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**5. Bearing Capacity of Shallow Footings**

CIV4249: Foundation Engineering Monash University

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**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

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**Limit State Design Limit state design equation: y F < f R**

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

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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

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**Geotechnical Design y F < f R 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

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**Applied Stress, F Fwall = 120 kN/m : Wwall = 20 kN/m : Wfoot = 10 kN/m**

What is the applied stress in these two situations? 1.0m 1.2m Applied stress = ( ) / 1.2 = 125 kPa in both cases

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**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/m3 What is the net applied stress in these two situations? 1.0m 1.2m Net applied stress for surface footing = 125 kPa Net applied stress for buried footing = *1.0 = 105 kPa qnet = 125 kPa qnet = 105 kPa

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**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

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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

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**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.

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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.

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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. 1 2 Weightless soil f = 0 qu = 4c 2c 2c 4c

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Upper Bound Solution Weightless soil f = 0 qu “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 = 2pc r O c e.g. slope stability - optimize failure surface; choose FoS

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**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 ¹ 0 : Rough punch passive active log spiral

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**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*

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**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.

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**Terzaghi Bearing Equation**

qu nett = c.Nc + p'o (Nq - 1) + 0.5B'N Solution for c and only soil Solution for D and only soil Solution for and only soil

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**Terzaghi Bearing Equation**

qu nett = c.Nc + p'o (Nq - 1) + 0.5B'N Overburden p'o = 'o D B Failure Zone (depth 2B) Generalized soil strength : c, (drainage as applicable) Soil unit weight : ' (total or effective as applicable) Adopt weighted average values !

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**Terzaghi Bearing Equation**

qu nett = c.Nc + p'o (Nq - 1) + 0.5B'N applies to strip footing Nc, Nq and N 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

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**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)

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**Example #1 1.0 1.7 x 2.3 Stiff Clay : cu = 75 kPa fu = 0o g = 18 kN/m3**

( *1.7/2.3) = 1.148 75 kPa 5.7 1.0 undefined 1.148*75*5.7*1.7*2.3 = 1919 kN 0.45 * 1919/1.33 = 649 kN = 166 kPa Shape Factor = c = Nc = Nq = Ng = Qu nett = / Qu nett =

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**Example #2 1.0 1.7 x 2.3 Medium Sand : c = 0 kPa f' = 35o g = 20 kN/m3**

40 10.2kN/m3 [20*(40-1)+0.5*0.852*1.7*10.2*40]*1.7*2.3 = 4205 kN 0.45 * 4205/1.33 = 1422 kN = 364 kPa c = p'o = Nq = g' = Ng = Qu nett = f/y Qu nett =

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**Footings with eccentric loads**

P e rigid e < B/6 : qmin qmax qmin = P (1-6e/B)/BL qmax = P (1+6e/B)/BL

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**Footings with eccentric loads**

P e rigid e > B/6 : qmin qmin = 0 qmax = P L(B-2e) qmax

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**Meyerhof Method for eccentric loads**

B P L' = L- 2e

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**Meyerhof Method for eccentric loads**

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2-way eccentricity L B 2e2 B' = B- 2e2 e2 e1 P 2e1 L' = L- 2e1

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**Footings with moments e = M P treat as equivalent eccentric load P M e**

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**Equivalent footing example**

Light tower 5.3x5.3 m Vertical Load = 500 kN Equiv Horizontal Load = 30 13m above base Determine: a) Maximum and minimum stresses under the footing b) Equivalent footing dimensions

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**Equivalent footing example**

Light tower 5.3x5.3 m Effective eccentricity = e/B = min = max = Effective area = 30*13/500 = 0.78m 0.78/5.3 = < 0.166B 500*(1- 6*0.147)/5.32 = 2.1 kPa 500*(1+ 6*0.147)/5.32 = 33.5 kPa 5.3 * ( *0.78) = 5.3 * 3.74m

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**Inclined Loads Fc = Fq = (1 - d / 90)2 Fg = (1 - d / f)2**

Correction Factors, Fc , Fq and Fg empirically determined from experiments

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**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

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**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

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**Example 4 - Bearing capacity after Hansen**

Ground inclination = 3.5o Load inclination = 10o 0.6 Firm Clay : cu = 40 kPa fu = 0o g = 17 kN/m3 1.0 1.7 x 2.3 Medium sand : f' = 34o g = 20 kN/m3 1.5 Grading dense : f' = 40o g = 21.5 kN/m3 Determine the ultimate bearing capacity (in kN)

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**Example 4 - Bearing capacity after Hansen**

1.5*34+1.9*40/3.4 = 37o 42.9 47.4 1.445 0.704 1.239 1.00 = Nc = Nq = N = sq = s = dq = d = iq = i = gq = g = bq = b = q = = 0.831 (1 = 2) 0.674 (2 = 3) 0.80 1.00 17*0.6+7*0.4 = 13 1.5* *11.7/3.4 = 11.0 kN/m3 Qu = 1.7*2.3*( ) = 3250 kN

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**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? Take strength of underlying stiffer material into account Approaches based on taking weighted average strength See Bowles, Das or other text B soft clay trending stronger 2B stiff clay or dense sand

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**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 dense sand soft clay

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**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

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Session 5 – 6 BEARING CAPACITY OF SHALLOW FOUNDATION

Session 5 – 6 BEARING CAPACITY OF SHALLOW FOUNDATION

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