6 Design of Constant Dia RCC Chimney and Foundation 23052014

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

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

0.1 1.00 19000

20295

P T O

-4.40

Weight of Lining per meter height =

4.50 - 2 ( 0.4 0.1 0.05

Allowable tebsile stress (Direct) N/mm2 -2.00 -2.80 -3.20 -3.60 -4.00

10.00

Allowable compressive stress (Bending) N/mm2 5.00 7.00 8.50 10.00 11.50 13.00

Allowable compressive stress (Direct) N/mm2 4.00 5.00 6.00 8.00 9.00

35 40

modular ratio m 18.67 13.33 10.98 9.33 8.11 7.18

Shape Factor 0.70

100As

bd

Permissible Shear Stress in Concrete Tc

N/mm2 for grade of concreteGrade of conc (N/mm2) 15 20 25 30

Grade of Steel (N/mm2) 250 415

Allowable tebsile stress N/mm2

140 230

Constant wind pressure intensity at middle

portion

1600 N/m2

Constant wind pressure intensity at bottom

portion

1400 N/m2

Lining Support Distance @ every 6.00 m

Constant wind pressure intensity at top portion 1800 N/m2

Thickness of chimney shell at bottom portion

of Chimney

400 mm Cross-Section of ChimneyCross-Section of ChimneyCross-Section of ChimneyCross-Section of Chimney

Height of balance bottom portion of Chimney 22.00 m

300 mm

100

Height of middle portion of Chimney 25.00 m fig 1

400

Thickness of chimney shell at top portion 200 mm

14

00

n

/m

2

22

.00

m Height of top portion of Chimney 25.00 m

Thickness of chimney shell at middle portion

of Chimney

Grade Concrete Mix M25 25 N/mm2

Modulus of Elasticity of Concrete Ec = 2.85E+04 N/mm2

Allowable tensile stress in steel 140 N/mm2

Modulus of Elasticity of steel Es = 2.05E+05 N/mm2

º C

25

.00

m

Coefficient of expansion of concrete and

Steel

1.1E-05 per deg C

Grade of Steel fy = ( 250 or 415) 250 N/mm2

Height of Fire Brick Lining above Ground

Level

48.00 m

16

00

3.90 m 300

The temperature of gases above

surrounding air

200

Unit weight of Fire Brick Lining 19000 N/m3 100 lining thickness

Fire Brick Lining 100 mm thk

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

4.10 m

Height of Chimney 72.00 m

25

.00

m

External Diameter of Chimney 4.50 m

1800 4.50 m

Dimensions of Chimney and Forces 200

Ref Calculation Output

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

Prepared by : Date : Job no : Sheet No :

cont'd :

Project :

Subject : Description : 4

Verified by : Date : Revision note :

Calculation SheetCalculation SheetCalculation SheetCalculation Sheet

Visit Abqconsultants.com

This program Designs and

Optimises RCC Chimney and

Foundation.

Written and programmed

by

:-

A B Quadri

www.abqconsultants.com

[email protected]

[email protected]

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=

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

6410.98 2150 2.08 2150

6.9073E+12 mm4

10.98 281492982693 mm2

4.50 4.10 1000

168860625 10.98

4.50 4.10 1000000

m below top

141750 12.5 1771875 N . m

1771875 1.049 1049 mm

0.7 1800 4.50 25.0 141750 12.5

400250 ok

25.00 67544 1688606

10012 1000 113 282.5 mm

2ΠR0.2 ok0.2 200 1000 400 mm2

2.25 0.1 2.15

28149 2.08 mm

201140 ok

28149 mm2 > 27018 mm2 ok

27018 mm216 mm Φ 27018 135

4.50 4.10 1000000100

128805

Stress at Section 25.00 m below top

1.00 ok50 mm ok

1.00 2500098960

400 mm thk 4.50 0.40 0.40 1.00 25000

67544

300 mm thk 4.50 0.30 0.30

Weight of Concrete per meter height

200 mm thk 4.50 0.20 0.20 1.00 25000

[email protected]

9959010210

9959010211

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 = * + * =



6890625 1.476 1476 mm4669977

141750 37.5 126000 12.5 6890625 N . m

0.7 1600 4.50 25.0141750 126000.00 267750.00

25.00 98960 4669977

0.7 1800 4.50 25.0

600180 ok

25.00 67544 25.00 20295

10012 1000 113 188.0 mm

2ΠR0.2 ok0.2 300 1000 600 mm2

2.25 0.15 2.10

40841 3.10 mm

126314

130 ok

40841 mm2 > 39584 mm2 ok

10039584 mm2

20 mm Φ 39584

50 mm ok

4.50 3.90 1000000


Thickness of shell = 300 mm

1.00 ok

N/mm2 allowable (Safe)Tensile stress = -0.011 < -0.8 N/mm2 allowable (Safe)

0.566 0.577

1.143 < 8.5

1049

1688606 1771875 1000 45002982693 6.9073E+12

6.9073E+122982693 4500

1029.2 mm

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 centre800

140 okP T O


10012 1000 113 141.0 mm

2ΠR0.2 ok0.2 400 1000 800 mm2

ok

2.25 0.15 2.10

58905 4.46 mm

105 nos491

120 ok

58905 mm2 > 51522 mm2

10051522 mm2

25 mm Φ 51522

50 mm ok

4.50 3.70 1000000


Thickness of shell = 400 mm

1.00 ok

N/mm2 allowable (Safe)Tensile stress = -0.533 < -0.8 N/mm2 allowable (Safe)

1.070 1.603

2.673 < 8.5

4669977 6890625 1000 45004365997 9.6716E+12

9.6716E+124365997 4500

984.5 mm1476

6410.98 2100 3.10 2100

9.6716E+12 mm4

10.98 408414365997 mm2

4.50 3.90 1000

25 10.98

4.50 3.90 1000000

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


(Tc-Ts) sinΦ (Π-Φ) cosΦ cosΦ ]} + ] fig 2

e = R4 2

[[ (Tc-Ts) {

sin2Φ+

Π-Φ

2.054.46 mm 1.742 m

10.98

25.00 50.00

400 mm 2.25 0.20

N/mm2 allowable (Safe)Tensile stress = -1.164 >= -0.8 N/mm2 allowable Check further

1.385 2.549

3.934 < 8.5

7950177 13848345 1000 45005740082 1.2225E+13

1.2225E+135740082 4500

946.6 mm1742

6410.98 2100 4.46 2100

1.2225E+13 mm4

10.98 589055740082 mm2

4.50 3.70 1000

795017725 10.98

4.50 3.70 1000000

13848345 N . m

13848345 1.742 1742 mm

141750 126000 97020 364770

141750 59.5 126000 34.5 97020 11.0

4.50 25.00.7 1400 4.50 22.000.7 1800 4.50 25.0 0.7 1600

9896022.00 20295 22.00 128805 795017725.00 67544 25.00 20295 25.00

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

Safe


140= 51.079 N/mm2 < 140

4420

4500 80 4420 mm

.: t1 = =364770

113

800140 ok

113.1 mm2 140 40 mm

12 mm Φ bar hoops 1000 113 141.0 mm

0.2 400 1000 800 mm2

100

140 N/mm2

safe(b) Stress in horizontal reinforcement

0.2 ok

[ ] = 43.96 N/mm2 <t1 = m * c1 *

N/mm2 < 8.5 N/mm2 safein Concrete 1726917

Compressive stressc1 =

7950177= 4.6037

4.46 * 0.0698 ]=

1726917

} + 10.98 * Π *{ 0.9976 + 1.6406 * 0.0698

400 - 4.46 ) *0.06976

Ts cosΦ ]1+cosΦ

7950177 =2050 c1 [ (

+ (Π-Φ) cosΦ } + mΠW =2Rc1 [ (Tc-Ts) { sinΦ

not ok which is slightly more than the actual value

86.00 º C 1.5010 radians

6922.40 1578.44

]Now adjust the value of angle Φ in such a way that the value of eccentricity e is >= 1.742

} + 10.98 * Π ** { sinΦ + (Π-Φ) cosΦ

0.00 º C

]4 2 2

[ ( 40.00 - 0.45 )

Π-Φ } +

10.98*

0.450.45 ) * {

sin2Φ+

0.45 * cosΦ

8500.8443.98 1.074 45.058

1.887

e = 205 * [ ( 40.00 -

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 = 0.8419


10.2329 14.60430.24509 0.21445

0.245 0.915

71.110.5 10.98 0.01116 0.875 1.867E+04

0.5

4.6037 10.98 0.01116 1.1E-05

- 1 ) * p ]c1 [ 1 + ( m

5 ok

160*

400 71.11 º C400 5 100

200 º C ok

80 ok200 0.8 160 º C

N/mm2

10.98

Ts 4.46 0.01116 1.10E-05 Concrete Temperature Co-efficient

Tc 400

4.6037 N/mm2 400

2.05E+05 N/mm2 2.05E+05 1.867E+04

350Cover to vertical steel = 50 mm

350 0.875

Thickness of steel Ts = 4.46 mmaTc 400 50

Thickness of shell Tc = 400 mm Thickness of lining Tl = 100 mm

fig 3

fig 4

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 * * *

+ * * *10.98 10.98 0.01116 0.01116


N/mm2

=10.98 0.01116 10.98 0.01116 0.875

0.01116 0.8750.000011 71.11 º C 18670

k2 = -mp + √2mpa + m2p2

10.98

95.36 N/mm2 < 140 N/mm2 ( safe )

fig 5

10.98 4.0937 0.875 0.28030.2803

0.10723 0.12254 0.5 0.0336

0.2803

Compressive 4.0937 N/mm2

or0.490581772

=14.604

0.107226563 0.122544643 0.5

0.01116 43.956 0.000011 71.1110.98 0.01116 0.875 0.5 18670

43.956 N/mm2 0.01116071 0.875 10.980.000011 71.11 º C 18670 N/mm2

0.842k 0.842

5.308 N/mm2

0.5

11.33 N/mm2

Thus the compressive stress more than the permissible--not ok

m c a - k 10.98 12.295 0.875

Since wind stresses are taken into account,

Permissible= * 8.5 =

0.842 1.1E-05 71.11 1.867E+04

12.295 N/mm2

* Tº * Ec1 +

k

Compressive c = = k * α

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


O.k

115.64 51.079 166.72 N/mm2

140 186.67 N/mm2 Safe

10.98 2.6118 0.900 0.17884

0.17884

115.64 N/mm2 < 140 N/mm2

10.98 10.98 0.00202 0.002020.17883981


=10.98 0.00202 10.98 0.00202 0.900

360 0.900400

k' = -mp' + √2mp'a + m2p'2

( safe )

AΦ 113.10 0.00202140 400

0.35649

83.15 N/mm2 < 140 N/mm2

0.35648596


10.98 5.2062 0.875 0.35649

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


6. Force acting at 0.00 level for Foundation Design :

795017713848345

3647700.00 level

P T O

1.52.0 ok

fig 6

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 * =

32P T O


64153.94 m2

OD3 m3269.39157

]P' 21391.43 8

Ok

OD2 OD4 m41885.74099

0.215380.38 Kn . M

M' 15380.38 0.7190 [ OD 1.75

M' 13848.34513848.345 364.77 2.30 1.70

7950.18 128.33 237.65 4223.83 8851.44

21391.43

14.00 4.50 18 1.70

14.00 2.30 25

1.70 0.2 254.10 18

fig 7

14000

OD

7950.18P'

7950.18 4.50 4.10

Footing Reinforcement dia Φ = 32 25 mm 14000Reinforcement cover c = 75 mm OD

Depth of Soil D = 1700 mm 4500Depth of Footing T = 2300 mm d

TDia of Footing OD = 14.00 m A = 200 mmLevel of footing below ground Totd = 4000 mm

4.50 mThickness of chimney wall t = 400 mm

2300

Kn . M

4000eccentricity e = M/P = 1.742 m

To

tdHorizontal load H = 364.77 KnOuter dia of chimney d =

1700

D

Density of soil Ws = 18 Kn/m3Axial Load P = 7950.18 KnMoment M = 13848.345

Steel Grade fy = 415 N/mm2S.B.C of Soil Qs = 200 Kn/m2

7. Design of Cirrcullar Chimney Foundation :

7950.18364.77 1742

Concrete Grade fc' = 25 N/mm2

200

A

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'


= 2.782 mfrom 'PQ'

fig 9

P T O

Pressure due to= 18.4 Kn / m2

Moment at 'PQ'

Area of Segment 'PQRS' 1.15028 m2

Uniform pressure = 139.0 Kn / m2

under area 'PQRS'Pressure due to

= 57.1 Kn / m2Moment at 'RS'

Radius of Foundation fro 7000Bar Spacing at ro 118Bar Spacing at fro 367

1295

3.00 0.0524Radius of Chimney ro 2250

1 Layer 32 12025 As

15380.38

9.74 > 1.5 safe

Design of Footing slab

Stabilising MomentOverturning Moment

M'21391.43 14.00

2

Ok

σ min 81.87 Kn / m2 > 0 Kn / m2 allowable Ok

σ max 196.05 Kn / m2 < 200 Kn / m2 allowable

21391.43 15380.38153.94 269.39

81.8

7

196.0

5

138.961 57.093

M'

Zxx

fig 8

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

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 = * = %*

0.494distribution steel 118 2225


> 1241 mm2

okProvide 16 200 p% 1295 100

1 dia bars.layer of Main radial reinforcement 1295 mm2

mm2

230 0.90378 2225

Provide 32 25 AΦ 32 25

2225 mm ok

M 573887856 1241

2096.081000 118 1.109

2300 mm 75 mm 2300 75

0.289

1.109

√ Mf √ 573.89 1000000

0.289 0.903783

fc'all 8.5 0.904

10.98 8.5 0.28910.98 8.5 230

415 N/mm2 230 N/mm2 10.9825 N/mm2 8.5 N/mm2

fig 10

Moment at 'PQ' Mf 503.33 70.56 573.89 Kn .m

1112.5 75

critical punching shear section

2225 critical shear Section

7000

1.61

c/L of Foundation

2300

90

0

32 120 200

2250 2500 2250

Straight portion Sloping portion

C / L of foundation

C / L

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


ok okok not ok ok ok

1 1

.: Provided spacing = mm 175 175

Minimum shear s = 2.5Asvfy/b mm 341 169 1 1

1

spacing mm 349 2638 245 193 142 91

1 1

Shear - MS bar dia fyall 140 12 12 1 1 1

Shear - Vs per main bar Kn 101 13 1 1

49 0

Shear Reinf Reqd Reqd Not Reqd Not Reqd Not Reqd Not Reqd

Shear actual Kn 181 133 123 90

0.23 0.23

Shear allowable Kn 80 120 127 116 97 69

Shear Stress tc N/mm2 0.31 0.23 0.23 0.23

ok ok

p% for Shear 0.50 0.25 0.25 0.25 0.25 0.25

ok ok ok ok

19 0

M allowable / bar Kn.m 599 599 599 367 270 138

M actual / bar Kn.m 575 194 157 73

0.260 0.000

CG of Segment 'PQxx' m 2.79 1.36 1.20 0.78 0.38 0.00

Area of Segment 'PQxx' m2 1.150 0.759 0.692 0.491

57.1 57.1

Pressure due to moment at section kn/m2 18.35 36.50 38.74 44.86 50.98 57.09

Pressure due to moment @ rs kn/m2 57.1 57.1 57.1 57.1

ok ok

Uniform pressure kn/m2 139.0 139.0 139.0 139.0 139.0 139.0

ok ok ok not ok

1005 804

p% 0.5018 0.2488 0.2347 0.1999 0.2387 0.2664

1292 825

Bar dia mm

As mm2 1295 1295 1295 1005

Effective depth mm 2225 2225 2225 1758

6250 7000

Reinforcement Spacing mm 116 234 248 286 326 366

1270 2225 OK

Shear :

distance from c/L mm 2250 4475 4750 5500

0.16 √ fck 176 0.8 176

√ 25 0.8 N/mm2

F 178831

Punching Shear at 'a1a2' 164.19 14.64 178.83 Kn

Allowable Punching Shear stress 0.16 √ fck 0.16

Kn / m2from 'a1a2'

Moment at 'RS'Pressure due to

= 27.4 Kn / m2Moment at 'PQ'

0.98682 m2under area 'PQRS' CG of Segment 'PQa1a2'

= 2.032 mPressure due to

= 57.1

Bar Spacing at fro 367 Length of segment 'RS'

Uniform pressure = 139.0 Kn / m2

Area of Segment 'PQa1a2'

Radius of Foundation fro 7000Bar Spacing at rps 176 Length of segment 'a1a2'

3363 mm

3.00 0.0524Radius of Punching shear rps 3363

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

2250 mm 1112.5 mm

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



17.812wrt bottom of middle portion

2807 < 200ok

2.8500E+04

2.4999E-08

4.2499E-04

3.7500E+04

1.7812E+01 (2)

25.00 m

1600

4365997

9.6716E+12

6.8906E+06

13332 > 200ok

Middle Portion :

1.1251E-04

1.6667E+04

1.8752E+00 (1)

1.875wrt bottom of top portion

1800

2982693

6.9073E+12

1.7719E+06

2.8500E+04

9.0008E-09

Check Deflection of Chimney :

Top Portion :25.00 m

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


ok


6.1000E+04

6.1256E+01 (3)

61.256wrt bottom of bottom portion

1175 < 200

5740082

1.2225E+13

1.3848E+07

2.8500E+04

3.9746E-08

7.1219E-04

Bottom Portion :22.00 m

1400

of 16

6 Design of Constant Dia RCC Chimney and Foundation 23052014

Documents

Transcript of 6 Design of Constant Dia RCC Chimney and Foundation 23052014