Final report

An Internship Report on
DESIGN AND DETAILING OF BOX CULVERT
Submitted in partial fulfillment for the award of the degree of
Master of Technology
In
Structural Engineering
Submitted By:
SUMEET DILIP DIVATAGI
USN: 1BI15CSE15
Internship Carried Out
at
STUP CONSULTANTS PVT. LTD
5th & 6th floor, Golden Enclave, Old Airport Road, Bengaluru-560017
Department of Civil Engineering
Bangalore Institute of Technology
K.R. Road, V.V. Puram Bengaluru- 560004
2016-17
INTERNAL GUIDES:
Mr. Madhan. S
Dr. P. M. Ravindra
Bangalore Institute of
Technology
EXTERNAL GUIDES:
Mr. Prabhanandan K
(Associate Principal Manager-Design)
Mr. Ashok Kumar G
(Senior Design Engineer)
STUP CONSULTANTS PVT. LTD.

BANGALORE INSTITUTE OF TECHNOLOGY
K. R. ROAD, V.V. PURAM, BENGALURU-560004
DEPARTMENT OF CIVIL ENGINEERING
(Post Graduate Studies)
Certificate
This is to certify that this internship report has been successfully carried out by SUMEET
DILIP DIVATAGI bearing USN: 1BI15CSE15 in partial fulfillment of the requirements for the
award of Master of Technology in Structural Engineering from Visvesvaraya Technological
University, Belagavi during the year 2016-2017. The internship report has been approved as it
satisfies the academic requirements in respect of internship work prescribed for the Masters of
Technology.
Examiners:
Name Signature
1.
2.
Mr. Madhan. S
(Asst. professor, Guide)
Dr. A. G. Nataraj
Principal, BIT
Dr. Aswath. M. U.
H.O.D
Department of Civil Engineering,
BIT
Dr. P. M. Ravindra
Co-ordinator, P.G. Studies

BANGALORE INSTITUTE OF TECHNOLOGY
K. R. ROAD, V.V. PURAM, BENGALURU-560004
DEPARTMENT OF CIVIL ENGINEERING
(Post Graduate Studies)
DECLARATION
I, the undersigned declare that this internship report is bonafide work carried out by me
during 2016-17 in partial fulfillment of the requirements for the award of Post-Graduation Degree of
Master of Technology in Structural Engineering of Visvesvaraya Technological University, Belagavi
and is based on the internship carried out in STUP CONSULTANTS PVT. LTD. Bengaluru under
the guidance of Mr. Madhan S, Asst. Professor and Dr. P.M. Ravindra, Professor, Department of
Civil Engineering, Bangalore Institute of Technology, Bengaluru and Mr. Prabhanandan K,
Associate Principal Manager, and Mr. Ashok Kumar G, Sr. Design Engineer, STUP consultants
Pvt. Ltd, Bengaluru.
I also declare that this internship report has not been submitted to any other University or Institute
for the award of any degree.
SUMEET DILIP DIVATAGI
USN: 1BI15CSE15
M. Tech (Structural Engineering)
Bangalore Institute of Technology
Bengaluru

ACKNOWLEDGEMENT
I express my gratitude to the Director of STUP CONSULTANTS PVT. LTD, Mr. A.
T. Samuel and the Management Team of STUP CONSULTANTS PVT. LTD. for providing an
opportunity to work as an intern in this deemed organization and their guidance throughout the
period of internship.
I express my sincere thanks to our internship guides, Mr. Prabhanandan K, Associate
Principal Manager-Design, and Mr. Ashok Kumar G, Senior Design Engineer for giving us an
insight about the Structural and Water Resource and Irrigation Design Industry and sharing their
knowledge and experiences in carrying out our design project in their busy schedule, without their
guidance and support my internship would not have been completed successfully.
I am also thankful to all the technical and non-technical staff of STUP CONSULTANTS
PVT. LTD, who have directly or indirectly helped me and supported me during my internship
program,
I’m grateful to Dr. A. G. Nataraj, Principal, Bangalore Institute of Technology, Prof. Dr.
P.M Ravindra, Professor & Coordinator- PG Studies, Department of Civil Engineering,
Bangalore Institute of Technology, and all the other faculties of Civil Engineering Department of
Bangalore Institute of Technology, Bengaluru, for their generous guidance, help and useful
suggestions.
I would like to place on record my deep sense of gratitude to Prof. Dr. Aswath M.U.,
Head of the Department, Department of Civil Engineering, Bangalore Institute of Technology,
Bengaluru for his extended support, generous guidance and encouragement for all our endeavors.
I would like to place on record my deep sense of gratitude to my internal guide Mr.
Madhan S, Asst. Professor Department of Civil Engineering, Bangalore Institute of Technology,
Bengaluru for his extended support, generous guidance and encouragement for all our endeavors.

TABLE OF CONTENTS
Certificate………………………………………………………………………………………i
Completion Certificate……………………………………………………………………….ii
Declaration……………………………………………………………………………………iii
Acknowledgement…………………………………………………………………………….iv
Table of Contents…………….………………………………………………………………v
List of Tables…………………………………………………………………………………viii
List of Figures………………………………………………………………………………..ix
Notations……………………………………………………………………………………...xi
Objectives of Internship……………………………………………………………………xiii
About the company………………………………………………………………………….xiv
CHAPTER 01: INTRODUCTION 01-02
1.0.Design and Detailing of Box Culvert 02
CHAPTER 02: HYDROLOGY 03-12
2.1. Hydraulic Particulars of the canal 04
2.2. Nalla Particulars 04
2.3. Calculation of Catchment Area 05
2.3.1. Grid Method 05
2.3.2. Planimeter 05
2.3.3. AutoCAD 06
2.4. Design Discharge Calculations 06
2.4.1. Empirical Formula Methods 06
2.4.1.1. Dicken's Formula 06
2.4.1.2. Ryve's Formula 06
2.4.1.3. Ingli’s Formula 07
2.4.2. Rational Formula 07
2.4.3. Modified Rational Formula 08
2.4.4. Area-Velocity Method 09
2.4.5. Conclusions 11
CHAPTER 03: HYDRAULICS 13-15
3.1. Vent Way Requirements 14
3.2. Scour Depth Calculations 15
3.3. Calculation of Afflux 16

CHAPTER 04: LOAD CALCULATIONS 16-33
4.1. Design Data 17
4.2. Load Calculations 19
4.2.1. Dead Load 19
4.2.2. Super Imposed Dead Load 19
4.2.3. Earth Pressure 20
4.2.4. Live Load Surcharge 20
4.2.5. Live Load 21
4.2.5.1. Class 70R Wheeled 21
4.2.5.2. Class 70R Maximum Bogie Load 25
4.2.5.3. Class 70R Tracked 27
4.2.5.4. Class A Single Lane 28
4.2.5.5. Class A Double Lane 31
CHAPTER 05: STRUCTURAL ANALYSIS OF BOX CULVERT 34-41
5.1. Design Section Forces 35
5.2. Combination of Loads for Limit State Design 39
CHAPTER 06: STRESS-BLOCK PARAMETERS 42-46
6.1. Calculation of Stress Block Parameters 43
CHAPTER 07: STRUCTURAL DESIGN OF BOX CULVERT 47-67
7.1. Center Wall Design 49
7.2. Typical long hand calculation for Top Slab Section 3 54
7.2.1. Ultimate Limit State 54
7.2.1.1. Flexural Design 54
7.2.1.2. Check for Shear 55
7.2.2. Serviceability Limit State 59
7.2.2.1. Permissible Stress Check 59
7.2.2.2. Check for Crack Width 60
7.3. Check for Bearing Pressure 64
7.3.1. Permanent Loads 65
7.3.2. Live Load 66
7.3.3. Pressure Calculations 67
CHAPTER 08: DESIGN OF WING WALL 68-122
8.0. Design of wing wall- data 69

8.1. Section 1-1 71
8.1.1. Dimensions of Section 1-1 71
8.1.2. Ultimate Limit State of Strength- Basic Combination 74
8.1.3. Limit State of Serviceability- Rare combination 82
8.1.4. Limit State of Serviceability- Quasi Permanent Combination 87
8.2. Section 2-2 92
8.3. Section 3-3 108
CHAPTER 09: CONCLUSIONS 123-124
ANNEXURE-I 125
ANNEXURE-II 149
REFERENCES 153

LIST OF TABLES
Table 2.1: Computation by Equivalent Slope Method 09
Table 2.2: Cross section at Box culvert site 10
Table 2.3: Design Discharge 11
Table 5.1: Load combination for Ultimate Limit State of Strength 39
Table 5.2: load Combination for Limit State of Serviceability 40
Table 5.3: Design forces from STAAD Pro. 41
Table 7.1: Design of sections for flexure- U.L.S 48
Table 7.2: Design of sections for Shear- U.L.S 50
Table 7.3: Check for maximum stress- S.L.S 51
Table 7.4: Check for crack width- S.L.S 52
Table 7.5: Check for Deflection- S.L.S 53

LIST OF FIGURES
Fig. 1.1: Location of Structure 02
Fig. 2.1: Trial Pit 04
Fig. 2.2: Catchment area 05
Fig. 2.3: Longitudinal section of Nalla 10
Fig. 2.4: Cross section at box culvert site 11
Fig. 3.1: Length of Barrel 15
Fig. 4.1: Dimensions of Box Culvert 17
Fig. 4.2: Earth Pressure 20
Fig. 4.3: Class 70R Wheeled 21
Fig. 4.4: Wheel arrangement- 70R Wheeled 22
Fig. 4.5: Dispersion of 70R Wheeled 22
Fig. 4.6: Class 70R Wheeled- Case 01 Dispersion 23
Fig. 4.9: Class 70R max bogie load 25
Fig. 4.10: Class 70R max bogie load- Case 01 Dispersion 25
Fig. 4.13: Class 70R Tracked- Wheel Configuration 27
Fig. 4.14: Class A Single lane- Wheel Configuration 28
Fig. 4.15: Class A Single lane- Case 01 Dispersion 29
Fig. 4.18: Class A Double lane- Case 01 Dispersion 31
Fig. 5.1: Box Culvert sections 35
Fig. 5.2: STAAD Model dimensions 35
Fig. 5.3: Node Numbers 36
Fig. 5.4: Beam Numbers 36
Fig. 5.5: Bending Moment Diagram due to Dead load 37
Fig. 5.6: Bending Moment Diagram due to SIDL 37

Fig. 5.7: Bending Moment Diagram due to Earth Pressure 38
Fig. 5.8: Bending Moment Diagram due to Live Load Surcharge 38
Fig. 5.9: Bending Moment Diagram due to Live Load (Class A 2 Lane) 39
Fig. 6.1: Stress Block Parameters 43
Fig. 6.2: Stress Block Parameters- values 43
Fig. 6.3: Stress Block Parameters- Balanced section 44
Fig. 7.1: Effective tension area 62
Fig. 7.2: Plan of Culvert 64
Fig. 7.3: Longitudinal section of Box Culvert 64
Fig. 7.4: Cross section of Box Culvert 64
Fig. 7.5: Live Load eccentricity 66
Fig. 8.1: Dimension nomenclature of Retaining wall 70
Fig. 8.2: Section 1-1 Dimensions 72
Fig. 8.3: Section 1-1 –Forces acting on stem- Basic combination 76
Fig. 8.4: Section 1-1 –Upward bearing pressure for footing- Basic Combination 80
Fig. 8.5: Section 1-1 –Forces acting on stem- Rare combination 83
Fig. 8.6: Section 1-1 –Upward bearing pressure for footing- Rare Combination 85
Fig. 8.7: Section 1-1 –Forces acting on stem- Quasi Permanent 88
Fig. 8.8: Section 1-1 –Upward bearing pressure for footing- Quasi Permanent 90

NOTATIONS
LATIN UPPER CASE LETTERS
A = Cross sectional area
Ac = Cross sectional area of concrete
As = Cross sectional area of reinforcement
Asw = Cross sectional area of shear reinforcement
As min = Minimum cross sectional area of reinforcement
As pro = Cross sectional area of reinforcement provided
D = Overall depth of cross section
Ec = Tangent modulus of elasticity of normal weight concrete at a stress
of σc=0
Ec eff = Effective modulus of elasticity of concrete
Es = Effective modulus of elasticity of steel
FOS = Factor of safety
Icr = Cracked moment of inertia of concrete section
M = Bending moment
MR = Resisting moment
MO = Overturning moment
NEd. = Design value of the applied axial force (tension or compression)
Pa = Active earth pressure
Pah = Horizontal component of active earth pressure
Pav = Vertical component of active earth pressure
S = Spacing
Sr max = Maximum crack spacing
SLS = Serviceability limit state
ULS = Ultimate limit state
V = Shear force
VEd. = Design value of the applied shear force
VRd.c = Design shear resistance
Wk = Crack width
Z = Sectional modulus

LATIN LOWER CASE LETTERS
bw = Width of the web
d = effective depth of the member
e = Eccentricity
fcd = Design value of concrete compressive strength
fck = Characteristic compressive cube strength of concrete at 28 days
fy = Yield strength of reinforcement
fctm = Mean value of axial tensile strength of concrete
h = Overall depth of cross section
kt = factor dependent on the duration of load
lo = Clear height of compression member between end restraints
xu = Neutral axis depth
z = Lever arm of internal forces
GREEK LOWER CASE LETTERS
σsc = Tensile stress in steel
σc = Compressive stress in concrete
σcp = Compressive stress in concrete from axial load
α = Angle; Ratio
β = Angle; Ratio; Coefficient
θ = Angle
Ꜫc = Compressive strain in concrete
Ꜫcu = Ultimate compressive strain in concrete
Ꜫs = Ultimate tensile strain in steel
μ = Coefficient of friction
ρ1 = Reinforcement ratio for longitudinal reinforcement
ρw = Reinforcement ratio for shear reinforcement
ϕ = Diameter of reinforcing bar
δ = Increment/Redistribution ratio
γm = Partial factors for a material property, taking account only of
uncertainties in the material property
ν = Strength reduction factor for concrete cracked in shear
Ꜫsm = Mean strain in the reinforcement
Ꜫcm = Mean strain in the concrete between cracks

OBJECTIVES OF INTERNSHIP
 Bridge gap between academics and industry
 Applicability of academics in industry
 To know the work flow.
 To learn the designs thoroughly.

ABOUT THE COMPANY
INTRODUCTION
 STUP is a full service project delivery consultancy company offering integrated planning,
architectural, engineering and project management services for transportation, marine, water,
power, telecommunications, commercial, institutional, recreational and manufacturing
facility infrastructure, and is an international firm with over 1200 professionals in more than
20 offices and global project locations.
 STUP, a French acronym for “Societe Technique pour l’Utilisation de la Precontrainte”
meaning “technical corporation for the utilization of prestressed concrete”
 STUP has served over 10,000 clients in 37 countries on projects of tremendous diversity
 Established in Paris in 1944 to spread knowledge of prestressed concrete and other inventions
of Mr. Eugene Freyssinet
 First global office was established by Mr. Yves Guyon
 STUP Consultants Pvt. Ltd. ("STUP") was established in India in 1963 and had been inspired
& led by C R Alimchandani for five decades.
 It has offices/served clients in: Afghanistan, Algeria, Bahrain, Bangladesh, Bhutan, Brunei,
Cambodia, Cyprus, France, Ghana, India, Indonesia, Iran, Iraq, Jordan, Kuwait, Laos, Libya,
Malaysia, Maldives, Nepal, Oman, Papua New Guinea, Philippines, Qatar, Russia, Sri Lanka,
Tanzania, U.A.E., United States, Vietnam, and Yemen.
 In India: Mumbai, Navi Mumbai, Bangalore, Chennai, Hyderabad, Kolkata, Delhi, Pune,
Ahmedabad
FIELD OF EXPERTISE
 AIRPORTS & AVIATION
 Master planning
 Airside Infrastructure
 Landside Infrastructure
 Runway Infrastructure & Taxiway
 Terminal Buildings
 ATC Towers
 Aircraft Manufacturing & Maintenance Unit
 Hangers & Maintenance Factory
 Maintenance Block
 Catering & Cargo Buildings

 URBAN, RURAL AND INDUSTRIAL DEVELOPMENT
 Master Planning & Urban Design
 Airports
 Corporate Headquarters & Commercial Complex
 High-tech Parks (IT, Bio-tech, Pharmaceutical, Apparels)
 Hospitality : Hotels & Resorts
 Universities & Institutes
 Industrial
 Residential & Mixed Use
 Healthcare & Hospitals
 SEZ and Integrated Townships
 Leisure & Sports
 Entertainment, Convention Centers & Retail
 Signature Public Buildings
 Interiors
 ENERGY, TELECOMMUNICATION AND SPACE INFRASTRUCTURE
 Containment for Nuclear Reactor Buildings
 Thermal & Hydro-electric Power Projects
 Thermal & Structural Design of Natural Draught Cooling Towers
 Thermal & Structural Design of Induced Draught Cooling Towers
 Functional & Structural Design of Tall Chimneys
 Cryogenic Tanks for Storage of LNG
 Special structure like tall pylons for supporting boilers etc.
 Material Conveyance Structures
 Structural and Civil Engineering for Energy related projects
 Water Intake and Circulation System
 ENVIRONMENTAL AND PUBLIC HEALTH ENGINEERING
 Water resources studies including design of systems
 Process design of water treatment and desalinization
 Collection, treatment and disposal of sewage, industrial effluent and solid waste
 Drainage Network and Discharge
 Specialized techniques for reservoir construction
 Environmental Consultancy Services

 ROADS, HIGHWAYS, EXPRESSWAYS
 Socio-techno-economic Feasibility and
 Traffic Studies
 Prioritization and Master plans
 Road Design, Strengthening, Widening and Expansion
 Urban and Rural Roads
 Expressways and Elevated Roads
 Flyovers and Interchange Systems
 Road Bridges
 Underpass/ Box-Pushing/ Tunneling
 Road Maintenance and Bridge Rehabilitation
 BRIDGES & FLYOVERS
 Cable Stayed Bridges
 Extra-dosed Bridges
 Suspension Bridges
 Segmental - Precast (Box) / Insitu (Box)
 Cantilever Construction / Balanced Cantilever
 Steel Girder Bridges-Through Type / Composite Deck Type Bridge/ Under Slung
 Arch Bridges
 Rail Cum Road Bridges
 Interchanges / Flyovers /T-Beam - Insitu / Precast T-Beam
 Incremental Launching / Nose Launching
 METROS
 Elevated Viaduct
 Elevated Station
 Underground Station
 Tunnel
 Underground Crossovers
 RAILWAYS
 Trackwork
 Railway Crossing Structures, Railway Station Building, Railway Plants and other
Infrastructure
 Railway Bridges
 Dedicated Freight Corridor

 OFFSHORE, HARBOR AND COASTAL ENGINEERING
 Ports and Harbor’s
 Mooring and Berthing Structures
 Jetties and Break Waters
 Ship lifts, Slipways and Dry Docks
 Offshore Yards
 Intake and Outfall
 Cargo Handling
 LPG / LNG / POL / Dry Bulk / Crude Oil Terminals
 Navigation Aids
 Rehabilitation of Marine Structures
 WATER RESOURCES AND AGRICULTURAL DEVELOPMENT
 Major and Minor Irrigation Projects & Command Area Development
 Aqueducts, Syphons, Canals and Canal Regulatory Works
 Intake Structures, Tunnels, Surge Shafts, Penstocks and Power Houses
 Engineering of Barrages, Major Dams and Irrigation Tanks
 Lift Irrigation Schemes
 Water Distribution Systems
 Water Resources Consolidation
 Flood Control
 Evaluation of the Safety of Dams
 Modernization of Canals
 CONSTRUCTION ENGINEERING, PROJECT MANAGEMENT AND TECHNOLOGY
TRANSFER
 Airport Projects
 Urban Infrastructure
 Building Design & Integrated Engineering
 Energy, Telecommunication and Space Infrastructure Projects
 Environmental and Public Health Engineering Projects
 Major Structures (Bridges & Flyovers)
 Highways (Roads, Highways & Expressways) / IE Engineering
 Metros & Railways
 Marine Projects
 Rehabilitation Projects

 Water Resources Projects
 Lender’s Engineer
 REHABILITATION OF STRUCTURES AND HERITAGE BUILDINGS
 Inspection and Surveys
 Tests (Destructive and Non-destructive)
 Rehabilitation Studies
 Restoration Studies
 Rehabilitation Schemes
 Restoration Schemes
 Residual Life Estimation
CLIENTS
Funding Agencies
 Asian Development Bank (ADB)
 African Development Bank (AFDB)
 World Bank (WB)
 Japan Bank of International Cooperation (JBIC)
 International Bank of Reconstruction and Development (IBRD)
 United Nations Development Programme (UNDP)
 World Health Organization (WHO)
 Department for International Development, UK (DFID)
 Kuwait Fund for Arab Economic Development (KFAED)
Government Bodies
 Govt. of United States
 Govt. of Marshall Island
 Sultanate of Oman
 Govt. of Laos PDR
 Govt. of Vietnam
 Govt. of Brunei
 Govt. of Iraq
 Govt. of U. A. E.
 Govt. of India
 Govt. of Ghana
 Govt. of Qatar

 Govt. of Malaysia
 Govt. of Indonesia
 Govt. of Bhutan
 Govt. of Kuwait
 Govt. of Algeria
 Govt. of Bangladesh
Contractors & Developers
 Sadbhav Engineering Ltd.
 Simplex Infrastructures Ltd.
 Essel Infrastructures Group
 Afcons Infrastructure Limited
 Innovative Technical Solutions Inc. (ITSI)
 Bechtel
 Degremont
 Alsthom
 Dumez
 Galfar
 Ideal Road Builders
 Gammon India Limited
 Larsen & Toubro Limited
 Consolidated Contractors Company (CCC)
 Six Construct
 Emaar
 Hindustan Construction Company (HCC)
Corporations
 Aeroport de Paris Ingenieurs
 Cognizant Software
 Marriot Hotels
 Reliance
 Kuwait Airways Corporation
 Hyatt Hotels & Resorts
 Birla Brothers
 Indian Oil Corporation
 Sterlite

 Oil and Natural Gas Company Limited
 Nuclear Power Corporation of India
 Ministry of Roads Transport and Highways
 National Highways Authority of India
 Central Public Works Department
 Ghaziabad Development Authority GDA)
 Thane Municipal Corporation (TMC)
 Municipal Corporation of Greater Mumbai (MCGM)
 Mumbai Metropolitan Region Development Authority (MMRDA)
EXTERNAL GUIDES:
1. Mr. Prabhanandan K
M.E. (Structures)
Associate Principal Manager (Design)
Experience: 17 years
2. Mr. Ashok Kumar. G.
M. Tech (Water Resource a& Hydrology)
Senior Design Engineer
Experience: 14 years

Design and Detailing of Box Culvert
Department Of Civil Engineering, BIT Page 1
CHAPTER 01
INTRODUCTION

1.0 DESIGN AND DETAILING OF BOX CULVERT
The Upper Krishna project constitutes of two dams:
 Almatti Dam
 Narayanpura Dam
Krishna Bhagya Jala Nigam Limited is implementing several lift irrigation scheme on the
Krishna basin to lift water and irrigate drought prone northern Karnataka districts.
Mulwad Lift Irrigation Scheme is taken on foreshore of Almatti reservoir:
 Scheme A consists of Stage I and Stage II required irrigating 30,850 hectares of lands up to
contour RL 560.00m and these works are already completed.
 Stage III required to irrigate 2,27,966 hectares of land up to contour RL 640.00m and the
work is in progress
Huvina Hipparagi Branch Canal
 The Stage III of MLIP is to lift water from RL 560m to RL 640m.
 It is the 3rd
lift at RL 560m and is called the Bijapur Main Canal.
Huvinu Hipparagi Branch Canal takes off from the Bijapur Main Canal at chainage 11.070km
and runs for a length of 63.88kms to irrigate about 23,676 hectares with discharge of 13.152 cumecs
at chainage 0.00 km.
A natural stream (nalla) crosses the canal at chainage 55.680km for which box culvert is
proposed.
Location: Longitude 76˚8’19.33” Latitude 16˚22’49.615”
Fig. 1.1: Location of Structure

CHAPTER 02
HYDROLOGY

2.1. HYDRAULIC PARTICULARS OF THE CANAL
Ground level = 577.794 m
Canal bed level = 580.928 m
Height of bed filing = 3.134 m
Design discharge in canal = 2.790 m3
/s
Bed width = 1.450 m
Full supply depth = 1.300 m
Free board = 0.450 m
Side slope = 1.5: 1
Bed fall = 1 in 5000
Velocity in trough = 0.631 m/s
Top width of canal at FSL = 5.350 m
Top width of canal at FBL = 6.700 m
Top width of canal at GL = 6.700 m
Lining thickness of canal = 0.080 m
Rear side slope = 1.5:1
Service road width = 5.500 m
Inspection path width = 3.000 m
2.2. NALLA PARTICULARS
Lowest nalla bed level = 577.794 m
Observed high flood level = 579.212 m
Width of nalla = 25.000 m
Trial Pit Details
All kinds of soil = 3.200 m
Soft Rock = 0.000 m
Hard Rock = 0.000 m
Fig. 2.1: Trial Pit

2.3. CALCULATION OF CATCHMENT AREA
2.3.1. Grid Method
Fig. 2.2: Catchment Area
No. of full squares = 85
No. of three quarter squares = 15
No. of half squares = 10
No. of quarter squares = 7
Total no. of squares = (85 × 1) + (15 ×3
4 ) + (10 ×1
2 ) + (7 ×1
4 )
= 103
Scale 1 cm = 15000 cm
1 cm = 0.150 cm
1 cm2
= 0.023 km2
Area = 103 x 0.023
= 2.318 km2
2.3.2. Planimeter
Least count of drum = 100 cm2
Least count of 1 division = 1 cm2
Least count of 1 vernier division = 0.1 cm2
Scale 1:15000
Box Culvert at
Chainage 55.680 km

No. of times the zero mark passes
the fixes dial (N) = 1
Initial reading (I R) = 0
Final reading (F R) = 4
Coinciding vernier division = 5
Area = (N ×LC Drum + (FR - IP) ×LC Div +Vernier ×LCVD)
= 104.500 cm2
Area to scale = Planimeter area × scale
= 2.351 km2
2.3.3. AutoCAD
The area calculated in AutoCAD= 2.338 km2
CONCLUSION
The area of catchment for further calculations = 2.351 km2
2.4. DESIGN DISCHARGE CALCULATIONS
2.4.1 EMPIRICAL FORMULA METHOD
2.4.1.1. Dicken's Formula
Q = C × M 3/4
(Cl. 4.2, IRC SP: 13-2004)
Q = Discharge in m3
/s
C = Dicken's Constant
= 11 - 14 where the annual rainfall is 60 - 120 cm
= 14 - 19 where the annual rainfall more than 120 cm
= 22 in Western Ghats
M = Catchment area km2
Q = 11 × 2.351 3/4
= 20.887 m3
/s
2.4.1.2. Ryve's Formula
Q = C × M 2/3
(Cl. 4.3, IRC SP: 13-2004)
Q = Discharge in m3
/s
C = Ryve's Constant
= 6.8 for areas within 25 km of the coast
= 8.5 for areas between 25 km and 160 km of the coast
= 15 for this case (Krishna River Basin) CWC Manual
= 10 for limited areas near the hills

Q = 15 × 2.351 2/3
= 26.523 m3
/s
2.4.1.3. Ingli's Formula
Q =
×
√
(Cl. 4.4, IRC SP: 13-2004)
Q = Discharge in m3
/s
Q =
× .
√ .
= 83.628 m3
/s
2.4.2. RATIONAL FORMULA
Q = λ × I0 × A (Cl. 4.7.9, Eq. 4.14, IRC SP: 13-2004)
λ =
. × ×
(Cl. 4.7.9, Eq. 4.14a, IRC SP: 13-2004)
tc =
. ×
.
(Cl. 4.7.5.2, Eq. 4.9, IRC:SP:13-2004)
Q = Discharge in m3
/s
λ = coefficient of runoff
f = fraction of rainfall
P = coefficient of runoff for catchment area
tc = time of concentration
L = distance from critical point to the structure (km)
H = the fall in level from critical point to the structure (m)
A = area in hectares
L = 1.680 km (Contour Map)
H = (601.000 - 577.794)
= 23.206 m (Contour Map)
A = 235.125 ha
F = 0.990 (fig 4.2, IRC-SP 13)
P = 0.600 (black cotton soil, Table 4.1 IRC-SP 13)
tc = 0.514 hrs
λ = 0.022
As per figure 6.2, page 44 of Flood Estimation Methods for Catchment Less than 25 km2,
Bridge and Flood Wings Report No. RBF – 16, Ministry of Railway, Government of India,
Ratio =

As per Plate 17, Atlas of State wise Genralised Isopluvial Maps of Southern India,
Indian Meteorological Department September 2007
50 year 24 hour rainfall = 200 mm
0.245 =
I0 = 4.9 cm/hr
Q = 0.022 × 4.9 × 235.125 = 25.308 m3
/s
2.4.3. MODIFIED RATIONAL FORMULA
This method is as per Flood Estimation Methods for Catchment Less than 25 km2, Bridge and
Flood Wings Report No. RBF – 16, Ministry of Railway, Government of India.
Q50 = 0.278 × C × I50 × A
Q50 = 50 year return flood peak m3
/s
C = Runoff coefficient
I50 = 50 year rainfall intensity (mm/hr) lasting for tc hour duration, where tc is the
time of concentration.
A = Catchment area in km2
= 2.351 km2
Runoff Coefficient [C]
From table 6.1,
C = 0.415 × (R ×F) 0.2
(Silt)
R = 50 year 24 hour point rainfall in cm
F = Areal reduction factor depending upon area and duration of rainfall
From table 6.2, for tc = 30.86 minutes and for catchment area less than 2.5 km2
F = 0.81
R = 20 cm from 50 years 24 hours Isopluvial map
C = 0.415 × (20 ×0.81)0.2
= 0.7244
Rainfall Intensity (I50)
Ratio = (Figure 6.2)
As per Plate 17, Atlas of State wise Genralised Isopluvial Maps of Southern India,
Indian Meteorological Department September 2007
50 year 24 hour rainfall = 200
0.245 =
I = 49 mm/hr

Q = 0.278 × 0.7244 × 49 × 2.35125
Q = 23.20 m3/s
2.4.4. AREA-VELOCITY METHOD
Calculation of Bed Slope
Table 2.1: Computation by Equivalent Slope Method
Chainage Distance Length
Lowest
NBL
Triangular
Area
Rectangular
Area
∑ of Area
m m m m m2 m2 m2
Up
stream of
Canal
100 0 0 580.792 - - -
80 20 20 580.454 3.380 29.080 32.460
60 20 40 580.280 1.740 25.600 27.340
40 20 60 580.034 2.460 20.680 23.140
30 10 70 580.634 -3.000 16.340 13.340
25 5 75 580.388 0.615 6.940 7.555
20 5 80 579.066 3.305 0.330 3.635
15 5 85 578.099 2.417 -4.505 -2.087
10 5 90 578.308 -0.522 -3.460 -3.982
5 5 95 578.692 -0.960 -1.540 -2.500
Center 0 5 100 577.794 2.245 -6.030 -3.785
Down
Stream
of Canal
-5 5 105 579.550 -4.390 2.750 -1.640
-10 5 110 579.129 1.052 0.645 1.697
-15 5 115 579.330 -0.503 1.650 1.148
-20 5 120 579.654 -0.810 3.270 2.460
-25 5 125 579.951 -0.743 4.755 4.013
-30 5 130 579.761 0.475 3.805 4.280
-40 10 140 579.016 3.725 0.160 3.885
-60 20 160 579.000 0.160 0.000 0.160
-80 20 180 579.000 0.000 0.000 0.000
-100 20 200 579.000 0.000 0.000 0.000
Total Area, A = 111.117
Level Difference, H =
×
= 1.111 m Fall = H/L = 0.006, i.e. = 1 in 180

Fig. 2.3: Longitudinal Section of Nalla
Cross Section
Table 2.2: Cross Section
Sl. No
HFL
(m)
Survey Data
Differenc
e in Bed
Level (m)
Breadth
of
Flow(m)
Depth
of
Flow
(m)
Area
(m2)
Wetted
Perimeter
(m)
Distanc
e (m)
Bed
Level
(m)
1 579.212 30 580.500 0.000 0 0.000 0.000 0.000
2 579.212 25 580.049 0.000 0 0.000 0.000 0.000
3 579.212 20 579.598 0.000 0 0.000 0.000 0.000
4 579.212 15 579.147 0.089 5 0.044 0.222 5.000
5 579.212 10 578.696 0.540 5 0.314 1.572 5.002
6 579.212 5 578.245 0.991 5 0.765 3.827 5.005
7 579.212 0 577.794 1.442 5 1.216 6.082 5.005
8 579.212 -5 578.245 0.991 5 1.216 6.082 5.000
9 579.212 -10 578.696 0.540 5 0.765 3.827 5.005
10 579.212 -15 579.147 0.089 5 0.314 1.572 5.005
11 579.212 -20 579.598 0.000 0 0.044 0.000 0.000
12 579.212 -25 580.049 0.000 0 0.000 0.000 0.000
13 579.212 -30 580.500 0.000 0 0.000 0.000 0.000
TOTAL = 23.187 35.022
577.500
578.000
578.500
579.000
579.500
580.000
580.500
581.000
0 20 40 60 80 100 120 140 160 180 200 220
NallaBedlevel(m)
Length (m)
LONGITUDINAL SECTION
L/S

Fig. 2.4: Cross Section
Cross sectional Area, A = 23.187 m2
Wetted perimeter, P = 35.022 m
Hydraulic mean radius, R = 0.662 m
Slope, S =
Velocity, V = × R1/3
× S1/2
=
.
× 0.6621/3
×
1
180
1/2
= 1.618 m/s
Discharge, Q = A × V = 37.511 m3
/s
2.4.5. CONCLUSIONS
Table 2.3: Design Discharge
Sl. No. Method
Discharge
(m3/s)
Remark
1 Dicken's 20.63 -
2 Ryve's 26.23 Madras Presidency
3 Ingli's 83.83 Bombay Presidency
4 Rational 25.31 -
5 Modified Rational 23.20 As per RBF 16
6 Area Velocity 37.51 -
From above Ingli’s formula is yielding more discharge, since it is used in Western Ghats
(Bombay Presidency) and it is comparatively high with respect to other empirical formula, hence it is
neglected.
577.500
578.000
578.500
579.000
579.500
580.000
580.500
581.000
-40 -30 -20 -10 0 10 20 30 40
ReducedLevel(m)
Chainage (m)
CROSS-SECTION
BEDLEVEL
HFL

As per article 6.2.1 page 21 of IRC:SP 13-2004, the maximum flood discharge to be
adopted for design should be higher of the above values as design discharge Q, provided it does not
exceed the next highest discharge by more than 50%.
As per above clause,
First maximum discharge = 37.51 m3
/s
Second maximum discharge = 26.23 m3
/s
Design flood discharge Q,
should not exceed = 1.5 × 26.23 = 39.345 m3
/s
From the above table,
Design flood discharge, Q = 37.51 m3
/s is adopted from area velocity method.

CHAPTER 03
HYDRALICS

3.1. VENT WAY REQUIREMENTS
Design flood discharge = 37.510 m3
/s
Observed high flood level = 579.212 m
Lowest nalla bed level = 577.794 m
Depth of water in nalla = 1.418 m
Canal bed level = 580.928 m
Depth below CBL,
i.e. available vent height = 2.654 m
Maximum allowable velocity = 2.700 m/s (Cl. 8.8.5, Pg 6, IS 10430-2000)
Area of flow required =
Q
V
(Q = A× V)
=
37.51
2.7
= 13.893 m2
Providing vent height = 2.654 m
Vent width required = 5.235 m
Say vent width required = 3 m in 2 Nos.
Nalla width at crossing = 25 m
Area of vent provide = 2 × 3 × 2.654
= 15.924 m2
Total area of flow provided is more than required, Hence OK
∴ Provide two vent of 3 m width x 2.654 m depth box culvert. Also, provide splayed wing
walls with returns on either side of the vents since the width of nalla at crossing is greater than the
vent way.
Check for velocity =
.
.
= 2.356 m/s
The velocity in the vent is less than the allowable maximum permissible limit, hence safe
Wetted perimeter of vents when full = 22.616 m
Hydraulic mean radius = 0.704
Longitudinal slope = 1 in 440
The longitudinal slope of culvert floor is flatter; hence make up the slope to 1:100.

3.2. SCOUR DEPTH CALCULATION
Normal scour depth, D =
. × /
/ (Clause 7.5, IRC SP: 13-2004)
Ksf = 0.35 (For silt, table 7.1, IRC SP: 13)
D =
. × . /
. /
= 2.247 m
Maximum scour depth = 1.27 x D (Cl. 10.4, IRC SP: 13– 2004)
= 2.854 m
Maximum scour level = H F L - Maximum scour depth
= 579.212 – 2.854
= 576.358 m
Depth of soft rock,
below nalla bed level = 3.2 m
Scour level = Nalla bed level – Top of soft rock
= 577.794 – 3.2
= 574.594 m
Hence provide cut off wall up to RL 574.594 m below lowest nalla be level.
Length of Barrel
Fig. 3.1: Length of Barrel
Width of head wall = 0.300m
FBL = 582.678m
RL of head wall = 581.748m
Side slope = 1.5:1
Banking width = (FBL – RL of head wall) × 1.5
= (582.678 -581.748) × 1.5
= 1.400m (one side)
Width of inspection path = 3.000m
Width of service road = 5.500m

Top width of canal = 6.700m
Barrel length = 2 × 0.3 + 2 × 1.4 + 3 + 5.5 + 6.7
= 18.600m
3.3. CALCULATION OF AFFLUX
Calculation of afflux is as per cl. 8.4.4.2 of IS 7784 (Part 1): 1993
h = [
.
+ 0.01524] × [ – 1]
A2
= c/s area before construction
= 13.893 m2
(from Cl. 3.1, pg. 14)
a2
= c/s area after construction
= (2.654 x 3.00) x 2
= 15.924 m2
= [
.
.
+ 0.01524] × [
.
.
– 1]
= -0.078 < 0
Hence no afflux
Top of Vent = Average Bed Level + Vent Height + Afflux +
Top Slab Thickness
Top of Vent = 577.794 + 2.654 + 0.000 + 0.400
= 580.848 m

CHAPTER 4
LOAD CALCULATIONS

4.1. DESIGN DATA
I. Box Details
1. Box clear width = 3.000 m
2. Box clear height = 2.654 m
3. Box barrel length = 18.600 m
4. No. of cell = 2.000 Nos.
5. Bottom slab thickness = 0.450 m
6. Top slab thickness = 0.400 m
7. Wall thickness = 0.400 m
8. Wall thickness (intermediate) = 0.200 m
9. Wall height (including slabs) = 3.504 m
10. Haunch horizontal (Bottom slab) = 0.600 m
11. Haunch vertical (Bottom slab) = 0.200 m
12. Haunch horizontal (Top slab) = 0.600 m
13. Haunch vertical (top slab) = 0.200 m
14. Height of soil on box = 1.830 m
Fig. 4.1: Dimensions of Box Culvert
II) MATERIALS
Grade of Concrete = M-25
Grade of Reinforcing Steel = Fe-500
III) DURABILITY (As per IRC: 112-2011)
Condition of exposure = Moderate(Cl.14.3.1 Table 14.1/ pg. 141)
Clear Cover = 75 mm (Cl.14.3.2.1 Table 14.2/ Note 7)
Minimum grade of Concrete = M-25 (Moderate condition)

IV) DESIGN PARAMATERS FOR RCC DESIGN
a) Reinforcement (Cl. 6.2 of IRC: 112-2011)
Grade of Steel = Fe-500
Characteristic Strength of Steel (fy) = 500 N/mm2
Material Factor (ϒs) = 1.15
Modulus of Elasticity (Es) = 200000N/mm2
b) Concrete (Cl. 6.4 of IRC: 112-2011)
Grade of Concrete = M-25
Characteristic Strength of Concrete (fck) = 25 N/mm2
Material Factor (ϒs) = 1.50
Coefficient of Friction (μ) = 0.50
Modulus of Elasticity (Ec) = 25000 N/mm2
Design value considered (0.446*fck) = 11.15 N/mm2
c) Constants
Modular ratio m =
Es
Ec
(1 + φ)
= 20.8
V) SOIL DATA AS PER SOIL TEST REPORT
1. Saturated density of soil γs = 20.000 kN/m3
2. Angle of internal friction of soil Φ = 30.000˚
3. Angle of wall friction δ = x Φ = 20.000 ˚
4. Angle which earth surface makes
with horizontal β = 0.000 ˚
5. Wall inclination to backfill α = 90.000 ˚
6. Co-efficient of earth pressure ko = 1-sin Φ = 0.500
7. Soil bearing capacity = 200.000 kN/m2
VI) REFERENCE CODES
IRC: 6-2014 Standard Specifications and Code of Practice for Road Bridges,
Section: II Loads and Stresses
IRC: 112-2011 Design Criteria for Concrete Road Bridges
IRC: 78-2014 Standard Specifications and Code of Practice for Road Bridges,
Section: VII Foundations and Substructures

4.2. LOAD CALCULATIONS
4.2.1. DEAD LOAD (Cl. 203, pg. 5, IRC: 6-2014)
Volume of top slab = 6.6 x 0.40 x 1
= 2.8 m3
Volume of bottom slab = 6.6 x 0.45 x 1
= 3.15 m3
Volume of side walls = 2 x 3.08 x 0.4 x 1
= 2.123 m3
Volume of center wall = 3.08 x 0.2 x 1
= 0.531 m3
Total volume = 8.690 m3
Therefore, total weight of concrete = 8.690 x 25
= 217.250 kN
Effective width = 0.2 + 3 + 0.2 + 3 + 0.2
= 6.600 m
∴ Base pressure due to self-weight =
217.25
6.6 × 1
= 32.91 kN/m
4.2.2. SUPER IMPOSED DEAD LOAD
a) At soil section
Soil depth = (FBL – CBL) + canal lining
= (582.678 - 580.925) + 0.08
= 1.83 m
Therefore, weight of soil on top of box = (1.83 x 20)
= 36.6 kN/m2
Therefore, base pressure due to soil weight = (36.6 x 1)
= 36.6 kN/m
b) At canal section:-
Depth of water = FBL – CBL
= 582.678 - 580.928
= 1.75 m
Therefore, weight of water = 1.75 x 10 x 1
= 17.5 kN/m2
Depth of canal lining = 0.08 m
Therefore, weight of canal lining = 0.08 x 25 x 1

= 2.0 kN/m2
Therefore, total weight at canal section = 17.5 + 2
= 19.5 kN/m2
Base pressure at canal section = 19.5 x 1
= 19.5 kN/m
4.2.3. EARTH PRESSURE (Cl. 214, pg. 41, IRC: 6-2014)
Fig. 4.2: Earth Pressure
Earth pressure at mid depth of top slab = k0 x γ x h
Earth pressure at rest k0 = 1- sin (ϕ)
= 1- sin (30)
= 0.5
At mid depth of top slab = 0.5 x 20 x (1.83 +
0.4
2
)
= 0.5 x 20 x 2.03
= 20.3 kN/m
At mid depth of bottom slab = 0.5 x 20 x (1.83+0.4+2.654+
0.45
2
)
= 0.5 x 20 x 5.109
= 50.28 kN/m
4.2.4. LIVE LOAD SURCHARGE
As per Cl. 214.1, IRC: 6-2014,
Surcharge due to live load equivalent to 1.2m
earth fill = 0.5 x 20 x 1.2
= 12.0 kN/m2
Top Slab
3.079
LLSEarth
PtressureBottom Slab

4.2.5. LIVE LOADS
4.2.5.1. CLASS 70R WHEELED
Fig. 4.3. Class 70R (Wheeled)
Maximum possible wheel load is in case of maximum
Single axle load = 5000 kg
Maximum tyre pressure = 5.273 kg/cm2
(Fig. 1, IRC: 6-2014)
Contact area =
5000
5.273
= 948.227 cm2
For 70R wheeled, tyre width = 41 cm
(Fig. 1, IRC: 6-2014)
For 70R wheeled, thread width = (41 – 5) = 36 cm
(Note 3, Annex A, IRC: 6-2014)
Contact length =
.
= 26.34 cm
Consider type "L" tyres:-
From Fig. 1, IRC: 6-2014, we have,
Diameter of tyre = 0.61 m
Effective tyre width = 0.86 m
Spacing between tyres = 0.86 – 2 x 0.41
= 0.04 m
Effective thread width = 2 x 0.36 + 0.04
= 0.76 m
Over all axle length = 2.79 m
Effective axle length = 2.79 – 0.76
= 1.93 m

Fig 4.4: Wheel Arrangement- 70R Wheeled
Dispersion angle = 45˚
(Cl. B3.4, Annex B3, IRC: 112-2011)
Dispersion dimension along road = 0.263 + 2 x 1.83
= 3.923 m
Dispersion dimension across road = (2.79 + 2 x 1.93) = 6.350 m
Fig 4.5: Dispersion of Load- 70R Wheeled
Therefore, Intensity =
Load × Impact factor
Dispersion area
Impact factor = 1.25
(Cl.208.3.a, IRC: 6-2014)
LOAD
(tonnes)
INTENSITY
(kN/m2
)
17.0 8.5
12.0 6.0
8.0 4.0

Case 1: Load on entry span only, first two axles of 17t concentrically placed on the span
Fig. 4.6: Class 70R (Wheeled) Case 1 Dispersion
Upward Bearing pressure σ =
P
±
Pe
P
A
=
. × . . × .
. ×
= 8.82 kN/m2
Z =
1×6.62
6
= 7.26 m3
Pe
z
=
( . × . × . ) – ( . × . × . )
.
=
-78.02
7.26
= -10.76 kN/m2
σmax = 8.82 + 10.76 = 19.57 kN/m2
σmin = 8.82 – 10.76 = -1.93 kN/m2
Case 2: Load on central wall, 2nd and 3rd axels placed equidistant from the central wall
Fig 4.7: Class 70R (Wheeled) Case 2 Dispersion

P
A
=
2×8.5×3.3+2×8.5×0.44+2×8.5×2.37
6.6×1
= 15.738 kN/m2
Z =
× .
= 7.260 m3
Pe
z
=
(8.5×2.37×2.115) – (8.5×3.74×1.43) + (8.5×2.37×2.115) + (8.5×3.74×1.43)
7.26
=
0
7.26
= 0 kN/m2
σmax = 15.74 + 0 = 15.74 kN/m2
σmin = 15.74 – 0 = 15.74 kN/m2
Case 3: The first two 17t axels placed concentrically on the second span
Fig. 4.8: Class 70R (Wheeled) Case 3 Dispersion
P
A
=
. × . . × . . × . . × .
6.6×1
= 15.738 kN/m2
Z =
× .
= 7.26 m3
Pe
z
=
(8.5×1.81×2.395) – (8.5×3.18×1.71) + (8.5×3.92×0.97) + (8.5×2.93×1.835)
7.26
=
-5.05
7.26
= -0.7 kN/m2
σmax = 15.25 + 0.7 = 15.95 kN/m2
σmin = 15.25 – 0.7 = 14.55 kN/m2

4.2.5.2. CLASS 70R MAXIMUM BOGIE LOAD
Fig. 4.9: Class 70R max bogie load
Intensity =
20×1.25
3.92×6.35
= 10.0 kN/m2
Case 1: Load on first span
Fig. 4.10: Class 70R max bogie load case 1 dispersion
P
A
=
10×3+10×3.92
6.6×1
= 10.50 kN/m2
Z =
1×6.62
6
= 7.26 m3
=
- (10×3×1.8)-(10×3.92×1.04)
7.26
=
-94.77
7.26
= -13.05 kN/m2
σmax = 10.5 + 13.05 = 23.535 kN/m2
σmin = 10.5 – 13.05 = -2.565 kN/m2

Case 2: Boogie placed concentrically on the center wall
P
A
=
× × .
. ×
= 11.88 kN/m2
Z =
1×6.62
6
= 7.26 m3
Pe
z
=
( × . × . ) ( × . × . )
7.26
=
0
7.26
= 0 kN/m2
σmax = 11.88 + 0 = 11.88 kN/m2
σmin = 11.88 – 0 = 11.88 kN/m2
Case 3: Load on second span

P
A
=
10×3.92+10×3
6.6×1
= 10.5 kN/m2
Z =
× .
= 7.26 m3
Pe
z
=
(10×3.92×1.04) + (10×3×1.8)
7.26
=
94.77
7.26
= 13.05 kN/m2
σmax = 10.5 + 13.05 = 23.55 kN/m2
σmin = 10.5 – 13.05 = -2.55 kN/m2
4.2.5.3. CLASS 70R TRACKED
Fig 4.13: Class 70R Tracked- Wheel Configuration
Dispersion along road = 4.57 + (2 x 1.83) = 8.23 m
Dispersion across road = 2.90 + (2 x 1.83) = 6.56 m
Intensity =
70 × 1.25
6.6 × 6.56
= 20.21 kN/mm2
P
A
=
20.21×6.6
6.6×1
= 20.21 kN/m2
Z =
× .
= 7.26 m3
Pe
z
=
-(20.21×3.3×1.65) + (20.21×3.3×1.65)
7.26
=
0
7.26
= 0 kN/m2
σmax = 20.21 + 0 = 20.21 kN/m2
σmin = 20.21 – 0 = 20.21 kN/m2

4.2.5.4. CLASS A SINGLE LANE
Fig. 4.14: Class A Single Lane- Wheel Configuration
Impact factor =
4.5
6+L
=
.
.
= 1.48

Axle
Load
(tonne)
Contact Area Dispersion Intensity
(kN/m2
)
B
(mm)
W
(mm)
Along
road(m)
Across
road(m)
Single
Lane
Double
Lane
11.4 250 500 3.91 5.96 7.25 14.50
6.8 200 380 3.86 5.84 4.50 9.00
2.7 150 200 3.81 5.66 1.85 3.70
Case 1: Two 11.4t axels placed equidistant from mid span of first span
Fig. 4.15: Class A Single Lane Case 1 dispersion
P
A
=
. × . . × . . × . × .
6.6×1
= 9.09 kN/m2
Z =
× .
= 7.26 m3
Pe
z
=
- (7.25×3.01×1.8) ( . × . × . ) ( . × . × . ) ( . × . × . )
7.26
=
-51.21
7.26
= -7.05 kN/m2
σmax = 9.09 + 7.05 = 16.14 kN/m2
σmin = 9.09 – 7.05 = 2.04 kN/m2

Case 2: The two 11.4t axels are placed equidistant from center support
P
A
=
. × . . × .
6.6×1
= 8.61 kN/m2
Z =
× .
= 7.26 m3
Pe
z
=
- (7.25×3.91×0.595) ( . × . × . )
7.26
=
0
7.26
= 0 kN/m2
σmax = 8.61 + 0 = 8.61 kN/m2
σmin = 8.61 – 0 = 8.61 kN/m2
Case 3: Two 11.4t axels placed equidistant from mid span of second span

P
A
=
. × . . × . . × .
6.6×1
= 8.952 kN/m2
Z =
× .
= 7.26 m3
Pe
z
=
( . × . × . ) (7.25×3.91×1.050) ( . × × . )
7.26
=
48.33
7.26
= 6.66 kN/m2
σmax = 8.952 + 6.66 = 15.61 kN/m2
σmin = 8.952 – 6.66 = 2.292 kN/m2
4.2.5.5. CLASS A DOUBLE LANE
Case 1: Two 22.8t axels placed equidistant from mid span of first span
Fig. 4.18: Class A Double Lane Case 1 dispersion
P
A
=
. × . . × . . × . × .
6.6×1
= 17.978 kN/m2
Z =
× .
= 7.26 m3
Pe
z
=
- (14.5×3.01×1.8) ( . × . × . ) ( . × . × . ) ( . × . × . )
7.26
=
-100.82
7.26
= -13.89 kN/m2
σmax = 17.978 + 13.887 = 31.865 kN/m2
σmin = 17.978 – 13.887 = 4.091 kN/m2

Case 2: The two 22.8t axels are placed equidistant from center support
P
A
=
. × . . × .
6.6×1
= 17.18 kN/m2
Z =
× .
= 7.26 m3
Pe
z
=
- (14.5×3.91×0.595) ( . × . × . )
7.26
=
0
7.26
= 0 kN/m2
σmax = 17.18 + 0 = 17.18 kN/m2
σmin = 17.81 – 0 = 17.18 kN/m2
Case 3: Two 22.8t axels placed equidistant from mid span of second span

P
A
=
× . . × . . × .
6.6×1
= 17.90 kN/m2
Z =
× .
= 7.260 m3
Pe
z
=
( × . × . ) (14.5×3.91×1.050) ( . × × . )
7.26
=
96.666
7.26
= 13.315 kN/m2
σmax = 17.903 + 13.315 = 31.218 kN/m2
σmin = 17.903 – 13.315 = 4.588 kN/m2

CHAPTER 5
STRUCTURAL ANALYSIS
OF
BOX CULVERT

The culvert is designed as a closed RCC structure. It is analyzed as plane frame of unit width
using standard STAAD.Pro software for DL+SIDL+EP+LL. The cross section is modeled with beam
members for 2D analysis. Since the bridge is resting on soil, the base slab is modeled considering
hinged support.
5.1. DESIGN SECTION FORCES:-
Section considered for design is as follows
Fig. 5.1: Sections
Fig. 5.2: STAAD Model Dimensions
1a 2a 3a
1 2 3
4
5
6

Fig. 5.3: Node Numbers
Fig. 5.4: Beam numbers

Fig. 5.5: Bending Moment Diagram due to Dead Load
Fig. 5.6: Bending Moment Diagram due to SIDL

Fig. 5.7: Bending Moment due to Lateral Earth Pressure
Fig. 5.8: Bending Moment due to Live Load Surcharge

Fig. 5.9: Bending Moment due to Live Load (CLASS A 2 Lane governing)
5.2. COMBINATION OF LOADS FOR LIMIT STATE DESIGN
a) Partial Safety Factor for verification of Structural Strength:
Only Basic Combination is applicable for the design of superstructure.
As per Amendment to IRC: 6-2014, Table 3.2, pg. 44
Table 5.1: Load combination for Ultimate Limit State of Strength
LOADS BASIC LOAD COMBINATION
Dead Load 1.35
Super Imposed Dead Load (SIDL) 1.35
Backfill Weight 1.50
Earth Pressure due to backfill
1.50 (Adding to Effect of Variable Load)
1.00 (Relieving to Effect of Variable Load)
Live Load Surcharge 1.20
Live Load 1.50
As per Cl 219.5.4 of IRC: 6, the additional earth pressure due to seismic need not be
considered.

b) Partial Safety Factor for verification of Serviceability Limit State:
As per Amendment to IRC: 6-2014, Table 3.3, pg. 46
Table 5.2: Load combination for Limit State of Serviceability
LOADS
RARE
COMBINATION
QUASI-
PERMANENT
Dead Load 1.00 1.00
Super Imposed Dead Load (SIDL) 1.00 1.00
Backfill Weight 1.00 1.00
Earth Pressure due to backfill 1.00 1.00
Live Load Surcharge 0.80 -
Live Load 1.00 -
As per Cl 219.5.4 of IRC: 6, the additional earth pressure due to seismic need not be
considered.
Rare Combination : To check for the stress limit in the member
Quasi-Permanent : To check for crack width and deflection in the member.

Table 5.3: Design Forces from STAAD Pro.
Member Section
Moment (kN-m)
Shear
Force
(kN)
Ultimate Moment
(Basic
Combination)
Serviceable
Moment (Rare
Combination)
Serviceable
Moment
(Quasi-
Permanent
Combination)
Top Slab
1 73.087 50.533 33.668 164.966
2 -67.6242 -46.6728 -25.714 -5.530
3 87.722 63.570 43.717 -173.138
Bottom Slab
1a -73.454 -50.356 -36.850 -186.531
2a 63.669 44.201 31.471 13.705
3a -113.576 -82.880 -60.580 196.619
Side Wall
4 73.547 49.432 33.670 -92.774
5 -44.103 -27.844 -23.170 -5.879
6 74.394 48.882 36.850 121.592
Center Wall
4a 2.745 0.000 0.000 -1.125
5a 0.000 0.000 0.000 -1.125
6a -0.720 0.000 0.000 -1.125

CHAPTER 06
STRESS-BLOCK PARAMETERS

6.1. CALCULATION OF STRESS BLOCK PARAMETER
Fig.6.1: Stress Block Parameters
Z = lever arm = (d – k2xu)
From similar triangles in strain diagram, we have
εcu
xu
=
εs
(d - xu)
εs =
(d - xu) × εcu
xu
εs xu + εcu xu = εcu d
εcu
εs+ εcu
=
εcu
εs+ εcu
d; where,
xu = neutral axis
d = effective depth of section
b = breadth of section
εcu = strain in concrete
εs = strain in steel
xu = depth of neutral axis in m
fck = grade of concrete in N/mm2
As per IRC: 112, εcu= 0.0035 and strain at which stress reaches design strength εo = 0.002
Fig.6.2: Stress Block Parameters-Values
εs
xu
C/S Strain diagram
Cu
Stress diagram
d
d-xu
εcu k1fck
xu
d-k2xu
Tu
d
xu
0.0035
d-0.42xu
0.42xu
Cu
Tu
0.446fck
d-xu
0.002

0.0035
xu
=
0.002
x1
x1 =
0.002 × xu
0.0035
x1 = 0.571 xu
=
4
7
xu
x2 = xu - x1
= xu - 0.571 xu
= 0.429 xu
=
3
7
xu
 Area of stress block,
A = A1+ A2
= (0.45 × fck × 0.429 × xu) + (2
3 × 0.45 × fck × 0.571 × xu)
A = 0.3645 × fck × xu
 Calculation of depth of Neutral Axis,
x =
ΣAi×xi
ΣAi
Σ (Aixi) = (2
3 × 0.45 × fck × 4
7 × xu) × (3
7 × xu + 3
8 × 4
7 × xu) +
0.45 × fck × 3
7 × xu × 3
7 ×
xu
2
= 0.1515 × fck × xu
2
x =
0.1515 × fck × xu
2
0.3645 × fck × xu
x = 0.42 xu
Centroid of compression force acts at a distance of 0.42 xu from compressive fiber.
Case 1: Balanced Section
In Balanced section, xu=xumax
Fig.6.3: Stress Block Parameters-Balanced section
xumax
0.0035
Z=d-0.42xu
0.42xu
Cu
d
Tu=0.87fyAst
0.446fck
d-xumax
0.002+ (0.87fy/Es)

At Equilibrium:
Cu = Tu
0.36×fck× xu max ×b = 0.87× fy ×Ast max
xu max =
0.87×fy×Ast max
0.36×fck×b
Dividing both sides by‘d’, we obtain
xu max
d
=
0.87×fy×Ast max
0.36×fck× b × d
But
Ast max
b×d
= pt max
pt max =
xu max
d
×
0.36×fck
0.87×fy
; where,
pt max = limiting percentage of steel
Applying initial triangles to strain diagram,
0.0035
xu max
=
0.002 +
0.87×fy
Ɛs
d - xu max
xu max
d
=
0.0035
0.0055+
0.87 ×fy
Ɛs
; where,
εs = 2×10 N/mm2
fy
xu max
d
250 0.53
415 0.48
500 0.46
Calculating Moment of Resistance:
Mu lim = Cu × Z
= 0.36×fck×xu max ×b× (d-0.42×xu max)
= 0.36×fck×
u max
d
×b× (d-0.42×
u max
d
) × d2
Case 2: Under Reinforced Section
In this section, tensile strain in steel attains its limiting value first and at this point the
strain in extreme compressive fiber is less than limiting strain.
εs < εcu
Neutral axis depth is obtained by equilibrium condition

0.36×fck× xu ×b = 0.87× fy ×Ast
xu =
0.87×fy×Ast
0.36×fck×b
…………………………………… (a)
Moment of Resistance
Mu = Tu x Z
= 0.87 × fy × Ast × (d - 0.42 xu)
= 0.87 × fy × Ast × (1 -
. u
d
) × d
From a,
u
d
=
0.87×fy×Ast
0.36×fck× b × d
Mu = 0.87 × fy × Ast × (1 -
0.42×2.417×fy×Ast
fck× b × d
) × d
= 0.87 × fy × Ast × (1 -
1.015×fy×Ast
fck× b × d
) × d
Mu = 0.87 × fy × Ast × (1 -
fy×Ast
fck× b × d
) × d

CHAPTER 07
STRUCTURAL DESIGN
OF
BOX CULVERT

Table 7.1: Design of sections for flexure (Ultimate Limit State of Strength)
Member Section
Moment
(kNm)
dreq
(mm)
D
(mm)
d
(mm)
Main Steel Distribution Steel
Ast
(mm2)
Astmin
(mm2)
Bar
Dia
(mm)
Spacingreq
(mm)
Spacingpr
(mm)
Ast(pr)
(mm2)
Astmin
(mm2)
Bar
Dia
(mm)
Spacing
(mm)
Top Slab
1 73.087 150 400 320 543.80 416 10 140 115 682.609 416 8 120
2 67.624 150 400 320 501.79 416 10 150 140 560.714 416 8 120
3 87.722 170 400 320 657.52 416 10 110 100 785.000 416 8 120
Bottom
Slab
1a 73.454 150 450 370 468.47 481 10 160 115 682.609 481 8 100
2a 63.669 140 450 370 404.63 481 10 160 130 603.846 481 8 100
3a 113.576 190 450 370 735.23 481 10 100 85 923.529 481 8 100
Side
Walls
4 73.547 150 400 320 547.34 416 10 140 115 682.609 416 8 120
5 44.103 120 400 320 323.54 416 10 180 150 523.333 416 8 120
6 74.394 150 400 320 553.88 416 10 140 115 682.609 416 8 120
Middle
Wall
4a 2.745 30 200 155 40.949 201.5 12 300 200 565.200 201.5 8 240
5a 0.000 0 200 155 0.000 201.5 12 300 200 565.200 201.5 8 240
6a 0.720 20 200 155 10.699 201.5 12 300 200 565.200 201.5 8 240
As per Cl. 7.6.4.1, pg. 57, IRC: 112-2011, axial force in side walls i.e. 221kN in Beam no. 5 and 172kN in Beam no. 7 is less than
0.1fcdAc = 0.1 x 11.15 x (0.4 x 1) = 446kN. In center walls the axial force (362kN) is exceeding 0.1fcdAc (223kN). Hence must be checked for
combined axial and bending compression member and is checked as per SP-16.

7.1. CENTER WALL DESIGN
Breadth of the column = 1000 mm
Overall depth of the column = 200 mm
Factored load Pu = 361.87 kN
Characteristic strength of concrete fck = 25 N/mm2
Characteristic strength of steel fy = 500 N/mm2
Clear height of compression member (lo) = 2654 mm
Effective length (0.7xlo) = 1857.8 mm
Factored moment Mu = 2.59 kNm
Assuming 10mm dia bars with 40mm clear cover
Effective cover d’
= (40 + (10/2))
= 45 mm
d’
/D = 0.23
Pu
fck bD
=
361.87
25 x 1000 x 200
= 0.07237
Mu
fckbD2 =
2.59
25 x 1000 x 2002
= 0.003
P
fck
= 0 (chart 38, SP-16)
Pt = 0
Minimum area of steel, Ast min (0.13*1000*155) = 201.5 mm2
As per Cl. 16.3.1, pg. 173, IRC: 112-2011,
 The diameter of bar should not be less than 12mm.
 The total area of the vertical reinforcement should be between 0.0024Ac and 0.04Ac outside
the locations of laps of vertical steel.
 This reinforcement should be provided at two faces taking into account the direct axial force
and biaxial bending, but shall not be less than 0.0012Ac on either face.
 The distance between two adjacent vertical bars shall not exceed 200.
∴ Provide 12mm dia bars at 200mm c/c
Area of steel provided, Ast pro =
π x 122
4
200
x 1000 = 565.416 mm2

Table 7.2: Design of sections for shear (Ultimate Limit State of Strength)
Member Section
Shear,
VNS
(kN)
Check for shear
z θ ρw Legs
Bar
Dia
(mm)
Asv
(mm2)
Spacing
(mm)ρ1 k VRd.c Requirement
Top Slab
1 164.966 0.002 1.791 118.842 Required 237.84 21.801 0.72 4 8 201.088 240.0
2 5.530 0.002 1.791 118.842 Not Required - - - 4 8 201.088 -
3 173.138 0.002 1.791 118.842 Required 232.83 21.801 0.72 4 8 201.088 240.0
Bottom
Slab
1 186.531 0.002 1.735 131.088 Required 280.73 21.801 0.72 4 8 201.088 270.0
2 13.705 0.002 1.735 131.088 Not Required - - - 4 8 201.088 -
3 196.619 0.002 1.735 131.088 Required 268.47 21.801 0.72 4 8 201.088 270.0
Side
Walls
4 92.774 0.002 1.791 118.842 Not Required - - - 4 8 201.088 -
5 5.879 0.002 1.791 118.842 Not Required - - - 4 8 201.088 -
6 121.592 0.002 1.791 118.842 Required 237.84 21.801 0.72 4 8 201.088 240.0
Middle
Wall
4 1.125 0.002 2.136 74.997 Not Required - - - 4 8 201.088 -
5 1.125 0.002 2.136 74.997 Not Required - - - 4 8 201.088 -
6 1.125 0.002 2.136 74.997 Not Required - - - 4 8 201.088 -

Table 7.3: Check for Serviceability (Maximum Stress)
Member Section
D
(mm)
Cover
(mm)
Dia
(mm)
Eff
Cover
(mm)
d (mm)
Ast
(mm2)
M
(kNm)
xu
(mm)
y (mm) Icr (mm4)
σsc
(N/mm2)
σc
(N/mm2)
Top Slab
1 400.000 75.000 10.000 80.000 320.000 682.609 50.533 82.160 237.840 9.880E+08 253.019 4.202
2 400.000 75.000 10.000 80.000 320.000 560.714 46.673 75.503 244.497 8.407E+08 282.344 4.192
3 400.000 75.000 10.000 80.000 320.000 785.000 63.570 87.170 232.830 1.106E+09 278.373 5.011
Bottom
Slab
1 450.000 75.000 10.000 80.000 370.000 682.609 50.356 89.265 280.735 1.356E+09 216.831 3.315
2 450.000 75.000 10.000 80.000 370.000 603.846 44.201 84.648 285.352 1.225E+09 214.183 3.055
3 450.000 75.000 10.000 80.000 370.000 923.529 82.880 101.528 268.472 1.733E+09 267.000 4.854
Side
Wall
4 400.000 75.000 10.000 80.000 320.000 682.609 49.432 82.160 237.840 9.880E+08 247.506 4.111
5 400.000 75.000 10.000 80.000 320.000 523.333 27.844 73.275 246.725 7.938E+08 180.017 2.570
6 400.000 75.000 10.000 80.000 320.000 682.609 48.882 82.160 237.840 9.880E+08 244.752 4.065
Middle
Wall
4 200.000 75.000 12.000 81.000 119.000 565.200 0.000 42.408 76.592 9.439E+07 0.000 0.000
5 200.000 75.000 12.000 81.000 119.000 565.200 0.000 42.408 76.592 9.439E+07 0.000 0.000
6 200.000 75.000 12.000 81.000 119.000 565.200 0.000 42.408 76.592 9.439E+07 0.000 0.000

Table 7.4: Check for Serviceability (Crack Width)
Member Section kt fct eff
hceff
(mm)
Aceff
(mm2)
ρpeff εsm-εcm k1 k2
Sr max
(mm)
Wk (mm)
Top Slab
1 0.5 2.9 105.95 105946.66 0.00644 0.000506 0.8 0.5 518.854 0.26
2 0.5 2.9 108.17 108165.74 0.00518 0.000467 0.8 0.5 582.942 0.27
3 0.5 2.9 104.28 104276.70 0.00753 0.000574 0.8 0.5 480.822 0.28
Bottom
Slab
1 0.5 2.9 120.24 120244.84 0.00568 0.000476 0.8 0.5 554.463 0.26
2 0.5 2.9 121.78 121784.05 0.00496 0.000457 0.8 0.5 597.857 0.27
3 0.5 2.9 116.16 116157.43 0.00795 0.000585 0.8 0.5 468.818 0.27
Side
Wall
4 0.5 2.9 105.95 105946.66 0.00644 0.000506 0.8 0.5 518.854 0.26
5 0.5 2.9 108.91 108908.23 0.00481 0.000449 0.8 0.5 608.778 0.27
6 0.5 2.9 105.95 105946.66 0.00644 0.000554 0.8 0.5 518.854 0.29
Middle
Wall
4 0.5 2.9 52.53 52530.74 0.01076 0.000000 0.8 0.5 444.601 0.00
5 0.5 2.9 52.53 52530.74 0.01076 0.000000 0.8 0.5 444.601 0.00
6 0.5 2.9 52.53 52530.74 0.01076 0.000000 0.8 0.5 444.601 0.00

CHECK FOR DEFLECTION
As per Cl.12.4.1, IRC: 112-2011,
Limiting values of deflection for vehicular loads =
Span
800
Table 7.5: Check for Serviceability (Deflection)
Member Span (m) Deflection (mm)
Permissible
deflection (mm)
Remark
Top Slab 3.300 0.660 4.125 OK
Bottom slab 3.300 0.362 4.125 OK
Hence OK

7.2. TYPICAL LONG HAND CALCULATION FOR TOP SLAB SECTION 3:
7.2.1. ULTIMATE LIMIT STATE
Ultimate moment Mu = 87.722 kN-m
Ultimate shear Vu = 173.138 kN
Depth required d required =
Mu
0.134 ×fck × b
=
. ∗
. ∗ ∗
= 161.69 mm
Diameter of the bar ϕ = 10 mm
Depth provided d provided = overall depth – clear cover –
ϕ
2
= 400 – 75 – 10
2
= 320 mm
∴ d provided > d required, hence OK.
Area of steel required Ast =
0.5×fck×b×d
fy
× [1 - 1-
4.6×Mu
fck×b×d2]
=
0.5×25×1000×320
500
× [1 - 1-
4.6×87.58×
25×1000×3202 ]
= 656.429 mm2
Minimum area of steel Ast min = 0.13% × b × d (Cl. 16.5.1.1, IRC: 112-2011)
=
0.13
100
× 1000 × 320
= 416 mm2
Spacing required = Least of
Area of one bar
Ast required
× 1000
2 × d
250
=
π × 102
4
656.429
× 1000
2 × 320
250
=
119.647
640
250
mm
∴ Spacing required = 119.66 mm
However provide spacing = 100 mm

Area of steel provided, Ast provided =
Area of one bar
Spacing provided
× 1000
=
π × 102
4
100
× 1000
= 785 mm2
Distribution Steel
Minimum area of steel Ast min = 0.13% × b × d (Cl. 16.5.1.1, IRC: 112-2011)
=
0.13
100
× 1000 × 320
= 416 mm2
Use diameter of bar = 8 mm
Spacing =
π × 82
4
416
× 1000
= 120.83 mm
Hence provide 8Ø @ 120 mm c/c
7.2.1.2 CHECK FOR SHEAR
As per Cl. 10.3.2, IRC: 112-2011, design shear resistance (VRd. c) must be greater than design
shear force acting at the section (VEd.)
VEd = 173.138 kN
VRd.c = [0.12×K×(80×ρ1×fck)0.33
+ 0.15×σcp]×b×d
VRd.c > vRd.c min
> (vmin + 0.15×σcp) ×b×d
> (0.031×K3/2
×fck
1/2
+ 0.15×σcp) ×b×d
Where,
K = 1 +
200
d
= 1 +
200
320
= 1.791
σcp =
NEd
Ac
< 0.2 fcd

=
0
Ac
= 0 (NEd = 0, no axial force)
ρ1 =
Ast
b×d
≤ 0.02
=
785
1000×320
= 0.00245
VRd.c = [0.12×1.791×(80×.00245×25)0.33
+0]×1000×320
= 116217 N
= 116.217 kN
vRd.c min = 0.031×1.7913/2
×251/2
×1000×320
= 118841.5 N
= 118.842 kN
VRd.c = vRd.c min
= 118.842 kN
VRd.c < VEd
Shear design required.
As per Eq. 10.8, IRC: 112-2011,
VRd.max= αcw×b×z×v1×
fcd
cot θ + tan θ
Where, αcw = 1 (Eq.10.9, IRC: 112-2011)
z = (d –xu)
Modular ratio, m =
E×s
Ec eff
Where,
Es = young's modulus of elasticity of steel in N/mm2
Ec eff = short term static modulus of elasticity of concrete in N/mm2
Ec eff =
Ec
1+ φ
=
5000× fck
1+ φ

=
5000×√25
1+ 1.6
= 9615.385 N/mm2
Modular ratio, m =
2 × 105
9615.385
= 20.8
Neutral axis 1000 ×
xu
2
2 = m × Ast × (d - xu)
1000 ×
xu
2
2 = 20.8 × 785 × (320 - xu)
xu = 87.17 mm
z = (320-87.17)
= 232.83 mm
v1 = 0.6 × 1 -
fck
310
(Eq.10.6, IRC: 112-2011)
= 0.6 × 1 -
25
310
= 0.6 × 0.919
= 0.5516
fcd = αcc ×
ck
γm
(Cl. 10.3.1, IRC: 112-2011)
= 0.67 ×
25
1.5
= 11.167 kN/mm2
173.138×103
= 1×1000×232.83×0.5516×
11.167
cot θ + tan θ
By trigonometric operations,
1
cot θ + tan θ
=
Sin 2θ
2
Sin 2θ =
173.138 × 103 × 2
1000 × 232.83 × 0.5516 × 11.167
= 0.242
2θ = Sin-1
(0.242)
θ =
.
2

θ = 6.986˚ < θmin (Cl.10.2.2. IRC: 112-2011)
∴ θ = 21.8˚
VRd.s =
Asw
s
×z×fywd×cot θ
Asw
s
=
. × 6
z×fywd×cot θ
=
. × 3
232.83×0.87× 500×cot 21.8
= 0.684
ρw =
Asw
s × b × sinα
(Eq. 16.4, IRC: 112-2011; α = 90°, vertical stirrups)
ρw min =
0.072× fck
fyk
(Eq. 16.5, IRC: 112-2011)
min =
0.072× fck ×b ×1
fyk
min =
0.072×√ ×1000 ×1
500
= 0.72 > 0.684
∴
Asw
s
= 0.72
Bar diameter = 8 mm
No. Legs = 4 Nos.
Asw = 4×
π × 82
4
= 201.06 mm2
s = Least of
0.75×d
= Least of
201.06
0.72
0.75×320
= Least of
279.289
240

= 240 mm
7.2.2. SERVICEABILITY LIMIT STATE
In serviceability limit state we check for:
 Permissible stress in concrete and steel for rare combination
 Crack width check for quasi permanent combination
 Deflection check
7.2.2.1 PERMISSIBLE STRESS CHECK
As per Cl. 12.2.1, pg. 120, IRC: 112-2011, the maximum compressive stress in concrete
under rare combinations of loads shall be limited to 0.48fck = 0.48 x 25
= 12.0 N/mm2
As per Cl. 12.2.2, pg. 120, IRC: 112-2011, the maximum tensile stress in steel under rare
combinations of loads shall be limited to 0.80fy = 0.8 x 500
= 400.00 N/mm2
We have,
Moment, M = 63.57 kNm
Modular ratio, m = 20.80
xu = 87.17 mm
To calculate cracked Moment of Inertia
Icr =
b×xu
3
12
+(A×h2
) + m × Ast × (d-xu)2
= [
1000×87.173
12
+1000×87.17×
. 2
]+20.8×785×(320-87.17)2
∴Icr = 1.106×109 mm4
Stress in Steel (σsc)
σsc =
63.57×106
1.106×109 × (320 – 87.17) × 20.80
= 278.373 N/mm2 < (Limiting σsc= 400N/mm2
)
Stress in Concrete
σc =
63.57×106
1.106×109 × 87.17
= 5.011 N/mm2 < (Limiting σc= 12 N/mm2
)
HENCE O.K

7.2.2.2 CRACK WIDTH CHECK
Serviceable moment M = 43.717 kN-m
Area of steel provided Ast = 785 mm2
Spacing provided S = 100 mm
Effective cover = 75+ 10
2 mm
= 80 mm
Modular ratio, m =
E×s
Ec eff
Where,
Es = young's modulus of elasticity of steel in N/mm2
Ec eff = short term static modulus of elasticity of concrete in N/mm2
Ec eff =
Ec
1+ φ
=
5000× fck
1+ φ
=
5000×√25
1+ 1.6
= 9615.385 N/mm2
Modular ratio, m =
2 × 105
9615.385
= 20.8
Neutral axis 1000 ×
xu
2
2 = m × Ast × (d - xu)
1000 ×
xu
2
2 = 20.8 × 785 × (320 - xu)
xu = 87.17 mm
Stress in reinforcement, σsc =
m × Mu × (d-xu)
I
Moment of inertia I = Ixx + Ah2
= [
1000×xu
3
12
+ 1000 × xu × xu
2
2 + m × Ast × (d-xu)]
= [
1000×87.173
12
+1000 × 87.17 × 87.172
2+20.8 × 785 ×(320 – 87.17)]
= 1.106 × 109 mm4
σsc =
20.8×43.717×106
×(320-87.21)
1.106 × 109
= 192.676 N/mm2

As per clause 12.3.4 of IRC: 112-2011,
Crack width, Wk = Sr max × (εsm - εcm)
Where,
Sr max = maximum crack spacing
εsm = mean strain in the reinforcement under the relevant combination of loads,
including the effect of imposed deformations, restrained thermal and shrinkage
effects and taking into account the effects of tension stiffening. For pre-
stressed members only the additional tensile strain beyond the state of zero
strain of the concrete at the same level is considered.
εcm = mean strain in concrete between cracks
εsm - εcm =
σsc -
kt × fct eff × (1+αe × ρp eff)
ρp eff
Es
≥
0.6×σsc
Es
(eq. 12.6, IRC:112-2011)
Where,
σsc = is the stress in the tension reinforcement assuming a cracked section
αe = m = 20.8
kt = factor dependent on the duration of the load which may be taken as 0.5
fct eff = is the mean of the tensile strength of the concrete effective at the time when
the cracks may first be expected to occur. In calculating the minimum
reinforcement to cater for shrinkage fcteff should be taken as the greater of 2.9
MPa or fctm (t).
= greater of
2.9
fctm (t)
fctm (t) = βcc (t)
α
× fctm (Eq. 6.7 of IRC:112-2011)
βcc (t) = exp S 1 − ⁄
(Eq. 6.3 of IRC:112-2011)
S = 0.25
t = age of concrete in days
t1 = 1 day
= exp 0.25 1 − 1⁄
= 1
fctm = 2.2 (Table 6.5 of IRC:112-2011)

fctm (t) = 1 × 2.2 = 2.2
fct eff = greater of
2.9
2.2
= 2.9
ρp eff = As
Ac eff
Ac eff = effective area of concrete in tension surrounding the reinforcement, of
depth hc eff
hc eff = Least of
⎩
⎪
⎨
⎪
⎧2.5×(h-d)
(h-xu)
3
h
2
= Least of
⎩
⎪
⎨
⎪
⎧2.5×(400-320)
(400-87.17)
3
400
2
= Least of
200
104.277
200
hc eff = 104.277 mm
Fig. 7.1: Effective tension area
Ac eff = hc eff × b
= 104.277 × 1000

= 104277 mm2
ρp eff = 785
104277
= 0.0075280
εsm - εcm =
192.676 -
0.5 × 2.9 × (1+20.8 ×0.00753)
0.00753
2 × 105
= -0.0001502
εsm - εcm =
0.6×192.676
2× 105
= 0.000578
εsm - εcm = Greater of -0.0001502
0.000578
= 0.000578
Sr max = 3.4×c +
0.425×k1×k2×φ
ρp eff
(Eq. 12.8, IRC: 112-2011)
Where,
c = clear cover
k1 = coefficient which takes account of the bond properties of the bonded
reinforcement
= 0.8 for deformed bars
= 1.6 for bars with an effectively plain surface
k2 = is a coefficient which takes into account of the distribution of strain
= 0.5 for bending
= 1.0 for pure tension
Φ = Diameter of bar
Sr max = 3.4×75 +
0.425×0.8×0.5×10
0.00753
= 480.822 mm
Wk = 480.822 × 0.000578
= 0.278 mm < 0.300 mm (Table 2.1, IRC: 112-2011)

7.3. CHECK FOR BEARING PRESSURE (Cl. 706, pg. 15, IRC: 78-2014)
Fig. 7.2: Plan of culvert
Fig. 7.3: Longitudinal Section of Box Culvert
Fig. 7.4: Cross Section of Box Culvert
X
Y

Width of culvert = 7 m
Length of barrel = 18.6 m
Section modulus:
Section modulus along x-direction Zx=
18.6×72
6
= 151.90 m3
Section modulus along y-direction Zy=
7×18.62
6
= 403.62 m3
7.3.1. PERMANENT LOADS
Item Description
Load (P)
(kN)
Eccentricity
(e) (m)
Pex Zy Pex/Zy
S1 0.5×1.4×0.93×7×20 91.000 -8.300 -755.300 403.620 -1.871
S2 1.4×0.82×7×20 161.000 -8.070 -1299.270 403.620 -3.219
S3 3×1.75×7×20 735.000 -6.100 -4483.500 403.620 -11.108
S4 0.5×2.625×1.75×7×20 322.000 -3.720 -1197.840 403.620 -2.968
W5 0.5×2.625×1.75×7×10 161.000 -2.850 -458.850 403.620 -1.137
W6 1.45×1.75×7×10 179.000 -1.250 -223.750 403.620 -0.554
W7 0.5×2.625×1.75×7×10 161.000 0.350 56.350 403.620 0.140
S8 0.5×2.625×1.75×7×20 322.000 1.220 392.840 403.620 0.973
S9 5.5×1.75×7×20 1347.500 4.850 6535.375 403.620 16.192
S10 1.4×0.82×7×20 161.000 8.070 1299.270 403.620 3.219
S11 0.5×1.4×0.93×7×20 91.000 8.300 755.300 403.620 1.871
S12 7×0.4×18.6×25 1302.000 0.000 0.000 403.620 0.000
S13 7×0.45×18.6×25 1465.000 0.000 0.000 403.620 0.000
S14 2×0.4×2.654×18.6×25 987.000 0.000 0.000 403.620 0.000
S15 0.2×2.654×18.6×25 247.000 0.000 0.000 403.620 0.000
W16 2×3×2.654×18.6×10 2962.000 0.000 0.000 403.620 0.000
Σ P = 10694.500 ΣPex/Zy = 1.538
P
A
=
10694.5
7×18.6
= 82.139 kN/m2

7.3.2. LIVE LOAD
Governing case for live load is Class A double lane
Fig. 7.5: Live Load eccentricity
Wheel No. P (kN) ex ey Pex Pex/zy Pey Pey/zx
1 114.000 2.625 3.250 299.250 0.741 370.500 2.439
2 114.000 4.425 3.250 504.450 1.250 370.500 2.439
3 114.000 5.275 3.250 601.350 1.490 370.500 2.439
4 114.000 7.075 3.250 806.550 1.998 370.500 2.439
5 114.000 2.625 2.050 299.250 0.741 233.700 1.539
6 114.000 4.425 2.050 504.450 1.250 233.700 1.539
7 114.000 5.275 2.050 601.350 1.490 233.700 1.539
8 114.000 7.075 2.050 806.550 1.998 233.700 1.539
9 27.000 2.625 -0.600 70.875 0.176 -16.200 -0.107
10 27.000 4.475 -0.600 120.825 0.299 -16.200 -0.107
11 27.000 5.225 -0.600 141.075 0.350 -16.200 -0.107
12 27.000 7.025 -0.600 189.675 0.470 -16.200 -0.107
13 27.000 2.625 -1.700 70.875 0.176 -45.900 -0.302
14 27.000 4.475 -1.700 120.825 0.299 -45.900 -0.302
15 27.000 5.225 -1.700 141.075 0.350 -45.900 -0.302
16 27.000 7.025 -1.700 189.675 0.470 -45.900 -0.302
ΣPex/Zy = 13.548 ΣPey/Zx = 14.275
X
Y

7.3.3 PRESSURE CALCULATIONS:
Case 1: Canal and culvert with full water
σ =
P
A
±
Pex
zy
±
Pey
zx
σ1 = 82.139 – (1.538+13.548) + 14.275
= 81.328 kN/m2
> 0 & < SBC = 200 kN/m2
σ2 = 82.139 + (1.538+13.548) + 14.275
= 111.500 kN/m2
> 0 & < SBC = 200 kN/m2
σ3 = 82.139 + (1.538+13.548) - 14.275
= 82.950 kN/m2
> 0 & < SBC = 200 kN/m2
σ4 = 82.139 - (1.538+13.548) - 14.275
= 52.778 kN/m2
> 0 & < SBC = 200 kN/m2
Case 2: Canal and culvert with no water
σ1 = 53.180 kN/m2
> 0 & < SBC = 200 kN/m2
σ2 = 86.454 kN/m2
> 0 & < SBC = 200 kN/m2
σ3 = 57.903 kN/m2
> 0 & < SBC = 200 kN/m2
σ4 = 24.629 kN/m2
> 0 & < SBC = 200 kN/m2
Case 3: Canal with full water and culvert with no water
σ1 = 58.579 kN/m2
> 0 & < SBC = 200 kN/m2
σ2 = 88.750 kN/m2
> 0 & < SBC = 200 kN/m2
σ3 = 60.200 kN/m2
> 0 & < SBC = 200 kN/m2
σ4 = 30.029 kN/m2
> 0 & < SBC = 200 kN/m2
Case 4: Canal with no water and culvert with full water
σ1 = 75.929 kN/m2
> 0 & < SBC = 200 kN/m2
σ2 = 109.203 kN/m2
> 0 & < SBC = 200 kN/m2
σ3 = 80.653 kN/m2
> 0 & < SBC = 200 kN/m2
σ4 = 47.379 kN/m2
> 0 & < SBC = 200 kN/m2

Department of Civil Engineering, B.I.T. Page 68
CHAPTER 08
DESIGN
OF
WING WALL

8.0 DESIGN OF WING WALL
 It is proposed to provide a Cantilever Retaining Wall for the Box Culvert as Wing Walls
 Limit State Method of Design as per IRC: 112-2011 is adopted with partial safety factors as
given in IRC: 6-2014
 For Stability check, factors as per IRC: 78 are used.
i) MATERIALS
Refer to pg. 16, chapter 4, Cl. 4.1 (II)
ii) DURABILITY
Refer to pg. 16, Chapter 4, Cl. 4.1. (III)
iii) DESIGN PARAMATERS FOR RCC DESIGN
a) Reinforcement (Cl. 6.2 of IRC: 112-2011)
Refer to pg. 17, Chapter 4, Cl. 4.1. (IVa)
b) Concrete (Cl. 6.4 of IRC: 112-2011)
Refer to pg. 17, Chapter 4, Cl. 4.1. (IVb)
iv) BACKFILL PROPOERTIES FOR DESIGN
Density of Compacted Backfill (ϒ) = 20 kN/m3
Angle of Internal Friction (φ) = 30 Deg
Angle between retaining wall & Backfill (α) = 0 Deg
Angle of Wall Friction (δ) = 20.00 Deg
Co-efficient of Active Earth Pressure (ka)
-For Infinite Backfill
Slope of Backfill surcharge (β) = 0 Deg
K =
Cos ( α)
Cos α Cos( )
× ( ) ( )
( ) ( )
.
= 0.4924
Co-efficient of Active Earth Pressure (ka)
-For Finite Backfill
Slope of Backfill surcharge (β) = 25.25 Deg
K =
Cos ( α)
Cos α Cos( )
× ( ) ( )
( ) ( )
.
= 0.2973

v) FOUNDATION PROPERTIES FOR DESIGN
Safe Bearing Capacity of Soil = 200 kN/m3
vi) REFERENCE CODES
Refer pg. 17, Chapter 4, cl. 4.1.
vii) COMBINATION OF LOADS FOR LIMIT STATE DESIGN
a) Partial Safety Factor for verification of Structural Strength:
Refer pg. 38, Chapter 5, cl. 5.2 (a)
b) Partial Safety Factor for verification of Serviceability Limit State:
Refer pg. 38, Chapter 5, cl. 5.2 (b)
Fig. 8.1: Dimension Nomenclature of Retaining Wall

8.1. SECTION 1-1
8.1.1 DIMENSIONS OF SECTION 1-1
F.B.L of the Canal = 582.680 m
R.L. at Top of the Wall = 581.750 m
R.L. at Foundation Level = 575.790 m
Height of the Wall (H1) = (581.750-575.790)
= 5.950 m
Allowable Surcharge height = 0.93 m
Thickness of Stem at top t1 = 0.30 m
Thickness of Stem at bottom t2 = 0.60 m (Min. 0.1xH)
Thickness of Base Slab at center D1 = 0.60 m (Min. 0.1xH)
Thickness of Base Slab at ends D2 = 0.30 m
Height of Stem h = Height of wall – Base slab thickness
= (5.950-0.60)
= 5.35 m
Width of Base Slab B = 5.10 m (0.4-0.7) x H
Width of Toe Slab a = 0.90 m
Width of Heel Slab b = B – t2 - a
= (5.10-0.60-0.90)
= 3.60 m
Surcharge Width b1 = Allowable Surcharge Height/ tan (β)
= 0.93/ tan (25.25)
= 1.976 m
Total Height including surcharge (H2) = H1 + [b1 x tan (β)]
= 6.89 m
As per Cl. 214.1/pg. 41/ IRC: 6-2014, Earth Pressure due to live load Surcharge (LLS)
Live Load Surcharge = 1.2 x ka x ϒ
= 1.2 x 0.297 x 20
= 7.128 kN/m2
Active Earth Pressure (Pa) = 0.5 x ka x ϒ x (H )
= 0.5 x 0.2973 x 20 x (6.89)2
= 140.98 kN/m2

As per Cl. 214.1, pg. 41, IRC: 6-2014, the Active Earth Pressure (AEP) is located at an
elevation of 0.42 of the height of the wall above the base.
Fig. 8.2: Section 1-1 Dimensions

STABILITY CHECK
Sl.
No
DESCRIPTION
FORCES (kN) LEVER
ARM
MOMENTS
(kN-m)
VERTICAL HORIZONTAL MR MO
a) SELF WEIGHT
1 S1 = 0.3x5.354x25 40.16 - 1.05 42.16 -
2 S2 = 0.5x0.3x5.404x25 20.08 - 1.30 26.10 -
3 S3 = 0.6x0.6x25 9.00 - 1.20 10.80 -
4 S4 = 0.5x0.3x3.6x25 3.38 - 0.60 2.03 -
5 S5 = 0.3x0.9x25 6.75 - 0.45 3.04 -
6 S6 = 0.5x0.3x3.6x25 13.50 - 2.70 36.45 -
7 S7 = 0.3x3.6x25 27.00 - 3.30 89.10 -
TOTAL 116.86
b) SOIL WEIGHTS
1 S8 = 0.5x1.976x0.93x3.6x20 18.42 - 2.52 46.36 -
2 S9 = 0.5x0.3x5.354x20 16.06 - 1.40 22.49 -
3 S10 = 0.932x1.924x20x1.50 35.86 - 4.14 148.40 -
4 S11 = 3.60x5.354x20 385.49 - 3.30 1272.11 -
5 S12 = 0.50x0.30x3.60x20 10.80 - 3.90 42.12 -
TOTAL 466.63
c) EARTH PRESSURES DUE TO BACKFILL AND SURCHARGE
1 Pa = 0.5x0.297x20x6.882
- - 2.50 0.00 0.00
2 PaH Pa=PaH - 140.98 2.89 0.00 407.72
3 PaV 0.00 0.00 - 0.00 0.00
4 LLS = 1.20x7.128x6.88 - 49.08 3.44 0.00 168.99
TOTAL ΣV=585.44 ΣH=190.06
ΣMR=
1741.15
ΣMO=
576.72
Total Vertical Load = 585.44 kN
Total Horizontal Load = 190.06 kN
Total Restoring Moment = 1741.15 kN-m
Total Overturning Moment = 576.72 kN-m

 = (ΣMR - ΣMO) / ΣV
= (1741.15 – 576.72) / 585.44
= 1.990
As per Cl. 706.3.4, pg. 18-19, IRC 78-2014, stability checks are carried out
F.O.S against Sliding = μ x ΣV / ΣH
= 0.5 x 585.44 / 190.06
= 1.54 >1.50 SAFE
F.O.S against Overturning = ΣMR / ΣMO
= 1741.15 / 576.72
= 3.02 >2.00 SAFE
Eccentricity = (B/2) - 
= (5.10/2) – 1.99
= 0.56 e<B/6 (0.85) SAFE
Base Pressure at Toe = B
× 1 +
×e
B
=
585.44
5.10
× 1 +
×0.56
5.10
= 191.38 kN/m2
< 200 kN/m2
SAFE
Base Pressure at Heel = B
× 1 −
×e
B
=
668.18
5.10
× 1 +
×0.56
5.10
= 38.62 kN/m2
> 0 kN/m2
SAFE
8.1.2. ULTIMATE LIMIT STATE DESIGN (U.L.S)- STRENGTH (BASIC
COMBINATION)
As per IRC: 6 -2014, Amendment, Table 3.2, pg. 44, the following Load Factors are
to be used for the Ultimate Limit State Design.
ϒself weight = 1.35
ϒSIDL = 1.35
ϒbackfill weight = 1.50
ϒearth pressure = 1.50
ϒLLS = 1.20

Sl.
No
DESCRIPTION
FORCES (kN)
LEVER
ARM
(m)
MOMENTS
(kN-m)
1 Self-Weight 161.81 0.00 283.06 0.00
2 Weight of Soil on heel 699.94 0.00 2297.22 0.00
3
Active Earth
Pressure
PaH 0.00 211.47 2.89 0.00 611.58
PaV 0.00 0.00 0.00 0.00 0.00
4 LLS 0.00 58.90 3.44 0.00 202.79
TOTAL 861.75 270.37 2580.28 814.38
 = 2.05
F.O.S against Sliding = 1.59
F.O.S against Overturning = 3.17
Eccentricity = 0.50
Base Pressure at Toe = 268.52 kN/m2
Base Pressure at Heel = 69.42 kN/m2
A) DESIGN OF STEM
Grade of Concrete = M-25 (Strength Class)
Characteristic Strength of Concrete (fck) = 25.00 N/mm2
Grade of Steel = Fe-500 (Strength Class)
Characteristic Strength of Steel (fy) = 500.00 N/mm2
Clear Cover = 75.00 mm
Diameter of Bar = 20 mm
Effective Depth (d) = 600-75-20/2
= 515 mm

Fig. 8.3: Section 1-1 -Forces acting on Stem- Basic Combination
 FORCES ACTING ON STEM:
 Force due to Active Earth Pressure (AEP) = 0.5 x ka x ϒ x h2
x ϒearth pressure
= 0.5 x 0.2 x 0.4924 x 5.42
x 1.50
= 191.48 kN/m2
 Lever arm for AEP = 0.42 x h
= 0.42 x 5.40
= 2.27 m
 Force due to Live Load Surcharge (LLS) = LLS x h x ϒLLS
= 7.128 x 5.4 x 1.20
= 45.80 kN/m2
 Lever arm for LLS = h /2
= 5.40 / 2
= 2.7 m
 BENDING MOMENT AND SHEAR FORCE
 Bending Moment (Mu) = (191.48 x 2.27) + (45.80 x 2.70)
= 553.17 kN-m
 Shear Force (Vu) = 195.07 + 46.22
= 237.27 kN
 CHECK FOR DEPTH
Breadth of Wall (b) = 1000 mm
Depth required (d) required =
Mu
. ×fck × b
(For Fe-500)
5.35m
0.3m
0.6m 237.27
kN/m2
AEP
45.80
kN/m2
LLS

=
553.17×106
0.134 ×25 × 1000
= 406.36 mm < (d) provided
Hence O.K.
 Ultimate Moment of Resistance (Mu)lim = 0.134 x fck x b x d2
= 0.134 x 25 x 1000 x (515)2
= 888.50 kN-m > Mu = 553.17 kN-m
Hence O.K.
 Tension Reinforcement for Stem (Ast):
Ast =
0.5×b×d×fck
fy
× 1 − 1 −
4.6 × Mu
fck×b×d
Ast =
0.5×1000×515×25
500
× 1 − 1 −
. ×241.29×
25×1000×515
= 2768.01 mm2
Considering 20 mm Ø bars,
c/c spacing =
×(Ast)reqd
Area of 1 bar
=
× .
×
= 110.20 mm
As per Cl. 16.6.1.1, max spacing must not exceed 2h (i.e. 600mm) or 250mm.
Hence, provide 20 mm Ø bars @ 85 mm c/c
Ast provided =
× ×
= 3695.99 mm2
Percentage of steel (pt) =
×(Ast)provided
b×d
=
×3695.99
1000×515
= 0.72%
 DISTRIBUTION STEEL
As per cl.16.5.1.1, pg. 175, IRC: 112-2011, minimum reinforcement to be provided
should be 0.0013bd
∴Ast min = 0.0013 x 1000 x 515 = 669.50 mm2
Hence, provide 10 Ø bars @ 115 mm c/c as distribution steel for Stem.
 DEVELOPMENT LENGTH
As per Cl. 15.2.3.3, pg. 150, IRC: 112-2011, the Development Length (Ld) is given
by

Ld=
∅fyd
4fbd
Where,
Ø = nominal diameter of the bar,
fyd = Design ultimate stress = fy/1.15
fbd = design values for favorable bond conditions given in Table 15.3.
From Table 15.3, pg. 150, IRC: 112-2011, the Design bond stress for M25 Concrete is 2.25
for deformed bars. Hence the value of Bond Stress is
fbd = 2.25 N/mm2
Ld =
20×0.87×500
4×2.25
Ld = 966.67 mm
 CURTAILMENT OF STEM REINFORCEMENT
The curtailment of main tension reinforcement has to be done at a section where the Area of
tension reinforcement required is 50%.
Steel provided for stem = 3695.99 mm2
i.e. 20mm Ø bars @ 85mm c/c
50% steel for stem = 1848.00 mm2
i.e. 20mm Ø bars @ 170mm c/c
B.M for 50% steel = 384.286 kN-m
Now, we need to calculate the height at which the BM is 384.286 kN-m.
M=(Pa×0.42h)+ LLS×
h
2
M= 0.5×ka×ϒ×h2
×0.42h + 1.2×ka×ϒ×h×
h
2
384.286= 0.5×0.492×20×0.42×h3
×1.5 + 1.2×0.297×20×
h2
2
×1.2
h = 4.565 m from top of stem and 0.789 from bottom of stem
But Actual Curtailment length = height of 50% Ast + Ld
= 0.789 + 0.967
= 1.756 m
Hence, curtail every alternate main reinforcement for stem at a height of 1.80 m from the
bottom.

 CHECK FOR SHEAR
As per Cl. 10.3.2, pg. 88, IRC: 112-2011, the design shear resistance (VRd.c) must be greater
or equal to the shear force acting at that section (VEd.)
VRd.c = [0.12K(80ρ1fck)0.33
+ 0.15σcp] bwd
Subjected to a minimum of VRd.c = (νmin + 0.15σcp]bwd
K=1+√(200/d) ≤ 2.0 where d is depth in mm.
νmin = 0.031K3/2
fck
1/2
σcp is limited to 0.2 fcd (N/mm2
) where σcp = NEd / Ac
ρ1 = Asl/(bwd) ≤ 0.02
d = 515mm, Ast pro = 3695.99 mm2
K = 1+
200
515
= 1.623
ρ1 = 3695.99/(1000x515) = 0.007177
σcp = 0 Since there is no axial force acting on the member
∴VRd.c = [0.12 x 1.23(80x0.007177x25)0.33
] 1000 x 515
= 241.63 kN
VRd min = (0.031 x 1.6233/2
x 251/2
) 1000 x 515
= 165.05 kN
Hence, VRd.c > VEd (237.27kN) HENCE O.K
Section
from
top
Breadth
(mm)
VEd. ρ1 σcp k VRd.c (VRd.c)min CHECK
3.554 1000.00 39.34 0.00446 0 1.6950 173.49 141.71 SAFE
5.350 1000.00 237.27 0.00717 0 1.6232 241.62 165.08 SAFE
HENCE O.K

B) DESIGN OF FOOTING
Fig. 8.4: Section 1-1 -Upward bearing pressure for footing- Basic Combination
 SECTION 1: FOR HEEL SLAB
BENDING MOMENT AND SHEAR FORCE CALCULATION
Sl.No Description Force
Lever
Arm
Moment
1
Self-Weight of Heel Slab
= 0.5x0.25x3.6x25x1.35 36.45 1.80 65.61 S6
= 0.5x3.60x0.25x1.35 18.23 1.20 21.87 S7
2
Weight of Soil Above Heel Slab
= 0.5x0.7905x1.676x20x1.50 19.87 1.12 22.21
S8
= 1.676x0.1415x20x1.50 7.11 0.84 5.96
= 0.932x1.9239x20x1.50 53.79 2.638 141.91 S10
= 3.60x5.404x20x1.50 578.23 1.80 1040.82 S11
= 0.50x0.25x3.60x20x1.50 16.20 2.40 38.88 S12
3 Pv 0.00 3.60 0.00
4 Base Pressure on Heel
-249.90 1.80 -449.82
-252.98 1.20 -303.58
TOTAL 227.00 583.85
0.515 m
0.9 m 0.6 m 3.6 m
0.55 m
268.52
N/mm2
69.542
N/mm2209.96
N/mm2
233.39
N/mm2
253.49
N/mm2
S/N 1S/N 2S/N 3
0.3 m
5.1 m

 SECTION 2: FOR TOE SLAB
Lever
Arm
Moment
1
Self-Weight of Toe Slab
= 0.5x0.25x0.90x25x1.35 4.56 0.30 1.37 S4
= 0.30x0.90x25x1.35 9.11 0.45 4.10 S5
2 Base Pressure on Toe
-210.05 0.45 -94.52
-15.81 0.60 -9.49
TOTAL -212.19 -98.56
 SECTION 3: FOR TOE SLAB AT CRITICAL SECTION
Lever
Arm
Moment
1
Self-Weight of Toe Slab
=0.50x0.431x0.121x25x1.35 0.83 0.127 0.10 S4
=0.30x0.431x25x1.35 4.36 0.19 0.73 S5
-108.20 0.19 -18.40
-3.65 0.25 -0.73
TOTAL -94.76 -18.29
 SECTION FORCES
Section Breadth (b)
Overall
Depth (D)
Effective
Depth (d)
B.M
(kN-m)
S.F
(kN)
1 1000 600.00 515.00 583.72 227.10
2 1000 600.00 519.00 -98.56 -212.19
3 1000 428.33 347.33 -18.29 -9

 TENSION REINFORCEMENT
Section
drequired
(mm)
dprovided
(mm)
Mu
(kN-m)
Mu lim
(kN-m)
Ast min
(mm2)
Ast req
(mm2)
c/c spacing Ast pro
(mm2)Required Provided
1 417.47 515.00 583.72 888.50 720.00 2944.10 106.71 90.00 3490.66
2 171.52 519.00 98.56 902.36 720.00 444.30 157.08 135.00 837.76
3 73.88 347.33 18.29 404.15 514.00 121.95 220.03 135.00 837.76
Hence provide,
 20 Ø bars at 90 mm c/c for Section -1
 10 Ø bars at 115 mm c/c as Distribution steel
 CHECK FOR SHEAR
Section
Breadth
(mm)
VED. ρ1 σcp k VRd.c (VRd.c)min CHECK
1 1000.00 227.00 0.006778 0.00 1.6232 237.11405 165.08 SAFE
3 1000.00 94.76 0.002412 0.00 1.7588 26.210005 125.58 SAFE
8.1.3. LIMIT STATE OF SERVICEABILITY (RARE COMBINATION)
As per IRC: 6 -2014, Amendment, Table 3.3, pg. 44, the following Load Factors are to be
used for the Ultimate Limit State of Serviceability- Rare Combination.
ϒself weight = ϒSIDL = ϒbackfill weight = ϒearth pressure = 1.00 and ϒLLS = 0.80
Sl. No DESCRIPTION
FORCES (kN) LEVER
ARM
MOMENTS
(kN-m)
1 Self-Weight 119.86 0.00 209.68 0.00
2 Weight of Soil on heel 466.63 0.00 1531.48 0.00
3
Active Earth
Pressure
PaH 0.00 140.98 2.89 0.00 407.72
PaV 0.00 0.00 5.10 0.00 0.00
4 LLS 0.00 39.27 3.44 0.00 135.20
TOTAL 585.44 180.24 1741.15 542.92

0.3m
 = 2.04
Eccentricity = 0.51
A) DESIGN OF STEM
 FORCES ACTING ON STEM
Fig. 8.5: Section 1-1 -Forces acting on Stem- Rare Combination
 DESIGN FORCES
Section
from
top
Wall
thickness
Breadth
b
(mm)
LLS
Lever
Arm
(m)
Active
Earth
Pressure
Lever
Arm
(m)
B.M
(kN-m)
S.F
(kN)
3.554 499.29 1000 20.54 1.80 63.75 1.51 133.30 84.289
5.350 600 1000 30.53 2.68 127.65 2.25 368.78 158.18
 CHECK FOR STRESS IN STEM
As per Cl. 12.2.1, pg. 120, IRC: 112-2011, the maximum compressive stress in concrete
under rare combinations of loads shall be limited to 0.48fck = 0.48 x 25
= 12.0 N/mm2
127.65
kN/m2
AEP
0.6m
30.53
kN/m2
LLS
5.35m

As per Cl. 12.2.2, pg. 120, IRC: 112-2011, the maximum tensile stress in steel under
rare combinations of loads shall be limited to 0.80fy = 0.8 x 500
= 400.00 N/mm2
Section
from
top
Effective
depth (d)
Bending
Moment
(M)
Area of
Steel Ast
pro
Neutral
Axis
(xu)
Moment
of Inertia
(Icr)
Stress in
Steel
σsc
(N/mm2)
Stress in
Concrete
σc
(N/mm2)
3.554 414.29 133.30 1848.00 144.62 3.86E+09 195.65 4.99
5.350 515.00 368.78 3695.99 214.65 1.02E+10 225.22 7.74
HENCE O.K
 To calculate Neutral axis:
We have,
Modulus of Elasticity of Steel (Es) = 200000 N/mm2
Modulus of Elasticity of Concrete (Ec) = 25000 N/mm2
As per Cl. 6.4.2.5, pg. 43, IRC: 112-2011
Creep Co-efficient (Փ) for 28 days = 1.60
Modular ratio (m) = Es / Ec eff
=
2×105
25000
1+1.60
= 20.80
Hence,
b×xu×(xu/2)=m×Ast×(d-xu)
1000×xu×(xu/2)=20.8×2855.99×(515-xu)
Solving for xu we get,
xu=194.99 mm
 To calculate cracked Moment of Inertia
Icr=
b×xu
3
12
+(A×h2
) +[m×Ast×(d-xu)2]
Icr=
1000×194.993
12
+(1000×194.99×(194.99/2)2
) + [20.8×2855.99×(515-194.99)2]
∴Icr = 8.550x109 mm4

 Stress in Steel (σsc)
σsc =
368.78×106
8.55×109 × (515-194.99) x20.80 = 286.94 N/mm2
< (Limiting σsc= 400N/mm2
)
HENCE O.K
 Stress in Concrete
σc =
378.43×106
8.55×109 ×(194.99) = 8.41 N/mm2
< (Limiting σc= 12 N/mm2
)
HENCE O.K
Fig. 8.6: Section 1-1 -Upward bearing pressure for footing- Rare Combination
0.3 m
143.24
N/mm2
46.41
N/mm2
5.1 m
3.6 m0.6 m0.9 m
0.60 m
S/N 1
0.515 m
183.58
N/mm2
159.37
N/mm2
173.33
N/mm2
S/N 2S/N 3

 BENDING MOMENTS AND SHEAR FORCES
Section 1
Sl. No Description Force Lever Arm Moment
1 Self-Weight of Heel Slab
27.00 1.80 48.60 S6
13.50 1.20 16.20 S7
2 Weight of Soil Above Heel Slab
13.25 1.12 14.81
S8
4.74 0.84 3.97
35.86 2.64 94.60 S10
385.49 1.80 693.88 S11
10.80 2.40 25.92 S12
3 Pv 0.00 3.60 0.00
-167.09 1.80 -300.77
-174.28 1.20 -209.14
TOTAL 149.27 388.08
Section 2
Sl. No. Force Lever Arm Moment
1 Self-Weight of Toe Slab
3.38 0.30 1.01 S4
6.75 0.45 3.04 S5
-143.44 0.45 -64.55
-10.89 0.60 -6.54
TOTAL -144.20 -67.03
Section 3
Sl.No Description Force Lever Arm Moment
0.611 0.13 0.08 S4
2.86 0.19 0.54 S5
-66.04 0.19 -12.58
-1.95 0.25 -0.50
TOTAL -64.52 -12.45

 SECTION FORCES
Section
Overall
Depth (D)
Breadth
(b)
Effective
Depth (d)
B.M
(kN-m)
S.F
(kN)
1 600.00 1000 515.00 388.02 149.27
2 600.00 1000 519.00 -67.03 -144.20
3 420.83 1000 347.33 -12.45 -64.52
 CHECK FOR STRESS
Section
Overall
Depth
(h)
Effective
Depth
(d)
Bending
Moment
(M)
Area of
Steel
As provided
Neutral
Axis
(xu)
Moment
of
Inertia
(Icr)
Stress in
Steel σsc
(N/mm2)
Stress in
Concrete
σc
(N/mm2)
1 600 515 388.02 3490.66 210.34 9.84E+09 249.90 8.29
2 600 519 67.03 837.76 118.19 3.35E+09 166.83 2.37
3 420.83 347.33 12.45 837.76 93.97 1.40E+09 47.04 0.84
8.1.4. LIMIT STATE OF SERVICEABILITY (QUASI PERMANENT
COMBINATION)
As per IRC: 6 -2014, Amendment, Table 3.3, pg. 44, the following Load Factors are to be
used for the Ultimate Limit State of Serviceability- Quasi Permanent Combination.
ϒself weight = ϒSIDL = ϒbackfill weight = ϒearth pressure = 1.00 and ϒLLS = 0
Sl. No DESCRIPTION
FORCES (kN) LEVER
ARM
MOMENTS
(kN-m)
1 Self-Weight 119.86 209.68
2 Weight of Soil on heel 466.63 1531.48
3
Active Earth
Pressure
PaH 0.00 140.98 2.89 0.00 407.72
PaV 0.00 0.00 5.10 0.00 0.00
4 LLS 0.00 0.00 3.44 0.00 0.00
TOTAL 586.49 140.98 1741.15 407.72

0.3m
 = 2.27
Eccentricity = 0.28
A) DESIGN OF STEM
 FORCES ACTING ON STEM
Fig. 8.7: Section 1-1 -Forces acting on Stem- Quasi Permanent
DESIGN FORCES
Section
from top
Wall
thickness
LLS
Lever
Arm (m)
Active
Earth
Pressure
Lever
Arm (m)
B.M
(kN-m)
S.F
(kN)
3.554 414.29 0 1.80 63.75 1.51 96.341 63.75
5.350 600 0 2.680 127.65 2.25 287.05 127.65
127.65
kN/m2
AEP
0.6m
5.35 m

CRACK WIDTH FOR STEM
Section
from
top
xu
σsc
(N/mm2)
σc
(N/mm2)
hc eff Ac eff ρp eff ϵs-ϵm Sr max Wk
3.554 144.62 141.405 3.61 119.13 119127.92 0.01551 0.000424 474.18 0.201
5.15 214.65 175.27 6.02 128.45 128449.54 0.02877 0.0005226 373.163 0.196
Crack width is calculated as per Cl. 12.3.4, pg. 125, IRC: 112-2011.
Wk=Sr.max(εsm-εcm)
Where,
hc eff is least of
2.5(h-d)
(h-x)/3
h/2
=
2.5(600-515)
(600-194.99)/3
600/2
=
212.50 mm
135.00 mm
300.00 mm
Hence, hc eff = 135 mm
Ac eff = b x hc eff = 1000 x 135 = 135000 mm2
ρp-eff = As/ Ac eff = (2855.99/135000) = 0.021150
Sr. max = 3.4c+
0.425k1k2ϕ
ρp-eff
= 3.4×75+
0.425×0.8×0.5
0.02115
= 415.72 mm
fct.eff = 0.7√0.446fck or 2.90 max Cl. 12.2.3, IRC: 112-2011
= 0.7√11.15 or 2.90
= 3.50 > 2.90
= 2.90
(εsm-εcm) =
σsc -kt
fct.eff
ρp-eff
1+αeρp-eff
Es
≥0.6
σsc
Es
=
. . ×
.
.
( . × . )
≥ 0.6
.
= 0.000654809 ≤ 0.000688991
(εsm-εcm) = 0.000688991
As per Cl. 12.3.2, Table 12.1, Pg. 112, IRC: 112-2011, the limiting crack width for moderate
exposure condition and reinforced member is 0.30 mm
∴ Wk = 415.72 x 0.000688991
= 0.28643 mm < 0.3mm
HENCE O.K.

Fig. 8.8: Section 1-1 -Upward bearing pressure for footing- Rare Combination
Section 1
1 Self-Weight of Heel Slab
27.00 1.80 48.60 S6
13.50 1.20 16.20 S7
2 Weight of Soil Above Heel Slab
13.25 1.12 14.81
S8
4.74 0.84 3.97
35.86 2.638 94.60 S10
385.49 1.80 693.88 S11
10.80 2.40 25.92 S12
3 Pv 0.00 3.60 0.00
-279.37 1.80 -502.86
-95.03 1.20 -114.03
TOTAL 116.25 281.09
Section 2
3.38 0.30 1.01 S4
6.75 0.45 3.04 S5
-125.27 0.45 -56.37
-5.94 0.60 -3.56
TOTAL -121.09 -55.89
0.60 m0.515 m
77.60
N/mm2130.40
N/mm2
139.19
N/mm2
146.81
N/mm2
5.1 m
0.3 m
3.6 m0.6 m0.9 m
S/N 3 S/N 2 S/N 1
152.39
N/mm2

Section 3
0.611 0.13 0.08 S4
2.86 0.19 0.54 S5
-55.93 0.19 -10.66
-1.06 0.25 -0.27
TOTAL -53.53 -10.30
 SECTION FORCES
Section
Overall Depth
(D)
Breadth
(b)
Effective
Depth (d)
B.M
(kN-m)
S.F
(kN)
1 600 1000 515 281.09 116.25
2 600 1000 519 -55.89 -121.09
3 428.33 1000 179.67 -10.30 -53.53
 CRACK WIDTH FOR FOOTING
Section xu
σsc
(N/mm2)
σc
(N/mm2)
hc eff Ac eff ρp eff ϵs-ϵm Sr max Wk
1 210.34 181.00 6.01 129.89 129888.32 0.02687 0.0005430 381.51 0.2072
2 118.19 139.10 1.97 160.60 160603.77 0.00522 0.0004173 646.08 0.2696
3 93.97 13.16 0.69 111.46 111455.14 0.00752 0.0000395 526.40 0.0208
HENCE O.K.

DEPARTMENT OF CIVIL ENGINEERING, B.I.T. Page 92
8.2. SECTION 2-2
8.2.1. DIMENSIONS OF SECTION 2-2
F.B.L of the Canal = 582.680 m
R.L. at Top of the Wall = 580.771 m
R.L. at Foundation Level = 575.794 m
Height of the Wall (H1) = 4.98 m
Allowable Surcharge height = 1.91 m
Surcharge Width = 4.05 m
Thickness of Stem t1 = 0.30 m
t2 = 0.60 m
Thickness of Base Slab D1 = 0.60 m
D2 = 0.30 m
Height of Stem h = 4.38 m
Width of Base Slab B = 4.90 m
Width of Toe Slab a = 0.50 m
Width of Heel Slab b = 3.80 m
Surcharge Width b1 = 4.10 m
Total Height including surcharge H2 = 6.91 m
Co-efficient of Earth Pressure ka = 0.4924
Coefficient of Friction μ = 0.50
Density of concrete = 25.00 kN/m3
Density of Compacted Backfill ϒ = 20.00 kN/m3
Live Load Surcharge = 0.000 kN/m2
As per Cl. 214.2, pg. 41, IRC 6-2014, the section 2-2 is at a distance greater than 3m from
the box culvert. Hence, the effect of LLS will not act upon the section and is ignored.

DEPARTMENT OF CIVIL ENGINEERING, B.I.T. Page 93
Fig. 8.9: Section 2-2 Dimensions

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