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SGI Varia; 75EXAMPLES FOR DESIGN OF FOUNDATIONSNguyen Truong TienSGI,LinkopingSweden, 1981 August

EXAMPLES FOR DESIGN OF FOUNDATIONSCONTENTExample 1.Design of a strip footingExample 2.Friction pile under vertical loadExample 3.Calculation of the settlement of ashallow footing embankment and apile foundationExample 4.Calculation of the settlement of a pile groupExample 5.Calculation of the primary consolidationsettlement of a road embankmentExample 6.Calculation of the bearing capacity ofa pile groupExample 7.Desing of foundation methodExample 8.Calculation of the settlement of a buildingExample 9.Design of piles by the CTH methodExample 1 0.Calculation of the stop driving criteriaExample 11.Design of a pile group subjected tohorizontal loadExample 1 2.Design of a footing subjected to horizontaland vertical loadExample 1 3.Calculation of the permissible load for arectangular footingExample 1 4 Calculation of the safety factorExample 1 5.Design of a pile group for a retaining wall

STATEN GEOTEKNISKA INSTITUTIExample 1Design a strip footing founded on a firm sandaccording toa) Swedish Building Code (SBN 75)b) the deformation methodc) ultimate bearing capacity formulasThe load on the slab 650 kN/m.QThe sand has p 1 . 8 t/m 3p, 1. 1 t/m 3 and cpI 36 .IThe ground water level is 2 m below ground surface,i.e. in the same level as the slab.Design also for a maximum total settlement accordingto Canadian manual (1978). Weight sounding 30 ht/0.2 m. Qc650KN/,,, . l . A « li!l-- &f--2,0.::.,l'l'l:C '' - i-0---.- ·' \D··.: '-c:r-- 4- . .'. GW.,Solution:a) SBN 75Allowable soil pressure:aSGI nr 198osS8'27ALLF185 1 001rn bn(1- b ) (1-tan 2 a )3123:5332

STATEN GEOTEKNISKA INSTITUT2For strip footing0oa b310, and a vertical load The ground water level is in the same level offooting :. h 0.The depth of footing under the ground surfaced 2.0 m, and for firm sand layer, from Table23:5332:n 0.22 MPa/mand0mmax 0.5 MPaSubstituting the value n in 23:5332,0.22 b MPaQ 0.650 MPa0actbb As00m act0.22 b then b 0.65b0.65/0.22 1.72 m (band0m1 . 7 m) 0.22 x 1.7 0.37 MPa 0.5 MPa.Comment: the foundation is designed with safety factorequal 3 according to SBN 75.b) The deformation methodAccording to Maslov's method, the allowable soilpressure0 m0mis calculated by the formula:TTgp' (b tan d p/p' c cot /gp') gpdTTcot 2substituting the values:0mSGInr198oa sas21ALLF 1as 79 001 TT 10 1.1 (b tan 36 2 0)1· 1 10·1.8·2cot 360- 36·TT2780

3STATENS GEOTEKNISKA INSTITUTAfter transformationCJas CJm 57.9 b 297actand CJm 650b CJthenactb 1 67 m(b 1. 70 m).Comment: The allowable soil pressure O-m, is the pressurethat don't develop the plastic zone below thedepth zmax, where zmax b tan jl '( see Fig 1 . 2) .Qc) By Canadian manual 1978The allowable bearing pressure is calculated b y :qa is(cNC pgd Nq ½ p 'gb Ny )For jl' 36 from Fig 3.4: Nq Ny 32 . 67(after Hansen)As qq CJ act 6501Jand assumed FS 3 then,-650-b 31 ( 1 . 8 · 1 0 · 2 · 3 2 . 6 7 ½ 1 1 . 1 0b · 3 2 . 6 7)After transformationb 2 6.54b -650 0b 1.37 m orb 1.40 mSGI nr 198DB 36527mr1651s 0 0 1

STATENS GEOTEKNISKA INSTITUT4therefore:0650 7:; act(Comment:0464 kPa 0.46 MPaact 0.5 MPa (SBN 75)).d) By DIN 4017The ultimate bearing capacity is calculated byAs strip foundation -vyFor p 36 :NqNyAssumed FS \) q1.39.4825.363 I therefore qa qu3.Substituting the values:650-b 13 ( 1 . 8 · 10 2 · 3 9 . 4 8 · 1 1 . 1 · 1 0 b · 2 5 . 3 6 . 1)After transformation:b 2 2.53b - 6.86 0b 1.64 m(b 1. 60 m)therefore0act650 T:"6 406 kPa(Comment: oact 0.406 MPa 0.5 MPa (SBN 75)).SGI nr 198oo as,21ALLF 1 ss 1 001

5STATEN GEOTEKNISKA INSTITUTe) Hungarian foundation specification (MSZ 15004)Assume the reduction factor to be a 1 0,7,a20.7, a3 0.6 a a1 a2 a3 0.29And the allowable pressure is:qa aquThe ultimate bearing capacity of the soil is cal culated by:qu p'gbNy0for cp 36 : Ny pgdNq 40.2 and Nq 37.8Substituting the values:qu 1.1 ·10-b·40.2 1.8·10·2·37.8qa 6 0 0.29(422b 1360.8);After transformation:b 2 3.23-5.31 0b 1,2 m.aact650 1 2 542 kPa 0.54 MPa 0.5 MPa (SBN 75)Comment: The a is 1/FS, as the value Ny is highaccording to c) and d), so the footing widthis only 1,2 m.SGI nr 198os ss,21ALLF I as ,s 001

6STATENS GEOTEKNISKA INSTITUTf) by Handbook Foundation Engineering (Kezdi)The allowable bearing pressure is obtained from theformula:1qa FS (cNc pdgNq ½ p' gbNy)as qa for '0650act 1 ) 'C 0 and FS 3.0 36 , bearing capacity factors (Table 3.1):Nq 37.75, Ny 56.31Substituting the values:650 {(10 1.8·2·37.75 ½ 1.1·10·56.31, b)bb 2 4.39b - 6.3 0b 1.14 m(b 1.20 m)Comments: The values of Ne are almost the same ford),e) and f) but Nyhas a big variation. Bydifferent solutions of ultimate bearing ca pacity, the variation of bis from 1.2 m to1.60 m, this is due to the variation of the Nyvalue. The value of b 1.2 m obtained bye)and f) will give a pressure not permittedaccording to SBN 75.g) Design for a total maximum settlementAccording to the Canadian manual for structures insand the allowable maximum total settlement is3.8 cm. The settlement may be estimated using therelationship:b qnetfCEswhere Es modulus of elasticity of the soilf C settlement coefficient, as given i Fig. 5.3(Kany)qn net design bearing pressuresSGI nr 198os ss521ALLF I a 1s 001

STATEN GEOTEKNISKA INSTITUTExample 2A friction pile 30 x 30 of precast concrete is drivento 12 m depth in firm sand. Calculate the bearingcapacity froma) static point of viewb) dynamic point of viewc) SW-bored pile instructions.The pile driving hammer has a free fall of 0.5 m.Allowable remaining settlement for the 6000 kN pileis 0.7 mm according to the Swedish Pile codes.The sand hasp 1.8 t/m 3,p' 1.1 t/m 3 and 34 .The ground water level is 2 m below the ground surface.7Young's modulus is 3x10 kPa.Also calculate the bearing capacity of the pile,· con sidering the effect of critical depth.Example 2firm sand 34 SGI nr 198oa sas21All, ,as 1s 001

STATENS GEOTEKNISKA INSTITUT7The value of qn is equal to q-pgd, where q 650/b,therefore:6506 0650 36qn 7:, - pgd - 1.s·10.2 7:,For strip footing a, and assumed the compressiblelayer z 2B, by Fig. 5.3, fc 1.07. The result ofweight sounding is given Mwcone static resistance qBored instructions).C 30 ht/0.2 m, which means the 10 MPa (q-c 0,33 M, SwWThe modulus of elasticity may bedetermined by the relationship:ES (1.5-:- 2) qC(Canadian manual, SAA code)if assumedE s 1.7 q, then-cEs 1,7·10 MPa 17 MPa 170.10 2 kPaSubstituting the values:3.8·10- 2 b(650 36)b1.07170-10 2After solution:b 1.37 m(b 1.40 m)Comment: the method require a approximation of the valueof the modulus of elasticity E, therefore itsis useful only for preliminary design purposes.SGI nr 198os as,21ALLF I as10 001

- 38 0. 7 4 B/20 .74 L/20. 1 00. 15L B0.20. . . . . . .0.30co.0 .40N0.50CO .M PRESS I BLE LAYE R.Q.LUzy ·. . . . .Es,:::cI .) 0.70C)LU 1 . 00 I (-'1 5 0LU0:::2.003.004.00SETTLEMENTCALCULATIONs uEfC . .s5.007.00WHERE q I q-yDf10.00L/B l .015.002 0. 00 -------.L.----.L--.Ll.-,-.1.--.l- J, ,1 . 01 .5 0.50SETT LEMEN T COE FF I Cl ENT, fFIG(After Kany)5. 3CURVESCRITICALCFOR CALCULATION OF SETTLEMENT ATPOINT , CTHE

STATENS GEOTEKNISKA INSTITUT2a) Statically:The bearing capacity of the pile: Pf PpP skin friction: For one pile element dz (Fig 2.2),fequilibrium gives:I·d P K O I 0 tan odzJSVLp fJ K o ' 0 tan odz - 0sVIltan o .Fig. 2.2IOv0 perimeter of pile andwhere K Isohtan o is the friction between thepile and the soil.Fig 2.3 shows the distribution of horizontal pressure,then the resulting horizontal load P is:/After Broms, the angle of friction atthe sand/concrete pile contacto and Ks ¾ · ¾34 25.5 2 for firm sandsubstituting the values: k:s( d 2 '- h:cz)& ,SGI nr 198oa sao212-3 '"" ,a 0012 x 4 x 0 3 (½ x 1 0x 1 .s xf 2x10x1.8;10x1 .1x8.8tan 25.50Pf 1090 kN

3STATEN GEOTEKNISKA INSTITUTThe point resistance is calculated by Caquot Kerisel:p-tanA-'p Ap gpL e' 'where: cross-sectional area of the pile tippgpL effective overburden pressure at the pile tip.ASubstituting the values:2pp 0.3pp 1470 kN10X(2x1 .8 1 0x1 .1)12Xe?tan /)'Then the bearing capacity of the pile isp P f P p 1090 14 70 2 5 6 0 kN .The Canadian ManualPoint resistance: Assume that L Dcri·t·ica 1 therefore:pg effective unit weight of the soilNq the bearing capacity factor for pile as derivedfrom Brezantsev (1961)For j) 34 Nq"" 75Substituting the values:2Pp 0.3 (1.8x10x2 1.1x10x10)x75P p 985 kNCalculation of the skin friction follows the sameprocedure that was presented, then:PSGInr198oB as,21ALLF 1 ss 10 001 1090 985 2075 kN

4STATENS GEOTEKNISKA INSTITUTb) DynamicallyThe bearing capacity of the pile is calculated by thefollowing pile driving formula:(SBN 75)n correction factor for the drop hammer ( 1 fora free falling hammer)Qr Qp h eCweight of the hammer (at least 3 tons for aconcrete pile). Here Qr 4 tons.total weight of the pile2Qp 2.4 X 0.3 X 12 2,59 tonsheight of drop of hammer: 0.5 mremaining penetration: 0.7 mm 0.7x10-3 mrebound of the soilThe rebound of the soil for a concrete pile can beassumed.1e Qrqp, where qp is the weight of the pile permetre, herele 42.4x0.3 2 18. 7 mlf length of the follower. Here lf 0Ep modulus of elasticity (Young's modulus):3x10 7 kPaSubstituting the values in the expression of c:C P11 X 18.7 6.9 X 10-6P0,32x3x107uSubstituting the values in the pile driving formula:Pu 0.8 x 1 x 0.5 x 4(1-0.2x2.59/4.0)/(0.7 0.0069p)u 0,0035Pu 1980 kNSGInr198os as521ALLF , as1 001 0.7 P·V- 14960 0

5STATENS GEOTEKNISKA INSTITUTc) by Sw.Bored Pile Instructions with NFor diagram 2.14211a, with 6: 3s ' 34 40, one obtains1the value N P /0 42 for a bored pile.qsf Vwhere Psf the failure pressure at the bored pile pointa' effective overburden pressure at the pileVpoint level. the bearing capacity factor.NqFor driven piles the bearing capacity is 3 times thatof bored piles, so N 3 x 42 146 for a driven pile.q01V 2 x 1.8 x 10 10 x 1.1 x 10 146 kPaThe failure pressure of the driven pile:Psf 146 x 146 18 396 kPaAccording to the diagram 2.14212p/0 V1m 12 for bored pilesExperience gives pm/0 V1 3 x 12 36 for driven pilesThe average overburden pressurea'V 0 146 73 kPap 36 x 73 2628 kPa2thenAspmmis the permissible shaft stress with the safetyfactor of 2, then the failure skin friction is assumed:Pmf 2628 x 2 5256 kPaand the bearing capacity of the pile:P Ap(psf pmf) 0 . 3 2 ( 1 8 3 9 6 5 2 5 6)SGInr198os sss21ALLF 1as 79 991 21 2 8 kN

6STATENS GEOTEKNISKA INSTITUTThe maximum shaft resistance can be calculated bythe formula:a'2l1-(1- u) (1-lQ.) jtan pP mf Ks 0v 4 l:.1Da'nVK-a sVa'uL1D0n earth pressure coefficient mean value of a'V 73 kPa 0 12 m 0.3 m 25 5 40 !:1.D1. 5Substituting the values:Pmf 1 5 x 7 3 x 4 x 4 0 [ 1 - ( 1-l ) 2 J 0. 4 8P mf 3700 kPaand the bearing capacity of the pileP 2 0 3 ( 1 8 3 9 6 3 7 0 0) mA ( P f p f)ps198 9This is almost the same value as obtained from thediagram 2.14212 and use of the safety factor of 2.0.Comment: The bearing capacity of the pile is calculatedby different methods, and is 2000 7 2500 kN. If a safetyfactor of 3 is assumed, the allowable load is 660 kN.The Sw. Bored Pile Instructions consider critical depth.SGI nr 198os sss21ALLF 1s, 10 001

7STATEN GEOTEKNISKA INSTITUTd) Calculation of the bearing capacity of the pile bythe Australian Code (SAA, 1978).The unit skin friction and point resistance reachlimiting values at a pile depth of about 8 to 20 timesthe pile diameter depending on the relative density.The value of critical depth D, the bearing capacityCfactor N,and the factor of correction for differentqtypes of pile and relative density are presented inTable A. 1 1 . 2 .In this case as ' 34 , the sand is medium dense andDCcl-F 1 . 0 and N 1 00.q 8,where d is the diameter of the pile. For equivalentarea, d 0.34 m and D C 8X0.34 2.72 m.The unit skin friction at critical depth is:F X 0 1V 1(1.8x10x2 1.1x10x0.72) 44 kPaThe unit point resistance:psf 01VN 44 x 100 4400 kPa.qThe bearing capacity of the pile is calculated by:PmfQ Qf Q p -2- 0 Dc p mf 0(L-D c ) Ap p. sfSubstituting the values:Q 4422x4x0.3x2.72 44x0.3x4(12-2.72) 0.3 x4400Q 958 kNComment: Considering the effect of critical depth, thebearing capacity is reduced to almost 50% of the valuesobtained by a), b) and c).SGInr198os ss521ALLF ,a, 10 001

STATEN GEOTEKNISKA INSTITUTExample 3Calculate the primary consolidation settlement in theclay in case a) to f)for one layer solution. The soilconsist of 10.0 m clay on sand. The clay is normally3consolidated, p 1.6 t/m , and the compression modulusnumber m 10.0. The ground water table is 1.0 m belowthe ground surface.a) Loading over a large areab) Lowering of the ground water table by 1 0 mc) Lowering of the pore pressure in the bottom ofthe clay layerd) Building 1 0e) Building 1 0surfacef)XX210 m10P.l2on the ground surfacefounded 2 m below the groundBuilding on piles.Solution: '10 lcP ,.JJ :.,X,.J- .,',-,e-,¼ !!e-r,: -r, "'/17.' I"\""-,A-F g. 3.1The settlement is caluclated byo0 1 60 1 h 1 ln . a,mif OCR 1.SGI nr 196Klintland Gratiska, Linkoping 19810'0

2STATEN GEOTEKNISKA INSTITUTwhere100is the effective overburden pressure and6 0 1 is the additional pressure due to the loading.Effective pressure at the middle of the layer 0 ' 1. 60 60As q110XX1 . 0 0.6X10X4 40 kPa 1 0 kPaThen0 1 000X40 101 22 cm7o ln 40b) Lowering of ground water table by 1.0 m I- -or , ,,1e1f--"- .::;.;.,,;. ,0 vnJ.:.:.::: .:.:.- ,1:.:. '.::.o vn ; -"-----'n.,."'1 GGV\JVV8 , 0mI Fig. 3.2Effective overburden pressure at