chapter 4 bearing capacity

CHAPTER THREE Shallow Foundations – Bearing Capacity

Contents

3.1 Introduction ........................... .......................................... ............................. ............................ ............................ ............................ ............................ ............................ ............................ ............................ ............................ ................. ... 2

3.2 Types of Bearing Capacity Failure ........................... ......................................... ............................ ............................. ............................. ............................ ............................ ............................ ................... ..... 3 3.2.2 Punching shear Failure....................................... ..................................................... ............................ ............................ ............................ ............................ ............................ ............................. ............................. ..............4 3.3

Basic Definitions ........................... .......................................... ............................. ............................ ............................ ............................ ............................ ............................ ............................ ............................ ................. ... 7

3.4 Factor of Safety F.S ............................ .......................................... ............................ ............................ ............................ ............................. ............................. ............................ ............................ ............................ ................... .....9 3.5

Method of Determining Bearing Capacity ............................ ........................................... ............................. ............................ ............................ ............................ ............................ ................. ... 10

3.5.1

Bearing Capacity Equations ............................ .......................................... ............................ ............................ ............................ ............................. ............................. ........................... ..................... ........10

3.5.1.1

Terzaghi Bearing Capacity Equation .......................... ......................................... ............................. ............................ ............................ ............................ ............................ ........................ ..........10

3.5.1.2

Meyerhof’s Bearing Capacity Equation ............................ ........................................... ............................. ............................ ............................ ............................ ............................ ................. ... 15

3.5.1.3

Hansen’s Bearing-Capacity Bearing-Capacity Equation............................ .......................................... ............................ ............................ ............................ ............................. ............................ ..................... ........17

3.5.1.4

Vesićs Bearing-Capacity Bearing -Capacity Equations .......................... ......................................... ............................. ............................ ............................ ............................ ............................ ........................ ..........19

3.5.1.5 Which Equations to be Use? ........................... ......................................... ............................ ............................ ............................ ............................ ............................ ............................ ........................... .............21 3.5.1.6

Ground Water Effects........................... ......................................... ............................ ............................. ............................. ............................ ............................ ............................ ............................ ................. ... 22

3.5.1.7 Footings with Inclined Loads............................. Loads........................................... ............................ ............................. ............................. ............................ ............................ ............................ ........................ ..........28 3.5.1.8 Footing with Eccentric Loading................................... ................................................. ............................ ............................ ............................ ............................. ............................ ........................... ................32 -

Ultimate Bearing Capacity under Eccentric Loading— Loading —One-Way Eccentricity ............................ ........................................... ............................. ................... .....34

Effective Area Method (Meyerhof, 1953) ............................ .......................................... ............................ ............................ ............................ ............................. ............................. .......................... ............ 34 -

Bearing Capacity— Capacity —Two-Way Eccentricity ............................ ........................................... ............................. ............................ ............................ ............................ ............................ ................. ... 37

3.5.1.9 Bearing Capacity for the Footing on Layard Soil ........................ ...................................... ............................ ............................ ............................ ............................ ........................... .............41 3.5.2 Bearing Capacity from SPT Test ........................... ......................................... ............................ ............................. ............................. ............................ ............................ ............................ ........................ .......... 43 3.5.3 Bearing Capacity from Filed Load Test ............... ............................. ............................ ............................ ............................ ............................ ............................. ............................. .......................... ............48 3.6 Building Codes Bearing Capacity Values .................................... .................................................. ............................ ............................ ............................ ............................. ............................. ................... ..... 54

Shallow Foundations – Bearing Capacity

3.1 Introduction Shallow foundations transmit the applied structural loads to the near-surface soils. In the process of doing so, they induce both compressive and shear stresses in these soils. The magnitudes of these stresses depend largely on the bearing pressure and the size of the footing. If the bearing pressure is large enough, or the footing is small enough, the shear stresses may exceed the shear strength of the soil or rock, resulting in a bearing capacity failure as shown in Fig.3.1. .

Figure 3.1 shear failure of foundation

Seldom has a structure collapsed or tilted over from a base shear failure in recent times. Most reported base failures ha h ave occurred under embankments or simi similar structur structures where a low facto factor of safety was deemed acceptable We should note that although our primary f ocus here is on estimating the ultimate bearing capacity f or framed structures and equipment f ound ations, the same principles apply to obtaining the bearing capacity f or other structures such as tower bases,, dams bases dams,, and fills. fills. It will be shown that the ultimate bearing capacity is m ore difficult to estimate for layered soils, foundations located on or near slopes , and f oundations subjected to tension loads.

AL-Qasim Green Green University University College Of Water Resources Engineering

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Foundation Engineering Mr. Wissam Nadir


3.2 Types of Bearing Capacity Failure 3.2.1 General shear Failure General shear failure is the most common mode. It occurs in soils that are relatively incompressible and reasonably strong, in rock, and in saturated, normally consolidated clays that are loaded rapidly enough. When the load increase to ultimate the surface is well defined and failure occurs quite suddenly, as illustrated by the load-displacement curve. A clearly formed bulge appears on the ground surface adjacent to the foundation. Although bulges may appear on both sides of the foundation, ultimate failure occurs on one side only, and it is often accompanied by rotations of the foundation.

Figure 3.2 General shear failure of foundation


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3.2.2 Punching shear Failure The opposite extreme is the punching shear failure. It occurs in very loose sands, in a thin crust of strong soil underlain by a very weak soil, or in weak clays loaded under slow, drained conditions. The high compressibility of such soil profiles causes large settlements and poorly defined vertical shear surfaces. As shown in Fig.3.3 little or no bulging occurs at the ground surface and failure develops gradually, as illustrated by the ever-increasing load settlement curve.

Figure 3.3 punching shear failure of foundation


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3.2.3 Local shear Failure Local shear failure is an intermediate case. The shear surfaces are well defined under the foundation, and then become vague near the ground surface as shown in Fig3.4 . A small bulge may occur, but considerable settlement, perhaps on the order of half the foundation width, is necessary before a clear shear surface forms near the ground. Even then, a sudden failure does not occur, as happens in the general shear case. The foundation just continues to sink ever deeper into the ground.

Figure 3.4 Local shear failure of foundation

Table 3.1 presents a summary of the type of bearing capacity failure that would most likely develop based on soil type and soil properties. Table 3.1


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As compared to the number of structures damaged, far fewer structures have bearing capacity failures. This is because of the following factors: 1. Settlement governs. The foundation design is based on several requirements and two of the main considerations are: 1. settlement due to the building loads must not exceed tolerable values, and 2. There must be an adequate factor of safety against a bearing capacity failure. In most cases, settlement governs and the foundation bearing pressures recommended by the geotechnical engineer are based on limiting the amount of settlement. 2. Extensive studies. There have been extensive studies of bearing capacity failures, which have led to the development of bearing capacity equations that are routinely used in practice to determine the ultimate bearing capacity of the foundation. 3. Factor of safety. In order to determine the allowable bearing pressure q all , the ultimate bearing capacity q ult is divided by a factor of safety. The normal factor of safety used for bearing capacity analyses is 3 4. Minimum footing sizes. Building codes often require minimum footing sizes and embedment depths. 5. Allowable bearing pressures.

In addition, building codes often have maximum

allowable bearing pressures for different soil and rock conditions .

AL-Qasim Green University College Of Water Resources Engineering

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3.3Basic Definitions There are various terms, especially those which are directly related to pressures intensity, involved in bearing capacity analysis which requires clear understanding without confusion. Therefore, it may be useful at this stage to write down their definitions and symbols, as follows:

1. Total overburden pressure



Is the intensity of total pressure or total stress due to the weights of both soil and soil water, on any horizontal plane at and below foundation level before construction operations come into action.

2. Gross foundation pressure (or gross contact pressure, or, gross loading intensity),



3. Gross effective foundation pressure,

 

is the gross foundation pressure less the uplift pressure due to height h of ground water table (W.T) above the foundation level, expressed as AL-Qasim Green University College Of Water Resources Engineering

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  = 



4. Net foundation pressure (or net contact pressure, or, net loading intensity),



,

Is the excess pressure or the difference between the gross foundation pressure



  =         =    

and the total overburden pressure

expressed as

directly beneath the foundation,

5. Gross ultimate bearing capacity (or gross ultimate soil pressure), gross

,

Is the minimum value of the gross effective contact pressure at which the supporting material fails in shear. It is often written as

only.

6. Net ultimate bearing capacity (or net ultimate soil pressure),

,

Is the minimum value of the excess effective contact pressure at which the supporting material fails in shear, expressed as

7. Net safe bearing capacity,

or

Is the net ultimate bearing capacity divided by a suitable safety factor SF, expressed as

    =    = 

8. Gross safe bearing capacity, gross qs or allowable bearing capacity


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



3.4 Factor of Safety F.S Most building codes do not specify design factors of safety. Therefore, engineers must use their own discretion and professional judgment when selecting F.S. Items to consider when selecting a design factor of safety include the following: 

Soil type. Shear strength in clays is less reliable than that in sands, and more failures

have occurred in clays than in sands. Therefore, use higher factors of safety in clays. 

Soil variability. Projects on sites with erratic soil profiles should use higher factors of

safety than those with uniform soil profiles. 

Importance of the structure and the consequences of a failure . Important projects,

such as hospitals, where foundation failure would be more catastrophic may use higher factors of safety than less important projects, such as agricultural storage buildings, where cost of construction is more important. Structures with large height-to-width ratios, such as chimneys or towers, could experience more catastrophic failure, and thus should be designed using higher factors of safety. 

The likelihood of the design load ever actually occurring . Some structures, such as

grain silos, are much more likely to actually experience their design loads, and thus might be designed using a higher factor of safety. Conversely, office buildings are much less likely to experience the design load, and might use a slightly lower factor of safety. Geotechnical engineers usually use factors of safety between 2.5 and 3.5 for bearing capacity analyses of shallow foundations. Occasionally we might use values as low as 2.0 or as high as 4.0.


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3.5Method of Determining Bearing Capacity 1- Bearing Capacity Equations 2- Bearing Capacity from SPT test 3- Bearing Capacity form CPT test 4- Bearing Capacity form Field Load Test

3.5.1 Bearing Capacity Equations 3.5.1.1 Terzaghi Bearing Capacity Equation Terzaghi (1943) was the first to present a comprehensive theory for the evaluation of the ultimate bearing capacity of shallow foundations. According to this theory, a foundation is shallow if its depth, is less than or equal to its width

 ≤

.

Later investigators, however, have suggested that foundations with equal to 3 to 4 times their width may be defined as shallow foundations.

Terzaghi suggested that for a continuous, or strip, foundation (i.e., one whose width to length ratio approaches zero), the failure surface in soil at ultimate load may be assumed to be similar to that shown in Figure 3.5. (Note that this is the case of general shear failure, as defined above)


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The effect of soil above the bottom of the foundation may also be assumed to be replaced by an equivalent surcharge,

=  (where

is a unit weight of soil).

The failure zone under the foundation can be separated into three parts (see Figure 3.5):

Figure 3.5

1- The triangular zone ACD immediately under the foundation 2- The radial shear zones ADF and CDE, with the curves DE and DF being arcs of a logarithmic spiral 3- Two triangular Rankine passive zones AFH and CEG The angles CAD and ACD are assumed to be equal to the soil friction angle Note that, with the replacement of the soil above the bottom of the foundation by an



equivalent surcharge , the shear resistance of the soil along the failure surfaces GI and HJ was neglected. Using equilibrium analysis, Terzaghi expressed the ultimate bearing

capacity in the form

 =  0.5


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   = 2 4 5∅2⁄ ℎ =.−∅⁄∅  =( 1)∅ → for ∅>0.0  =5.7  → for ∅=0.0  = ∅ ∅ 1 

Donald P. Coduto 2001 developed an equation below by fitting Terzaghi curve to

match. It produces

values within about 10 percent of Terzaghi’s values. Alternatively,

Kurnbhojkar (1993) provides a more precise. but more complex, formula for

  +∅   = +.∅   ,   ∅ =   



.

Where

= ultImate bearing capacity

= effective cohesion for soil beneath foundation =Terzaghi’s bearing capacity factors

= shape factor

=∅

= effective friction angle for soil beneath foundation

= vertical effective stress at depth

surface



below the ground

For strip round square Rectangular



1.0

1.3

1.3

1.0

0.6

0.8

⁄ 10. 3  10. 2 ⁄

= passive pressure

= effective unit weight of the soil = depth of foundation below ground surface

B = width (or diameter) of foundation


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Note

  ∅

-

The first and third term of equation above related to the soil properties (

-

The second term of equation related to the surcharge load above the base of foundation

-

If saturated undrained conditions exist (clay), we may conduct a total stress analysis

= 

with the shear strength defined as and

=0.0 as shown in Table 3. 2


 = ∅ and

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=0

In this case,

)

  = 5.7,

= 1.0,



Table 3.2 Terzaghi’s bearing capacity factors


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3.5.1.2

Meyerhof’s Bearing Capacity F Equation

Meyerhof (1951, 1963) proposed a bearing capacity equation similar to that of Terzaghi but included



-

A shape factor

-

He also included depth factors

  d

with the depth term

.

and inclination factors

footing load is inclined from the vertical)

i

(For cases where the

The shape, depth, and inclination factors in Table 33 are from Meyerhof (1963). The shape factors do not greatly differ from those given by Terzaghi except for the addition of

s

.

D≈B q  :  =  0.5  :  =  0.5  =  0.5 ,    =    = ∅ 45 ∅2  =( 1)∅  =( 1)tan1.4∅ Up to a depth of

, the Meyerhof

is not greatly different from the Terzaghi

value. The difference becomes more pronounced at larger D/B ratios.

Meyerhof (1963) suggested the following form of the general bearing capacity equation:

Where:


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,,   =ℎ    = ℎ     ,    = 


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3.5.1.3

Hansen’s Bearing-Capacity Equation

Hansen (1970) proposed the general bearing capacity case and N factor equations shown below. This equation is readily seen to be a further extension of the earlier Meyerhof 1951 work.

    =  0. 5 ℎ ∅=0   =5.∅ 14 1        = ∅ 45 2  =( 1)∅  =1.5( 1)tan∅

The extensions include base factors for situations in which the footing is tilted from the horizontal

and for the possibility of a slope

give ground factors

of the ground supporting the footing to

·

The Hansen equation implicitly allows any D/B and thus can be used for both shallow (footings) and deep (piles, drilled caissons) bases.


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Shallow Foundations – Bearing Capacity Table 3.4

Table 3.5 Table 3.6


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3.5.1.4

Vesićs Bearing-Capacity Equations

N N  N s  q =cNsdigb qNsdigb 0. 5γB Nsdigb when ∅=0 use q =5.∅ 14 s1 s d i b g q  = ∅ 45 2  =( 1)∅  =2( 1)tan ∅

The Vesić (1973, 1975) procedure is essentially the same as the method of Hansen (1961) with select changes. The

and

different . There are differences in the ,

terms are those of Hansen but , and

is slightly

terms as in Table 3.6.

The Vesić equation is somewhat easier to use than Hansen's because Hansen uses the i terms in computing shape factors


.

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Table 3.6


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3.5.1.5 Which Equations to be Use? The Terzaghi equations, being the first proposed, have been very widely used. Because of their greater ease of use (one does not need to compute all the extra shape, depth, and other factors) they are still used probably more than they should be. They are only suitable for: A concentrically loaded footing on horizontal ground. They are not applicable for footings carrying a horizontal shear and/or a moment or for tilted bases. Both the Meyerhof and Hansen methods are widely used. The Vesic’method has not been much used. As previously noted there is very little difference between the Hansen and Vesic’ methods. From these observations its suggest the following equation use: Use

Best f or

Terzaghi



Very cohesive soils



where D/B



For a quick estimate of qult , to compare with other methods.



Do not use f or f ootin gs with moments and/or horizontal f orces or

≤1

for tilted bases and/or sloping ground.

Hansen, Meyerhof ,

Any situation that applies, depending on user pref erence or f amiliarity

Vesic

with a part icular method.

Hansen, Vesic



when f ootin g is on a slope or



When base is tilted



when

D /B


>

1

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3.5.1.6

Ground Water Effects

The presence of shallow groundwater affects shear strength in two ways: 1- the reduction of apparent cohesion, and 2- The increase in pore water pressure. When exploring the subsurface conditions, we determine the current location of the groundwater table and worst-case (highest) location that might reasonably be expected during the life of the proposed structure. We then determine which of the following three cases describes the worst-case field conditions as shown in Figure 3.6: Case 1: if the water table located so that

 =  

0≤ ≤

The factor in the bearing capacity equations takes the form

Where:

 =

= saturated unit weight of soil Unit weight of water

Also, the value of in the last term of the equations has to be replaced by

 = 

.

0≤≤  ̅ =      

Case 2. For a water table located so that

In this case, the factor ( ) in the last term of the bearing capacity equations must be replaced by the factor

Case 3. When the water table is located so that d > B, the water will have no effect on

the ultimate bearing capacity.


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Example: Compute the allowable bearing pressure using the Terzaghi equation for the

square footing of B=2 m and soil parameters shown in Figure 3.7. Use a safety factor of

q = =.−∅  ⁄∅0.=2.5212    = 2 4 5∅2⁄ =7.43  =( 1)∅ → for ∅>0.0 →  =17.7 3 to obtain

.

Solution:

 ⁄ =17. 3   ° ∅=20 =20 

 = +.+∅ ∅ =4.4 Or go to table 3.2

=  =1.3     =0. 8   =17.3×1.2=20.76/  =20×17.7×1.320.7×7.430.5×17. 3×2×4.4×0.8  =460.2153.860.9=674.9⁄  = . = 674.3 9 =225 ⁄ AL-Qasim Green University College Of Water Resources Engineering

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Example: A mat foundation 22.5 × 60.0 m supports a silo, which completely covers the

mat area. The dead weight W of the complete structure, unloaded, is assumed to be 200 MN. The foundation level is located at 3 m below ground surface. The soil profile consists of a uniform saturated clay deposit with average undrained shear strength of 75 kPa and unit weight of 17.5 kN/m . The water table is located at the ground surface. If the safety factor SF against bearing capacity failure is to be not less than three and neglecting soil adhesion on the walls of the silo: Determine the maximum vertical load V which the silo may carry, considering bearing capacity failure only and using Terzaghi bearing capacity equation.

  = 200000   ℎ     ℎ  −    = 22. 5×60   3 ×10 (3× 17.510) =7.4×10 95.65 .,=∅=0  =5.7 0.5 =1.0  =0.0  =10. 3  =1.1125    ⁄ . =75×5. 7 ×1. 1 1253 1 7. 5 10 ×1=498. 1  . =.  =498. 1 7. 5 ×3=475. 6 ⁄    = .. = 475.3 6 =158.5 ⁄    =7.4×10 −95.65=158.5 →=85000 =8500 Solution:


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Example: A square footing is required to carry a gross load of1500 kN. The base of the

footing is to be 4.5 m below the ground level. The ground water table will rise to the ground surface. The soil is saturated clay having the following properties: apparent cohesion c = 57.5 kPa, Ø = 0 and γ = 19 .2kN/ m3. Using the Terzaghi bearing capacity equation with a safety factor SF =3.0, find a suitable size for the footing. Solution:

.,=∅=0 →=5. 70.,5=1.0  =0 .=. =    =  ⁄  . =57. 5 ×5. 7 ×1. 3 =426. 1    = .. = 327.375 =142⁄     = 1500  ℎ     1500 ℎ      =   4.5×10 (4.5× 19.210) =  86.4   =  142= 1500  86.4 →=6.6  Use square footing with dimension of 2.6mX2.6m


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Example: A strip footing is to be designed to carry a gross foundation load equals

800kN/m run at a depth of 0.7 m in a gravelly sand stratum. Appropriate shear strength parameters are c= 0 and Ø = 40 . Assume that the water table exists at the foundation level. The sand unit weights above and below the water table are 17 and 20 kN/m , respectively. Using Terzaghi bearing capacity equation with a safety factor of three, considering shear failure only, determine the width of the footing. Solution:

,. = =0,ℎ 0.5     .== 0. 50.5  ( ) . °     ∅=40 ,   =81.3   =121.5 . =17×0. 7 ×81. 3 0. 52 010 ×121.5  17×0.7  . =955. 5 7607. 5  5    = .. = 955.57607. =318.52202.5 3     =×1         ℎ       =800 0.7×17   =  800  318.52202. 5 = 0 . 7 ×17   202.5 330.42B800=0 →B=1.33 m use B=1.4 m AL-Qasim Green University College Of Water Resources Engineering

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Example 3.2: A 30 m by 50 m mat foundation is to be built as shown in Figure 3.8, Compute the ultimate bearing capacity. Solution: 1- By Terzaghi equation Take the effect of water table

̅=<  → 2    ̅==8.72 18.=18.58.579.=9.8 =8.35 7 ⁄⁄  30  =  0.5   ∅=30 →  =22. 5   =20. 1   30 ⁄ =10. 2 × =0. 8 8 =10.2 50 =010×18.5×22.50.5×9.35×30×20.1×0.88=6643.2 ⁄   = ∅ 0.5  = ∅ 45 2=18.4  =( 1)tan1.4∅ =15.65   = =10.1  4 5∅2⁄  =1. 18 ⁄  = =010×18.  445∅2 =1. 0 6  =10. 1  5×18.  ×1. 1 8×1. 0 60. 5 ×9. 3 5×30×15. 6 5×1. 1 8×1. 0 6  =7003 ⁄

Since C = 0 there is no need to compute any of the other factors in the first term of the bearing capacity equation. Go to table 3.2

2- By Meyerhof equation


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3.5.1.7 Footings with Inclined Loads Inclined loads are produced when the footing is loaded with both a vertical (V) and a horizontal component (H); of loading. This loading is common for many industrial process footings where: 

Horizontal wind loads are in combination with the gravity loads.



The retaining wall base design is a classic case of a foundation with both a horizontal (the lateral earth pressure) and vertical loading.



Eccentricity results from the vertical load not being initially located at B/2.



And from the lateral earth pressure effects and a number of other types of industrial foundations are subjected to horizontal loads.

In any case, the load inclination results in a bearing capacity reduction over that of a foundation subjected to a vertical load only. The inclination factors of Tables 3.3, 3.4, 3.5 and 3.6 can be used with the Meyerhof, Hansen, or Vesic bearing capacity equations. The Terzaghi equations have no direct provision for a reduction in cases where the load is inclined. The Meyerhof inclination factors




are reasonably selfexplanatory.

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Example: A square column foundation shown in Figure below is to be constructed on a sand deposit. The allowable load (P) will be inclined at an angle

=20°

with the

vertical. The standard penetration numbers N 60 obtained from the field are as follows. Depth (m) N 60 1.5

3

3

6

4.5

9

6

10

7.5

10

9

8

=18 ⁄ C=0

∅=27.10.3 0.00054 ∅ ∅

Determine P. use FS=3 and Solution

1. Estimate the angle of friction Depth (m) N 60 1.5

3

28

3

6

29

4.5

9

30

6

10

30

7.5

10

30

9

8

30

Average

30


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2. Use Meyerhof’s Bearing Capacity Equation for inclined load

 =  0.5 =  = 0.5  =18×0.7 =12.6 ⁄  = ∅ 45 ∅2=4515 =18.4  =( 1)tan1.4∅=18.41tan1.4×30=15.67  = =10.1   , ℎ:  =4 5∅2⁄=3  = =10. 1 ×3×1=1. 3  = =10. 1 0.7  = =10.°  1√ 3 1.25 =1.097  =1 90 ° =0.605  °   =1 ∅° =0.11  =12.6×18.×0.141=227. ×1.3×1.68097×0. 6 050. 5 ×18×1. 2 5×1. 3 ×15. 6 7×1. 0 97 ⁄  Since C=0, the first term of equation will be zero



Calculate bearing capacity factor



Calculate shape factor



Calculate depth factor


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3. Calculate the safe bearing capacity

  6    ⁄  =   = 227.6812. =71. 7   3  =  →= ×=71.7×1.25×1.25=112  =12.6×18.=5824×1.⁄3×1. 0970.5×18×1.25×1.3×15.67×1.097  >500 ⁄   =500 ⁄   50012. 6     =  = 3 =162.46 ⁄   =  →= ×=162.46×1.25×1.25=253.85 4. Find the allowable load P

Example: Resolve the example above by using vertical load

Note: if


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Shallow Foundations – Bearing Capacity 3.5.1.8 Footing

with Eccentric Loading

In several instances, as with the base of a retaining wall, foundations are subjected to moments in addition to the vertical load, as shown in Figure 3.10. In such cases, the distribution of pressure by the foundation on the soil is not uniform.


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 6 . =   6 . =   =  . =  1 6  . =  1 6

The nominal distribution of pressure is

Note that, in these equations, -

when the eccentricity (e) becomes B/6, q min is zero.

-

For e >B/6, qmin will be negative, which means that tension will develop. Because soil cannot take any tension, there will then be a separation between the foundation and the soil underlying it. The nature of the pressure distribution on the soil will be as shown in Figure 3.10a.


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-

Ultimate Bearing Capacity under Eccentric Loading —One-Way Eccentricity

Effective Area Method (Meyerhof, 1953)


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The following is a step-by-step procedure for determining the ultimate load that the soil can support and the factor of safety against bearing capacity failure: Step 1. Determine the effective dimensions of the foundation (Figure 3.11a):

 =2  =    =  0.5 , ,      = × × and

Note that if the eccentricity were in the direction of the length of the foundation, the value of

= L- 2e. The value of

would equal B.

Step 2. Use the equation shown below to calculate the ultimate bearing capacity

Note that -

Evaluate the shape factor

-

To determine the depth factor

with

and

instead of L and B

do not replace B with

Step 3. The total ultimate load can be resisted by the footing is:


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



Example: A continuous foundation is shown in Figure below. If the load eccentricity is 0.2 m, determine the ultimate load, Pu, per unit length of the foundation. Use Meyerhof’s effective area method. Solution:

 =  0.5  = 0.5 = =16.5×1.5=24.75 ⁄   = ∅ 45 ∅2=4520 =64.2  =( 1)tan1.4∅=64.21tan1.4×40=93.7  =0  = =1  = =10. 1  1.5, ℎ:  =4 5∅2⁄ =4.6  = =10. 1√ 4.6 2 =1.16  ==2=22×0.  =1 2 =1. 6  =24.75×64.2 ×1×1.16×10.5 ×16.5 ×1.6 ×93.7 ×1×1.16×1 =3277.91 ⁄ Since C= 0 , the first term will be equal to zero.



Calculate bearing capacity factor



Calculate shape factor

For strip foundation



Calculate depth factor


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7 5  ⁄  =  = 3277.9124. =1084. 4       3 =500 ⁄   =  →= ×=500×1.6×1.0=800 -

Bearing Capacity—Two-Way Eccentricity


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A footing may be subjected to eccentric loading and such condition leads to a reduction in bearing capacity. The estimate of ultimate bearing capacity for footings with eccentricity may be obtained by the following method: The effective area method (Meyerhof, 1953, 1963; Hansen, 1970).

(a) Determine the effective footing dimensions

=2 =2

using

Where eB is the eccentricity parallel to B Where eL is the eccentricity parallel to L

Note that

-

 

  A = BB aLnd L

The smaller of the two dimensions (i.e.

) is the effective width of the

footing used in the Nγ term of the bearing capacity equations. -

Effective area of the footing

.

(b) Use the effective footing dimensions

in computing the shape factors.

(c) Use actual B and L dimensions in computing the depth factors. (d) Use a general bearing capacity equation to obtain q ult. The ultimate load that the foundation can support is

=. ×A

. AL-Qasim Green University College Of Water Resources Engineering

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Example 3.4: A square footing is 1.8 X 1.8 m with a 0.4 X 0.4 m square column. It is loaded with an axial load of 1800 kN and



= 450 kN.m and

 γ

= 360 kN m.

and c = 20 kPa. The footing depth D = 1.8 m; the soil unit weight

∅= 36

°

= 18.00 kN/m3; the

water table is at a depth of 6.1 m from the ground surface. What is the allowable soil pressure, if SF = 3.0, using Meyerhof’s equation.

 =  = 1800360 =0.20   =  = 1800450 =0.25  ℎ   <⁄6 =0.3     ==2 =1. 8 2×0. 2 0=1. 4   2 =1. 8 2×0. 2 5=1. 3     =1.4    =1.3  Solution:

Find


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 =  0.5  =51  =38  =44.4    ⁄   =1.715  =10.2 45∅2  = =10.14 5∅2⁄  =1.357  =10. 2 4 5∅2⁄  =1. 39   =10. 1  4 5∅2⁄   =1. 2  =  0.5     =20×51×1. 7 15×1. 3 9 1 8×1. 8 ×38×1. 3 57×1. 2 0. 5 ×18×1. 3 ×44. 4 ×1.357×1.2=5282.33 ⁄2      = .  1800= 5282.331.3 8×182 =1749.97 ⁄2 2   = 500 ⁄2  = 1.3×1.4 =989 ⁄ >  =500 ⁄   By Meyerhof Equation

-

Find the bearing capacity factors

-

Find the shape factors

-

Find the depth factors


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3.5.1.9 Bearing Capacity for the Footing on Layard Soil Thus far, the analyses in this chapter have considered only the condition where C, Ø and



are constant with depth. However, many soil profiles are not that uniform. Therefore,

we need to have a method of computing the bearing capacity of foundations on soils where C, Ø and



are vary with depth. There are three primary ways to do this:

1- Evaluate the bearing capacity using the lowest values of C, Ø and



in the zone

between the bottom of the foundation and a depth H below the bottom.

H =0.5Btan 45∅⁄2

Where:

This is the zone in which bearing capacity failures occur and thus is the only zone in which we need to assess the soil parameters. This method is conservative, since some of the shearing occurs in the other, stronger layers. However, many design problems are controlled by settlement anyway, so a conservative bearing capacity analysis may be the simplest and easiest solution. In other words, if bearing capacity does not control the design even with a conservative analysis, there is no need to conduct a more detailed analysis. 2- Use weighted avenge values C, Ø and



based on the relative thicknesses of each

stratum in the zone between the bottom of the footing and a depth B below the bottom. This method could be conservative or unconservative, but should provide acceptable results so long as the differences in the strength parameters are not too great. 3- Consider a series of trial failure surfaces beneath the footing and evaluate the stresses on each surface using methods similar to those employed in slope stability analyses. The surface that produces the lowest value of q is the critical failure surface. This method is the most precise of the three, but also requires the most effort to implement. It would be appropriate only for critical projects on complex soil profiles. AL-Qasim Green University College Of Water Resources Engineering

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In this course, we will use the second method in estimate the ultimate bearing capacity

Example. A footing of 3 X 6 m is to be placed on a twolayer clay deposit as in Figure. Estimate the ultimate bearing capacity. Solution: Important note

If the depth

 H >



 H  ⁄ H=ℎ=0.5B tan45∅ 2    ℎ ℎ  ⁄  HWeight =of0.la5yerB t1=an 45∅2 =0. 5 ×3×tan45=1. 5 > =1. 2 2  1. 2 2 = 1.5 1.=0.51.81322 Weight of layer 2=  = 1.5 =0.186  =0.813×770. 186×115=84. 0 7  =0. = 25.1=0. 41 1 =0. 4  =0. 24  =5.14×84.07× 10.10.24 1.83×17.26=610.62 kN/ Then the failure will occur in the first layer (use C, Ø and If

<

of first layer).

The failure will penetrate the second layer (use the average C, Ø and Where:

By using Hansen equation

Example: For the square footing of (2m*2m) dimension shown in Figure. Determine the safety factor agents bearing capacity failure. AL-Qasim Green University College Of Water Resources Engineering

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Solution. 1- Check is the footing on layered soil or no?

⁄ =0. HH <=0.5=2.Bta5n4ℎ5∅2 5 ×2×tan45=1. 0              ℎ     . ,ℎ        =   0.5   =0  ∅=0 , = =60×5.7×1.3=444.6⁄  =  = 2×2130 =32.5 ⁄ .  =  = 444.32.46 =13.68

2- Use Terzaghi equation or any other equation discussed above to find ultimate bearing capacity

3- Find allowable bearing capacity

4- Find the factor of safety

3.5.2 Bearing Capacity from SPT Test Mayerhof (1956, 1974) published equations for computing the net allowable bearing capacity (net qa) for 25 mm settlement. Considering the accumulation of field observations and the stated opinions of the author and others, Bowles (1977) adjusted AL-Qasim Green University College Of Water Resources Engineering

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the Mayerhof equations for an approximate 50% increase in net qa to obtain the

   =     ≤        =        >   =      =ℎ =[10.33]≤1.33 0. 5   =       ≅0.5    following:

Where:

= depth of W.T below foundation level ≤ B

D = depth of foundation B = width of footing

F1, F2, F3 and F4 = factors depend on the SPT hammer energy ratio Er, as given below

Note that, for complete saturation (submergence) conditions, that is when W.T is above

the foundation level or when

  ≅0.5  =1  =0,

; and

for

≥ B.

In these equations N55 is the statistical average corrected value for the footing influence zone of about 0.5 B above the foundation level to at least 2 B below. AL-Qasim Green University College Of Water Resources Engineering

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In the above three equations the allowable soil pressure is for an assumed 25-mm settlement. In general, for cohesionless soils, it is possible to assume that settlement S is proportional to net soil pressure net q. Based on this assumption, the settlement Si caused by any given net soil pressure net q Si is

 = × 

Example: A footing foundation 3.5 m square is to be constructed at a depth of 2 m. At

the site of a proposed structure the variation of the SPT number (N60) with depth in a deposit of normally consolidated sand is as given below: Depth , m

1.5

3


4.5 Page 45 of

6 54

7.5

9





6

8

9

8

13

14

If the settlement is not to exceed 30 mm, determine the net allowable bearing capacity using: (a) The bearing capacity from SPT test, (b) The Meyerhof bearing capacity equation with a safety factor = 3.

Solution: 1. Correct the SPT number. The corrected SPT number



for granular soil can be calculated as shown below:


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1 =  . 1  =100′°  , ° =×  =18×1.′°5=27  1.5

6

3

8

54

1.36

11.54≈12

4.5

9

81

1.11

10

6

8

108

0.962

7.5

13

118.2*

0.92

9

14

142.5

0.837

Depth m

1.924

31.5

*

10.88≈11 7.11.976≈12 ≈8 11.72≈12

° =× =6.5×181.5× 20.210 =118. 2 1 =∑  ℎ ℎ 1 1. 5 3 4. 5 6 7. 5 9 1 .×12=11 = 31.5 ×12 31.5 ×11 31.5 ×10 31.5 ×8 31.5 ×12 31.5

2. Calculate the allowable bearing capacity for 25 mm settlement

       =        >  =[10.33]=[10.333.25]=1.188 ≤1.33 


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 =6.22=4.5 >=3.5 →  =1.0  60 =0.05, 60 =0.08,  =0.3,  =1.2  =55 =55×11=12   .  +.    = .  .  × 1.19×1=210.4 ⁄    = ×  =30× 210.254 =252.5 ⁄ 

3.5.3 Bearing Capacity from Filed Load Test


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The procedure has been standardized as ASTM D 1194, which is essentially as follows: 1. Decide on the type of load application. If it is to be a reaction against piles, they should be driven or installed first to avoid excessive vibration and loosening of the soil in the excavation where the load test will be performed. 2. Excavate a pit to the depth the test is to be performed. The test pit should be at least four times as wide as the plate and to the depth the foundation is to be

placed. 3. A load is placed on the plate, and settlements are recorded from a dial gauge accurate to 0.25 mm. Load increments should be approximately one-fifth of the estimated bearing capacity of the soil. Time intervals of loading should not be less than 1 h and should be approximately of the same duration for all the load increments. 4. The test should continue until a total settlement of 25 mm is obtained, or until AL-Qasim Green University College Of Water Resources Engineering

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the capacity of the testing apparatus is reached. After the load is released, the elastic rebound of the soil should be recorded for a period of time at least equal to the time duration of a load increment. The Figure below is a typical semilog plot of time versus settlement (as for the consolidation test).

The following points are related to results of a plate-load test:


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

For clay soils it might be said that q ult is independent of footing size, giving:

, =,  , =, 



For sand soils, practically, extrapolation may be justified using



For clayey soils (cohesive soils) and for the same qult of the footing and plate, the following empirical approximate relationship regarding footing settlement may be used:



  , =,       0. 3         , =,  0.3

For sandy soils (cohesionless soils) and for the same qult of the footing and plate, the following empirical relationship regarding footing settlement may be used:

According to Housel, size of a footing to carry a given load for a given safe settlement may be established by using data from at least two PLT in the following equation AL-Qasim Green University College Of Water Resources Engineering

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Where:

,,  

 ==            

= total load on a bearing plate area

respectively

= perimeter of plate No.1 and No.2 respectively

M and n = constant corresponding to the bearing pressure and parameter shear respectively. After finding m and n resolve the equation above to calculate the dimension of footing.

==             =  =    

Example: The results of two-plate load are given in the following table. AL-Qasim Green University College Of Water Resources Engineering

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Plate dimeter, D (m) Total load kN Settlement (mm) 0.305 32.2 20 0.610 71.8 20 Use the test data shown in table above to find the dimension of square foundation, that

will construct to support square column carry a total load of 715 kN. The tolerable settlement is 20 mm. Solution:

 ==         32.2 = 4 ×0.305×0.305……………1 71.8 = 4 ×0.61×0.61 ……………2  == 29.51.696 715==51.9×  29.66× 4 715 =51.9×  118.64 Solve the equation above to find m and n -

For the foundation to be designed

B= 2.75 m


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chapter 4 bearing capacity

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