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2 ENCE 4610 Foundation Analysis and Design Lecture 3 Bearing Capacity of Shallow Foundations: Eccentric Loading of Foundations Effect of Water Table and Layered Soils

3 Brinch-Hansen s Complete Formula Similar in basic format to Terzaghi's Method, but takes into account a larger number of factors Some variations in the way it is implemented

4 Shape Factors From Verruijt; Fellenius' factors slightly different

Load Inclination Factors Verruijt Fellenius The problem with load inclination factors such as this is that they require knowing the load on the foundation before it is analysed for bearing capacity.

5 Load Inclination Factors Verruijt Fellenius The problem with load inclination factors such as this is that they require knowing the load on the foundation before it is analysed for bearing capacity. In some cases this is OK, in some it is not. For this course, the preferred values are those of Fellenius Only problem is that iγ is singular for purely cohesive soils Fellenius also gives values for AASHTO code

6 Allowable Bearing Capacity Factors when considering selection of a factor of safety Most foundations designed by ASD for geotechnical strength (LRFD to be discussed later) q ult r r =q = a a F = u F Foundation is then designed so that the allowable bearing pressure is not exceeded

7 Bearing Capacity Example q = (2)(121) = 242 psf φ = 31 degrees Shape Factors sc = 1.2 Inclination factors unity for vertical load Unit load on foundation = (76)/((5)(5)) = 4.75 ksf sq = 1+sin(31) = s γ = 0.7

8 Bearing Capacity Example p = (1)(1.2)(0)(32.671) + (1) (1.515)(0.242)(20.631) + (1)(0.7)(0.5)(0.121)(5) (23.591) = ksf FS = ( )/4.75 Effective stress at the base of the foundation is customarily included in this way

9 Concentric vs. Vertical Loading

10 Expressing Load Eccentricity and Inclination Load Divided by Inclination Angle Total Load Q Total Vertical Load Qv Total Horizontal Load Qh Angle of Inclination α = arctan(q h /Q v ) Load can be concentric or eccentric Load and Eccentricity Total Vertical Load Q Eccentricity from centroid of foundation e Horizontal Load (if any) not included Moment and Eccentric Load Total eccentric vertical Load Q with eccentricity e Replace with concentric vertical load Q and eccentric moment M=Qe Continuous Foundations Moments, loads expressed as per unit length of foundation, thus Q/b or M/b

11 One-Way Loading One-way loading is loading along one of the centre axes of the foundation Three cases to consider Resultant loads outside the middle third result in foundation lift-off and are thus not permitted at all

12 Equations for One-Way Pressures with Eccentric/Moment Loads W f is foundation weight If q at any point is less than zero, resultant is outside the middle third

13 Given o o o Find o o Example of One Way Eccentricity Continuous Foundation as shown Groundwater table at great depth Weight of foundation (concrete) not included in load shown Whether resultant force acts in middle third Minimum and maximum bearing pressures

14 Example of One Way Eccentricity Compute Weight of Foundation W f /b = (5)(1.5)(150) = 1125 lb/ft Compute eccentricity e= ( M /b) Q /b = 8000 =0.61 ft B 6 =5 =0.833 ft.>0.61 ft. 6 Thus, eccentricity is within the middle third of the foundation and foundation can be analysed further without enlargement at this point

15 Example of One Way Eccentricity Compute minimum and maximum bearing pressures

16 Two-Way Eccentricity Eccentricity in both B and L directions produces a planar distribution of stress Kern of Stability Foundation stable against overturn only if resultant falls in the kern in the centre of the foundation Resultant in the kern if 6 e B B + 6e L L 1 e B, e L = eccentricity in B, L directions

17 Bearing Pressure at Corners Two-Way Eccentricity Helpful hint to prevent confusion of eccentricity of finite vs. infinite (continuous) foundations o o o Always use one-way eccentricity equations for continuous foundations Always use two-way eccentricity equations for finite foundations Two-way equations will reduce to one-way equations if one of the eccentricities (e B, e L ) is zero

18 Two-Way Eccentricity 12 m Example 12 m Given Grain silo design as shown Each silo has an empty weight of 29 MN; can hold up to 110 MN of grain Weight of mat = 60 MN Silos can be loaded independently of each other Find Whether or not eccentricity will be met with the various loading conditions possible Eccentricity can be oneway or two-way

19 Two-Way Eccentricity Example One-Way Eccentricity Largest Loading: two adjacent silos full and the rest empty Q = (4)(29) + 2(110) + 60 = 396 MN M = (2)(110)(12) = 2640 MN-m B 6 =50 6 =8.33m>6. 67m e= M Q e= e=6.67 m Eccentricity OK for one-way eccentricity

20 Two-Way Eccentricity Example Two-Way Eccentricity Largest Loading: one silo full and the rest empty P = (4)(29) = 286 MN M B = M L = (110)(12) = 1320 MN-m e B =e L = M Q = =4.62m 6 e B B + 6e L L =2 ( ( 6 ) ( 4.62) 50 ) =1.11>1 Not acceptable

21 Two-Way Eccentricity Example Two-Way Eccentricity Solution to Eccentricity Problem: increase the size of the mat 6 e B B + 6e L L =2 ( ( 6 ) ( 4.62) B ) =1 B=L=55. 4 m Necessary to also take other considerations into account (bearing failure, settlement, etc.)

22 Equivalent Footing Procedure Structural Geotechnical (NAVFAC DM 7.02)

23 Equivalent Footing Using Two- Way Eccentricity Example Largest Loading: one silo full and the rest empty Result of Two-Way Eccentricity Analysis e B = e L = 4.62 m B = L = 55.4 m (expanded foundation) Equivalent Footing Dimensions B = B 2e B = 55.4 (2)(4.62) B = 45.8 m = L (as B = L and e B = e L )

24 Equivalent Footing Using Two- Way Eccentricity Example B 6 =50 6 =8.33m>6. 67m One-Way Eccentricity Largest Loading: two adjacent silos full and the rest empty B = L = 55.4 m (expanded foundation) e B = 6.67m e L = 0 m B = B -2e B = 55.4 (2) (6.67) = 42.1 m L = L = 55.4 m

25 Groundwater and Layered Soil Effects Layered Soils are virtually unavoidable in real geotechnical situations Softer layers below the surface can and do significantly affect both the bearing capacity and settlement of foundations Shallow groundwater affects shear strength in two ways: o o Reduces apparent cohesion that takes place when soils are not saturated; may necessitate reducing the cohesion measured in the laboratory Pore water pressure increases; reduces both effective stress and shear strength in the soil (same problem as is experienced with unsupported slopes)

26 Solution for Groundwater and Layered Soil Bearing Capacity Weighted average is, overall, the best way of handling both of these situations Groundwater creates additional soil layer Valid unless soil strengths have major variations Three ways to analyze layered soil profiles: 1. Use the lowest of values of shear strength, friction angle and unit weight below the foundation. Simplest but most conservative. 2. Use weighted average of these parameters based on relative thicknesses below the foundation. Best balance of conservatism and computational effort. 1. Use width of foundation B as depth for weighted average 3. Consider series of trial surfaces beneath the footing and evaluate the stresses on each surface (similar to slope failure analysis.) Most accurate but calculations are tedious; use only when quality of soil data justify the effort

27 Example with Layered Soils and Groundwater Find o Check adequacy against bearing capacity failure using weighted average method, Brinch- Hansen Formula and FS = 3 Given o Square spread footing as shown 2.5 m

28 Layered Soil Example Failure zone exists for a distance B (1.8 m) below the foundation, i.e., from 1.9 m to 2.7 m below the ground surface (an assumption of the method) o Unsubmerged silty sand layer is 0.6 m deep o Submerged silty sand layer is 0.5 m deep o Fine-to-medium sand layer is 0.7 m in the failure zone Weighting factors o o o Unsubmerged silty sand layer: 0.6/1.8 = 0.33 Submerged silty sand layer: 0.5/1.8 = 0.28 Fine-to-medium sand layer: 0.7/1.8 = 0.39 Weighted Values of Soil Parameters o c' = ( )(5 kpa) + (0.39) (0) = 3 kpa o φ' = ( )(32) + (0.39)(38) = 34º o γ = (0.33)(18.2) + (0.28)( ) +(0.39)( ) = 12.4 kn/m3 Note that submerged unit weights were used for submerged layer This avoids the need to then compute an average buoyant weight In the event a layer is not submerged, the moist unit weight can be used for that layer and a buoyant weight for the submerged layers

29 Questions?

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