Bearing Capacity Of Shallow Foundation.pdf 3m6m53

Bearing Capacity of Shallow Foundation

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Bearing Capacity is the ability of a soil to a load from foundation without causing a shear failure or excessive settlement. The sign of Bearing Capacity (B.C) and this units as pressure's unit ton/m2, KN/ m2, Kg/cm2, lb/ft2 etc… so can called Bearing Pressure Definitions: 1. Ultimate B.C ( q

ult

)

It's the gross pressure at the base of foundation at which the soil fails in shear. It's not used for design because it has a big value 2. Net ultimate B.C ( q

u net

)

It's net increase in pressure at the base of foundation cause the failure… u net

=

ult

-

DF

Where: DF ult u net

=

Over burden pressure at foundation level

=

Ultimate B.C

=

Net ultimate B.C

3. Net safe Bearing Capacity ( q

n.s

):

It's the pressure at which foundation designed.

q

n.s

=

Where: F.S

= Factor of safety equal from(2 to 5)

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4. Safe Bearing Capacity ( q ): s

It means the gross safe Bearing Capacity which used in design. q

s

=

q

gs

=

q

ns

=

+ γD

+ γD

5. Net safe settlement pressure q

n.p

It's the net pressure which the soil carry without increase in allowable settlement. 6. Net allowable B.C q

n.all

It's the net pressure which can be used for the design of foundation, which ensure that there is not shearing failure, or the settlement within reach the limit, to choose the allowable B.C ( q

n.all

).

If the net safe settlement pressure more than net safe B.c q q

n.p all

>

q

=

q

n.s

n.s

If the net safe B.C more than the net safe settlement pressure the the allowable B.C equal the net safe settlement pressure. q q

n.s all

>

q

=

q

=

Net safe B.C

=

Net settlement pressure

n.p

n.p

where: q q

n.s n.p

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q

all

=

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Net allowable B.C (Design allowable B.C)

Shear Failure: Failure soil at foundation level due to shear strength happened when increased the foundation load or decreased the resistance of soil for shear. Shear failure happened on many stages: I)

Stage I: The soil in the elastic case and behave as the part of foundation it still that, and by increasing the load performed the region I which called active zone.

II)

Stage II: At this stage the foundation load effect on the active zone and neighboring soil so perform the region which called arc of logarithmic spiral zone.

III)

Stage III: By increasing the load performed the third part curve in which part the soil became in the ive case it make to resist the failure. The soil fails when load on foundation increase and became more than the soil resist or B.C of soil. See Fig. (1).

In this case there are three component produced to resist the failure of soil. I)

(P )

pγ

= Component produced by the weight of shear zone II, III.

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II)

(P )

= Component produced by the cohesive stress.

III)

(P ) = Component produced by the surcharge q.

pc

pq

Terzaghi's Bearing Capacity Theory: Terzaghi Assumptions: The failure for Terzaghi's theory as shown in Fig (2):

1. The base of foot is rough, to prevent the shear displacement. 2.

The foot is shallow foundation, i.e. the depth of foundation is less than the width of foot… D

f

≤

B

3. Shear strength above the level of the base of foot is negligible. i.e. above ( F.L ). 4. Consider only the surcharge which produced as uniform pressure

C=0

q = γD

F

at foundation level. 5. The load on foundation is vertical and uniform.

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6. The foot is long strip foot. As mentioned before …. qult

=

(Pp)γ + (Pp)c + (Pp)q

(P )

=

Component produced by cohesive stress.

(P )

=

Component produced by surcharge q = γD

pγ

pc

(P ) = pq

q

F

Component produced by the weight of soil in zone II, III.

=CN

ult

N , N , Nγ c q

c

+qN =

q

+ 0.50 γ B Nγ

Bearing Capacity factor which are dimensionless depend on

angle of shear resistance φ.

N

=

q

a

=

N

=

N

=

K

=

c

γ

p

= coefficient of ive earth pressure.

A.R.E Bearing Capacity Equation:

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‫‪Bearing Capacity of Shallow Foundation‬‬

‫‪http://osp.mans.edu.eg/geotechnical/Ch3.htm‬‬

‫من كتاب المواصفات المصرية يحدث االنھيار تحت األساس لعدت أسباب منھا ‪:‬‬ ‫‪ .1‬زيادة حمل األساس‪.‬‬ ‫‪ .2‬تناقص مقاومة تربة األساس‬ ‫‪ .3‬حفر التربة المجاورة لألساس‬ ‫‪ .4‬تقدر قدرة تحمل التربة القصوى إما بالحساب أو بيانيا ً وھو بدراسة الجزء الموضح في الرسم وباستخدام‬ ‫المعادالت‪.‬‬ ‫وذلك في حالة تربة متجانسة إلى العمق ‪ d‬ومنسوب رسوب المياه الجوفية أوطي من أي أسفل المنسوب ‪ d‬أي أن‬ ‫التربة جافة فإنه ‪...‬‬ ‫… ‪I. Under concentrated vertical central load‬‬ ‫‪+ qN λ + γ BN λ‬‬

‫‪γ γ‬‬

‫‪2‬‬

‫‪= CN λ‬‬

‫‪c c‬‬

‫‪q q‬‬

‫‪Where:‬‬ ‫‪Ultimate Bearing Capacity.‬‬

‫=‬

‫‪Cohesive stress.‬‬

‫=‬

‫‪C‬‬

‫)‪Over burden pressure above (F-L‬‬

‫=‬

‫‪q‬‬

‫‪γ D‬‬

‫=‬

‫)‪Unit weight of soil above (F-L‬‬

‫=‬

‫‪1 F‬‬

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‫‪q‬‬

‫‪ult‬‬

‫‪γ‬‬

‫‪1‬‬

‫‪q‬‬

‫‪ult‬‬

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γ

=

Unit weight of soil at the base of foundation

B

=

Width of foundation

2

N , N , N = Bearing Capacity (B.C) factors depend on φ (angle of internal c

q

γ

friction) πtanφ

2

N

=e

NC

= ( Nq - 1 ) cotφ

Nγ

= ( Nq - 1 ) tanφ

q

. tan ( 45 +

Nc, Nq, Nγ = F (φ )

λ , λ , λγ = C q

)

see table (1)

factors depend on the shape of foundation Shape dimension [B, L].

Foundation

λC - λq

λγ

Strip

1.0

1.0

Rectangular Square & Circle

1 + 0.3 B/L 1.3

1 – 0.3 B/L 0.7

Table (2) The value of shape Factor

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D

F

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= Depth of foundation , Show Fig (4)

Fig(4) , Cases of depth of foundation

II.

Eccentric Vertical Load: 1. Eccentricity in direction L = eL

A = area = The shape factors become -

λ , λ , λγ = F ( B / L ) C q 2. Eccentricity in direction B = e as the mentioned before .. B

-

B = B – 2e

B

-

A = B .L And the shape factor became λ , λ , λγ = F ( B / L) C q -

qu = CN λ + qN λ + γB Nγ λγ c C

q q

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3. Eccentricity in direction (B, L). -

B = B – 2e

B

-

L = L – 2e

L

And the shape factors become -

-

λ , λ , λγ = F ( B / L ) C q -

-

A=B .L

III.

Central inclined… In the case of inclined load R the resultant can analyses at two components H & V where : H= V=

1.

By increasing the angle of δ the value of Bearing Capacity decreased where δ = tan

2. H

-1

H/V.

≤

Where A = Area of foundation 3. B.C equation become q

-

ult

= C N λ i + q N λ i + γB Nγλγiγ C C C

q q q

Where: i , i , iγ = Inclination Factors C q

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i

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=

q

iγ

=

i

=i -

C

q

H

= Horizontal component of load

V

= Vertical component of foot

φ

= Angle of Internal Friction

Special Case: When à

When

φ

=0

cotφ = α i

q

= iγ = i = 1 C

C

=0

i

=

q

=

iγ

= =

i

C

IV.

=i q

Eccentric inclined on foot is inclined and eccentricity so we make it as

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mentioned before in II & III. V.

Effect of Ground surface inclination:

The Bearing capacity of soil decrease when the foot lies near from inclination of ground surface. See Fig.(4).

From fig (5) note that … -

-

1. The surcharge decrease from q to q so that value of Nq becomes Nq . -

2. The surface which produced to resist the failure L decrease and become L -

so Nc decrease to Nc .

N γ = as before Where: ,

= B.C factors in case of inclined G.S this function

F (b/B, D/B, β, φ). See table (3) and B.C equation become q

ult

=C

λ +q C

-

λ + γB Nγ λγ q

Table (3) From Code

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Table (3)

Example: Calculate the allowable B.C (q ) for surcharge foundation 36 m wheares depth all

o

2

of foundation 1.5 m for C – φ soil where φ = 10 ,C = 4 t/m and unit weight of soil 1.8 3

o

t/m , and compare the results if there back fill inclinations with β = 60 , b = 0. Solve: 1. φ = 10

o

from table

.................»

N

c

= 8.5

N

= 2.5

Nγ

= 0.5

q

2. 3. q = γD = 1.8 ´ 1.5 = 2.7 ton/m

2

F

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4. q

ult

http://osp.mans.edu.eg/geotechnical/Ch3.htm

-

= CN λ + qN λ + γB Nγλγ C C

q q

= 4 ×8.5 × 1.15 + 2.7 × 2.5× 1.15+ 1.8× 3 ×0.5 × 0.85 2

= 49.16 t/m q

n ult

q

all

= 49.16 – 2.7 = 46.46 t/m

2

=

2

= 15.48 t/m

2

= 1.5 Kg/cm For Inclination: Φ

= 10

β

= 60

o

=

= 0.5

.............................»

From table

=0 = 6.33 = 0.5 = 0.5 q

ult

=C

λ +q C

λ + γBNγλγ q

= 4 × 6.33 ×1.15 + 2.7 × 0.5 ×1.15 + 1.8 × 3× 0.5 × 0.85 = 32.96 ton/m q

net ult

q

all

2

= 32.96 – 2.7 = 30.27 = 30.3 / 3

2

= 10.1 ton/m

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2

= 1 Kg / cm % decrease

=

VI . Effect of ground water table ( G. W. T): q

ult

= CN λ + qN λ + γBN λ C C

q q

γ γ

1.) G.W.T under G.S and above the base q = γDF = γ γB

=γ

sat

sub

. dw + γ

sub

. h1

.B

2.) G.W.T under foundation level:

a.

, it means that the water is far

If dw >

from the shear failure plan and his effect so … q = γD γB

F

à as the case of soil dry or bulk.

b. If dw < , and the ( G.W.T ) lies between the base and shear failure plan. q =γ D

1 F

γD = [γ

+ Fw (γ - γ

sub

1

sub

)]

where: Fw = Coefficient depend on φ and in fig (6).

as shown

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Fig (6)

VII. Effect of Multi Layer: To calculate B.C for Multi layer soil for foot ( B . L ). 1. Calculate the B.C for first layer by using the properties of this soil (q ). 1all

2. Calculate the B.C for second layer (q ) by using the properties of second 2all

layer γ2, φ2, C2 where B- become ( B + h ) and

= DF + h1. After that calculate the

equavelent B.C q

=

2equ

3. q

Compare the values of q

1all

and

2equ

If Design If

q

1all

q q

all

1all

>q

2equ

=q

2equ



2equ

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Design

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qall = q1all

Example: As shown in fig the B.C at F.L for clay layer = 1.0 kg/cm2 under clay soil lies organic clay soil at 3.0 m under (F.L) wheares B.C = 0.2 kg/cm2 determine the allowable B.C if the foot … a. Strip ( B = 2 ). b. Square foot ( 2 2× ) m

Solve : a. for strip foot: q

=

equ

2

=

= 0.5 kg/cm 2

1all

q

= 1 kg/cm

q

equ

2

= 0.5 kg/cm

2

all

design = 0.5 kg/cm

b. For square: q

equ

=

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2

= q q

= 1.25 kg/cm 2

all

= 1 kg/cm

q

equ

= 1.25 kg/cm

2

2

all

design = 1 kg/cm

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