17-retaining wall design
TRANSCRIPT
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
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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-
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.
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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
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
=
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Braja M. Das (2007). Principle of Foundation Engineering, sixth edition.
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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
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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⋅+=
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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
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⋅
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Check for bearing capacity failure
Braja M. Das (2007). Principle of Foundation Engineering, sixth edition.
qΣV
B 1m( )⋅±
ΣV( ) e⋅B2
⋅
112
B3( )⋅⋅=
The maximum and minimum pressure
qmax qtoe= ΣVB
16 e⋅B
+⎛⎜⎝
⎞⎟⎠
⋅=
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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
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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.
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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⋅≥=
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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:=
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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=
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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
⋅=
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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
⋅=:=
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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=:=
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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=:=
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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⋅=:=
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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
"need required shear reinforcement" otherwise
:=
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
"need required shear reinforcement" otherwise
:=
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-