38
Foundation Engineering Foundation Engineering Lecture #04 Lecture #04 The Bearing Capacity of Soils -Terzaghi’s Ultimate Bearing Capacity -Terzaghi’s Ultimate Bearing Capacity -Meyerhof’s Method -Brinch Hansen’ Method -Vesic’s Method Luis A. Prieto-Portar, 2009

General Bearing Capacity

Embed Size (px)

DESCRIPTION

General Bearing Capacity

Citation preview

Page 1: General Bearing Capacity

Foundation EngineeringFoundation Engineering

Lecture #04Lecture #04

The Bearing Capacity of Soils

-Terzaghi’s Ultimate Bearing Capacity-Terzaghi’s Ultimate Bearing Capacity-Meyerhof’s Method

-Brinch Hansen’ Method-Vesic’s Method

Luis A. Prieto-Portar, 2009

Page 2: General Bearing Capacity

These wheat storage silos in Canada are an example of the bearing capacity failure of a large foundation. This failure alerted engineers to the mechanism of how surface loads may

exceeded the shear strength of the soils beneath the foundation.

Page 3: General Bearing Capacity

Some foundations in congested cities must consider how they transfer loads onto adjacent utilities and the adjacent building’s foundations, in addition to the underlying soil.

This is a typical excavation to repair a utility in a New York City street.

Page 4: General Bearing Capacity

Toronto’s Queen and Yonge Street intersection.

Page 5: General Bearing Capacity

A shallow foundation must:1. be safe against an overall shear failure in the soil that supports it. 2. cannot experience excessive displacement (in other words, settlement).

The definitions of bearing capacity are,qo is the actual contact pressure of the soil at the footing’s invert;qult is the load per unit area of the foundation at which the shear failure in soil occurs and is called the ultimate bearing capacity of the foundation; andqall is the load per unit area of the foundation that is supported without an unsafe movement of the soil, and is called the allowable bearing capacity.

CONCENTRIC LOAD ( NO ECCENTRICITY)CONCENTRIC LOAD ( NO ECCENTRICITY)

B X LB

q

Q

y

xFigure 1.

Page 6: General Bearing Capacity

A continuous footing resting on the surface of a dense sand or a stiff cohesive soil is shown in Figure 2a with a width of B. If a load is gradually applied to the footing, its settlement will increase. When the load per unit area equals qult a sudden failure in the soil supporting the foundation will take place, with the failure surface in the soil extending to the ground surface. This type of sudden failure is called a general shear failure.

If the foundation rests on send or clayey soil of medium compaction (Figure 2b), an increase of load on the foundation will increase the settlement and the failure surface will gradually extend outward from the foundation (as shown by the solid line). When the load per unit area on the foundation equals qult, the foundation movement will be like sudden jerks. A considerable movement of the foundation is required for the failure surface in soil to extend to the ground surface (as shown by the broken lines). The load per unit area at which extend to the ground surface (as shown by the broken lines). The load per unit area at which this happens is the ultimate bearing capacity qult. Beyond this point, an increase of the load will be accompanied by a large increase of footing’s settlement. The load per unit area of the footing qult, is referred to as the first failure load (Vesic 1963). Note that the peak value of qis not realized in this type of failure, which is called the local shear failure in soil.

If the foundation is supported by a fairly loose soil, the load-settlement plot will be like the one in Figure 2c. In this case, the failure surface in soil will not extend to the ground surface. Past the value qult, the load-to-settlement plot will be steep and practically linear. This type of failure is called the punching shear failure.

Page 7: General Bearing Capacity

S e t t l e m e n t

L o a d / u n i t a r e a , q

q u

B L o a d / u n i t a r e a , q

q u

q u ( 1 )

B

( a ) F a i l u r e

i n s o i ls u r f a c e

Dense sands and stiff cohesive soils.

S e t t l e m e n t

L o a d / u n i t a r e a , q

q uq u

S u r f a c ef o o t i n g

B

( c )

s u r f a c e F a i l u r e

( b ) F a i l u r es u r f a c e

Figure 2. The nature of bearing capacity failures in soils: (a) the general shear failure; (b) the local shear failure; and (c) the punching shear failure. (Redrawn from Vesic, 1973).

Soils of medium density or stiffness.

Loose and soft soils.

Page 8: General Bearing Capacity

R e l a t i v e d e n s i t y D0 . 6

5 0 0

1 0 02

and

1

u

B

8 0 7 0

9 0

2 0 0

3 0 0

4 0 0

B

uq1

P u n c h i n g S h e a r

0 . 3

7 0 06 0 0

0 . 2 0 . 4

L o c a l S h e a r

0 . 5

G e n e r a l S h e a r

r

0 . 7 0 . 8

2

u(1) 6 0

S m a l l s i g n s i n d i c a t ef i r s t f a i l u r e l o a d

8 i n . C i r c u l a r p l a t e6 i n . C i r c u l a r p l a t e4 i n . C i r c u l a r p l a t e2 i n . C i r c u l a r p l a t e2 X 1 2 i n . R e c t a n g u l a r p l a t e( r e d u c e d b y 0 . 6 )

D r y u n i t w e i g h t ( l b / f t )

8 5

2 0

1 0

3 0

21

4 0

5 0

12 B

9 0

u ( 1 )q

3

9 5

and

B

7 0 B1

L e g e n d

Figure 3. Variation of qu(1) / ½ �B and qu / ½ �B for circular and rectangular plates on the surface of sand. (Adapted from Vesic, 1963).

Page 9: General Bearing Capacity

Based on experimental results from Vesic (1963), a relation for the mode of bearing capacity failure of foundations can be proposed (Figure 4), where

Dr is the relative density in sand,Df is the depth of the footing measured from the ground surface,B is the width and L is the length of the footing (Note: L is always greater than B).

0 . 4R e l a t i v e d e n s i t y D

P u n c h i n g S h e a r f a i l u r e

1

00 . 2

r

G e n e r a l S h e a rL o c a l S h e a rf a i l u r e f a i l u r e

0 . 6 0 . 8 1 . 0

B5

4 D f

3

Df/B

°

f a i l u r e

2

f a i l u r e f a i l u r e

Figure 4. Modes of foundation failure in sand (after Vesic, 1963).

Page 10: General Bearing Capacity

Figure 5. Range of settlement of circular and rectangular plates at ultimate loads for Df / B = 0 in sand (after Vesic, 1973).

Page 11: General Bearing Capacity

Terzaghi’s Ultimate Bearing Capacity Theory:

Using an equilibrium analysis, Karl Terzaghi expressed in 1943 the ultimate bearing capacity qu of a particular soil to be of the form,

(for strip footings, such as wall foundations)

(for square footings, typical of interior columns)

(for circular footings, such as towers, chimneys)

= c + + 0.5 B

= 1.3c + + 0.4 B

u c q

u c q

q N q N N

q N q N N

γγγγ

γγγγ

γγγγ

γγγγ

γγγγ (for circular footings, such as towers, chimneys)

where, q = �Df is the removed pressure from the soil to place the footing, Nc, N�, and Nq

are the soil-bearing capacity factors, dimensionless terms, whose values relate to the angle of internal friction �. These values can be calculated when � is known or they can be looked up in Terzaghi’s Bearing Capacity Factor Table.

Terzaghi’s bearing capacity equations have been modified to take into account the effects of the foundation shape (B/L), depth (Df) and the load inclination (i). This will be discussed further in the next section for the General Bearing Capacity equation.

= c + + 0.3 B u c qq N q N Nγγγγγγγγ

Page 12: General Bearing Capacity

(1970)

(1951, 1963)

Page 13: General Bearing Capacity

The factor of safety FS against a bearing capacity failure is thus defined as,

where qall is the gross allowable load-bearing capacity and qnet is the net ultimate bearing capacity.

The factor of safety is chosen according the function of the structure, but never less than 3 in all cases.

The net ultimate bearing capacity is defined as the ultimate pressure per unit area of

= ultall

qq

FS

The net ultimate bearing capacity is defined as the ultimate pressure per unit area of the footing that can be supported by the soil in excess of the pressure caused by the surrounding soil at the foundation level.

Therefore,

A footing will obviously not settle at all if the footing is placed at a depth where the weight of the soil removed is equal to the weight of the column’s load plus the footing’s weight.

= − = −net ult ult fq q q q Dγγγγ

Page 14: General Bearing Capacity

Choosing the Bearing Capacity Factors.

Based on laboratory and field studies of bearing capacities, the basic failure surface in soil suggested by Terzaghi, has been modified by Meyerhof, Hansen and Vesic. The relation for Nq was presented by Reissner (1924),

whereas the equation for Nc was originally derived by Prandtl in 1921 to explain the behavior of hot steel being extruded through dies,

2 tan tan tan 45 e e2

� �= ° + =� �� �

q PN Kπ φ π φφ

Nc = (Nq – 1) cot �

and finally, Caquot and Kerisel (1953) and Vesic (1973) proposed a relation for N� , latermodified by Coduto as,

2( 1) tan '

1 0.4sin(4 ')qN

φφ

+=

+

Page 15: General Bearing Capacity

A) Terzaghi’s Bearing Capacity Factors.

2 0 75 2

22 45 21

5 7 0 0

2 1

=° +

−= = = >

+

' '( . / )tan

q '

q' 'c c '

'

eNcos ( / )

NN . for or N for

tan( N )tan

π φ φπ φ φπ φ φπ φ φ

ϕϕϕϕ

ϕ ϕϕ ϕϕ ϕϕ ϕφφφφ

φφφφ2 1

1 0 4 4

+=

+

'q

'

( N )tanN

. sin( )γγγγ

φφφφφφφφ

Page 16: General Bearing Capacity

B) Meyerhof’s Bearing Capacity Factors.

C) Brinch Hansen’s Bearing Capacity Factors.

2 45 2

1

1 1 4

'tan 'q

'c q

'q

N e tan ( / )

N ( N )cot

N ( N )tan( . )

π φπ φπ φπ φ

γγγγ

ϕϕϕϕ

ϕϕϕϕ

ϕϕϕϕ

= ° +

= −

= −

2 45 2

1

1 5 1

'tan 'q

'c q

'q

N e tan ( / )

N ( N )cot

N . ( N )tan

π φπ φπ φπ φ

γγγγ

ϕϕϕϕ

ϕϕϕϕ

ϕϕϕϕ

= ° +

= −

= −

Page 17: General Bearing Capacity

ANGLE OF FRICTION TERZAGHI MEYERHOF HANSEN� (DEGREES) Nc Nq N� Nc Nq N� Nc Nq N�

0 5.70 1.00 0.00 5.10 1.00 0.00 5.10 1.00 0.002 6.30 1.22 0.18 5.63 1.20 0.01 5.63 1.20 0.014 6.97 1.49 0.38 6.19 1.43 0.04 6.19 1.43 0.055 7.34 1.64 0.50 6.49 1.57 0.07 6.49 1.57 0.076 7.73 1.81 0.62 6.81 1.72 0.11 6.81 1.72 0.118 8.60 2.21 0.91 7.53 2.06 0.21 7.53 2.06 0.2210 9.60 2.69 1.21 8.34 2.47 0.37 8.34 2.47 0.3912 10.76 3.29 1.70 9.28 2.97 0.60 9.28 2.97 0.6314 12.11 4.02 2.23 10.37 3.59 0.92 10.37 3.59 0.9715 12.86 4.45 2.50 10.98 3.94 1.13 10.98 3.94 1.1816 13.68 4.92 2.94 11.63 4.34 1.37 11.63 4.34 1.4318 15.52 6.04 3.87 13.10 5.26 2.00 13.10 5.26 2.0820 17.69 7.44 4.97 14.83 6.40 2.87 14.83 6.40 2.9522 20.27 9.19 6.61 16.88 7.82 4.07 16.88 7.82 4.13

BEARING CAPACITY FACTORS FOR GENERAL SHEAR

22 20.27 9.19 6.61 16.88 7.82 4.07 16.88 7.82 4.1324 23.36 11.40 8.58 19.32 9.60 5.72 19.32 9.60 5.7525 25.13 12.72 9.70 20.72 10.66 6.77 20.72 10.66 6.7626 27.09 14.21 11.35 22.25 11.85 8.00 22.25 11.85 7.9428 31.61 17.81 15.15 25.80 14.72 11.19 25.80 14.72 10.9430 37.16 22.46 19.73 30.14 18.40 15.67 30.14 18.40 15.0732 44.04 28.52 27.49 35.49 23.18 22.02 35.49 23.18 20.7934 52.64 36.50 36.96 42.16 29.44 31.15 42.16 29.44 28.7735 57.75 41.44 42.40 46.12 33.30 37.15 46.12 33.30 33.9236 63.53 47.16 51.70 50.59 37.75 44.43 50.59 37.75 40.0538 77.50 61.55 73.47 61.35 48.93 64.07 61.35 48.93 56.1740 95.66 81.27 100.39 75.31 64.20 93.69 75.31 64.20 79.5442 119.67 108.75 165.69 93.71 85.37 139.32 93.71 85.37 113.9644 151.95 147.74 248.29 118.37 115.31 211.41 118.37 115.31 165.5845 172.29 173.29 294.50 133.87 134.87 262.74 133.87 134.87 200.8146 196.22 204.19 426.96 152.10 158.50 328.73 152.10 158.50 244.6548 258.29 287.85 742.61 199.26 222.30 526.45 199.26 222.30 368.6750 347.51 415.15 1153.15 266.88 319.06 873.86 266.88 319.06 568.57

Page 18: General Bearing Capacity

Example 1.A square foundation is 5 feet by 5 feet in plan. The soil supporting the foundation has a friction angle of � = 20° and c = 320 lb/ft². The unit weight of the soil � = 115 lb/ft3. Determine the allowable gross load on the foundation with a factor of safety of 4. Assume that the depth of the foundation Df is 3 feet and that a general shear failure occurs in the soil. Use Terzaghi’ Method.

Solution: Terzaghi’s formula for a square footing is,

u c q

c q

q = 1.3c N + q N + 0.4 B N

where N =17.69, N =7.44 and N =4.97γγγγ

γγγγ

γγγγ

7 44 5

114

u

u

uall

q = 1.3(0.320)(17.69) + (0.115)(3 )( . ) + 0.4 (0.115 )( )(4.97)q = 11 ksf

q ksfq = 2.77 ksfFS

Thus, the total gross load allowed fr

∴ = =

2 2all all

om the column will be,

Q =q (B ) = (2.77 ksf)(5 70ft) = kips

Page 19: General Bearing Capacity

The General Bearing Capacity Equation.

The Terzaghi ultimate bearing capacity equations presented previously are for continuous, square, and circular footings only. They do not include rectangular footings (0 < B/L < 1), or take into account the shearing resistance along the failure surface in the soil above the bottom of the foundation, or the inclination of the footing or the load. To account for all of these, Meyerhof (1963) suggested the following General Bearing Capacity equation:

u

1q

2= + +

c cs cd ci q qs qd qi s d ic N F F F q N F F F B N F F Fγ γ γ γγ

where c = the cohesion;q = the excavated soil’s pressure at the footing’s invert (its bottom);� = the unit weight of the soil;B = width of foundation ( equal to the diameter for a circular foundation);

Nc, Nq, N� are the bearing capacity factors;

Fcs, Fqs, F�s are the shape factors;Fcd, Fqd, F�d are the depth factors; andFci, Fqi, F�i are the load inclination factors.

u 2c cs cd ci q qs qd qi s d iγ γ γ γ

Page 20: General Bearing Capacity

Shape Factors.

Depth Factors for Df / B ≤≤≤≤ 1.

Depth Factors for Df/B >1.

1 qcs

c

NBF

L N= + 1 tanqs

BF

Lφ= + 1 0.4s

BF

Lγ = −

1 0.4 fcd

DF

B= + 21 2 tan (1 sin ) f

qd

DF

Bφ φ= + − 1dFγ =

Depth Factors for Df/B >1.

Inclination Factors with ββββ is the inclination of load with respect to the vertical.

11 0.4 tan fcd

DF

B− � �

= + � �� �

2 11 2tan (1 sin ) tan fqd

DF

Bφ φ −= + − 1dFγ =

2

190

o

ci qi oF Fβ� �

= = −� �� �

2

1iFγβφ

� �= −� �� �

Page 21: General Bearing Capacity

Example 2.Compare the predicted ultimate vertical footing load Pu using both the Hansen and Meyerhof methods, and compare their results with the actual field measured load of 1863 kPa, given that L = 2.0 m.

The soil is a saturated sand, because the water table is close to the surface. The effective unit weight is �’ = 9.31 kN/m3.

The angle of internal friction was determined from a triaxial test to be 42�. determined from a triaxial test to be 42�. The corrected field value is therefore,

1.5( ) 17

1.5(42 ) 17 46

= − °

= ° − ° =field triaxial

ofield

φ φ

φ

Page 22: General Bearing Capacity

a) For the Brinch Hanson method,For � = 46°, Nq = 158.51 Nc = 0 and Nγγγγ = 244.65Fqs = 1 + (B/L) tan φφφφ = 1 + (0.5/2) tan46� = 1.26Fγγγγs = 1 - 0.4 (B/L) = 1 – 0.4 (0.5/2) = 0.9Fqd = 1 + 2 tan φφφφ (1 - sin φφφφ ) = 1 + 0.155(0.5/0.5) = 1.155F�d = 1.0

qu = 0 + γγγγ’Df Nq Fsq Fdq + ½ γγγγ’ B Nγγγγ Fsγγγγ Fdγγγγ= (9.31)(0.5)(158.51)(1.26)(1.155) + (0.5)(9.31)(0.5)(244.65)(0.9)(1) = 1,586 kPa

Hansen predicts 1,586 kPa versus 1,863 kPa (measured) or a + 15% under-predicted.

b) For the Meyerhof method,For � = 46°, Nq = 158.51 Nc = 0 and Nγγγγ = 329.74Kp = tan2(45+�/2) = 6.13 therefore �Kp = 2.48Fqs = Fγγγγ s = 1 + 0.1 Kp (B/L) = 1 + (0.1)(6.13)(0.5/0.5) = 1.61Fqd = Fγγγγ d = 1 + 0.1 �Kp (B/L) = 1+ (0.1)(2.48)(0.5/0.5) = 1.25

qu = 0 + γγγγ’ Df Nq Fqs Fqd + ½ γγγγ’ B Nγγγγ Fγγγγ s Fγγγγ d

= (9.31)(0.5)(158.51)(1.61)(1.25) +(0.5)(9.31)(0.5)(329.74)(1.61)(1.25) = 3,030 kPa

Meyerhof predicts 3,030 kPa versus 1,863 kPa (measured) or a +63% over-predicted.

Page 23: General Bearing Capacity

Example 3.Determine the size of the square footing, if the soil has a � = 105 pcf, �sat = 118 pcf, Df = 4 feet, D1 = 2 ft. The gross design load Qall is 150 kips with a FS = 3. The field SPTs are as follows,

Depth(ft) N (blows/ft)5 4

10 615 620 10 25 5 D 1

W a te rta b le

γ φ c = 0γ s a t

φ

150 kips

Solution. Using Ncor = CN Nfield

where Ncor is the corrected N value to a standard of �’v [95.6 kN/m² or 2,000 lb/ft²];CN is the correction factor, Nfield is the N value obtained in the field and the Liao and

Whitman relationship is

D f

B X B

φc = 0

aN '

v

p C

σσσσ=

Page 24: General Bearing Capacity

The corrected standard penetration number Ncor can now be determined,Depth Nfield

v'= a

cor field

pN N

σ( ) ( )( ) ( )( )

( ) ( )( ) ( )( )

( ) ( )( ) ( )( )

0.5

0.5

0.5

2 0.105 3 0.118 0.0624 5 4 4 = 1.7

2

2 0.105 8 0.118 0.062410 6 6 = 3.4

2

2 0.105 13 0.118 0.062415 6 6 = 4.1

2

+ −� � � �

+ −� � � �

+ −� � � �

The average Ncor seems to be about 4.3, and using the Hatanaka-Uchida (1997) equation to find the averaged angle of internal friction �,

( ) ( )2

220 10 10

� �

( ) ( )( )

( ) ( )( ) ( )( )

0.5

0.5

0.105 18 0.118 0.0624= 7.8

2

2 0.105 23 0.118 0.062425 5 5 = 4.3

2

+ −� � � �

+ −� � � �

20 + 20 = 20(4.3) 20 29= + ≈ °corNφ

Page 25: General Bearing Capacity
Page 26: General Bearing Capacity
Page 27: General Bearing Capacity

Example 4.Using the data of Example 1, if the factor of safety is increased to 5, what is the new net allowable load for the footing?

Solution :

( )

2 2 52

11 from Example 1, and = (0.115)(3) 0.

345 ksf

11 - 0.3452.1 ksf

5(2.1)( 70 5) =

= = =

= = ≈

∴ = =

u f

uall net

q ksf q D

qq

FSkips versusQ q B kips

γ

2 2( ) 52 (2.1)( 70 5) = ∴ = =all net all kips versusQ q B kips

Page 28: General Bearing Capacity

Example 5.Again using the data of Example 1, and using the equation for a square footing and a factor of safety against shear failure of 1.5, determine the net allowable load for this footing. Use Hansen’s bearing capacity factors, without modifiers. Apply the factor of safety to both parameters, c and �.Solution :

-1 -1d

320 psf 213 psf

1.5

tan tan 20 tan tan 13.64

1.5

= = ≈

� � °� �= = = ° � �� �

dshear

cc

FS

FSφφd

( )

(

1.5

Therefore, 1.3 0.4 B

For 13.64 , 2.29, 3.59, and 10.4

� �� �

= + +

= ° ≈ ≈ ≈

shear

all net d c q

q c

all ne

FS

q c N q N N

N N N

q

γ

γ

γφ

( ))

( )

( )

(1.3)(0.213)(10.4) (0.115) 3 (3.59) (0.4)(0.115)(5)(2.29)

4.6 ksf

a 115nd (4.6)(5)(5 57 .) 5

= + +

=

= = =

t

all net

all net kips

q

Q tonsNOTE: There appears to be a large discrepancy between the results of Examples 4 and 5. The use of trial and error shows that, when the FS (shear) is about 1.2 the results are approximately equal.

Page 29: General Bearing Capacity

Modification of the Bearing Capacity Equations for the Water Table.

Case I: When 0 ≤≤≤≤ D1 ≤≤≤≤ Df .

where γγγγsat is the saturated unit weight soil and γγγγw is the unit weight of water. The general bearing capacity equation is now,

( )1 2e sat wq D Dγ γ γ= + −

qu = c Nc Fcs Fcd Fci + qe Nq Fqs Fqd Fqi + (γγγγ sat – �w) B Nγγγγ Fγγγγs Fγγγγd Fγγγγi

Page 30: General Bearing Capacity

Modification of the Bearing Capacity Equations for the Water Table.

Case II. When 0 ≤≤≤≤ d ≤≤≤≤ B .

where γγγγsat is the saturated unit weight soil and γγγγw is the unit weight of water. The general bearing capacity equation is now,

qu = c Nc Fcs Fcd Fci + q Nq Fqs Fqd Fqi + [�’ + (� - �’)d/B] B Nγγγγ Fγγγγs Fγγγγd Fγγγγi

fq Dγ=

Page 31: General Bearing Capacity

Modification of the Bearing Capacity Equations for the Water Table.

Case III. When d ≥≥≥≥ B, the water table will have no effect on the ultimate bearing capacity.

qu = c Nc Fcs Fcd Fci + q Nq Fqs Fqd Fqi + [�’ + (� - �’)d/B ] B Nγγγγ Fγγγγs Fγγγγd Fγγγγi

Page 32: General Bearing Capacity

Example 6.

Calculate the bearing capacity for a square footing, given c = 350 lb/ft2, γ = 105 lb/ft3, γsat = 118 lb/ft3, Df = 4.0 feet, Dw = 2.0 feet, φ = 25�, B = 4.0 feet, Qrequired = 50,000 lb and FS = 3.

Page 33: General Bearing Capacity

Bearing capacity factors: For � = 25�, Nq = 6.4Nc = 14.8 and Nγγγγ = 5.4

Shape factors:Fqs = 1 + (1) tan 25� = 1.40 Fcs = 1 + (1) 6.4/14.8) = 1.43Fγγγγs = 1 – 0.4(1) = 0.6

Depth factors: Df /B = (4/4) � 1 Fqd = 1 + 2 tan φ (1 - sinφ )2 tan-1 (4/4) = 1.28Fcd = 1 + 0.4 tan-1(4/4) = 1.35Fγγγγd = (1 – �/�)2 = 1

Inclination Factors: Inclination Factors: Fci = Fqi = F�i = 1

Consideration of water table: 0 < Dw < Df therefore,

q = D1 γγγγ + D2 (γγγγsat - γγγγw) = (2.0)(105) + (2.0)(118 - 62.4) = 321 lb/ft2

So,qu = c Nc Fcs Fcd Fci + q Nq Fqs Fqd Fqi + γγγγ B Nγγγγ Fγγγγs Fγγγγd Fγγγγi

= (350)(14.8)(1.43)(1.35)(1)+(321)(6.4)(1.40)(1.28)(1)+(55.6)(3.3)(5.4)(0.6)(1)(1) qu = 14,000 lb/ft2

(FS)(Qall) = (qu)(A) therefore Qall = (14,000)(4)(4)/3 = 50,700 lb > 50000 lb Satisfactory.

Page 34: General Bearing Capacity

70 1 0.33 1.20.04

fall

DNq if B m

B� �

= + ≤� �� �

Bearing Capacity from SPT.

Several papers have been published that provide statistical data that predict the bearing capacity of footings whilst controlling their settlement to 1 inch. The data is based on the results of SPTs with a correction to 70%, that is N70. The allowable bearing capacity can be provided on a preliminary basis from,

270 0.3

1 0.33 1.20.06

fall

DN Bq if B m

B B� �+� �= + ≥� �� �

� � � �

Alternatively, a chart is provided in the next slide that also predict allowable bearing capacities for surface loaded footings with a settlement limited to approximately 1” (25 mm). Equations used are shown on figure. Depth-factor adjustment are also shown. The N provided is approximately N55.

Page 35: General Bearing Capacity

400

500

600

700

800

qa, kPa

Bearing Capacity from SPT.

55

0.05all

Nq =

255 0.3

1 0.330.08

0 1.2

fall

f

DN Bq

B B

for D B and B m

� �+� �= +� �� �� � � �

≤ ≤ >

N=25

N=30

N=35

N=40

0

100

200

300

0 1 2 3 4 5 6

B, m

1.2 m

N=15

N=5

N=10

N=20

N=25

Page 36: General Bearing Capacity

γNNq qc 8.0~8.0~

( ) 25.13000052.028

cmkg

qq cult −−=

( ) 25.1300009.048

cmkg

qq cult −−=

Bearing Capacity using CPT.

For Granular Soils:

strip footings

square footings

228.02cmkg

qq cult +=

234.05cmkg

qq cult +=

For Cohesive Soils:

strip footings

square footings

Page 37: General Bearing Capacity

Props for stability when using dead weights.Jack

Dead weight of a truck or a beam attached to anchor piles

Bearing Capacity for Field Load Tests.

Several dial gauges attached to an independent suspension system to record plate settlements with each increment of the jack load.

Short block

Steel plateanchors or pilesto provide a reactive force

Page 38: General Bearing Capacity

References:

1. J. Bowles, “Foundation Analysis and Design”, McGraw-Hill;2. B. Das, “Principles of foundation Engineering”, Thompson;3. Liu, Everett, “Soils and Foundations”, Prentice Hall;