Design of Chimneys Introduction Chimneys are tall slender structures. Chimneys are used to convey and large combustion gases away from the operating area of an industry. Chimneys are scientifically designed to take cognizance of gas temperature and velocity acts, corrosion effects etc.A chimney is by which waste gases are discharged at high elevation so that after dilution due to a atmospheric turbulence, their concentration and that of their entrained solid particles is within acceptable limits on reaching the ground. A chimney achieves simultaneous reduction in concentration of a number of pollutants (SO2, Fly ash etc.) and being highly reliable it does not require a stand by. Classification of chimney 1. Based on height a) Tall – more than 150 m height b) Short 2. Number of flues a) Single b) Multi 3.Structural support a) guyed b) self supporting 4. Lining a) lined b) unlined 5. Shape a) circular b) rectangular c) elliptical d) triangular 6. Based on materials of construction a) Brick b) RCC i) Insitu
c) Steel i) mild steel
ii) precast
ii) stainless steel
iii)Prestressed concrete
iii) weathering steel
d) Fibre reinforced plastic
.
Physical dimensions a. Shape and base dimensions b. Exit size c. Physical height Shape and Base dimensions: The base dimensions of a chimney are largely governed by structural considerations and these dimensions are so chosen that stresses due to dead load together with those due to wind, temperature, earth quake and other effects are safely withstood by the chimney fabric. From the flow considerations a cylindrical chimney is preferred, but for a concrete chimney a tapper of about 1:50 to 1:100 is usually provided depending on its height and geographical location. Exit size: Generally the top dimension of a flux are fixed such that a given volume of gases can be discharged of a design exit velocity. In practice these dimensions have to be marginally adjusted to cater for a diminished velocity along the walls if a chimney has to handle a range of gas volumes, then the exit velocity should be high enough at a maximum load and at the same time should exceed the limiting values when operating at full load.
Physical height: The following considerations to dictate the physical height of chimney. i) ii)
To generate a draft which will cause gases to flow out with the desired exit velocity To satisfy the local regulations in respect of permissible GLC (Ground Level Concentrations) of pollutants.
Parts of Chimney: A RCC chimney is generally in circular in shape with a rigid concrete shell. The concrete is made of M25 or M30 grade and provided with vertical and horizontal reinforcements. A fire brick lining 100to 150mm thick is provided inside the concrete shell with an air gap of 100 t0180mm to reduce the temperature gradient from the interior surface of fire brick lining to the exterior surface of the concrete shell. Reinforced concrete brackets are provided at regular intervals to support the fire brick lining. At the bottom of the chimney, provision is made for a flue opening. The chimney rests on a circular raft foundation. The various parts of the chimney are shown in fig. Design factors:
The chimneys are designed to with stand the stresses developed due to 1. Self weight of chimney ( The dead weight of the concrete and weight of fire brick lining together with the brackets) 2. Wind pressure ( the wind pressure depends upon the velocity of wind at a given place as per IS875. The design wind loads on the chimney depends upon the cross sectional shape of the chimney. For circular 0.7 may be taken. The windward side concrete surface will be subjected to tension and the leeward side will be in compression) 3. Temperature stresses ( due to temperature gradient between the inside and outside faces of the chimney, temperature stresses are induced in the chimney walls both in the vertical and horizontal planes. The inner surface of the shell being at a higher temperature, tends to expand more than the outer surface . This restrains the expansion of the inner fibres to a certain extent. This restrained expansion results in compression of the inner fibres and tension of the outer fibres. This will causes the BM in the vertical plane.)
STRESSES IN R.C SHAFTS DUE TO SELF WEIGHT AND WIND LOADS
Vertical Steel Air Gap Compression
Hot Gases
Outside Face R
Q Inside Face
S
Moment
Tension Compression
Fig 1 Temperature stresses in R.C Chimney The following notations are used in the analysis of stresses in R.C shafts. W = Total weight of the shaft above the considered section P = Resultant wind force acting at a distance h from the section As = Area of reinforcement assumed to be in the form of a ring at the center of thickness of shell ts = Thickness of steel ring R = Radius of centre of thickness ts = (
As 2πR
)
d = Outside diameter of the shell tc = thickness of concrete shell n = Co-efficient of neutral axis depth σc = Compressive stress of concrete at the centre of thickness of shell σs = Tensile stress in steel
α = Angle subtended by the neutral axis at the centre m = modular ratio M = Bending moment at the section N P X
𝑡𝑐′
X
X
X
θ
P X
P
X
Wind force
dθ P
X
A (𝑅 +Rcosα) (𝑅 𝑐𝑜𝑠𝜃 +Rcosα ) 𝜎𝑐
𝜎𝑐′
P X
(𝑅 −Rcosα)
Fig 2 stresses in chimney shafts
Referring to Fig. 2, consider a strip R.dθ at an angle θ from xx. Stress in concrete at the level of elementary strip, 𝜎𝑐 ′ = 𝜎𝑐 [
𝑅𝑐𝑜𝑠𝜃 + 𝑅𝑐𝑜𝑠𝛼 𝑐𝑜𝑠𝜃 + 𝑐𝑜𝑠𝛼 ] = 𝜎𝑐 [ ] 𝑅 + 𝑅𝑐𝑜𝑠𝛼 1 + 𝑐𝑜𝑠𝛼
Area of strips = (R.dθ.tc) Area of steel in strip = (R.dθ.ts) Total compressive force in concrete and steel
(𝜋−𝛼)
C = 2 ∫0
2𝑅𝜎 𝑡
(𝑅. 𝑑𝜃 )𝑡𝑐 𝜎𝑐 [ (𝜋−𝛼)
𝑐 𝑐 ] ∫0 C = [1+𝑐𝑜𝑠𝛼
𝑐𝑜𝑠𝜃+𝑐𝑜𝑠𝛼 1+𝑐𝑜𝑠𝛼
(𝜋−𝑎)
] + 2 ∫0
(𝑐𝑜𝑠𝜃 + 𝑐𝑜𝑠𝛼 )𝑑𝜃 + [
(𝑅. 𝑑𝜃)𝑡𝑐 (𝑚 − 1) 𝜎𝑐 [
2𝑅𝜎𝑐 𝑡𝑠 (𝑚−1) 1+𝑐𝑜𝑠𝛼
(𝜋−𝛼)
] ∫0
𝑐𝑜𝑠𝜃+𝑐𝑜𝑠𝛼 1+𝑐𝑜𝑠𝛼
]
(𝑐𝑜𝑠𝜃 + 𝑐𝑜𝑠𝛼)𝑑𝜃
Integrating and simplifying
𝐶= [
2𝑅𝜎𝑐 ] [{𝑡𝑐 + (𝑚 − 1)𝑡𝑠 }[𝑠𝑖𝑛𝛼 + (𝜋 − 𝛼)𝑐𝑜𝑠𝛼]] 1 + 𝑐𝑜𝑠𝛼
Similarly total tension in steel is given by 𝛼
T = 2 ∫0 (𝑅𝑑𝜃 )𝑡𝑠 . 𝑚𝜎𝑐 (
𝑅 𝑐𝑜𝑠𝜃−𝑅 𝑐𝑜𝑠𝛼 𝑅+𝑅 𝑐𝑜𝑠𝛼
)
2𝑅𝜎 𝑚𝑡
𝑐 𝑠 ] [sin 𝛼 − 𝛼 cos 𝛼 ] = [ 1+cos 𝛼
Equating the sum of the internal forces to external load W W = (C-T) 2𝑅𝜎
𝑐 ] [(𝑡𝑐 − 𝑡𝑠 ){𝑠𝑖𝑛𝛼 + (𝜋 − 𝛼)𝑐𝑜𝑠𝛼 } + 𝜋𝑚𝑡𝑠 . cos 𝛼 ] 𝑊 = [1+𝑐𝑜𝑠𝛼
(1)
Equating the element of external forces to the sum of the moments of the internal forces we have (𝜋−𝛼)
𝑀=∫
𝜋
𝐶. 𝑅. 𝑐𝑜𝑠𝜃 + ∫ 𝑇. 𝑅. 𝑐𝑜𝑠𝜃
0
𝑀= [
0
(𝜋−𝛼) 2𝑅2 𝜎𝑐 (𝑐𝑜𝑠 2𝜃 + cos 𝛼 cos 𝜃 )𝑑𝜃 + ] [𝑡𝑐 + (𝑚 − 1)𝑡𝑠 ] ∫ 1 + cos 𝛼 0
2𝑅 2 𝜎𝑐 . 𝑚. 𝑡𝑠 𝜋 [ ] ∫ (cos 𝜃 − cos 𝜃 cos 𝛼)𝑑𝜃 1 + cos 𝛼 0 Integrating and simplifying we have the final equation for the moment M, as 𝑀= [
2𝑅2 𝜎𝑐 sin 2𝛼 (𝜋 − 𝛼) 2𝑅2 𝜎𝑐 . 𝑚𝑡𝑠 𝛼 sin 2𝛼 ] [(𝑡𝑐 + (𝑚 − 1)𝑡𝑠 )] [ ]+ [ ][ − ] + 1 + cos 𝛼 4 2 1 + cos 𝛼 2 4 2𝑅2 𝜎
𝑀 = [1+cos𝑐𝛼] [(𝑡𝑐 − 𝑡𝑠 )] [ Eccentricity e = (M/W)
sin 2𝛼 4
+
(𝜋−𝛼) 2
]+ [
𝑚𝑡𝑠 .𝜋 2
]
(2)
𝑒 = 𝑅 {( 𝑡
[(𝑡𝑐 −𝑡𝑠 )(
sin 2𝛼 𝜋−𝛼 𝑚.𝑡 .𝜋 + 2 )+ 2𝑠 ] 4
𝑐 −𝑡𝑠 )[sin 𝛼+(𝜋−𝛼 ) cos 𝛼 ]+𝜋𝑚𝑡𝑠 .cos 𝛼
}
(3)
The value of α which satisfies the Eq. 3 is determined by Trial and Error. Knowing α the stresses in concrete and steel can be evaluated using Eq.1. STRESSES IN HORIZONTAL REINFORCEMENT DUE TO SHEAR FORCE If H = Horizontal shear force at the section d = diameter of the chimney S = pitch of hoop bars At = Area of hoop bar in one pitch length Area of steel resisting shear in one metre height = ( 2.𝐴𝑡 .1000𝑥𝜎𝑠
Shear force resisted = (
𝑠
2 𝐴𝑡 1000 𝑠
) if σs = stress in steel
)
(a)
If horizontal distance between reinforcement on both sides is assumed as 0.8d, shear/meter = (
𝐻𝑋1000
𝐿𝑒𝑣𝑒𝑟 𝑎𝑟𝑚
)= (
1000 𝐻 0.8 𝑑
)
(b)
Equating (a) and (b) (
2𝐴𝑋1000𝑋𝜎𝑠 1000 𝐻 )=( ) 𝑠 0.8 𝑑 𝐻.𝑠
𝜎𝑠 = (1.6 𝐴 𝑑) 𝑡
(4)
Where d and s are expressed in mm units At = is in mm2 units. STRESSES DUE TO TEMPERATURE DIFFERENCE In the walls of a reinforced concrete chimney, stresses are developed due to the temperature gradient between the inner and outer surface of the walls. This temperature drop from inside to outside surface tends to expand the inner surface relative to outer one. Due to the monolithic action of the entire wall, differential expansion is not possible and hence equal expansion takes place so that the shell is compressed on its inside surface and pulled on the outside surface. As a whole there is an average increase in the length of the shell due to the temperature gradient. Effect of Temperature only on stresses in chimney walls
Let To = Temperature difference betweem inside and outside with a linear temperature gradient α = Coefficient of expansion of steel and concrete e = strain due to temperature difference m = modular ratio ts = Area of reinforcement per unit width tc = Area of concrete per unit width σct = stress in concrete due to temperature σst = stress in steel due to temperature p = (ts/tc) 𝑎𝑡𝑐
k = neutral axis depth constant
𝑡𝑐
Air Gap
Lining
To (Temperature Difference)
ktc
σst
σct
(Tα-e) = [e-(1-a)Tα}
Tα e
(1-a)Tα
= Net strain in concrete (compressive)
Fig. 3 Temperature stresses in Chimney walls Referring to Fig. 3 and considering the force equilibrium we have the following relation (1/2)𝜎𝑐𝑡 𝑘𝑡𝑐 = 𝑡𝑠 . 𝜎𝑠𝑡 = 𝑝. 𝑡𝑐 . 𝜎𝑠𝑡 𝜎𝑠𝑡 = (
𝜎𝑐𝑡 𝑘 𝛼𝑡𝑐 − 𝑘𝑡𝑐 𝑎−𝑘 ) = 𝑚. 𝜎𝑐𝑡 ( ) = 𝑚𝜎𝑐𝑡 2𝑝 𝑘𝑡𝑐 𝑘 𝑘 2 = 2𝑝𝑚(𝑎 − 𝑘 )𝑎𝑛𝑑
𝑘 = −𝑚𝑝 + √2𝑚𝑝𝑎 + 𝑝2 𝑚2
(5)
Rise in temperature in reinforcement = (1-a)T Free expansion of steel = (1-a)αT Thetensile stress in steel is due to the difference in between that due to strain e and due to temperature rise (1-a)T Stress in steel σst = Es[(e-(1-a))αT] At the neutral axis , there is free expansion due to strain e e = (1-k)αT Stress in steel σst = Es[(1-k)αT-(1-a)αT]
(6)
σst = Es.α.T(a-k) Stress in concrete σct = Ec[(Tα-e) = Ec[Tα-(1-k)αT]
(7)
σct = Ec.α.k.T
COMBINED EFECT OF WIND LOAD SELF-WEIGHT AND TEMPERATURE ON STRESSES IN CHIMNEY WALLS The stresses developed at the neutral axis, compression zone (Leeward side) and tension zone (Wind ward side) of the chimney due to combined effect of wind loads, self weight amd temperature will be examined separately. Case 1. Stresses at Neutral axis There are no stresses at the neutral axis due to the external loads. But there are stresses developed due to temperature difference Stress in steel σst = Es[(1-k)αT-(1-a)αT]
(6)
Stress in concrete σct = Ec[(Tα-e) = Ec[Tα-(1-k)αT]
(7)
Case 2. Stresses in Compression zone (Leeward side) The Leeward side of the chimney is under compression due to the effect of wind loads and selfweight. If te temperature stresses are are superposed, the final effect will be to increase the compressive stress in concrete and decrease the stress in steel. 𝑡𝑐 𝑎𝑡𝑐
Inside
Outside
Inside
Stresses due to wind and self-weight
atc σ’s
k’tc σ’c
Fig 4 Temperature stresses, Compression zone (Leeward side) Referring to Fig. 4 and using the following notations σc = Compressive stress in concrete assumed uniform due to the effect of self-weight and wind loads k’tc = Position of neutral axis 𝜎𝑐′ = Compressive stress in concrete due to combined effect 𝜎𝑠′ = Stress in steel due to combined effect e = strain due to temperature difference of T o Total compression will remain unchanged in the section
Considering the force equilibrium, we have 1 𝜎𝑐 . 𝑡 + (𝑚 − 1)𝑡. 𝜎 = ( 𝜎𝑐′ . 𝑘. 𝑡𝑐 ) − 𝑡𝑠 . 𝜎𝑠′ 2 ts = p.tc 1 𝑎 − 𝑘′ 𝜎𝑐 . 𝑡𝑐 + (𝑚 − 1)𝑡𝑠 . 𝜎𝑐 = ( 𝜎𝑐′ . 𝑘 , . 𝑡𝑐 ) − 𝑝. 𝑡𝑐 . 𝑚𝑐 𝜎𝑐′ ( ′ ) 2 𝑘 𝜎[1+(𝑚−1)𝑝]
𝜎 = { 𝑘′ 2
𝑎−𝑘′ ) 𝑘′
−𝑚𝑝(
}
(8)
Change in stress in concrete at inside face = (𝜎𝑐′ -σc)
(a)
(𝜎𝑐′ -σc) = Ec(Tα-e) Change of stress in steel = (𝜎𝑠′ + 𝑚𝜎𝑐 )
(𝜎𝑠′ + 𝑚𝜎𝑐 ) = 𝐸𝑠 [𝑒 − (1 − 𝑎)𝛼𝑇] But 𝜎𝑠′
[
𝜎𝑐′𝑚 (
𝜎𝑐′
=
𝑚
𝑎−𝑘 ′ ( 𝑘′ )
𝑎−𝑘′ )+𝑚.𝜎𝑐 𝑘′
𝐸𝑠
(b)
] + (1 − 𝑎)𝛼𝑇 = 𝑒
(9)
Also from Eq. (a), we have
𝑇𝛼 −
(𝜎𝑐′−𝜎𝑐 ) 𝐸𝑐
=𝑒
(d)
Equating, Eqs. (c) and (d), we get
𝜎𝑐′ = [
𝐸.𝑎.𝑇𝛼 (𝑎−𝑘′ ) 𝑘′
1+
]
(9)
Equating Eqs. (8) and (9) 𝜎𝑐 [1+(𝑚−1)𝑝] 𝑘′
𝑎−𝑘′ −𝑚𝑝( ) 2 𝑘′
= [
𝐸𝑐 .𝑎.𝑇.𝛼
]
(𝑎−𝑘′ ) 1+ ′ 𝑘
(10)
For given value of the various variables, we can evaluate k’ from Eq. (10) and then compute 𝜎𝑐′ from Eq. (9) and 𝜎𝑠′ , the stress in steel. If k’ is more than unity, the whole thickness of concrete t c will be in compression and the stresses can be analysed using the same procedure. Case 3. Stresses in Tension zone (Wind ward side) The chimney section in the wind ward zone is in tension due to the effect of self-weight and wind loads. Concrete is assumed to take negligible tension and hence the whole tension is resisted by steel. tc
atc
Outside
Inside
σs σ’s
k’tc σ’c
Fig 5 Temperature stresses, Tension zone (Wind ward side) Referring to Fig.5 and using the following notations: σs = Tensile stress in steel due to wind load and self-weight k’tc = position of neutral axis σc = compressive stress in concrete σs = stress in steel due to combined effect The effect of temperature is to develop compressive stress in concrete and increase the tensile stress in steel. The total force in the section remains unchanged, considering the equilibrium of forces 1 𝑡𝑠 . 𝜎𝑠 = (𝑡𝑠 𝜎𝑠′ − 𝜎𝑐′ 𝑘 ′𝑡𝑐 ) 2
𝑎−𝑘 ′
𝜎𝑠′ = 𝑚. 𝜎𝑐′ (
But
𝑘′
) and ts = p.tc
𝑎 − 𝑘′ 1 𝑝. 𝑡𝑐 . 𝜎𝑠 = 𝑝. 𝑡𝑐 . 𝑚. 𝜎𝑐′′ ( ′ ) − . 𝜎𝑐′ . 𝑘 ′𝑡𝑐 𝑘 2
𝜎𝑐′ = {
𝑝.𝜎𝑠
[𝑝𝑚(
𝑎−𝑘′ 𝑘′ )− ] ′ 2 𝑘
}
Change in strain in concrete at inside face is =
(11)
𝜎
𝜎𝑐′
𝐸𝑠
𝐸𝑐
[ 𝑠+
]
𝜎𝑠 𝜎𝑐′ [ + ] = (𝑇. 𝛼 − 𝑒) 𝐸𝑠 𝐸𝑐 𝜎𝑠 𝜎𝑐′ 𝑒 = [𝑇𝛼 − − ] 𝐸𝑠 𝐸𝑐
(𝑎)
Change of strain in steel is given by 𝜎𝑠′ − 𝜎𝑠 ( ) = [𝑒 − (1 − 𝑎)𝛼𝑇] 𝐸𝑠 𝜎𝑠′ − 𝜎𝑠 𝑒= ( ) + (1 − 𝑎)𝛼𝑇 𝐸𝑠
(𝑏)
Equating the Eqs. (a) and (b) we have the relation, 𝜎𝑐′ = [𝑎. 𝛼𝑇𝐸𝑐 −
𝜎𝑠′ ] 𝑚
Substituting for 𝜎𝑠′ and simplifying
𝜎𝑐′ = 𝛼𝑇𝐸𝑐 𝑘 ′
(12)
Equating the Eqs. (11) and (12), we have
[
𝑝.𝜎𝑠 𝑎−𝑘′ 𝑘′ 𝑝𝑚( ′ )− 2 𝑘
] = 𝛼. 𝑇. 𝐸𝑐 𝑘 ′
(13)
The value of k’ can be evaluated using the known values of the other variables and then the values of stress in concrete 𝜎𝑐′ and the stress in steel 𝜎𝑠′ can be estimated using relevant equations. STRESSES IN HORIZONTAL REINFORCEMENT DUE TO TEMPERATURE DIFFERENCE At high temperatures, the inner surface of the chimney is prevented from expansion and therefore gets compressed. The outer surface will expand more than natural expansion and will be in tension. Due to temperature stresses, generally the hoop tries to expand and consequently tensile stress will develop in the hoop reinforcement. tc
atc σ’s
K’tc
σ’c
Fig. 6 Stresses in horizontal reinforcement due to temperature difference Taking unit height of wall and referring to Fig. 6 and using the following notations. ′ 𝑘𝑡𝑐 = position of neutral axis
𝜎𝑐′ = compressive stress in concrete 𝜎𝑠′ = tensile stress in steel 𝐴′𝑠 = Area of hoop reinforcement /unit height As = cross sectional area of horizontal steel S = spacing 𝐴𝑠′ =
𝐴𝑠 = 𝑝. 𝑡𝑐 𝑠
𝜎𝑠′ = 𝑚. 𝜎𝑐′ (𝑎 − 𝑘 ′)/𝑘 ′
(14)
Consider the force equilibrium of the section, compressive force in concrete on the inner side = tensile force in horizontal reinforcement 1 ′ ′ 𝑎 − 𝑘′ ′ ′ 𝜎 𝑘 𝑡 = 𝐴𝑠 𝜎𝑠 = 𝑝. 𝑡𝑐 𝑚𝜎𝑐 ( ′ ) 2 𝑐 𝑐 𝑘 𝑘′ 𝑎 − 𝑘′ = 𝑚𝑝 ( ′ ) 2 𝑘 𝑘 ′ = √2𝑝𝑚𝑎 + 𝑝2 𝑚2 − 𝑝𝑚
(15)
Using this equation the position of neutral axis is determined. Let
e = actual strain
Stress in concrete = 𝜎𝑐′ = (αT-e)Ec Stress in steel = 𝜎𝑠′ = [e-(1-a)αT]Es From these two relations, 𝑒 = [𝛼𝑇 − 𝜎𝑐′ /𝐸𝑐 ]
𝜎′
and 𝑒 = [𝐸𝑠 − 𝛼𝑇(1 − 𝑎)] 𝑠
𝜎𝑐′ 𝜎𝑠′ [𝛼𝑇 − ] = [ + 𝛼𝑇(1 − 𝑎)] 𝐸𝑐 𝐸𝑠 [𝜎𝑠′ + 𝑚𝜎𝑐′ ] = 𝐸𝑠 𝛼. 𝑇. 𝑎
(16)
Knowing the value of k’, the stresses in steel and concrete 𝜎𝑐′ and 𝜎𝑠′ can be obtained by solving Eqs. (14) and (16).
