Design of cantilever and counter fort retaining walls : Retaining wall - Retains Earth - when level difference exists between two surfaces

Retaining
wall -Retains Earth -when level difference exists between two surfaces

A) Gravity
wall (h<3m) -Masonry or Stone

B) Cantilever
wall (h>3m and h<6m)

C) Counterfort
wall (h>8m)

D) Buttress wall [Transverse stem support
provided on front side]

E)E) Bridge abutment [Additional horizontal restraint from bridge deck]

A) Gravity wall (h<3m) -Masonry or Stone

B) Cantilever wall (h>3m and h<6m)

C) Counterfort wall (h>8m)

D) Buttress wall [Transverse stem support provided on front side]

E)E) Bridge abutment [Additional horizontal restraint from bridge deck]

Stability -Overturning and Sliding -Avoided by providing sufficient base width.

__Earth
pressure and stability requirements__:

Pressure,
*P
*=*C**g**Z*

Where, Z = depth, *g* = Unit weight

C_{a} = 1
-*Sin**f / *1
+*Sin**f *

*Cp *= 1 +*Sin**f / *1 -*Sin**f *

Always, Cp > Ca.

Eg: If^{0}, Ca?= 1/3=and30Cp = 3.

In sloped backfill,

1.__Effect of surcharge on
level backfill__:

Pa = P_{a1} + P_{a2},
where,

__Note__ : Purpose of retaining wall
is to retain earth and not water. Therefore, submerged condition should be
avoided by providing and maintaining proper drainage facilities [including
provision of weep holes].

2. __Effect of water in the
backfill__:

__Stability__ requirements

ii)
__Sliding:__ [Friction between base slab and supporting soil]

F = ?.[where,R R = W]

R
-> Resultant soil pressure at footing base

?->
Coefficient of static friction [0.35 -Silt & 0.60 -Rough rock]

FOSsliding
= 0.9*F
/ * _{Pa}_{Cos}* _{q}* ?
1.4

When P_{a} is very high, shear key
projection can be provided below footing base [Produces passive resistance P_{ps},
which is generally neglected, otherwise].

Sliding is reduced by
providing shear key [like a plug, anchors inside]

*P _{ps} *=

X_{sk}
-> Flexural reinforcement from stem is extended straight into shear key near
the

toe.

__Note__: For economical design,
soil pressure resultant(R) must be in line with front face of wall.

__Preliminary
proportioning of cantilever retaining wall:__

1. The
thickness of base slab is h/12 or 8% of the height of wall + surcharge.

2. The
base thickness of stem should be greater than the thickness of base slab

3. The
top thickness of stem should not be less than 150mm.

4. Clear
cover for stem is 50mm and base slab is
75mm

5. Minimum
length of base slab is given by

where,_{R}=Coefficient?
depending on the pressure distribution

?_{R}
= 0.5 for rectangular pressure distribution & 0.67 for trapezoidal pr.dist.

6. Minimum
length of heel slab is given by

__Notes__:

1. The
critical section for moment is at front face of stem.

2. The critical
section for shear
is at „d? f

3. The
stem, heel and toe slabs are designed as cantilever slabs for the resultant
pressure.

4. Temperature
and shrinkage reinforcement is provided as 0.12% of cross section along the
transverse direction to the main reinforcement and front face of the stem.

1. Determine suitable
dimensions of a cantilever retaining wall, which is required to support a bank
of earth 4.0m high above ground level on the toe side of the wall. Consider the
backfill surface to be inclined at an angle of 15^{o} with the
horizontal. Assume good soil for foundation at a depth of 1.25m below ground
level with SBC of 160kN/m^{2}. Further assume the backfill to comprise
of granular soil with unit weight of 16kN/m^{3} and an angle of
shearing resistance of 30^{o}. Assume the coefficient of friction
between soil and concrete to be 0.5.

Given: h = 4.0 + 1.25m

? =o 15 ? = o 30

re = 16kN/m3

qa = 160kN/m^{2}

? = 0.5

Minimum depth of
foundation (Rankine?s),

__Preliminary
proportioning__:

Thickness
of footing base slab = 0.08h = 0.08 x 5.25 = 0.42m

Provide
a base thickness of 420mm for base slab.

Assume
stem thickness of 450mm at base of stem tapering to 150mm at top of wall.

For economical proportioning
of length „L?, a be in line with front face of the stem.

height
above wall]

Assuming a triangular base
pressure distribution, L = 1.5X = 3.0m

Preliminary
proportions are shown in figure.

For the assumed proportions, the retaining
wall is checked for stability against overturning and sliding.

__Stability against
overturning__:

Pa = Active pressure exerted by retained
earth on wall [both wall and earth move in same direction]

Pp
= Passive pressure exerted by wall on retained earth [both move in opposite
direction]

Ca -> same for dry and
submerged condition, si significantly.

Force due to active pressure,

P_{a} = 0.373(16)(5.786)^{2}/2 = 99.9kN [per
m length of wall]

FOS = 0.9
x Stabilising force
or moment / Destabilising force
or moment

Therefore,
FOS_{(overturning)} = 0.9*Mr / Mo *? 1.4

Overturning moment, M_{o}
= (P_{a}Cos?)h?/3 = 96.5(5.786/3) =
186.

To find the distance of
resultant vertical force from heel,

Distance
of resultant vertical force from heel,

X_{w}
= M_{w}/W = 230.6/232.9 = 0.99m

Stabilising
moment (about toe),

M_{r} = W(L -X_{w})
+ P_{a}Sin?(L) =-0 .99)232+77.69(3 = 468.1kNm [per m length of wall]

FOS(overturning) = 0.9*Mr
/ Mo = *0.9*x*(468.1 +77.6) / 186.1 = 2.26 > 1.40

__Soil
pressure at footing base__:

Resultant vertical reaction,
R = W = 232.9kN [per m length of wall]

Distance
of R from heel, L_{R} = (M_{w} + M_{o}) / R = (230.6 +
186.1)/232.9 = 1.789m

Eccentricity,
e = L_{R} -L/2 = 1.789 -3/2 = 0.289m < L/6 ->[0.5m]

Hence
the resultant lies within the middle third of the base, which is desirable.

Maximum pressure as base,

__Stability
against sliding__:

Sliding force, P_{a}Cos? =
96.5kN

Resisting force, F
= ?R =[Ignoring0 .passive5 pressurex232ontoe.side]9 =
116.4

FOS(Sliding) = 0.9*F
/ P _{a} Cos*

Hence
a shear key may be provided.

Assume a shear key 300mm x
300mm at a distance of 1300mm from toe as shown in figure.

h_{2} = 950 + 300 + 1300tan30^{o} = 2001mm

Pp = C_{p}._{e}?.(h_{2}^{2}
-h_{1}^{2})/2 = 3 x 16 x (2.001^{2} -0.950^{2})/2
= 74.11kN.

FOS (Sliding) = 0.9(116.4
+74.44) / 96.5 =1.78 >1.4
[SAFE]

__Design of toe slab__:

Assuming a clear
cover of 75mm
and 16mm ?
use

d
= 420 -75 -8 = 337mm

V_{u}
= 1.5 [112 +81.9 / 2] x(1 - 0.337) =
96.42kN [Vu is
design shear at
„d? from fac

M_{u} = 1.5{(81.9 x
1^{2}/2) + (112 - 81.9) x 1/2 x 1^{2} x

2/3} = 76.48 kNm/m length

Nominal shear stress, *t* = *V/ bd = *96.42*x*10^{3}
/ 1000*x*337 =0.286*N*
/ *mm*^{2}

Using M20
concrete,

For a *t* =0.29N/mm_{2},
p_{t} (required) = 0.2%

K = *M _{u }*

For p_{t} = 0.2%, A_{st}
= 0.2/100 x 1000 x 337 = 674 mm^{2} / m

Spacing
= ( 1000*x**p**x*16^{2} / 4 )
/ 674 =
298mm

Provide
16mm*f* @ 290mm c/c at bottom of
toe slab

Ld
= *fs *_{s }* / *4*t *_{bd}* = *(16).0.87
*f* * _{y }/ *4

Since
length available is 1m, no curtailment is sorted.

__Design
of heel slab__:

V_{u} = =128.06*kN*

M_{u}
= 1.5{(82.54 x 1.55^{2}/2) + (128.6 - 82.54) x 1/2 x 1.55^{2} x

2/3}
= 203.96 kNm/m length

Nominal
shear stress, *t* = *V / bd = *128.06*x*10^{3
} / 1000*x*337 = =0.38*N* / *mm*^{2}

Using M20 concrete,

For a *t* =0.39N/mm_{2},
p_{t} (required) = 0.3%

K = *M _{u }/ *

For p_{t} = 0.565%,
A_{st} = 0.565/100 x 1000 x 337 = 1904.05 mm^{2} / m

Spacing
= (1000*x**p**x*16^{2} / 4) / 1904.05
=
105.61mm

Provide
16mm*f* @ 100mm c/c at bottom of
toe slab

L_{d}
= *fs *_{s}*
/ *4*t *_{bd}*
= *(16).0.87 *f* * _{y} / *4

Since
length available is 1.55m, no curtailment is sorted.

__Design
of vertical stem__:

Height
of cantilever above base = 5250 -420 = 4830mm

Assume a clear cover of 50mm and 16mm? bar, d_{at
base} = 450 -50 -16/2 = 392mm

M_{u}
= 1.5(C_{a}._{e}?.h^{3}/6) = 1.5(1/3)(16 x 4.92^{3}/6)
= 150.24kNm.

K
= *Mu
/ bd ^{2} = *150.24

p_{t} = 0.3%, A_{st} =
0.295/100 x 1000 x 392 = 1200mm2. Spacing = 1000 x 201/1200 = 160mm

Provide
16mm*f* @ 160mm
c/c in the
stem, extending= int

752mm.

__Check for shear__: [at „d?
from base]

V_{u} (stem) = 1.5(C_{a}._{e}?.Z^{2}/2)
= 1.5(1/3 x 16 x (4.83 -0.392)^{2}/2) = 53.83kN

*t*_{v}*=
Vu / bd = *53.83*x*10^{3} / 1000*x*392
= =0.135*N* / *mm*^{2} <*t**c*

* *[where,_{c}=0.39N/mm?^{2}
for p_{t} = 0.3%))

Hence,
SAFE.

__Curtailment
of bars__:

Curtailments
of bars in stem are done in two stages:

At
1/3^{rd} and 2/3^{rd} height of the stem above base.

__Temperature
and shrinkage reinforcement__:

A_{st} = 0.12/100 x
10 x 450 = 540mm^{2}

For I 1/3^{rd} height, provide 2/3^{rd}
of bar near front face (exposed to weather) and 1/3^{rd} near rear
face.

For II 1/3^{rd} height, provide 1/2
the above. For III 1/3^{rd} height, provide 1/3^{rd} of I case.

Provide nominal bars of 10mm
@ 300mm c/c vertically near front face.

__DESIGN OF COUNTERFORT
RETAINING WALL__

__Preliminary
proportioning of counterfort retaining wall__:

1. Thickness
of heel slab and stem = 5% of Height of wall

2. Thickness
of toe slab [buttress not provided] = 8% of Height of wall

3. Thickness
of counterfort = 6% of height of wall

4. In
no case thickness of any component be less than 300mm.

5. Spacing
of counterforts = 1/3^{rd} to 1/2
of Height of wall

6. Each
panel of stem and heel slab are designed as two way slab with one edge free
(one way continuous slab).

7. The
toe slab is designed as

a. Cantilever
slab when buttress is not provided

b. One
way continuous slab, when buttress is provided

8. Counterfort
is a triangular shaped structure designed similar to a T-Beam as vertical
cantilever with varying depth (stem acts as flange). The main reinforcement is
along the sloping side. Stirrups are provided in the counterfort to secure them
firmly with the stem. Additional ties are provided to securely tie the
counterfort to the heel slab.

1) Design a suitable counterfort retaining
wall to support a leveled backfill of height 7.5m above ground level on the toe
side. Assume good soil for the foundation at a depth of 1.5m below ground
level. The SBC of soil is 170kN/m^{2} with unit weight as 16kN/m^{3}.
The angle of internal
friction.Thecoefficientof isfriction ?between=30thesoil and concrete is 0.5.
Use M25 concrete and Fe415 steel.

Depth of foundation = 1.5m

Height of wall =
7.5 + 1.5 = 9m

Thickness of heel
and stem =
5% of 9m
= 0.45m

Thickness of toe slab = 8%
of 9m = 0.72m

L_{min}
= 1.5 x 3 = 4.5m

Thickness
of counterfort = 6% of 9 = 0.54m

Stability Condition:

Earth pressure calculations:

Xw = 874.69/510.75 = 1.713m

FOS (overturning) = 0.9Mr/Mo

Where

Mo = P_{a}.h/3 = C_{a}._{e}?.h^{3}/6
= 0.33x16x93/6 = 647.35kNm.

Mr = (L -Xw).W = 510.75(4.5
-1.713) = 1423.6kNm.

FOS (overturning) = 1.98
> 1.4

Hence, section is safe against overturning.

__Sliding__:

FOS _{(sliding)}
= 0.9(?R)/P_{a}Cos?

F = ?R
= 0.5 x
510.75 = 255.375kN

P_{a} = C_{a}._{e}?.h^{2}/2
= 215.784

Where, L_{R} = (M + M_{o})/R,
e = L_{R} -L/2, where, L_{R} = (874.688 + 647.352)/510.75 =
2.98m

&
e = L_{R} -L/2 = 2.98 -(4.5/2) = 0.73 < L/6 ^{à}
(0.75m)

Since the maximum earth pressure is greater
than SBC of soil, the length of base slab has to be increased preferably along
the toe side. Increase the toe slab by 0.5m in length.

?W = 510.75
+ 0.5 x
25 x 0.72
= 519.75kN

Additional load due to increase in toe slab
by 0.5m is, Moment = 0.5/2 + 4.5 = 4.75m

?M = 874.69
+ 42.75 =
917.438kNm.

L_{R} = (Mo + M) / R = (917.438 +
647.352)/519.75 = 3.011m

e= L_{R} -L/2 = 3.011 -(5/2) = 0.511m
< L/6 - > (0.833m)

FOS _{(Sliding)} = 0.9(?R)/P_{a}=0.9(0.5x519.75)/215.784
= 1.08 < 1.4.

Hence the section is not safe against
sliding. Shear key is provided to resist sliding. Assume shear key of size 300
x 300mm.

P_{ps} = C_{p}._{e}?.(h2^{2}
-h1^{2})/2 = 164.5kN/m

FOS _{(sliding)} = 0.9(?R_{ps})/P_{a} =
+1.77P> 1.4 [where,
h1 = 1.2m, h2 = 1.2 + 0.3 + 1.39 = 2.88m]

Hence, section is safe in sliding with shear
key 300 x 300mm.

__Design of Toe Slab__:

Effective cover = 75 + 20/2 = 85mm

Toe slab is designed similar to cantilever
slab with maximum moment at front face of the stem and maximum shear
at„d?fromfront face of stem.

d = 720 -85 = 635mm.

M = 80.38x2^{2}/2 + ½ x 2 x 49.94 x
2/3 x 2 = 160.76 + 122.76 = 227.35kNm.

SF at 0.635m, = 49.94/2 x 0.635 = 15.606kN

Area of trapezoid = ½.h.(a + b) = ½(2
-0.635)(130.32 + 95.98) = 154.44kN

Factored SF = 231.66kN; Factored Moment =
341.02kNm.

K = M_{u}/bd^{2}
^{à} A_{st} = 1551.25mm^{2 }^{à} Spacing
= 1000a_{st}/A_{st} ^{à} 16mm @125mmc/c.

Transverse reinforcement: = 0.12% of c/s

= 0.12/100 x 1000 x 720 = 864mm^{2}

Provide 10mm @100mm c/c.

__Design of heel slab__:

The heel slab is designed as an one way
continuous slab with moment wl^{2}/12 at the support and wl^{2}/16
at the midspan. The maximum shear at the support is w(l/2 -d). The maximum
pressure at the heel slab is considered for the design.

Moment at the support, M_{sup} = wl^{2}/12
= 106.92 x 2.5^{2}/12 = 55.688kNm.

Moment at the midspan, M_{mid} = wl^{2}/16
= 41.76kNm

The maximum
pressure acting on
the heel_{st}requiredslabat

midspan and support are found.

Factored M_{sup} =
83.53kNm ^{à} Ast = 570.7mm^{2}

Factored M_{mid} =
62.64kNm ^{à} Ast = 425.4mm^{2}

Using
16mm ? bar,
Spacing ^{à}=Provide1000ast16mm@110mm/Astc/c

At midspan, spacing = 156.72mm ^{à} Provide 16mm @ 150mm c/c

Transverse reinforcement = 0.12% of c/s =
0.12/100 x 1000 x 500 = 600mm^{2}

For 8mm bar, Spacing = 83.775mm ^{à}
Provide 8mm @80mm c/c.

__Check for shear__:

Maximum shear = w (l/2 -d) = 107 (2.5/2
-0.415) = 89.345kN

Factored shear force = 134.0175kN

?v = 0.33N/mm , ?c
= 0,.29N/mm?cmax =
3.1N/mm

Depth has to be increased. __Design of stem__:

The stem is also designed as
one way continuous slab with support moment wl^{2}/12 and midspan
moment wl^{2}/16. For the negative moment at the support, reinforcement
is provided at the rear side and for positive moment at midspan, reinforcement
is provided at front face of the stem.

The maximum moment varies from a base intensity of K_{a}._{e}?.h=1/3x16x(9-0.5)=
45.33kN/m M_{sup} = wl^{2}/12 = 1.5 x 45.33 x 3.54^{2}/12
= 71kNm

M_{mid} = wl^{2}/16 = 1.5 x 45.33 x 3.54^{2}/16
= 53.26kNm Effective depth = 500 -(50 + 20/2) = 440mm

A_{st} at support = 1058mm^{2}, For
16mm ?,190mmSpacing.Provide16mm
=@ 190mm c/c

A_{st} at midspan =718mm^{2}, For 16mm
?,280mmSpacing.Provide16mm@280mm= c/c Max. SF = w (l/2 -d) = 60.29kN, Factored
SF = 90.44kN

Transverse reinforcement = 0.12% of c/s ^{à} 8mm @ 80mmc/c

?_{v} = 0.188N/mm^{2},_{c}
=?0.65N/mm^{2},_{cmax}?= 3.1N/mm^{2} Safe in Shear. __Design
of Counterfort__:

The counterfort is designed as a cantilever beam whose
depth is equal to the length of the heel slab at the base and reduces to the
thickness of the stem at the top. Maximum moment at the base of counterfort, M_{max}
= K_{a}._{e}?.h^{3}/6 x L_{e}

Where, L_{e} ^{à} c/c
distance from counterfort

M_{max} = 1932.5kNm, Factored Mmax = 2898.75kNm

A_{st} = 2755.5mm^{2}, Assume 25mm?bar,
No. of bars required = 2755.5/491.5 = 5.61 ~ 6 The main reinforcement is
provided along the slanting face of the counterfort. __Curtailment of
reinforcement__:

Not all the 6 bars need to be taken to the
free end. Three bars are taken straight to the entire span of the beam. One bar
is cut at a distance of, *n* -1/8.5 = h1^{2}/8.5^{2} , where
n is the total number of bars and h1 is the distance from top. *N*

When n = 6, h1 = 7.75m [from bottom]

The second part is cut at a distance of,

*n *-2/* n = h*2/8.5^{2} , h2=6.94m [from bottom]

The third part is cut at a distance of,

*n *-3/* n = h*3^{2}/8.5^{2} , h3=6. 01m [from bottom]

Vertical ties and horizontal
ties are provided to connect the counterfort with the vertical stem and the
heel slab.

__Design of horizontal ties__:

Closed stirrups are provided
to the vertical stem and the counterfort. Considering 1m strip, the tension
resisted by reinforcement is given by lateral pressure on the wall multiplied
by contributing area.

T = Ca.e?.h x h, where, Ast = 1/ 0.87_{
fy}

T = 1/3 x 16 x (9 -0.5) x 3.54 = 160.48kN Factored force,
T = 1.5 x 160.48kN

A_{st} = 666.72mm^{2}. For 10mm ?,
Spacing110mm. =

Provide 10mm@110mm c/c closed stirrups as horizontal
ties. __Design of vertical ties__:

The vertical stirrup
connects the counterfort and the heel slab. Considering 1m strip, the tensile
force is the product of the average downward pressure and the spacing between
the counterforts. T = Avg(43.56 & 107) x Le = 266.49 kN

Factored
T = 399.74 kN

Ast = 1107.15mm^{2}. For 10mm
?, Spacing70.93mm. =

Provide 10mm @ 70mm c/c.

Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail

Civil : Design Of Reinforced Concrete And Brick Masonry Structures- Retaining Walls: Design of cantilever and counter fort retaining walls : Design Of Retaining Walls |

**Related Topics **

Privacy Policy, Terms and Conditions, DMCA Policy and Compliant

Copyright Â© 2018-2023 BrainKart.com; All Rights Reserved. Developed by Therithal info, Chennai.