design of const dia rcc chimney
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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 - em-abquadri@yahoo.com-12-2-826/2 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
Prep By : A B Quadri- Abq Consultants - 9959010210 - em-abquadri@yahoo.com-12-2-826/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-500028
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)
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 20mm 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
.: M = * + * =
.: Eccentricity e = M = = m =W
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
Stress at Section 50.00 m below top
<
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
Prep By : A B Quadri- Abq Consultants - 9959010210 - em-abquadri@yahoo.com-12-2-826/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-500028
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=
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
3)
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 25mm 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 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
Stress at Section 60.00 m below top
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
Prep By : A B Quadri- Abq Consultants - 9959010210 - em-abquadri@yahoo.com-12-2-826/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-500028
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 = * + * + *=
.: 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=
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
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
Prep By : A B Quadri- Abq Consultants - 9959010210 - em-abquadri@yahoo.com-12-2-826/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-500028
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 +
Prep By : A B Quadri- Abq Consultants - 9959010210 - em-abquadri@yahoo.com-12-2-826/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-500028
=
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
Prep By : A B Quadri- Abq Consultants - 9959010210 - em-abquadri@yahoo.com-12-2-826/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-500028
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
Prep By : A B Quadri- Abq Consultants - 9959010210 - em-abquadri@yahoo.com-12-2-826/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-500028
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
Prep By : A B Quadri- Abq Consultants - 9959010210 - em-abquadri@yahoo.com-12-2-826/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-500028
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
Prep By : A B Quadri- Abq Consultants - 9959010210 - em-abquadri@yahoo.com-12-2-826/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-500028
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
Prep By : A B Quadri- Abq Consultants - 9959010210 - em-abquadri@yahoo.com-12-2-826/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-500028
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
Prep By : A B Quadri- Abq Consultants - 9959010210 - em-abquadri@yahoo.com-12-2-826/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-500028
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
Prep By : A B Quadri- Abq Consultants - 9959010210 - em-abquadri@yahoo.com-12-2-826/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-500028
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
Prep By : A B Quadri- Abq Consultants - 9959010210 - em-abquadri@yahoo.com-12-2-826/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-500028
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
Modulus of Elasticity of Concrete = N/mm2
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
Ratio L / δ = L / L /
wrt bottom of middle portion
P T O Prep By : A B Quadri- Abq Consultants - 9959010210 - em-abquadri@yahoo.com-12-2-826/2 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
Modulus of Elasticity of Concrete = N/mm2
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
Ratio L / δ = L / L /
wrt bottom of bottom portion
P T O Prep By : A B Quadri- Abq Consultants - 9959010210 - em-abquadri@yahoo.com-12-2-826/2 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
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