Counter+Fort++Reatining+WAll+With+HORZONTAL+Bach+Fill (1)

7/28/2019 Counter+Fort++Reatining+WAll+With+HORZONTAL+Bach+Fill (1)

1/35

RCC design B.C.Punmia

18.2 TYPE OF RETAINING WALLS

1 Gravity walls

2 Cantilever retaining walls a. T- shaped b. L- shaped

3 Counterfort retainig walls.

4 Buttresssed walls.

The cantilever retaining wall resist the horizontal earth pressure as well as other vertical pressure by w

A gravity retaining wall shown in fig 1 is the one in which the earth pressure exrted by the

back fill is resisted by dead weight of wall, which is either made of masonry or of mass concrete . The

stress devlop in the wall is very low ,These walls are no proportioned that no tension is devloped any

where, and the resultant of forces remain withen the middle third of the base.

A retaining wall or retaining structure is used for maintaining the ground surfgaces at defrent

elevations on either side of it. Whenever embankments are involed in construction ,retaining wall are

usually necessary. In the construction of buildins having basements, retaining walls are mandatory.

Similsrly in bridge work, the wing walls and abutments etc. are designed as retaining walls , to resist

earth pressure along with superimposed loads. The material retained or supported by a retaining wall

is called backfill lying above the horizontal plane at the elevation of the top of a wall is called the

surcharge , and its inclination to horizontal is called the surcharge angle b

In the design of retaining walls or other retaining structures, it is necessary to compute the

lateral earth pressure exerted bythe retaining mass of soil. The equation of finding out the lateral earth

pressure against retaining wall is one of the oldest in Civil Engineering field. The plastic state of

strees, when the failure is imminent, was invetigated by Rankine in1860. A Lot of theoretical

experiment work has been done in this field and many theory and hypothesis heve benn proposed.

RETAINING WALL

Retaining walls may be classified according to their mode of resisting the earth pressure,and

according to their shape. Following are some of commen types of retaining walls (Fig)


2/35

y of beending of varios components acting as cantilever s.A coomon form of cantilever retaining waal


3/35

Hieght of cantilever wall from ground level = 7.00 mUnit weight of Earth = 18 KN/m

3

Angle of repose = 30 DegreeSafe Bearing capacity of soil = 180 KN/m

3

Coffiecent of friction = 0.5Concrete M- 20

wt. of concrete = 25000 N/m

3

Steel fe 415 N/mm sst 230 N/mm2m 13.33 scbc 7 N/mm2

Nominal cover = 30 mmFoundation depth = 1.00 m

Counter forts width = 0.50 m

Stem thickness At footing #REF! mm At top #REF! mm

Heel width 2000 mm Toe width 1700 mm

Footing width 4100 mm Key #REF! x #REF! mm

STEM : -

Main#REF! 10 mm F@ 90 mm c/c

#REF! 10 mm F@ #REF! mm c/c

Top 10 mm F@ #REF! mm c/c

Distribution 8 mm F@ #REF! mm c/c

Tamprecture 8 mm F@ #REF! mm c/c

TOE : -

Main 10 mm F@ 100 mm c/c

Distribution 8 mm F@ 0 mm c/c

HEEL : -

Main 10 mm F@ #REF! mm c/c

Distribution 8 mm F@ 180 mm c/c

DESIGN SUMMARY

100% Reinforcement upto m



DESIGN OF COUNTOR FORT RETAINING WALL with

horizontal back fill

Reinforcement Summary

[email protected]


4/35

mm F

mm F @ c/c

@ c/c

mm F

mm F @ c/c

@ c/c

mm F

mm F @ c/c

@ c/c

mm F

@ c/c

mm F

@ c/c

mm F

@ c/c

mm F @ c/c mm F

@ c/c

## F mm F

@ c/c

mm F

@ c/c

mm F

@ c/c

1700

4100

####

####

#REF!

####

####

####

####

#REF!

####

1700

Toe

####

####

#### ####

Out side

####

####

Earth side

####

####

2000

400

10

####

####

####

10

####

10

####

####

####

####

####

Heel

2000

####

####

####

####

####

####

400

####

####

#REF!

####

####

####

####

#REF!

####

####

0

7000

[email protected]


5/35

Hieght of cantilever wall from ground level = 7.00 m

Unit weight of Earth g = 18 kN/m = N/m

Angle of repose = 30 Degree

Safe Bearing capacity of soil q0 = 180 kN/m

Coffiecent of friction m = 0.5

Concrete = M 20Steel fe = 415

Nominal cover = 30 mm effective cover 40 mm

Foundation depth = 1.00 m

Counter forts width assume = 0.50 m 500 mm

1 Des ign Constan ts :For HYSD Bars = 20

sst = = 230 mm = #### N/mm2 = 25 kN/mm2scbc = = 7 mm

m = 13.33

x

13.33 x 7 + 230

j=1-k/3 = 1 - 0.289 / 3 = 0.904R=1/2xc x j x k = 0.5 x 7 x 0.904 x 0.289 = 0.913

1 - 0.5 1

1 + 0.5 Ka

2Diamension of var ious parts : -

= 7.00 + 1.00 = 8.00 m

The ratio of length of slabe (DE) to base width b is given by eq.

q0

2.2 y H 2.2 x 18 x 8.00

Keep a = 0.43 Eq (1)The width of base is given by Eq.

( 1 - 0.43 )x( 1 + 1.29 )

x 8.00 x 0.33

( 1 - 0.43 )x 0.5

0.5 b = 0.50 x 8.00 = m

Hence Provided b = m

Width of toe slab = a x b = 0.43 x 4.00 = 1.72 m m

taking the uniform thickness of stem = 300 mm = 0.30 m

= 4.00 - 0.30 - 1.70 = 2.00 m

Let the thickness of base slab = 300 mm = 0.3 m

H 1/4 8.00 1/4

y 18

b

6.55

This width is excessive. Normal practice is to provide b between 0.5 to 0.6 H .

m

0.33

Provided toe slab =

3.50

The wall will be unsafe against sliding. This will be made safe by providing a shear Key at base .

1-sin F

1+sinF

=

for design purpose

Kp= =

180=

== 30 \ 3

for design purpose

3.84

.

m

1.70

m

3.5 x x

4.00

4.00

Hence width of heel slab

0.7HKa 0.7

=

Taking maximum value of H =

b =

2.86=

Ka

m*c+sst

=x

(1-a) m=

0.432

=

0.95 H

= 0.95

The base width from the considration of sliding is given by Eq.

b

m*c

=

Ka = =For soil, F

DESIGN OF COUNTERFORT RETAINING WALL

Cocrete M

wt. of concrete

13.33 7

k=

18000

= 0.289

a = 1 - = -1

8.00

(1- a)x(1+3 a)

0.333

Hence height of wall above base H

x

Clear spacing of counter fort =

[email protected]


6/35

keep them at 3.00 m apart. Let us provide counterfort over toe slab, upto ground level at

3.00 m clear distance.

3

Full dimension wall is shown in fig 1a

Let w1 = weight of rectangular portion of stem

w2 = weight of base slab

w3 = weight of soil on heel slab.

The calculation are arrenged in Tableforce(kN) Moment about toe (KN-m)

w1 1 x 0.30 x 7.70 x 25 =

w2 1 x 0.30 x 4.00 x 25 =

w3 1 x 2.00 x 7.70 x 18 =

Sw =

Total resisting moment = kN-m ..(1)

0.33 x 18 x( 8.00 )2

2 2

8

3

mSw 0.50 x 364.95

PH

- 512 = kN-m

\ Distance x of the point of application of resultant, from toe is

SM 486.44 b 4.00

Sw 364.95 6 6

b 4.00

2 2SW 6 e 364.95 6x 0.67 183 > 180

b b 4.00

m, So that total wiodth is = 4.10

force(kN) Moment about toe (KN-m)

w1 1 x 0.30 x 7.70 x 25 =

w2 1 x 0.30 x 4.10 x 25 =

w3 1 x 2.00 x 7.70 x 18 =

Sw =

- = kN-m

\ Distance x of the point of application of resultant, from toe is

SM 523.00 b 4.10

Sw 365.70 6 6

b 4.10

2 2

SW 6 e 365.70 6x 0.62 170.1 < 180

b b 4.10

SW 6 e 365.70 6x 0.62 8.30 < 180

b b 4.10

1.50.95

kN -m2

==

=

Hence safe=4.10

Pressure p2 at Heel

The revised computations are arranged in table

Detail

1 + =

0.67

4.00

1.80

859

lever arm

58 1.95 113

31

277

1035.00

2.05 63

0.667

486.44

> 2Hence

safe

3.1

total MR365.70

x = = =

512.00 523.00

= 0.683

= Hence safe

Eccent ic i ty e =

kN -m2

< 0.683

Pressure p1 at toe= 1 +

Hence safe

[email protected]


7/35

170.10 - 8.30

170.10 - 8.30

4

= 3.00 m

= 8.30 kN/m2

witch is minimum at C.

= 7.70 x 1 x 1.00 x 18 = kN-m2

= 0.30 x 1 x 1.00 x 25 = kN-m2

= 138.60 + 7.50 - 8.30 = N-m2

Maximum negative bending moment in heel slab.at counter fort is

Pl2 137.8 x 3.00

12

BM x 106

Rxb x

137.8 x 3.00

= 0.72 % tc = 0.33

V 207 x 1000

tc x b 0.33 x 1000

500 60 mm d = 500 - 60 = 440 mm

207 x 1000

1000 x 440

x

230 x 0.904 x 440

3.14xdia2 = 3.14 x 12 x 12 = 113 mm2

4 4

\ Spacing A x1000 / Ast = 113 x 1000 / 1130 = mm

Hence Provided 12 mm F bar, @ mm c/cLet us check this reinforcement for development length at point of contraflexure is situated at distance

of 0.211.L In over case, the slab is continuous, but we will assume the same position of contraflexure

i.e. at 0.211 x 3.00 = 0.63 m from the face of conunterforts.

pL l L l

2 2 2 2

3.00

2

Assuming that all the bars will avilable at point of contraflexure,

M = sst x Ast x j x d = 230 x 1130 x 0.904 x 440 = N-mmLo = witch ever is more = 440 mm

Ld = 45x F = 45 x 12 = 540 mm

M

V

440 mm beyond the point of contraflexure. After that, curtail

length of each bar available = 630 + 440 = 1070 > Ld = 540 m

PL2

3

16 4

3 3

4 4

M1

mm providing effective cover =

Hence depth required from shear point of veiw d

However keep =

=

These bars will be provide at the top face of heel slab. Maximum Passive B.M. =

- x

or d,

0.33

0.63

100

tv = N/mm2= 0.47 Shear reinforcement required

2'=

Effective depth required =

207 kN

N/mm2

= 626 mm=

1.80

119886

=

Design of Heel slab:-

The Pressure intencity p1 under E is p1

=

2

this is

excessive

103.35 kN-m

=

p1 =

For balance section , having P

mm0.913

>

=103.35

1000

Shear force V

336

Area of steel at supports is given by Ast

Shear force at this point is given

= 137.8

\ + Lo

99.07

138.6

87.20

=

The pressure distribution on the heel slab is shown in fig 1b .consider a strip 1 meter

170.10

137.8

170.10 -4.10

x

Down ward weight of slab per unit area

kN-m2

The Pressure intencity p2 under B is p2

p

Hence net pressure intensities will be P

- x

7.5

Clear spacing b etween co unter fort

kN-m2

wide.Near the outer edge C. The upward pressure intencity

Down ward load due to weight of Earth.

=M1

4.10=2.10=

x

Hence safe

N

=

- x + = p

=

=

12

-

100

mm210'

6103.35

=

=

103350000

+119886

103350000

using 12 mm bars

1130

A =

=

x 1130

540>1302

Hence safe

= mm2847

=

=

440

Cotinue these bars by a distance lo = d =

half bars, and continue the remaining half throughout the length. At the point of curtailment,

mm

\ Area of Bottom steel Ast2 = x Ast1

[email protected]


8/35

3.14xdia2 = 3.14 x 12 x 12 = 113 mm2

4 4


Hence Provided 12 mm F bar, @ mm c/c

1000 x 113

Let us check this reinforcement for development length crierion at point of contraflexur,

M Where V = Shear at point of contraflexure= N

V = 0.63 mAssuming that all bars are available at point of contraflexure,


Ld = 45x F = 45 x 12 = 540 mm as before

M

V

Thus continue all bottom bars to a point distance Lo = 440 mm from the point of contraflexure,

630 - 440 = 190 mm from the center of sports.

At this point half bars can be discontinued. Since this distance is quite small,

it is better to continue these bars upto center of counterfors.

Reinforcement near B :- The c/c spacing of reinforcement near B may be increased, because P decrease

due to increase in upward soil reaction. Consider a strip 1 m wide near B

kN/m2 As found earlier.

weight of earth + weight of counterforts - upward soil reaction

\ Net downward load p = 138.60 +( 0.50 x 25 ) - 87.2 = 63.9 kN/m2

This is about = 63.90 / 137.8 = 0.464 of load intencity at C

= 100 / 0.464 = 200 mm c/c at the top face, near supports

Spacing of steel bars at the bottom face, at mid span= 133 / 0.464 = 300 mm c/c

0.12

100

P D2

3.14 x ( 12 )'2

41000 x 113

shear stress at C = tv = 0.47

100 x 1130

1000 x 440

0.26 % = 0.21 (See Table 3.1)

Safe if tv< tc Here 0.47 > 0.21 Hence shear reinforcement required

Vc = tc b x d= 0.21 x 1000 x 440 = #### N or 92.4 kN

= 92.4 kN

92.4 1.50 - x1

207

= 0.80 m on either side of each counterforts.

The requirement is there form a strip of unit width paassing through C,

such that shear force at the counterforts isd = 92.4 kN

Net down ward pressure at C = 137.80 kN/m2, 63.90 kN/m

2

Lt net down ward pressure at B1=w1 x 3/2=1.5w1 This is equal to = 92.4 kN

92.40

1.50

137.80 - 6461.6= 137.8 =y1 - 36.95 Y1w1= -137.80

2.00

1.50

Net down ward pressure at B =

w= = 61.6 kN/m2 However at Y1 from C,

0.80 mThe position is given by= =

Permissible shear stress for

Hence shear stirrups are required upto distance

steel provided tc

=

500

Shear reinforcement.

=

\ Spacing = = 188

\ +Lo>Ld

i.e. upto a distance =

119886+ 440

+Lo>LdInherent in criterion :

>

Distribution steel = x 1000

Hence spacing of steel bars

Net downward load p' =

x

=79531705

119886

Distance from face of supports

Hence safe5401103=

=130

=

79531705

870 mm2

using 12 mm bars A =

% of steel provided =

Actual Ast

mm2

133

130

or d,

600

Upward soil reaction at B is = 87.2

mm2113

N/mm2

or x1 -

Consider a section distance x1 from face of counterfort, where shear force is

=1.50

=

Using 12 mm F bars, Area ==

600

=

4

mm say = 180 mm c/c

0.70

0.26 %

[email protected]


9/35

\ y1 = 2.10 m

Hence shear reinforcement is required in triangular portion on the other side of counterforts shown hatched in fig .

However, we will provide shear strirrups in reangular portion x1 x y1= 0.80 x 2.10 = 1.68 m on

either side of counterforts.

Let us provide 4 8 mm F wire

P D2

3.14 x ( 8 )'2

4

201 x 230 x 440

207 - 92.40 )x 1000Hence provided the 8 mm F 4 lgd strirrups @ 170 mm c/c either side of each counterforts.

5Design of toe slab :- Since the toe slab is also large, provide counterforts over it, upto ground level at

3.00 m clear distance face to face. The toe slab will thus bend like a contious slab.

= 500 mm or 0.5 m

= 0.50 x 1 x 1 x 25 = kN//m2

= 170.10 - 12.50 = kN//m2

= 99.07 - 12.50 = kN//m2

Cosidering strip of unit width at D.

wl2 157.60 x 3.00

2

12

BM x 10 6

Rxb x

157.6 x 3.00

= 0.30 N/mm2 assuming % steel 0.5 % table 3.1

236 x 1000

1000 x 0.300

500 60 mm d = 500 - 60 = 440 mm

x230 x 0.904 x 440

3.14xdia2 = 3.14 x 12 x 12

4 4


Hence Provided 12 mm F bar, @ 80 mm c/c

1000 x 113


M

V

Hence shear force at the point of contraflexure is V =

w L 3.00

2 2

M = 230 x 1413 x 0.904 x 440 = NmmLo = witch ever is more = 440 mm

Ld = 45x F = 45 x 12 = 540 mm

M

V

Hence satisfied , continue these bars, at the bottom of toe slab, beyond the point of contraflexure

440.0 mm i.e. by a distance of 630 + 440.0 = 1070 mm

mm359.8

1292

mm2=

mm=

=

m

178Sv=Asv.ssv.d

V - Vc

=

= = 201 mm2

4

legged stirrups of

Using 8 mm F bars, Area =

1000

The depth of slab required from shear point of view is given by d= V / (b x tc)

d =

=

Total weight of toe slab

=

= 236.4 kN

Taking a permissible stress tc

kN/m2

Similarly Net upward pressure intencity at E

= 10'6

86.57

118.2 x

-

10'6

This is excessive ,However we will keep the same depth as that of heel


Max. negative B.M.

Shear force V =

Assume total depth of toe slab

Net upward pressure intencity at D 157.60

12.50

2

118.20

0.913

12=

mmArea of steel at supports, at bottom is Ast =

=

However keep =

using 12 mm bars A = 113

118.20

1413 mm2

mm providing effective cover =

and provide shear strirrups to take up excessive shearing stress.

788 mm

N

Inherent in criterion :

or d,

\ +Lo>Ld =129239020.21

by a distance of Lo=

> 540 Hence safe137112

+ 440 = 1383

+Lo>Ld

87.5

Actual Ast = =80

0.63=Where the point of contraflexure occure at

distance x rom supports

V =

129239020

137112.0- x 157.6= x 0.63 =

[email protected]


10/35

from the face of counterforts

3 3

4 4

3 3

4 4

3.14xdia2 = 3.14 x 12 x 12 = 113 mm2

4 4

\ A x1000 / Ast = 113 x 1000 / 969 = mm

Hence Provided 12 mm F bar, @ mm c/c1000 x 113


M

V

Assuming that all bars provided at top face,are available at point of contraflexure,


Ld = 45x F = 45 x 12 = 540 mm

M

V

Thus continue all bottom bars to a point distance Lo = 440 mm from the point of contraflexure,

630 - 440 = 190 mm from the center of sports.

At this point half bars can be discontinued. Since this distance is quite small,

it is better to continue these bars upto center of counterfors.

Reinforcement at E :- At a section distance 1 meter from E,

170.1 - 8.30

\ Net upward pressure = 138.5 - 12.50 = kN/m2

This is about = 126.0 / 157.60 = 0.80 ofw at D

\ Spacing of bottom steel = 87.5 / 0.80 = mm say = 100 mm

Spacing of top steel = 117 / 0.80 = mm say = 140 mm

0.12100

P D2

3.14 x ( 12 )'2

4

1000 x 113

Shear reinforcement shear force at D = kN

236.40 x 1000

Beam Ht.x beam wt. 1000 x 440

100 x 1413 table

1000 x 440 3.1

0.32 % = 0.24 (See Table 3.1)

Safe if tv< tc Here 0.54 > 0.24 Shear reinforcement required

Vc = tc b x d= 0.24 x 1000 x 440 = N or 105.6 kN

= 105.6 kN

105.6 1.50 - x2

236.40

= 0.80 m on either side of each counterforts.

The requirement is there form a strip of unit width paassing through D, Let us consider a strip through E1,

distance y2 from D, such that shear force at the counterforts is kN. To find the position ofY2

consider the net pressure distribution below the toe.

= 969

> 540 Hence safe

as before

N

137112

or d,

93992015+ 440 = 1126

110

137112

236.4

146

600

x 0.80 =

x

170.1

mm2

93992015

mm2

N-m

mm2

Where V = Shear at point of contraflexure=

kN/m2

126.0

109

-4.10

Inherent in criterion : +Lo>Ld

i.e. upto a distance =

upward soil pressure =

117

110Actual Ast = = 1028

= x

\ +Lo>Ld =

Spacing =

using 12 mm bars A =

\ Area of Bottom steel Ast2 = x Ast1 = x 1292

106x 118.20 x 10'

6

0.70


1.50=

Hence shear stirrups are required upto distance

N/mm2

N/mm2

The position is given by= = or x2 = 1.50

180

= x 500

-

4= 113

600=

Permissible shear stress tc for

% of steel provided =

Consider a section distance x1 from face of counterfort, where shear force is

105600

tc

shear force=

N/mm2

=

=

Shear stress tv =

\ Spacing = 188 mm say

=

Distribution steel

\ =

1000

steel provided tc

= 0.32 %

= 0.54

138.50

mm c/c

0.24

0.80 m

mm2

105.6

88.7Again, positive B.M. x M1 =

[email protected]


11/35

= 12.50 Hence net pressure intencity below D an dE are

below D = 170.1 - 12.50 = 157.6 kN/m2,

and below E 99.1 - 12.50 = 86.6 kN/m

Let the net pressure intencity at E1 = w2 x 3/2 = 1.5 w2 kN/m2

\ Shear force at the counterforts at E1 = w2 x 3/2 =1.5m w2 kN/M2

This is equal to = 105.6 kN

105.60

1.50

157.60 - 86.6

= 157.60 - 39.46 Y2 = \ y2 = 2.20 mThis is > than DE DE = 1.80 m

= kN/m2

1.50 x 86.57 = kN/m2

Considered a section distence Z from the face of dounterforts (Point E), where S.F. is 105.6 kN

105.60 1.50 - Z

129.85

= 0.30 m only.

However, we will provide shear strirrups for whole of rectangular area (shown dotted),

for width DD 1 ,= x2 = 0.80 m and length DE = 0.30 m

Let us provide 8 8 mm F wire

P D2

3.14 x ( 8 )'2

4

402 x 230 x 440

236.4 - 105.60 )x 1000

Hence provided the 8 mm F 8 lgd strirrups @ 300 mm c/c either side of each counterforts.

6Design of stem (vert ical slab)

The stem acts as a continuous slab. Considred 1 m strip at B .

The intencity of earth pressure is given by. ph = KayH1 = 0.33 x 18 x 7.50 = 45.0 kN/m2

Hence revised H1= 8.00 - 0.50 = 7.50 m

L2 45.00 x( 3 )2

12

BMRxb x

60 mm, so total depth = 192 + 60 = 252 mm

300 mm and effective thickness = 300 - 60 = 240 mm

this increased thickness will keep the shear stress within limit so that additional shear

45.0 x 3 67.50 x 1000 N

1000 x 240 mm2

this is less than = tc = 0.3

230 x 0.904 x 240

Reinforcement corresponding to p = 0.50 %

0.50 x 1000 x 240

P D2

3.14 x ( 12 )'2

4

1000 x 113

113 100 x 1256

90 1000 x 240%1000 = = 0.52

0.28

N/mm2

at 0.5% reinfocement (see Table 3.1)

kN \ =

=

mm c/c

= mm2

=

=

=

=0.913

Shear force V =2

=


w2= = 70.4

reinforcement not required.

Providing effective cover =

However provide total depth d =

12

Sv=

Actual shear forceat E =

=

(2)

Equating the two we get,

1.80Y2157.60However at Y2 from D, w2=

70.4

39.46157.6

mmV - Vc

Asv.ssv.d= =

= 402 mm2

4

129.8

The position of Z is given = 1.22

-

=1.50

=

- y2 =

1.50 -

is = pbd/100

100

=

mm say =

4

90

kN/m2

\

Hence shear stirrups are to be procided for a region DEE2D 1 , where EE2

B.M. ph x

=

legged stirrups of

311

=

mm2

Area of steel near conuterforts is

Actual AS provided= x 1256

Using 12 mm F bars, Area

1200

=

Spacing = = 94

m

N-mm

.(1)

= 0.30

= 677

113

mm2

= 1200

100As

bdmm

2and

=

Self weight of toe slab

105.6Hence shear force at E is more than

Using

or Z

8 mm F bars, Area

=

33750000

=

=

33750000

33750000 mm192

tc

1000

67.5

[email protected]


12/35

Let us check these bars for devlopment length, near points of contraflexure, so as to satisfy the criterion

M

V

0.211 x 3.00 = 0.63 m from the face of counterforts ,

shear force at this point given by V= pL/2(l/2-x)+(L/2)

pL l L l 3

2 2 2 2 2

= 58.7 kN

M = = 230 x 1256 x 0.904 x 240 = N-mmLo = 12 F or D , whichever is more = 240 mm

Ld = 45 F = 45 x 12 = 540 mm

M

V 58.70 x 1000

It is thus essencial to continue all the bars upto a point distance= 240 mm beyond

240 + 630 = 870 mm say = 900 mm from the

face of counterforts. These bars are to be provided at the inner face of stem slab.

3 x M1 3

4 4

3x Ast 3

4 4

P D2 3.14 x ( 12 )'2

4

1000 x 113

113 100 x 942

120 1000 x 942

Let us check these bars for devlopment length, near points of contraflexure, so as to satisfy the criterion

M

V

M = = 230 x 942 x 0.904 x 240 = N-mm

Lo = 12 F or D , whichever is more = 240 mm

Ld = 45 F = 45 x 12 = 540 mm V = 58.7 As before\ M

V 58.70 x 1000

The spacing of reinforcement at B , found above can be increased with height .

The pressurep h and hence the bending moment decreaases linearly with height.

Hence the spacing of bars can be increase gradually to say 300 mm c/c near top.

0.12

100

P D2

3.14 x ( 10 )'2

4

1000 x 79

7Design of main counterfort.

Let us assuming thickness of counterforts is = 500 mm. The counterfort will thus

be spaced @ 300 x 50 = 350 cm c/c. They will thus receive earth pressure from

3.5 3.5 m

At any section at depth h below the top A, the eerth pressure acting on each counter forts will be

1

3

62661343

1307 mm > 540 Hence safe+ Lo =62661343

sst xAst x jc xd

- x

240 =

= 67.5=

= + Lo > Ld

942

\ Ast

\ Spacing = = 120

540

=

mm2

+

For fixed beam or slab carrying U.D.L. , the point of

contraflexure is at a distance of 0.211 L

x - 0.63

+

Spacing =

= x

+46996007

a width of m and down ward reaction from heel slab for width of

240

79

1041

= 200

contraflexure point =

point of contraflexure, i.e. upto a point

Maximum positive B.M. = =

V = - x

Assuming that all the bars will be available at the point of contrflexure,

p x

N-mm

Area of steel = = = 1256 =

25312500x 33750000

942mm say = 120 mm c/c

Using 12 mm F bars, Area = = = 113 mm2

4

Actual AS provided= 1000 x = 942 mm2

and100As

= = 0.1 %bd

= + Lo > LdAssuming that all reinforcement is extended upto poin

of contraflexure.

mm2

4

sst xAst x jc xd 46996007

= 360

Using 10 mm F bars, Area = = =

mm2

Lo mm >=+ Hence safe

h

300

=

\ Distribution reinforcement = x 1000 x

mm c/c360

mm say= 218

x 3.5 = 21 h kN/m18.00 x h

[email protected]


13/35

similarly, net down ward pressure on heel at c is

= 7.50 x 18 + 0.50 x 25 - 8.30 = kN/m2

and that at B = 7.50 x 18 + 0.50 x 25 - 87.2 = kN/m2

Hence reaction transferrred to each counterfort are will be

At C, = 139.2 x 3.50 = kN/m

At B, = 60.3 x 3.50 = kN/m

The variations of horizontal and vertical forces on counterfort are shown in fig.

The critical section for the counterfort will be F, since below this, enormous depth will be available to resist bending.

7.00 m is = 21 x 7.00 = 147 kN/m1

2

B.M. = 514.5 x 7 / 3 = kN-m or N-mm

Conterforts act as a T beam. However, even as a reactangular beam, depth required

BM

Rxb x



Angle F of faceAC is given by Tan F = 2.00 / 7.50 = 0.3 \ F = 15

\ sin F = 0.259 and cos F = 0.966

Depth F1 C1 = AF1 sin F = 7 x 0.259 = 1.82 m or mm

\ Depth FG = 1820 + 300 = 2120

Asssuming that the steel reinforcement is provided in 2 layer with mm space

between them and providing a nominal cover 30 mm and main bars of mm F dia

the effective depth will be = 2120 -( 30 + 20 + 12 + 10 = 2048 mm

230 x 0.904 x 2048

3.14xdia2 = 3.14 x 20 x 20

4 4

\ No. of bars 2820 / 314 = 9 2 layers

M d 2048

d' cos F 0.966

500 x 2120

100xAs 100 x 9 x 314

b xd x

the height h where half of the reinforcement can curtailed will be equal to HH = 8.00 = 2.8 m

below A, i.e. at point H. To locate the position of point of curtailmenton AC, draw Hl parallel to FG.

mm

beyond l, i.e. extented upto l1. The location of H corresponding to l1 can be locate by drawing line l1H1

parallel FG. It should be noted that l1G should not less than 45 F = 900 mm similarly, other bars can be

curtailed, if desired,

Design of Hor izontal t ies:-

The vertical stem has a tendency to saprate outfrom the counterforts, and hence

should be tie to it by horizontalties. At any depth h below the top, force causing sepration

1

3

here h = 7 \ force = 18 x 7 = 126 kN/m

x 18 h kN/mh x 3.00 = 18



Effective shear force

60.3

139.2

487.20

211.05

Pressure intencity at h =

Shear force at F = x 147 x 7.00 = 514.5 kN

1820

1200.5 1200500000

Effective depth required = =1200500000

= 1622 mm0.913 500

20

20

Area of steel at supports, at bottom is Ast = 2820 mm

using 20 mm bars A = = 314

1200500000=

=2120 mm

mm2

No. provode these in

where d' = ==

361606.1321\ Effective shear force = 514500 -

-Q tan F

N

tv = N/mm2

2120x 0.27 =1200500000

\361606.1321

= 0.341138

= 0.3 %

240Thus half bars can be curtailed at l. However these should be extent by a distance 12 F =

area of steel =500 2120

thus the shear stress tv is more than permisssible shear stress tc. However, the vertical

and horizontal ties provided in counterforts will bear the excess shear stress.

\ tc = 0.21 N/mm2

[email protected]


14/35

126 x 1000

2x3.14

4

1000 x 157

200 300 mm at top

Design of vert ical t ies:-

Similar to the stem slab, heel slab has also tendency to seprate out from counterforts,due to net down ward force, unless tied properly by vertical ties.

487.20 x 3.00

211.05 x 3.00

417.6 x 1000

2x3.14

4

1000 x 226

180.9 x 1000

2x3.14

4

1000 x 226

120 280 mm

8Design of front counter for ts :-

170.1 99.07 kN/m2

at E.

500 mm thick toe slab = 0.5 x 25 = kN/m2

= 170.1 - 12.5 =kN/m

= 99.07 - 12.5 = kN/m2

The center to center spacing of counterforts, 500 mm wide is 3.50 m.Hence upward force transmitted

to counterforts at D 157.6 x 3.50 = kN/m and at E x 3.50 = 303 kN/m

The counterforts bent up as cantilever about face FE. Hence DF will be in compression while D 1 E1 will be

in tension, and main reinforcement will be provide at bottom face D 1 E1

( 551.6 + 303.0 )x

303 + 2 x 551.6 1.80

+ 3

\ B.M. = 769 x 0.99 = KN-m OR N-m

500 x 0.913



Thus project the counterforts 400 mm above groud level,to point F1 as shown in fig 4

230 x 0.904 x 1320

P D2

3.14 x ( 25 )'2

4

2767 provide these in one layer and continue by a=

=Using 25 mm F bars, Area = = 491 mm2

4

Area of steel near conuterforts is =759000000

= 2767 mm2



759 759000000

\ d =

759000000

= 1289 mm

551.63030.99 m from Eacting at x = x =

and at E 86.6

551.6

1.80 = 769

Down ward weight of 12.5

hence net w at D 157.6

Refer fig 1 The upward pressure intencity varies from kN/m2

at D, to

)2= 226 mm

2

kN/m

86.6

2.0Total upward force

mm

mm c / c at C toThus the spacing of vertical tie can be increase gradually from

\ spacing of ties = = 287 mm say 280

legged ties, As = x( 12

787

using 12 mm f 2

mm

steel required at B = = 787 mm230

mm1816 say 120\ spacing of ties = = 124

12 )2= 226 mm

2

1816 mm230

using 12 mm f 2 legged ties, As

steel required at C =

= x(

3.50417.6The down wars force ar C will be

The down wars force ar B will be 180.9

Near end C, the heel slab is tied to counterforts with the help of main reinforcement of counterforts.

kN/m3.50

548=spacing

however provide mm c/c at bottom and gradualy increase to

= 286 mm

=\ steel required

x( 10using 10 mm f 2 legged ties, As =

230

)2= 157 mm

2

= mm548

kN/m

see fig.

=

[email protected]


15/35

491

M 0.90

d' 1.80

759 0.90

1.32 1.80

482 x 1000

500 x 1320

100xAs 100 x 6 x 491

b xd xtv > tc shear reinforcement is required

2x3.14

4

Vc = tcxbxd = 0.280 x 500 x 1320 =

V1 = V - Vc = - = N

230 x 226 x 1320

230 mm c/c provide 2 x 12 mm f bars on top for holding.

9Fixing effect in stem, toe and heel :-

At the junction of stem, toe and heel slab fixing moment are included,which

normal reinforcement given below may be provided.

(I) In stem@ 0.8x0.3 =0.24% of cross section, to be provided at inner face,

in vertical direction,for a length 45 F

0.24

100

P D2

3.14 x ( 10 )'2

4

1000 x 79

Length embedment in stem, above heel slab = 45 x 10 = 450 mm

(II) In toe slab @ 0.12% to be provided at the lowae face

0.12100

P D2

3.14 x ( 10 )'2

4

1000 x 79

Length embedment in stem, above toe slab = 45 x 10 = 450 mm

(III) In heel slab @ 0.12% to be provided in upper face

0.12

100

P D2

3.14 x ( 10 )'2

4

1000 x 79

Length embedment in stem, above heel slab = 45 x 10 = 450 mm

Each of above reinforcement should anchored properly in adjoining slab, as shown in fig 5

10. Design o f shear key:-

However providethese @

and tv = =

\ svVs

ssv.Asv.d= =

296700

0.730

= x =769 - kN

N/mm2

481500

% \ tc

12

= 0.280area of steel = = 0.45

500 1320

2 = x(

since

using 12 mm f

N

481500 184800 296700

)2= 226 mm

2

184800

table

3.1

legged ties, As

231

distance of 45 F beyond E

Effective shear force = Q - tan F

\

V

mmsubject to a maxi.

300 mm

N/mm2

From fig 4 tanF =

481.5

=

are at right angles to their normal direction of bending. These moment are not determine , but

Using 10 =

= 720=Ast x 1000

500 =

\ mm2x 300

= x 1000 x

mm F bars, Area = mm2

4= 79

= 100Spacing = = 109720

mm say mm c/c

600 mm2

Using 10 mm F bars, Area = = =

\

79

Ast

79 mm2

4

\ Spacing = = 131 mm say

\ Ast = x

= =

mm c/c600

1000 x 500 = 600 mm2

130=

= 130Spacing = = 131 mm say


mm c/c600

The wall is in unsafe in sliding, and hence shear key will have to be provided, as shown in fig. 6

mm2

4

\

[email protected]


16/35

Pp = KpP = 3.00 x 99.07 = kN/m2

Pp x a = a

18

2.00

or PH = 3.00 x( 8 + a )2

= 4.10 ax 18 = 73.80 a

\ SW = 364.95 + 73.80 a Refer force calculation table=

mSw+Pp 0.5 x ( 364.95 + 73.80 a) 297.20 a

3 x( 8.00 + a2 )

0.5 x ( 364.95 + 73.80 a) 297.20 a

1.5

( 8 + a)2 = 182.48 + 36.9 a+ 297.20 a

4.5

( 8 + a)2 = 40.55 + 74.24 a

64 + 16 a + a2 = 74.24 a- 16 a + 40.55 - 64

a2 = 58.24 a -23.45

or a = a2 - 58.24 a +

or a = 0.405 m say = 410 mm

= 400 mm. and width of key 400 mm

Now size of key = 400 x 400 mm

PH = 3.00 x( 8.00 + a )2

= 3.00 x( 8.00 + 0.40 )2

PH = kN

= a = x

= kN Hence

SW = 364.95 + 73.80 a

= 364.95 + 73.80 x 0.40

= kN

Actual force to be resisted by the key at F.S. 1.5 is = 1.5PH - mSW= 1.5 x 211.7 - 0.5 x 394.5 = kN

120.29 x 1000

400 x 1000

120.29 x 200 x 1000

1/6 x 1000 x( 400 )2

= N/mm2 Hence safe

The details reinforcement shown in fig 7

297.20 0.40

211.68

Sliding force at level D1C1 =

Hence keep depth of key

23.45

297.20

120.29

= 0.30 N/mm2

=

x3

0.90

394.5

118.88

Bending stress =

\ shear stress =

.(2)

a )2

297.20

x( 8 +1.5

+

Hence equilibrium of wall, permitting F.S. 1.5 against sliding we have

8

=a)2

PH=1.5

x(

Weight of the soil between bottom of the base and D 1C1

x0.33

Let the depth of key =a intensity of passive pressure Pp devloped in front of key depend upon

the soil pressure P in front of the key

297.20

\ total passive pressure Pp =

[email protected]


17/35

0.3

`

0.3

7.50

7.00

3.00

0.3

2.00 D1

D E B C 1.00

2.00 E1 E B B

4.10 D y2

y2

w2

x2

63.9

0

1.80

FIG. 2

DESIGN OF COUNTOR FORT RETAINING WALL with horizontal back fill

170.1

99.0

7

1.70

1.80

x2

8.3

0

FIG. 1

0.50

4.00

87.2

0

86.5

7

157.6

0


18/35

0.30 A

F1

h

G.Level F

21 h kn/m

8.00 D F E7.50

G D1 E1

G1

F 147.00 kN/m

F1 F2 551.6

1.00 E B kN/m

C

168.00 kN/m

211.05 487.2kN/m 300

D E

mm F

mm c/c 450 D1

Pp

10 mm F

130 mm c/c

2.00

7.00

0.5


1.80

1

0.40

0.50

a

4.10FIG. 3

FIG. 4

0.51.00

0.50

FIG. 6FIG. 5

170.1

0

99.0

7

10

100

450


19/35

300 300

12 mm F 10 mm F

300 mm c/c 300 mm c/c

12 mm F 1 10 mm F

240 mm c/c 265 mm c/c

12 mm F 10 mm F

180 mm c/c 230 mm c/c

8000

12 mm F 10 mm F

120 mm c/c 200 mm c/c

12 mm F 0 mm F

110 mm c/c 12 mm F 0 mm c/c

3 900 180 mm c/c 2x12 mm F

2 2

12 mm F2 lgd

230 mm c/c

3 12 mm F 12 mm F

12 mm F 130 mm c/c 80 mm c/c

130 mm c/c 400 1 25 mm F 400

0 mm F 6 No.Bars 0 mm

0 mm c/c 0 mm

400


Cross -section mid way between co

2000

400

Holding bars

Cross -section mid way between counterfoorts

5001800


20/35

mm

mm

Fron

tcounterforts

mm


21/35


22/35


23/35


24/35


25/35


26/35


27/35


28/35


29/35


30/35


31/35


32/35


33/35

M-10 M-15 M-20 M-25 M-30 M-35 M-40

(N/mm2) Kg/m2

(N/mm2) Kg/m2

M 10 3.0 300 2.5 250

M 15 5.0 500 4.0 400

M 20 7.0 700 5.0 500

M 25 8.5 850 6.0 600

M 30 10.0 1000 8.0 800

M 35 11.5 1150 9.0 900

M 40 13.0 1300 10.0 1000

M 45 14.5 1450 11.0 1100M 50 16.0 1600 12.0 1200

M-10 M-15 M-20 M-25 M-30 M-35 M-40

Grade of concrete M-15 M-20 M-25 M-30 M-35 M-40

Modular Ratio 18.67 13.33 10.98 9.33 8.11 7.18

scbc N/mm2 5 7 8.5 10 11.5 13

m scbc 93.33 93.33 93.33 93.33 93.33 93.33

kc 0.4 0.4 0.4 0.4 0.4 0.4

jc 0.867 0.867 0.867 0.867 0.867 0.867

Rc 0.867 1.214 1.474 1.734 1.994 2.254

Pc (%) 0.714 1 1.214 1.429 1.643 1.857

kc 0.329 0.329 0.329 0.329 0.329 0.329

jc 0.89 0.89 0.89 0.89 0.89 0.89

Rc 0.732 1.025 1.244 1.464 1.684 1.903

Pc (%) 0.433 0.606 0.736 0.866 0.997 1.127

kc 0.289 0.289 0.289 0.289 0.289 0.289

jc 0.904 0.904 0.904 0.904 0.904 0.904Rc 0.653 0.914 1.11 1.306 1.502 1.698

Pc (%) 0.314 0.44 0.534 0.628 0.722 0.816

kc 0.253 0.253 0.253 0.253 0.253 0.253

jc 0.916 0.916 0.916 0.914 0.916 0.916

Rc 0.579 0.811 0.985 1.159 1.332 1.506

Pc (%) 0.23 0.322 0.391 0.46 0.53 0.599

Table 1.15. PERMISSIBLE DIRECT TENSILE STRESS

Grade of concrete

2.8 3.2 3.6 4.0 4.4

Table 1.16.. Permissible stress in concrete (IS : 456-2000)

Tensile stress N/mm2 1.2 2.0

Grade of

concrete

Permission stress in compression (N/mm2) Permissible stress in bond (Average) for

plain bars in tention (N/mm2)Bending acbc Direct (acc)

(N/mm2) in kg/m2

-- --

0.6 60

0.8 80

0.9 90

1.0 100

1.1 110

1.2 120

1.3 1301.4 140

Table 1.18. MODULAR RATIO

Grade of concrete

Modular ratio m31

(31.11)

19

(18.67)

13

(13.33)

11

(10.98)

9

(9.33)

8

(8.11)

7

(7.18)

Table 2.1. VALUES OF DESIGN CONSTANTS

(a) sst =

140

N/mm2

(Fe 250)

(b) sst =

190

N/mm2

(d) sst =

275

N/mm2

(Fe 500)

(c ) sst =

230N/mm2

(Fe 415)


34/35

M-15 M-20 M-25 M-30 M-35 M-40

0.18 0.18 0.19 0.20 0.20 0.20

0.22 0.22 0.23 0.23 0.23 0.23

0.29 0.30 0.31 0.31 0.31 0.32

0.34 0.35 0.36 0.37 0.37 0.38

0.37 0.39 0.40 0.41 0.42 0.42

0.40 0.42 0.44 0.45 0.45 0.46

0.42 0.45 0.46 0.48 0.49 0.49

0.44 0.47 0.49 0.50 0.52 0.52

0.44 0.49 0.51 0.53 0.54 0.55

0.44 0.51 0.53 0.55 0.56 0.57

0.44 0.51 0.55 0.57 0.58 0.60

0.44 0.51 0.56 0.58 0.60 0.62

0.44 0.51 0.57 0.6 0.62 0.63

300 or more 275 250 225 200 175 50 or less

1.00 1.05 1.10 1.15 1.20 1.25 1.30

M-15 M-20 M-25 M-30 M-35 M-40

1.6 1.8 1.9 2.2 2.3 2.5

Grade of concrete M-10 M-15 M-20 M-25 M-30 M-35 M-40 M-45 M-50

tbd (N / mm2) -- 0.6 0.8 0.9 1 1.1 1.2 1.3 1.4

tbd (N / mm2) tbd (N / mm2)

M 15

M 20

M 25

M 30

M 35

M 40M 45

M 50

Table 3.1. Permissible shear stress Table tc in concrete (IS : 456-2000)100As Permissible shear stress in concrete tc N/mm

2

bd

< 0.15

0.25

0.50

0.75

1.00

1.25

1.50

1.75

2.00

2.25

2.50

2.75

3.00 and above

Table 3.2. Facork

Over all depth of slab

k

Table 3.3. Maximum shear stress tc.max in concrete (IS : 456-2000)

Grade of concrete

tc.max

Table 3.4. Permissible Bond stress Table tbd in concrete (IS : 456-2000)

Table 3.5. Development Length in tension

Grade of

concrete

Plain M.S. Bars H.Y.S.D. Bars

kd = LdF kd = LdF

0.6 58 0.96

0.9 39 1.44

1 35 1.6

60

0.8 44 1.28 45

40

36

33

1.2 29 1.92 30

1.1 32 1.76

28

1.4 25 2.24 26

1.3 27 2.08


35/35

Degree sin cos tan tan Degree sin cos

10 0.17 0.98 0.18 0.18 10 0.17 0.98

11 0.19 0.98 0.19 0.19 11 0.19 0.98

12 0.21 0.98 0.21 0.21 12 0.21 0.98

13 0.23 0.97 0.23 0.23 13 0.23 0.97

14 0.24 0.97 0.25 0.25 14 0.24 0.97

15 0.26 0.97 0.27 0.27 15 0.26 0.97

16 0.28 0.96 0.29 0.29 16 0.28 0.96

17 0.29 0.96 0.31 0.31 17 0.29 0.96

18 0.31 0.95 0.32 0.32 18 0.31 0.95

19 0.33 0.95 0.34 0.34 19 0.33 0.95

20 0.34 0.94 0.36 0.36 20 0.34 0.94

21 0.36 0.93 0.38 0.38 21 0.36 0.93

22 0.37 0.93 0.40 0.40 22 0.37 0.93

23 0.39 0.92 0.42 0.42 23 0.39 0.92

24 0.41 0.92 0.45 0.45 24 0.41 0.92

25 0.42 0.91 0.47 0.47 25 0.42 0.91

30 0.50 0.87 0.58 0.58 30 0.50 0.87

35 0.57 0.82 0.70 0.70 35 0.57 0.82

40 0.64 0.77 0.84 0.84 40 0.64 0.77

45 0.71 0.71 1.00 1.00 45 0.71 0.71

50 0.77 0.64 1.19 1.19 50 0.77 0.64

55 0.82 0.57 1.43 1.43 55 0.82 0.57

60 0.87 0.50 1.73 1.73 60 0.87 0.50

65 0.91 0.42 2.14 2.14 65 0.91 0.42

Value of angle Value of angle

Documents

Counter+Fort++Reatining+WAll+With+HORZONTAL+Bach+Fill (1)