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

Similar documents
DESIGN AND DETAILING OF COUNTERFORT RETAINING WALL

Name :. Roll No. :... Invigilator s Signature :.. CS/B.TECH (CE-NEW)/SEM-3/CE-301/ SOLID MECHANICS

1. ARRANGEMENT. a. Frame A1-P3. L 1 = 20 m H = 5.23 m L 2 = 20 m H 1 = 8.29 m L 3 = 20 m H 2 = 8.29 m H 3 = 8.39 m. b. Frame P3-P6

CHAPTER 4: BENDING OF BEAMS

UNIT II SHALLOW FOUNDATION

CIVIL DEPARTMENT MECHANICS OF STRUCTURES- ASSIGNMENT NO 1. Brach: CE YEAR:

Delhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web: Ph:

ε t increases from the compressioncontrolled Figure 9.15: Adjusted interaction diagram

MAHALAKSHMI ENGINEERING COLLEGE

Sub. Code:

twenty one concrete construction: shear & deflection ARCHITECTURAL STRUCTURES: FORM, BEHAVIOR, AND DESIGN DR. ANNE NICHOLS SUMMER 2014 lecture

3 Hours/100 Marks Seat No.

CHAPTER 4. Design of R C Beams

Job No. Sheet No. Rev. CONSULTING Engineering Calculation Sheet

RETAINING WALL LOADS: Horizontal Equivalent Fluid Pressure = pcf. (Load Case = Soil)

KINGS COLLEGE OF ENGINEERING DEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK. Subject code/name: ME2254/STRENGTH OF MATERIALS Year/Sem:II / IV

Stress Analysis Lecture 3 ME 276 Spring Dr./ Ahmed Mohamed Nagib Elmekawy

QUESTION BANK SEMESTER: III SUBJECT NAME: MECHANICS OF SOLIDS

Module 14. Tension Members. Version 2 CE IIT, Kharagpur

R13. II B. Tech I Semester Regular Examinations, Jan MECHANICS OF SOLIDS (Com. to ME, AME, AE, MTE) PART-A

TABLE OF CONTANINET 1. Design criteria. 2. Lateral loads. 3. 3D finite element model (SAP2000, Ver.16). 4. Design of vertical elements (CSI, Ver.9).

QUESTION BANK DEPARTMENT: CIVIL SEMESTER: III SUBJECT CODE: CE2201 SUBJECT NAME: MECHANICS OF SOLIDS UNIT 1- STRESS AND STRAIN PART A

Annex - R C Design Formulae and Data

9.5 Compression Members

2012 MECHANICS OF SOLIDS

INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad

VTU EDUSAT PROGRAMME Lecture Notes on Design of Columns

Downloaded from Downloaded from / 1

Design of Beams (Unit - 8)

OUTCOME 1 - TUTORIAL 3 BENDING MOMENTS. You should judge your progress by completing the self assessment exercises. CONTENTS

Entrance exam Master Course

Project Name Structural Calculation for Feature Pressing

Mechanics of Structure

Civil Engineering Design (1) Design of Reinforced Concrete Columns 2006/7

Unit III Theory of columns. Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE, Sriperumbudir

CHAPTER 4. Stresses in Beams


Slenderness Effects for Concrete Columns in Sway Frame - Moment Magnification Method (CSA A )

Mechanics of Materials MENG 270 Fall 2003 Exam 3 Time allowed: 90min. Q.1(a) Q.1 (b) Q.2 Q.3 Q.4 Total

SRI CHANDRASEKHARENDRA SARASWATHI VISWA MAHAVIDHYALAYA


Associate Professor. Tel:

DESIGN OF STAIRCASE. Dr. Izni Syahrizal bin Ibrahim. Faculty of Civil Engineering Universiti Teknologi Malaysia

BE Semester- I ( ) Question Bank (MECHANICS OF SOLIDS)

3.2 Reinforced Concrete Slabs Slabs are divided into suspended slabs. Suspended slabs may be divided into two groups:

Roadway Grade = m, amsl HWM = Roadway grade dictates elevation of superstructure and not minimum free board requirement.

18.Define the term modulus of resilience. May/June Define Principal Stress. 20. Define Hydrostatic Pressure.

PURE BENDING. If a simply supported beam carries two point loads of 10 kn as shown in the following figure, pure bending occurs at segment BC.

Pre-stressed concrete = Pre-compression concrete Pre-compression stresses is applied at the place when tensile stress occur Concrete weak in tension

PERIYAR CENTENARY POLYTECHNIC COLLEGE PERIYAR NAGAR - VALLAM THANJAVUR. DEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK

UNIT 1 STRESS STRAIN AND DEFORMATION OF SOLIDS, STATES OF STRESS 1. Define stress. When an external force acts on a body, it undergoes deformation.

Sabah Shawkat Cabinet of Structural Engineering Walls carrying vertical loads should be designed as columns. Basically walls are designed in

Job No. Sheet No. Rev. CONSULTING Engineering Calculation Sheet. Member Design - Steel Composite Beam XX 22/09/2016

Design of AAC wall panel according to EN 12602

Reinforced Concrete Structures

CONSULTING Engineering Calculation Sheet. Job Title Member Design - Reinforced Concrete Column BS8110

Design of a Balanced-Cantilever Bridge

Stress Analysis Lecture 4 ME 276 Spring Dr./ Ahmed Mohamed Nagib Elmekawy

N = Shear stress / Shear strain

Lecture-04 Design of RC Members for Shear and Torsion

External Pressure... Thermal Expansion in un-restrained pipeline... The critical (buckling) pressure is calculated as follows:

Only for Reference Page 1 of 18

CE6306 STRENGTH OF MATERIALS TWO MARK QUESTIONS WITH ANSWERS ACADEMIC YEAR

UNIT-I STRESS, STRAIN. 1. A Member A B C D is subjected to loading as shown in fig determine the total elongation. Take E= 2 x10 5 N/mm 2

STRENGTH OF MATERIALS-I. Unit-1. Simple stresses and strains

Solution: The moment of inertia for the cross-section is: ANS: ANS: Problem 15.6 The material of the beam in Problem

APPENDIX 1 MODEL CALCULATION OF VARIOUS CODES

International Construction Consulting, LLC

7.2 Design of minaret: Geometry:

Structural Steelwork Eurocodes Development of A Trans-national Approach

Appendix J. Example of Proposed Changes

Neutral Axis Depth for a Reinforced Concrete Section. Under Eccentric Axial Load

STRESS STRAIN AND DEFORMATION OF SOLIDS, STATES OF STRESS

3.5 Reinforced Concrete Section Properties

CHAPTER 6: ULTIMATE LIMIT STATE

R.C. FOUNDATION ANALYSIS, DETAILS & DETAILING

SPECIFIC VERIFICATION Chapter 5

UNIVERSITY OF SASKATCHEWAN ME MECHANICS OF MATERIALS I FINAL EXAM DECEMBER 13, 2008 Professor A. Dolovich

Lecture-03 Design of Reinforced Concrete Members for Flexure and Axial Loads

ENG1001 Engineering Design 1

PUNCHING SHEAR CALCULATIONS 1 ACI 318; ADAPT-PT

Detailing. Lecture 9 16 th November Reinforced Concrete Detailing to Eurocode 2

This procedure covers the determination of the moment of inertia about the neutral axis.

The University of Melbourne Engineering Mechanics

Design of Reinforced Concrete Beam for Shear

Sample Question Paper

Software Verification

Bending and Shear in Beams

ENGINEERING COUNCIL DIPLOMA LEVEL MECHANICS OF SOLIDS D209 TUTORIAL 3 - SHEAR FORCE AND BENDING MOMENTS IN BEAMS

FLOW CHART FOR DESIGN OF BEAMS

Flexure: Behavior and Nominal Strength of Beam Sections

REVIEW FOR EXAM II. Dr. Ibrahim A. Assakkaf SPRING 2002

FIXED BEAMS IN BENDING

BOOK OF COURSE WORKS ON STRENGTH OF MATERIALS FOR THE 2 ND YEAR STUDENTS OF THE UACEG

Purpose of this Guide: To thoroughly prepare students for the exact types of problems that will be on Exam 3.

/ / W ./- \...\..S/. i x ju L \sixu. o/xiyjx[a-.xj; wnere, x - Case 2- If NA lies below the flange, with variable stress in the flange, Df<xu<7/3 Df

Design of reinforced concrete sections according to EN and EN

Strength of Material. Shear Strain. Dr. Attaullah Shah

DEPARTMENT OF CIVIL ENGINEERING

PES Institute of Technology

Transcription:

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 19000 N/m3 100 Height of Fire Brick Lining above Ground 48.00 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 25.00 m Thickness of chimney shell at middle portion 300 mm of Chimney Height of middle portion of Chimney 25.00 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 22.00 m Lining Support Distance @ every 6.00 m Constant wind pressure intensity at top portion Calculation Sheet Height of Chimney 72.00 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/m2 1600 1400 n/m2 Job no : Revision note : 1800 4.50 m 25.00 m 25.00 m 22.00 m 100 4.10 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". 3.90 m 300 400 Constant wind pressure intensity at middle portion Constant wind pressure intensity at bottom portion Shape Factor 0.70 1600 N/m2 1400 N/m2 Grade of Steel (N/mm2) 250 415 Allowable tebsile stress N/mm2 140 230 Visit Abqconsultants.com Grade of conc (N/mm2) 15 20 25 30 Allowable compressive stress (Direct) N/mm2 4.00 5.00 6.00 8.00 9.00 Allowable compressive stress (Bending) N/mm2 5.00 7.00 8.50 10.00 11.50 13.00 Allowable tebsile stress (Direct) N/mm2-2.00-2.80-3.20-3.60-4.00 Weight of Lining per meter height 35 40 modular ratio m 18.67 13.33 10.98 9.33 8.11 7.18 This program Designs and Optimises RCC Chimney and Foundation. Written and programmed 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 *( 4.50-2 ( 0.4 + + )) * 1.75 0.44 0.47 0.49 0.50 0.52 0.52 by 0.1 0.05 0.1 * 1.00 * 19000 2.00 0.44 0.49 0.51 0.53 0.54 0.55 :- A B Quadri www.abqconsultants.com abquadri@yahoo.com abquadri@gmail.com 2.25 0.44 0.51 0.53 0.55 0.56 0.57 20295 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 10.00-4.40 100As bd Permissible Shear Stress in Concrete Tc N/mm2 for grade of concrete Page 1 of 16

9959010210 9959010211 Weight of Concrete per meter height For 200 mm thk shell, w Π [ 4.50-0.20 ] * 0.20 * 1.00 * 25000 67544 N / m For 300 mm thk shell, w Π [ 4.50-0.30 ] * 0.30 * 1.00 * 25000 98960 N / m For 400 mm thk shell, w Π [ 4.50-0.40 ] * 0.40 * 1.00 * 25000 128805 N / m 2) Stress at Section 25.00 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 ^2-4.10 ^2 ) * 1000000 100 4 27018 mm2 Nos of 16 mm Φ bars 27018 135 201 Hence provide 140 bars of 16mm Φ suitably placed along the circumference Actual As 28149 mm2 > 27018 mm2 Equivalent thickness of steel ring placed at the centre of the shell thickness ( R 2.25-0.1 2.15 m ) is Ts 28149 2.08 mm 2ΠR Horizontal steel (hoops) may be provided @ 0.2 % of sectional area Area of steel per metre height of chimney 0.2 * 200 * 1000 400 mm2 100 Hence pitch s of 12 mm Φ bar hoops 1000 * 113 282.5 mm 400 Provide these at 250 mm centre W 25.00 * 67544 1688606 N P1 0.7 * 1800 ( 4.50 * 25.0 ) 141750 N acting at 12.5 m below top.: M 141750 * 12.5 1771875 N. m.: Eccentricity e M 1771875 1.049 m 1049 mm W 1688606 For M 25 concrete, m 10.98.: Eqivalent area A Π/4 * ( 4.50 ^2-4.10 ^2 ) * 1000000 + ( 10.98-1 )* 28149 2982693 mm2 Eqivalent moment of inertia I (Π / 64) ( D 4 -d 4 )+(m-1) Π R ts (R) 2 Π * ( 4.50 ^ 4-4.10 ^ 4 ) * 1000 ^ 4 + 64 ( 10.98-1 ) * Π * 2150 * 2.08 * 2150 ^ 2 6.9073E+12 mm 4 Page 2 of 16

For no tension to develop, allowable eccentricity 2 I 2 * 6.9073E+12 AD 2982693 * 4500 1029.2 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 1688606 ± 1771875 * 1000 * 4500 2982693 2 * 6.9073E+12 0.566 0.577 ± A 2I Compressive stress 1.143 N/mm2 < 8.5 N/mm2 allowable (Safe) Tensile stress -0.011 N/mm2 < -0.8 N/mm2 allowable (Safe) 2) Stress at Section 50.00 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 ^2-3.90 ^2 ) * 1000000 100 4 39584 mm2 Nos of 20 mm Φ bars 39584 126 314 Hence provide 130 bars of 20mm Φ suitably placed along the circumference Actual As 40841 mm2 > 39584 mm2 Equivalent thickness of steel ring placed at the centre of the shell thickness ( R 2.25-0.15 2.10 m ) is Ts 40841 3.10 mm 2ΠR Horizontal steel (hoops) may be provided @ 0.2 % of sectional area Area of steel per metre height of chimney 0.2 * 300 * 1000 600 mm2 100 Hence pitch s of 12 mm Φ bar hoops 1000 * 113 188.0 mm 600 Provide these at 180 mm centre W 25.00 * 67544 + 25.00 * 20295 + 25.00 * 98960 4669977 N P1 0.7 * 1800 ( 4.50 * 25.0 ) + 0.7 * 1600 ( 4.50 * 25.0 ) 141750 + 126000.00 267750.00 N.: M 141750 * 37.5 + 126000 * 12.5 6890625 N. m.: Eccentricity e M 6890625 1.476 m 1476 mm W 4669977 Page 3 of 16

For M 25 concrete, m 10.98.: Eqivalent area A Π/4 * ( 4.50 ^2-3.90 ^2 ) * 1000000 + ( 10.98-1 )* 40841 4365997 mm2 Eqivalent moment of inertia I ( Π/64 ) ( D 4 -d 4 ) + ( m-1 ) Π R ts (R) 2 Π * ( 4.50 ^ 4-3.90 ^ 4 ) * 1000 ^ 4 + 64 ( 10.98-1 ) * Π * 2100 * 3.10 * 2100 ^ 2 9.6716E+12 mm 4 For no tension to develop, allowable eccentricity 2 I 2 * 9.6716E+12 AD 4365997 * 4500 984.5 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 4669977 ± 6890625 * 1000 * 4500 4365997 2 * 9.6716E+12 1.070 1.603 ± A 2I Compressive stress 2.673 N/mm2 < 8.5 N/mm2 allowable (Safe) Tensile stress -0.533 N/mm2 < -0.8 N/mm2 allowable (Safe) 3) Stress at Section 72.00 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 ^2-3.70 ^2 ) * 1000000 100 4 51522 mm2 Nos of 25 mm Φ bars 51522 105 nos 491 Hence provide 120 bars of 25mm Φ suitably placed along the circumference Actual As 58905 mm2 > 51522 mm2 Equivalent thickness of steel ring placed at the centre of the shell thickness ( R 2.25-0.15 2.10 m ) is Ts 58905 4.46 mm 2ΠR Horizontal steel (hoops) may be provided @ 0.2 % of sectional area Area of steel per metre height of chimney 0.2 * 400 * 1000 800 mm2 100 Hence pitch s of 12 mm Φ bar hoops 1000 * 113 141.0 mm 800 Provide these at 140 mm centre Page 4 of 16

W 25.00 * 67544 + 25.00 * 20295 + 25.00 * 98960 + 22.00 * 20295 + 22.00 * 128805 7950177 N P1 0.7 * 1800 ( 4.50 * 25.0 ) + 0.7 * 1600 ( 4.50 * 25.0 ) 0.7 * 1400 ( 4.50 * 22.00 ) 141750 + 126000 + 97020 364770 N.: M 141750 * 59.5 + 126000 * 34.5 + 97020 * 11.0 13848345 N. m.: Eccentricity e M 13848345 1.742 m 1742 mm W 7950177 For M 25 concrete, m 10.98.: Eqivalent area A Π/4 * ( 4.50 ^2-3.70 ^2 ) * 1000000 + ( 10.98-1 )* 58905 5740082 mm2 Eqivalent moment of inertia I ( Π/64 ) ( D 4 -d 4 ) + ( m-1 ) Π R ts (R) 2 Π * ( 4.50 ^ 4-3.70 ^ 4 ) * 1000 ^ 4 + 64 ( 10.98-1 ) * Π * 2100 * 4.46 * 2100 ^ 2 1.2225E+13 mm 4 For no tension to develop, allowable eccentricity 2 I 2 * 1.2225E+13 AD 5740082 * 4500 946.6 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 7950177 ± 13848345 * 1000 * 4500 5740082 2 * 1.2225E+13 1.385 2.549 ± Compressive stress 3.934 N/mm2 < 8.5 N/mm 2 allowable (Safe) Tensile stress -1.164 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 25.00 m and 50.00 m will not be applicable. We shall analyse the section for stresses by method discussed in 8.3. Tc 400 mm R 2.25-0.20 2.05 m Ts 4.46 mm eccentricity e 1.742 m m 10.98 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 + 4 2 (Tc-Ts) { sinφ + (Π-Φ) cosφ } mπ Ts cosφ [ ] fig 2 Page 5 of 16

e 205 * [ ( 40.00 - [ ( 40.00-0.45 ) Π-Φ * Π 0.45 ) * { sin2φ + 4 2 } + 10.98 * 0.45 2 * { sinφ + (Π-Φ) cosφ } + 10.98 * Π * 0.45 * cosφ ] ] Now adjust the value of angle Φ in such a way that the value of eccentricity e is > 1.742 m Assume Φ 86.00 º C 1.5010 radians.: e 6922.40 + 1578.44 8500.84 43.98 + 1.074 45.058 1.887 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 * c1 7950177 1 + 0.06976 Compressive stress 7950177 c1 in Concrete 1726917 { 0.9976 + 1.6406 * 0.0698 1726917 c1 [ ( 4.6037 400-4.46 ) * } + 10.98 * Π * 4.46 * 0.0698 ] 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 [ ] 43.96 N/mm2 < 140 N/mm 2 safe Horizontal steel (hoops) may be provided @ 0.2 % of sectional area Area of steel per metre height of chimney 0.2 * 400 * 1000 800 mm 2 100 Hence pitch s of 12 mm Φ bar hoops 1000 * 113 141.0 mm 800 Provide these at 140 mm centre As 113.1 mm 2 in pitch s 140 mm centre, if the cover is 40 mm then D1 4500-80 4420 mm.: t1 p * s 364770 * 140 2 * As * D1 2 * 113 * 4420 51.079 N/mm 2 < 140 N/mm 2 allowable Safe Page 6 of 16

(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 400-50 350 mm Cover to vertical steel 50 mm a 350 0.875 c1 4.6037 N/mm 2 400 Es 2.05E+05 N/mm 2 Ec 2.05E+05 1.867E+04 10.98 N/mm 2 p Ts 4.46 0.01116 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 * 0.8 160 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 160 400 71.11 º C * 400 + 5 * 100 To locate -neutral axis in the shell thickness, use Eq. 8.10 c1 [ - 1 ) * p ] 1 + ( m 0.5 * k 2 - m * p * (a - k) α * T * Ec.: 4.6037 * [ 1 + ( 10.98-1 ) * 0.01116 ) ] 1.1E-05 * 71.11 * 0.5 * k 2-10.98 * 0.01116 * ( 0.875 - k ) 1.867E+04 or 10.2329 14.6043 k 2 + 0.24509 * k - 0.21445 k 2 + 0.245 * k - 0.915 0 solving for k k 0.8419 Page 7 of 16

.: Compressive a * α * Tº * Ec c k * α * Tº * Ec Stress in Concrete a - k 1 + k 0.842 * 1.1E-05 * 71.11 * 1.867E+04 12.295 N/mm 2 Since wind stresses are taken into account, Permissible 4 * 8.5 Stress in Concrete 3 11.33 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 10.98 * 12.295 * ( 0.875-0.842 ) k 0.842 5.308 N/mm 2 (d) Stresses on windward side, due to temperature gradient p * t1 α * Tº * Ec m * p * ( a - k ) - 0.5 * k 2 where t1 43.956 N/mm 2 p 0.01116071 a 0.875 m 10.98 α 0.000011 Tº 71.11 º C Ec 18670 N/mm 2.: 0.01116 * 43.956 0 0.000011 * 71.11 10.98 * 0.01116 * ( 0.875 - k ) - 0.5 * k 2 * 18670 or 0.490581772 0.107226563-0.122544643 k - 0.5 * k 2 14.604 0.10723-0.12254 k - 0.5 * k 2 0.0336 solving k 0.2803.: Compressive c α * Tº * Ec * k 4.0937 N/mm 2 Stress in Concrete Tensile stress in Steel, assuming concrete to be fully cracked. fig 5 t m c a - k 10.98 * 4.0937 * ( 0.875-0.2803 ) k 0.2803 95.36 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 10.98 p 0.01116 a 0.875 α 0.000011 Tº 71.11 º C Ec 18670 N/mm 2 or k 2-10.98 * 0.01116 + 2 * 10.98 * 0.01116 * 0.875 + 10.98 * 10.98 * 0.01116 * 0.01116 Page 8 of 16

k 2 0.35648596.: Compressive c 2 α * Tº * Ec * k 2 5.2062 N/mm 2 Stress in Concrete Tensile stress in concrete, assuming concrete to be fully cracked. t 2 m c 2 a - k 2 10.98 * 5.2062 * ( 0.875-0.35649 ) 0.35649 k 2 83.15 N/mm 2 < 140 N/mm 2 ( safe ) (b) Stress in horizontal reinforcement due to temperature : p' A Φ 113.10 0.00202 S T c 140 * 400 a' 360 0.900 400 From Eq. 8.13. k' -mp' + 2mp'a + m 2 p' 2 or k' - 10.98 * 0.00202 + 2 * 10.98 * 0.00202 * 0.900 + 10.98 * 10.98 * 0.00202 * 0.00202 k' 0.17883981.: Compressive c' α * Tº * Ec * k' 2.6118 N/mm 2 Stress in Concrete Tensile stress in concrete, assuming concrete to be fully cracked. t 2 m c' a' - k' 10.98 * 2.6118 * ( 0.900-0.17884 ) k' 0.17884 115.64 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 115.64 + 51.079 166.72 N/mm 2 Since wind is also acting, permissible t 4 * 140 186.67 N/mm 2 Safe allowable tensile stress in steel 3 Page 9 of 16

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 7950177 N M 13848345 N. m V M V 364770 N 0.00 level Page 10 of 16

7. Design of Cirrcullar Chimney Foundation : e 364.77 1742 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 7950.18 Kn Moment M 13848.345 Kn. M D eccentricity e M/P 1.742 m Horizontal load H 364.77 Kn Outer dia of chimney d 4.50 m Thickness of chimney wall t 400 mm Dia of Footing OD 14.00 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 32 25 mm 14000 Reinforcement cover c 75 mm OD 200 1700 2300 P 7950.18 Soil filling inside 4000 Totd Axial load at the base of footing 14000 OD 7950.18 P' P + Weight of Chimney Wall + Soil Filling inside of wall + Weight of soil + Self weight of footing 7950.18 + Π ( 4.50 ^2-4.10 ^2 ) 4 * ( 1.70 + 0.2 ) * 25 + Π ( 4.10 ^2 ) * 18 4 fig 7 + Π ( 14.00 ^2-4.50 ^2 ) * 18 * 1.70 4 + Π ( 14.00 ^2 ) * 2.30 * 25 4 7950.18 + 128.33 + 237.65 + 4223.83 + 8851.44 21391.43 kn M' 13848.345 + H * ( D + T + A ) 13848.345 + 364.77 * ( 2.30 + 1.70 + 0.2 ) 15380.38 Kn. M.: e' M' 15380.38 0.7190 < [ OD 1.75 m P' 21391.43 8 Ok ] Axx Π * OD 2 Ixx Π * OD 4 1885.74099 m 4 4 64 153.94 m 2 Zxx Π * OD 3 269.39157 m 3 32 Page 11 of 16

The maximum and minimum base pressures are given by σ P' ± M' 21391.43 15380.38 153.94 269.39 ± 138.961 57.093 ± 81.87 A Zxx fig 8 196.05 σ max 196.05 Kn / m2 < 200 Kn / m2 allowable Ok σ min 81.87 Kn / m2 > 0 Kn / m2 allowable Ok Factor of Safety against overturning Stabilising Moment Overturning Moment P' * OD 2 M' 21391.43 * 14.00 2 15380.38 9.74 > 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 0.0524 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' 139.0 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' 1.15028 m2 CG of Segment 'PQRS' from 'PQ' 2.782 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

C / L of foundation 2250 2500 2250 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 16 Φ @ 200 c/c 1 900 2300 1112.5 75 cover critical punching shear section 2225 critical shear Section C / L 7000 Section A - A fig 10.: Moment at 'PQ' Mf 503.33 + 70.56 573.89 Kn.m fy 415 N/mm2.: fyall 230 N/mm2 m 10.98 fc' 25 N/mm2.: fc'all 8.5 N/mm2.: k 10.98 * 8.5 0.289 10.98 * 8.5 + 230 j 1 - k 1-0.289 0.90378 3 3 R 1 fc'all j k 1 * 8.5 * 0.904 * 2 2 1.109 0.289 Hence d Mf 573.89 * 1000000 2096.08 mm 1000 * R 118 * 1.109.: adopt T 2300 mm cover 75 mm d 2300-75.: d 2225 mm effective depth As M 573887856 1241 Fyall * j * d 230 * 0.90378 * 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 16 Φ @ 200 c/c p% 1295 * 100 0.494 % distribution steel 118 * 2225 Page 13 of 16

Punching Shear : Check Punching shear at ro + d/2 from the c/l r 2250 mm d 1112.5 mm r + d 2 2 Φ 3.00 º 0.0524 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' 139.0 Kn / m2 57.1 Kn / m2 27.4 Kn / m2 Area of Segment 'PQa1a2' CG of Segment 'PQa1a2' from 'a1a2' 0.98682 m2 2.032 m.: Punching Shear at 'a1a2' F 164.19 + 14.64 178.83 Kn Allowable Punching Shear stress 0.16 fck 0.16 * 25 0.8 N/mm2.: Depth required for punching shear do F 178831 0.16 fck * 176 0.8 * 176 1270 < 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 2250 4475 4750 5500 Reinforcement Spacing mm 116 234 248 286 326 366 Effective depth mm 2225 2225 2225 1758 1292 825 Bar dia mm 32 + 25 32 + 25 32 + 25 32 + 16 32 + 16 32 + 0 As mm 2 1295 1295 1295 1005 1005 804 p% 0.5018 0.2488 0.2347 0.1999 0.2387 0.2664 not Uniform pressure kn/m2 139.0 139.0 139.0 139.0 139.0 139.0 Pressure due to moment @ rs kn/m2 57.1 57.1 57.1 57.1 Area of Segment 'PQxx' m2 1.150 0.759 0.692 0.491 M actual / bar Kn.m 575 194 157 73 19 0 M allowable / bar Kn.m 599 599 599 367 270 138 p% for Shear 0.50 0.25 0.25 0.25 0.25 0.25 Shear Stress tc N/mm2 0.31 0.23 0.23 0.23 Shear actual Kn 181 133 123 90 Shear - Vs per main bar Kn 101 13 1 1 Shear - MS bar dia fyall 140 12 12 1 1 1 spacing mm 349 2638 245 193 142 91 Minimum shear s 2.5Asvfy/b mm 341 169 1 1.: Provided spacing mm 175 175 not 6250 7000 57.1 57.1 Pressure due to moment at section kn/m2 18.35 36.50 38.74 44.86 50.98 57.09 0.260 0.000 CG of Segment 'PQxx' m 2.79 1.36 1.20 0.78 0.38 0.00 0.23 0.23 Shear allowable Kn 80 120 127 116 97 69 49 0 Shear Reinf Reqd Reqd Not Reqd Not Reqd Not Reqd Not Reqd 1 1 1 1 1 Page 14 of 16

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 2982693 mm2 Moment of Inertia of Top portion of Chimney 6.9073E+12 mm4 Wind Moment at the base of Top Portion 1.7719E+06 N.m Modulus of Elasticity of Concrete 2.8500E+04 N/mm2 M / Ei 9.0008E-09 1/mm Area of M / Ei of top portion 1.1251E-04 C.g of Area of M / Ei of top portion 1.6667E+04 mm Moment of Area of M / Ei from top portion 1.8752E+00 (1) Partial Deflection of Top Portion δtop 1.875 mm wrt bottom of top portion Ratio L / δ L / 13332 > L / 200 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 4365997 mm2 Moment of Inertia of Middle portion of Chimney 9.6716E+12 mm4 Wind Moment at the base of Middle Portion 6.8906E+06 N.m Modulus of Elasticity of Concrete 2.8500E+04 N/mm2 M / Ei 2.4999E-08 1/mm Area of M / Ei of Middle portion 4.2499E-04 C.g of Area of M / Ei of Middle portion from top 3.7500E+04 mm Moment of Area of M / Ei of Middle portion 1.7812E+01 (2) Partial Deflection of Top Portion δtop 17.812 mm wrt bottom of middle portion Ratio L / δ L / 2807 < L / 200 Page 15 of 16

Bottom Portion : Height of Bottom portion of Chimney 22.00 m Wind intensity of Bottom portion of Chimney 1400 N/m2 Area of Bottom Portion of chimney 5740082 mm2 Moment of Inertia of Bottom portion of Chimney 1.2225E+13 mm4 Wind Moment at the base of Bottom Portion 1.3848E+07 N.m Modulus of Elasticity of Concrete 2.8500E+04 N/mm2 M / Ei 3.9746E-08 1/mm Area of M / Ei of Bottom portion 7.1219E-04 C.g of Area of M / Ei of Bottom portion from top 6.1000E+04 mm Moment of Area of M / Ei of Bottom portion 6.1256E+01 (3) Total Deflection of Top Portion δtop 61.256 mm wrt bottom of bottom portion Ratio L / δ L / 1175 < L / 200 Page 16 of 16