Foundations with D f equal to 3 to 4 times the width may be defined as shallow foundations. TWO MAIN CHARACTERISTICS ULTIMATE BEARING CAPACITY

Transcription

1 oundation Analysis

2 oundations with D f eual to 3 to 4 times the width may be defined as shallow foundations. TWO MAI CHARACTERISTICS o Safe against overall shear failure o Cannot undergo excessive displacement, or settlement ULTIMATE EARIG CAPACITY o the load per unit area of the foundation at which shear failure in soil occurs

3 GEERAL SHEAR AILURE Long rectangular model footing of width at the surface of a dense sand The triangular wedge-shaped zone of soil marked I is pushed down and in turn presses the zones marked II and III sideways. The soil on the both sides of the foundation will bulge out and the slip surface will extend to the ground surface.

4 LOCAL SHEAR AILURE Medium dense sand or clayey soil of medium compaction Movement of the foundation will be accompanied by sudden jerks The triangular wedge-shaped zone of soil marked I is pushed down but unlike in general shear failure, the slip surface end somewhere inside the soil.

5 PUCHIG SHEAR AILURE airly loose soil Soil will not extend to the ground surface

6 GEERAL SHEAR AILURE Applies to dense granular soil and to firmer saturated cohesive soils subject to undrained loading (the UU and CU shearing conditions apply) PUCHIG SHEAR AILURE Applies to compressible soils, such as sands having lowto-medium relative density, and for cohesive soils subject to slow loading (the CD shearing condition apply)

7 Vesic, 963 Laboratory plate loadbearing tests on circular and rectangular plates supported by a sand at various relative densities of compaction, D r. D r about 70% (general shear failure)

8 Vesic, 973 Relationship for the mode of bearing capacity failure of foundations resting on sands.

9 Vesic, 973 D r, relative density of sand D f, depth of foundation measured from the ground surface = L + L where,, width of foundation L, length of foundation

10 General range of S/ with the relative density of compaction of sand. General Shear ailure ultimate load may occur at settlements of 4 to 0% of. Local Shear or Punching Shear ailure ultimate load may occur at settlements of 5 to 5% of.

11 earing capacity failure in soil under a rough rigid continuous (strip) foundation. The triangular zone ACD immediately under the foundation. The radial shear zones AD and CDE, with the curves DE and D being arcs of a logarithmic spiral 3. Two triangular Rankine passive zones AH and CEG

12 Continuous or Strip oundation where, u = c C + + γ γ c is the cohesion is the unit weight of soil is the euivalent surcharge load eual to γd f C,, γ are bearing capacity factors that are nondimensional and are functions only of the soil friction angle ɸ.

13 where, K pγ is the passive pressure coefficient Modified for: Suare oundation u =.3c C γ γ Circular oundation u =.3c C γ γ

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15 LOCAL SHEAR AILURE Strip oundation u = 3 c C + + γ γ Suare oundation u = 0.867c C γ γ Circular oundation u = 0.867c C γ γ ɸ = tan ( 3 tan ɸ )

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17 all = u S et stress increase on soil = net ultimate bearing capacity S net(u) = u where, net(u) is the net ultimate bearing capacity = γd f So, all(net) = u S The factor of safety should be at least 3 in all cases.

18 . A suare foundation is m x m in plan. The soil supporting the foundation has a friction angle of ɸ = 5 and c = 0 k/m. The unit weight of soil, γ, is 6.5 k/m 3. Determine the allowable gross load on the foundation with a factor of safety (S) of 3. Assume that the depth of the foundation (D f ) is.5 m and that general shear failure occurs in the soil. Solution u =.3c C γ γ rom Table 3., for ɸ = 5, C = 5.3 =.7 γ = 8.34 Thus, u =.3 0 (5.3) + (.5x6.5)(.7) + 0.4(6.5)()(8.34) = k/m

19 So, the allowable load per unit area of the foundation is all = u S = Thus, the total allowable gross load is = k/m Q = all ( ) = 359.5(x) = 438 k

20 The bearing capacity euation is modified when the water table is in the proximity of the foundation. earing Capacity Euation Modified earing Capacity Euation Case I Case II Case III

21 GEERAL SHEAR AILURE u u u c' c.3c'.3c' c c (Continuous or Strip oundation) (Suare oundation) (Circular oundation)

22 If 0 D D f, = D γ + D (γ sat - γ w ) u c' c = γd f where, γ sat = sat. unit wt. of soil γ w = unit wt. of water γ in ½γ γ becomes γ where γ = γ sat - γ w

23 If 0 d, = γd f u c' c γ in the last term is _ ' d ( ') * The preceding modifications are based on the assumption that there is no seepage force in the soil.

24 If d, u c' c *The water will have no effect on the ultimate bearing capacity.

25 SHAPE: The bearing capacity ens do not address the case of rectangular foundations (0 < /L < ). Wherein L >. DEPTH: The ens also do not take into account the shearing resistance along the failure surface in soil above the bottom of the foundation. LOAD ICLIATIO: The load on the foundation may be inclined.

26 u c' c cs cd ci s d i s d i where, c is the cohesion is the effective stress at the level of the bottom of the foundation γ is the unit weight of soil is the width of foundation (or diameter for circular foundation) cs, s, γs are shape factors cd, d, γd are depth factors ci, i, γi are load inclination factors c,, γ are bearing capacity factors

27 u c' α = 45 + ϕ / c cs cd ci s d i s d i = tan (45 + ϕ /) eπtan ϕ Reissner (94) c = ( ) cot ϕ Prandtl (9) γ = ( + ) tan ϕ Cauot and Kerisel (953), Vesic (973)

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29 Shape actors Reference: Deeer (970) L L L s s c cs 0.4 ' tan

30 Depth actors Reference: Hansen (970) ') sin '( tan ' tan ' 0.4 d f d c d d cd d d f cd f D or D or D tan ') sin '( tan ' tan ' 0.4 tan d f d c d d cd d d f cd f radians D or radians D or D

31 Inclination actors Reference: Meyerhof (963); Hanna and Meyerhof (98) ci i i ' 90 inclination of the load on the foundation with respect to the vertical

32 . Solve Problem using the general bearing capacity euation. Suare foundation m x m ' c' 6.5k / m S 3 D 5 0k / m f. 5 m General shear failure Red: Allowable gross load 3

33 u c' c cs cd Solution: earing Capacity actors: ci s d i s d i c tan ( ( (45 ' / ) e )cot' ) tan' tan ' tan (45 5/ ) e (0.66 )cot tan 5 (0.66 ) tan *Table 3.3 can also be used. Load Inclination actors Since load is vertical, ci, i, γi =

34 Shape actors cs s s L L 0.4 L c tan' cs =+ (/)(0.66/0.7) =.54 s = + (/)tan5 =.466 γs =-0.4(/)= 0.6 Depth actors (D f / =.5/ = 0.75) D f or' cd d d tan'( sin ') d d.33 c tan' D f tan 5 ( tan 5)( sin 5).5.33

35 u c' c cs cd ci s d i s d i u = x u =,373. k/m all = u S =, = k/m Q = x =,830.8 k

36 3. A suare foundation ( x ) has to be constructed as shown below. Assume that γ = 6.5 k/m 3, γ sat = 8.55 k/m 3, = 34, D f =. m, and D = 0.6 m. The gross allowable load, Q all, with S = 3 is 667. k. Determine the size of the footing using the general bearing capacity euation.

37 Since there is no cohesion, ecomes En. En. u c' c cs u all all cd ci Q all u S s 3 d s d i ' 667. k / m s d s ' d s s d d i earing Capacity actors rom Table 3.3, for ϕ = 34 = 9.44 γ = 4.06

38 s s d d 0.4 tan'( sin') L tan' L 0.4 tan D f tan34( sin 34) 4.05 Case I = D γ + D (γ sat - γ w ) = (0.6)(6.5) + (0.6)( )= 5.4 k/m

39 m error and y trial S u all all Q all Ens Combine x x S u all m k x d s d s S u all En.3, & : (4.06)(0.6)() 9.8) ( / ) 0.6( ) ' ( 3.