Retaining wall

Retaining Walls
By
Engr. Mudassir Abbas

Retaining walls are used to hold back masses of earth or other
loose material where conditions make it impossible to let those
masses assume their natural slopes.
Function of retaining wall
 Such conditions occur when the width of an excavation, cut,
or embankment is restricted by conditions of ownership, use of
the structure, or economy. For example, in railway or highway
construction the width of the right of way is fixed, and the cut or
embankment must be contained within that width.
Similarly, the basement walls of the buildings must be located
within the property and must retain the soil surrounding the
base.

Free standing retaining walls, as distinct from those that form
parts of structures, such as basement walls, are of various types.
Types of retaining walls
 The gravity retaining wall retains the
earth entirely by its own weight and
generally contains no reinforcement. It
is used up to 10 ft height.
 The reinforced concrete cantilever
retaining wall consists of the vertical arm
that retains the earth and is held in position
by a footing or base slab. In this case, the
weight of the fill on top of the heel, in
addition to the weight of the wall,
contributes to the stability of the structure.
Since the arm represents a vertical
cantilever, its required thickness increase
rapidly, with increasing height. It is used in
the range of 10 to 20 ft height.

In the counterfort wall the
stem and base slab are tied
together by counterforts which
are transverse walls spaced
at intervals and act as tension
ties to support the stem wall.
Counterforts are of half or
larger heights. Counterfort
walls are economical for
heights over 25 ft.
Property rights or other restrictions sometimes make it necessary
to place the wall at the forward edge of the base slab, i.e. to omit
the toe. Whenever it is possible, toe extensions of one-third to
one-fourth of the width of the base provide a more economical
solution.

A buttress wall is similar to a
counterfort wall except that
the transverse support walls
are located on the side of the
stem opposite to the retained
material and act as
compression struts. Buttress,
as compression elements,
are more efficient than the
tension counterforts and are
economical in the same
height range.
A counterfort is more widely used than a buttress because the
counterfort is hidden beneath the retained material, whereas the
buttress occupies what may otherwise be usable space in front of
the wall.

This is an free standing wall category. A wall type bridge abutment
acts similarly to a cantilever retaining wall except that the bridge
deck provides an additional horizontal restraint at the top of the
stem. Thus this abutment is designed as a beam fixed at the
bottom and simply supported or partially restrained at the top.

 w is unit weight of the soil
Earth Pressure
 C0 is a constant known as the coefficient of earth pressure at
rest According to Rankin, the coefficient for active and passive
earth pressure are
φ−δ+δ
φ−δ−δ
δ= 22
22
CosCosCos
CosCosCos
CosCa
φ−δ−δ
φ−δ+δ
δ= 22
22
CosCosCos
CosCosCos
CosCp
 For the case of horizontal surface δ=0
φ+
φ−
=
sin1
sin1
Cah
φ−
φ+
=
sin1
sin1
Cph
 For liquid P=wwh, ww is the
unit weight of water.
Soil retaining structure Ph=Cowh

Earth pressure for common condition of loading
2
whC
2
1
P
3
h
y
ah=
=
φ=φ=δ
=
=
CosCFor
Cawh
2
1
p
3
h
y
a
2

Earth pressure for common condition of loading
)h2h(whC
2
1
P
)h2h(3
hh3h
y
ah
2
′+=
′+
′+
=

1. Individual parts should be strong enough to resist the applied
forces
Stability Requirement
2. The
wall as
a whole
should
be
stable
against
(i)
Settlem
ent (ii)
Sliding
(iii)
Overtur
ning

It is necessary to ensure that the pressure under the footing does
not exceed the “permissible bearing pressure” for the particular
soil.
By the formula
Settlement
I
MC
A
N
q
min
max
±=
 If , compression will act throughout the section
3
a

>

( )[ ]
[ ] [ ]
















v
21
2
v
22
v
1
2
v
2
vvv
2,1
3
v
2
v
v
3
v
2
v
v
2,1
3
v
v
2,1
R
qq
2
awhen
2a6
R
q&a64
R
q
a63
RaR6R3R
q
12
4
aR2R
R
12
2
aR
4
R
R
q
12
2
Ra
2R
q
===
−=−=
−±=



−±=
−
±=
−
±=






−
±=
Unit dimension in the direction perpendicular to the paper.
Settlement

0q&
R2
q
11
Rq
2
R
6
12
1
R
2
R
6
23
12
1
Rq
12
2
R
32R
q
2
v
1
v2,1
v
v
v
v2,1
v
v
2,1
33
3
==






±=












×





±=












×




 −
±=
×





−
±=










Settlement

( )( )
a3
R2
q
a3q
2
1
R
v
v
=






=
Settlement

Sliding
F = µRv
5.1
P
F
h
≥
Overturning
2
momentgOverturnin
momentgStabilizin
≥

1. Lateral earth pressure will be considered to be live loads and a
factor of 1.7 applied.
Basis of Structural Design
2. In general, the reactive pressure of the soil under the structure
at the service load stage will be taken equal to 1.7 times the soil
pressure found for service load conditions in the stability
analysis.
3. For cantilever retaining walls, the calculated dead load of the
toe slab, which causes moments acting in the opposite sense to
those produced by the upward soil reaction, will be multiplied by
a factor of 0.9.

4. For the heel slab, the required moment capacity will be based
on the dead load of the heel slab itself, plus the earth directly
above it, both multiplied by 1.4.
5. Surcharge, if present, will be treated as live load with load
factor of 1.7.
6. The upward pressure of the soil under the heel slab will be
taken equal to zero, recognizing that for severe over load stage
a non linear pressure distribution will probably be obtained, with
most of the reaction concentrated near the toe.
Basis of Structural Design

Drainage
 Reduction in bearing capacity.
 Hydrostatic pressure.
 Ice pressure
 Drainage can be provided in various ways
i. Weep holes, 6 to 8 in. 5 to 10 ft horizontally spaced. 1 ft3
stone at rear end at the bottom weep holes to facilitate
drainage and to prevent clogging.
ii. Longitudinal drains embedded in crushed stone or gravel,
along rear face of the wall.
iii. Continuous back drain consisting of a layer of gravel or
crushed stone covering the entire rear face of the wall with
discharge at the ends.

Problem
 A gravity retaining wall consisting of plain concrete w=144
lb/ft3
is shown in fig. The bank of supported earth is
assumed to weigh 110 lb/ft3
ø=300
and to have a coefficient
of friction against sliding on soil of 0.5.
Solution
333.0
30sin1
30sin1
sin1
sin1
Ca =
+
−
⇒
φ+
φ−
=
3
30sin1
30sin1
sin1
sin1
Cp =
−
+
⇒
φ−
φ+
=
lb36.2637)12)(110)(333.0(
2
1
whC
2
1
P 22
aah =





==
lb.ft44.10549
3
12
36.2637momentgOverturnin =×=

 Restoring moment:
 Moment about toe
Component
Weights
Force Moment arm Moment
W1 (6)(1)(144) 864 x 3 2592 lb.ft
W2 (1)(11)(144) 1584 x 1 1584 lb.ft
W3 (4)(144)(11) 8976 lb.ft
W4 (4)(11)(110) 10083.33 lb.ft
W5 (0.5)(11)(110) 605 x 5.75 3478.75 lb.ft
Rv = 8641 lb M=26714.08 lb.ft






2
1






×+× 4
3
1
5.13168
Solution






×+× 4
3
2
5.12420





2
1
Safety factor against overturning = O.K253.2
44.10549
08.26714
>=

 Distance of resultant from the toe
 Thus resultant is out of kern distance
 The max. soil pressure will be
)87.1)(3(
)8641)(2(
a3
R2
q v
⇒=
 q = 3080.57 lb/ft2
< 4000 psf
Solution
ft87.1
8641
44.1054908.26714
a =
−
=

Sliding
 Assuming that soil above footing toe has eroded and thus
the passive pressure is only due to soil depth equal to
footing thickness.
Solution
lb516)1)(110)(3(
2
1
whC
2
1
P 22
pph =





==
 Friction force between footing concrete and soil.
 F.O.S. against sliding = O.K
lb5.4320)8641)(5.0(RF v ==µ=
5.17.1
36.2637
1655.4320
>=
+

Estimating size of cantilever retaining wall
 The base of footing should be below frost penetration about 3’
or 4’.
Height of Wall
 Stem is thickest at its base. They have thickness in the range
of 8 to 12% of overall height of the retaining wall. The minimum
thickness at the top is 10” but preferably 12”.
Stem Thickness.
 Preferably, total thickness of base fall between 7 and 10% of
the overall wall height. Minimum thickness is at least 10” to 12”
used.
Base Thickness

Estimating size of cantilever retaining wall
 For preliminary estimates, the base length can be taken
about 40 to 60% of the overall wall height.
Base Length
 Another method refer to fig. W is assumed to be equal to
weight of the material within area abcd.
 Take moments about toe and solve for x.

Problem
 Design a cantilever retaining wall to support a bank of earth of
16 ft height above the final level of earth at the toe of the wall.
The backfill is to be level, but a building is to be built on the fill.
 Assume that an 8’ surcharge will approximate the lateral
earth pressure effect.
Weight of retained material = 130 lb/ft3
Angle of internal friction = 35o
Coefficient of friction b/w
concrete and soil = 0.4
fc’=3000 psi
fy=40,000 psi
Maximum soil
pressure
= 5 k/ft2

Solution
 Allowing 4’ for frost penetration to the bottom of footing in
front of the wall, the total height becomes.
Height of Wall
 h = 16 + 4 = 20 ft.
 At this stage, it may be assumed 7 to 10% of the overall
height h.
Thickness of Base
 Assume a uniform thickness t = 2’ ( 10% of h )
 h = 20’ h’ = 8’
Base Length

( )( )( )
( )( )( )( )( )
( )
( ) ( )( )( )
( )
ft148.8
82203
820320
h2h3
hh3h
y
lb2.11707
3620120271.05.0
822020120
sin1
sin1
2
1
)h2h(whC
2
1
P
22
ah
=
×+
+
=
′+
′+
=
=
=
×+





φ+
φ−






=
′+=
Solution

 Moments about point a
 W = (120)(x)(20+8) = 3360 x lb
 yP)
2
x
)(W( ×=
 )148.8)(2.11707()
2
x
)(x3360( =
 x = 7.54 ft
 So base length = 1.5 x x = 11.31 ft
Use 11 ft 4” with x = 7’-8” and 3’-8’ toe length
Solution

Stem Thickness
 Prior computing stability factors, a more accurate
knowledge of the concrete dimensions is necessary.
Solution
 The thickness of the base of the stem is selected with the
regard for bending and shear requirements.
 P for 18’ height and h’ = 8’
( )( )( )
lb12.9951
)8218(18120271.0
2
1
)h2h(whC
2
1
P ah
=
×+





=
′+





=
ft412.7
)8218(3
)8)(18)(3()18(
)h2h(3
hh3h
y
22
'
'
=
×+
+
=
+
+
=

 Mu = (1.7) Py = (1.7) (9951.12) (7.412)
= 125388.09 lb.ft
Solution
02784.075.0
03712.0
000,4087000
87000
000,40
)3000)(85.0)(85.0(
f87000
87000
f
f)85.0(
bmax
yy
c
b
1
=ρ=ρ
=






+
=








+
′β
=ρ
For adequate deflection control, choose
686.15
300085.0
000,40
f85.0
f
m
)m
2
1
1(fRthen
c
y
yn
=
×
=
′
=
ρ−ρ=
01392.0
2
1
max =ρ=ρ

Solution
( )( )
67.16d
65.3370
4969.0
1209.125388
R
M
R
MquiredRe
bdquiredRe
496
)686.1501392.0
2
1
1)(000,40)(01392.0(
m
2
1
1fR
n
n
n
n2
yn
′′=
=
×
=
φ
=
=
=
××−=






ρ−ρ=
 Total thickness = 16.7 + 0.5 + 3 = 20.26”
Try 21” thickness of base of stem and select 12” for top of the wall

Shear at d
d used now = 17.5” = 1.458’
 At 18’ – 1.458’ = 16.542’ from top
( )( )( )( )
.requiredisorcementinfreshearnoSo,VVSince
lb7.19553
5.17123000285.0
bdf2V
lb96.17879
P7.1V
lb92.8752
82542.16542.16120271.0
2
1
)h2h(whC
2
1
P
uu
cu
u
ah
>φ
=
××××=
′×φ=φ
=
=
=
×+





=
′+





=
Solution

F.O.S Against Overturning
Component Force Arm Moment
W1 (5.92)(18)(120)=12787.2 107156.74
W2 (0.75)(18)(150)=1012.5 3.67+1+0.25=4.92 4981.50
W3 (18)(1)(150)=2700 3.67+0.5=4.17 11259
W4 (11.33)(2)(150)=3399.00 19255.34
W5 (18)(0.75)(120)=810 3.67+1+0.5=5.17 4187.7
W6 (6.67)(8)(120)=6403.2 51257.62
Total 27111.9 198097.9
)
2
1
(
)
2
1
(
38.8
2
92.5
2167.3 "
=++
665.5
2
33.11
=
005.8
2
67.6
167.3 =++
Solution

 P = 11707.2 lb y= 8.148 ft
 Overturning Moment = 11707.2 x 8.148 = 95390.27 lb.ft
F.O.S. against overturning O.K2077.2
27.95390
9.198097
>==
Location of Resultant & Footing Soil Pressure
 Distance of the resultant from the front edge is

7883.3
9.27111
27.953909.198097
a
)righting(loadTotal
momentgOverturninmomentRighting
a
=
−
=
−
=
Middle third = 3.7778 ft, So resultant is within the middle third.
Solution
( )
( )7883.3633.114
33.11
9.27111
q
)a64(
R
q
21
v
21
×−×=
−= 


2
22
ft/lb74.14)2a6(
R
q v
=−= 

q1= 4771.12 lb/ft2
< 5 k/ft2
So O.K against bearing pressure.
Solution

 Passive earth pressure against 2’ height of footing aphCwh
2
1 2
=
Solution
.requirediskeySo5.1964.0
2.11707
44276.10844
.S.O.F
lb82.442
sin1
sin1
)2)(120(
2
1
<=
+
=
=






φ−
φ+






=
F.O.S. against sliding
 force causing sliding = P = 11707.2 lb
 Frictional resistance = µR
 = (0.4) (27111.9)
 = 10844.76 lb

 The front of key is 4” in front of back face of the stem. This
will permit anchoring the stem reinforcement in the key..
ft/lb58.2635ordinateTotal
ft/lb84.2620x
33.11
38.4756
243.6
x
=
=
 Frictional resistance between soil to soil = µR
lb17.13191
)087.5(
2
58.263512.4771
)(tan
=





 +
φ=
Solution

 Frictional resistance between heel concrete to soil = µR
lb19.3309
)243.6(
2
74.1458.2635
)4.0(
=





 +
=
( )( ) ( ) lbh4.22169.3h120
2
1
Cwh
2
1
pressureearthPassive
22
ph
2
=





=
=
Solution
 F.O.S. against sliding = 1.5
ft19.2h
2.11707
h4.22119.330917.13191
5.1
2
=
++
=
So use key of height = 2’-3” = 2.25’

Design of Heel Cantilever
Wu = (1.7) (120) (8) + (1.4) [18 x 120 + 2 x 150 ]
= 5076 lb/ft
( )( )
ft.lb76.88947
92.576.5
2
1
2
W
M
2
u
2
=






=
=

Vu = Factored shear a joint of stem and heel
Solution
When the support reaction introduces compression into
the end region, then critical shear is at a distance d
from face of support. However, the support is not
producing compression, therefore, critical shear is at
joint of stem and heel.

Vu = Factored shear a joint of stem
and heel
= (5.92) (5076)
= 30049.92 lb
Solution
u
cc
V11.24023
5.21123000285.0
bdf2V
<=
××××=
′φ=φ
 So depth is required to be increased.
98.26
)12)(3000)(2)(85.0(
92.30049
bf2
V
d
c
u
′′==
′φ
=
 Therefore heel thickness 30”, d = 27.5”

Now Wu=(1.7)(120)(8)+(1.4)[(17.5)(120)+(2.5)(150)] = 5097 lb/ft
.ft.lb75.89315)92.5)(5097(
2
1
M 2
u =





=
005.0
f
200
00337.0
f
mR2
11
m
1
psi23.131
)5.27)(12)(9.0(
1289315
bd
M
RquiredRe
y
min
y
n
u
n 22
==ρ
=





−−=ρ
=
×
=
φ
=
Solution

 As = ρmin bd = 1.8 in2
 Use # 8 @ 5” c/c (As = 1.88 in2
)
 Dev. Length required = 23” top bars so 23 × 1.3 = 29.9”
 Available = 5.92’-3” = 68.04” O.K
Design of Toe Slab
33.11
66.7
38.4756
x
=
x = 3215.7
3215.7 + 14.74 = 3230.44
(6801.33)lb/ft



 +
=
2
12.477144.3230
)7.1(Wu
Wu = 6801.33 lb/ft – 270 = 6531.33
Solution
Overload factor = 0.9
{d=24”-3.5”=20.5”
=1.71 ft
Self load=(0.9)(1×2×150)=270 lb/ft

( )( )( )
3.116
5.20129.0
1292.43984
bd
M
RquiredRe
ft.lb92.43984
2
)67.3)(33.6531(
2
W
M
22
u
n
u
u
22
=
×
=
φ
=
===

So ρmin will control
As = (0.005)(12)(20.5) = 1.23 in2
Use # 8 @ c/c (As = 1.26 in2
)
2
1
7
 Available dev. Length = 3.76’ – 3” = 41.01” Required = 23”
At a distance d= 20.5” = 1.71’
3.67’ – 1.71’ = 1.96’
3.394874.1456.3933
56.3933x
33.11
37.9
38.4756
x
=+
=
=
Solution

Earth pressure lb55.14526)7.1)(96.1(
2
12.47713.3948
=




 +
=
Vu = 14526.55 – 270 x 1.96 = 13997.37 lb
u
cc
Vlb76.229055.20123000285.0
bd)f2(V
>=××××=
′φ=φ
 So no shear reinforcement is required
Reinforcement for stem
Solution
( )( )( )( )
( ) ( )
32.7
5.100
25.726
825.173
85.173)5.17(
h2h3
hh3h
y
lb43.9532
825.175.17120271.0
2
1
)h2h(whC
2
1
P
22
ah
′==
×+
××+
=
′+
′+
=
=
×+





=
′+=

( )( )( )
)in67.2A(c/c
2
1
4@bars9#Use
in457.25.17120117.0bdA
0117.0
000,40
08.425686.152
11
686.15
1
686.15
300085.0
000,40
f85.0
f
m
f
mR2
11
m
1
psi08.425
)5.17(129.0
121.117163
bd
M
R
ft.lb1.11716323.743.95327.1M
2
s
2
s
c
y
y
n
u
n
u
22
=
″
=××=ρ=
=





 ××
−−=ρ
=
×
=
′
=






−−=ρ
=
××
×
=
φ
=
==
Reinforcement for stem
Solution

At 5’ from top
P=1707.3lb
y=2.302’ Mu=6681.34 lb.ft
=6.68 k.ft
At 10’ from top
P=4227.6 lb
y=4.36’ Mu=31334.97 lb.ft
=31.33 k.ft
At 15’ from top
P=7560.9 lb
y=6.29’ Mu=80848.7 lb.ft
=80.85 k.ft
Solution
ft.k16.117
in.lb2.1405957ft.lb1.117163baseatMu
=
=

With Full Reinforcement
 C = 0.85fc’ba = 0.85 × 3000 × 12 × a = 30600a lb
 T = Asfy = 2.67 x 40000 = 106800
 a = 3.49 in
 At top of wall d = 8.5”
ft.k11.54.in.lb6.649290
2
49.3
5.8)106800)(9.0(
2
a
dfA9.0M ysn
==






−=





−=φ
At base of stem d=17.5”
Solution
ft.k2.126.in.lb6.1514370
2
49.3
5.17)106800)(9.0(Mn
==






−=φ

With half Reinforcement
 C = 0.85fc’ba = 0.85 × 3000 × 12 × a = 30600a lb
 T = Asfy = 1.335 × 40,000 = 53400 for #9@9” c/c(As=1.33 in2
)
 a = 1.745 in
ft.k55.30.in.lb65.366577
2
745.1
5.8)53400)(9.0(Mn
==






−=φ
Solution
ft.k59.66.in.lb65.799117
2
745.1
5.17)53400)(9.0(Mn
==






−=φ

With one-fourth Reinforcement
 C = 0.85fc’ba = 0.85 × 3000 × 12 × a = 30600a lb
 T = Asfy = 0.67 × 40,000 = 26800 for #9@18” c/c(As=0.67 in2
)
 a = 0.88 in
ft.k2.16.in.lb2.194407
2
88.0
5.8)26800)(9.0(Mn
==






−=φ
Solution
ft.k29.34.in.lb2.411487
2
88.0
5.17)26800)(9.0(Mn
==






−=φ

 The actual termination point is found by extending beyond the
intersection of capacity moment line with the factored moment
diagram a distance of either the effective depth d or 12 bar
diameters, whichever is greater.
Solution
 Thus half bars should be cut at 4’-8” distance from bottom and
further half bars should be cut 8’-8” theoretically from bottom.

36.6y
5.17
9
x
125.17
9
y12
x
′′=×=
×
=
×
d at 4.6’ from bottom (y=12.9’)
= 8.5 + 6.63 = 15.13”
15”≅
85.4y
5.17
9
x ′′=×=
d at 8.6’ from bottom (y=8.9’)
= 8.5 + 4.58 = 13.08”
13”
Solution
≅

 bar used # 9 of diameter = 1.128”
 12 db = 13.54” = 14”
 Therefore half bars should be terminated actually at
4’-8”+15”=15’-11” from bottom
 and further half bars should be terminated at
8’-8”+14” =9’10”
 For tension bars to be terminated in the tension zone, one
of the following condition must be satisfied.
2. Continuing bars must provide at last twice the area required
for bending moment at the cut off point.
1. Vu at the cut-off point must not exceed two-thirds of the
shear strength øVn.
3. Excess shear reinforcement is provided.
 at 12.9’ from topbd)f2(V cc
′φ=φ
Solution

( )
kips68.9V
3
2
kips14.53
1000
1
13123000285.0V
.satisfiedis)1(Condition
kips31.10
1000
1
h2hwhC
2
1
7.1V
kips11.17V
3
2
kips16.76
1000
1
15123000285.0V
c
c
ahu
c
c
=φ×
=
×××××=φ
=
×


 ′+=
=φ×
=
×××××=φ
Solution

( )
( )
kips13.6
1000
1
829.89.8120271.0
2
1
7.1
1000
1
h2hwhC
2
1
7.1V ahu
=
×



×+××××=
×


 ′+=
 Condition (1) satisfied so bars can be terminated.
The above condition are imposed as a check stress concentration.
 Shear at bottom = Vu = 1.7 x 9.53243 = 16.21 kips
Solution
kips56.19
1000
1
5.17123000285.0Vc =×××××=φ
Since øVc > Vu so no need of shear reinforcement.

 The ρ used should not be less than at any point. This
minimum limit, strictly speaking, does not apply to retaining
walls. However, because the integrity of retaining wall
depends absolutely on the vertical walls, it appear prudent to
use this limit un such cases.
yf
200
 First termination point is 5’-11” from bottom
where d=14.46” As =1.335 in2
 Second termination point is 9’-10” form bottom where
d=12.44” As=0.6675 in2
Solution
005.0
f
200
0077.0
y
=>=ρ
ρ = 0.0045 ≈ 0.005 Therefore the above condition is satisfied.

 Another requirement is that maximum spacing of the primary
flexural reinforcement exceed neither 3 times the wall thickness
nor 18 in. These restrictions are satisfied as well.
 For splices of deformed bars in tension, at sections where the
ratio of steel provided to steel required is less than 2 and where
no more than 50% of the steel is spliced, the ACI code requires
a class-B splice of length 1.3 ld.
 ld for # 9 bars =29”
 Splice length = 1.3 × 29 = 37.7” or 3’-2” O.K
Solution

Temperature & shrinkage reinforcement
 Total amount of horizontal bars (h is average thickness)
ft/in39.0
2
50.2012
12002.0bh002.0A 2
s =
+
××==
 Since front face is more exposed for temperature changes
therefore two third of this amount is placed in front face and one
third in rear face.
Solution
Accordingly # 4 @ 9 in. c/c As=0.26 in.ft/in26.0A
3
2 2
s =
 Use # 3 @ 10 in. c/c As=0.13 in2
.ft/in13.0A
3
1 2
s =
 For vertical reinforcement on the front face, use any nominal
amount. Use # 3 @ 18 in. c/c
Since base is not subjected to extreme temperature changes,
therefore # 4@ 12” c/c just for spacers will be sufficient.

Retaining wall

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