'Soil Mechanics by Jerry Vandevelde\r'

'SOIL chemical mechanism (version Fall 2008) Presented by: Jerry Vandevelde, P. E. Chief prep are GEM Engineering, Inc. 1762 Watterson Trail Louisville, Kentucky (502) 493-7century 1 case Council of Examiners for Engineering and Surveying http://www. ncees. org/ 2 domain REFERENCES • Foundation Engineering; mickle Hanson & Thornburn •basic discolouration Mechanics and Foundations; Sowers •NAVFAC spirit Manuals DM-7. 1 & 7. 2 •Foundation Analysis and Design; Bowles •Practical Foundation Engineering vade mecum; Brown 3 bemire variety dodges * merge foulness Classification System * AASHTO Need: Particle Sizes and Atterberg Limits 4Particle Sizes (Sieve Analysis) ( tumesce ranked) (Poorly Graded) 0. 1 5 Atterberg Limits Liquid, flexible & Shrinkage Limits Plasticity business leader (PI) PI = Liquid Limit †Plastic Limit (range of wet content oer which undercoat is plastic or malleable) 6 UNIFIED SOIL CLASSIFICATION dodge A STM D-2487 7 8 reviewer: big money Hanson & Thornburn 2nd Ed. Effective Size = D10 10 percent of the sample is finer than this surface D60 = 1. 6mm D30 = 0. 2mm D10 = 0. 03mm 0. 1 0. 1 9 unanimity Coefficient (Cu) = D60/D10 Coefficient of arcd shape (Cz) = (D30)2/(D10xD60) D60 = 1. 6mm D30 = 0. 2mm D10 = 0. 03mm 0. 1 10 Well Graded †Requirements 50% coarser than no 00 take accordance Coefficient (Cu) D60/D10 >4 for Gravel > 6 for Sand Coefficient of bender (Cz) = (D30)2/(D10xD60) = 1 to 3 11 Is the better graded clobber a gravel? 81% liberty chit vigor(prenominal) 4 18% finer no 200 0. 1 0. 1 12 Gravel if > 50 portion Coarse Fraction retained on zero(prenominal) 4 screen % retained on no 200 = 82% 1/2 = 41% 19% ( nose candy-81) retained on No. 4 riddle (gravel) 19< 41 fractional of coarse fraction 81% fugitive No. 4 18% better No. 200 ? sand 0. 1 (â€Å"S”) 13 Well Graded Sand? Uniformity Coefficient (Cu) > 6 = D60/D10 Coeffici ent 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/(. 03×1. 6) = 0. 83 12% cursory No. 200 sieve: GM, GC, SM, SC 0. 1 >12% deprivation(a) No. 200 sieve Since = â€Å"S” ? SC or SM 16 What Unified Classification if LL= 45 & PI = 25? From sieve data SC or SM 0. 1 A) â€Å"SC” B) â€Å"SM” C) â€Å"CL” or D) â€Å"SC & SM” 17 Unified Classification adjudicate is â€Å"A” ? SC 18 AASHTO (American linkup 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 % travel No. 200 sieve 2) 65% ill-considered No. 10 sieve 3) 40% passing No. 40 sieve 4) mint LL = 45 & PI = 25 20 18 percent passing No. 200 sieve; 65 p ercent 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 great power 23 potty- muckle (Phase Diagram) • building block book of account of smut contains: gibe flashiness Va broadcast tot Vt Vv Vw Vs irrigate Ww Ws burthen Wt Soil †telephone line (gases) †body of water system (fluid) †Solid Particles 24 wet national = ? eight of water/ freight of dry soil ? = Ww/Wd water firing/(moist soil fish †water loss) ? = Ww/(Wm-Ww) and ? =(Wm-Wd)/Wd 25 Mass †multitude Relationships stringency or social unit lean = Moist social unit encumbrance = ? m ? ?m = Wm/Vt = ? d + ? ?d ? = (? m †? d )/ ? d ? ?d + ? d = ? m ? m= (1+ ? ) ? d ? d = ?m/(1+ ? ) b 26 rack up batch = ? Volume (solid + water + air) = Vs+Vw+Va ? Va = Vt †Vs- Vw sub stance Volume Va blood line rack up Vt Vv Vw Vs body of water Ww Ws Weight Wt Soil 27 Relationship Between Mass & Volume Volume = Mass/(Specific gravitation x Unit Weight of water supply) = Ws/(SGxWw) Va summarize Volume Air score Vt Vv Vw Vs Water Ww Ws Weight Wt Soil 28Specific sedateness = weight of material/ weight of same ledger of water Soil Specific Gravity Typical Range 2. 65 to 2. 70 Specific Gravity of Water = 1 29 intensiveness = S verbalised as percent S = flashiness of water/ masses of voids x ascorbic acid fundamental Volume Va Air Total S = Vw/Vv x 100 Ww Ws Weight Vt Vv Vw Vs Water Wt Soil Always ? 100 30 Porosity n = rule book of voids/ bestow volume n = Vv/Vt debase 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 storey of strength 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 volume 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 situration 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 resolve is â€Å"C” 34 Ref: NAVFAC DM-7 35 resume remove Adjustments follow Material Properties: ?m = cx pcf & ? = 10% placed Fill Properties: ? d = cv pcf & ? = 20% How much take on is inevitable to produce 30,000 cy of remove? How much water must be added or upstage from each cf of involve? Total Volume Va Air Total Vt Vv Vw Vs Water Ww Ws Weight Wt Soil 36 take Fill Adjustments Borrow Material Properties: ?m = 110 pcf & ? = 10% ?d = ? m /(1+? ) = 110/(1. 10) =100 pcf; Ww = 110-100=10 lbs pose 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% primed(p) Fill Properties: ? d = 105 pcf & ? = 20% Since bear ? d =100pcf & fit ? 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 lb s 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 monitoring device: wet parsimoniousness Relationships Establishes the unique relationship of wet to dry density for each specialised soil at a undertake compaction energy MOISTURE-DENSITY alliance 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 wet gist (%) 40 Proctor: wet Density Relationships • 4” mold 25 blows • 6” mold 56 blows measuring †5. 5 lb power hammer †dropped 12 in †3 layers Standard: ASTM D-698 AASHTO T-99 limited: ASTM D-1557 AASHTO T-150 • Modified †10 lb hammer †dropped 18 in †5 layers 41 PROCTOR COMPACTION TEST utmost run dry Density †Highest density for that degree of compactive political campaign Optimum wet Content †moisture conten t at which uttermost dry density is achieved for 42 that compactive effort Proctor: wet Density Relationships MOISTURE-DENSITY human relationship 108. 0 106. 0 104. 0 modify 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 indispensable 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 reduce A to B (14 to 24 %) 44 Proctor: Zero Air Voids air travel Relationship of density to moisture at saturation for constant specific sedateness (SG) Can’t achieve fill in zone right of zero air voids line ZMOISTURE-DENSITY RELATI ONSHIP 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. 65×62. 4/(1+24×2. 65/100) ? d=101. 1 pcf state is â€Å"B” 47 Ref: Peck Hanson & Thornburn Static judgment 48 take care strong tension at efflo rescence x Ref: Peck Hanson & Thornburn unadulterated Unit Weight ? sat 5’ ? sat = cxxv pcf Moist Unit Weight ? M Dry Unit Weight ? Dry 7’ go under (buoyant) Unit Weight = ? sat †62. 4 x 49 augur effective stress at point x Ref: Peck Hanson & ThornburnTotal Stress at X 5’ ? sat = cxxv pcf = 5 x 62. 4+ 7x 125= 1187psf concentre 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 down(prenominal) electric current Gradient 51 down(prenominal) hunt 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 prior problem 52 Upward Flow Gradient Ref: Peck Hanson & Thornburn 53 One Dimensional Consolidation ?e/pn 54 Primary Phase colonization (e log p) ? H = (H x ? )/(1+eo) e o ? H H 55 Consolidation turn up Pre-consolidation Pressure Cc = slope of e log p gross(a) curve est. Cc = 0. 009(LL-10%) Skempton Rebound or recompression curves 56 56 e- l o g p Calculate Compression major power; 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 dish out is â€Å"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 repoint Conditions • Q=kiAt • Q= k (h/L)At • k=QL/(Ath) 59 If Q =15cc & t = 30 atomic descend 16 what is the permeability k=QL/(Ath) 10cm 5cm A) 0. 01 cm/ unsweet B) 0. 01×10-2 cm/ due south 25cm2 C) 0. 1 cm/sec 60 Constant inte rrogative Permeability Calculate k Q =15cc & t = 30 sec • k=QL/(Ath) • k= 15(5)/(25(30)10) • k= 0. 01 cm/sec respond is â€Å"A” 10cm 5cm 25cm2 61 Falling Head Permeability • k=QL/(Ath) (but h varies) • k=2. 3aL/(At) log (h1/h2) • where a = pipette subject area • h1 = initial take aim • h2 = final charge 62 If t = 30 sec; h1= 30 cm; h2 = 15 cm L= 5 cm; a= 0. cm2; A= 30 cm2; calculate k A) 2. 3×10-3 cm/sec B) 8. 1×10-6 cm/sec C) 7. 7×10-4 cm/sec 63 Falling Head Permeability k=2. 3aL/(At) log (h1/h2) k= 2. 3 (0. 2) 5 /(30×30) log (30/15) k= 7. 7×10-4 cm/sec Answer is â€Å"C” 64 •Flow lines & head drop lines must intersect at right angles •All areas must be square •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 width of sectionalization t=time Flow Nets 6ft 66 What flow/day? as sume k= 1×10-5 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 â€Å"quick conditions” pc =2( one hundred twenty)= 240 psf (total stress) Flow Nets Below water level theatrical role 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 p’c = pc -(? + ?? ) p’c = 52. 8 psf >0, soil is not quick ?sat=120 pcf 68 Stress Change fix (1H:2V) For square footing ?? z=Q/(B+z)2 69 If Q= 20 kips, Calculate the steep stress increase at 7 feet below the footing stinker 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 & nonresilient material) If B= 6. 3’ in a square footing with 20 kips load, what is the vertical stress i ncrease 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 participate in the PE review course on Soil Mechanics! More questions or comments? You can email me at: [email&#clx;protected] com 77\r\n'