lecture 12- earth pressure and sturucture rankine theory [1]
TRANSCRIPT
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Earth pressure and structure (Rankine theory)
Dr. Md Mizanur Rahman
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
Acknowledge: Dr. D. A. CameronPrevious course coordinator
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Earth pressures on retaining structures
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
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Earth pressures on retaining structures
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
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Earth pressures on retaining structures
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
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Earth pressures on retaining structures
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
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Two Methods
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
- to estimate the earth pressures on structures
1. RankinePlausible stress states
2. CoulombPlausible failure mechanisms
Relative merits of approaches?
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Rankine earth pressures
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
A lower bound estimate
Effective horizontal stress,
H = Kz
where, K = earth pressure coefficient
z = effective vertical stress
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Earth pressure states (retaining walls)
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
“At rest” wall is not moving, so the soilan intermediate state
PassiveActive
Both are failure states
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Earth pressure at rest
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
“AT REST” PRESSURE
The intermediate state
K = Ko = fn(soil type, density, OCR)
The soil is unable to move laterally - can’t expand, OR contract
e.g soil confined in a large body of soil
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Active state (stress relaxation)
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
Normal stress
Shear stress
3f 3o 1
At rest stateActive state
failure envelope
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Passive state (stress intensification)
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
Normal stress
Shear stress
1f3o 1
Passive stateAt rest state
failure envelope
3f
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All three states
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
Normal stress
Shear stress
1o
Active state
Passive state
At Rest failure envelope
Note:Active state: stress relaxation
Passive state: stress intensification
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The 3 States (consider a vertical retaining wall)
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
H/z
Wall movement
Kp
Ka
NB: Passive needs LARGE strains
KO
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Equations for Rankine States
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
(Can be derived from Geometry of Mohr’s circles)
For ACTIVE STATE
Case 1A: c = 0
H = Kaz
and Ka =
[Ka max 0.333 for loose sand]
)sin(1
)sin(1
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Active state
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
Normal stress
Shea
r str
ess
NB: Active state = a failure state
Failure, f , nf (
1 - 3 )/2
(1 + 3)/2
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Active state
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
31
31sin
)sin1()sin1( 31
)sin1(
)sin1(
1
3
aK
From the geometry,
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Active state (with cohesion)
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
Case 2A: c 0
H = Kaz - 2cKa
Notes:
• the 2nd term is a constant!
• z = (z) + z
i.e. stress due to self weight + extra due to surface load
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Can soil undergo tension?
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
If z = 0, then H 0
Now if z = z,
At what depth will H = 0?
H = Kaz - 2cKa
This depth is called the depth of
cracking, zc, & defines the potential
tension zone
z
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Depth of Cracking
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
By definition:
At zc, H = 0
Therefore,
H = 0 = Ka zc - 2cKa
Therefore,
zc = [2cKa][ Ka ]
Or a
cKγ'
c2z
z
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The tension zone
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
The pulling power of cohesive soil is ignored in calculations of pressures behind retaining walls over the depth zc because:
- tension is unsustainable
i.e. short term only!
However, no compressive pressures exist in this zone = a dead zone
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Evidence of a tension zone
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
How can unsupported, vertical-sided trenches be cut to metres depth in clay soils?
What depth is possible?
What happens if it rains?
Warning: people laying pipes have died in collapsed trenches!
OH&S???
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Summary of active state
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
• Stresses relaxedcommon retaining wall situation
• Ka = (1 – sin)/(1+ sin )clean sand, Ka 0.33 usually
• Theoretical tension or crack zone from cohesive strength (c)
may be applied to slope stability
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Passive state
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
Again, from Geometry of Mohr’s circles
Case 1P: c = 0
H = Kpz
and Kp =
[Kp min 3 for loose sand]
aK
1
)sin(1
)sin(1
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Passive state (with cohesion)
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
Case 2P: c 0
H = Kpz + 2cKp
Note:
1. the 2nd term provides greater constant passive pressure component
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Orientation of Failure Planes
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
From Mohr’s circlesActive state:
(45 + /2) to horizontalPassive state:
(45 - /2) to horizontal
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Orientation of Failure Planes
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
Sliding surfaces?
ACTIVE
PASSIVE
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School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
Tension is ignored!
ACTIVE-2cKa
zc
+2cKp PASSIVE
Typical Lateral stresses, c 0
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The Influence of Pore Water
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
Steady state (and seepage) pressures add to lateral stresses on walls
Should Ka be applied to the pore water pressure?
NO WAY!
Hydrostatic means K = 1
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The Influence of Pore Water
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
No Water
c = 0
H = H = Ka z
Water
c = 0
H = Ka z u = wz
+
TOTAL LATERAL STRESS
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The Influence of Pore Water
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
In the previous example
uniform soil, no surface load and with or without a Water Table at
ground level,
Almost twice the total lateral pressure is
experienced with the high Water Table
Effective lateral stresses are halved, BUT full pwp
is exerted!
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Importance of Drainage for Retaining Walls (Drains, Filters & Weep holes)
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
Weep holes
Granular zone or geofabric drain
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Examples
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
From Whitlow - modified
Find the total resultant thrust and its point of action behind vertical-backed retaining walls of height, 12 m, resulting from earth and water pressures given the following situations
1. Surface horizontal; no surcharge; single soil layer, c = 0, = 30, = 18 kN/m3
2. Surface horizontal; uniform surcharge of 10 kPa; single soil layer: c = 0, = 30, = 18 kN/m3
3. Surface horizontal; no surcharge; two soil layer:
0-5 m depth, c = 0, = 30, = 18 kN/m3
> 5 m depth, c = 0, = 36, = 20 kN/m3
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Examples
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
Thrust = lateral pressure x area
= average pressure x height over which it acts, per m length of wall
9 m
40 kPa
360 kN
20 kPa
360 kN
60 kPa3 m
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Examples
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
Resultant Thrust = Resultant, Pa = (all thrusts)
Point of action found by summing moments about a point and dividing by Pa
10 kPa12
m
360 kN
60 kPa4 m
120 kN
X mLocation of
resultant force
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Examples
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
AnswerResultant, Pa = (120 + 360)
Pa = 480 kN per m length of wall
Point of action found by summing moments about the base and dividing by Pa
120 x 6 + 360 x 4 = Pa x X
X = (720 +1440)/480 = 4.5 m
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Examples
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
Q1. c = 0, = 30, = 18 kN/m3
Ka = 0.333
72 kPa
12 m
432 kN
8 m
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Examples
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
Q2. As for 1 but 10 kPa surcharge
Ka = 0.3333.33 + 72 kPa
432 kN
8 m
40 kN
10 kPa
ANSWER 472 kN/m, 4.17 m
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Examples
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
Q3. Two granular soils
Ka1 = 0.333
12 m
5 m
Ka2 = 0.26
At z = 5 m, z = 90 kPaAt z = 12 m, z = 140 kPa
ANSWER
366 kN/m, 4.15 m
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Example
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
From Whitlow, cont’d6. Surface horizontal; no surcharge; single soil layer, cu = 45 kPa, u = 0,
= 18 kN/m3
7. Surface horizontal; no surcharge; single soil layer, c = 15 kPa, = 20, = 18 kN/m3
11. Surface horizontal; no surcharge; two soil layer,
0-4 m depth, c = 0, = 30, = 19.6 kN/m3
> 4 m depth, c = 25 kPa, = 15, = 18.2 kN/m3
ANSWERSQ6 441 kN/m, 2.33 mQ7 408 kN/m, 3.21 mQ11 458 kN/m, 3.61 m
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Example
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
Q7. c = 15 kPa, = 20, = 18 kN/m3
(0.49x216 - 300.49 kPa
OR 84.8 kPa
12 m
408 kN
zc
Ka = 0.49
zc = 2.38 m 21 = 0.49x18xzcANSWER: 408 kN/m, 3.21 m
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Limitations of Rankine
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
1. Vertical backs of walls only
2. Backfill surface must be regular
– a solution exists for a sloping backfill, provided slope angle, <
– BUT pressures act parallel to the slope - theoretically wrong!
3. Backfill loads / surcharge effects approximated
4. Wall friction ignored!
– friction is beneficial
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Summary
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
1) Earth pressures are needed for design of retaining walls & excavations
2) Three major states: at rest, active and passive
─ Last 2 are failure states
3) Earth pressure coefficients are based on effective stresses
4) Water pressures are important
− total lateral stresses
5) Cohesion leads to potential cracked zone for Active state
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Excavation Bracing
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
Trench
Support systems:soldier beams (vertical)& shuttering between them or steel sheeting
Strut
Possiblefailure shape
Steelsheeting
Wale
PLAN
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Example
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
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Design of Bracing
School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)
Earth pressures are not simple
- propping forces from struts
- progressive construction
Empirical design earth pressures
- struts designed for thrust
Refer to Notes for guidance
Information Only