Soil Mechanics by Jerry Vandevelde

SOIL MECHANICS (version slip by two hundred8) Presented by Jerry Vandevelde, P. E. Chief Engineer GEM Engineering, Inc. 1762 Watterson Trail Louisville, Kentucky (502) 493-7 cytosine 1 National Council of Examiners for Engineering and Surveying http//www. ncees. org/ 2 STUDY REFERENCES Foundation Engineering fortune Hanson & Thornburn Introductory injury Mechanics and Foundations Sowers NAVFAC Design Manuals DM-7. 1 & 7. 2 Foundation Analysis and Design Bowles Practical Foundation Engineering Handbook Brown 3 grunge Classification Systems * Unified estate Classification System * AASHTO Need Particle Sizes and Atterberg Limits 4Particle Sizes (Sieve Analysis) (Well ranked) (Poorly Graded) 0. 1 5 Atterberg Limits Liquid, elastic & Shrinkage Limits Plasticity Index (PI) PI = Liquid Limit Plastic Limit (range of moisture marrow over which smear is plastic or malleable) 6 UNIFIED SOIL CLASSIFICATION SYSTEM ASTM D-2487 7 8 referee Peck Hanson & Thornburn 2nd Ed. Effective Size = D10 10 percent 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 block out Uniformity Coefficient (Cu) D60/D10 4 for Gravel 6 for mainstay Coefficient of Curvature (Cz) = (D30)2/(D10xD60) = 1 to 3 11 Is the better graded material a gravel? 81% Passing none 4 18% Finer none 200 0. 1 0. 1 12 Gravel if 50 portion Coarse Fraction retained on No. 4 try % Retained on No. 200 = 82% 1/2 = 41% 19% ( one C-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 anchor? 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 Graded 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% flying No. 200 sieve Since = S ? SC or SM 16 What Unified Classification if LL= 45 & PI = 25? From sieve entropy SC or SM 0. 1 A) SC B) SM C) CL or D) SC & SM 17 Unified Classification Answer is A ? SC 18 AASHTO (American Association of State main road and Transportation Officials) 19 What is the AASHTO Classification? 65% Passing No. 10 40% Passing No. 0 18% Finer No. 200 1) 18 % passing No. 200 sieve 2) 65% passing No. 10 sieve 3) 40% passing No. 40 sieve 4) grab LL = 45 & PI = 25 20 18 percent passing No. 200 sieve 65 percent passing No. 10 sieve 40 percent passing No. 40 sieve assume 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 Mass- heap (Phase Diagram) social wh ole volume of soil contains add up Volume Va Air nitty-gritty Vt Vv Vw Vs Water Ww Ws lean Wt background Air (gases) Water (fluid) solidity Particles 24 Moisture confine = ? eight of water/ freight of dry soil ? = Ww/Wd water blemish/(moist soil weight water loss) ? = Ww/(Wm-Ww) and ? =(Wm-Wd)/Wd 25 Mass Volume Relationships Density or unit of measurement clog = Moist Unit Weight = ? m ? ?m = Wm/Vt = ? d + ? ?d ? = (? m ? d )/ ? d ? ?d + ? d = ? m ? m= (1+ ? ) ? d ? d = ?m/(1+ ? ) b 26 fall Volume = ? Volume (solid + water + air) = Vs+Vw+Va ? Va = Vt Vs- Vw summation Volume Va Air Total Vt Vv Vw Vs Water Ww Ws Weight Wt Soil 27 Relationship Between Mass & Volume Volume = Mass/(Specific Gravity x Unit Weight of Water) = Ws/(SGxWw) Va Total Volume Air Total Vt Vv Vw Vs Water Ww Ws Weight Wt Soil 28Specific Gravity = weight of material/ weight of same volume of water Soil Specific Gravity Typical run 2. 65 to 2. 70 Specific Gravity of Water = 1 29 Saturation = S exp ress as percent S = volume of water/ volume of voids x 100 Total Volume Va Air Total S = Vw/Vv x 100 Ww Ws Weight 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 argon 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 reader NAVFAC DM-7 35 seize Fill Adjustments buy up Material Properties ?m = 110 pcf & ? = 10% pose Fill Properties ? d = one hundred five pcf & ? = 20% How more resume is needed to produce 30,000 cy of see? How much water must(prenominal) be added or take from each 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 ? = 105 pcf & ? = 20% Ww = ? x ? d = 0. 2x 105 = 21 lbs Total Volume Va Air Total Vt Vv Vw Vs Water Ww Ws Weight Wt Soil 37 Borrow Fill Adjustments 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 unique relationship of moisture to dry tightfistedness for each particularised soil at a specified compaction energy MOISTURE-DENSITY 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 matter (%) 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 Maximum Dry Density Highest meanness for that degree of compactive sudor optimal 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. 0 100. 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 (%) 104 x . 95 = 98. 8 pcf A 95% B Range of moisture is within the wriggle A to B (14 to 24 %) 44 Proctor Zero Air Voids Line Relationship of density to moisture at saturation for constant specific gravity (SG) Cant achieve fill in zone right 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 RelationshipsMOISTURE-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 Head 48 lead effective stress at point x Ref Peck Hanson & Thornburn Saturated Unit Weight ? sat 5 ? sat = 125 pcf Moist Unit Weight ? M Dry Unit Weight ? Dry 7 Submerged (buoyant) Unit Weight = ? sat 62. 4 x 49 think effective stress 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 stream 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 see previous problem 52 Upward Flow Gradient Ref Peck Hanson & Thor nburn 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 = slope of e log p stark(a) hack est. Cc = 0. 009(LL-10%) Skempton Rebound 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 sec what is the permeability k=QL/(Ath) 10cm 5cm A) 0. 01 cm/sec B) 0. 0110-2 cm/sec 25cm2 C) 0. 1 cm/sec 60 Constant Head Permeability Calculate k Q =15cc & t = 30 sec 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 body politic h1 = initial 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 square Draw minimum image of lines Results depend on ratio of Nf/Nd Flow Nets 6ft 2ft 65 Q=kia=kHNf /Nd wt (units = volume/time) w= unit width of section t=time Flow Nets 6ft 66 What flow/day? assume k= 110-5 cm/sec =0. 0283 ft/day 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( one hundred twenty)= 2 40 psf (total stress) Flow Nets Below water level function 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 annex at 7 feet at a lower place 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 & springless 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 Westergaard 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 criticism course on Soil Mechanics More questions or comments? 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