Week 7 Eccentricity and Bearing Capacity of Foundations

Effect of eccentric footing on bearing capacity

Resultantof

superstructurepressure

DD

B

Concentric

DD

e

Eccentric

B′

B′ = B-2e why????????????

Effect of eccentricity on foundation base

The bearing capacity equation is

developed with the idealization that the

load on the foundation is concentric.

However, the forces on the foundation

may be eccentric (i.e., foundation

subjected to additional moment). In

such situations, if the loads are eccentric

in both directions, then; the base of

foundation (B x L) shall be considered

as:

Further, area of foundation to be

considered for safe load carried by

foundation is not the actual area, but the

effective area is:

eB=MB/P , eL=ML/PB′ = B – 2eB L′ = L – 2eL

A′ = B′ x L′

eB must be ≤B/6 otherwise!!!!!!!

Eccentrically Loaded Foundations

Bearing pressures at corners-Two way eccentricity

Verify stability of footing for the effect of one-way bending moment

maxqqs

)6

1(max B

e

BL

Pq B)

61(min B

e

BL

Pq B

MB

PWhen eB<B/6

Verify stability of footing for the effect of one-way bending moment

P

MB

When eB >B/6

What happens?

There will be separation of foundation from soil beneath and stresses will be redistributed; Not recommended

Effect of two-way bending moment

P

MB

B

We change B to B′ to be used in the bearing capacity calculation:

B′=B-2eB L′=L-2eL

eB=MB/P

ML

L

ML

MBP

qmax = )

qmin = )

maxqqs

To calculate the bearing capacity we have to change: B to B′ =B-2eB

and L to L′ = L-2eL

A′ =B′ x L′

eB=MB/P

eL=ML/P

ML

MB

B

L

P

Verify stability of footing for the effect of two-way bending moment

Example1: Eccentrically loaded Foundation/one-way moment For the rectangular footing shown, draw the pressure distribution below the footing base for the following conditions:

B=3m, L=5m, Q=1600kN, M = 800 kN-m

Solution:MB

QeB=MB/P =800/1600 =0.5m

qmin = )

qmax = )

eB=MB/P =800/1600 =0.5m

q min =1600/3x5 (1-6x0.5/3 - 0) = 0.0kN

q max =1600/3x5 (1+6x0.5/3+0) =213.33kN

213.33kN

0.0kN

A square footing (1.8x1.8)m with a (0.4x0.4)m square column. It is loaded with an axial load of 1800 kN and Mx =450 kN.m; My =360 kN.m. The soil has Φ =36° and C=20kN/m2 . The footing depth D =1.8m; the soil unit weight =18.0 kN/m³; the water table is at a depth of 6.1 m from the ground surface. Find the effective footing dimensions to contain the applied moments.

Example 2: (Eccentrically loaded foundation/two-way moment)

Solution:1- Refer to the red and orange boxes to the right.2- In this footing we should switch the footing dimensions L’=1.3m to become the smaller dimension and B’=1.4m to become the longer dimension.

For the mat

foundation and

loading shown, find

the pressure

intensity values (q)

at points A to F. All

columns measure

(0.5x0.5)m

(Eccentrically loaded Mat/Raft foundation)

Solution

Example: One & Two‐Way Eccentricity

Given:- Grain silos as shown; - Proposed raft dimensions = 50m x

50m- Each silo has a diameter of12m and an

empty weight of 29 MN; can hold up to 110 MN of grain.

- Silos are 24m apart (c/c) from each other

- Weight of raft = 60 MN- Silos can be loaded independently of

each other

Find: Design the raft by checking:1. Whether or not eccentricity will be

met with the various loading conditions possible

2. Eccentricity can be one way or two-way

One-Way Eccentricity

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

e = M/Q = 2640/396 = 6.67m

B /6 = 50/6= 8.33 m > 6.67 m

Eccentricity OK for one-way

eccentricity, i.e., (No negative

pressures)

Two-Way Eccentricity

Largest Loading:

one silo full and the rest empty

Result of Two-Way - Eccentricity

Analysis

• eB = eL = 4.62 m

• B = L = 50.0 m (proposed

foundation)

Equivalent Footing Dimensions

• B’ = B – 2eB = 50.0 – (2)(4.62)

• B’ = 40.8 m = L’ (as B = L and eB =

eL)

End of Week 7