and γ d (levee ) = ( 0. Ws (borrow) = Ws (levee ) Ws But γ d = or Ws = γ dV or γ d (borrow)Vborrow = γ d (levee )Vlevee V γ 112.700cy γ d (borrow) ⎣ 95 pcf ⎦ 73 .95)(106 ) = 100.47.7 pcf ⎤ ∴ Vborrow = Vlevee d (levee ) = ( 23. (Revision: Aug-08) Your company has won a contract to provide and compact the fill material for an earth levee.300 cy ⎣⎝ 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. Your borrow pit has a silty clay with an in-situ moist density of 112. and a Gs = 2. (28) Determine the required number of truckloads. (26) Determine the volume of fill required for the levee. 68 )( 62. 700 yd 3 ⎜γ ⎟ ⎝ 67.314 ⎝ truck capacity ⎠ ⎝ 15 yd / truck − load ⎠ 3 74 .7 pcf and γ d ( hauled ) = = = 67. 700 yd 3 ⎞ Number of truck-loads = ⎜ = ⎟ ⎜ ⎟ = 2.7 pcf ⎞ Vhauled = Vlevee ⎜ d (levee ) ⎟ = ( 23.4 ) but γ d (levee ) = 100.7 pcf ⎠ ⎝ d ( hauled ) ⎠ ⎛ Vhauled ⎞ ⎛ 34.47 ⎛ γ ⎞ ⎛ 100. Ws ( hauled ) = Ws (levee ) γ d ( hauled )Vhauled = γ d (levee )Vlevee ⎛ γ d (levee ) ⎞ ∴ Vhauled = Vlevee ⎜ ⎟⎟ ⎜γ ⎝ d ( hauled ) ⎠ GS γ w ( 2.Number of truck-loads required is based on.300 yd 3 ) ⎜ ⎟ = 34.7 pcf 1 + e 1 + 1. rectangularly shaped in plan. The dam is to be built with a silty clay soil. The borrow suppliers quoted the following: Pit Price ($/yd3) Gs S (%) w (%) A 1.71 49 22 C 0. with a specific gravity of 2.69 65 22 B 0.**Compaction-13: Multiple choice compaction problem of a levee. Specifications call for a compacted soil at the dam with a dry unit weight of 97 pcf at an OMC of 31 percent.700 ft2 e) 2.200 ft long by 1.70.78 2. What is the dam’s cross-sectional area? a) 675 ft2 b) 2.91 2. (Revision: Aug-08) A town’s new reservoir will impound its fresh water with a small earth dam.100 ft2 c) 2. and its cross-section is shown below in figure B. Assume all voids are totally devoid of any gas (or air).550 ft2 75 .05 2.66 41 18 Questions: 1.350 ft2 d) 2.750 ft wide. available from three different local sources. The perimeter of the dam will be 2. 81 d) 1.2 6.01 e) 1.000 yd3 c) 900.93 b) 1.75 Myd3 3.2.1 b) 0.550 ft3 b) 300.00 76 .22 c) 0.000 yd3 d) 1. What is the voids ratio e of the material in pit A? a) 0.000 ft3 b) 134 acre-ft c) 7 Mft3 d) 154 acre-ft e) 45 · 106 gallons 4.91 c) 1.79 e) 1.98 b) 0.84 d) 0.72 7. What is the designed voids ratio e of the compacted soil in the dam? a) 1. What is the approximate volume of water impounded (stored) in the dam? a) 6. What is the approximate total volume V of soil required for the dam? a) 2.300.1 e) 0.02 d) 1.22 Myd3 e) 0. What is the voids ratio e of the material in pit B? a) 0. 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 5.92 c) 0. 98.23 Myd3 e) 1.25 Myd3 d) 1.00 Myd3 b) 1.95 Myd3 d) 0.07 Myd3 11.05 e) 1. Assume that the voids ratio e of pit A is 0.18 Myd3 d) 1.29 Myd3 b) 1.02 Myd3 c) 1.17 9.21 Myd3 12. Assume that the voids ratio e of pit B is 0. 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 77 . What is the equivalent volume required of pit A to place 1.2 million cubic yards of compacted soil in the dam? a) 1.2 million cubic yards of compacted soil in the dam? a) 1.12 c) 1. what is the equivalent volume required of pit B to place 1.37 Myd3 c) 1.10.34 Myd3 b) 1.2 million cubic yards of compacted soil in the dam? a) 1.81. What is the voids ratio e of the material in pit C? a) 1.05 Myd3 e) 1.96 Myd3 10.10 b) 1.20 Myd3 c) 0. Assume that the voids ratio e of pit C is 1.97 Myd3 e) 0. 8.08 d) 1. what is the equivalent volume required of pit C to place 1. γS →Unit weight of solids. C→ Hazen’s coefficient. q→ Flow rate (ft3 per second per foot of width). Sc → Seepage capacity. TS → Surface tension (typically given as 0. Neq→ The number of equipotentials drops in Forheimer’s equation. H→ Thickness of the soil layer. umax → Maximum pore water pressure. Ludgeon→ Standard unit for permeability ( 10−4 mm ). σV→ Vertical effective stress. 78 . Q→ Total seepage (total flow). kH → Coefficient of horizontal permeability. k→ coefficient of permeability in D’Arcy’s equation .073 N per meter). γW →Unit weight of water . po →In-situ vertical pressure at any depth . d→ Diameter of a capillary tube Dx → Diameter of a soil % finer (represents % finer by weight) e →The voids ratio. kV → Coefficient of vertical permeability L→ Distance of the hydraulic head loss.. sec Nf→ The number of flow channels in Forheimer’s equation. u → pore water pressure. h→ Thickness of the aquifer. Chapter 5 Permeability of Soils Symbols for Permeability A→ area of a seepage surface. GS → Specific gravity of the solids of a soil. hC → Height of the rising capillary i→ Hydraulic gradient. whereas the falling head test is used for fine-grained (cohesive) soils (silts M. 79 . the “standard” unit is the Ludgeon (10. (b) In Europe.*Permeability–01: Types of permeability tests and common units. 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. and in the USA the standard unit is the Meinzer. (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 clays C). and sands S).4 mm/sec). 45gal ⎞⎛ 8ft ⎞ gallons q = k A=⎜0. and showed its phreatic surface at elevation 942 feet MSL. Δh ⎛ mm⎞⎛ 3600s ⎞⎛ 1in ⎞⎛ 1ft ⎞⎛ 7.*Permeability-02: Use of Hazen’s formula to estimate the k of an aquifer. and a sample of its soil showed the grain size distribution below. Estimate the permeability using Hazen’s formula with the coefficient C = 12. An aquifer stratum was identified. Solution: mm Allen Hazen’s (1893) formula for the permeability k = C ( D10 ) 2 = (12 ) (0.16 mm GRAIN SIZE DISTRIBUTION DIAGRAM 80 . If the thickness of the aquifer was a uniform 12 feet between both points.05 per ft width L ⎝ sec ⎠⎝ hour ⎠⎝ 25. A piezometer (measures the location of the WT) was installed 2500 feet downstream from the boring.500 ft ⎠ hour D10 D10 = 0. estimate the quantity of flow per foot of width in gallons/hour (1 ft 3 ≈ 7.16mm) 2 = 0.31 ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ 3 ⎟⎜ ⎟(12 ft)(1ft) =1. and it found the phreatic surface (water table) 5 feet below the ground surface.45 gallons ). (Revision: Aug-08) A test boring was performed at an elevation 955 feet MSL.4mm⎠⎝12in⎠⎝ 1ft ⎠⎝ 2.31 s The hydraulic head drop is (955 ft – 5 ft) – (942 ft) = 8 feet Applying D’Arcy formula. The sand is 6 feet thick. +1050’ ELEV. Δh ft ⎛ 86. 400 s ⎞ ⎡ (1050 − 1021) ft ⎤ ( 6. Compute the seepage loss q from the canal in if the permeability of day − mile ft the sand is 2 x10-3 . A stratum of sand intersects both the river and the canal below their water levels.*Permeability-03: Flow in a sand layer from a canal to a river. The elevation of the water surface in the canal is at +1050 feet and in the river at +1021 feet. sec 250’ ELEV. (Revision: Aug-08) A canal and a river run parallel an average of 250 feet apart.0 ft ) ⎛⎜ 5. and is sandwiched between strata ft 3 of impervious clay. 000 day − mile 81 . +1021’ CLAY 6’ SAND CLAY Solution: D’Arcy’s formula for q yields. 280 ft ⎞ q=k A = ( 2 ) (10) −3 ⎜ ⎟⎢ ⎥ ⎟ L s ⎝ day ⎠ ⎣ 250 ft ⎦ ⎝ mile ⎠ ft 3 q = 635. 5. What is the combined kHcomb for the upper 4 m in cm/sec? Sand 1m . kH1 = 5kV1 = 5(CD102 ) = (5)(1. 1 meter thick. k1H1 + k2 H 2 (1. Use the grain-size distribution curve of Permeability-02.920(10-6 ) m / s The formula for combining several horizontal layers is. where k is in cm/s if C ranges from 0.16 mm)2 = 1.5)(0. Its vertical permeability kV can be estimated using Hazen’s formula with C = 1.*Permeability-04: Find the equivalent horizontal permeability of two layers. 3 meters thick.8-1. Its kH is known to be approximately 500% of the kV. (Revision: Aug-08) The topmost layer is loose.8 ×10−4 cm / s H1 + H 2 1m + 3 m 82 .920 x10 m / s ) (1 m ) + (10 m / s ) ( 3 m ) −6 −6 kHcomb = = = 4. D10 is in mm.5 (to over-estimate) and the sieve analysis shown here. Marl 3m Solution: Use Hazen’s formula to find the permeability.92 cm / s = 1. Below the sand stratum is a marine marl. with a kV = kH = 10-6 m/s. clean sand. 4 cm/sec Fort Thompson Formation Solution: H (1+ 5 + 3 +12) m kV (EQ) = = H1 H2 H3 H4 1m 5m 3m 12 m + + + + + + k1 k2 k3 k4 10−1cm / s 2×10−3 cm / s 10−3 cm / s 2×10−4 cm / s = ( 2.. (Revision: Aug-08) The soil profile shown below is typical of Miami-Dade County.100 cm) = 0. 83 . Estimate the equivalent permeabilities kV(eq) and kH(eq) in cm/sec.8×10−3 cm / s Therefore = = 15 kV 0. *Permeability-05: Equivalent vertical and horizontal permeabilities. and the ratio of kH(eq) / kV(eq).551x10 )s 1 1 kH (EQ) = [ H1k1 + H2k2 + H3k3 + H4k4 ] = ⎡⎣(1)(10−1) + (5)(2×10−3 ) + (3)(10−3 ) + (12)(2×10−4 )⎤⎦ H ( 21 m) kH (EQ) = 4.8×10−3 cm / s kH 4.32×10−3 cm / s The horizontal permeability is 15 times larger than the vertical permeability. 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 = 2x10.32×10−3 cm / s 3 (6. 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: 1 ( ft )( cm/ s) ⎡ 3 10−3 + 3 2 10−4 + 3 10−5 + 3 2 10−3 ⎤ kH(eq) = (k1H1 + k2H2 + k3H3 + k4H4 ) = ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( )⎦ H (12 ft ) ⎣ kH(eq) = 8x10−4 cm/sec The equivalent vertical permeability of all four layers is: H 12 ft kVeq = = = 0.37 x10 − 4 cm / sec H1 H 2 H 3 H 4 3' 3' 3' 3' + + + −3 + −4 + −5 + kV1 kV2 kV3 kV4 10 2 x10 10 2 x10 − 3 Therefore the ratio of the horizontal to the vertical permeability is: k H eq 8 x1 0 − 4 c m / s = = 22 kV eq 0 .*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.3 7 x 1 0 − 4 c m / s 84 . it can not “bypass” any layer.0012 cm / s ) ⎜ ⎟ (10 cm )(10 cm ) ⎜ ⎟ = 291 cm / hour 3 H ⎝ 450 mm ⎠ ⎝ 1 hr ⎠ 85 .*Permeability–07: Do not confuse a horizontal with a vertical permeability.9 x 10-4 cm/sec. Therefore. 600sec ⎞ ∴ q = keq iA = keq A = (0. every soil layer has the same flow v = v1 = v2 = v3 and the total head Δh = Δh1 + Δh2 + Δh3. Find the rate of water supply in cm3/hr.9 x10−4 cm / s −2 −3 Δh ⎛ 300 mm ⎞ ⎛ 3. (Revision: Aug-08) The soil layers below have a cross section of 100 mm x 100 mm each. kC = 4. H 450mm kV ( eq ) = = = 1. Δh=300 mm hA hB A B C v H1=150 mm H2=150 mm H3=150 mm Solution: This is a trick drawing: it “looks” like a horizontal flow. kB =3 x 10-3 cm/sec.. The permeability of each soil is: kA =10-2 cm/sec. but in reality it is a vertical flow because the flow has to cross through every layer.2 (10−3 ) cm / sec H1 H 2 H 3 150mm 150mm 150mm + + + + k1 k2 k3 1x10 cm / s 3x10 cm / s 4. 544 Notice that since k1 = 0. e3 k= (1 + e ) e13 ( 0.544 ( 0.57. Estimate the increased permeability using the Kozeny-Carman formula of this same sand when its voids ratio has increased to 0.*Permeability-08: Permeability as a function of the voids ratio e.012 ∴ k2 = = 0.72 ) 3 k2 k2 e2 1 + e2 1 + 0.72. Solution: Using the Kozeny-Carman formula.72 0. A 26% increase of the voids ratio has effected a doubling the permeability.012 cm/s the permeability has almost doubled.012 cm/sec at a voids ratio of 0. (Revision: Aug-08) The coefficient of permeability of fine sand is 0.012 1 + e1 ∴ 1 = = = 1 + 0. 86 .573 = 0.57 ) 3 k 0.022 cm / s 0. *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. Δh = 4 feet H1=2’ H2=6’ Clayey sand Solution: a) The hydraulic gradient i and the voids ratio e are: Δh 4 ft Gsγ w (2.63)(62.4 pcf ) i= = = 0.667 and e= −1 = − 1 = 0.465 H 2 6 ft γd 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 q = kiA = ⎜ 0.238 ⎟ (−0.667)(36 ft ) = −5.74 2 ⎝ min ⎠ min b) The flow direction is UP. 87 *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. 88 -4 (T s ) (c o s α ) − 4 ( 0 .0 7 5 N / m ) ( c o s 3 ° ) hc = = dγ w d ( 9 .8 1 k N / m 3 ) − 4 ( 7 .5 x 1 0 − 2 N / m ) (c o s 3 ° ) = - 4 1 h1 = m ( 7 .5 x 1 0 m ) ( 9 .8 1 k N / m ) −7 3 − 4 ( 7 .5 x 1 0 N / m ) ( c o s 3 ° ) −2 h2 = = - 0 .4 1 m = 410m m ( 7 .5 x 1 0 m ) ( 9 .8 1 k N / m ) −5 3 − 4 ( 7 .5 x 1 0 N / m ) ( c o s 3 ° ) −2 h3 = = - 0 .0 4 1 m = 41 m m ( 7 .5 x 1 0 m ) ( 9 .8 1 k N / m ) −4 3 89 *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: −4To cos α −4 ( 0.064 N / m ) cos 3° hc = = = −0.26 m dγ w (10 −4 m )( 9.81 kN / m 3 ) Discontinuous moisture zone Vapor flow Capillary fringe zone Capillary flow Capillary saturation zone hc 100% Saturation zone z 90 *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: 0.0306 0.0306 1. The Hazen empirical formula for capillary rise is hC = = = 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 Seγ ω (1)( 0.40 )( 9.81) Se = wGS ∴ w = = = = 0.148 GS γS 26.5 ⎛ 1 + w ⎞ ( 26.5 )(1.148 ) kN ∴ γ SAT = γ S ⎜ ⎟= = 21.8 3 ⎝ 1+ e ⎠ 1.40 m For 50% saturation, w= Seγ ω = ( 0.5 )( 0.40 )( 9.81) = 0.074 γS 26.5 ⎛ 1 + w ⎞ ( 26.5 )(1.074 ) kN ∴γ = γ S ⎜ ⎟= = 20.3 3 ⎝ 1+ e ⎠ 1.40 m 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 91 *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 Q = = 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 The seepage capacity of the surface stratum is Sc , Q 154 gallons / min ( ft 3 ) ( min ) Sc = = = 8.8 x10−3 cfs / ft / ft C ⋅ L '⋅ H (15 ft )( 2.6 ft ) ( 7.45 gallons )( 60 sec ) 92 **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) Δh 10’ e = 0.217 (Sample 2) Geomembrane 15’ e = 0.048 (Sample 3) L 20’ e = 0.015 (Sample 4) 25’ e = 0.006 (Sample 5) (29) Determine the permeability of sample #1; A trial test of a vibroflotation (densification) probe was taken to a depth of 5 feet and showed a densified voids ratio e = 0.55. If this densification ratio is attained to the full depth of 25 feet, at 93 therefore k= = iA iπ d 2 Since 50 grams of water is equivalent to 50 cm3 (for γw = 1g/cm3) 50cm3 cm3 Δh 27in but .5π ( 6in ) ⎜ 2.066 mm / sec 2 ea2 Therefore kb = ka * 2 = 0.5 5sec sec L 6in ( 4 ) ⎛⎜10 cm ⎞ 4q ⎟ (10mm / cm) mm ⎝ in ⎠ ∴ ka = = = 0.54 ∴ Vibroflotation has reduced the permeability by half.122 ( 0. q = = 10 . q 4q From Darcy.066 = = 0. Solution: Step 1: Determine the permeability k of sample # 1.what depth (to the nearest 5 feet) would you place the bottom of the pond in order to keep the total seepage Q below 50.4k0.75) 2 eb Step 3: The ratio of densified permeability to in-situ permeability kb 0. Q= k i A.85e 2 ( 0.000 gal/min? Assume seepage only through the bottom of the pond.122 Step 4: Find the densified permeability at each sample depth. Using Casagrande’s relation k = 1. and that the pond is kept filled.122 iπ d 2 2⎛ cm ⎞ 2 sec 4. 94 . ka 0. and i= = = 4.54 ⎟ ⎝ in ⎠ Step 2: Determine the new permeability of the sand due to vibroflotation densification.55) = 0. 4(0.015 5.005 2.56x10-5 Step 5: Estimate the required depth h of the pond.281×10 ft ⎠ = = 2.4e 1.217 0. k Δh Q From Darcy’s equation Q = kiA or ki = = L A ⎛ ft 3 ⎞⎛ 1mm ⎞ ⎜ 6. 680 min ⎟ ⎜ −3 ⎟ Q ⎝ ⎠ ⎝ 3.005 -15 0.155 mm / sec 1.750 0. Consider this rate to be constant.4 k 0 .006 1.4mile2 ) 5280 ft mile (1min ) 60 sec min 95 .122 0.85 = 2 = = 0.122mm / sec) k = 1 .79 × 10−4 mm ( ) ( ) sec 2 A ( 4. -5 0.4 k 0.752 ) The corresponding densified permeability k = 1.066 -10 0. k (0.74x10-5 -25 0.010 0.48 gal) = 6.85 e 2 (x ratio) Depth (feet) Original eo Original k (mm/sec) Densified k (mm/sec) 0. Q = 50.680 ft3/min.000 gallons/min (1ft3 / 7.048 0.08x10-5 2. 0.8 5 e 2 therefore k0.03x10-5 0.76x10-4 -20 0. 0. 0. place the bottom of pond at -19 feet.5 6 x1 0 ⎟ = 0 . Q A (× 10 −4 mm sec ) 0 0 1 2 3 4 5 6 -5 -10 Depth ft -15 18.76 -25 -30 96 .7 ft -20 2.2 2 4 x1 0 L 5 ⎝ sec ⎠ se c Therefore.14 x10 mm / sec L 10 ⎝ sec ⎠ or at a depth of: -20 ft. Δh 15 ⎛ −4 mm ⎞ −4 k = ⎜ 2.76 x10 ⎟ = 4. Δh 20 ⎛ −5 m m ⎞ −4 m m k = ⎜ 0 . Consider the depth -15 ft. Chapter 6 Seepage and Flow-nets Symbols for Seepage and Flow-nets 97 . 98 . because two equipotential lines intersect each other (equipotential lines and flowlines must intersect orthogonally to each other).*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. b) Incorrectly drawn mesh. c) The well should be at the center of the net (a sink or a source point). because two flow-lines intersect each other (same as above). 0 m ψ2 17. and (b) would the dam be more stable if the cutoff wall was placed under its tail-water side? 15 m Δh = 6.3 × 10 m / sec/ per m of dam width 2 N eq ⎝ sec ⎠ ⎝ 100 cm ⎠ ⎝ 10 ⎠ Since the dam is 500 meters wide (shore-to-shore) the total flow Q under the dam is.3 × 10−6 m3 / sec ⎤⎦ ⎜ 3 ⎟ ⎜ 31. (a) the total seepage loss under the dam in liters per year. Placing the cutoff wall at the toe would allow higher uplift hydrostatic pressures to develop beneath the dam. the number of flow channels Nf = 3 and the equipotentials Neq = 10.5 × 10 6 ⎟ = 100 ⎝ 1 m ⎠⎝ year ⎠ year b) No.0 m.5 × 10 −4 ⎟⎜ −6 ⎟ (6.0 m ψ3 IMPERVIOUS STRATUM (CLAY OR ROCK) Solution: (a) Notice that ∆h = 6. *Flow-nets-02: A flow-net beneath a dam with a partial cutoff wall.0 m ψ1 10. 99 .0 m) ⎜ ⎟ = 6.5 x 10-4 cm/s. The dam is half a kilometer in width (shore to shore) and the permeability of the silty sand stratum is 3.0 m 2. thereby decreasing the dam’s stability against sliding toward the right (down-stream). ⎛ 103 liters ⎞ ⎛ sec ⎞ million liters Q = Lq = 500 m ⎡⎣ 6. Nf ⎛ cm ⎞ ⎛ m ⎞ ⎛ 3⎞ q = k Δh = ⎜ 3. Find. (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. Using Forcheimer’s equation. 0015 ⎟⎜ ⎟ ⎜ ⎟ ( 30 '− 5') ⎜ ⎟ = 0.185 ft 3 ⎞ q ⎜ 5 hr ⎟ ft v= =⎜ ⎟ ≈ 0. Overestimate the flow by using Hazen’s coefficient C = 15 to determine the permeability k.000 foot wide concrete dam. and (2) the velocity at point “a” in feet/hour.4 mm ⎠ ⎢⎣12 in ⎥⎦ ⎝ 1 hour ⎠ ⎝ 12 ⎠ hr − ft of dam ⎛ ft 2 ⎞ ft 3 ∴ Q = Lq = (1. The soil has a GS = 2.67. Δh = 30 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. (Revision: 12 Oct.*Flow-nets-03: The velocity of the flow at any point under a dam.002 A ⎜ (19 ft high )(1 ft wide ) ⎟ hr ⎜ ⎟ ⎝ ⎠ 100 .0015 sec Using Forheimer's equation with flow lines N f = 5 and equipotentials N eq = 12. (1) determine the seepage underneath the 1.185 ⎟ = 185 ⎝ hr ⎠ hr The velocity at "a" has a flow q in only that channel. where the height of the net’s square is 19 feet.01 mm.185 N eq ⎝ sec ⎠ ⎝ 25. Nf ⎛ mm ⎞ ⎛ 1 in ⎞ ⎡ 1 ft ⎤ ⎛ 3. or q/5. D10 = 0. 600 sec ⎞ ⎛ 5⎞ ft 3 q = k Δh = ⎜ 0.-08) Using the flow net shown below. ⎛ 0. 000 ft ) ⎜ 0. 30 m/day. 070 ⎝ N eq ⎠ ⎝ da y ⎠ ⎝ 10 ⎠ day 101 .7 m Cross-section of levee looking north.3 ⎟ ( 23 m ) ⎜ ⎟ = 2. Solution: Using Forheimer’s equation. Laboratory tests indicate that the permeability of the 80-year old levee is 0. 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 B 23 m 2:1 2:1 D A E DRAINAGE 112 m 2. the Everglades are contained with levees. ⎛ N ⎞ ⎛ m ⎞ ⎛ 3⎞ m3 Q = Lq = L ⎜ k Δh f ⎜ ⎟⎟ = (1. Levee #111 runs North-South about 2 kilometers west of Krome Avenue and its cross section is show below. (Revision: Aug-08) In western Miami-Dade County. 000 m ) ⎜ 0.*Flow-nets-04: Flow through an earth levee. 400sec ⎞ Q = Lq = L ⎜ kΔh ⎟ = (1.8 feet 102 .200 ft ) ⎜ 0.200 feet wide. 40’ rock toe 7 1 2 A3 4 5 6 0 25’ Clay (an aquiclude) Solution: ⎛ Nf ⎞ ⎛ mm ⎞ ⎛ 1inch ⎞⎛ 1 ft ⎞⎛ 7.1 feet ⎝ 7 ⎠ 2 The static head at "A" is approximately ( 40 ft ) = 26.17 mm ) 2 = 0.4 ⎞ At point "A" the dynamic pressure head is ⎜ ⎟ ( 40 ft ) = 25. use C =15.8×106 day ⎛ 4.5 gallons ⎞ ⎛ 3 ⎞ ⎛ 86.1 ft + 26. (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.43 ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ( 40 ft ) ⎜ ⎟ ⎜ ⎟ ⎜ Neq ⎟⎠ ⎝ sec ⎠ ⎝ 25. static and dynamic heads in a dam.7 ft = 51. to overestimate flows.43 sec Note: 8 ≤ C ≤ 15 for D10 mm.4 mm ⎠⎝ 12inches ⎠⎝ 1 ft 3 ⎠ ⎝ 7 ⎠ ⎝ 1day ⎠ ⎝ gallons Q = 18. and the dam is 1.*Flow-nets-05: Finding the total. the total head = static + dynamic = 25.7 feet 3 Therefore. 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 mm Using Hazen's formula k = CD 102 = 15 D102 = 15(0. 000 feet wide is shown below. and (c) the pore pressures along a trial failure surface along the line ED.**Flow nets-06: Hydraulic gradient profile within an earth levee. 103 . (b) the hydraulic gradient in square I. (Revision: Aug-08) The cross-section of an earth dam 5. in ft3 / minute. Determine (a) the seepage flow through the dam. 4 mm ⎠ ⎝ 12 inches ⎠ ⎝ day ⎠ ⎝9⎠ ft 3 Q = 454.04 mm.6 ( 9 ) 62. with C = 15 to overestimate the permeability of the dam.2 (c) The pore pressures along ED are approximately.Solution: (a) From graph D10 = 0. Δh 40/9 iI = = = 0. 400 sec ⎞ ⎛3⎞ Q = Lq = L ⎜ k Δh f ⎜ ⎟⎟ = ( 5.7 ( 9 ) 62.83 ksf 40 6 u= 40' .4 = 1109 psf = 1.024 mm / sec 2 2 ⎛ N ⎞ ⎛ mm ⎞ ⎛ 1inch ⎞ ⎛ 1 ft ⎞ ⎛ 86.4 = 277 psf = 0.28 ksf 104 . 000 ft ) ⎜ 0.4 ( 9 ) 62. at pore pressure E 0 0 40 1 u= 40' .66 ksf 3 40 u= 40' . k = C ( D10 ) = (15 )( 0.40 lI 11.04 mm ) = 0.5 ( 9 ) 62.55 ksf 40 D u= 40' .4 = 1803 psf = 1.3 ( 9 ) 62.80 ksf 40 2 u= 40' . 000 day (b) The gradient in square I is.39 ksf 40 4 u= 40' .4 = 555 psf = 0.11 ksf 40 5 u= 40' .4 = 832 psf = 0.4 = 1664 psf = 1.024 ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ( 40 ft ) ⎜ ⎟ ⎝ N eq ⎠ ⎝ sec ⎠ ⎝ 25.5 ( 9 ) 62.2. Using Hazen’s relation.8 ( 9 ) 62.4 = 1387 psf = 1. The specific gravity of the solids was found to be 2.**Flow-net-07: Flow into a cofferdam and pump size. with 80 % passing the # 200 sieve. (30) What type of soil was the sample? (31) Will a large 3 m3 per minute pump be adequate to maintain a 1 m draw down below the bay bottom? Use FS > 2. The dynamic viscosity of water is 10- 2 Poise (dynes-sec/cm2) at 20oC. 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. (Revision: Aug-08) A cofferdam is to be built in the middle of a bay to place the foundations of a tall television tower.65. The soil is uniform in size. Pump CL 7m 1m 15 m Clay stratum 105 . the hydrometer dropped 16 cm. After 1 hour of precipitation. 106 . 002 mm).074 sec sec Use Forheimer’s formula to estimate the total flow Q into the cofferdam: Nf ⎛ mm ⎞ ⎛ 4 ⎞⎛ m ⎞ −5 m 3 q = k Δh = ⎜ 7.6 x10 N eq ⎝ sec ⎠ ⎝ 8 ⎠ ⎝ 10 m m ⎠ m−s ⎛ m 3 60 s ⎞ m3 Q = ( perimeter )( q ) = 80 m ⎜ 29.42 ⎝ m − s min ⎠ min 3 m 3 / min Pump FS = = 2 .4 x10 −2 ⎟ ( 8 m ) ⎜ ⎟⎜ 3 ⎟ = 29.07 mm ) 2 = 0.6 x10 −5 . b) Hazen’s formula permits us to estimate the permeability k of the soil: mm mm k = CD102 = 15(0.075 mm to 0.42 m 3 / min 107 .81dynes x 3600sec ) ⎟ ⎠⎝ ⎠ d = 0. ⎟ = 1.11 > 2 OKAY Pump is adeq uate ! 1.65 − 1) ( 9.070 mm Therefore the soil is silt (0.Solution: a) Use Stoke’s formula to find diameter of the bay bottom particles: 18η L ⎛ 18η ⎞ ⎛ 18 x10−2 dynes − sec x cm3 x 16cm ⎞ D10 = d = . =⎜ ⎟⎜ ⎟ γ S − γ W t ⎜⎝ γ W ( GS − 1) t ⎟ ⎜ cm 2 ( 2. a) At what depth of the excavation will the limestone (shear strength = 0. b) What size pump do you need (m3/minute) with a factor of safety of 3? Pump Elevation +0 m kN Sand γ = 1 7 m3 Elevation –1. The plan size of the site is 100 x 80 meters. Some of the soil properties are shown below.1 MN/m2) have a punching shear failure? Suggest using a 1m x 1m plug as a model.*Flow-nets-08: Drainage of deep excavations for buildings.0 m Anchor Limestone kN γ = 19 m3 Nf = 3 A Neq = 8 Elevation –7 m 0 8 F2 F1 F3 Elevation –9 m 1 2 7 3 4 5 6 kN 108 Sandstone γ = 21 Elevation –11 m m3 . (Revision: Aug-08) A new office building will require a two-level underground parking garage. 6 m / m i n ≈ 5 m / m i n 3 3 ⎝ sec ⎠ ⎝ 8 ⎠ ⎝ m ⎠⎝ mm ⎠ m3 F or a F S = 3 use a 15 pum p.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 li f t .8 1 kN ⎞ F u p lift = u A = (Δ hγ w ) A = ⎟ (1 m )= 2 59 kN ⎝ m3 ⎠ MN kN F s h e a r = τ A = 0 .6 m .5 ) ( 0 . ⎛ a ⎞⎛ N f ⎞ Q = qL = Lk ⎜ ⎟ ⎜ ⎟⎟ Δ h ⎝ b ⎠ ⎜⎝ N q ⎠ mm k = C D 120 = (1 2 . m in 109 . 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 .F s h e a r r e s i s ta n c e = 0 ( 6 m ) ⎛⎜ 9 .1 1 3 sec Δh = 5m ⎛ mm ⎞⎛ 3 ⎞ ⎛ 60 ⎞ ⎛ m ⎞ ∴ Q = ( 3 6 0 m ) ⎜ 0 .1 (4 x ) m 2 = 4 0 0 x m2 m 59 ∴ x = = 0 .1 1 3 ⎟ ⎜ ⎟ (5 m ) ⎜ ⎟ ⎜10 −3 ⎟ = 4 .1 4 7 m w i t h F S = 7 ∴ x =1m 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 .0 9 5 m m ) 2 w here = 0 . T he perm eability k = C ( D10 ) = (15 )( 0. N f = 3.006 m m / s 2 2 ⎛ N ⎞ ⎛ mm ⎞ ⎛ 3 ⎞ ⎛ m ⎞ ⎛ 60 s ⎞ m3 Q = L q = L k Δh ⎜ f ⎜N ⎟⎟ ( = 200 + 800 ) ⎜ m 0.*Flow-nets-09: Dewatering a construction site.45 gal ⎞ g a l lons Q = ⎜ 2.200 gallons per minute pump or two 600 gallons per minute pumps.16 ⎝ eq ⎠ ⎝ s ⎠ ⎝ 4 ⎠ ⎝ 10 m m ⎠ ⎝ 1 m in ⎠ m in 3 ⎛ m3 ⎞ ⎛ ft ⎞ ⎛ 7. for a factor of safety of 2 use at least a 1.02 m m ) = 0.16 ⎟⎜ ⎟ ⎜ 3 ⎟ = 60 0 ⎝ m in ⎠ ⎝ 0 .006 ⎟( 8 m )⎜ ⎟ ⎜ 3 ⎟⎜ ⎟ = 2. 000 m . 110 . 30 m ⎠ ⎝ ft ⎠ m in Therefore. N eq = 4 and L = perim eter = 1. (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. What size pump do you need (gpm) with a Factor of Safety = 2? Dewatering wells Marshy soils Δh = 2m Excavated site CL Δh = 2m 1 Δh = 2m 2 3 Δh = 2m Lowered WT Neq = 4 Nf = 3 Solution: N otice that Δ h = 8 m .02 mm. The plan area of the excavation is 400 m long by 100 m wide. The soil has a D10 of 0. to prevent the excavation from flooding. Estimate the dewatering capacity requirements. (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. CL 40 m 40 m 10m 10m 80 m Phreatic surface condition is CL fulfilled along the top flow-line GWL Sheet-pile wall Datum for h (½ square) h=0 h=1(1/3)m h=2(2/3)m h=4m h=5(1/3)m h=6(2/3)m (½ square) h=8m h=9(1/3)m (½ square) (½ square) Assumed recharged boundary h = 10 4m Scale 111 . in m3/hour. The value of Δh is to scale. but you may use 10 m.*Flow-net-10: Dewatering in layered strata. N eq = 8 k = kxk y = ( 5 x1 0 −4 m / s )( 5 x1 0 − 5 m / s ) = 1 . N f = 4. 6 0 0 s ⎞ m3 Q = 0 .2 5 3 ⎜ ⎟⎜ ⎟ = 911 ⎝ s ⎠ ⎝ hr ⎠ hr 112 .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 ⎞⎤ ⎡ ⎛ −4 m ⎞ ⎛ 4 ⎞⎤ Q = qp = ⎢k Δh ⎜ ⎜ N ⎟⎟ ⎥ p = ⎢ (1 .Solution: Δh = 10m .6 ) ⎜ 1 0 ⎟ (1 0 m ) ⎜ ⎟ ⎥ ( 3 2 0 m ) ⎢⎣ ⎝ eq ⎠ ⎥⎦ ⎣ ⎝ s ⎠ ⎝ 8 ⎠⎦ ⎛ m 3 ⎞ ⎛ 3. Reservoir surface inclined filter A 10 ft H=100 ft shell dam permeability k1= 0. The material should be relatively impermeable (clay) and should not shrink or swell excessively. 15% size = 1/8”. (32) State the function and properties of the core shell and drains. 113 .001 ft/day core L = 150 ft grout h1=5 ft horizontal drain B Note: The grain size of core = 100% passes 1”. and 85% size = 0. strong and durable. (35) What grading requirements should be specified for the inclined filter A? (36) What minimum permeability k is required in the horizontal drain B to prevent saturation from rising into the random fill zone? Give the results of k in ft/day. to resist seepage. (Revision: Aug-08) An earth dam on a pervious but strong earth foundation has the cross-section shown in the figure below. The shell provides the structural strength to support and protect the core.001 in. Foundation layer permeability k2 = 0.1 ft/day Solution: (a) The core is used to retain water within the dam. **Flownets-11: Flow through the clay core of an earth dam. that is. through the foundation. (33) What is the function of the grout between the core and foundation? Under what conditions is it most important? (34) Calculate the seepage quantity per foot of length of the dam through the dam. and the total seepage quantity. The material must be more permeable than the core material. The core of the dam is sealed from the jointed rock foundation with a thin layer of 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(Δh/L)b in ft3/day/ft where b is the normal distance between streamlines. The flow net divides the core into 4 zones (#1 at the bottom, #4 at the top). 4 4 Q = k1 ∑ ij bj = k1 ∑ (Δh)j (b)j j=1 j=1 In zone #1, the flow net is nearly rectangular, so (b)1 = 2; for zone # 4, (b )j = 1 l l The average head loss hl across the core in each zone (p) + Zu = constant = 100’ on the upstream face γ On down stream face of core (p) = 0 is assumed in the drain, γ so that ZL + hL = constant = 100’ on the downstream face Using an average ZL for each zone by scaling, Zone#1 ZL = 2’ hL = 98’ ΔhL = 98 6 6 Zone#2 ZL = 10’ hL = 90’ ΔhL = 90 5.5 5.5 Zone#3 ZL = 25’ hL = 75’ ΔhL = 75 4.5 4.5 Zone#4 ZL = 55’ hL = 45’ ΔhL = 45 3 3 4 Q1 = k1 ∑Δhj (b/L)j = 0.001 [98’(2) + 90’ + 75’ + 45’] = 0.081 ft3/day/ft of dam j=1 6 5.5 4.5 3 2) Through the foundation Q2 = NF kh 114 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. 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: 115 kmin = 2LQ = 2(150)(3.83) = 46 ft/day h1 2 52 116 Chapter 7 Effective Stresses and Pore Water Pressure Symbols for Effective Stresses and Pore Water Pressure 117 *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? 118 *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' =⎢ ⎥ − γW 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) ⎧⎪ ⎡ ( GS + e ) γ W ⎤ ⎫⎪ ⎡ ⎡⎣ 2.78 + ( 0.54 ) (2.78) ⎤⎦ (9.81) ⎤ σ ' = γ ' hsoil = ⎨⎢ − γ ⎥ W ⎬ soil ⎢ h = − 9.81 ⎥ (15 m ) ⎩⎪⎣ 1 + e ⎦ ⎭⎪ ⎣⎢ 1 + ( ) 0.54 (2.78) ⎦⎥ σ ' = 105 kPa 119 *Effective Stress–03: The concept of effective stress with multiple strata. (Revision: Aug-08) The City of Houston, Texas has been experiencing a rapid lowering of its phreatic surface (draw- downs) during the past 49 years due to large volumes of water pumped out of the ground by industrial users. a) What was the effective vertical stress at a depth of 15 m in 1960? b) What is the effective stress at the same depth in 2009? c) What happens to the ground surface as a result of the draw-downs? Solution: a) 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. 120 with numerical values at each interface.4) x (6) = 562 Δh = (110. Pay heed to the capillarity in the upper clay.4) x (4) = -250 Δh = (62.4) x (4)] + 664} = 854 Δh = {[(117-62.7) x (6) = 664 Δh = {[(110-62. show a plot of the pore water pressure and the effective stress along the right margin of the figure. Assume S = 50%.4) x (9)] + 854} = 854 Δh = 664 – 250 = 414 Δh = 0 + 854 = 854 Δh = 562 + 1346 = 1908 Depth (ft) u + σ' σ 0 0 0 0 6 -250 664 414 10 0 854 854 19 562 1346 1908 121 .Effective Stress-03B Revision In the soil profile shown below. Solution: Δh = (62. Chapter 8 Dams and Levees Symbols for Dams and Levees 122 . 5) = 35. (Revision: Sept.3 ft Pressure head at B = 30 ft + 8 ft . Solution: The dynamic head drop per equipotential is. ⎛ p + pB ⎞ ⎛ 35. HA − HB 30 ft − 5 ft Δ (Δh) = = = 1.Δ h (1.8 ft / drop N eq 14 drops Pressure head at A = 30 ft + 8 ft .*Dams-01: Find the uplift pressure under a small concrete levee.4 ) 3 = 169. Points A and B are at the corners of the concrete levee. 000 = 16 9 ⎝ 2 ⎠ ⎝ 2 ⎠ ft ft ft 123 .-08) Calculate the uplift force at the base of the weir.Δ h (10) = 20.1 ft The upli fting force F is.3 + 20.1 ⎞ lb lb kip F = L⎜ A γ w ⎟ = 98 ' ⎜ ⎟ ( 62. per foot of width. *Dams-02: Determine the uplift forces acting upon a concrete dam. to 100% of the tail-water at the toe. Assume γ Concrete = 145 pcf a) Determine the Factor of Safety against overturning. o 124 . if the sand that underlay the dam has ϕ = 3 7 . (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. and b) Determine the FS against sliding. 7 2 ⎟⎟(300 ft )⎜ (300 ft )⎟ ft ⎠ ⎝ 2 ⎠ 2⎝ ft ft ⎠ ⎝3 ⎠ x3 .340 kips / ft 2 ft ⎛ k ⎞ ⎛ 300 ft ⎞ 1 ⎛ k k ⎞ ⎛2 ⎞ ⎜⎜ 3.7 ) 2 ( 300 ft ) = 2. 3 Step 2: The hydrostatic uplift at the base of the dam.0624 )( 60 ft ) = 3.9 2 − 3.3 ft left of toe ⎛ k ⎞ 1⎛ k k ⎞ ⎜⎜ 3.9 + 3.534 kips / ft 2 ⎝2 ft ⎠ 1 y1 . from toe O = (52 ft ) = 17. from toe O = (60 ft ) = 20 ft above the toe.9 2 − 3.0624 3 ⎟⎟(285 ft ) = 2.534 kips / ft 2 where the pressure pLEFT = ( 0.145 3 ⎟⎟⎜⎜ (300 ft )(40 ft ) + ⎜ ⎟(260 ft )(300 ft )⎟⎟ = 7.9ksf and the pressure pRIGHT = (1.395 kips / ft ⎝ ft ⎠⎝ ⎝2⎠ ⎠ ∑M0 = 0 (1740 k )(280 ft ) + (5655 k )⎛⎜ 2 (260 ft )⎞⎟ x1 . from toe O = ⎝ = 176.Solution: Step 1: Determine all forces on the dam ΣFV = 0 ⎛ k ⎞⎛ ⎛1⎞ ⎞ V1 = weight of dam = ⎜⎜ 0.0624 3 ⎟⎟(60 ft ) = 112 kips / ft 2 ⎝ 2 ft ⎠ 1 y 2 .0624 )( 285 ft ) = 11. from toe O = ⎝3 ⎠ = 198 ft left of toe (1740 k + 5655 k ) k ⎛1 ⎞ V2 : vertical weight of water upon toe sec tion = 0. 3 ⎛1 k ⎞ H 2 : lateral force from tailwater = ⎜⎜ 0.3 ft left of toe 3 ⎛1 k ⎞ H 1 : lateral force from headwater = ⎜⎜ 0.0624 ⎜ (60 ft )(52 ft )⎟ = 97 kips / ft ft 3 ⎝2 ⎠ 1 x 2 .7 2 ⎟⎟(300 ft ) ⎝ ft ⎠ 2⎝ ft ft ⎠ 125 .7 2 ⎟⎟(300 ft ) + ⎜⎜11. 1 V3 : uplift force = ( pLEFT + pRIGHT )( 300 ft ) = 2.67 ( 0.00 ) γ w h = 1.67 ) γ w h = 0. from toe O = (285 ft ) = 95 ft above the toe.7 ksf 1 k ∴ V3 = (11.7 2 ⎟⎟(300 ft )⎜ ⎟ + ⎜⎜11.0 ( 0. S.3) + (112 )( 20 ) = 2. against sliding. resisting forces H 2 + (V1 + V2 − V3 ) tan φ FS sliding = = driving forces H1 FS sliding = (112 ) + ( 7395 + 97 − 2340 ) tan 37° = 1.3) Step 4: The F.2 > 2 GOOD ( 2534 )( 95 ) + ( 2340 )(176. resisting moments V x + V2 x 2 + H 2 y 2 FS = = 1 1 overturning moments V3 x3 + H 1 y1 FS overturning = ( 7395 )(198 ) + ( 97 )(17.58 < 2 NOT GOOD ENOUGH 2534 126 .Step 3: The factor of safety (FS) against overturning (taken about the toe). L→ Length of the loaded selected region. z→ Depth of the soil at the point of interest. C→ consolidation. Dp→ Increased stress on the soil from a surface loaded area. σn→ Normal stress. p→ Stress of the loaded area. τn→ Shear stress. σx→ stress at an specific point (x). z→ Depth of the stratum stat of the soil. n→ The ratio (L/Z). γ→ Unit weight of the soil. AB→ Scale to the depth of interest to determine the size of the surface structure graph the Newmark’s graph. po → The effective stress at the point of interest.01. a chart divided into 100 areas. Symbols for Newmark IV → The influence value in the Newmark’ chart (for example. qu→ Ultimate shear strength of a soil. ф → The angle of internal friction of the soil. σθ→ The normal stress on a plane with an angle θ with respect to the major principal stress plane (σ1). τminx→ Minimum shear stress. GS→ Specific gravity of the solids of a soil. 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). m→ The ratio (B/Z). Symbols for Boussinesq Stresses B→ Width of the loaded selected region. γ→ Unit weight of the soil. each is IV=0. 127 . N→ Normal load carried by a foundation. q→ The load of the footing. M→ Nomber of squares (enclosed in the Newmark’ chart). τmax→ maximum shear stress. qu→ Ultimate shear strength of a soil. σmin→ Minimum normal axial stress. σmax→ Maximum normal axial stress. τθ→ The shear stress on aplane with an angle θ respect to the major principal stress plane (σ1). 128 . (a) The graphical solution.2 2 ⎝ 2 ⎠ ⎝ 2 ⎠ m ⎛ σ −σ ⎞ kN τ maximum = ⎜ 1 3 ⎟ = 9.6 + 4.18 ⎞ kN τ θ = ⎜ 1 3 ⎟ sin 2θ = ⎜ ⎟ sin 2 ( 50° ) = 9.6 + 4. (Revision: Jan-09) A soil particle is found to be subjected to a maximum stress of 14. Find the σ and τ on the plane of θ = 50° with respect to the major principal stresses. (b) The calculated solution.6 − 4.18 kN/m2.18 ⎞ kN σθ = ⎜ ⎟+⎜ ⎟ cos 2θ = ⎜ ⎟+⎜ ⎟ cos 2 ( 50° ) = 3.6 kN/m2. and a minimum stress of – 4.18 ⎞ ⎛ 14.4 2 ⎝ 2 ⎠ m 129 .6 2 ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ m ⎛ σ −σ ⎞ ⎛ 14. and also find τmax. ⎛ σ1 + σ 3 ⎞ ⎛ σ1 − σ 3 ⎞ ⎛ 14.*Mohr-01: Simple transformation from principal to general stress state. 57° = 41. The direction of the major δ principal stress bisects the angle α.77 σ = σ1 + σ3 2 σ .63° if q = 432.57° β = 85° 68.-08) Equations for the principal stresses in the elastic half-space shown below for a uniformly loaded strip footing are as follows: σ1 = Β /2 Β /2 q/π(α + sin α) and σ 3 = q/π(α .R = 100 .5 kPa then σ1 = (σx + σy)/2 + R = 100 + 70.38 σ = β θ σ = Ο 2θ = 94.26.20° .7 = 29.20° 2θ + β = 180° δ = arc tan 0.sin α).*Mohr – 02: Find the principal stresses and their orientation. Solution: σ = σ1 + σ2 2 σ .75B and y = 0.70.7 = 170.τ σ = Ο β = 85.5 = 68. α Calculate the vertical stress σy. (Revision: Sept.τ τ = α + δ = arc tan 2.3 kPa 130 .5B using Mohr’s diagram.7 kPa σ3 = (σx + σy)/2 .5 = 26. and τxy at point A if x = 0. the horizontal stress σx.23 σ = Ο 47. (Revision: Sept-09) Given the general stresses at a point in a soil.*Mohr – 03: Find the principal stresses and their orientation. determine the principal stresses and show them on a properly oriented element. 131 . If the angle θ between the vertical stress and the principal stress is 60°.6 kN/m2. The sample had been under a vertical load of 150 kN/m2. and a shear stress of 86.*Mohr – 04: (Revision: Sept-09) A sample of clean sand was retrieved from 7 m below the surface. a horizontal load of 250 kN/m2. what is the angle of internal friction φ of this sample? Solution: 132 . √ 90-125 + (-40) 2 2 2 τn θ 30° σ1 = 151 psf σ3 = 64 psf B σx σn 125 lb/ft2 τxy .*Mohr – 05: Normal and shear stress at a chosen plane.64 cos60° 2 2 σn = 129 psf τn = σ1 . Solution: σy = 90 lb/ft2 (a) σ1 = σv + σx +√ σv .σ3 cos 2θ 2 2 = 151 + 64 + 151.40 lb/ft 2 A σ3 = 90 + 150 .40 lb/ft2 (b) σn = σ1 + σ3 + σ1 .σx 2 + τ xy 2 2 2 2 .σ3 sin 2θ 2 = 151 – 64 sin60° 2 τn = 38 psf 133 . determine the normal and shear stresses on the plane AB. (Revision: Sept-09) Using a Mohr circle. (c) the angle theta. B 20 kN/m2 σ3 = 0 10 kN/m2 A β β C A Graphical Solution: OR B Analytical Solution: τ τmax = 12.4˚ (b) α in the figure is in the angle 2θ between (σ. (b) the angle alpha.2 sin 2θ + cos 2θ = 0 (a) σ1 = 25 kN/m 2 and σ3 = 0 …… θ = 63. σ1 = = 25 KN / m2 .5 kN/m2 …… τmax= 12. σ3 = 0 …… α = 2θ = 126.9˚ sin 126.. as shown. τ) and σ3. find (a) the maximum and minimum principle stresses. -10) ∴ 1 .**Mohr – 07: Back figure the failure angle (Revised: Sept-09) From the stress triangle shown below.5 1 σ1 +σ1 σ1 +σ1 σ1 = + COS 2θ 2 2 σ1 σ1 40 20 = + COS 2θ ∴ σ1 = 2θ 2 2 1 + cos 2θ σ1=25 1+cos2θ σ3=0 2 σ1 −σ3 τn = SIN 2θ 2 2θ 2 σ1 20 (σ.5 kN/m2 134 .90 (c) …… θ = 63.4˚ and τmax= 12. τ) 10 = SIN 2θ ∴ ϑ1 = 2 SIN 2θ (20. and (d) the value for the maximum shear stress. 20 …. Every morning. a misguided foreman tightens the screw mechanism on the strut “just to be safe”.*Mohr – 08: find the Principle pressure using Mohr (Revised Sept-09) The temporary excavation shown below is braced with a steel tube strut. just behind the wall. has been measured with a pressure sensor installed by the Engineer. estimate the normal and shear stresses at that point A along a potential failure 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. If the potential failure planes in the soil behind the wall sustain 60° angles with respect to the vertical wall. The stress on a soil particle at point A.7 kN/m2 135 . Since θ = 60° is with respect to the major principal stress (σ1) plane. It now measures 40 kN/m2. Solution: At point A: σv = hγ = (1.25 m) (16 kN/m3) = 20 kN/m2 ∴ σv is the minor principal stress at A. prove that θ = 45°+ φ/2 using Mohr’s circle. c=0) shows that σ1=11.e.2 ksf at failure. (Revised Sept-09) For a clean sand. Find the angle φ for this sand. By inspection in ΔOAB (180°-2θ)+90°+φ=180° ∴ 2θ=90°+φ ∴ θ = 45° + φ/2 A failure test on a clean sand (i. 136 . Solution: For sand c=0.*Mohr – 09: Relation between θ and φ.5 ksf and σ3=3. 52 + 552 = 103 psf σ1 = 212.5) = 32° σn = 212.5 psf 0102 = (300-125)/2 = 87.5 + 103 = 315.5 psf σ3 = 212.5 – 103 = 109.5 psf πC0102 = tan-1 (55/87.5 psf 01B = 87.*Mohr – 10: (Revised Sept-09) Determine the normal and shear stresses on the plane AB. 001 = (300+125)/2 = 212.5 – 103 cos (32+40) = 181 psf τn = 103 sin (32+40) = 98 psf 137 . (39) Determine the angle θ between the major principal stress and the state of stress shown in the figure above.*Mohr–11: (Revised Sept-09) (37) Derive the equation that transforms a general state of stress to the principal state of stress. (38) Determine the value of the major principal stress. 138 . (Hint: Use Mohr’s circle for a graphical solution). Maximum and minimum principal stresses: Normal and shear stresses: 139 . and the normal and shear stresses on plane AB.*Mohr – 12: (Revised Sept-09) Determine the maximum and minimum principal stresses. 5° qu = -123 Pa 140 .) σ1f = qu (4. (Revised Sept-09) A soil sample has been tested and when plotted developed the Mohr-Coulomb envelope of failure shown below.) σf = σθ +τ τf = τθ (3. (2) the normal and shear stresses on the failure plane.*Mohr – 13: Data from Mohr-Coulomb failure envelope. (3) the angle of failure with respect to the principal axis. and (4) the soil tensile strength.) qu Solution: σ1f = qu = 300 Pa σf = σθ = 86 Pa τf = τθ = 136 Pa θ = 115°/2 = 57. +τ (2.) θ 25 ° 2θ = 115° 96 kN/m2 +σ σ1 +σ (1. Find (1) axial stresses at failure. 200kPa 2 H 141 . and cohesion of 12 kPa.sin φ ) σ3 = d − R = −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 36° ∴σ1 = (300 kPa) tan2 (45°+ ) = 1160 kPa 2 τ b) H = c(cot φ ) φ R σ1 + H = (σ 3 + H ) tan2 (45°+ ) 2 c Φ σ3 σ1 σ H = (12 kPa)(cot 36°) ≈ 17 kPa d 36° σ1 = (300 +17) tan2 (45° + ) −17 = 1. 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. The angle of internal friction was found to be 36°. If the minor principal stress was 300 kPa.**Mohr – 14: (Revised Sept-09) A dry sample of sand was tested in a triaxial test. R = sin φ d R (1 . What was the new maximum principal stress? Solution: τ τ Φ = 36° T R Φ Φ σ 0 σ 0 σ3 d τ’ a) Failure will occur when the Mohr circle becomes tangent to the failure envelopes. σ 3 ∴ (c) sin φ cot φ = σ1 -σ3 − σ1 +σ3 sin φ 2 2 2 2 ∴ 2 ((c ) sin φ cot φ ) = σ 1 .σ 3 + (1 + sin φ ) ∴ 2 (c ) cos φ + σ 1 + sin φ = σ1 sin φ ⎟⎠ 1 − sin φ 1 − sin φ 3 ⎝ cos φ 1 + sin φ φ⎞ Since 1 − sin φ ≈ tan 45 0 + φ( ) and 1 − sin φ ⎛ ≈ tan 2 ⎜ 45 0 + ⎟ 2⎠ ⎝ ⎛ φ⎞ ⎛ φ⎞ σ 1 = σ 3 tan 2 ⎜ 45 0 + ⎟ + 2c tan⎜ 45 0 + ⎟ ⎝ 2⎠ ⎝ 2⎠ 142 .(sin φ ( σ 1 ) . Solution: θ θ = 45° + ∴ 2θ = 90° + φ 2 σ1 −σ3 ad = 2 ad sinφ ≈ (1) af σ1 +σ3 af = af + af = (c) cotθ + (2) 2 c of tan θ = ∴ = cot θ ∴ of = (c) cot θ (3) of c According to the Mohr’s circle properties: σ1 -σ3 σ1 -σ3 σ1 -σ3 oa = σ 3 + ∴ oa = ∴ af = (c) cot φ + (4)From (1) & (4): 2 2 2 σ1 -σ3 ⎛ σ +σ3 ⎞ σ1 -σ3 sin φ = 2 ∴ sinφ ⎜ (c )cot φ + 1 ⎟= (c )cot φ + σ 1 + σ 3 ⎝ 2 ⎠ 2 2 (c ) sin φ cot φ + σ 1 + σ 3 sin φ = σ 1 . (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.sin φ ) .*Mohr – 15: Derive the general formula for horizontal stress.(σ 3 + sin φ ( σ 3 ) ) ⎛ ∴ 2 ⎜⎜ (c ) sin φ cot φ ⎞ ⎟ = σ 1 (1 . 005 y The increase in the vertical stress is found from Δp = po M (IV) where M is the number of “squares” enclosed in the Newmark chart and (IV) is the influence value. 10 m x 10 m in plan. 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.005) Æ Δp2 = 84 kN/m² (which represents a 78% reduction in the vertical stress).-08) A construction site has a surface layer of Aeolic sand 2 m thick. (Revision: Sept. AB = 12 m Æ Δp2 = po M(IV) = (400 kN/m²) (42) (0. For Δp1. *Newmark–01: Stress beneath a tank at different depths. Solution: BB == 1100 m m B = 10 m for point #1 SSA ANND D 22 m m 1 C CL C LA LAAY Y 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. underlain by a 10 m thick clay stratum. AB = 2 m Æ Δp1 = po M(IV) = (400 kN/m²) (190) (0. with a contact pressure po = 400 kN/m². The project involves placing a wastewater treatment tank.005) Æ Δp1 = 380 kN/m² For Δp2. 143 . at a depth of 3 m below the edge of the footing.5 k N / m 2 ⎣ ( 3 m )( 3 m ) ⎦ The number of elements inside the outline of the plan is about M = 45. Hence. (Revision: Aug-08) Find the stress at the point A shown below. ⎡ 6 6 0 kN ⎤ Δ p = q M ( IV ) = ⎢ ⎥ ( 4 5 )( 0 . 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.*Newmark-02: The stress below the center of the edge of a footing.5 k N / m 2 ⎣ ( 3 m )( 3 m ) ⎦ 144 .0 0 5 ) = 1 6 .0 0 5 ) = 1 6 . ⎡ 6 6 0 kN ⎤ Δ p = q M ( IV ) = ⎢ ⎥ ( 4 5 )( 0 . Hence. Solution: The number of elements inside the outline of the plan is about M = 45. at the point C.*Newmark-03: Stress at a point distant from the loaded footing.005 Depth Point = Z Solution: Set AB = 5 and draw the footing to that scale. Influence Value (IV) = 0.8 ksf. (Revision: Aug-08) The footing shown below has a load q = 1. therefore Δp = q M (IV) = (1.800 psf)(8)(0. The number of affected areas M =8. Find the stress at a depth of 5 feet below the footing invert.005) = 72 psf 145 . (Revision: Aug-08) A small but heavy utility building will be placed over a 2 m thick sand stratum.*Newmark-04: Stresses coming from complex shaped foundations.005 ) = 68 m2 kN Point B (bottom of clay) q B = q M ( IV ) = (100 )(100 )( 0. 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 Point A (top of clay stratum) q A = q M ( IV ) = (100 )(136 )( 0.005 ) = 50 2 m q + qB ( 70 + 50 ) kN qaverage = A = = 59 2 2 2 m 146 . 001 Solution: The contact stress is qo = (0. (Revision: Aug-08) A circular oil storage tank will be built at the shore of Tampa Bay.*Newmark-05: Stress beneath a circular oil tank.81)(10 ) ⎤⎦ ( 680 )( 0. Find the stress increase from a fully loaded tank.5 m2 147 . at mid-clay stratum. and 15 m high. Set AB 10 m Influence Value (IV) = 0.001) = 28.95)(9. using the Newmark influence chart shown below. It will be 20 m in diameter.81 kN/m3)(15m) = 140 kN / m 2 At mid-clay depth along the centerline of the tank (depth = 10 m) ∴ OQ = 10 m kN σ v' = ( qo − γ w h ) M ( IV ) = ⎡⎣140 − ( 9. The water table is at practically at the surface. The tank sits upon a 2 m thick sand deposit that rests upon a clay stratum 16 m thick. and (b) at its outer edge. (a) directly under the center of the tank. The soil profile consists of 3 m of a dry sand.5 kN / m 2 The stress at point C has drop ped to 50% of the stress at E. what are the new stresses at the top and bottom of the clay stratum due to the building’s loading? (41) What is the expected differential settlement between the building’s center and one of its corners.5 kN/m3.5 kN/m3 under laid by a 5 m thick clay layer with a γ = 18.30 and a Gs = 2.005 ) = 87. (Revision: Aug-08) A small but heavily loaded utility building has dimensions of 20 m x 20 m. ∴ Δ σ E = 100 kN / m 2 Set AB = 7 m and the number of units M = 175 ∴ Δ σ F = qM ( IV ) = (100 )(175 )( 0. in mm? (42) If a laboratory sample 4” thick of the field clay attained 50% consolidation in 5 hours. what time will the clay layer in the field attain 60% consolidation? q =100 kN/m2 1m SAND 3m 2m C E • • CLAY γ = 18. (40) Using the Newmark method. and the phreatic surface coincides with the top of the clay stratum.**Newmark-06: Use Newmark with a settlement problem. It applies a uniform load on its mat foundation of 100 kN/m2. Its mat foundation sits 1 m below the surface. The clay stratum is under laid by another sand stratum. Cc = 0.96 )( 50 ) = 48 kN / m 2 Average stress in the clay stratum beneath the center = (100 + 87.70.5 kN/m3 w = 22 % 5m •D F • Solution: (1) Set AB = 2 m and observe that the building's foot-print covers the entire graph.8 kN 2 m2 Average stress in the clay stratum beneath the corner = ( 50 + 48 ) = 49 kN 2 m2 148 . ∴ Δ σ C = 50 kN / m 2 and ∴ Δ σ D = ( 0. with γ = 17. a moisture content of 22%.5 ) = 93. 59 S 1 The in-situ stress at mid-clay stratum before the building is built is. ( 0.8) = 74.26)( 2.3) + ( 93.22)( 2.8) ⎞ the settlement at the center ΔH = log ⎜ o ⎟= log ⎜⎜ ⎟⎟ = 0.5) + ( 2.4 mm) 2 2 T H2 cv = σ drained = t tF ( 5 hours ) Solving for the time of settlement in the field t f .33 m − 0.3) + ( 49) ⎞ the settlement at the corner ΔH = log ⎜ o ⎟= log ⎜⎜ ⎟⎟ = 0.70 wGs ( 0.3 m2 Therefore.12 m = 120 mm (3) The time required to attain 60% consolidation in the field is.21m 1+ e ⎝ po ⎠ 1+ ( 0.(2) The differential settlement between the center and a corner of the building is.59) ⎝ ( 74.30 and GS = 2.33 m 1+ e ⎝ po ⎠ 1 + ( 0. Cc = 0. ( 0.500 mm) = ( 0.18)( 2x25.21 m = 0.3) ⎠ the differential settlement is Δ ( ΔH ) = 0.4 mm) 2 1 ⎝ 24 hours ⎠⎝ 365 days ⎠ 149 .5 m)(18. CC H ⎛ p + Δσ ⎞ ( 0.5 − 9.30)( 5m) ⎛ ( 74.59 ) ⎝ ( 74.26)( 2. kN po = γ sand hsand + γ clay hat mid −clay = ( 3 m)(17.70) Seo = wGS ∴ eo = = = 0.3) ⎠ CC H ⎛ p + Δσ ⎞ ( 0.500 mm) ( 5 hours ) ⎛ 1day ⎞⎛ 1 year ⎞ ≈ 2 years 2 tF = ⎜ ⎟⎜ ⎟ ( 0.30)( 5m) ⎛ ( 74.18)( 2x25. 150 .0626 2.2 psf.2733 16. 4000.*Stress–01: Stress increase at a point from several surface point loads.50 0. B and C respectively.000 5 10 0.12 0. the vertical stress increase at D from the three loads A.2 psf Therefore.18 (102+52) 1/2 = Δp from B 4. the vertical increase in stress contributed by each at a depth z =10 feet is found by. as shown below.50 11. Using the Boussinesq (1883) table on page 202 for vertical point loads.18 Δp from C 6.000 10 1. and 6000 lbs act at points A. A 10 feet B 10 feet C 5 feet D Solution.0626 1. Determine the increase in vertical stress at a depth of 10 feet below point D.000 10 1. (Revision: Aug-08) Point loads of 2000. B and C is 20.12 0. ⎧ ⎫ P ⎪ 3 1 ⎪ P Δpz = 2 ⎨ ⎬ = 2 I1 z ⎪ 2π 5/ 2 ⎡( r / z ) + 1⎤ ⎪ z 2 ⎩ ⎣ ⎦ ⎭ Increase in P r (ft) z (ft) Δp r/z I1 the load at: (lbs) (psf) (102+52) 1/2 = Δp from A 2.25 11.40 Total = 20. 000 kN ⎞ Δqz = qo (4 I 4 ) = ⎜ ⎟ (4 I 4 ) = ⎜⎜ ⎟⎟ ( 4 )( 0.5m L 2.0m z 6. (Revision: Aug-08) Determine the vertical stress increase in a point at a depth of 6 m below the center of the invert of a newly built spread footing. placed on the ground surface carrying a columnar axial load of N = 2.5 m and L1 = 2.000 kN.25 and n = 1 = = 0.000 kN B=3m L=4m Depth = 6 m B1 1. Solution: The Boussinesq solution for a rectangular loaded area only admits finding stresses below a corner of the loaded area. the footing must be cut so that the load is at a “corner” (shown as the quarter of the area).0345 ⎛ N ⎞ ⎛ 2. 3 m by 4 m in area. N = 2.0 m.33 z 6.0m m= = = 0. Therefore.*Stress-02: Find the stress under a rectangular footing. where the reduced footing dimensions for the shaded area are B1 = 1.0m Use the table and extrapolate and find I 4 = 0.0345 ) = 23 kN / m 2 ⎝ BL ⎠ ⎝ ( 3 m )( 4 m ) ⎠ 151 . 2 kN / m 2 which is a 48% increase in stress. Δpo = q (4 I 4 ) = ( 300 )( 4 )( 0.5 m L 2.076 ) = 91.8 kN / m 2 b) When the WT drops from -1 m to -5 m below the footing. Δpo = 91. N = 4.076 a) The total stress increase from the footing is.625 A 15 m 2 m2 z 4m z 4m ∴ I 4 = 0.8 ) = 61. Δpo' = Δpo − u = ( 91.375 and n = = = 0. the effective stress is identical to the total stress.5 m q= = = 300 m= = = 0. 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. 152 .500 kN kN B 1. (Revision: Aug-08) Find the effective stress increase in the soil at a depth of 4 m below the footing. Therefore the effective stress increase is.*Stress-03: The effect of the WT on the stress below a rectangular footing.500 kN B=3m WT L=5m WT-1 Depth = 4 m WT-2 Solution: N 4.2 ) − ( 3 m )( 9.2 kN / m 2 and the effective stress when the WT is 1 m below the footing is. Solution: The stress increase.12) = 18 kN / m 2 Similarly.*Stress–04: Finding the stress outside the footing area.5 and n = = = 1. For the loaded area shown in (b): B 2 L 4 m = = = 0.2 kN / m 2 Therefore.2 = 10. Δp .5 Z 4 Z 4 Δp2 = (150 )( 0. can be written as : Δp = Δp1 − Δp2 where Δp1 = stress increase due to the loaded area shown in (b). Δp = Δp1 − Δp2 = 18 − 7. Δp2 = stress increase due to the loaded area shown in (c).0 Z 4 Z 4 Δp1 = qI 4 = (150)(0.25 and n = = = 0.048 ) = 7.8 kN / m 2 153 . for the loaded area show in (c): B 1 L 2 m = = = 0. (Revision: Aug-08) Find the vertical stress increase Δp below the point A at a depth z = 4 m. Determine the increase in the vertical stress Δp at the depth of the pipe. 5 ft 4 ft 5 ft 2 ft B A C C 10 ft 3 ft 10 ft 3 ft PLAN VIEW SECTION Solution: For the expanded 5’ x 13’ area.6 and n= = = 1 therefore.800 . . I 4 = 0.800 2 ⎟ (0.-08) A clay sanitary pipe is located at a point C below the footing shown below. (Revision: Sept.200 Z 5 Z 5 For virtual 3’ x 5’ area B 3 L 5 m= = = 0.136 Z 5 Z 5 The increase in stress at point C from the footing is therefore.*Stress-05: Stress below a footing at different points.136) = 115 psf ⎝ ft ⎠ 154 .6 therefore. ⎛ lb ⎞ Δp = q ( I 4 . B 5 L 13 m= = = 1 and n= = = 2. . I 4 = 0.800 psf. which is z = 5 feet below the footing invert. The footing has a uniformly distributed load q = 1. q = 1.200 . and 3 feet away from its edge. B A .0. .I 4' ) = ⎜ 1. . 2 = 0.5' ) 2z 2(15' ) Δp 2 Δp2 ⇒ = = −1 = = 1.332 psf 155 . (Revision: Aug-08) Calculate the stress increase at the point A due to the new road embankment.25)(15') ⎜120 3 ⎟ = 450 psf ⎝ ft ⎠ ------------------------------------------------------- 2x2 2(−12. 2x1 2(15' ) 2z 2(15' ) Δp1 Δp1 ⇒ = =2 = =2 = 0. Using Boussinesq.02 B3 15' B3 15' q ⎛ lb ⎞ ∴Δp 3 = (0. whereas the contribution from the left and right hand slopes are Δp1 and Δp 3 respectively.47)(15') ⎜120 3 ⎟ = 846 psf ⎝ ft ⎠ ------------------------------------------------------------ 2 x3 2(40' ) 2z 2(15' ) Δp 3 Δp3 ⇒ = = 5.47 B2 25' B2 25' q ⎛ lb ⎞ ∴Δp 2 = (0.02)(15') ⎜120 3 ⎟ = 306 psf ⎝ ft ⎠ Δp = Δp1 + Δp 2 + Δp 3 = 450 + 846 + 36 = 1.3 = =2 = 0.*Stress-06: Stress increase from a surcharge load of limited width.25 B1 15' B1 15' q ⎛ lb ⎞ ∴Δp1 = (0. Δp2 25m 15 ft 1 1 Δp1 1 Δp3 1 Z 15 ft Solution: The contribution from the central portion of the fill is Δp2 . B 1.*Stress-07: Finding a stress increase from a surface load of limited width.038 For I4(H1): m′ = B / H1 = 1.3.6 kN / m 2 156 .5 m 1. between z = 3 m and z = 5 m. I4(H2) = 0.3 n′ = L / H2 = 1.5 For m′ = n′ = 0.5 / 3 = 0.5.4 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: 4Δpavg ( H 2 / H1 ) = (4)(3. 2 q = 100 kN/m z L .086 Therefore: (5)(0. A 3m SECTION 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 )) ⎤ Δpavg ( H 2 / H1 ) = q ⎢ 2 4 2 ⎥ = 100 ⎢ ⎥⎦ ⎣ H 2 − H1 ⎦ ⎣ 5−3 For I4(H2): m′ = B / H2 = 1.038) − (3)(0.086) Δp av(H 2/H 1) = 100 × = 3 . I4(H1) = 0.5 / 3 = 0.3 For m′ = n′ = 0.4) = 13.5 / 5 = 0.5 / 5 = 0. .5 n′ = L / H1 = 1. (Revision: Aug-08) Determine the average stress increase below the center of the loaded area.5 m 3m 3m 5m A . 0007σ 0 v ) dz If a particular stratum has a function γ = 95 + 0. find the vertical stress at a depth of 100 feet below the surface. and integrating by parts.**Stress-08: Stress increase as a function of depth.0007 σ v = 135.0007 σv . Z 100 σv = ∫ γ (σ ) dz 0 v = ∫ (95 + 0. where γ is in pcf and σv is in psf.0007 z − 1) 100 = 9.0007σ V ) = ∫ dz 0 1 ln ( 95 + 0. σv dσ v 100 ∫ 0 ( 95 + 0.840 psf 157 .840 psf 0 At Z = 100 σ v = 135. (Revision: Aug-08) The vertical stress σv in a soil at any depth below the surface can be estimated as a function of the soil unit weight γ by the equation.0725 − 1) ∴ σ v = 9. Solution: Rearranging.800 (1.0007σ V )0 V = z100 σ 0 0.800 ( e0. ES → Soil elastic modulus. I→ Influence factor. Eeq→ Equivalent modulus. MT→ Transverse moment. essentially equivalent to the strain ε in the soil. v→ Poisson’s ratio. ΔHi→ Elastic settlement. Δ(Δe) → Differential settlement between adjacent foundation. IZ→ Simplifying influence factor. γ→ Unit weight of the soil. 158 . C1 → Embedment coefficient. ε→ Strain at mid stratum. Chapter 10 Elastic Settlements Symbols for Elastic Settlements N → Raw value of the STP (obtained in the field). C2→ Creep correction factor. qo → Contact pressure. the average value of the strain is about 0.30 ⎛ 144 in 2 ⎞⎛ k ⎞ Δ = C1C2 qo ⎜ ⎟ dz = (1)(1)(80 2 )( 2 )(24. A surface SPT test shows an N = 12 and the tire’s bearing area is roughly square. the rutting is. 000 lb ⎠ ∴Δ ≈ 0. The front wheels carry 20% of the load on tires inflated with 80 psi air pressure. Therefore. Since the crane loads are on the surface and only for a few days. Therefore. it is permissible to assume that there is no creep and therefore C1 = C2 = 1. The strain reaches a maximum value of 0.2 inches) = 24.*Elastic Settlement-01: Settlement (rutting) of a truck tire.5)(20%)(120. Use the Schmertmann method to estimate the rutting. Calculate the possible rutting depth to your temporary jobsite road built from in-situ compacted medium sand. 159 . ⎛ ε ⎞ lb 0. and the influence factor Iz is equivalent to the strain ε. 000 lb) B2 = = 2 = 150 in 2 ∴ B = 12.2 inches tire pressure q o (80 lb / in ) A rough estimate of the soil’s elastic modulus is Es = 14N = 14(12) = 168 ksf. it is a dense sand. (Revision: Jan-08) You are required to move a 60-ton truck-mounted crane onto your construction site.4 in) × ⎜ 2 ⎟⎜ ⎟ ⎝ Es ⎠ in 168 k / ft ⎝ ft ⎠ ⎝ 1.6. Solution: Each front tire has a square bearing area of BxB such that: tire's load (0.4 inches.5 inches of rutting. Since the sand is compacted. for the single layer of soil.3 throughout its depth to 2B = 2(12. 60 ksf B 49 ft 2 The coefficients for the Schmertmann method are C1 and C2 : ⎛D γ ⎞ ⎛ 0.48 ⎞ Depth factor C1 = 1.60)(0.60 ⎠ Creep factor C2 = 1.0 − 0. for long-term use.281) = 1.281 Q 200 k The contact pressure on the soil is. Es Assume ≅ 14 where Es is in ksf.35)(3. embedded 4 ft below the ground surface.5 ⎜ f ⎟ = 1.5 ⎜ ⎟ = 0.46 0.090 2 60 210 0.0 − 0.35 for a five year period . (Revision: Aug-08) Estimate the settlement of a square footing placed on a fine.16 0.120 kcf ) = 3.*Elastic Settlement-02: Schmertmann method used for granular soils.130 3 66 168 0. Use the Schmertmann method. used for fine medium sands.93 ⎝ qo ⎠ ⎝ 3.27 inches 0 E 160 . N Solution: I z Δz Layer ∆z Es Iz=ε Es (in) (ksf) 1 42 140 0. medium dense sand.93)(1. qo = 2 − Df γ = − ( 4 ft ) (0.30 0. 2B ε ∴Δ = C1C2 qo ∑ Δz = (0.061 ∑= 0. The Schmertmann formula for the elastic settlement Δ is. The modulus E for loose sand can be calculated using the following formula: ES ≈ 10 ( N + 15 ) ksf 161 . q0 = = = 7. x 3 ft. Solution: Q 64 kips The contact pressure on the soil is. footing when it is placed on a uniform clean sand with γ = 110 pcf.11 ksf B 2 ( 3 ft )2 The SPT value indicates that the soil is a loose sand.*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. 35 0.5 )( 0.0025 2 4.07 ft = 0.11−( 0.110 )( 4 ) ⎞ 1.0086 The correction factors are as follows: ⎛ 0. C2 = 1 + 0.0061 Σ = 0.110)( 4) ⎤⎦( 0. Creep factor.The following table summarizes the data and calculations: Layer’s SPT Soil’s Average strain at εZΔZ/E Layer elastic thickness (average) mid number modulus ΔZ (feet) N E (ksf) stratum ε (ft3/kip) 1 1.5γ D f ⎞ ⎛ ( 0.97)(1.97 ⎜ q0 − γ D f ⎝ ⎠ ⎝ 7.1 ⎠ The total elastic settlement is. ⎛t ⎞ ⎛ 5 ⎞ 2.110 × 4 ) ⎠ Since C1 > 0. 2B ε 1 2 ( qo −γ Df ) ∑ dz = ( 0.5 6 210 0. Depth factor.5.34) ⎡ Δ= CC ⎣7.30 0.11 − ( 0.2 ) log ⎜ ⎟ = 1.84in 0 Es 162 .5 7 220 0.34 ⎝ 0. C1 = 1 − ⎜ ⎟⎟ = 1 − ⎜⎜ ⎟⎟ = 0.2log ⎜ yr ⎟ = 1 + ( 0.0086) = 0.1 ⎠ ⎝ 0. this is FINE. 3. (Revision: Aug-08) Use the 2:1 method to find the average stress increase (Δq) due to the applied load Qu in the 5-foot sand stratum directly beneath the footing. 1 H2 Qu Δq = H ∫ ( B + Z ) dz H1 2 where H1 = 0 feet (the footing’s invert) to H2 = 5 fee (bottom of the sand stratum). 163 .*Elastic Settlement-04: The 2:1 method to calculate settlement. 1 − v2 ΔH = qo B Iw Es For square and flexible footings the influence factor is about IW = 0. The stress increase can be found by integrating the above equation. If ES = 400 ksf and v = 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. what is the expected immediate settlement ΔHi? Qu=120 kips Solution: The settlement ΔH of an elastic media (the 5 foot thick sand stratum in this case) can be found from the theory of elasticity as. 32 ⎞ ⎛ 12 in ⎞ ΔH = qo B Iw = ⎜ 2 ⎟ ( 6 ft ) ⎜ ⎟ ( 0.52 inches Es ⎝ 3 6 ft ⎠ ⎝ 400 ksf ⎠ ⎝ ft ⎠ 164 .95 ) ⎜ ⎟ = 0. 5 1 ⎡ Q ⎤ 1 ⎛ 120kip 120kip ⎞ Δq = ⎢ = ⎜ − ⎟ = 1.82ksf H ⎣ B + z ⎥⎦ 0 5 ft ⎝ 6 ft 11 ft ⎠ but 1 − v2 ⎛ 120 kip ⎞ ⎛ 1 − 0. Then. medium dense sand stratum (with φ = 36o and γ = 112 pcf) is 3 ksf.5 ΔH B 8 ft Therefore if the settlement at column B is ΔHB = 1 in. Column A has a design load of 430 kips and column B has a design load of 190 kips. 165 .. Δ (ΔH ) = 1. since Δ(ΔH) should be < 1" and θ < 0.*Elastic Settlement-05: Differential settlement between two columns.0 in = 0.0014 L ⎛ 12in ⎞ ( 30 ft ) ⎜ ⎟ ⎝ ft ⎠ Both of these values [Δ(ΔH) and θ] are acceptable.5 in. then the settlement at column A will be ΔHA = 1. Is this Δ(ΔH) acceptable for columns spaced 30 ft apart? Solution: Footing size for column A: QA 430 kips BA = = = 12 feet qall 3 ksf Footing size for column B: QB 190 kips BB = = = 8 feet qall 3 ksf ΔH A 12 ft A quick estimate of the ratio of settlement to the proportionality is = = 1. Select footing sizes and determine the differential settlement Δ(ΔH) between them.5 inches and the rotation between the two columns is θ= Δ ( ΔH ) = ( 0.5 in − 1. (Revision: Aug-08) The allowable bearing capacity of a 30-ft thick.0033.5 in ) = 0. Also.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 = 0.05 Z s where Zs is the thickness of the granular stratum beneath the footing in feet.11 5’ CLAY Cc = 0.52 ∴ q = ( 0. Westergaard. Solution: Assume an initial value of B = 10 feet. (b) Westergaard's method.4ksf ) = 1.25 ksf qo 166 .*Elastic Settlement-06: Compare the settlements predicted by the Boussinesq. determine the size of a square spread footing. and (c) the 2:1 method. Estimate the initial settlement Δ H i = 0. for the values shown below. and the 2:1 methods. using: (a) Boussinesq's method.5 inches. (Revision: Aug-08) Compute the average stress Δq at mid-clay stratum. Q = 240 kips 4’ γS= 110 pcf 5’ SAND WT γ = 120 pcf eo = 1. to give ΔHi in inches.42 ROCK The contact pressure qo of the footing is: Q 240 kips qo = = = 2.52 )( 2. in order to limit total settlement ΔH = ΔHi + ΔHc + ΔHS to only 1. it is the least conservative method. Therefore. Since this would predict faster consolidation rates.05 Z s = 0. qo B+z Q 240 kips q= = = 0. and the secondary settlement ΔHs.11 in The in-situ effective stress qo' at mid-clay layer. ΔH = ΔH i + ΔH C + ΔH s = 1.50 inches.5 ft )2 Note that the Boussinesq method provides the highest predicted stress. the total settlement ΔH is equal to the immediate ΔHi.120 3 − 0.05 5 ft = 0.33 ∴ q = ( 0. The instantaneous settlement (ΔHi): ΔH i = 0. plus the consolidation ΔHc.5 feet: B = 10 ft. use the 2:1 method's stress of 0.78 ksf. but limited to no more than 1.33)( 2.4 ksf ) = 0. for this problem.(b) Stress at mid-clay stratum using Westergaard’s method for a square footing: q = 0.5 ft ⎥ = 1.78 ksf AB + Z (10 ft + 7.5 ft.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 7.78 ksf from the 2:1 method. z = 7.13 ksf ⎝ ft ⎠ ⎣⎝ ft ft ⎠ ⎦ Using q = 0. before placing the footing is: ⎛ kip ⎞ ⎡⎛ kip kip ⎞ ⎤ qo ' = γ s hs + γ c hc = ⎜ 0.11 3 × 9 ft ⎟ + ⎢⎜ 0.0624 3 ⎟ × 2.50 inches 167 . Therefore the maximum permissible consolidation settlement ΔHc is limited to: ΔH c = 1.34 ksf Therefore B = 19 feet Since the initial B = 10 feet.39 in = log ⎜ ⎟ = 1.6 ft AB + Z Δq 0.39 ksf Using the 2:1 method: Q Q 240 kips Δq = B+Z = = = 26.But ΔHi = 0.11 in. 168 . the new value of 19 feet should be used to re-iterate towards a better solution that converges.42 ) ⎜ 5 ft × Cc H q + Δq ' ⎝ ft ⎟⎠ ⎛ 1.39 in or ⎛ 12 in ⎞ ( 0. and ΔHs is negligible for this problem.50 in − ΔH i = 1.11 ⎝ 1.50 in − 0.11 in = 1.13 ksf + Δq ⎞ ΔH c = log o = 1.13 ksf ⎠ Therefore Δq = 0.39 in 1 + eo qo ' 1 + 1. 2 mm Es ⎝ m ⎠ ⎝ 20. ⎝ E s ⎠ Q = 2.5 and C1 = C2 = 1. qv2 = 163. qv3 = 77. Δq ⎛ 1000 mm ⎞ ⎛ 104 kPa ⎞ ΔH = C1C2 (0.95. ⎛ 1 − v2 ⎞ ⎛ 1 − 0.0kPa ⎥ H ⎣⎝ 2 ⎠ ⎦ 6m ⎣⎝ 2 ⎠ ⎦ Δqv = 104 kPa The elastic settlement via Schmertmann is.6B) = (1)(1)( 0.3 kPa )( 3 m) ⎜ ⎟ ( 0.400 kPa ⎠ The alternative method from the theory of elasticity would yield.0 kPa.0kPa + 44. qv4 =44. ΔH ⎡⎛ q1 + qn ⎞ ⎤ 1.0).5 m 1. where v = 0.0kPa ⎞ ⎤ Δqv = ⎢⎜ ⎟ + q2 + q3 + K + qn −1 ⎥ = ⎢⎜ ⎟ + 163.3 kPa.3 kPa.5 m A 3m B SAND 5m ROCK Solution: The average stress from multiple layers is solved via this formula.400 kPa.*Elastic Settlement-07: Schmertmann versus the strain methods.3kPa + 28.3 and Iw = 0.400 kPa ⎠ ⎝ m ⎠ 169 .100 kN B = 3m x 3m 1.5m ⎡⎛ 233. (Revision: Aug-08) Compute the immediate settlement ΔHi using the Schmertmann formula using an average Δq value (qv1 = 233.95) ⎜ ⎟ = 10 mm ⎝ Es ⎠ ⎝ 20. Compare the results with an alternate method B ⎛1− v2 ⎞ using ΔH = qo B⎜⎜ ⎟⎟ I w .6)( 3 m) ⎜ ⎟⎜ ⎟ = 9.3kPa + 77.32 ⎞ ⎛ 1000 mm ⎞ ΔH = qv1B ⎜ ⎟ I w = ( 233. = 0.0 kPa and qv5 = 28. Es at point A is Df 20. where ES is in ksf and N is the corrected SPT value. (Revision: Aug-08) Determine the elastic settlement using the Schmertmann method of the 10'x 10' footing as shown below.0 260 0.5 560 0. (ksf) (average strain) ΔZ.00962 3 2.0 360 0.00263 Σ = 2B = 20 ft Σ= 0.50 0.25 0.35 0. Solution: The data from these strata are placed into a table below.00486 2 5.*Elastic Settlement-08: The Schmertmann method in multiple strata. Estimate the elastic modulus using ES = 10(N + 15). (feet) 1 5.02115 170 .5 300 0. Layer Soil Modulus IZ = ε I Z ΔZ (ft/kp) Layer No.10 0.00112 5 5. Thickness E ES.00292 4 2.0 190 0.35 0. 02115 ) = 0.⎢ ⎥ = 0.042 ft = 0.5( )]= 1.(0.5(0.5 .35 2B ε ΔH i = C1C2 (q .[0. γ Df ⎡ 0.γ D f )∑ dz = ( 0.1)(5) ⎤ The depth coefficient C1 =1.35 ) ⎡⎣ 2.1)( 5 ) ⎤⎦ ( 0.1)(5) ⎦ The time coefficient C2 = 1.88 )(1.05 in 0 E 171 .The Schmertmann coefficients are.75 q -γ D f ⎣ 1.5 − ( 0. 54 0.0 2B ε Δ = C1C2 ( qo − γ D f ) ∑ dz = ( 0. and carries a uniform load of 6 ksf.26 2 300 3.0 ) ⎡⎣6 − ( 0.95+0.12(6) = 0.95 For clay 2: qo = 5.75 B = 3.96.47 0.33 0.35 30.20(25x12) 5.54 Δ1 = log 10 = 1. The structure is of reinforced concrete with column spacing at 25 ft.95+0.09(6) = 0. For the soil profile conditions shown below.**Elastic Settlement-09: Settlement of a mat foundation.09q = 0. 87.92 )(1.0 − 0. (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.23 3 300 8.29 γ Df ⎡ ( 0. Set eo1 = 1.100 )( 8 ) ⎦ Thetime factor (creep) C2 = 1.30 10.80 Σ = 84.29 ) = 3.5 = 1. eo2 = 0.125 + 0.5 = 7.110 .20(25x12) 5.0624)12.5 = 5.13 in 2 5.0.5 ft below the surface.72 ksf CcH 1 q + Δq 0.95 + ( 0.100)(50) + (0.50 43.0.95 ksf At the mid clay stratum is at 87.5/50=1.130 .5 ⎢ ⎥ = 0.00.92 ( q − γ D f ) ⎣ 6 − ( 0.95 At corner Δq = 0.89.100 )( 8 ) ⎤ The depth factor C1 = 1. determine the total settlement at the center and a corner of the foundation. Layer Δz (in) Es (ksi) Iz =ε (Iz/Es)Δz (in3/kip) 1 300 3.78 in 0 Es b) Consolidation settlement Method A (take each layer at a time). (ie.0624)12. Is the calculated differential settlement acceptable? Solution: a) Using the Schmertmann method. and from e = 125/γ.48 in 1+ e o qo 1+1 5.0 − 0.47 0.125 .5 B/2) Δq = 0. eo3 = 0.58 ksf 172 .72 Δ1 = log 10 o = log 10 = 1.0624)12.5 + (0.0624)25 + (0. For clay 1: qo = (0.12q = 0.100 )( 8 ) ⎤⎦ ( 84.0. 35(25x12) 7.04q = 0. Δedge = 3.65 Method B: the equivalent layer equation Δ Δe q o + Δq = where Δe = C c log H 1+ e o qo The total settlement on the centerline is.03q = 0.30 0.65+0.5 = 8. 0. 173 .5 B/2) Δq = 0.70 inches The allowable for reinforced concrete buildings Δ(Δ)< 0.15(25x12) 8.78 + 1.midlayer at 112 ft = 4.35(25x12) 7.08 inches The differential settlement is Δ(Δ) = 0.003(span) = 0.24 ksf 0.130 + 0.65+0.5 + (0.78 + 1.24 Δ3 = log 10 = 0.89 8.90 inches Therefore.78 inches and along the foundation edge. 0.30 Δ2 = log 10 = 0.0.89 8.624)12.58 Δq . the design is acceptable.003(25x12 inches) = 0.65 ksf at midlayer (137.58 +0.140 .05q = 0.13 + 0. Δ6 = 3.58 +0.96in 1+0.96 + 0.58 For clay 3: qo = 7.21 = 6.96 7.48 + 1.25 in at the centerline 1+0.18 ksf 0.21in 1+0.15(25x12) 8. Δq .5' / 5.42 Δ2 = log 10 = 1.28 = 6.58 + (0.42 0.65 at corner Δq = 0.18 Δ3 = log 10 = 0.0624)12.07q = 0.96 7.5 B/2.28in 1+0.25 + 0. DHs→ Second consolidation settlement. OCR → Over-consolidation ratio (ratio of in-situ stress divided by the overburden stress). GS→ Specific gravity of the solids of a soil. u→ Pore water pressure. 174 . DHtotal→ Total settlement of a structure. Pe→ Pre-consolidation pressure of a specimen. Dp → Increasing pressure on the surface. U→ Degree of consolidation. DHc→ Plastic settlement (also called primary consolidation). γW → Unit weight of water. H→ Depth of zone influence. CC→ Compression index. Chapter 11 Plastic Settlements Symbols for Plastic Settlements e→ Voids ratio. ES → Soil elastic modulus. γSAT → Saturated unit weight of the soil. (Revision: Oct. when h =15 ft.*Plastic Settlement–01: Porewater pressure in a compressible soil.? c. How high will the water rise in the piezometer immediately after the application of the surface load of 3 ksf? b.000 The pore water pressure is u0 = Δp = γ wh = 3.000 lb / ft 2 ∴ h = = = 48.4 175 . Find h when the degree of consolidation at A is 60%. What is the degree of consolidation from the 3 ksf at point A. Δ p = 3 ksf h Ground water table 20 feet 15 feet Sand 4 feet Clay 10 feet A Rock Solution: a) Assume a uniform increase of the initial excess pore water pressure throughout the 10-foot thickness of the clay layer: Δp 3.1 feet γw 62.-08) a. 4 pcf ) ⎠ c) When UA = 60 %.6 )( 3.1 ft )(62.b) The degree of consolidation at A is UA (%) when h = 15 feet: ⎛ u ⎞ ⎛ (15 ft )(62. 000 psf ⎠ ∴ u A = (1 − 0. 000 psf ) = 1.4 pcf ) ⎞ U A % = ⎜1 − A ⎟100 = ⎜1 − ⎟100 = 69% ⎝ u0 ⎠ ⎝ (48. 200 psf ) = 19.4 pcf ) 176 . what is the value of h? ⎛ u ⎞ ⎛ uA ⎞ U A = 0. 200 psf h= uA = (1.2 feet γw ( 62.6 = ⎜1 − A ⎟ = ⎜1 − ⎟ ⎝ u0 ⎠ ⎝ 3. 74 ksf The consolidation settlement is .009( LL − 10) = 0.009(50 − 10) = 0.*Plastic Settlement-02: Total settlement of a single layer.34 )( 2.0624) kcf (7 ft ) + (0. HCc p + Δ p (17 )(12 ) (0. Δp = 1 ksf Sand 8 feet γ = 110 pcf. (Revision: Aug-08) A new building is planned upon the site shown below.120 − 0.67.74 + 1 ΔH = log 10 o = log 10 = 7. LL = 50%.115 − 0.91 1.6 i nches 1 + eo po 1 + 0.36 Se = wG s but S = 1 ∴ e = wG s = ( 0.91 The stress of the clay at its mid − stratum before the building was built is . Find the primary consolidation settlement if the clay is normally consolidated.36) 1.5 ft ) p o = 1. Water Table 7 feet γsat = 115 pcf Clay 17 feet w = 34%.67 ) = 0. Assume that the clay solids have a specific gravity of 2.0624) kcf (8.74 177 . γsat = 120 pcf Solution: Skempton formula for the C c = 0. 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. Apply Boussinesq's formula to find the reduction of the vertical stress with depth. po = γ d −sand Hdry−sand + (γ sat − γ w )Hsat −sand + (γ sat − γ w )Hmid −clay po = (0.10kcf )(10 ft) + (0.009(40 −10) = 0.*Plastic Settlement-03: Boussinesq to reduce the stress with depth.110 − 0.5 feet 2 178 . (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.0624)kcf (5 ft) + (0.120 − 0.n' = ⎥ ⎣ H 2 − H1 ⎦ ⎣ H2 H2 ⎦ 5 B = L = = 2.0624)kcf (5 ft) po = 1. 200 kips 5’ 10 feet 5’ x 5’ footing Sand γdry = 100 lb/ft3 Ground Water Table 120 lb/ft3 5 feet Clay 10 feet γsat =110 lb/ft3.27 The stress of the clay at its mid − stratumbefore the building was built is.53 ksf The Δpavg below the center of the footing between z1 = 15 feet to z2 = 25 feet is given as. The clay is normally consolidated. LL = 40 Solution: Skempton formula Cc = 0.0.009(LL −10) = 0. e = 1. ⎡ H 2 I 4( H 2 ) − H1 I 4( H1 ) ⎤ ⎡ B L ⎤ Δpavg = 4q ⎢ ⎥ where I 4( H 2 ) = f ⎢ m′ = . 167 and n′ = = = 0.167 H1 15 H1 15 From Boussinesq's Table for I 4 I 4( H 2 ) = 0.075 The average pressure at mid-clay layer is thus given by.6 inches 1 + eo po 1+1 1. ⎡ H 2 I 4( H 2 ) − H1 I 4( H1 ) ⎤ ⎛ 200 kips ⎞ ⎡ (25)(0.05 and I 4( H1 ) = 0.40 ΔH = log10 = log10 = 1.53 + 0.27) 1. B 2.05) − (15)(0.53 179 .5 L 2. HCc po + Δpavg (10 )(12 ) (0.5 m′ = = = 0.075) ⎤ Δpavg = 4q ⎢ H − H ⎥ = 4 ⎜⎜ ′ ′ ⎟⎟ ⎢ ( 5 )( 5 ) ⎣ 25′ − 15′ ⎥⎦ = 0.4 ksf = 400 psf ⎣ 2 1 ⎦ ⎝ ⎠ The consolidation settlement is. 11 Therefore.11)] ( 9. σ ′ = (4.*Plastic Settlement -04: Surface loads with different units.11) The effective unit weight for the clay is. the clay unit weight is γ sat = ( Gs + eo ) γ w = [(2.11 175 180 .78) = 1. γ’ = 10.40(2. this value must be determined through the other information provided. q = 1.4 kN m3 ) + (7.2 daN/cm2 4.78) + (1.4 kN/m3 Normally consolidated clay 7.6 m Ground Water Table Sand 6.0 m γ = 17.3 kN m3 ) = 175kN / m 2 The consolidation settlement is.78.6m)(0.1 kN m3 1 + eo 1 + (1.6 m Gs =2.6m)(17. the Skempton relation is.81) = 18.26 m = 260 mm 1 + eo σ o′ 1 + 1. PL = 30% Solution: Notice that the data provided does not include a unit weight for the clay stratum. HCc σ ′ + Δσ (7.6 kN/m3. γ s = γ wGs = 27. Cc = 0.6 kN m3 ) + (6.6m 2)(8.32) 175 + 120 ΔH = log10 = log10 = 0.32 The stress at the mid-clay stratum. Therefore.8 kN m3 Se = (1)e = wGs = 0. γ ′ = γ sat − γ w = 18.009(45 − 10) = 0.009( wL − 10) = 0.0m)(10. w = 40%.1 − 9.3 kN m3 Also. (Revision: Aug-08) Find the total settlement under a building that applies the load shown below.81 = 8. PI = 15%. 08 0. Solution: 181 .850 766.52 0.94 1. p (kN/m2) Void ratio. e 23.080 191. Cc.105 95. (Revision: Aug-08) The results of a laboratory consolidation test on a clay sample are given below: Pressure.04 0. pc .112 47.*Plastic Settlement-05: Pre-consolidation pressure pc and index Cc.76 1.731 (43) a) Draw an e-log p plot. (44) b) Determine the pre-consolidation pressure.985 383.88 1. (45) c) Find the compression index. 8 p1 = 300 kN m 2 and e1 = 0. pc.451 ⎛ p2 ⎞ ⎛ 500 ⎞ log ⎜ ⎟ log ⎜ ⎟ ⎝ p1 ⎠ ⎝ 300 ⎠ 182 . e1 − e2 0.8 Cc = = = 0. Cc. p2 = 500 kN m 2 and e2 = 0. From the e-log p plot.5 kN / m 2 c) Find the compression index.9 ∴ pc = 117. From the slope of the graph.b) Determine the pre-consolidation pressure.9 − 0. and (b) the voids ratio of the clay if the total pressure at its mid-height is 1000 kN/m2. 760 ft. Clay Elev. Find (a) the effective overburden pressure at mid-height of the compressible clay layer. 752 ft.81.98.*Plastic Settlement-06: Final voids ratio after consolidation.427 log( p2 p1 ) log(500 200) 0. 710 ft. Ground Water Table Sand and Gravel Unit Weight = 132 lb/ft3 Elev.4) lb ft 3 (11 ft ) po = 3. Sand and Gravel Unit Weight = 132 lb/ft3 Elev.81 b) Cc = = = 0.4) lb ft 3 (20 ft ) + (125. Elev.14 ksf e1 − e2 0. 732 ft.4 − 62. Unit Weight = 125. 721 ft.427 = ∴ e2 = 0.4 lb/ft3 Elev.68 log(1000 200) 183 . (Revision: Aug-08) The clay stratum shown in the profile below has a total vertical stress of 200 kN/m2 at its mid- height with a voids ratio of 0. 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. When the vertical stress increases to 500 kN/m2 the voids ratio decreases to 0.98 − e2 0.98 − 0. 0’ using the boring report shown below. The rising WT may reduce settlement. 184 . (Revision: Aug-08) Find the settlement due to lowering of the phreatic surface from elevation 349.032 ft 1 + 0.71ksf Δq Δp = + Δp due to the lowering WT H Δq2 = Δp − Δq1 = 1.0624 pcf )( 3 ft ) + ( 0. Solution: po = ( 0.96 3 versus 0.030(10) ΔH1 = log 3 + 1.110 pcf )( 6 ft ) + ( 0.110 pcf )( 7 ft ) + ( 0.5’ to 344.0624 pcf )( 3 ft ) po = 1.96 3 0.693 C H ΔH1 = c log po + Δq1 1+ e po 0 0.853 − 1.110 − 0.083ft the additional settlement.*Plastic Settlement-07: Settlement due to a lowered WT.034(10) ΔH 2 = log 3 + 0.157 ft 1 + 0.110 − 0.16 = 0.693 = 0.16 = 0. 1 kN/m3 Clay 3m γb = 9.5m 2m WT 4mx4m 4m Sand γb = 9.*Plastic Settlement-08: The over-consolidation ratio (OCR).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. Given that Gs = 2. (Revision: Sept-08) Oedometer (consolidation) tests of several samples from the clay stratum yields the consolidation curve shown below.65) γd = = = 14.1kN / m3 )⎤⎦ + ⎡⎣(1.65.2kN / m3 )⎤⎦ = 79. γ bGs (9.6 kN / m3 Gs −1 1. Solution: 1200 kN 1.4 kPa 185 .5m)(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. find (a) the value of po.1kN / m3 )(2. (b) The value of pc and (c) the OCR of the clay. we need to find the unit weight of the sand stratum.6kN / m3 )⎤⎦ + ⎡⎣(4m)(9. 8 1.80 9.6 3.8 1 1. 2 3H dr Coefficient of consolidation(C v ) = 4t Time (minutes) 0 0.32 0.2 S ettlem en t (m m .86 0.64 1. The initial sample thickness is 20 mm.96 (mm. The results are shown below.19 3.) 0. (3) Estimate what could be a good estimate of the elastic modulus of this soil Eo (kPa) (4) What sort of permeability k (in m/s) could you estimate for this soil (from Eo)? Solution: (1) Time vs. (Revision: Aug-08) An odometer test was performed on a peat soil sample from an FDOT project in the Homestead area.**Plastic Settlement-09: Coefficient of consolidation Cv.10 4.2 186 . Estimate of the coefficient of consolidation cv as.23 0.) ( 12 ) 0 0.40 4. with two-way drainage through porous stones. The vertical stress increment is 10 kPa.60 16.4 0.45 0. Settlement Time (min)^(1/2) 0 1 2 3 4 5 0 0. simulating field conditions.40 0.33 0.0 Settlement 0 0.0 time = (min) (1) Plot a graph of settlement against the square root of time.28 2.65 0.13 1.57 0.55 2. (2) Determine the value of the coefficient of consolidation cv (in m2/s).16 0. 30σ z ' k ⎤ ⎡1 + e ⎤ ⎛ σ z ⎞⎛ k ⎞ ⎛ k ⎞ e≈ ∴ cv = ⎢ ⎥⎢ ⎥ ∴ cv = ⎜ ⎟ ⎜ ⎟ = Eo ⎜ ⎟ 1 + eo ⎣ γw ⎦ ⎣ Cc ⎦ ⎝ ez ⎠ ⎝ γ w ⎠ ⎝γw ⎠ k= cvγ w = ( 1.08 × 10−7 m 2 s )( 9.08 ×10 4t x 4(11.81 kN 3 ) m = 5.05 m (4) Cc ⎡ 2.56 min)(60sec)(1000mm)2 sec (3) 1 mm Voids − ratio (ev ) = = 0.05 ∴ ev = 5% 20 mm σv 10 ×103 N Eo = = 2 = 200 × 103 Pa = 200 kPa ev 0.3 × 10−9 m Eo ( 200 kN 2 m ) sec 187 .56 min 3d 2 3(10mm)2 (1min)(1m)2 −7 m 2 cv = = = 1.(2) vt x = 3.40 min1/ 2 and time(t x ) = 11. 18 )(12in / ft ) ⎛ 5 ⎞ ∴ ΔH s = α log ⎜ 2 ⎟ = log ⎜ ⎟ = 0.8 − ( 0.02.76 ) ⎝ 1. The time of completion of the primary settlement is approximately 18 months.02 )( 8.59 = 2.15 inches 1 + eo ⎝ po ' ⎠ 1 + ( 0. by finding the change in the voids ratio during primary consolidation.8 ) ⎝ 2.650 psf.5 ⎠ The total consolidation settlement is thus =ΔH p + ΔH s = 2. Cα H ⎛t ⎞ ΔH s = log ⎜ 2 ⎟ 1 + ep ⎝ t1 ⎠ We must find e p first. 650 + 970 ⎞ e p = eo − Δe = eo − Cc log ⎜ ' ⎟ = 0. The in-situ stress at mid-clay layer is po = 2.5 foot clay stratum 5 years after the primary consolidation? Solution: The primary consolidation ΔH p is. 650 ⎠ The secondary consolidation ΔH s is. and it found that the clay’s initial voids ratio was eo = 0.*Plastic Settlement -10: Secondary rate of consolidation.74 inches 188 . 650 + 970 ⎞ ΔH p = log ⎜ ⎟= log ⎜ ⎟ = 2.28 ) log ⎜ ⎟ = 0. What is the total consolidation of the 8. The secondary compression index Cα = 0. (Revision: Aug-08) An oedometer (consolidation) test is performed on a normally consolidated clay stratum that is 8.59 inches 1 + ep t ⎝ 1⎠ 1 + ( 0.5 ft )(12in / ft ) ⎛ 2. Cc H ⎛ po' + Δp ' ⎞ ( 0.76 ⎝ po ⎠ ⎝ 2. 650 ⎠ C H ⎛ t ⎞ ( 0.28.15 + 0.5 feet thick. and the building exerts a pressure through its mat foundation of 970 psf.8 and its primary compression index is Cc = 0.28 )( 8.5 − 0. ⎛ po' + Δp ' ⎞ ⎛ 2. 2 0. Tv H dr2 (0.3 0. and Tv = 0.7 0.4 0.*Plastic Settlement-11: Using the Time factor Tv.2 Time factor .00294 cm2/sec t (75 days × 24 × 60 × 60) 0 10 Percent consolidation 20 30 40 50 60 70 80 90 100 0 0.5 0. (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.9 1 1.848 for 90% consolidation. Solution: The clay layer has two-way drainage. Find the coefficient of consolidation cv of this clay in cm2/sec.848)(150 cm) 2 cv = = = 0.6 0.8 0.1 0.1 1. Tv 189 . *Plastic Settlement-12: The time rate of consolidation. Determine the time required to attain 70 percent consolidation (Tv = 0. It was observed that 45 percent consolidation (Tv = 0.40)(12.15) (2 in) 2 (0. (Revision: Aug-08) An oedometer (consolidation) test is performed on a 4” thick specimen. t Tv H dr 2 ⎛ Tv H dr 2 ⎞ ⎛ Tv H dr 2 ⎞ cv = =⎜ ⎟ = ⎜ ⎟ t ⎝ t ⎠laboratory ⎝ t ⎠ field ( 0.5 ft × 12in / ft ) ⎞ 2 ∴ t field = ( 78 hours ) ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ 24hours ⎠ ⎝ 365days ⎠ ⎝ 0. Building’s load q = 4 ksf 20’ WT SAND 25’ CLAY SAND Solution: TvH dr2 The coefficient of consolidation cv = is the same for the lab and field samples.5 ft ×12 in / ft ) 2 ∴ = ( 78 hours ) t field ⎛ 1day ⎞ ⎛ 1 year ⎞ ⎛ 0.40 ⎞ ⎛ (12.15 ⎠ ⎝ (2in) 2 ⎠ ∴ t field = 134 years to attain 70% consolidation 190 . drained on top and bottom.15) was attained in 78 hours.40) in a job site where the clay stratum is shown below. 500mm) 2 191 . therefore. ⎡ ⎤ ⎡ ⎤ cv t90( field ) ⎢ c (75days × 24 × 60 × 60) ⎥ ⎢ cv t90 ⎥ T90 = =⎢ v ⎥ =⎢ H dr2 ( field ) 3. 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. 000mm 2 30mm 2 ⎥ ⎢ ( ) ⎥ ⎢( ) ⎥ ⎣ 2 ⎦ field ⎣ 2 ⎦ laboratory (75days × 24 × 60 × 60) (15mm ) 2 ∴ t90(lab ) = = 648 seconds =10 minutes (1.*Plastic Settlement-13: Time of consolidation t.-08) Using the information derived from Problem 11. (Revision: Oct. 5 min) T65 H dr2 (0. Solution: (1) tlab t( field ) = H dr2 (lab ) H dr2 ( field ) (3. to undergo 50% consolidation? (2) Find the time required for the clay layer in the field as described in part (a) above. 3. (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 8.34)(3. (1) How long will it take for a similar clay layer in the field.5 min) ∴ t( field ) = 2 = 557.*Plastic Settlement-14: Laboratory versus field time rates of settlement.36 ×10−5 m 2 / min 192 .2 m thick. 400 min = 668 days cv 0. 000 min = 387 days ⎛ 25mm ⎞ ⎜ ⎟ ⎝ 2 × 1000mm ⎠ (2) 2 ⎛ 25 ⎞ (0.197) ⎜ ⎟ Tv H dr2 ⎝ 2 × 1000 ⎠ = 0.36 ×10−5 cv = = t50 (8. but drained at the top only.2m) 2 ∴ t( field 65) = = = 961.2m) 2 (8. to reach a 65% consolidation.5 minutes. Tv 193 .1 1.1 0.3 0.197) ⎡⎣(10 ft )( 30.197) ⎡⎣( 20 ft )( 30.5 0.6 0.7 0.2 0. *Plastic Settlement-15: Different degrees of consolidation.2 Time factor .9 ⎠ (0.5 0.7 0.1 0.002 cm2/sec.2 Time factor .6 0.5% ⎝ 8.9 1 1.8 0.8 0.2 0.4 0. (Revision: Aug-08) A clay layer 20 feet thick sitting on top of granite bedrock.002 cm / sec ) ⎜⎝ 360sec ⎟⎠ ⎜⎝ 24h ⎟⎠ 2 (0.3 0.197 and t= H 2 dr = = 424 days ⎛ 1hr ⎞ ⎛ days ⎞ cv ( 0.48cm / ft ) ⎤⎦ 2 Tv (b) T50 = 0. experiences a primary consolidation of 8.002 cm 2 / sec ) ⎜⎝ 360sec ⎟⎠ ⎜⎝ 24h ⎟⎠ 0 10 Percent consolidation 20 30 40 50 60 70 80 90 100 0 0. Tv 0 10 Percent consolidation 20 30 40 50 60 70 80 90 100 0 0.9 1 1. (b) The time to reach 50% settlement if cv is 0.4 0.9 inches.48cm / ft ) ⎤⎦ 2 T (c) t = v H dr2 = = 106 days ⎛ 1hr ⎞ ⎛ days ⎞ cv ( 0. (c) The time for 50% consolidation if the clay stratum is doubly-drained? Solution: ⎛ 2 ⎞ (a) U % = ⎜ ⎟100 = 22. Find: (a) The degree of consolidation when the settlement reaches 2 inches.1 1. and point B (at the edge of the tank) are.4 ) = 2. The stress at point A ⇒ 0.2 ksf point B ⇒ 0.4 ksf Using the Boussinesq pressure diagram (next two pages) for B = 75' .5 minutes.43(2.-08) An oedometer test on a 1” thick. 194 .43Δp = 0.120 − 0. provides the stress levels at any point in the soil mass.110 − 0.5' )( 0. doubly-drained sample from the clay stratum (shown below) attained 50% consolidation in 6.062 ) = 1. Solution: a) The surface load from the oil tank (neglecting the weight of the tank) is Δp: Δp = hγ oil = (40 ft )(60 pcf ) = 2. **Plastic Settlement-16: Excavate to reduce the settlement. (b) The time required for 75% consolidation in the field.91Δp = 0. (c) The depth of excavation for minimal settlement.062 ) + ( 28. before the tank was built: po = (10 ' )( 0.90 ( 2. (Revision: Oct. thus.4 ) = 1. Find: (a) The total differential settlement of the fully loaded tank.0 ksf The in-situ effective stress at point A was po.95 ksf The settlements at point A (below the center of the tank). 95 ⎠ CH ⎛ p '+ Δp ⎞ 0. using the relationship between field and lab conditions.5 min t75 ∴ t75 = 13.120 ) + x ( 0. through the coefficient of consolidation.0 ⎞ ΔH B = c log ⎜ o ⎟= log ⎜ ⎟ = 1. since the water is also removed). Cc H ⎛ p '+ Δp ⎞ 0. c) The settlement can be reduced by excavating weight of soil equal to the weight of the structure (note: total. and Tv for 75% = 0.8 years for 75% consolidation.2 ) ⎢0. Tv for 50% = 0. σ total = (10 ')( 0.95 + 1.49 ft = 18inches b) Since the time required for consolidation is.5 ft ) 2 Tv field H 2 field Tvlab H 2 field ⎣ ⎝ 12in ⎠ ⎦ = ∴ cv = = t75 tlab 6. 2 ⎡ ft ⎞ ⎤ ( 0.34 ft − 1.27 ) ⎝ 1.4 ( 57 ') ⎛ 1.5in ⎛⎜ ⎟⎥ ( 0.34 ft 1 + eo ⎝ po ' ⎠ (1 + 1.2 ⎞ ΔH A = log ⎜ o ⎟= log ⎜ ⎟ = 3.9 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 195 .85 ft 1 + eo ⎝ po ' ⎠ (1 + 1.197.95 ⎠ ∴ The differential settlement between A and B is = 3.110 ) = 2.27 ) ⎝ 1.480.4 ksf Solving for x: ∴ x = 10.4 ( 57 ') ⎛ 1. Assume need to excavate “x” feet from the clay layer.480 )( 28.85 ft = 1.95 + 2. not effective. e. The sand surcharge has a γ = 115 pcf. This is the lead time required to place the surcharge before construction.70.30). 196 . The surcharge will have to weigh the same. TV = 0. (f) The SPT in the sand stratum is N = 15.000 ft2. (e) The time required to attain 60% consolidation of the clay stratum (i. the settlement from the clay stratum is a consolidation settlement. an initial void ratio of 0. spread over a square foundation 200 ft by 200 ft. This method minimizes the settlement of the structure. using a fill with a unit weight of γ = 115 pcf. **Plastic Settlement-17: Lead time required for consolidation of surcharge. The total weight of the building is 136. 000 ft 2 lb but . h ft Sand Surcharge (Area = 200’ x 200’) 15 ft WT 30 ft Sand 20 ft Clay Impermeable rock Solution: (a) The weight of the new building is estimated at Q = 136. as shown below.01. spread over an area = 200’ x 200’ = 40. The sand stratum has a CC = 0.4 ksf A 40. a γ = 125 pcf and an initial void ratio of 0. The unit pressure of the surcharge γsurcharge is. a γ = 130 pcf and a consolidation coefficient of 10-3 in2/second. Field tests showed that the clay stratum has a liquid limit of 28 percent.000 kips. (Revision: Aug-08) A common method used to accelerate the consolidation of a clay stratum is a sand surcharge. 000kips qsur = = = 3.95. Q 136.4 ksf ∴ h = 30 feet high ft Both the sand and clay strata contribute to the settlement. The surcharge load will force the clay to attain a large part of its settlement before the structure with the same load is built. such that. However. (d) Determine the total settlement at mid-clay under the center of the surcharge. qsur = γ sur h = 115 3 (h) = 3. An office building is planned to be built on the site shown below.000 kips. 3 inches 197 .95) ⎝ 3.30 Δ = C1C2 qo ⎜ ⎟ dz = (1)(1)(3. depth factor C1 = 1 and creep factor C2 = 1 ⎛ε ⎞ 0.94in + 1. Total Time = ΔH S + (60%)ΔH C = 1.162 Also eo = 0. Δp = γ h = (0.30) ⎡ 20 ft 12in /1 ft ⎤ ⎛ 1min ⎞ ⎛ 1hr ⎞ ⎛ 1day ⎞ t= v = ⎣ ⎦ ⎜ ⎟⎜ ⎟⎜ ⎟ = 200 days for 60% consolidation c v (10 in / sec ) −3 2 ⎝ 60sec ⎠ ⎝ 60 min ⎠ ⎝ 24hr ⎠ The total settlement that has taken place at 200 days (60% consolidation) is.0624)(10) = 3.130 − 0.115kcf )(30 ft ) = 3. ( )( ) 2 T H dr2 (0.009(28 -10) = 0.69in b) The lead time t required for the surcharge to accomplish its task is. HCc ⎛ p + Δp ⎞ (240in)(0.162) ⎛ 3.6)(5.4ksf )( )(360 in) = 1.94in) = 5. The initial stress at mid-clay stratum is.75in ⎝E⎠ 210 ksf The total settlement = 5.75in = 7.95 and H clay stratum = 20 ft = 240 in.5 ksf The increased load from the surcharge Δp is.5 + 3. p0 = γ s h + γ sb h = γ cb h = (0.009( LL -10) = 0.94 inches (1 + e0 ) ⎝ p0 ⎠ (1 + 0. H sand = 30 ft = 360in ES ≈ 14 N = (14 )(15 ) = 210k / ft 2 .73in + (0. Using Skempton's formula Cc = 0.5 ⎠ The settlement from the sand stratum is an elastic settlement.45 ⎞ ΔH c = log10 ⎜ 0 ⎟= log10 ⎜ ⎟ = 5.0624)(15) + (0.125kcf )(15 ft ) + (0.125 − 0.45 ksf The settlement created by the surcharge is. 59. and its index of compression can be found from a relation proposed by Sowers as Cc = 0.75 (eo – 0.2 0.**Plastic Settlement-18: Settlement of a canal levee. The in-situ voids ratio of the marl is 0.2 Time factor .9 × 10 −4 ft 2 / lb . and a v = 2.30).0 × 10 −5 cms / s . Tv 198 .8 0.6 0.4 0.3 0. in order to preconsolidate that layer for a future building.1 0. Find the time required (in days) for the marl to attain 50% consolidation.9 1 1. k (1 + eo ) 2 cv = ft /day avγ w where the permeability k = 1. (Revision: Aug-08) A uniform surcharge of sand 20 feet in height will be placed over the marl stratum as shown below.1 1. Find the total settlement of the surcharge at its point A.5 0. 0 10 Percent consolidation 20 30 40 50 60 70 80 90 100 0 0. The coefficient of consolidation cv for the marl can be found from the relation.7 0. ⎛ 1 + eo ⎞ cv = k ⎜ (1.75 ( 0.0 ×10−5 cm / sec ) (1 + 0.59 − 0. the in-situ stress σ o is.5 ft 2 / day 199 .2 )( 40 ft ) 2 Tv H dr2 t= = = 128 days cv 2.5 ⎞ ∴Δ = log10 ⎜ o ⎟= log10 ⎜ ⎟ = 20.4 ×103 sec ⎞ ⎟= ⎟ = 2.218 At midpoint of the marl.2. 400 psf ⎝ ft 3 ⎠ The increase in stress due to the surcharge Δσ is.218 ) ⎛ 2.75 ( eo − 0.500 psf ⎝ ft ⎠ Therefore.59 ) ⎛ 1in ⎞⎛ 1 ft ⎞ ⎛ 86. the consolidation settlement of the marl is.5 ft / day 2 3 ⎜ ⎟⎜ ⎟⎜ ⎝ avγ w ⎠ ( 2.59 ) ⎝ 2. ⎛ lb ⎞ σ o = γ h = ⎜120 ⎟ ( 20 ft ) = 2.4 + 2.4 ⎠ The time required for 50% of consolidation to take place is found through the coefficient of consolidation cv .Solution. ⎛ lb ⎞ Δσ = γΔh = ⎜ 125 3 ⎟ ( 20 ft ) = 2. ( 0. cc = 0.30 ) = 0.4lb / ft ) ⎝ 2. HCc ⎛ σ + Δσ ⎞ ( 40 ft )(12in / ft )( 0.4inches 1 + eo ⎝ σo ⎠ (1 + 0. therefore the time t required is.9 × 10 ft / lb )( 62.30 ) = 0.54cm ⎠⎝ 12in ⎠ ⎝ −4 2 day ⎠ The time factor for 50% consolidation is Tv = 0. 7 ( 0. (Revision: Aug-08) An oedometer (consolidation) test was performed on a clay sample 3 cm high.7Cc H ⎛ p '+ Δp ' ⎞ 0. drained on both sides. **Plastic Settlement-19: Differential settlements under a levee.67 min ) ⎜ ⎟ = 1.70 ) ⎨ c log10 ⎜ o ⎟⎬ ⎪⎩1 + eo ⎝ po ' ⎠ ⎪⎭ The in-situ stress at mid-clay stratum before the surcharge was applied was. ⎪⎧ C H ⎛ p '+ Δp ' ⎞ ⎪⎫ ΔH 70% = ( 0.7 + 72 ⎞ ∴ 70%ΔH = log10 ⎜ o ⎟= log10 ⎜ ⎟ = 14.67 minutes.5m ) = 71.7 kN / m 2 i and Δp ' = 72 kN / m 2 0.76 years 6 HL ⎝ 1. (c) The magnitude of that settlement in that time.81) kN / m3 ( 3. Seventy percent consolidation was attained in 6. Find: (b) The time required to attain 70% consolidation of the clay stratum.20 )( 700cm ) ⎛ 71.8 cms 1 + eo ⎝ p o ' ⎠ (1 + 1) ⎝ 71. Solution.45 × 10 min = 2.7 ⎠ 200 . and both are 70% consolidation. and taken from mid-stratum shown below. ∴ po ' = ∑ γ i hi = (18kN / m3 ) ( 2m ) + ( 20 − 9.5cm ⎠ (b) The amount of settlement that takes place at 70% consolidation is. Tv H dr 2 ⎡ Tv H dr 2 ⎤ ⎡ Tv H dr 2 ⎤ t field H 2field cv = =⎢ ⎥ =⎢ ⎥ ∴ = 2 t ⎣ t ⎦ laboratory ⎣ t ⎦ field tlab H lab 2 tlab H F 2 ⎛ 700cm ⎞ or t field = 2 = ( 6. (a) Since the soil is the same clay in the laboratory and the field. Vertical stresses induced by a uniform load on a circular area 201 . 60 0.69 1.28 2.20 Settlement Delta H (mm) 0.80 9. s (c) Estimate what could be a good estimate of the elastic modulus of this soil. E o (in kPa).33 0.20 202 .00 0.2503x + 0. m (d) What sort of permeability k ( ) could you estimate for this soil (from E o )? s Solution: (a) Settlement vs.00 0. 3H dr2 cv = 4t Given Data Time(min) 0.23 0.00 5.64 1.00 4. The results are shown below.80 1.86 0.13 1. ***Plastic Settlement-20: Estimate of the coefficient of consolidation cv. m2 (b) Determine the value of cv ( ).00 3. The initial sample thickness was 20 mm with two way drainage through porous stones.00 Settlement (mm) 0.60 16. The vertical stress increment is 10 kPa.32 0.40 y = 0.10 4. simulating field conditions.45 0.00 1.3. Use an estimate of the coefficient of consolidation cv as.55 2.16 0.57 0. (Revised Oct-09) An oedometer test was performed on a peat soil sample from an FDOT project in the Homestead area.00 0.00 0.0376 0.19 3.96.00 0.00 0.80 1. Time^1/2 Time^1/2(min)^1/2 0.65 0.00 2.00 (a) Plot a graph of settlement against the square root of time.96 Time^1/2(min)^1/2 0.40 4. 178 * 10 ⎟⎜ 9 . 81 ⎟⎟ cv γ w ⎜ ⎜⎝ sec ⎟⎠⎝ m3 ⎠ ⎟ m k= =⎜ ⎟ = 4. use 1.05 or 5% L (20mm) σ v 10 * 10 3 N Eo = = = 200kPa Vertical Modulus εv 0.62 min 3d 2 3(10 min) 2 (1 min)(1m) 2 −8 m 2 cv = = = 9.(b) t x = 3.30σ z k ⎛ 1 + e0 ⎞ σ ⎛ k ⎞ k cv = ⎜⎜ ⎟⎟ or cv = z ⎜⎜ ⎟⎟ = E o γ w ⎝ Cc ⎠ εz ⎝γw ⎠ γw *note: (2.05m 2 Cc From Coduto (pages 390 and 391) ε v ≈ but Coduto (page 424) 1 + eo 2.62 min)(60 sec)(1000mm) 2 sec (c) The strain at the end of the loading is: Δ (1mm) εv = = = 0.178 *10 4t x 4(13.0 for log) ⎛⎛ −8 m ⎞⎛ 2 kN ⎞ ⎞ ⎜ ⎜ 9.30 is for ln.69 min 1 / 2 → t x =13.50 * 10 −9 Eo ⎜ ⎛ kN ⎞ ⎟ s ⎜ 200 2 ⎟ ⎜ ⎝ m ⎠ ⎟ ⎝ ⎠ 203 . 739] = 33.279)(2. and a specific gravity of 2. (Revised Oct-09) Find the Cc and the “apparent” OCR for a superficial clay stratum that is 4m thick.**Plastic Settlement-21: The apparent optimum moisture content. A sample of the clay has a water content of 27.65 + 0.65.123 Se = wGs (Saturation means S = 1) e = wGs = (0.600]/[log(930)-log(230)] = 0.739 γ = [(Gs + e) γw]/[1 + e] = [(2.13 204 .12) = 5.25/1.81)]/[1 + 0.675 230 0. An oedometer test showed the following results: Voids Ratio Vertical Effective Stress 0.73935 = 0.638 480 0.733 28 0.739)(9.6 930 Solution: Cc = [Ca-Cb]/[log(O’z b)-log(O’z a)] = [0. The water is at the surface.73 60 0.9%.65) = 0.675-0.739 = 19.736 15 0.12 OCR = (98)/(19. or the differential settlement due to unequal pressures.) Given: P = 150. For clay: Gs = 2. Assume that each building’s mat rotates as a rigid plate. The clay has a specific gravity of 2. They also have similar foundations. (Revised Oct-09) Two tall buildings sit next to each other in the downtown area of Boston.000 kips. The dry unit weight of improved soil is 110 pcf. that has an allowable bearing capacity of 17 ksf. The total load (dead + live + wind) of each building is 150.L = 12%. P. P. The mats are sitting upon a thick prepared stratum of carefully improved soil 40 feet thick. The clay has a consolidation coefficient of 10-4 in2/sec. The second building “B” was finished by early 1995. 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.68. a PI of 52% and a PL of 12% before building “A” was built in 1975. Each mat is a thick reinforced concrete slab. The water table coincides with the interface between the improved soil and the clay. which consist of a simple mat foundation. w = 32%. cv = 10-4 in2/sec For the improved soil: γd =110 pcf 205 .**Plastic Settlement-22: The differential settlement between two buildings.68. They are separated by a narrow 5 foot alley and are both 50 stories high (550 feet tall).000 kips qa = 17 ksf. γ =121 pcf.I = 52%. Also a moisture content of 32%. 5 feet thick and 100 feet by 100 feet in plan view. How much do you predict will building ‘A” drift towards building “B” by early 1996. whereas the in-situ unit weight of the clay is 121 pcf. 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. Diagram 1 PLAN VIEW PLAN VIEW A 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 E Fill Clay Sand 206 . 344 lb/ ft2 Induced Stresses: To solve the induced stresses we will use the Boussinesq’s Method: (σ z )induced = [( ( q 2BLz f B 2 + L2 + z 2f ) 1/ 2 )] × B 2 + L2 + 2 z 2f −1 + Π − sin − ( 2 BLz f B 2 + L2 + 2 z 2f ) 1/ 2 ( ( ) 4 Π z 2f B 2 + L2 + z 2f + B 2 L2 ) B 2 + L2 + z 2f ( ) z 2f B 2 + L2 + z 2f + B 2 L2 p 150.256.256.533 lb/ ft2 • The total final stresses at corner “D” (σ’zf )D = σ’zo + (σz) induced + (σz) fill = 2344 lb/ ft2+ 1059. 1059.532. Time Factor Determining the time factor for building A: Tv = _4Cvt_ H2dr For 21 years: t = 662.933] x 100% The degree of consolidation at TvA = 165.000 × 10 3 q= = = 15.79 lb/ ft2 + 4180 lb/ ft2 = 7.180 lb/ ft2 • The total final stresses at corner “C” (σ’zf )C = σ’zo + (σz) induced + (σz) fill = 2344 lb/ ft2+ 8008.4) = 2.65 lb/ ft2 . loadings and geostatic factors). L and zf into the Boussinesq’s equation yields: (σz )induced = 8008.000 sec ) Tv A = = 165.79 lb/ ft2 ≈ 7.Solution: Initial Condition: Geostatic Stresses σ’zo = ∑Hγ – u = (40)(121) – (40)(62.56 is: 207 .583.000 sec 4 (10 -4 in2 / sec)(662.085+Tv)/0. Stresses due to fill: (σz )fill = H fill γ fill = (110) (38) = 4.56 (40)2 Degree of Consolidation U = [1-10 –(0. B.79 lb/ ft2 Exists differential stresses at the corners of the foundation.000 lbs A (100 )(100 ) zf = 58 ft (at mid-clay) Replacing the values of q.65 lb/ ft2 + 4180 lb/ ft2 = 14.584 lb/ ft2 (Note: Both buildings A and B will experience the same final stress due to fill.65 lb/ ft2 ≈ 14. 344 lb/ ft2 208 . Therefore. Therefore.344 ⎠ At corner D: H= 40 ft (σ’zf )D = 7584 lb/ ft2 σ’zo = 2.32 91.46 ft = 101.009 (64 – 10) = 0. δc U= (δ c )ult δ c = (δ c )ult Assuming normally consolidated soils Cc ⎛σzf ⎞ δc = ∑ H log ⎜ ⎟ (Eqn 1-1) 1 + e0 ⎜σz ⎟ ⎝ 0 ⎠ 1 + eo σ’zo Cc = 0. σ’zo. At corner C: H= 40 ft (σ’zf )C = 14.68 )(62.533 lb/ ft2 σ’zo = 2.82 ⎝ 2.486 (40 ft ) log ⎛⎜ 14.085+166)/0. we will have differential settlement. (σ’zf )C .4 ) . Cc and eo into (Eqn 1-1): δc = 0.67 Recall that uniform stress does not exist beneath the corners of the mat foundation.533 ⎞⎟ = 8.U = [1-10 –(0.486 Gsγ w e0 = −1 γd γd = γ = 121 = 91.82 1+ w 1 + 0.933] x 100% =100% By early 1996. Cc = 0.1 = 0.67 ∴ e0 = (2.55 in 1 + 0.344 lb/ ft2 Replacing values of H. the soils are completely consolidated. 21 yrs after the load was applied.009 (LL – 10) LL = PL + PL = 52 + 12 = 64 % Therefore. 209 .344 ⎠ The Δδc = 8.52 ft = 200 inches.45 ft = 65.01 ft C' 100 ft Δ = tan -1 (3.35 in 1 + 0.72 = 16.46 ft D' ? 3.82 ⎝ 2. Cc and eo into (Eqn 1-1): δc = 0.Replacing values of H. σ’zo.486 (40 ft ) log ⎛⎜ 7.01/ 100) =1.01 ft Diagram 2 A DRIFT = x A 550 ft ' ? D C 5. (σ’zf )D .584 ⎞⎟ = 5.45 = 3.72º Drift = x = 550 tan 1.46 -5.45 ft 8. (Revision: Aug-08) Estimate the average settlement from primary consolidation of the clay stratum under the center of the bridge pier.2 m m2 Determine eo : From chart pc ~ 150 < po normally consolidated. eo = 0.8 0 3 = 6 8 . Solution: Stress at mid-clay stratum: kN kN 3 m x 1 9 .9 m m2 kN kN kN 7 m x 9 .6 0 3 = 1 9 .6 2 3 = 5 8 .81.7 m m2 m2 kN kN 2 m x 9 .**Plastic Settlement-23: Settlement of a bridge pier. Determine Δσ or Δp : 210 .6 for a total ρ o = 1 4 6 . 7 + 99.5 ⎞ ΔΗ = Η ⎜ c ⎟ log10 ⎜ o ⎟ = 4m ⎜ ⎟ log10 ⎜ ⎟ = 0.5 2 A ⎝ 8 m x 10 m ⎠ m Q Determine the settlement ΔΗ : ⎛ C ⎞ ⎛ p + Δp ⎞ ⎛ 0.15 m ⎝ 1 + eo ⎠ ⎝ po ⎠ ⎝ 1 + 0.7 ⎠ 211 .071 ) ⎜ ⎟ = 99.286 and n= = 0. 4m 5m m = = 0.81 ⎠ ⎝ 14 6.071 10 m 10 m Q ⎛ 28 x10 3 kN ⎞ kN Δσ = 4 I4 = 4 ( 0.31 ⎞ ⎛ 146.500 ∴ I 4 = 0. τf→ The normal shear stress at a failure. σ1→ stress σ1’→ stress σ3’→ stress σ3→ Confirming pressure σd → τ→ The shear stress. φ→ The angle of inclination of the plane of failure caused by the failure shear stress. u → Pore water pressure. σ’→ Effective stress. c u→ p→ p’→ q→ q’→ qu→ Ultimate shear strength of a soil. 212 . ф→ The angle of internal friction of the soil. uc→ ud→ σ→ The normal axial stress. Chapter 12 Shear Strength of Soils Symbols for Shear Strength of Soils c→ The cohesion of a soil particle. 8 2 .8 + 2. u = 1. At failure.0 = 4.*Shear strength–01: Maximum shear on the failure plane.8 cm 2 kg σ 3 = 2.8 − 2 ⎞ kg Shear stress τ= sin 2θ = ⎜ ⎟ sin114° = 1. cm kg kg σ 1 − σ 3 = 2.4 at θ = 45° 2 ⎝ 2 ⎠ cm 2 213 .8 − 2 ⎞ kg Maximum shear τ MAX = sin 2θ = ⎜ ⎟ = 1.27 2 ⎝ 2 ⎠ cm 2 ⎛ σ1 + σ 3 ⎞ ⎛ σ1 − σ 3 ⎞ ⎛ 4.83 ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ cm 2 σ1 − σ 3 ⎛ 4.8 + 2 ⎞ ⎛ 4. (Revision: Oct-08) A consolidated un-drained triaxial test was performed on a specimen of saturated clay with a kg chamber pressure σ 3 = 2.8 − 2 ⎞ kg Normal stress σ =⎜ ⎟+⎜ ⎟ cos 2θ = ⎜ ⎟+⎜ ⎟ cos114° = 2. 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.0 2 .0 cm 2 σ1 − σ 3 ⎛ 4. Solution: kg σ 1 = 2.8 2 and the failure plane angle θ = 57° . and note that they are equal.27 cm 2 Compare that to τ = 1.03 cm2 kg S57° = c '+ σ ' tan φ = 0.60 tan 24°) = 1. therefore failure does not occur.4 − 1. Now at the plane of maximum shear stress θ = 4 5 ° ⎛ 4.03 tan 24°) = 1. and so for both.8 + (1.8 + (1. and φ = 24°. .4 2 ⎝ 2 ⎠ ⎝ 2 ⎠ cm kg σ ' = 3. show why the sample cm 2 failed at 57 grades instead of the plane of maximum shear stress.8 − 2 ⎞ kg σ =⎜ ⎟+⎜ ⎟ cos 90° = 3.51 cm 2 The shear strength at 45º is much larger than at 57º.*Shear strength–02: Why is the maximum shear not the failure shear? (Revised Oct-09) kg Using the results of the previous Problem 01. c ' = 0.8 = 1.83 − 1.6 2 cm kg s45° = c '+ σ ' tan φ = 0.80 . Solution: On failure plane kg σ ' = (σ − u ) = 2. 214 .8 + 2 ⎞ ⎛ 4.27 kg/cm from the previous problem.8 = 1. S 57 ° = τ 57 ° failed. on the failure plane θ = 57° .*Shear strength–03: Find the maximum principal stress σ1. (Revised Oct-09) Continuing with the data from the two previous problems.31 2 . at failure σ '3 = σ 3 = 2 . that is u = 0.457 ) σ '1 − ( 0.132 ) σ '1 ⎣⎝ 2 ⎠ ⎝ 2 ⎠ ⎦ ⎛σ ' −2 ⎞ and τ = ⎜ 1 ⎟ sin114° = ( 0.80 + ⎢⎜ 1 ⎟ + ⎜ 1 ⎟ cos114°⎥ tan 24° = (1. in a drained test u = 0. with σ 3 = 2. 215 . What will be the major principal cm stress at failure? Solution: kg a) Analytically.38 2 cm cm cm b) Graphically.426 ) + ( 0. the same soil specimen is now loaded kg slowly to failure in a drained test.6 2 and τ θ = 2. cm 2 ⎡⎛ σ ' + 2 ⎞ ⎛ σ ' − 2 ⎞ ⎤ s = c '+ σ ' tan φ = 0.914 ) ⎝ 2 ⎠ kg kg kg At failure s = τ ∴ σ '1 = 7.0 2 . σ θ = 3. ⎛ φ1 ⎞ ⎡ 2⎛ 26 ⎞ ⎤ kN σ 1 ' = σ 3 ' tan 2 ⎜ 45° + ⎟ = (103.5 ) ⎢ tan ⎜ 45° + ⎟ ⎥ = 265 2 ⎝ 2⎠ ⎣ ⎝ 2 ⎠⎦ m 216 . what was the major principal stress at failure? Solution: φ φ θ = 45° + ∴ 58° = 45° + ∴ φ = 26° 2 2 Using the equation that relates the major principal stress σ1 to the minor principal stress σ3. (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. and with c = 0 (the value of cohesion for a normally consolidated clay).*Shear strength–04: Find the effective principal stress. If the sample was tested with a chamber confining pressure of 103.5 kN/m2. q = 30.5 ta n α ≈ = = 0 .5.5 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: q4 2 4 2 .5 psi (1) σ1 = 148 psi for σ3 = 30 psi p = 89.6 6 8 ∴ φ = 42° 217 .*Shear strength–05: Using the p-q diagram. q = 59.0 psi (3) σ1 = 605 psi for σ3 = 120 psi p = 362. q = 242. (Revised Oct-09) Triaxial tests performed on samples from our Miami Pamlico formation aeolian sand.0. showed the peak stresses listed below.6 6 8 p4 3 6 2 . Plot these values on a p-q diagram to find the value of the internal angle of friction.5 s i n φ = t a n α = 0 .0. q = 126.0 psi (2) σ1 = 312 psi for σ3 = 60 psi p = 186. σ1 = 76 psi for σ3 = 15 psi p = 45.5. 4 5 ° OA σ '1 + σ '3 σ '1 + σ '3 552 + 276 2 218 . from which: σ '1 − σ '3 AB 2 σ '1 − σ '3 552 − 276 sin φ = = = = = 0 . The results kN kN are as follows: σ '3 = 276 2 ( Δ σ d ) f = 276 2 m m Determine: (a) The angle of friction φ. The Mohr circle and the failure envelope are shown in the figure below. 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. (Revised Oct-09) A consolidated-drained triaxial test was conducted on a normally consolidated clay. (b) The angle θ that the failure plane makes with the major principal plane. **Shear strength–06: Consolidated-drained triaxial test. the effective major and minor principal stresses at failure are as follows: kN kN σ '1 = σ 1 = σ 3 + ( Δ σ d ) f = 276 + 276 = 552 and σ '3 = σ 3 = 276 m2 m2 Part A.3 3 3 ∴ φ = 1 9 . Part B.7 ) = 130 2 2 m 219 .7 ° 2 2 Part C. φ 1 9 . σ '1 + σ '3 σ '1 − σ '3 σ' ( on the failure plane ) = + cos 2θ 2 2 σ '1 − σ '3 and τ f = sin 2θ 2 kN kN Substituting the values of σ '1 = 552 .7 ) = 36 8 2 2 2 m 552 − 276 kN and τf = sin 2 ( 54 .7 ° in the preceding m2 m2 equations. σ '3 = 276 and θ = 5 4 . 552 + 276 552 − 276 kN σ' = + cos 2 ( 54.4 5 ° θ = 45° + = 45° + = 5 4 . kN The tests were taken with pore water pressure measurements.5m ) = 173 2 ⎝ m ⎠ m kN kN kN ∴ S = c '+ σ '1 tan φ ' = 20 2 + 173 3 tan 24° = 96. and c ' = 20 2 . **Shear strength–07: Triaxial un-drained tests.5 ⎜17. 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 γ sat = γ d + nγ w = (16 + 0.5) = 297 − ⎜10 3 ⎟ (12.6 3 ⎟ = 297 2 ⎝ m ⎠ ⎝ m ⎠ ⎝ m ⎠ m ⎛ kN ⎞ kN ∴ For σ '1 = σ 1 − u = σ 1 − γ w (9 + 3.3x10 ) = 19 m3 ⎛ kN ⎞ ⎛ kN ⎞ ⎛ kN ⎞ kN ∴ For σ 1 = 4m ⎜16 3 ⎟ + 9m ⎜19 3 ⎟ = 3.6 2 m m m 220 . (Revised Oct-09) Triaxial un-drained tests were performed on clay samples taken from the stratum shown below. 5 (173) = 86 m2 kN kN kN and σ 3 = σ '3 + u = 86 2 + (13m + 3.5m )10 3 = 251 2 m m m 221 . kN σ '3 = K oσ '1 = 0.5σ '1 = 0.(b) Since the clay is saturated. 6 (Δσ d ) f = 384.37 m2 m2 Find the values of the cohesion c and the angle of internal friction φ.6 m2 m2 Similarly. Consolidated-drained triaxial tests were conducted on these two specimens. the principal stresses at failure are. the principal stresses at failure for specimen 2 are kN kN σ 3 ' = σ 3 = 50 and σ 1 ' = σ 1 = σ 3 + ( Δσ d ) f = 50 + 384.6 = 510. The following are the results of the tests: Specimen 1: Specimen 2: kN σ 3 = 100 σ 3 = 50 kN m2 m2 kN kN ( Δσ d ) f = 410. (Revised Oct-09) Two similar clay soil samples were pre-consolidated in triaxial equipment with a chamber pressure of 600 kN/m2.4 = 434.4 m2 m2 222 . kN kN σ 3 ' = σ 3 = 100 and σ 1 ' = σ 1 = σ 3 + ( Δσ d ) f = 100 + 410.**Shear strength-08: Consolidated and drained triaxial test. Solution: For Specimen 1. 6 ) = (100 ) tan 2 ⎢ 45° + ⎛⎜ ⎟ ⎥ + 2c tan ⎢ 45° + ⎜ ⎟ ⎥ ⎣ ⎝ 2 ⎠⎦ ⎣ ⎝ 2 ⎠⎦ kN 510.4 ) = ( 50 ) tan 2 ⎛⎜ 45° + ⎛ ⎟ + 2c tan ⎜ 45° + ⎟ ⎝ 2⎠ ⎝ 2⎠ φ1 ⎞ Subtracting both equations ( 76. for specimen 1 φ1 ⎞ φ1 ⎞ ( 510. ⎡ 12 ⎞ ⎤ ⎡ ⎛ 12 ⎞ ⎤ ( 510.2 ) = ( 50 ) ta n 2 ⎛⎜ 45 ° + ⎟ ∴ φ1 = 1 2 ° ⎝ 2 ⎠ Substituting φ 1 into the equation.These two samples are over-consolidated.6 ) = (100 ) tan 2 ⎛⎜ 45° + ⎛ ⎟ + 2c tan ⎜ 45° + ⎟ ⎝ 2⎠ ⎝ 2⎠ and for specimen 2 φ1 ⎞ φ1 ⎞ ( 434.5 + 2.47c ∴ c = 145 m2 223 .6 = 152. Using the relationship given by equation ⎛ φ1 ⎞ ⎛ φ1 ⎞ σ 1 ' = σ 3 ' tan 2 ⎜ 45° + ⎟ + 2c tan ⎜ 45° + ⎟ ⎝ 2⎠ ⎝ 2⎠ Thus. 00 -0. . The center of the circle must lie on the σ’ axis. - e 10. DISPLACEMENTS(mm) σ’ XX σ’YY τxy τyx Stage x y (kpa) (kpa) (kpa) (kpa) a 0 0 30 70 0 0 b 0.2 0. τ xy ) and (σ ' yy .5 31.***Shear strength-09: Plots of the progressive failure in a shear-box.25˚ e 70 32 0.50 90.44 0˚ c (peak) 103 60 0.0 d 3.60 145 70 43.58 5. (Revised Oct-09) A soil test is performed in the shear-box shown below. Use a compass to locate by trial and error the center of the Mohr circle.6 70 24. TABLE OF VALUES σ 1 '+ σ 3 ' σ 1 '− σ 3 ' S'= τ= τ Change in angle between STAGE 2 2 S' x + y plan (in degrees) kPa kPa a 50 20 0. Assign positive normal stresses to compression and positive shear stresses are counter-clockwise.0 -31.40 0˚ b 70.5 -32.50 71 70 31. Plot the Mohr circles of stress for each stage. the x and y planes remain perpendicular. and it must be equidistant from the two stress points (σ ' xx .82 . τ yx ) .30 -0.50 -0. The test data lists the stresses and displacements.0 Solution: For small displacements.00 1.3 c (peak) 2.46 21˚ 224 . .3 =49. 225 . 226 . **Shear strength-10: Shear strength along a potential failure plane. The unit weight of the soil is 120 lb/ft3 above the WT and 123 lb/ft3 below. then compute the new shear strength if it rose to level C.4 3 ⎟⎟ (20 ft ) = 1248 2 When the groundwater table is at B: ⎝ ft ⎠ ft ⎛ 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. The soil has an angle φ ' = 30° and no cohesive strength.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 227 . Potential shearing surface C▼ B▼ ●A Solution: ⎛ lb ⎞ lb u = γ z z w = ⎜⎜ 62. Compute the shear strength at point A along this surface when the groundwater table is at level B. 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. (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. Vertical plane: σ ' z = Kσ ' z = (0. Solution: Point A .Horizontal plane: kN kN kN σ ' = ∑ γH − u = (17. Using this data.5kPa) tan 28° = 41.0 3 )(3.1 kPa s = c' + σ ' tan φ ' = 10 kPa + (32.6 kPa Point A . (Revised Oct-09) Samples have been obtained from both soil strata shown in the figure below.1m) − (9.5 3 )(1. The c’ and φ’ values obtained from these diagrams are shown in the figure below.8 3 )(1.8 kPa s = c' + σ ' tan φ ' = 10kPa + (59.1 kPa) tan 28° = 27.1 kPa 228 .54)(59. and C.1m) m m m σ ' z = 59. A series of shear strength tests were then performed on both samples and plotted in diagrams below. B.0m) + (17. compute the shear strength on the horizontal and vertical planes at points A.***Shear strength-11: Use of the Mohr-Coulomb failure envelope.5 kPa) = 32. but change suddenly at the boundary between the two strata.Using similar computations: Point B vertical plane s = 57.4 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 strength would increase gradually with depth within each stratum. 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. and K values. φ’. The strength at point C is even higher than at point B because it is in a new strata with different c’. Thus. Assume the principal stresses act vertically and horizontally. the shear strength at point B is greater than that at point A.5 kPa Point C vertical plane s = 68. because the effective strength is greater.2 kPa Point B horizontal plane s = 35.1 kPa Point C horizontal plane s = 54. Failure envelope and Mohr’s circle 229 . 0 m ) + (17 .1 kPa) tan 28° = 27. 5 )(1 . 8 )(1.8 kPa s = c' + σ ' tan φ ' = 10kPa + (59.1 kPa s = c' + σ ' tan φ ' = 10 kPa + (32. and C.5 kPa) = 32. Using this data. compute the shear strength on the horizontal and vertical planes at points A. 1m ) − (9 . (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.***Shear strength-11b: Use of the Mohr-Coulomb failure envelope. B.1 kPa 230 . Solution: Point A .1m) m3 m3 m3 σ ' z = 59.5kPa) tan 28° = 41.54)(59.6 kPa Point A .Horizontal plane: kN kN kN σ ' = ∑ γH − u = (17.0 )(3 .Vertical plane: σ ' z = Kσ ' z = (0. 2 kPa Point B horizontal plane s = 35. the shear strength at point B is greater than that at point A. 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. The strength at point C is even higher than at point B because it is in a new strata with different c’. the strength would increase gradually with depth within each stratum. φ’. but change suddenly at the boundary between the two strata. Assume the principal stresses act vertically and horizontally.Using similar computations: Point B vertical plane s = 57.1 kPa Point C horizontal plane s = 54. Thus.4 kPa Commentary At each point the shear strength on a vertical plane is less than that on a horizontal plane because K < 1. and K values.5 kPa Point C vertical plane s = 68. because the effective strength is greater. Failure envelope and Mohr’s circle 231 . In addition. S= 96.6 kN/m3 )= 297 kN/m3 σ1’ = σ1 − u = σ1 − γw (9 + 3.6 NM/m3 Since the clay is saturated.3)10] = 19 kN/m3 For the clay: σ1 = (4m)(16 kN/m3 )+ (9m)(19 kN/m3 )+ (3.5m] = 172 kN/m3 S = c’ + σ1’ tan φ = (20 KN/m3 + 172 kN/m3 tan 24ο) = 96. and φ= 24º.5) = 297 . (Revised Oct-09) Triaxial un-drained tests were performed on clay samples taken from the stratum shown below.Find (1) the clay shear strength at mid-stratum.5m)(17. The test were taken with pure water pressure measurements and yield a c` = 20 kN/m3.30 σ1 γ =16 kN/m3 H= 7m σ3 clay γ =17. 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.6 kN/m3 KN C` = 20 /m3 σ` 232 .**Shear strength-12: Triaxial un-drained tests. and (2) the effective and total stresses at that same level acting on a vertical face of a small element.6 kN/m3 Impermeable rock stratum Solution: For the gravel: γsat = γsat + nγw = [16 + (0. H=4 sandy gravel H= 13m n= 0.10 kN/m3 [12. 6 kN/m3 )= 297 kN/m3 σ1’ = σ1 − u = σ1 − γw (9 + 3.**Shear strength-12b: Triaxial un-drained tests.Find (1) the clay shear strength at mid-stratum. and φ= 24º. σ3’ = k σ1’ = 0.3)10] = 19 kN/m3 For the clay: σ1 = (4m)(16 kN/m3 )+ (9m)(19 kN/m3 )+ (3. H=4 Sandy gravel H= n= 0.6 NM/m3 Since the clay is saturated. S= 96.6 kN/m3 C` = 20 KN/m3 σ` 233 .10 kN/m3 [12.6 kN/m3 σ3 Solution: For the gravel: γsat = γsat + nγw = [16 + (0.5(172) = 86 kN/m3 σ3 = σ3’ + u = 86 kN/m3 + (86m + 86m) 10 kN/m3 = 251 kN/m3 τ φ = 24ο.5m] = 172 kN/m3 S = c’ + σ1’ tan φ = (20 KN/m3 + 172 kN/m3 tan 24ο) = 96. The test were taken with pure water pressure measurements and yield a c` = 20 kN/m3.30 σ1 γ =16 kN/m3 H= σ3 Clay γ =17.5) = 297 . (Revised Oct-09) Triaxial un-drained tests were performed on clay samples taken from the stratum shown below. and (2) the effective and total stresses at that same level acting on a vertical face of a small element.5m)(17. 36 +0.429 *Assume that the clay was normally consolidated to find σ' at midlevel in the clay (point A). it is required to know the γsat of the sand.9 psi.30 γ (1+ e) 103(1+0.70 γW 62. that is c = 0. **Shear strength-13: Determine the principal stresses of a sample. d=13' H1=40' γd = 103 pcf n = 30% H2=20' •A c = 2. 20 feet thick is covered by a 40 foot sandy gravel stratum with a porosity of 30%.4 ( G S + e)γ W (2.4 γ sat = = = 122pcf 1+ e 1.429)62.30 0. Tests on the un-drained samples of the clay gave c = 2.9 psi γsat = 112 pcf φ' = 24o Solution: (a) In order to find s.30 0.36 1. and a dry unit weight of 103 pcf. n 0. γSAT = 112 psf and φ' = 24o. and (b) the effective and total stress acting on the vertical face of a soil element at the clay midlevel (point A). Find: (a) the soil shear strength s = c + σ'tanφ' at the clay's midlevel (point A).n 1. (Revised Oct-09) A clay layer. 234 .0.429 and G S = d = = 2.429) e= = = = 0. u = (10 + 27 )( 0. σ’ A = γ h S + γ’ h S + γ’ h C = ( 0.122 − 0.95 x2 = = = 4.95ksf in 10 lb ft ft The pore water pressure u is. we find θ through a graphical solution as follows.3 = 5.9 2 x3 x 2 + 3. σ A = σ A' + u = 3.7ksf (b) To find the stress on the vertical face of the soil element at A.38ksf tan φ’ tan 24 o o o x 3 = s( tan 24 )= 1. the toatl stress is.0624 )( 27 ) ⎤⎦ + ⎡⎣( 0.103)(13) + ⎡⎣( 0.44 2 ( tan 24 o )= 0.53ksfs = 1.4ksf and s = c + σ’ A tan φ’ lbkip 144in 2 k s = 2.42ksf +1.4 + 2.42 x1 = = = 0.122 − 0. φ o θ = 45 o + = 45 o + 24 = 57 o_2θ = 114 o 2 2 c 0.95( tan 24 )= 0.0624 )(10 ) ⎤⎦ = 3.3ksf Therefore.939ksf tan φ’ tan 24 o s 1.868ksf 235 .0624 ) = 2. 868 .18 + 2.13ksf cos φ’ cos 24 o σΝ3 = x2 + x3 .18 + 2(2.939 .95 R= = = 2.38 + 0. s 1.0.18 + 2R = 2.13) = 6.R = 4.2.5 ksf 236 .13= 2.18 ksf σΝ1 = 2.44 ksf σ3 = σΝ3 + u = 2.31 = 4.x1 . that is c = 0. Tests on the un-drained samples of the clay gave c = 2.**Shear strength-13b: Determine the principal stresses of a sample. γSAT = 112 psf and φ' = 24o. (Revised Oct-09) A clay layer.429 and G S = d = = 2. it is required to know the γsat of the sand. Find (1) the soil shear strength s = c + σ'tanφ' at the clay's midlevel (point A). 237 .n 1.4 ( G S + e)γ W (2. and (2) the effective and total stress acting on the vertical face of a soil element at the clay midlevel (point A). n 0. 20 feet thick is covered by a 40 foot sandy gravel stratum with a porosity of 30%.30 0.429) e= = = = 0.9 psi.70 γW 62.4 γ sat = = = 122pcf 1+ e 1.0.429)62. and a dry unit weight of 103 pcf. D=13’ H=40’ Sand γd= 103 pct n= 30% γsat= 103 pct • A Clay c = 2.30 0.9 psi H= 20’ φ' = 24o Rock Solution: In order to find s.429 * Assume that the clay was normally consolidated to find σ' at midlevel in the clay (point A).36 1.30 γ (1+ e) 103(1+0.36 +0. we find θ through a graphical solution as follows.95ksf in 10 lb ft ft The pore water pressure u is.939ksf tan φ’ tan 24 o s 1.0624 )(10 ) ⎤⎦ = 3.38ksf tan φ’ tan 24 o o o x 3 = s( tan 24 )= 1.95 x2 = = = 4.42ksf +1.868ksf 238 .4 + 2.4ksf and s = c + σ’ A tan φ’ kip 144in 2 lb k s = 2.53ksfs = 1.44 2 ( tan 24 o )= 0.95( tan 24 )= 0.3 = 5.0624 )( 27 ) ⎤⎦ + ⎡⎣( 0.7ksf To find the stress on the vertical face of the soil element at A.42 x1 = = = 0. the toatl stress is.122 − 0.0624 ) = 2.3ksf Therefore.σ’ A = γ h S + γ’ h S + γ’ h C = ( 0.103)(13) + ⎡⎣( 0. φ o θ = 45 o + = 45 o + 24 = 57 o_2θ = 114 o 2 2 c 0.122 − 0.9 2 x 3 x 2 + 3. u = (10 + 27 )( 0. σ A = σ A' + u = 3. s 1.13= 2.939 .13ksf cos φ’ cos 24 o σΝ3 = x2 + x3 .5 ksf 239 .95 R= = = 2.18 + 2.18 + 2R = 2.18 ksf σΝ1 = 2.x1 .44 ksf σ3 = σΝ3 + u = 2.13) = 6.868 .31 = 4.38 + 0.R = 4.0.2.18 + 2(2. **Shear strength-14: Formula to find the maximum principal stress. Solution: h φ θ = 45° + → 2 θ = 90 + φ 2 d σ1 − σ 3 ad = 2 φ ф 2 2θ f c o τ3 a τ1 σ3 σ1 45° From the figure. the cohesion c and the angle of internal friction φ. ad sin φ = fa (1) σ1 + σ 3 fa = fo + oa = (c) cot φ + 2 fo tan φ = c → = cot φ → fo = (c)(cot φ) fo c Using the properties of the Mohr circle. (σ 1 − σ 3 ) (σ 1 + σ 3 ) (σ 1 + σ 3 ) oa = σ3 + → oa = → fa = (c) cot φ + (2) 2 2 2 (σ 1 − σ 3 ) σ1 + σ 3 σ1 − σ 3 Introducing (1) into (2): sin φ = 2 (σ + σ 3 ) → (sin φ)[(c) cot φ + ] = 2 2 (c) cot φ + 1 2 240 . (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. →[(c) sin φ cot φ] + [ σ 1 +σ3 σ −σ3 sin φ] = 1 → (c) sin φ cot φ = [σ 1 −σ3 ] .[σ 1 +σ3 sin φ] 2 2 2 2 → (2) [(c) sin φ cot φ] = [σ1 – (sin φ)( σ 1)] – [σ3 + (sin φ)( σ 3)] cos φ → (2) [(c) sin φ ]= σ1 [1 – (sin φ)] – σ3[1 + (sin φ)] sin φ 2(c) cos φ (1 + sin φ ) + (σ 3 ) = σ1 (1 − sin φ ) (1 − sin φ ) cos φ φ (1 + sin φ ) φ Since ≈ tan (45° + ) and ≈ tan2 (45° + ) (1 − sin φ 2 (1 − sin φ ) 2 ⎛ ϕ⎞ ⎛ ϕ⎞ σ 1 = σ 3 tan 2 ⎜ 45° + ⎟ + 2 c tan ⎜ 45 ° + ⎟ ⎝ 2⎠ ⎝ 2⎠ 241 . Chapter 13 Slope Stability Symbols for Slope Stability 242 . 3 m ) ⎛⎜ 20.-08) A slope cut to 1.0 m )(11.*Slope-01: Factor of Safety of a straight line slope failure. 650 ⎟ cos16 ° ⎥ tan 30 ° ⎝ m ⎠ ⎣⎝ m ⎠ ⎦ FS = = 2.1 kN/m3. Solution: T he traingule of rock above the potential slip plane has a w eight W per unit width. 650 1 kN kN W = 2 ⎝ m ⎠ m T he length L of the slip plane is.4 m ) + ⎢ ⎜ 9. is this slope stable? Use a unit weight of 20. (Revision: Oct. ( 85.4 m cos16 ° T herefore.1 3 ⎞⎟ = 9.5H:1V will be made in a shale rock stratum that has bedding planes that have an apparent dip of 16˚ (see the figure below).7 > 2 OK ⎛ kN ⎞ ⎜ 9. Resisting For ces cL + ⎡⎣ (W ) cos α ⎤⎦ tan φ FS = = = D riving Forces (W ) sin α ⎛ kN ⎞ ⎡⎛ kN ⎞ ⎤ ⎜ 22 2 ⎟ ( 88. 85.0 m L= = 88. If the acceptable factor of safety against failure is at least 2 along the lower-most bedding plane. and bedding strength parameters of c = 22 kPa and φ = 30˚. 650 ⎟ sin 1 6 ° ⎝ m ⎠ 243 . 6 5 0 m ⎟ c o s 1 6 ° − ( 3 1 . 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 . 4 m . 2 m ) = 3 1 . 4 k P a w ⎝ m ⎠ R e s is tin g F o rc e s c ' L + ⎡⎣ (W ) c o s α − u ⎤⎦ t a n φ FS = = = D riv in g F o rc e s (W ) s i n α ⎛ kN ⎞ ⎡⎛ kN ⎞ ⎤ ⎜15 ⎟ (8 8 . 2 m . c o n s e r v a t i v e ly . (Revision: Oct. 8 1 3 ⎟ (3 . What happens to that factor of safety if the water table rises to the level shown below? Use a unit weight of 20.-08) In the previous problem the slope appeared to be stable with a factor of safety = 2.7. ⎛ kN ⎞ u = γz w = ⎜ 9 . 6 5 0 m T h e le n g t h L o f t h e s li p p la n e i s s t i ll 8 8 . 244 . and bedding strength parameters are reduced by the effective parameters of c’ = 15 kPa and φ’ = 20˚. 7 6 i s le s s t h a n t h e m i n i m u m a c c e p t a b le v a lu e o f 2 .7 6 < 2 NG ⎛ kN ⎞ ⎜ 9 . 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. 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 . 4 m )+ ⎢ ⎜ 9 . a t w a t e r d e p t h z w a b o v e t h e p la n e t h a t r a n g e f r o m 0 t o 3 . 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 le n g t h L . Solution: kN T h e w e i g h t W o f t h e r o c k t r i a n g le p e r u n i t w i d t h i s s t i ll 9 .1 kN/m3.*Slope-02: Same as Slope-01 but with a raising WT. 4 k P a )⎥ t a n 2 0 ° ⎝ m2 ⎠ ⎣⎝ ⎠ ⎦ FS = = 1 . 130 – 0.125(8’)(tan40)22’ + (1.WWV (b2 ) + Vb3 + ae 2 g = 205(40)–(1/2)(0.7.064)[30(15)+(1/2)(40)(15)]15+70(55)+205(0.(0.900 = 1.500/12.200 .WwH (d1 ) .050 k-ft Therefore: FS = Mr/Mo = 19.15 Î GOOD! 245 .2 ksf B φ C Wsoil 15’ Sandy clay c = 1.064)(15)² .064)(21)(tan15)(102)] = 80[23k + 221k] = 19.2+0. The site is located in a seismic zone with a seismic coefficient of 0. 20’ 15’ 40’ 15’ 15’ A 45’ Clayey sand Wcrane SEA c = 0.500 k-ft W (d ) M 0 = Wb1 .8 ksf φ Solution: Mr = R[S1(AB)+S2(BC)] = R[(C1’+σ1’tanφ1)AB + (C2’+σ2’tanφ2)BC] = 80’[(0.*Slope-03: Is a river embankment safe with a large crane? (Revision: Oct.15)(50) Mo = 8.8+(0. The total weight of the soil per unit width are Wsoil = 205 kips and Wcrane =70 kips.-08) Determine if the work site shown below is safe.500/9050 = 2.900 k-ft Therefore: FS = Mr/Mo = 19. provided you consider the minimum acceptable factor of safety for the man-made waterfront slope shown below to be 2. the circular lengths are AB = 22 feet and BC = 102 feet. Assume the arc radius is 80 feet.2 –720 +3850 +1540 = 12.15.51 ÎNot Good! Removing Crane Î Mo =9. 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.86 < 2 NG ∑ W sin α 7.751 lb/ft ∑ = 20. the soil parameters are c = 225 lb/ft2 and φ = 15º.-08) The stability of a slope was analyzed by the method of slices.*Slope-04: Simple method of slices to find the FS.166 plf FS = = = 1. Solution: 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 ∑ = 7.166 lb/ft cL + ∑ (W cos α ) tan nφ (225 psf )(40 ft ) + 20. 751 plf 246 . (Revision: Oct. Determine the factor of safety along this trial surface. compute the factor of safety along the trial circle.3 ⎞ lb = 1 0 .3 ⎞ lb = 7.-08) lb A 30 ft tall.0 ⎞ lb = 2 .9 ⎜ ⎟ 1 7 .4 ⎜ ⎟ 1 2 3 = 1 5. 8 0 0 b ⎝ 2 ⎠ ⎝ 2 ⎠ ft W3 ⎛ 1 2 .6 ⎜ ⎟ 119 + 7.0 ⎞ lb = 1 2 .2 ⎞ lb = 9 . The soil is homogeneous.6 ⎞ ⎛ 5 .0 ⎞ ⎛ 1 2 .5 + 1 4 . Solution: Weights: W1 ⎛ 1 0 .9 ⎞ ⎛ 7 .8 + 9.0 ⎜ ⎟ 119 = 2.8 ⎞ ⎛ 10 . 6 2 0 b ⎝ 2 ⎠ ft W2 ⎛ 1 0 .3 ⎜ ⎟ 119 + 9.6 ⎜ ⎟ 123 = 26.**Slope-05: Method of slices to find the factor of safety of a slope with a WT. The unit weight is 119 pcf above the groundwater table.0 + 7 . 8 0 0 b ⎝ 2 ⎠ ⎝ 2 ⎠ ft W4 ⎛ 5 .3 ⎜ ⎟ 123 = 39. (Revision: Oct. and 123 pcf below.8 = 1 6 2 0 b ⎝ 2 ⎠ ⎝ 2 ⎠ ft W5 ⎛ 16.1 ⎜ ⎟ 1 1 9 + 1 2 .1 ⎜ ⎟ 1 7 .3 + 1 2 .4 ⎜ ⎟ 1 1 9 + 9 .5H:1V slope is to be built as shown below.9 ⎞ lb = 4. 1. 4 00 b ⎝ 2 ⎠ ft 247 .1 ⎜ ⎟ 1 2 3 = 3 0.3 ⎞ lb = 9.8 + 12.7 + 7. 900 b ⎝ 2 ⎠ ⎝ 2 ⎠ ft W6 ⎛ 12.8 ⎜ ⎟ 1 1 9 = 6.9 + 8 . with c ' = 400 ft 2 and φ ' = 29° .5 ⎞ ⎛ 5 . 70 0 b ⎝ 2 ⎠ ⎝ 2 ⎠ ft W7 ⎛ 9. Using the ordinary method of slices.2 + 1 0 . 2 + 1 0 .800 -7 400 29 160 9.4 = 2 3 0 ⎝ 2 ⎠ f t2 u7 = 0 c 'l + W α ⎛ lb ⎞ φ ⎛ lb ⎞ W Slice (lb) c '⎜ 2 ⎟ u⎜ 2 ⎟ l ( ft ) ⎛W ⎞ ( lb ) b ( Deg ) ⎝ ft ⎠ ( Deg ) ⎝ ft ⎠ ⎜ ⎟ ( cosα − ul ) tanφ ' b ⎝b⎠ 1 6620 -18 400 29 0 11.Average pore water pressure at base of each slice: u1 = 0 ⎛ 5 .2 ⎞ lb u2 = ⎜ ⎟ 6 2 .700 53 400 29 230 12.900 3 30.700 16. 400 FS = = = 1.900 7 24.8 12. ⎛W ⎞ c ' l + ⎜ ⎟ ( cos α − ul ) tan φ ' ⎝b⎠ 84.0 + 1 0 .4 = 6 5 0 ⎝ 2 ⎠ f t2 ⎛ 1 0 .800 8 400 29 470 12.200 Σ = 84.4 = 4 7 0 ⎝ 2 ⎠ ft 2 ⎛ 1 0 .100 ⎜ ⎟ ⎝b⎠ Note how slices # 1 and 2 have a negative α because they are inclined backwards.600 2.300 4 39.7 ⎞ lb u4 = ⎜ ⎟ 6 2 . 248 .100 Therefore.4 = 1 6 0 ⎝ 2 ⎠ ft 2 ⎛ 5 .000 2 15.2 18.83 < 2 NG ⎛W ⎞ 46.7 + 7 .0 ⎞ lb u3 = ⎜ ⎟ 6 2 .000 -2.900 24 400 29 650 13.400 6 13.00 67 400 29 0 10.000 10.9 20.800 16.6 8.2 4.600 4.4 8.200 5 26. the factor of safety is.4 = 5 6 0 ⎝ 2 ⎠ ft 2 ⎛ 7 .700 -1.3 ⎞ lb u6 = ⎜ ⎟ 6 2 . 400 Σ = 46.3 ⎞ lb u5 = ⎜ ⎟ 6 2 .700 38 400 29 560 11.5 11. 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. the border between slices 2 and 3). Then.-08) Using the Swedish slip circle method. (Revision: Oct. compute the factor of safety along the trial circle shown in the figure below.**Slope-06: Swedish slip circle solution of a slope stability. One of the slice borders should be directly below the center of the circle (in this case. compute the weight and moment arm for each slide using simplified computations as follows: 249 . Solution: Divide the slide mass into vertical slices as shown. 0 + 9 .5 m 2 7 .8 ⎜ ⎟ 1 7 .1 + 7 .0 d2 = = − 3 .0 = 4 2 0 b ⎝ 2 ⎠ m Moment arms: 4 .0 ⎞ kN : = 2 .8 = 5 9 0 b ⎝ 2 ⎠ m W4 ⎛ 5 .9 m 2 0 .8 = 1 6 2 0 b ⎝ 2 ⎠ ⎝ 2 ⎠ m W5 ⎛ 5 .9 + 7 .9 + 7 .0 ⎜ ⎟ 1 7 .7 m 3 250 .0 − = − 8 .3 ⎞ ⎛ 8 .9 ⎜ ⎟ 1 7 .8 = 1 3 0 b ⎝ 2 ⎠ m W3 ⎛ 9 .6 ⎜ ⎟ 1 7 .1 ⎜ ⎟ 1 7 .8 + 1 2 .8 = 8 0 b ⎝ 2 ⎠ m W2 ⎛ 2 .0 + 7 .9 + = 6 .0 ⎞ kN = 7 .6 m 2 5 .0 ⎞ ⎛ 1 2 .0 + 1 0 .1 + = 1 0 .8 ⎞ kN = 0 .9 + 7 .1 d5 = 2 .8 ⎞ kN = 7 .0 = 1 4 0 b ⎝ 2 ⎠ m W7 ⎛ 9 .1 + 7 .0 + 7 .Solution: Weights W1 ⎛ 2 .2 + 0 .6 d 1 = − 7 .5 m 2 2 .9 d3 = = 1 .2 ⎜ ⎟ 1 7 .8 = 1 4 5 0 b ⎝ 2 ⎠ ⎝ 2 ⎠ m W 6 ⎛ 1 0 .8 d6 = 2 .1 d4 = 2 .9 ⎜ ⎟ 1 7 .5 m 2 7 .8 ⎞ kN = 5 .0 ⎞ kN = 4 .2 ⎜ ⎟ 1 7 .3 + 9 .1 ⎜ ⎟ 1 7 .2 + = 1 7 .9 ⎞ kN = 2 .9 + 8 .1 d7 = 2 .5 m 3 − 7 .8 + = 1 9 . Slice Su ( kPa ) θ ( Deg ) Su θ W ⎛ kN ⎞ d ( m) ⎛W ⎞ ⎜ ⎟ ⎜ ⎟ d b ⎝ m ⎠ ⎝b⎠ 1 80 -8. 280 FS = u = = 1. 8 3 0 ∑ ⎜⎝ b ⎟⎠ d 251 .460 40 30 1200 7 420 19.5 10.830 ∑Sθ π ( 23.7 8.5 -690 2 130 -3.6 2.9 2 < 2 Not Goo d 180 ⎛W ⎞ 180 36.530 5 1450 10.280 Σ = 7280 Σ = 36.9 15.5 -450 3 80 76 6080 390 1.800 6 140 17.5 890 4 1620 6.6 ) 2 π R2 7. Chapter 14 Statistical Analysis of Soils Symbols for the Statistical Analysis of Soils 252 . σ’→ Effective stress. kH → Horizontal permeability. kV → Vertical permeability. w→ water content. icritical→ Critical hydraulic gradient. Chapter 15 Lateral Pressures from Soils Symbols for Lateral Pressures from Soils Dx → Diameter of the grains distributed (represent % finer by weight). GS→ Specific gravity of the solids of a soil. γSAT →Saturated unit weight of a soil γW →Unit weight of water. u → pore water pressure. e →The voids ratio. g’→ Bouyant unit weight of a soil. σV’→ Vertical Effective stress. hsoil→ depth of the soil. H→ Maximum depth of excavation or thickness of a soil layer. VW → Volume of water. 253 . Figure for symbols used in the Coulomb earth pressures.θ) Ka = 2 ⎡ sin(δ + φ )sin(φ − α ) ⎤ cos2θ cos(δ + θ ) ⎢1 + ⎥ ⎣ cos(δ + θ )cos(θ − α ) ⎦ cos2 (φ + θ) Kp = 2 ⎡ sin(φ − δ )sin(φ + α ) ⎤ cos2θ cos(δ − θ ) ⎢1 − ⎥ ⎣ cos(δ − θ )cos(α − θ ) ⎦ 254 . Coulomb’s lateral pressure coefficients Ka and Kp.Formulas and Figures for Lateral Stresses. cos2 (φ . 255 . Ka for the case where δ = 2/3 φ.Ka for the case of θ = 0º and α = 0º. 256 . Failure modes for flexible walls (sheet-piling).Kp for θ = 0º and α = 0º. 33 ) = 0. The wall will not have to retain water. K a = tan 2 ⎜ 45° − ⎟ = tan 2 ⎜ 45° − ⎟ = 0.38 )(10 ) = 1.150 kcf )( tan 30° ) Factor of Safety ( FS ) = = = 1.07 < 2 NG overturning moment (1.9 kips ) 257 .38 ksf 1 The force against the wall is Fa = ( pa ) h = ( 0. Estimate.115 kcf )(10 ft )( 0. Solution: ⎛ φ⎞ ⎛ 30° ⎞ (a) The Rankine active earth pressure coefficient is.9 kips )(10 / 3 ft ) (c) The stability of the wall against sliding towards the left is found by. Factor of Safety ( FS ) = resisting moment = ( 3')(10 ')(1')( 0. and (c) its stability against sliding (use a Factor of Safety ≥ 2).5 ft ) = 1. (a) the lateral force on the wall from the backfill in an active pressure condition.150 kcf )(1. (b) its stability against overturning.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. 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.5 )( 0. resisting force ( 3' )(10 ' )(1' )( 0.37 < 2 NG driving force (1. *Lateral-01: A simple wall subjected to an active pressure condition.33 ⎝ 2⎠ ⎝ 2 ⎠ The lateral pressure at the bottom of the wall is pa = γ hK a = ( 0. (Revision: Sept.-08) Consider a small 10-foot tall and 3 feet thick concrete retaining 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). δ = 12°. Rankine ' s Fp = 0.517 = 3.323 ⎡ sin(δ + φ )sin(φ − α ) ⎤ ⎡ sin(12 + 35)sin(35 − 20) ⎤ cos θ cos(δ + θ ) ⎢1 + 2 ⎥ cos 0 cos(12 + 0) ⎢1 + 2 ⎥ ⎣ cos(δ + θ )cos(θ − α ) ⎦ ⎣ cos(12 + 0)cos(0 − 20) ⎦ cos2 (φ + θ) cos2 (35 + 0) Kp = 2 = 2 = 3. the total passive force Fp on the wall per unit length is.690 versus 3. *Lateral–02: Compare the Rankine and Coulomb lateral coefficients. Using these values.517 ⎡ sin(φ − δ )sin(φ + α ) ⎤ ⎡ sin(35 −12)sin(35 + 20) ⎤ cos2θ cos(δ − θ ) ⎢1 − ⎥ cos 0 cos(12 − 0) ⎢1 − 2 ⎥ ⎣ cos(δ − θ )cos(α − θ ) ⎦ ⎣ cos(12 − 0)cos(20 − 0) ⎦ When α = 0º.1)(10) ( 3. θ = 0º and α = 20°.θ) cos2 (35 . cos2 (φ .690) + 2( 9)(10) 3.271 2 2 φ 35° 1 K p = tan 2 (45° + ) = tan 2 (45° + ) = 3.1)(10) ( 3. (Note: δ is the angle of friction between the soil and the backside of the wall. the Rankine coefficient is 3. (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°.690 = 3.1 kN/m3 and c = 9 kN/m2? Solution: (a) Rankine’s active and passive earth pressure coefficients.520 kN / m2 2 258 .0) Ka = 2 = 2 = 0. γ = 18.690 Note that Ka = 2 2 KP Coulomb’s active and passive earth pressure coefficients. θ = 0º and δ = 0º the Coulomb formula becomes identical to Rankine’s.685 kN / m2 2 Coulomb ' s Fp = 0.5)(18.517 for Coulomb’s. (b) What is the passive earth force on the wall at failure if the wall is 10 m high. ) = 0.5γ h2 K p + 2ch K p = ( 0. φ 35° K a = tan 2 (45° .) = tan 2 (45° .5γ h2 K p + 2ch K p = ( 0.5)(18. (b) Therefore.517) + 2( 9)(10) 3. (Revision: Sept-08) Using the Rankine method.*Lateral-03: Passive pressures using the Rankine theory. exerted upon a temporary retaining wall by a large jacking system (which is not shown in the figure). find the magnitude and location of the passive pressure force Fp with respect to the heel of the wall (point B). Solution: 259 . 63)( 6. K o = 1 − sin φ ' = 1 − sin 30° = 0.5 ) ⎡⎣( 0. because there is no surcharge loading upon the surface of Sand #1. Solution: From Jaky's empirical relation.*Lateral-04: The “at-rest” pressure upon an unyielding wall.105kcf )(10 ft ) = 0.302 )(10 ) + ( 0.8 ft from the top of the wall.12 = 12. c = 0 psf and φ = 30º.25 )(15) + (1.12 )(16.624 ksf 1 1 1 Fo = ∑ fi = F 1 + F 2 + F 3 + F 4 = ( 0.624 )(10 ) i 2 2 2 kip Fo = 2.67 ) + ( 3.63 + 5. Given: Sand #1 has a unit weight of 105 pcf.67 ) = 173.122 − 0. (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.525)(10 ) + ( 0.105 )(10 ) + ( 0.5 kip 12. 260 .525)(10 ) + ( 0.1 kip − ft 12. Sand #2 has a unit weight of 122 pcf.49 + 3.25 + 1.5 )( 0. at z = 10 feet σ 'h = K oσ 'v = ( 0.5 ft z= ( 2.525 ksf at z = 20 feet σ 'h = ( 0.5 kip z = 13.49 )(16. c = 0 psf and φ = 30º.823 ksf σ w = γ w h = ( 0.50 at z = 0 feet σ ' = 0 ksf .0624 pcf )(10 ft ) = 0.67 ) + ( 5.0624 )10 ⎤⎦ = 0. and (c) find the lateral force upon the wall whilst considering the clay separation. 2c K a Solution: a) The coefficient of active earth pressure is.630ksf ) ∴ hcrack = = = = 11. pa = 0.113kcf )( 21 ft ) (1) − 2 ( 0.2c K a ∴ γ hK a = 2c K a 2c K a 2c 2 ( 0. ⎛ 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.2 feet γ Ka γ K a ( 0.26 = 1. (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. 21 ft γ zK a . Find: (a) The Rankine active earth pressure on the wall. pa = γ hK a .11 ksf b) The crack stops where the pressure is zero.48 k / ft of wall 2 ( 0.113 kcf )( 21 ft )(1) − 2 ( 0.113kcf )(1) c) The total (Rankine) active earth force upn the wall Fa is.630ksf ) 1 = 2.63ksf )( 21 ft )(1) + 2 = 5.63ksf ) 2 1 Fa = ( 0.37 − 1. pa = σ 3 = γ hK a − 2c K a = ( 0. *Lateral-05: The contribution of cohesion to reduce the force on the wall.113kcf ) 261 . (b) Estimate the depth of separation of the clay from the wall. 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 ⎛ 2c ⎞ 1 Fa = 1 2 (γ HK a − 2c K a ) ⎜H − ⎜ ⎟ = γ H − 2cH K a + γ K a ⎟⎠ 2 2 2c 2 γ ⎝ 2 ( 0. 5 m3 The stress at point "a" is σ a =0.3 3 ⎟ ( 4 m ) (0.41 ⎛ kN ⎞ The forc e Fo = ½ γ d h 2 K o = ½ ⎜ 18.3 + (0.26) = 38 kN per meter of wall 2 ⎝ m ⎠ 3) The buoyant weight γ ' of the flooded sand is. 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.26) = 4.81) − ( 9. (Revision: Sept-08) A 4 m wall retains a dry sand backfill with a unit weight of 18.26 ⎝ 2⎠ ⎝ 2 ⎠ ⎛ kN ⎞ The force Fa = ½ γ d h 2 K a = ½ ⎜ 18. kN γ ' = γ sat − γ w = γ d − nγ w − γ w = 18. kN σ b = γ d hK a = (18. and at "b" which is 1 meter below the surface.3 )(1m )(0. an angle of internal friction of 36˚ and a porosity of 31%.3 kN/m3.8 kN / m 2 ∴ F1 = ½(4. and its location from its heel.41) = 60 kN per meter of wall 2 ⎝ m ⎠ 2) When th e wall deflects to the left sufficiently to develop an active pressure condition. if the wall does deflect enough to develop a Rankine active earth pressure condition? 3) What is the new force on the wall.3 3 ⎟ ( 4 m ) (0.**Lateral-06: The effect of a rising WT upon a wall’s stability. The backfill is fully drained through weep holes. 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 .81) = 11.31) ( 9.4 kN / m m3 262 .sin 36 ° = 0. ⎛ φ⎞ ⎛ 36 ° ⎞ K a = tan 2 ⎜ 45 ° − ⎟ = tan 2 ⎜ 45 ° − ⎟ = 0.8 kN / m 2 )(1m ) = 2.sin φ = 1 . 5 kN / m m The water pressure and force.1)(1) y= = R 74. 38kN 263 .26) = 9.4 kN / m 2 ∴ F4 = ½(29. ΔF = ( 74.1 kN / m m 4 Therefore R = ∑F i =1 = 74. F1d1 + F2 d 2 + F3 d3 + F4 d 4 ( 2.5m ) + (13.81 3 )(3m) = 29. kN σ bc = γ d hK a = (18.8kN / m 2 )(3m) = 14.3 3 )(1m)(0.5 3 )(3m)(0.4kN / m 2 )(3m) = 44.4 kN / m The location of the resultant is y.26) = 4. kN σ w = γ w h = (9.4 y = 1.17 m from the bottom of the wall. The percent increase in load upon the wall due to flooding is.5 )(1m ) + ( 44.33m ) + (14.4 )( 3.4 kN / m m kN σ c = γ ' hK a = (11.8 kN / m 2 ∴ F2 = (4.4kN − 38kN ) = 96% increase.0kN / m 2 )(3m) = 13.4 )(1.0 kN / m 2 ∴ F3 = ½(9. 2kN / m3 )(7 m) 2 (0.0 m high retaining wall has a horizontal backfill of dry sand with a unit weight of 17.*Lateral-07: The effects of soil-wall friction upon the lateral pressure. Solution: The force applied to the wall first requires the coefficient of active earth pressure.307) = 129. (Revision: Sept-08) A 7. FH 129.5 m = (138 ) (3.307 2 2 The horizontal force FH per unit width of wall is. find the magnitude of the active earth force upon a length of wall equal to 3. The wall is cast-in-place concrete.2 kN/m3 and an angle of internal friction φ = 32˚. Ignoring the effect of the passive pressure upon the toe of the footing.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 .5 m. φ 32° K a = tan 2 (45° − ) = tan 2 (45° − ) = 0.5m) = 482kN FH δ FV R 264 .5 m assuming Rankine conditions. with a friction angle δ = 20˚. FH = ½γ h 2 K a = ½(17. Total Active Force every 3.5kN / m FH = R cos δ ∴ R = = = 138 kN / m cos δ cos 20° We are asked what is the total force every 3. 5 ) ( 8 .8 ) ( 6 m ) = 5 8 .1m f ro m A . 3 3 ) = 8 . 5 . 2 2 T h e φ = 9 0 ° is r e a lly a c o m b in a tio n o f s h e a r a n d c o h e s io n ( " c e m e n ta tio n " ) .2 kN/m3 3m φ = 90° P li A 0 S o lu tio n .5 k N / m T h e lo c a tio n y = y 1 F1 + y 2 F 2 = (4 m ) (1 3 .1 k N / m F 2 = ½ p 2 H = ( 0 .8 ) ( 6 m ) = 1 7 6 . F to ta l (1 8 9 . for the worst case (clogged weep holes). 7 ) ( 3 m ) = 1 3 . 4 ) = 2 .9 . p 1 = γ ' h1 K a = (γ SAT -γ w ) h1 K a = (1 8 . 8 ) ( 3 ) ( 0 . ϕ 30° K a sand = ta n 2 ( 4 5 ° − ) = ta n 2 ( 4 5 ° − ) = 0 . T h e w o r s t a c tiv e p r e s s u r e lo a d o c c u r s w h e n th e w a te r ta b le r a is e s to th e to p o f th e w a ll.5 kN/m3 3m φ = 30° H=6m Weep holes γ = 21.4 k N / m F to ta l = 1 8 9 . 1 ) + ( 2 m ) (1 7 6 .*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).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 ∴ th e lim e s to n e d o e s n o t lo a d th e w a ll.7 k N / m 2 p2 = γ w H = (9 . 5 ) h1 + H F1 Ftotal p1 h2 F2 ŷ A p2 265 .8 k N / m 2 F1 = ½ p 1 h1 = ( 0 .5 )(5 8 . 1m WT worst load case Medium dense sand γsat = 18. 25 ksf Step 3 266 .0624) (20’) = 1.115 .125 – 0.0624) (10’) + (0.t. (Revised Oct-09) Draw the pressure diagram on the wall in an active pressure condition.217] = 0.13 Solution: Step 1 Ka1 = tan2 (45°.333) = 0.5 + (0.0624) (10’)] [0.217] = 0.115 – 0. and find the resultant Ftotal on the wall and its location with respect to the top of the wall.40°/2) = 0.30°/2) = 0.5 ksf w.0624) (10’)] [0.*Lateral-09: Strata with different parameters.333] = 1.5 + (0.18 5 γ = 125 pcf 10’ ф = 40° 3 4 c 1.83 + c=0 0.115 – 0.66 0. a 0.217 Step 2 The stress on the wall at point a is: pa = q Ka 1 = (2.83 c=0 γ = 115 pcf 10’ 1 ф = 30° H = 20’ 2 b 0.66 0.66 ksf The stress at point c is: pc’ = [q + (γ’h)1 + (γ’h)2] Ka 2 = [2.333 Ka 2 = tan2 (45°.5) (0.25 0.83 ksf The stress at b (within the top stratum) is: pb’+ = (q + γ’h) Ka 1 = [2.= (q + γ’h) Ka 2 = [2.79 ksf The pressure of the water upon the wall is: pw = γwh = (0. q = 2.0624)(10’)] [0.01 ksf The stress at b (within bottom stratum) is: pb’ .0.5 + (0. 25) (20’) = 12.30 kips/ft F2 = ½ (10’)(0.83) = 8.3 + 20 ⋅ 0.0 kips/ft Step 4 The location of forces ŷ is at: 5 ⋅ 8.90 kips/ft F3 = (10’) (0.60 kips/ft F4 = ½ (10’)(0.65 + 40 ⋅ 12.5 kips/ft Ftotal = 29.5 yˆ = 3 3 3 = 0. The forces from each area: F1 = (10’) (0.9 + 15 ⋅ 6.13) = 0.6 + 50 ⋅ 0.66) = 6.65 kips/ft F5 = ½ (1.2 feet from top of wall 267 .66 ksf 29 The stress at point c is: ŷ = 11.18) = 0. 11) (20’) – (2) (0.59 ksf σc.84 c+ 0 1 3.29 10’ c=0 3 4 A φ = 40° a γ = 130 pcf +0 48 +0 77 Lateral load from the surcharge σc+ = Ka1 q = (0. Calculate the magnitude of the resultant Ftotal of the active force above the point “A” upon the wall.70 = -0.25 -0.48 Dense sand +1.0.59 . Solution: Notice that the vertical pressure diagram will always increase in magnitude.84 ksf -0. which may increase or decrease the pressures on the wall. Surcharge q = 0.84 ksf ∴ ∑ σc = 0.25’ no water present 20’ Sandy clay 16.29 ksf 268 .50) 0.75’ c = 500 psf 2 φ = 10 ° b+ b.25 ksf σb+ = Ka 1 γ h – 2c K a 1 + q Ka = (0.70)(0.*Lateral-10: The effects of a clay stratum at the surface. Assume Rankine conditions.7) (0.= -2c K a1 = -2(0. +0.84 = -0.84 kcf) = 0.70 = 1. but the horizontal pressures are governed by the Ka coefficient. The sheet pile wall shown below is flexible enough to permit the retained soil to develop an active earth pressure condition.5) 0. 40° / 2) = tan2 25° = 0.29)(10’) = + 1.22 F2 = ½ (1.75’) = +10.48 + (0.41 k/ft (tension).= Ka 2 γ h – 0 = (0.11) (20’) = 0.48 + Ka 2 γ h = 0.48 + 0.29)(16.48 ksf σa = 0.13)(10’) = 0.0. σb.φ /2) = tan2 40° = 0.77 ksf Ka 1 = tan2 (45° . Ka 2 = tan2 (45° .80 k/ft F4 = ½ (0.22)(0.45 k/ft Ftotal = +16.48)(10’) = + 4.29 = 0.80 k/ft F3 = (0.70 F1 = ½ (-0.22) (0.25’) = .6 kip/ft 269 .25)(3. 15)(24) 3 = 103 kip/ft 270 .-08) The wall shown below will be used to retain the sides of an excavation for the foundations of a large building.2(0. (1) What is the minimum distance x from the anchor to behind the wall? (2) What is your recommended factor of safety for the anchor? What is an economical load for the anchor? x 5’ Grouted anchor A 24’ φ = 30° c = 150 psf O Solution: (1) The anchor must be beyond the passive slip plane. (Revision: Sept. or (x) tan 30º = 19’ or x = 33 feet. (2) Ka = tan2(45º .33 = 5.33) .**Lateral-11: Anchoring to help support a wall.84 kip/ft with the force located at ŷ = ⅓(19’) = 6.15)(24) 0.33 and Kp = tan2(45º + φ/2) = 3.φ/2) = 0. The potential passive failure force (from the anchor) on the wall Fp is: Fp = ½γH2Kp + 2cH K p = ½ (0.33’ above point O (note that the tensile portion does not load the wall).105)(24)2(3) + 2(0.0 The active force upon the wall per unit width Fa is: Fa = ½γH2Ka -2cH K a = ½(0.105)(24’)2(0. The engineer has decided to use earth anchors in lieu of braces or rakers to stabilize the wall. Fa ( FS ) = or (FS)2 = p = =17.2) = 24.2 ( FS ) Fa 5.The factor of safety should be the same for an active failure as a passive failure. A common spacing would be 10 feet.2) = 24. which means A = 245 kips. a simple Fp F 103 kips equation could be written as. Therefore.5 kips/ft).5 kips/ft (which is the same as using the passive force = (103)/(4. and depends on cost factors.84 kips Note that this corresponds to a load in the anchor of (5. The horizontal spacing of the anchors is not influenced by this analysis.6 ∴ FS = 4.84)(4. 271 . 66 + 0.11)6’ + (0.31+0.46 ksf ∴ 1.0624)(5) (0.**Lateral-12: The effect of five strata have upon a wall.0624)](0.0.110)(6) (0.2c√Ka = [2 + (0.0.203 ksf at h = -8’ Δp3 = (γ2 .4)(0.0624)(8)(1) = 0.γw)h K2a = (0.704) – 2(0.0626) (9)](1) –2(0.49) = 0.22 ksf at h = -(25 + dh)’ = [2 + 0.35 ksf at h = -(17 + dh)’ = [2 + 0.704) = 0.6)(0.13 ksf at h = -30’ Δp6 = (γ5-γw)h K5a = (0.126-0.γw)h K3a = (0. (Revision Oct-09) Plot the pressure diagram and find the resultant force F and its location under an active pressure condition.95 ksf at h= -17’ Δp4 = (γ3 .333) = 0.49) – 2(0.0624)(2)(0.γw) 2’] K3a – 2c √(K3a from p = γh Ka .45 ksf 272 .76 ksf at h = -25’ Δp5 = (γ4 .84) = 0.46 = 2.120 – 0.125 + (0.307) = 0.8)(1) = 1.γw)h K4a = (0.141 ksf ∴1.76 + 0.417 ksf at h = -(8+dh)’ = [q + (γ1) 6’ + (γ2 .125 – 0.14 = 1.120 .403 = 1. At h=0’ p1 = q K1a = (2) (0.120-0.66 + 0.95+0.0624)(9)(0.0624)2’](0.572 + 8(0.403 ksf ∴0.125 .307) = 0.614 ksf at h = -6’ Δp2 = γ1h K1a = (0.7) = 1.125 + 0. F1 = (0.31)(5) = 6.61)(26) = (1.614)(6) = 3.35 kips F11 = 0.40(9) = 1.0 kips ∴ y = 611 / 57.042)(2) = 0.80 kip F7 = (1.5(0.5(0.1 = 10.203)(6) = 0.61 kips F3 = (0.461)(8) = 1.68 kips The resultant R is.04 The location of R is…….141)(5) = 0.51(1.84 kips F9 = (1.63 kips F4 = 0.63)(23) F6 = 0.55 kips F10 = 0.817)(2) = 1.09(y) = (3.95)(9) = 8.5(0.758)(8) = 14.∑M0 = 0 (about 0) F5 = (0.5(0. R = ∑ Fi = 57.1 kips F8 = 0.68)(27) + (0.5(0.1 kips 273 .55 kips 57.1 kips F2 = 0.50)(24) = 18.7 feet above “0” 57. (a) overturning.6 φ2 = 28 c2 = 30 3 2 m m 274 .6 φ1 = 32 c1 = 0 m 3 m kN o kN γ2 = 17.5 m 0.8 kN o 3 γ conc = 23.62m 4 H = 8 m 1 H = 9.58m 2 1.**Lateral-13: The stability of a reinforced concrete wall.4 α = 10ο 0. and (c) bearing capacity failures. (Revised Oct-09) Calculate the Factor of Safety against.5 m 0. 0.75 m 0.96m 0 yφc kN γ1 = 16. (b) sliding.6 m 3 3. 63m) + (0.6)(2/3)(30) = 355 kN/m FSS = Fh / FR = 1.1 kN/m Fh = Fa cos10° = (248 kN/m)(0. K1 = K2 = 2/3 Kp = tan2( 45° + 28°/2 ) = 2. MR = 23.58 m)2(16.4m)(8m)(1.75 m)2 + (2)(30kN/m2) 2.5)(2.77)(17.617m)(3.8m)] (1m) + 16.322 cos β + cos 2 β − cos 2 φ ' cos10o + cos 2 10o − cos 2 32o Fa = (1/2) H2 γ1 Ka = (1/2)(9.43m)] (1m) + 43.46 c) the factor of safety (FSBC) against a bearing capacity failure.6 kN/m3)(1.5m)(4.8 kN/m3 [(3. 275 .77 (1.75 m) = 249 kN/m the driving force = Fh = 244 kN/m the resisting force = FR = ΣV tan(2/3)(28) + (5.85m) + (1/2)(0. cos β − cos 2 β − cos 2 φ ' cos10o − cos 2 10o − cos 2 32o Ka = cos α = cos10o = 0.985) = 244 kN/m a) The factor of safety against overturning is found by taking moments about point “O”.96m)(5.6m)(2. The resisting moment against overturning is MR.8 kN/m3)(0.322) = 248 kN/m Fv = Fa sin10° = (248 kN/m)(0.42 b) The factor of safety (FSS) against sliding failure.6m)(1m) = 2661 kN-m and the overturning moment is MO = Fh (1/3) H’ = 244 kN/m (9.2m)(8m)(1.5m)(8m)(3.1 kN/m (5.58m)(1/3) = 777 kN-m FSO = MR / MO = 3 .174) = 43.6 kN/m3[(0.90m) + (1/2)(0.77 Fp = (1/2) γ2 H2 Kp + 2 c2 H Kp = (0. 6m – 2 (0.6)) = ⎝ B ⎠ = 1.8)(1.00 = 9.7 Nγ = 16.08 qu = (1/2)(17.7 Df 1.98 ⎛ Ph ⎞ ψ = tan −1 ⎜ -1 ⎟ = tan (244 / 749) = 18.4 = 1 + 0.96 ⎝ 90 ⎠ 2 2 ⎛ ψ ⎞ ⎛ 18.8 Nq = 14.7)(1)(0.6 ⎢⎣1 + 5.08)(0.58) + (30)(25.75/5.04° ⎝ ΣV ⎠ ⎛ ψ ⎞ Fqi = Fci = ⎜1 − ⎟ 2 = 0.8 − = 0.58 ⎝ φ ⎠ ⎝ 28 ⎠ Fγd = 1 2 ⎛ Df ⎞ Fqd = 1 + 2 tan φ (1 − sin φ ) ⎜ 2 ⎟ = 1 + 2 tan28°(1-sin28°) (1.8)(14.98 m qu = (1/2) γ2 B’ Nγ Fγds Fγid + C2 Nc Fcds Fcid + q Nq Fqds Fqid q = γ2 D = (17.75) = 30.31m) = 4.31 m 2 ΣV 749 ΣV ⎡ 6e ⎤ 749 ⎡ 6(0.7)(1.14 B' 4.14)(0.96) + (30.6 ⎥⎦ = 178 kN/m B’ = B – 2e = 5.75 Fcd = 1 + 0.78 276 .98)(16.6)(4.6)(1.4 = 1.8 kN/m2 using φ2 = 28° → Nc = 25.96) = = 1740 kN/m2 FSBC = qu / qtoe = 1740 / 178.31) ⎤ 2 qtoe = B ⎢⎣1 + B ⎥⎦ = 5.04 ⎞ Fγi = ⎜⎜1 − ⎟⎟ = ⎜1 − ⎟ = 0. B ΣMr − ΣMo 2661 − 777 e= − = e = 2 . ***Lateral-14: Derive a formula that provides K and σH as a function of σv. Solution: The definition of an “active pressure” condition is when σh decreases until it touches point D on the Mohr-Coulomb failure envelope. when the soil mass is on the verge of failure). AO and OC gives: 277 . CD CD sin φ = = AC AO + OC ⎛σ −σ3 ⎞ ⎛σv −σa ⎞ CD = radius of failure circle = ⎜ 1 ⎟=⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ AO = c cot φ ⎛σ +σ3 ⎞ ⎛σv −σa ⎞ OC = ⎜ 1 ⎟=⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ Substituting values into the equation for CD. (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. Find σa: From the figure. β).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° 90° + φ φ α= = 45° – 2 2 1 + sin φ = tan (α + φ) cos φ φ = tan (45° − +φ ) 2 ⎛ φ⎞ = tan⎜ 45° + ⎟ ⎝ 2⎠ ⎛ φ⎞ sin ⎜ 45° + ⎟ ⎝ 2⎠ = ⎛ φ⎞ cos⎜ 45° + ⎟ ⎝ 2⎠ ⎛ φ⎞ cos⎜ 45° + ⎟ cos φ ⎝ 2⎠ ∴ = 1 + sin φ ⎛ φ⎞ sin ⎜ 45° + ⎟ ⎝ 2⎠ But for any complementary angles β and (90° . φ φ φ φ Thus. cos (45° sin (45° - and sin (45° cos (45° 2 2 2 2 278 . cos β = sin (90° .β).cos φ + ⎜ v ⎟ sin φ = v ⎝ 2 ⎠ 2 Τhis then gives: σa (sin φ + 1) = σv (1 – sin φ) − 2c. ⎛σv −σa ⎞ ⎜ ⎟ ⎝ 2 ⎠ sin φ = ⎡ ⎛σv +σa ⎞⎤ ⎢c cot φ + ⎜ 2 ⎟⎥ ⎣ ⎝ ⎠⎦ Rearrange the equation to make σa the subject: This gives: ⎛σ +σa ⎞ σ −σa 2c. 2c ( Ka ) ½ . ) 2 2 Rankine’s expression gives the effective φ ∴ σa = (γz) Ka . ) . where Ka = tan2 (45° . ⎛ φ⎞ sin ⎜ 45° − ⎟ cos φ ⎝ 2⎠ ⎛ φ⎞ ∴ = = tan⎜ 45° − ⎟ 1 + sin φ ⎛ 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° − ⎟ 1 + sin φ ⎝ 1 + sin φ ⎠ ⎝ 1 + sin φ ⎠ (1 + sin φ ) 2 (1 + sin φ ) 2 ⎝ 2⎠ φ φ σa = σv tan2 (45° . the slip planes can be described by the grid of lines shown below: Δ 279 .2c tan (45° . ) 2 Using this equation. 588) ⎤ sin ( 66° ) ⎢1 + ⎥ ( 0. (a) The magnitude of the active earth force Pa on the wall. Calculate the coefficient of active earth pressure (Coulomb) K a using. (c) The location of the resultant of both forces.33H Dense sand γ = 18 kN/m3 φ = 36° 0 Solution. α = 0°. Determine.866 )( 0.914 ) ⎢1 + ⎥ ⎢⎣ sin ( 66° ) ⎥⎦ ⎢⎣ ( 0. (Revision: Sept-08) The reinforced concrete retaining wall shown below will be subjected to a horizontal seismic load of 0. (b) The magnitude of the earthquake active earth force Pae on the wall. 1m ΔPae increase due to the earthquake load located at 0. β = 90° and δ = 2 / 3φ = 24°.654 Ka = 2 = 2 = 0.2346 ⎡ sin ( 60° ) sin ( 36° ) ⎤ ⎡ ( 0. sin 2 ( β + φ ') Ka = 2 ⎡ sin (φ '+ δ ) sin (φ '− α ) ⎤ sin β sin ( β − δ ) ⎢1 + 2 ⎥ ⎢⎣ sin ( β − δ ) sin (α + β ) ⎥⎦ sin 2 ( 90° + 36° ) Ka = 2 ⎡ sin ( 36° + 24° ) sin ( 36° − 0° ) ⎤ sin 2 ( 90° ) sin ( 90° − 24° ) ⎢1 + ⎥ ⎢⎣ sin ( 90° − 24° ) sin ( 0° + 90° ) ⎥⎦ sin 2 (126° ) 0. φ = 36°.914 ) ⎥⎦ 280 . **Lateral-15: The magnitude and location of a seismic load upon a retaining wall.6H Pae the earthquake load H=5m Pa lateral load from the soil located at 0.2 g without a vertical component. 7kN / m ) 2 281 . δ = 2 / 3φ = 2 / 3 ( 36° ) = 24° and θ ' = tan −1 ⎢ h ⎥ = tan −1 ( 0.3° ⎣1 − kV ⎦ sin 2 (φ + β − θ ') ∴ K ae = 2 ⎡ cos θ 'sin 2 β sin ( β − θ '− δ ) ⎢1 + sin ( φ + δ ) sin ( φ − θ '− α ) ⎤ ⎥ ⎢⎣ sin ( β − δ − θ ') sin (α + β ) ⎥⎦ sin 2 ( 36° + 90° − 11.6H above the base of the wall. found by locating the force Pae at a height 0.6 H )( ΔPae ) + (1/ 3) H ( Pa ) ( 0.7kN / m ) + (1/ 3)( 5m ) ( 53kN / m ) 2 2 z= = = 2. the active earth force Pa is obviously ( 0. kV = 0. 1 Pa = γ H 2 K a = ( 0.7 kN / m 2 2 The earthquake force is ΔPae = Pae − Pa = 83. 1 Pae = γ H 2 (1 − kV ) K ae = ( 0.2346 ) = 53 kN / m 2 2 Calculate the earthquake coefficient of active earth pressure (Coulomb) K ae .2.372 ⎡ 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.33) H above the base. ( 0. ⎡ k ⎤ kh = 0.5 ) (18kN / m3 ) ( 5m ) ( 0.3° ) K ae = 2 = 0.5 ) (18kN / m3 ) ( 5m ) (1 − 0) ( 0.1 m Pae (83.372 ) = 83.6 )( 5m ) ( 30.7 − 53 = 30.The active earth force Pa is.2 ) = 11.7 kN / m The location of the resultant earthquake force z is. 282 . Assume no vertical seismic component (kv=0). (Revision: Aug-08) The reinforced concrete retaining wall shown below will be designed to a horizontal seismic loading of 0.2 g. **Lateral-16: Seismic loading upon a retaining wall. for zero lateral displacement. (b) The weight of the wall under seismic conditions. 1m H=5m Dense sand γ = 16 kN/m3 φ = 36° δ = 2/3 φ = 24º 0 Solution.5 inches. for a lateral displacement = 1. Determine. (c) The weight of the wall under seismic conditions. (a) The weight of the wall Ww under static conditions. Chapter 16 Braced Cuts for Excavations Symbols for Braced Cuts for Excavations 283 . e. and the sample reached failure with a deviator stress of 90 kN/m2. as shown below. (N. if they are spaced every 4 m horizontally.65). s-1 to s-3). 284 .46 at a moisture of 34% (assume Gs = 2. Two triaxial laboratory tests were performed on samples of the clayey sand. In order to design the steel horizontal strut shown. The deviator stress needed to attain failure was 160 kN/m2. The first sample had a confining pressure of 0 kN/m2.: the deviator stress is the additional vertical stress required to reach failure.*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.B. Further laboratory tests show that this clayey sand had an in-situ voids ratio of 0. Show all your calculations. i. you must first find the force and moment on one of them. The second sample had its confining stress increased to 30 kN/m2. we can get that φ2 = 32 o (Gs )(γ W ) (2.65)(9810) Gs = 2.307 2 285 . Effective Stress Mohr’s Circle for failure Angle τ (kN/m2) σ (kN/m2) From the Mohr’s Circle.65 .46 o φ K A 1 = tan 2 (45o − 25 ) = 0.406 K A = tan2 (45o − ) 2 2 o KA 2 = tan 2 ( 45 o − 32 ) = 0. γW = 9810 N/m2 Î γs = = Î γS = γ2 = 17.8 kN/m2 1+ e 1 + 0. 54 kN/m2 Pb’+ = [KA1 (q + γ1h1)] = [(.4 kN/m 286 .27 3m 3 F3 F4 F5 34.54 2m 1 F1 F2 2 18.307 [90 + (15)(3) + (17.406) (15kN/m2 x 3m)] Î 54.0m F3: 3m + 2m (1/2) = 4.99 kN/m2 Pc = [(q + γ1h1 + (γ2-γW) h2] KA2 = .8-9.8-9.36 19.62 kN/m2 36.81) (3)] Î 34.307) (90 + (17.35 kN/m2 PW = γW h W = (9.5m F2: 3m (2/3) = 2.62 Location of the Forces (with respect to the top datum): F1: 3m (1/2) = 1.81 kN/m2 Pb’.54)(3/2) = 27.6 kN/m F2 = (Pb+. Pa = (q) (KA1) = (90kN/m2) (.33m F5: 3m + 2m (2/3) = 4.406) Î 36.Pa)( h1/2) = (54.= [KA2 (q + (γ2-γW) h] = [(.0m F4: 3m + 2m (2/3) = 4.99 4 5 11.33m Magnitude of the Forces: F1 = (Pa)( h1) = (36.81-36.81)(2)]Î 46.81)(2) Î 19.54 kN/m2)(3m) = 109. 96 ∑M c = 0 Where C is located at the bottom of the trench along with RA RB is located at the end of the strut.84 kN (2.96 kN/m Ftot = (∑ F )(space b/t struts) = (237.33) Located at y f = = 2.84 kN 109.98(4) + 11.Pb-)( h2/2) = (46. Î RB (3m) .5) + 27.33) + 19.98 kN/m F4 = (Pc . F3 = (Pb-)( h2) = (34.35-34.99)(2/2) = 11.6(1.96kN/m)(4m) Î 951.36 kN/m F5 = (PW ) (h W/2) = (19.62(4.62)(2/2) = 19.44 kN ÎRA = 209.62 kN/m ∑F = F 1 + F2 + F3 + F4 + F5 = 237.4(2) + 69.36(4.34m) = 0 Î RB = 742.66m 237.951.99 kN/m2) (2m) = 69.40 kN 287 . 34 m 288 .44 kN 0 kN -209.0kN-m 0 kN-m 0 kN-m 0.40 kN Moment Diagram 490.Shear Diagram 742.66 m 2. which is a reinforced concrete pipe with a diameter of 3 m. Solution: Use Stroud’s relation to estimate the un-drained cohesion of the soil (the previous problem provided the shear strength): cu = KN = (3. γH if > 4 the clay is soft to medium cu γH if ≤ 4 the clay is stiff cu In this problem. Therefore. (Revision: Sept-08) Design a braced excavation for a large sanitary sewer force-main. 289 .21 < 4 ∴ this is a stiff clay cu ( 70 ) Also. Assume φ = 0.**Braced cuts-02: A 5 m deep excavation with two struts for support. γH = (17 )( 5) = 1. and γ = 17 kN/m³. The trench should be 5 m deep and 5 m wide. The SPT for the silty clay is Navg = 20. since γH/ cu < 6. the sheet-piling should extend at least 1. The phreatic surface is below bottom of excavation.5 kN/m²) (20) = 70 kN/m².5 m below bottom. we must choose the larger of.51)(26).25/3]-(0.51)(26). Since φ = 0.2c Ka therefore zc = 2c/γ = 2(70 kN/m²)/ 17 kN/m³ = 8.51/2] = 0 therefore.25m)(26)[0. by symmetry F22 = F12 = 13.51)(26)[0.25)(26)+(0.6 m OK Step 2: Determine the lateral loads at strut locations and excavation bottom.6 kN/m 290 .6 kN/m ∑ Fy = 0 = .3γ H = 0.9 + 1/2 (1.9 kN/m ∑ Fy = 0 = -15.16m)-(0. Isolation the left portion between the surface and strut #2.5)(1.Step 1.6 kN/m Isolating the right portion between strut #2 and the trench bottom. ⎡ ⎛ 4c ⎞ ⎤ (1) pa = γ H ⎢1 − ⎜ u ⎟ ⎥ ⎣ ⎝ γ H ⎠⎦ (2) pa = 0.F22 + (3. Establish the lateral earth pressure distribution. Using Peck's (1967) apparent pressure envelope.3 (17 )( 5 ) = 25. F1 = 15.51+1.F3 therefore. F’2 = 13. ∑ MF’2 = 0 = F1(1.2 m >> 0.5 kN / m 2 The location of the top strut should be less then the depth of the tensile crack zc.F’2 = 0 therefore. Ka → Ka = 1.75-0. therefore σ3 = σa = (γ)(zc)Ka . F3 = 70. Mmax = 96 kN-m/m Step 4: Select the steel-piling .25)(26)(1. Step 6: Select the struts.46 m³/ m-105.2 kN/m. Design the steel pipe wales and the struts. Design the thickness of the slab (diaphragm).00056m³ = 56 m³/ m-105 Choose a PDA-27 section. The bottom of the trench has the highest lateral load.000 kN/m² = 18.56 kN-m/m MC = (2.9(0. therefore σallow = 50%fy = 172 MN/m² The required section modulus S S = MMax/ σall = 96 kN-m/ 172. which provides 57. the areas under the shear diagram): MA = ½(0.6 m³/ m-105 At strut level #2 the load is 27.71/2) = 96 kN-m/m Obviously. (May use 3 m to reduce steel size. Step 5: Select the horizontal waler at each strut level. calculated in Step 6 below. Swale at 1 = Mmax/σallow = 32 kN-m/ 172. Swale at 2 = Mmax/σallow = 54. Step 7: Check for possible heave of the excavation bottom.000 kN/m² = 31.71)(26)(2.6 m³/ m-105 Notes: 1. Wales are commonly channels or WF beams.4 kN-m Therefore.4 kN-m/ 172. Select the horizontal spacings to be 4 m. 291 . Propose to cast a concrete “mud” slab at the bottom of the trench. B.25/3)-15. with 70.2)(4)²/8 = 54.75 kN-m/m MB = ½(1. Mmax = F1s²/8 = (16)(4)²/8 = 32 kN-m (where s is the spacing) therefore.60)(12. Finding moments at A. 2.60/3) = 0.48)(0. Mmax= F2s²/8 = (27. the spacing s is = 4 m.6 kN per every meter. but increases the difficulty of placing the concrete pipes).2 kN/m)(4m) = 109 kN (Design the steel for the struts).65) = 3.Step 3: Find the maximum moment Mmax in the sheet-piling. & C (that is. Level # 1 strut = F1s = (16 kN/m)(4m) = 64 kN Level # 2 strut = 2 F2s = (27. Assume fy = 50 ksi = 345 MN/m².000 kN/m² = 0. At strut level #1 the load F1 is 16 kN/m. 84 + 0.K.4 > 2 O. δγ from 1 to 5 cms.4)(0.16 B/L)]/ γH = (70)(6.84)/(17)(5) = 4. Expect δh from 5 to 10 cms. FSagainst heaving = [cNc(0.Braced cuts in clay may become unstable if the bottom heaves upward and fails a section of wall. Step 8: Expected lateral yielding of the sheet-piling and ground settlement behind the wall. 292 . 2γ H to 0.4γ H cu 293 . (Revision: Sept-08) A four-strut braced sheet pile installation is designed for an open cut in a clay stratum. as shown below. γH ⎛ 4c ⎞ if > 4 the clay is soft to medium then σ a = γ H ⎜1 − u ⎟ cu ⎝ γH ⎠ γH if ≤ 4 the clay is stiff then σ a = 0. Find: 1. B. The struts are spaced longitudinally (in plan view) at 4. medium or stiff. 2. The lateral earth pressure diagram for the braced sheet pile system. Assume that the sheet piles are pinned or hinged at strut levels B and C.0 m center to center. C. and D.*Braced cuts-03: Four-struts bracing a 12 m excavation in a soft clay. Solution: From Terzaghi and Peck (1967). The loads on struts A. a clay is soft. 3 kN / m 3 )(12 m ) ⎢ (17.48 kN / m ) ( 4.0 m ) − 100.3 kN / m ) (12 m ) = 4.5 m ) (15.48 kN / m ) ( 4. In the free body diagram.0 m )( 4.2 kN 2 In the free body diagram.0 m ) = 92 kN 2 In the free body diagram.5 m ) (15.6 kN From ∑H = 0 FB1 = 1 ( 2 ) (1.03 m ⎞⎟⎠ + (1.0 m ) ⎛⎜⎝ 1. 2.52 m ⎞⎟⎠ − ( F 2 2 A )( 3.0 m ) = 0 ∴ FA = 100. ⎛ 4c ⎞ ⎡ (4)(48 kN / m 2 ) ⎤ pa = γ H ⎜1 − u ⎟ = (17. ∑M C =0 ( FD )( 3.0 kN 294 .0 m ) − ( 4.33 > 4 3 ∴ this is a soft to medium clay cu ( 48 kN / m ) 2 Peck (1969) provided a criterion for soft to medium clays.6 kN = 85.0 m ) (15. part (b) FB 2 = FC1 = 1 ( 2 ) ( 3.48 kN / m ) ( 3.48 kN / m 2 ) ( 4.5 m ) (15.0 m ) ⎛⎜ 4.5 m + 4.5 m 3. part (a).5 m ⎞ ⎟=0 ⎝ 2 ⎠ ∴ FD = 209.48 kN / m 1 − 2 ⎝ γH ⎠ ⎣ ⎦ The lateral earth pressure diagram for the braced sheet pile system in soft clays is.48 kN / m ) ( 4.3 kN / m3 )(12 m) ⎥ = 15. qu 96 kN / m 2 Determine the cohesion from Mohr's circle cu = = = 48 kN / m 2 2 2 ∴ γH = (17. part (c). ∑M B =0 ( 1 2 ) (15.0 m ) ⎛⎜⎝ 1. 0 kN 295 .2 kN + 92.0 m ) − 209.6 kN Therefore.5 m ) (15. FA = 100.0 m ) = 0 FC 2 = ( 4.5 m ) (15.9 kN + 69.48 kN / m 2 ) ( 4.From ∑H = 0 FC 2 + FD − ( 4.6 kN = 162.48 kN / m2 ) ( 4.9 kN = 178.0 kN = 69.6 kN FB = 85.1 kN FC = 92.5 kN FD = 209. Chapter 17 Bearing Capacity of Soils Symbols for the Bearing Capacity of Soils 296 . 00 0.29 173.90 19.54 41 0.25 11.71 0.18 20.4712 41.70 4.07 6.99 134.60 2.12 33.3491 26.05 42 0.44 13.56 32.34 4.07 133.87 55.49 0.41 118.81 46 0.16 8 0.31 0.00 61.34 1.1396 14.88 0.85 115.82 4.34 1.07 3.30 368.49 23.70 1.58 8.49 1.43 17 0.58 45 0.94 1.20 79.37 8.0524 11.37 113.92 265.86 73.4014 32.72 10.01 5.25 0.85 31.80 8.88 24 0.38 1.18 5.2618 20.73 18 0.32 28 0.4363 36.63 17.50 674.47 0.44 41.59 85.19 11.14 18.35 0.50 22 0.50 328.78 14 0.09 0.98 1.69 32 0.00 105.93 64.48 20 0.45 2.6283 87.86 16.64 819.60 5.92 8.44 31.49 70.80 14.21 0.59 37.10 5.02 4 0.53 298.35 48.07 6.19 1.64 0.46 19.88 319.2094 17.33 38.51 415.62 55.32 93.7156 151.40 48.57 196.00 0.44 5.03 25.18 20.24 27.11 6.08 46.63 12.50 173.83 18.5236 52.77 1.85 600.84 9.16 29.92 15.47 0.96 77.40 50 0.24 19.69 1.65 93.80 17.72 0.3665 28.98 17.81 0.76 3.28 2.10 1.26 0.38 11.31 64.96 7.72 10.0698 12.26 2.00 5.66 12.73 152.39 11 0.05 37.37 115.14 105.63 4.34 1.6807 119.15 199.52 6.33 63.34 14.71 85.81 1.92 53.40 2.61 17.00 2 0.81 3.63 42.17 38.75 40.15 23.66 152.00 1 0.24 15.95 21 0.81 7.95 347.86 266.19 346.10 158.79 33 0.87 9.08 5.75 41.65 10.06 0.31 211.64 187.90 95.94 224.76 40 0.21 7.07 15.89 16.20 9.74 9.26 34.16 1.77 35 0.00 22.06 873.05 8.55 67.00 0.7854 271.37 115.31 229.40 13.63 18.13 13.07 6 0.3840 30.10 21.95 147.33 67.49 1.89 106.04 28.92 10.8029 323.49 10.11 7 0.57 297 .21 414.0349 11.05 8.99 83.34 258.66 6.13 9.0175 10.86 4.40 15.45 199.20 0.80 2.06 568.13 10.07 61.64 26.22 204.18 5.49 23.00 1.01 27.26 0.02 35.1222 13.16 2.90 1.09 119.52 48 0.92 47.00 5.13 77.43 50.81 1.21 27.88 319.35 48.61 75.20 0.31 64.75 44.6109 79.8378 484.75 25 0.81 7.63 1.10 158.23 30.34 4.74 215.04 6.94 27 0.56 6.36 11.59 37.96 66.53 47.50 12 0.73 1.10 0.64 187.14 18.61 6.1571 15.37 3.27 55.43 0.19 6.44 12.80 2.86 16.88 7.5061 48.13 12.45 3.28 5.97 15 0.81 15.6632 107.13 23 0.16 118.2793 21.44 8.79 10.63 2.02 2.39 6.87 200.50 61.40 15.53 22.72 344.0000 10.92 265.25 0.69 75.80 14.1745 15.30 16.01 171.28 7.7330 173.90 13.26 6.85 8.37 139.07 53.82 18.38 1.87 134.60 2.94 1.67 20.18 16 0.87 55.63 42.12 9.10 1.74 133.44 28.16 29.28 2.07 3.33 12.93 5.19 23.16 22.67 49 0.12 33.34 2.05 37 0.41 25.66 81.11 50.4538 39.51 6.5411 56.30 526.34 2.90 113.11 99.18 22.66 6.48 52.85 7.31 165.40 8.94 13.5934 72.44 10.36 12.6458 96.57 0. Bearing Capacity Factors for General Shear Angle φ Angle φ Terzaghi Meyerhof Hansen (Degrees) (Radians) Kpγ Nc Nq Nγ Nc Nq Nγ Nc Nq Nγ 0 0.15 2.92 3.10 151.23 7.46 13.43 0.4887 44.66 4.98 3.0873 12.83 6.00 0.32 9.17 39 0.88 7.01 137.61 79.65 47 0.09 0.19 25.47 83.30 37.65 6.80 2.65 14.29 40.07 16.30 10 0.94 13.53 2.57 0.72 10.20 9.86 73.40 57.30 1.28 22.21 299.61 7.75 10.97 70.04 44.71 0.20 35.72 11.02 5.37 11.42 18.80 451.3316 25.26 222.81 3.15 7.09 14.75 140.29 1.30 33.09 32.06 0.64 26.41 3.80 56.29 271.29 287.63 4.7679 231.52 26.66 4.22 9 0.58 126.42 15.15 46.84 30 0.23 42.8552 613.19 1.75 30.57 172.67 108.87 262.15 7.2269 18.11 4.27 9.69 7.47 8.53 2.63 1.16 48.92 2.87 134.92 2.7505 198.5760 66.72 0.90 1.01 3 0.15 1155.92 36 0.71 32.26 222.80 2.56 6.08 19 0.87 14.09 24.97 0.76 26 0.93 5.09 26.59 0.50 456.81 93.25 11.94 29 0.97 1.33 173.2967 22.37 3.72 19.78 29.68 4.8203 391.11 99.63 13 0.5585 61.98 3.50 35.83 6.19 7.21 11.70 13.3142 23.60 5.56 5.09 2.4189 34.67 30.31 0.77 67.6981 134.40 2.20 93.72 10.55 241.59 0.67 20.00 13.00 5.22 0.27 95.8727 801.96 43 0.93 56.46 23.31 95.07 31 0.60 9.49 25.35 8.10 44 0.10 5.05 5 0.97 0.71 85.28 173.04 4.50 244.77 20.1047 13.38 38 0.16 1.77 1.80 7.44 1.1920 16.15 42.26 2.07 6.12 20.88 0.92 229.97 266.32 9.82 4.62 1.44 34 0.2443 19.64 36. qult = cN c Fsc Fdc + qN q Fsq Fdq + 0.The bearing capacity of a soil is its ability to carry loads without failing in shear.4γ BNγ Fsγ Fdγ Fiγ and the factors are. Brinch Hansen in Denmark (1970) and Vesic in the USA modified these factor to a greater refinement. Field tests in Canada by Meyerhof (1963) lead to modification factors.5 ( N q − 1) tan φ 298 .4φ ) Brinch Hansen (1970): The general equation. N q = eπ tan φ tan 2 ( 45° − φ / 2 ) N c = ( N q − 1) cot φ Nγ = 1.5γ BNγ where.75π −φ / 2) tan φ a cos 2 ( 45° − ϕ / 2 ) N c = ( N q − 1) cot φ tan φ ⎛ K pγ ⎞ Nγ = ⎜ − 1⎟ 2 ⎝ cos φ ⎠ 2 Meyerhof (1963): For vertical loads. a2 Nq = where a = e( 0. qult = cN c Fsc Fdc Fic + qN q Fsq Fdq Fiq + 0. Finally.3c ' N c + qN q + 0. qult = cN c Fic Fdc + qN q Fiq Fdq + 0. These bearing capacity factors are based on these three authors: Terzaghi (1943): For square footings. N q = eπ tan φ tan 2 ( 45° − φ / 2 ) N c = ( N q − 1) cot φ Nγ = ( N q − 1) tan (1.4γ BN γ Fiγ Fd γ and the factors are.4γ BNγ For continuous or wall footings qult = c ' N c + qN q + 0. The fist method was developed by Karl Terzaghi in 1943. qult = 1.4γ BNγ Fsγ Fd γ and for inclined loads. There are four major methods to predict failure. q = γ D f and the factors are. 1) ⎤⎦ B2 3 ⎣ 294 = 250.4γ BNγ where q = D f γ The allowable bearing capacity qall with the factor of safety of 3 is. and Nγ =41. N c =57. qult = 1.5 + 99. N q =41.4) (18.3c ' N c + qN q + 0.8.96 = 0 ∴ B = 0.3c ' N c + qN q + 0.3c ' N c + qN q + 0. (Revision: Sept-08) The square footing shown below must be designed to carry a 294 kN load. qall = qult 1 3 3 ( = 1. W = 294 kN Df = 1 m γ = 18.4.90 m 299 . Use Terzaghi’s bearing capacity formula to determine B of the square footing with a Factor of Safety =3.4) + (0.15) B(41.*Bearing–01: Terzaghi’s bearing capacity formula for a square footing.5B B2 B 3 + 2.15 kN/m3 φ = 35° W c =0 B Solution: Terzaghi's formula for the ultimate bearing capacity qult of a square footing is.4γ BNγ ) and qall = W 294 kN B2 = B2 or 294 1 B2 3 ( = 1. Substituting these values into Terzaghi's equation.1.52 B 2 − 2. we get 294 1 = ⎡( 0 ) + (18.15)(1) (41.4γ BNγ ) For φ =35º. 15) B (37.15 )( 0. and N γ = 37.30.4γ BNγ Fsγ Fd γ ) For φ = 35º. qall = qult 1 ' 3 3 ( = c N c Fsc Fdc + qN q Fsq Fdq + 0.6 ⎝L⎠ ⎝1⎠ ⎝L⎠ 2 ⎛ Df ⎞ 2⎛ 1 ⎞ Fdq = 1 + 2 tan φ (1 − sin φ ) ⎜ ⎟ = 1 + 2 ( tan 35° ) (1 − sin 35°) ⎜ ⎟ ≈ 1.94 B − 5.45 = 0 ∴ B = 0.4γ BNγ Fsγ Fd γ ) and qall = W 294 kN B2 = B2 or 294 1 ' B2 3 ( = c N c Fsc Fdc + qN q Fsq Fdq + 0.65 m B 300 .12.3) (1.94 B or B 3 + 7.15. W = 294 kN Df = 1 m γ = 18.25 ) + (0. we get 294 1 = ⎡( 0 ) + (18.4γ BNγ Fsγ Fd γ Fiγ where q = D f γ Since the load is vertical.7 )(1. all three inclination factors Fic =Fiq =Fiγ =1.4) (18.15)(1) (33. N c = 46. Use Meyerhof’s bearing capacity formula to determine B with a factor of safety =3. qult = c ' N c Fsc Fdc Fic + qN q Fsq Fdq Fiq + 0.1 + 53. ⎛B⎞ ⎛1⎞ ⎛B⎞ Fsq = 1 + ⎜ ⎟ tan φ = 1 + ⎜ ⎟ tan 35° = 1.70 and Fsγ = 1 − 0.4(1) = 0. (Revision: Sept-08) The square footing shown below must be designed to carry a 294 kN load.4 ⎜ ⎟ = 1 − 0.6 )(1) ⎤⎦ B2 3 ⎣ 294 2 = 428.15 kN/m3 φ = 35° W c =0 B Solution: Meyerhof's formula for the ultimate bearing capacity qult of a square footing is. Substituting these values into Meyerhof''s equation.*Bearing–02: Meyerhof’s bearing capacity formula for a square footing. N q = 33.25 and Fd γ = 1 ⎝ B ⎠ ⎝B⎠ The allowable bearing capacity qall with the factor of safety of 3 is. 3) (1.73B − 5.15) B (33. all three inclination factors Fic =Fiq =Fiγ =1.255 ) + (0.7 and Fsγ = 1 − 0. Substituting these values into Hansen's equation.25 B or B 3 + 8.4 ⎜ ⎟ = 1 − 0.4γ BN γ Fsγ Fd γ Fiγ where q = D f γ Since the load is vertical.97 = 0 ∴ B = 0.4γ BN γ Fsγ Fdγ ) and qall = W 294 kN B2 = B2 or 294 1 ' B2 3 ( = c N c Fsc Fdc + qN q Fsq Fdq + 0.255 and Fd γ = 1 ⎝ B ⎠ ⎝B⎠ The allowable bearing capacity qall with the factor of safety of 3 is. qall = qult 1 ' 3 3 ( = c N c Fsc Fdc + qN q Fsq Fdq + 0.15 kN/m3 φ = 35° W c =0 B Solution: Hansen's formula for the ultimate bearing capacity qult of a square footing is.4γ BN γ Fsγ Fd γ ) For φ = 35º.4) (18.92 )( 0. qult = c ' N c Fsc Fdc Fic + qN q Fsq Fdq Fiq + 0. W = 294 kN Df = 1 m γ = 18.7 )(1. (Revision: Sept-08) The square footing shown below must be designed to carry a 294 kN load.30.12.92.6 )(1) ⎤⎦ B2 3 ⎣ 294 2 = 429. we get 294 1 = ⎡( 0 ) + (18.8 + 49.*Bearing–03: Hansen’s bearing capacity formula for a square footing.15)(1) (33. ⎛B⎞ ⎛1⎞ ⎛B⎞ Fsq = 1 + ⎜ ⎟ tan φ = 1 + ⎜ ⎟ tan 35° = 1. and N γ = 33. Use Brinch Hansen’s bearing capacity formula to determine B with a factor of safety =3.6 ⎝L⎠ ⎝1⎠ ⎝L⎠ 2 ⎛ Df ⎞ 2⎛ 1 ⎞ Fdq = 1 + 2 tan φ (1 − sin φ ) ⎜ ⎟ = 1 + 2 ( tan 35° ) (1 − sin 35°) ⎜ ⎟ ≈ 1. N q = 33. N c = 46.4(1) = 0.70 m B 301 . 3c ' N c + qN q + 0.15) B (41.850 kg/m3 φ = 35° W c =0 B Solution: The soil density ρ = 1.4. (Revision: Sept-08) The square footing shown below must be designed to a load of 30. determine the size B of the square footing.000 kgm.3c ' N c + qN q + 0.4) + (0. ⎛ kg m ⎞ ⎛ m⎞ ⎜ 1. 294 1 2 = ⎡⎣( 0 ) + (18.90 m 302 .4γ BNγ ) For φ = 35º.52 B 2 − 2. 000 kg m ) ⎛⎜ 9.850 kgm / m3 converts to a unit weight via γ = ρ g ( like F = ma ) .850 m3 ⎟ ⎜ 9. N c = 57.000 kgm Df = 1 m ρ = 1. N q = 41. Using a factor of safety of 3 and using Terzaghi’s method. (1. m = 30.3c ' N c + qN q + 0.8. and N γ = 41.15)(1) (41. qult = 1.4) (18.96 = 0 B 3 B = 0.1.81 s 2 ⎟ γ = ρg = ⎝ ⎠⎝ ⎠ = 18.4γ BN γ ) P 294 and qall = 2 = 2 B B or 294 1 B 2 3 ( = 1.15 kN / m3 and the load to be supported by the footing is. 000 N / kN ) ( 30.4γ BN γ ∴ qall = qult 1 3 3 ( = 1.81 m⎞ ⎟ ⎝ s2 ⎠ W = ma = = 294 kN (1. 000 N / kN ) Terzaghi's ultimate bearing capacity of a square footing is given by.1) ⎤⎦ ∴ B 3 + 2.*Bearing–04: Same as #01 but requiring conversion from metric units. Nγ = 15.2 ksf ( Almost a three − to − one) 303 .4)(0. (Revision: Sept-08) Using Terzaghi’s method.23)(17.2)(16.5)(4.115)(2. N q' = 6.5) + (0.8) + 0.30)(31. qult = 1.4(0.2 ksf ' ' ' ' qult 3 3 2 2 and ϕ ' = (ϕ ) = (28°) = 18.5)(15.4γ BNγ For φ = 28° N c = 31.30) = 0.6) + (0. the cohesion and angle of internal friction are reduced by two-thirds.115)(2.115)(2) = 0.1 ksf versus qult −local failure = 6. qult = 1.1 ksf To find the value of the bearing capacity of a local shear failure.3c ' N c + qN q + 0.*Bearing–05: General versus local bearing capacity failures.0) = 18.4γ BNγ where c ' = c = (0.7° which give N c' = 16.52 3 3 ∴ qult −local = (1.6.5 and Nγ' = 4.23 ksf Therefore.52) = 6.3(0. 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.2 ksf ' qult − general failure = 18.2.3) (0.8.3c ' N C + qN q + 0.0 and q = γ D f = (0.23)(6.2) + (0. 2 2 −local = 1. N q = 17. 5 )(158.0 m ) (1) The Hansen formula predicts an ultimate bearing capacity of. φtriaxial = 42° and γ’ = 9. qult = 0 + qN q Fqs Fqd + 0.9)(1. BL ( 0.0) ∴ qult = 1.5 ⎞ Fqs = 1 + ⎜ ⎟ tan ϕ = 1 + ⎜ ⎟ tan 46 ° = 1.5 and N γ = 244. 863 kPa ( Meyerhof underestimates by 4 % ) . c = 0.0 m.73)(0. qult = 0 + qN q Fqs Fqd + 0. (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. L = 2.5)(159)(1. 863 kPa measured ( Hansen underestimates by 20%) (2) The Meyerhof formula with φ = 46º.5 ⎞ Fγ s = 1 − 0.31 kN/m3 (the WT is at the surface).31) ( 0.*Bearing–06: Comparing the Hansen and Meyerhof bearing capacities.5 ⎞ Fqd = 1 + 2 tan ϕ (1 − sin ϕ ) 2 ⎜ f ⎟ = 1 + 2 tan 46°(1 − sin 46°) 2 ⎜ ⎟ = 1.5)(9. 863 kN qult = = = 1. 782 kPa versus 1.27) (1. 304 .16) + (0. Given B = 0.5γ BN γ Fγ s Fγ d Lee ' s adjustment formula is φ ps = 1.5 ⎠ Fγ d = 1.863 kPa was the field measured failure load.5 m )( 2.5 and N γ = 328.863 kN.5 ) (1.0) ∴ qult = 1.5 m.7 3.4 ⎜ ⎟ = 1 − 0.863 kN ▼ WT Df = 0.31)(0.5φtriaxial − 17 ° = 1.65 ⎛B⎞ ⎛ 0.16 ⎝ B ⎠ ⎝ 0. N q = 158.5)(245)(0.5 m Solution: Pult 1.26 ⎝L⎠ ⎝ 2 ⎠ ⎛B⎞ ⎛ 0.5γ BN γ Fγ s Fγ d ∴ qult = 0 + (9.9 ⎝L⎠ ⎝ 2 ⎠ ⎛D ⎞ ⎛ 0.5 ( 42 ° ) − 17° = 46° For ϕ = 46 °.0 ∴ qult = 0 + (9. 485 kPa versus 1.5)(9.27)(1. Pult = 1. N q = 158.31)(0.9)(1.5)(328.16 ) + (0.31)(0.5 m B = 0.4 ⎜ ⎟ = 0. 10 G sγ w G sγ w 101.7 Fqs = Fγ s = 1 + (0 .61 = 0.8 pcf and V s = = 1 + wN 1. q ult = c ' N c ( Fcs Fcd Fci ) + qN q ( Fqs Fqd Fqi ) + 0.2) ⎜ f ⎟ K p = 1 + (0.7 305 .46.68.1) ⎜ ⎟ K p = 1 + (0. and (b) if the water table rises to the ground surface? The soil has a unit weight of 112 pcf.8 set V = 1 ft 3 ∴ V s = = 0.2) ⎜ ⎟ K p = 1 + (0.2) ⎜ ⎟ (2.49 ⎝L⎠ ⎝ 5.39) ( 62.1) ⎜ f ⎟ K p = 1 + (0. a cohesion cu = 240 psf and a specific gravity of solids of Gs = 2.7 feet with M eyerhof's equation.7 ⎠ ⎛D ⎞ ⎛ 4 ⎞ Fqd = Fγ d = 1 + (0.2) ⎜ ⎟ 2.46) = 1.46 = 1.61 ft 3 ∴ V v = V − V s = 1 − 0.2 pcf ⎝V ⎠ and γ ' = γ sat − γ w = 126.7 ⎠ ⎛D ⎞ ⎛ 4 ⎞ Fcd = 1 + (0.4 ) = 126. Solution: (a) Find γ SAT to determine γ '.*Bearing–07: Increase a footing’s width if the WT is expected to rise.4 = 63. Fqi and Fγ i = 1 ⎛ ϕ⎞ ⎛ 25 ° ⎞ For ϕ ≥ 10 ° K p = tan 2 ⎜ 45 ° + ⎟ = tan 2 ⎜ 45 ° + ⎟ = 2.1)( )(2.46) = 1.8 + (0. a moisture of 10%.7 ⎞ Fcs = 1 + (0. γ 112 Ws γ dry γ dry = = = 101.8 pcf T ry B = 5.46 = 1. therefore ⎝ 2⎠ ⎝ 2 ⎠ ⎛B⎞ ⎛ 5.1) ⎜ ⎟ 2.22 ⎝ B ⎠ ⎝ 5.68 ( 62.5γ BN γ ( Fγ s Fγ d Fγ i ) w here the load inclination factors Fci .11 ⎝ B ⎠ ⎝ 5.2 − 62.25 ⎝L⎠ 5.4 ) ⎛V ⎞ but γ sat = γ dry + nγ w = γ dry + ⎜ v ⎟ γ w ∴ γ sat = 101. (Revision: Sept-08) Use Meyerhof’s bearing capacity formula (with a factor of safety = 3) to select a footing’s width B if.39 ft 3 2.7 ⎠ ⎛B⎞ 5. (a) the water table is as shown below. φ = 25º. 46) = 1.09 )(1.5γ BN γ ( Fγ s Fγ d Fγ i ) qult = (0.7 ft FS 3 qall 6.5)(0.25 ) qult = 16.49 ⎝L⎠ ⎝7⎠ Fcs = 1.25)(1.2 ⎜ ⎟ (2.49 )(1.7)(1.11)(1) + (0.22)(1) + (0. 306 .18 ) + (0.54 Iterate once more. and N γ = 6.67 )(1.0 feet.54 ksf and B 2 = = = 36.77 qult = c ' N c ( Fcs Fcd Fci ) + qN q ( Fqs Fqd Fqi ) + 0.49 same as above ⎛D⎞ ⎛4⎞ Fqd = Fγ d = 1 + 0.09 ) + (0.7)(1.2 ⎜ ⎟ K p = 1 + 0.6 ksf qult 18.7 ft was a good choice.67)(1.7.01 ft FS 3 qall 5.2 ksf therefore B 2 = = = 32. (b) When the water table rises to the ground surface.7) (1.09 ⎝B⎠ ⎝7⎠ Fqs = Fγ s = 1.25)(1.1⎜ ⎟ K p = 1 + 0.1 ft 2 ∴ B = 6.1⎜ ⎟ 2.2 Therefore the choice of B = 5.62 Q 200 qall = = = 5.062) ( 4 )(10.25 )(1.The Meyerhof bearing capacity factors for φ = 25° are N c = 20.5)(0.112)(4)(10. need a larger footing. ⎛B⎞ ⎛7⎞ Fcd = 1 + 0.24)(20.46 = 1.25 ft 2 ∴ B = 5.62 ksf qult 16.7.7)(6.112)(5.49)(1.11)(1) qult = 18. try B = 7.5 feet.6 Q 200 qall = = = 6.7 )(1. N q = 10. and find B = 7.25 same as above qult = (0.24)(20.062)(7) ( 6. 5 γ dry = = = 16.6 3 2 ⎥+⎢ ⎢⎣ ( 2.8 ) = 20.37) ( 9.10 m G s γ wet (2.10 kN γ dry 16.1 m3 ⎡ 0.5 3 and Vs = = = 0.4 ) ⎥⎦ 2 2 m Using Hansen’s method with φ = 35º.85 (18.68) ( 9. The average effective weight γ e of the soil can also be given by the formula: dw γ' γ e = (2H − dw ) γ wet + 2 ( H − d w ) 2 2 H H ⎛ ϕ⎞ ⎛ 35 ° ⎞ where H = (0.8 ⎤ kN ∴ γ e = ( (2)(2.5) tan ⎜ 45 ° + ⎟ = 2.40 − 0.5 + (0.3 and Nγ = 33.4 ) ⎥⎦ ⎢⎣ ( 2.63 = 0.40 − 0.0 − V s = 1 − 0. 307 .**Bearing–08: The effect of the WT upon the bearing capacity.10 ) ⎤ ⎡ 20.85 ) = 12.5)(2.85 m Set the total volume V = 1 m 3 γ wet 18. the bearing capacity factors are Nq = 33.37 m 3 and γ sat = γ dry + nγ wet = 16.63 m 3 1+ w 1 + 0. 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. (Revision: Sept-08) Using the Hansen method.40 m ⎝ 2⎠ ⎝ 2 ⎠ and d w = depth to the WT below the footing invert = 0.1 − 9.85) ) ⎢ ⎥ ( 2.8 ) kN Vv = 1.92.5) B tan ⎜ 45 ° + ⎟ = (0. 4 9 7 q a ll = = 7 4 9 kP a 2 308 .1 0 ⎝ B ⎠ ⎝ 2 .7 0 L ⎝ 2 .5 Fγ d = 1 .5 γ e B N γ ( Fγ s Fγ d ) q u lt = (1 8 . B ⎛ 2 .5 ⎞ Fq s = 1 + tan φ = 1 + ⎜ ⎟ tan 3 5 ° = 1 .5 ⎠ B 2 . 4 9 7 kP a 1.6 )(1 ) q u lt = 1.0 )( 3 3 )(1 .4 ( ) = 0 .4 = 1 − 0 . th e u ltim ate an d allo w ab le b earin g cap acities are.1) (1 .0 T h erefo re.5 Fγ s = 1 − 0 .6 )( 2 .5) (1 2 .7 0 )(1 .5 ⎠ 2 ⎛ D ⎞ 2 ⎛ 1 ⎞ Fq d = 1 + 2 tan φ (1 − sin φ ) ⎜ f ⎟ = 1 + 2 tan 3 5 o (1 − sin 3 5 o ) ⎜ ⎟ = 1 .5 )( 3 4 )( 0 .6 L 2 .1 0 ) + (0 . q u lt = 0 + q N q ( F q s F q d ) + 0 . 1 kN/m3 0. For φ = 32° . qult = cN c Fcs Fcd + qN q Fqs Fqd + 0.20 / 35.292 ⎢⎣ N q tan φ ⎥⎦ ⎣⎝ 23.62 = 1.273 ⎞ ⎤ Fcd = Fqd − ⎢ ⎥ = 1.22)(1) = 1.5γ BN y Fys Fyd The inclination factors Fci .50 ) = 1.61 m φ =32° 0. N c = 35.65 ⎝ Nc ⎠ Fqs = 1 + tan φ = 1 + 0. Fqi .20 × 0.273 − ⎢⎜ ⎟ ⎥ = 1.49.62 ⎛B⎞ Fys = 1 − 0.22 m Solution: The Hansen formula for a footing is.62 ⎠ ⎦ 309 .*Bearing–09: Finding the gross load capacity. and Fγ i are all equal to 1 because the load is vertical.60 ⎝L⎠ 2⎛ D ⎞ Fqd = 1 + 2 tan φ (1 − sin φ ) ⎜ f ⎟ = 1 + (2)(0.18 and N γ = 20.79 and B / L = 1 ⎛N ⎞ Fcs = 1 + ⎜ q ⎟ = 1 + ( 23.61 m γ t = 21 07 kN/m3 1. N γ =18.4 ⎜ ⎟ = 1 − 0.273 for D f / B ≤ 1 ⎝ B ⎠ Fyd = 1 ⎡ (1 − Fqd ) ⎤ ⎡⎛ 1 − 1. (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 q = 23.62)(0.4 = 0. 07 − 9. the total gross load N is.81) = 17.81)(1.62)(1.22m) 2 = 487 kN 310 . N = qall B 2 = (327 kN / m 2 )(1.61m ) (18.5 )( 0.The WT is located above the footing.9 kN / m 2 ∴ qult = (17.22 )( 20.1 kN / m3 ) + ( 0. therefore.273)(23.07 − 9.20) + ( 0. ⎛ q ⎞ ⎛ 981kN / m ⎞ 2 qall = ⎜ ult ⎟ = ⎜ ⎟ = 327 kN / m 2 ⎝ 3 ⎠ ⎝ 3 ⎠ Hence. q = ( 0.9)(1.6 )( 21.8 )(1) = 981 kN / m 2 Therefore.61m )( 21. 47 )( 2.4) = 4. q ult = 0 + qN q Fqs Fqd + 0.2(0. Qult = q ult ( B ′L ′ ) = ( 8.8 ⎝ L′ ⎠ ⎝ 4.73 kips 311 .58 ) = 1.4 ⎜ ⎟ = 1 − 0. N q = 18.2 ⎠ 2 ⎛ 2 ⎞ Fqd = 1 + 2 ( tan 30 ° )(1 − sin 30 ° ) ⎜ ⎟ = 1.**Bearing–10: The effect of an eccentric load upon bearing capacity. Determine the gross ultimate load Qult applied eccentrically upon the footing.5 feet.1 ⎞ F ys = 1 − 0. given that γ = 115 pcf.4) (1.2 ( 0. (Revision: Sept-08) A rectangular footing measures 5 feet by 2.2 ) = 74.275 ) + (0.2 e x = ( 2.11 5)(2.67 ⎛ B' ⎞ ⎛ 2.67) ( 0.2 ⎠ F yd = 1 q ult = ( 2)(0.275 ⎝ 2.29 ⎝ L ⎠ ⎝ 4.1)(4.5γ B ′N γ F ys Fγ d For φ = 30 o .2 e y = ( 5 ) .4 and N γ = 15.1 ⎞ Fqs = 1 + ⎜ ' ⎟ tan φ = 1 + ⎜ ⎟ ( 0.47 ksf Henc e .5 ) . and the ultimate bearing capacity of the soil qult. c = 0 and φ = 30°.5)(0. M eyerhof's ultimate bearing capacity formula with c = 0 is.8 )(1 ) = 8.29 )(1.1)(15.1 ⎠ ⎛ B′ ⎞ ⎛ 2.1 ft and the effective length L ' = L .2 ft. Solution: The effective width footing width B ' = B .115 ) (18.4 ⎜ ⎟ = 0.2 ) = 2. 4° = 44.34 ⎝7⎠ Since φ > 10° .5γ BN γ S γ 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 312 .73 ⎝ Lr ⎠ 7. Hansen’s qult = 0.8 ) tan 36° = 40. and the WT was not found.85 ) = 1. The footing depth Df = 6.1 ⎛B ⎞ 7 ∴ Sc = 1 + 0.6 ' = 7.5 feet Q 400 Q 400 ∴ Br = B − 2ey = 8'− 1' = 7 feet and Lr = L − 2ex = 8'− 0.1φtr = 1.**Bearing–11: The effect of an inclined load upon the bearing capacity.4 ⎛D ⎞ and d c = 1 + 0.7° ) = 36° ⎛ 36° ⎞ N q = eπ tan 36° tan 2 ⎜ 45° + ⎟ = 37.4φ ) = ( 36. Un-drained triaxial tests (the soil is not saturated) gave φ = 33º and c = 200 psf.6 Nγ = ( N q − 1) tan (1.8 ⎝ 2 ⎠ Nc = (N q − 1) cot φ = ( 36. 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: My 120 ft − k M x 200 ft − k Eccentricities ex = = = 0. My = 120 ft- kips.85 ⎜ ⎟ = 1.2 ⎠ ( )⎛6⎞ 3.8 ) tan 50.2 K p ⎜ r ⎟ = 1 + 0.5 ( 36.4 feet (ie.0 and d q = dγ = 1. the soil unit weight is 115 pcf. Lr > Br) Adjusting the φ from triaxial (φtr ) to a plane-strain value (φ ps ) via Lee’s formulation. S q = Sγ ≅ 1.0 feet.2 ( 3. φ ps ≅ 1.8 ) cot 36° = 50.3 feet and ey = = = 0.4 Nγ = 1.1 ( 32.2 K p ⎜ ⎝ Br ⎟ = 1 + 0. (Revision: Sept-08) A square 8’ x 8’ footing is loaded with an axial load of 400 kips and Mx = 200 ft-kips.0 .5 ( N q − 1) tan φ = 1. 5 + 26.115 )( 7 )( 40.75 ) = 851 kips Qall ⎛ 851 ⎞ qall = =⎜ ⎟ = 13.Also i = g = b = 1.5 ( 0.200 )( 50.9 ksf FS 3 1 1 ⎛ e ⎞2 ⎛ 0.5 ⎞ 2 Rey = 1− ⎜ ⎟ = 1− ⎜ ⎟ = 0.0 for this problem.1)(1) + ( 0.81)( 0.73 )(1.6 )(1.34 ) + ( 0.3 ksf B 2 ⎝ 64 ⎠ ⎛ 400 ⎞ (The contact load qo = 13 ⎜ ⎟ = 6.8 )(1) = qult = 16.7 q all = = = 21.9 ( 8 x8 )( 0.115 )( 6 )( 37.3 ⎞ 2 R ex = 1− ⎜ x ⎟ = 1− ⎜ ⎟ = 0.1 = 65.7 ksf qult 65.81 ⎝ B ⎠ ⎝ 8 ⎠ 1 1 ⎛ ey ⎞ 2 ⎛ 0. 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 = 0.75 ⎝ B ⎠ ⎝ 8 ⎠ Qall = qall ( B 2 ) ( R e x ) ( R e y ) = 21.1 + 23.1 ksf ) ⎝ 851 ⎠ 313 . (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. 270 4. 3. 74 Solution: It is presumed that all the building’s footings will be placed at roughly elevation 284 or thereabouts.4 Elevation 296.6 292. Meyerhof has proposed formulas for the allowable bearing capacity adjusted so that the settlement is limited to 1-inch.**Bearing-12: Interpretation of borings to estimate a bearing capacity.trace of coarse sand Boring No. Figures to the right of each boring log indicate the numbeer of blows required 69 to drive the 2-in. These formulas are: N qall = ( K D ) for B ≤ 4 ft 4 314 .8 13 Firmer 4 14 290 Elevation 288.No water encountered in any of the borings.6 295 Elevation Got 295. boring No.0 Firmer 7 Elevation Elevation 5 Got 290. -OD split spoon. This is fine for the building area covered by borings # 3. State your reasons. Topsoil Fine brown silty sand . Borings were made using standard procedures with 2-in.small gravel Fine to medium brown silty sand -some small to medium gravel Brown silty clay Fine brown silty sand .0 Dark 25 6 7 brown 71 29 Got Sandy 10 Firmer Hard 8 16 46 285 Got 71 Got Firmer Got Firmer 25 51 27 Firmer 47 Hard Got Firmer 69 280 22 36 6 in. using a 140 lb weight falling 30 in. 38 boulder 38 62 Cohesive 34 Got 34 67 firmer 39 275 32 Notes: 1.3 Boring No.-OD split spoon 12 in. All elevations are in accordance with plot furnished by architect.2 Boring No. 4 and 5 because they have good SPT values. Hard 2.5 Boring No. 2? The building is a four-story (five on the low side) office building with column loads around 160 kips. 5 feet is excessive since 6 ⎝ 45 ⎠ Q 160kips q0 = = = 7.06 ≅ 13 ksf ≤ 14 ksf OK B 2 (3.9 ksf This suggests that a B = 4. and use formula N 56 ⎛ 0. That places the shallow foundation at elevation 274 ft.64ksf < 13 ksf NOT GOOD ⎛Q⎞ Let’s use B= 3.33 32 ⎛ 0.33Df /B ≤ 1.5 + 1 ⎞ qall = ⎜ ⎟ (1) = 13.25sf Assume B < 4 ft. they will be deeper by 1-story (ie.64 ksf ⎝B ⎠ Use B = 3.5 ⎞ Kd = ⎜1 + ⎟ = 1. This area will have bearing in the same strata. 2 N ⎛ B +1⎞ qall = ⎜ ⎟ ( K D ) for B> 4ft 6⎝ B ⎠ ⎛ Df ⎞ where K D = 1 + 0. and 5) 3 Let’s assume B=4.33 4 4 ⎝ 3.33 x 4.33Df ⎞ qall= ( KD ) kd =1+ 0. 315 .33 56 (#3. 4.90 feet. for 5-story building). say B~ 3.50’ and Df = 4.5 ft .90 feet qall = 10.50 ⎠ qall =10.5 ft and Df =0 2 56 ⎛ 4.5’ qall= N kd kd =1+0.9 qall = 13.33Df /B ⎜1 + ⎟ and Df = 0 qall = 14 ksf 4 4 ⎝ B ⎠ Q 160 kips ∴ qo = = = 13.33 ⎝ B ⎠ 47 + 51 + 71 For the silty sand use N= = 56.64 ksf q0 = ⎜ 2 ⎟ = 10.5) 2 For footings in area of borings # 1 and #2. N= 32 and using B= 3.9 ksf B2 20.51 ksf ≤ 10.33 ⎜ ⎟ ≤ 1. Chapter 18 Shallow Foundations Symbols for Shallow Foundations 316 . Properties of Reinforcing Steel (British and SI units). 317 . (Revision: Sept-08) Design a square reinforced concrete footing for a column 15”x15” with 4 # 8 rebars.5 ksf < 10 ksf for qult …GOOD ( 7. 15” B = 7.5 feet. D L = 100 kips f c' = 3.*Footings–01: Analyze a simple square footing.2 DL + 1. 000 psi q allowable = 4 ksf from qult = 10 ksf and FS = 2. 000 psi = 164 psi ≈ 23.5 feet 15” + d 318 . QULT = 1.42 feet therefore use B = 7.6 (120 ) = 120 + 192 = 312 kips QU 312 ∴ qO = = = 5.2 (100 ) + 1. qa 4 ksf 2) Check ultimate parameters: that is the actual soil pressure qo under Qult. the effective depth. and.7 ksf 4) Find d. vC = vall = 4 ( 0.5 ft ) 2 2 B 3) Compute the allowable concrete shear strength vc ( allowable ) vall = 4φ f 'C where ϕ = 0.5 Solution: 1) Footing size for service loads: Q 220 kips B= = = 7.6 LL = 1.75 for shear and torsion . 000 psi LL = 120 kips f y = 60. (in this case two-way shear governs) in feet.75 ) 3. 25d − 3. round-out to d = 15 inches.5 ⎞⎛ 15 ⎞ ⎡ 2 ⎛ 15 ⎞ ⎤ ⎛ 5.22 ft Which yield to two solutions for d = and using the modified formula equation 2a: −2.26 = 0 which yields d = +0.5 ⎞ ⎛ d ⎜ 23.5 ft 15” 1 ft w L q0 319 .25d − 3.0 = 0 +1.5 − ⎜ ⎟ ⎥ ⎜ ⎟=0 ⎝ 4 ⎠ ⎝ 4 ⎠⎝ 12 ⎠ ⎣⎢ ⎝ 12 ⎠ ⎦⎥ ⎝ 4 ⎠ d 2 + 1. 5) Compute the area of steel AS for flexure. ⎛ BLqo ⎞ ⎛ 15 15 ⎞ ⎡ ( 7.7 + 2 ⎟ + d ⎜ 23.6 inches.5 − ⎜ 12 ⎟ ⎟ ⎛ B−w⎞ ⎝ ⎠⎠ Unit strip of the cantilever arm = ⎜ ⎟=L ∴ L=⎝ = 3.5 )2 ( 5.7 + ⎟⎜ ⎟ − ⎢7. ⎛ q ⎞ ⎛ qo ⎞ 2 ⎛ qo ⎞ d 2 ⎜ vc + o ⎟ + d ⎜ vc + ⎟ w − ( B − w ) ⎜ ⎟ = 0 2 ⎝ 4 ⎠ ⎝ 4⎠ ⎝ 4⎠ 5. ⎛ ⎛ 15 ⎞ ⎞ ⎜ 7.7 ⎥⎦ d 2 + 1. the formula is.39 ft Use the largest d = 1.5 ) ⎤ 4d + 2 ( b + c ) d − ⎜ 2 ⎟ = 4d + 2 ⎜ + ⎟ d − ⎢ 2 ⎥=0 ⎝ vc ⎠ ⎝ 12 12 ⎠ ⎢⎣ 23. It is not necessary to check for wide-beam shear on a square footing.22 feet = 14.47 ft When the column has a rectangular area bxc.13 ft ⎝ 2 ⎠ 2 B = 7.5 ⎞ 2 ⎛ 5. 5 ft (0. AS = 7.0018bh or fy 200 ρ min = = 0.99 = 0 As2 − 15.0023)(12)(15) = 3.11 = 0 AS = 0. The cantilever moment: ⎡⎛ k ⎞ ⎤ ( ) ( ) 2 ⎢ ⎜ 5.59 in2 per foot of footing For B = 90” (7. 320 .0018 fy Therefore.13 ft 12 in ⎥ ⎛ q o L2 ⎞ ⎣ ⎝ ft ⎠ ⎦ = 359 in − kips MU =⎜ ⎟= ⎝ 2 ⎠ 2 ⎛d −a⎞ M U = φ AS f y ⎜ ⎟ where φ = 0. AS 0.08 in2 (Check ACI 10.9 ) AS f y ⎜ ⎟ ∴ = AS ⎜ ⎟ ⎝ 2 ⎠ ( 0.12 for temperature and shrinkage.3).5.3 As + 6.41 ρ= = = 0.0018 bd 12 (15 ) 3 f c' ρ min = = 0.41 in2/ft) = 3.9 )( 60 ksi ) ⎝ 2 ⎠ 15 AS − 0.0023 > 0.5 (0.9 for tension ⎝ 2 ⎠ AS f y AS ( 60 ) a= = = 1.9 ) f y ⎝ 2 ⎠ 359 in − kip ⎛ A ⎞ = AS ⎜ 15 in − 1.18 in2 ) @ 12 inches on center or 5 # 8 bars ( AS = 3.1) for minimum steel and ACI 7.5 2 ⎟ 3. and the embedment length of the dowels.85 f 'C b 0.85 ( 3 )(12 in ) Substituting a into the M U equation above.1 in2 or 0.96 S ⎟ ( 0.5’) use 6 # 7 bars ( AS = 3.96 AS 0.41 in 2 per foot of footing The total steel required across the footing is AS = 7.95 in2 ) @ 12 inches on center Check for Development length Ld (ACI-318-08.98 AS 2 − 5. ⎛d −a⎞ MU ⎛d −a⎞ M U = ( 0.12. 321 . 02 ⎜ ⎟ = 4.5) = 6 ksf. M P = 165 kips P M = 20 ft-kips DL = 100 kips LL = 65 kips d 3’-d 15” T 7 feet d) Determine the flexural moment for a strip of footing 1 foot wide.6(65) = 224 kips ⎛Q ⎞ ⎛ 224 ⎞ ⎛Q ⎞ ⎛ 224 ⎞ qall = qmax ⎜ u ⎟ = 3. GOOD.121 ft P 165 ⎛ P ⎞⎛ 6 e ⎞ ⎛ 165 ⎞ ⎛ 6 ( 0.02 ksf b) Check bearing pressure at ultimate load: Qu = 1.72 ksf and q min = 3.72 ⎜ ⎟ = 5.1 ksf ⎝ P ⎠ ⎝ 165 ⎠ ⎝ P ⎠ ⎝ 165 ⎠ Then consider qall = 5.2(100) + 1. b) Check the bearing pressure at ultimate load. c) Calculate the wide beam shear. 322 .121' ) ⎞ q = ⎜ ⎟ ⎜1 ± ⎟=⎜ ⎟ ⎜1 ± ⎟= ⎝ A ⎠⎝ B ⎠ ⎝ 7 ' x7 ' ⎠ ⎝ 7' ⎠ q max = 3.*Footings–02: Add a moment to the load on a footing.05 ksf and qall = qmin ⎜ u ⎟ = 3. qall = 4 ksf ( with a FS = 1. a) Compute the load’s eccentricity e. (Revision: Sept-08) The footing shown below is a square footing with the dimensions and loads as shown.5) . P = DL + LL Solution: M 20 a) Compute the load eccentricity: e = = = 0. f 'C = 3 ksi . f y = 60 ksi .1 ksf < qall (FS) = 4 (1. 1 c) Calculate the allowable one-way (wide beam) shear.3 inches ≈ 12 in assumed 2ϕ f c' b (0. 5.75)(110)(12) V U 3’ - d 4.1 ksf d) Determine the flexural moment for a strip 1 foot wide: x 3 ⎛ 5 .9 x − 0 .2 kips Equation (1) 2 2 2 2 The allowable shear = 2φ f c' = 110 psi Vu (9. q = = 4.1 ksf 5. assume a d = 1 ft.1 5 7 x 2 ⎞ ft − k ip M = ∫ Vdx = 0 ∫ 0 ⎜ ⎝ 2 ⎟d x = 2 5 .6 2 B c 7 1 VU = q ( − T − ) = 4.2)(1000) Equation (2) The required d = = = 9.1 + 4.9 ⎠ ft 323 .6( − 1 − ) = 9. *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 7’ My x bd 3 ( 7 ' )( 7 ' ) 3 N Mc qo = ± where I= = = 200 ft 4 A I 12 12 qo = 165k ± ( 20k . ft )( 3.5 ft ) = ( 7 ' x7 ') 200 ft 4 qo max = 3.7 ksf and qo min = 3.0 ksf < 4 ksf allowable GOOD 324 My N x 3.0 ksf 3.7 ksf b) The ultimate load on the soil: Qu = 1.2 DL + 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 M u c 238 kips ( 32 ft .kips )( 3.5 ft ) qu = ± = ± = A I ( 7 ')( 7 ' ) 200 ft 4 qu max = 5.9 ksf and qu min = 4.7 ksf My Q x 4.7 ksf 5.9 ksf 325 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 My 7’ w =12” 12” Critical d 2’ Section Vu d From ∑f y = 0 , the ultimate shear is Vu = 82, 600 lbs , but Vu = φ 2 f 'c bd , Vu 82, 600 lbs d = = = 10.6 in < 12 in assumed. φ 2 f 'c b ( 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’ 326 ⎛ ⎛d ⎞⎞ Vu = φ 4 f 'c bo d , where bo is the perimeter of the critical section; bo = 4 ⎜ c + 2 ⎜ ⎟⎟ ⎝ ⎝2 ⎠⎠ ∴ d = Vu = ( 5.9 ksf ( 7 ft x 7 ft − 2 ft x 2 ft )1000 ) = 14.8 in assumed; φ 4 f 'c b ( 0.85 )( 4 ) ( 3000 ) ( 4 )( 24 in ) must increase d. Recalculate by trying d = 14.0 in. ∴ d = Vu = ( 5.9 ksf ( 7 ft x 7 ft − 2.17 ft x 2.17 ft )1000 ) = 13.5 in φ 4 f 'c b ( 0.85 )( 4 ) ( 3000 ) ( 4 )( 26 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 )( 7 ft )( 3 ft ) 2 Mu = s = = 186 kip . ft 2 2 As f y 0.85 f 'c Ba 0.85 ( 3000 )( 7 ft )( a ) a= ( where b = B) then As = = = 0.3a 0.85 f 'c b 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 2 ⎟⎜ 1442 2 ⎟⎜ 1.16 − ⎟ ft = ⎝ in ⎠ ⎝ ft ⎠ ⎝ 2⎠ a = 0.07 ft and ∴ As = 0.3 a = 0.021 As A 0.021 ft 2 Percentage of steel p= = s = = 0.00257 > pmin = 0.0018 bd Bd ( 7 ft )(1.167 ft ) 327 As min = 0.021 ft 2 = 3.024 in 2 Therefore, use 6 # 7 bars at 12 inches. ⎛ f ⎞ e) Check development length, Ld (ACI 12.2.2) Ld = 0.04 Ab ⎜ y ⎟ > 0.0004db f y ⎜ f' ⎟ ⎝ c ⎠ ⎛ 0.44 ( 60, 000 ) ⎞ Ld = 0.04 Ab ⎜ ⎟ > 0.0004 ( 0.75 )( 60, 000 ) ⎝ 3000 ⎠ Ld = 19.3 in > 18 in c ⎛ 12in ⎞ Actual Ld ( provided ) = B− − ( cover ) = ⎜ 7 ft ⎟ − 6 in − 3 in = 75 in 2 ⎝ 1 ft ⎠ f) Check bearing strength on concrete (ACI 318-02 10.15) ⎛ A ⎞ ⎛ A ⎞ Bearing strength = 0.85φ f 'c A g ⎜ 2 ⎟ where ⎜⎜ 2 ⎟⎟ ≤ 2 ⎜ A ⎟ ⎝ g ⎠ ⎝ Ag ⎠ Ag is the area of the column and A2 is the area of the footing. ⎛ A ⎞ ⎛ 3 kips ⎞ 0.85φ f 'c Ag ⎜ 2 ⎟ = 0.85 ( 0.70 ) ⎜ ⎟ (144 in ) ( 2 ) = 514 kips 2 Then ⎜A ⎟ ⎝ in 2 ⎠ ⎝ 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 AS = 0.20 in 2 ( 4 ) = 0.80 in 2 h) Final design 328 *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 ) B2 = = = 8 m2 ∴ B = 2.82 m qa ⎛ kN ⎞ ⎜ 200 2 ⎟ ⎝ m ⎠ M 800 e= = = 0.5 m ∴ L ≥ 6 ( 0.5 ) = 3 m N 1600 If L = 3 m, try for qmax ≅ 2qavg iterate by trying footing: 2.5 m x 4 m = B x L. N 1600 kN q avg = = = 160 kPa A 10 m 2 N 1600 ⎡ 6 ( 0.5 ) ⎤ q max = = ⎢1 + ⎥ = 280 kPa ⎛ 6e ⎞ 10 ⎣ 4 ⎦ BL ⎜ 1 + ⎟ ⎝ L ⎠ Note that qmax exceeds qavg by 33 %, therefore increase the area. Try using 2.75 m x 4.5 m dimensions; 329 N 1600 kN q avg = = = 130 kPa A 2.75 ( 4.5 ) m 2 N ⎡ 6 ( 0.5 ) ⎤ q max = = 130 ⎢1 + ⎥ = 217 kPa ⎛ 6e ⎞ ⎣ 4.5 ⎦ BL ⎜ 1 + ⎟ ⎝ L ⎠ Iterate again, with B = 3.0 m and L = 5.0 m N 1 6 0 0 kN q a vg = = = 1 0 7 kP a A 3 m (5 m ) N 1600 ⎡ 6 ( 0 .5 ) ⎤ q m ax = = ⎢1 + ⎥ = 1 7 1 kP a GOOD ⎛ 6e ⎞ 15 ⎣ 5 ⎦ BL ⎜1 + ⎟ ⎝ L ⎠ The footing dimensions are B = 3 m by L = 5 m. 330 5 m N 1600 Nu ⎛ 2240 ⎞ ⎡ 6 ( 0.*Footings–05: Design the steel for the previous problem.5 ) ⎤ qmin = =⎜ ⎟ ⎢1 − ⎥ = 60 kPa ⎛ 6e ⎞ ⎝ 15 ⎠ ⎣ 5 ⎦ A ⎜1 − ⎟ ⎝ L⎠ q max = 239 kPa and q min = 60 kPa 331 .2 ( 800 ) + 1. Solution. (Revision: Sept-08) Design the previous footing using f 'c = 21 MPa and f y = 415 MPa .6 ( 800 ) = 2240 kN M 800 e= = = 0. 1) Check the ultimate pressures: N u = 1.6 ( LL ) = 1.5 ) ⎤ qmax = =⎜ ⎟ ⎢1 + ⎥ = 239 kPa ⎛ 6e ⎞ ⎝ 15 ⎠ ⎣ 5 ⎦ A ⎜1 + ⎟ ⎝ L⎠ N ⎛ 2240 ⎞ ⎡ 6 ( 0.2 ( DL ) + 1. 5 m + 0.25 = 0 ∴ T ≅ 0.225 – d ) : dv = qdx 2.1 ( 2.60 m 2T 2T We will use the highest of (2a) or (2b).7 = 0 T = 0.2 x 2 ⎞ ⎫⎪ V = ∫ qdx = ∫ ( 266 − 40.8T − 24. therefore T = 0.25 − T x x ⎧⎪ ⎛ 40. for a f 'c = 21 MPa . Using the simplified equation for a square footing: 4T 2 + 2 ( 0.60 m. 332 . 2240 kN qavg = = 149 kPa < qall = 200 kPa Good 15 m 2 2a) Calculate footing depth T based on punching shear. and Vc = 1.5T − 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 = 2.2 x ) dx = ⎨ 266 x − ⎜ ⎟⎬ 0 0 ⎩⎪ ⎝ 2 ⎠ ⎭⎪ 0 V = 598 − 266T − 20.5 m ) T − (15x 149 ) =0 2240 T 2 + 0.29 MPa .25 − T ) = 2 Vc 1290 V = = ∴ T 2 + 40. 0012 = 12 m m 4) Check footing: cm2 1m 2 Longitudinal steel: 28.m ⎛ a⎞ ⎛ a ⎞ Mu M u = φ As f y ⎜ d − ⎟ ∴ As ⎜ d − ⎟ = ⎝ 2⎠ ⎝ 2 ⎠ φ fy ⎛ 23.002 bT 1 ( 0.25 2.m 2 2 2 ⎛ a⎞ M cm 2 and As ⎜ d − ⎟ = u ∴ As 2 − 0.2 x 2 ⎞ ⎡ 266 x 2 40.0515 As + 0.3 As 0.002 (1)( 0.3 ⎞ 597 As ⎜ 0. 0 0 0 c m Therefore use 17 # 25 mm bars at 17.85 ( 21)(1) 2.0515 As + 1.6 − ⎟= = As 2 − 0.002 ( b )(T ) = 0.37 x10 − 2 = 0 ⎝ 2 ⎠ 0.61 ⎝ 2 ⎠ φ fy m As 7.9 ( ) 415 cm 2 As = 28.6 cm2 2 = 8 .2 x 3 ⎤ Mu = ∫ Vdx = ∫ 0 0 ⎜ ⎝ 266 x − 2 ⎟ ⎠ dx = ⎢ ⎣ 2 − 6 ⎦0 ⎥ = 597 kN .5 ) ⎞ ⎞ 2 ⎡ ( B − 0.4 6 x 1 0 − 3 m m 1 0.6 ) 3) b) Find AS (transverse): Use the high average.37 kPa ⎛ ⎛ ( 3 − 0.000386 = 0 ∴ As = 7. As f y 415 As 3) a) Find AS ( Longitudinal ): a= = = 23.25 2.61 x 10 −4 m 2 Check for ρ= = = 0.0047 > 0.25 ⎛ 40.00282 Check p= = 0.6 ) m2 cm 2 ∴ Use minimum As = 0.2 (3 m) = 84.2 m As 0.6 cms o − c 333 .00126 < 0. q= ( q ave + q max ) = (149 + 266 ) = 20.6 ) = 0.002 GOOD bd 1 ( 0.85 f 'c b 0.5 ) ⎤ 2 q⎢ ⎜ 215 ⎜ ⎟ ⎟⎟ ⎥ ⎜ 2 wl 2 ⎣ 2 ⎦ =⎝ ⎝ ⎠ ⎠ Mu = = = 168 kN . Sketch: 334 . 75” 12 in T=12 in B 335 . the minimum cover is 3” from the steel to the footing invert. (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 7.0. d = 12” .3” . Therefore. *Footings–06: Design a continuous footing for a pre-cast warehouse wall.2. Assume # 4 bars.7.5” / 2 = 8. Ultimate load.52 ksf Estimate B = Q/ qo = 4.15 k/ ft3 ) – ( 3 ft )( 0.0 + 2.76 ft.6 ( 1.33 = 1.0 kips/ ( 3 ft )( 1 ft ) = 2.0 ksf < qult = 4 ksf Æ GOOD 2) Check the shear strength of the footing: The critical section for shear section occurs at a distance d from the face of the wall.110 k/ ft3 ) = 2 – 0. assume B = 3 ft. Therefore. b =12 in a d T B Find the ultimate soil pressure qult.d in d Vu T qo = 2.2 ) = 4.1 ).1 ksf 336 . qo = 2.2 k/ 1.11.6 LL = 1.0 = 6. (ACI 15. qo = qall – qconcrete – qsoil-above footing qo = 2 ksf – (1 ft )( 0.1 ksf B = 3 ft Vu = (12 − d ) qo (1) 12” 12” b =12” 12 in .15 – 0.2 ( 3 ) + 1.2 DL +1.0 kips Therefore. per unit length.5 & ACI 11. the soil pressure at ultimate loads qu is: qu = U / ( B )( 1 ) = 6. U = 1.52 ksf = 2. The actual soil pressure from the structure is. 85 ⋅ 2 ⋅ 3000 ⋅ 12in ⋅ 6.05 kip-ft / ft of wall But.6 in2 per ft.5in = 7.1 k/ ft2 = 0.3 kips / ft of wall → GOOD Therefore. a = Asfy / [ 0.8 kips / ft of wall >> 0. Mu = Ø Mn = 0.05 = 0 Two possible answers: As ( 1 ) = 6. Therefore. 1.4 Rechecking.8 As ) 53 As2 – 351As – 1.5 / 2 = 6. As ( ACI 15.75in = 9.12.96 As (inches) and. the total thickness T = d + bar diameter + 3” = 10” + 1” + 3” = 14” 3) Design the flexural reinforcement.85 ⋅ 2 ⋅ 3000 ⋅ 12in ⋅ 8. d = 10” – 3” – 0. of wall As ( 2 ) = 0.0 ft to say 0. we can reduce the thickness of the footing T from 1.85 f’c ( b ) ] = [ (60 ksi ) As ) ] / [ 0.4): Mu = qo ℓ2/ 2 where ℓ = 12” Mu = (2. of wall The percentages of steel with As ( 1 ) & As ( 2 ) ( ACI 7.75”) ( 1 ft / 12 in ) x 2.05 kips-in / in = 0. Mn = Asfy ( d – a/2) = As ( 60 k / in2 ) ( d – a / 2 ) = 60 As ( 6.1 ksf x 1 ft2 )/ 2 = 1. Vu = 0.2.57 kips / ft Since Vu << ØVc . The ultimate shear Vu = (12” – 8.9 Mn Therefore.85 ft (~10”).85 ( 3 ksi )( 12 in ) ] = 1.003 in2 per ft.5 in – 9.5 in – 9.57 kips / ft of wall The concrete shear strength must be: Vu ≤ φ Vc = φ 2 f' c bd Vu = 0.9 (60 k / in2 ) As ( 6.8 As ) But.5” > 6” = dmin from ACI 15.1 ) 337 . of wall ( As = 0. As = ρmin b d = ( 0.14 in2 per ft.85 ( 3 ) / 60 ) 0.20 in2 ) 4) Check the development length.5 in) = 0.75 ρb = 0. we are missing 3” on each side. Ld ( ACI 12.000 / ( 87.8” or 12” Æ Clearly.0018 Therefore.5 in ) = 0.000 Ld = 0.04 Ab (But not less than 0.021 therefore.6 in2 / (12 in )( 6.0004 < 0.75 ρb.2 ): fy 60. use ρmin = 0. ρ1 = As(1) / bd = 6.000 ) = 0.0018 minimum steel The maximum steel percentages allowed ρmax = 0.0004 d b f y ) = 0. Increase the footing width B by 6” to B = 3.021 ) = 0.0018 )( 12 in )( 6.5 in) = 0.085 > ρmax = 0.085 ρ2 = As(2) / bd = 0. of wall Æ use 1 # 4 every ft.85 ( 87.000 + fy ) ) = ( 0.5 ft. and the design could be further optimized.016 Note that ρ1 = 0.) 338 .003 in2 / (12 in )( 6. ρmax = 0. where ρb = ( 0. reduces qo.000 / ( 87.016 therefore.000 + 60.20 ⋅ = 8.75 ( 0. 12” controls.8" f' c 3000 Ld = 8.04 ⋅ 0. Presently we have 12” – 3” cover = 9” < 12”.85 f’c / fy ) β ( 87. Therefore. (Note that increasing B. 5 in ) = 0. ( As = 0. to offset shrinkage and temperature effects ( ACI 7.0018 ( 42 in )( 6.49 in2 Provide 3 # 4 bars at 12” o.5 ft T = 13” As = 1 # 4 @ 12” along the wall Use minimum steel in longitudinal direction.c.12 ): As = ( 0. B = 3.0018 )( b )( d ) = 0.Therefore.60 in2 ) 339 . 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 7-02.26 Cf = 1. qz = 0. assuming an Exposure C. (Revised Oct-09) Design a spread footing for the billboard sign shown below using FBC-2004 and ASCE 7.2 M 32 The sign shape factor is = = 1. Category I.2) (32 ft x 20 ft) Therefore: F= = 32.98 I = 1.4 kips 1000 340 . and the water table.02.00256 Kz (IV)2 Kz = 0.**Footings–07: Design the footings of a large billboard sign.05 F = qz Gh Cf Af ∴ qz = 52 psf V = 146 mph Gh = 1.26) (1. Ignore the torsion and the wind load on the column.6 N 20 (34 psf) (1. 24 Fdγ = (depth factor for width) = 1.4 (Df / B) = 1.150 ksf q = γDf = (embedment pressure) = (0.19 Fdc = (depth factor for cohesion) = 1.39 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) = 6.867 Fdq = (depth factor for embedment) = 1 + 2 tan φ(1 – sin φ)2 (Df/B) = 1.5 (Nq – 1)tan φ = 2.4 (B/L) = 0.14 Fsγ = (shape factor for width) = 1.130 ksf)(3 ft) = 0.0) c Fiq = [ 1 – (0.7 H) / (V +Af Ca cot φ)]α qult = c’ Nc Fsc Fdc Fic +⎯q Nq Fsq Fdq Fiq + 0.4 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. ⎜ ⎟ ⎝ (Af Ca) ⎠ where ca = (0.0 + (Nq / Nc) (B/L) = 1.95 Fsq = (shape factor for embedment) = 1.40 Nc (factor for cohesion at φ = 20°) = (Nq – 1)cot φ = 14.6 to 1.0 ⎛ 1-H ⎞ Fic = (inclination factor) = 0. .0 + (B/L) sin φ = 1.5 H) / (V +Af Ca cot φ)]d where 2 ≤ d ≤ 5 Fiγ = [1 – (0.11 Fsc = (shape factor for cohesion) = 1.0 + 0.5 .49 Mx = 10 kips x 15’ = 150 k-ft My = 32.83 Nγ (factor for width at φ = 20°) = 1. Step #2: Calculate loads on footing Weight of steel column = γs L A = 0.4 k x 30’ = 972 k-ft Mz = 32.0 – 0. c (cohesion) = 0.5 γ B’ Nγ Fsγ Fdγ Fiγ 341 . 2) = 0.2 sin 20 = 1.275 x 972 kip-ft = 1239 kip-ft ⎛ 336 ⎞ ⎛ 6 x 336 x 0.4 >> 1.09 qmax.33) + 0. = 5750 / 1069 = 5.05 x 240 Kips = 252 kips Mu-x = 1.4 kip per feet Punching shear will not govern by observation.min = P ⎛ . 342 .0 + (0.625) ⎞ ⎛ .89 ksf < 2.92 DQ = 1 + (0.6P e ⎞ + ⎜ 2 y⎟ + ⎜ ⎛ .989 – 0.625' P 240 ⎛ 240kips ⎞ ⎛ .4)(3/10) = 1.6L + 1.0 + (0.431 x 0.min = ⎜ ⎟+⎜ ⎟ +⎜ ⎟ = 0.05 x 150 kip-ft = 158 kip-ft Mu-y = 1.MIN = ⎜ ⎟±⎜ ⎟±⎜ ⎟ = 0.0 – 0.33)(0.019) = 14.5 ksf GOOD ⎝ 500 ⎠ ⎝ 102 x 50 ⎠ ⎝ 10 x 502 ⎠ Step #5: Check out (long direction).7W) FACTORED LOADS: Pu = 1. 0.315)(3/10) = 1.0 + 0.625 ⎞ ⎛ 6 x 336 x 4.15(50-21.4 Kips (30+3) = 1069 kip-ft MR = 5 Kips x 25’ + 225 Kips x 25’ = 5750 kip-ft F.2 FS = 3.4 kips #3 LOAD COMBINATION = 0.09 DC = 1.9 ⎞ ∴ Q MAX.S. MOT = 32.05) ⎞ qmax.2) = 1.019 ksf ⎝ 10 x 50 ⎠ ⎝ 10 2 x 50 ⎠ ⎝ 10 x 50 2 ⎠ Check beam shear: D = 36” – 4” = 32” VU = 0.019x(50 – 21.4(0.150) ) = 240 kips 2 ⎟ BL ⎝ BL ⎠ ⎝ BL ⎠ FTG WT My 972 ex = = = 4.6P e x ⎞ where P = 15 kips + ( (3x10x50x0.05' P 240 Mx 150 ey = = = 0.Step #4: Assume: B = 10’ L = 50’ D = 3’ B/L = 0.2D + 1.75 (1.6(240)(4.0 SC = 1.6(240)(0.989 ksf.5 GOOD Check Sliding RS = Σ V tan f + CB = 240 tan f + 150 (10) = 1587 kips >> 32.07 SJ = 1.12 Q = (130)x3 = 390 SQ = 1. each way. a short drilled shaft would be an efficient and inexpensive foundation. #3 Design for flexure in long direction f’C = 3000 PSI and fY = 60000 PSI A = AS x FY / 0. That alternative will be covered in the drilled shaft section later in this course.85 F’C B AS = MU x 12 / F FY (D – A/2) A = 1.83 ⎞ R= = ⎜ ⎟ = 0.9 )( 60 ) ⎝ 2 ⎠ Therefore. NOTE: In lieu of such a large and expensive footing.0022 > 0. = 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.64 in. AS = 0.64 ⎞ ⎜ 132 .90 in2 OK ⎛ 1. ⎟ M U = ( 0. As = 0.0018 OK BD ⎝ 12 x 32 ⎠ USE # 7 @ 8” O/C.83 in2 AS ⎛ 0. 343 . Chapter 19 Combined Footings Symbols for Combined Footings 344 . Chapter 20 Mat Foundations Symbols for Mat Foundations 345 . 4 D f qult(net) = 5.14(1.14 cu (1+ )(1+ ) L B 0.95 ksf ) [1+ ] [1 + ] = 12 ksf ( 45 ft ) ( 30 ft ) 346 . Solution: Mat foundations in purely cohesive soils have the following ultimate bearing capacity: 0.5 feet below the surface and resting upon a saturated clay stratum with cu = 1.5 ft ) qult(net) = 5.*Mat Foundations–01: Ultimate bearing capacity in a pure cohesive soil.195B 0.40 ( 6.195 ( 30 ft ) 0.950 lb/ft2 and φ = 0º. (Revision: Sept-08) Determine the ultimate bearing capacity of a mat foundation measuring 45 feet long by 30 feet wide placed 6. Chapter 21 Deep Foundations .Single Piles Symbols for Single Piles of Deep Foundations 347 . *Single-Pile–01: Pile capacity in a cohesive soil.-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. 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.97) + ( 0. The clay has γsat = 18. Q 1. (Revision: Oct. Solution: The ultimate capacity of the pile Qult is given by the simple formula. ∴ Qult = ( 0.890 kN 2 The allowable capacity is. φ=0º and cu = 70 kN/m2.890kN Qall = ult = = 630 kN 3 3 348 .38m ) (70 kN / m 2 )(10.75 ) (70 kN / m2 )4(0. The water table lies below the tip of the pile.38m)(20m) = 1.5 kN/m3. Determine the allowable capacity of the pile for a FS = 3 using the α- method. Chapter 22 Deep Foundations .Pile Groups and Caps Symbols for Pile Groups and Caps of Deep Foundations 349 . 4 kips < 423 kips. and a service pile load of 42 kips per pile.85 = 69.0 kips/ pile. Solution.0 kips/pile. which is less than the maximum ultimate load. Since the footing weight will be about 3 kips/pile. say 59. (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.0/0. The permissible shear force around the pile will be. The shear perimeter is bo = π(12 + d) = 99. and the load per pile is Pu = 530/9 = 58. Pn = Pu/φ = 59. assume d = 19. The number of piles required in N=W/P = 340/39 = 8.000 psi.**Pile-caps–01: Design a pile cap for a 9-pile cluster.0 = 39. the pile will not punch through the pile cap (footing). Figure 1 Figure 2 350 . or 9 piles. the net service load per pile is 42. ƒy = 40. we calculate the punching shear stress. Punching shear around a single pile often governs the footing depth determination.7.4) (160) + (1. and 12” diameter piles.7) (180) = 530 kips. Referring to Fig.0″. thus Wu = (1. 2. except in cases in which the loads are small. 70 kips/ pile.7 bars. In this case. 1.5″. After several trials.0-3. Use ƒ׳c = 3000 psi. Use a pile pattern as shown in Fig. The net ultimate load is used to design the footing. it will be shown that beam shear governs. Vc = 4√f׳c bod = 4√3000 (99) (19.9.5) / 1000 = 423 kips Since the actual shear force is the nominal pile reaction. The vertical steel in the column consists of 12 No. 9 kips The force to be resisted is Vn = Vou/φ = 472/0.0 kips.3 ft-kips φ Ru = Mu / bd2 = 398. Refer to Fig. 4.625)2 = 121. Since b = B= 8′-6″ = 102″. The permissible punching shear force (βc = 18/18 < 2) is given by (6. Figure 3 Figure 4 Perimeter shear (punching shear) must now be checked around the column in a similar manner. Vou = 472 kips on 8 piles outside of the critical section as shown on Fig. Beam shear must now be checked. For practical reasons use 29″.3 x 12. Hence the footing is satisfactory for beam shear. Refer to Fig. 3. Three piles exist beyond the critical section.0) = 177. Thus c = a + b = 18.625″ and bo = 4(37. The total dept required will be 28.625) / 1000 = 219.85 = 555. the permissible beam shear (one-way shear) force on the critical section is Vc = 2√ƒ′c bd = 2 √3000 (102) (19.0″. 4 Mu = (177) (27/12) = 398.67 psi 351 . clearance above the pile butts will be 3″ and embedment of the piles will be 6″.85 = 208. 6 bars will be used.12) as Vc = 4 √3000 (150) (19.000 / (102)(19. In this case. so Vu = (3) (59. 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. Assuming No.625) / 1000 = 644.625″. Hence. Refer to Figure 3.75″.0 + 19.625) = 150.3 kips. therefore the pile cap (footing) is satisfactory for punching shear.2 kips. this furnishes an effective depth d = 19.625 = 37.3 kips The force to be resisted is the nominal shear force. Vn = Vu / φ = 177/ 0. The bending moment about the face of the column must now be investigated. it will still be less than ρ min. Thus. the minimum are of steel will be provided. The total weight is WF = (8. However. Since any section having less than minimum reinforcement is usually considered to be un-reinforced.000) (102) (19. Further. The assumed weight of 3. 7 bars (As = 10. ρ min = 200 / ƒy = 0.2 kips And the weight per pile is 26. The assumed footing weight must finally be checked. discloses the fact that the steel ration required is less than the minimum steel ration.0018bh if ƒy = 60.5) (8.000 psi and 0.2/ 9 = 2.925) = 10.0in.91 kips / pile.2 Use seventeen No. if the steel ration required is increased by 1/3. Hence.0 kips / pile is most satisfactory. The 1983 ACI Code is not explicit concerning minimum steel for footings.5) (29) (12.5) / 1000 = 26.002bh for minimum steel area if ƒy 40. the Code does not permit un-reinforced (plain concrete) pile caps.2 in.005. It would appear that 4/3 times the required steel ration would satisfy the 1983 ACI Code. This corresponds to temperature and shrinkage reinforcement requirements.2 for ƒ ′c = 3000 psi and ƒy = 40. some structural engineers use 0.2).000 psi. The final details are shown below 352 .Table 5.000 psi. As = (200/ƒy) bd = (200/ 40. Chapter 23 Deep Foundations: Lateral Loads Symbols for Lateral Loads on Deep Foundations 353 . medium 12 to 18 3.dense 55 to 65 15.800 to 2. (Revision: Oct-08) Determine the lateral load capacity Qg of a steel H-pile (HP 250 x 0.000 to 1. The top end of the pile is allowed to deflect laterally 8 mm.000 to 18.loose 3. The modulus of horizontal subgrade reaction nh is a function of ks at any depth z. Mg = 0).medium 20 to 25 5. For simplicity assume that there is no moment applied to the top of the pile (that is.500 to 7.dense 32 to 45 9.500 .5 to 8. k z = nh z nh modulus of horizontal subgrade reaction Type of soil lb/in3 kN/m3 Dry or moist sand .000 to 12.400 . Solution: The subgrade modulus ks is a description of the reaction of the soil mass to vertical loads.loose 6.200 .000 354 .000 .500 to 4.**Lateral loads on piles-01: Find the lateral load capacity of a steel pile.000 Submerged sand .5 to 5.0 1.834) fully embedded to a depth of 25 m in very dense submerged sand.0 1. Now choose the parameters for the steel H-pile. this is the same table in British units.254 m. the moment of inertia about the strong axis is Ip = 123 x 10-6 m4. For the steel HP 250 x 0.From this table and the soil conditions noted above. choose nh = 12 MN/m3 for the modulus. the steel’s yield strength is Fy = 248 MN/m2 and the pile depth d1 = 0. For future reference.834 pile. its modulus of elasticity is Ep = 207 x 106 kN/m2. 355 . The characteristic length T of a pile-soil system is given by. 000 ) Therefore. Q gT 3 M gT 2 Δ = Az + Bz but M g = 0 EpIp EpIp In this problem we are given this value of Δ = 8 mm.16 m = 21. at a depth z = 0. 356 . T= 5 EpI p = 5 ( 207 x10 )(123 x10 ) = 1.16 m 6 −6 nh (12 . where the coefficient Az is taken from a table of coefficients kz = nh z for long piles. so this is a long pile. and we want to find the allowable lateral load Qg.6 > 5. the ratio L / T = 25 m / 1. The formula for the pile’s top end lateral deflection Δ at any depth z is given by. 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. Ip (123 x10 −6 ) m4 M max = Fy d1 / 2 ( = 248 MN / m 2 ) ( 0.435)(1.The magnitude of the lateral load Qg limited by the displacement condition only is. and ignores that the pile has a moment capacity. 357 .254 m / 2 ) = 240 kN − m ∴ Qg = M max = ( 240 kN − m ) = 268 kN AmT ( 0.772. Δ(E p I p ) ( 0.772 )(1.16 ) 3 Since the value of the allowable lateral load Qg found above is based on the limiting displacement conditions only.008 m ) ( 207 x106 kN / m 2 )(123 x10−6 m 4 ) ∴ Qg = = = 54 kN AzT 3 ( 2. Qg = 54 kN. The maximum allowable moment that the pile can carry is.16 m ) This last value of Qg emanating from the moment capacity is much larger than the value of Qg = 54 kN found for the deflection criterion. Therefore use. Chapter 24 Reinforced Concrete Retaining Walls and Bridge Abutments Symbols for Reinforced Concrete Retaining Walls 358 . an allowable soil bearing capacity of qall = 3 ksf. of retaining wall).**RC Retaining Walls–01: Design a RC wall for a sloped backfill.33' The forces o 15' 16'-6" Pv = ½(12)(18 Pv Ph = ½(39)(1 Ph Step 2: Stabi 4 6. (Revision: Oct-08) Design a reinforced concrete wall with a backfill γ = 125 pcf. shears and moments are per linear ft. and a friction at the base of φ = 30º.11' 6 2' Moment About A 1'-6" 5 3' 1'-6" 5'-6" 10' Area Area Force Arm Moment 359 . Design the wall and check for it’s stability under working loads. 3 1 1. (Note: All loads.83' 1' Solutio 1 n: Step 1: Find The active la ⎛ 3 2 K a = tan 2 ⎜ 4 ⎝ The presure pb = γ HK a = 18. 875 (1 + 0.9 = 1.0 = 3. F = 10.50 6.0 x 15.0 x 0.5 x 15.5 x 15.125) (3.67 3.2 and2 5.k.5 4 ½ x 0. 104.75 then e = 5.25 5.5)² = 4.7 5 10.6 360 .63 1.5 x 0.75 kips x 0.6 ft o.75 8.31 2.150 = 0.8 (0.4 kips OK without Key 6.55 ksf OK < 3 ksf 10 10 Check F against Sliding Shear available along base = 18.125 = 0.03 x 0.6 6.75 (1 + 6 x 0.7 kips 2 (0.0 x 0. Max F = 10.75 ΣM=104.6 – 10 = 0.4 ΣV=18.0 x 0.00 Ph ΣH = 6.6 ) = 1.5 = 15.9 = 5.0 = 82.0 x 2.83 1.9 + 4.36) = 2.0 = 6. < B 2 6 Soil Pressure at Toe of Base qmax = 18.58 = 10.00 13.6 ft 18.7 kips .25 6.940) Min.56 6.6 = 2. (kip) (ft) (kip-ft) 1 ½ x 5.60 6.8 kips ½ γ H² Pp = 5.125 = 0.6 6.00 11.150 = 2.0 x 1.5x 1.75 28.75 x 0.3 6 3.83 = 5.11 40.9 kips Passive force at toe Use S = 2/3 (30º) = 20º .7 = 15.4 3 1.125 = 10.4 Pv 2.0 = 15.9 Location of Resultant From point A. Pp = ( cos δ ) = 5.150 = 2. 005 .000 psi .06/(14 x 12) = 0.85 = 2. ρshrinkage = 0.000 Step 3: Design the stem of the wall.96 in²/ft At bottom of wall Use #6 @ 5” ctrs 5' As = 1.90) = 0.000 = 110 psi fy = 40.06 in²/ft ρ=1.9 (top bars) √3. Load Factors Stem – Use 1.000) = 29.5 / (40 x 0.7 Ph Base (toe and heel) – distribute ΣV uniformly over front B/3 Concrete and Steel Data Capacity reduction factors: 0.0278 OK 14" 361 .0063 >0.04 Ab (40.Step 2: Design parameters.005 and <0.85(shear) F’c = 3.5 kip Ph As = 34.90(flexure).7 ½ 39 (15)² = 7.46 M = 7. 12" Vertical Reinforcement Ph = 1.46 x 5 x 12 = 448 kip-in Use 6” batter on front.550 psi (for stress block) Vc = 2 √3.2 Ab (bottom bars) x 1. ρmin = 0. 0.002 ld = 0.000psi x 0.0278 .4 = 40. ρmax = 0. then t = 12 + 6 = 18” Use d = 18 – 4 = 14” 15' Assume arm = d – a/2 = 13” Pv T = 448/13 = 34. 06 (40) (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.90) = αmin = 1.44) = 1' . . As = 1.25 in OK < 2. hook 362 .6" 10' #6 @ 5"ctrs 6' top of base 15' 0 448 kip-in 496 kip-in 248 kip-in 14" plus std.06 in/ft At bottom : Mr = 1.5 / (2.5) = 286 kip-in 286 kip-in 143 kip-in 2" 0 #6 @ 10"ctrs 17 kip-in 8'-10" 5' 133 kip-in Ld = 40.90) (7.46 / (14 x 12 x 0.Check compressive stress block: C = T = 34.90) (13) = 496 kip-in At top : Mr = 106 (40) (0.9 (0.55 x 12 x 0.0 Check Shear: 7. 0035 > 002. cutoff and develop bars above level where: M = 0.0.75 kips = qmax = 5.53 (40) (±11) (0.59 in²/ft (40 x 0.0.15 x (3.wt of concrete base = 5.005 Therefore.0) = 15.9) Check ρ = 0.23 ksf = 5. shrinkage OK 363 .45 kips 5.23 ksf Net toe pressure for design = qmax .63 ksf (10 / 3) wt.63ksf .004 > 0.5 (0. ρ = 0. : As = 0.53 / (±11 x 12) = 0.150) = 0.25 ksf wt of concrete base = 1.15 ksf front face of stem 3' d =14" 1'-6" V = 5.15 kip M = 15.53 in²/ft .25ksf .0 (0.wt.125) = 0.90) = 158 kip-in 1.4 kips = 0. of soil over toe . Assume t = 18” . d =14” ΣV = 18.Compute level of cutoff for #6 @ 10” ctrs.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.45 x (3/2) x 12 =278 kip-in Assume arm = 13 in T = 278/13 = 21.59 = 0. Distribute ΣV over front B/3. of soil over toe = 2.002 < 0. 14 x 12 Ρ< 0.88 in²/ft Compressive stress block and shear OK by inspection after stem computations Development length : Extend full base width. so increase As by 1/3 .005. As = 0.33 = 0.80 in²/ft use #6 @ 6” ctrs. then As = 0. therefore ld OK 364 .59 x 1. 41 (2/3) (5.150) = 0.5) = 6. As = 0.88 + 0.5 kip As = 31.88 in²/ft ρ > 0.002 and >0.125) = 1.80 kips V2 = ½ (2.10 ksf Weight of concrete base = 0.5) (12) = 128 kip-in M2 = 6.83 (0.33) (5.8 (1/3) (5.5) (12)=282 kip-in Total M = 128 + 282 = 410 kip-in Assume arm = 13” T= 410/13 = 31.9) Use #6 @ 6” ctrs.41 kips d =14" 1'-6" Total V = 5.41 = 12.11 ksf V1 = ½ (2.88 in²/ft (40 x 0.33 ksf 2.0278 OK Compressive stress block and shear OK Development length: Extend full base width.5) = 5.23 ksf Net pressure for design = 2.125) = 2.11 ksf At Back: Weight of soil above hell at back = 16.33 ksf 2.5 (0.005 and < 0.0 (0.11) (5.23 ksf Net pressure for design = 1.5kip = 0. therefore ld OK Horizontal shrinkage steel in stem: 365 . At Stem: Weight of soil above heel at back face of stem = 15.88 ksf Weight of concrete base = 1.80 + 6.21 kips 5'-6" M1= 5.23 = 2. Step 5: Design the heel of the base of the wall. 002 (±15) (12) = 0.13 in²/ft Total As = 0. bars.27 + 0.40 in²/ft Step 6: Finished sketch of the wall. Required: 0.36 in²/ft of height Use #4 @ 9” ctrs. top and bottom As = 0. bars. 3 1 12" #4@ 9" front face #4@18" back face Provide 3" clear to all bars except 112 " Vert.13 = 0. back face #6 @ 5" #6 @ 6" 2' 3' 1'-6" 5'-6" 1'-6" #6 @ 6" 2"x6" key #4 @ 12" 366 .27 in²/ft Use #4 @ 18” ctrs back As = 0. front As = 0.40 in²/ft Horizontal shrinkage steel in base: Use #4 @ 12 ctrs. back face clear to bars in front face of stem #6 @ 10" #4 @ 2'-0" 16'-6" 6" drains @ 10'-0" Vert. Chapter 25 Steel Sheet Pile Retaining Walls Symbols for Steel Sheet Pile Retaining Walls 367 . 368 .**Sheet-pile Wall-01: Free-Earth for cantilevered walls in granular soils. Determine the pressures on the cantilevered wall.70 Solution: Step 1. γ = 115 pcf γ ' = 65 pcf φ = 35° therefore K a = 0.27 and K p = 3. (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. 860 psf 369 J pE = 3 . find the dimension z shown in the figure of previous page.5 − 14 435 .Step 2.860 psf 3 .6 D ( ) pE = γ D K p − K a − p A1 = 65 D ( 6. d) Use z to check if ΣM = 0. b) Assume a value for D. 860 − 435 10.27 ) D = 435 + 17. The procedure will follow the following steps: a) From statics.56 ) = 409 D + 10 . p A 2 = 620 psf . p = 14 .5 feet . c) Calculate z. Determine the depth of embedment D. adjust D for convergence.29 ) − 435 = 409 D − 435 ( ) pJ = γ D K p − K a + γ HK p = 65 D ( 6.29 ) + (115 )(14 )( 6. 560 Try D = 10.27 ) = 435 psf pA 2 = pA1 + γ DK a = ( 435 ) + ( 65 )( 0. p A1 = 435 psf . pA1 = γ HK a = (115 )(14 )( 0. 040 lb 2 2 1 1 P2 = p A1 y = ( 435 )(1. 030 − 4 . 360 = 26 . 0 ⎞ M max = P1l1 + P2 l2 − P3 l3 = ( 3. 040 + 218 ) ∴ x2 = = = 16 ∴ x = 4. z= ( pE − pA1 ) D − Hp1 pE + pJ ( 2) ΣM = 0 about any po int. Determine the maximum moment Mmax (point of zero shear).0 ) ⎞ ⎛ 4.0 + 4. say F .06 feet γ ' ( K p − Ka ) ( )( ) 65 6 .(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 . 29 1 1 P1 = pA1 H = ( 435 )(14 ) = 3 . 280 lb 2 2 M max = P1l1 + P2 l2 − P3 l3 where . 000 ft − lb = 312 in − kips . 040 ) ⎜ + 1. 1 ⎛ H⎞ ⎛ D2 ⎞ ⎛ z2 ⎞ ⎛ D2 ⎞ ⎛ D2 ⎞ ( H ) pA1 ⎜ D + ⎟ + ( pA1 ) ⎜ ⎟ + ( pE + pJ ) ⎜ ⎟ − ( pE + pA2 ) ⎜ ⎟ + ( pA2 − pA1 ) ⎜ ⎟ = 0 2 ⎝ 3 ⎠ ⎝ 2 ⎠ ⎝ 6⎠ ⎝ 6 ⎠ ⎝ 6 ⎠ Step 3. H l1 = + y+ x 3 2y l2 = +x 3 x l3 = 3 ⎛ 14 ⎞ ⎛ 2 (1. pA1 435 y= = = 1.29 ) 1 ( ) P3 = γ ' K p − K a ( 4 ) = 3 .0 ⎟ − ( 3. 300 + 1. 280 ) ⎜ ⎟ 370 ⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠ M max = 29 . 1 ' 2 ( ) γ K p − K a x 2 = P1 + P2 2 ( P1 + P2 ) 2 ( 3 .0 ⎟ + ( 218 ) ⎜ + 4.0 ) = 218 lb 2 2 also.0 feet γ ( K p − Ka ) ' ( 65)( 6. 371 . 000 )(12 ) The required sec tion mod ulus S = = = 10.5 = 27. The total sheet-pile length L = H + D = 14.5 feet 372 . Determine the sheet-pile length L. This is shown plotted with the red lines. Step 5. 000 ) ∴ Use a MP − 116 that provides a S = 10. an Ex − Ten − 45 steel sec tion is used .7 in3 at a lower cos t .If instead .8 in3 fs ( 29 .0 + 13. f s = 29 ksi M max ( 26 . Chapter 26 MSE (Mechanically Stabilized Earth) Walls Symbols for Mechanically Stabilized Earth Walls 373 . (Revision: Oct-08) Determine the length L of a geotextile-reinforcing for the 16 foot high temporary retaining wall shown below. etc.**MSE Walls-01: Design the length L of geotextiles for a 16 ft wall. Comment: These geotextile walls are usually used for temporary civil works. such as detour roadways. and the available granular backfill has a unit weight of γ = 110 lb/ft3 and the angle of internal friction is φ = 36°. Also determine the required vertical spacing of the reinforcing layers Sv and the required lap length ll. then the face is stabilized and the plastic geotextile is protected from ultra-violet light through a layer of shotcrete. If the wall must become permanent. σ a = K aσ v = K a ( γ z ) where K a = tan2 ( 45° − ϕ / 2 ) = tan2 ( 45° − 36° / 2 ) = 0.26 374 . temporary abutments or excavation walls. The geotextile chosen has an allowable strength of σG of 80 lb/inch. 16 feet Solution: The active pressure on the wall is. SV = σG = (80 x12 lb / ft ) = 2.26 )(1.5.5) At a depth of z = 16 feet from the top. SV = σG = (80 x12 lb / ft ) = 1.3 to 1. Permanent walls should use at least FS ≥ 2.4 feet ≈ 17 inches (γ zK a ) FS (110 )(16 )( 0. 375 . below z = 8 feet use SV = 16 inches throughout. as shown in the figure.5) Notice how the spacing becomes denser the deeper we go below the surface.8 feet ≈ 34 inches (γ zK a ) FS (110 )(8)( 0. For this problem choose FS = 1.5) At a depth of z = 12 feet from the top.5 for temporary walls. choose SV = 20 inches from z = 0 to z = 8 feet. σG σG SV = = σ a FS ( γ zK a ) FS At a depth of z = 8 feet from the top.9 feet ≈ 22 inches (γ zK a ) FS (110 )(12 )( 0. where the Factor of Safety (FS) is generally chosen between 1.26 )(1.26 )(1.Step 1: Find the vertical spacing SV of each layer of geotextile. SV = σG = (80 x12 lb / ft ) = 1. 68 0.48 0.78 0.731 8. 376 .81 112 9.00 1.438SV) L inches feet feet feet feet feet 16 1.5 ) + V tan ( 45° − φ / 2 ) 2 tan ϕ F tan ( 45° − 36° / 2 ) 2 ( 0.582 3. we can now prepare a table with the required lengths.34 1.5. which is usually assumed to be 2/3 φ of the soil.33 3.33 2. Therefore.582 2. page 408). Again use FS = 1. and the angle φF is the soil-to-geotextile angle of friction.67 1. but never smaller than 3 feet.66 96 8.34 1.Step 2: Find the length of each layer of geotextile L. L = l R + le = ( H − z ) + SV K a ( FS ) = ( H − z) S ( 0.26 Based on this table. the length L of the geotextile layer is.67 1. Step 3: Find lap length ll for the geotextile.04 0.731 6.99 144 12.51)(H-z) (0. z SV (0. and a few are shown in this table.33 1.26 )(1.40 0.00 1.731 5.445 ) L = ( 0.731 4.582 1. use L = 8.67 4.51)( H − z ) + 0.51 76 6.5 feet for z ≤ 8 feet and L = 4.67 7. which is composed of two parts.33 0.08 0.21 56 4.0 feet for z > 8 feet.67 5. and le which is the effective length of the geotextile beyond the failure zone (see the first figure. lR which is the length of the geotextile within the Rankine failure zone. Other values can be used.66 176 14.93 0.67 4.438 SV From this equation. 219 SV = ( 0. use ll = 3 feet Comment: These MSE problems commonly use Rankine’s active pressure coefficient.365 feet < 3 feet ⎝ 12 in / ft ⎠ Therefore. However.26 )(1. as shown in this figure: 377 .219 ) ⎜ ⎟ = 0.5 ) ll = = = 0.219 SV 4σ v tan ϕ F 4 tan ⎡⎣( 23 ) ( 36° ) ⎤⎦ For a depth of z=16 inches. the actual value of K must depend on the degree of restraint of the type of reinforcement. ⎛ 20 inches ⎞ l l = 0. SV σ a ( FS ) SV ( 0.