Visit Abqconsultants.com. This program Designs and Optimises RCC Chimney and Foundation. Written and programmed

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Jocelin Hines
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1 Prepared by : Date : Verified by : Date : Project : Ref Calculation Output Design of RCC Chimney :- 1) Dimensions of Chimney and Forces 200 Unit weight of Fire Brick Lining N/m3 100 Height of Fire Brick Lining above Ground m Level The temperature of gases above 200 º C surrounding air Coefficient of expansion of concrete and 1.1E-05 per deg C Steel Grade of Steel fy ( 250 or 415) 250 N/mm2 Allowable tensile stress in steel 140 N/mm2 Modulus of Elasticity of steel Es 2.05E+05 N/mm2 Grade Concrete Mix M25 25 N/mm2 Modulus of Elasticity of Concrete Ec 2.85E+04 N/mm2 Thickness of chimney shell at top portion 200 mm Height of top portion of Chimney m Thickness of chimney shell at middle portion 300 mm of Chimney Height of middle portion of Chimney m fig 1 Thickness of chimney shell at bottom portion 400 mm Cross-Section of Chimney of Chimney Height of balance bottom portion of Chimney m Lining Support every 6.00 m Constant wind pressure intensity at top portion Calculation Sheet Height of Chimney m External Diameter of Chimney 4.50 m Fire Brick Lining 100 mm thk Air Gap Between Wall & Fire Brick Lining (min) 100 mm 1800 N/m n/m2 Job no : Revision note : m m m m m lining thickness Sheet No : cont'd : Subject : Description : 4 Note : Input data in yellow cells only and ensure all check boxes are displaying "" or "safe" m Constant wind pressure intensity at middle portion Constant wind pressure intensity at bottom portion Shape Factor N/m N/m2 Grade of Steel (N/mm2) Allowable tebsile stress N/mm Visit Abqconsultants.com Grade of conc (N/mm2) Allowable compressive stress (Direct) N/mm Allowable compressive stress (Bending) N/mm Allowable tebsile stress (Direct) N/mm Weight of Lining per meter height modular ratio m This program Designs and Optimises RCC Chimney and Foundation. Written and programmed Π *( ( )) * by * 1.00 * :- A B Quadri abquadri@yahoo.com abquadri@gmail.com N As bd Permissible Shear Stress in Concrete Tc N/mm2 for grade of concrete Page 1 of 16

2 Weight of Concrete per meter height For 200 mm thk shell, w Π [ ] * 0.20 * 1.00 * N / m For 300 mm thk shell, w Π [ ] * 0.30 * 1.00 * N / m For 400 mm thk shell, w Π [ ] * 0.40 * 1.00 * N / m 2) Stress at Section m below top Let the vertical reinforcement be 1.00 % of the concrete area place at a cover of 50 mm As 1 * Π * ( 4.50 ^ ^2 ) * mm2 Nos of 16 mm Φ bars Hence provide 140 bars of 16mm Φ suitably placed along the circumference Actual As mm2 > mm2 Equivalent thickness of steel ring placed at the centre of the shell thickness ( R m ) is Ts mm 2ΠR Horizontal steel (hoops) may be 0.2 % of sectional area Area of steel per metre height of chimney 0.2 * 200 * mm2 100 Hence pitch s of 12 mm Φ bar hoops 1000 * mm 400 Provide these at 250 mm centre W * N P1 0.7 * 1800 ( 4.50 * 25.0 ) N acting at 12.5 m below top.: M * N. m.: Eccentricity e M m 1049 mm W For M 25 concrete, m : Eqivalent area A Π/4 * ( 4.50 ^ ^2 ) * ( )* mm2 Eqivalent moment of inertia I (Π / 64) ( D 4 -d 4 )+(m-1) Π R ts (R) 2 Π * ( 4.50 ^ ^ 4 ) * 1000 ^ ( ) * Π * 2150 * 2.08 * 2150 ^ E+12 mm 4 Page 2 of 16

3 For no tension to develop, allowable eccentricity 2 I 2 * E+12 AD * mm The actual eccentricity is 1049 mm. Hence some tension will be developed in the leeward side. The maximum and minimum stresses are given by σ W ± MD ± * 1000 * * E ± A 2I Compressive stress N/mm2 < 8.5 N/mm2 allowable (Safe) Tensile stress N/mm2 < -0.8 N/mm2 allowable (Safe) 2) Stress at Section m below top Thickness of shell 300 mm Let the vertical reinforcement be 1.00 % of the concrete area place at a cover of 50 mm As 1 * Π * ( 4.50 ^ ^2 ) * mm2 Nos of 20 mm Φ bars Hence provide 130 bars of 20mm Φ suitably placed along the circumference Actual As mm2 > mm2 Equivalent thickness of steel ring placed at the centre of the shell thickness ( R m ) is Ts mm 2ΠR Horizontal steel (hoops) may be 0.2 % of sectional area Area of steel per metre height of chimney 0.2 * 300 * mm2 100 Hence pitch s of 12 mm Φ bar hoops 1000 * mm 600 Provide these at 180 mm centre W * * * N P1 0.7 * 1800 ( 4.50 * 25.0 ) * 1600 ( 4.50 * 25.0 ) N.: M * * N. m.: Eccentricity e M m 1476 mm W Page 3 of 16

4 For M 25 concrete, m : Eqivalent area A Π/4 * ( 4.50 ^ ^2 ) * ( )* mm2 Eqivalent moment of inertia I ( Π/64 ) ( D 4 -d 4 ) + ( m-1 ) Π R ts (R) 2 Π * ( 4.50 ^ ^ 4 ) * 1000 ^ ( ) * Π * 2100 * 3.10 * 2100 ^ E+12 mm 4 For no tension to develop, allowable eccentricity 2 I 2 * E+12 AD * mm The actual eccentricity is 1476 mm. Hence some tension will be developed in the leeward side. The maximum and minimum stresses are given by σ W ± MD ± * 1000 * * E ± A 2I Compressive stress N/mm2 < 8.5 N/mm2 allowable (Safe) Tensile stress N/mm2 < -0.8 N/mm2 allowable (Safe) 3) Stress at Section m below top Thickness of shell 400 mm Let the vertical reinforcement be 1.00 % of the concrete area place at a cover of 50 mm As 1 * Π * ( 4.50 ^ ^2 ) * mm2 Nos of 25 mm Φ bars nos 491 Hence provide 120 bars of 25mm Φ suitably placed along the circumference Actual As mm2 > mm2 Equivalent thickness of steel ring placed at the centre of the shell thickness ( R m ) is Ts mm 2ΠR Horizontal steel (hoops) may be 0.2 % of sectional area Area of steel per metre height of chimney 0.2 * 400 * mm2 100 Hence pitch s of 12 mm Φ bar hoops 1000 * mm 800 Provide these at 140 mm centre Page 4 of 16

5 W * * * * * N P1 0.7 * 1800 ( 4.50 * 25.0 ) * 1600 ( 4.50 * 25.0 ) 0.7 * 1400 ( 4.50 * ) N.: M * * * N. m.: Eccentricity e M m 1742 mm W For M 25 concrete, m : Eqivalent area A Π/4 * ( 4.50 ^ ^2 ) * ( )* mm2 Eqivalent moment of inertia I ( Π/64 ) ( D 4 -d 4 ) + ( m-1 ) Π R ts (R) 2 Π * ( 4.50 ^ ^ 4 ) * 1000 ^ ( ) * Π * 2100 * 4.46 * 2100 ^ E+13 mm 4 For no tension to develop, allowable eccentricity 2 I 2 * E+13 AD * mm The actual eccentricity is 1742 mm. Hence some tension will be developed in the leeward side. The maximum and minimum stresses are given by σ W ± MD ± * 1000 * * E ± Compressive stress N/mm2 < 8.5 N/mm 2 allowable (Safe) Tensile stress N/mm2 > -0.8 N/mm 2 allowable Check further The eccentricity is quite high. Due to this, tensile stresses in the windward side are expected to be greather than 0.8 N/mm 2 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 m and m will not be applicable. We shall analyse the section for stresses by method discussed in 8.3. Tc 400 mm R m Ts 4.46 mm eccentricity e m m A 2I In order to find the position of N.A., use equation 8.3 : [ mπ Ts } 2 ] + e R (Tc-Ts) { sin2φ Π-Φ (Tc-Ts) { sinφ + (Π-Φ) cosφ } mπ Ts cosφ [ ] fig 2 Page 5 of 16

6 e 205 * [ ( [ ( ) Π-Φ * Π 0.45 ) * { sin2φ } * * { sinφ + (Π-Φ) cosφ } * Π * 0.45 * cosφ ] ] Now adjust the value of angle Φ in such a way that the value of eccentricity e is > m Assume Φ º C radians.: e m not which is slightly more than the actual value.: consider Φ 0.00 º C The maximum stress c1 in concrete is found from Eq.8.1 2Rc1 W 1+cosΦ [ (Tc-Ts) { sinφ + (Π-Φ) cosφ } + mπ Ts cosφ ].:.: 2 * 2050 * c Compressive stress c1 in Concrete { * c1 [ ( ) * } * Π * 4.46 * ] N/mm 2 < 8.5 N/mm 2 safe Tensile stress in Steel, assuming concrete to be fully cracked. 1 - cosφ t1 m * c1 * 1 + cosφ (b) Stress in horizontal reinforcement [ ] N/mm2 < 140 N/mm 2 safe Horizontal steel (hoops) may be 0.2 % of sectional area Area of steel per metre height of chimney 0.2 * 400 * mm Hence pitch s of 12 mm Φ bar hoops 1000 * mm 800 Provide these at 140 mm centre As mm 2 in pitch s 140 mm centre, if the cover is 40 mm then D mm.: t1 p * s * * As * D1 2 * 113 * N/mm 2 < 140 N/mm 2 allowable Safe Page 6 of 16

7 (c) Stress on leeward side due to temperature gradient fig 3 fig 4 Thickness of shell Tc 400 mm Thickness of lining Tl 100 mm Thickness of steel Ts 4.46 mm.: atc mm Cover to vertical steel 50 mm a c N/mm Es 2.05E+05 N/mm 2 Ec 2.05E E N/mm 2 p Ts α 1.10E-05 per º C Concrete Tc 400 Temperature Co-efficient Temperature difference 200 º C Let us assume that 80 % of temperature drops through the lining and shell. Drop in temperature 200 * º C Asssuming that drop in lining is 5 times more than that in shell, per unit thickness, the drop of temperature through concrete is given by, Tº º C * * 100 To locate -neutral axis in the shell thickness, use Eq c1 [ - 1 ) * p ] 1 + ( m 0.5 * k 2 - m * p * (a - k) α * T * Ec.: * [ 1 + ( ) * ) ] 1.1E-05 * * 0.5 * k * * ( k ) 1.867E+04 or k * k k * k solving for k k Page 7 of 16

8 .: Compressive a * α * Tº * Ec c k * α * Tº * Ec Stress in Concrete a - k 1 + k * 1.1E-05 * * 1.867E N/mm 2 Since wind stresses are taken into account, Permissible 4 * 8.5 Stress in Concrete N/mm 2 Thus the compressive stress more than the permissible--not 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 m c a - k * * ( ) k N/mm 2 (d) Stresses on windward side, due to temperature gradient p * t1 α * Tº * Ec m * p * ( a - k ) * k 2 where t N/mm 2 p a m α Tº º C Ec N/mm 2.: * * * * ( k ) * k 2 * or k * k k * k solving k : Compressive c α * Tº * Ec * k N/mm 2 Stress in Concrete Tensile stress in Steel, assuming concrete to be fully cracked. fig 5 t m c a - k * * ( ) k N/mm 2 < 140 N/mm 2 ( safe ) (e) Stresses on the Neutral axis.(i.e. temperature effect alone) k 2 -mp + 2mpa + m 2 p 2 where m p a α Tº º C Ec N/mm 2 or k * * * * * * * Page 8 of 16

9 k : Compressive c 2 α * Tº * Ec * k N/mm 2 Stress in Concrete Tensile stress in concrete, assuming concrete to be fully cracked. t 2 m c 2 a - k * * ( ) k N/mm 2 < 140 N/mm 2 ( safe ) (b) Stress in horizontal reinforcement due to temperature : p' A Φ S T c 140 * 400 a' From Eq k' -mp' + 2mp'a + m 2 p' 2 or k' * * * * * * * k' : Compressive c' α * Tº * Ec * k' N/mm 2 Stress in Concrete Tensile stress in concrete, assuming concrete to be fully cracked. t 2 m c' a' - k' * * ( ) k' N/mm 2 < 140 N/mm 2 O.k These stresses are due to temperature effect alone. To this we must add the stresses due to wind. Hence total stress in steel N/mm 2 Since wind is also acting, permissible t 4 * N/mm 2 Safe allowable tensile stress in steel 3 Page 9 of 16

10 5. Flue Opening : Provide a flue opening 1.5 m wide and 2.0 m high at bottom. 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 P N M N. m V M V N 0.00 level Page 10 of 16

11 7. Design of Cirrcullar Chimney Foundation : e Data FFL H Concrete Grade fc' 25 N/mm2 A Steel Grade fy 415 N/mm2 FGL S.B.C of Soil Qs 200 Kn/m2 Density of soil Ws 18 Kn/m3 Axial Load P Kn Moment M Kn. M D eccentricity e M/P m Horizontal load H Kn Outer dia of chimney d 4.50 m Thickness of chimney wall t 400 mm Dia of Footing OD m A 200 mm T Level of footing below ground Totd 4000 mm Depth of Soil D 1700 mm 4500 Depth of Footing T 2300 mm d Footing Reinforcement dia Φ mm Reinforcement cover c 75 mm OD P Soil filling inside 4000 Totd Axial load at the base of footing OD P' P + Weight of Chimney Wall + Soil Filling inside of wall + Weight of soil + Self weight of footing Π ( 4.50 ^ ^2 ) 4 * ( ) * 25 + Π ( 4.10 ^2 ) * 18 4 fig 7 + Π ( ^ ^2 ) * 18 * Π ( ^2 ) * 2.30 * kn M' H * ( D + T + A ) * ( ) Kn. M.: e' M' < [ OD 1.75 m P' Ok ] Axx Π * OD 2 Ixx Π * OD m m 2 Zxx Π * OD m 3 32 Page 11 of 16

12 The maximum and minimum base pressures are given by σ P' ± M' ± ± A Zxx fig σ max Kn / m2 < 200 Kn / m2 allowable Ok σ min Kn / m2 > 0 Kn / m2 allowable Ok Factor of Safety against overturning Stabilising Moment Overturning Moment P' * OD 2 M' * > 1.5 safe Design of Footing slab Assume initially 1 Layer of 32 mm Φ bars 120 nos spaced radially along the + 25 mm Φ bars As 1295 mm2 circumferance Φ 3.00 º radians Radius of Chimney ro 2250 mm Radius of Foundation fro 7000 mm Bar Spacing at ro 118 mm Length of segment 'PQ' Bar Spacing at fro 367 mm Length of segment 'RS' Uniform pressure under area 'PQRS' Kn / m2 Pressure due to Moment at 'RS' 57.1 Kn / m2 Pressure due to Moment at 'PQ' 18.4 Kn / m2 Area of Segment 'PQRS' m2 CG of Segment 'PQRS' from 'PQ' m fro A ro P b1 a1 Φ rp Q R a2 b2 Main Reinforcement S Area covered by one unit of Main Reinforcement Critical Section for Moment Footing Outer Dia A rs Chimney Outer Dia Line of Punching Shear Line of Shear fig 9 Page 12 of 16

13 C / L of foundation r Straight portion Sloping portion 16 Φ top radial reinforcement 1.61 c/l of Foundation Shear Stirrups (if required) Main Radial reinforcement Circullar reinf 32 Φ 120 nos c/c cover critical punching shear section 2225 critical shear Section C / L 7000 Section A - A fig 10.: Moment at 'PQ' Mf Kn.m fy 415 N/mm2.: fyall 230 N/mm2 m fc' 25 N/mm2.: fc'all 8.5 N/mm2.: k * * j 1 - k R 1 fc'all j k 1 * 8.5 * * Hence d Mf * mm 1000 * R 118 * : adopt T 2300 mm cover 75 mm d : d 2225 mm effective depth As M Fyall * j * d 230 * * 2225 mm 2.: Provide 32 Φ + 25 Φ AΦ Π * ( 32 ² + 25 ² ) 1 dia bars. 4 layer of Main radial reinforcement 1295 mm 2 > 1241 mm 2 Provide c/c p% 1295 * % distribution steel 118 * 2225 Page 13 of 16

14 Punching Shear : Check Punching shear at ro + d/2 from the c/l r 2250 mm d mm r + d 2 2 Φ 3.00 º radians Radius of Punching shear rps 3363 mm Radius of Foundation fro 7000 mm Bar Spacing at rps 176 mm Length of segment 'a1a2' Bar Spacing at fro 367 mm Length of segment 'RS' 3363 mm Uniform pressure under area 'PQRS' Pressure due to Moment at 'RS' Pressure due to Moment at 'PQ' Kn / m Kn / m Kn / m2 Area of Segment 'PQa1a2' CG of Segment 'PQa1a2' from 'a1a2' m m.: Punching Shear at 'a1a2' F Kn Allowable Punching Shear stress 0.16 fck 0.16 * N/mm2.: Depth required for punching shear do F fck * * < 2225 mm provided OK Shear : Check shear at r + d from the c/l, End of Straight portion, and at three points at sloping portion. distance from c/l mm Reinforcement Spacing mm Effective depth mm Bar dia mm As mm p% not Uniform pressure kn/m Pressure due to rs kn/m Area of Segment 'PQxx' m M actual / bar Kn.m M allowable / bar Kn.m p% for Shear Shear Stress tc N/mm Shear actual Kn Shear - Vs per main bar Kn Shear - MS bar dia fyall spacing mm Minimum shear s 2.5Asvfy/b mm : Provided spacing mm not Pressure due to moment at section kn/m CG of Segment 'PQxx' m Shear allowable Kn Shear Reinf Reqd Reqd Not Reqd Not Reqd Not Reqd Not Reqd Page 14 of 16

15 Check Deflection of Chimney : Data : Top Portion : Height of Top portion of Chimney m Wind intensity of top portion of Chimney 1800 N/m2 Concrete Area of Top Portion of chimney mm2 Moment of Inertia of Top portion of Chimney E+12 mm4 Wind Moment at the base of Top Portion E+06 N.m Modulus of Elasticity of Concrete E+04 N/mm2 M / Ei E-09 1/mm Area of M / Ei of top portion E-04 C.g of Area of M / Ei of top portion E+04 mm Moment of Area of M / Ei from top portion E+00 (1) Partial Deflection of Top Portion δtop mm wrt bottom of top portion Ratio L / δ L / > L / 200 Middle Portion : Height of Middle portion of Chimney m Wind intensity of Middle portion of Chimney 1600 N/m2 Area of Middle Portion of chimney mm2 Moment of Inertia of Middle portion of Chimney E+12 mm4 Wind Moment at the base of Middle Portion E+06 N.m Modulus of Elasticity of Concrete E+04 N/mm2 M / Ei E-08 1/mm Area of M / Ei of Middle portion E-04 C.g of Area of M / Ei of Middle portion from top E+04 mm Moment of Area of M / Ei of Middle portion E+01 (2) Partial Deflection of Top Portion δtop mm wrt bottom of middle portion Ratio L / δ L / 2807 < L / 200 Page 15 of 16

16 Bottom Portion : Height of Bottom portion of Chimney m Wind intensity of Bottom portion of Chimney 1400 N/m2 Area of Bottom Portion of chimney mm2 Moment of Inertia of Bottom portion of Chimney E+13 mm4 Wind Moment at the base of Bottom Portion E+07 N.m Modulus of Elasticity of Concrete E+04 N/mm2 M / Ei E-08 1/mm Area of M / Ei of Bottom portion E-04 C.g of Area of M / Ei of Bottom portion from top E+04 mm Moment of Area of M / Ei of Bottom portion E+01 (3) Total Deflection of Top Portion δtop mm wrt bottom of bottom portion Ratio L / δ L / 1175 < L / 200 Page 16 of 16

DESIGN AND DETAILING OF COUNTERFORT RETAINING WALL

DESIGN AND DETAILING OF COUNTERFORT RETAINING WALL When the height of the retaining wall exceeds about 6 m, the thickness of the stem and heel slab works out to be sufficiently large and the design becomes