design and detailing of retaining walls learning outcomes...
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DESIGN AND DETAILING OF RETAINING WALLS
Learning Outcomes:g
•• AfterAfter thisthis presentpresent youyou willwill bebe ableable toto dodo thetheAfterAfter thisthis presentpresent youyou willwill bebe ableable toto dodo thethecompletecomplete designdesign andand detailingdetailing ofof differentdifferent typestypes ofofretainingretaining wallswalls..
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Cantilever Retaining wall with shear keywith shear key
Batter
Drainage HoleToeToe
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Classification of Classification of Retaining wallsRetaining walls
• Cantilever retaining wall-RCC(Inverted T and L)(Inverted T and L)
• Counterfort retaining wall-RCC• Buttress wall-RCCButtress wall RCC
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Classification of Retaining Classification of Retaining Classification of Retaining Classification of Retaining wallswalls
BackfillBackfill
Gravity RWT-Shaped RW
L-Shaped RWTile drain
Counterfort ButtressBackfill
Weep hole
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Counterfort RW Buttress RW
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Earth Pressure (P)Earth Pressure (P)
• Earth pressure is the pressure exerted by the retaining material on the retaining wall This GLon the retaining wall. This pressure tends to deflect the wall outward.
GL
• Types of earth pressure : Pa
• Active earth pressure or earth pressure (Pa) and
• Passive earth pressure (Pp).
• Active earth pressure tends to
Variation of Earth pressure
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• Active earth pressure tends to deflect the wall away from the backfill.
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Factors affecting earth Factors affecting earth pressurepressure
• Earth pressure depends on type of backfill the height of wall and the backfill, the height of wall and the soil conditions
Soil conditions: The different soil conditions are
• Dry leveled back fill• Moist leveled backfill• Submerged leveled backfill• Leveled backfill with uniform
surcharge6
surcharge• Backfill with sloping surface
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Analysis for dry back Analysis for dry back fillsfillsfillsfills
Maximum pressure at any height, p=kaγh GLMaximum pressure at any height, p kaγh Total pressure at any height from top,
pa=1/2[kaγh]h = [kaγh2]/2 h
GL
Bending moment at any height M=p xh/3= [k γh3]/6
Pa
HGL
M=paxh/3= [kaγh3]/6
∴ Total pressure, Pa= [kaγH2]/2
M
∴ Total pressure, Pa [kaγH ]/2 ∴Total Bending moment at bottom,
M = [kaγH3]/6
kaγH
H=stem height
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• Where, ka= Coefficient of active earth pressure• = (1 sinφ)/(1+sinφ)=tan2φ• = (1-sinφ)/(1+sinφ)=tan2φ• = 1/kp, coefficient of passive earth
pressurepressure• φ= Angle of internal friction
γ=Unit weigh or density of backfillγ g y
• If φ= 30°, ka=1/3 and kp=3. Thus ka is 9 times kpp p
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Backfill with sloping surface
• pa= ka γH at the bottom pa a γand is parallel to inclined surface of backfill
GL
• ka= ⎥⎥⎦
⎤
⎢⎢⎣
⎡ −−
φθθφθθθ22
22 coscoscoscos
• Where θ=Angle of
⎥⎦⎢⎣ −+ φθθ 22 coscoscos
surcharge∴ Total pressure at bottom
9=Pa= ka γH2/2
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Stability requirements of RWStability requirements of RW
• Following conditions must be f d f b l f llsatisfied for stability of wall
It h ld t t• It should not overturn• It should not slide
• i.e Max. pressure at the toe should not exceed the safe bearing not exceed the safe bearing capacity of the soil under working condition
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Check against overturningCheck against overturning
Factor of safety against overturning g
= MR / MO ≥ 1.55 (=1.4/0.9)Where,
M S bili i MR =Stabilising moment or restoring momentMO =overturning momentMO overturning moment
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Check against SlidingCheck against Sliding
• FOS against sliding • FOS against sliding • = Resisting force to
sliding/Horizontal force g/causing sliding
• = μ∑W/Pa ≥ 1.55 (=1.4/0.9)
Friction μ ∑W
SLIDING OF WALL
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SLIDING OF WALL
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Design of Shear key• In case the wall is
unsafe against sliding
• pp= p tan2 (45 +φ/2) H
• = p kp• where pp= Unit
passive pressure il b
PAH+a
H
on soil above shearing plane AB
• p= Earth pressure at BCapp
RAC
at BC
• R=Total passive resistance=p xa
θ=45 + φ/2
app
Bθ
ΣμW kaγ(H+a)
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resistance=ppxa
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Design of Shear keyDesign of Shear key--Contd.,Contd.,
• If ∑W= Total vertical force acting at the key base
,,
baseφ= shearing angle of passive resistanceR T t l i f • R= Total passive force = pp x a
• PA=Active horizontal pressure at key base for H+aH+aμ∑W=Total frictional force under flat base
• For equilibrium, R + μ∑W =FOS x PA
14• FOS= (R + μ∑W)/ PA ≥ 1.55
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M i h M i h Maximum pressure at the toeMaximum pressure at the toe
W Hx1
x
W1
W2
W4
PΣW
Hh
x2
W3
H/3
Pa
R
TW3
b/2b/6e
xb
Pressure below the Retaining WallPmax
Pmin.
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Retaining Wall
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• Let the resultant R due to ∑W and Pa• lie at a distance x from the toe.e at a d sta ce o t e toe• X = ∑M/∑W, ∑M = sum of all moments about toe.
• Eccentricity of the load = e = (b/2-x) < b/6
• Minimum pressure at heel= >Zero. ⎥⎦
⎤⎢⎣⎡ −
∑=
be
bW 61Pmin
• For zero pressure, e=b/6, resultant should cut the base within the middle thirdthe base within the middle third.
• Maximum pressure at toe= < SBC of soil.⎤⎡∑ eW 6
16⎥⎦⎤
⎢⎣⎡ +
∑=
be
bW 61Pmax
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Depth of foundationDepth of foundationDepth of foundationDepth of foundation
R ki ’ f l • Rankine’s formula: 2
sin1 ⎤⎡ − φSBC• Df = sin1sin1
⎥⎦
⎤⎢⎣
⎡+ φ
φγSBC
•• =
2akγ
SBC Df
•γ
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Preliminary Proportioning Preliminary Proportioning (T shaped wall)(T shaped wall)(T shaped wall)(T shaped wall)
• Stem: Top width 200 mm 200
to 400 mm• Base slab width b= 0.4H to
f0.6H, 0.6H to 0.75H for surcharged wallB l b thi k H/10
H
H/10 –• Base slab thickness= H/10 to H/14
• Toe projection= (1/3 1/4) b 0 4H t 0 6H
tp= (1/3-1/4)b H/14
• Toe projection= (1/3-1/4) Base width
b= 0.4H to 0.6H
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Behaviour or structural action
• Behaviour or • Behaviour or structural action and design of stem, heel and toe slabs are same as that of any that of any cantilever slab.
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DistDist.fromtop Every
alternateh1
alternate bar cutAst/2 h2
L
h1c
AA /2h2
Ast
Ldt
AstProvided
Ast/2
Ast
C ti C t il t20
Cross section Curtailment curve
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Design of Heel and ToeDesign of Heel and ToeDesign of Heel and ToeDesign of Heel and Toe
1 Heel slab and toe slab should also be1. Heel slab and toe slab should also bedesigned as cantilever. and determinethe maximum bending moments at thethe maximum bending moments at thejunction.
2 D t i th i f t 2. Determine the reinforcement. 3. Also check for shear at the junction. 4. Provide enough development length.
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Design of stem
• As(min)=14/fy bd
g
• As(min)=14/fy bd• 1.34As =• Development length (Stem • Development length (Stem
steel)
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•Check for shear
x x
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x1 W1
W
W4
ΣW
Hh
x2W2
WH/3
Pa
R
TW3
b/2b/6e
xb Forces acting
Pmax
Pmin.
0.75m 0.45m 1.8m
30.16 T/m2
Forces acting on the wall
and the 120.6 T/m2
T/m224.1
97.99 22.6
P b l th R t i i W ll
pressure below the wall
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Pressure below the Retaining Wall
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Design of heelDesign of heel
x
x
120.6 T/m2
30.16 T/m2
25
T/m
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Ch k f h t d f j ti
Design of toe slabDesign of toe slab--Contd.,Contd.,
• Check for shear: at d from junction (at xx as wall is in compression)
xd
xd
Ldt
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Other deatailsOther deatails
• Construction joint• DrainageDrainage
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Design and DetailingDesign and Detailing of of g gg gCounterfort Retaining Counterfort Retaining
llllwallwall
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Counterfort Retaining wall
• When H exceeds about 6m,
g
When H exceeds about 6m,• Stem and heel thickness is more• More bending and more steel
Cantilever T type Uneconomical
CF
• Cantilever-T type-Uneconomical• Counterforts-Trapezoidal section• 1.5m -3m c/c Stem
Base Slab
CRW
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Parts of CRWParts of CRWSame as that of Cantilever Retaining wall • Same as that of Cantilever Retaining wall Plus Counterfort
Stem Counterforts
Toe Heel ToeBase slab
C ti Pl31
Cross section Plan
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Design of Stem• The stem acts as a continuous slab • Soil pressure acts as the load on the
slab. BF• Earth pressure varies linearly over the
height• The slab deflects away from the earth
f b t th t f t
BF
face between the counterforts• The bending moment in the stem is
maximum at the base and reduces towards top towards top.
p=Kaγh
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Maximum Bending moments for Maximum Bending moments for stemstem
• Maximum + B.M= pl2/16 • (occurring mid-way between
counterforts)• and l• and• Maximum - B.M= pl2/12 • (occurring at inner face of
f )-
counterforts)
• Where ‘l’ is the clear distance
p+
• Where l is the clear distance between the counterforts
• and ‘p’ is the intensity of soil
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Design of Toe SlabDesign of Toe Slab• The base width=b =0.6 H to 0.7 H• The projection=1/3 to 1/4 of base
width.• The toe slab is subjected to an upward
soil reaction and is designed as acantilever slab fixed at the front face ofthe stem
H
the stem.• Reinforcement is provided on earth
face along the length of the toe slab.In case the toe slab projection is large• In case the toe slab projection is largei.e. > b/3, front counterforts areprovided above the toe slab and theslab is designed as a continuous bslab is designed as a continuoushorizontal slab spanning between thefront counterforts.
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Design of Heel Slab
• The heel slab is designed as a continuousslab spanning over the counterforts and issubjected to downward forces due to weight ofsubjected to downward forces due to weight ofsoil plus self weight of slab and an upward forcedue to soil reaction.
• Maximum +ve B.M= pl2/16• (mid-way between counterforts)• And
BF• And• Maximum -ve B.M= pl2/12• (occurring at counterforts)
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Design of CounterfortsDesign of Counterforts• The counterforts are subjected to j
outward reaction from the stem. • This produces tension along the outer
sloping face of the counterforts. • The inner face supporting the stem is
in compression. Thus counterforts are designed as a T-beam of varying depth depth.
• The main steel provided along the sloping face shall be anchored properly at both ends
TC
at both ends. • The depth of the counterfort is
measured perpendicular to the sloping side.
d
side.
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Behaviour of Counterfort Behaviour of Counterfort Behaviour of Counterfort Behaviour of Counterfort RWRW
-M
+M Important points
COUNTERFORT
•Loads on Wall
•Deflected shapeCOUNTERFORTSTEM
•Deflected shape
•Nature of BMs
-M•Position of steel
•Counterfort details
TOE +MHEEL SLAB
Counterfort details
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a. Proportioning of Wall a. Proportioning of Wall p gp gComponentsComponents
• Coefficient of active pressure = ka = 1/3• Coefficient of passive pressure= kp = 3Coefficient of passive pressure kp 3• The height of the wall above the base • = H = 7 + 1.25 = 8.25 m. Hh1=
7 m• Base width = 0.6 H to 0.7 H • (4.95 m to 5.78 m), Say b = 5.5 m
1 25 m
7 m
• Toe projection = b/4 = 5.5/4 = say 1 .2 m • Assume thickness of vertical wall = 250 mm • Thickness of base slab = 450 mm
1.25 m
• Thickness of base slab = 450 mm b=5.5 m
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S i f fS i f fSpacing of counterfortsSpacing of counterforts
l = l = 33..5 5 (H/γ)(H/γ)00..2525 = = 33..5 5 ((88..2525//1818))00..2525 = = 22..88 88 m m
∴∴ c/c spacing = c/c spacing = 22 88 88 + + 00 40 40 = = 33 28 28 m m l
∴∴ c/c spacing = c/c spacing = 22..88 88 + + 00..40 40 = = 33..28 28 m m say say 3 3 mm
∴∴ Provide counterforts at Provide counterforts at 3 3 m c/c. m c/c. Assume width of counterfort = Assume width of counterfort = 400 400 mmmm∴∴ clear spacing provided = l = clear spacing provided = l = 3 3 -- 00..4 4 = =
22..6 6 m m
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250
Details of wall250 mm
CF 3 /
h 7 8
CF: 3m c/c, 400 mm
h=7.8 mH=8.25 mh1=7 m
θ
d
4.05m θ1.2 m
b=5 5 m
1.25m
T40
b=5.5 mT
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Di f M
b. Check Stability of Wall
Sr.No.
Description of loads Loads in kN
Dist. ofe.g. from
T in m
Moment about
T in kN-m
1 Weight of stemW1
25x0.25x1x7.8= 48.75
1.2 + 0.25/2=1.325 64.59
2 Weight of baseslab W2
25x5.5x1x0.45= 61.88 5.5/2 =2.75 170.17
3 Weight of earthover heel slab W3
18x4.05x1x7.8= 568.62
1.45 +4.05/2= 3.475 1975.95
Total ΣW = 679.25 ΣW =2210.71
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250 mm
W1
H8250
W3W1
h1= 70008250ΣW
PA
R
A B C D1200 mm 4050 mmDf= 1250
PA PA
H/3450
eX b/2TW2
b/3
kaγH
Pressure distribution
Cross section of wall-Stability analysis
b/3
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Cross section of wall Stability analysis
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Stability of Stability of Stability of Stability of wallswalls
• Horizontal earth pressure on full heightf llof wall
• = Ph = kaγH2 /2 =18 x 8.252/(3 x 2) =204 19 kN204.19 kN
• Overturning moment = M0Overturning moment M0• = Ph x H/3 = 204.19 x 8.25/3 = 561.52
kN.m.• Factor of safety against overturning• = ∑ M / M0 = 2210.71/561.52 = 3.94 >
1 5543
1.55∴ safe.
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CheckCheck forfor slidingsliding
TotalTotal horizontalhorizontal forceforce tendingtending toto slideslide thethewallwallwallwall
== PPhh == 204204..1919 kNkNResistingResisting forceforce == ∑µ∑µ WW == 00 5858 xx 679679 2525ResistingResisting forceforce == ∑µ∑µ..WW == 00..5858 xx 679679..2525== 393393..9797 kNkN∴∴FactorFactor ofof safetysafety againstagainst slidingsliding∴∴FactorFactor ofof safetysafety againstagainst slidingsliding== ∑µ∑µ..WW // PPhh == 393393..9797//204204..1919== 11..9393 >> 11..5555 ...... safesafe..
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CheckCheck forfor pressurepressure distributiondistribution atat basebase
LetLet xx bebe thethe distancedistance ofof RR fromfrom toetoe (T),(T),∴∴ xx == ∑∑ MM // ∑∑ WW∑∑ // ∑∑Eccentricity=eEccentricity=e == b/b/22 -- xx << b/b/66∴∴WholeWhole basebase isis underunder compressioncompression..
MaximumMaximum pressurepressure atat toetoe== pp == ∑W∑W // bb (( 11++66e/b)e/b) ==== ppAA == ∑W∑W // bb (( 11++66e/b)e/b) ==
MinimumMinimum pressurepressure atat heelheeluu p essu ep essu e atat eeee
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I t it f t j ti f t ith t I t it f t j ti f t ith t Intensity of pressure at junction of stem with toe Intensity of pressure at junction of stem with toe i.e. under Bi.e. under B
Intensity of pressure at junction of stem with heel Intensity of pressure at junction of stem with heel i.e. under Ci.e. under C
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250 mm
H8250ΣW R
PA
R
A B C D1200 mm 4050 mm
450
1250
T
80.39kN/ 2166 61 143 9147 8153 9
eX b/2T
kN/m2166.61kN/m2
143.9147.8153.9
475500 mm
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b) Design of Toe slabb) Design of Toe slab• Max. BMB = psf x (moment due to soil pressure -
moment due to wt. of slab TB]
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Check for ShearCheck for ShearCritical section for shear: At distance d ( • Critical section for shear: At distance d (= 390 mm) from the face of the toe
d
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-MCounterfort RW+M
COUNTERFORTSTEMSTEM
-M
TOE +MHEEL SLAB
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(c) Design of Heel Slab( ) g
80.39kN/m2166.61
kN/m2143.9147.8153.9
kN/m
5500 mm
71.26
7.75 kN/m Forces on heel slab71.26 kN/m
DC
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A B C1200 4050 1250
R
A B Cmm mm
450
eX b/2 Area for4050 mm
C DTOE
Area for stirrups
26003000
TOE
HEEL
SFD139
x1
1
Net down force dia
71.28 kN/m
7.75 kN/m
y1 Shear analysis and Zone of shear steel
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Net down force dia.
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(d) Design of Stem (Vertical (d) Design of Stem (Vertical (d) Design of Stem (Vertical (d) Design of Stem (Vertical Slab).Slab).
C ti l b i b t th • Continuous slab spanning between the counterforts and subjected to earth pressure.
• The intensity of earth pressure• The intensity of earth pressure• = ph = ka γh• Area of steel on earth side near counterforts : Area of steel on earth side near counterforts : • Maximum -ve ultimate moment,• Mu = ph l2/12 u ph /
• Required d
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( ) D i f C t f tAt any section at any depth h below the top, the At any section at any depth h below the top, the
(e) Design of CounterfortAt any section at any depth h below the top, the At any section at any depth h below the top, the
total horizontal earth pressure acting on the total horizontal earth pressure acting on the counterfortcounterfort
= = 11//2 2 kk yy hh22x c/c distance between counterfortx c/c distance between counterfort= = 11//2 2 kkaayy hh22x c/c distance between counterfortx c/c distance between counterfort
Counterfort acts as a TCounterfort acts as a T--beam. beam.
Even assuming rectangular section,Even assuming rectangular section,
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The effective depth is taken at The effective depth is taken at right angle to the reinforcement.
h =7.8 m
d
4.05m θ
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Design of Horizontal TiesDesign of Horizontal TiesTh di t ll b th ll t f t f 1 • The direct pull by the wall on counterfort for 1 m height at base
• = k γh x c/c distance =1/3x18 x 7 8 x 3 = 140 4 • = kaγh x c/c distance =1/3x18 x 7.8 x 3 = 140.4 kN
• Area of steel required to resist the direct pullq p• = 140.4 x 103/(0.5fy) = 583 mm2 per m height.• Using 8 mm 2-legged stirrups, Ast = 100 mm2
• spacing = 1000 x 100/583 = 170 mm c/c.• Since the horizontal pressure decreases with h,
th i f ti b i dthe spacing of stirrups can be increased
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Design of Vertical TiesDesign of Vertical TiesDesign of Vertical TiesDesign of Vertical Ties
• The maximum pull will be exerted at the end of heel slab where the net downward force = 71.26 kN/m.kN/m.
• Total downward force at D• = 71.26 x c/c distance. • Required Ast = 1.5 x 213.78 x 103/(0.87 x 415) =
888 mm2888 mm• Using 8 mm 2-legged stirrups , Ast = 100 mm2
• spacing = 1000 x 100/888 = 110 mm c/c.p g / /
• Increase the spacing of vertical stirrups towards the end C
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the end C
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250 mm 0-200mm
STEM COUNTERFORT
7000
STEM COUNTERFORT
12@200 8250 mm
12@200
1200 mm 4050 mm450
1250
16@120 12@200 12@200
Cross section between counterforts
TOE HEEL
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Cross section between counterforts
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250 mm
12@400 8-22 1.77m
8 - # 22
#8@110-450, VS#12@
#12@200 8250
8 # 22
#8@170-450, HS
@110-300
1200 mm450
1250
#16@120 #12@200 #12@200
Cross section through counterforts59
Cross section through counterforts
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STEMSTRAIGHT BARS STEMBARS
Backfill B kfillBackfill
0.3l
Backfill
0.25 l
Section through stem at the junction of Base slab
With straight barsWith cranked
bars
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Section through stem at the junction of Base slab.
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Lateral pressure against retaining wall d t h l d due to surcharge loads
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