wk7 raft foundations + eccentricity
DESCRIPTION
Geotechnical engineeringTRANSCRIPT
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