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5. Bearing Capacity of Shallow Footings CIV4249: Foundation Engineering Monash University CIV4249: Foundation Engineering Monash University.

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Presentation on theme: "5. Bearing Capacity of Shallow Footings CIV4249: Foundation Engineering Monash University CIV4249: Foundation Engineering Monash University."— Presentation transcript:

1 5. Bearing Capacity of Shallow Footings CIV4249: Foundation Engineering Monash University CIV4249: Foundation Engineering Monash 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: –Loads –Capacity

3 Limit State Design Limit state design equation:  F <  R F = action (kN or kPa)  = load factor –(AS1170) - Loading Code –Dead Load = 1.25 –Live Load = 1.50 –Hydrostatic = 1.00 Typical value –2/3 dead + 1/3 live –  = 1.33 R = capacity (kN or kPa)  = capacity red n 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

4 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 q allow = q ult  FoS I want you to apply limit state design principles Equivalent “Factor of safety” =  /  For  = 1.33 implies  = 0.44 to 0.53

5 Geotechnical Design Generally working with stresses applied stressOn LHS concerned only with that applied stress which acts to cause rupture available strengthOn RHS concerned with the available strength which acts to prevent rupture  F <  R

6 F wall = 120 kN/m : W wall = 20 kN/m : W foot = 10 kN/m What is the applied stress in these two situations? Applied Stress, F 1.2m 1.0m

7 Net Applied Stress, F F wall = 120 kN/m : W wall = 20 kN/m : W foot = 10 kN/m :  = 20 kN/m 3 What is the net applied stress in these two situations? 1.2m 1.0m q net = 125 kPa q net = 105 kPa

8 Net Applied Stress Rule For bearing capacity: q net applied =  ' below -  ' beside ALWAYS WORK WITH NET APPLIED STRESSES NEVER WORK WITH GROSS APPLIED STRESSES

9 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

10 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.

11 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.

12 will not fail“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. Lower Bound Solution 1 2 Weightless soil  = 0 ququ 0 2c 4c = 4c

13 Upper Bound Solution must“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.” Weightless soil  = 0 ququ r O c c q u r. r/2 =  r.c.r = 2  c e.g. slope stability - optimize failure surface; choose FoS

14 Other classic analytical solutions for weightless soils: Solutions with  = 0 : –Prandtl smooth punch : q u = 5.14c –Prandtl rough punch : q u = 5.7c Solutions with   0 : –Rough punch passive active log spiral

15 Solutions for real soils There is no rigorous mathematical solution for a soil which contains cohesion, c, and angle of friction, , and weight, . EmpiricalnumericalEmpirical or numerical approaches must be used to provide methods of estimating bearing capacity in practical situations. finite elementboundary elementNumerical approaches include finite element and boundary element methods and would rarely be used in practice *

16 Terzaghi Approximate Analysis Solution for soil with c, ,  and D > 0 Solution is based on superposition of 3 separate analytical cases: –Soil with  and  but c = D = 0 : q u = N .f(  ) –Soil with  and D but c =  = 0 : q u = N q f(D) –Soil with  and c but  = D = 0 : q u = N c f(c) Each case has a different failure surface, so superposition is not theoretically valid.

17 Terzaghi Bearing Equation Solution for c and  only soil q u nett = c.N c + p' o (N q - 1) + 0.5B  'N  Solution for D and  only soil Solution for  and  only soil

18 Terzaghi Bearing Equation B q u nett = c.N c + p' o (N q - 1) + 0.5B  'N  p' o =  ' o D Generalized soil strength : c,  (drainage as applicable) Soil unit weight :  ' (total or effective as applicable) Overburden Failure Zone (depth  2B) Adopt weighted average values !

19 Terzaghi Bearing Equation –applies to strip footing –N c, N q and N  are functions of , and are usually given in graphical form –c,  and  ' 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 q u nett = c.N c + p' o (N q - 1) + 0.5B  'N 

20 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: –N c multiplier is (1+ 0.2B/L)

21 Example #1 Stiff Clay : c u = 75 kPa  u = 0 o  = 18 kN/m x 2.3 Shape Factor = c = N c = N q = N  = Q u nett =  Q u nett = ( *1.7/2.3) = kPa undefined 1.148*75*5.7*1.7*2.3 = 1919 kN 0.45 * 1919/1.33 = 649 kN = 166 kPa

22 Example #2 Medium Sand : c = 0 kPa  ' = 35 o  = 20 kN/m x 2.3 c = p ' o = N q =  ' = N  = Q u nett =  Q u nett = 0 kPa 1.0*20 = 20 kPa kN/m 3 40 [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

23 q min q max e P e < B/6 : q min = P (1-6e/B)/BL q max = P (1+6e/B)/BL rigid Footings with eccentric loads

24 q min = 0 q max = 4P. 3L(B-2e) q min q max e P e > B/6 : rigid Footings with eccentric loads

25 Meyerhof Method for eccentric loads P e L 2e2e LL- 2e L' = L- 2e B

26 Meyerhof Method for eccentric loads

27 2-way eccentricity P e1e1e1e1 L 2e12e12e12e1 LL- 2e 1 L' = L- 2e 1 B e2e2e2e2 2e22e22e22e2 BB- 2e 2 B' = B- 2e 2

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

29 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

30 Effective eccentricity = e/B =  min =  max = Effective area = Light tower 5.3x5.3 m 30*13/500 = 0.78m 0.78/5.3 = < 0.166B 500*(1- 6*0.147)/5.3 2 = 2.1 kPa 500*(1+ 6*0.147)/5.3 2 = 33.5 kPa 5.3 * ( *0.78) = 5.3 * 3.74m Equivalent footing example

31 Inclined Loads Correction Factors, F c, F q and F  empirically determined from experiments F c = F q = (1 -  / 90  2 F   = (1 -  /  ) 2

32 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 q u = cN c s c d c i c + qN q s q d q i q  'BN  s  d  i  s, d, and i are shape, depth and load inclination factors

33 Analyses by Hansen, Vesic q u = cN c s c d c i c g c b c + qN q s q d q i q g q b q  'BN  s  d  i  g  b  N c,N q,N   : Meyerhof bearing capacity factors s c,s q,s  : shape factors d c,d q,d  : depth factors i c,i q,i  : load inclination factors g c,g q,g  : ground inclination factors b c,b q,b  : base inclination factors

34 Example 4 - Bearing capacity after Hansen x 2.3 Medium sand :  ' = 34 o  = 20 kN/m 3 Grading dense :  ' = 40 o  = 21.5 kN/m Load inclination = 10 o Ground inclination = 3.5 o Firm Clay : c u = 40 kPa  u = 0 o  = 17 kN/m 3 Determine the ultimate bearing capacity (in kN)

35 0.831 (  1 = 2) (  2 = 3) *0.6+7*0.4 = * *11.7/3.4 = 11.0 kN/m 3  = N c = N q = N  = s q = s  = d q = d  = Example 4 - Bearing capacity after Hansen i q = i  = g q = g  = b q = b  = q =  = Q u = 1.7*2.3*( ) = 3250 kN 1.5*34+1.9*40/3.4 = 37 o

36 Stratified Deposits - 1 soft clay stiff clay or dense sand B 2B2B2B2B trendingstronger 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

37 Stratified Deposits - 2 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 soft clay dense sand

38 Terminology Ultimate Bearing Pressure, q uUltimate Bearing Pressure, q u –as computed by any number of methods Maximum Safe Bearing Pressure, q sMaximum Safe Bearing Pressure, q s –q s = q u  FoS Allowable Bearing Pressure, q aAllowable Bearing Pressure, q a –take settlement into consideration : q a  q s Design Pressure, q dDesign Pressure, q d –construction practicalities/ standardization may dictate larger footings : q d  q a


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