Soil Mechanics by Jerry Vandevelde

SOIL MECHANICS (version Fall two hundred8) Presented by Jerry Vandevelde, P. E. Chief get up GEM Engineering, Inc. 1762 Watterson Trail Louisville, Kentucky (502) 493-7 deoxycytidine monophosphate 1 National Council of Examiners for Engineering and Surveying http//www. ncees. org/ 2 STUDY REFERENCES Foundation Engineering Peck Hanson & Thornburn Introductory flaw mechanism and Foundations Sowers NAVFAC Design Manuals DM-7. 1 & 7. 2 Foundation Analysis and Design Bowles mulish Foundation Engineering Handbook Brown 3 Soil motley Systems * Unified Soil miscellanea System * AASHTO Need Particle sizes and Atterberg Limits 4Particle Sizes (Sieve Analysis) (Well judge) (Poorly Graded) 0. 1 5 Atterberg Limits irrigatey, Plastic & Shrinkage Limits Plasticity Index (PI) PI = Liquid Limit Plastic Limit (range of moisture content over which dirty is plastic or malleable) 6 UNIFIED SOIL CLASSIFICATION dodging ASTM D-2487 7 8 referee Peck Hanson & Thornburn 2nd Ed. Effective Size = D10 10 portion of the sample is finer than this size D60 = 1. 6mm D30 = 0. 2mm D10 = 0. 03mm 0. 1 0. 1 9 Uniformity Coefficient (Cu) = D60/D10 Coefficient of Curvature (Cz) = (D30)2/(D10xD60) D60 = 1. 6mm D30 = 0. 2mm D10 = 0. 03mm 0. 1 10 Well Graded Requirements 50% coarser than no. 00 concealment Uniformity Coefficient (Cu) D60/D10 4 for discombobulate 6 for moxie Coefficient of Curvature (Cz) = (D30)2/(D10xD60) = 1 to 3 11 Is the better graded real(a) a gravel? 81% Passing zero(prenominal) 4 18% Finer zero(prenominal) two hundred 0. 1 0. 1 12 Gravel if 50 Percent Coarse Fraction retained on zero(prenominal) 4 sieve % Retained on No. 200 = 82% 1/2 = 41% 19% ( atomic number 6-81) retained on No. 4 sieve (gravel) 19 41 half of coarse fraction 81% Passing No. 4 18% Finer No. 200 ? sand 0. 1 (S) 13 Well Graded guts? Uniformity Coefficient (Cu) 6 = D60/D10 Coefficient of Curvature (Cz) = 1 to 3 = (D30)2/(D10xD60) 14 D60 = 1. 6mm D30 = 0. 2mm D10 = 0. 3mm 0. 1 Well Grad ed Sand? Uniformity Coefficient (Cu) D60/D10 = 1. 6/. 03 = 53 6 D60 = 1. 6mm D30 = 0. 2mm D10 = 0. 03mm Coefficient of Curvature (Cz) = (D30)2/(D10xD60) = 0. 22/(. 031. 6) = 0. 83 12% Passing No. 200 sieve GM, GC, SM, SC 0. 1 12% transitory No. 200 sieve Since = S ? SC or SM 16 What Unified assortment if LL= 45 & PI = 25? From sieve data SC or SM 0. 1 A) SC B) SM C) CL or D) SC & SM 17 Unified compartmentalization Answer is A ? SC 18 AASHTO (American Association of State Highway and Transportation Officials) 19 What is the AASHTO Classification? 65% Passing No. 10 40% Passing No. 0 18% Finer No. 200 1) 18 % button No. 200 sieve 2) 65% overtaking No. 10 sieve 3) 40% passing No. 40 sieve 4) read LL = 45 & PI = 25 20 18 percent passing No. 200 sieve 65 percent passing No. 10 sieve 40 percent passing No. 40 sieve take for granted LL = 45 & PI = 25 21 AASHTO Classification 1 2 3 4 4 1) 18 % passing No. 200 sieve 2) 65% passing No. 10 sieve 3) 40% passing No. 40 sieve 4) assume LL = 45 & PI = 25 22 AASHTO Group Index 23 trade-Volume (Phase Diagram) unit of measurement volume of acres contains append Volume Va tonal pattern check Vt Vv Vw Vs wet Ww Ws load Wt Soil Air (gases) wet (fluid) Solid Particles 24 wet Content = ? eight of water/ weight of dry blur ? = Ww/Wd water loss/(moist speck weight water loss) ? = Ww/(Wm-Ww) and ? =(Wm-Wd)/Wd 25 potful Volume kindreds Density or Unit freight = Moist Unit pitch = ? m ? ?m = Wm/Vt = ? d + ? ?d ? = (? m ? d )/ ? d ? ?d + ? d = ? m ? m= (1+ ? ) ? d ? d = ?m/(1+ ? ) b 26 entireness Volume = ? Volume (solid + water + air) = Vs+Vw+Va ? Va = Vt Vs- Vw tally Volume Va Air positive Vt Vv Vw Vs Water Ww Ws weightiness Wt Soil 27 Relationship Between Mass & Volume Volume = Mass/(Specific dryness x Unit heaviness of Water) = Ws/(SGxWw) Va Total Volume Air Total Vt Vv Vw Vs Water Ww Ws tilt Wt Soil 28Specific Gravity = weight of significant/ weight of same volume of water Soil Specific Gravity Typical Range 2. 65 to 2. 70 Specific Gravity of Water = 1 29 Saturation = S expressed as percent S = volume of water/ volume of voids x 100 Total Volume Va Air Total S = Vw/Vv x 100 Ww Ws incubus Vt Vv Vw Vs Water Wt Soil Always ? 100 30 Porosity n = volume of voids/ total volume n = Vv/Vt Void Ratio e = volume of voids/ volume of solids e = Vv/Vs Total Volume Va Air Total Vt Vv Vw Vs Water Ww Ws Weight Wt Soil 31 What is the degree of saturation for a soil with SG = 2. 68, ? m = 127. 2 pcf & ? = 18. 6 percent A) 88. 4 Total Volume VaAir Total Vt Vv Vw Vs Water Ww Ws Weight B) 100. 0 Wt Soil C) 89. 1 32 What are the porosity and degree of saturation for a soil with SG = 2. 68, ? m = 127. 2 pcf & ? = 18. 6 percent = 107. 3pcf ?d = ? m/(1+ ? ) = 127. 2/(1. 186) Total Volume Va Air Total Vt Vv Vw Vs Water Soil Ww Weight Wt Ws Ww = ? m- ? d = 19. 9 pcf Vw = Ww/62. 4 = 0. 319 cf Vs = ? d /(SGx62. 4) = 0. 642 cf Va = Vt Vw Vs = 1- 0. 319 0. 642 = 0. 039 cf Vv = Vw + Va = 0. 358 cf 33 What are the porosity and degree of saturation for a soil with SG = 2. 68, ? m = 127. 2 pcf & ? = 18. 6 percent Vw = 0. 319 cf, Vs = 0. 642 cf, Vv = 0. 358 cf Total VolumeVa Air Total Degree of Saturation = Vw/Vv x 100 Ww Weight Wt Ws Vt Vv Vw Vs Water = 0. 319/0. 358 x 100 = 89. 1% Soil Answer is C 34 Ref NAVFAC DM-7 35 Borrow Fill Adjustments Borrow hooey Properties ?m = cx pcf & ? = 10% Placed Fill Properties ? d = one hundred five pcf & ? = 20% How a great deal borrow is needed to produce 30,000 cy of fill? How much water essential be added or removed from all(prenominal) cf of fill? Total Volume Va Air Total Vt Vv Vw Vs Water Ww Ws Weight Wt Soil 36 Borrow Fill Adjustments Borrow Material Properties ?m = 110 pcf & ? = 10% ?d = ? m /(1+? ) = 110/(1. 10) =100 pcf Ww = 110-100=10 lbs Placed Fill Properties ? = one hundred five pcf & ? = 20% Ww = ? x ? d = 0. 2x one hundred five = 21 lbs Total Volume Va Air Total Vt Vv Vw Vs Water Ww Ws Weight Wt Soil 37 Borrow Fill Adjustm ents Borrow Properties ? m = 110 pcf, ? d =100 & ? = 10% Placed Fill Properties ? d = 105 pcf & ? = 20% Since borrow ? d =100pcf & fill ? d =105pcf, 105/100 =1. 05 It takes 1. 05 cf of borrow to make 1. 0 cf of fill For 30,000 cy, 30,000 x 1. 05 = 31,500 cy of borrow Total Volume Va Air Total Vt Vv Vw Vs Water Ww Ws Weight Wt Soil 38 Borrow Fill Adjustments Borrow Material Properties Ww = 10 lbs Placed Fill Properties Ww = 21 lbs Water supplied from borrow in each cf of fill = 10 x 1. 5 = 10. 5 lbs 21 lbs 10. 5 = 10. 5 lbs short/1. 05 cf 10. 5lbs/1. 05 cy = 10 lbs of water to be added per cf borrow Total Volume Va Air Total Vt Vv Vw Vs Water Ww Ws Weight Wt Soil 39 Proctor moisture Density Relationships Establishes the unequaled relationship of moisture to dry tightness for each specific soil at a specified compaction energy MOISTURE-DENSITY family relationship 108. 0 106. 0 104. 0 D ry D ensity (pcf) 102. 0 100. 0 98. 0 96. 0 94. 0 92. 0 90. 0 88. 0 8. 0 10. 0 12. 0 14. 0 16. 0 18. 0 20. 0 22. 0 24. 0 26. 0 28. 0 Moisture Content (%) 40 Proctor Moisture Density Relationships 4 mold 25 blows 6 mold 56 blows Standard 5. 5 lb hammer dropped 12 in 3 layers Standard ASTM D-698 AASHTO T-99 Modified ASTM D-1557 AASHTO T-150 Modified 10 lb hammer dropped 18 in 5 layers 41 PROCTOR COMPACTION TEST upper limit teetotal Density Highest density for that degree of compactive effort Optimum Moisture Content Moisture content at which maximum dry density is achieved for 42 that compactive effort Proctor Moisture Density Relationships MOISTURE-DENSITY RELATIONSHIP 108. 0 106. 0 104. 0 Dry Density (pcf) 102. 0 100. 0 98. 0 96. 0 94. 0 92. 0 90. 0 88. 0 8. 0 10. 0 12. 0 14. 0 16. 0 18. 0 20. 0 22. 0 24. 0 26. 0 28. 0 Moisture Content (%)What density is required for 95% Compaction? What range of moisture would facilitate achieving 95% compaction? 43 Proctor Moisture Density Relationships MOISTURE-DENSITY RELATIONSHIP 108. 0 106. 0 104. 0 Dry Density (pcf) 102. 0 1 00. 0 98. 0 96. 0 94. 0 92. 0 90. 0 88. 0 8. 0 10. 0 12. 0 14. 0 16. 0 18. 0 20. 0 22. 0 24. 0 26. 0 28. 0 Moisture Content (%) 104 x . 95 = 98. 8 pcf A 95% B Range of moisture is within the curve A to B (14 to 24 %) 44 Proctor Zero Air Voids Line Relationship of density to moisture at saturation for constant specific sedateness (SG) Cant achieve fill in zone even out of zero air voids line ZMOISTURE-DENSITY RELATIONSHIP 108. 0 106. 0 104. 0 Dry Density (pcf) 102. 0 100. 0 98. 0 96. 0 94. 0 92. 0 90. 0 88. 0 8. 0 10. 0 12. 0 14. 0 16. 0 18. 0 20. 0 22. 0 24. 0 26. 0 28. 0 Moisture Content (%) 45 Proctor Moisture Density Relationships MOISTURE-DENSITY RELATIONSHIP 108. 0 106. 0 104. 0 Dry Density (pcf) 102. 0 100. 0 98. 0 96. 0 94. 0 92. 0 90. 0 88. 0 8. 0 10. 0 12. 0 14. 0 16. 0 18. 0 20. 0 22. 0 24. 0 26. 0 28. 0 Moisture Content (%) If SG = 2. 65 & moisture content is 24% What dry density achieves 100% saturation? A) 100. 0 pcf B) 101. 1 pcf 46 Proctor Moisture Density Relations hipsMOISTURE-DENSITY RELATIONSHIP 108. 0 106. 0 104. 0 Dry Density (pcf) 102. 0 100. 0 98. 0 96. 0 94. 0 92. 0 90. 0 88. 0 8. 0 10. 0 12. 0 14. 0 16. 0 18. 0 20. 0 22. 0 24. 0 26. 0 28. 0 Moisture Content (%) X ?d=SG62. 4/(1+? SG/100) ? d=2. 6562. 4/(1+242. 65/100) ? d=101. 1 pcf Answer is B 47 Ref Peck Hanson & Thornburn Static coping 48 Calculate effective test at point x Ref Peck Hanson & Thornburn Saturated Unit Weight ? sat 5 ? sat = one hundred twenty-five pcf Moist Unit Weight ? M Dry Unit Weight ? Dry 7 Submerged (buoyant) Unit Weight = ? sat 62. 4 x 49 Calculate effective line at point x Ref Peck Hanson & ThornburnTotal Stress at X 5 ? sat = 125 pcf = 5 x 62. 4+ 7x 125= 1187psf Pore Pressure at X 7 = 12 x 62. 4 = 749 psf Effective Stress at X = 1187-749= 438 psf x or (125-62. 4) x 7=438 psf 50 Ref Peck Hanson & Thornburn downward Flow Gradient 51 Downward Flow Gradient 3 Total Stress at X = 5 x 62. 4+ 7x 125= 1187psf Pore Pressure at X ? sat = 125 pcf 7 = (12-3) x 62. 4 = 562 psf Effective Stress at X = 1187-562 = 625 psf 5 x or 438 + 3 x 62. 4 = 625psf capture previous problem 52 Upward Flow Gradient Ref Peck Hanson & Thornburn 53 One Dimensional Consolidation ?e/pn 54 Primary Phase Settlement (e log p) ? H = (H x ? )/(1+eo) eo ? H H 55 Consolidation Test Pre-consolidation Pressure Cc = side of e log p virgin curve est. Cc = 0. 009(LL-10%) Skempton recant or recompression curves 56 56 e- l o g p Calculate Compression Index Cc 1. 50 1. 40 1. 30 Void Ratio (e) 1. 20 1. 10 ksf 0. 1 1 4 8 16 32 (e) 1. 404 1. 404 1. 375 1. 227 1. 08 0. 932 1. 00 0. 90 A) 0. 21 B) 0. 49 57 0. 80 0. 1 1 10 100 Pr essur e ( ksf ) Cc is the slope of the virgin e-log p e- l o g p Cc = -(e1-e2)/log (p1/p2) 1. 50 Cc=-(1. 375-1. 227)/log(4/8) Cc = 0. 49 Answer is B ksf 0. 1 1 4 8 16 32 (e) 1. 404 1. 404 1. 375 1. 227 1. 08 0. 932 1. 40 Cc Void Ratio (e) . 30 1. 20 1. 10 1. 00 0. 90 0. 80 0. 1 1 10 100 Pr essur e ( ksf ) 58 Permeability Constant Head Conditions Q=kiAt Q = k (h/L)At k=QL/(Ath) 59 If Q =15cc & t = 30 randomness what is the permeability k=QL/(Ath) 10cm 5cm A) 0. 01 cm/ south B) 0. 0110-2 cm/ southward 25cm2 C) 0. 1 cm/ irregular 60 Constant Head Permeability Calculate k Q =15cc & t = 30 south k=QL/(Ath) k= 15(5)/(25(30)10) k= 0. 01 cm/sec Answer is A 10cm 5cm 25cm2 61 Falling Head Permeability k=QL/(Ath) (but h varies) k=2. 3aL/(At) log (h1/h2) where a = pipette area h1 = sign head h2 = final head 62 If t = 30 sec h1= 30 cm h2 = 15 cm L= 5 cm a= 0. cm2 A= 30 cm2 calculate k A) 2. 310-3 cm/sec B) 8. 110-6 cm/sec C) 7. 710-4 cm/sec 63 Falling Head Permeability k=2. 3aL/(At) log (h1/h2) k= 2. 3 (0. 2) 5 /(3030) log (30/15) k= 7. 710-4 cm/sec Answer is C 64 Flow lines & head drop lines must intersect at right angles All areas must be neat Draw minimum number of lines Results depend on ratio of Nf/Nd Flow Nets 6ft 2ft 65 Q=kia=kHNf /Nd wt (units = volume/time) w= unit comprehensiveness of section t=time Flow Nets 6ft 66 What flow/ twenty-four hours? assume k= 110-5 cm/sec =0. 0283 ft/ twenty-four hour period Q= kH (Nf /Nd) wt Q= 0. 0283x8x(4. 4/8)x1x1 Q= 0. 12 cf/day 2ft Flow Nets ft 67 Check for quick conditions pc =2( cxx)= 240 psf (total stress) Flow Nets Below water level use saturated unit weight for total stress ?= 2(62. 4) = 124. 8 (static pressure) = 1/8(8)(62. 4)= 62. 4 (flow gradient) = 240-(124. 8+62. 4) 2ft 2ft 6ft pc = pc -(? + ) pc = 52. 8 psf 0, soil is not quick ?sat=120 pcf 68 Stress Change Influence (1H2V) For square footing z=Q/(B+z)2 69 If Q= 20 kips, Calculate the vertical stress increase at 7 feet below the footing bottom 5 8 7 70 If Q= 20 kips, Calculate the vertical stress increase at 7 feet below the footing bottom 5 8 z = 0000 (8+7)(5+7) 7 z = 111 psf 71 Westergaard (layered elastic & inelastic material) If B= 6. 3 in a square footing with 20 kips load, what is the vertical stress increase at 7 below the footing bottom? 72 Westergaard Q = 20 kips B = 6. 3 Z = 7 z = ? 73 W estergaard 7/6. 3 = 1. 1B z = 0. 18 x 20000/6. 32 = 90. 7 psf 74 Boussinesq (homogeneous elastic) Q = 20 kips B = 6. 3 Z = 7 z = ? 75 Boussinesq Z/B = 1. 1 z = 0. 3 x 20000/6. 32 = 151 psf 76 Thanks for participating in the PE review passage on Soil Mechanics More questions or comments? 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