Reatining WAll With HORZONTAL Bach Fill

7/29/2019 Reatining WAll With HORZONTAL Bach Fill

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RCC design B.C.Punmia

18.2 TYPE OF RETAINING WALLS

1 Gravity walls

2 Cantilever retaining walls a. T- shaped b. L- shaped

3 Counterfort retainig walls.

4 Buttresssed walls.

The cantilever retaining wall resist the horizontal earth pressure as well as other vertical pressure by w

A gravity retaining wall shown in fig 1 is the one in which the earth pressure exrted by the

back fill is resisted by dead weight of wall, which is either made of masonry or of mass concrete . The

stress devlop in the wall is very low ,These walls are no proportioned that no tension is devloped any

where, and the resultant of forces remain withen the middle third of the base.

A retaining wall or retaining structure is used for maintaining the ground surfgaces at defrent

elevations on either side of it. Whenever embankments are involed in construction ,retaining wall are

usually necessary. In the construction of buildins having basements, retaining walls are mandatory.

Similsrly in bridge work, the wing walls and abutments etc. are designed as retaining walls , to resist

earth pressure along with superimposed loads. The material retained or supported by a retaining wall

is called backfill lying above the horizontal plane at the elevation of the top of a wall is called the

surcharge, and its inclination to horizontal is called the surcharge angle b

In the design of retaining walls or other retaining structures, it is necessary to compute the

lateral earth pressure exerted bythe retaining mass of soil. The equation of finding out the lateral earth

pressure against retaining wall is one of the oldest in Civil Engineering field. The plastic state of

strees, when the failure is imminent, was invetigated by Rankine in1860. A Lot of theoretical

experiment work has been done in this field and many theory and hypothesis heve benn proposed.

RETAINING WALL

Retaining walls may be classified according to their mode of resisting the earth pressure,and

according to their shape. Following are some of commen types of retaining walls (Fig)


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y of beending of varios components acting as cantilever s.A coomon form of cantilever retaining waal


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12 mm f@

400 mm c/c

1 Hieght of cantilever wall from ground level = 3.00 m 8 mm f @ 12 mm f@

2 Unit weight of Earth = 18 KN/m3 300 mm c/c 200 mm c/c

3 Angle of repose = 30 Degree4 Safe Bearing capacity of soil = 100 KN/m

3 8 mm f @ 12 mm f@

5 Coffiecent of friction = 0.5 300 mm c/c 100 mm c/c

6 Concrete M- 20

m = 13.33 wt. of concrete 25000 N/m3 8 mm f @ 10 mm f@7 Steel fe = 415 N/mm st = 230 N/mm 160 mm c/c 120 mm c/c

cbc = 7 N/mm2

8 Nominal cover = 30 mm 10 mm f@

9 Foundation depth = 1.00 m 310 130 mm c/c

10 Stem thickness At footing 300 mm At top 200 mmToe width 900 mm Heel width 1200 mm 900

11 Footing width 2400 mm Key 300 x 300 mm 8 mm f @

180 mm c/c 300

12 Reinforcement Summary 8 mm F @ 160 `

STEM :-Main mm f@

2.58 12 mm F@ 100 mm c/c mm c/c1.94 12 mm F@ 200 mm c/cTop 12 mm F@ 400 mm c/c

mm f@Distribution 8 mm F@ 160 mm c/c 2.58 mm c/cTamprecture 8 mm F@ 300 mm c/c

TOE :-Main 10 mm F@ 130 mm c/c mm f@Distribution 8 mm F@ 180 mm c/c mm c/c

HEEL:-Main 10 mm F@ 120 mm c/cDistribution 8 mm F@ 180 mm c/c Out side Earth side

DESIGN SUMMARY

100% Reinforcement upto m



DESIGN OF T SHAPED CANTILEVER RETAINING WALL

Horizontal back fill

1.94

200

1000

200

1200

200

940

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Hieght of cantilever wall from ground level = 3.00 m

Unit weight of Earth g = 18 kN/m = N/m

Angle of repose = 30 Degree

Safe Bearing capacity of soil q0 = 100 kN/m3

Coffiecent of friction m = 0.5 = 25 N/mmConcrete = M 20

Steel fe = 415

Nominal cover = 30 mm

Foundation depth = 1.00 m

1 Design Constants:-For HYSD Bars = 20

st = = 230 N/mm2

= #### N/mm2

cbc = = 7 N/mm3

m = 13.33x

13.33 x 7 + 230

j=1-k/3 = 1 - 0.289 / 3 = 0.904

R=1/2xc x j x k = 0.5 x 7 x 0.904 x 0.289 = 0.913

2 Diamension of base:-

height of wall above base, H = 3.00 + 1.00 = meter

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

q0

2.2 y H 2.2 x 18 x 4.00

Keep = 0.37 Eq (1)

The width of base is given by Eq. sin = 0.5

1 - 0.5

1 + 0.5

( 1 - 0.37 )x( 1 + 1.11 )

x 4.00 x 0.33

( 1 - 0.37 )x 0.5

0.6 b = 0.60 x 4.00 = m

Hence Provided b = m

Width of toe slab = a x b = 0.37 x 2.40 = 0.89 m mLet the thickness of stem = H/12 = 4.00 / 12 = 0.33 or say = 0.30 m

= 2.40 - 0.30 - 0.90 = 1.20 m

4 Thickness of stem:-

= 4.00 - 0.30 = 3.70 m consider 1 m length of retaining wall

K x y x H13 0.33 x 18 x( 3.70 )

3

BM 50.65 x 10 6

Rxb 0.913 x 1000

= 240 mm and total thickness = 240 + 60 = 300 mm

Assuming that 12 mm F bar will be used. a nominal cover of = 60 - 6 = 54 mm= 200 mm at top so that effective depth of = 140 mm

DESIGN OF T SHAPED CANTILEVER RETAINING WALL with horizontal back fill

30

=

m*c+sst

=x

(1-a) m=

0.369

=0.7HKa

=

=

k=

b 2.96

This width is excessive. Normal practice is to provide b between 0.4 to 0.6 H .

Taking maximum value of H = 2.40

=

m*c

Provided toe slab =

1.90

Cocrete M

wt. of concrete

13.33 7

b

=

Keep d

0.7

2.40

mm

0.333

=

=Effective depth required

1-sin F0.95 H

6

Hence width of heel slab

The wall will be unsafe against sliding. This will be made safe by providing a shear Key at base .

Ka =

-1=

18000

100=

= 0.289

-

= 0.95

=

4.00

Ka

4.00

(1- a)x(1+3 a) 1+sinF

m

=

The base width from the considration of sliding is given by Eq.

.

Reduce the total thickness

Heigth AB

Maximum Bending momentat B =

236=

=

m

0.90

for design

purpose

50.65 Kn-m

a

6

b

x

1

0.33

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5

Full dimension wall is shown in fig 1a

Let W1 = weight of rectangular portion of stem

w2 = weight of triangular portion of stem

w3 = weight of base slab

w4 = weight of soil on heel slab.

The calculation are arrenged in Table

force(kN) Moment about toe (KN-m)

w1 1 x 0.20 x 3.70 x 25 =w2 1/2 x 0.10 x 3.70 x 25 =

w3 1 x 2.40 x 0.30 x 25 =

w4 1 x 1.20 x 3.70 x 18 =

Sw =Total resisting moment = kN-m ..(1)

0.33 x 18 x( 4.00 )2

2 2

4

3

- 64 = kN-m

\ Distance x of the point of application of resultant, from toe isSM 126.20 b 2.40Sw 121.05 6 6

b 2.40

2 2

SW 6 e 121.05 6x 0.16 70.28 < 100b b 2.40

SW 6 e 121.05 6x 0.16 30.59 < 100b b 2.40

70.28 - 30.59

70.28 - 30.59

6

(1) (2) Down ward weight of slab

= 0.30 x 1 x 1.00 x 25 = 7.50 kN-m2

= 70.28 - 7.50 = 62.8 kN-m2

underD

55.40 - 7.50 = 47.90 kN-m2

= 0.50 x( 62.78 + 47.9 ) x 0.90 = kN

47.90 + 2.00 x 62.78 0.9047.90 + 3

\ B.M. at E = 49.81 x 0.47 = kN-m

BM 23.42 x 10 6

Rxb 0.913 x 1000


= 200 mm or m at edge say = 0.26 mx

230 x 0.904 x 200

= 1/2 x 1131 = mm2

=

1.04 ==

x 1 +

Stability of wall:-

Design of toe slab:-

190.20

21.6

143.86

4.39

Hence safe

Hence safe

0.4

160

x

Pressure p at the junction of stem with toe slab is

p =

And at underE =

Total force = S.F. at E

lever arm

Hence net pressure intensities will be

Pressure distribution net moment SM =

Over turning

\ F.S. against over turning

Eccenticity e

= =x

Pressure p1 at

Heel= 1 -

Earth pressure

p=

Ka x y x H2

23.42

= =

Reduce the total thickness to

1.2

1.8

=

190.20

>

126.20

1.1

Detail

18.54.63 0.95

20.35

190.20

total MR121.05

18

79.92

kN ..(2)

Over turning moment Mo = 64 kN-m

= 48

x48

==

2Hence

safe

To make safe against sliding. Ew will have provid a shea Key .

190.20

64= 2.97=

= - x 0.16

1.04 m=

- 0.4

2.40=

m 450 Hence safe

0.12 260 + 200100

P D2 3.14 x ( 8 )4

1000 x 50

7 Design of heel slab :-

3.70 m high 2 eight of heel slab 3 upward soil pressure

1.20 x 3.70 x 1 x 18 = 79.9 = kN

Acting at 0.60 m from B.

= 1.20 x 0.26 x 1 x 25 = kN

Acting at 0.60 m from B.

= 1/2 x( 50 + 31 )x 1.20 = kN50 + 2 x 31 1.20

+ 3

\ Total force = = 80 + 7.80 - 49 = kN

\ B.M.at B =( 80 x 0.60 )+( 8 x 0.60 )-( 49 x 0.55 )

= 25.9 x 10'6

BM x 10 6

Rxb x


= 200 mm or m at edge (as that of toe slab)

x

230 x 0.904 x 200

P D2 3.14 x ( 10 )'2

4

1000 x 79

120 mm c/c at the top of keep slab. Take the reinforcement into toe

45 x D = 45 x 10 = 450 mm to the left ofB

0.12 260 + 200100

P D2 3.14 x ( 8 )'2

4

1000 x 50

Ast/A = 623 / 79 = 7.94 say = 8 No.8 bars of mm F at Bottom

8 x 79

1000 x 260

39 x 1000

Beam Ht.x beam wt. 1000 x 200

0.24 % = 0.21 (See Table 3.1)

Safe if tv< tc Here 0.20 < 0.21

8 Reinforcement in the stem:-

We had earliar assume the thickeness of heel slab as = 0.30 m

Total weight of soil =

Hence safe

Total weight of heel slab

Acting at

Effective depth required =

Hence provided these @

from a distance of

Using

Distribution steel

=Ausing 10 mm bars

39.18S.F. at B

Total upward soil pressure 48.62

7.80

Three force act on it

1. down ward weight of soil =

80.00

130

= 50

=50.44 30.59

KN say

1000

Reduce the total thickness to 0.20

=

N-mm2

0.55x = m from B

130

168=25.89

0.913 1000

Keep effective depth d

mm

Ast =25.89 10

6

sst x j x DBM x 10

6

= = 623 mm2

mm2

623= 126 mm say =

=4

mm F bars, Area = = 79

Spacing =

10

= x 1000 x 276

= 182 mm say =

=Using 8 mm F bars, Area =

2=

Spacing =

mm2

4= 50

276

Nomber of Bars =Hence Provided 10

% of steel provided =

mm c/c

mm c/c180

120

mm2

x 100

= 0.20 N/mm2

%= 0.24

Permissible shear stress for steel provided tc N/mm2

Shear stress tv =shear force

=

180 mm c/c

x = 276 mm2

2

276

=mm F bars, Area = mm24

= 182 mm say =

Distribution steel = x

Spacing =

Using 8

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0.26 m only. Hence revised H1= 4.00 - 0.26 = 3.74 m

H1 0.33 x 18 x( 3.74 )3

6

BM xRxb x


= 200 mm or m at edge

x

230 x 0.904 x 250P D2 3.14 x ( 12 )'2

4

1000 x 113

113

100

Continue alternate bars in the toe slabto serve as tensile reinforcement ther.Discontinue the remaining

45 x 12 = 540 mm beyond B, In th etoe slab.

250 - 140

250 - 140

H3

d'

h Ast' d'1/3

H1 Ast d

where Ast' = reinforcement at depth h Ast = reinforcement at depth H1

d' = effective depthat depth h d = effective depthat depth H1Ast' 1 h 1 d'

Ast 2 H1 2 d

Subsituting d = 250 =( 140 + 29.4 x h ) we get

140 + 29.4 x h 1/3

x

140 x 29.4 x h 1/3

x

h = 0.471 x ( 140 + 29.4 x h )1/3 ..(3)

h = 2.83 m 0.471 x ( 140 + 29.4 x h)

- h =

H1 H1

(2)1/3

2.83 m.Thus half bars can be curtailed can be curtailed at this point

.Howerver, the bars should be extented by a distance of 12 F = 12 x 12 = 144 mmOr d = 250 mm whichever is more beyond the point.

\ h = 2.83 - 0.25 = 2.58 m. Hence curtailed half bars at at height of2.58 m below the top . If we wish to curtailed half of the remaining bars so that remaining

Ast' 1

Ast 4

h 1 x d' 140 + 29.4 x h 1/3

H1 4 d x

x 140 x 29.4 x h4 x

h = 0.374 x ( 140 + 29.4 x h )1/3 ..(4)

h = 2.19 m 0.374 x ( 140 + 29.4 x h) - h =

H1

(4)1/3

+

1000

106

kN-m

= = 239 mm

=6

=

52.31

= = 0.63H1

x=250

x 3.741.59

This can be solved bytrial and error, Noting that if the effective thickness of stem w=are constant,

h would have been equal to

Hence from .(2)

0.79 x

2.36

\

h = 3.74

\ h

=

h = H1

m

1.26

Solving (3) by trial, we get h =

=

2.97 m

4

3.74 =

1/2 Ast

Effective depth required =0.913

than

2

=

x

if Ast =

h = 3.74

=\

2 250

mm and d'

x

250

x

52.31=M Kay x

While it has now been fixed as

Keep effective depth d

Reduce the total thickness to 0.20

Ast =52.31 10

6

=sst x j x D

BM x 100= 1007 mm2

Using 12 mm F bars, Area = = = 113 mm24

Spacing = = 112 mm say = 100 mm c/c1007

Actual AS provided = 1000 x x 1131 mm2

H =( Ast dNow As Ast

=

half bars after a distance of 45 F =Between A and B some of bars can be curtailed. Cosider a section at depth below the top of stem

)1/3

,,,'(1)

(where h In meter)

29.41

hx

or

= 140

Hence

d' +

This can be solved by trial and error, Noting that if the effective thickness of stem are constant,

h would have been equal to = =

he effective depth d' at section is 140

x h )=( 140 +

H

3.74 x h

0.02

0.04

250

= H1

remaining reinforcement is one forth of that provided arB, we have

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2.19 m.Thus half bars can be curtailed can be curtailed at this point

.Howerver, the bars should be extented by a distance of 12 F = 12 x 12 = 144 mmOr d = 250 mm whichever is more beyond the point.

\ h = 2.19 - 0.25 = 1.94 m. Hence stop half bars the remaining barsby 1.94 m below the top of the stem . Continue rest of the bars to the top of the stem

Check for shear:-

Shear force = 182

41.96 x 1000

1000 x 250

Distribution and temprechure reinforcement:-

= 310 + 200

0.12

100

P D2 3.14 x ( 8 )4

1000 x 50

= 8 = 300 mm c/cboth way in outer face

9 Design of shear key:-

Pp = KpP = 3.00 x 55.40 = kN/m3 Kp=1/ka = 3.00

Pp x a = a

18

2.00

or PH = 3.00 x( 4 + a )2

= 2.40 ax 18 = 43.20 a

\ W = 121.05 + 43.20 a Refer force calculation table=

mSw+Pp 0.5 x ( 121.05 + 43.20 a) 166.19 a3 x( 4 + a

2 )

0.5 x ( 121.05 + 43.20 a) 166.19 a

1.5

( 4 + a)2 = 60.52 + 21.6 a+ 166.19 a

4.5

( 4 + a)2 = 13.45 + 41.7 a

16 + 8 a + a2 = 41.73 a- 8 a + 13.45 - 16

a2 = 33.73 a -2.55

or a = a2 - 33.73 a +

or a = 0.076 m say = 80 mm

= 300 mm. Keep width of key 300 mm (equal to stem width)

F where (45 + F/2) =2 shearing angle of passive resistance

\ a1 = 0.3 x ( 3.00 )1/2

a1 = 0.52 m = DE = 0.90 m

Hence satisfactory.

Now size of key = 300 x 300 mm

PH = 3.00 x( 4.00 + a )2

= 3.00 x( 4.00 + 0.30 )2

PH = kN

0.17 < Permissible shear stress table 3.1

45 +a tan F = a tan x

55.47

it should be noted that passive pressure taken into account above will be devloped only when lengtha1 given below is avilable in front of key ;

However, provided minimum value of a

2.55

Actual length of the slab available

= a kpa1 =

1.5

+

Hence equilibrium of wall, permitting F.S. 1.5 against sliding we have

PH=1.5

x3

=

.(2)Weight of the soil between bottom of the base and D 1C1

0.33 x x( 4Sliding force at level D1C1 = a )2

mm2

Using 8 mm F bars, Area

164 mm say =

\ total passive pressure Pp = 166.19

166.19

mm bars

mm c/cat the inner face of

wall,along its length160

for tempreture reinforcement provide

\ spacing306

= = = 50.24

The wall is in unsafe in sliding, and hence shear key will be provided below the stem as shown in fig.

Let the depth of key =a intensity of passive pressure Pp devloped in front of key depend upon

the soil pressure P in front of the key

=

mm2

4

mm

\ Distribution reinforcement = x 1000 x 255 = 306

Average thickness of stem

2= 255

tv\

p =kayH2

1.8=

kN3.742

=41.96

2

=

x= 0.33 x

Solving (4) by trial, we get h =

=x( 4 + a)2

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= a = x

= kN Hence

SW = 121.05 + 43.20 a

= 121.05 + 43.20 x 0.30

= kN

Actual force to be resisted by the key at F.S. 1.5 is = 1.5PH - mSW= 1.5 x 55.5 - 0.5 x

= kN

16.20 x 1000

300 x 1000

16.20 x 150 x 1000

1/6 x 1000 x( 300 )2

= N/mm2 Hence safe

Bending stress =

\ shear stress =

166.19 0.30

49.86

0.16

134.01

166.19

134.01

16.20

= 0.054 N/mm2

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A A A

m H1 m m H1= m m

toe heel

D E B C D E B C D E B

Toe Toe a1

m m

D1 e

p = pp

70.28

55.40

50.4

4

30.5

9

b = 2.40

2.40

0.90W2 1.20

W1

W20.00

2.40

70.2

8

0.30 0.30

3.00 3.70

4.00H=

1

0.20

1.20

W1

3.00

0.20 0.20

3.00 3.70

W1

P=

1.00

0.90

a b

b =

55.4

0

2.40

W2

P=


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A

Outer side face

mm F Earth side Face@ c/c `

mm F

@ C/Cm

mm F@

mm F@ c/c

mm Fmm F @

@ c/c

mm FN.S.L. @

mm F mm F

@ c/c @

Toe Heel

Reinforcement Detail Reinforcement

Foundation level

mm F mm F@ c/c @ c/c `

1000

300

1.9

4

2.5

8

10

C/C

8

160

200

C/C

740

3000

8

8

8

300

160

3740

200

4.00H=

180

60

0.20

Earth side Face

130

8

10

C/C

C/C

12

8

300

300

300

450

Outer side

180

120

400

12

200

12

100


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13/16


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M-10 M-15 M-20 M-25 M-30 M-35 M-40

(N/mm2) Kg/m2

(N/mm2) Kg/m2

M 10 3.0 300 2.5 250

M 15 5.0 500 4.0 400

M 20 7.0 700 5.0 500

M 25 8.5 850 6.0 600

M 30 10.0 1000 8.0 800

M 35 11.5 1150 9.0 900

M 40 13.0 1300 10.0 1000

M 45 14.5 1450 11.0 1100M 50 16.0 1600 12.0 1200

M-10 M-15 M-20 M-25 M-30 M-35 M-40

Grade of concrete M-15 M-20 M-25 M-30 M-35 M-40

Modular Ratio 18.67 13.33 10.98 9.33 8.11 7.18

scbc N/mm2 5 7 8.5 10 11.5 13m scbc 93.33 93.33 93.33 93.33 93.33 93.33

kc 0.4 0.4 0.4 0.4 0.4 0.4

jc 0.867 0.867 0.867 0.867 0.867 0.867

Rc 0.867 1.214 1.474 1.734 1.994 2.254

Pc (%) 0.714 1 1.214 1.429 1.643 1.857

kc 0.329 0.329 0.329 0.329 0.329 0.329

jc 0.89 0.89 0.89 0.89 0.89 0.89

Rc 0.732 1.025 1.244 1.464 1.684 1.903

Pc (%) 0.433 0.606 0.736 0.866 0.997 1.127

kc 0.289 0.289 0.289 0.289 0.289 0.289

jc 0.904 0.904 0.904 0.904 0.904 0.904Rc 0.653 0.914 1.11 1.306 1.502 1.698

Pc (%) 0.314 0.44 0.534 0.628 0.722 0.816

kc 0.253 0.253 0.253 0.253 0.253 0.253

jc 0.916 0.916 0.916 0.914 0.916 0.916

Rc 0.579 0.811 0.985 1.159 1.332 1.506

Pc (%) 0.23 0.322 0.391 0.46 0.53 0.599

Table 1.15. PERMISSIBLE DIRECT TENSILE STRESS

Grade of concrete

Tensile stress N/mm2 1.2 4.0 4.42.0 2.8 3.2 3.6

Table 1.16.. Permissible stress in concrete (IS : 456-2000)

Grade of

concrete

Permission stress in compression (N/mm2) Permissible stress in bond (Average) for

plain bars in tention (N/mm2)Bending acbc Direct (acc)

(N/mm2) in kg/m2

-- --

0.6 60

0.8 80

0.9 90

1.0 100

1.4 140

1.1 110

1.2 120

31

(31.11)

19

(18.67)

13

(13.33)

11

(10.98)

1.3 130

9

(9.33)

8

(8.11)

7

(7.18)

Table 1.18. MODULAR RATIO

Grade of concrete

Modular ratio m

Table 2.1. VALUES OF DESIGN CONSTANTS

(a) sst =140

N/mm2

(Fe 250)

(b) sst =190

N/mm2

(d) sst =275

N/mm2

(Fe 500)

(c ) sst =

230N/mm2

(Fe 415)


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

0.18 0.18 0.19 0.20 0.20 0.20

0.22 0.22 0.23 0.23 0.23 0.23

0.29 0.30 0.31 0.31 0.31 0.32

0.34 0.35 0.36 0.37 0.37 0.38

0.37 0.39 0.40 0.41 0.42 0.42

0.40 0.42 0.44 0.45 0.45 0.46

0.42 0.45 0.46 0.48 0.49 0.49

0.44 0.47 0.49 0.50 0.52 0.52

0.44 0.49 0.51 0.53 0.54 0.55

0.44 0.51 0.53 0.55 0.56 0.57

0.44 0.51 0.55 0.57 0.58 0.60

0.44 0.51 0.56 0.58 0.60 0.62

0.44 0.51 0.57 0.6 0.62 0.63

300 or more 275 250 225 200 175 50 or less

1.00 1.05 1.10 1.15 1.20 1.25 1.30

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

1.6 1.8 1.9 2.2 2.3 2.5

Grade of concrete M-10 M-15 M-20 M-25 M-30 M-35 M-40 M-45 M-50

tbd (N / mm2) -- 0.6 0.8 0.9 1 1.1 1.2 1.3 1.4

tbd (N / mm2) tbd (N / mm2)

M 15

M 20

M 25

M 30

M 35

M 40M 45

M 50

Table 3.1. Permissible shear stress Table c in concrete (IS : 456-2000)

100As Permissible shear stress in concrete tc N/mm2

bd

< 0.15

0.25

0.50

0.75

1.00

1.25

1.50

1.75

2.00

2.25

2.50

2.75

Table 3.2. Facor k

3.00 and above

Grade of concrete

Over all depth of slab

k

Table 3.3. Maximum shear stress tc.max in concrete (IS : 456-2000)

tc.max

Table 3.4. Permissible Bond stress Table bd in concrete (IS : 456-2000)

Table 3.5. Development Length in tension

Grade of

concrete

Plain M.S. Bars H.Y.S.D. Bars

kd = LdF kd = LdF

0.6 58 0.96

0.9 39 1.44 40

1 35 1.6 36

60

0.8 44 1.28 45

27 2.08

33

1.2 29 1.92 30

1.1 32 1.76

28

1.4 25 2.24 26

1.3


16/16

Degree sin cos tan

10 0.174 0.985 0.176

15 0.259 0.966 0.268

16 0.276 0.961 0.287

17 0.292 0.956 0.306

18 0.309 0.951 0.325

19 0.326 0.946 0.344

20 0.342 0.940 0.364

21 0.358 0.934 0.384

22 0.375 0.927 0.404

23 0.391 0.921 0.424

24 0.407 0.924 0.445

25 0.422 0.906 0.466

30 0.500 0.866 0.577

35 0.573 0.819 0.700

40 0.643 0.766 0.839

45 0.707 0.707 1.000

50 0.766 0.643 1.192

55 0.819 0.574 1.428

60 0.866 0.500 1.732

65 0.906 0.423 2.145

Value of angle

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Reatining WAll With HORZONTAL Bach Fill

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