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5. Bearing Capacity of Shallow Footings CIV4249: Foundation Engineering Monash University 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: – Loads – Capacity Limit State Design • Limit state design equation: yF<fR • F = action (kN or kPa) • y = load factor – – – – (AS1170) - Loading Code Dead Load = 1.25 Live Load = 1.50 Hydrostatic = 1.00 • Typical value – 2/3 dead + 1/3 live – y = 1.33 • R = capacity (kN or kPa) • f = capacity redn factor – – – – (AS2159) - Piling Code Static test = 0.70 to 0.90 CPT design = 0.45 to 0.65 SPT design = 0.40 to 0.55 • Why a range? – variability in site conditions and in quality or quantity of exploration Factor of Safety • Working or Allowable stress method is currently used in practice • No Australian Standard • By convention Factor of Safety = 2.5 to 3.0 • qallow = qult FoS • I want you to apply limit state design principles • Equivalent “Factor of safety” = y/f • For y = 1.33 implies f = 0.44 to 0.53 Geotechnical Design yF<fR • Generally working with stresses • On LHS concerned only with that applied stress which acts to cause rupture • On RHS concerned with the available strength which acts to prevent rupture Applied Stress, F • Fwall = 120 kN/m : Wwall = 20 kN/m : Wfoot = 10 kN/m • What is the applied stress in these two situations? 1.2m 1.0m Net Applied Stress, F • Fwall = 120 kN/m : Wwall = 20 kN/m : Wfoot = 10 kN/m : g = 20 kN/m3 • What is the net applied stress in these two situations? 1.2m qnet = 125 kPa 1.0m qnet = 105 kPa Net Applied Stress Rule • For bearing capacity: qnet applied = s 'below - s 'beside ALWAYS WORK WITH NET APPLIED STRESSES NEVER WORK WITH GROSS APPLIED STRESSES Available Strength, R • Methods that can be used to determine available strength: 1. Historical / experience : • Building Codes may specify allowable values in particular formations 2. Field loading tests • Plate loading tests for very large projects 3. Analytical solutions • Upper and lower bound solutions for special cases 4. Approximate solutions • Solutions for general cases Field (Plate) Loading Tests - + 0.3m 1.2m • Testing footing under actual soil conditions • Measure load-deflection behaviour • Expensive mobilization and testing • Need to apply scaling laws • Different zone of influence • Affected by fabric – fissuring, partings etc. Analytical Solutions • The 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. Lower Bound Solution u = 4c • “If an equilibrium distribution of qstresses can be found which balances the applied load, and nowhere violates the yield criterion, the soil mass will not fail or will 1 2 - i.e. it will be just at the point of failure” Weightless be a lower-bound estimate of capacity. soil f = 0 0 2c 2c 4c Upper Bound Solution • “If a solution is kinematically admissible qu r. r/2 = p r.c.r and simultaneously satisfies equilibrium qu = 2pc considerations, failure must result - i.e. it O estimate of will be an upper-bound capacity.” c r c Weightless soil f = 0 e.g. slope stability - optimize failure surface; choose FoS Other classic analytical solutions for weightless soils: • Solutions with f = 0 : – Prandtl smooth punch : qu = 5.14c – Prandtl rough punch : qu = 5.7c • Solutions with f 0 : – Rough punch passive active log spiral 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* Terzaghi Approximate Analysis • Solution for soil with c, f, g and D > 0 • Solution 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. Terzaghi Bearing Equation qu nett = c.Nc + p'o (Nq - 1) + 0.5Bg'Ng Solution for c and f only soil Solution for D and f only soil Solution for g and f only soil Terzaghi Bearing Equation qu nett = c.Nc + p'o (Nq - 1) + 0.5Bg'Ng Overburden p'o = g'o D B Failure Zone (depth 2B) Generalized soil strength : c, f Soil unit weight : g' (total or effective as applicable) (drainage as applicable) Adopt weighted average values ! Terzaghi Bearing Equation qu nett = c.Nc + p'o (Nq - 1) + 0.5Bg'Ng – applies to strip footing – Nc, Nq and Ng are functions of f, and are usually given in graphical form – c, f and g' refer to soil properties in the failure zone below the footing – p'o is the effective overburden pressure at the founding level – shear strength contribution above footing level is ignored : conservative for deeper footings Application to other than strip footings • Strip footings represent a plane-strain case • What is different for a rectangular footing? • Correction factors applied - e.g. Schultz: – Nc multiplier is (1+ 0.2B/L) Example #1 1.0 1.7 x 2.3 Stiff Clay : cu = 75 kPa fu = 0o g = 18 kN/m3 • • • • • • • Shape Factor = (1 + 0.2*1.7/2.3) = 1.148 c = 75 kPa Nc = 5.7 Nq = 1.0 Ng = undefined Qu nett = 1.148*75*5.7*1.7*2.3 = 1919 kN f / y Qu nett = 0.45 * 1919/1.33 = 649 kN = 166 kPa Example #2 1.0 1.7 x 2.3 Medium Sand : c = 0 kPa f' = 35o g = 20 kN/m3 • c = 0 kPa • p'o = 1.0*20 = 20 kPa • Nq = 40 • g' = 10.2kN/m3 • Ng = 40 • Qu nett = [20*(40-1)+0.5*0.852*1.7*10.2*40]*1.7*2.3 = 4205 kN • f/y Qu nett = 0.45 * 4205/1.33 = 1422 kN = 364 kPa Footings with eccentric loads P e e < B/6 : rigid qmin qmin = P (1-6e/B)/BL qmax = P (1+6e/B)/BL qmax Footings with eccentric loads P e e > B/6 : rigid qmin qmin = 0 qmax = 4P . 3L(B-2e) qmax Meyerhof Method for eccentric loads L e B P 2e L' = L- 2e Meyerhof Method for eccentric loads 2.50 PB/L 2.00 q(min) Meyerhof q(max) average 1.50 1.00 0.50 0.00 0.00 0.05 0.10 0.15 0.20 e/B 2-way eccentricity L 2e2 B e2 P 2e1 L' = L- 2e1 B' = B- 2e2 e1 Footings with moments M P e P e=M P treat as equivalent eccentric load Equivalent footing example Light tower 5.3x5.3 m Vertical Load = 500 kN Equiv Horizontal Load = 30 kN @ 13m above base Determine: a) b) Maximum and minimum stresses under the footing Equivalent footing dimensions Equivalent footing example Light tower 5.3x5.3 m • • • • • Effective eccentricity = 30*13/500 = 0.78m e/B = 0.78/5.3 = 0.147 < 0.166B smin = 500*(1- 6*0.147)/5.32 = 2.1 kPa smax = 500*(1+ 6*0.147)/5.32 = 33.5 kPa Effective area = 5.3 * (5.3 - 2*0.78) = 5.3 * 3.74m Inclined Loads Fc = Fq = (1 - d / 90)2 Fg = (1 - d / f)2 • Correction Factors, Fc , Fq and Fg empirically determined from experiments Meyerhof Approx Analysis • differs from Terzaghi analysis particularly for buried footings – soil above footing base provides not only surcharge but also strength – more realistic i.e. less conservative qu = cNcscdcic + qNqsqdqiq + 0.5g'BNgsgdgig • s, d, and i are shape, depth and load inclination factors Analyses by Hansen, Vesic qu = cNcscdcicgcbc + qNqsqdqiqgqbq + 0.5g'BNgsgdgigggbg Nc ,Nq ,Ng : Meyerhof bearing capacity factors sc ,sq ,sg : shape factors dc ,dq ,dg : depth factors ic ,iq ,ig : load inclination factors gc ,gq ,gg : ground inclination factors bc ,bq ,bg : base inclination factors Example 4 - Bearing capacity after Hansen Ground inclination = 3.5o Load inclination = 10o 1.0 Firm Clay : cu = 40 kPa fu = 0o g = 17 kN/m3 1.7 x 2.3 1.5 Medium sand : f' = 34o g = 20 kN/m3 Grading dense : f' = 40o g = 21.5 kN/m3 Determine the ultimate bearing capacity (in kN) 0.6 Example 4 - Bearing capacity after Hansen • f = 1.5*34+1.9*40/3.4 = 37o • Nc = 0 • Nq = 42.9 • Ng = 47.4 • sq = 1.445 • sg = 0.704 • dq = 1.239 • dg = 1.00 • • • • • • • • iq = 0.831 (1 = 2) ig = 0.674 (2 = 3) gq = 0.80 gg = 0.80 bq = 1.00 bg = 1.00 q = 17*0.6+7*0.4 = 13 g= 1.5*10.2+1.9*11.7/3.4 = 11.0 kN/m3 Qu = 1.7*2.3*(664+168) = 3250 kN Stratified Deposits - 1 B soft clay 2B stiff clay or dense sand • For uniform soils, zone of influence typically ~ 2B • Failure surface will tend to be more shallow • Ignore strength increase? • Place footing deeper? trendingof underlying • Take strength stiffer stronger material into account • Approaches based on taking weighted average strength • See Bowles, Das or other text Stratified Deposits - 2 dense sand soft clay • Ignoring underlying layer unconservative • compute load spread and analyze as larger footing with reduced stress on underlying soil • use parameters of underlying soil in bearing equation • again, look at texts for different approaches Terminology • Ultimate Bearing Pressure, qu – as computed by any number of methods • Maximum Safe Bearing Pressure, qs – qs = qu FoS • Allowable Bearing Pressure, qa – take settlement into consideration : qa qs • Design Pressure, qd – construction practicalities/ standardization may dictate larger footings : qd qa