UNIT 4

Earth Retaining Structures

4.1 Types of Retaining Walls

In order to retain earth in filling having a greater height, retaining walls are constructed. The retaining walls are usually preferred in following construction:

(1) Rail and road projects where earth filling is required.

(2) If basements are provided in buildings.

(3) Wing walls and abutments act as retaining walls.

There are following types of retaining walls:

(1) Cantilever or T-shaped retaining wall

(2) Counterfort retaining wall

(3) Buttress retaining walls

4.2 Components of retaining wall 

The cantilever walls may prove to be uneconomical if the retaining wall’s height is more than a certain limit. In order to achieve economy, counterfort retaining walls are provided.

These retaining walls are vertical beams which are connected to the stem as well as heel slab by reinforcement.

The design of heel slab and vertical stem may be treated as a continuous slab and not as a cantilever slab.

If the toe side is provided with the counterforts, the retaining wall is known as buttress wall.

Cantilever, counterfort and buttress retaining walls are shown in Figs. 3.1 and 3.2.

(a) Cantilever retaining wall without toe projection

(b) Cantilever retaining wall with fillets

                                                            Fig. 4.1

(a) Counterfort Retaining Wall :

Fig. 4.2

4.3 Behavior and application of retaining walls

If a wall retains the earth level upto the top of retaining wall, it is known as retaining wall without surcharge.

If the earth on the retained side is not level or the earth carries load, the earth is said to have surcharge.

In this case, the pressure exerted by earth on wall is more.

Fig.  4.3

4.4 Stability criteria for Cantilever Retaining Wall                                                                                                                                                                                               

Fig. 4.4               Fig. 4.5

• Active earth pressure is defined as horizontal pressure exerted by retained earth on retaining wall.
• Passive earth pressure is defined as the resisting pressure applied by wall on the retained earth.
• The variation of this earth pressure is linear along the depth of retained earth.
• The total earth pressure upto the depth H (Fig. 3.4) is the area of pressure diagram which is 1/2 KaWH2 and acts at a height of H/3 from bottom. Hence, moment due to horizontal  pressure = KawH3              ...(1)

Where Ka =  and  is the angle of repose of soil.

• In case of a sloping surcharge at an angle

 to the horizontal pressure at a depth of

H1 due to earth= Ka WH1 and total earth

Pressure upto a depth of H1 = KaW

Which acts at a height of H1/3 and parallel

To the ground surface.

Moment about the base= KaWcos …(2)

Where, Ka = cos                     Fig. 4.6

• If the earth has a level surcharge of ws/unit run, then pressure at all depths is some of magnitude KaWs and the surcharge pressure at depth H is KaWs H acting at H/2 from bottom (Fig. 3.6)

Moment due to this at bottom =KaWH2 …(3)

4.5 Design of T-shaped or Cantilever Retaining Wall

• The components of cantilever retaining wall are shown in Fig. 4.7.
• Stem is designed as a cantilever retaining earth.
• The toe and heel slabs are also designed as

Cantilevers resisting upward soil pressure and

Downward earth pressure.

• The retaining wall is provided with or without

Shear key at the bottom of the base slab to protect

The wall against sliding.

4.5.1 The general guidelines to decide the preliminary dimensions of the retaining wall:

Following guidelines should be adopted while deciding the dimensions of a cantilever retaining wall:

(1) Top width of stem, b2= 200 mm

(2) Bottom width – Design for the maximum bending moment.

(3) Height of retaining wall = H

(4) Width of base slab = b

(5) b is usually kept between 0.5H to 0.6 H for walls

Without surcharge and 0.7H for walls with surcharge.

(6) Toe projection = b/3       

Fig. 4.8

(7) The thickness of base slab is normally kept as same as the bottom width of   

Stem i.e. b1 =H1

(8) The wall is to be checked for horizontal sliding. The factor of safety against  

Sliding is > 1.5,

Where W is total vertical load,  is coefficient of friction and p is horizontal

Earth pressure.

(9) If necessary, a key is provided.

The ratio of overturning moment to the restoring moment should be less                                               than 0.87 or the ratio of restoring moment to the overturning moment should be more than 1.15.

Problems based on Design of Cantilever Retaining Walls

Example 4.5.2 : Design a cantilever retaining wall to retain earth for a height of 4m above ground level. The earth density is 18kN/m3 and its angle of repose is 30. The embankment is horizontal at its top. The SBC of soil is 200 kN/m3 and coefficient of friction between soil and concrete is 0.5. Take M20 grade concrete and Fe415 steel.

Ans:

H2  =  3.5m;=18 kN/m3 ;=30 ;SBC qo=200 kN/m2

 = 0.5 ; fck=20N/mm2 ; fy=415 N/mm2

Step 1 : Coefficient of active earth pressure:

Ka =  = =

 Minimum depth of foundation is

ymin =  = = 1.23 m

Provide depth of foundation =1.25 m

 Height of retaining wall =4 + 1.25 = 5.25 m

Step 2 : Preliminary dimensions of retaining wall

b =0.5 H to 0.6 H = 0.5  5.25 to 0.6  5.25 = 2.625 m to 3.15 m

Say b =3m 

Toe projection =b/3 = 3/3 = 1 m

Thickness of base slab = Thickness of stem =  =  = 0.4375 m say 0.45 m

Assume the top width of stem as 0.2 m. Fig. 3.9 shows dimensions of retaining wall. Various forces on retaining wall are shown in Fig. 1-Q. 3.7.

(a)                                                                                                                                                      (b)
 

Fig. 4.9

Step 3: Check for Stability :

Horizontal pressure pH =  Ka H2 =  18  5.252= 81.86kN

Overturning moment M0 = PH= 81.86   = 143.26 kN-m

Factor of safety for overturning F1 = = 2.24 > 1.15.

Hence OK

F2 = = = 0.9 < 1.5

Hence, shear key is to be provided.

Step 4: Pressure under base slab

Total moment about point O = M – M0  = 369.63 – 143.26 = 226.37 kN-m

Total vertical load =186.3kN

Horizontal distance from O where resultant intersects the base line,

= = 1.215 m 

 Eccentricity, e  = – 1.215 = 0.285 m

 Maximum pressure p1 = = = 97.497 kN/m2

Minimum pressure p2 = = 26.703 kN/m2

Thus p1< SBC of soil, hence ok.

Step 5 : Design of stem :

Stem acts as a cantilever of height 4.80 m subjected to a uniformly varying load of Kah

Maximum moment at the base of cantilever

= Kah2 =  Kah3 =  18  4.803 = 110.592 kNm

 Mu = 1.5  110.592 = 165.888 kN-m

For M20 and Fe415, Mu(lim) = 0.138 bd2 fck

 165.888  106 = 0.138  1000  d2 20

d2 = 60.104  103 

d = 245.16 mm

Depth  d = 350 mm  and overall depth D = 400 mm

Main steel :

Mu = 0.87 fy Ast d

165.888  106 = 0.87  415  Ast 350 

 1312.74 = Ast

 Ast2– 16867.5Ast + 1312.74 16867.5 = 0 

 Ast = 1434.78 mm2

Use 12mm  bars

 s =  1000 = 86.15mm

Provide 12 mm bars @ 80 mm c/c

Distribution steel :

Average thickness of wall =  = 325 mm

Ast = 1000  325 = 390 mm2

Providing 195 mm2 on each face and using 8 mm bars.

S =  1000 = 257.7 mm

Provide 8 mm bars @ 250 mm c/c on tension face.

A mesh of 8 mm bars @ 250 mm is given on compression face of the wall.

Step 6 :Curtailment of vertical bars  : 

One third of vertical bars are curtailed at a height of 1.6 m from base and another 1/3 at a height of 3.2 m from base.

Check for shear : V = pH = 87.86 kN

Vu = 1.5  81.86 = 122.79 kN 

v =  = 0.273 N/mm2

P =  = 0.20  

­e = 0.4 N/mm2

 No shear R/F is required.

Step 7 :Design of Toe slab

Pressure at face of toe = 26.703 + (97.497 – 26.703)=73.90 kN/m2

M = 73.90 + 1 (97.497 – 26.703)  1=84.17 kN.m

Mu= 1.5 84.17 = 126.25 kN.m

d = 350 mm

 126.25  106 = 0.87  415  Ast 350 

 999.10 = Ast

A– 16867.5 Ast + 16867.5  999.10=0

Ast = 1066.54 mm2

Using 12 mm bars

s = 1000 = 106.04 mm                       Fig. 4.10

Providing 12 mm bars @ 300 mm c/c in both directions.

Step 8: Design of Heel slab

Width of heel slab = 1.55 m

Pressure at the face of column = 26.703 +   (97.497 – 26.703)=63.28 kN/m2

Weight of back fill = γH1 = 18  4.80 = 86.4 kN/m

Self weight = 0.45  1  25 = 11.25 kN/m

 Total load = 86.4 + 11.25= 97.65 kN/m

 Mmax = 97.65 – 26.703  – (63.28 – 26.703)  1.55 1.55

= 70.725 kN-m

Mu = 1.5 70.725 = 106.08 kN-m

Main steel :

106.08  106 = 0.87  415  Ast 350

 839.52 = Ast

 A– 16867.5 Ast + 839.52  16867.5 = 0 

 Ast = 886.05 mm2

Use 12mm bars, 

 s =  = 127.6 mm

provide 12 mm  bars @ 120 mm c/c in both directions.

Step 9 :Design of shear key

Pressure at face of shear key = 73.90 kN/m2

Coefficient of passive earth pressure

Kp = = 3

If ‘a’ is the projection of shear key, resistance offered by passive earth pressure.

= Kp vertical pressure = 3  73.9  a = 221.7a kN

  Factor of safety against sliding F2  =  = 1.5

 = 1.4  83.835 + 221.7a = 114.604  a = 0.138 m

  Provide 200 mm deep shear key.

Fig. 4.11

Example 4.5.3: Design the stem of a RC cantilever retaining wall, retaining leveled earth 5 m above the base level. Take the density of earth as 18 kN/m3 and angle of repose as 30. Toe projection 1.8 m, heel projection 1.7m, thickness of base slab as 450 mm.

Ans.:

 = 18 kN/m3

Ka =  = =  

Stem acts as a cantilever of height 5 m subjected

To a uniformly varying load of Kah.

Maximum moment at the base of cantilever

=  Kah2 =  Kah3 =  18  53

= 125 kN-m

Assume M20 concrete grade and Fe245 steel.                          Fig. 4.12

Mu = 0.87 fy Ast d

125  106 = 0.87  415  Ast 450

 769.36 = Ast

 A– 21686.7 Ast + 769.36  21686.7 =0 

Ast = 798.78 mm2

Using 12 mm bars s =  = 141.58 mm

 Provide 12 mm  @ 140 mmc/c

Distribution steel :

Average thickness of well =  = 325 mm

Ast =  1000  325 = 390 mm2

Providing 195 mm2 on each face and using 8mm bars.

s =  = 257.77 mm

 Provide 8 mm bars @ 250 mm c/c on tension face.

A mesh of 8 mm bars @ 250 mm is given on compression face of the wall.

Curtailment of vertical bars :

One third of vertical bars may be curtailed at a height of 1.67 m from base and another one third at a height of 3.33 m from base as shown in Fig.3.13.

Check for shear:

V = PH = KaH2 =  18  52 = 75 kN 

Vu = 1.5 75  = 112.5 kN,  V =  = 0.25 N/mm2

P =  = 0.25  

c =0.4N/mm2 No shear R/F is required.

Fig. 4.13

Example 4.5.4: Determine the dimensions of a T-shaped retaining wall for a height of 5 m above the ground level. The top of the earth is surcharged at 20  with horizontal. The angle of repose of earth is 30 and its density is 20kN/m3. The safe bearing capacity of soil is 50 kN/m2 and coefficient of friction between concrete and soil is 0.55.             

Ans.:

H2 = 5m;        =20 kN/m3;       =30 ;     SBC q0=90 kN/m2

  = 0.55;        fck=20 N/mm2 ;          fy=415 N/mm2

Ka =  = =  

Minimum depth of foundation

= =  = 1.5 m

 Provide a minimum depth of 1.5 m below ground level.

Total height of retaining wall  = 5 + 1.5 = 6.5 m

Use 500 mm thick base slab

Height of stem = 6.5 – 0.5 = 6 m

Let the top width of stem = 200 mm

Width of base slab= 0.7H = 0.7  6.5 = 4.55 m  4.6 m

Width of toe = 4.6/3 = 1.53 m      

Fig. 4.14

Example 4.5.5: Design the vertical stem of a T-shaped retaining wall for a height of 3.5 m above the ground level. The top of earth retained is horizontal. The angle of repose of earth is 30 and its density is 20 kN/m3. The safe bearing capacity is 100 kN/m2. Use M25 grade concrete and Fe415 grade steel.             

Ans.:

H2 = 3.5 m ;       =20 kN/m3 ;     =30;        q0=100 kN/m2 ; 

 = 0.5 (assume);  fck=25 N/mm2;  fy=415 N/mm2

Coefficient of active earth pressure Ka =  = =

Minimum depth of foundation = = = 0.55 m

Provide depth of foundation = 1m

Height of retaining wall = 3.5 + 1 = 4m

Preliminary dimensions of Retaining wall :

b = 0.5H to 0.6H = 0.5 4 to 0.6  4 = 2 m to 2.4 m  

Provide b = 2.4m Toe projection = = = 0.8 m

Thickness of base slab=thickness of stem =  =  = 0.33 m

Say 0.4 m

Let top of width of stem = 0.2 m

Fig. 4.15

Stability check:

Horizontal pressure PH = KaH2 =   20  42 = 53.33kN

Overturning moment M0 = PH= 53.33  = 71.1 kN.m 

F1 =  = 2.68 > 1.4  …OK

F2 =  = = 1.159 < 1.4

Hence shear key is to be provided.

Design of stem:

Maximum moment at the base of cantilever

=  Kah2 =   Kah3 =   20  43 = 71.11kN.m

 Mu = 1.5  71.11 = 106.67 kN-m=0.87 fy Ast d

 106.67  106 = 0.87  415 Ast 350

 844.125 = Ast

A– 21084.33Ast + 844.125  21084.33  = 0

Ast = 880.93 mm2

Using 12 mm bars, s =  = 128.38 mm

Provide 12mm @ 120 mm c/c.

Distribution steel:

Average thickness of wall =  = 300 mm

Ast =  1000  300 = 360 mm2

Providing 180 mm2 on each face and using 8 mm bars.

s =  = 279 mm

Provide 8mm  @ 270 mm c/c on tension side.

A mesh of 8mm  @ 270 mm c/c is given on compression side of the wall.

Curtailment of vertical bars:

One third of vertical bars may be curtailed at a, height of 1.2 m from base and another one third at a height of 2.4 m from the base.

Fig. 4.16

Example 4.5.6: A cantilever retaining wall retains an earth embankment with a horizontal top 3.75m above ground level. Design the stem for following data: Density of earth = 19 kN/m3, Angle of internal friction = 30, SBC of soil = 180 kN/m2 Coefficient of internal friction between soil and concrete = 0.5, M20 concrete grade and Fe 415 grade steel.

Ans.:

Given data:

H2 = 3.75m, r = 19 kN/m3,  = 30, SBC q0 = 180 kN/ m2 , = 0.5, fck = 20 N/mm2,

fy = 415 N/mm2 .

Coefficient of active earth pressure = ka =  =  =

Minimum foundation depth ymin = = = 1.05m

Provide foundation depth = 1.1m

 Height of retaining wall = 3.75 + 1.1 = 4.85 m

Preliminary dimensions:

B = 0.5 H to 0.6 H = 0.5  4.85 to 0.6  4.85

= 2.425 m to 2.91m

Assume b = 2.5 m

Toe projection = b/3 = 2.5/3 = 0.83

Provide 0.75 m

Thickness of base slab = Thickness of stem

=  = = 0.4m

Assume top width of stem = 0.2m.

Fig.4.17: Preliminary dimensions

Fig. 4.18: Forces on retaining wall

Stability check:

Horizontal pressure PH = karH2 =  19  4.852 = 74.49 kN

Overturning moment Mo = PH   = 74.49 = 120.42 kNm

Factor of safety for overturning F1 =  = 1.96 > 1.4

Hence O.K.

F2 = =  = 0.99 < 1.4

Design of stem:

Maximum moment at the base of cantilever

= karh2  =   19  = 93.02 kNm

Mu = 1.5  93.02 = 139.525 kNm

Mu(lim) = 0.138 bd2fck 

139.525  106 =  0.138  1000  d2 20

d = 224.84 mm

Provided  = 350mm and D = 400 mm

Ast =  0.5 

=  0.5  20 

= 1188.40mm2

Use 12 mm    bars.

S =  1000 = 95.16 mm

 Provide 12 mm              @ 90mm c/c

Distribution steel:

Average thickness of wall = = 300 mm

st =  1000  300 = 360 mm2 

Provide 180 mm3 on each face.

Use 8mm bars 

 S =  1000 = 279 mm

 Provide 8mm bars @ 270 mm c/c on tension face and compression face.

Fig.4.19: Reinforcement details

Example 4.5.7: Design of heel slab and stability analysis of a cantilever retaining wall is to be carried out for the following data: Height of horizontal backfill = 3.5m, Concrete grade = M20, Steel = Fe415, Backfill density = 18kN/m3, Angle of friction  = 30 coefficient of friction between soil and concrete  = 0.5, SBC of soil = 150 kN/m3 , depth of foundation = 1.1 m.

Ans.:

H2 = 3.5m, r = 18kN/m3,  = 30, SBC q0 = 150 kN/m2, M = 0.5, fck = 20N/mm2
fy= 415 N/ mm2  

Coefficient of active earth pressure = ka  =   =   =

Depth of foundation = 1.1 m

 Height of retaining wall = 3.5 + 1.1 = 4.6m

Preliminary Dimensions of Retaining Wall :

B  = 0.5 H to 0.6H = 0.5  4.6 to 0.6  4.6

= 2.3 m to 2.76m

Say b = 2.5m

Toe projection = b/3 = 0.75m

Thickness of base slab = Thickness of stem

=  =

= 0.38 m say 0.4m.

Let top width of stem = 0.4 m 

Fig. 4.20: Forces on retaining wall

Stability check:

Horizontal pressure  PH   =  karH2  = 18  4.62 = 63.48 kN

Overturning moment  M0  = PH  = 63.48 = 97.336 kNm

Factor of safety for overturning F1 =   = 2.30 > 1.4   Hence O.K.

F2  =  =  = 1.12 < 1.4

Design of heel slab :

Total moment about ‘o’ = M – M0 = 248.8 – 97.336 = 151.46 KN.m

Total vertical load = 158.56 kW

Horizontal distance from o where resultant intersects base line.

=  = 0.955 m

 Eccentricity  e =  – 0.955 = 0.295 m

Maximum pressure  

P1 =  =

=  108.32 kN/m2

Maximum pressure P2 =

= 18.52 kN/m2

Fig. 4.21

Fig. 4.22: Base pressure

0  –  18.52 1.35 –   x

2.5  –  108.32 2.5 – 89.8 

 Pressure intensity at1.35m  1.35 – x x  = = 48.492

=   18.52 + 48.492=67.012 kN/m2  

Weight of backfill = rH1 = 18  4.6 = 82.8 kN/m

Self weight = 0.4  1  25 = 10 kN/m

Total downward load w = 82.8 + 10= 92.8 kN/m

Mmax   = 92.8  – 18.52  –   (67.012 – 18.52)  1.35  1.35

=  52.96 kN-m

Mu =   1.5  52.96 = 79.43 kNm

Ast =    bd

=    1000   350=654.25 mm2

Provide minimum reinforcement of 12 mm bars @ 225 mm c/c in both directions.

Example 4.5.8: Design cantilever retaining wall for following data: Height of wall = 5.5m, Depth of foundation 1.2 m, Density of soil = 19 kN/m3, Angle of internal friction = 34,
SBC of soil = 225 kN/m2 coefficient of friction base and soil = 0.5. At the top level of backfill, a rectangular GSR is constructed 2m away from wall whose plan details are as below:

(i)Plan area = 15m  50 m (ii) Material M25 and Fe 415, (iii) Total weight of GSR under
     full water condition = 7500 kN. 

Ans. :

Fig. 4.23: Retaining wall

 = 34, H = 5.5m, r = 19kN/m3, SBC q0 = 225 kN/ m2 ,=0.5, fck = 20 N/mm2, fy = 415 N/mm2

Coefficient of active earth pressure Ka = =  =

Preliminary dimensions of Retaining wall:

B = 0.5H to 0.6 H = 0.5  5.5 to 0.6  5.5 = 2.75m to 3.3m

 Provide b = 3m

Toe projection = b/3 = 3/3 = 1m

Thickness of base slab = Thickness of stem =  =  = 0.46 m  0.5m

Let top width of stem is 0.2 m

Fig. 4.24: Forces on retaining wall

Stability check:

Horizontal pressure PH =  KarH2 =  19  5.52 = 81.18 kN

Overturning moment M0 = PH   = 81.18 = 148.83 kN-m

Factor of safety for overturning

F1 =  = 3.67 > 1.4  Hence O.K.

F2 =  =  = 1.65 > 1.4

Hence, shear key need not be provided.

Pressure under Base slab:

Total moment about O = 607.5 – 148.83 = 458.67 kNm

Total vertical load  =   298.75 kN 

=  = 1.53 m

Eccentricity e =  – 1.53 = – 0.03 mm  

e < b/

0.03  < 0.5

 Maximum pressure  = = =105.56 kN/m2

Minimum pressure =   = 93.6 kN/m2

Design of stem:

Maximum moment at base of cantilever

= ka  rh2 =  19  = 148.8 kNm

  Mu = 1.5  148.8 = 223.24 kNm 

Mu(lim)= 0.138 bd2fck

  223.24  106   =  0.138  1000   d2  20 

 d = 284.40 mm

   Provide d  =  450mm and overall depth D = 500 mm

Ast =  

=   =1475 mm2

Use 12m m    bars, S    = 1000 = 76.67 mm

 Provide 12 mm     bars @ 75mm c/c

Distribution steel:

Average thickness of wall =  = 350 mm

Ast =  1000   350 = 420 mm2 (Provide half on each face) 

Use 8mm bars 

S =  = 239.3mm

 Provide 8mm bars @ 230 mm c/c on tension face and compression face of the wall.

Curtailment of vertical bars:

One third of vertical bars are curtailed at a height of 1.67m from base and another one third at 3.33m from base.

Check for shear:

V =  PH  = 81.18 kN Vu = 1.5  81.18 = 121.77kN

v  =  = 0.27 N/mm2 

Pt  =  =  = 0.25 %

c = 0.4 N/mm2  v < v <cmax

No shear reinforcement is required.

Design of Toe slab :

0 – 105.56 2  –  ?

3  – 93.6 3  –  11.96

2  –?

Pressure at face of toe = 97.58 x = 7.97

M =  97.58  +  1   (97.58 – 93.6 ) 1

=  48.79 + 1.32=50.11 kNm Fig. 4.25 : Base pressure

  Mu =  1.5  50.11 = 75.175 kN.m

d =  450 mm

 Ast = 

=  =473.25mm2

Astmin =  1000  450= 540 mm2

Use 12mm bars  S = (/4)  122 1000/540 = 209.44 mm

Provide 12mm     @ 300mm c/c in both directions 

Design of steel slab:

0  –  105.56 

1.5  –  A

3  –  93.6

 3  –  11.96

1.5  –  x

x =  = 5.98

  A = 105.56 – 5.98 = 99.58 kN/m2      

Fig. 4.26: Base pressure

Weight of backfill = rH1 = 19  5 = 95 kN/m

Self weight = 0.5  3  25 = 37.5 kN/m

Weight of water tank = 10  1.5  5 = 75 kN/m

Total downward load = 207.5 kN/m

 Max. Bm =  – 105.56  –   (105.56 – 99.58)1.5 

= 233.44 – 118.755 – 2.242=112.44 kNm

 Mu = 1.5  112.44 = 168.66 kN-m

Ast = =1093.79 mm2

Use 12mm  bars.

S =  = 103.39 mm

 Provide 12 mm     @ 100 mm c/c in both directions.

Example 4.5.9: Design an RCC cantilever retaining wall to retain earth embankment 4 m height above the G.L. The angle of surcharge is at an angle of 15 to horizontal weight of soil and its angle of repose is 18 kN/10m3 and 30 respective. The good foundation is available at depth of
1 m below GL, the SBC of soil = 160 kN/m3 coefficient of friction -  – 0.62. Use M20 and Fe415.

Ans.:

Given: Df    = 1 m   ;      H  =  4 + 1 = 5 m

1. Preliminary s/c.:

(a)  Assuming top width = 120 mm

(b) Width of base slab = 0.55 H to 0.75 H

=0.55  5 to 0.75  5

= 2.75 to 3.75 M

Assuming B = 3.5 m

(c)Toe projection=  or =  or    = 0.83 or 1.167

Assuming Toe projection  = 1 m

(d)  Thickness of base =  to  = 0.33 to 0.5 m

Fig. 4.27

Assuming thickness of base = 0.5 m

Thickness of vertical wall

Fa = H2 kah2

Ka =  cos  ,

 =  15 ,  = 30

Ka = 0.373

Pa =  0.373  18  (4.5)2 = 67.97 kN

PaH = Pa cos   = 67.97 cos 15 =  65.65 kN    

Moment   =  PaH = 65.65   (at base max) = 98.48 kN-m

98.48  106 =  0.138 fck bd2,  

Factored   Mu= 98.48  1.5 = 147.71  106 N-mm

Mu = Mulim ,147.71  106 =0.138 fck bd2

 147.71   106=0.138  20  1000  d2,   d=231.34 mm

D = d + cover = 231.34 + 40 = 271.34

AssumingD=400 mm

d    = 400 – 40 = 360 mm

Since the thickness or toe and heel slab

The height of vertical wall. So the design

Of v wall should be taken after toe and heel

1. Width of heel slab  :

= 3.5 – 1 – 0.4  = 2.1 m

2. Force on the body:

Fig. 4.28: Geometry of retaining wall

(i)     Earth pressure:

 Pa =  Ka w = 0.373  18  (5 + 0.56)2

= 103.78 kN

PaH = Pa cos   = 103.78  cos 15 

= 100.24 kN                                              

(ii) Self weight:

(a) Weight of earth upto top

w1 = Area × Density of soil; 

w1 = 2.1 × 4.5 × 18 = 170.1

x1 =  + 0.4 + 1 = 2.45 m

(b) Weight of soil above top

 w2 =  × 2.1 × 0.56 × 18 = 10.58 kN 

x2 =   × 2.1 + 0.4 + 1 = 2.8 m

(c)   Weight of vertical wall   

w3 = 0.2 × 4.5 × 25 = 22.5 kN; x3 

= 1 + (0.4 – 0.2) +  = 1.3 m

Weight of triangular portion of vertical wall

w4 =  × 0.2 × 4.5 × 25 = 11.25 kN 

x4 =  × 0.2 + 1 = 1.13 m                                        Fig. 4.29

(d) Weight of base slab

w5 = 3.5 × 0.5 × 25 = 43.75 m; x5 =   1.75 m

w = w1 + w2 + w3 + w4 + w5  = 258.18 kN

wx = (w1x1 + w1x2 + w3x3 + w4x4 + w5x5) = 564.89 kN.m

3. Check against stability:

(a) Check for overturning, overturning moment

=  1.2 × 1 +  = 1.2 × 100.24 ×  =222.93 kN-m

Restoring moment =  0.9 × wx = 0.9 × 564.89 = 508.4 kN-m

Restoring moment > overturning moment

508.4 > 222.93 safe.

(b) Check for sliding :

Sliding force = 1.4 × PaH = 1.4 × 100.24 = 140.34 kN

Restoring force = 0.9 w = 0.9 × 0.62 × 258.18 = 144.06 kN

Restoring force > sliding force   Safe.

(c) Check for base pressure:

1. Considering Pav :

MToe = wx – PaH ×  + Pav × B

 v =  w + PaH = 258.18 + 26.86 = 285.04 kN

MToe = 564.89 – 100.24  + 26.86  3.5 = 473.12 kN.m

Z =  =  = 1.66 e  =  – Z =  – 1.66 = 0.09

Pmax =  =  = 94.00 ≯ SBC ok.

(b) Neglecting Pav :

MToe = wx –  =564.89 – 100.24  = 379.11 kN-m

 v =  w = 258.18 kN Z =  =  = 1.47

e =  – z = 0.28 Pmax =   =

= 109.17 kN≯ SBC   safe

Pmin =  =  = 38.36 > 0  safe

4.  Design of toe slab:

Fig.  4.31: Base pressure   Fig. 4.32: Resultant pressure

From similar 

=  y2 = 50.58

Total  = y1 + y2 = 38.36 + 50.58=88.94

=  y4 = 42.486 

Total  = y3 + y4   = 38.36 + 42.486 =80.85  

Resultant pressure

At A = 109.17 – 12.5 = 96.67 kN/m3

At B = 88.94 – 12.5  = 76.44 kN/m3

Self weight = 0.5  25 = 12.5 

Shear force  = Area of  + Area of rectangle = (1) (96.67 – 76.44) + (1)  76.44

= 86.55 kN

Factored SF = 1.5  86.55 = 129.825 kN

BM = (Area of  C.G. Distance) + (Area of   C.G. Distance)

=  + 76.44  1  0.5 = 44.96 kN-m

Mu   = 1.5  45.96 = 68.94 kN-m

Check for depth:

Mu = Mulim 67.44  106 = 0.138 fck bd2

67.44  106 = 0.138  20  1000  d2 

d = 156.32

D = 156.32 + 40=196.32 mm <Dassumed 

Provide D = 500 m

d = 460 mm

Area of steel:

Ast =   bd

=  1000  460 = 423.39 mm2

Astmin =  b  D = 600 mm2 as Ast<Astmin

 Providing Ast = 600 mm2 Ld   + lo

47    + 46020.84

Assuming 12 mm    bars

Spacing  =   = 188.49 < 3 d or 300  ok.

Providing 12 mm   bar @ 180 mm c/c

Area of distribution steel:

=  b D =   1000  460 = 552 mm2
    Provide Astmin= 600mm2

Assuming 8 mm  bars

Spacing  =  = 83.77 ≅ 80 mm  ok.

Check for shear:

v =   =   = 0.28 N/mm2

Point % =   =  = 0.136 %

c = 0.28 N/mm2; v = c

Design of shear reinforcement is not required

(6) Design of heel slab :

Resultant pressure

(a) At  c =   80.85 – 12.5 – 72

=   – 3.65

(b) At D  =   38.36  – 12.5 – 80.96

=   – 55.1

Max S.F =    Area of  + Area of 

=    2.1  (55.1 – 3.65) + 3.65  2.1

=   61.69 kN

Factored SF  = 1.5  61.69 = 92.53 kN

BM= (Area of  cg. Dist) + (Area of  cg. Dist)

=   + 3.65  2.1 

= 83.68 kN-m

Factored B.M. = 1.5  83.68 = 125.52

Check for depth:

Mu = Mulim 

125.52  106 = 0.138  1000  d2 20  

d = 213.25 

D = 213.25 + 40  Fig.                            4.33: Pressure diagrams

= 253.25 mu < depress  o.k.

Provide D =  500 mm d = 460 mm

Area of steel:

Ast = 1000  460=783.86 mm2

Ld =  + ld 47  =  + 460

 = 38.65 mm

Assuming 16 mm bars

Spacing =  = 256.5 < 3d or 300 mm

 Provide 16 mm @ 250 mm.

Distribution steel:

Ast =  b  D =  1000  500 = 600 mm2

Assuming 8 mm   bars

Spacing  =  = 83.77

Providing 8     mm

Check for shear:

v =  =  = 0.2 N/mm2

pt =  =  = 0.17

c = 0.28 +  (0.17 – 0.15) = 0.296 N/mm2

k = 1

v < c 0.2 < 0.296

Shear reinforcement is not required.

(7) Design of vertical wall

Pa =  Kaw =  0.373  16   (4.5 + 0.56)2=76.40 kN

PaH = Pa cos  = 76.4 cos 15 = 73.79

pav = Pa sin  = 19.77

BM at base = PaH = 73.79  = 124.46 kNm

Mu = 1.5  124.46 = 186.69

Check for depth:

Mu =   Mulim 186.69  106 =   0.138  1000  d2 20

d =   260.07      D =   300.07 < Dassumed (400) ok.

 providing D = 400 mm  d = 360 mm

Area of steel   =   bd

=  1000  360 = 1581.13

ld   + lo

Shear force, v = PaH = 73.79 vu = 1.5  73.79 = 110 .68

ld   + 360

47   2046.75   43.54

Providing 20 mm     bars

Spacing = =198.69 mm < 3d or 300 mm 

Providing 20 mm     190 mm c/c

Curtailment: pressure at half height

Pressure at  = 2.25 from Top

Pa =  0.373  16  (2.25 + 0.56)2 = 23.56 kN

PaH = Pa cos  = 23.56 cos 15 = 22.75 kN

BM = PaH = 22.75  = 21.31 kN-m

Mu = 1.5  21.3 = 31.965 kN-m D =  = 0.3 m

d = 300  – 40 = 260 mm

Ast = = 260  1000

= 350.48 mm2

(Ast)min =  1000  300 = 360 mm

Providing Ast 360 mm2

Assuming 12 mm bar

Spacing =  = 314 < 3d or 300 ≅ 300 mm

Check for shear :

SF =  PaH = 73.79 kN 

Factored SF = vu = 1.5  73.79 = 110.685 kN

v =  =  = 0.307 

pt =     =   = 0.465

pt  c

0.25   0.36

0.5   0.48

c = 0.36 +  (0.465 – 0.25)  = 0.46

k = 1   c = 0.46 

 v < c safe.

Fig.4.34: Reinforcement details

Example 4.5.10: For the cantilever wall of height 3.5m, fix the basic dimensions of various elements. The angle of repose of soil is 30. SBC of soil is 200 kN/m3 and density of soil is 18 kN/m3. Friction coefficient between soil and concrete is 0.55. Do the check for stability, sliding and overturning. Design the stem of the retaining wall.

Ans.:

H2 = 3.5m, γ = 18kN/m3,  = 30, SBC q0 = 200 kN/m2, μ = 0.5, fck = 20N/mm2
fy= 415 N/ mm2    

Coefficient of active earth pressure = ka  =   =   =

 Minimum depth of foundation is

ymin =  = = 1.23 m

Provide depth of foundation  = 1.25 m

 Height of retaining wall =3.5 + 1.25 = 4.75 m

Preliminary Dimensions of Retaining Wall:

B  =  0.5 H to 0.6H = 0.5 4.75 to 0.6 4.75

= 2.375 m to 2.85m

Say b = 2.7m

Toe projection = 2.7/3 = 0.9m

Thickness of base slab = Thickness of stem

= 

= 0.395 m say 0.4m.

Let top width of stem = 0.2 m

4.6 Concept of Counter fort Retaining Wall

4.6.1 Various parts of a counter fort retaining wall:

A counter fort retaining wall is to be provided where the height of earth fill is greater than 6m.

Fig. **

Fig. 4.35: Parts of counter fort retaining wall

4.6.2 Design steps for a counter fort retaining wall:

Design steps:

1. Tentative trial section.  

2. Forces

3. Stability check

(i) Overturning  (ii) Sliding  (iii) Base pressure

4. Design of section

(i) Vertical wall 

 (ii) Toe slab  

(iii) Heel slab

(iv) Counter fort

(I) Trial section:

1. Base width = 0.6 H to 0.7 H (H = total height of wall)

2. Toe width =  to   or 

3. Thickness of vertical walls @ 0.3 m

4. Counter fort thickness @ 0.4 m

5. Thickness of base slab = 2l  or 4l

l  – Spacing of  counter fort in m

H – Height in m

4.6.3 Various forces acting on a counter fort retaining wall:

Forces:

(1)    Horizontal back fill :

Intensity of earth pressure  = Ka wH

Ka  =  Coefficient of active earth pressure.

Ka  =  

w   =  unit weight of soil

    = angle of repose           Fig. 4.37 : Pressure distribution diagram (Horizontal backfill)

Total pressure   =  Ka wH H Pa  =    KawH2

(2) Inclined backfill (with surcharge) :

Fig. 4.38: Pressure distribution diagram (Inclined backfill)

Active earth pressure,  Pa   =   Ka w 

Ka = cos 

Where,   = angle of surcharge  = angle of repose

(3) Passive earth pressure:

The passive earth pressure is exerted when it has tendency to move towards the backfill such a case may occur when the retaining wall supports soil of different depths on both.

PP =  Kp wH2  Kp  =  

(4) Self weight of wall

(i) Weight of earth (without surcharge)

 w1 =  width  height  density of soil

x1  =  distance of load w1 from toe i.e. (A)

(ii) Weight of earth (with surcharge)

w2  =  Area of triangle  density of soil

x2 =  distance of load w2 from A

(iii) Weight of wall

w3 =   Thickness  height  density of conc.

x3 =   distance of  load w3 from A 

Fig. 4.39: Loads of different part

In retaining wall

(iv) weight of base slab

w4  = Thickness  width  Density of conc

x4  = distance  of  load w4 from A

w = w1 + w2 + w3 + w4 M  @ A

MToe = w1 x1 + w2 x2  + w3 x3 + w4 x4 

4.6.4 Stability check for a counter fort retaining wall:

(i) Check against overturning : (Pg 33, cl-20.1)

(1) Overturning moment = Pa

(2) Restoring moment = wx = w1 x1 +w2 x2 + w3 x3  ….

0.9  Restoring moment  1.2  overturning moment

0.9 wx 1.2  Pa ( safe)

If unsafe change the dimensions

(ii) Check against sliding :  (clause 20.2)

(1) Sliding force  = 1.4  Pa

(2) Restoring force = 0.9  w 

Restoring force  sliding force

0.9  w  1.4  Pa                                    …Ok

If it is unsafe there is no need to change the dimensions but provide shear key.

(iii) Check for base pressure :

(1) Maximum pressure = Pmax =

Eccentricity = e =  – z z = 

Pmax  S.B.C. Of soil

(2) Maximum pressure = Pmin =  0

Problems based on Design of Counterfort Retaining Walls

Example 4.6.5: Design a retaining wall for the following data.

(1) Height above GL = 6M.  (2) Depth of foundation below GL = 1m

(3) The density of soil = 16 kN/m3  (4) Angle of repose = 30

(5) Surcharge = 10   (6) SBC of foundation = 150 kpa

(7) Coefficient of friction = 0.6.

Use M20 and Fe 415. The counter fort is on earth side only.

Ans.:

(1) Preliminary section:

Total height of wall = H = 6 + 1 = 7m

(i) Base width = 0.6H to 0.7H=0.6  7 to 0.7  7=4.2 to 4.9 m

Assuming B = 4.5 m

(ii) Toe width =  to  or = to  or  = 1.5 to 1.125 or 1.17

Assuming toe width = 1.2 m

(iii)  Assuming thickness of vertical wall = 0.5 m

(iv)  width of heel slab =  4.5 – 1.2 – 0.25 = 3.05 m

(v) Thickness of base slab = 2lH or 45

Assuming l = spacing of counter fort = 3.5 m

Thickness of base slab = 37 or 49

Assuming  = 40 cm

(2) Forces on retaining wall:

(i) Earth Pressure H1 = H + y1    

Fig. 4.40: Geometry of retaining wall

H1 = 7 + 0.54=7.54 m        

Pa  =     Ka w          Fig. 4.41: Height of surcharge

w =    Specific weight of soil

=   16 kN/m3  (given)

Ka =   cos 

Tan 10  = 

y1 = 0.54 m

 – angle of surcharge

= 10 and  – angle of repose

= 30 

Ka = 0.35                                            Fig. 4.42: Loads and pressure

Distribution diagram

Pa =  0.35  16  (7.54)2=159.18 kN

Acting at H1/3 from base   =  = 2.51m from base

Horizontal component PaH = Pa cos =159.18 cos 10=156.76 kN

Vertical comp Pav = Pa Sin = 27.64 kN

(3) Self weight:

(i) Weight of soil (up to top)

w1 = area  density of soil = 3.05  (7 – 0.4)  16 = 322

x1 = 3.05/2 + 0.25 + 1.2=2.975 m

(ii) Weight of surcharge,

w2  = ½  3.05  0.54  16 = 13.2 kN

x2 = 2/3  3.05 + 0.25 + 1.20 = 3.48 m

(iii) Weight of vertical wall (stem)

w3 = 0.25  6.6  25 = 41.25 kN x3  = 1.2 +  = 1.325 m

(iv) Weight of base slab,

w4 =  0.4  4.5  25 = 45 kN                 x4 =   = 2.25

w =  w1 + w2 + w3 + w4 = 421.55 wx = w1 x1 + w2 x2 + w2 x3 + w4 x4

wx = 1158.48 kN – m

(4) Check for stability:

(i) Check against overturning:

Overturning moment =  1.2  PaH=1.2  156.76  = 472.78 kN – m

Restoring moment = 0.9  Wx=0.9 (1158.48)=1042.63

Restoring moment > overturning moment

 safe against overturning

(ii) Check against sliding:

Sliding force  =  1.4  PaH = 1.4  156.76=219.46 kN

Restoring force =  0.9  w=0.9  0.6  421.55=227.64 kN

(iii) Restoring force is greater than sliding force   safe

Check against base pressure:

(a)     Considering Pav :

 Moment @ Toe = MToe  = w1 x1  + w2 x2 + w3 x3 + w4 x4  + Pav B – PaH

= 1158.48 + 27.64  4.5 – 156.76 

MToe = 888.86 –m              Z =    =

v =   421.55 + 27.64 = 449.19 kN     Z ==1.978 1.98

e =    – z =  – 1.98 = 0.27

Pmax =     =

≯S.B.C.(150 kN/m2)=135.74  ≯150

Pmin = => 0

= 63.88 > 0   OK.

(b) Neglecting Pav :

MToe  = w1 x1  + w2 x2  + w3 x3  + w4 x4 – PaH=1158.48 – 156.76 

= 766 kN – m

v = w = 421.45 kN     Z =   =  = 1.81

e =  – Z =  – 1.81 = 0.44

Pmax = ≯ S.B.C.=

≯150 kN/m2 =148.6 ≯150

  OK

Pmin = > 0=> 0=38.72 > 0  OK

(5) Design of vertical wall:

(i) A vertical wall is designed as vertical continuous slab spanning between counter forts

Span = spacing of counter fort = 3.5 m

Fig. 4.43: Shear force and Bending moment distribution diagram

Horizontal component of base pressure w1 = ka w  H1 cos 

Ka – coefficient of active earth pressure

w – Density of soil

w1   = 0.35  16   (7.54 – 0.4) cos 10=39.37

Maximum BM =  = = 40.19 kN.m

l  = spacing of counter fort

Ultimate BM  = Mu  = 1.5  40.19=60.29 kN.m

Maximum S.F = 0.6 w1l

V  = 0.6 × 39.37 × 3.5=82.67 kN

Vu  = 1.5 × V = 1.5 × 82.67 = 124.01 kN

(ii) Check for depth:

Maximum BM = Mu lim Mu = Mulim 

60.29 × 106  = 0.138 bd2 fck 60.29 × 106 =  0.138 × 1000 × d2 × 20

d  = 147 mm

Assuming cover 40 mm.

Dread  = d + cover = 147 + 40

= 187 mm  < Dassumed

< 250 mm

Providing D = 250 mm and d = 210 mm

(iii) Area of steel:

(a) Area of main steel,

Ast = 

Bd =     100 × 2

= 870.42 mm2 ld   + lo

ld =  =  = 47  ….  For M20 and Fe415.

47    + 210   14.81

Providing 12 mm dia bar.

Spacing  =  = 129.93 < 3d or 300 ok.

Providing 12 mm diameter bar @ 120 mm c/c from base upto 3.3. m.

Curtailment:

Providing 12 mm diameter bar @ 240 mm c/c at 3.3 m from base and upto top of wall.

(b) Area of distribution steel =  × b × D= × 1000 × 250 = 300 mm2

Assuming 8 mm diameter bar.

Spacing  =  = 167.59 < 5 d or 450 mm

Providing 8 mm @ 160 mm c/c.

(iv) Check for shear                                      IS : 456-2000/Page 73/table 19

v =   =    = 0.59 N/mm2

pt % =   = 

pt % = 0.44

c = 0.36 +  (0.44 – 0.25) = 0.45 N/mm2

v > c

Design of shear reinforcement is required

Design shear force = Vus

Vus = u – c bd   = 124 × 103 – 0.45 × 1000 × 210 = 29.51 × 103 N

sv = 

Assume 8 mm  5 legged stirrups.

(1) sv =  = 645.7 mm

(2) sv =     =   =226 mm

(3) sv = 0.75 d or 300 mm = 0.75 × 210 or 300  = 157.5 mm

Taking least value of the three

= 157.5  mm≅150 mm

 Providing 8 mm   stirrups @ 150 mm c/c

Final provision: 

(1) Main steel = 12 mm @ 120 mm c/c upto 3.3 m from base.

 = 12 mm @ 240 mm c/c from 3.3 to 6.6 m from base.

(2) Distribution = 8 mm @ 160 mm c/c steel

(3) Stirrups = 8 mm legged stirrups @ 150 mm c/c

Design of toe slab:

Designing as cantilever slab fixed at the face of vertical wall.

From similar  property

= 

y2 = 80.58

Total ordinate = y1 + y2 = 38.72 + 80.58 = 119.3                

Fig.  4.44: Upward pressure distribution diagram

 Resultant pressure

(1) At A = 148.6 – 10 = 138.6 kN/m2 ()

(2) At B =119.3 – 10 = 109.3 kN/m2 ()

Shear force = Area of triangle + Area of rectangle

  =  (1.2) (138.6 – 109.3) + 1.2 × 109.3

= 148.7 kN

Factored shear force

Vu = 1.5 × 148.7 = 223.12 kN

BM = (Area of rectangle × c.g. Dist.)

+ (Area of triangle × c.g.dist)

M =   1.2 × 109.3 ×  +  (1.2)

× (138.6 – 109.3)  × 1.2

=   92.76 kN.m

Mu = 1.5 × 92.76= 139.14 kN-m

Check for depth:

Maximum BM = 139.14 kNm 

Mulim  = 0.138 fck bd2

Mu lim  = BM max 

0.138 fck bd2  = 139.14 × 106

0.138 × 20 × 1000 × d2 = 139.14 × 106

d  = 224.5 mm

Assuming cover =40 mm                      Fig.  4.45: Resultant pressure

Diagram

 Drequired  = 224.5 + 40 = 264.5 mm <Dassumed (400 mm) 

 Providing D = 400 mm

d  = 400 – 40 = 3600 mm

Area of steel reinforcement:

Ast = b.d

=   × 360 × 1000=1146 mm2

Check for Ld

Ld =  + L0 For M20 and Fe 415 Ld=

Ld = 47  L0 = d = 360 mm

  47    + 360   20.92 mm

Assuming 16 mm  bars

Spacing  =   = 175.31 ≅ 170 mm < 3d or 300.

Providing 16 mm   bars @ 170 mm c/c

Provide at bottom because

Area of distribution steel =  b × D= × 1000 × 400 = 480 mm2

Assuming 8 mm  bars

Spacing  = < 3 d or 300=104 mm ≅ 100 mm

Providing 8 mm  bars @ 100 mm c/c

Check for shear

v =   =   = 0.62

 Pt =  Ast(pro      = 

=   = 1182.72 mm

= 

= 0.33

From IS page 73 table (19)

c = 0.36 +

= 0.3984 N/mm2

v > c

 Design of shear reinforcement is required.

c ≯ cmax

Design of shear reinforcement:

 vus = vu – c bd = 223.12 × 103 – 0.39 × 1000 × 360 = 82.7 × 103 N

Assuming 8 mm 5 legged stirrups.

(1) sv = = = 4.28mm (Not feasible)

(2) sv =  = =226 mm

(3) sv = 0.75 d or 300 mm

Taking least value = 226 mm 

 Spacing = 220 mm

 Providing 8 mm   5 lugged stirrups @ 220 mm c/c.

Final provision:

(1) Main steel  =   16 mm  @ 170 mm c/c 

(2) Distribution steel =   8 mm  @ 100 mm c/c 

(3) Stirrups  =   8 mm  5 legged stirrups @ 220 mm c/c.

Design of Heel slab:

(1) Base press

(2) Self weight = 0.4 × 25 = 10

Fig. 4.46             Fig. 4.47 

(3) Earth press

(a) At left edge = 6.6 × 16 = 105.6 (b) At right edge = 7.15 × 16 = 114.64

(4) Resultant press

(a) At left edge = 113.19 – 10 – 1056 = – 2.49 kN/m2

(b) At right edge = 38.72 – 10 – 11464 = – 85.52 kN/m2

Taking maximum value

w2  = 85.52 kN/m2

Maximum SF, v = 0.6 w2l=0.6 × 85.52 × 3.5  

l   = spacing of counter fort = 3.5 m

= 179.59 

Factored S.F.

vu = 179.59 × 1.5 = 269.38 kN

Max BM = 

M = 85.52 =87.3 kNm

Factored B.M.,   Mu = 1.5  87.3 = 130.95 kNm

Check for depth:

Mu = Mu(lim) 130.95  106 = 0.138 bd2 fck

130.95  106 = 0.138  1000  d2  20

d = 217.84 mm D   = 217.84 + 40=257.84  Dassumed (400 mm)

Providing D = 400 mm d = 400 – 40 = 360 mm

Area of steel:

Ast =  bd

= 1000 360 mm=1074.5 mm2

Check for Ld :

Ld   + L0

For Mw and Fe415  Ld = 47 

L0 = d = 360 mm 47    + 360

  18 mm

Assuming 16 mm  bar

Spacing =  =187 mm  3d or 300

Providing 16 mm  @ 180 mm c/c

Distribution steel:

Ast =  b  D= 1000  400=480 mm2

Assuming 8 mm  bars

Spacing = =104 ≅ 100

Providing 8 mm  @ 100 mm c/c

Check for shear:

V =  = =0.74 N/mm2

pt = ,  Ast(prov) =   = 1117 mm2

= =0.31

c = 0.36 +  (0.31 – 0.25)=0.38 N/m2