17-Retaining Wall Design

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.

Page-254-

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

15.2 Proportioning Retaining Walls

Page-255-

In designing retaining walls, an engineer must assume some of their dimensions. Calledproportioning, such assumptions allow the engineer to check trial sections of the walls forstability.

In figure below shows the general proportions of various retaining wall components thatcan be used for initial checks.

Bowles,J. E. (1977).

Jack C. McCormac, James K. Nelson (2006). Design of Reinforced Concrete ACI 318-05Code Edition, seventh edition.

Page-256-


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.

Page-257-

Failure of retaining wall: (a) by overturning; (b) by sliding; (c) by bearing capacity failure; (d) by deep-seated shear failure.


1. Checiking for Overturning

The Rankine active pressure

Pa12γ1⋅ H2⋅ Ka⋅= Ka

1 sinϕ1−

1 sinϕ1+=

The horizontal active force

Ph Pa cos α( )⋅=

The vertical active force

Pv Pa sin α( )⋅=

The Rankine passive pressure

Pp12

Kp⋅ γ2⋅ D2⋅ 2 c2⋅ Kp⋅ D⋅+= Kp tan 45degϕ22

+⎛⎜⎝

⎞⎟⎠

2

=

Page-258-


Page-259-

Procedure for Calculating ΣM R

Section Area Weight/unit

length of wallMoment arm

measured from CMomentabout C

1 A1 w1 γ1 A1⋅= x1 M12 A2 w2 γ2 A2⋅= x2 M23 A3 w3 γc A3⋅= x3 M34 A4 w4 γc A4⋅= x4 M45 A5 w5 γc A5⋅= x5 M56 A6 w6 γc A6⋅= x6 M6

Pv B MvΣV ΣMR

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

FSoverturingΣMRM0

=M1 M2+ M3+ M4+ M5+ M6+ Mv+

Pa cos α( )⋅H3

⎛⎜⎝

⎞⎟⎠

⋅=

2. Checking for Sliding along the Base

The factor of safety against sliding can be calculated as

FSslidingΣFRΣFd

=ΣV tan δ'( )⋅ B c'a⋅+ Pp+

Pa cos α( )⋅=

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

Page-260-

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 12

to 23

. Thus,

FSslidingΣV tan k1 ϕ2⋅( )⋅ B k2⋅ c2⋅+ Pp+

Pa cos α( )⋅=

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 lengthof the wall becomes

Pp12γ2⋅ D1

2⋅ Kp⋅ 2c2 D1⋅ Kp⋅+=

Page-261-

Alternatives for increasing the factor of safety with respect to sliding


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

XMnetΣV

=

The ecectricity of the resultant R

eB2

X−=

The pressure distribution under the base slab may be determined by using simple principlefrom the mechanics of materials. First, we have

qΣVA

±Mnet y⋅

I⋅=

where Mnet = moment = ΣV( ) e⋅

I = moment of inertia per unit length of the base section = 12

1m( )⋅ B2⋅

Page-262-

Check for bearing capacity failure


qΣV

B 1m( )⋅±

ΣV( ) e⋅B2

⋅

112

B3( )⋅⋅=

The maximum and minimum pressure

qmax qtoe= ΣVB

16 e⋅B

+⎛⎜⎝

⎞⎟⎠

⋅=

Page-263-

qmin qheel= ΣVB

16 e⋅B

−⎛⎜⎝

⎞⎟⎠

⋅=

The ultimate bearing capacity of a shallow foundation

qu c2 Nc⋅ Fcd⋅ Fci⋅ q Nq⋅ Fqd⋅ Fqi⋅+12γ2⋅ B'⋅ Nγ⋅ Fγd⋅ Fγi⋅+=

where

q γ2 D⋅= B' B 2 e⋅−= Fcd 1 0.4DB'

⋅+=

Fqd 1 2 tanϕ2⋅ 1 sinϕ2−( )2⋅DB'

⋅+= Fγd 1=

Fci Fqi= 1ψ

90−⎛⎜

⎝⎞⎟⎠

2= Fγi 1

ψ

ϕ2−⎛

⎜⎝

⎞⎟⎠

2= ψ tan 1− Pa cosα⋅

ΣV

⎛⎜⎝

⎞⎟⎠

=

The factor of safety against bearing capacity failure can be calculated

FSbearingqu

qmax=

Generally, a factor of safety of 3 is required.

15.4 Effect of Surcharge

Differnt type of loads are often imposed on the surface of the backfill behind aretaining 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 horizontalpressure due to the surcharge is constant throughout the depth of the retaining wall.

H/3

H

DD/3

ws

H/3

H

DD/3

ws

Ph2Ph1Pp

surcharge effect under a uniform load

hswsw

=

where hs = equivalent height of soil

ws = pressure of the surcharge

w = unit weight of soil

Page-264-

The total pressure is

Ph Ph1 Ph2+= Ka w⋅H2

2H hs⋅+

⎛⎜⎝

⎞⎟⎠

⋅=

In the case of a partial uniform load acting at a distance from the wall, only a portion of thetotal surcharge pressure affects the wall as in Figure below.

ws

H/3

H

DD/3

45o

h’

ws

H/3

H

DD/3

45o

h’ Ph1Ph2Pp

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 partialsurcharge 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 125

the supported height or lenght,

whichever is shorter, but not less than 100mm.

tmin125

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⋅ diameter 16mm>if

0.0020 b⋅ h⋅ diameter 16mm≤if

= for fy 400MPa≥

3. The minimum area of the vertical reinforcement is 0.0015 b⋅ h⋅ , but it may be reducedto 0.0012 b⋅ h⋅ if diameter 16mm or smaller deformed bars with fy 400MPa≥ are used.

Av.min 0.0015 b⋅ h⋅ diameter 16mm>if

0.0012 b⋅ h⋅ diameter 16mm≤if

= for fy 400MPa≥

4. The maximum spacing of the vertical or the horzontal reinforcing bars is the smaller of450mm or three times the wall thickness.

Page-265-

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

23

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

the wall thickness. As.exterior

23

As= and covermin 50mm13

hwall≤=

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 wallthickness.

covermin 25mm13

hwall≤=

c/ The minimum steel area in the wall footing (heel or toe) according to ACI Codeis 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= .Because this minimum steel area is relately small, it is a common practice to increase itto that minimumn As.min required for flexure:

As.shrinkage 0.0018 b⋅ h⋅ fy 400MPa=if

0.002 b⋅ h⋅ fy 350MPa= fy 275MPa=∨if

=

As.min0.25 f'c⋅

fyb⋅ d⋅

1.4fy

b⋅ d⋅≥=

Page-266-

Example15.1

b ws

γ1

ϕ1H1c1

configuration of retainingwall

γ2D ϕ2 H2c2

B1 B2 B3

B

Geometry of wall:

base slab width B1 1m:= B2 0.6m:= B3 1.9m:=

B B1 B2+ B3+ 3.5m=:=

stem width of upper wall b 0.3m:=

wall height H1 5.4m:= H2 0.6m:=

H H1 H2+ 6m=:= Soil properties:

backfill soil γ1 18kN

m3:= ϕ1 35deg:=

soil in font γ2 19.5kN

m3:= ϕ2 20deg:= c2 20

kN

m2:=

allowable soil bearing capacity qa 190kN

m2:=

depth of soil in font D 0.9m:=

Page-267-

Material:

concrete compressive strength f'c 25MPa:=

steel yield strength fy 400MPa:=

unit weight of concrete γc 24kN

m3:=

surcharge uniform load ws 12kN

m2:=

Solution

1. Checking of wall against overturning

a/ the moment resisting

A

B3 H1⋅

b H1⋅

B2 b−( ) H1⋅

2

B H2⋅

⎡⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎦

:= A

10.26

1.62

0.81

2.1

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

m2=

Page-268-

ws

A1 Ph1A2

Ph2 H2A3 H

3Pp A4

qimnqmax

ΣR CL

x e0.5B

forces acting on the wall

γ

γ1

γc

γc

γc

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:= γ

18

24

24

24

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

kN

m3⋅=

ORIGIN 1:= i 1 rows γ( )..:=

Wi Ai γi⋅:= W

184.68

38.88

19.44

50.4

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

kNm

⋅=

Page-269-

X

BB32

−

B1 B2+b2

−

B1B2 b−( )

2+

B2

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

:= X

2.55

1.45

1.15

1.75

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

m=

Mi Wi Xi⋅:= M

470.934

56.376

22.356

88.2

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

kN m⋅1m

⋅=

ΣR W∑ 293.4kNm

⋅=:=

ΣMR M∑ 637.866kN m⋅1m

⋅=:=

Section No. Area (m2) Unit weight (kN/m) Moment arm (m) Moment(kNm/m)1 10.260 184.680 2.550 470.9342 1.620 38.880 1.450 56.3763 0.810 19.440 1.150 22.3564 2.100 50.400 1.750 88.200

293.400 637.866

b/ the unfactor force acting on the wall

using Rankine equation

Ka1 sin ϕ1( )−

1 sin ϕ1( )+0.271=:= hs

wsγ1

0.667m=:=

Ph112

Ka⋅ γ1⋅ H2⋅ 87.801kNm

⋅=:=

Ph2 Ka γ1⋅ hs⋅ H⋅ 19.511kNm

⋅=:=

c/ the overturning moment

Mo Ph1H3

⎛⎜⎝

⎞⎟⎠

⋅ Ph2H2

⎛⎜⎝

⎞⎟⎠

⋅+ 234.135kN m⋅1m

⋅=:=

Page-270-

the safety of factor against overturning

FSoverturningΣMRMo

2.724=:=

wall "is not overturning" FSoverturning 2≥if

"is overturning" otherwise

:=

wall "is not overturning"=

2. Checking of wall against sliding

Rankine passive force per unit length

let k123

0.667=:= k2 k1 0.667=:=

Kp tan 45degϕ22

+⎛⎜⎝

⎞⎟⎠

2

2.04=:=

Pp12

Kp⋅ γ2⋅ D2⋅ 2c2 Kp⋅ D⋅+ 67.521kNm

⋅=:=

Ph Ph1 Ph2+ 107.312kNm

⋅=:=

the safety of factor against sliding

FSslidingΣR tan k1 ϕ2⋅( )⋅ B k2⋅ c2⋅+ Pp+

Ph1.712=:=

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

"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

xΣMR Mo−

ΣR1.376m=:=

eB2

x− 0.374m=:=

Page-271-

footing "is not upward" eB6

≤if

"is upward" otherwise

:=

footing "is not upward"=

b/ the maximum and minimum pressure

qmaxΣRB

16 e⋅B

+⎛⎜⎝

⎞⎟⎠

⋅ 137.569kN

m2⋅=:=

qminΣRB

16 e⋅B

−⎛⎜⎝

⎞⎟⎠

⋅ 30.088kN

m2⋅=:=

wall "is not failure" qmax qa≤if

"is failure" otherwise

:=

wall "is not failure"=

4. Design of stema/ main reinforcementThe 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⋅ 140.481kNm

⋅=:=

Pu2 1.6 Ph2⋅ 31.218kNm

⋅=:=

Mu1 1.6 Pu1H13

⋅ Pu2H12

⋅+⎛⎜⎝

⎞⎟⎠

⋅ 539.448kN m⋅1m

⋅=:=

h B2 0.6m=:= d1 h 30mm20mm

2+⎛⎜

⎝⎞⎟⎠

− 560 mm⋅=:=

RMu1

0.9 d12⋅

1.911 MPa⋅=:=

ρ 0.85f'cfy

⋅ 1 12 R⋅

0.85 f'c⋅−−

⎛⎜⎝

⎞⎟⎠

⋅ 0.00501=:=

Page-272-

ρmin max0.25MPa

f'cMPa

⋅

fy

1.4MPafy

,

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

0.0035=:=

As1 max ρ ρmin, ( ) d1⋅ 2.808mm2

mm⋅=:=

As0π 20mm( )2⋅

4314.159 mm2⋅=:=

s1 FloorAs0As1

10mm, ⎛⎜⎝

⎞⎟⎠

110 mm⋅=:=

Because the moment decrease along the height of the wall, the reinforcement area maybe reduced according to the moment requirement. It is practical to use one As1 or

spacing 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 1.6Pu12

H13

⋅ Pu2H12

⋅+⎛⎜⎝

⎞⎟⎠

⋅ 337.155kN m⋅1m

⋅=:=

hB2 b+

2450 mm⋅=:= d2 h 30mm

20mm2

+⎛⎜⎝

⎞⎟⎠

− 410 mm⋅=:=

RMu2

0.9 d22⋅

2.229 MPa⋅=:=

ρ 0.85f'cfy

⋅ 1 12 R⋅

0.85 f'c⋅−−

⎛⎜⎝

⎞⎟⎠

⋅ 0.0059=:=

As2 max ρ ρmin, ( ) d2⋅ 2.419mm2

mm⋅=:=

s2 FloorAs0As2

10mm, ⎛⎜⎝

⎞⎟⎠

120 mm⋅=:=

Page-273-

b/ Temperature and shrinkage reinforcement

the minimum horizontal reinforcement at the base of wall is:

As.min 0.002 B2⋅ 12cm2

1m⋅=:=

for the bottom third H13

1.8 m⋅=

As.min_0.33 0.002B2 2⋅

3⋅ 8

cm2

1m⋅=:=

for the upper two-thirds 2H1

33.6m=

because the front face of the wall is mostly exposed to temperture changes, use halfto two-thirds of the horizontal bars at the external face of wall and place the balanceat the internal face.

As.tem 0.5 As.min⋅ 6cm2

1m⋅=:=

use diameter of shrinkage bar ds 14mm:=

sshrinkage Floor

π ds2⋅

4

As.tem10mm,

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

250 mm⋅=:=

c/ Design for shear the critical section for shear is at a distance d d1 560 mm⋅=:= from the bottomof the stem, at this section the distance for the top equal Hd H1 d− 4.84m=:=

Hu 1.612

Ka⋅ γ1⋅ Hd2⋅ Ka γ1⋅ hs⋅ Hd⋅+⎛⎜

⎝⎞⎟⎠

⋅ 116.595kN1m

⋅=:=

ϕVc 0.75 0.17⋅ MPaf'c

MPa⋅ d⋅ 357

kN1m

⋅=:=

stem_wall "no need shear reinforcement" ϕVc Hu≥if

"need required shear reinforcement" otherwise

:=

stem_wall "no need shear reinforcement"=

Page-274-

5/ Design of the heel

Vu 1.2 B3 H1⋅ γ1⋅ B3 H2⋅ γc⋅+( )⋅ 1.6 B3 hs⋅ γ1⋅( )⋅+ 290.928kN1m

⋅=:=

h H2 600 mm⋅=:= d h 50mm25mm

2−⎛⎜

⎝⎞⎟⎠

− 562.5 mm⋅=:=

ϕVc 0.75 0.17⋅ MPaf'c

MPa⋅ d⋅ 358.594

kN1m

⋅=:=

heel "no need shear reinforcement" ϕVc Vu≥if


:=

heel "no need shear reinforcement"=

Mu VuH12

⋅ 785.506kN m⋅1m

⋅=:=

RMu

0.9 d2⋅2.758 MPa⋅=:=

ρ 0.85f'cfy

⋅ 1 12R

0.85 f'c⋅−−

⎛⎜⎝

⎞⎟⎠

⋅ 0.00741=:=

As max ρ ρmin, ( ) d⋅ 4.17mm2

mm⋅=:=

sheel Floor

π 25mm( )2⋅4

As10mm,

⎡⎢⎢⎣

⎤⎥⎥⎦

110 mm⋅=:=

As.dis 0.0018 d⋅ 1.012mm2

mm⋅=:=

sdis Floor

π ds2⋅

4

As.dis10mm,

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

150 mm⋅=:=

Page-275-

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 distanced 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

qminq1 q2

B1 d− d

detail stress of soil under the base

q1 qmax 137.569kN

m2⋅=:= q2

q1 B B1− d−( )⋅

B76.154

kN

m2⋅=:=

Vu 1.6q1 q2+

2

⎛⎜⎝

⎞⎟⎠

⋅ B1 d−( )⋅ 1.2 H1 γc⋅( )⋅ B1 d−( )⋅− 6.763kN1m

⋅=:=

toe "no need shear reinforcement" ϕVc Vu≥if


:=

toe "no need shear reinforcement"=

Mu 1.6 q2B1

2

2⋅ q1 q2−( ) B1⋅

2B13

⋅+⎡⎢⎢⎣

⎤⎥⎥⎦

⋅ 1.2 H1 γc⋅( ) B1 d−( )⋅B12

⋅−:=

Mu 92.412kN m⋅1m

=

Page-276-

RMu

0.9 d2⋅0.325 MPa⋅=:=

ρ 0.85f'cfy

⋅ 1 12R

0.85 f'c⋅−−

⎛⎜⎝

⎞⎟⎠

⋅ 0.00082=:=

As max ρ ρmin, ( ) d⋅ 1.969mm2

mm⋅=:=

stoe Floor

π 20mm( )2⋅4

As10mm,

⎡⎢⎢⎣

⎤⎥⎥⎦

150 mm⋅=:=

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

Page-277-

17-Retaining Wall Design

Documents

Transcript of 17-Retaining Wall Design