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Visit Abqconsultants.comThis program Designs andOptimises Constant Dia RCC Chimney and Foundation.Written and programmed byA B Quadriwww.abqconsultants.comabquadri@yahoo.comabquadri@gmail.com99590102109959010212
Citation preview
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 - em-abquadri@yahoo.com-12-2-826/2 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
abquadri@yahoo.com
abquadri@gmail.com
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=
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-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
abquadri@gmail.com
9959010210
9959010211
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
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-www.abqconsultants.com
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
Stress at Section 50.00 m below top
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
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 centre800
140 okP 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-www.abqconsultants.com
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
Stress at Section 72.00 m below top
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
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
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-www.abqconsultants.com
(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
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
Safe
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-www.abqconsultants.com
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 -
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 = 0.8419
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-www.abqconsultants.com
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
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 * * *
+ * * *10.98 10.98 0.01116 0.01116
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-www.abqconsultants.com
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
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
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-www.abqconsultants.com
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
Compressive 2.6118 N/mm2
=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
Compressive 5.2062 N/mm2
10.98 5.2062 0.875 0.35649
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
Prep By : A B Quadri- Abq Consultants - 9959010210 - em-abquadri@yahoo.com-12-2-826/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-500028-www.abqconsultants.com
6. Force acting at 0.00 level for Foundation Design :
795017713848345
3647700.00 level
P T O
1.52.0 ok
fig 6
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 * =
32P 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-www.abqconsultants.com
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
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'
Prep By : A B Quadri- Abq Consultants - 9959010210 - em-abquadri@yahoo.com-12-2-826/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-500028-www.abqconsultants.com
= 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
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 = * = %*
0.494distribution steel 118 2225
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-www.abqconsultants.com
> 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
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
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-www.abqconsultants.com
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
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 /
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-www.abqconsultants.com
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
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 /
ok
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-www.abqconsultants.com
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
Page 16 of 16
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