BEARING CAPACITY OF SHALLOW FOUNDATION DUE TO INCLINED LOAD AND ECCENTRIC LOADING

Lecture-3

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References Braja M Das, Shallow Foundation Bearing Capacity

and Settlement, 2nd ed, 2009 Muni Budu, Soil Mechanics and Foundations, 2nd

edition, John Wiley & Sons, 2007, USA Bowles, Foundation Analysis and design, 5th edition,

Mc Graw-Hill, 1997 R.F. Craig, Soil Mechanics, (English & terjemahan

Prof.Dr.Ir Budi S Supandji) D.P. Coduto, Foundation design 1994

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Due to bending moments and horizontal thrusts transferred from the superstructure, shallow foundations are often subjected to eccentric and inclined loads.

Eccentric loading 3

Stress distribution 4

Eccentric and moment loading 5

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Eccentrically loaded footing with (a) Linearly varying pressure distribution (structural design), (b) Equivalent uniform pressure distribution (sizing the footing).

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INCLINED LOADING - MEYERHOF

CONTINUOUS FOUNDATIONS 10

Meyerhofs Theory (Continuous Foundation) 11

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Note that the horizontal component of the inclined load per unit area on the foundation qh cannot exceed the shearing resistance at the base, or

where c = unit base adhesion = unit base friction angle

where Ncq, N q = bearing capacity factors that are functions of the soil friction angle and the depth of the foundation Df For a purely cohesive soil ( = 0),

For cohesionless soils c = 0; hence

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Meyerhofs bearing capacity factor Ncq for purely cohesive soil ( = 0). 14

Meyerhofs bearing capacity factor Nq for cohesionless soil ( = 0, = ).

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ECCENTRIC LOADING

CONTINUOUS FOUNDATIONS 19

ECCENTRIC LOADING MEYERHOF

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Continuous foundation with eccentric loading

B = B - 2e

Q u= qu A where A = effective area = B 1 = B

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ECCENTRIC LOADING PRAKASH & SARAN

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Derivation of the bearing capacity theory of Prakash and Saran for eccentrically loaded rough continuous foundation.

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Prakash and Sarans bearing capacity factor Nc(e) . 24

where N (e), Nq(e), Nc(e) = bearing capacity factors for an eccentrically loaded continuous foundation

Prakash and Sarans bearing capacity factor Nq(e) . Prakash and Sarans bearing capacity factor N (e) 25

RECTANGULAR FOUNDATIONS 26

ECCENTRIC LOADING, ONE WAY MEYERHOF

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Ultimate Load on Rectangular Foundation

Eccentric load on rectangular foundation

el = MB /Q eB = ML /Q

where eL, eB = load eccentricities, respectively, in the directions of the long and short axes of the foundation MB, ML = moment components about the short and long axes of the foundation,respectively

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According to Meyerhof,

where A = effective area = BL B = effective width L = effective length

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Other book use factor F instead of

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B = B 2 eB ; L = L; A = BL if e=0 calculate L 2eL The effective area is A = B( L eL)

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ECCENTRIC LOADING, ONE-WAY PRAKASH & SARAN

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Prakash and Saran

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ECCENTRIC LOADING, TWO-WAY

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CASE I

Rectangular foundation with one-way eccentricity

Effective area for the case of eL/L 1/6 and eB/B 1/6.

Case I (eL/L 1/6 and eB/B 1/6)

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Effective area for the case of eL/L < 0.5 and eB/B < 1/6.

CASE II

Case II (eL/L < 0.5 and 0 < eB/B < 1/6)

A = ( L1 + L2) B

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Case III (eL/L < 1/6 and 0 < eB/B < 0.5) Figure on the right shows the case under consideration. Knowing the magnitudes of eL/L and eB/B, the magnitudes of B1 and B2 can be obtained from Figures 3.20 and 3.21. So, the effective area can be obtained as A = ( B1 + B2 ) L In this case, the effective length is equal to L = L The effective width can be given as B = A/L

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Case IV (eL/L < 1/6 and eB/B < 1/6) The eccentrically loaded plan of the foundation for this condition is shown in Figure on right. For this case, the eL/L curves sloping upward in Figure below represent the values of B2/B on the abscissa. Similarly, in Figure below the families of eL/L

Effective area for the case of eL/L < 1/6 and eB/B < 1/6

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Case V (Circular Foundation)

In the case of circular foundations under eccentric loading (Fig. a), the eccentricity is always one way. The ulimate load for the foundation can be used Meyerhoff equation concidering: 1. The bearing capacity factors for a given friction angle are to be determined from those presented in Tables previously given 2. The shape factor is determined by using the relationships given by replacing B for B and L for L whenever they appear. 3. The depth factors are determined from the relationships given in Table However, for calculating the depth factor, the term B is not replaced by B.

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EXAMPLE 45

Example A shallow foundation measuring 2 m 3 m in a plan is subjected to a centric load and a moment. If eB = 0.2 m, eL = 0.6 m, and the depth of the foundation is 1.5 m ,determine the allowable load the foundation can carry. Use a factor of safety of 4. For the soil, given: unit weight g = 18 kN/m3; friction angle f = 35; cohesion c = 0. Use Vesics Ng (Table 2.4), DeBeers shape factors (Table 2.6), and Hansens depth factors (Table 2.6). Solution For this case,

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ECCENTRIC OBLIQUE LOADING

CONTINUOUS FOUNDATIONS 49

Ultimate Bearing Capacity of Eccentrically Obliquely Loaded Foundations

Saran and Agarwal.

where Nc(ei), Nq(ei), Ng(ei) = bearing capacity factors q = gDf

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Example

For a continuous foundation, given: B = 1.5 m; Df = 1 m; = 16 kN/m3; eccentricity e = 0.15 m; load inclination = 20. Estimate the ultimate load Qult.

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Solution with c = 0,

B = 1.5 m, q = Df = (1)(16) = 16 kN/m2, e/B = 0.15/1.5 = 0.1, and = 20. From Figures 3.27c and 3.28c, Nq(ei) = 14.2 and N(ei) = 20. Hence

Qult= (1.5) [(16)(14.2) 1+(16)(1.5)(20) ]= 700.8 kN/m

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MAT FOUNDATION 53

Mat foundation 54

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Clay soil 58

Granular soil 59

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Bearing capacity of shallow foundation due to INCLINEd LOAD and eccentric loadingSlide Number 2Eccentric loadingStress distributionEccentric and moment loadingSlide Number 6Slide Number 7Slide Number 8Slide Number 9CONTINUOUS FOUNDATIONSMeyerhofs Theory (Continuous Foundation)Slide Number 12Slide Number 13Slide Number 14Slide Number 15Slide Number 16Slide Number 17Slide Number 18CONTINUOUS FOUNDATIONSSlide Number 20Continuous foundation with eccentric loadingSlide Number 22Slide Number 23Slide Number 24Slide Number 25RECTANGULAR FOUNDATIONSSlide Number 27Ultimate Load on Rectangular FoundationSlide Number 29Slide Number 30Slide Number 31Slide Number 32Slide Number 33Slide Number 34Slide Number 35Slide Number 36CASE ISlide Number 38CASE IISlide Number 40Slide Number 41Slide Number 42Slide Number 43Slide Number 44EXAMPLESlide Number 46Slide Number 47Slide Number 48CONTINUOUS FOUNDATIONSUltimate Bearing Capacity of EccentricallyObliquely Loaded FoundationsExample Slide Number 52MAT FOUNDATIONMat foundationSlide Number 55Slide Number 56Slide Number 57Clay soilGranular soil Slide Number 60