Design of Const Dia RCC Chimney

1

3

1)

15 20 25 30 35 40

0.25 0.22 0.22 0.23 0.23 0.23 0.23

0.50 0.29 0.30 0.31 0.31 0.31 0.32

0.75 0.34 0.35 0.36 0.37 0.37 0.38

1.00 0.37 0.39 0.40 0.41 0.42 0.42

1.25 0.40 0.42 0.44 0.45 0.45 0.46

1.50 0.42 0.45 0.46 0.48 0.49 0.49

= Π *( + + )) * 1.75 0.44 0.47 0.49 0.50 0.52 0.52

* * 2.00 0.44 0.49 0.51 0.53 0.54 0.55

2.25 0.44 0.51 0.53 0.55 0.56 0.57

= N 2.50 0.44 0.51 0.55 0.57 0.58 0.60

2.75 0.44 0.51 0.56 0.58 0.60 0.62

3.00 0.44 0.51 0.57 0.60 0.62 0.63

fig 1

Modulus of Elasticity of steel Es =

Unit weight of Fire Brick Lining 19000

0.1

m

400

mm

mm

º C

N/mm2

N/mm2

N/mm2

2.05E+05

48.00

per deg C

200

7.00

Height of middle portion of Chimney

Constant wind pressure intensity at bottom

portion

Constant wind pressure intensity at top portion

8.50

Grade of Steel fy = ( 250 or 415)

Allowable tensile stress in steel 140

Weight of Lining per meter height =

Shape Factor

Constant wind pressure intensity at middle

portion

13.33

6.00 m

-2.00 -2.80

Design of RCC Chimney :-Design of RCC Chimney :-Design of RCC Chimney :-Design of RCC Chimney :-

Modulus of Elasticity of Concrete Ec = 2.85E+04

Dimensions of Chimney and Forces

Height of Chimney

Height of Fire Brick Lining above Ground

Level

Height of top portion of Chimney

Thickness of chimney shell at top portion

1.1E-05

Grade Concrete Mix M25 25

Coefficient of expansion of concrete and

Steel

190000.4- 2 (

Thickness of chimney shell at bottom portion

of Chimney

Thickness of chimney shell at middle portion

of Chimney

Lining Support Distance @ every

The temperature of gases above

surrounding air

1800

mHeight of balance bottom portion of Chimney

Air Gap Between Wall & Fire Brick Lining (min) 100

N/mm2

25.00 m

mm

External Diameter of Chimney

4

Subject :

Fire Brick Lining 100 mm thk

Ref Calculation

200

4.00

4.00 m

Output

200

mm

300

10.00

25.00

m

m

60.00

m

1001

60

0

Prepared by : Date :ABQ Consultants

A B Quadri

4

Date :

Project :

4 Description :

4

4

Verified by :

Job no :

4

1

4

Sheet No :

cont'd :

Revision note :

4

Calculation SheetCalculation SheetCalculation SheetCalculation Sheet

Engineers Planners & Valuers - Civil / Structural design engineers

100

300

3.60 m

lining thickness

1600

5.00

N/mm2

Cross-Section of ChimneyCross-Section of ChimneyCross-Section of ChimneyCross-Section of Chimney

25

.00

m

25

.00

m

10

.00

m 14

00

n

/m

2

3.40 m

400

10.98 9.33

20

6.00 8.00

140

10.00

1800

1400 N/m2

N/m2

N/m2

4.00

250

N/m3

0.70

18.67

10.005.00

4.000.1 1.00

13.00

-3.20 -3.60

17310

Allowable compressive stress (Direct) N/mm2

Allowable compressive stress (Bending) N/mm2

Allowable tebsile stress (Direct) N/mm2

Prep By : A B Quadri- Abq Consultants - 9959010210 - [email protected]/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-500028

-4.00 -4.40

0.05

250 415Grade of Steel (N/mm2)

Allowable tebsile stress N/mm2

35

8.11

9.00

11.50

230

Grade of conc (N/mm2)

modular ratio m 7.18

15 40

Permissible Shear Stress in Concrete Tc

N/mm2 for grade of concrete100As

bd

25 30

P T O

Note : Input data in yellow cells only and ensure all check boxes are displaying "ok" or "safe".

Page 1 of 16

For shell, w = Π [ - ] * * *= N / m

For shell, w = Π [ - ] * * *= N / m

For shell, w = Π [ - ] * * *= N / m

2)

Let the vertical reinforcement be % of the concrete area place at a cover of

As = 1 * Π * ( ^2 - ^2 ) *4

=Nos of bars = =

Hence provide bars of 16mm Φ suitably placed along the circumference

Actual As =

Equivalent thickness of steel ring placed at the centre of the shell thickness ( R = - = m ) is

Ts = =

Horizontal steel (hoops) may be provided @ % of sectional areaArea of steel per metre height of chimney = * * =

Hence pitch s of mm Φ bar hoops = * =

Provide these at mm centreW = * = N

P1 = * ( * ) = N acting at

.: M = * =

.: Eccentricity e = M = = m =W

For M concrete , m =

.: Eqivalent area = A = Π/4 * ( ^2 - ^2 ) * +( - 1 )*

=Eqivalent moment of inertia = I = (Π / 64) ( D 4 -d 4 )+(m-1) Π R ts (R) 2

= Π * ( ^ 4 - ^ 4 ) * ^ 4 +

( - 1 ) * Π * * * ^ 2=

1000

28149

19002.36mm4

3.60

4.8286E+12

ok

282.5 mm

250400

0.3087179

4.00

1000000

25000

113

2.00

28149 mm2

2.36

12

0.7

ok

4.00 25.0

mm

12.5

1000000

mm2mm Φ

3.60100

16

140

4.00

126000

2000.2

mm thk

0.1

14922571055

400 mm21000

1.055

12.5 1575000 N . m

ok

m below top

250001.00

1.00 25000

2ΠR

119

1575000

1.90

100

0.20

1130974.00 0.40

Stress at Section 25.00 m below top

1.00mm50

59690200

4.00 0.30

0.40

Weight of Concrete per meter height

0.20mm thk

23876

10.98

400 mm thk

300


1.00

10.98

64

3.60

126000

28149

20123876

1800

25.00 59690

4.00

4.00

10.98

1900

25

2668534 mm2

1492257

0.2

mm

1000

P T O

okok

> 23876 mm2

ok

Page 2 of 16

For no tension to develop, allowable eccentricity = 2 I = 2 *AD *

=

The actual eccentricity is mm. Hence some tension will be developed in the leeward side.

The maximum and minimum stresses are given by σ = W ± MD

A 2I

= ± * *2 *

= ±

Compressive stress = N/mm2 N/mm2

2)


As = 1 * Π * ( ^2 - ^2 ) *4

=Nos of bars = =


Actual As =


Ts = =



Provide these at mm centreW = * + * +

* = N

P1 = * ( * ) + * ( * )= + = N

.: M = * + * =


4000

N/mm2 allowable (Safe)

mm

N/mm2

904.7

26685344.8286E+12

mm

-0.8 allowable

4.00

314ok

0.15 1.85

40001575000 1000

111

1055

1492257

Tensile stress = (Safe)

10034872

12.5 6125000 N . m

1492 mm

< 8.5-0.093

3.40 1000000


<

50

130

40841 mm2

20 mm Φ 34872

0.559

4.8286E+122668534

0.652

1.212

1000 600 mm2100

2ΠR0.20.2 300

mm600

18017310

ok

12 1000 113 188.0

25.0059690

1.492

126000 37.5

25.00

mm2

Thickness of shell = 300 mm

mm

25.0

1.00

2.00

40841 3.51

4104491


P T O

0.7 1800 4.00

25.00 87179

25.00.7 1600 4.00238000.00126000 112000.00

112000

61250004104491

okok

> 34872 mm2 ok

ok

Page 3 of 16

For M concrete , m =

.: Eqivalent area = A = Π/4 * ( ^2 - ^2 ) * +( - 1 )*

=Eqivalent moment of inertia = I = ( Π/64 ) ( D 4 -d 4 ) + ( m-1 ) Π R ts (R) 2

= Π * ( ^ 4 - ^ 4 ) * ^ 4 +

( - 1 ) * Π * * * ^ 2=


=The actual eccentricity is mm. Hence some tension will be developed in the leeward side.


A 2I

= ± * *2 *

= ±


3)


As = 1 * Π * ( ^2 - ^2 ) *4

=Nos of bars = =


Actual As =


Ts = =



Provide these at mm centre

(Safe)allowable (Safe)

1000

3.40 100000010.98 40841

25 10.98

4.00

6410.98 1850 3.51

3894758 mm2

4.00 3.40

6.7041E+12

N/mm2

1492

3894758

4104491

1.054 1.827

4000860.7 mm

1000 4000

2.881 < 8.5

3894758 6.7041E+12

Tensile stress = -0.773 < -0.8


Thickness of shell = 400 mm

4.00 3.20 1000000

1.0050 mm

1.85

45239491

120

93 nos

10045239 mm2

25 mm Φ

5.07 mm2ΠR

58905 mm2

2.00 0.15

800100

12 1000 113 141.0 mm

0.2 4000.2


P T O

800140 ok

mm2

okok

> 45239 mm2 ok

ok

ok

N/mm2 allowable

6125000

18506.7041E+12 mm4

1000

58905

Page 4 of 16

W = * + * + * +* + * = N

P1 = * ( * ) + * ( * )* ( * )

= + + = N

.: M = * + * + *=


For M concrete , m =.: Eqivalent area = A = Π/4 * ( ^2 - ^2 ) * +

( - 1 )*=

Eqivalent moment of inertia = I = ( Π/64 ) ( D 4 -d 4 ) + ( m-1 ) Π R ts (R) 2

= Π * ( ^ 4 - ^ 4 ) * ^ 4 +

( - 1 ) * Π * * * ^ 2=


=The actual eccentricity is mm. Hence some tension will be developed in the leeward side.


A 2I

= ± * *2 *

= ±


The eccentricity is quite high. Due to this, tensile stresses in

the windward side are expected to be greather than 0.8 N/mm2

resultingin cracking of concrete. Hence it is assumed that onlysteel will take the tensile stresses and concrete in the tensile zone will be ignored. Thus, the method of analysis used at

m and m will not be applicable. We shall analyse the section for stresses by method discussed in § 8.3.

Tc = R = - = mTs = eccentricity e =m =

In order to find the position of N.A., use equation 8.3 :

mΠ Ts

2{ + } mΠ Ts

fig 2

25.00 50.00

10.98

1.609

[ (Tc-Ts)

sin2Φ

sinΦ

Π-Φ

(Π-Φ)

[ ](Tc-Ts)

e = R{ +

cosΦ cosΦ

10.98

1850 5.07

]} +

0.202.00

4 2

1.80m

5.0

27720039200

39200

126000 112000

5111764 mm2

4.00 3.20

18508.4252E+12 mm4

8.4252E+12

100064

5111764

25.00 59690 25.00 17310 25.00 8717910.00 17310 5408566

0.7 1800 4.00 25.0

10.00 113097

0.7 25.00.7 1400 4.00 10.00

126000 47.5 112000 22.5

1600 4.00

540856625 10.98

8701000 N . m

8701000

5890510000003.20

1609 mm1.609

4.0010.98

4000

1609

5408566 8701000 1000 4000

824.1 mm

3.124 < 8.5

5111764 8.4252E+12

1.058 2.065

N/mm2 allowable (Safe)Tensile stress = -1.007 >= -0.8 N/mm2 allowable Check further

400 mm5.07 mm


P T O

Page 5 of 16

* Π

m

Assume Φ = =

.: e = + =+

= m.: consider Φ =

The maximum stress c1 in concrete is found from Eq.8.1

.: 2 * *1 +

c1

.:

Tensile stress in Steel, assuming concrete to be fully cracked.1 - cosΦ1 + cosΦ

Horizontal steel (hoops) may be provided @ % of sectional area

Area of steel per metre height of chimney = * * =

Hence pitch s of = * =

Provide these at mm centre

As = in pitch s = mm centre, if the cover is thenD1 = - =

p * s *2 * As * D1 2 * *

N/mm2

allowable

not ok0.00 º C

{ 0.9976

140

safe

]

]0.0698*

34.00* [ ] =m

<

1.609

º C radians86.00 1.5010

>=

10.98 * Π cosΦ* *0.51

+ 1.6406 * +0.0698 }

=1518945

=

12 1000 113

(b) Stress in horizontal reinforcement

N/mm2 safe

mm

mm2

< 140 N/mm2

= N/mm2 43.7676140

141.0

3.5607=54085661518945

Safe

Compressive stressc1

in Concrete

t1

.: t1 = =3920113

277200

)

]* * ++ *( 40.00Π-Φ 0.51

- 0.51 )sin2Φ[ 10.98

{

c1*

e = 180

W

*40.00 0.51( -

{ 4 2 } 2

=

18005408566 =

1+cosΦ

2Rc1 [ (Tc-Ts)

0.06976

7651.4545.136

}sinΦ + (Π-Φ) cosΦ

1573.241.219

[

{

10.98 Π

< 8.5

c1

* *

- 5.07

]*

} + mΠ(Π-Φ)

400 800

N/mm2

5.07

Ts cosΦ

400 )

sinΦ +


=

mm

140 ok

1000

40

mm800

113.1 mm2 1404000 80 3920

N/mm2

6078.2143.92

1000.2

mm Φ bar hoops

1.695

+

Now adjust the value of angle Φ in such a way that the value of eccentricity e is

ok

which is slightly more than the actual value

P T O

[ (

cosΦ

0.2

Page 6 of 16

(c) Stress on leeward side due to temperature gradient

.: = - = mm

a = =c1 =

Es = Ec = =

p = = = α = per º C

Temperature difference =

Let us assume that % of temperature drops through the lining and shell.Drop in temperature = * =Asssuming that drop in lining is times more than that in shell, per unit thickness,the drop of temperature through concrete is given by,

Tº = =+ *

To locate -neutral axis in the shell thickness, use Eq. 8.10

= α * T * Ec* k2 - m * p * (a - k)

.: * [ 1 + ( - 1 ) * ) ] = * ** k2

- * * ( - k )

or =k2 + * k -

k2 + * k - = 0

solving for k k =

fig 3

fig 4

100

100

Thickness of lining Tl =

160

400mm

mm

N/mm2

5.07

N/mm2

350

mm

10.98

0.875

50

Concrete Temperature Co-efficient

0.793

2.05E+05

0.01267

2.05E+05

0.24343

0.278

350

aTc 400

400

º C0.8

1.867E+04

8.0219

( 1

200

0.27821

3.5607

+ pm -1

80

0.7620

400 5

mmThickness of shell Tc =

Tc 400

3.5607

5.07Thickness of steel Ts =

Cover to vertical steel =

Ts

50

5

1.10E-05

[0.5

º C

º C

) *

200

N/mm2


71.1110.98 1.1E-050.01267

]

0.5 10.98 0.01267 0.875

160*

400 71.11

c1

14.6043

P T O

1.867E+04

ok

ok

ok

Page 7 of 16

.: a * α * Tº * EcStress in Concrete a - k

= * * *

=

4Stress in Concrete 3

The above analysis is based on the assumption that the tension caused by temperature variation cannot be taken by concrete, and it is taken entirely by steel.

Stress in Steel = t = = * * ( - )

=

(d) Stresses on windward side, due to temperature gradient

p * t1 = α * Tº * Ecm * p * ( a - k ) - * k2

where t1 = p = a = m =α = Tº = Ec =

.: * 0 = ** * ( - k ) - * k2 *

- k - * k2

- k - * k2 =

solving k =

.: c = α * Tº * Ec * k =Stress in Concrete

Tensile stress in Steel, assuming concrete to be fully cracked.

t = m c a - k = * * ( - )k

=

(e) Stresses on the Neutral axis .(i.e. temperature effect alone)

where m = p = a =α = Tº = Ec =

or k2 - * + √ 2 * * *

+ * * *

N/mm2

140 N/mm2 <

º C 18670

0.312340.31234

+

0.5=

0.875

N/mm2

0.01267

0.7620.762

fig 5

10.98

71.11

10.98 0.01267

0.000011

4.5615

11.129

-mp

14.604

=10.98 0.01267

18670

0.01267 0.875

0.3123

º C

k2 =

10.98

√2mpa + m2p2

10.98 0.01267 0.875

=

α

71.11

10.98 0.01267

10.98

71.11

18.113 N/mm2

0.8750.01266892

0.0295

k

0.875 0.5

90.23 N/mm2

0.13910473

0.430718713

Compressive

0.12172 0.13910

Compressive =c =

1 +

0.121716639

* Tº

( safe )

k

0.5

0.5

33.998 N/mm2

71.110.00001118670

4.5615

0.875

Thus the compressive stress less than the permissible

m c

k

10.98

Since wind stresses are taken into account,

Permissible= 11.33

0.762 1.1E-05

8.5*

11.129 N/mm2

10.98

0.01267

a - k

0.000011

33.998

N/mm2

* Ec*

or

1.867E+04

N/mm2


P T O

Page 8 of 16

k2 =

.: c2 = α * Tº * Ec * k2 =Stress in Concrete

Tensile stress in concrete, assuming concrete to be fully cracked.

t2 = m c2 a - k2= * * ( - )

k2

=

(b) Stress in horizontal reinforcement due to temperature :

p' = = =S Tc *

a' = =

From Eq. 8.13.

or k' - * + √ 2 * * *

+ * * *k' =

.: c' = α * Tº * Ec * k' =Stress in Concrete

Tensile stress in concrete, assuming concrete to be fully cracked.

t2 = m c' a' - k' = * * ( - )

k'

=

These stresses are due to temperature effect alone. To this we must add the stresses due to wind.Hence total stress in steel= + = Since wind is also acting, permissible t = 4 * =

allowable tensile stress in steel 3

( safe )

N/mm2

0.00202

0.900

0.00202

0.875 0.37352

140

5.4550 N/mm2

O.k

43.768 159.41 N/mm2

< 140 N/mm2

Safe

<

0.17884

115.64

0.37352

N/mm2

Compressive

115.64

80.42 N/mm2

k' = -mp' + √2mp'a + m2p'2

0.37351931

0.00202

10.98 5.4550

10.98

0.900400360

Compressive

=0.00202

0.1788398110.98 10.98

10.98 0.00202

140 186.67 N/mm2

N/mm2

0.17884

2.6118

AΦ

140 400113.10


P T O

10.98 2.6118 0.900

Page 9 of 16

5. Flue Opening :

Provide a flue opening m wide andm high at bottom.

The boundary of the opening is thickened andreinforced as shown in Fig A. The vertical steel bars are bent on either side of theopening as shown

P P = NM = N . Mm

V M V = N

5408566

fig 6

1.52.0

6. Force acting at 0.00 level for Foundation Design :

0.00 level

8701000277200


P T O

ok

Page 10 of 16

Pe

Data FFL H

FGLSoil filling inside

Axial load at the base of footing

== P + Weight of Chimney Wall + Soil Filling inside

of wall + Weight of soil + Self weight of footing= + Π ( ^2 - ^2 )

4* ( + ) *+ Π ( ^2 ) *

4+ Π ( ^2 - ^2 ) * *

4+ Π ( ^2 ) * *

4= + + + +

= kn

= + H * ( D + T + A )= + * ( + + )=

.: e' = = = < = m

Axx = Π * Ixx = Π * =4

=Zxx = Π * =

fig 7

25

25

1.70

183.22 4325.97

0.2

8851.44

Reinforcement cover c = 75 mm

5408.57

18

4.00

14000

OD

8701

9865.2418882.61

0.5225

9865.24 Kn . M

M'

2.30

8OD

OD

18882.61

18

2.30

1.70

1700

4000

14000d

T

D

Level of footing below ground Totd = 4000 mm

2300Dia of Footing OD = 14.00 m

A =

Outer dia of chimney d = 4.00 mThickness of chimney wall t = 400 mm

Depth of Footing T = 2300 mm

200 mm

Depth of Soil D = 1700 mm

1609

7. Design of Cirrcullar Chimney Foundation :

Concrete Grade fc' = 25 N/mm2

200

Steel Grade fy = 415 N/mm2

18

Moment M = 8701 Kn . M

A

277.2

Horizontal load H = 277.2 Kneccentricity e = M/P = 1.609 m

S.B.C of Soil Qs = 200 Kn/m2Kn/m3

Axial Load P = 5408.57 KnDensity of soil Ws =

14.00

5408.57 113.41

14.00

0.23.60

25

P'

Footing Reinforcement dia Φ = mm

M' 8701

3.605408.57 4.00

1.70

Ok

153.94 m264

OD2 OD4 1885.74099

OD3 m3

32


269.39157

]1.75

m4

32

[

277.2

P'

P T O

5408.57

To

td4000

Page 11 of 16

The maximum and minimum base pressures are given by σ = P' ±

A

= ±

= ±

=

=

Factor of Safety against overturning = = P' * OD2

= *

=

Assume initially of mm Φ bars nos spaced radially along the + mm Φ bars = mm2 circumferance

Φ = º = radians= = mm= = mm= mm Length of segment 'PQ'

= mm Length of segment 'RS'

=CG of Segment 'PQRS'

fig 8

fig 9

Design of Footing slab

1 Layer 32

from 'PQ'=

120

0.0524

1295

1.5 safe>

Stabilising Moment

25

M'

> 0

7000


Kn / m2 allowable

122.664

Zxx

σ min 86.04 Kn / m2

<

36.620

Ok

18882.61153.94 269.39

σ max Kn / m2 allowable200

Uniform pressure

Pressure due toMoment at 'RS'

under area 'PQRS'

Ok

86.0

4

10.5

Radius of ChimneyRadius of FoundationBar Spacing at ro 105

rofro

Bar Spacing at fro 367

=

=

Kn / m2

Kn / m2

2.963 m

=

M'

2000

Kn / m2

122.7

159.28

9865.24

13.40

36.6

Area of Segment 'PQRS' 1.1781 m2

As

3.00

Overturning Moment

9865.24

Kn / m2

Pressure due toMoment at 'PQ'

P T O

159.2

8

18882.61 14.002

ro

froΦ

Main Reinforcement

Area covered by one unit of Main Reinforcement

Critical Section for Moment

Footing Outer Dia

Chimney Outer Dia

A A

P

R

S

Q

Line of Punching Shear

Line of Shear

Sa1

a2

b2

b1

rs

rp

Page 12 of 16

r16 Φ top radial reinforcement

1

Shear Stirrups (if required)

Main Radial reinforcement Circullar reinf

Φ nos 16 Φ @ c/c

cover

Section A - A

.: = + =

fy = .: fyall = m =fc' = .: fc'all =

.: k = * =* +

j = 1 - k = 1 - =3

R = 1 j k = 1 * * *2 2

=

Hence d = = * = mm* R *

.: adopt T = cover = d = -

.: d = effective depth

As = = =Fyall * j * d * *

.: Φ + Φ = Π * ( ² + ² )4

=

Φ @ c/c = * = %*distribution steel

Straight portion

2300

90

0

32 120 200

ok1116

1000000

0.556

AΦ

mm2

105 2225100

25

Main radial reinforcement

16 200

7000

415 N/mm2

464.70

8.5 0.904

Moment at 'PQ' Mf

fig 10

1.109

230

2500 2500

1000

fc'all

Mf 516.06

25 N/mm2 8.5

10.98 8.510.98

10.98

mm

√

1.109

8.5 230

0.289

1.79

1112.5

0.90378

516.06

N/mm2

0.289

75

1116

75

2225

mm

>

ok

32

mm2

516059761

51.36

N/mm2

0.289

Sloping portion

critical punching shear section

2225

c/L of Foundation

critical shear Section

32

230 0.90378

Provide

1 dia bars.Provide

p%

M

mm

75

layer of


22252300

√105

2000

Kn .m

1295

1295

25

3

P T O

2300

2108.25

mm2

C / L of foundation

C / L

Page 13 of 16

r = d = r + d =2 2

Φ = º = radians= = mm= = mm= mm= mm

=

.: F = + =

= = * =

.: Depth required for punching shear do = =* *

= < mm provided

Check shear at r + d from the c/L, End of Straight portion, and at three points at sloping portion.

32 + 25 32 + 25 32 + 25 32 + 16 32 + 16 32 + 0

245

ok

1

193

M allowable / bar Kn.m

ok

Area of Segment 'PQxx'

CG of Segment 'PQxx'

M actual / bar

10.46Pressure due to moment at section

1292

1005

22.10

Kn.m 516

m

1.178m2

36.6

kn/m2

mm

mm2

p%

ok okok ok

.: Provided spacing = mm 175 175

1 1Minimum shear s = 2.5Asvfy/b mm 381 180 1

ok

1

142 91spacing mm 414 6433

1

Shear actual

Shear - Vs per main bar

Shear Stress tc

140Shear - MS bar dia fyall

Shear Reinf

1 1

11

p% for Shear

N/mm2

Kn

1

0.25

0.23

12

0.23

ok

113

Reqd

Moment at 'PQ'

from 'a1a2'

Pressure due to=

36.6 Kn / m2Moment at 'RS'

Reqd

0.25

ok

0.23

27.90

163

61675333

36.6

0.23

2225

0.80.16

mm

fck0.16

F

1177

√ fck

2000

2225

2225

220 234

1295 1295

1758

4225 4500

1.50 1.34

mm

0.8√

163367

2000

As

599 367

85

distance from c/L

OK

Check Punching shear at ro + d/2 from the c/L

0.250.25

Kn

Kn

118

5

120Shear allowable

81 45110

112 96

0.538

Uniform pressure

kn/m2

104 322

1005

0.2417

ok ok ok ok

0.2057

122.7

ok

Reinforcement Spacing

Effective depth

Bar dia

278

2225

1295

kn/m2

mm

0.815

32.26

36.6 36.6

0

7000

366

825

804

122.7

36.62

√

Kn / m2

m

1.02919Area of Segment 'PQa1a2'

Kn

250.16

CG of Segment 'PQa1a2'm2

N/mm2

ok

0

ok

153469163

153.47143.00 10.47

73

36.6Pressure due to moment @ rs

Not Reqd

mm

0.2488

2.96

599

0.5597 0.26640.2646

122.7 122.7 122.7 122.7

ok

0.753

23.54

156188

599

0.87

Length of segment 'a1a2'

Length of segment 'RS'

36.6

12

0.50

0.31

72

157

Not Reqd

Radius of Foundationrps

3.00

1112.5

Bar Spacing at fro

Radius of Punching shear

Punching Shear at 'a1a2'

Kn / m2

Allowable Punching Shear stress

Punching Shear :

0.287

0.00

270 138

19

0.000

0.43

Uniform pressure = 122.7

Pressure due to=

7000

Shear :

mm

16.3

under area 'PQRS'

Bar Spacing at rps

0.0524

fro3113

3113

P T O

= 2.193

mm

1

0.25

0.23

69

Not Reqd Not Reqd

ok


Page 14 of 16

Data :

Height of Top portion of Chimney =

Wind intensity of top portion of Chimney = N/m2

Concrete Area of Top Portion of chimney = mm2

Moment of Inertia of Top portion of Chimney = mm4

Wind Moment at the base of Top Portion = N.m

Modulus of Elasticity of Concrete = N/mm2

M / Ei = 1/mm

Area of M / Ei of top portion =

C.g of Area of M / Ei of top portion = mm

Moment of Area of M / Ei from top portion =

Partial Deflection of Top Portion δtop = mm

Ratio L / δ = L / L /

Height of Middle portion of Chimney =

Wind intensity of Middle portion of Chimney = N/m2

Area of Middle Portion of chimney = mm2

Moment of Inertia of Middle portion of Chimney = mm4

Wind Moment at the base of Middle Portion = N.m


M / Ei = 1/mm

Area of M / Ei of Middle portion =

C.g of Area of M / Ei of Middle portion from top = mm

Moment of Area of M / Ei of Middle portion =

Partial Deflection of Top Portion δtop = mm


wrt bottom of middle portion

P T O Prep By : A B Quadri- Abq Consultants - 9959010210 - [email protected]/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-500028

1600

3894758

6.7041E+12

6.1250E+06

2.8500E+04

3.2057E-08

5.4377E-04

3.7500E+04

2.2776E+01 (2)

25.00

1800

2668534

4.8286E+12

1.5750E+06

m

2.384

10485 200ok

Check Deflection of Chimney :

2.8500E+04

1.1445E-08

1.4306E-04

1.6667E+04

2.3843E+00

Top Portion :

(1)

25.00 m

wrt bottom of top portion

22.776

2195 < 200ok

Middle Portion :

>

Page 15 of 16

Height of Bottom portion of Chimney =

Wind intensity of Bottom portion of Chimney = N/m2

Area of Bottom Portion of chimney = mm2

Moment of Inertia of Bottom portion of Chimney = mm4

Wind Moment at the base of Bottom Portion = N.m


M / Ei = 1/mm

Area of M / Ei of Bottom portion =

C.g of Area of M / Ei of Bottom portion from top = mm

Moment of Area of M / Ei of Bottom portion =

Total Deflection of Top Portion δtop = mm


wrt bottom of bottom portion

P T O Prep By : A B Quadri- Abq Consultants - 9959010210 - [email protected]/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-500028

3.6236E-08

3.4147E-04

5.5000E+04

4.1556E+01 (3)

41.556

1444 < 200ok

Bottom Portion :10.00 m

1400

5111764

8.4252E+12

8.7010E+06

2.8500E+04

Page 16 of 16

Documents

Design of Const Dia RCC Chimney