Presentation on theme: "5. Bearing Capacity of Shallow Footings"— Presentation transcript:
1 5. Bearing Capacity of Shallow Footings CIV4249: Foundation EngineeringMonash University
2 Bearing Capacity Ultimate or serviceability limit state? “What is the maximum pressure which the soils can withstand for a given foundation before the soil will fail?”Design for less but how much less?Uncertainty with respect to:LoadsCapacity
3 Limit State Design Limit state design equation: y F < f R F = action (kN or kPa)y = load factor(AS1170) - Loading CodeDead Load = 1.25Live Load = 1.50Hydrostatic = 1.00Typical value2/3 dead + 1/3 livey = 1.33R = capacity (kN or kPa)f = capacity redn factor(AS2159) - Piling CodeStatic test = 0.70 to 0.90CPT design = 0.45 to 0.65SPT design = 0.40 to 0.55Why a range?variability in site conditions and in quality or quantity of exploration
4 Factor of SafetyWorking or Allowable stress method is currently used in practiceNo Australian StandardBy convention Factor of Safety = 2.5 to 3.0qallow = qult ¸ FoSI want you to apply limit state design principlesEquivalent “Factor of safety” =y / fFor y = 1.33 implies f =0.44 to 0.53
5 Geotechnical Design y F < f R Generally working with stresses On LHS concerned only with that applied stress which acts to cause ruptureOn RHS concerned with the available strength which acts to prevent rupture
6 Applied Stress, F Fwall = 120 kN/m : Wwall = 20 kN/m : Wfoot = 10 kN/m What is the applied stress in these two situations?1.0m1.2mApplied stress = ( ) / 1.2 = 125 kPa in both cases
7 Net Applied Stress, F qnet = 125 kPa qnet = 105 kPa Fwall = 120 kN/m : Wwall = 20 kN/m : Wfoot = 10 kN/m : g = 20 kN/m3What is the net applied stress in these two situations?1.0m1.2mNet applied stress for surface footing = 125 kPaNet applied stress for buried footing = *1.0 = 105 kPaqnet = 125 kPaqnet = 105 kPa
8 Net Applied Stress Rule For bearing capacity:qnet applied = s 'below - s 'besideALWAYS WORK WITH NET APPLIED STRESSESNEVER WORK WITH GROSS APPLIED STRESSES
9 Available Strength, RMethods that can be used to determine available strength:1. Historical / experience :Building Codes may specify allowable values in particular formations2. Field loading testsPlate loading tests for very large projects3. Analytical solutionsUpper and lower bound solutions for special cases4. Approximate solutionsSolutions for general cases
10 Field (Plate) Loading Tests -+0.3m1.2mTesting footing under actual soil conditionsMeasure load-deflection behaviourExpensive mobilization and testingNeed to apply scaling lawsDifferent zone of influenceAffected by fabric -fissuring, partings etc.
11 Analytical SolutionsThe failure of real soils with weight, cohesion and friction is a complex phenomenon, not amenable to simple theoretical solutions.If simplifying assumptions are made, it is possible to develop particular analytical solutions.These analytical solutions must be based either on principles of equilibrium or kinematic admissibility.
12 Lower Bound Solution“If an equilibrium distribution of stresses can be found which balances the applied load, and nowhere violates the yield criterion, the soil mass will not fail or will be just at the point of failure” - i.e. it will be a lower-bound estimate of capacity.12Weightlesssoil f = 0qu= 4c2c2c4c
13 Upper Bound SolutionWeightlesssoil f = 0qu“If a solution is kinematically admissible and simultaneously satisfies equilibrium considerations, failure must result - i.e. it will be an upper-bound estimate of capacity.”qu r. r/2 = p r.c.r= 2pcrOce.g. slope stability - optimize failure surface; choose FoS
14 Other classic analytical solutions for weightless soils: Solutions with f = 0 :Prandtl smooth punch : qu = 5.14cPrandtl rough punch : qu = 5.7cSolutions with f ¹ 0 :Rough punchpassiveactivelog spiral
15 Solutions for real soils There is no rigorous mathematical solution for a soil which contains cohesion, c, and angle of friction, f, and weight, g.Empirical or numerical approaches must be used to provide methods of estimating bearing capacity in practical situations.Numerical approaches include finite element and boundary element methods and would rarely be used in practice*
16 Terzaghi Approximate Analysis Solution for soil with c, f, g and D > 0Solution is based on superposition of 3 separate analytical cases:Soil with f and g but c = D = 0 : qu = Ng.f(g)Soil with f and D but c = g = 0 : qu = Nq f(D)Soil with f and c but g = D = 0 : qu = Nc f(c)Each case has a different failure surface, so superposition is not theoretically valid.
17 Terzaghi Bearing Equation qu nett = c.Nc + p'o (Nq - 1) + 0.5B'NSolution for c and only soilSolution for D and only soilSolution for and only soil
18 Terzaghi Bearing Equation qu nett = c.Nc + p'o (Nq - 1) + 0.5B'NOverburdenp'o = 'o DBFailure Zone (depth 2B)Generalized soil strength : c, (drainage as applicable)Soil unit weight : ' (total oreffective as applicable)Adopt weighted average values !
19 Terzaghi Bearing Equation qu nett = c.Nc + p'o (Nq - 1) + 0.5B'Napplies to strip footingNc, Nq and N are functions of f, and are usually given in graphical formc, f and g' refer to soil properties in the failure zone below the footingp'o is the effective overburden pressure at the founding levelshear strength contribution above footing level is ignored : conservative for deeper footings
20 Application to other than strip footings Strip footings represent a plane-strain caseWhat is different for a rectangular footing?Correction factors applied - e.g. Schultz:Nc multiplier is (1+ 0.2B/L)
21 Example #1 1.0 1.7 x 2.3 Stiff Clay : cu = 75 kPa fu = 0o g = 18 kN/m3 ( *1.7/2.3) = 1.14875 kPa5.71.0undefined1.148*75*5.7*1.7*2.3 = 1919 kN0.45 * 1919/1.33 = 649 kN = 166 kPaShape Factor =c =Nc =Nq =Ng =Qu nett = / Qu nett =
22 Example #2 1.0 1.7 x 2.3 Medium Sand : c = 0 kPa f' = 35o g = 20 kN/m3 4010.2kN/m3[20*(40-1)+0.5*0.852*1.7*10.2*40]*1.7*2.3 = 4205 kN0.45 * 4205/1.33 = 1422 kN = 364 kPac =p'o =Nq =g' =Ng =Qu nett =f/y Qu nett =
23 Footings with eccentric loads Perigide < B/6 :qminqmaxqmin = P (1-6e/B)/BLqmax = P (1+6e/B)/BL
24 Footings with eccentric loads Perigide > B/6 :qminqmin = 0qmax = P L(B-2e)qmax
25 Meyerhof Method for eccentric loads BPL' = L- 2e
31 Inclined Loads Fc = Fq = (1 - d / 90)2 Fg = (1 - d / f)2 Correction Factors, Fc , Fq and Fg empirically determined from experiments
32 Meyerhof Approx Analysis differs from Terzaghi analysis particularly for buried footingssoil above footing base provides not only surcharge but also strengthmore realistic i.e. less conservativequ = cNcscdcic + qNqsqdqiq + 0.5g'BNgsgdgigs, d, and i are shape, depth and load inclination factors
34 Example 4 - Bearing capacity after Hansen Ground inclination = 3.5oLoad inclination = 10o0.6Firm Clay : cu = 40 kPa fu = 0o g = 17 kN/m31.01.7 x 2.3Medium sand : f' = 34o g = 20 kN/m31.5Grading dense : f' = 40o g = 21.5 kN/m3Determine the ultimate bearing capacity (in kN)
36 Stratified Deposits - 1 B soft clay 2B stiff clay or dense sand For uniform soils, zone of influence typically ~ 2BFailure surface will tend to be more shallowIgnore strength increase?Place footing deeper?Take strength of underlying stiffer material into accountApproaches based on taking weighted average strengthSee Bowles, Das or other textBsoft claytrendingstronger2Bstiff clay ordense sand
37 Stratified Deposits - 2 dense sand soft clay Ignoring underlying layer unconservativecompute load spread and analyze as larger footing with reduced stress on underlying soiluse parameters of underlying soil in bearing equationagain, look at texts for different approachesdense sandsoft clay
38 Terminology Ultimate Bearing Pressure, qu as computed by any number of methodsMaximum Safe Bearing Pressure, qsqs = qu ¸ FoSAllowable Bearing Pressure, qatake settlement into consideration : qa £ qsDesign Pressure, qdconstruction practicalities/ standardization may dictate larger footings : qd £ qa
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