Retaining Counterfort Wall Design

Document No.ISBT_MAIN_01 Rev. No. 0

Project Title

Client

Name of Unit

Drawing Ref.

Designed by: G.C. Checked by: G.C. Approved by: Date: 30.11.06

Reference

A)

= 205.50 M= 205.14 M= 200.70 M= 0.00 M= -0.36 M= 4.80 M= 0.30 M= 0.36 M= 4.44 M= 5.10 M= 4.00 M= 2.00 M= 2.45 M= 0.60 M= 1.00 M= 0.60 M= 0.30 M= 0.75 M= 0.45 M= 0.42 M= 0.50= 2.00 T/M2

= 1.94 T/M3

= 1.00 T/M3

= 0.94 T/M3= M30= 5.00 Cm.

Overall Depth of Counterforts at bottom

Surcharge Load at Formed Ground Level on soil, qUnit weight of Saturated Soil, sat (Max.)Unit weight of Water, w Unit weight of Saturated Soil, sub (Max.)Grade of Concrete considered Clear cover to reinforcement provided

Overall Depth of Horizontal beam at top of Wall

Thickness of Counterfort assumed Overall Thk. of Wall spanning between Counterforts

Coefficient of Earth Pressure at Rest, k0

Overall Width of Horizontal beam at top of Wall

Overall Thk. of Base Slab spanning between Counterforts

Kolkata - 700019

Variation of Depth of Counterforts from -4.8 M starts at

Overall Clear Depth of Counterforts at bottom

Formed Ground Level at 205.5M corresponds to

Height of Retaining Wall above Formed Ground Level

High Flood Level (HFL) at 205.14M corresponds to

Overall Height of Retaining Wall

Overall Depth of Counterforts at top

DESIGN PARAMETERS FOR RCC RETAINING WALL:

Centre - to - centre distance of Counterfort (Span of Wall)

Height of Retaining Wall below Formed Ground Level

Formed Ground Level High Flood Level (HFL)

Height of Retaining Wall for saturated soil pressure, H1Height of Retaining Wall for submerged soil pressure, H2

Top of Slab at Basement Level

Fig. 1: Section showing Counterfort retaining Wall for Bus Parking Area at -4.8M Level

Design of Peripheral Retaining Wall at basement level for Bus Parking Area at -4.8 M level:

RCC DESIGN OF RETAINING WALL (-4.8M) FOR TERMINAL BUILDING OF MAIN ISBT

INTERSTATE BUS TERMINAL, SARAI KALE KHAN, NEW DELHI

COMMISSIONER (TRANSPORT), GOVERNMENT OF DELHI

MAIN TERMINAL BUILDING FOR ISBT

Institute For Steel Development & Growth

ISPAT NIKETAN', 1st. Floor

52 / 1A Ballygunge Circular Road

CALCULATIONS

0.30

0.60

0.45

2.0

1.0

5.10

0.30

0.42

Formed Ground Level

(EL. +205.50M)

High Flood Level (HFL)(EL. +205.14M)

Lateral Pressure on Walldue to Surcharge Load q(k )0x

Lateral Pressure on Wall

due to Saturated Soil(k H1)x sat0

H1 = 0.36 M

H2 = 4.44 M

x

Lateral Pressure on Walldue to Submerged Soil

(k H2)0 x subx


due to Water ( H2)w x

Counterfort Beam

EL. +200.70M

Retaining Wall

Basement Slab

Page 1 of 7


Project Title

Client

Name of Unit

Drawing Ref.


Reference

Kolkata - 700019









CALCULATIONS

i)

Since, the major portion of the wall will be in contact with water, the same has been designed based on the principles of uncracked design as per working stress method

Load Cases:

i) =k0satH1= 0.35 T/M2 (Triangular)ii) =k0subH2= 2.09 T/M2 (Triangular)iii) =k0q = 1.00 T/M2 (Uniform)iv) =wH2 = 4.44 T/M2 (Triangular)

Now for all practical purposes, there will be two types of combined load cases as follows:1.00 T/M2 = f16.88 T/M2 = f2

Here, Shorter Span, lx = 4.00 MLonger Span, ly = 5.16 M

ly/lx = 5.16/4 = 1.29 i.e. 1.3

Which gives from IS: 456 - 2000,

x (+ve) = 0.036 y (+ve) = 0.024

x (-ve) = 0.047 y (-ve) = 0.032

Mx (+ve) = Positive BM for Horizontal Span = x (+ve) x f1 x lx2 = 0.58 T-m/m

Mx (-ve) = Negative BM for Horizontal Span = x (-ve) x f1 x lx2 = 0.75 T-m/m

My (+ve) = Positive BM for Vertical Span = y (+ve) x f1 x lx2 = 0.38 T-m/m

My (-ve) = Negative BM for Vertical Span = y (-ve) x f1 x lx2 = 0.51 T-m/m

Here, Horizontal Span, lx = 4.00 MVertical Span, lz = 5.16 M

lx/lz = 4/5.16= 0.78

Considering total length of wall to be divided equally @ 4.0 M c./c. between Counterforts and a Longitudinal Beam running throughout at the top of wall, the wall panel closely follow case 2 (figure 2) of Chart 53 of page 185 of Reynold's Handbook, freely supported at top edge and fixed at other three sides

(Refer "Reinforced Concrete Designer's Handbook", Tenth Edition, Charles E. Reynold & James C. Steedman)

For which,

1 = 0.027

Design of Wall Slab (Uncracked Design):

Lateral Pressure on Wall for Surcharge

Coeff. for Maximum -ve Vertical BM at base i.e.at junction of base slab =

Lateral Pressure on Wall for Water

Load Case II, Linearly varying Lateral Pressure for Soil & Water =Load Case I, Uniform Lateral Pressure due to Surcharge =

Analysis for Triangular Pressure Loading:

Analysis for Uniform Pressure Loading:

Lateral Pressure on Wall for Submerged Soil Lateral Pressure on Wall for Saturated Soil

Page 2 of 7


Project Title

Client

Name of Unit

Drawing Ref.


Reference

Kolkata - 700019









CALCULATIONS

2 = 0.0095

3 = 0.033

4 = 0.015

Mx (+ve) = Positive BM for Horizontal Span = 4 x f2 x lx2 = 1.65 T-m/m

Mx (-ve) = Negative BM for Horizontal Span = 3 x f2 x lx2 = 3.63 T-m/m

My (+ve) = Positive BM for Vertical Span = 2 x f2 x lz2 = 1.74 T-m/m

My (-ve) = Negative BM for Vertical Span = 1 x f2 x lz2 = 4.94 T-m/m

Now, considering the bending moments arising due to uniform and triangular laoding are additive irrespective of their position of occurrence, the values of Maximum Bending Moments are:

Mx (+ve) = Positive Maximum BM for Hor. Span = 0.58 + 1.65 = 2.23 T-m/m

Mx (-ve) = Negative Maximum BM for Hor. Span = 0.75 + 3.63 = 4.38 T-m/m

My (+ve) = Positive Maximum BM for Ver. Span = 0.38 + 1.74 = 2.12 T-m/m

My (-ve) = Negative Maximum BM for Ver. Span = 0.51 + 4.94 = 5.46 T-m/m

Now, using Tor steel rebar and considering Asc = As i.e. cross-sectional area of compression rebar is equal to the cross-sectional area of tension rebar and k = d1/d = 0.39/0.45 = 0.87 say 0.85Where, d1 = effective depth and d = overall depth of wall

Now, considering uncracked section for M30 grade of concrete & k = 0.85, the value of M/bd12 = 3.57(Refer Table 2.6 (page 15) of "Handbook of Tor Steel Research Foundation" - Design of Water Retaining Structure with Torsteel)

Hence, Effective depth, d1 required per meter length of wall,

d1 = deff. = (5.46 x 105) / (3.57 x 100) = 39.090 Cm.

Hence, overall depth d, required = 39.09 + 0.8 + 5 = 44.89 Cm.(Considering Maximum dia of reinforcement used as 16 mm and Clear cover of 50 mm)

Overall depth provided = 45 Cm., Hence OK,

Now, percentage of reinforcement required to be provided = 100 P = 0.3

Area of reinforcement required = 11.81 Cm2 per m

Provide 16 Tor reinforcement @ 160 c/c on both faces of wall vertically at junction of base slab

At junction of counterfort and wall, effective depth required = (4.38 x 105) / (3.57 x 100) = 35.037

Area of reinforcement required = 10.58 Cm2 per m

Provide 16 Tor reinforcement @ 160 c/c on both faces of wall horizontally at junction of counterfort & wall

(Considering no -ve BM at top of Vertical Span allowing top edge beam to rotate)

Coeff. for Maximum +ve Vertical BM at around 0.4 x Vertical Span =

Coeff. for Maximum -ve Horizontal BM at supports for Horizontal Span =

Coeff. for Maximum +ve Horizontal BM at center of Horizontal Span =

Page 3 of 7


Project Title

Client

Name of Unit

Drawing Ref.


Reference

Kolkata - 700019









CALCULATIONS

ii)

Since, the major portion of the counterfort will be in contact with water, the same has been designed based on the principles of uncracked design as per working stress method

Considering the total horizontal load on a wall panel to be carried by each counterforts for an effective command area of 4.8 M x 4.0 M,

Maximum Cantilever Moment at Section A - A,= 0.5 x 6.88 x 4.0 x 4.8 x 4.8 / 3 + 1 x 4.0 x 4.8 x 4.8 / 2 = 151.70 T-m/m

Hence, Effective depth, d1 required for Counterfort at Section A - A,

d1 = deff. = (151.70 x 105) / (3.57 x 75) = 238.024 Cm.

Overall Depth required = 238.024 + 1.25 + 5.0 = 244.3 Cm.

Overall Depth provided = 2.45 M = 245.0 Cm., Hence OK

Now, Various pressures at Section B - B, i.e. for H3 = H2 - 1.0 M = 3.44 M= 3.80 M

i) =k0satH1= 0.35 T/M2 (Triangular)ii) =k0subH3= 1.62 T/M2 (Triangular)iii) =k0q = 1.00 T/M2 (Uniform)iv) =wH3 = 3.44 T/M2 (Triangular)

Now for all practical purposes, there will be two types of combined load cases as follows:1.00 T/M2 = f1'5.41 T/M2 = f2'

Hence,Maximum Cantilever Moment at Section B - B,= 0.5 x 5.41 x 4.0 x 3.8 x 3.8 / 3 + 1 x 4.0 x 3.8 x 3.8 / 2 = 80.92 T-m/m

Hence, Effective depth, d1 required for Counterfort at Section B - B,

d1 = deff. = (80.92 x 105) / (3.57 x 75) = 173.85 Cm.

Load Case II, Linearly varying Lateral Pressure for Soil & Water =

Height of Pressure Diagram below Formed Ground Level

Lateral Pressure on Wall for Saturated Soil Lateral Pressure on Wall for Submerged Soil Lateral Pressure on Wall for Surcharge

Design of Counterfort (Uncraked Design):

Fig. 2: Section showing Pressure Diagram for Retaining Wall of Bus Parking Area at -4.8M Level

Load Case I, Uniform Lateral Pressure due to Surcharge =

Lateral Pressure on Wall for Water

0.30

0.60

0.45

2.0

1.0

5.10

0.30

0.42

Formed Ground Level

(EL. +205.50M)

High Flood Level (HFL)(EL. +205.14M)


due to Surcharge Load

H1 = 0.36 M

H2 = 4.44 M


due to Soil & Water

= 6.88 T/m

Counterfort Beam

EL. +200.70M

Retaining Wall

Basement Slab

= 1.0 T/m2

2

Page 4 of 7


Project Title

Client

Name of Unit

Drawing Ref.


Reference

Kolkata - 700019









CALCULATIONS

Overall Depth required = 238.024 + 1.0 + 5.0 = 179.8 Cm.

Overall Depth provided = 2.45 M = 245.0 Cm., Hence OK

Now, Maximum Shear at Section A - A,VAA = 0.5 x 6.88 x 4.0 x 4.8 + 1.0 x 4.0 x 4.8 = 85.21 T

Maximum Shear at Section B - B,VBB = 0.5 x 5.41 x 4.0 x 3.8 + 1.0 x 4.0 x 3.8 = 56.29 T

Now, reinforcement required for Maximum Bending Moment at Section A - A,

Ast. reqd. = Asc. reqd. = (0.302 x 238.024 x 75)/100 = 53.91 Cm2

For 25 dia. Tor reinforcement bar, Area available = 4.91 Cm 2

Using 25 dia. Tor reinforcement bar, Numbers required on both faces = 11

and reinforcement required for Maximum Bending Moment at Section B - B,

Ast. reqd. = Asc. reqd. = (0.302 x 173.85 x 75)/100 = 39.38 Cm2

Using 25 dia. Tor reinforcement bar, Numbers required on both faces = 8

Now, for maximum shear force, VAA at section A - A,

100Ast/bd1 = (100 x 11 x 4.91) / (75 x 238.024) = 0.30255 %

For which, c = 2.46 Kg/Cm2

VCAA = 2.46 x 75 x 238.024 = 43915.5 Kg = 43.92 T

Hence, VSAA = VAA - VCAA = 41.29 T = 41289.6 Kg


Spacing of 10 dia. Tor rebar = (2 x 0.79 x 1500 x 238.024) / 41290 = 13.662

Provide 2 L , 10 Tor reinforcement bar @ 125 c/c. upto 1.0M height from -4.8M level

Now, for maximum shear force, VBB at section B - B,

100Ast/bd1 = (100 x 8 x 4.91) / (75 x 238.024) = 0.22003 %

For which, c = 2.20 Kg/Cm2

VCBB = 2.20 x 75 x 238.024 = 39274 Kg = 39.27 T

Hence, VSBB = VBB - VCBB = 17.02 T = 17015.6 Kg


Spacing of 10 dia. Tor rebar = (2 x 0.79 x 1500 x 238.024) / 17016 = 33.153

Provide 2 L , 10 Tor reinforcement bar @ 250 c/c. from -3.8M level to top of counterfort

Provide 10 Tor reinforcement bar @ 250 c/c. from -4.8M level as side face reinforcement on both sides

Page 5 of 7


Project Title

Client

Name of Unit

Drawing Ref.


Reference

Kolkata - 700019









CALCULATIONS

iii)

Considering the total horizontal freely supported moments are to be effective on each top edge beams, themoment coefficients are as follows:

O/A Depth of Beam = 0.5 M, Width of Beam = 0.4 M,

Clear cover to main rebar = 5.0 Cm., Modular Ratio, m = 13.04

5 = 0.0205

6 = 0.0290

Bending Moment due to combined triangular loading:

Mx (+ve) = Positive BM for Horizontal Span = 6 x f2 x lx2 = 3.19 T-m/m

Bending Moment due to uniform loading:

Mx (+ve) = Positive BM for Horizontal Span = x (+ve) x f1 x lx2 = 0.58 T-m/m

Total +ve Horizontal Bending Moment = 3.77 T-m/m

Now, for maximum vertical Bending Moment which will impart Torsion in the edge beam:

Torsional Moment due to combined triangular loading:

My (-ve) = Torsional Moment for Vertical Span = 5 x f2 x lz2 = 3.75 T-m/m

Torsional Moment due to uniform loading:

My (-ve) = Torsional Moment for Vertical Span = y (-ve) x f1 x lx2 = 0.51 T-m/m

Total Torsional Moment = 4.27 T-m/m

Equivalent Bending Moment = [T x (1 + D/b)]/1.7 = 5.64 T-m/m

Total Effective Design Bending Moment = 9.41 T-m/m

Now, effective depth required, d1 reqd. = (9.41 x 105) / (13.04 x 40) = 42.48 Cm.

Considering clear cover of 50 mm and maximum dia of main reinforcemnt as 20 mm and link dia as 6 mm,Overall Depth reqd. = 42.48 + 5.0 + 1.0 + 0.6 = 49.1 Cm. < 50.0 Cm.

Hence OKAst reqd. = (9.41 x 105) / (2300 x 0.9 x 43.4) = 10.48 Cm.2

Using 4 Nos. 20 Tor reinforcement bar, total area provided = 12.56 Cm.2 > 10.48 Cm.2

Hence OK

Since the edge beam is freely supported, Provide nominal Shear Reinforcement, 8 dia. Tor @ 200 c/c.

Minimum spacing of shear reinforcement to be provided = (Asv x 0.87 x fy) / (0.4 x b) = 22.566 Cm.

Hence OK

Coeff. for Maximum Freely supported vertical Bending Moment =

Coeff. for Maximum Freely supported horizontal Bending Moment =

Design of Horizontal Beam at Top (Cracked Design):

Page 6 of 7


Project Title

Client

Name of Unit

Drawing Ref.


Reference

Kolkata - 700019









CALCULATIONS

iv)

Considering overall depth of Pile Cap = 1.00 MFor Maximum Overturning Moment at - 5.8 M level,

i) k0satH1= 0.35 T/M2 (Triangular)ii) k0sub(H2+1)= 2.56 T/M2 (Triangular)iii) k0q = 1.00 T/M2 (Uniform)iv) w(H2+1) = 5.44 T/M2 (Triangular)

Now for all practical purposes, there will be two types of combined load cases as follows:1.00 T/M2 = f1''8.35 T/M2 = f2''

Considering length of Pile Cap = 3.20 M, and width of Pile Cap = 0.90 M

Dia of Pile Cap = 600 mm, and Clear edge distance from pile cap = 150 mm

Maximum Overturning Bending Moment:

= 0.5 x 8.35 x 4.0 x 5.8 x 5.8 / 3 + 1 x 4.0 x 5.8 x 5.8 / 2 = 254.45 T-m

Balancing Moment,

=2x4x4.8x2x2.2 + 0.45x4x5.1x2.5x(0.225+0.75) + 4x2x0.42x2.5x2.2 + 1x3.2x0.9x2.5x1.6 = 221.34 T-m (Weight of Soil) (Weight of Wall) (Weight of Base Slab) (Weight of Pile Cap)

Additional Bending Moment (Couple) to be taken by Piles = 134.90 T-m

Using 2 nos 600 dia piles, lever arm available = 2.3 M

Load on each pile= 135.62/2.3 = + 58.65 T

Total vertical Load on pile cap,

= 2x4x4.8x2 + 0.45x4x5.1x2.5 + 4x2x0.42x2.5 + 1x3.2x0.9x2.5 = 115.35

Vertical laod on each pile = 57.68 T

Maximum Compressive load on pile = 116.33 T (Using 600 dia pile, capacity = 128.5 T)

Maximum Tensile load on pile = -0.98 T (For 600 dia pile, Tension Capacity = 64.25 T)

Hence OK

Load Case I, Uniform Lateral Pressure due to Surcharge =Load Case II, Linearly varying Lateral Pressure for Soil & Water =

Check for Overturning & Load on piles:

Lateral Pressure on Wall for Saturated Soil = Lateral Pressure on Wall for Submerged Soil = Lateral Pressure on Wall for Surcharge = Lateral Pressure on Wall for Water =

Page 7 of 7

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Retaining Counterfort Wall Design