Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

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Design Of Counter Fort Retaining Wall Reference: Example 18.4, R.C.C. Design Vol. - I, B.C. Punamia 1 Design Constants Hieght of cantilever wall from ground leve = = Unit Weight of water (ϒw) = Angle of repose (φ) = = Co-eff of friction (µ) = = = Cover = Foundation Depth = Width of counterfort = = = Neuteral axis constant (k) = Lever arm constant (j) = Moment of resistance constant ® = = 2 Dimension of Various Parts Height of wall above base (H) = = The ratio of length of slabe (DE) to base width b α = = α = Adopting α = The width of base is given by Eq. b = b = = Normal practice is to provide b between 0.7 to 0.8 Taking maximum value of H = b = Width of toe slab = = Provide toe slab = Saturated weight of Earth (γ sat ) SBC of Soil (qo) Unit Weight of Concrete (fck) fy σ cbc σst Coefficient of earth pressure, K a

Transcript of Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

Page 1: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

Design Of Counter Fort Retaining WallReference: Example 18.4, R.C.C. Design Vol. - I, B.C. Punamia

1 Design ConstantsHieght of cantilever wall from ground level = 11.5 m

= 18

Unit Weight of water (ϒw) = 10Angle of repose (φ) = 30

= 125Co-eff of friction (µ) = 0.5

= 25 MPa

= 415 MPaCover = 40 mmFoundation Depth = 1 mWidth of counterfort = 0.5 m

= 8.5

= 230Neuteral axis constant (k) = 0.289Lever arm constant (j) = 0.904Moment of resistance constant ® = 1.109

= 0.333

2 Dimension of Various PartsHeight of wall above base (H) = 11.5 +

= 12.5 mThe ratio of length of slabe (DE) to base width b is given by eq.

α = 1 -

= 1 -

α = 0.75Adopting α = 0.40 ….

The width of base is given by Eq.

b = 0.95 H x

b = 0.95 x

= 5.97Normal practice is to provide b between 0.7 to 0.8 H Taking maximum value of H = 0.7

b = 8.75 mWidth of toe slab = α x b

= 3.50 m

Provide toe slab = 3.50

Saturated weight of Earth (γsat ) kN/m3

kN/m3

SBC of Soil (qo) kN/m2

Unit Weight of Concrete (fck)

fy

σcbc N/mm2

σst N/mm2

Coefficient of earth pressure, Ka

Page 2: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

Taking the uniform thickness of stem = 0.6 mHence width of heel slab = 4.65 mLet thickness of base slab = 0.8 m

Clear spacing of counter fort = 3.5 xH

= 3.20 m

Providing spacing of counterfort = 3.00 m

3 Stability of Wall

= Weight of rectangular portion of stem

= Weight of base slab

= Weight of soil on heel slab.

Detail

1 x 0.60

1 x 0.80

1 x 4.65

Total resisting moment = 7724 kNm

Earth pressure (p) == 158.33 kN

Moment == 4123.26 kNm

F.O.S. against overturning = 1.87 >

F.O.S. against sliding == 4.20 >

Pressure DistributionNet Moment (SM) = 3601 kNmDistance x of the point of application of resultant, from toe is

x == 2.71 m

e = b/2 - x= 1.67

b/6 = 1.46Hence un safe as e > b/6

= 1 +b

=1329.79

x8.75

= 325.69

γsat

Let w1

w2

w3

w1

w2

w3

(Ka x γ' x H)+(γw x H)

p*H2/6

µSw/p

SM/Sw

Pressure p1 at toeSW

kN/m2

Page 3: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

Hence un safe as p1 > 125

= 1 -b

1329.79x

8.75

= -21.74

Hence un safe as p2 < 0

= 325.69 -

= 186.72

= 325.69 -

= 162.90

4 Design of Heel SlabClear spacing b/w counterforts = 3.00 mThe pressure distribution on the heel slab is shown in fig 2. Consider a strip 1 meter wide.

Upward pressure intensity (near outer edge, C) = -21.74

Down ward load due to weight of Earth. = 210.6

Down ward weight of slab per unit area = 15

= 203.86

== 152.89 kNm

Effective depth required (d) =BMRxb

= 371.39 mmProviding overall depth (D) = 500 mmEffective depth (d) = 460 mm

= 800

= 1598.94

= 1598.94

Dia of bar = 20 mmSpacing of bar calculated = 196 mmSpacing of bar provided = 150 mm

Shear force (V) = PL/2

Pressure p2 at heelSW

kN/m2

The Pressure intencity p1 under E,

p1

kN/m2

The Pressure intencity p2 under B,

p2

kN/m2

kN/m2

kN/m2

kN/m2

Hence net pressure intensities will be P kN/m2

M1 Pl2/12

Minimum steel required 0.2% of b D ― further 20% [IS :3370,7.1]

mm2

Ast

M=Ast σst j dmm2

Provide Ast1 mm2

Page 4: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

= 305.79 kN

= 0.35

= 0.26Refernce: Pg. 84, Table 23, IS 456 : 2000

= V/(d x b)= 0.66 664.7609

Hence shear reinforcement is required as τv > τcHence depth required from shear point of veiw (d) = V/(τc x b)

= 1176.12 mmProviding overall depth (D) = 1200 mmHence effective depth (d) = 1160 mm

= 1920

= 634.06

= 1920

Dia of bar = 20 mmSpacing of bar calculated = 163 mmSpacing of bar provided = 150 mm

Distribution Steel = 1920Dia of bar = 20 mmSpacing of bar calculated = 163 mmSpacing of bar provided = 150 mm

Hence point of contraflexure is at = 0.633 mmShear force at this point is given = P(L/2 - x)

= 176.75 kN= 176746.62 N

Assuming that all the bars will avilable at point of contraflexure,

M == 462979286.112 Nmm

= 12φ or d, whichever is more= 1160 mm

= 45φ= 900 mm

= 3779.45 mmHence safe as M/V+Lo > Ld

= 1160 mm beyond the point of contraflexure. After that, curtail half bars, and continue the remaining half throughout the length.

100Ast/bd

τc N/mm2

τv

Minimum steel required 0.2% of b D ― further 20% [IS :3370,7.1]

mm2

Ast

M=Ast σst j dmm2

Provide Ast1 mm2

mm2

Let us check this reinforcement for development length at point of contraflexure which is situated at distance of 0.211 x L. In this case, the slab is continuous, but we will assume the same position of contraflexure.

Ast σst j d

Lo

Ld

M/V + Lo

Cotinue these bars by a distance Lo = d

Page 5: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

At the point of curtailment, length of each bar availa = 1793 mmHence safe

These bars will be provide at the top face of heel slab.

Maximum passive B.M. =

=

=

= 1440

= 1920

= 1920

Dia of bar = 20 mmSpacing of bar calculated = 163 mmSpacing of bar provided = 150 mm

Let us check this reinforcement for development length crierion at point of contraflexur,

M == 462979286.112 Nmm

= 12φ or d, whichever is more= 1160 mm

= 45φ= 900 mm

= 3779.45 mmHence safe as M/V+Lo > Ld

= 1160 mm from the point of contraflexure.

i.e. upto distance = -527 mm from the centre of support. At this point half bars can be discontinued. Since this distance is quite small, it is better to continue these bars upto center of counterfors.

5 Design of Toe SlabProviding counterfort over toe slab upto ground level.Assuming total depth of toe slab = 500 mm

Total weight of toe slab = 12.5

Net upward intensity at D = 325.69-12.5

= 313.19

Net upward intensity at E = 186.72-12.5

= 174.22Cosidering strip of unit width at D.

=

= 234.8925 kNm

Effective depth required (d) =BMRxb

= 460.33 mmProviding overall depth (D) = 500 mm

PL2/16

3/4 x M1

Area of bottom steel (Ast2) 3/4 x Ast1

mm2

Minimum steel required 0.2% of b D ― further 20% [IS :3370,7.1]

mm2

Provide Ast2 mm2

Ast σst j d

Lo

Ld

M/V + Lo

Cotinue these bars by a distance Lo = d

kN/m2

kN/m2

kN/m2

kN/m2

kN/m2

Max. negative B.M. (M1) wL2/12

Page 6: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

Effective depth (d) = 460 mm

= 800

= 2456.46

= 2456.46

Dia of bar = 20 mmSpacing of bar calculated = 127 mmSpacing of bar provided = 100 mm

Shear force (V) = PL/2= 469.78 kN

= 0.53

= 0.32Refernce: Pg. 84, Table 23, IS 456 : 2000

= V/(d x b)= 1.02

Hence shear reinforcement is required as τv > τcHence depth required from shear point of veiw (d) = V/(τc x b)

= 1468.08 mmProviding overall depth (D) = 1500 mmHence effective depth (d) = 1460 mm

= 2400

= 773.95

= 2400

Dia of bar = 20 mmSpacing of bar calculated = 130 mmSpacing of bar provided = 100 mm

Distribution Steel = 2400Dia of bar = 20 mmSpacing of bar calculated = 130 mmSpacing of bar provided = 100 mm

Let us check this reinforcement for development length crierion at point of contraflexur.Point of contraflexure is at = 0.633 mmShear force at this point is given = P(L/2 - x)

= 271.54 kN= 271535.73 N

M == 728394135.478 Nmm

Minimum steel required 0.2% of b D ― further 20% [IS :3370,7.1]

mm2

Ast

M=Ast σst j dmm2

Provide Ast1 mm2

100Ast/bd

τc N/mm2

τv

Minimum steel required 0.2% of b D ― further 20% [IS :3370,7.1]

mm2

Ast

M=Ast σst j dmm2

Provide Ast1 mm2

mm2

Ast σst j d

Page 7: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

= 12φ or d, whichever is more= 1460 mm

= 45φ= 900 mm

= 4142.50Hence safe as M/V+Lo > Ld

= 1460 mm beyond the point of contraflexure. After that, curtail half bars, and continue the remaining half throughout the length.

At the point of curtailment, length of each bar availa = 2093 mmHence safe

These bars will be provide at the top face of toe slab.

Maximum passive B.M. =

=

=

= 1800.00

= 2400

= 2400

Dia of bar = 20 mmSpacing of bar calculated = 130 mmSpacing of bar provided = 100 mm

Let us check this reinforcement for development length crierion at point of contraflexur,

M == 728394135.478 Nmm

= 12φ or d, whichever is more= 1460 mm

= 45φ= 900 mm

= 4142.50 mmHence safe as M/V+Lo > Ld

= 1460 mm from the point of contraflexure.

i.e. upto distance = -827 mm from the centre of support.

6 Design of StemThe stem acts as a continuous slab. Considred 1 m strip at B .

=

= 67.80

= 11.30 m

== 50.85 kNm

Effective depth required (d) =BM

Lo

Ld

M/V + Lo

Cotinue these bars by a distance Lo = d

PL2/16

3/4 x M1

Area of bottom steel (Ast2) 3/4 x Ast1

mm2

Minimum steel required 0.2% of b D ― further 20% [IS :3370,7.1]

mm2

Provide Ast2 mm2

Ast σst j d

Lo

Ld

M/V + Lo

Cotinue these bars by a distance Lo = d

The intencity of earth pressure is given by (ph) KaγH1

kN/m2

Hence revised H1

Negative B.M. in slab, (M1) phL2/12

Page 8: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

Effective depth required (d) =Rxb

= 214.18 mmProviding overall depth (D) = 600 mmEffective depth (d) = 560 mm

= 960

= 436.82

= 960.00

Dia of bar = 16 mmSpacing of bar calculated = 209 mmSpacing of bar provided = 200 mm

Distribution Steel = 960Dia of bar = 16 mmSpacing of bar calculated = 209 mmSpacing of bar provided = 200 mm

Shear force (V) = PL/2= 101.70 kN

= 0.17

= 0.20Refernce: Pg. 84, Table 23, IS 456 : 2000

= V/(d x b)

= 0.18Hence no shear reinforcement is required as τv < τc

Let us check this reinforcement for development length crierion at point of contraflexur.Point of contraflexure is at = 0.633 mmShear force at this point is given (V) = V(L/2 - x)

= 88.17 kN= 88173.90 N

M == 111753620.786 Nmm

= 12φ or d, whichever is more= 560 mm

= 45φ= 720 mm

= 1827.42Hence safe as M/V+Lo > Ld

= 560 mm beyond the point of contraflexure. After that, curtail half bars, and continue the remaining half throughout the length.

At the point of curtailment, length of each bar availa = 1193 mmHence safe

These bars are to be provided at the inner face of the stem.

Minimum steel required 0.2% of b D ― further 20% [IS :3370,7.1]

mm2

Ast

M=Ast σst j dmm2

Provide Ast1 mm2

mm2

100Ast/bd

τc N/mm2

τv

N/mm2

Ast σst j d

Lo

Ld

M/V + Lo

Cotinue these bars by a distance Lo = d

Page 9: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

Maximum passive B.M. =

=

=

= 720.00

= 960

= 960

Dia of bar = 16 mmSpacing of bar calculated = 209 mmSpacing of bar provided = 200 mm

Let us check this reinforcement for development length crierion at point of contraflexur,

M == 111753620.786 Nmm

= 12φ or d, whichever is more= 560 mm

= 45φ= 720 mm

= 1827.42 mmHence safe as M/V+Lo > Ld

7 Design of Main CounterfortLet us assuming thickness of counterforts is = 500 mmSpacing of counterforts = 350 cm c/c

=

= 21 h kN/mSimilarly, net down ward pressure on heel at C is = (11.3 x 18) + (1200/1000) x 25 - -21.74

= 255.14Similarly, net down ward pressure on heel at B is = (11.3 x 18) + (1200/1000) x 25 - 162.9

= 70.5Hence reaction transferrred to each counterfort will be,At C = 255.14 x (350/100)

= 892.99 kN/mAt B = 70.5 x (350/100)

= 246.75 kN/mPressure intencity at h = 11.5 m

= 241.5 kN/mShear force at F (Q) = 0.5 x 241.5 x 11.5

= 1388.63 kNB.M. = h/3 x 1388.625

= 5323.06 kNm= 5323062500 Nmm

PL2/16

3/4 x M1

Area of steel (Ast2) 3/4 x Ast1

mm2

Minimum steel required 0.2% of b D ― further 20% [IS :3370,7.1]

mm2

Provide Ast2 mm2

Ast σst j d

Lo

Ld

M/V + Lo

At any section at depth h below the top A, the earth pressure acting on each counter forts will be Ka x γsat x h x L

kN/m2

kN/m2

Page 10: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

Effective depth required (d) =BMRxb

= 3099.03 mmProviding total depth (D) = 3150 mmEffective depth (d) = 3110 mmAngle θ of the face AC is given by, =

tan θ = 4.65/11.3tan θ = 0.41150442478

θ = 22.37 DegreesSin θ = 0.3805Cos θ = 0.9248

== 4.38 m= 4376.26 mm

FG = 4977.00 mmAsssuming that the steel reinforcement is provided in two layers and providing a nominal cover of 40 mm and 20 mm dia bar.Effective depth (d) = 4977 - (40 + 20 + 10)

= 4907.00 mm

= 5218.47

Using dia. of bar = 20 mm

= 314.16No. of bars = 17 barsProvide the bars in two layers.Effective shear force = Q-(M/d')tanθ

d' = d/cosθ= 5306.23 mm

Effective shear force = 975815.19 N= V/(d x b)

= 0.37

= 0.20

= 0.21Hence shear reinforcement is required as τv > τc

However, the vertical and horizontal ties provided in counterforts will bear the excess shear stress.

= h m

= 3.54 m below A, i.e. at point H.To locate the position of point of curtailmenton AC, drawing Hl parallel to FG.

= 240

= 900 mm

Design of Horizontal Ties

F1G1 AF1 x sinθ

Area of steel at supports, at bottom (Ast ) M=Ast σst j d mm2

Aφ mm2

τv

N/mm2

100Ast/bd

τc N/mm2

The height h where half of the reinforcement can curtailed (H)

Thus half bars can be curtailed at l. However these should be extent by a distance 12φ

mm beyond I, i.e. extented upto I1

The location of H corresponding to I1 can be locate by drawing line I1H1 parallel FG

It should be noted that I1G should not less than 45φ

Page 11: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

At any depth h below the top, force causing sepratio == 207 kN/m

= Force/Stress

= 900Using 2 legged ties of dia. = 10 mm

=

= 157.08Spacing required = 174.53 mmSpacing provided = 150 mm

Design of Vertical TiesThe downward force at C = (892.99 x 3)/(350/100)

= 765.42 kN/mThe downward force at B = (246.75 x 3)/(350/100)

= 211.5 kN/m

= Force/Stress

= 3327.91Using 2 legged ties of dia. = 20 mm

=

= 628.32Spacing required = 188.80 mmSpacing provided = 150 mm

= Force/Stress

= 919.57Using 2 legged ties of dia. = 10 mm

=

= 157.08Spacing required = 170.82 mmSpacing provided = 150 mm

8 Design of Front Counterfort

The upward pressure intensity varies from 325.69 186.72

Downward weight of toe slab = 37.5

Net weight at D = 288.19

Net weight at E = 149.22The center to center spacing of counterforts = 3.5 mHence upward force transmitted to counterforts at D = 288.19 x 3.5 kN/m

= 1008.67 kN/mUpward force transmitted to counterforts at E = 149.22 x 3.5 kN/m

= 522.27 kN/mTotal upward force = 1/2 x (1008.665 + 522.27) x 3.5

= 2679.14 kNForce will be acting at x = ((522.27 + 2 x 1008.665)/(1008.665 + 522.27)) x (3.5/3)

Ka x γsat x h x L

Ast required

mm2

Aφ 2Π/4 x D2

mm2

Ast required at C

mm2

Aφ 2Π/4 x D2

mm2

Ast required at B

mm2

Aφ 2Π/4 x D2

mm2

kN/m2 at D to

kN/m2

kN/m2

kN/m2

Page 12: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

= 1.94 m from EB.M. = 2679.14 x 1.94 kNm

= 5197.53 kNm

Effective depth required (d) =BMRxb

= 3062.27Providing total depth (D) = 4000 mmEffective depth (d) = 3960 mmThus height of counterfort above ground level = 3000 mm

= 6313.93

= 6400

= 6400.00

Dia of bar = 32 mm

= 804.25No. of bars = 8 barsEffective shear force (V) = Q-(M/d)tanθ

tan θ = 1.14Effective shear force (V) = 1179130.88 N

= V/(d x b)

= 0.60

= 0.11

= 0.19Hence shear reinforcement is required as τv > τc

Using 2 legged ties of dia. = 10 mm

=

= 157.08

== 376200 N

== 802930.88 N

== 178.18

= 150 mm c/c Provide 2 x 10 mm holding bar at top

9 Fixing Effect in Stem, Toe and HeelAt the junction of stem, toe and heel slab fixing moment are induced, which are at right angles to their normal direction of bending. These moment are not determine , but normal reinforcement given below may be provided.

(i)

i.e. upto point F3.

Area of steel at supports, at bottom (Ast ) M=Ast σst j d mm2

Minimum steel required 0.2% of b D ― further 20% [IS :3370,7.1]

mm2

Provide Ast1 mm2

Aφ mm2

τv

N/mm2

100Ast/bd

τc N/mm2

Aφ 2Π/4 x D2

mm2

Vc τc x b x d

Vs V - Vc

Spacing required (sv) (σst x Aφ x d)/Vs

Spacing required (sv)

In stem @0.3% of cross section, to be provided at inner face, in vertical direction,for a length 58φ

Page 13: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

= (0.3/100) x 1000 x 600

= 1800Dia. of bar = 16 mm

= 201.06Spacing required = (201.06 x 1000)/1800

= 111.70 mmSpacing provided = 100 mmLength of embedment in slab above heel slab = 928 mm

(ii) In toe slab @0.15% to be provided at the lowar face

= (0.15/100) x 1000 x 1500

= 2250Dia. of bar = 20 mm

= 314.16Spacing required = (314.16 x 1000)/2250

= 139.63 mmSpacing provided = 100 mmLength of embedment in slab above heel slab = 1160 mm

(iii) In heel slab @ 0.15% to be provided in upper face

= (0.15/100) x 1000 x 1200

= 1800Dia. of bar = 16 mm

= 201.06Spacing required = (201.06 x 1000)/1800

= 111.70 mmSpacing provided = 100 mmLength of embedment in slab above heel slab = 928 mm

Ast

mm2

Aφ mm2

Ast

mm2

Aφ mm2

Ast

mm2

Aφ mm2

Page 14: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

1

2.2 y H125

2.2 x 18 x 12.50

Eq (1)

(1-α) x (1+3α)

12.500.333

( 1 - 0.40 )x( 1

q0

Ka

Page 15: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

Weight of rectangular portion of stem

Weight of soil on heel slab.

Detail force(kN) lever arm

x 11.70 x 25 = 175.5 3.8

x 8.75 x 25 = 175 4.38

x 11.70 x 18 = 979.29 6.43

= 1329.79

1.5 Safe against overturning

1.5 Safe against sliding

6 e

b

1 +6 x 1.67

8.75

1/4

Sw Total MR

Page 16: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

6 e

b

1 +6 x 1.67

8.75

325.69 - -21.74x 3.50

8.75

325.69 - -21.74x 4.10

8.75

Page 17: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

0.66Hence shear reinforcement is required as τv > τc

12φ or d, whichever is more

Hence safe as M/V+Lo > Ld

mm beyond the point of contraflexure. After that, curtail half bars, and continue the remaining half throughout the length.

Page 18: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

12φ or d, whichever is more

Hence safe as M/V+Lo > Ld

mm from the point of contraflexure.

mm from the centre of support. At this point half bars can be discontinued. Since this distance is quite small, it is better to continue these bars upto center of counterfors.

Page 19: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

Hence shear reinforcement is required as τv > τc

Page 20: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

12φ or d, whichever is more

Hence safe as M/V+Lo > Ld

mm beyond the point of contraflexure. After that, curtail half bars, and continue the remaining half throughout the length.

12φ or d, whichever is more

Hence safe as M/V+Lo > Ld

mm from the point of contraflexure.

mm from the centre of support.

Page 21: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

Hence no shear reinforcement is required as τv < τc

12φ or d, whichever is more

Hence safe as M/V+Lo > Ld

mm beyond the point of contraflexure. After that, curtail half bars, and continue the remaining half throughout the length.

Page 22: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

12φ or d, whichever is more

Hence safe as M/V+Lo > Ld

(11.3 x 18) + (1200/1000) x 25 - -21.74

(11.3 x 18) + (1200/1000) x 25 - 162.9

Page 23: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

Asssuming that the steel reinforcement is provided in two layers and providing a nominal cover of 40 mm and 20 mm dia bar.

Hence shear reinforcement is required as τv > τcHowever, the vertical and horizontal ties provided in counterforts will bear the excess shear stress.

m below A, i.e. at point H.

mm beyond I, i.e. extented upto I1

Page 24: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

1/2 x (1008.665 + 522.27) x 3.5

((522.27 + 2 x 1008.665)/(1008.665 + 522.27)) x (3.5/3)

kN/m2 at E.

Page 25: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

m from E

Hence shear reinforcement is required as τv > τc

At the junction of stem, toe and heel slab fixing moment are induced, which are at right angles to their normal direction of bending. These moment are not determine , but normal reinforcement given below may be provided.

Page 26: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012
Page 27: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

0.333+ 1.20 )

A 0.6

θ

II1

H1

Page 28: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

H12.50

11.5 G

3.00F

3.50 1.00 4.65 D E B C

8.75

FIGURE 1Moment about toe (KN-m)

666.90

765.63

6291.94

325.

697724

FIGURE 2

F3 G1

F1 F2

kN/m

2

Page 29: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

mm beyond the point of contraflexure. After that, curtail half bars, and continue the remaining half throughout the length.

Page 30: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

mm beyond the point of contraflexure. After that, curtail half bars, and continue the remaining half throughout the length.

Page 31: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

mm beyond the point of contraflexure. After that, curtail half bars, and continue the remaining half throughout the length.

Page 32: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

At the junction of stem, toe and heel slab fixing moment are induced, which are at right angles to their normal direction of bending. These moment are not determine , but normal reinforcement given below may be provided.

Page 33: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

16 mm φ

0.928100 mm c/c

0.928

20 mm φ 20 mm φ100 mm c/c 100 mm c/c

1.16

Page 34: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

1.50

-21.

74kN

/m2

Page 35: Counterfort Retaining Wall B.C. Punamia 05-Feb-2012

M-15 M-20 M-25 M-30 M-35 M-40

0.15 0.18 0.18 0.19 0.20 0.20 0.20 0.210.25 0.22 0.22 0.23 0.23 0.23 0.23 0.260.50 0.29 0.30 0.31 0.31 0.31 0.32 0.320.75 0.34 0.35 0.36 0.37 0.37 0.381.00 0.37 0.39 0.40 0.41 0.42 0.421.25 0.40 0.42 0.44 0.45 0.45 0.461.50 0.42 0.45 0.46 0.48 0.49 0.491.75 0.44 0.47 0.49 0.50 0.52 0.522.00 0.44 0.49 0.51 0.53 0.54 0.552.25 0.44 0.51 0.53 0.55 0.56 0.572.50 0.44 0.51 0.55 0.57 0.58 0.60