CANTILEVER RETAINING 

WALL


What is a retaining wall?

A Retaining wall is a structure used to retain Earth or other materials and to maintain ground surface at different elevations on either side of it.


The main components of a retaining wall are :

Stem

Toe slab

Heel slab 

Shear key

 

Difference between cantilever retaining wall and counterfort retaining wall :

 

 

CANTILEVER RETAINING WALL

COUNTERFORT RETAINING WALL

·         It consists of Stem, Heel slab, Toe slab and Shear key

           It consists of stem, Toe slab, Heel slab and Counterforts

 

·         Stability is maintained by weight of retaining wall and weight of earth on heel slab.

           The vertical stem as well as heel slab act as continuous slab because of provision of counterforts.

·         It resists the horizontal Earth pressure as well as vertical pressure by way of bending of various components acting as cantilever.

          Counterfort reduces bending moment in stem and heel slab by providing support to them.

·         It is used when the height of wall is up to 6 m.

           It is used when the height of wall is above 6 m.

 

                           

Figure shows the deflection profile of a cantilever retaining wall

 

From the figure it is seen that the tension face of stem is near to the back fill, and the main reinforcement should be provided on tension face in vertical direction in stem.

For heel slab the top portion acts as tension face hence main reinforcement is provided on top face of  Heel. For Toe slab bottom portion of the toe slab is the tension face so main reinforcement is provided at bottom in Toe slab.

 

These are the simple steps for the design of a cantilever retaining wall when the following data has been provided in the question.

DESIGN STEPS

GIVEN DATA : HEIGHT , SBC OF SOIL, ø , µ , Unit weight of soil( ),GRADE OF CONCRETE AND GRADE OF STEEL

STEP 1 :  CO-EFFICIENT OF EARTH PRESSURE

·        Ka =  (1-sin∅)/(1+sin∅)

·        Kp =  (1+sin∅)/(1-sin∅)

STEP 2 : PRELIMINARY DIMENSIONS

·         dmin =  {qo/(γ )} * ((1-sin∅)/(1+sin∅))2

·         Overall Height = Given Height + dmin

·         Base Width B = √(3P/2γ)     where P = 0.5 Kγ  H2

·         Toe Width = B/3

·         Total Base Width = B + (B/3)

·         Thickness of Base Slab =  H/12   to  H/15

·         Thickness of Stem at Base

Mu = 0.5 K γ h2 x (h/3)   x 1.5                                                                                  (i)

Mu = 0.138 fck b d2                                                                                                                           (ii)

Comparing (i) & (ii) , value of d is obtained.

D = d + Effective Cover (generally 50mm)

Top width of stem is to be assumed (generally 0.2m).

Also provide shear key of suitable size.

STEP 3: STABILITY CALCULATIONS

                                  

Sliding Force Pah = 0.5 Ka γH2

Overturning Moment  Mo = 0.5 Ka γ H2 x  H/3 

LOAD TYPE

(col. 1)

VERTICAL LOAD

(col. 2)

PERPENDICULAR DISTANCE  (col. 3)

MOMENT ABOUT A

(KN m) (col. 4)

W1

W2

W3

W4

W5

Find the loads and add them ∑W  will be obtained

Distance is measured from toe

( col. 2 x col. 3)

Adding values of rows 1 to 5 of col. 4

MR will be obtained.

 

Find  , by using the equation ,    = MR Mo

Find eccentricity , e = 

Find Max. Pressure at toe σmax =    which must < SBC (hence safe)

Find Min. Pressure at Heel σmin =  which must be > 1.55 (hence safe)

Find Factor Of Safety against overturning = MR / Mo

Find Factor of Safety against sliding =

STEP 4: DESIGN OF STEM

·         Find factored Bending Moment Mu= 0.5 Ka  h2 x   x 1.5

·         Find Mu/Bd2and from SP-16 find pt  where d = D – Effective cover

·         Find Ast = ptBd/100

·         Provide suitable dia bars at suitable spacing by the use of Steel Table or SP-16

·         For Distribution Steel, Find Avg. thickness of stem =

·         Find Ast(min) = 0.12% x 1000 x (Avg. thickness of stem)

·         Provide suitable dia bars at suitable spacing

·         On Outer face use 0.06% steel (i.e Ast(min) / 2 )bothways.

·         Check for shear :

Find Vu =0.5 Ka  h2 x 1.5

τv= Vu/ bd , then find pt & by use of Table 19 on pg 73 of IS 456 -2000 find τc. (If τv  < τc  hence safe).

 STEP 5 : DESIGN OF HEEL SLAB

·         From Pressure Distribution Below Base diagram find,

·         Pressure at junction of stem with Heel Slab

·         Pressure at junction of stem with toe Slab

·         Total downward pressure on heel slab = weight of backfill + self weight of heel slab

·         Plot diagram showing Net Pressure on heel

·         Find factored shear force and Bending Moment

·         Find Mu/Bd2and from SP-16 find pt

                                   (where d = Base slab thickness – effective cover)

·         Find Ast = ptBd/100

Provide suitable dia bars at suitable spacing on top face of heel

·         For Distribution Steel

Find Ast(min) = 0.12% x 1000 x Base Slab thickness

Provide suitable dia bars at suitable spacing on bottom face

·         For crack control use 0.06% steel (i.e Ast(min) / 2 )bothways on bottom face

·         Check for shear :

Find Vu =0.5 Ka  h2 x 1.5

τv= Vu/ bd , then find pt & by use of Table 19 on pg 73 of IS 456 -2000 find τc. (If τv  < τc  hence safe).

STEP 6: DESIGN OF TOE SLAB

·         Total downward pressure on toe slab = self weight of toe slab

·         Plot diagram showing Net Pressure on toe

·         Find factored shear force and Bending Moment

·         Find Mu/Bd2and from SP-16 find pt

                                         (where d = Base slab thickness – effective cover)

·         Find Ast = ptBd/100

Provide suitable dia bars at suitable spacing.

Half RF of stem anchored in toe will serve as toe RF.

·         For Distribution Steel

Find Ast(min) = 0.12% x 1000 x Base Slab thickness

Provide suitable dia bars at suitable spacing.

STEP 7: DESIGN OF SHEAR KEY

·         Find Ast(min) = 0.12% x 1000 x D where D is width of shear key

·         Total Ast in key = half of main RF in stem + (Ast(min) / 2 value which found  in step 4) 

·         If Total Ast in key > Ast(min)                          (hence O.K)

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