Prepared by by :
Date :
A B Quadri Verified by :
Date :
4 Project :
4
Subject :
4
Job no :
ABQ Consultants
1
4 Revision note :
Engineers Planners & Valuers - Civil / Structural design engineers engineers
Sheet No :
1
cont'd 4
3
:
4
Calculation Sheet
4 Description :
Ref
4
Calculation
Output
Note : Input data in yellow cells only and ensure all check boxes are displaying "ok" or "safe".
Design of RCC Chimney :4.00 m
1800
1) Dimensions of Chimney and Forces
200 3.60 m
Height of Chimney External Diameter of Chimney Fire Brick Lining
60.00 4.00 100 100 19000 48.00
Air Gap Gap Between Between Wall & Fire Fire Brick Lining (min)
Unit weight of Fire Brick Lining Height of Fire Brick Lining above Ground Level
200
The temperature of gases above surrounding air
m m 0 0 . 5 2
m mm thk mm m
250 140 2.05E+05 25 2.85E+04 200 25.00 300
Thickness of chimney shell at top portion
Height of top portion of Chimney Thickness of chimney shell at middle portion of Chimney
Height of middle portion of Chimney
25.00 400
Thickness of chimney shell at bottom portion of Chimney
N/mm2 N/mm2 N/mm2 mm
0 2 0 m 4 / 1 n
m mm
1600
N/m2
Constant wind pressure intensity at bottom portion
1400
N/m2
fig 1
m m
Grade of Steel (N/mm2) Allowable tebsile tebsile stress N/mm2
15
20
25
30
35
40
18.67
13.33
10. 98
9.33
8.11
7.18
4.00
5.00
6.00
8.00
9.00
10.00
5.00
7.00
8.50
10.00
11.50
13.00
-2 . 0 0
-2 . 8 0
-3 . 2 0
-3.60
-4. 00
-4 . 4 0
Weight of Lining per per meter meter height = = Π *( 4.00 - 2 ( 0.4 + 0.1 0.1 * 1.00 * 19000 17310
N
100
Cross-Section of Chimney
0.70
Shape Factor
m 0 0 . 0 1
m mm
Constant wind pressure intensity at middle portion
Constant wind pressure intensity at top portion
400
N/mm2
N/m2
Lining Support Distance @ every
m 0 0 . 5 2
N/mm2
10.00 6.00 1800
Height of balance bottom portion of Chimney
=
300
3.40 m
0 0 6 1
ºC
1.1E-05 per deg C
Coefficient of expansion of concrete and Steel Grade of Steel Steel fy = ( 250 or 415) Allowable Allowab le tensile tensile stress in steel Modulus of Elasticity Elasticity of steel steel Es = Grade Concrete Mix M25 Modulus of Elasticity of Concrete Ec =
Grade of conc (N/mm2) modular ratio m Allowable compressive compressive stress stress (Direct) N/mm2 Allowable compressive compressive stress stress (Bending) (Bending) N/mm2 Allowable tebsile tebsile stress (Direct) (Direct) N/mm2
lining thickness
100
N/m3
+ 0.05 )) *
100As bd
250
41 5
140
230
Permissible ble Shear S tress in Concrete Tc N/mm2 for grade of concrete
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
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
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 00 0. 44 44
0.51 0.57
0.60
0.62 0.63
PTO Prep By : A B Quadri- Abq Consultants - 9959010210 -
[email protected]/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-500028
Weight of Concrete per meter height For
200 mm thk
shell, w =
Π
[ 4.00 - 0.20 ] * 0.20 * 1.00 * = 59690 N/m
25000
For
300 mm thk
shell, w =
Π
[ 4.00 - 0.30 ] * 0.30 * 1.00 * = 87179 N/m
25000
For
400 mm thk
shell, w =
Π
[ 4.00 - 0.40 ] * 0.40 * 1.00 * = 113097 N / m
25000
2) Stress at Section 25.00 m below top 1.00 % of the concrete area
Let the vertical reinforcement be mm place at a cover of 50
ok ok
As =
1 * Π * ( 4.00 ^2 - 3.60 ^2 ) * 1000000 4 100 = 23876 mm2 Nos of = 119 16 mm Φ bars = 23876 201 Hence provide 140 bars of 16mm Φ suitably placed along the circumference Actual As =
28149
mm2
23876
>
mm2
ok
ok
Equivalent thickness of steel ring placed at the centre of the shell thickness ( R = 2.00 0.1 = 1.90 m ) is Ts =
28149 = 2.36 mm 2ΠR Horizontal steel (hoops) may be provided @ Area of steel per metre height of chimney
0.2 % of sectional area 0.2 * 200 * 1000 = 400 100 282.5 mm 12 mm Φ bar hoops = 1000 * 113 = 400 Provide these at 250 mm centre 59690 = 1492257 N
Hence pitch s of W =
25.00
*
P1 =
0.7 *
1800 (
.: M =
126000
4.00 *
12.5
.: Eccentricity e = M = 25
concrete , m
.: Eqivalent area
= A =
=
) =
=
126000 N acting at
1575000
1575000 = 1492257
W
For M
* 25.0
=
12.5
ok mm2
ok m below top
N.m
1.055
m =
1055
mm
10.98
4.00 ^2 - 3.60 ^2 ) * 1000000 + ( 10.98 1 )* 28149 = 2668534 mm2 Eqivalent moment of inertia = I = (Π / 64) ( D 4 -d 4 )+(m-1) Π R ts (R) = Π * ( 4.00 ^ 4 - 3.60 ^ 4 ) * 1000 ^ 4 + 64 ( 10.98 1) * Π * 1900 * 2.36 * 1900 ^ 2 = 4.8286E+12 mm4 Π/4
* (
PTO Prep By : A B Quadri- Abq Consultants - 9959010210 -
[email protected]/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-500028
For no tension to develop, allowable eccentricity
=
2 I AD
= The actual eccentricity is
1055
= 2 * 4.8286E+12 2668534 * 4000 904.7
mm
mm. Hence some tension will be developed in the leeward side.
The maximum and minimum stresses are given by
σ
=
W
± MD
A
=
1492257 2668534
=
0.559
1575000 1000 * 4.8286E+12 2*
±
2I
4000
*
0.652
±
Compressive stress = Tensile stress =
1.212 -0.093
N/mm2 N/mm2
< <
8.5 -0.8
(Safe) (Safe)
N/mm2 allowable N/mm2 allowable
2) Stress at Section 50.00 m below top Thickness of shell =
300
mm
Let the vertical reinforcement be mm place at a cover of 50
1.00 % of the concrete area
ok ok
As =
1 * Π * ( 4.00 ^2 - 3.40 ^2 ) * 1000000 4 100 = 34872 mm2 Nos of bars = 34872 = 111 20 mm Φ 314 Hence provide 130 bars of 20mm Φ suitably placed along the circumference Actual As =
40841 mm2
34872
>
mm2
ok
ok
Equivalent thickness of steel ring placed at the centre of the shell thickness ( R = 2.00 0.15 = 1.85 m ) is Ts =
40841 = 3.51 mm 2ΠR Horizontal steel (hoops) may be provided @ Area of steel per metre height of chimney Hence pitch s of W =
25.00 25.00
P1 = =
0.7 * 1800 ( 126000 +
.: M =
* *
0.2 % of sectional area 0.2 * 300 * 1000 = 600 100 188.0 mm 12 mm Φ bar hoops = 1000 * 113 = 600 Provide these at 180 mm centre 59690 + 25.00 * 17310 + 87179 = 4104491 N
126000
4.00 * 25.0 112000.00 *
.: Eccentricity e = M = W
37.5 +
ok mm2
=
) + =
0.7 * 1600 ( 238000.00 N
112000
6125000 = 4104491
*
12.5
1.492
m =
4.00
= 1492
* 25.0
6125000
ok
) N.m
mm
PTO Prep By : A B Quadri- Abq Consultants - 9959010210 -
[email protected]/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-500028
For M
25
concrete , m
.: Eqivalent area
=
= A =
10.98
Π/4
* (
(
10.98 -
=
4.00 ^2 1 )* 3894758 mm2
3.40 ^2 ) * 1000000 + 40841
Eqivalent moment of inertia = I = ( Π/64 ) ( D 4 -d 4 ) + ( m-1 ) Π R ts (R) = Π * ( 4.00 ^ 4 - 3.40 ^ 4 ) * 1000 ^ 4 + 64 ( 10.98 1) * Π * 1850 * 3.51 * 1850 ^ 2 = 6.7041E+12 mm4 For no tension to develop, allowable eccentricity = 2 I = 2 * 6.7041E+12 AD 3894758 * 4000 = 860.7 mm The actual eccentricity is 1492 mm. Hence some tension will be developed in the leeward side. The maximum and minimum stresses are given by
σ
=
W A
=
4104491 3894758
=
1.054
±
6125000 1000 * 6.7041E+12 2*
± MD
2I
4000
*
1.827
±
Compressive stress = Tensile stress =
2.881 -0.773
N/mm2 N/mm2
< <
8.5 -0.8
N/mm2 allowable N/mm2 allowable
(Safe) (Safe)
3) Stress at Section 60.00 m below top Thickness of shell =
400
mm
Let the vertical reinforcement be mm place at a cover of 50
1.00 % of the concrete area
ok ok
As =
1 * Π * ( 4.00 ^2 - 3.20 ^2 ) * 1000000 4 100 = 45239 mm2 Nos of = 93 nos 25 mm Φ bars = 45239 491 Hence provide 120 bars of 25mm Φ suitably placed along the circumference Actual As =
58905 mm2
45239
>
mm2
ok
ok
Equivalent thickness of steel ring placed at the centre of the shell thickness ( R = 2.00 0.15 = 1.85 m ) is Ts =
58905 = 5.07 mm 2ΠR Horizontal steel (hoops) may be provided @ Area of steel per metre height of chimney Hence pitch s of
12
mm Φ bar hoops
0.2 % of sectional area 0.2 * 400 * 1000 = 800 100 = 1000 * 113 = 141.0 mm 800 Provide these at 140 mm centre
=
ok mm2
ok
PTO Prep By : A B Quadri- Abq Consultants - 9959010210 -
[email protected]/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-500028
W =
25.00 10.00
P1 =
0.7 * 1800 ( 0.7 * 1400 ( 126000 +
= .: M = =
* *
59690 17310
126000 8701000
+ +
4.00 4.00
25.00 10.00
* *
17310 + 113097 =
* 25.0 ) + * 10.00 ) 112000 +
* 47.5 + N.m
.: Eccentricity e = M =
0.7 *
1600 (
39200
112000
*
8701000 = 5408566
W
25.00 * 5408566
22.5 +
1.609
87179 N
4.00
* 25.0
=
277200
N
*
5.0
39200
m =
+
1609
)
mm
For M 25 concrete , m = 10.98 .: Eqivalent area = A = Π/4 * ( 4.00 ^2 - 3.20 ^2 ) * 1000000 + ( 10.98 1 )* 58905 = 5111764 mm2 Eqivalent moment of inertia = I = ( Π/64 ) ( D 4 -d 4 ) + ( m-1 ) Π R ts (R) = Π * ( 4.00 ^ 4 - 3.20 ^ 4 ) * 1000 ^ 4 + 64 ( 10.98 1) * Π * 1850 * 5.07 * 1850 ^ 2 = 8.4252E+12 mm4 For no tension to develop, allowable eccentricity = 2 I = 2 * 8.4252E+12 AD 5111764 * 4000 = 824.1 mm The actual eccentricity is 1609 mm. Hence some tension will be developed in the leeward side. The maximum and minimum stresses are given by
σ
=
W
± MD
A
=
5408566 5111764
=
1.058
8701000 1000 * 8.4252E+12 2*
±
2I
4000
*
2.065
±
Compressive stress = Tensile stress =
3.124 -1.007
N/mm2 N/mm2
< >=
8.5 -0.8
N/mm N/mm
allowable (Safe) allowable Check further
The eccentricity is quite high. Due to this, tensile stresses in the windward side are ex ected to be reather than 0.8 N mm2 resultingin cracking of concrete. Hence it is assumed that only steel will take the tensile stresses and concrete in the tensile zone will be ignored. Thus, the method of analysis used at 25.00 m and 50.00 m will not be applicable. We shall analyse the section for stresses by method discussed in § 8.3. 400 Tc = Ts = 5.07 m = 10.98
mm mm
R= 2.00 - 0.20 = 1.80 eccentricity e = 1.609 m
m
In order to find the position of N.A., use equation 8.3 :
e
=
[ R [
(Tc-Ts)
{
(Tc-Ts) {
sin2Φ
4
+
sinΦ +
Π-Φ
2 (Π-Φ)
}
+
mΠ Ts
2
cosΦ }
]
mΠ Ts
fig 2 cosΦ
] PTO
Prep By : A B Quadri- Abq Consultants - 9959010210 -
[email protected]/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-500028
e
=
180
*
[(
[(
40.00 -
40.00 0.51
sin2Φ
0.51
) *
{
{
sinΦ
+ (Π-Φ)
) *
4
+
Π-Φ
}
2
cosΦ
}
+
10.98 * Π 2
86.00
Φ =
.: e =
ºC =
]
+ 10.98 * Π * 0.51 * >=
Now adjust the value of angle Φ in such a way that the value of eccentricity e is
Assume
0.51
*
cosΦ
1.609
]
m
1.5010 radians
6078.21 + 1573.24 = 7651.45 43.92 + 1.219 45.136 1.695 m not ok which is slightly more than the actual value 0.00 ºC
= .: consider Φ =
The maximum stress c1 in concrete is found from Eq.8.1 W = .:
2Rc1 1+cosΦ
5408566
[
(Tc-Ts)
=
2 * 1800 * 1 + 0.06976
{ =
.: Compressive stress c1 = in Concrete
{
sinΦ
+ (Π-Φ)
0.9976 + 1.6406
c1
cosΦ
[(
* 0.0698
}
}
+ mΠ Ts cosΦ
400
-
5.07
]
) *
+ 10.98 * Π * 5.07 * 0.0698
]
1518945 c1 5408566 1518945
=
N/mm2
3.5607
Tensile stress in Steel, assuming concrete to be fully cracked. 1 - cosΦ t1 = m * c1 * = 1 + cosΦ
[
]
34.00
<
8.5
N/mm2
N/mm2
safe
140
N/mm2
<
safe
(b) Stress in horizontal reinforcement Horizontal steel (hoops) may be provided @ Area of steel per metre height of chimney
Hence pitch s of
As = D1 =
.:
12
=
0.2 0.2
*
ok
% of sectional area 400 100
*
1000
=
800
mm Φ bar hoops = 1000 * 113 = 141.0 mm 800 Provide these at 140 mm centre
mm2
ok
in pitch s = 140 mm centre, if the cover is 40 mm then 80 = 3920 mm p * s 277200 * 140 t1 = = = 43.7676 N/mm2 < 140 2 * As * D1 2 * 113 * 3920
113.1 4000
mm2
N/mm2 allowable
Safe
PTO Prep By : A B Quadri- Abq Consultants - 9959010210 -
[email protected]/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-500028
(c) Stress on leeward side due to temperature gradient fig 3
fig 4
Thickness of shell Tc = Thickness of steel Ts =
400 5.07
mm mm
Thickness of lining Tl =
.: Cover to vertical steel = c1 =
3.5607
Es = p =
50
Ts Tc
400
-
50
=
a =
350 400
=
0.875
Ec =
N mm2
5.07 400
=
0.01267
Temperature difference
=
200
2.05E+05 10.98
α =
=
per º C
1.10E-05
ºC
1.867E+04
400
160 5 *
+
mm
ok
400
*
100
N mm2
oncre e Temperature Co-efficient
80 % of temperature drops through the lining and shell. Let us assume that ºC = 200 * 0.8 = 160 Drop in temperature 5 times more than that in shell, per unit thickness, Asssuming that drop in lining is the dro of tem erature throu h concrete is iven b , Tº =
350
mm
N mm
=
mm
aTc =
2
2.05E+05
100
=
71.11
ok ok
ºC
To locate -neutral axis in the shell thickness, use Eq. 8.10 c1
[1
+ ( m - 1 ) * p
0.5 * k -
.:
=
α
* T * Ec
m * p * (a - k)
3.5607 * [ 0.5 * k -
1+ ( 10.98
or
10.98 1) * * 0.01267 * (
k +
0.27821
8.0219 * k -
k +
0.278
* k -
solving for k
]
k =
0.01267 ) 0.875 =
0.24343 0.793
=
] k )
=
1.1E-05 *
71.11 * 1.867E+04
14.6043
0
0.7620
PTO Prep By : A B Quadri- Abq Consultants - 9959010210 -
[email protected]/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-500028
a * α * Tº * Ec = k * α * Tº * Ec a - k 1 + k = 0.762 * 1.1E-05 * 71.11 * 1.867E+04
.: Compressive Stress in Concrete
c =
= 11.129 Since wind stresses are taken into account, Permissible Stress in Concrete
4 * 3
=
8.5
N mm2
=
N/mm2
11.33
Thus the compressive stress less than the permissible 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
= mc
a - k k
= 10.98 * =
11.129 * (
18.113
0.875
0.762 0.762
)
N mm2
(d) Stresses on windward side, due to temperature gradient m * p * ( where
-
t1 = 33.998 α = 0.000011
.: 10.98 or
= α * Tº * Ec
p * t1 a - k )
*
solving
-
k =
.: Compressive Stress in Concrete
p = 0.01266892 a = N mm2 ºC Tº = 71.11 Ec =
0.01267 * 0.01267 * (
0.121716639 0.12172
0.5 * k
33.998 0.875 -
0.430718713 0.13910473 k 0.13910
k
0 k )
-
0.875 18670
m = 10.98 2 N mm
0.000011 *
*
=
0.5 * k 14.604
=
-
0.5
*
k
-
0.5
*
k =
71.11 18670
0.0295
0.3123 c = α * Tº * Ec * k =
4.5615
N mm2
Tensile stress in Steel, assuming concrete to be fully cracked. t
= m c a - k = k =
10.98
*
90.23
4.5615
* ( <
N mm2
fig 5
0.875
- 0.31234 ) 0.31234
140
N mm2
( safe )
(e) Stresses on the Neutral axis .(i.e. temperature effect alone) k 2 = -mp + √2mpa + m2p2 where or
k 2 = -
m = 10.98 α = 0.000011 10.98
*
p =
0.01267 a = 0.875 ºC Tº = 71.11 Ec =
0.01267 + √
2* +
10.98 10.98
*
*
18670
0.01267 * 10.98
*
N mm2
0.875
0.01267 *
0.01267
PTO Prep By : A B Quadri- Abq Consultants - 9959010210 -
[email protected]/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-500028
k 2 = 0.37351931 c2 = α * Tº * Ec * k 2 =
.: Compressive Stress in Concrete
5.4550
N mm2
Tensile stress in concrete, assuming concrete to be fully cracked. 2
= m c2 a - k 2 = k 2
10.98
=
5.4550
80.42
0.875 - 0.37352 0.37352 <
N mm2
140
N mm2
0.00202 *
0.900
( safe )
(b) Stress in horizontal reinforcement due to temperature : p' =
Φ
113.10 = 140 * 400
=
S Tc 360 400 From Eq. 8.13. a' =
=
0.00202
0.900
k' = -mp' + √2mp'a + m2p'2
or
k'
=
-
10.98
*
0.00202 + √
2* +
10.98 10.98
*
*
10.98
*
0.00202 *
2.6118
N mm2
0.00202
k' = 0.17883981
.: Compressive Stress in Concrete
c' = α * Tº * Ec * k' =
Tensile stress in concrete, assuming concrete to be fully cracked. 2
= m c' a' - k' = k' =
10.98
115.64
*
2.6118
N mm2
* (
<
0.900 - 0.17884 ) 0.17884 140
N mm2
O.k
These stresses are due to temperature effect alone. To this we must add the str esses due to wind. Hence total stress in steel = 159.41 = 115.64 + 43.768 N mm2 140 = 186.67 N mm2 Safe Since wind is also acting, permissible t = 4 * 3 allowable tensile stress in steel
PTO Prep By : A B Quadri- Abq Consultants - 9959010210 -
[email protected]/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-500028
5. Flue O enin : 1.5 m wide and Provide a flue opening 2.0 m high at bottom. ok The boundary of the opening is thickened and reinforced as shown in Fig A. The vertical steel bars are bent on either side of the opening as shown
fig 6
6. Force acting at 0.00 level for Foundation Design : P V
M
0.00 level
P = M = V =
5408566 8701000 277200
N N .m N
PTO Prep By : A B Quadri- Abq Consultants - 9959010210 -
[email protected]/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-500028
7. Design of Cirrcullar Chimney Foundation : P Data Concrete Grade Steel Grade S.B.C of Soil Density of soil Axial Load Moment eccentricity Horizontal load
Outer dia of chimney Thickness of chimney wall Dia of Footing Level of footing below ground
Depth of Soil Depth of Footing Footing Reinforcement dia Reinforcement cover
d= t= OD = A= Totd =
D= T= Φ= c =
25 415 200 18 5408.57 8701 1.609 277.2 4.00 400 14.00 200 4000 1700 2300 32 25 75
N/mm2 N/mm2 Kn/m2 Kn/m3 Kn Kn . M m Kn m mm m mm mm mm mm mm mm
0 0 2
M'
A
FGL Soil filling inside
0 0 7 1
0 0 3 2
D 0 0 0 4
T
0 0 0 4 1
D O
= 5408.57 = P + Weight of Chimney Wall + Soil Filling inside of wall + Weight of soil + Self weight of footing = 5408.57 + Π ( 4.00 ^2 3.60 ^2 ) 4 * ( 1.70 + 0.2 ) * 25 + Π ( 3.60 ^2 ) * 18 4 + Π ( 14.00 ^2 4.00 ^2 ) * 18 * 4 + Π ( 14.00 ^2 ) * 2.30 * 25 4 = 5408.57 + 113.41 + 183.22 + =
18882.61
kn
= = =
8701 8701 9865.24
+ H * ( D + T + A ) + 277.2 * ( 2.30 + Kn . M
.: e' =
Axx
M' P'
=
9865.24 18882.61
= Π * OD2 4 = 153.94 m
d t o T
4000 d 14000 OD
Axial load at the base of footing P'
277.2 H
FFL fc' = fy = Qs = Ws = P= M= e = M/P = H=
5408.57
e 1609
fig 7
1.70
4325.97
1.70
+
8851.44
0.2
)
1.75
m
+
=
0.5225
<
Ixx
= Π *
OD4 =
1885.74099
m4
OD3 =
269.39157
m3
[
OD = 8 Ok
]
64 Zxx
= Π * 32
PTO Prep By : A B Quadri- Abq Consultants - 9959010210 -
[email protected]/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-500028
The maximum and minimum base pressures are given by
σ
=
P'
M'
±
A
Zxx
fig 8
= =
18882.61 153.94
9865.24 269.39
±
122.664
36.620
±
8 2 . 9 5 1
4 0 . 6 8
σ max
=
159.28
Kn / m2
<
200
Kn / m2 allowable
Ok
σ min
=
86.04
Kn / m2
>
0
Kn / m2 allowable
Ok
Factor of Safety against overturning
=
Stabilising Moment = P' * OD Overturning Moment 2 M' 18882.61 * 14.00 2 9865.24
=
=
13.40
>
safe
1.5
Design of Footing slab Assume initially
1
Φ = 3.00 º = Radius of Chimney =
Layer
of +
32 25
mm mm
Φ bars Φ bars
Radius of Foundation
=
Bar Spacing at ro Bar Spacing at fro
= =
0.0524 radians ro = 2000 mm fro = 7000 mm 105 mm Length of segment 'PQ' 367 mm Length of segment 'RS'
=
122.7
Kn / m2
=
36.6
Kn / m2
Uniform pressure under area 'PQRS' Pressure due to Moment at 'RS' Pressure due to Moment at 'PQ'
120 nos spaced radially along the As = 1295 mm2 circumferance
R
=
10.5
Kn / m2
Main Reinforcement
b1
Area of Segment 'PQRS' =
CG of Segment 'PQRS' from 'PQ'
=
1.1781 2.963
m2 m
S
a1
P fr o
b2 Critical Section for Moment
a2
Φ
ro
Area covered by one unit of Main Reinforcement
Q Footing Outer Dia
A
A
rp
Chimney Outer Dia
rs
Line of Punching Shear Line of Shear
fig 9
PTO Prep By : A B Quadri- Abq Consultants - 9959010210 -
[email protected]/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-500028
C / L of foundation
2000
2500
r
2500
Straight portion
Sloping portion 16 Φ top radial reinforcement
1.79 1
c/L of Foundation
Shear Stirrups (if required) Main Radial reinforcement 32
120
Φ
Circullar reinf
nos
16 Φ
@
1112.5
200
75
0 0 3 2
0 0 9
c/c
cover
critical punching shear section
2225
7000 Section A - A
C/L
.: Moment at 'PQ' Mf =
464.70
fy = fc' =
415 25
.: fyall = .: fc'all =
.:
k = j
N/mm2 N/mm2
=
R =
10.98 10.98 * 11 2
+
* 8.5
d =
k 3
=
1-
fc'all
j
k =
.: adopt T = .: d = As = .: Provide 1 layer of Provide
√
M Fyall * j
=
230 8.5
516.06
N/mm2 N/mm2 =
Kn .m
m =
10.98
0.289
230
= * d
230
32 Φ + 25 Φ AΦ dia bars. Main radial reinforcement
*
0.289
516.06 * 1000000 105 * 1.109
√
mm cover = 75 mm effective depth
16 Φ @ 200 c/c distribution steel
fig 10
0.289 = 0.90378 3 1* 8.5 * 0.904 2 1.109
Mf = 1000 * R
2300 2225
51.36
8.5 +
= Hence
critical shear Section
mm
d =
2300
2108.25 mm -
75
ok
516059761 * 0.90378 *
= Π * ( 4 = 1295
=
32
²
=
mm 2
1116
2225 +
mm 2
25 >
²
) mm2
1116
ok p% =
1295 * 100 105 * 2225
=
0.556
%
PTO Prep By : A B Quadri- Abq Consultants - 9959010210 -
[email protected]/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-500028
Punching Shear : Check Punching shear at ro + d/2 from the c/L r
=
2000
d = 1112.5 mm r + d = 2 2 Φ = 3.00 º = 0.0524 radians Radius of Punching shear = rps = 3113 mm = Radius of Foundation fro = 7000 mm Bar Spacing at rps = 163 mm Length of segment 'a1a2' Bar Spacing at fro = 367 mm Length of segment 'RS' Uniform pressure under area 'PQRS' Pressure due to Moment at 'RS' Pressure due to Moment at 'PQ' .:
mm
=
122.7
Kn / m2
=
36.6
Kn / m2
=
16.3
Kn / m2
Punching Shear at 'a1a2'
0.16
CG of Segment 'PQa1a2'
143.00 fck
+ =
=
10.47
0.16 *
do = 0.16
Shear :
=
from 'a1a2'
√
.: Depth required for punching shear
mm
Area of Segment 'PQa1a2' = 1.02919
F =
Allowable Punching Shear stress =
3113
√
1177
√
F fck * <
2.193
= 25
m
153.47 =
0.8 =
163 2225
m2
Kn N/mm2
153469 0.8 * 163
mm provided
OK
Check shear at r + d from the c/L, End of Straight portion, and at three points at sloping portion.
distance from c/L Reinforcement Spacing Effective depth Bar dia As p% Uniform pressure Pressure due to moment @ rs Pressure due to moment at section Area of Segment 'PQxx' CG of Segment 'PQxx' M actual / bar M allowable / bar p% for Shear Shear Stress tc Shear allowable Shear actual Shear Reinf Shear - Vs per main bar Shear - MS bar dia fyall 140 spacing Minimum shear s = 2.5Asvfy/b .: Provided spacing =
mm 2000 4225 4500 5333 6167 7000 mm 104 220 234 278 322 366 mm 2225 2225 2225 1758 1292 825 mm 32 + 25 32 + 25 32 + 25 32 + 16 32 + 16 32 + 0 mm2 1295 1295 1295 1005 1005 804 0.5597 0.2646 0.2488 0.2057 0.2417 0.2664 ok ok ok ok ok ok kn/m2 122.7 122.7 122.7 122.7 122.7 122.7 kn/m2 36.6 36.6 36.6 36.6 36.6 36.6 kn/m2 10.46 22.10 23.54 27.90 32.26 36.62 m2 1.178 0.815 0.753 0.538 0.287 0.000 m 2.96 1.50 1.34 0.87 0.43 0.00 Kn.m 516 188 156 73 19 0 Kn.m 599 599 599 367 270 138 ok ok ok ok ok ok 0.50 0.25 0.25 0.25 0.25 0.25 N/mm2 0.31 0.23 0.23 0.23 0.23 0.23 Kn 72 113 120 112 96 69 Kn 157 118 110 81 45 0 Reqd Reqd Not Reqd Not Reqd Not Reqd Not Reqd Kn 85 5 1 1 1 1 12 12 1 1 1 1 mm 414 6433 245 193 142 91 mm 381 180 1 1 1 1 mm 175 175 ok ok ok ok ok ok
PTO Prep By : A B Quadri- Abq Consultants - 9959010210 -
[email protected]/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-500028
Check Deflection of Chimney : Data :
Top Portion : Height of Top portion of Chimney
=
25.00
m
Wind intensity of top portion of Chimney
=
1800
N/m2
Concrete Area of Top Portion of chimney
=
2668534
mm2
Moment of Inertia of Top portion of Chimney
=
4.8286E+12
mm4
Wind Moment at the base of Top Portion
=
1.5750E+06
N.m
Modulus of Elasticity of Concrete
=
2.8500E+04
N/mm2
M / Ei
=
1.1445E-08
1/mm
Area of M / Ei of top portion
=
1.4306E-04
C.g of Area of M / Ei of top portion
=
1.6667E+04
Moment of Area of M / Ei from top portion
=
2.3843E+00
Partial Deflection of Top Portion wrt bottom of top portion Ratio L / δ
=
2.384
δtop
=
L
/
10485
mm (1) mm >
L
/
200 ok
Middle Portion : Height of Middle portion of Chimney
=
25.00
m
Wind intensity of Middle portion of Chimney
=
1600
N/m2
Area of Middle Portion of chimney
=
3894758
mm2
Moment of Inertia of Middle portion of Chimney
=
6.7041E+12
mm4
Wind Moment at the base of Middle Portion
=
6.1250E+06
N.m
Modulus of Elasticity of Concrete
=
2.8500E+04
N/mm2
M / Ei
=
3.2057E-08
1/mm
Area of M / Ei of Middle portion
=
5.4377E-04
C.g of Area of M / Ei of Middle portion from top
=
3.7500E+04
Moment of Area of M / Ei of Middle portion
=
2.2776E+01
Partial Deflection of Top Portion wrt bottom of middle portion Ratio L / δ
=
22.776
δtop
=
L
/
2195
mm (2) mm <
L
/
200 ok
PTO Prep By : A B Quadri- Abq Consultants - 9959010210 -
[email protected]/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-500028
Bottom Portion : Height of Bottom portion of Chimney
=
10.00
m
Wind intensity of Bottom portion of Chimney
=
1400
N/m2
Area of Bottom Portion of chimney
=
5111764
mm2
Moment of Inertia of Bottom portion of Chimney
=
8.4252E+12
mm4
Wind Moment at the base of Bottom Portion
=
8.7010E+06
N.m
Modulus of Elasticity of Concrete
=
2.8500E+04
N/mm2
M / Ei
=
3.6236E-08
1/mm
Area of M / Ei of Bottom portion
=
3.4147E-04
C.g of Area of M / Ei of Bottom portion from top
=
5.5000E+04
Moment of Area of M / Ei of Bottom portion
=
4.1556E+01
To a De ec ion o Top Por ion wrt bottom of bottom portion
=
41.556
Ratio L /
δ
δtop
=
L
/
1444
mm (3) mm
<
L
/
200 ok
PTO Prep By : A B Quadri- Abq Consultants - 9959010210 -
[email protected]/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-500028
