Height of Retaining Earth = 4.5 m Depth of Foundation below Ground Level = 1.5 m Unit Weight of Soil ( ϒ ) = 18 KN / m ³ 0 24 Angle of internal Friction in Degree φ ( ) = Coefficient of friction between Earth & Concrete = 0.5 Allowable Bearing Pressure on Soil = 200.0 KN / m ² Sandy clay loam (clay +30% sand), moist clay Concrete Grade : M 15 X 0.48 Steel Grade : Fe 415 d Unit Weight of Concrete = 25 KN / m ³ u
Hence Total Height of Retaining Wall ( H ) = ( 4.5 + 1.5 ) = 6.0 m Therefore Retaining Wall Designed as a Counterfort Type. Stability Considerations : Coefficient of Active Earth Pressure is K a
1 Sin 1 Sin
( 1 - Sin 24 ) / ( 1 + Sin 24 )
= 0.422
Total force due to Earth pressure on the Wall :
P a
1 2
K a H 2
( 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.
Considerr 1 m Le Conside L engt ngth h of Wall Wall.. Weight of Wall ( W 1 ) =
0.5 x ( 0.3 + 0.6 ) x 5.4 x 25 = 0.6 + ( 0.3 + 0.6 ) / 2 =
Distance of its Line of action from Point O ( X 1 ) =
60.75 KN 1.950 m
Weight of the base Slab ( W 2 ) =
( 3.6 x 0.6 x 25 ) = 54.00 KN Distance of its Line of action from Point O ( X 2 ) = ( 3.6 / 2 ) = 1.80 m ( W ) Weight of the earth over the heel slab is 3 = 0.5 x ( 2.4 + 2.7 ) x 5.4 x 18 = 247. 247.90 90 KN KN ( X 3 ) Distance of its Line of action from Point O = 3.6 - (2/3) x ( 3.6 - 1.5 - 0.5 x ( 0.6 + 0.3 ) ) = 1.90 1.90 m
Overturning Moment due to dead Load effect ( M od ) = ( 136.73 x 2 ) = Overturning Moment due to live Load effect ( M os ) = 0
273.5 KN-m
( 1.2 M od + 1.4 M os ) =
( 1.2 x 273.46 + 1.4 x 0 ) = 328.2 KN-m Restoring moment ( M r ) = 0.9 x ( W 1 X 1 + W 2 X 2 + W 3 X 3 ) 0.9 x ( 60.75 x 1.95 + 54 x 1.8 + 247.9 x 1.9 ) = Mr = It is seen that M r ( 1.2 M od + 1.4 M os ) > ( Greater than ) Hence the Structure is safe against in Overturning. Sliding Force ( F S ) = Active earth Pressure ( P a ) = 136.73 KN Resisting force available from friction between The Base and Soil : F r = μ x 0.9 x ( W 1 + W 2 + W 3 ) = 0.5 x 0.9 x ( 60.75 + 54 + 247.9 ) = 163.19 KN
618.0 KN-m
F r F S
( 163.19 / 136.73 ) = 1.19 , Which is always less than 1.4. Hence, the structure is unsafe in Sliding. 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. Try a key projecting 0.05 m below the base, Which will reach up to 1.55 m below the G.L. 1 Sin K p 1 Sin
P p
1
( 1 + Sin 24 ) / ( 1 - Sin 24 ) =
2.37 0.900 m
2
K p H 2 Available Passive Earth Pressure Force on B = 0.5 x 2.37 x 18 x 1.55 x 1.55 = 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
A
1.550 m
B
Net available Passive Earth Pressure = ( 51.25 - 17.28 ) = 33.97 KN Hence, total resisting force against Sliding is now ( F r ) = = ( 163.19 + 33.97 ) = 197.16 KN F r ( 197.16 / 136.73 ) = Which is greater than 1.4. 1.442 F S Hence, the structure has now become safe against Sliding. Total vertical Force on Soil ( R ) = ( W 1 + W 2 + W 3 ) = ( 60.75 + 54 + 247.9 )
= 362.70 KN
Distance of its line of action from the Point O is : W X 1 1
x
W X 2 W X 3 P a 2 3
R
H
3
( 60.75 x 1.95 + 54 x 1.8 + 247.9 x 1.9 - 136.73 x 6 / 3 )/ 362.7
= 1.14 m Take 1.15 m x̅ = Eccentricity of R from the centre of the base is ( e ) = ( 1.8 - 1.15 ) = 0.65 m towards point O Hence, pressure on the soil below the base is as follows :Below the Toe ( q 1 ) = ( 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 = Below the Heel ( q 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 = 2 Since ( q 1 ) is ( < ) less than the allowable bearing pressure on the Soil i.e. KN / m The Structure is safe against subsidence. Net upward or downward pressure on the base : The Retaining wall will be designed by the limit state method for which factored loads should be considered. Hence all the loads should be multiplied by factor 1.5 The pressure below the base will be as f ollows : ( At the Toe = 1.5 x 182.61 ) = ( 1.5 x q 1 ) = 273.92 KN / m ²
At the Heel =
( 1.5 x q 2 ) =
(
1.5 x 18.89 ) =
28.34 KN / m ²
0.6 m
2.400 m
A
B
C
D
0.600 m
0.600 m
28.34 KN-m ² 273.915 KN / m
232.99
192.06
2
Pressure Distribution below the Base 67.32 KN-m ² 9 2 KN / m
Load intensity from above 39.00 KN-m ²
-124.7 KN / m 2 223.99 KN / m 2 2 Resultant Load Intensity on Base Slab KN / m Factored load intensity on the base due to self weight of base and the earth above is as follows : 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 ² 264.915
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 m The Heel Slab is a continuous slab sp anning 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 ) 175.00 mm of Clear Span = ( 0.083 ) x 2.1 x 1000 = It is seen that the supports are wider than ( 1 / 12 ) of clear span. Hence, effective span = ( Clear Span ) = 2.1 m Bending moment in the strip of slab, near counterforts is
M u
w l 12
2
( 39 x 2.1 x 2.1 ) / 12 =
14.3 KN-m
Effective depth for balanced condition is found as : 2
0.36 x 15 x 0.48 x ( 1 - 0.42 x 0.48 ) x 1000 x (d )= 83.13 mm , Say 90 mm d=
14.3 x 1000000
Tension Reinforcement near top surface 14.3 x 1000000 / ( 0.87 x 415 x (1-0.42 x 0.48) x 90 ) A st = = 551.2 mm 2 / m Critical section for shear occurs at the face where the slab joints with the counterforts. Shear force ( V u ) : Vu= 0.5 x ( 2.4 - 0.3 ) x 39 = 41.0 KN
τV=
41 x 1000 / ( 1000 x 90 ) =
( 100 x A st ) / ( b x d ) =
0.46 N / mm ²
100 x 551.2 / ( 1000 x 90 ) =
0.612
From Design Shear Strength of Concrete Table :-
For ( 100 x A st ) / ( b x d ) = Since τ V
< less than
τC=
0.61 , τ
the slab is safe in Shear.
, hence
C
0.496 N / mm ²
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 ) = 90 mm 41 x 1000 / ( 1000 x 90 ) = τV= 0.456 N / mm ² Now τ V
< ( less than )
τC
the slab is safe in Shear.
, hence
Tension Reinforcement :( 90 / 90 ) x 551.2 = A st =
551.20 mm 2 / m near the top surface. The spacing may be
dia. Bars - @ Provide 12 mm 45 mm C / C, 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 / m dia. Bars - @ 200 mm C / C, at right angles to the main reinforcement. Provide 12 mm Bending moment at the mid-span :
M u
w l 2 16
( 39 x 2.1 x 2.1 ) / 16 =
10.7 KN-m
Tension Reinforcement near bottom surface 0.87 x 415
x A st x ( 90 x
90 - 415 / ( 15 x 1000 ) x 2
A st - ( A st ) x
a = 0.028
Taking ( - ) sign, hence A st = 12 mm
0.028 =
10.7 x 1000000 29635.78
b = -90.000 A st =
Provide
A st ) =
dia. Bars - @
c = 29635.78
2
2841.84 mm / m ( When taking + sign) 372.44 mm 2 / m ( When taking - sign)
2 372.44 mm / m 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.
Average intensity of earth pressure on the strip = ( K a x ϒ x h )
= 1.5 x 0.422 x 18 x ( 5.4 + 4.4 ) / 2 Bending moment in the strip of slab, near counterforts is M u
w
l 2
12
= 55.8 KN / m ²
( 55.83 x 2.1 x 2.1 ) / 12 = 20.50 KN-m/m
Effective depth for balanced condition is found as : 0.36 x 15 x 0.48 x ( 1 - 0.42 x 0.48 ) x 1000 x
2
(d )=
20.5 x 1000000
d = 99.5 mm Tension Reinforcement near top surface 20.5 x 1000000 / ( 0.87 x 415 x (1-0.42 x 0.48) x 99.5 ) A st = 2 = 714.73 mm / m ( 100 x A st ) / ( b x d ) = 100 x 714.73 / ( 1000 x 99.5 ) =
0.718
From Design Shear Strength of Concrete Table :-
For ( 100 x A st ) / ( b x d ) =
τC= 0.718 , 0.53 N / mm ² Since thickness of slab will be kept at least 90 mm, the effective depth will be If the % of reinforcement is kept unchange, than τ C = ( 90 - 35 ) = 55 mm. 0.53 N / mm 2 as before. Hence tension reinforcement : ( 55 / 99.5 ) x 714.73 = A st = 395.08 mm 2 / m dia. Bars - @ Provide 12 mm 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. Shear force ( V u ) : Vu= 0.5 x ( 2.4 - 0.3 ) x 55.83 = 58.6 KN
Therefore, τ V = Now τ V
> ( greater than )
1.065 N / mm ² 58.6 x 1000 / ( 1000 x 55 ) = the vertical slab is unsafe in Shear. hence τC ,
Bending moment at the mid-span :
M u
w l
2
16
( 55.83 x 2.1 x 2.1 ) / 16 =
15.39 KN-m/m
Tension Reinforcement 0.87 x 415
x A st x (
55 - 415 / ( 15 x 1000 ) x
55 x A st - ( A st ) a = 0.028 A st =
Taking ( - ) sign, hence A st =
2
x
A st ) =
15.39 x 1000000
0.028 =
42625.68
c = 42625.68 b = -55.000 2 #NUM! mm / m ( When taking + sign) #NUM! mm 2 / m ( When taking - sign)
#NUM!
dia. Bars - @ 200 mm C / C, Provide 12 mm the spacing may be increased to 300 mm C / C, Temperature & Shrinkage Reinforcement = dia. Bars - @ 300 mm C / C, Provide 10 mm
2
mm / m on the front face. towards the top. ( 0.12 / 100 ) x 90 x 1000 = 108 mm 2 / m may be provided in vertical direction.
The toe bends upward as a cantilever due to upward soil reaction. Design of Toe : Bending moment ( M u ) : Mu =
( 2 x 264.915 + 223.985 ) x 0.6 x 0.6 / 6 =
Effective depth for balanced condition is found as : 0.36 x 15 x 0.48 x ( 1 - 0.42 x 0.48 ) x 1000 x d = 147.8 mm
, Say
2
(d )= 150 mm
45.2 x 1000000
45.20 KN-m/m
Tension Reinforcement : A st =
= Provide
16 mm
45.2 x 1000000 / ( 0.87 x 415 x (1-0.42 x 0.48) x 150 ) 2
1045.34
mm / m dia. Bars - @ 85 mm C / C.
( 100 x A st ) / ( b x d ) =
100 x 1045.34 / ( 1000 x 150 ) =
0.697
From Design Shear Strength of Concrete Table :-
For ( 100 x A st ) / ( b x d ) =
τC=
0.697 ,
Overall depth of toe slab = ( 150 + 35 ) =
185 mm
Critical section for shear occurs at an effective depth i.e. of the vertical wall. Shear force ( V u ) :
Vu=
0.52 N / mm ² 0.150 m away from the front
0.5 x ( 264.915 - 151.96 ) x 1.35 x 10 =
Therefore, τ V =
762.463 x 1000 / ( 1000 x 150 ) = 5.080 N / mm > ( greater than ) , the toe slab is unsafe in Shear. τ C , hence
As τ V
V uc =
Now
dia. Bars - @ Provide 10 mm Shear Strength of inclined bars is :
V us
0.87 f y A sv Sin(450 )
0.529 x 1000 x 150 / 1000 =
762.46 KN/m
²
79.35 KN/m
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
Hence, strength of toe slab in shear is V u = ( V uc + V us )
Vu =
( 79.35 + 20.04 ) =
99.39 KN/m
< (less than the . which is 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 dia. Bars - @ 180 mm C / C, at right angles to main reinforcement bars. Provide 10 mm Design of Shear Key : The key projects 0.05 m below the base, and its lowest point is below the G.L. in the front. Factored bending moment on the key is ( M u ) : 1.76
M u 1.5
K
1.76
p
1.5
= 63.99 x
1.55 m
h h 1.5 dh
1.5 x
2
2.37 x 18 x (
h - 1.5 h ) dh
1.5
1.76
h3 h2 1 . 5 3 2 1.5
= 0.12 KN-m/m
This is a small bending moment. A shear key of thickness 100 mm and reinforcement of 10 mm dia. bars - @ 200 mm C / C will be sufficient.
having vertical
Design of Counterforts : For taking up tension, steel reinforcement should be provided alon the sloping edge and the °. One counterfort supports earth pressure from 3 m width of vertical is α = 21.45 the vertical slab. Maximum bending moment occurs where the counterforts meets the b ase. At this point depth of earth = ( 6 - 0.6 ) = 5.4 m . Factored bending moment ( M u ) : -
M u 1.5
1 6
K a h 3
3 = 1.5 x ( 1 / 6 ) x ( 0.422 ) x 18 x 3 x ( 5.4 ) = 897.07 KN-m/m
³
Even for a rectangular beam of width = balanced condition, is given by -
300 mm,
0.36 x 15 x 0.48 x ( 1 - 0.42 x 0.48 ) x 300 x
(d )=
d = 1202.1 mm
the required effective depth for
2
897.07 x 1000000
1100 mm Overall depth of counterfort available at the base = 3000.0 mm, there will be at least = ( 3000 - 100 ) = 2900.0 mm, Hence, the dimensions of counterfort are alright. Tension reinforcement required in the counterfort is 0.87 x 415
x A st x (
, Say
2900 - 415 / ( 15 x 300 ) x 2
2900 x
and effective depth available which is more than required.
A st ) =
897.07 x 1000000
0.092 =
A st - ( A st ) x
a = 0.092
2484614.32 c = 2484614.32
b = -2900.0 A st =
Taking ( - ) sign, hence A st =
2
30640.33 mm / m ( When taking + sign) 881.41 mm 2 / m ( When taking - sign) 881.41
2
mm / m 5 bars in each row. Provide 16 mm dia. Bars 10 Nos. in two layers of The force which tries to separate the strip from the counterfort is ( 1.5 x 0.422 x 18 x 4.9 x 2.7 ) = 1.5 x K a x ϒ x h x ( 3 - 0.3 ) = = 150.74 KN/m Area of two legged ties ( 150.74 x 1000 ) / ( 0.87 x 415 ) = 417.50 mm ² / m 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.
V u 1.5
1
K a h 2 3 =
1.5 x ( 1/2 ) x 0.422 x 18 x 3 x ( 6 - 0.6 ) ² = 498.37 KN V u tan( ) 498.37 - ( 897.07 / 2.9 ) x tan ( 21.45 ° ) = 376.9 KN d τ v = 376.9 x 1000 / ( 300 x 2900 ) = 0.433 N / mm ² 100 A st ( ( 100 x 10 x 200.96 ) / ( 300 x 2900 ) ) x Cos ( 21.45 ) ° = b d = 0.22
2 M u
From Design Shear Strength of Concrete Table :-
For ( 100 x A st ) / ( b x d ) =
V uc =
τC=
0.215 ,
( 0.335 x 300 x 2900 / 1000)
0.335 N / mm ²
= 291.45 KN
For the Ties already provided :V us = 0.87 x 415 x 2 x ( 3.141 / 4 ) x 10 x 10 x 2900 / ( 300 x 1000 )
= 548.23 KN ( V uc + V us ) = ( 291.45 + 548.23 ) = It is seen that
a
( V uc + V us )
839.68 KN
> (greater than ) V u
M u d
tan( )
Hence, the counterfort section is safe against shear. Bending moment in the counterfort decrease towards the top. Tension reinforcement can, therefore, be curtailed accordingly. Hence, depth from top where 5 bars out of are no longer required. 10 bars h= ( 5.4 / 2 ) = 2.700 m It is proposed to extend of the inner layer by 5 bars 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.
At this point, Shear Force is -
V u 1.5
1 2
K a h 2 3 =
( 1 / 2 ) x 1.5 x 0.422 x 18 x 2.508 x 2.508 x 3
Effective depth at this point = ( 0.3 + 2.508 x Tan 21.45° - 0.1 ) = Available tension reinforcement after cut-off is A st =
( 5 x 201.056 ) =
( 100 x A st ) / ( b x d ) =
= 107.50 KN 1.185 m
1006.4 mm ²
100 x 1006.4 / ( 1100 x 300 ) =
0.305
From Design Shear Strength of Concrete Table :-
For ( 100 x A st ) / ( b x d ) =
For the
V uc =
10 mm
τC=
0.305 ,
( 0.35 x 300 x 1100 / 1000) dia. Two legged ties spaced at
0.350 N / mm ²
= 115.50 KN 300 mm C / C
V us = 0.87 x 415 x 2 x ( 3.141 / 4 ) x 10 x 10 x 1100 / ( 300 x 1000 )
= 207.95 KN ( V uc + V us ) = ( 115.5 + 207.95 ) = 2 ( V + V ) = ( 0.667 x 323.45 ) = uc us 3 Vu It is seen that < (less than )
323.45 KN 215.74 KN 2 ( V uc + V us ) 3
Hence, the can be safely curtailed at this point. 5 bars 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 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 / m Sectional area of two legged vertical ties required here is = ( 105.3 x 1000 ) / ( 0.87 x 415 ) = 291.65 mm ² / m Spacing of 10 mm dia. Two legged vertical ties is = ( 2 x 78.5375 / 291.65 ) x 1000 = 538.60 mm C / C Provide a spacing of 150.00 mm C / C at this point. 2 Towards the point C , the vertical load intensity is only -124.74 KN/m . Total downward force 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 ² / m Adopt a spacing of C / C at this point. 300.0 mm Joint Reinforcement :At the junction of the vertical wall with the base slab, a rigid connection can b e ensured by providing additional joint reinforcement. of concrete area may be provided as 0.30 % 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 ² / m dia. Bars - @ 180 mm C / C, as joint reinforcement. Provide 12 mm
0.3 m
4.5 m W1=
X1=
1.05 m
X3=
1.90 m
60.75 KN W3=
6.0 m 247.90 KN
Pa =
136.73 KN
0.600 m 1.5 m
W2=
54.00 KN
2.00 m
0.600 m
1.55 m O
0.6 m
0.05 m 3.6 m X 2 = 1.80 m
0.3 m 12 mm dia. Bars @ 300 mm C/C
10 mm - 2 Legged ties @ 300 mm C/C
2.508 m 16 mm - 5 Nos.
Cut off point
6.0 m
16 mm - 10 Nos. 10 mm - 2 Legged Ties, 300 mm C/C 10 mm - 425 mm C/C
10 mm - 2 Legged Ties, 150 mm C/C
0.3 m 12 mm dia. Bars @ 300 mm C/C
10 mm - 2 Legged ties @ 300 mm C/C
2.508 m 16 mm - 5 Nos.
Cut off point
6.0 m
16 mm - 10 Nos. 10 mm - 2 Legged Ties, 300 mm C/C
10 mm - 2 Legged Ties, 150 mm C/C
10 mm - 425 mm C/C 12 mm dia. Bars @ 95 mm C/C 10 mm - 180 mm C/C
0.19 m
0.05 m
16 mm bars @ 85 mm C/C
12 mm - @ 200 mm C/C
10 mm - 200 mm C/C
12 mm @ 45 mm C/C
3.6 m 0.6 m
0.1 m Side View
2.4 m 0.30 m Counterforts
1.25 m
12 mm - 45 mm C/C, increased to 200 mm C/C towards wall 12 mm - 200 mm C/C
0.600 m 12 mm - 200 mm C/C
End View
1.25 m
12 mm dia. - 200 mm C/C, increased to 300 mm C/C at Top
12 mm dia. - 95 mm C/C, Increased to 300 mm C/C towards Top
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
Sample Location Angle of internal Cohesion (c) Number friction (F) 1 93º42' E 26º40' N 2 93º43' E 26º39' N 3 93º44' E 26º37' N 4 93º46' E 26º38' N 5 93º47' E 26º37' N 6 93º50' E 26º34' N 7 93º53' E 26º34' N 8 93º55' E 26º31' N 9 93º57' E 26º30' N 10 93º57' E 26º28' N 11 93º57' E 26º26' N 12 93º55' E 26º24' N 13 93º54' E 26º22' N 14 93º53' E 26º19' N 15 93º52' E 26º16' N 16 93º50' E 26º13' N 17 93º50' E 26º08' N 18 93º47' E 26º03' N 19 93º48' E 26º02' N 20 93º48' E 25º58' N
22 17 33 28 27 30 36 17 36 28 18 20 18 17 16 23 15 18 26 18 23.15
0.5 6 3.03 1.6 2.6 1.05 0.42 1.32 2.2 2 1.32 0.64 0.34 0.63 0.51 1.03 0.8 0.67 0.45 0.31 1.371
N/cm2 S Alluvial Sandy S D
The geote bank stability provided val Textura subordinat 5x10-4 to 1x10 with
Different geotechnical parameters and critical height for a few locations along the bank o Sample Location
γ
β
cu
ɸu
Barguriagaon Butalikhowa Golaghat Dachmuagaon
1.79 1.87 1.54 1.53
41.25º 31.63º 41.25º 41.87º
2.6 2.2 1.05 2
27º 36º 30º 28º
c' = 2/3 cu Ø' = tan-1 173 147 70 133
18.76º 25.84º 21.05º 19.52º
Hc 31.74 290.91 18.19 29.12
Parameters in relation to bank stability at two different conditions for the Dhansi b 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 57 30° 0° 14.93 0.91 0.019 714.98 21.29° 161.59 1.31 0.0172 12 35° 0° 12.69 0.91 0.019 607.61 21.29° 075.03 1.31 0.0172 57 40° 0° 10.99 0.91 0.019 526.36 21.29° 045.33 1.31 0.0172 34 45° 0° 09.66 0.91 0.019 462.51 21.29° 031.22 1.31 0.0172 23 50° 0° 08.58 0.91 0.019 410.85 21.29° 023.22 1.31 0.0172 17 55° 0° 07.68 0.91 0.019 368.03 21.29° 018.16 1.31 0.0172 13 60° 0° 06.93 0.91 0.019 331.84 21.29° 014.69 1.31 0.0172 11 65° 0° 06.28 0.91 0.019 300.72 21.29° 012.19 1.31 0.0172 09
70° 0° 05.71 0.91 0.019 273.61 21.29° 010.30 1.31 0.0172 07 75° 0° 05.21 0.91 0.019 249.67 21.29° 008.82 1.31 0.0172 06 80° 0° 04.77 0.91 0.019 228.32 21.29° 007.64 1.31 0.0172 05 85° 0° 04.37 0.91 0.019 209.07 21.29° 006.67 1.31 0.0172 5 90° 0° 04.00 0.91 0.019 191.58 21.29° 005.85 1.31 0.0172 4
Soil Bearing Capacities Soil Type oft, wet, pasty or muddy soil soil, loam, sandy loam (clay +40 to lay loam (clay +30% sand), moist Compact clay, nearly dry olid clay with very fine sand ry compact clay (thick layer) Loose sand Compact sand Red earth Murram Compact gravel Rock
kN/m² 27 - 35 80 - 160 215 - 270 215 - 270 -430 320 - 540 160 - 270 215 - 320 -320 -430 750 - 970 -1700
hnical properties of the bank sediments and analysis along the Dhansiri River channel have uable information in land resource evaluation. lly dominant Sandy-Clay-Loam (76%) with Clay-Loam (24%), ranging permeability from -6 cm/sec has characterized the bank sediments oor stability susceptible to liquefaction
f the Dhansiri River Zone Risk Unstable Risk Risk i River channel sediments
44.25 06.94 14.38 52.73 77.98 68.79 82.90 19.04 28.50
84.24 71.81 81.63 7.60 5.55
