pondasi-1-6

7/23/2019 PONDASI-1-6

http://slidepdf.com/reader/full/pondasi-1-6 1/18

Foundation Engineering Foundation Engineering

Lecture #05 Lecture #05

Bearing Capacity of Soils

on

Eccentrically Loaded Footings

L. Prieto-Portar, 2008

7/23/2019 PONDASI-1-6


Foundations with a One-Way Eccentricity.

In most instances, foundations are subjected to moments in addition to the vertical load

as shown below. In such cases the distribution of pressure by the foundation upon the soil

is not uniform.

Q M

B B

B X L

e

F o r e < B /6

q m in

q m a x

F o r e > B /6

q m a x

e

(b )( a )

Figure 6. An

eccentrically

loaded

footing.

The effectivewidth is now,

B’ = B - 2e

whereas the

effective length is

Still,

L’ = L

7/23/2019 PONDASI-1-6


The distribution of the nominal pressure is :

where Q is the total vertical load and M is the moment on the footing in one axis.

Figure 6b shows a force system equivalent to that in Figure 6a. The distance e is the

eccentricity of the load, or

Substituting equation 16 in equations 14 and 15 yields:

max 2

min 2

6 (14)

and

6 q (15)

Q M q BL B L

Q M

BL B L

= +

= −

M e (16)Q

=

max

min

6

1 (17 a)

6

q 1 (17 b)

Q e

q BL B

and

Q e

BL B

= +

= −

7/23/2019 PONDASI-1-6


Note that in these equations, when the eccentricity e becomes B/6, q min is zero. For e >

B/6, q min will be negative, which means that tension will develop. Because soils can sustain

very little tension, there will be a separation between the footing and the soil under it.Also note that the eccentricity tends to decrease the load bearing capacity of a

foundation. In such cases, placing foundation column off-center, as shown in Figure 7 is

probably advantageous. Doing so in effect, produces a centrally loaded foundation with a

uniformly distributed pressure.

M

e

Q

Figure 7. A footing with the column off-center to preserve a uniform pressure on the soil.

7/23/2019 PONDASI-1-6


The general bearing capacity equation is therefore modified to,

qu’ = c N c F cs F cd F ci + q N q Fqs Fqd Fqi + γ γγ γ B’ N γ γγ γ Fγ γγ γ s Fγ γγ γ d Fγ γγ γ i

and Qu = qu’ B’ L’

7/23/2019 PONDASI-1-6


Foundations with Two-way Eccentricities.

Consider a footing subject to a vertical ultimate load Qult and a moment M as shown in

Figures 8a and 8b. For this case, the components of the moment M about the x and y axis

are M x and M y respectively. This condition is equivalent to a load Q placed eccentrically

on the footing with x = e B and y = e L as shown in Figure 8d.

M

B X L

Q u lt

B(a )

Qult

B

L

Qult

My

x

y

My

Qult

Mx eL

eB

(b) (c) (d)

Figure 8. Analysis of a footing with a two-way eccentricity.

7/23/2019 PONDASI-1-6


Note that,

ult

c q

ult u

(18)

If Q is needed, it can be obtained as ' ' ,

1' cN qN B ' N (19)2

A' is the effective area B'L'

Finally Q = q (

y x B L

ult ult

ult u

u cs cd ci qs qd qi s d i

M M e and eQ Q

Q q A where

q F F F F F F F F F

and

γ γ γ

= =

=

= + +

cs qs s

A')

As before, to evaluate F , F , and F , use the effective length (L') and the effective width

(B') dimensions instead of L and B, respectively. To calculate F , ,and ,do not replaceB with B

cd qd d F F

γ

γ

'. In determining the effective area (A'), effective width (B'), and the effective (L'),

four possible cases may arise (Highter and Anders, 1985).

7/23/2019 PONDASI-1-6


B1

B

Qult

L1

Effectivearea

eL

eB

L

1 1

ffective area for the case of and .6 6

L Be e

L B≥ ≥Figure 9. Case I. The e

7/23/2019 PONDASI-1-6


1 1

1 1

1 1 and .

6 6

1The effective area for this condition is A' (20)

2

3 3where B 1.5 and L 1.5 (21 a and b)

The effective length L

L B

B L

e e

L B

B L

e e B L

B L

≥ ≥

=

= − = −

Case I :

1 1', is the larger of the two dimensions B L .

Therefore, the effective width is:

' ' (22)

'

L

or

A B

L

e

=

Case II :

1 2

1 2

1 1 and 0 . The effective area for this case is shown in Figure 10a,

2 6

1 ' ( ) (23)

2The magnitude of L and L can be det

Be

L B

A L L B

< < <

= +

1 2

1 2

ermined from Figure 10b. The effective width is,

' ' (24)

( arg )

The effective length is ' ( arg ) (25)

A B

L or L whichever is l er

L L or L whichever is l er

=

=

7/23/2019 PONDASI-1-6


0.4

0.1

For

obtaining

L2 /L

0.60

0

0.2 0.4

0 .

1 6

0.2

0.1

0.3

e L / L

0 . 1 0

eB /B =

0 . 1 2

0 . 1 4 0

. 0 6

0 . 0 8

0 . 0

4

For

obtaining

L1 /L

0.8 1.0

0 . 0

1

0 . 0

2

0 . 0 2

0 . 0 1

0.080.06

0.04

0.5

eB /B =0.167

L1 /L , L2 /L

(b)

L

(a)

Qult

L2eB

B

L1

eL

Effective

area

1 1Effective area for the case of and 0 (After Highter and Anders, 1985)

2 6

'The effective width is ' and the effective length is ' .

L Be e

L B

A B L L

L

< < <

= =

Figure 10. Case II.

L1 and L2 can be determined from the figure,

1 2

'' B

L orL

=

using in the denominator the largest

value of L, that is,

L’ = L1 or L2 whichever is larger

7/23/2019 PONDASI-1-6


1 2

1 1 and 0 . The effective area is shown in Figure 11a.

6 2

1 ' ( ) (26)

2

The effective width is,

'

L Be e

L B

A B B L

B

< < <

= +

=

Case III :

1 2

' (27)

The effective length is equal to L' L (28)

The magnitudes of B and B can be determined

A

L =

22

2

fron Figure 11b.

1 1

and . The effective area is shown in Figure 12a.6 6

The ratio and thus B can be determined by using the curves that slope upward.

Similarely, the ratio

L B

L

e e

L B

B e

B L

L

L

< <Case IV :

2

2 2 2

and thus L can be determined by using the curves that slope downward.

The effective area is then,

1 ' ( )( ) (29)

2

Le

L

A L B B B L L= + + −

7/23/2019 PONDASI-1-6


00 0.2 0.4 0.6 0.8 1.0

0.1

0.2

0.3

0.4

0.5

B1 /B , B2 /B

e B / B

Effectivearea

Qult

0 . 1

6

For

obtaining

B2 /B

For

obtaining

B1 /B

0 . 1 4

eL /L =

0 . 1 2

0 . 1 0

0 . 0 8

0 . 0

6

0 . 0

4

0 . 0

2

0 . 0

1

eL

eB

B

L

eL /L =

0 . 0 2

0 . 0 1

0.167

0.10.08

0.060.04

(a)

(b)

B1

B2

Foundation with Two-way Eccentricity, Case III

1 1Effective area for the case of and 0

6 2(After Highter ans Anders, 1985).

L Be e

L B

< < <Figure 11.

7/23/2019 PONDASI-1-6


00 0.2 0.4 0.6 0.8 1.0

0.05

0.10

0.15

0.20

B2 /B , L2 /L

e B / B

Effectivearea

Qult

eL /L = 0.02

eL

eB

B

L

0.02 = eL /L

(a)

(b)

B2

For obtaining L2 /L

For obtaining B2 /B

L2

0 .0 4

0 .0 6

0 . 0

8 0 .1

0 .1 4 0.04

0.060.080.10

0 . 1

2 0 . 1

6

0 . 1

4

Foundation with Two-way Eccentricity, Case IV

1 1Effective area for the case of and

6 2

(After Highter ans Anders, 1985)

L Be e

L B< <Figure 12.

7/23/2019 PONDASI-1-6


Example 7.

A square foundation is shown in Figure 13. Assume that the one- way load eccentricity eis 0.15 m. Determine the ultimate load Qult.

Solution : With c = 0, equation 19 becomes:

γ sat =18kN/m

φ = 30°c = 0

1.5 m x 1.5 m

0.7 m

sand3

( ) ( )

1q 2

2

f

' qN B'N

and D 0.7 18 12.6 kN m

u qs qd qi s d iq F F F F F F

q

γ γ γ

γ

= +

= = =

Figure 13.

For = 30º, N q = 18.4 and N = 22.4

B’ = 1.5 – 2(0.15) = 1.2 m

L’ = 1.5 m

7/23/2019 PONDASI-1-6


( ) ( ) ( )2

B ' 1 .2 1 tan 1 tan 3 0 1 .4 6 2

' 1 .5

0 .289 0 .7 1 2 tan 1 - s in 1 1 .135

1 .5

B ' 1 .2 1 - 0 .4 1 - 0 .4 0 .68

L ' 1 .5

q s

f

q d

s

F L

DF

B

F γ

φ

φ φ

= + = + ° =

= + = + =

= = =

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

12

2

1

' 1 2 .6 1 8 .4 1 .4 6 2 1 .1 3 5 1 8 1 .2 2 2 .4 0 .6 8 1

384 .7 164 .50 549 .2 kN m

B ' ' ' 1 .2 1 .5 5 4 9 .2 9 8 8

d

u

u lt u

F

S o

q

H en ce

Q L q kN

γ =

= +

= + =

= = ≈

7/23/2019 PONDASI-1-6


Example 8.

Using the data from Example 5, and having a two-way eccentricity, for Figure 14,determine the ultimate load Qult.

1.5m

e =0.3 m

1.5m

L

e =0.15mB

2.0

5.1

3.0

L

e :SolutionL ==

1.05.1

15.0

B

e B ==

2

21

1.193m)0.315)(1.5(1.275

2

1

)LL(2

1 'A

)51.3.(FromEq

=+

=+=

Figure 14.

7/23/2019 PONDASI-1-6


1

1q 2

2

q

qs

A' 1.193From equation 24 B' 0.936

L' 1.275

From equation 25, L' L

Since c 0

' qN B'N

(0.7)(18) 12.6 kN /m

For 30 , N 18.4 and N 22.4

'F 1 tan'

u qs qd qi s d i

f

q F F F F F F

and q D

B L

γ γ γ

γ

γ

φ

= = =

=

=

= +

= = =

= ° = =

= +

s

2

qd

d

0.9361 tan 30 1.4241.275

' 0.936F 1 0.4 1 0.4 0.706

' 1.275

(0.289)(0.7)

F 1 2 tan (1 sin ) 1 1.1351.5

F 1

1 ' ' '( ' )

2

(1.193)[(12.6)

f

ult u q qs qd s d

B

L

D

B

Q A q A qN F F B N F F

γ

γ

γ γ γ

φ

φ φ

γ

= + ° =

= − = − =

= + − = + =

=

∴ = = + =

= (18.4)(1.424)(1.135) (0.5)(18)(0.936)(22.4)(0.706)(1)]

606 kN

+

=

7/23/2019 PONDASI-1-6


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;

pondasi-1-6

Documents