coulomb earth pressures 2009
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coulomb earth pressureTRANSCRIPT
Earth Pressures by the Coulomb Approach
Source Material Provide By : D. A. Cameron Rock & Soils 2009
Dr. I Wayan Sengara
Lateral Earth Pressure
Retaining structures such as retaining walls, basement walls, and bulkheads commonly are used in foundation engineering.
These structures are also subjected to lateral loads, coming from either a soil mass, or from seismic forces operating upon these soil masses.
Lateral Earth Pressure
Some of the examples of these lateral loads covered in this lecture include:
1. Soil stability problems;
2. The temporary of permanent bracing of excavations;
3. The design of retaining walls for highways (abutments) or buildings (walls), and
4. The various types of anchorages that transfer loads from the walls to foundations.
Lateral Earth Pressure
Lateral Earth Pressure
Lateral Earth Pressure
In order to be able to design the structures that can carry these lateral loads, a theory of how these lateral loads behave is reviewed. Specifically, this lecture will present,
1. Define the different types of lateral pressures: at rest, active and passive earth conditions,
2. Distinguish between the theories developed by Charles Coulomb and William Rankine.
Coulomb (1736-1806) Charles-Augustine de
Coulomb (1736-1806) was a military engineer and a famous French physicist that discovered the force between two electrical charges.
Less known was his development of the first thoroughly analytical study of lateral earth pressures which he published in 1776.
That theory remains the standard choice of analysis for lateral forces upon structures in soils.
Rankine (1820-1872)
William J.M. Rankine (1820-1872), the famous Scot engineer and physicist is best known as one of the founders of the science of thermodynamics.
He held the Queen Victoria Chair of civil engineering at the University of Glasgow.
In soil mechanics, he simplified Coulomb’s theory for cases when the surface of the backfill is horizontal, the friction between the wall and the backfill is negligible and the retaining wall is vertical.
COULOMB APPROACH
Upper bound theorem
(Rankine is a lower bound)
Plausible collapse mechanism
assumed
Equilibrium of forces, moments or
energy
Predicted force > collapse load
NOTE: when an upper bound and a
lower bound agree = the true
solution!
COULOMB APPROACH
HANDLES: Irregular backfill surfaces Sloping backs of walls Sloping backfill Surface surcharges Wall friction
Interface friction angle,
Although failure surfaces were
known to be curved, a planar
approximation was adopted
(OK for Active earth pressures)
Influence of Friction
Fa
ilure
line
Pa
Active
Pp
Failu
re lin
e
Passive
Coulomb Wedge Analysis
Case 1: Uniform soil, c = 0,
vertical back of retaining wall
- no wall friction
W
N
T
Pa
a R
FORCES
W = weight of soil wedge
needs support
N = force normal to sliding plane from underlying supporting soil
T = tangential force along sliding plane = Ntan
R = resultant reaction from supporting soil forces, T & N
Pa = maxm reaction from wall required for
equilibrium for “critical” wedge angle, a
VECTOR DIAGRAMSCase 1: no wall friction
W
Pa
R
90 - a +
a -
a
N
R
90Sliding
plane
SOLUTIONBut
a
2
tanθ2
HW
0.322
0.324
0.326
0.328
0.33
0.332
0.334
50 55 60 65 70
trial angle, a (degrees)
2Pa/( H
2)
)tan(θtanθ2HP aa
2
a
)Wtan(θP aa
Note: for this simple
problem, Rankine and
Coulomb expressions
coincide – failure plane is at
(45 + /2) = 60
i.e. where the maximum
value of Pa is obtained
AND Ka = 0.333 = 2Pa/(H2)
for this simple problem
Case 2: Uniform soil, c = 0,
vertical backed retaining wall,
wall friction (tan)
R
W
N
T
a
Pa
VECTOR DIAGRAMS
Case 2: with wall friction
RW
Pa
90 - a +
a -
SOLUTION
(as before)
From the sine rule:
)θsin(90
W
)sin(θ
P
aa
a
))tan(θθsin(90
)sin(θ
2
HP
aa
a2
a
a
2
tanθ2
HW
0.286
0.288
0.29
0.292
0.294
0.296
0.298
50 55 60 65
trial angle, a (degrees)
2Pa/( H
2)
4 flatter failure plane
‘Ka’ 11% less
Influence of = 2/3 = 20- simple case study 2
Consider - - -
natural soil
natural soil
slidi
ng s
urfa
ce?backfill
tension crack
FORCES
natural soil
R
Pa
W
cwh
cl
R
Pa
W
cwh
cl
FORCE VECTOR DIAGRAM
Consider water in crack
natural soil
natural soil
slidi
ng s
urfa
ce?backfill
tension crack
FORCES
natural soil
R
Pa
W
cwh
clU = 0.5wz2
z
Pa
W
cwh
cl
FORCE VECTOR DIAGRAM
U
R
SEEPAGE FORCES
R
Pa
W
cwh
clU
So what’s Coulomb method all about? Must appreciate all the forces acting on
the potentially unstable wedge of backfill soil, including shear against the wall
The forces includes a base reaction, the magnitude of which is unknown
Do know the direction which is useful in finding the active thrust – force vectors
Numerous trial wedges are needed to find the maximum thrust analagous to finding the critical
sliding surface in slope stability
OVERALL SUMMARY
Earth pressures are needed for design
of retaining walls & excavations
3 major states: at rest, Active &
Passive
Earth pressure coefficients are based
on effective stresses
Water pressures are important
Cohesion leads to potential cracked
zone for Active earth pressure state
Coulomb a superior approach as wall
interface forces considered
Masonry Gravity walls
Crib gravitywall(concrete,timber)
Gabiongravitywall
DESIGN OF RETAINING WALLS
Generally rely on mass - needs to be substantial
Concrete Cantilever Walls
Flexural strength BUT also mass of soil above footing, which
moves with the wall
Active thrust
Dead Weights
Design Requirements
a) The wall is structurally sound
flexure, compression & tension
b) The “foundation” can carry the loads
will not fail - “bearing capacity”
c) The wall is safe against overturning
and translation
toppling or sliding
d) Settlement and tilting are minimized
remains “serviceable”
Design Steps
Determine:
1. earth pressures
2. resultant thrust behind wall
3. soil reactions at base of wall (footing)
4. location of resultant soil reaction on
base
take moments of all forces about
toe of wall
Consider the loadssubscript w = “wall”
N
T
“toe”
x
Tw
Pw
H
B
W
xw
y
and the Reactions
EXAMPLE
• Evaluate earth pressures
• Pw and Tw, horizontal and vertical
components of Pa
• Soil reaction: N = (W + Tw)
• Moments of all forces about toe
of wall to find x
(Nx + Pwy) = (TwB + Wxw)
i.e. overturning = restoring moments
“Moment equilibrium”
Design Steps, cont’d.
Checks
1. Base reaction is in MIDDLE third?
If not, then widen footing to avoid “lift
off”
2. Resistance to potential overturning?
1.5moment goverturnin
moment restoring
3. Bearing capacity of soil
foundation?
4. Base sliding resistance?
EXAMPLE cont’d5. 0.67B x 0.33B?
6. Overturning?
7. Bearing capacity = function of
soil shear strength parameters
(later in the course)
8. Base sliding?
1.5yPNx
BTWx
w
ww
NtanBc T where1.5,P
Ta
w
Adhesion & wall friction
Adhesion, ca, between soil foundation and
wall footing
─ natural soil and concrete?
─ sliding resistance of interface for zero
normal force
Wall or interface friction, tan
Typically assume:
─ ca = 0.67c'
─ tan = 0.67tan'
but can vary with roughness of footing
material
Soil base reactionsN Centreline
x
B/2
Average pressure, N/(LB)
+ Pressure due to moment, Ne
m 1LLB
6Ne2
e =(B/2 –x)
Point of lift off at zero contact pressure
6
Be
i.e._when
0B
6e1
B
N
or
0B
6Ne
B
N2
i.e. “the middle third rule”
Design Steps, cont’d.
Finally, check:
9. Global stability
Slope stability
10. Shear & moment capacity of
wall & base
Designers of Retaining Walls:
AS 4678 – 2002
Some Features:
1) 3 levels of classification of structure
2) characteristic strengths factored for
ultimate or serviceability limit state
3) reduction factors depend on how well
controlled the backfill compaction is
4) Load factors 1.25DL,1.25'h, 1.5LL
5) Minm. LL of 5 kPa on backfill surface
6) good information on reinforced soil
systems
SMITH’S EXERCISES
7.2 Gravity retaining wall: Check FoS for sliding and overturning Answers 3.53 and 1.85
2.5 m
1.8 m
4.0 m
1.0 m
c = 0, = 35,
= 18 kN/m3
wall = 23.5 kN/m3
Passive pressure at front of wall?