DESIGN OF RETAINING WALLS

DESIGN OF RETAINING WALLS

Retaining Walls - Applications
highway

High-rise building basement wall

Metros and Subways Road Train

Tunnel E E Dock Abutment E

RETAINING WALLS TYPES GRAVITY WALLS

RETAINING WALLS TYPES CANTILEVER

RETAINING WALLS TYPES COUNTERFORT

TYPES RETAINING WALLS COUNTERFORT SIIT-Thammasat University
School of Civil Engineering-AIT

TYPES RETAINING WALLS BUTTRESS SIIT-Thammasat University

PARTS CANTILEVER RETAINING WALLS STEM or Wall Slab BACKFILL FRONT HEEL
SIIT-Thammasat University CANTILEVER RETAINING WALLS PARTS STEM or Wall Slab BACKFILL FRONT TOE HEEL KEY School of Civil Engineering-AIT

Lateral Earth Pressure
(R.P. Weber) ?? (R.P. Weber)

Water Pressure and Soil Pressure
Consider hydrostatic condition Consider “at-rest” (geostatic) condition Anisotropic sz Isotropic sx sz ≠ sx > sx

SIIT-Thammasat University
gy School of Civil Engineering-AIT

PASSIVE EARTH PRESSURE
SIIT-Thammasat University EARTH PRESSURES PRESSURE AT REST ACTIVE EARTH PRESSURE PASSIVE EARTH PRESSURE School of Civil Engineering-AIT

PRESSURE AT REST RIGID SIIT-Thammasat University

At Rest Earth Pressure One common earth pressure coefficient for the “at rest” condition in granular soil is: Ko = 1 – sin(φ) Where: Ko is the “at rest” earth pressure coefficient and φ is the soil friction value. h z z K0z h/3 K0h

EARTH PRESSURES SIIT-Thammasat University

movement Active Failure

ACTIVE EARTH PRESSURE RANKINE ACTIVE EARTH PRESSURE
3 = 1 . tan2 (45-/2)-2c.tan (45-/2)

3 = 1 . tan2 (45-/2)-2c.tan (45-/2)
RANKINE ACTIVE EARTH PRESSURE 3 = 1 . tan2 (45-/2)-2c.tan (45-/2) a = v . tan2 (45-/2)-2c.tan (45-/2) a = v . Ka – 2cKa Ka = tan2 (45 - /2)

Active Stress Distribution (c ≠ 0)
zo γ c ≠ 0 Φ dry soil H _ - = Ka γH 2 c (Ka)1/2 Ka γH – 2 c (Ka)1/2 Find zo: Ka γzo – 2 c (Ka)1/2 = 0 Zo = 2c / γ (Ka)1/2 Pa = ?

zc = depth of tensile crack
ACTIVE EARTH PRESSURE Note : z = 0  v = 0 ; a = -2cKa z = H  v = H The tensile stress decreases with depth and becomes zero at a depth z = zc or zcKa – 2cKa = 0 and zc = depth of tensile crack

(for granular soil, c = 0) For c- soil ACTIVE EARTH PRESSURE
RANKINE ACTIVE EARTH PRESSURE FOR INCLINED BACKFILL (for granular soil, c = 0) For c- soil

SIIT-Thammasat University PASSIVE EARTH PRESSURE School of Civil Engineering-AIT

movement Passive Failure

RANKINE PASSIVE EARTH PRESSURE

RANKINE PASSIVE EARTH PRESSURE p= v . tan2(45+/2) + 2c . tan (45+/2)

RANKINE PASSIVE EARTH PRESSURE Kp = tan2 (45 + /2) h = v . Kp + 2cKp

Passive Stress Distribution (c ≠ 0)
γ c ≠ 0 Φ dry soil H + - = Kp γH 2 c (Kp)1/2 Kp γH + 2 c (Kp)1/2 h = v . Kp + 2cKp Pp = ? Kp = tan2 (45 + /2)

Ka < K0< Kp

o -△ +△ +△ -△ E Ep Eo Ea △a △p Relation among three earth pressures
Lateral Earth Pressure -△ +△ +△ -△ E o Ep Eo Ea △a △p Relation among three earth pressures

Example 1

Example 2 A h1 =2m B h=5m h2 =3m C 1=17kN/m3 c1=0 1=34o 2=19kN/m3
Lateral Earth Pressure Example 2 h=5m 1=17kN/m3 c1=0 1=34o 2=19kN/m3 c2=10kPa 2=16o h1 =2m h2 =3m A B C

Solution: A B C h1=2m 10.4kPa h=5m 4.2kPa h2=3m 36.6kPa
Lateral Earth Pressure Solution: A B C h=5m h1=2m h2=3m 10.4kPa 4.2kPa 36.6kPa

Active Stress Distribution (c = 0)
γ c = 0 Φ dry soil H Pa = ? ? - What is this value σa‘ = Ka σv’ – 2 c (Ka)1/2 σa‘ = Ka σv’ σa‘ is the stress distribution Pa is the force on the wall (per foot of wall) How is Pa found?

Passive Stress Distribution (c = 0)
γ c = 0 Φ dry soil H Pp = ? ? - What is this value σp‘ = Kp σv’ – 2 c (Kp)1/2 σp‘ = Kp σv’ σp‘ is the stress distribution Pp is the force on the wall (per foot of wall) How is Pp found?

Stress Distribution - Water Table (c = 0)
H1 Effective Stress Pore Water Pressure Ka γ H1 H2 Pa Ka γ H1 Ka γ’ H2 γw H2 or Ka (γ H1 + γ’ H 2) Pa = Σ areas = ½ Ka γH12 + Ka γH1H2 + ½ Ka γ’H22 + 1/2γwH22

Stress Distribution With Water Table
Why is the water pressure considered separately? (K) H1 Effective Stress Pore Water Pressure Ka γ H1 H2 Pa Ka γ H1 Ka γ’ H2 γw H2 or Ka (γ H1 + γ’ H 2)

Fill material is granular soil
Assumptions: Fill material is granular soil Friction of wall and fill material is considered Soil failure shape is plane (BC1, BC2 …) ACTIVE EARTH PRESSURE COULOMB ACTIVE EARTH PRESSURE Pa = ½ Ka .  . H2

ACTIVE EARTH PRESSURE COULOMB’S ACTIVE EARTH PRESSURE WITH A SURCHARGE ON THE BACKFILL

COULOMB PASSIVE EARTH PRESSURE Pp = ½ Kp .  . H2

STABILITY OVERTURNING SLIDING BEARING SIIT-Thammasat University

OVERTURNING Highway Loading (Surcharge) SIIT-Thammasat University

Active Pressure Soil+Surcharge
SIIT-Thammasat University OVERTURNING Overturning Forces Full Surcharge Here No Surcharge Here Active Pressure Soil+Surcharge School of Civil Engineering-AIT

OVERTURNING Restoring Forces Weight of Wall No Passive Pressure
SIIT-Thammasat University OVERTURNING Restoring Forces Weight of Wall No Passive Pressure Weight of Soil (with care) Weight of Soil School of Civil Engineering-AIT

OVERTURNING Restoring Moment FOS vs OT = Overturning Moment
SIIT-Thammasat University OVERTURNING Restoring Moment Overturning Moment FOS vs OT = A FOS = 2 is considered sufficient School of Civil Engineering-AIT

SIIT-Thammasat University SLIDING Sliding Forces Full Surcharge Here No Surcharge Here Active Pressure Soil+Surcharge H1 School of Civil Engineering-AIT

SLIDING a=Coeff of Friction Resisting Forces No Surcharge Here
SIIT-Thammasat University SLIDING Resisting Forces No Surcharge Here Resisting Forces H2 + a S V a=Coeff of Friction Vs1 Vc1 Vs2 H2 Vc2 Vc3 School of Civil Engineering-AIT

A FOS = 1.5 is considered sufficient
SIIT-Thammasat University SLIDING without KEY Passive Earth Pressure Force+a S V Active Earth Pressure Force FOS vs Sliding = A FOS = is considered sufficient School of Civil Engineering-AIT

SIIT-Thammasat University SLIDING with KEY Sliding Forces No Surcharge Here Active Pressure Soil+Surcharge School of Civil Engineering-AIT

SLIDING with KEY Resisting Forces No Surcharge Here Vc1 Vs2 Vs1 H Vc2
SIIT-Thammasat University SLIDING with KEY Resisting Forces No Surcharge Here Vc1 Vs2 Vs1 H Vc2 Vc3 School of Civil Engineering-AIT

SIIT-Thammasat University SLIDING with KEY Find Vertical forces acting in front and back of key No Surcharge Here Active Pressure Soil+Surcharge RESULTANT Vc1 Vs2 Vs1 Vc2 Vc3 School of Civil Engineering-AIT

SLIDING with KEY Determine Pressure Distribution Under Base A=B e
SIIT-Thammasat University SLIDING with KEY Determine Pressure Distribution Under Base A=B S=B2/6 e x V B B/2 School of Civil Engineering-AIT

SLIDING with KEY Determine Force in Front of KEY y2 y3 P2 P1 y1
SIIT-Thammasat University SLIDING with KEY Determine Force in Front of KEY B x1 P1 P2 y1 y2 y3 y3=y2+(y1-y2) (B-x1)/B P1=(y1+y3) x1/2 P2=V-P1 School of Civil Engineering-AIT

SIIT-Thammasat University
SLIDING with KEY When Pressure Distribution Under Base is Partially Negative e V B B/2 School of Civil Engineering-AIT

SLIDING with KEY e V B x 3x Determine P1 and P2 once again 2V P2 3x P1
SIIT-Thammasat University SLIDING with KEY e V B x 3x Determine P1 and P2 once again 2V 3x P2 P1 School of Civil Engineering-AIT

Passive Earth Pressure Force
SIIT-Thammasat University SLIDING with KEY Active Earth Pressure Force Total Sliding Force = H1 Total Resisting Force = P1 tan f + a P2 + H2 Force on and Back of Key Friction b/w Soil, Concrete Force in Front of Key Internal Friction of Soil Passive Earth Pressure Force School of Civil Engineering-AIT

BEARING There are two possible critical conditions
SIIT-Thammasat University BEARING There are two possible critical conditions 1. No surcharge on heel 2. Surcharge on heel School of Civil Engineering-AIT

SIIT-Thammasat University BEARING This case has been dealt already No Surcharge on Heel Active Pressure Soil+Surcharge RESULTANT Vc1 Vs2 Vs1 Vc2 Vc3 School of Civil Engineering-AIT

SIIT-Thammasat University BEARING DETERMINE THE PRESSURE DISTRIBUTION UNDER BASE SLAB Surcharge on Heel Vs Active Pressure Soil+Surcharge RESULTANT Vc1 Vs2 Vs1 Vc2 Vc3 School of Civil Engineering-AIT

Determine Pressure Distribution Under Base
SIIT-Thammasat University Determine Pressure Distribution Under Base A=B S=B2/6 e x V B B/2 School of Civil Engineering-AIT

Allowable Bearing FOS vs Bearing = Max Bearing Pressure
SIIT-Thammasat University Compare Pressure with Bearing Capacity B FOS vs Bearing = Allowable Bearing Max Bearing Pressure School of Civil Engineering-AIT

2V/3x 3x Allowable Bearing FOS vs Bearing = Max Bearing Pressure 2V/3x
SIIT-Thammasat University ALTERNATELY B 2V/3x 3x FOS vs Bearing = Allowable Bearing Max Bearing Pressure 2V/3x School of Civil Engineering-AIT

END OF PART I

BENDING OF WALL SIIT-Thammasat University

Critical Section Moment
SIIT-Thammasat University DESIGN OF STEM CRITICAL SECTIONS Critical Section Shear d Critical Section Moment Active Pressure Soil+Surcharge School of Civil Engineering-AIT

h H1=Ca s h y1 H2=0.5 Ca gs h2 y2 DESIGN OF STEM Design Moment
SIIT-Thammasat University DESIGN OF STEM Design Moment =1.6 (H1 y1 + H2 y2) Surcharge = s N/m2 h H1=Ca s h y1 H2=0.5 Ca gs h2 y2 School of Civil Engineering-AIT

h H'1=Ca s (h-d) H'2=0.5 Ca gs (h-d)2 d DESIGN OF STEM
SIIT-Thammasat University DESIGN OF STEM Design Shear=1.7(H '1+H '2) Surcharge = s N/m2 h H'1=Ca s (h-d) H'2=0.5 Ca gs (h-d)2 d School of Civil Engineering-AIT

Critical Section Moment
SIIT-Thammasat University DESIGN OF TOE SLAB CRITICAL SECTIONS d Critical Section (Shear) Critical Section Moment School of Civil Engineering-AIT

DESIGN OF TOE SLAB Design Loads 1.6Soil Pressure 0.9 Self Wt
SIIT-Thammasat University DESIGN OF TOE SLAB Design Loads 1.6Soil Pressure 0.9 Self Wt 0.9 Soil in Front (may be neglected) School of Civil Engineering-AIT

TOE : DESIGN MOMENT 1.6(0.5 T y3) T/3 +1.6(0.5 T y1) 2T/3 -0.9 wc T2/2
SIIT-Thammasat University TOE : DESIGN MOMENT 1.6(0.5 T y3) T/3 +1.6(0.5 T y1) 2T/3 -0.9 wc T2/2 -0.9 ws T2/2 y3 y1 T School of Civil Engineering-AIT

TOE : DESIGN SHEAR 1.6(0.5 Ts) y3 Ts/T +1.6(0.5 T y1-0.5 d [y1/T] d)
SIIT-Thammasat University TOE : DESIGN SHEAR 1.6(0.5 Ts) y3 Ts/T +1.6(0.5 T y1-0.5 d [y1/T] d) -0.9 wc Ts -0.9 ws Ts y3 y1 Ts=T-d School of Civil Engineering-AIT

Critical Section Moment & Shear
SIIT-Thammasat University DESIGN OF HEEL SLAB CRITICAL SECTIONS Critical Section Moment & Shear TENSION FACES School of Civil Engineering-AIT

Soil Pressure Neglected
SIIT-Thammasat University DESIGN OF HEEL SLAB DESIGN LOADS 1.6s gs +1.2 gc Soil Pressure Neglected School of Civil Engineering-AIT

BENDING OF WALL SIIT-Thammasat University

MAIN REINFORCEMENT Minimum 75 mm Clear Cover SIIT-Thammasat University

ACI CODE SECONDARY STEELS
SIIT-Thammasat University ACI CODE SECONDARY STEELS ACI ACI ACI Minimum SLAB School of Civil Engineering-AIT

END OF PART II

Weepers Or Weep Holes Sand + Stone Filter DRAINAGE
SIIT-Thammasat University DRAINAGE Weepers Or Weep Holes Sand + Stone Filter School of Civil Engineering-AIT

Drainage Pipes f 100-200 mm @ 2.5 to 4 m
SIIT-Thammasat University DRAINAGE Drainage Pipes f to 4 m School of Civil Engineering-AIT

Perforated Pipe Suited for short walls DRAINAGE (Alternate)
SIIT-Thammasat University DRAINAGE (Alternate) Perforated Pipe Suited for short walls School of Civil Engineering-AIT

End of Part III END OF PART III

Active and Passive Limit Conditions
Ka = Coefficient of Active Earth Pressure (Wall Moving Away from Backfill) (small sx) Active Failure Condition movement Active Failure Wedge (45+f/2) Kp = Coefficient of Passive Earth Pressure (Wall Moving Toward Backfill) (large sx) Passive Failure Condition movement Passive Failure Wedge (45 -f/2)

Rankine Active Failure Surface
Pole Point

Rankine Passive Failure Surface
Pole Point

Consider Mohr’s Circles…
sx decreases until failure sx increases until failure movement Passive Failure

Evolution of lateral stress with wall movement…
Stationary (at rest) Movement away from backfill Movement toward backfill Passive Failure at Kp Active Failure at Ka Ka < K0< Kp

σv’ H σh’ We can calculate σv’ Now, calculate σh’ which is the horizontal stress σh‘/ σv‘ = K Therefore, σh‘ = Kσv‘ (σV‘ is what?)

There are 3 states of lateral earth pressure Ko = At Rest Ka = Active Earth Pressure (wall moves away from soil) Kp = Passive Earth Pressure (wall moves into soil) Passive is more like a resistance σv z H σh

At Rest Earth Pressure σv H σh
At rest earth pressure occur when there is no wall rotation such as in a braced wall (basement wall for example) Ko can be calculated as follows: Ko = 1 – sin φ for coarse grained soils Ko = [PI / 100] for NC soils Ko (oc) = Ko (NC) (OCR)1/2 for OC soils σv z H σh

Sand, normally consolidated clay  M = 1
AT REST EARTH PRESSURE q Jaky, Broker and Ireland  Ko = M – sin ’ Sand, normally consolidated clay  M = 1 Clay with OCR > 2  M = 0.95 z v =  . z + q Broker and Ireland Ko = PI , 0  PI  40 Ko = PI , 40  PI  80 v h Sherif and Ishibashi  Ko =  +  (OCR – 1)  = (LL – 20) = (LL – 20) LL > 110%   = 1.0 ;  = 0.19 At rest, K = Ko

DESIGN OF RETAINING WALLS

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