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Retaining Walls
1
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.
2
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 20ft height.
3
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.
Types of retaining walls
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.
4
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.
Types of retaining walls
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.
5
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.
Types of retaining walls
6
w is unit weight of the soil
Earth Pressure
C0 is a constant known as the coefficient of earth pressure at rest According to Rankine, the coefficient for active and passive earth pressure are
22
22
CosCosCos
CosCosCosCosCa
22
22
CosCosCos
CosCosCosCosCp
For the case of horizontal surface =0
sin1sin1
Cah
sin1sin1
Cph
For liquid P=wwh, ww is the unit weight of water.
Soil retaining structure Ph=Cowh
7
Earth pressure for common condition of loading
2whC21
P
3h
y
ah
CosCFor
Cawh21
p
3h
y
a
2
8
Earth pressure for common condition of loading
)h2h(whC21
P
)h2h(3hh3h
y
ah
2
9
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) Settlement (ii) Sliding (iii) Overturning
10
Stability Requirement
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
IMC
AN
qminmax
If , compression will act throughout the section3
a
11
v
vv
Rqqawhen
aR
qaR
q
21
2221
2
26&64
Stability Requirement Settlement
12
0&2
21 qR
q v
Stability Requirement Settlement
13
a
Rq v
3
2
Stability Requirement Settlement
14
Sliding
F = Rv
Stability Requirement
5.1PF
h
Overturning
2moment gOverturnin
moment gStabilizin
15
1. Lateral earth pressure will be considered to be live loads and a factor of 1.6 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 1.6 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.
16
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 1.2.
5. Surcharge, if resent, will be treated as live load with load factor of 1.6.
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
17
Drainage
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.
18
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
19
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.20
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 = 120 lb/ft3
Angle of internal friction = 35o
Coefficient of friction b/w concrete and soil = 0.4
fc’=3000 psi
fy=60,000 psi
Maximum soil pressure
= 5 k/ft221
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
22
ft148.8
82203820320
h2h3hh3h
y
lb2.11707
3620120271.05.0
822020120sin1sin1
21
)h2h(whC21
P
22
ah
Solution
23
Moments about point a
W = (120)(x)(20+8) = 3360 x lb
yP)2x
)(W(
)148.8)(2.11707()2x
)(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
24
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(18120271.021
)h2h(whC21
P ah
ft412.7
)8218(3)8)(18)(3()18(
)h2h(3hh3h
y22
'
'
25
Mu = (1.6) Py = (1.6) (9951.12) (7.412)
= 118004 lb.ft
Solution
),4(0203.0
005.0003.0
003.0
000,60
)3000)(85.0)(85.0(
005.0003.0
003.0)85.0( 1max
NilsonTableAOR
f
f
y
c
For economy and ease of bar placement,
008.04.0 max
26
Solution
"75.20
304129.0
12118004
ReRe
304
588.01
22
'
d
bd
M
R
Mquiredbdquired
psif
ffR
u
n
n
c
yyn
Total thickness = 20.75 + 0.5 + 2.5 = 23.75”
Try 24” thickness of base of stem and select 12” for top of the wall27
Shear at dd used now = 24-0.5-2.5=21”=1.75’ At 18’ – 1.75’ = 16.25’ from top
.inf,
20704
21123000285.0
2
13634
6.1
8521
8225.1625.16120271.02
1
)2(2
1
requiredisorcementreshearnoSoVVSince
lb
bdfV
lb
PV
lb
hhwhCP
uu
cu
u
ah
Solution
28
F.O.S Against Overturning
Component Force Arm Moment
W1 (5.667)(18)(120)=12239.7 3.67+2+5.667/2 104037.9
W2 (1)(18)(150)=1350.1 3.67+1+1/3 6750.6
W3 (18)(1)(150)=2700 3.67+0.5=4.17 11250
W4 (11.33)(2)(150)=3399.00 11.33/2 19267
W5 (18)(1)(120)=1080 3.67+1+2/3 4187.7
W6 (6.67)(8)(120)=6400 3.67+1+6.67/2 51200
Total 27170.1 198266.6
)2
1(
)2
1(
Solution
29
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.K2 08.227.95390
6.198266
Location of Resultant & Footing Soil Pressure Distance of the resultant from the front edge is
786.3
1.27170
27.953906.198266
Re
a
loadTotal
momentgOverturninmomentsistinga
Middle third=L/3= 3.77 ft, a>L/3,So resultant is within the middle third.
Solution
786.3633.114
33.11
1.27170
)64(
21
21
q
aR
q v
30
222 / 11)26( ftlbaR
q v
q1= 4783.7 lb/ft2 < 5 k/ft2
So O.K against bearing pressure.
Solution
31
Passive earth pressure against 2’ height of footing aphCwh21 2
Solution
.5.1966.02.11707
44210868...
82.442
sin1
sin1)2)(120(
2
1
requirediskeySoSOF
lb
F.O.S. against sliding force causing sliding = P = 11707.2 lb
Frictional resistance = R
= (0.4) (27170.1)
= 10868 lb
32
The front of key is 4” in front of back face of the stem. This will permit anchoring the stem reinforcement in the key..
ftlbordinateTotal
ftlbx
x
/5.2538
/5.252733.11
)117.4783(
6
Frictional resistance between soil to soil = R
lb4.13773
)34.5(2
5.25387.4783)(tan
Solution
33
Frictional resistance between heel concrete to soil = R
lb4.3059
)6(2
115.2538)4.0(
lbh4.22169.3h12021
Cwh21
pressureearthPassive
22
ph2
Solution
F.O.S. against sliding = 1.5
fth
h
81.12.11707
4.2214.30594.137735.1
2
So use key of height = 2’34
Design of Heel CantileverWu = (1.6) (120) (8) + (1.2) [18 x 120 + 2 x 150 ]
= 4488 lb/ft
ftlb
WM u
.72055
67.544882
12
2
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.
35
Design of Heel CantileverVu = Factored shear a joint of stem
and heel= (5.67) (4488)
= 25432 lb
Solution
u
cc
V
bdfV
20211
5.20123000275.0
2
So depth is required to be increased.
"26)12)(3000)(2)(75.0(
25432
2
bf
Vd
c
u
Therefore heel thickness 30”, d = 26.5”36
Now Wu=(1.6)(120)(8)+(1.2)[(17.5)(120)+(2.5)(150)] = 4506 lb/ft
.. 72344)67.5)(4506(2
1 2 ftlbM u
0033.0200
002.02
111
53.2385.0
14.51)5.26)(12)(9.0(
1272344Re
min
'
22
y
y
n
c
y
un
f
f
mR
m
f
fm
psibd
MRquired
SolutionDesign of Heel Cantilever
37
As = min bd = 1.06 in2
Bar area of #8=3.14(1)^2/4=0.79, No of bars required=1.06/0.79=1.35, Take=2, S=12inch/No. of bars,
Use # 8 @ 6” c/c (As = 1.57 in2 )
Design of Toe Slab
33.11
66.7
7.4772
x
x = 3226.7
3226.7 + 11 = 3237.7
Wu=(6417.15)lb/ft
2
7.47837.3237)6.1(uW
Wu = 6417.15 lb/ft – 270 = 6147.15
Solution
Overload factor = 0.9{ d=24”-3.5”=20.5”
=1.71 ft
Self load=(0.9)(1×2×150)=270 lb/ft
38
109
5.20129.0
1241332Re
. 413322
)67.3)(15.6147(
2
22
22
bd
MRquired
ftlbW
M
un
uu
So min will control
As = (0.0033)(12)(20.5) = 0.82 in2
Use # 8 @ c/c (As = 1.26 in2) Table A-3 of Nilson21
7
At a distance d= 20.5” = 1.71’
3.67’ – 1.71’ = 1.96’
1.3958111.3947
1.394733.11
37.9
7.4772
x
x
Solution
39
Earth pressure lb 1.13707)6.1)(96.1(2
7.47831.3958
Vu = 13707.1 – 270 x 1.96 = 13177.9lb
u
cc
Vlb
bdfV
202115.20123000275.0
)2(
So no shear reinforcement is required
Reinforcement for stem
Solution
32.75.10025.726
825.17385.173)5.17(
h2h3hh3h
y
lb43.9532
825.175.17120271.021
)h2h(whC21
P
22
ah
40
)57.1(/6@8#
26.12112005.0
005.0
000,60
27858.23211
58.23
1
53.23300085.0
000,60
85.0
211
1
278)21(129.0
12110271
.15.11027123.743.95326.1
2
2
22
inAccbarsUse
inbdA
f
fm
f
mR
m
psibd
MR
ftlbM
s
s
c
y
y
n
un
u
Reinforcement for stemSolution
41
Temperature & shrinkage reinforcement
Total amount of horizontal bars (h is average thickness)
ftinbhAs /43.02
241212002.0002.0 2
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 @ 8 in. c/c As=0.29 in2.ftinAs /28.03
2 2
Use # 3 @ 9 in. c/c As=0.15 in2. ftinAs / 14.03
1 2
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. 42
Solution
43