Counterfort Retaing Wall-Original

25
Height of Retaining Earth = 4.5 m Depth of Foundation below Ground Level = 1.5 m 18 KN / m ³ 24 Coefficient of friction between Earth & Concrete = 0.5 Allowable Bearing Pressure on Soil = 200.0 KN / m Concrete Grade : M 15 0.48 Steel Grade : Fe 415 Unit Weight of Concrete = 25 KN / m ³ 6.0 m Therefore Retaining Wall Designed as a Counterfort Type. Stability Considerations : = 0.422 Total force due to Earth pressure on the Wall : ( 0.5 x 0.422 x 18 x 6 x 6 ) = 136.73 KN This force acts horizontally at a point 2 above the foundation Level. Consider 1 m Length of Wall. 0.5 x ( 0.3 + 0.6 ) x 5.4 x 25 = 0.6 + ( 0.3 + 0.6 ) / 2 = ( 3.6 x 0.6 x 25 ) = 54.00 KN 1.80 m 0.5 x ( 2.4 + 2.7 ) x 5.4 = 247.90 KN 3.6 - (2/3) x ( 3.6 - 1.5 - = 1.90 m 273.5 KN-m 0 328.2 KN-m 0.9 x ( 60.75 x 1.95 + 54 x 1.8 + 247.9 x 1.9 ) = > ( Greater than ) Hence the Structure is safe against in Overturning. 136.73 KN Resisting force available from friction between The Base and Soil : = 0.5 x 0.9 x ( 60.75 + 54 + 247.9 ) = 163.19 KN Unit Weight of Soil ( ϒ ) = Angle of internal Friction in Degree φ ( 0 ) = ² Sandy clay lo sand), moist cl Hence Total Height of Retaining Wall ( H ) = ( 4.5 + 1.5 ) = Coefficient of Active Earth Pressure is ( 1 - Sin 24 ) / ( 1 + Sin 24 ) Weight of Wall ( W 1 ) = Distance of its Line of action from Point O ( X 1 ) = Weight of the base Slab ( W 2 ) = Distance of its Line of action from Point O ( X 2 ) = ( 3.6 / 2 ) = Weight of the earth over the heel slab is ( W 3 ) = Distance of its Line of action from Point O ( X 3 ) = Overturning Moment due to dead Load effect ( M od ) = ( 136.73 x 2 ) = Overturning Moment due to live Load effect ( M os ) = ( 1.2 M od + 1.4 M os ) = ( 1.2 x 273.46 + 1.4 x 0 ) = Restoring moment ( M r ) = 0.9 x ( W 1 X 1 + W 2 X 2 + W 3 X 3 ) M r = It is seen that M r ( 1.2 M od + 1.4 M os ) Sliding Force ( F S ) = Active earth Pressure ( P a ) = F r = μ x 0.9 x ( W 1 + W 2 + W 3 ) Sin Sin K a 1 1 2 2 1 H K P a a d X u

description

Counterfort Retaing Wall-Original

Transcript of Counterfort Retaing Wall-Original

Page 1: Counterfort Retaing Wall-Original

Height of Retaining Earth = 4.5 mDepth of Foundation below Ground Level = 1.5 m

18 KN / m ³

24

Coefficient of friction between Earth & Concrete = 0.5Allowable Bearing Pressure on Soil = 200.0 KN / m

Concrete Grade : M 150.48

Steel Grade : Fe 415Unit Weight of Concrete = 25 KN / m ³

( 4.5 + 1.5 ) = 6.0 mTherefore Retaining Wall Designed as a Counterfort Type.

Stability Considerations :

( 1 - Sin 24 ) / ( 1 + Sin 24 ) = 0.422

Total force due to Earth pressure on the Wall :

( 0.5 x 0.422 x 18 x 6 x 6 ) = 136.73 KN

This force acts horizontally at a point 2 above the foundation Level.

Consider 1 m Length of Wall. 0.5 x ( 0.3 + 0.6 ) x 5.4 x 25 = 60.75 KN

0.6 + ( 0.3 + 0.6 ) / 2 = 1.950 m

( 3.6 x 0.6 x 25 ) = 54.00 KN ( 3.6 / 2 ) = 1.80 m

0.5 x ( 2.4 + 2.7 ) x 5.4 x 18= 247.90 KN

3.6 - (2/3) x ( 3.6 - 1.5 - 0.5 x ( 0.6 + 0.3 ) )= 1.90 m

( 136.73 x 2 ) = 273.5 KN-m 0

( 1.2 x 273.46 + 1.4 x 0 ) = 328.2 KN-m

0.9 x ( 60.75 x 1.95 + 54 x 1.8 + 247.9 x 1.9 ) = 618.0 KN-m > ( Greater than )

Hence the Structure is safe against in Overturning.

136.73 KNResisting force available from friction between The Base and Soil :

= 0.5 x 0.9 x ( 60.75 + 54 + 247.9 ) = 163.19 KN

Unit Weight of Soil ( ϒ ) =Angle of internal Friction in Degree φ ( 0 ) =

² Sandy clay loam (clay +30% sand), moist clay

Hence Total Height of Retaining Wall ( H ) =

Coefficient of Active Earth Pressure is

Weight of Wall ( W 1 ) =Distance of its Line of action from Point O ( X 1 ) =

Weight of the base Slab ( W 2 ) =Distance of its Line of action from Point O ( X 2 ) =

Weight of the earth over the heel slab is ( W 3 ) =

Distance of its Line of action from Point O ( X 3 ) =

Overturning Moment due to dead Load effect ( M od ) =

Overturning Moment due to live Load effect ( M os ) =

( 1.2 M od + 1.4 M os ) =

Restoring moment ( M r ) = 0.9 x ( W 1 X 1 + W 2 X 2 + W 3 X 3 )

M r =It is seen that M r ( 1.2 M od + 1.4 M os )

Sliding Force ( F S ) = Active earth Pressure ( P a ) =

F r = μ x 0.9 x ( W 1 + W 2 + W 3 )

Sin

SinK a 1

1

2

2

1HKP aa

d

X u

Page 2: Counterfort Retaing Wall-Original

( 163.19 / 136.73 ) = 1.19 , Which is always less than 1.4. Hence, the structure is unsafe in Sliding.

Try a key projecting 0.05 m below the base, Which will reach up to 1.55 m below the G.L.

( 1 + Sin 24 ) / ( 1 - Sin 24 ) = 2.370.900 m

A

Available Passive Earth Pressure Force on B = 0.5 x 2.37 x 18 x 1.55 x 1.55 1.550 m= 51.25 KN

Available Passive Earth Pressure Force on A = = 0.5 x 2.37 x 18 x 0.9 x 0.9 = 17.28 KN

BNet available Passive Earth Pressure = ( 51.25 - 17.28 ) = 33.97 KN

= ( 163.19 + 33.97 ) = 197.16 KN ( 197.16 / 136.73 ) = 1.442 Which is greater than 1.4.

ð Hence, the structure has now become safe against Sliding.

= ( 60.75 + 54 + 247.9 ) = 362.70 KNDistance of its line of action from the Point O is :

( 60.75 x 1.95 + 54 x 1.8 + 247.9 x 1.9 - 136.73 x 6 / 3 )/ 362.7

= 1.14 m 1.15 m ( 1.8 - 1.15 ) = 0.65 m towards point O

Hence, pressure on the soil below the base is as follows :- ( 362.7 / 3.6 ) + ( 362.7 x 0.65 x 4.5 / ( 3.6 x 3.6 ) )

= ( 100.75 + 81.86 ) = 182.61 KN / m ² ( 362.7 / 3.6 ) - ( 362.7 x 0.65 x 4.5 / ( 3.6 x 3.6 ) )

= ( 100.75 - 81.86 ) = 18.89 KN / m ²

The Structure is safe against subsidence.

Net upward or downward pressure on the base :

considered. Hence all the loads should be multiplied by factor 1.5The pressure below the base will be as follows :

At the Toe = ( 1.5 x 1.5 x 182.61 ) = 273.92 KN / m ²

At the Heel = ( 1.5 x 1.5 x 18.89 ) = 28.34 KN / m ²

A key may be provided below the base to obtain additional resisting force due to Passive earth pressure in front. The soil above the toe will be neglected from consideration as it is likely to be excavated.

Remaining depth of soil up to the bottom of the key will be considered to be provide passive resistance.

Hence, total resisting force against Sliding is now ( F r ) =

Total vertical Force on Soil ( R ) = ( W 1 + W 2 + W 3 )

Take x ̅ =Eccentricity of R from the centre of the base is ( e ) =

Below the Toe ( q 1 ) =

Below the Heel ( q 2 ) =

Since ( q 1 ) is ( < ) less than the allowable bearing pressure on the Soil i.e.

200.00 KN / m 2

The Retaining wall will be designed by the limit state method for which factored loads should be

q 1 ) = (

q 2 ) = (

S

r

F

F

Sin

SinK p 1

1

2

2

1HKP pp

S

r

F

F

R

HPXWXWXW

xa 3332211

Page 3: Counterfort Retaing Wall-Original

0.6 m 2.400 mA B C D

0.600 m 0.600 m

28.34 KN-m ²273.915 232.99 192.06

Pressure Distribution below the Base

67.32 KN-m ²9

Load intensity from above

39.00 KN-m ²

-124.7

264.915 223.99

Resultant Load Intensity on Base Slab

At Point A = ( 0.6 x 0.6 x 25 ) = 9.00 KN-m ² At Point B = ( 0.6 x 0.6 x 25 ) = 9.00 KN-m ² At Point C = 0.6 x ( 0.6 x 25 + 5.4 x 18 ) = 67.32 KN-m ² At Point D = 0.6 x ( 0.6 x 25 + 5.4 x 18 ) = 67.32 KN-m ²

Design of Heel Slab :Let the spacing of the Counterforts be kept = ( 0.4 x H ) = ( 0.4 x 6 ) = 2.4 m C / C

Let the thickness of the Counterforts be kept = 0.30 mThe Heel Slab is a continuous slab spanning horizontally over the counterforts & bearing a net

vertical load intensity. Consider a 1.0 m ( 1000 mm ) wide strip of slab.Clear Span = ( 2.4 - 0.3 ) = 2.10 m

( 1/12 ) of Clear Span = ( 0.083 ) x 2.1 x 1000 = 175.00 mmIt is seen that the supports are wider than ( 1 / 12 ) of clear span.

Hence, effective span = ( Clear Span ) = 2.1 m

( 39 x 2.1 x 2.1 ) / 12 = 14.3 KN-m

0.36 x 15 x 0.48 x ( 1 - 0.42 x 0.48 ) x 1000 x 14.3 x 1000000 d = 83.13 mm , Say 90 mm

KN / m 2

KN / m 2

KN / m 2

KN / m 2

KN / m 2

Factored load intensity on the base due to self weight of base and the earth above is as follows :

Bending moment in the strip of slab, near counterforts is

Effective depth for balanced condition is found as :( d 2 ) =

12

2lwM u

Page 4: Counterfort Retaing Wall-Original

14.3 x 1000000 / ( 0.87 x 415 x (1-0.42 x 0.48) x 90 )

= 551.2

Critical section for shear occurs at the face where the slab joints with the counterforts.

0.5 x ( 2.4 - 0.3 ) x 39 = 41.0 KN

41 x 1000 / ( 1000 x 90 ) = 0.46 N / mm ²

100 x 551.2 / ( 1000 x 90 ) = 0.612From Design Shear Strength of Concrete Table :-

0.61 , 0.496 N / mm ²

< less than the slab is safe in Shear.

90 mm

41 x 1000 / ( 1000 x 90 ) = 0.456 N / mm ²

< ( less than ) the slab is safe in Shear. Tension Reinforcement :-

( 90 / 90 ) x 551.2 = 551.20 mm

Provide 12 mm dia. Bars - @ 45 mm C / C, near the top surface. The spacing may be increased towards the stem to 200 mm C / C

Keep overall thickness = ( 90 + 35 ) = 125.00 mm( Keeping End Cover or end spacing = 35 mm )

Temperature & Shrinkage Reinforcement = ( 0.12 / 100 ) x 125 x 1000 = 150 mm 2 / mProvide 12 mm dia. Bars - @ 200 mm C / C, at right angles to the main reinforcement.

( 39 x 2.1 x 2.1 ) / 16 = 10.7 KN-m

0.87 x 415 90 - 415 / ( 15 x 1000 ) x 10.7 x 1000000

90 x 0.028 = 29635.78

a = 0.028 b = -90.000 c = 29635.78

2841.84

372.44

372.44Provide 12 mm dia. Bars - @ 200 mm C / C, near the bottom surface.

Design of Vertical Slab :Clear height of vertical slab over the heel = ( 6 - 0.6 ) = 5.400 m

Considering 1 m high strip of slab spanning horizontally & continuous over the counterforts.Effecyive Span = 2.100 m as for the heel slab.

Tension Reinforcement near top surface -A st =

mm 2 / m

Shear force ( V u ) : V u =

ð τ V =

( 100 x A st ) / ( b x d ) =

For ( 100 x A st ) / ( b x d ) = ð τ C =

Since τ V τ C , hence

Hence, it is proposed to increase the effective depth of the slab while keeping the percentage ( % ) of tension reinforcement is same as before.

Try an effective depth ( d ) =τ V =

Now τ V τ C , hence

A st = 2 / m

Bending moment at the mid-span :

Tension Reinforcement near bottom surface -

x A st x ( A st ) =

A st - ( A st ) 2 x

A st =mm 2 / m ( When taking + sign)

mm 2 / m ( When taking - sign)

Taking ( - ) sign, hence A st = mm 2 / m

16

2lwM u

Page 5: Counterfort Retaing Wall-Original

= 1.5 x 0.422 x 18 x ( 5.4 + 4.4 ) / 2 = 55.8 KN / m ²

( 55.83 x 2.1 x 2.1 ) / 12 = 20.50 KN-m/m

0.36 x 15 x 0.48 x ( 1 - 0.42 x 0.48 ) x 1000 x 20.5 x 1000000 d = 99.5 mm

20.5 x 1000000 / ( 0.87 x 415 x (1-0.42 x 0.48) x 99.5 )

= 714.73 100 x 714.73 / ( 1000 x 99.5 ) = 0.718

From Design Shear Strength of Concrete Table :-

0.718 , 0.53 N / mm ²Since thickness of slab will be kept at least 90 mm, the effective depth will be

( 90 - 35 ) = 55 mm. 0.53 N / mm 2

( 55 / 99.5 ) x 714.73 = 395.08 mm

Provide 12 mm dia. Bars - @ 95 mm C / C, near the back face.the spacing may be increased to 300 mm C / C, towards the top.

Critical section for shear occurs at the face where the slab joins with counterforts. 0.5 x ( 2.4 - 0.3 ) x 55.83 = 58.6 KN

58.6 x 1000 / ( 1000 x 55 ) = 1.065 N / mm ²

> ( greater than ) the vertical slab is unsafe in Shear.

( 55.83 x 2.1 x 2.1 ) / 16 = 15.39 KN-m/m

0.87 x 415 55 - 415 / ( 15 x 1000 ) x 15.39 x 1000000

55 x 0.028 = 42625.68a = 0.028 b = -55.000 c = 42625.68

#NUM!

#NUM!

#NUM!Provide 12 mm dia. Bars - @ 200 mm C / C, on the front face.the spacing may be increased to 300 mm C / C, towards the top.

Temperature & Shrinkage Reinforcement = ( 0.12 / 100 ) x 90 x 1000 = 108 mm 2 / mProvide 10 mm dia. Bars - @ 300 mm C / C, may be provided in vertical direction.

Design of Toe : The toe bends upward as a cantilever due to upward soil reaction.

( 2 x 264.915 + 223.985 ) x 0.6 x 0.6 / 6 = 45.20 KN-m/m

0.36 x 15 x 0.48 x ( 1 - 0.42 x 0.48 ) x 1000 x 45.2 x 1000000 d = 147.8 mm , Say 150 mm

Average intensity of earth pressure on the strip = ( K a x ϒ x h )

Bending moment in the strip of slab, near counterforts is

Effective depth for balanced condition is found as :( d 2 ) =

Tension Reinforcement near top surface -

A st =

mm 2 / m

( 100 x A st ) / ( b x d ) =

For ( 100 x A st ) / ( b x d ) = ð τ C =

If the % of reinforcement is kept unchange, than τ C =as before. Hence tension reinforcement :

A st = 2 / m

Shear force ( V u ) : V u =

Therefore, τ V =

Now τ V τ C , hence

Bending moment at the mid-span :

Tension Reinforcement -x A st x ( A st ) =

A st - ( A st ) 2 x

A st =mm 2 / m ( When taking + sign)

mm 2 / m ( When taking - sign)

Taking ( - ) sign, hence A st = mm 2 / m

Bending moment ( M u ) :M u =

Effective depth for balanced condition is found as :( d 2 ) =

12

2lwM u

16

2lwM u

Page 6: Counterfort Retaing Wall-Original

Tension Reinforcement :

45.2 x 1000000 / ( 0.87 x 415 x (1-0.42 x 0.48) x 150 )

= 1045.34Provide 16 mm dia. Bars - @ 85 mm C / C.

100 x 1045.34 / ( 1000 x 150 ) = 0.697From Design Shear Strength of Concrete Table :-

0.697 , 0.52 N / mm ²Overall depth of toe slab = ( 150 + 35 ) = 185 mm

Critical section for shear occurs at an effective depth i.e. 0.150 m away from the front of the vertical wall.

0.5 x ( 264.915 - 151.96 ) x 1.35 x 10 = 762.46 KN/m

762.463 x 1000 / ( 1000 x 150 ) = 5.080 N / mm ² > ( greater than ) , the toe slab is unsafe in Shear.

Now 0.529 x 1000 x 150 / 1000 = 79.35 KN/m

Provide 10 mm dia. Bars - @ inclined at 45 ° to the horizontal at a spacing of 425 mm C / C.

( 0.87 x 415 x 78.5 x Sin 45 / 1000 ) = 20.04 KN/m

( 79.35 + 20.04 ) = 99.39 KN/m which is < (less than the . 762.46 KN/m

The toe slab is therefore safe in shear.Temperature & Shrinkage Reinforcement = ( 0.12 / 100 ) x 185 x 1000 = 222 mm 2 / m

Provide 10 mm dia. Bars - @ 180 mm C / C, at right angles to main reinforcement bars.

Design of Shear Key :The key projects 0.05 m below the base, and its lowest point is 1.55 m

below the G.L. in the front.

1.5 x 2.37 x 18 x (

= 63.99 x = 0.12 KN-m/m

This is a small bending moment. A shear key of thickness 100 mm and having vertical reinforcement of 10 mm dia. bars - @ 200 mm C / C will be sufficient.

Design of Counterforts :For taking up tension, steel reinforcement should be provided alon the sloping edge and the

= 21.45 °. One counterfort supports earth pressure from 3 m width of the vertical slab. Maximum bending moment occurs where the counterforts meets the base.

At this point depth of earth = ( 6 - 0.6 ) = 5.4 m .

3 = 1.5 x ( 1 / 6 ) x ( 0.422 ) x 18 x 3 x ( 5.4 ) ³ = 897.07 KN-m/m

A st =

mm 2 / m

( 100 x A st ) / ( b x d ) =

For ( 100 x A st ) / ( b x d ) = ð τ C =

Shear force ( V u ) : V u =Therefore, τ V =

As τ V τ C , hence

V uc =

Shear Strength of inclined bars is :

Hence, strength of toe slab in shear is V u = ( V uc + V us )

V u =

Factored bending moment on the key is ( M u ) :

h 2 - 1.5 h ) dh

vertical is α

Factored bending moment ( M u ) : -

)45(87.0 0SinAfV svyus

dhhhKM pu

76.1

5.1

5.15.1 76.1

5.176.1

5.1

23

25.1

3

hh

3

6

15.1 hKM au

Page 7: Counterfort Retaing Wall-Original

Even for a rectangular beam of width = 300 mm, the required effective depth for

0.36 x 15 x 0.48 x ( 1 - 0.42 x 0.48 ) x 300 x 897.07 x 1000000 d = 1202.1 mm , Say 1100 mm

Overall depth of counterfort available at the base = 3000.0 mm, and effective depth availablethere will be at least = ( 3000 - 100 ) = 2900.0 mm, which is more than required.

Hence, the dimensions of counterfort are alright.

0.87 x 415 2900 - 415 / ( 15 x 300 ) x 897.07 x 1000000

2900 x 0.092 = 2484614.32

a = 0.092 b = -2900.0 c = 2484614.32

30640.33

881.41

881.41Provide 16 mm dia. Bars - 10 Nos. in two layers of 5 bars in each row.

The force which tries to separate the strip from the counterfort is = 1.5 x ( 3 - 0.3 ) = ( 1.5 x 0.422 x 18 x 4.9 x 2.7 )

= 150.74 KN/m ( 150.74 x 1000 ) / ( 0.87 x 415 ) = 417.50 mm

Provide 10 mm dia. Two legged ties in horizontal direction, with a vertical spacing of 300 mm C / C. This spacing may be maintained up to the top.

Critical section for shear occurs where counterfort joins with the base slab.

3 = 1.5 x ( 1/2 ) x 0.422 x 18 x 3 x ( 6 - 0.6 ) ²= 498.37 KN

498.37 - ( 897.07 / 2.9 ) x tan ( 21.45 ° ) = 376.9 KN 376.9 x 1000 / ( 300 x 2900 ) = 0.433 N / mm ²

= ( ( 100 x 10 x 200.96 ) / ( 300 x 2900 ) ) x Cos ( 21.45 ) °= 0.22

From Design Shear Strength of Concrete Table :-

0.215 , 0.335 N / mm ²è ( 0.335 x 300 x 2900 / 1000) = 291.45 KN

For the Ties already provided :- 0.87 x 415 x 2 x ( 3.141 / 4 ) x 10 x 10 x 2900 / ( 300 x 1000 )

= 548.23 KN ( 291.45 + 548.23 ) = 839.68 KN

> (greater than )

Hence, the counterfort section is safe against shear.

Hence, depth from top where 5 bars out of 10 bars are no longer required.h = ( 5.4 / 2 ) = 2.700 m

It is proposed to extend 5 bars of the inner layer by 12 times the bar dia.beyond this point and then curtail them. Hence, actual point of cut-off from top -

= ( 2.7 - 12 x 0.016 ) = 2.508 m from top.

balanced condition, is given by -( d 2 ) =

Tension reinforcement required in the counterfort is -x A st x ( A st ) =

A st - ( A st ) 2 x

A st =mm 2 / m ( When taking + sign)

mm 2 / m ( When taking - sign)

Taking ( - ) sign, hence A st = mm 2 / m

K a x ϒ x h x

Area of two legged ties =

² / m

τ v =

For ( 100 x A st ) / ( b x d ) = ð τ C =V uc =

V us =

 ( V uc + V us ) =

It is seen that  ( V uc + V us )

Bending moment in the counterfort decrease towards the top. Tension reinforcement can, therefore, be curtailed accordingly.

°

2

2

15.1 hKV au

)tan(d

MV u

u

db

Ast100

)tan(

d

MV u

u

Page 8: Counterfort Retaing Wall-Original

3 = ( 1 / 2 ) x 1.5 x 0.422 x 18 x 2.508 x 2.508 x 3 = 107.50 KN

Effective depth at this point = ( 0.3 + 2.508 x Tan 21.45° - 0.1 ) = 1.185 m Available tension reinforcement after cut-off is -

( 5 x 201.056 ) = 1006.4 mm ²

100 x 1006.4 / ( 1100 x 300 ) = 0.305From Design Shear Strength of Concrete Table :-

0.305 , 0.350 N / mm ²è ( 0.35 x 300 x 1100 / 1000) = 115.50 KN

For the 10 mm dia. Two legged ties spaced at 300 mm C / C

0.87 x 415 x 2 x ( 3.141 / 4 ) x 10 x 10 x 1100 / ( 300 x 1000 ) = 207.95 KN

( 115.5 + 207.95 ) = 323.45 KN ( 0.667 x 323.45 ) = 215.74 KN

It is seen that < (less than )

Hence, the 5 bars can be safely curtailed at this point.

Counterforts = ( 3 - 0.3 ) = 2.70 m . Total downward force trying to separate the strip from the counterforts at the point D is -

= ( 39 x 2.7 ) = 105.3 KN / mSectional area of two legged vertical ties required here is -

= ( 105.3 x 1000 ) / ( 0.87 x 415 ) = 291.65 mm Spacing of 10 mm dia. Two legged vertical ties is -

= ( 2 x 78.5375 / 291.65 ) x 1000 = 538.60 mm C / CProvide a spacing of 150.00 mm C / C at this point.

Towards the point C , the vertical load intensity is only -124.74 at the point C is - = ( -124.74 x 2.7 ) = -336.8 KN / m

Sectional area of two legged vertical ties required here is - = ( -336.798 x 1000 ) / ( 0.87 x 415 ) = -932.83 mm

Adopt a spacing of 300.0 mm C / C at this point.

Joint Reinforcement :-At the junction of the vertical wall with the base slab, a rigid connection can be ensured by providing

additional joint reinforcement. 0.30 % of concrete area may be provided as reinforcement in the vertical slab, extending for a height of 2.50 m. These bars

should be embedded into the heel slab also.Area of reinforcement = 0.3 % of ( 1000 x 90 ) = 270.0 mm

Provide 12 mm dia. Bars - @ 180 mm C / C, as joint reinforcement.

At this point, Shear Force is -

A st =

( 100 x A st ) / ( b x d ) =

For ( 100 x A st ) / ( b x d ) = ð τ C =V uc =

V us =

 ( V uc + V us ) = ( V uc + V us ) =

V u ( V uc + V us )

Vertical ties are needed to connect the heel slab to the counterfort. Consider a strip of unit width of slab, in horizontal direction and supported over the counterforts. Clear width of slab between two

² / m

KN/m 2. Total downward force

² / m

² / m

2

2

15.1 hKV au

3

2

3

2

Page 9: Counterfort Retaing Wall-Original

0.3 m

4.5 m

60.75 KN 6.0 m

1.05 m

247.90 KN

1.90 m 136.73 KN

0.600 m

1.5 m 54.00 KN 2.00 m

0.600 m1.55 m

O0.6 m 0.05 m

3.6 m

1.80 m

W 1 =

X 1 =

W 3 =

X 3 = P a =

W 2 =

X 2 =

Page 10: Counterfort Retaing Wall-Original

0.3 m

2.508 m

16 mm - 5 Nos.

Cut off point6.0 m

16 mm - 10 Nos.

10 mm - 425 mm C/C

10 mm - 180 mm C/C

0.19 m

0.05 m 16 mm bars @ 85 mm C/C 12 mm @ 45 mm C/C

10 mm - 200 mm C/C 3.6 m

0.6 m 0.1 mSide View

2.4 m

0.30 m

Counterforts

1.25 m12 mm - 200 mm C/C

0.600 m

12 mm - 200 mm C/C End View

12 mm dia. Bars @ 300 mm C/C 10 mm - 2 Legged ties -

@ 300 mm C/C

10 mm - 2 Legged Ties, 300 mm C/C 10 mm - 2 Legged Ties,

150 mm C/C

12 mm dia. Bars @ 95 mm C/C

12 mm - @ 200 mm C/C

12 mm - 45 mm C/C, increased to 200 mm C/C towards wall

0 0 00

0

000

0

000

0 0 000

0 0 000 0

Page 11: Counterfort Retaing Wall-Original

1.25 m

10 mm dia. - 300 mm C/C 2.4 m

1.25 m

0.30 m

0.3 m

Sectional Plan

12 mm dia. Bars @ 180 mm C/C

2.5 m

1.25 m

Joint Reinforcement

12 mm dia. - 95 mm C/C, Increased to 300 mm C/C towards Top

12 mm dia. - 200 mm C/C, increased to 300

mm C/C at Top

00

00

00 0

0

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Sample Location Angle of internal Cohesion (c) Soil Bearing CapacitiesNumber friction (F)1 93º42' E 26º40' N 22 0.5 N/cm2 Soil Type2 93º43' E 26º39' N 17 6 Soft, wet, pasty or muddy soil3 93º44' E 26º37' N 33 3.034 93º46' E 26º38' N 28 1.65 93º47' E 26º37' N 27 2.6 Compact clay, nearly dry6 93º50' E 26º34' N 30 1.05 Solid clay with very fine sand7 93º53' E 26º34' N 36 0.42 Dry compact clay (thick layer)8 93º55' E 26º31' N 17 1.32 Loose sand9 93º57' E 26º30' N 36 2.2 Compact sand

10 93º57' E 26º28' N 28 2 Red earth11 93º57' E 26º26' N 18 1.32 Murram12 93º55' E 26º24' N 20 0.64 Compact gravel13 93º54' E 26º22' N 18 0.34 Rock14 93º53' E 26º19' N 17 0.6315 93º52' E 26º16' N 16 0.51 The geotechnical properties of the bank sediments and16 93º50' E 26º13' N 23 1.03 bank stability analysis along the Dhansiri River channel have17 93º50' E 26º08' N 15 0.8 provided valuable information in land resource evaluation.18 93º47' E 26º03' N 18 0.67 Texturally dominant Sandy-Clay-Loam (76%) with19 93º48' E 26º02' N 26 0.45 subordinate Clay-Loam (24%), ranging permeability from20 93º48' E 25º58' N 18 0.31 5x10-4 to 1x10-6 cm/sec has characterized the bank sediments

23.15 1.371 with poor stability susceptible to liquefaction

Different geotechnical parameters and critical height for a few locations along the bank of the Dhansiri River

Sample Location γ β Ø' = tan-1

Barguriagaon 1.79 41.25º 2.6 27º 173 18.76º 31.74Butalikhowa 1.87 31.63º 2.2 36º 147 25.84º 290.91

Golaghat 1.54 41.25º 1.05 30º 70 21.05º 18.19Dachmuagaon 1.53 41.87º 2 28º 133 19.52º 29.12

Parameters in relation to bank stability at two different conditions for the Dhansiri River channel sedimentsb Saturated Unsaturated

F' Sn c' gsat Hc Fm Sn cm g Hc(N/cm2) (N/CC) (cm) (N/cm2) (N/CC) (cm)

25° 0° 18.04 0.91 0.019 864.16 21.29° 751.60 1.31 0.0172 57244.2530° 0° 14.93 0.91 0.019 714.98 21.29° 161.59 1.31 0.0172 12306.9435° 0° 12.69 0.91 0.019 607.61 21.29° 075.03 1.31 0.0172 5714.3840° 0° 10.99 0.91 0.019 526.36 21.29° 045.33 1.31 0.0172 3452.7345° 0° 09.66 0.91 0.019 462.51 21.29° 031.22 1.31 0.0172 2377.9850° 0° 08.58 0.91 0.019 410.85 21.29° 023.22 1.31 0.0172 1768.7955° 0° 07.68 0.91 0.019 368.03 21.29° 018.16 1.31 0.0172 1382.9060° 0° 06.93 0.91 0.019 331.84 21.29° 014.69 1.31 0.0172 1119.0465° 0° 06.28 0.91 0.019 300.72 21.29° 012.19 1.31 0.0172 0928.5070° 0° 05.71 0.91 0.019 273.61 21.29° 010.30 1.31 0.0172 0784.2475° 0° 05.21 0.91 0.019 249.67 21.29° 008.82 1.31 0.0172 0671.8180° 0° 04.77 0.91 0.019 228.32 21.29° 007.64 1.31 0.0172 0581.6385° 0° 04.37 0.91 0.019 209.07 21.29° 006.67 1.31 0.0172 507.60

Alluvial soil, loam, sandy loam (clay +40 to 70% sand)Sandy clay loam (clay +30% sand), moist

clay

cu ɸu c' = 2/3 cu Hc

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90° 0° 04.00 0.91 0.019 191.58 21.29° 005.85 1.31 0.0172 445.55

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Soil Bearing Capacities

Soil Type kN/m²Soft, wet, pasty or muddy soil 27 - 35

80 - 160215 - 270

Compact clay, nearly dry 215 - 270Solid clay with very fine sand -430Dry compact clay (thick layer) 320 - 540

Loose sand 160 - 270Compact sand 215 - 320

Red earth -320Murram -430

Compact gravel 750 - 970Rock -1700

The geotechnical properties of the bank sediments andbank stability analysis along the Dhansiri River channel haveprovided valuable information in land resource evaluation.

Texturally dominant Sandy-Clay-Loam (76%) withsubordinate Clay-Loam (24%), ranging permeability from

5x10-4 to 1x10-6 cm/sec has characterized the bank sedimentswith poor stability susceptible to liquefaction

Different geotechnical parameters and critical height for a few locations along the bank of the Dhansiri River

Zone

RiskUnstable

RiskRisk

Parameters in relation to bank stability at two different conditions for the Dhansiri River channel sediments

25° 0° 18.04 0.91 0.019 864.16 21.29° 751.60 1.31 0.0172 57244.2530° 0° 14.93 0.91 0.019 714.98 21.29° 161.59 1.31 0.0172 12306.9435° 0° 12.69 0.91 0.019 607.61 21.29° 075.03 1.31 0.0172 5714.3840° 0° 10.99 0.91 0.019 526.36 21.29° 045.33 1.31 0.0172 3452.7345° 0° 09.66 0.91 0.019 462.51 21.29° 031.22 1.31 0.0172 2377.9850° 0° 08.58 0.91 0.019 410.85 21.29° 023.22 1.31 0.0172 1768.7955° 0° 07.68 0.91 0.019 368.03 21.29° 018.16 1.31 0.0172 1382.9060° 0° 06.93 0.91 0.019 331.84 21.29° 014.69 1.31 0.0172 1119.0465° 0° 06.28 0.91 0.019 300.72 21.29° 012.19 1.31 0.0172 0928.5070° 0° 05.71 0.91 0.019 273.61 21.29° 010.30 1.31 0.0172 0784.2475° 0° 05.21 0.91 0.019 249.67 21.29° 008.82 1.31 0.0172 0671.8180° 0° 04.77 0.91 0.019 228.32 21.29° 007.64 1.31 0.0172 0581.6385° 0° 04.37 0.91 0.019 209.07 21.29° 006.67 1.31 0.0172 507.60

Alluvial soil, loam, sandy loam (clay +40 to 70% sand)Sandy clay loam (clay +30% sand), moist

clay

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90° 0° 04.00 0.91 0.019 191.58 21.29° 005.85 1.31 0.0172 445.55