17 retaining wall design

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15 Retaining Walls Design

15.1 Introduction

Conventional retaining walls can generally be classified into four varieties:

1 Gravity retaining walls

2 Semigravity retaining walls

3 Cantilever retaining walls

4 Counterfort retaining walls

Braja M Das (2007) Principle of Foundation Engineering, sixth edition

Jack C McCormac, James K Nelson (2006) Design of Reinforced Concrete ACI318-05 Code Edition, seventh edition

In designing retaining walls, an engineer must assume some of their dimensions Called

proportioning, such assumptions allow the engineer to check trial sections of the walls for

Braja M Das (2007) Principle of Foundation Engineering, sixth edition

1.9-3 Stability of Retaining Walls

A retaining wall may fail in any of the following ways:

- It may oveturn about its toe

- It may slide along its base

- It may fail due to the loss of bearing capacity of the soil supporting the base

- It may undergo deep-seated shear failure

- It may go through excessive settlement

Failure of retaining wall: (a) by overturning; (b) by sliding; (c) by bearing capacity failure;

(d) by deep-seated shear failure

Braja M Das (2007) Principle of Foundation Engineering, sixth edition

1 Checiking for Overturning

The Rankine active pressure

=

Braja M Das (2007) Principle of Foundation Engineering, sixth edition

Procedure for Calculating ΣM R

length of wall

Moment arm measured from C

Momentabout C

Note: γ1 = unit weight of backfill

γc = unit weight of concrete = 24kN

m3

The factor of safety against overturning can be calculated as

The usaul minimum desirable value or the factor of safety with respect to overturning is 2 to 3

2 Checking for Sliding along the Base

The factor of safety against sliding can be calculated as

where ΣFR = sum of the horizontal resisting forces

ΣFd = sum of the horizontal driving forces

δ' = angle of friction between the soil and the base slab

c'a = adhesion between the soil and the base slab

A minimum factor of safety of 1.5 against sliding is generally required

Check for sliding along thebase

Braja M Das (2007) Principle of Foundation Engineering, sixth edition

In many cased, the passive force Pp is ignored in calculating the factor of safety with

respect to sliding In general, we can write δ' = k1 ϕ2⋅ and c'a k2 c2= ⋅ In most cases, k1a

nd k2 are in range from 1

If the desired value of FSsliding is not achieved, several alternatives may be investigated:

1 Increase the width of the base slab (i.e., the heel of the footing)

2 Use a key to the base slab If a key is include, the passive force per unit length

of the wall becomes

Pp = 12⋅γ2⋅D12⋅Kp+2c2 D1⋅ ⋅ Kp

Alternatives for increasing the factor of safety with respect to sliding

Braja M Das (2007) Principle of Foundation Engineering, sixth edition

3 Checking for Bearing Capacity Failure

The net moment of these forces about point C

Mnet ΣMR ΣM0= −

The distance from point C to E

X MnetΣV

Check for bearing capacity failure Braja M Das (2007) Principle of Foundation Engineering, sixth edition

q ΣV

B 1m⋅( )±

ΣV( ) e⋅ B

2

⋅1

12 B3

qmin qheel= ΣV

B 1

6 e⋅B

=

where

q = γ2 D⋅ B' = B 2 e− ⋅ Fcd 1 0.4= + ⋅DB'Fqd 1 2 tanϕ2⋅ ⋅(1 sin− ϕ2)2 D

B'

⋅+

Differnt type of loads are often imposed on the surface of the backfill behind a

retaining wall If the load is uniform, an equivalent height of soil, hs may be assumed acting on the wall to account for the increased pressure For the wall shown in Figure below, the horizontal pressure due to the surcharge is constant throughout the depth of the retaining wall

H/3

H

D D/3

ws

H/3

H

D D/3

ws

Ph2Ph1Pp

surcharge effect under a uniform load

hs = wsw

where hs = equivalent height of soil

ws = pressure of the surcharge

w = unit weight of soil

The total pressure is

D/3

45 o

Ph2Pp

surcharge effect under a partial uniform load at a distance from the wall

It is common practice to assume that the effective height of pressure due to partial

surcharge is h' measured from point B to the base of retaining wall The line AB form an angle of

45deg with horizontal

15.5 Design Requirements

The ACI Code provides methods for bearing wall design The main requirements are as folllow

1 The minimum thickness of bearing wall is 1

25 the supported height or lenght,

whichever is shorter, but not less than 100mm

tmin = 251 ⋅hwall where hwall = the height of wall

2 The minimum area of the horzontal reinforcement in the wall is 0.0025 b⋅ h⋅ , but thisvalue may be reduced to 0.0020b h⋅ if diameter 16mm or smaller deformed bars with

fy 400MPa≥ are used

Ah.min 0.0025 b⋅ h⋅ if diameter > 16mm

0.0020 b⋅ h⋅ if diameter ≤ 16mm

3 The minimum area of the vertical reinforcement is 0.0015 b⋅ h⋅ , but it may be reduced

to 0.0012 b⋅ h⋅ if diameter 16mm or smaller deformed bars with fy 400MPa≥ are used

smax min 3t 450mm= ( , ) where t = the wall thickness

5 If the wall thickness exceed 250mm, the vertical and horzontal reinforcement should beplaced in two layers parallel to the exterior and interior wall surface, as follows:

a/ For exterior wall surface, at leat 0.5 of the reinforcement As ( but not more than

2

3As ) should have a minimum concrete cover of 50mm but not more than 1/3 ofthe wall thickness

As.exterior = 23As and covermin 50mm= ≤ 13hwall

b/ For interior wall surface, the balance of the reinforcement in each direction shouldhave a minimum concrete cover of 25mm but not more than 1/3 of the wall

thickness

covermin 25mm= ≤ 13hwall

c/ The minimum steel area in the wall footing (heel or toe) according to ACI Code

is that required for shrinkage and temperature reinforcement, which is 0.0018 b⋅ h⋅

when fy 400MPa= and 0.002 b⋅ h⋅ when fy 350MPa= or fy 275MPa= Becau

se this minimum steel area is relately small, it is a common practice to increase it

to that minimumn As.min required for flexure:

Example15.1

b

ws

γ1ϕ1

H1c1

m3

:= ϕ2 20deg:= c2 20kN

m2:=

allowable soil bearing capacity qa 190kN

m2:=

Material:

concrete compressive strength f'c := 25MPa

m3:=

m2

:=

Solution

1 Checking of wall against overturning

a/ the moment resisting

A

B3 H1⋅

b H1⋅B2 b−

( )⋅H12

A2

Ph2 H2

3

qimnqmax

⋅

=

B B32

ΣMR ∑M 637.866 kN m⋅

1m

⋅

=:=

Section No Area (m2) Unit weight (kN/m) Moment arm (m) Moment(kNm/m)

b/ the unfactor force acting on the wall

using Rankine equation

Ka 1 sin− ( )ϕ1

1 sin+ ( )ϕ1 = 0.271

γ1 = 0.667 m:=

Ph1 12⋅Ka⋅γ1⋅H2 87.801 kN

m

⋅

=:=

Ph2 Ka γ1⋅ ⋅hs⋅H 19.511 kN

m

⋅

=:=

c/ the overturning moment

1m

⋅

=:=

the safety of factor against overturning

FSoverturning ΣMR

Mo = 2.724:=

wall "is not overturning" if FSoverturning 2≥

"is overturning" otherwise:=

wall = "is not overturning"

2 Checking of wall against sliding

Rankine passive force per unit length

let k1 := 23 = 0.667 k2 := k1 0.667=

Kp tan 45deg ϕ2

2+

=:=

Pp 12⋅Kp⋅γ2⋅D2+2c2 Kp⋅ ⋅D 67.521 kN

m

⋅

=:=

Ph Ph1 Ph2+ 107.312 kN

m

⋅

=:=

the safety of factor against sliding

FSsliding ΣR tan k1 ϕ2⋅ ( ⋅ ) +B k2⋅ ⋅c2+Pp

:=

wall "is not sliding along the base" if FSoverturning 1.5≥

"is sliding along the base" otherwise:=

wall = "is not sliding along the base"

3 Checking of wall against bearing capacity failure

a/ the eccentricity of the resultant

x ΣMR Mo−

ΣR = 1.376 m:=

B

footing "is not upward" e B

6

≤if

"is upward" otherwise:=

footing = "is not upward"

b/ the maximum and minimum pressure

qmax ΣRB 1 6 e⋅

B+

wall "is not failure" if qmax qa≤

"is failure" otherwise:=

wall = "is not failure"

4 Design of stem

a/ main reinforcement

The lateral forces applied to the wall are calculate using a load factor of 1.6 Thecritical section for bending moment is at the bottom of the wall Calculate the appliedultimate forces:

Pu1 1.6 Ph1⋅ 140.481 kN

m

⋅

=:=

Pu2 1.6 Ph2⋅ 31.218 kN

m

⋅

=:=

Mu1 1.6 Pu1⎛ ⋅H13 +Pu2⋅H12

h := B2 0.6m= d1 h 30mm 20mm

2+

ρmin max

0.25MPa f'c

MPa

⋅fy

1.4MPafy,

As1 max ρ ρmin( , )⋅d1 2.808 mm

2mm

⋅

=:=

Because the moment decrease along the height of the wall, the reinforcement area may

be reduced according to the moment requirement It is practical to use one As1 orspacing s1 for the lower half and a second As2 or spacing s2 for the upper half of thewall To calculate the moment at midheight of wall from the top:

Mu2 1.6 Pu1

2

H13

h B2 b+

2 = 450 mm⋅

2+

⋅

=:=

b/ Temperature and shrinkage reinforcement

the minimum horizontal reinforcement at the base of wall is:

As.min 0.002 B2⋅ 12 cm

21m

⋅

=:=

for the bottom third H1

3 = 1.8 m⋅As.min_0.33 0.002⋅B2 23⋅ 8 cm

21m

⋅

=:=

for the upper two-thirds 2H1

3 = 3.6 m

because the front face of the wall is mostly exposed to temperture changes, use half

to two-thirds of the horizontal bars at the external face of wall and place the balance

at the internal face

As.tem 0.5 As.min⋅ 6 cm

21m

⋅

=:=

use diameter of shrinkage bar ds 14mm:=

sshrinkage Floor

π ds ⋅ 24As.tem , 10mm

c/ Design for shear

the critical section for shear is at a distance d := d1 560 mm= ⋅ from the bottom

of the stem, at this section the distance for the top equal

stem_wall "no need shear reinforcement" if ϕVc Hu≥

"need required shear reinforcement" otherwise:=

stem_wall = "no need shear reinforcement"

5/ Design of the heel

Vu 1.2 B3 H1⋅( ⋅ ⋅γ1+B3 H2⋅ ⋅γc) +1.6 B3 hs⋅( ⋅ ⋅γ1) 290.928 kN

1m

⋅

=:=

heel "no need shear reinforcement" if ϕVc Vu≥

"need required shear reinforcement" otherwise:=

heel = "no need shear reinforcement"

Mu Vu⋅H12 785.506 kN m⋅

1m

⋅

=:=

⋅

=:=

sheel Floor

π 25mm ⋅ ( )24

As.dis 0.0018 d⋅ 1.012 mm

2mm

⋅

=:=

sdis Floor

π ds ⋅ 24As.dis , 10mm

6 Design of the toe

The toe of base acts as a cantiveler beam subject to upward pressure The critical sectionfor the bending moment is at the front face of the stem The critical for shear is at a distance

d from the front face of stem

The toe is subject to an upward pressure from the soil and downward pressure due to selfweight of the toe slab

critical sectionfor shear in toe

toe "no need shear reinforcement" if ϕVc Vu≥

"need required shear reinforcement" otherwise:=

toe = "no need shear reinforcement"

Mu 1.6 q2 B1

22

⋅ (q1 q2− )⋅B1 2B1

3

⋅+

Mu 92.412= kN m1m⋅

R Mu

0.9 d⋅ 2

0.325 MPa⋅

=:=

⋅

=:=

stoe Floor

π 20mm ⋅ ( )24

upper stem wall steelbar

shrinkage steel bardistibution steel bar

stem base wall steel bar

heel steel bar

toe steel bar

steel bar sketch for retaining wall