Tribhuwan University
Institute of Engineering
Pulchowk Campus
DEPARTMENT OF CIVIL ENGINEERING
FINAL REPORT ON
DESIGN OF BRIDGE OVER KERUNGA KHOLA,
CHITWAN
Supervisor
Asso. Prof. Nabin Chandra Sharma
Department of Civil Engineering
Prepared By
ANJAN LUITEL (069/BCE/012)
BIBEK KUMAR KHATRI (069/BCE/031)
BIGYA GYAWALI (069/BCE/034)
BIKAL SHAKYA (069/BCE/035)
BIKESH LAGE (069/BCE/037)
BUDDHIMAN TAMANG (069/BCE/047)
October, 2016
Tribhuwan University
Institute of Engineering
Pulchowk Campus
Department of Civil Engineering
Lalitpur, Nepal
Certificate
This is to certify that the final year project entitled “DESIGN OF BRIDGE OVER
KERUNGA KHOLA, CHITWAN” was submitted by the students to the DEPARTMENT
OF CIVIL
ENGINEERING in partial fulfillment of requirement for the degree of Bachelor of
Engineering in Civil Engineering. The project was carried out under special supervision and
within the time frame prescribed by the syllabus.
…………………………….
Asso. Prof. Nabin Chandra Sharma
Project Supervisor
…………………………….
…………………………….
Mr. Dinesh Gupta
Mr. Rajan Suwal
Internal Examiner
External Examiner
…………………………….
Dr. Kamal Bahadur Thapa
Head of Department
Department of Civil Engineering
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
ACKNOWLEDGEMENT
We’re grateful to our Department of Civil Engineering, Pulchowk Campus for providing us
with this project to further enhance our knowledge in the field of civil engineering and its
application.
Bridge construction has been a predominant feature of human progress and evolution. It
has always been a symbol of peace and development. Hence, in the context of our country
which has more than 6000 rivers and rivulets, construction of bridge plays a very important
role in development and making it an undoubted reason for choosing this project.
We’re indebted to our advisor and supervisor Assoc. Prof. Dr. Nabin Chandra Sharma for his
valuable time, instructions and guidance throughout the project.
We’re heartily thankful to Mr. Chuman Babu Shrestha, capacity building specialist, LRBSU
for arranging us the field visit and providing the design data. We’re so deeply obliged to site
engineer of both bridge sites for their supervision during our site visit.
i
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
ABSTRACT
We aim to analysis and design a combination of T-Girder and Truss Bridge by
using the different theories of civil engineering. The combination of structural mechanics and
foundation engineering forms the base of our design.
The knowledge of influence line diagram, shear force diagram and bending moment
diagram are the essentials of this project. IRC codes were used as the guidelines in the design
of the bridge.
The superstructure components for truss (slab, stringer, crossbeam and truss girder)
and t-girder (slab, t-girder, cross beam) have been designed under class AA and class A
loading as prescribed by IRC. Pigeaud’s method was used in analysis of the slab and HendryJaeger method in the analysis of t-girders.
The substructure components include bearing, abutments, piers and foundation.
Considering the various forces acting on the substructure, spread foundation is provided and
the abutment is gravity type.
Hence with the help of our supervisor and the available materials and data, we have
made an effort to analyse and design The Bridge.
Salient Features
Particulars
Name of the Project
Location
Zone
District
Village/ Town
Name of Road
Chainage of the Bridge Site
Geographical Location
Easting
Northing
Classification of Road
Type of the Road Surface
Terrain/ Geology
Information on the Structure
Total Length of the Bridge
Span Arrangement
Total Width of the Bridge
No. of Lanes
Width of:
Carriageway
Footpaths
Kerbs
Required Information/ Range/ Values
Design of Bridge over Kerunga Khola
Narayani
Chitwan
Kalyanpur V.D.C.
531972.828 (Longitude – 84β°19’26”)
3050960.78 (Latitude – 27β°34’34”)
Local Road
Gravel
Plain
60 m
12 + 36 + 12 m
7.2 m
Two
6m
Not provided
0.6 m
ii
Design of Bridge Over Kerunga Khola, Chitwan
Particulars
Type of Superstructure
Type of Bearings
Type of Abutments
Type of Piers
Type and Depth of Foundation
Design Data
Live Load
Net Bearing Capacity of Soil
Design Discharge
Linear Waterway
SUMMARY OF QUANTITIES
Grade and Quantity of Concrete
In Superstructure
In Substructure
Grade and Quantity of Reinforced Steel
In Superstructure
In Substructure
Grade and Quantity of Structural Steel
In Superstructure
In Substructure
In Foundation
Quantities of other materials
Brick Masonry
Gabion Masonry
Formworks
Timber
Stone Masonry
SUMMARY OF COST
Superstructure
Substructure
Foundation Works
Approach Road
River Training Works
Miscellaneous Works
Grand Total
Cost per m run
2069 – AB Bridge
Required Information/ Range/ Values
RCC T-Girder + Steel Truss
Elastomeric Pad Bearings
RCC rectangular abutment
RCC Solid Piers
Spread Footing for Abutments
Well Foundation for Piers
IRC Class AA (Wheeled and Tracked)
IRC Class A
300 kN/m2
95.54 m3/sec
46.43 m
M25 – 154.4 m3
M20 – 548.136 m3
Fe415 – 12.12 MT
Fe415 – 64.328 MT
E250 – 76.066 MT
411.018 m2
-
Rs. 1,30,22,728 /Rs. 1,26,88,792 /Rs. 2,57,11,520 /(Truss) Rs. 3,05,578/(T-Girder) Rs. 84,248 /-
iii
Design of Bridge Over Kerunga Khola, Chitwan
NOTATION
∅
πuv
γm
Ag
Ah
Ast
Asv
bf
bw
d
d’
D
E
fck
fy
I
Ip
Ld
pc
pt
R
Sa/g
Sv
Xu
Xul
Z
Diameter of Bar
Shear Stress
Partial Safety Factor
Gross Area
Horizontal Seismic Coefficient
Area of Steel in Tension
Area of Stirrups
Flange width
Web width
Effective depth
Effective Cover
Overall Depth
Young’s modulus of Elasticity
Characteristic Strength of Concrete
Characteristic Strength of Steel
Importance Factor
Polar Moment of Inertia
Development Length
Percentage of Steel in Compression
Percentage of Steel in Tension
Response Reduction Factor
Average Response Acceleration Coefficient
Spacing of Stirrups
Actual depth of Neutral Axis
Ultimate depth of Neutral Axis
Zone Factor
iv
2069 – AB Bridge
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
TABLE OF CONTENTS
INTRODUCTION ....................................................................................................................................... 1
1.1
Introduction ............................................................................................................................ 1
1.2
Title of project work................................................................................................................ 1
1.3
Project assignment.................................................................................................................. 1
1.4
Objectives................................................................................................................................ 2
1.5
Client Requirements ............................................................................................................... 2
METHODOLOGY ...................................................................................................................................... 3
2.1.
Acquisition of Data .................................................................................................................. 3
2.2.
Structural Planning and Preliminary Design ........................................................................... 7
2.3.
Idealization and Analysis of bridge structure........................................................................ 13
ACQUISITION OF DATA FOR DESIGN ..................................................................................................... 17
3.1
Topographical Survey............................................................................................................ 17
3.2
Geology and Topography ...................................................................................................... 17
3.3
Hydrology .............................................................................................................................. 18
3.4
Observation visit and Verification of data ............................................................................ 21
SELECTION OF BRIDGE TYPE ................................................................................................................. 24
STRUCTURAL PLANNING AND PRELIMINARY DESIGN .......................................................................... 25
T-Beam Bridge ................................................................................................................................... 25
Truss Bridge....................................................................................................................................... 27
ABUTMENT/PIER HEIGHT CALCULATION: ......................................................................................... 28
STRUCTURAL ANALYSIS AND DESIGN OF BRIDGE COMPONENTS ........................................................ 29
A.
Design of RCC T-Girder (12 m span) bridge .............................................................................. 29
I.
Analysis and design of Deck slab ....................................................................................... 29
II.
Analysis and Design of Main Girder .................................................................................. 44
III.
Analysis of Cross Beam ..................................................................................................... 65
IV.
Design of Elastomeric bearing .......................................................................................... 68
B.
Design of Steel Truss (36 m span) bridge .................................................................................. 75
I.
Design of Deck Slab ........................................................................................................... 75
II.
Design of stringer beam .................................................................................................... 76
III.
Design of cross girders ...................................................................................................... 78
v
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
IV.
Design of Steel Trusses ..................................................................................................... 80
V.
Design of Truss Joints ........................................................................................................ 90
VI.
Consideration of Wind Load on Member Stresses ........................................................... 91
VII.
Consideration of Seismic Force ......................................................................................... 94
VIII.
Design of Bracings ............................................................................................................. 95
IX.
Design of Connections .................................................................................................... 100
X.
Design of Elastomeric bearing for truss .......................................................................... 107
C.
Design of RCC Abutment ......................................................................................................... 116
I.
Planning and Preliminary Design ..................................................................................... 116
II.
Analysis and Design of Abutment Cap ............................................................................ 118
III.
Analysis and Design of Abutment Stem .......................................................................... 119
IV.
Design of Dirt Wall .......................................................................................................... 126
V.
Analysis and Design of Spread Footing ........................................................................... 130
D.
Design of RCC Solid Pier .......................................................................................................... 134
I.
Planning and preliminary design ..................................................................................... 134
II.
Load Calculation .............................................................................................................. 135
III.
Analysis and Design of Pier Cap ...................................................................................... 139
IV.
Analysis and Design of Pier Stem .................................................................................... 140
DRAWINGS .......................................................................................................................................... 150
APPROXIMATE ESTIMATES ................................................................................................................. 151
CONCLUSION AND RECOMMENDATIONS........................................................................................... 155
BIBLIOGRAPHY .................................................................................................................................... 156
ANNEXES ............................................................................................................................................. 157
Detail Design of Connections at Truss Joints .................................................................................. 157
vi
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
INTRODUCTION
1.1
Introduction
Bridges has been a predominant feature of human progress and evolution. Bridges
are used by people and vehicles for crossing rivers, lakes, ravines, canyons, railroads and
highways. Bridges must be strong enough to safely support their weight as well as the
weight of the vehicle that pass over it.
As Nepal is a mountainous country with a lot of river and rivulets, we need a lot of
bridges just to join one part of the country to another. So we need to construct a lot of
bridges to ease the extension of road network as well as to carry out other development
works in an efficient way. So there is a huge potential of bridge engineering in Nepal.
In this project we are assigned to design a bridge over Kerunga Khola joining Devi
Mandir Chowk of Madhi V.D.C. to Kerunga Chowk of Kalyanpur V.D.C. of Chitwan
district. As it is a local road so a single lane bridge shall suffice. But as per the provisions
in Nepal Road Standards (a minimum of 6 m carriageway for bridges longer than 50 m),
the bridge will be designed as a two-lane bridge. We are supposed to design the most
economic bridge for this section based of the various data provided by LRBSU. This
report is also prepared as a part of project work for the fulfilment of the Project-II as per
the syllabus of Bachelor of Civil Engineering fourth year second part.
In Nepal, mostly RCC T-beam superstructure is preferred as the resources to design
and construct are readily available in Nepal. But for understanding of various other
superstructures, we chose to use a Steel Truss bridge along with the T-girder over the
river. The variation in design procedures for the two types and a need to accommodate
these superstructures over the pier has helped to enhance our understanding of the
essentials of Bridge Engineering.
1.2
Title of project work
The main objective of this project is to design a bridge over Kerunga Khola by using
the Limit State approach of design. So this project is entitled as “Design of Bridge over
Kerunga Khola, Chitwan”.
1.3
Project assignment
Following assignments were completed during the completion of the project:
1.1.1. Study of topographic, geological, hydrological, geotechnical and traffic
condition of bridge site from previous reports.
1.1.2. Visit of bridge site and preparation of site observation report including
verification of data required.
1.1.3. Finding the different alternate span arrangements of bridge for the site and
comparing them.
1.1.4. Carryout design and detailing of selected bridge type.
1.1.5. Design of bearing.
1.1.6. Design of abutment and pier.
1.1.7. Design of foundation.
1
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
1.1.8. Preparation of detail drawing of bridge superstructures with its all
components, abutments, pier, bearing and footing required for the construction
of selected bridge type.
1.4
Objectives
As the title suggests the main objective of this project is to analyse and design the
Steel Truss Bridge by Limit State Method. In addition to this following objectives were
set while fulfilling the above maintained assignments:
a. To be familiar with the bridge and its design.
b. To recognize various types of bridges.
c. To know about various type of loading and their forms of application.
d. To understand various methods used in the design of the structural component of
loading of bridge and their limitations.
e. To be familiar with the standard specification regarding the design of bridge.
Thus the objective of this project is to obtain the basic ideas of bridge building.
1.5
Client Requirements
The bridge was designed for LRBSU as per their requirement based on the present
need of the site and possible future need. The requirements of the bridge to be
designed are as follows:
ο· Type of Road
Local Road
ο· Footway
No
ο· Wearing surface
Asphalt Concrete
ο· Bridge type
Permanent
2
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
METHODOLOGY
2.1. Acquisition of Data
2.1.1. Preliminary Data
For the design of our bridge the preliminary data needed was acquired from the report prepared by
consultancy but in actual practice it is done by following methods.
Site selection survey
Site selection survey is done by a team of bridge engineer, geotechnical engineer, surveyor and hydrologist.
After consultation with local residents, technical personnel of Divisional Road Office of the site, proposed
bridge alignment is fixed.
Topographical survey
Tachometric survey is carried for detailed engineering survey of the proposed bridge site. Theodolites, level
machines, staffs and measuring tape are usually used for detailed survey.
After consultation with the technical personnel and the local villagers and as directed by the river morphology;
an axis joining line joining left bank and right bank is fixed. Temporary Benchmark is also fixed.
The bridge site detailing area covers a suitable region along the length of river both upstream and
downstream. It also covers left and right banks along the existing approach roads. L-section of approach road
and river measured and also the benchmarks and bridge points are shown in contour maps.
Geotechnical Investigation
Geotechnical investigation is one of the major parts of the project work for the design of the proposed bridge
at Kerunga Khola in Chitwan district. Geotechnical investigation works includes core drilling, test pitting, visual
investigation at site. Detail report and test results with bore logs are included in the actual project report.
For our project the site and its contour map, hydrological data and geotechnical data were provided by LRBP
(Local Road and Bridge Programme).
2.1.2.
Hydrology
Methods
The maximum discharge which a bridge across a natural stream is to be designed to pass can be estimated by
the following methods:
a) By using one of the empirical formulae applicable to the region;
b) By using the rational method involving the rainfall and other characteristics for the area;
c) By the area velocity method, using the hydraulic characteristics of the stream such as cross sectional
area, and the slope of the stream;
d) From any available records of the flood discharges observed at the or at any other site at the vicinity.
It is desirable to estimate the flood discharge by all or at least two of the above methods. These methods are
briefly discussed here.
A. Empirical formulae
Empirical formulae for flood discharge from a catchment have been proposed of the form:
Q=CAn
Where, Q=maximum flood discharge in m3 per second.
A= catchment area in Km2.
C= constant depending on the nature of the catchment and the location.
n= constant.
A popular empirical formulae of the above type is Ryve’s formulae given by equation
Q=CA2/3
3
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
The value of C is taken as 6.5 for the tracts near the coast, 8.5 for the areas between 25 and 50 km of
the coast and 10 for the limited areas near hills.
B.
Rational formulae
A rational formula for the flood discharge should take into account the intensity, distribution and
duration of rainfall as well as the area, shape, slope, permeability and initial wetness of catchment.
Many complicated formulae are available in treatises on hydrology. A typical rational formula is:
Q=AIoλ
Where, Q= maximum flood discharge in m3 per second
A= catchment area in Km2.
Io= peak intensity of rainfall in mm per hour.
λ= a function depending upon characteristics of catchment in producing the peak run-off.
λ=0.56Pf/ (to+Io)
to= concentration time in hours.
to= (0.88 L3/H)0.35
L= distance from the critical point to the bridge site in kilometres.
H= difference in elevation between the critical point and the bridge site in meters.
P= percentage coefficient of run-off for the catchment characteristics.
F= a factor to correct for the variation of intensity of rainfall I o over the area of the catchment.
C.
Area-velocity method
The area-velocity method based on the hydraulic characteristics of the stream is probably the most
reliable among the methods of determining the flood discharge. The velocity obtaining in the stream
under the flood conditions is calculated by Manning’s or similar formula: Manning’s formula is used
here. The discharge is given by equation:
Q= AV
Where, Q= discharge in m3 per second.
A= wetted area in m2.
V= velocity of flow in m/s
= (1/n) ο΄R2/3ο΄S1/2
n= coefficient of roughness.
S= slope of stream.
R= hydraulic mean radius in m.
= wetted area in m2/wetted area in meters.
The wetted area and wetted perimeter are obtained from the cross-section of the stream at site
drawn to scale with the floods levels marks therein. The quantity S in equation denotes the slope of
stream.
D. Estimation from flood marks
If flood marks can be observed on an existing bridge structure near the proposed site, the flood
discharge passed by the structure can be estimated reasonably well, by applying appropriate
formulae available in treatises on hydraulics. It is possible by inspection to ascertain the flood levels
soon after a flood. Sometimes, these flood marks can be identified even years after a flood, but it is
desirable to locate these as soon after the flood as possible.
Design discharge
The design discharge may be taken as the maximum values obtained from the least two method
mentioned. If the values obtained by two methods differ by 50%, then the maximum design discharge
is limited to 1.5 times the lower estimate. Freak discharge of high intensity due to failure of dam or
tank constructed upstream of the bridge site need not to be catered for. From consideration of
4
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
economy, it is not desirable to aim to provide for the passage of the very extraordinary flood that may
ever happen at a particular site. It may be adequate to design for a flood occurring once in a 20 years
in case of culverts and once in 100 years for bridges and to ensure that the rarer floods be passed
without excessive damage to structure.
Linear waterway
When the watercourse to be crossed in an artificial channel for irrigation or navigation, or when the
banks are well defined for artificial streams, the linear waterway should be full width of channel or
stream.
For large alluvial stream with undefined banks, the required linear waterway should be
determined using Lacey’s formula given by equation
L = C√Q
Where, L= the linear waterway in m.
Q= the designed maximum discharge in m3 per second.
C= a constant, usually taken as 4.8 for regime channels, but may vary from 4.5 to 6.3
according to local conditions.
It is not desirable to reduce the linear waterway below that for regime condition. If a
reduction is affected, special attention should be given to afflux and velocity of water under the
bridge. With reduced waterway, velocity would increase and greater scour depth would be involved,
requiring deeper foundations. Thus, any possible saving from a smaller linear waterway will be offset
by extra expenditure on deeper foundations and protective works.
Afflux is the heading up of water over the flood level caused by the constriction of waterway
bridge site.it can be computed from equation
X = (v2/2g) ο΄ ((L2/c2.L12)-1)
Where, X= afflux
V= velocity of normal flow in stream
g= acceleration due to gravity
L= width of stream in at H.F.L
L1= linear waterway under the bridge
c= coefficient of discharge through the bridge, taken as 0.7 for sharp entry and 0.9 for bell
mouth entry.
The afflux should be kept minimum and limited to 300mm. afflux causes increase in velocity in
downstream side, leading to greater scour depth and requiring deeper foundations. The road
formation level and the top level of the guide bunds are dependent on the maximum water level on
upstream side including afflux.
Economical span
Considering only the variable items for the given linear waterway, the total cost of the
superstructure increases with the increase or decrease in span length. The most economical span is
that for which the cost of superstructure equals the cost of sub-structure. This condition may be
derived as below:
Let,
A= cost of approaches
B= cost of two abutments including foundations
L= total linear waterway
l= length of one span
n= number of spans
P= cost of one pier, including foundation
C= total cost of bridge.
Assuming that the cost of superstructure of one span is proportional to the square of the span length,
total cost of superstructure equals n.kl2, where k Is a constant.
5
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
The cost of railing, flooring, etc., is proportional to the total length of bridge and can be taken as k’L.
C= A + B + (n-1) P + nkl2 +K’L
For minimum cost, dC/dL should be zero, we get
P= k.l2
Therefore, for an economic span, the cost of superstructure of one span is equal to the cost of
substructure of same span.
The economical span (le) can then be computed from the equation
Le= √(P/k)
P and k are to be evaluated as average over a range of possible span lengths.
Location of pier and abutments
Pier and abutments should be as to make the use of best the foundation conditions
available. As far as possible the most economical span as above should be adopted. If navigational or
aesthetic requirements are to be considered, the span may be suitably modified. As a rule, the
number of span should be kept low, as pier obstructed water flow. If pier are necessary, odd number
of span should be preferred.
For small bridges with open foundations and solid masonry piers and abutments, the
economical span is approximately 1.5 times the height of pier or abutments, while that for masonry
arch bridges it is about 2 times the height of keystone above the foundation, the question has to be
examined in detail.
The alignment of pier and abutments should be, as far as possible, parallel to the mean
direction of flow in the stream. If any temporary variation in the direction and velocity of stream
current is anticipated, suitable protective works should be provided to protect the sub-structure
against the harmful effects on stability of the bridge structure.
Vertical clearance above H.F.L
For the high level bridges, a vertical clearance should be allowed between the H.F.L, and the
lowest point of the superstructure. This is required to allow for any possible error in the estimation of
the H.F.L., and the design discharge. It also allows floating debris to pass under the bridge without
damaging the structure.
The difference between the vertical clearance and the free-board is sometimes not clearly
understood. While vertical clearance is the difference in level between H.F.L. and the lowest point of
the superstructure, freeboard is associated with the approaches and guides bunds. The freeboard at
any point is the difference between the highest flood level after allowing afflux, if any, and the
formation level of the embankment on the approaches or the top level of guide bunds at that point,
for high level bridges, the freeboard should not be less than 600 mm.
Scour Depth
Scour of stream bed occurs during the passage of a flood discharge, when the velocity of
stream exceeds the limiting velocity that can be withstand by the particles of the bed material. The
scour should be measured with reference to existing structures near the proposed bridge site, if this is
possible. Due allowance should be made in the observed value for additional scour that may occur
due to the designed discharge being greater than the flood discharge for which the scour was
observed, and also due to increased velocity due to obstruction to flow caused by the construction of
bridge.
Where the above practical method is not possible, the normal depth of scour may be
computed by equation for natural streams in alluvial beds.
d=0.473(Q/f).33
Where,
d= normal depth of scour below H.F.L. for regime conditions in a stable
channel in meters.
Q=designed discharge in m3 per second
f=Lacey’s silt factor for a representative sample of bed material
The minimum depth of foundation below H.F.L. is kept at 1.33d for erodible strata
6
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
2.2. Structural Planning and Preliminary Design
Considering the fact that we had very little background in the field of bridge design, this project
turned out to be more strenuous than we had ever imagined. The fine guidance from our supervisor and
the books recommended by him combined with immense hard work from each one of us are the reasons
for the successful completion of this project. The books available in the library were also of great help.
Although they were few in number, the old reports on steel truss bridges and R.C bridges facilitated us
with plenty of necessary materials. Finally, the internet served us with the resources required for the final
touch up of this report.
a.
Types of bridge
i. According to Function
ο· Aqueduct
ο· Viaduct
ο· Highway
ο· Railway
ο· Road-cum-rail or Pipeline Bridge
ii. According to the material of construction
ο· Timber
ο· Masonry
ο· Iron
ο· Steel
ο· Reinforced concrete
ο· Prestressed concrete
ο· Composite
iii. According to the type of superstructure
ο· Slab
ο· Beam
ο· Truss
ο· Arch
ο· Suspension
iv. According to the position of bridge floor relative to superstructure
ο· Deck
ο· Through
ο· Half through
ο· Suspended
v. According to the inter-span relations
ο· Simple
ο· Continuous
ο· Cantilever
vi. According to the method of connection of different parts of superstructure
ο· Pin connected
ο· Welded
ο· Riveted
vii. According to the road level relative to the high flood level of the river
ο· High-level
ο· Submersible
viii. According to the method of clearance for navigation
ο· High-level
ο· Movable-Bascule
ο· Movable-swing
ix. According to anticipated type of service and duration of use
ο· Permanent
7
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
ο· Temporary
ο· Military
b. Selection of Bridge Site
It may not be always possible to have a wide choice of sites for a bridge. This is particularly so in case of
bridges in urban areas and flyovers. For rivers bridges, in rural areas, usually a wider choice may be
available. The characteristics of an ideal site for a bridge across a river are:
ο·
ο·
ο·
ο·
ο·
ο·
ο·
ο·
ο·
A straight reach of river.
Steady river flow without whirls and across currents.
A narrow channel with firm banks.
Sustainable high banks above high flood level on each side.
Rock or other hard in-erodible strata close to the river bed level.
Proximity to a direct alignment of the road to be connected.
Absence of sharp curves in the approaches.
Absence of expensive river training works.
Avoidance of excessive underwater construction.
In selection a site a care should be taken to investigate a number of probable alternative sites and then
decide on the site which is likely to serve the needs of the bridges at the least cost.
c.
Loading
IRC loads for the bridge design:
According to IRC: 6-2000, road bridges and culverts are classified on the basis of loadings that they are
designed to carry.
IRC class AA loading:
This loading is to be adopted within certain limits, in certain existing or contemplated industrial areas, and
along certain specified highways and areas. Bridges designed for class AA loading should be checked for
class A loading is considered in each lane.
IRC class A loading:
This loading is normally considered on all in which dominant bridges and culverts are constructed. One
train of class A loading is considered in each lane.
IRC class B loading:
This loading is normally considered when the structure is temporary and for bridges in specified area.
Structures with timber spans are to be regarded as temporary structures.
d. Superstructure
The basic function of bridge superstructure is to permit the uninterrupted smooth passage of traffic
over it and to transmit the loads and forces to the substructure safely through the bearings. Although it is
difficult to stipulate the aesthetic requirements, it should, however, be ensured that the type of
superstructure adopted is simple, pleasing to the eye, and blends with the environment.
The superstructure of any bridge must be designed such that it satisfies geometric and load carrying
requirements set forth by its owner. This geometric requirement depends upon the number and widths of
traffic lanes and footpaths that have to be carried across. They also depend on overall alignment and
various horizontal and vertical clearances required above and below the roadway. The superstructure
designed has to meet various structural design requirements such as strength, stiffness and stability.
The horizontal and vertical alignment of a bridge is governed by the geometrics of the highway,
roadway or channel, it is crossing. For girder type bridges, the girders may either be curved or straight,
and may be aligned on chords between supports with the deck slab built on the curve. The following
points require close examination when girders are aligned on a chord:
ο·
ο·
ο·
ο·
Non symmetric deck cross section
Deck finish of the warped surface
Vertical alignment of the curbs and railings, to preclude visible discontinuities
Proper development of super elevation
The components of superstructure are as follows:
Lighting
8
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
The lighting of the bridge is generally in accordance with the provisions of the authority having jurisdiction
on that area.
Drainage
The transverse drainage of the roadway is usually accomplished by providing suitable crown in the
roadway surface, and the longitudinal drainage is accomplished by camber or gradient.
Traffic lane
Roads designed for traffic flow can be single lane, double lane or more. Road width in meters should be
divided by 3.65 and the quotient approximated to the nearest whole number of design traffic lanes. We
have designed our bridge with two traffic lane.
Road width
Road width is the distance between the roadside faces of the kerbs which depends on the number and
width of traffic lanes and the width of the bounding hard shoulders. For our project we have designed
road width of 7.5 m.
Footpaths
Footpaths or walkways are generally provided where pedestrian traffic is anticipated, but not on major
arteries or in country sides. Its width is 1.5 m generally, but may be as narrow as 0.6 m and as wide as 2.5
m depending on the requirements. For our project we have designed footpath of 1 m wide and 300 mm
deep.
Road kerb
The road kerb is either surmountable type or insurmountable type. In the absence of walkways, a road
kerb is combined with parapet.
Parapets
Parapets can be of many shapes and of variable sturdiness. They are designed to prevent a fast moving
vehicle of a given mass from shooting off the roadway in the event of an accidental hit. Their height varies,
but it should be at least 700 mm. For our project we have designed parapet of 1000 mm depth and 150
mm width.
Handrails
The parapets are usually mounted by metal hand rail, about 350 mm high. Their roadside face is double
sloped. For our project we have designed handrail of size 50 mmο΄50 mm.
Crash barriers
Sometimes walkways are protected from the erring vehicular traffic by crash barriers which act as
insurmountable kerbs and deflect the hitting vehicles back into the traffic lane.
Expansion and roadway joints
To provide for expansion and contraction, joints should be provided at the expansion end of spans, at
other points, where they may be desirable. Joints are preferably sealed to prevent erosion and filling of
debris.
Medians
On expressways and freeways, the opposing traffic flows are separated by median strips. These reduce the
possibility of accidents due to head on collisions.
Super-elevation
The super-elevation of the surface of a bridge on a horizontal curve is provided in accordance with the
applicable standard. This should preferably not exceed 0.06 m per meter, and never exceed 0.08 m per
meter.
e.
Substructure
Substructure of a bridge refers to that part of it which supports the structure that carries the roadway or
the superstructure. Thus substructure covers pier and abutment bodies together with their foundations,
and also the arrangements above the piers and abutments through which the superstructure bears on the
structure. The latter are called bearings.
i. Foundation
9
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
A foundation is that part of the structure which is in direct contact with the ground and transmits loads
to it. A footing is that part of the foundation that transmits the loads directly to the soil.
Types of Foundations:
A. Deep Foundations
Deep foundations generally have depth greater than the width. They are constructed by various
special means. They are of following types:
ο·
Piles
Piles are essentially giant sized nails that are driven into the subsoil or are placed in after
boring holes in the subsoil. The giant sized nails that are driven into the subsoil or are placed
in after boring holes in the subsoil. The giant sized nails are made of concrete, steel or timber
and can be square, rectangular, circular or H-shaped in section. A group of piles is capped
together at top, usually by a reinforced concrete cap, to support the pier of crapped together
at top, usually by a reinforced concrete cap, to support the pier or abutment body above.
ο·
Caissons or wells
Caisson is constructed at open surface level in portions and sunk downwards mechanically by
excavating soil from within the dredge hole all the way till its cutting edge reaches the
desired founding level. The well is then effectively scaled at bottom and at least partly filled
by sand. The surface level and the portions near it are capped. The pier or abutment is then
constructed on the cap.
B. Shallow Foundations
A foundation is shallow if its depth is less than or equal to its width. These are generally placed after
open exaction, and are called open foundations. The design of open foundations is based on complete
subsoil investigations. But in case of low safe bearing capacity of soil, such foundations have to be
disallowed. The selection of the appropriate type of open foundation normally depends upon the
magnitude and disposition of structural loads, requirements of structures (settlement characteristics,
etc.), type of soil or rock encountered, allowable bearing pressures, etc. Where rocky stratum is
encountered at shallow depths, it may be preferable to adopt open foundations because of its
advantage in permitting proper seating over rock and speed of construction work. They are of
following types:
ο·
Spread Footing (Isolated footing, combined footing, strip footing)
An isolated footing is a type of shallow foundation used to transmit the load of an isolated
column to the subsoil. This is the most common type of foundation. The base of the column
is enlarged or spread to provide individual support for the load.
A spread footing which supports two or more columns is termed as combined footing. The
combined footing may be rectangular in shape if both the columns carry equal loads, or may
be trapezoidal if they carry unequal loads. If the independent spread footings of two columns
are connected by a beam, it is called strap footing. A strap footing may be used where the
distance between the columns is so great that a combined trapezoidal footing becomes quite
narrow.
The strap footing consists of single continuous R.C. slab as foundation of two or three or
more columns in a row. It is suitable at locations liable to earthquake activities. It also
prevents differential settlement. In order to have better stability a deeper beam is
constructed in between the columns. It is also known as continuous footing.
ο·
Mat or Raft Footing
A raft or mast is a footing that covers the entire area beneath a structure and supports all the
walls and columns. When the allowable soil pressure is low or the loads are heavy the use of
spread footings would cover more than one half of the area and it may prove more
economical to use mat or raft foundation. The mat or raft tends to bridge over the erratic
deposits and eliminates the differential settlement. It is also used to reduce settlement
above highly compressive soils, by making the weight of structure and raft approximately to
weight of soil excavated.
ii. Bearing
10
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Bearings are provided in bridges at the junction of the girders or slabs and the top of pier and
abutments. Bearings transmit the load from the superstructure to substructure in such a way that the
bearing stresses developed are within the safe permissible limits. The bearings also provide for small
movements of the superstructure. The movements are induced due to various reasons such as:
ο·
ο·
ο·
ο·
ο·
Movement of the girders in the longitudinal direction due to variations in the temperature
The deflection of the girder causes rotations at the supports
Due to sinking of the supports the vertical movements are developed
Movements due to shrinkage and creep of concrete
In the case of prestressed girders, prestressing the girders cause movements of girders in the
longitudinal direction
Types of Bearings
a.
Fixed Bearings
Fixed bearings permit rotations while preventing expansion. They are of the following types:
ο· Steel Rocker bearing
ο· R.C. Hinge bearing
b. Expansion Bearings
Expansion bearings accommodate both horizontal movements and rotations, they are of
following types:
ο· Sliding Plate bearing
ο· Sliding cum Rocker bearing
ο· Steel Roller cum Rocker bearing
ο· R.C Rocker cum Roller bearing
ο· Elastomeric bearing
Elastomeric Bearing
Elastomeric bearings are widely used in present times as they have less initial and maintenance cost.
Besides occupying a smaller space, elastomeric bearings are easy to maintain and also to replace
when damaged, chloroprene rubber termed as neoprene is the most commonly used type of
elastomer in bridge bearings. Neoprene pad bearings are compact, weather resistant and flame
resistant. Hence, nowadays elastomeric bearings have more or less completely replaced steel rocker
and roller bearings.
iii. Pier
The bridge supports in between the abutment supports are referred to as piers. The choice of
construction of the bridge deck will dictate the choice of the type of pier. If support is required at
intervals across the full width of the bridge deck, then some form of supporting wall or portal frame is
made for the pier. However when deck has some capacity within itself to span transversely at an
intermediate support positions by means of a diaphragm within the depth of the deck, there is wider
choice available for pier.
Simplicity in the formation of a pier not only has the merit of providing easier and more economical
construction, but it is also likely to produce more attractive result. But for some special cases,
complex shapes may be adopted. In this case the bearings are placed at the heads or the feet of the
piers.
Types of Pier
Depending on the type, size and dimensions of the superstructure, the following types of piers are in
general use:
a)
Solid type pier
The solid type pier is generally built using brick or stone masonry or concrete. This type with
cut ease water is widely used for river bridges.
b) Trestle type pier
The trestle type pier comprises of a number of reinforced concrete columns with a
concreting cap at the top. The trestle type of pier finds wide applicability in the case of
flyovers and elevated roadways generally used for crossing in city roads.
c) Hammer-head type pier
11
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Hammer-head type of pier consists of a massive single pier with cantilever caps on opposite
sides resembling the head of a hammer. This type of pier is generally suitable for elevated
roadways and when used in river bridges, there is minimum restriction of waterway.
d) Cellular type pier
For the construction of massive piers carrying multilane traffic, it is economical to use cellular
type reinforced concrete piers which results in the savings of concrete. However cellular type
piers require costly shuttering and additional labor for placing of reinforcements. For tall
piers, slip forming work can be adopted for rapid construction.
e) Framed type pier
R.C. type piers are aesthetically superior and rigid due to monolithic joints between the
vertical, inclined and horizontal members. These type of piers are ideally suited to reduce the
span length of main girders on either side of center line of the pier resulting in savings in the
cost of superstructure. However this type of construction requires two expansion joints at
close intervals with increase of maintenance cost.
Forces acting on piers
The various forces to be considered in the design of piers are as follows:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
Dead load of superstructure and pier.
Live load of vehicles moving on the bridge.
Effect of eccentric live loads.
Impact effect for different classes of loads.
Effect of buoyancy on the submerged part of the pier.
Effect of wind loads acting on the moving vehicles and the superstructure.
Forces due to water current.
Forces due to wave action.
Longitudinal forces due to tractive effort of vehicles.
Longitudinal forces due to braking of vehicles.
Longitudinal forces due to resistance in bearings.
Effect of earthquake forces.
Forces due to collision for piers in navigable rivers.
The stability analysis for the piers is generally made by considering some of the critical forces which
will have significant effect on the stresses developed in the piers.
Design of pier
The salient dimensions of pier like the height, pier width and batter are determined as follows:
a)
Height
The top level of pier is fixed to 1 to 1.5m above the high flood level, depending upon the
depth of water on the upstream side. Sufficient gap between the high flood level and top of
pier is essential to protect the bearings from flooding.
b) Pier Width
The top of pier should be sufficient to accommodate the two bearings. It is usually kept at a
minimum of 600 mm more than the outer to outer dimension of the bearing plates.
c)
Pier Batter
Generally the sides are provided with a batter of 1 in 20 to 1 in 24. Short piers have vertical
sides. The increased bottom width is required to restrict the stresses developed under loads
within safe permissible values.
d) Cut and Ease Waters
The pier ends are shaped for streamlining the passage of water. Normally the cut and ease
waters are either shaped circular or triangular.
iv. Abutment
Abutments are end supports to the superstructure of a bridge. Abutments are generally built using
solid stone, brick masonry or concrete. An abutment has three distinct structural components:
12
Design of Bridge Over Kerunga Khola, Chitwan
a.
b.
c.
2069 – AB Bridge
Breast wall
Wing wall
Back wall
The design of abutment is done precisely in the same manner as the design of pier. The dimensions
are first determined from the practical point of view and its stability is subsequently tested. The
important additional force which the abutment has to withstand is the earth pressure of the earth
filling behind the abutment. The minimum top width of the abutment should be 3 to 4 feet with the
front batter of 1 in 24 and back batter of 1 in 6. Eddies erode the toes of the bank behind the
abutment and thus the cost of maintenance of the road is increased. In order to overcome this defect
and give the smooth entry and exit to the water, splayed wing walls to the abutment are constructed.
Function
ο·
ο·
To finish up the bridge and retain the earth filling
To transmit the reaction of the superstructure to the foundation.
Design
ο·
ο·
ο·
ο·
ο·
Height: Height is kept equal to that of piers.
Abutment batter: The water face is kept vertical or a small batter of 1 in 24 to 1 in 12 is given. The
earth face is provided with a batter of 1 in 3 to 1 in 6 or it may be stepped down.
Abutment width: The top width should provide enough space for bridge bearings and bottom
width is dimensioned as 0.4 to 0.5 times the height of the abutment.
Length of abutment: The length of abutment must be at least equal to the width of the bridge.
Abutment cap: The bed block over the abutment is similar to the pier cap with a thickness of 450
to 600 mm.
Forces acting on abutment
ο·
ο·
ο·
ο·
ο·
ο·
Dead load due to superstructure
Live load due to superstructure
Self weight of the abutment
Longitudinal force due to tractive effort and braking
Forces due to temperature variation
Earth pressure due to backfill
Abutment should be designed in such a way that it can resist the forces mentioned above.
2.3. Idealization and Analysis of bridge structure
2.3.1. Influence Line Diagram
Usually the structures are analysed for loads which do not change their points of application on the
structure. Very often structures have to be analysed for a number of parallel moving loads which keep on
changing their positions on the structure. In such cases the internal stresses in the structure at any given point,
which depend on the positions of the loads, keep on varying as the loads take up different positions on the
structure.
A typical instance is a bridge loaded with a number of moving vehicles, which are then said to constitute a train
of wheel loads. In order to design such structures it is not enough to analyse the structure for a given position
of loads and calculate the stress resultants namely: bending moments, radial and normal shear forces at any
section in a member of the structure. The engineer must know the maximum values of stress resultants, both
positive and negatives, at any section of the structure. Sometimes the designer would even like to know the
maximum deflection at a given point when a structure is subjected to moving loads.
The maximum value of the stress resultants or the deflection at a given section could be found by taking a
number of trial positions of the loads. Such a procedures apart from being time consuming is also uncertain.
The task is very much simplified by using the concept of influence line.
13
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
An influence line is a graph or curve showing the variation of any function such as reaction, bending moment,
shearing force, deflection etc. at a given point of a structure, as a unit load parallel to a given direction, crosses
the structure.
An influence line is thus a relation between the required function at any given point of the structure and the
position of a moving unit load on the structure.
The direction of the moving unit load depends on the nature of loading to be expected in the structure.
Use of Influence Line Diagrams
Using the principle of superposition, the following two types of problems can be solved with the help of
influence lines:
ο·
ο·
First, if the influence line for a function is known, its value for a given position of a number of parallel
moving loads can be found.
The second application is of far more practical importance, influence lines can be used to locate very
easily that particular position of a number of parallel moving loads on a structure, which will give the
maximum positive or maximum negative value of a function at a given point on the structure.
2.3.2. Pigeaud’s Method
Pigeaud’s method is used for the analysis of slabs spanning in two directions.
Analysis of Slab Decks
i) Slabs spanning in one direction
For slabs spanning in one direction, the dead load moments can directly be computed assuming the slab to be
simply supported between the supports. Bridges deck slabs have to be designed for I.R.C. loads, specified as
class AA or A depending on the importance of the bridge. For slabs supported on two sides, the maximum
bending moment caused by a wheel load may be assumed to be resisted by an effective width of slab
measured parallel to the supporting edges,
For a single concentrated load the effective width of dispersion may be calculated by the equation,
be=Kο΄x(1-x/L)+bw
Where, be=Effective width of slab on which load acts
L=Effective span
x=Distance of centre of gravity from nearer support
bw=Breadth of concentration of load
K=a constant depending on the ratio (B/L) and is compiled in IRC:21-2000 for simply supported and
continuous slabs.
ii) Slabs spanning in two directions
In the case of bridge decks with tee beams and cross girders, the deck slab is supported on all four sides and is
spanning in two directions. The moments in two directions can be computed by using the design curves
developed by M. Pigeaud.
The method developed by Pigeaud is applicable to rectangular slabs supported freely on all four sides and
subjected to a symmetrically placed concentrated load as shown in the figure below.
The notations used are as follows:
L=Long span length
B=Short span length
u,v=Dimensions of the load spread after allowing for dispersion through the deck slab
K=Ratio of short to long span=(B/L)
M1=Moment in the span direction
M2=Moment in the long span direction
m1 and m2=coefficient for moment along short and long direction
14
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
µ=poison’s ratio for concrete generally assumed as 0.15
W=Load from the wheel under consideration
The dispersion of the load may be assumed to be at 45β° through the wearing coat and deck slab according to
IRC:21-2000 code specifications. Consequently, the effect of contact of wheel or track load in the direction of
span shall be taken as equal to the dimension of the tyre contact area over the wearing surface of the slab in
the direction of slab plus twice the overall depth of the slab inclusive of the thickness of the wearing surface. It
is sometimes assumed to be at 45β° through the wearing coat but at steeper angle through the deck slab.
The bending moments are computed as
M1=(m1 + µο΄m2) ο΄W
M2=(m2 + µο΄m1) ο΄W
Fig: Dispersion of wheel load through deck slab
Fig: Dispersion of wheel load through wearing coat
The values of the moment coefficients m1 and m2, depend upon parameters (u/B), (v/L) and K.
Curve to compute moment coefficients of slabs completely loaded uniformly distributed load or dead load of
slab for different values of K and 1/K is also given below. The Pigeaud’s curves used for the estimation of the
moment coefficients m1 and m2 for value of k= 0.9 used in our design also follows.
15
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Fig: Moment coefficients for slabs completely loaded with uniformly distributed load,
coefficient is m1 for K and m2 for 1/K
2.3.3. Hendry-Jaeger Method
In this method, the cross beam can be replaced in the analysis by a uniform continuous transverse medium of
equivalent stiffness. According to this method, the distribution of loading in an interconnected bridge deck
system depends on the following three dimensionless parameters.
12 πΏ
ππΈπΌπ‘
α =A= 4 ( )3 ο΄
π β
π2 β πΆπ½
πΈπΌ
β=F= ( )
c=
2π πΏ πΈπΌπ‘
πΈπΌ1
πΈπΌ2
where L = span of the bridge
h = spacing of longitudinal girders
n = no. of cross beams
EI,CJ = flexural and torsional rigidities respectively of one girder
EI1,EI2 = flexural rigidities of the outer and inner longitudinal girders, where these are different
EIt = flexural rigidities of one cross beam
In case of beam and slab bridge without cross beams, nEIt in above equation is to be replaced by L.EIt, where
the latter gives the total flexural rigidity of the slab deck. Normally, for reinforced concrete T-beam bridges,
the flexural rigidities of the outer and inner longitudinal girders will be nearly equal.
The parameter A is the most important of the above three parameters. It is a function of the ratio of the
span to the spacing of longitudinals and the ratio of transverse to longitudinal rigidities. The second parameter
F is a measure of the relative torsional rigidity of the longitudinals and is difficult to determine accurately, due
to uncertainties surrounding the CJ values for practical girder sections.
Graphs giving the values of the distribution coefficients (m) for different conditions of number of
longitudinals (2-6) and two extreme values of F i.e. zero and infinity are available.
mF = m0+(m∞-m0)√
πΉ√π΄
3+πΉ√π΄
where mF is the required distribution coefficient and m0 and m∞ are respective coefficients for F=0 and F=∞.
Using the graph this method is applied.
16
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
ACQUISITION OF DATA FOR DESIGN
For the purpose of design we adopted the design data provided by LRBP. The
acquisition of data as given in the report prepared by A-One consultancy P .Ltd. is as
below:
3.1 Topographical
Survey
Topographical survey was carried out to prepare topographical map for pertinent
information that may be required for design, construction and maintenance.
Centre line of proposed bridge site
The bridge axis as established during site selection was surveyed. The elevation and
co-ordinates of the bore holes along the axis were surveyed.
Bench marks
The reference bench mark was established to start with the survey works. The suitable
and convenient place for starting bench mark was marked as BM1 on the permanent
concrete pillar which is situated near by the bridge site on left bank of the river.
3.2 Geology
and Topography
This report on soil investigation of Kerunga Khola Bridge in Chitwan discussed the sub-surface
exploration works carried out at the proposed site and presented the findings of the
investigation. The report is based on two boreholes:
The investigation work included boring, SPT Test, Laboratory tests and Analysis of
various test results to predict the allowable bearing capacity of sub-soil at the proposed
bridge site. The details of the investigation work as well as that of findings of the
analysis carried out are presented.
17
Design of Bridge Over Kerunga Khola, Chitwan
3.3
2069 – AB Bridge
Hydrology
Hydro-Meteorological Data
There was no gauging station in Kerunga Khola. The stream flow data for this river were
not available. Hence the estimation of the flow for this river had to be determined from
other approaches. To determine Design discharge, following methods are used.
1. Rational Method
A typical rational formula is:
Q=A.Í.l
Where, Q= Maximum flood discharge in m3 per second
A= catchment area km2
Í= peak intensity of rainfall in mm per hour
l = a function depending on the characteristics of the characteristics of
the catchment in producing the peak run-off
=
0.56ππ
tc+1
tc= concentration time in hours
= (0.87×L3/H)0.385
L= distance from the critical point to the bridge site in kilometers
H= difference in elevation between the critical point and the bridge site
in meters
P= coefficient of run-off for the catchment characteristics
f= a factor to correct for the variation of intensity of rainfall Í over the area of
catchment
Here, A= 56.3 km2
Í= .2496 mm/hr
L= 39.8 km
H= 1152.89 m
tc=(0.87×L3/H)0.385
= (0.87×39.83/1152.89)0.385
=4.42 hr
P= 0.20
f= 0.813
l=
0.56×0.20×0.813
4.41+1
=0.0168
Q=A.Í.l
=0.0168×53.6×0.2496
=0.8267 m3/sec
18
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
2. Area Velocity Method
A. From Available Field data, cross- section of River is shown figure below.
Average A=49.485m2
Average velocity = 0.203 m/s
Discharge (Q) = 49.485ο΄2ο΄0.203 =20.09m3/sec
Since it is such a small value, we go in our field data
B. From our site Visit & levelling Data, cross-section of River is shown figure below.
Average A=224.58m2
Average velocity = 0.203 m/s
Discharge (Q) =224.58ο΄2ο΄0.203 =91.18m3/sec
3. Empirical Formula
A. Ryve’s formula
Q= CA2/3
Where, Q= maximum flood discharge in m3 per second
A= catchment area in square kilometers
C= constant depending on the nature of the catchment and location
19
Design of Bridge Over Kerunga Khola, Chitwan
For flat tracts, take C= 6.8
Now, Q = CA2/3
= 6.8 × (56.3) 2/3
= 99.89 m3/sec
B. Modified Dicken’s Formula
QT= CTA0.75
1185
CT= 2.342 log(0.6T)log(
π
)+4
π+6
P=100× π΄+6 , where a= perpetual snow area in sq. km = 0
T= return period in years = 50
A= Catchment area in sq. km= 53.6 sq. km
0+6
P= 100×53.6+0 = 11.19
1185
CT= 2.342log(0.6×50)log(11.19)+4 = 11.00
QT= CTA0.75
= 11×53.60.75
= 217.90 m3/sec
C. Fuller’s Method
π΄
Qmax= QT (1+2(2.59)-0.3 )
QT = Qav (1+ 0.8 logT)
Qav= Cf A0.8
Cf =1.03 for Nepal
Qav= Cf A0.8
= 1.03×53.60.8
= 24.898 m3/sec
QT = Qav (1+ 0.8 logT)
=24.898(1+0.8log50)
=58.74 m3/sec
π΄
Qmax= QT (1+2(2.59)-0.3 )
53.6
= 58.74 (1+2(2.59)-0.3 )
= 106.07 m3/sec
20
2069 – AB Bridge
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Discharge Calculation table
S.No
Method
Discharge
1
Rational Method
0.8267 m3/sec
2
Area Velocity Method
91.18m3/sec
3
Ryve’s formula
99.89 m3/sec
4
Modified Dicken’s Formula
217.90 m3/sec
5
Fuller’s Method
106.07 m3/sec
As discharges from Area-velocity method and Ryve’s method are close to each other, we
consider discharges from these two methods only.
Hence, Design Discharge = (91.18+99.89)/2 = 95.54m3/sec
Calculation of Linear Waterway & Scour Depth:
Linear Waterway
= 4.75√π
Db = 95.54/46.43
= 2.058
= 4.75√95.54
= 46.43m
Db2
Scour Depth = 1.34( ππ π )^0.33
= 1.34(
2.0582
1.25
)^0.33
= 2m
The river at proposed bridge site is in regime condition and water way width is about
46.43m.
3.4 Observation
visit and Verification of data
Site visit was arranged at Kerunga Khola Bridge located in Kalyanpur VDC, Chitwan to
verify the data provided during the design.
Survey Observations
We first located the available benchmarks at site and the bridge axis points (Axis Left and
Axis Right). We transferred RL from the benchmark and then carried out leveling work along
the proposed bridge axis to obtain a tentative cross section. We also identified the high
flood marks after consulting with the local people and then transferred RL for the high flood
marks. The linear waterway at the bridge axis c/s was also found out by taping. The
observed data for both bridge sites are as follows:
21
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
BRIDGE SITE SURVEY
LEVELLING DATA
LOCATION: KERUNGA KHOLA
Hydrological Observations
STATIONS
CHAINAGE
BM1
AR
IS
FS
RISE
FALL
0.73
137.45
0+000
0.991
0.261
137.189
0+004
1.304
0.313
136.876
0+005.3
1.735
0.431
136.445
0+009
1.701
0.034
136.479
0+012
1.61
0.091
136.57
0+016
1.652
0.042
136.528
0+019
1.755
0.103
136.425
0+024.82
3.075
1.32
135.105
0+026.52
3.431
0.356
134.749
0+030.52
3.37
TP-1
AL
BS
REDUCED
LEVEL(m)
1.685
0.362
0.061
134.81
3.008
137.818
0+035.08
4.06
0+038.22
3.895
0.165
135.608
0+041.12
3.65
0.245
135.853
0+043.92
3.17
0.48
136.333
0+047.32
2.79
0.38
136.713
0+051.32
2.175
0.615
137.328
0+053.82
2.07
0.105
137.433
0+057.02
1.61
0.46
137.893
0+060.12
1.2
0.41
138.303
0.648
0.552
138.855
1.188
0.492
139.347
TP-2
1.68
HFL(2059)
SUM
2.375
4.095
2.198
7.098
∑BS-∑FS=4.095-2.198=1.897
∑Rise-∑Fall=7.098-5.201=1.897
135.443
5.201
We determined the approximate flow velocity by float method and also measured the linear
waterway 25m u/s and 25m d/s.
KERUNGA KHOLA
Date: 2072/10/16
Time: 11:40 am
Average velocity measurement by Float Method:
Length of stretch: 5.0m
22
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Time taken:
Obs. 1: 23 secs , Obs 2: 26 secs and Obs 3: 25 secs
Average time taken = 24.67 secs
Thus, average velocity = 5/24.67 = 0.203 m/s
Geotechnical Observations
We got to observe geotechnical investigation going on at site. Wash boring method of
investigation was adopted to determine soil profile as well as SPT value. The soil type of site
was also found out.
SITE
TYPE OF SOIL
KERUNGA
SILTY SAND
23
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
SELECTION OF BRIDGE TYPE
The choice of an appropriate type of bridge and planning of its basic features usually
constitutes a crucial decision to be taken by the bridge engineer. The designer must
consider all the preliminary data made available to him from the detailed investigation
before arriving at a solution. The entire completed structure should be the most suitable to
carry the desired traffic, adequately strong to support the incident loads, economical in first
cost and maintenance and aesthetically pleasing.
There are no hard-and-fast rules that can be used for the choice of bridge types in all cases.
Good design result from the serious search for the best solution for the given situation.
Hence we are proposing two alternative bridge types for each site.
Kerunga khola site
Proposed bridge axis alignment is curved w.r.t the river axis and the H.F.L level is quite
higher in this site.
For this site, we proposed
1) Prestressed box girder bridge of 60 m span
ο· With an intent of providing a single span without any intermediate piers in
the river i.e. not disturbing the natural waterway, prestressed box girder
bridge is suitable.
ο· Box girder has relatively high torsional rigidity as compared to the simple Tgirder.
ο· With the use of prestressed material and due to full section being effective,
the dead load is reduced and also the quantity of reinforcement necessary
for the bridge structure is reduced.
2) Steel truss mid span(36 m) with T-girder side spans(12 m each)
ο· Two different materials can be adopted to economize the bridge spans.
ο· Steel truss is economical for the span between 25-300 m.
ο· Prefabrication of steel truss is possible which helps in easy construction.
ο· Simple monolithic construction is possible with T-girder.
Alternative 2 will be adopted.
Though suitable for the given site, a prestressed box girder bridge is deemed to be
more complex in terms of analysis and design. Considering it to be more advanced for our
level of study and as the basics are yet to be comprehended, it shall be looked into only
when we are through with our adopted alternative.
As suggested by our supervisor, a truss superstructure for 36 m span may prove to be
uneconomical and a composite bridge superstructure would have been more appropriate.
But for learning purposes, we stayed on our decision and are implementing the
alternative 2 – as proposed – in our site.
24
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
STRUCTURAL PLANNING AND
PRELIMINARY DESIGN
For the given span of 60 m, a steel truss of 36 m span will be provided for the
central span and two t-girders of 12 m span each will be provided as side
spans. The abutment and pier shall be provided for depth beyond maximum
scour level and adequate vertical clearance off HFL shall be made available
underneath the decks.
Preliminary Design
T-Beam Bridge
1.
2.
3.
4.
Carriageway width = 6 m
Width of kerb = 0.6 m on each side
Height of kerb = 300 mm = 0.3 m
Wearing coat = Take Asphalt Concrete for wearing coat of bridge. Thickness of
wearing coat is taken 50 mm at edge and 92.5 mm at crown of carriage way to give
about 2% camber.
5. RC post of 225ο΄225ο΄1100 mm no: = 8 posts @ 1.5 m c/c
6. Size of RC slab = 200 mm thickness depth and 150 mm at tip of cantilever slab
7. Main girder:
Width of web (bw) = 300 mm > 250 mm (minimum)
Depth (D) = effective span/12 = 1 m
Number = 3 @ 2 m c/c
8. Fillet size = 150 x 300 mm and angle of inclination = 110β°
9. Spacing of cross beams = 2.5 m
10. No. of cross beams = 5
25
Design of Bridge Over Kerunga Khola, Chitwan
26
2069 – AB Bridge
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Truss Bridge
1.
2.
3.
4.
5.
6.
7.
8.
9.
Height of bridge = L/10 to L/6 = 3.6 to 6 m = Take 6 m
Depth of cross beam = B/(15 to 20) = 480 to 360 mm = Take 400 mm (B=7200 mm)
Single span of truss member = 4.5 m
Depth of stinger (D’) = span/20 = 4.5/20 = 0.225 m
No. of stringers = 4
Spacing of stringers = 2 m
Depth of slab = 200 mm
Angle of inclination of truss member = tan-1(6/4.5) = 53.13β° (preferable 45β° to 55β°)
Horizontal clearance = 6 m
27
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
ABUTMENT/PIER HEIGHT CALCULATION:
We have,
River bed level = 134.749 m
High flood level = 139.347 m (from field observation)
Maximum scour depth level = 130.749 m
Minimum foundation level=134.749-1.33*4=129.429 m
Calculation:
ο·
High flood depth = high flood level – river bed level = 139.347 – 134.749m = 4.598 m
ο·
Maximum scour depth up to foundation level= river bed level – Minimum foundation
level = 134.749 –129.429= 5.32 m
ο·
Clearance above HFL = 0.9m (min.)
Take 1.5m considering afflux condition.
ο·
Total depth of main girder = 1m
ο·
Height of abutment = 4.598 + 5.32+ 1.5 +1 = 12.418 m
ο·
Adopted height of abutment = 12.5m
28
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
STRUCTURAL ANALYSIS AND DESIGN
OF BRIDGE COMPONENTS
A. Design of RCC T-Girder (12 m span) bridge
I.
Analysis and design of Deck slab
1. Cantilever slab
0.15m
0.70m
Wheel of 114 KN
axle of Class A load
0.3m
0.17m
0.35m
0.6m
0.85m
Fig. Cantilever slab with dead and live load
Calculation of dead loads
ο§ Weight of railing = (9 × 0.225 × 0.225 × 1.1 × 25 + 12 × 3 × 0.0437) × 1.35/12 = 1.59 KN/m
(load acts 0.1625m from tip)
ο§ Weight of W.C. = 0.08 × 0.85 × 22 × 1.75= 2.618 KN/m (load acts at 0.425 m from support)
ο§ Weight of kerb = 0.6 × 0.3 × 25 × 1.35= 6.075 KN/m (loads acts at 0.25m from tip)
ο§ Weight of slab
= 0.17 × 1.45 × 25 × 1.35 = 8.32KN/m (rectangular portion) (0.725m from support)
= 0.5 × (0.35 - 0.17) × 0.85 × 25 × 1.35 = 2.582 KN/m (triangular part) (0.283m from
support)
Calculation of live loads
Live load per unit width of slab is calculated placing a wheel of 114 KN Axle at 0.15 m from
the face of the kerb. Live load per unit width of slab is found by dividing live load by
effective width of slab ‘bef’.
bef = 1.2a + b1 ≤ π⁄3
Where a = 0.2 + 0.5/2 = 0.45 m and b1 = 0.25 + 2 × 0.08 = 0.45 m
Thus, bef = 0.589 m and π⁄3 = 1.45/3 = 0.483 m
Adopt, bef = 0.483 m
57×0.483×1.5×1.25
Live load per unit width with impact =
= 213.75 KN/m (0.45 m from
0.5×0.483
support)
Where Impact factor = 1.25 and partial safety factor = 1.5
Design longitudinal bending moment at the face of main girder
Design BM = Max. BM due to dead load + Max. BM due to live load
29
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Mu = (1.59 × (1.45 - 0.1625) + 2.618 × 0.425 + 6.075 × 1.15 + 8.32 × 0.725 + 2.582 × 0.283) +
213.75 × 0.45
= 16.91 + 96.188 = 113.098 KN-m/m
Maximum transverse bending moment in direction of traffic
Mu = 0.3 BM due to live load + 0.2 BM due to dead load
= 0.3 × 96.188 + 0.2 × 19.91
= 32.84 KN-m/m
Design of slab
Check the depth of slab
Depth provided = 350-25-10/2 = 320 mm
ππ’
113.098 × 10^6
Balanced depth = √ ππ = = √
3.45 × 1000
= 195.85 mm
Where, Q = 0.36 fck × 0.48 × (1 -0.416 × 0.48) = 3.45
Depth provided > Balanced depth so, OK.
Main Reinforcement
Mu,l = 0.36 ο΄ fck ο΄ b ο΄ xu,lο΄ (d - 0.416 ο΄ xu,l)
= 0.36 x 25 x 1000 x 0.48 x 320(320 - 0.416 x 0.48 x 320)
= 354.04 KN
Since Mu,l>Mu , it is designed as SRURS.
0.87∗415∗π΄π π‘
Xu = 0.36∗25∗1000
= 0.0401Ast
113,09×106
& Ast = 0.87×415(320−0.416×0.0401Ast)
Solving, Ast = 1034mm2 > min. 0.12%
Provide 12mm ∅ bars @100mm c/c
Transverse Reinforcement
ππ’
π×d2
32.84∗10^6
= 1000×3202 = 0.32
Using SP16, Pt = 0.0915% <0.12%
So, provide minimum reinforcement.
Astmin = 0.12x1000x350/100
= 420mm2
Provide 10mm ∅ bars @180mmc/c
30
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
2. Restrained Slab
Analysis of slab
Direction
of traffic
Effective span in transverse direction of bridge = 2 – 0.3 = 1.7 m
Effective span in longitudinal direction of bridge = 3 - 0.25 = 2.75 m
Calculation of Bending Moment due to Dead Load
Bending moment is calculated by using PIGEAUD’S METHOD
Dead load due to W.C. and self-wt. of slab (W) = (0.08 × 22 × 1.75+ 0.2 × 25 × 1.35) × 1.7 ×
2.75 + 2 × 0.5 × 0.15 × 0.3 × 2.75 × 1.35 = 52.7 KN
BM in shorter span of slab = (m1 + µm2) × W × 0.8 = (0.048+0.15 × 0.016) × 52.7 ×0.8 =
2.125 kN-m
BM in longer span of slab = (m2 + µm1) × W × 0.8 = (0.016+0.15 × 0.048) × 52.7 ×0.8 = 0.978
kN-m
Shorter Span
Where, m1 = 0.048, m2 = 0.016 for K = Longer Span = 0.62 and 1⁄πΎ = 1.613, µ = 0.15
Calculation of Shear Force due to dead load
Max. SF due to dead load in shorter span of slab
= SF due to self-weight of WC & slab with fillet
=
ππ’π
3
+ Vufillet
= (0.2×25×1.35+0.08×22×1.75) ×
1.70
3
+ 0.25×0.15×0.3×1/2×2.75×25
= 5.957 KN (25% of weight fillet)
31
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Calculation of Bending Moment due to IRC Class A Loading
Bending moment is calculated by using PIGEAUD’S METHOD
BM due to Wheel I
BM in shorter span = (m1 + µm2) × W × 0.8 = 14.702 KN-m
BM in longer span = (m2 + µm1) × W × 0.8 = 11.649 KN-m
Where, m1 = 0.16, m2 = 0.11 for K =0.6 and m1 = 0.155, m2 = 0.125 for k =0.7
We have, k =1.7/2.75 = 0.62
By interpolation, m1 = 0.159 and m2 = 0.113
π’
Also, π΅ =
π+2π
π΅
=
0.5+2 × 0.08
1.7
π£
= 0.388 and, πΏ =
π+2π
πΏ
W= 57 × IF × κ©f = 57 × 1.25 × 1.5 = 106.875 KN
BM due to Wheel II
BM in shorter span = ½ × (BM of patch I – BM of patch II)
BM in longer span = ½ × (BM of patch I – BM of patch II)
For patch – I,
32
=
0.25+2 × 0.08
2.75
= 0.149
Design of Bridge Over Kerunga Khola, Chitwan
π’
π΅
=
π+2π
π΅
=
0.5+2 × 0.08
1.7
= 0.388 and
π£
πΏ
=
2069 – AB Bridge
π+2π
πΏ
=
2.65+2 × 0.08
2.75
= 1.022
For k = 0.6, m1 = 0.082, m2 = 0.023 and for k = 0.7, m1 = 0.082, m2 = 0.033
Thus, for k = 0.62, m1 = 0.082 and m2 = 0.025, W = 57 × 1.25 × 1.5 = 106.875 KN
BM of patch – I in shorter span = (m1 + µm2) × W × 0.8 × 0.5 × 2.65 / (0.5 × 0.25) = 77.715
KN-m
BM of patch – I in longer span = (m2 + µm1) × W × 0.8 × 0.5 × 2.65 / (0.5 × 0.25) = 33.805
KN-m
For patch – II,
π’
π΅
=
π+2π
π΅
=
0.5+2 × 0.08
1.7
= 0.388 and
π£
πΏ
=
π+2π
πΏ
=
2.15+2 × 0.08
2.75
= 0.84
For k = 0.6, m1= 0.095, m2 = 0.03 and for k = 0.7, m1= 0.098, m2= 0.04
Thus, for k = 0.62, m1= 0.0956 and m2= 0.032, W = 57 × 1.25 × 1.5 =106.875 KN
33
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
BM of patch – II in shorter span = (m1 + µm2) × W × 0.8 × 0.5 × 2.15 / (0.5 × 0.25) = 73.824
KN-m
BM of patch – II in longer span = (m2 + µm1) × W × 0.8 × 0.5 × 2.15 / (0.5 × 0.25) = 34.074
KN-m
Thus, due to wheel load II,
BM in shorter span = ½ × (BM of patch I – BM of patch II) = 0.5 x (77.715 – 73.824) = 1.945
KN-m
BM in longer span = ½ × (BM of patch I – BM of patch II) = 0.5 x (33.805 – 34.074) = 0.1345 KN-m
Calculation of Shear Force due to IRC Class A Loading
Shear force is calculated by effective width method.
For wheel I,
a = 0.25 + 0.2 + 0.08 = 0.53m
π΅
πΏ
= 0.62 so, α = 1.86 (for continuous slab)
b1 = W+2h = 0.25 + 2×0.08 = 0.41
Then,
bef 1 = α × a (1- a/l) + b1
= 1.86×0.53(1-0.53/1.70) + 0.41
= 1.09m
57
Load due to wheel I = maximum of (1.09 ππ
RA = 219.57 KN
57
1.09 0.25
+
2
2
) × 1.25 × 1.5 × 2
= 319.03 KN/m
RB= 99.46 KN
Design shear force due to DL & LL in shorter span = 5.957 + 219.57 × 0.8 = 181.613 KN
34
Design of Bridge Over Kerunga Khola, Chitwan
Calculation of Bending Moment due to IRC Class AA Loading
(Tracked Load)
π’
1.01
π£
2.75
=
= 0.594 for u = 0.85 + 2×0.08 = 1.01 m
π΅ 1.70
=
= 1 for v = 3.6 + 2×0.08 = 3.76 m (limited to 2.75 m)
πΏ 2.75
For k = 0.62,
m1=0.07 and m2=0.019
Effective load including impact = 350×1.25×1.5×2.75/3.76= 479.97 KN
So,
BM in shorter span = (0.07+0.15×0.019) ×479.97 = 34.97 KNm
BM in longer span = (0.019 + 0.15×0.07) ×479.97 = 14.16 KNm
(Wheeled Load)
35
2069 – AB Bridge
Design of Bridge Over Kerunga Khola, Chitwan
BM due to wheel I
u = 0.3+ 2×0.08 = 0.46
v = 0.15+ 2×0.08 = 0.31
π’
π΅
=
0.46
1.7
π£
0.31
= 0.27 and πΏ = 2.75 = 0.113
From Pigeaud’s curve for k = 0.6,
m1=0.195 and m2=0.16
w = 62.5 × IF × gf = 62.5 × 1.25 × 1.5 = 117.18KN
So,
BM in shorter span = (m1+µm2) × w × 0.8 =
(0.195+0.16×0.15) ×117.18×0.8 = 20.53KNm
BM in longer span = (µm1+m2) × w × 0.8 =
(0.16+0.15×0.195) ×117.18×0.8 = 17.74KNm
BM due to wheel 2
36
2069 – AB Bridge
Design of Bridge Over Kerunga Khola, Chitwan
For patch I,
π’
π΅
=
1.5+2×0.08
π£
= 0.976 and πΏ =
1.7
0.15+2×0.08
= 0.112
2.75
For k = 0.62 ≅ 0.6, m1 = 0.09 and m2 = 0.08
So,
BM of patch I in shorter span
1
= (m1+μm2) ×w×0.8×0.3×0.15×0.15×1.5
1
= (0.09+0.15×0.08) ×37.5×0.8×0.3×0.15×0.15×1.5
= 15.3 KNm
BM of patch I in longer span
= (μm1+m2) ×w×0.8×
1
×0.15×1.5
0.3×0.15
1
= (0.15×0.09+0.08) ×37.5×0.8×0.3×0.15×0.15×1.5
=14.025 KNm
For patch II,
π’
π΅
=
0.9+2×0.08
1.7
π£
= 0.62 and πΏ =
0.15+2×0.08
2.75
= 0.113
For k = 0.62 ≅ 0.6, m1 = 0.12 and m2 = 0.11
So,
BM of patch I in shorter span
1
= (m1+μm2) ×w×0.8×0.3×0.15×0.15×1.5
1
= (0.12+0.15×0.11) ×37.5×0.8×0.3×0.15×0.15×0.9
=12.29 KNm
BM of patch I in longer span
1
= (μm1+m2) ×w×0.8×0.3×0.15×0.15×1.5
1
= (0.15×0.12+0.11) ×37.5×0.8×0.3×0.15×0.15×0.9
= 11.52 KNm
Thus,
π
BM of wheel load 2 in shorter span =π× (15.3-12.29) = 1.505 KNm
π
BM of wheel load 2 in longer span = π× (14.025-11.52) = 1.25 KNm
BM due to wheel load 4
37
2069 – AB Bridge
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
For patch I,
π’
=
π΅
0.3+2×0.08
1.7
π£
= 0.27 and πΏ =
2.55+2×0.08
2.75
= 0.98
For k = 0.62 ≅ 0.6, m1 = 0.095 and m2 = 0.025
So,
BM of patch I in shorter span
1
= (m1+μm2) ×w×0.8×0.3×0.15×0.15×1.5
= (0.095+0.15×0.025)
1
×62.5×0.8×0.3×0.15×0.30×2.55
= 83.94 KNm
BM of patch I in longer span
1
= (μm1+m2) ×w×0.8×0.3×0.15×0.15×1.5
1
= (0.15×0.095+0.025) ×62.5×0.8×0.3×0.15×0.30×2.55
= 33.36 KNm
For patch II,
π’
π΅
=
0.3+2×0.08
1.7
π£
= 0.27 and πΏ =
2.25+2×0.08
2.75
= 0.88
For k = 0.62 ~ 0.6, m1 = 0.10 and m2 = 0.03
So,
BM of patch I in shorter span
1
= (m1+μm2) ×w×0.8×0.3×0.15×0.3×2.25
= (0.10+0.15×0.03)
1
×62.5×0.8×0.3×0.15×0.3×2.25
= 78.38 KNm
1
BM of patch I in longer span = (μm1+m2) ×w×0.8×0.3×0.15×0.30×2.25
1
= (0.15×0.10+0.03) ×62.5×0.8×0.3×0.15×0.3×2.25
38
Design of Bridge Over Kerunga Khola, Chitwan
=33.75 KNm
Thus,
π
BM of wheel load 4 in shorter span = π× (83.94-78.38) = 2.78 KNm
π
BM of wheel load 4 in longer span = π× (33.36-33.75) = -0.195 KNm
BM due to wheel load 3
39
2069 – AB Bridge
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
For patch I,
π’
π΅
=
1.5+2×0.08
π£
= 0.98 and πΏ =
1.7
2.55+2×0.08
2.75
= 0.99
For k = 0.62 ≅ 0.6, m1 = 0.049 and m2 = 0.012
So,
1
BM of patch I in shorter span= (m1+μm2) ×w×0.8×0.3×0.15×2.55×1.5
1
= (0.049+0.15×0.012) ×37.5×0.8×0.3×0.15×1.5×2.55
=129.54 KNm
1
BM of patch I in longer span= (μm1+m2) ×w×0.8×0.3×0.15×2.55×1.5
1
= (0.15×0.049+0.012) ×37.5×0.8×0.3×0.15×1.5×2.55
=49.34 KNm
For patch II,
π’
π΅
=
0.9+2×0.08
1.7
π£
= 0.62 and πΏ =
2.55+2×0.08
2.75
= 0.99
For k = 0.62 ~ 0.6, m1 = 0.069 and m2 = 0.019
So,
1
BM of patch II in shorter span= (m1+μm2) ×w×0.8×0.3×0.15×0.9×2.55
1
= (0.069+0.15×0.019) ×37.5×0.8×0.3×0.15×0.9×2.55
=109.93 KNm
1
BM of patch II in longer span= (μm1+m2) ×w×0.8×0.3×0.15×0.9×2.55
1
= (0.15×0.069+0.019) ×37.5×0.8×0.3×0.15×0.9×2.55
=44.91 KNm
40
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
For patch III,
π’
π΅
=
1.5+2×0.08
π£
= 0.98 and πΏ =
1.7
2.25+2×0.08
= 0.88
2.75
For k = 0.62 ≅ 0.6, m1 = 0.055 and m2 = 0.018
So,
1
BM of patch III in shorter span= (m1+μm2) ×w×0.8×0.3×0.15×1.5×2.25
1
= (0.055+0.15×0.018) ×37.5×0.8×0.3×0.15×1.5×2.25
=129.83 KNm
1
BM of patch III in longer span= (μm1+m2) ×w×0.8×0.3×0.15×1.5×2.25
1
= (0.15×0.055+0.018) ×37.5×0.8×0.3×0.15×1.5×2.25
=59.06 KNm
For patch IV,
π’
π΅
=
0.9+2×0.08
1.7
π£
= 0.62 and πΏ =
2.25+2×0.08
2.75
= 0.88
For k = 0.62 ~ 0.6, m1 = 0.076 and m2 = 0.025
So,
1
BM of patch IV in shorter span= (m1+μm2) ×w×0.8×0.3×0.15×0.9×2.25
1
= (0.076+0.15×0.025) ×37.5×0.8×0.3×0.15×0.9×2.25
=107.66 KNm
1
BM of patch IV in longer span= (μm1+m2) ×w×0.8×0.3×0.15×0.9×2.25
1
= (0.15×0.076+0.025) ×37.5×0.8×0.3×0.15×0.9×2.25
=49.14 KNm
Thus,
π
BM of wheel load 3 in shorter span = π× (BM of patch I - BM of patch II – BM of patch III +
BM of patch IV ) = ¼ × (129.54 – 109.93- 129.83 + 107.66) = -0.64 KNm
π
BM of wheel load 3 in longer span = π× (BM of patch I BM of patch II – BM of patch III + BM of patch IV ) = ¼ ×
(49.34 – 44.91 – 59.06 + 49.14) = -1.37 KNm
Calculation of Shear Force due to IRC Class AA Loading
(Tracked wheel)
beff = 3.76m ( limited to 2.75 m)
41
Design of Bridge Over Kerunga Khola, Chitwan
Effective load including impact
2.75
= 350 ×1.25×1.5×3.76
= 479.97 KN
RA=280.93KN
RB=199.04KN
Design SF due to DL and LL in shorter span
= 280.93×0.8+5.957
= 230.7 KN/m
(Wheeled Load)
For wheel I,
a= 0.15+0.2+0.08=0.43
π΅
α =1.86 for πΏ ≅ 1
b1= 0.15+2×0.08=0.31m
Then, beffI = αxa (1- a/l) + b1
= 1.86×0.43(1- 0.43/1.7)+0.31
= 0.908m
2×62.5
Effective load due to wheel I = (0.908
2
+
0.15
2
) ×1.25×1.5
= 443.05 KN/m
For wheel II,
a = 0.67
π΅
α =1.86 for πΏ ≅ 1
b1= 0.15+2×0.08=0.31m
Then, beffII = αxa (1- a/l) + b1
= 1.86×0.67(1- 0.67/1.7)+0.31
= 1.065m
2×37.5
Effective load due to wheel I = (1.065
2
+
0.15
2
) ×1.25×1.5
= 231.5 KN/m
RA=422.22KN
RB=252.33KN
Design SF due to DL and LL in shorter span
= 422.22×0.8+5.957
42
2069 – AB Bridge
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
= 343.73 KN/m
Summary
The calculated BMs and SFs may be tabulated as follows:
Load
Bending Moment
Shear Force
Shorter Span
Longer Span
Dead Load
2.125
0.978
5.957
Live Load
IRC Class A
16.647
0.1345
219.57
Loading
IRC Class
34.97
14.16
280.93
AA(Tracked)
IRC Class
24.175
17.425
422.22
AA(Wheeled)
Total BM
37.095
15.138
Combined Shear
(DL+0.8xLL)
181.613
230.7
343.73
Design of slab
Check depth of slab
Effective depth of main reinforcement (d1) = 200-25-10/2 = 170mm
Effective depth of secondary reinforcement (d2) = 200-25-10-10/2
= 160mm
ππ’
37.095×106
dbal = √π×b = √ 3.45×1000 = 103.7 mm < 170mm and < 160mm
I.e. dbal < dprov OK
Find reinforcing bars:
Since dbal < dprov , section of slab is designed as singly reinforced under-reinforced section (
SRURS) . In the example, section design has been done by using SP16.
a) Reinforcing bars in shorter and longer direction of slab
In short span,
ππ’
π×d2
37.095×106
= 1000×1702 = 1.28
By interpolation (Table 2 of SP16)
0.392−0.376
1.3−1.25
ππ‘−0.376
= 1.28−1.25
Pt = 0.3856 % > Ptmin = 0.12%
Astreq= 0.3856/100×100×170
= 655.52 mm2
Provide 10mm ∅ bar @ 110mm c/c.
Astprovided = 706.85 mm2
In long span,
43
Design of Bridge Over Kerunga Khola, Chitwan
ππ’
π×d2
2069 – AB Bridge
15.136×106
= 1000×1602 = 0.6
Pt = 0.172 % > Ptmin = 0.12%
Astreq = 0.172/100×100×170
= 275.2 mm2
Provide 10mm ∅ bar @ 280mm c/c.
Astprovided = 314.16 mm2
b) Temperature reinforcement
Provide 10 mm ∅ bars @300 mm c/c in both direction at top of slab.
Check for shear: ππ’π£ ≤ Kππ’π
ππ’ 343.73×1000
ππ’π£ = ππ =
1000×170
= 2.02 N/mm2
Shear strength of concrete section ππ’π = 0.46 N/mm2 for M25 and Pt = 0.433 %
Limiting value of Shear Stress ππ’π, πππ₯ = 3.1 N/mm2 [Refer IS 456 table 19, 20]
Depth factor (K) = 1
Since ππ’π£>ππ’π, design shear reinforcement.
0.87 ππ¦×π΄π π£×π
Sv = (τuv−τuc)bw×d=
0.87×415×9×π×8^2/4×170
(2.02−0.46)×1000×170
= 100 < 0.75d and <300 mm
Adopt Sv = 100mm.
Provide 8mm ∅ 9-legged vertical stirrups @100 mm c/c.
II.
Analysis and Design of Main Girder
Calculation of dead load on a main girder per running meter of span
ο·
Weight of wearing coat = 6×0.08×22×1.75/3 = 6.16KN/m
ο·
Weight of railing = (9×0.225×0.225×1.1×25+12×3×0.0437) ×1.35/12 = 1.59KN/m
ο·
Weight of kerb = 0.6×0.3×25×1.35 = 6.075 KN/m
ο·
Weight of slab:
a) Middle portion=4.3×0.2×12×25×1.35
=348.30KN
1
1
b) Fillet=(2 × 0.15×0.3×12×4×25+2 × 0.15×0.3×1.1×8×25×2) ×1.35 = 49.815 KN
c) Cantilever part =(0.6×0.17×12×2×25+0.85×0.26×12×2×25) ×1.35 = 261.63KN
Weight of slab =
π‘ππ‘ππ π€πππβπ‘ ππ π πππ
ππ.ππ ππππππ×π πππ
=
659.745
3×12
= 18.33 KN/m
ο· Weight of web of main beam= 0.3×0.8×25×1.35 = 8.1KN/m
Total design dead load per main girder
= 6.16 + (1.59+6.075) ×2/3 + 18.33 + 8.1
= 37.70 KN/m
ο·
Self-wt. of Cross girder acts as a point load on main girder.
44
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
a. Self-wt. of web of internal cross girder on a girder = 0.25 × 0.55 × 1.7 × 25 × 1.35
2
× 3 = 5.259 KN
2
b. Self-wt. of web of end cross girder on a girder = 0.25 × 0.3 × 1.7 × 25 × 1.35 × 3
=2.869 KN
Dead load on main girder
Calculation of maximum BM and SF at critical sections due to DL
Reaction at support (RA) = (5.259×3+2.869×2+37.7×12)/2 = 236.96KN
BM at Mid span = RA×6 - 2.869×6 - 5.259×3-37.7×6×6/2
= 236.96×6 - 711.59 = 710.17 KNm
BM at quarter span = RA×3 - 2.869×3 - 37.7×3×3/2
= 236.96×3 - 178.26 = 532.62 KNm
SF at supports = 236.96 kN
Calculation of maximum BM and SF at critical sections due to LL
(Hendry Jaeger method of Lateral Load Distribution)
End Main Girder
CG of the section(y)
=
0.17
0.2 1
0.18 1
0.15
)+0.85×0.2×(1− )+ ×0.18×0.85×(1−0.17−
)+ ×0.15×0.30×(1−0.20−
)
2
2
2
3
2
3
1
1
0.3×1+1.45×0.17+0.85×0.20+ ×0.18×0.85+ ×0.15×0.30
2
2
0.3×1×0.5+1.45×0.17×(1−
= 0.741m
1.45×0.173
0.85×0.203
0.30×13 0.85×0.183 0.30×0.153
I=
+ 12 + 12 + 36 + 36 +1.45×0.17(1-0.17/2-0.741)2+0.20×0.85×
12
(1-0.20/2-0.741)2+0.3×1×(0.741-1/2)2+0.5×0.18×0.85(0.11-0.741)2+0.9×0.15×0.30×(0.750.741)2
45
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
= 0.05557741 m4
Interior Main Girder
CG of the section (y)
=
0.2×2×(0.8+
0.2
0.8
1
0.15
)+0.3×0.8× +2× ×0.15×0.30×(0.8−
)
2
2
2
3
1
0.2×2+0.3×0.8+2× ×0.15×0.30
2
= 0.715 m
I=
2×0.23
1
12
+
0.3×0.83
12
0.30×0.153
+2×
36
+0.2×2× (1 -
0.2
2
- 0.715)2 + 0.3×0.8× (0.715 -
0.8 2
)
2
+2×2×0.15×0.30×(0.8-0.15/3-0.715)2= 0.0517487m4
Cross girder
End cross girder
CG of the section (y)
=
0.25×0.40×
0.40
0.20 1
0.15
+1.625×.20×(0.60−
)+ ×0.30×0.15×(0.60−0.20−
)
2
2
2
3
1
0.25×0.40+1.625×0.2+ ×0.30×0.15
2
= 0.425m
0.25×0.43
1.625×0.23
0.3×0.153
IT = 12 +
+ 36 + 0.25×0.40× (0.425-0.40/2)2 + 1.625×0.20× (0.6-0.2/212
0.425)2+1/2×0.15×0.30× (0.425-0.35)2 = 0.009462 m4
Intermediate cross girder
Centroid of the section(y)
=
0.55
0.20
1
0.15
+3.0×.02×(0.75−
)+2× ×0.30×0.15×(0.75−0.20−
)
2
2
2
3
1
0.25×0.55+3.0×0.2+2× ×0.30×0.15
2
0.25×0.55×
= 0.575m
0.25×0.553
3.0×0.23
0.3×0.153
IT =
+ 12 +2× 36 + 0.25×0.55 × (0.575-0.55/2)2 + 3.0×0.20× (0.75-0.2/212
0.575)2+2×1/2×0.15×0.30× (0.575-0.5)2 = 0.021526 m4
∑EIT =E (3×0.021526 +2×0.009462) = E×0.083502 m4
EI = E× 0.05557741 m4 (Taking larger of longitudinal flexural rigidity)
Then,
12
πΏ
12
12
α = π4 ×(β)3×∑EIT/EI = π4 ×( 2 )3×( E×0.083502)/( E× 0.05557741) = 39.98
β=
π2 β
2
×πΏ ×CJ/∑EIT
Take β = ∞ (for T-beam bridges, having 3 longitudinal girders with a number of cross beams)
Distribution coefficients:
p11= 0.40, p12= 0.32, p13= 0.28
p21= 0.32, p22= 0.34, p23= 0.34
p31= 0.28, p32= 0.34, p33= 0.38
And,
46
Design of Bridge Over Kerunga Khola, Chitwan
IF = 1.25 and γF = 1.5
1. Class A load
47
2069 – AB Bridge
Design of Bridge Over Kerunga Khola, Chitwan
H1=1.7W
H2=0.5W
H3=1.8W
Reactions:
R1=p11×H1+p12×H2+p13×H3=1.344W
R2=p21×H1+p22×H2+p23×H3=1.326W
R3=p31×H1+p32×H2+p33×H3=1.33W
Reaction Factors:
RF1=1.344
RF2=1.326
RF3=1.330
BM at mid span due to LL
Max. BM= (13.5(0.8+0.25) + 57(3+2.4) + 34(0.85))RF × IF × γF
= 350.875×RF×1.25×1.5
= 657.89×RF KNm
For End girder, max. BM = 657.89×RF3 (max (RF1, RF3))
= 657.89×1.344
= 884.20 KNm
For Interior girder, max. BM = 657.89 × RF2
= 657.89×1.326
48
2069 – AB Bridge
Design of Bridge Over Kerunga Khola, Chitwan
= 872.36 KNm
BM at quarter span due to LL
Max. BM = (57(2.25+1.95) + 34(0.875+0.125)) × RF × IF × γF
= 273.4×RF×1.25×1.5
= 512.625×RF
For End girder, max. BM = 512.625×RF3 (max (RF1, RF3))
= 512.645×1.344
= 688.97 KNm
For Interior girder, max. BM= 512.625×RF2
= 512.625×1.326
= 679.74 KNm
SF at support due to LL
Max. SF = (57(1+0.9) + 34(0.54+0.292+0.042)) × RF × IF × γF
= 138.016×RF×1.25×1.5
=258.78×RF
For End girder, max. SF= 258.78×RF3 (max (RF1, RF3))
= 258.78×1.344 = 347.80 KNm
For Interior girder, max. SF= 258.78×RF2
= 258.78×1.326
= 343.14 KNm
SF at quarter span due to LL
49
2069 – AB Bridge
Design of Bridge Over Kerunga Khola, Chitwan
Max. SF = (57(0.75+0.6) + 34(0.292+0.042)) × RF × IF × γF
= 88.306×RF×1.25×1.5
=165.57×RF
For End girder, max. SF= 165.57×RF3 (max (RF1, RF3))
= 165.57×1.344 = 222.53 KNm
For Interior girder, max. SF= 165.57×RF2
= 165.57×1.326
= 219.55 KNm
2. Class AA wheeled load
50
2069 – AB Bridge
Design of Bridge Over Kerunga Khola, Chitwan
H1=65.31 KN
H2=124.38KN
H3=10.31KN
51
2069 – AB Bridge
Design of Bridge Over Kerunga Khola, Chitwan
Reactions:
R1=p11×H1+p12×H2+p13×H3=68.81 KN = 0.344 W (for W= 200KN)
R2=p21×H1+p22×H2+p23×H3=66.69 KN = 0.333 W (for W= 200KN)
R3=p31×H1+p32×H2+p33×H3=64.49 KN = 0.322 W (for W= 200KN)
Reaction Factors:
RF1=0.344
RF2=0.333
RF3=0.322
BM at mid span due to LL
Max. BM= (200(2.7+2.7)) RF × IF × γF
= 200×5.4×RF×1.25×1.5
=2025×RF KNm
For End girder, max. BM= 2025×RF1 (max (RF1, RF3))
= 2025×0.344= 696.60 KNm
For Interior girder, max. BM= 2025×RF2
= 2025×0.333
= 674.33 KNm
BM at quarter span due to LL
52
2069 – AB Bridge
Design of Bridge Over Kerunga Khola, Chitwan
Max. BM = (200(2.25+1.95)) RF × IF × γF
= 200×4.2×RF×1.25×1.5
=1575×RF KNm
For End girder, max. BM= 1575×RF1 (max (RF1, RF3))
= 1575×0.344= 541.80 KNm
For Interior girder, max. BM= 1575×RF2
= 1575×0.333
= 524.48 KNm
SF at support due to LL
Max. SF = 200(1+0.9) × RF × IF × γF
= 200(1+0.9) × RF×1.25×1.5
=712.5×RF
For End girder, max. SF= 712.5×RF1 (max (RF1, RF3))
= 712.5×0.344= 245.10 KNm
For Interior girder, max. SF= 712.5×RF2
= 712.5×0.333
= 237.26 KNm
SF at quarter span due to LL
Max. SF = 200(0.75+0.65) × RF × IF × γF
= 200×1.4×RF×1.25×1.5
=525×RF
For End girder, max. SF= 525×RF1 (max (RF1, RF3))
53
2069 – AB Bridge
Design of Bridge Over Kerunga Khola, Chitwan
= 525×0.344= 180.60 KNm
For Interior girder, max. SF= 525×RF2
= 525×0.333
= 174.83 KNm
3. Class AA track load
54
2069 – AB Bridge
Design of Bridge Over Kerunga Khola, Chitwan
H1=240.625 KN
H2=192.5 KN
H3=266.875 KN
Reactions:
R1=p11×H1+p12×H2+p13×H3=232.58 KN = 0.665W (for W= 350 KN)
R2=p21×H1+p22×H2+p23×H3=233.19 KN = 0.666W (for W= 350 KN)
R3=p31×H1+p32×H2+p33×H3=234.24 KN = 0.669W (for W= 350 KN)
Reaction Factors:
RF1=0.665
RF2=0.666
RF3=0.669
BM at mid span due to LL
55
2069 – AB Bridge
Design of Bridge Over Kerunga Khola, Chitwan
Max. BM= 97.22(2.1×3.6 + (3-2.1) ×1/2×3.6) RF × IF × γF
= 97.22(2.1×3.6 + (3-2.1) ×1/2×3.6)×RF×1.25×1.5
=1673.40×RF KNm
For End girder, max. BM= 1673.40×RF3 (max (RF1, RF3))
= 1673.40×0.669= 1119.50 KNm
For Interior girder, max. BM= 1673.40×RF2
= 1673.40×0.666
= 1114.48 KNm
BM at quarter span due to LL
Max. BM = 97.22(2.1×3.6 + (3-2.1) ×1/2×3.6) RF × IF × γF
= 97.22(2.1×3.6 + (3-2.1) ×1/2×3.6)×RF×1.25×1.5
=1673.40×RF KNm
For End girder, max. BM= 1673.40×RF3 (max (RF1, RF3))
= 1673.40×0.669 = 1119.50 KNm
For Interior girder, max. BM= 1673.40×RF2
56
2069 – AB Bridge
Design of Bridge Over Kerunga Khola, Chitwan
= 1673.40×0.666
= 1114.48 KNm
SF at support due to LL
Max. SF =97.22(1/2 × (1+0.7) ×3.6) × RF × IF × γF
= 297.4932×RF×1.25×1.5
=557.80 ×RF
For End girder, max. SF = 557.80×RF3 (max (RF1, RF3))
= 557.80×0.669 = 373.17 KNm
For Interior girder, max. SF= 557.80×RF2
= 557.80×0.666
= 371.49 KNm
SF at quarter span due to LL
Max. SF =97.22(1/2 × (0.45+0.75) ×3.6) × RF × IF × γF
= 209.9952×RF×1.25×1.5
= 393.74 ×RF
For End girder, max. SF= 393.74×RF3 (max (RF1, RF3))
= 393.74×0.669= 263.41 KNm
For Interior girder, max. SF= 393.74×RF2
= 393.74×0.666 = 262.23 KNm
Summary
The calculated BMs and SFs may be tabulated as follows:
57
2069 – AB Bridge
Design of Bridge Over Kerunga Khola, Chitwan
Load
2069 – AB Bridge
Bending Moment (kNm)
Mid-Span
Quarter-Span
710.17
532.62
Exterior Girder
Dead Load
Live Load
IRC Class A
Loading
IRC Class
AA(Tracked)
IRC Class
AA(Wheeled)
Shear Force (kN)
Supports
Quarter-Span
236.96
119.01
884.20
688.97
347.80
222.53
1119.50
1119.50
373.17
263.41
696.60
541.80
245.10
180.60
Interior Girder
Live Load
IRC Class A
Loading
IRC Class
AA(Tracked)
IRC Class
AA(Wheeled)
872.36
679.74
343.14
219.55
1114.48
1114.48
371.49
262.23
674.33
524.48
237.26
174.83
1. Design of End Main Girder
Design Section
Thickness of flange (Df)
Average thickness of left part (t1) =
Average thickness of left part (t2) =
Df =
π‘1+π‘2
2
=
0.223+0.226
2
0.17×0.6+0.26×0.85
1.45
0.2×0.55+0.3×0.275
0.85
= 0.223 m
= 0.226 m
= 0.224 m
Effective width of flange (bef)
π
bef = 5 + ππ€ =
12
5
+ 0.3 = 2.7m > bact = 2.6 m
Actual width on right side = 1 m
Adopt bef = 2×1 = 2 m
CG of section
Y=
0.224
0.776
)+0.3×0.776×
2
2
0.224×2×(0.776+
0.224×2+0.776×0.3
= 0.717 m
MOI of section
Ixx =
2×0.2243
12
+
0.3×0.7763
12
+ 0.224×2×(1 -
0.224
2
- 0.717)2 + 0.3×0.776×(0.717 -
MOI of actual section of girder about XX axis
Ixxgross = 0.05557741 m4
For End Girder,
58
0.776 2
)
2
= 0.0519 m4
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Design BM due to LL at mid-span = 1119.50 kNm (Class AA tracked load)
Design BM due to LL at quarter-span = 1119.50 kNm (Class AA tracked load)
Design SF due to LL at mid-span = 373.17 kN (Class AA tracked load)
Design SF due to LL at quarter-span = 263.41 kN (Class AA tracked load)
Design BM due to DL at mid-span = 702.24 kNm
Design BM due to DL at quarter-span = 526.68 kNm
Design SF due to DL at support = 232.11 kN
Design SF due to DL at quarter-span = 119.01 kN
Therefore,
Design BM at mid-span = 1119.50 + 702.24 = 1821.74 kNm
Design BM at quarter-span = 1119.50 + 526.68 = 1646.18 kNm
Design SF at support = 373.17 + 232.11 = 605.28 kN
Design SF at quarter-span = 263.41 + 119.01 = 382.42 kN
Design of section for Bending
Mid-span section
Xu,l = 0.48 d = 0.48×912 = 437.76 mm
Where, d = 1000 – 40 – 32 – 32/2 = 912 mm
Since Xu,l > Df, Neutral axis lies in web.
(Considering 40mm clear cover and 3 layers of 32 mm ∅ tension bars in main girder)
Find Mu,l = 0.36×fck×b×Xu,l×(d – 0.416×Xu,l) + 0.446×fck×(bef – bw)×Df×(d – Df/2 ) = 4259.43
kNm
Mu = 1821.74 kNm
Since Mu < Mu,l, the section is designed as SRURS.
Find Xu considering Xu < Df.
Mu = 0.36×fck×bef×Xu×(d – 0.416×Xu)
Or, 1821.74×106 = 0.36×25×2000×Xu×(912-0.416×Xu)
∴ Xu = 117.24 mm < Df i.e. N. Axis lies in flange.
Find area of steel for SRUR flanged section when N.A. lies in flange.
ππ’
Ast,req = 0.87×fy×(d−0.416×Xu) = 5845.12 mm2 > Ast,min
Where, Ast,min = 0.2% of bd = 0.002 × 300 × 912 = 547.2 mm2 [Refer Cl.305.1, IRC21]
Provide 8-32 mm ∅ bar.
Ast,prov = 6433.98 mm2 and Pt =
6433.98
912×300
× 100 = 2.35 % < Pt,max = 2.5 %
Quarter-span section
Mu = 1646.18 kNm
Since Mu < Mu,l, the section is designed as SRURS.
Find Xu considering Xu < Df.
Mu = 0.36×fck×bef×Xu×(d – 0.416×Xu)
59
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Or, 1646.18×106 = 0.36×25×2000×Xu×(912-0.416×Xu)
∴ Xu = 105.34 mm < Df i.e. N. Axis lies in flange.
Find area of steel for SRUR flanged section when N.A. lies in flange.
ππ’
Ast,req = 0.87×fy×(d−0.416×Xu) = 5251.71 mm2 > Ast,min
Where, Ast,min = 0.2% of bd = 0.002 × 300 × 912 = 547.2 mm2 [Refer Cl.305.1, IRC21]
Provide 8-32 mm ∅ bar.
6433.98
Ast,prov = 6433.98 mm2 and Pt = 912×300 × 100 = 2.35 % < Pt,max = 2.5 %
Design of section for Shear
Support section
ππ’
ππ’π£ = bwd =
605.28×1000
300×912
= 2.21 N/mm2
Where d = 1000 – 40 – 32 - 32/2 = 912 mm
For M25 and Pt = 2.35 %, ππ’π = 0.862 N/mm2 and ππ’π, = 3.1 N/mm2
Since ππ’π£ > ππ’π, design shear reinforcement.
Take 10 mm∅ 4-legged vertical stirrups for shear reinforcement.
912
Sv = 0.87×ππ¦×π΄π π£× π⁄Vu, net = 0.87×415×4×π×102/4×(2.21−0.862)×300×912
= 280.48 mm < 0.75d (= 684 mm) and < 300 mm
Adopt Sv = 270 mm
Provide 10 mm ∅ 4-legged vertical stirrups @ 270 mm c/c from support to quarter span.
Quarter-span section
ππ’
ππ’π£ = bwd =
382.42×1000
300×912
= 1.40 N/mm2
Where d = 1000 – 40 – 32 - 32/2 = 912 mm
For M25 and Pt = 2.35 %, ππ’π = 0.862 N/mm2 and ππ’π, = 3.1 N/mm2
Since ππ’π£ > ππ’π, design shear reinforcement.
Take 10 mm∅ 2-legged vertical stirrups for shear reinforcement.
912
Sv = 0.87×ππ¦×π΄π π£× π⁄Vu, net = 0.87×415×2×π×102/4×(1.40−0.862)×300×912
= 351.39 mm < 0.75d (= 684 mm) and > 300 mm
Adopt Sv = 300 mm
Provide 10 mm ∅ 2-legged vertical stirrups @ 300 mm c/c from quarter to quarter span of
the other side.
Detailing of reinforcement
Since the no. of bars required is same at both mid-span and quarter-span, no curtailment is
done.
Anchorage of main reinforcing bars at supports
Extension of bar beyond the face of support = 23ld
π΄π π‘,πππ
Now, ld = 0.7×l0×π΄π π‘,ππππ£ for bars with hook end
5251.71
= 0.7×(46×32)×6433.98 where l0 = nφ
60
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
= 841.06 mm
Extension of bar beyond the face of support = 23ld = 23 × 841.06 = 560.71 mm ≈ 561 mm
Side Face Reinforcement
Take 0.1% of web area.
As = 0.001×(1000 – 224)×300 = 232.80 mm2
232.80
No. of 8 mm ∅ bars required on each face =
π
4
2× ×8×8
≈3
Provide 3-8 mm ∅ on each face.
Check for Limit State of Serviceability in Deflection
(Method of Sufficient Stiffness)
π
Check π ≤ αβγδλ
Where,
π
π
12×1000
=
912
= 13.16
α = 20 (simply supported)
10
10
β = π πππ = 12
π΄π π‘,πππ
5845.12
fs = 0.58×fy×π΄π π‘,ππππ£ = 0.58×415×6433.98 = 218.67 N/mm2
For Pt = 2.35 %, γ = 0.86
For 2-32 mm ∅ compression bars,
π
4
2× ×322
Pc = 300×912 = 0.588 %
For Pc = 0.588 %, δ = 1.16
For flange, λ = 0.8 since
π€ππ π€πππ‘β
ππππππ π€πππ‘β
=
300
2000
= 0.15
10
So, αβγδλ = 20×12×0.86×1.16×0.8 = 13.30
π
i.e.
π
≤ αβγδλ
OK.
Hence, deflection of the girder is under control.
2. Design of Intermediate Main Girder
Design Section
Thickness of flange (Df)
Average thickness of left part (t1) =
0.35+0.2
×0.3
2
0.2×0.55+
0.85
Average thickness of left part (t2) = 0.226 m
Df =
π‘1+π‘2
2
=
0.226+0.226
2
= 0.226 m
Effective width of flange (bef)
π
bef = 5 + ππ€ =
12
5
+ 0.3 = 2.7m > bact = 2 m
Actual width on each side = 1 m
61
= 0.226 m
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Adopt bef = 2×1 = 2 m
CG of section
Y=
0.226
)+0.3×0.774×0.774/2
2
0.226×2×(0.774+
0.226×2+0.774×0.3
= 0.717 m
MOI of section
2×0.226^3
0.3×0.774^3
0.226
Ixx =
+
+ 0.226×2×(1 - 2 - 0.717)2 + 0.3×0.774×(0.717 12
12
m4
MOI of actual section of girder about XX axis
Ixxgross = 0.0517487 m4
For End Girder,
Design BM due to LL at mid-span = 1114.48 kNm (Class AA tracked load)
Design BM due to LL at quarter-span = 1114.48 kNm
Design SF due to LL at mid-span = 371.49 kN
Design SF due to LL at quarter-span = 262.23 kN
Design BM due to DL at mid-span = 725.88 kNm
Design BM due to DL at quarter-span = 544.41 kNm
Design SF due to DL at support = 238.02 kN
Design SF due to DL at quarter-span = 124.92 kN
Therefore,
Design BM at mid-span = 1114.48 + 725.88 = 1840.36 kNm
Design BM at quarter-span = 1114.48 + 544.41 = 1658.89 kNm
Design SF at support = 371.49 + 238.02 = 609.51 kN
Design SF at quarter-span = 262.23 + 124.92 = 387.15 kN
0.774 2
)
2
= 0.0519
Design of section for Bending
Mid-span section
Xu,l = 0.48 d = 0.48×912 = 437.76 mm
Where, d = 1000 – 40 – 32 – 32/2 = 912 mm
Since Xu,l > Df, Neutral axis lies in web.
(Considering 40mm clear cover and 3 layers of 32 mm ∅ tension bars in main girder)
Find Mu,l = 0.36×fck×b×Xu,l×(d – 0.416×Xu,l) + 0.446×fck×(bef – bw)×Df×(d – Df/2 ) = 4285.48
kNm
Mu = 1840.36 kNm
Since Mu < Mu,l, the section is designed as SRURS.
Find Xu considering Xu < Df.
Mu = 0.36×fck×bef×Xu×(d – 0.416×Xu)
Or, 1840.36×106 = 0.36×25×2000×Xu×(912-0.416×Xu)
∴ Xu = 118.51 mm < Df i.e. N. Axis lies in flange.
Find area of steel for SRUR flanged section when N.A. lies in flange.
62
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
ππ’
Ast,req = 0.87×fy×(d−0.416×Xu) = 5908.48 mm2 > Ast,min
Where, Ast,min = 0.2% of bd = 0.002 × 300 × 912 = 547.2 mm2 [Refer Cl.305.1, IRC21]
Provide 8-32 mm ∅ bar.
6433.98
Ast,prov = 6433.98 mm2 and Pt = 912×300 × 100 = 2.35 % < Pt,max = 2.5 %
Quarter-span section
Mu = 1658.89 kNm
Since Mu < Mu,l, the section is designed as SRURS.
Find Xu considering Xu < Df.
Mu = 0.36×fck×bef×Xu×(d – 0.416×Xu)
Or, 1658.89×106 = 0.36×25×2000×Xu×(912-0.416×Xu)
∴ Xu = 106.20 mm < Df i.e. N. Axis lies in flange.
Find area of steel for SRUR flanged section when N.A. lies in flange.
Ast,req =
ππ’
0.87×fy×(d−0.416×Xu)
= 5294.44 mm2 > Ast,min
Where, Ast,min = 0.2% of bd = 0.002 × 300 × 912 = 547.2 mm2 [Refer Cl.305.1, IRC21]
Provide 8-32 mm ∅ bar.
6433.98
Ast,prov = 6433.98 mm2 and Pt = 912×300 × 100 = 2.35 % < Pt,max = 2.5 %
Design of section for Shear
Support section
ππ’
ππ’π£ = bwd =
609.51×1000
300×912
= 2.23 N/mm2
Where d = 1000 – 40 – 32 - 32/2 = 912 mm
For M25 and Pt = 2.35 %, ππ’π = 0.862 N/mm2 and ππ’π, = 3.1 N/mm2
Since ππ’π£ > ππ’π, design shear reinforcement.
Take 10 mm∅ 4-legged vertical stirrups for shear reinforcement.
912
Sv = 0.87×ππ¦×π΄π π£× π⁄Vu, net = 0.87×415×4×π×102/4×(2.23−0.862)×300×912
= 276.38 mm < 0.75d (= 684 mm) and < 300 mm
Adopt Sv = 270 mm
Provide 10 mm ∅ 4-legged vertical stirrups @ 270 mm c/c from support to quarter span.
Quarter-span section
ππ’π£ =
ππ’
bwd
=
387.15×1000
300×912
= 1.415 N/mm2
Where d = 1000 – 40 – 32 - 32/2 = 912 mm
For M25 and Pt = 2.35 %, ππ’π = 0.862 N/mm2 and ππ’π, = 3.1 N/mm2
Since ππ’π£ > ππ’π, design shear reinforcement.
Take 10 mm∅ 2-legged vertical stirrups for shear reinforcement.
912
Sv = 0.87×ππ¦×π΄π π£× π⁄Vu, net = 0.87×415×2×π×102/4×(1.415−0.802)×300×912
63
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
= 309.91 mm < 0.75d (= 684 mm) and > 300 mm
Adopt Sv = 300 mm
Provide 10 mm ∅ 2-legged vertical stirrups @ 300 mm c/c from quarter to quarter span of
the other side.
Detailing of reinforcement
Since the no. of bars required is same at both mid-span and quarter-span, no curtailment is
done.
Anchorage of main reinforcing bars at supports
Extension of bar beyond the face of support = 23ld
π΄π π‘,πππ
Now, ld = 0.7×l0×π΄π π‘,ππππ£ for bars with hook end
5294.44
= 0.7×(46×32)×
6433.98
where l0 = nφ
= 847.90 mm
Extension of bar beyond the face of support = 23ld = 23 × 847.90 = 565.27 ≈ 566 mm
Side Face Reinforcement
Take 0.1% of web area.
As = 0.001×(1000 – 224)×300 = 232.80 mm2
No. of 8 mm ∅ bars required on each face =
232.80
π
4
2× ×8×8
≈3
Provide 3-8 mm ∅ on each face.
Check for Limit State of Serviceability in Deflection
(Method of Sufficient Stiffness)
π
Check π ≤ αβγδλ
Where,
π
π
=
12×1000
912
= 13.16
α = 20 (simply supported)
10
10
β = π πππ = 12
π΄π π‘,πππ
5908.48
fs = 0.58×fy×π΄π π‘,ππππ£ = 0.58×415×6433.98 = 221.04 N/mm2
For Pt = 2.35 %, γ = 0.86
For 2-32 mm ∅ compression bars,
π
4
2× ×322
Pc = 300×912 = 0.588 %
For Pc = 0.588 %, δ = 1.16
π€ππ π€πππ‘β
300
For flange, λ = 0.8 since ππππππ π€πππ‘β = 2000 = 0.15
10
π
So, αβγδλ = 20×12×0.86×1.16×0.8 = 13.30 > π
OK
Hence, deflection of the girder is under control.
64
Design of Bridge Over Kerunga Khola, Chitwan
III.
2069 – AB Bridge
Analysis of Cross Beam
1. Intermediate Cross Beam
Self-weight of wearing coat = 0.08x22x1x2x1.75 = 6.16 kN/m
Self-weight of slab = 0.2x25x1x2x1.35 = 13.50 kN/m
Total = 6.16 + 13.50 = 19.66 kN/m
Self-weight of fillet = ½x0.15x0.3x25x2x1.35 = 1.52 kN/m
Self-weight of cross beam = 0.55x0.25x25x1.35 = 4.64 kN/m
Total = 1.52+4.64 = 6.16 kN/m
19.66 kN/m
6.16 kN/m
Dead Load on Intermediate Cross Beam
Longitudinal Position of Class A Load for Maximum BM and SF
Calculation of max BM at mid-span due to DL and LL
Mu = 6.16x2x½ - 6.16x1x½ + 19.66x½x2x½x1 - ½x19.66x1x1/3 + 106.875/2x1
= 60.22 kNm
Calculation of max SF at support due to DL and LL
Vu = 6.16x2/2 + ½x19.66x½x2 + 106.875x0.2/2 + 106.875 = 133.553 kN
2. End Cross Beam
Self-weight of wearing coat = ½x6.16 = 3.08 kN/m
Self-weight of slab = ½x13.50 = 6.75 kN/m
Total = 3.08 + 6.75 = 9.83 kN/m
Self-weight of fillet = 0.76 kN/m
Self-weight of cross beam = 0.4x0.25x25x1.35 = 3.375 kN/m
Total = 0.76 + 3.375 = 4.135 kN/m
65
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
9.83 kN/m
4.135 kN/m
Dead Load on End Cross Beam
Longitudinal Position of Class A Load for Maximum BM and SF
Mu = 4.135/2x1 - 4.135x1x½ + 9.83x½x2x½x1 - ½x9.83x1x1/3 + 106.875/2x1
= 56.714 kNm
Vu = 4.135x2/2 + ½x9.83x½x2 + 106.875x0.2/2 + 106.875 = 126.613 kN
Design of Cross beams
1. Intermediate Cross beam
Design section
bef = π/5 + bw = 2/5 + 0.25 = 0.65m ≤ bact = 3 m
d = 750 – 40 – 16/2 = 702 mm
Design section for bending
Xu,l = 0.48 d = 336.96 mm
Since Xu,l > Df and Df > 0.43Xu,l, when NA lies in web, we have,
Mu,l = 0.36xfckxbxXu,lx(d – 0.416xXu,l ) + 0.446fckx(bef – bw)x yf,l x(d – yf,l/2 )
Where yf,l = 0.15Xu,l+0.65Df = 180.54mm
Mu,l = 918.523kNm
We have, Mu= 60.22 kNm
Since Mu < Mu,l, section is designed as SRURS.
Mu = 0.36xfckxbefxXu (d – 0.416xXu)
Or, 60.22×106 = 0.36x25x650xXu (702 - 0.416xXu)
∴ Xu = 14.79 mm < Df
ππ’
60.22×10^6
Ast= 0.87ππ¦(π−0.416ππ’) = 0.87×415(702−0.416×14.79) =239.70 mm2 < Astmin
Where Astmin = 0.2% of bwd = 0.002 ×250 ×702=351.00 mm2
Provide 3-16mm ∅ bars as longitudinal bars.
66
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Astprov=603.19mm2, Pt=0.347%
Design for shear
Vu
ππ’π£ = bw×d =
133.553 × 1000
250 × 702
= 0.76 N/mm2
Where d = 750 – 40– 16/2 = 702 mm
ππ’π = 0.41 N/mm2 for M25 and Pt= 0.347 %,
ππ’π, πππ₯ = 3.1 N/mm2 for M25
Since ππ’π£>ππ’π, design shear reinforcement.
Take 8mm ∅ 2-legged vertical stirrups for shear reinforcement.
0.87ππ¦×π΄π π£×d
Sv = (τuv−τuc)×bw×d =
0.87×415×(2×π/4×82 )×702
(0.76−0.41)×250×702
= 414.82mm > 300mm
< 0.75d = 526.50mm
Adopt Sv = 300mm
Provide 8 mm ∅ 2-legged vertical stirrups @ 300 mm c/c.
2. End Cross beam
Design section
bef = π/10 + bw = 2/10 + 0.25 = 0.45m ≤ bact = 3 m
d = 500 – 40 – 16/2 = 452 mm
Design section for bending
Xu,l = 0.48 d = 216.96 mm
Since Xu,l > Df and Df > 0.43Xu,l, when NA lies in web, we have,
Mu,l = 0.36 x fck x b x Xu,l x (d – 0.416 x Xu,l) + 0.446 x fck x (bef – bw) x yf,l x (d – yf,l/2 )
Where yf,l = 0.15Xu,l + 0.65Df = 162.54 mm
Mu,l = 310.97 kNm
We have, Mu= 56.714 kNm
Since Mu < Mu,l, section is designed as SRURS.
Mu = 0.36 x fck x bef x Xu (d – 0.416Xu)
Or, 56.714×106 = 0.36 x 25 x 450 x Xu (452 - 0.416Xu)
∴ Xu = 31.92 mm < Df
56.714×106
ππ’
Ast = 0.87ππ¦(π−0.416ππ’) = 0.87×415(452−0.416×31.92) = 358.04 mm2 > Astmin
Where Astmin = 0.2%of bwd = 0.002 ×250 ×452 = 226.00 mm2
Provide 3-16mm ∅ bars as long bars.
Astprov=603.19mm2, Pt=0.347%
Design for shear
Vu
ππ’π£ = bw×d =
126.613 × 1000
250 × 452
= 1.12 N/mm2
67
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Where d = 500 – 40– 16/2 = 452 mm
ππ’π = 0.41N/mm2 for M25 and Pt = 0.347 %,
ππ’π, πππ₯ = 3.1 N/mm2 for M25
Since ππ’π£>ππ’π, design shear reinforcement.
Taking 8mm ∅ 2-legged vertical stirrups for shear reinforcement.
0.87ππ¦×π΄π π£×d
Sv = (τuv−τuc)×bw×d =
0.87×415×(2×π/4×82 )×452
(1.12−0.41)×250×452
= 204.48 mm < 300mm
< 0.75d=526.50mm
Adopt Sv = 200mm
Provide 8 mm ∅ 2-legged vertical stirrups @ 200 mm c/c.
IV.
Design of Elastomeric bearing
Calculation of Loads on Bearing
1. DL from Superstructure
ο· Weight of wearing coat = 6.0 × 0.08 × 22 × 12.25 = 129.36 KN
ο· Weight of railing = 2× 9× 0.225 × 0.225 × 1.1 × 25 + 2 × 12 × 3 × 0.0437 = 28.21 KN
ο· Weight of kerb = 0.3 × 0.6 × 12.25 × 2 × 25 = 110.25 KN
ο· Weight of slab = 488.7 + 0.2 × 7.2 × 0.25 × 25 = 497.7 KN
Where
a) Middle portion = 4.3 × 0.2 × 12 × 25 =258 KN
b) Fillet = 1/2× 0.15 × 0.3 × 12 × 4 × 25 + 1/2×0.15 × 0.3 × 1.1 × 8 × 25 ×2 = 36.9
KN
c) Cantilever part = 0.6 × 0.17 × 12 × 2 × 25 + 0.85 × 0.26 × 12 × 2 × 25 = 193.80
KN
ο· Weight of web of main girder = 0.3 × 0.8 × 12.25 × 25 × 3 = 220.5 KN
ο· Weight of web of cross girder = (0.25 × 0.55 × 1.7 × 3 × 2 + 0.25 × 0.40 × 1.7 × 2 ×2)
× 25 = 52.06 KN
Total DL from super structure (Wu) = 1038.08 KN
DL from superstructure on a bearing (DLsup) = 1038.08/6 = 173.01 KN
2. LL from Superstructure
Maximum LL on a bearing (LL) = Maximum reaction of a main girder = 373.17/1.5 =
250.27 kN
3. Load due to braking effort [Refer IRC 21: Cl 211.2]
ο·
Class A load
Braking load = 0.2 × (2 × 114 + 3 × 68) × 2 = 172.8 KN
ο·
Class AA track load
68
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Braking load = 0.2 × 700 = 140 KN
ο·
Class AA
Braking load = 0.2 × 400 = 80 KN
Taking braking effort due to Class A loading,
Horizontal Braking load on a bearing (FbrH) = 172.8/6 = 28.8 KN
Braking loads acts at 1.2m above wearing coat.
Point of application of braking load is 2.28 m (1.2+0.08+0.2+0.8) from bearing. It
induces vertical reaction on bearing.
172.8 × 2.28
Vertical reaction on a bearing due to braking load= 12 × 3 = 10.94 KN
4. Wind load [Refer IRC 6: Cl 209.3.5]
ο· Wind load in transverse direction of bridge (FWT) = PZ × A × G × CD = 71.90 KN
Take, Ht. of bridge = 7 m, Basic wind speed = 47 m/s and Plain Terrain
Where, Vz = 27.80 × 47/33 = 39.59 m/s [Refer Cl. 209, IRC 6]
PZ = 463.70 × 472/332 = 940.60 N/m2
G = 2 (for span up to 150 m)
CD = 1.3 × 1.5 = 1.95 [B/π· = 7.2, π΅/π·≥ 6 and more than single girder]
A = (1 + 0.3) × 12.25 + 0.225 × 1.1 × 9 + 0.0483 × (12-9 × 0.225) × 3 = 19.60 m2
Wind load in transverse direction on a bearing (FWT) = 71.90/6 =11.98 KN
ο· Wind load in longitudinal direction of bridge (FWL) = 0.25 × FTW = 17.975 KN
Wind load in longitudinal direction on a bearing (FWL) = 17.975/6 = 3.00 KN
ο
ο· Wind load in vert. dir. of bridge (FWV) = PZ × A3 × G × CL = (940.60 × 10-3)× 7.2 × 12.25 ×
2 × 0.75 = 124.44 KN
Wind load in vertical direction on a bearing (FWV) = 124.44/6 = 20.74 KN
5. Seismic Load
π
πΌ
Seismic load (FSh) = 2 × π
×
ππ
π
×π
[Refer Cl. 219, IRC 6]
Take, Seismic Zone - V, Soil Strata - Medium, Damping - 5 %, Bridge Class - Normal
Where,
π
πΌ
Ah = πΌh = 2 × π
×
ππ
π
= 0.225; Z = 0.36, I = 1, R = 2, ππ/π = 2.5
W = 1038.08 KN in longitudinal direction
W = 1038.08 + 0.2 × 2 × (2 × 114 + 3 × 68) = 1208.63 KN in transverse direction
So,
69
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
FShL = 0.225 × 1035.83= 233.06 KN in longitudinal direction of bridge
FShT = 0.225 × 1208.63= 271.94 KN in transverse direction of bridge
Then,
ο§ Seismic load in transverse direction on a bearing (FShT) = 271.94/6 = 45.32 KN
ο§ Seismic load in longitudinal direction on a bearing (FShL) = 233.06/6 = 38.84 KN
ο§ Vertical reaction due to seismic load on support of bridge (FSv)
Seismic loads acts on c. g. of seismic weight. It creates additional vertical load on bearing.
Consider c. g. of seismic weight = 0.9 m from bearing.
V. reaction on a bearing when s. load acts in tr. dir. (FSvT) = 172.97/2 = 86.485 KN
Vertical reaction on a bearing when seismic Load acts in long dirn (FSVL) =
= 8.74 KN
233.06 × 0.9
12
×½
6. Load due to temperature variation, creep and shrinkage effect
Maximum horizontal force on a bearing (Fcst) = Δ/ππ × G × A = 5.44 KN
where,
ο¨ Strain due to temp., creep and shrinkage = 5 × 10-4 [Refer IRC 83 Part II Cl. 916.3.4]
ο¨ Horizontal deformation of bearing ( Δ ) = 5×10-4×12.25 × 103 × ½=3.0625 mm
ο¨ Shear modulus of elastomer (G) = 1 N/mm2 [Refer IRC 83 Part II, Cl. 915.2.1]
ο¨ Preliminary height of bearing (h0) = 52 mm
ο¨ Preliminary effective sectional area of bearing (A) = b × l = 238 × 388 = 92344 mm 2
Load Combinations [Refer IRC 6 Table 1]
Vertical load
Combination Along
Across
of load
Traffic
Traffic
I
[N]
II(A)
[N+T]
Dl
LL
sup
Dl
LL
sup
Permissible
stress (%)
sup
100%
FbrV
Dl
LL
Horizontal Load
Along
Across
Traffic
Traffic
FbrH
Dl
LL
sup
FbrV
115%
FbrH
Fcst
III(A)
[N+ T+ W]
Dl
LL
sup
FbrV
FbrH
133%
Fcst
FWV
sup
VI
[N+T+S]
Dl
LL
sup
Dl
0.2LL
F WV
FWL
F WT
sup
Dl
0.2LL
0.5FbrV
0.5FbrH
150%
Fcst
FS VL
FS VT
FS hL
FS hT
70
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Calculation of Loads on Bearing According to Combination of Loads
Vertical and horizontal loads subjected to bearing in the direction of traffic are only taken for
design.
Combination I [N]
Total Vertical load = DLSup + LL + FbrV = 173.01 + 250.27 + 10.94 = 434.22 KN
Total Horizontal load = FbrH = 28.8 KN
Combination II (A) [N+T]
Total Vertical load = DLSup + LL + FbrV = 173.01 + 250.27 + 10.94 = 434.22 KN
Total Horizontal load = FbrH + Fcst = 28.8 + 5.44 = 34.24 KN
Combination III (A) [N+T+W]
Total Vertical load = DLSup + LL+ FbrV + FWV = 173.01 + 250.27 + 10.94 + 20.74 = 454.96
KN
Total Horizontal load = FbrH + Fcst + FWL = 28.8 + 5.44 + 3.0 = 37.24 KN
Combination VI [N+T+S]
Total Vertical load = DLSup + 0.2 × LL + 0.2 FbrV + FsvL
= 173.01 + 0.2 × 250.27 + 0.2 × 10.94 + 8.74= 233.99 KN
Total Horizontal load = 0.2 × FbrH + Fcst + FshL = 0.2 × 28.8 + 5.44 + 38.84 = 50.04 KN
Design of Elastomeric Pad Bearing for Combination I [N]
Effective Span = 12 m, Total Vertical Load = 434.22 kN, Total Horizontal Load = 28.8 kN
1. Geometrical Design
Nmax = 434.22 kN
Nmin = 173.01 kN
Try Standard plan dimension 250x400 mm for elastomeric bearing (IRC 83 – Part II)
From table,
Loaded Area = 92344 mm2
Thickness of individual elastomer layer hi = 12 mm
Thickness of outer layer he = 6 mm
Thickness of steel laminate hs = 4 mm
Adopt 3 internal layers and 4 steel laminates.
Overall Thickness = 3x12+4x4+2x6 = 64 mm
Effective Thickness of elastomer h = 3x12+2x6 = 48 mm
Adopt side cover c = 6 mm
Check for Geometry
1. l0/b0 = 400/250 =1.6 ≤ 2
2. h = 48 < b0/5 = 50 mm
And > b0/10 = 25 mm
92344
3. Shape factor S = 2(238+388)×12 = 6.146 (between 6 and 12)
71
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
434.22×1000
Nmax
4. Bearing Stress in concrete (σm) = Loaded Area = 92344 = 4.70 MPa
A1/A2 is limited to 2.
A1
Allowable BS = 0.25 × πππ × √A2 = 0.25 × 25 × √2 = 8.84 N/mm2
So,
Bearing stress in concrete ≤ Allowable Bearing stress
OK
2. Structural Design
1. Check for Translation
Design strain in bearing (γd) < 0.7
i.e. γd = γbd = Δππ/h + τmd = 0.0625+0.312 = 0.3745 < 0.7
where
Shear strain per bearing due to shrinkage, creep and temperature= Δππ/h
=
5×10−4 ×12×103
2×48
OK
= 0.0625
π»
28.8×103
Shear strain due to longitudinal forces = π΄×πΊ = 92344×1 =0.312
2. Check for Rotation
Design rotation in bearing (αd) ≤ βnαbi, max
1
αd = αdDL+ αdLL = 400 × MDL × L × 10-3 × πΈπ
2
× Igr
1
+ 400 × MLL × L × 10-3 × πΈπ × Igr
= 0.00629
where
L =12 m
MDL = 725.88/1.35 = 537.69 KN-m
MLL = 1119.50/1.5 = 746.33 KN-m
Ec = 5000 √fck = 25000 N/mm2
πΌππ= 0.05557741 m4
β = 0.1σm= 0.1× 439.70× 1000/(92344) = 0.476
n=6
αbi,max = 0.5 πππππ₯×hπ/(π×π 2) = 0.5×10×12/(238×6.1462) = 0.00667
βnαbi, max= 0.0190
αd ≤ βnαbi, max
3. Check for Friction
Design strain in bearing (γd) ≤ 0.2 + 0.1σm
i.e. 0.2 + 0.1σm = 0.2 + 0.1x4.76 = 0.676 > 0.3745
And
Normal stress in bearing (σm) > 2 MPa and ≤ 10 MPa
4. Check for Shear Stress
Total shear stress ≤ 5 MPa
i.e. τc + τh + τα = 2.849 MPa < 5 MPa
where
Shear stress due to compression = 1.5xσm/S = 1.5x4.76/6.146 = 1.162 MPa
Shear stress due to horizontal deformation = 0.3745x1 = 0.3745 MPa
Shear stress due to rotation = 0.5x(b/hi)2 x αbi,max = 0.5× (238/12)2 ×0.00667
72
OK
OK
OK
OK
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
= 1.312 MPa
Check of Elastomeric Pad Bearing for Combination VI [N + T + S]
Effective Span = 12 m, Total Vertical Load = 233.99/1.50 = 155.99 kN, Total Horizontal Load
= 50.04/1.50 = 33.36 kN
1. Geometrical Design
Nmax = 155.99 kN
Nmin = 173.01/1.50 = 115.34 kN
Designed bearing dimension 250x400 mm
Loaded Area = 92344 mm2
Thickness of individual elastomer layer hi = 12 mm
Thickness of outer layer he = 6 mm
Thickness of steel laminate hs = 4 mm
Adopt 3 internal layers and 4 steel laminates.
Overall Thickness = 3x12+4x4+2x6 = 64 mm
Effective Thickness of elastomer h = 3x12+2x6 = 48 mm
Adopt side cover c = 6 mm
Check for Geometry
1. l0/b0 = 400/250 =1.6 ≤ 2
2. h = 48 < b0/5 = 50 mm
And > b0/10 = 25 mm
92344
3. Shape factor S = 2(238+388)×12 = 6.146 (between 6 and 12)
155.99×1000
Nmax
4. Bearing Stress in concrete (σm) =
=
= 1.69 MPa
Loaded Area
92344
A1/A2 is limited to 2.
A1
Allowable BS = 0.25 × πππ × √A2 = 0.25 × 25 × √2 = 8.84 N/mm2
So,
Bearing stress in concrete ≤ Allowable Bearing stress
OK
2. Structural Design
1. Check for Translation
Design strain in bearing (γd) < 0.7
i.e. γd = γbd = Δππ/h + τmd = 0.00625+0.361 = 0.36725 < 0.7
where
Shear strain per bearing due to shrinkage, creep and temperature= Δππ/h
=
5×10−4 ×12×103
2×48
= 0.00625
π»
33.36×103
Shear strain due to longitudinal forces = π΄×πΊ = 92344×1 =0.361
2. Check for Rotation
Design rotation in bearing (αd) ≤ βnαbi, max
1
1
αd = αdDL+ αdLL = 400 × MDL × L × 10-3 × πΈπ
+ 400 × MLL × L × 10-3 × πΈπ × Igr
2
73
× Igr
OK
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
= 0.00629
where
L =12 m
MDL = 725.88/1.35 = 537.69 KN-m
MLL = 1119.50/1.5 = 746.33 KN-m
Ec = 5000 √fck = 25000 N/mm2
πΌππ= 0.05557741 m4
β = 0.1σm= 0.1× 155.99× 1000/(92344) = 0.169
n=6
αbi,max = 0.5 πππππ₯×hπ/(π×π 2) = 0.5×10×12/(238×6.1462) = 0.00667
βnαbi, max= 0.00676
αd ≤ βnαbi, max
3. Check for Friction
Design strain in bearing (γd) ≤ 0.2 + 0.1σm
i.e. 0.2 + 0.1σm = 0.2 + 0.1x1.69 = 0.369 > 0.36725
And
Normal stress in bearing (σm) > 2 MPa and ≤ 10 MPa
4. Check for Shear Stress
Total shear stress ≤ 5 MPa
i.e. τc + τh + τα = 2.091 MPa < 5 MPa
where
Shear stress due to compression = 1.5xσm/S = 1.5x1.69/6.146 = 0.412 MPa
Shear stress due to horizontal deformation = 0.36725x1 = 0.36725 MPa
Shear stress due to rotation = 0.5x(b/hi)2 x αbi,max = 0.5× (238/12)2 ×0.00667
= 1.312 MPa
Summary: Bearing Design
Provide standard 250x400 mm elastomeric pad bearing.
Overall Thickness = 64 mm
Thickness of individual elastomer layer = 12 mm
No of internal elastomer layers = 3
Thickness of each laminate = 4 mm
No of laminates = 4
Thickness of outer layer = 5 mm
74
OK
OK
OK
OK
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
B. Design of Steel Truss (36 m span) bridge
Effective span = 8@4.5m = 36 m
Roadway width = 6 m
Kerb = 0.6 m
Stringer Spacing = 1.5 m
Adopt IRC Class AA Tracked & Class A vehicles
M25 grade concrete & Fe415 steel
I.
Design of Deck Slab
Deck slab shall be designed as a one-way slab.
Analysis may be simplified by considering a single 1.5m span between stringers for mid-span
moment and two spans for positive and negative bending moments as in figures below.
Dead Load
Self-wt. of Deck slab = 0.2x25 = 4.8kN/m2
Self-wt. of wearing coat =.08x22 = 1.76KN/m2
Total dead load = 4.8x1.35 + 1.76x1.75 = 9.56KN/m2
Live Load
Here, only class AA tracked vehicle has been adopted for calculations as it is evidently seen
from the figures that larger moments cannot be induced by other vehicles.
Load = W x γf x IF = 350 x 1.5 x 1.1 = 577.5 kN
From figure, Maximum positive moment = 216.563 kNm
And Maximum negative moment = 130.29 kNm
Effective width for the tracked vehicle may be calculated as
beff = (3.6 +2 x 0.08) = 3.76 m
Then, for unit m strip, positive moment = 57.6 kNm/m
And negative moment = 34.652 kNm/m
Design of slab section
Total design moments are
75
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Positive moment = 2.688+57.6 = 60.288 KNm
Negative moment = 2.49+34.652 = 37.142 KNm
Q = 0.36 x 25 x 0.48(1 - 0.416 x 0.48) = 3.46
Effective depth of slab is given by
π
d = √ππ
60.288×1000
=√
3.46
= 132 mm
Adopt overall depth = 200mm
Effective depth = 200-10/2-25 = 170mm
For SRURS (positive moment),
π΄π
ππ
π
=
ππ.πππ×ππππ
πππ∗πππ
For SRURS (positive moment),
π΄π
= 2.086
ππ
π
Pt = 0.548%
Ast = 0.548x200x1000/100 = 1096 mm2
Use 12mm ∅ bars @100mm c/c.
=
ππ.πππ×ππππ
πππ∗πππ
= 1.285
Pt = 0.406%
Ast = 0.406x200x1000/100 = 812 mm2
Use 12mm ∅ bars @130mm c/c.
And use 10mm ∅ bars @300mm c/c (Astmin) as distribution bars on top and bottom along
longer direction of slab.
II.
Design of stringer beam
Dead load due to self-wt. of slab and wearing coat = 9.56x1.5 = 14.34kN/m
Self-wt. of stringer (assumed) = 1KN/m
Total Dead load = 15.34KN/m
Max. BM = 15.34x4.52/8 = 38.83KNm
Max. SF = 15.34x4.5/2 = 34.52KN
For live load,
Figure: position for Max. BM
Max. BM due to live load = (175x2.25-350x0.5x0.5x1.8)1.5x1.1 = 389.81KNm
76
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Figure: position for Max. SF
Max. SF due to live load = (350x2.7/4.5)x1.5x1.1 = 346.5KN
Designed BM = 389.81+38.83 = 428.64KNm
Designed SF = 346.5+34.52 = 381.02KN
[IS 800:2007 steel code, clause 8.2.1.2]
Md = βbZpfy/γmo
ο Zp = 428.64x106x1.1/250 = 1.87x106 mm3
Use ISHB450 with Zp = 1.955x106 mm3.
Section Classification
250
π
250
∈ = √ ππ¦ = 1, π‘π = 2∗13.7 = 9.12 < 9.4 ∈,
π
=
π‘π€
(450−2∗13.7)
9.8
= 43.12 < 84 ∈
Section is plastic i.e β = 1.0
Check for Moment capacity
Md = βbZpfy/γmo < 1.2Zefy/rmo
= 1x1955.03x1000x250/(1.1x106) <1.2x1740x250/(1.1x106 )
= 444.32KNm < 474.54KNm
OK
Check for Shear capacity
Vd=
=
Av∗fy
√3∗ππ0
450∗9.8∗250
√3∗1.1
K
=578.66KN >381.02KN
OK
Check for Deflection
5 × ππΏ3
πΏπππ₯ =
384πΈπΌ
W=15.34x4.5+350=419.03KN
5×419.03×1000×45003
πΏπππ₯ = 384×2×105 ×39200×104 < 4500/240
77
Design of Bridge Over Kerunga Khola, Chitwan
=6.34mm < 18.75
Check for Web Buckling
πΎπΏ
450
= 2.5 ×
= 114.8
π
9.8
For buckling class ‘c’, fy = 250
Fcd = 89.04 MPa
Fcdw = (b1+n1) x tw x fcd
= (300+225) x 9.8 x 89.04
= 458.11KN > 381.02
OK
b1 = 300mm assumed
OK
Check for Web Crippling
Fcdc=(b1+n2) x tw x fyw/rmo
Where n2 = 2.5(13.7+15) = 71.75
Fcdc = (300+71.75) x 9.8 x 250/1.1 = 827.99KN > 381.02
III.
2069 – AB Bridge
Design of cross girders
Span of cross girder=6+2x0.6=7.2m
Impact factor=10%
Dead load due to slab and wearing coat=9.56x4.5=43.02KN/m
Dead load due to stringer beam=0.872x4.5x1.35=5.3KN
Add load due to connectors=0.375KN
Total load=5.68KN
Self-wt. of cross girder = (0.2L+1) KN/m = (0.2x7.2+1) x 1.35 = 3.3KN/m
Total UDL = (43.02+3.3) = 46.32 KN/m
Figure b
Max. BM = (180.95x3.6-5.68x3-5.68x1.5-46.32x3.6x3.6x0.5) = 325.71KNm
Max. SF = 180.95KN
For live load class AA tracked
Figure: position for Max. BM
78
OK
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Max. BM = (350x3.6-350x1.025)x1.5x1.1=1487.06KNm
Figure: position for Max. SF
Max. SF = (350x5.4/7.2+350x3.35/7.2)x1.5x1.1= 701.83KN
Designed BM=325.71+1487.06=1812.77KNm
Designed SF=180.95+701.83=882.78KN
Md = βbZpfy/γmo
Zp = 1812.77x106x1.1/1x250
= 7.98x106mm3
Use IS WPB 600x300x285.47
250
∈= √ ππ¦ =1,
π
305
π
= 2∗40 = 3.81 < 9.4 ∈ π‘π€ =
π‘π
(620−2∗40)
21
= 25.71 < 84 ∈
Section is plastic i.e. β=1.0
Check for Moment capacity
Md = βbZpfy/γmo <1.2Zefy/rmo
=1x8772.31x1000x250/(1.1x106) <1.2x7659.6x250/(1.1x106 )
=1993.7KNm <2088.98KNm ok
Check for Shear capacity
Vd =
=
Av∗fy
√3∗ππ0
620∗21∗250
√3∗1.1
K
=1708.43KN >882.78 ok
Check for Deflection
5 × ππ 3
πΏπππ₯ =
384πΈπΌ
79
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
W = 46.32x7.2+5.68x5+30x2x2x1.1x1.5 = 1516.904KN
πΏπππ₯ =
5×1516.904×1000×72003
384×2×105 ×237447×104
< 7200/240
=15.52mm< 30
Check for Web Buckling
πΎπΏ
600
= 2.5 ×
= 71.43
π
21
For buckling class ‘c’, fy = 250
Fcd = 133.76MPa
Fcdw = (b1+n1)x tw x fcd
= (300+310) x 21 x 133.76
= 1713.47KN > 882.78
OK
b1 = 300mm assumed
OK
Check for Web Crippling
Fcdc = (b1+n2)xtwxfyw/rmo
Where n2 = 2.5(40+27) = 167.5
Fcdc = (300+167.5)x21x250/1.1 = 2231.25KN > 882.78
IV.
OK
Design of Steel Trusses
Pratt truss of 8 panels of 4.5m each
Span of truss = 36m
Height of truss = 6m
Dead load due to deck slab, wearing coat, stringer beam and cross girder acting at each
node
=
46.32∗7.2+5.68∗5
2
= 181KN
Self-wt. of truss = (0.15L+5.5)=0.15x36+5.5=10.9KN/m
Self-wt. at each node point = 4.5x10.9=49.05KN
Total dead load = 181+49.05 = 230.05KN
Live loads: IRC class AA loading, maximum BM is produced when the class AA vehicle is
closer to main girder
80
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Max load transferred at the edge of kerb = (350x4.975/7.2+350x2.925/7.2)x1.5x1.1
= 633.65KN
Average UDL = 633.65/3.6 = 176KN/m
Forces in truss member
1) ILD for U1L0
Load on Right Side (between L1 and L8)
FL0U1xsin 53.13β° + R1 = 0
π
1
ο FL0U1 = − π ππ 53.13°
Load at L1; R1 = 7⁄8
So, FL0U1 = -1.094 kN
Now,
Force due to Dead Load
= 230.05(1.094 + 0.938 + 0.78 + 0.625 + 0.469 + 0.313 + 0.156)
= 1006.7 kN (Comp)
Force due to Live Load
= 176x[ ½ (0.985 + 1.094)x0.45 + ½ (0.985 + 1.094)x3.15]
= 658.63 kN (Comp)
2) ILD for U1U2
81
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Load on right side
FU1U2x6 + R1x9 = 0
9
ο FU1U2 = − 6 R1
Load at L2; R1 = 6⁄8
So, FU1U2 = -1.125 kN
Load on left side
FU1U2x6 + R2x4.5 = 0
ο FU1U2 = -4.5R2
Load at L2; R2 = 2⁄8 So, FU1U2 = -1.125 kN
Load at L1; R2 = 1⁄8 So, FU1U2 = -0.563 kN
Now,
Force due to Dead Load
= 230.05(0.563 + 1.125 + 0.938 + 0.75 + 0.563 + 0.375 + 0.188)
= 1035.69 kN (Comp)
Force due to Live Load
= 176x[ ½ (1.013 + 1.125)x3.6]
= 677.32 kN (Comp)
3) ILD for U2U3
Now,
Force due to Dead Load
= 1293.8 kN (Comp)
Force due to Live Load
= 846.17 kN (Comp)
4) ILD for U3U4
82
Design of Bridge Over Kerunga Khola, Chitwan
Now,
Force due to Dead Load
= 1380.3 kN (Comp)
Force due to Live Load
= 902.88 kN (Comp)
5) ILD for L0L1
∑Fx = 0
ο FL0L1 + FL0U1 cos 53.13β° = 0
ο FL0L1 = R1 cot 53.13β°
Load at L1; R1 = 7⁄8
So, FU1U2 = 0.656 kN
Now,
Force due to Dead Load
= 603.65 kN (Tens)
Force due to Live Load
= 394.73 kN (Tens)
6) ILD for L1L2
Now,
Force due to Dead Load
83
2069 – AB Bridge
Design of Bridge Over Kerunga Khola, Chitwan
= 603.65 kN (Tens)
Force due to Live Load
= 394.73 kN (Tens)
7) ILD for L2L3
Now,
Force due to Dead Load
= 1035.69 kN (Tens)
Force due to Live Load
= 677.16 kN (Tens)
8) ILD for L3L4
Now,
Force due to Dead Load
= 1293.801 kN (Tens)
Force due to Live Load
= 846.173 kN (Tens)
(Diagonal Members)
9) ILD for U1L2
Load on Right Side
FU1L2xsin 53.13β° - R1 = 0
84
2069 – AB Bridge
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
π
1
ο FU1L2 = π ππ 53.13°
Load at L2; R1 = 6⁄8
So, FL0U1 = 0.938 kN (Tens)
Load on left Side
FU1L2xsin 53.13β° + R2 = 0
π
2
ο FU1L2 = − π ππ 53.13°
Load at L1; R2 = 1⁄8
So, FL0U1 = - 0.156 kN
Now,
Force due to Dead Load
= 230.05(-0.156 + 0.938 + 0.782 + 0.625 + 0.469 + 0.313 + 0.156)
= 503.579 kN (Tens)
Force due to Live Load
Maximum Compression
= 176x[ ½ (0.05 + 0.156)x3.6
= 65.261 kN
Maximum Tension
= 559.79 kN
10) ILD for U2L3
Now,
Force due to Dead Load
= 230.05(- 0.157 – 0.313 + 0.781 + 0.625 + 0.469 + 0.312 + 0.156)
= 430.884 kN (Tens)
Force due to Live Load
Maximum Compression
= 163.66 kN
Maximum Tension
= 460.31 kN
85
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
11) ILD for U3L4
Now,
Force due to Dead Load
= 230.05(- 0.156 – 0.313 – 0.469 + 0.625 + 0.469 + 0.313 + 0.156)
= 143.78 kN (Tens)
Force due to Live Load
Maximum Compression
= 262.63 kN
Maximum Tension
= 361.47 kN
(Vertical Members)
12) ILD for U1L1
Force due to Dead Load
= 230.05x1 = 230.05 kN (Tens)
Force due to Live Load
= 176x ½ (0.6 + 1)x3.6 = 506.88 kN (Tens)
13) ILD for U2L2
86
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Consider section (1)-(1)
Load on left side (between L0 and L2)
Considering right portion
FU2L2 = R2 (Tens)
Load at L0; R2 = 0; FU2L2 = 0
Load at L2; R2 = 2⁄8 ; FU2L2 = 0.25 (Tens)
Load on right side (between L3 and L8)
Considering left portion
R1 + FL2U2 = 0
ο FL2U2 = - R2
Load at L3; R1 = 5⁄8 ; FL2U2 = - 0.625 (Tens)
Load at L8; R1 = 0 ; FL2U1 = 0
Now,
Force due to Dead Load
= 230.05 (0.125 + 0.25 – 0.625 – 0.5 – 0.375 – 0.25 – 0.125)
= 345.075 kN (Comp)
Force due to Live Load
Maximum Tension
= 176x ½ (0.163 + 0.25)x3.6 = 130.84 kN
Maximum Compression
= 176x ½ (0.533 + 0.615)x3.6 = 366.85 kN
14) ILD for U3L3
Now,
Force due to Dead Load
87
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
= 230.05x0.5
= 115.025 kN (Comp)
Force due to Live Load
Maximum Tension
= 176x ½ (0.288 + 0.375)x3.6 = 210.04 kN
Maximum Compression
= 176x ½ [(0.533 + 0.615)x0.31 + (0.5 + 0.409)x3.29] = 287.92 kN
15) ILD for U4L4
FU4L4 = 0
Class A Load
FIG. Transverse Positioning Of Class A load
Response Due To Class A loading
Ordinates Of ILD
Class A
load
L0-U1
U1-U2
U2-U3
U3-U4
L0-L1
L1-L2
L2-L3
L3-L4
U1-L2
U2L3
U3-L4
U1-L1
U2-L2
13.5
13.5
57
57
34
Force
Nature
Forcex2.3
0.048 0.316 1.094 1.05 0.903 0.79 0.69
0.58 0.72 1.125 1.07 0.89 0.77 0.64
0.95 1.07 1.406 1.33 1.06 0.87 0.69
1.125 1.23
1.5
1.4
1.04 0.79 0.54
0
0.19 0.656 0.61 0.54 0.48 0.42
0
0.19 0.656 0.61 0.54 0.48 0.42
0.563 0.72 1.125 1.07 0.89 0.76 0.64
0.96 1.07 1.406 1.33 1.06 0.87 0.7
0
0.16 0.938 0.9
0.74 0.64 0.4
0
0.05 0.156
0
0
0
0
0
0
0.781 0.474 0.59 0.48 0.38
0.16 0.157 0.31 0.04
0
0
0
0
0
0.625 0.58 0.43 0.33 0.23
0.31 0.36 0.469 0.18
0
0
0
0.44 0.28
1
0.77
0
0
0
0.59
0.52
0.5
0.29
0.35
0.35
0.52
0.5
0.43
0
0.28
0
0.12
0
0
228.204
238.545
289.302
287.5325
135.587
135.587
237.9755
289.777
182.066
9.567
130.355
24.2295
106.425
46.038
110.61
c
c
c
c
t
t
t
t
t
c
t
c
t
c
t
524.8692
548.6535
665.3946
661.32475
311.8501
311.8501
547.34365
666.4871
418.7518
22.0041
299.8165
55.72785
244.7775
105.8874
254.403
0
0.12
0.22
0
116.515
27.3615
c
t
267.9845
62.93145
0
0.625 0.59
0.163 0.25 0.163
34
0.48
0
34
0.39
0
88
34
0.3
0
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Ordinates Of ILD
Class A
load
13.5
13.5
57
57
34
U3-L3
0.26
0
0.28
0
0.375
0.5
0.14
0.46
0
0.35
34
34
34
0
0
0
0.26 0.18 0.097
Force
Nature
Forcex2.3
36.645
84.878
t
c
84.2835
195.2194
The maximum axial loads in each member of the truss are tabulated below:
Members
DL kN
LL kN
DL+LL kN
L0-U1
1006.7 (C)
658.63 (C)
0
1665.33 (C)
U1-U2
1035.69 (C)
677.32 (C)
0
1713.01 (C)
U2-U3
1293.8 (C)
846.17 (C)
0
2139.97 (C)
U3-U4
1380.3 (C)
902.88 (C)
0
2283.18 (C)
L0-L1
603.65 (T)
394.73 (T)
0
998.38
(T)
L1-L2
603.65 (T)
394.73 (T)
0
998.38
(T)
L2-L3
1035.69 (T)
677.16 (T)
0
1712.85 (T)
L3-L4
1293.80 (T)
846.17 (T)
0
2139.97 (T)
U1-L2
503.579 (T)
559.79 (T)
65.262 (C)
1063.369 (T)
U2-L3
430.884 (T)
460.31 (T)
163.66 (C)
891.194 (T)
U3-L4
143.78 (T)
361.47 (T)
262.63 (C)
505.25
(T)
U1-L1
230.05 (C)
506.88 (C)
0
736.93
(C)
U2-L2
345.075 (C)
366.85 (C)
130.84 (T)
711.925 (C)
U3-L3
115.025 (C)
287.92 (C)
210.04 (T)
402.945 (C)
U4-L4
0
0
0
0
Properties of Bolt and Gusset Plate
BOLT
Diameter of bolt = M20 bolts of the product grade C
Property class = 10.9 (for steel truss bridge)
From table,
Ultimate tensile strength of bolt material (fub) =1040 Mpa
Yield strength of bolt material (fyb) =940 Mpa.
GUSSET PLATE
Adopt 20 mm thickness.
Ultimate tensile strength of plate (fu) = 410 Mpa.
Yield strength of plate (fy) =250 Mpa.
89
Design of Bridge Over Kerunga Khola, Chitwan
V.
2069 – AB Bridge
Design of Truss Joints
In structures like bridges where the load on connections can undergo many cycles of
reversal, fatigue of bolts can become critical if the connection is allowed to slip with each
reversal. So slip resistant connections is designed. Therefore, slip is not permitted at
ultimate load.
Slip Resistance
The design frictional force produced by a bolt at the interface of the connecting parts is
given by,
Vdsf = Vnsf/ϒmf
where Vnsf = the nominal frictional capacity produced by a bolt
= µf x nc x Kh x Fo
In which,
µf = the coefficient of friction or slip factor, usually 0.55
nc = the number of effective interfaces offering frictional resistance = 1
Kh = 1.0 for standard clearance
ϒmf = 1.25 for slip resistance designed at the ultimate load
Fo = the minimum bolt tension (proof load) =Anb x fo
Anb = the net tensile stress area of the bolt
For M20 bolt, Anb = 245 mm^2
fo = the proof stress = 0.7xfub
fub = the ultimate tensile strength of the bolt material
fo = 0.7xfub = 0.7x1040
= 728 MPa
Then, Vnsf = 0.55 x 1 x 1.0 x(245x728)
= 98098 N
Vdsf = Vnsf/ ϒmf
= 98098/1.25
= 78478.4 N
The detailed calculations for connections design at each joint has been provided in the
Annex.
Summary: Truss Joint Connections
The connection details have been tabulated below:
Members
Total Load kN
Section Provided
Bolt Arrangement
L0-U1
1665.33 (C)
DC250
3x4
U1-U2
1713.01 (C)
DC250
3x4
U2-U3
2139.97 (C)
DC300
3x5
U3-U4
2283.18 (C)
DC300
3x6
L0-L1
998.38 (T)
DC250
3x3
L1-L2
998.38 (T)
DC250
3x3
90
Design of Bridge Over Kerunga Khola, Chitwan
VI.
2069 – AB Bridge
L2-L3
1712.85 (T)
DC300
3x4
L3-L4
2139.97 (T)
DC300
3x5
U1-L2
1063.37 (T)
4-ISA 70x70x10
2x4
U2-L3
891.19 (T)
4-ISA 65x65x10
2x3
U3-L4
505.25 (T)
4-ISA 55x55x8
2x2
U1-L1
736.93 (C)
4-ISA 100x100x10
2x3
U2-L2
711.93 (C)
4-ISA 100x100x10
2x5
U3-L3
402.95 (C)
4-ISA 80x80x10
2x2
U4-L4
0
4-ISA 80x80x10
2x2
Consideration of Wind Load on Member Stresses
No.
Depth(mm)
Width(mm)
Face(m)
Area(m2)
4
316
300
0.316
5.688
2
266
300
0.266
2.394
4
316
300
0.316
5.688
4
266
300
0.266
4.788
3. End posts
2
266
300
0.266
3.99
4. Verticals
7
300
212
0.212
8.904
5. Diagonals
2
300
152
0.152
2.28
2
300
142
0.142
2.13
2
300
122
0.122
1.93
7×0.5 m2 @ top
-
-
-
-
3.5
9×0.5m2
-
-
-
-
4.5
-
-
-
-
29.844
Members
1. Top chords
2. Bottom chords
6. Gusset plates
area @ 0.5m2 for
@ bottom
7. Deck structure
Case (I) Bridge is unloaded
1
οΆ Wind load acting on top chord = Wind load acting on top chord + 2 wind load on
1
verticals + 2 wind load on diagonals and end posts + wind load on top gusset
1
= Pz × G × CD × (Area of top chord + 2 (areas of end post + verticals + diagonals) +
1
top gusset) + Pz × G ×CD × η (Area of top chord + 2 (areas of end post + verticals + diagonals)
+ top gusset)
1
= 940.60 × 2(5.668+2.394 + 2(3.99+5.544+2.28+2.13+1.83)+3.5)(1.776+0.876× 1.776)
= 121.90 KN
91
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
1
οΆ Wind load acting on bottom chord = Wind load acting on bottom chord + 2 wind load on
1
verticals + 2 wind load on diagonals and end posts + wind load on bottom gusset
1
= Pz × G × CD×( Area of bottom chord + 2 (areas of end post + verticals + diagonals) +
1
bottom gusset + deck structure area) + Pz × G ×CD × η (Area of bottom chord + 2 (areas of
end post + verticals + diagonals )+ bottom gusset)
= 940.60 x2 (1.776x (5.688+4.788+0.5(3.99+5.544+2.28)+4.5+29.844 )+ 0.876x
1.776(5.688+4.788+0.5(3.99+5.544+2.28)+4.5)
= 230.598 KN
[Chandra, 1981]
(a) Overturning effect due to wind when the bridge is unloaded
Taking moment about the level of bearings,
2Rx7.5 = 121.90 x6.158+ 230.598 x 0.158
Or, R = 52.47 KN
Thus, 2R = 104.95 KN
Due to the overturning effect, a thrust of 104.95 KN acts downward on leeward
girder,
92
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Increase in stress in central top chord member U3U4 of leeward girder is given by,
1
104.95
= 2 x 1.5x36x 36
= 78.71 KN (C)
Increase in stress in central bottom chord member L3L4 of the leeward girder is given
by,
1
= 2 x 1.406x36x
104.95
36
= 73.78 KN (T)
(b) (i) Lateral effect of top chord bracing when bridge is unloaded
Wind pressure acting on top lateral bracing is shown in figure.
The top chord of leeward girder are subjected to tension due to lateral effect of top
bracing.
Therefore, the force in these members decrease.
Decrease in force in central top chord member due to top lateral bracing is,
=
121.90∗27
8
x
1
7.5
= 54.855 KN (T)
(ii) Lateral effect of bottom chord bracing when the bridge is unloaded
Wind load acting on bottom bracing is shown in figure.
93
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
The bottom chord members of leeward girder are subjected to tension.
Therefore, the forces in these members increase.
Increase in force in central bottom chord member due to bottom lateral bracing is,
=
230.598∗36
8
x
1
7.5
= 138.36 KN (Tension)
Case (II) Bridge is loaded
When the bridge is loaded with the vehicles, then the weight of the vehicle
counter act the wind load acting on the truss members. So, this case is not shown here.
Thus,
Total T on L3L4 = 2139.97 + 73.78 + 138.36 = 2352.1 kN < 3165.45 kN
OK
And,
Total C on U3U4 = 2283.18 + 78.71 – 54.855 = 2307 kN < 2952.6 kN
OK
VII. Consideration of Seismic Force
Sections
Weight (kN/m)
No.
Length (m)
Total (kN)
Deck
-
-
-
2015.47
Stringer
-
-
-
156.96
Cross girder
-
-
-
184.66
DC 300
1.092
8
4.5
39.31
DC250
0.984
8
4.5
17.71
4
3.75
14.76
Bottom Part
ISA 70x70x10
1.004
4
3.75
15.06
ISA 65x65x10
0.94
4
3.75
14.1
ISA 55x55x8
0.7
4
3.75
10.5
ISA 60x60x10
0.876
14
3
36.792
ISA 65x65x8
(bracing)
0.07
16
8.75
9.8
Live Load
1108
Total
3626.7
Top Part
DC 300
1.092
8
4.5
39.31
DC250
0.984
4
4.5
17.71
4
3.75
14.76
ISA 70x70x10
1.004
4
3.75
15.06
ISA 65x65x10
0.94
4
3.75
14.1
94
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
ISA 55x55x8
0.7
4
3.75
10.5
ISA 60x60x10
0.876
14
3
36.792
ISA 45x45x5
(bracing)
0.034
12
8.75
3.57
Total
151.8
Zone factor (z) = 0.36 (for zone V)
D = 3626.7+151.8 = 3778.5 KN
F = k = 115.89 KN/m (from conjugate beam analysis)
Time period (T) = 2(D/1000F)0.5
= 0.3611 sec
From code,
Sa/g = 2.5 (for soil type II and T= 0.3611 sec <0.4sec)
Importance factor (I) = 1 (for normal bridge)
Ah = Z/2xI/Rx Sa/g
0.36
=
2
1
∗ 4 ∗ 2.5
= 0.1125
For Bottom part,
Feq = Ah x Seismic Weight
= 0.1125 x 3626.7
= 408 KN
For Top part,
Feq = Ah x Seismic Weight
= 0.1125 x 151.8
= 17.1 KN
VIII. Design of Bracings
Design of Top Bracing
Load due to Wind effect will be considered.
95
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Design Load=54.24KN
π
Based on yielding, Ag = ππ¦/ππ0
=
54.24∗1000
250/1.1
= 238mm2
Adopt <45x45x5, Ag = 428 mm2 (It may be difficult in bolting for smaller sections.)
Check for Rupture
Gusset Thickness =12mm, M20 bolts of grade 10.9
fus=410MPa , fub=1040MPa
fys=250MPa,
fyb=940MPa
Slip Resistance=78478.4N
No. of Bolts =
54.24∗103
78478.4
=1
Adopt 2 Bolts.
π½ = 1.4 − 0.076 ∗
π€
π‘
ππ¦
∗ ππ’
= 1.4 - 0.076x45/5x250/410x(45+25-5)/50
= 0.858
Tdn =
0.9∗π΄ππ∗ππ’ π½∗π΄ππ∗ππ¦
ππ1
+
ππ0
=
0.9∗(45−2.5)∗5∗410 0.858∗(45−2.5)∗5∗250
1.25
+
1.1
Design of Bottom Bracing
Load due to Seismic effect will be considered.
96
= 104.2 kN > 54.24 kN
OK
Design of Bridge Over Kerunga Khola, Chitwan
Designed AF=208.16KN
π
208.16
Based on yielding, Ag=ππ¦/ππ0 =250/1.1 =916mm2
Adopt <65x65x8,Ag=976mm2
Check For Rupture
208.16∗10^3
No. of Bolt=
= 2.65
78478.4
Adopt 3 Bolts
β = 1.4-0.076x65/8x250/410x(65+35-8)/100
= 1.05>0.7
Tdn =
=
0.9∗π΄ππ∗ππ’ π½∗π΄ππ∗ππ¦
+
ππ1
ππ0
0.9∗390∗410 1.05∗610∗250
+
1.25
1.1
= 260.7KN>208.16KN
Where Anc=(65-4-22)x10=390
Ago=(65-4)x10 =610
Block Shear
Avg=150x10=1500 mm2
Avn=(150-1.5x22)x10=1170mm2
Atg=30x10=300mm2
Atn=(30-0.5-22)x10=75mm2
Tdb=
1500∗250 0.9∗75∗410
√3∗1.1
+
1.25
or
0.9∗1170∗410
√3∗1.25
+
300∗250
1.1
=218.96 or 267.58
=218.96KN>208.16KN ok
97
2069 – AB Bridge
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Design of Portal Bracing
Design of members bc, de
Design tension, T = 250.89 kN
Ag required = ππ¦
π
⁄πΎπ0
=
250.89×103
250⁄
1.1
= 1103.92 mm2
Use <65x65x10 with Ag = 1200 mm2
For M20 bolts of property class 10.9,
Vnsf = 78478.4 kN
∴ No of Bolts =
250.89×10^3
78478.4
= 3.2
Take 4 bolts.
Check for Rupture
65
250
β = 1.4 – 0.076× 10 × 410 ×
Tdn =
0.9∗π΄ππ∗ππ’
ππ1
+
π½∗π΄ππ∗ππ¦
ππ0
=
65+35−10
150
= 1.22 > 0.7 and <1.4
0.9∗10∗(60−22−5)∗410
1.25
+
1.22∗10∗(65−5)∗250
= 264.676 kN > 250.89 kN
1.1
OK
Check for Block Failure
Avg = 200x10 = 2000 mm2
Avn = (200 – 3.5x22)x10 = 1230 mm2
Avg = 30x10 = 300 mm2
Avg = (30 – 0.5x22)x10 = 190 mm2
Tdb =
2000×250
1.1×√3
+
0.9×190×410
1.25
or
0.9×1230×410
1.25×√3
+
300×410
1.1
= 318.52 or 277.82 kN
= 277.82 kN > 250.89 kN
OK
Design of member bf
Design Compression = 247.84 kN
98
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Assume fcd = 60 MPa
Ag required =
247.84×103
60
= 4130.67 mm2
Try <150x150x20 with Ag = 5620 mm2 and rv = 29.3 mm
Now,
πΏ
=
7200
= 245.73
ππ£
29.3
π1+π2 150+150
=
2π‘
= 7.5
2×20
250
ε = ( ππ¦ )0.5 = 1
π2 πΈ
ε( 250 )0.5 = (
π 2 ×2×105 0.5
)
250
λvv =
πΏ
ππ£
π2 πΈ 0.5
ε(
)
250
=
λφ =
π1+π2
2π‘
π2 πΈ 0.5
ε(
)
250
= 0.08
245.73
88.86
= 88.86
= 2.76
From Table 12 of IS code,
k1 = 0.20, k2 = 0.35, k3 = 20
λe = √π1 + π2 × ππ£π£ 2 + π3 × ππ 2 = 1.73
From Table 7 of IS Code, α = 0.49 for buckling class ‘c’
Then,
Φ = 0.5[1 + πΌ(ππ − 0.2) + ππ 2 ] = 2.37
1
χ = [π+(π2 −ππ 2 )0.5 ] = 0.25
πππ
fcd = χxπΎπ0 = 56.96 Mpa
Strength = fcdxAg = 56.96x5620 = 320.1 kN >247.84 kN
No of Bolts =
247.84×10^3
78478.4
OK
= 3.2
Take 4 bolts.
Check for Rupture
β = 1.4 – 0.076×
Tdn =
0.9∗π΄ππ∗ππ’
ππ1
150
+
20
250
× 410 ×
π½∗π΄ππ∗ππ¦
ππ0
=
150+75−20
150
= 0.9 > 0.7 and <1.4
0.9∗20∗(150−22−10)∗410
1.25
+
0.9∗20∗(150−10)∗250
= 1269.4 kN >> 250.89 kN
Check for Block Failure
Avg = 200x20 = 4000 mm2
Avn = (200 – 3.5x22)x20 = 2460 mm2
Avg = 75x20 = 1500 mm2
99
1.1
OK
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Avg = (75 – 0.5x22)x20 = 1280 mm2
Tdb =
4000×250
1.1×√3
+
0.9×1280×410
1.25
or
0.9×2460×410
1.25×√3
+
1500×410
1.1
= 902.7 or 978.3 kN
= 902.7 kN β« 250.89 kN
OK
Design of Sway Bracing
Load on sway braces = 33.97 kN (T/C)
Use <45x45x5 with two bolts at each end as sway bracing.
IX.
Design of Connections
Design of connection (cross beam and stringer connection)
Different plane connection:
Shear force from one stringer (Vu) = 381.02 KN
Total shear for connection design = shear from two stringers = 2x381.02 = 762.04 kN
Moment from stringer (Mu) = 428.64 KNm
Design Moment = 60% of Mu = 0.6x428.64 = 257.184KNm
For M20 bolts of grade 10.9, we know the following properties:
fyb = 940 N/mm2
fub = 1040 N/mm2
Nominal Frictional strength of bolt (Vnsf) = μfxnexkhxFo
Here, μf = the coefficient of friction or slip
ne = the number of effective interfaces offering frictional resistance(i.e 2)
kh = 1.0 for standard clearance
Fo = the minimum bolt tension (proof load = Anbxfo)
Anb = the net area of the bolt
fo = the proof stress = 0.7xfub = 0.7x1040 = 728
fub = the ultimate tensile strength of the bolt material
100
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
So, Fo = 245 mm2 x 728 N/mm2 = 178360 N
Then, Vnsf = 0.52x2x1x178360
= 185.49 KN
Design Strength of bolt (Vdsf) = Vnsf/ϒmf = 185.49/1.25 = 148.39 KN
Now,
6π
No. of bolts required (n) = √π×π×πππ
Where,
Tdb= Tdf = Tnf/ ϒmf
ϒπ1
Tnf = nominal tensile strength of bolt = 0.9xfubxAnb ≤ fybxAsbxϒππ
= 0.9x1040x245 ≤ 940x314x1.25/1.1
= 229.32 KN ≤ 335.41 KN
So, Tdf = 229.32/1.25 = 183.456 KN
p = pitch distance = 2.5d = 2.5x20 = 50mm
No. of vertical rows (m) = 2
6×257.184×106
Therefore, n= √2×50×183.456×103 = 9.17≈ 10
To arrange 10 no. of bolts in a single line is not adjustable and not okay with our problem, so
we try again with M27 bolts.
Frictional strength of bolt (Vnsf) = μfxnexkhxFo
where Fo = Anbxfo = 459 mm2 x 728 N/mm2 = 334152 N
So, Vnsf = 0.52x2x1x334152
= 347.52 KN
Design strength of bolt (Vdsf) = Vnsf/ϒmf = 347.52/1.25 = 278.01 KN
6π
Now, No. of bolts required (n) = √π×π×πππ
Tdb= Tdf = Tnf/ ϒmf
ϒπ1
Tnf = nominal tensile strength of bolt = 0.9xfubxAnb ≤ fybxAsbxϒππ
= 0.9x1040x459 ≤ 940x573x1.25/1.1
= 429.624 KN ≤ 612.07 KN
So, Tdf = 429.624/1.25 = 343.699 KN
p = pitch distance = 2.5d = 2.5x27 = 67.5mm
Adopt p = 70mm
No. of vertical rows (m) = 2
6×257.184×106
Therefore, n= √2×70×343.699×103 = 5.66 ≈ 6 bolts
Total no. of bolts in same plane = 2x6 = 12
Check:
Shear force in extreme critical bolt, Vb =
ππ’
π
=
762.04
2∗12
101
= 31075 KN
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
∑yi2 = 2(352+1052+1752)
= 85750 mm2
M’ = M/2 = 257.184/2 = 128.59 KNm
Yn = 175 mm
Tensile force in extreme critical bolt (Tb) =
π′ × π¦π
∑yi2
=
128.59∗106 ∗175
85750
= 262.43 KN
Combined shear and tension equation:
ππ
2
ππ
2
(πππ) + (πππ) ≤ 1
63.50 2
262.43 2
= (148.39) + (343.699)
= 0.766 < 1
OK
Above value 0.766 which is less than 1 shows that the design connection we designed is okay.
Stringer and angle section connection (same plane connection)
Adopt 24mm HSFG bolt of grade 12.9 having following properties:
fyb = 1100 N/mm2
fub = 1220 N/mm2
Then,
Frictional strength of bolt (Vnsf) = μfxnexkhxFo
Where Fo =Anbxfo = 353 mm2 x (0.7x1220) N/mm2 = 301462 N
Vnsf = 0.52x2x1x301462
= 313.52 KN
Design Strength of bolt (Vdsf) = Vnsf/ϒmf = 313.52/1.25 = 250.82 KN
6π
No. of bolts required (n) = √π×π×πππ π
Tdb= Tdf = Tnf/ ϒmf
ϒπ1
Tnf = nominal tensile strength of bolt = 0.9xfubxAnb ≤ fybxAsbxϒππ
= 0.9x1220x353 ≤ 940x452x1.25/1.1
102
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
= 387.594 KN ≤ 428.82 KN
So, Tdb = 387.594/1.25 = 310.07 KN
p = pitch distance = 2.5d = 2.5x24 = 60mm
No. of vertical rows (m) = 2
6×257.184∗106
Therefore, n= √2×60×250.82×103 = 7.16 ≈ 8 bolts
The arrangement of 8 bolts in a single vertical line is not adjustable as required spacing
between the bolts is not sufficient.
So, trying with M36 HSFG bolts,
Frictional strength of bolt (Vnsf) = μfxnexkhxFo
Where Fo =Anbxfo = 817 mm2 x (0.7x1220) N/mm2 = 697718 N
So, Vnsf = 0.52x2x1x697718
= 725.63 KN
Design Strength of bolt (Vdsf) = Vnsf/ϒmf = 725.63/1.25 = 580.5 KN
6π
No. of bolts required (n) = √π×π×πππ π
Tdb= Tdf = Tnf/ ϒmf
ϒπ1
Tnf = nominal tensile strength of bolt = 0.9xfubxAnb ≤ fybxAsbxϒππ
= 0.9x1040x817 ≤ 940x1018x1.25/1.1
= 7640412 KN ≤ 1087.41 KN
Tdf = 764.712/1.25 = 611.77 KN
p = pitch distance = 2.5d = 2.5x36 = 90mm
No. of vertical rows (m) = 2
6×257.184×106
Therefore, n= √2×90×580.5×103 = 3.84 ≈ 4 bolts
Hence, 4 bolts in each vertical line can be arranged with required spacing.
103
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Design of connection (cross beam and truss)
Different plane connection:
Shear force from one cross beam (Vu) = 882.78KN
Total shear for connection design = shear from one cross beam = 1x882.78 =
882.78kN
Moment from s (Mu) = 1812.77 KNm
Design Moment = 60% of Mu = 0.6x1812.77 = 1087.662 KNm
For M20 bolts of grade 10.9, we know the following properties:
fyb = 940 N/mm2
fub = 1040 N/mm2
Nominal Frictional strength of bolt (Vnsf) = μfxnexkhxFo
Here, μf = the coefficient of friction or slip
ne = the number of effective interfaces offering frictional resistance(i.e 2)
kh = 1.0 for standard clearance
Fo = the minimum bolt tension (proof load = Anbxfo)
Anb = the net area of the bolt
fo = the proof stress = 0.7xfub = 0.7x1040 = 728
fub = the ultimate tensile strength of the bolt material
So, Fo = 245 mm2x728 N/mm2 = 178360 N
Then, Vnsf = 0.52x1x1x178360
= 92.745 KN
Design Strength of bolt (Vdsf) = Vnsf/ϒmf = 92.745/1.25 = 74.196 KN
Now,
6π
No. of bolts required (n) = √π×π×πππ
Where,
Tdb= Tdf = Tnf/ϒmf
ϒπ1
Tnf = nominal tensile strength of bolt = 0.9xfubxAnb≤fybxAsbxϒππ
= 0.9x1040x245 ≤ 940x314x1.25/1.1
= 229.32 KN ≤ 335.41 KN
So, Tdf = 229.32/1.25 = 183.456 KN
p = pitch distance = 2.5d = 2.5x20 = 50mm
No. of vertical rows (m) = 4
6×1087.662×106
Therefore, n= √4×50×183.456×103 = 13.336≈ 14
To arrange 14 no. of bolts in a single line is not adjustable and not okay with our problem, so
we try again with M27 bolts.
Frictional strength of bolt (Vnsf) = μfxnexkhxFo
Where Fo =Anbxfo= 459 mm2x728 N/mm2 = 334152 N
104
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
So, Vnsf = 0.52x1x1x334152
= 173.76 KN
Design strength of bolt (Vdsf) = Vnsf/ϒmf = 173.76/1.25 = 139.008 KN
6π
Now, No. of bolts required (n) = √π×π×πππ
Tdb= Tdf = Tnf/ϒmf
ϒπ1
Tnf = nominal tensile strength of bolt = 0.9xfubxAnb≤fybxAsbxϒππ
= 0.9x1040x459 ≤ 940x573x1.25/1.1
= 429.624 KN ≤ 612.07 KN
So, Tdf = 429.624/1.25 = 343.699 KN
p = pitch distance = 2.5d = 2.5x27 = 67.5mm
Adopt p = 70mm
No. of vertical rows (m) = 4
6×1087.662×106
Therefore, n= √4×70×343.699×103 = 8.234 ≈ 9 bolts
To arrange 9 no. of bolts in a single line is not adjustable and not okay with our problem, so we
try again with M30 bolts.
Frictional strength of bolt (Vnsf) = μfxnexkhxFo
whereFo =Anbxfo= 0.78x706.85 mm2x728 N/mm2 = 401382.44 N
So, Vnsf = 0.52x1x1x401382.44
= 208.72 KN
Design strength of bolt (Vdsf) = Vnsf/ϒmf = 208.72/1.25 = 166.976 KN
6π
Now, No. of bolts required (n) = √π×π×πππ
Tdb= Tdf = Tnf/ϒmf
ϒπ1
Tnf = nominal tensile strength of bolt = 0.9xfubxAnb≤fybxAsbxϒππ
= 0.9x1040x0.78x706.85 ≤ 940x706.85x1.25/1.1
= 516.06 KN ≤ 755.044 KN
So, Tdf = 755.044/1.25 = 604.035 KN
p = pitch distance = 2.5d = 2.5x30 = 75m
No. of vertical rows (m) = 4
6×1087.662×106
Therefore, n= √4×75×604.035×103 = 6 bolts
Total no. of bolts in same plane = 4x6 = 24
Check:
Shear force in extreme critical bolt,Vb =
ππ’
π
882.78
=
24
= 36.78 KN
∑yi2 = 4(37.52+112.52+187.52)
= 196875 mm2
M’ = M/2 = 1087.662/2 = 543.831 KNm
105
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Yn = 187.5 mm
π′ ×π¦π
Tensile force in extreme critical bolt (Tb) =
∑yi2
=
543.831∗106 ∗187.5
196875
= 517.93 KN
Combined shear and tension equation:
ππ
2
ππ
2
(πππ) + (πππ) ≤ 1
36.78 2
517.93 2
= (74.196) + (604.35)
= 0.98 < 1
OK
Above value 0.98 which is less than 1 shows that the design connection we designed is okay.
Cross beam and angle section connection (same plane connection)
Adopt 24mm HSFG bolt of grade 12.9 having following properties:
fyb = 1100 N/mm2
fub = 1220 N/mm2
Then,
Frictional strength of bolt (Vnsf) = μfxnexkhxFo
Where Fo =Anbxfo= 353 mm2x(0.7x1220) N/mm2 = 301462 N
Vnsf = 0.52x2x1x301462
= 313.52 KN
Design Strength of bolt (Vdsf) = Vnsf/ϒmf = 313.52/1.25 = 250.82 KN
6π
No. of bolts required (n) = √π×π×πππ π
Tdb= Tdf = Tnf/ϒmf
ϒπ1
Tnf = nominal tensile strength of bolt = 0.9xfubxAnb≤fybxAsbxϒππ
= 0.9x1220x353 ≤ 940x452x1.25/1.1
= 387.594 KN ≤ 428.82 KN
So, Tdb = 387.594/1.25 = 310.07 KN
p = pitch distance = 2.5d = 2.5x24 = 60mm
No. of vertical rows (m) = 2
106
Design of Bridge Over Kerunga Khola, Chitwan
Therefore, n= √
6×1087.662∗106
2×60×250.82×103
2069 – AB Bridge
= 14.72≈ 15 bolts
The arrangement of 15 bolts in a single vertical line is not adjustable as required spacing
between the bolts is not sufficient.
So, trying with M36 HSFG bolts,
Frictional strength of bolt (Vnsf) = μfxnexkhxFo
Where Fo =Anbxfo= 817 mm2x(0.7x1220) N/mm2 = 697718 N
So, Vnsf = 0.52x2x1x697718
= 725.63 KN
Design Strength of bolt (Vdsf) = Vnsf/ϒmf = 725.63/1.25 = 580.5 KN
6π
No. of bolts required (n) = √π×π×πππ π
Tdb= Tdf = Tnf/ϒmf
ϒπ1
Tnf = nominal tensile strength of bolt = 0.9xfubxAnb≤fybxAsbxϒππ
= 0.9x1040x817 ≤ 940x1018x1.25/1.1
= 7640412 KN ≤ 1087.41 KN
Tdf = 764.712/1.25 = 611.77 KN
p = pitch distance = 2.5d = 2.5x36 = 90mm
No. of vertical rows (m) = 2
6×1087.662×106
Therefore, n= √2×90×580.5×103 = 7.9≈ 8 bolts
Hence, 8 bolts in each vertical line are to be arranged but our dimension of cross beam doesn’t
fit all the bolts. So we are trying 3 vertical lines.
6×1087.662×106
n= √4×90×580.5×103 = 5.58≈ 6 bolts
Hence, 6 bolts in each vertical line can be arranged with required spacing.
X.
Design of Elastomeric bearing for truss
Calculation of Loads on Bearing
107
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
3. DL from Superstructure
From calculation table seismic force in bracing
Total DL from super structure (Wu) = 2776.2 KN
DL from superstructure on a bearing (DLsup) = 2776.2/4 = 694.05 KN
4. LL from Superstructure
Transverse positioning of class A vehicle:
Distribution factor for one bearing = 0.88W+0.64W+0.52W+0.28W
=2.32W
Longitudinal positioning of class A vehicle:
Total load = 2.32W
= 2.32[57(1+0.966)+34(0.85+0.764+0.68+0.6)+13.5x0.042]
= 2.32x211.025
= 489.58 KN
5. Load due to braking effort
ο·
Class A load
Braking load = 0.2 × (2 × 114 + 4 × 68 + 2 × 27) × 2 = 221.6 KN
ο·
Class AA track load
Braking load = 0.2 × 700 = 140 KN
ο·
Class AA
Braking load = 0.2 × 400 = 80 KN
Taking braking effort due to Class A loading,
Horizontal Braking load on a bearing (FbrH) = 221.6/4= 55.4 KN
Braking loads acts at 1.2m above wearing coat.
Point of application of braking load is 2.1 m (1.2+0.08+0.2+0.62) from bearing. It
induces vertical reaction on bearing.
221.6 × 2.1
Vertical reaction on a bearing due to braking load= 36× 2 = 6.46 KN
6. Wind load
From the wind calculation of bracing,
Wind load acting on the top chord = 121.90 KN
Wind load acting on the bottom chord = 230.598 KN
Total wind load acting on the truss = 121.90 + 230.598 = 352.498 KN
Wind load in transverse direction on a bearing (FWT) = 352.498/4 = 88.1245 KN
108
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
ο§ Wind load in longitudinal direction of bridge (FWL) = 0.25 × FTW= 88.1245 KN
Wind load in longitudinal direction on a bearing (FWL) = 88.1245/4 = 22.031 KN
ο
ο§ Wind load in vert. dir. of bridge (FWV) = PZ × A3 × G × CL = 940.6 × 7.8 × 36.3 × 2 × 0.75
= 399.48 KN
Wind load in vertical direction on a bearing (FWV) = 399.48/4 = 99.87 KN
7. Seismic Load
π
πΌ
Seismic load (FSh) = 2 × π
×
ππ
π
×π
[Refer Cl. 219, IRC 6]
Take, Seismic Zone - V, Soil Strata - Medium, Damping - 5 %, Bridge Class - Normal
Where,
π
πΌ
Ah = πΌh= 2 × π
×
ππ
π
= 0.225; Z = 0.36, I = 1, R = 2, ππ/π= 2.5
W = 2670.5 KN in longitudinal direction
W = 2670.5 + 0.2 × 2 × (2 × 114 + 4 × 68 + 2 ×27) = 2892.10 KN in transverse direction
So,
FShL = 0.225 x 2670.5= 600.86 KN in longitudinal direction of bridge
FShT = 0.225 × 2892.10= 650.72 KN in transverse direction of bridge
Then,
ο§ Seismic load in transverse direction on a bearing (FShT) =600.86/4 =150.22 KN
ο§ Seismic load in longitudinal direction on a bearing (FShL) = 650.72/4 =162.68 KN
ο§ Vertical reaction due to seismic load on support of bridge (FSv)
Seismic loads acts on c. g. of seismic weight. It creates additional vertical load on bearing.
Consider c. g. of seismic weight = 0.5 m from bearing.
V. reaction on a bearing when s. load acts in tr. dir. (FSvT) =
600.86× 0.5
7.5
× ½ =20.03 KN
Vertical reaction on a bearing when seismic Load acts in long dirn(FSVL) =
=4.52 KN
650.72× 0.5
36
×½
8. Load due to temperature variation, creep and shrinkage effect
Maximum horizontal force on a bearing (Fcst) = Δ/ππ× G × A = 26.23 KN
where,
ο¨ Strain due to temp., creep and shrinkage = 5 × 10-4 [Refer IRC 83 Part II Cl. 916.3.4]
ο¨ Horizontal deformation of bearing ( Δ )=5×10-4×36.3 × 103 × ½=9.075 mm
ο¨ Shear modulus of elastomer (G) = 1 N/mm2 [Refer IRC 83 Part II, Cl. 915.2.1]
ο¨ Preliminary height of bearing (h0) = 52 mm
ο¨ Preliminary effective sectional area of bearing (A) = b × l = 308 × 388 = 150304 mm 2
Load Combinations [Refer IRC 6 Table 1]
109
Design of Bridge Over Kerunga Khola, Chitwan
Vertical load
Combination Along
Across
of load
Traffic
Traffic
I
[N]
II(A)
[N+T]
Dl
LL
sup
Dl
LL
sup
Permissible
stress (%)
sup
100%
FbrV
Dl
LL
Horizontal Load
Along
Across
Traffic
Traffic
2069 – AB Bridge
FbrH
Dl
LL
sup
FbrV
115%
FbrH
Fcst
III(A)
[N+ T+ W]
Dl
LL
sup
FbrV
FbrH
133%
Fcst
FWV
sup
VI
[N+T+S]
Dl
LL
sup
Dl
0.2LL
F WV
FWL
F WT
sup
Dl
0.2LL
0.5FbrV
0.5FbrH
150%
Fcst
FS VL
FS VT
FS hL
FS hT
Calculation of Loads on Bearing According to Combination of Loads
Vertical and horizontal loads subjected to bearing in the direction of traffic are only taken for
design.
Combination I [N]
Total Vertical load = DLSup + LL + FbrV = 694.05 + 489.58 + 6.46 = 1190.09 KN
Total Horizontal load = FbrH = 55.4 KN
Combination II (A) [N+T]
Total Vertical load = DLSup + LL + FbrV = 694.05 + 489.58 + 6.46 = 1190.09 KN
Total Horizontal load = FbrH + Fcst = 55.4 + 26.23 = 81.63 KN
Combination III (A) [N+T+W]
Total Vertical load = DLSup + LL+ FbrV + FWV = 694.05 + 489.58 + 6.46 + 99.87 = 1289.96
KN
Total Horizontal load = FbrH + Fcst + FWL = 55.4 + 26.23 + 22.03= 103.66 KN
Combination VI [N+T+S]
Total Vertical load = DLSup + 0.2 × LL + 0.5 FbrV + FsvL
= 694.05 + 0.2 × 489.58 + 0.5 × 6.46 + 4.52= 795.20 KN
Total Horizontal load = 0.5 × FbrH + Fcst + FshL = 0.5 × 55.4 + 26.23 + 162.68 = 216.61 KN
Design of Elastomeric Pad Bearing for Combination I [N]
Effective Span = 36 m, Total Vertical Load = 1190.09 kN, Total Horizontal Load = 55.4 kN
110
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Geometrical Design
Nmax = 1190.09 kN
Nmin = 55.4 kN
Try Standard plan dimension 320x500 mm for elastomeric bearing (IRC 83 – Part II)
From table,
Loaded Area = 150304 mm2
Thickness of individual elastomer layer hi = 10 mm
Thickness of outer layer he = 5 mm
Thickness of steel laminate hs = 3 mm
Adopt 3 internal layers and 4 steel laminates.
Overall Thickness = 3x10+4x3+2x5 = 52 mm
Effective Thickness of elastomer h = 3x10+2x5 = 40 mm
Adopt side cover c = 6 mm
Check for Geometry
1. l0/b0 = 500/320 =1.5625 ≤ 2
2. h = 40 < b0/5 = 64 mm
And > b0/10 = 32 mm
150304
3. Shape factor S = 2(308+488)×10 = 9.44 (between 6 and 12)
1163.665×1000
Nmax
4. Bearing Stress in concrete (σm) = Loaded Area = 150304
= 7.74 MPa
A1/A2 is limited to 2.
A1
Allowable BS = 0.25 × πππ × √A2 = 0.25 × 30 × √2 = 10.61 N/mm2
So,
Bearing stress in concrete ≤ Allowable Bearing stress
Structural Design
1. Check for Translation
Design strain in bearing (γd) < 0.7
i.e. γd = γbd = Δππ/h + τmd = 0.225+0.369 = 0.594 < 0.7
where
Shear strain per bearing due to shrinkage, creep and temperature= Δππ/h
=
5×10−4 ×36×103
2×40
OK
OK
= 0.225
π»
55.4×103
Shear strain due to longitudinal forces = π΄×πΊ =150304×1 =0.369
2. Check for Rotation
Design rotation in bearing (αd) ≤ βnαbi, max
3. Check for Friction
Design strain in bearing (γd) ≤ 0.2 + 0.1σm
i.e. 0.2 + 0.1σm = 0.2 + 0.1x7.74 = 0.974 > 0.594
And
111
OK
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Normal stress in bearing (σm) > 2 MPa and ≤ 10 MPa
4. Check for Shear Stress
Total shear stress ≤ 5 MPa
i.e. τc + τh + τα = 2.687 MPa < 5 MPa
where
Shear stress due to compression = 1.5xσm/S = 1.5x7.74/9.44 = 1.23 MPa
Shear stress due to horizontal deformation = 0.594x1 = 0.594 MPa
Shear stress due to rotation = 0.5x(b/hi)2 x αbi,max = 0.5× (308/10)2 ×0.00182
= 0.863 MPa
OK
OK
Check for Combination VI [N + T + S]
Effective Span = 36 m, Total Vertical Load = 795.20 kN, Total Horizontal Load = 216.61 kN
Geometrical Design
Nmax = 795.20/1.5 = 530.133 kN
Nmin = 216.61/1.5 = 144.41 kN
Try Standard plan dimension 320x500 mm for elastomeric bearing, as for Combination I [N]
From table,
Loaded Area = 150304 mm2
Thickness of individual elastomer layer hi = 10 mm
Thickness of outer layer he = 5 mm
Thickness of steel laminate hs = 3 mm
Adopt 3 internal layers and 4 steel laminates.
Overall Thickness = 3x10+4x3+2x5 = 52 mm
Effective Thickness of elastomer h = 3x10+2x5 = 40 mm
Adopt side cover c = 6 mm
Check for Geometry
1. l0/b0 = 500/320 =1.5625 ≤ 2
2. h = 40 < b0/5 = 64 mm
And > b0/10 = 32 mm
150304
3. Shape factor S = 2(308+488)×10 = 9.44 (between 6 and 12)
Nmax
530.133×1000
4. Bearing Stress in concrete (σm) = Loaded Area = 150304 = 3.53 MPa
A1/A2 is limited to 2.
A1
Allowable BS = 0.25 × πππ × √A2 = 0.25 × 30 × √2 = 10.61 N/mm2
So,
Bearing stress in concrete ≤ Allowable Bearing stress
Structural Design
1. Check for Translation
112
OK
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Shear strain per bearing due to shrinkage, creep and temperature= Δππ/h
=
5×10−4 ×36×103
2×40
= 0.225
π»
144.41×103
Shear strain due to longitudinal forces, τmd = π΄×πΊ = 150304×1 =0.961
Design strain in bearing (γd) < 0.7
i.e. γd = γbd = Δππ/h + τmd = 0.225+0.961 = 1.186 < 0.7
NOT OK
2. Check for Rotation
Design rotation in bearing (αd) ≤ βnαbi, max
3. Check for Friction
Design strain in bearing (γd) ≤ 0.2 + 0.1σm
i.e. 0.2 + 0.1σm = 0.2 + 0.1x3.53 = 0.553 > 1.186
And
Normal stress in bearing (σm) > 2 MPa and ≤ 10 MPa
NOT OK
OK
4. Check for Shear Stress
Total shear stress ≤ 5 MPa
i.e. τc + τh + τα = 2.61 MPa < 5 MPa
OK
where
Shear stress due to compression = 1.5xσm/S = 1.5x3.53/9.44 = 0.561 MPa
Shear stress due to horizontal deformation = 1.186x1 = 1.186 MPa
Shear stress due to rotation = 0.5x(b/hi)2 x αbi,max = 0.5× (308/10)2 ×0.00182
= 0.863 MPa
Hence, the bearing fails. Try 400x800 elastomeric pad bearing to fulfill the failed criteria.
Loaded Area = 305744 mm2
Thickness of individual elastomer layer hi = 12 mm
Thickness of outer layer he = 6 mm
Thickness of steel laminate hs = 4 mm
Adopt 3 internal layers and 4 steel laminates.
Overall Thickness = 3x12+4x4+2x6 = 64 mm
Effective Thickness of elastomer h = 3x12+2x6 = 48 mm
Adopt side cover c = 6 mm
Check for Geometry
1. l0/b0 = 800/400 =2 ≤ 2
2. h = 48 < b0/5 = 80 mm
And > b0/10 = 40 mm
305744
3. Shape factor S = 2(388+788)×12 = 10.83 (between 6 and 12)
Nmax
530.133×1000
4. Bearing Stress in concrete (σm) = Loaded Area = 305744 = 1.734 MPa
A1/A2 is limited to 2.
A1
Allowable BS = 0.25 × πππ × √A2 = 0.25 × 30 × √2 = 10.61 N/mm2
113
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
So,
Bearing stress in concrete ≤ Allowable Bearing stress
OK
Structural Design
5. Check for Translation
Shear strain per bearing due to shrinkage, creep and temperature= Δππ/h
=
5×10−4 ×36×103
2×48
= 0.188
π»
144.41×103
Shear strain due to longitudinal forces, τmd = π΄×πΊ = 305744×1 =0.472
Design strain in bearing (γd) < 0.7
i.e. γd = γbd = Δππ/h + τmd = 0.188+0.472 = 0.66 < 0.7
OK
6. Check for Rotation
Design rotation in bearing (αd) ≤ βnαbi, max
7. Check for Friction
Design strain in bearing (γd) ≤ 0.2 + 0.1σm
i.e. 0.2 + 0.1σm = 0.2 + 0.1x1.734 = 0.3734 > 0.66
And
Normal stress in bearing (σm) > 2 MPa and ≤ 10 MPa
8. Check for Shear Stress
Total shear stress ≤ 5 MPa
i.e. τc + τh + τα = 1.59 MPa < 5 MPa
where
Shear stress due to compression = 1.5xσm/S = 1.5x1.734/10.83 = 0.240 MPa
Shear stress due to horizontal deformation = 0.66x1 = 0.66 MPa
Shear stress due to rotation = 0.5x(b/hi)2 x αbi,max = 0.5× (388/12)2 ×0.00132
= 0.69 MPa
114
OK
OK
OK
Design of Bridge Over Kerunga Khola, Chitwan
Summary: Bearing Design
Provide standard 400x800 mm elastomeric pad bearing.
Overall Thickness = 64 mm
Thickness of individual elastomer layer = 12 mm
No of internal elastomer layers = 3
Thickness of each laminate = 4 mm
No of laminates = 4
Thickness of outer layer = 6 mm
115
2069 – AB Bridge
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
C. Design of RCC Abutment
I.
Planning and Preliminary Design
A. Selection of Type of Abutment
Abutment may be of masonry or reinforced cement concrete. Masonry is technically /
economically feasible up to 5m height of abutment. In the particular case, abutment is
greater than 5 m height. So reinforced concrete wall type abutment has been selected.
B. Material Selection
Following materials are to be taken for the construction of the abutment:
οΆο M20 grade of concrete for abutment stem and dirt wall
οΆο M25 grade of concrete for abutment cap
οΆο Fe 415 HYSD bars for all RC work
C. Geometry of Abutment
οΆο Seating width = 0.515 m
Minimum seating width = 305 + 2.5 × span + 10× Ht. of abutment [Ref. Cl. 219.9, IRC 6]
= 305 + 2.5 × 12 + 10 × 12.5 = 460 mm
Seating width ≥ Bearing width + 150 mm + Projection of cap + Width of Expansion Joint
≥ 0.25 + 0.15 + 0.075 + 0.04 = 0.515 m
Width of Expansion joint ≥ 12 × 103 × 0.000011/ ΜC × 50 ΜC × ½ = 3.3 mm
≥ 5 × 10-4 × 12 × 103 × ½ = 3 mm and
≥ 20 mm
Adopt 40mm
οΆο Height of dirt wall = Depth of girder + height of bearing – thickness of approach slab
= 1 + 0.052 – 0.3
= 0.752 m
οΆο Thickness of dirt wall = 0.25 m
Where, Thickness of dirt wall, tdw ≥ 200 mm and Height of Dirt wall/7= 0.752/7 = 0.107 m
οΆο Width of stem of abutment = 1.25 m
Width of stem ≈12.5/10=1.25 π
Width of stem ≥thickness of dirt wall + seating width – projection = 0.515 + 0.25 -0.075 =
0.69 m
οΆο Thickness of footing = 2 m
116
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Thickness of footing ≈ H/8 = 12.5/8 = 1.56 m
οΆο Width of footing (B) = 9.5 m
B≈ 0.75×H = 0.75×12.5 = 9.37 m
B ≥ Area of footing / length
πππ‘ππ ππππ‘ππππ πΏππππ
≈ 1.5 × π΄ππππ€ππππ πππππππ πππππππ‘π¦ ππ π πππ×Length of footing
= 1.5 ×
434.22∗3
300×5.2
= 1.25 m
ο
οΆο Thickness of abutment cap = 300 mm
Minimum thickness of cap = 200mm
οΆο Length of abutment = 5.2 m
Length of abutment ≥ C/C distance between external girders + Width of bearing + 2 ×
Clearance
= 4 m + 0.4 m + 2 × 0.4 m = 5.2 m
οΆο Size of Approach slab = 3.5 m × 6 m × 0.3 m
117
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Cross section of Abutment
II.
Analysis and Design of Abutment Cap
Check thickness of abutment cap for punching shear
ππ’π£≤ ππ ππ
434.22 × 1.5 × 1000
434.22× 1.5 × 1000
Where, = (2 × 400+2 × d+2 × 250+2 × d) × d = (2 × 400+2 × 252+2 × 250+2 × 252) × 252 = 1.12 N/mm2
ππ ππ =1.25 N/mm2, ππ =1+π½π=1, ππ =0.25 √πππ = 0.25√25 = 1.25
d= 300 – 40 – 16/2 = 252 mm
Take area of steel AS = 1 % of area of cap and distribute these bars equally at top and
bottom of cap. [Refer IRC 78 CL. 716.2]
οΆο As in longitudinal direction of abutment
= 1% of 300 × 1325 = 3975 mm2,
As on one side = 3975/2 = 1987.5 mm2
Take 12 mm ∅ bar, n = 17.57, Adopt n =18
ο
οΆο As in transverse direction of abutment
= 1% of 300 × 5200 = 15600 mm2
As on one side = 15600/2 = 7800 mm2
Take 12 mm ∅ bar n = 68.96, Adopt n = 69
In transverse direction bars are provided in the forms of stirrups.
In addition, two layers of mesh reinforcement, each consisting 6 mm ∅ @ 75 mm c/c in both
directions one at 20 mm and other at 100 mm from the top of cap are provided directly
under the bearing.
118
Design of Bridge Over Kerunga Khola, Chitwan
III.
2069 – AB Bridge
Analysis and Design of Abutment Stem
Load Calculation
1. DL from superstructure [Refer bearing design]
Total DL from superstructure = 1038.08 KN
Load on an abutment per unit length (DLSS) =
1038.08
5.2×2
= 99.81 KN/m
2. Weight of approach slab [Take half of total weight of approach slab]
= 0.3 × 3.5 × 6× 25 × 1/2 = 78.75 KN
Load on an abutment per unit length (DL Ap.S) = 78.75/5.2 = 15.14 KN/m
3. LL from superstructure
Maximum LL = (LLexterior girder × 2 + LLinterior girder)/1.5 =
373.17 ×2+371.49
1.5
= 745.22 KN
Load on an abutment per unit length (LL) = 745.22/5.2 = 143.375 KN/m
4. Load from Braking Effort [Refer bearing design]
Horizontal braking load per unit length (FbrH) = 172.8 / (5.2 × 2) = 16.635 KN/m
Vertical reaction due to braking load per unit length (FbrV) = 10.94x3/5.2 = 6.314 KN/m
5. Wind Load [Refer bearing design]
Transverse Wind load per unit length, FWT = 71.90 / (5.2 × 2) = 6.913 KN/m
Longitudinal Wind load per unit length, FWL = 17.975/ (5.2 × 2) = 1.728 KN/m
Vertical Wind load per unit length, FWV = 124.44 / (5.2 × 2) = 11.965 KN/m
6. Seismic Load due to the DL and LL from superstructure
119
Design of Bridge Over Kerunga Khola, Chitwan
π
πΌ
Seismic load (FSh) = 2 × π
×
ππ
π
2069 – AB Bridge
× W [Refer Cl. 219, IRC 6]
FShL = 233.06/ 1.5 = 155.373 KN in longitudinal direction of bridge
FShT = 271.94/1.5 = 181.293 KN in transverse direction of bridge
Take, Seismic Zone - V, Soil Strata - Medium, Damping - 5 %, Bridge Class - Normal
Where, Ah = πΌh =
Z
2
I
×R×
Sa
g
= 0.15; Z = 0.36, I = 1, R = 3,
ππ
π
= 2.5
W = 1035.83 KN in longitudinal direction
W = 1208.63 KN in transverse direction
Seismic load in transverse direction per unit length, FShT = 181.293/ (5.2 × 2) = 17.432 KN/m
Seismic load in longitudinal direction per unit length, FShL = 155.373/(5.2 × 2) = 14.94 KN/m
Vertical reaction due to seismic load on support of bridge (FSVT) = 172.97/ (5.2 × 2) =16.632
KN/m
Vertical reaction on abutment per unit length when seismic load acts in longitudinal
direction, (FSVL) = 155.373 × 0.9 / (12 × 5.2) = 2.241 KN /m
7. Load due to temperature variation, creep & shrinkage effect
Δ
Load on three bearings due to CST = βo × G × A
=
3.0625
52
× 1 × 250 × 400
= 17.67 KN
Load per unit length FCST = 17.67/5.2 = 3.398 KN/m
8. Self-Weight of abutment
Self-weight = (0.752 × 0.25 + 0.3 × 1.325 + 1.25 × 9.148) × 5.2 × 25
= 1562.67 KN
Load per unit length, DLAb = 1562.67/5.2 = 300.51 KN/m
9. Seismic load due to the self-weight of abutment
Z
I
FS Abt hT = FS Abt hL = 2 × R ×
=
0.36
2
Sa
g
×W
1
× 3 × 2.5 × 1562.67
= 234.4 KN
Load per unit length, FS Abt hL = 234.4/5.2 = 45.08 KN/m
10. Load due to static earth pressure
Load due to static earth pressure is found by Coulomb’s theory.
PA= 0.5 × gsoil × H2 × KA
= 0.5 × 20 × (10.5)2 × 0.226
= 249.17 KN/m
(KA = 0.226)
120
Design of Bridge Over Kerunga Khola, Chitwan
cos2 ( φ− π)
where, KA = cos2 π×cos2(δ + π) ×
2069 – AB Bridge
1
1 2
sin (φ+δ)×sin( φ−i ) 2
(1+(
) )
cos(α−i)×cos( α+δ)
= 0.226
π=37β°, i = 0β°, πΏ = 2/3 × 37β°≈ 24.67β°, πΌ = 0β°, πΎsoil = 20 KN/m3, H= 12.5 m
Horizontal component of load per unit length PEPH(s) = PA cos (24.67°) = 226.42 KN/m
Vertical component of load per unit length PEPV(s) = PA sin (24.67°) = 104 KN/m
11. Load due to dynamic earth pressure
Load due to active earth pressure has been found by Mononobe Okabe Theory
PA= 0.5 × gsoil × H2 × KAdyn = 403.52 KN
where , KAdyn = (1 ± aV) ×
cos2 ( φ− π−ψ )
cosψ × cos2 π×cos2 (δ + π+ψ )
×
1
1 2
sin (φ+δ)×sin( φ−i−ψ) 2
(1+(
) )
cos(α−i)×cos( α+δ+ψ)
= 0.366
π=37β°, i = 0β°, πΏ = 2/3 × 37β°≈ 24.67β°, πΌ = 0β°
ah = 0.15, aV = 0.15 × 2/3 = 0.10
αh
π = tan-1( 1±αv ) = 7.77β° & 9.47β°
Horizontal component of load per unit length PEPH(D) = PA cos (24.67°) =366.68 KN/m
Vertical component of load per unit length PEPV(D) = PA sin (24.67°) = 168.42 KN/m
12. Surcharge load
1.2 m earth fill from road surface is taken as surcharge load.
Psur = KA × πΎsoil × h × W = 0.226 × 20 ×1.2 × 10.5 = 56.95 KN/m
Horizontal component of load per unit length PsurH = Psur cos (24.67°) = 51.75 KN/m
Vertical component of load per unit length PsurV = Psur sin (24.67°) = 23.77 KN/m
13. Backfill weight on heel slab of footing
WBF = (12.5 – 2 – 0.3) × 6.25 × 5.2 ×20
= 6630 KN
Load per unit length WBF = 6630/5.2 = 1275 KN/m
14. Weight of footing
WFooting = 2× 9.5× 5.2 × 25 = 2470 KN
Load per unit length WFooting = 2470/5.2 = 475 KN/m
Analysis of Abutment Stem
Responses of abutment at bottom and at 4 m from the bottom for basic combination and
seismic combination of loads have been calculated. Loads taken are vertical and longitudinal
loads. Although seismic and wind load in transverse direction are greater than seismic and
121
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
wind load in longitudinal direction, reduction in earth pressure and other loads in transverse
directions make the longitudinal direction’s load critical.
[Annex B, Table 3.2, IRC 6-2010]
Responses of abutment at its bottom in basic combination of loads
Load
(KN)
Distanc
e from
bottom
(m)
ϒf
Eccentricity
x
(m)
Pu
(KN)
Mux
(KN-m)
y (m)
DLss
87.38
1.35
-0.210
117.963
-24.77
DLwc
12.43
1.75
-0.210
21.7525
-4.57
DLAp.s
15.14
1.35
-0.5
17.8875
-5.93
143.375
1.5
-0.210 215.0625
-45.16
LL
12.5-12=
9.5
Muy
(KN
-m)
Hx
(KN)
Hy
(KN)
FbrH
16.635
1.15
FbrV
6.314
1.15
FwL
1.728
1.5
9.5
0
24.624
2.592
Fcst
3.398
1.5
9.5
0
48.42
5.097
DLAb
300.51
1.35
0
405.69
0
104
1.5
-0.625
156
-97.5
0
0
1200.12
-0.625
30.156
-16.5858
PEPV(S)
-0.210
0.42*1
0.5=4.2
8.4
PEPH(S)
226.42
1.5
PsurV
23.77
1.2
51.75
12.5/2 1.2 2=4.25
PsurH
0
181.74
7.2611
0
Total
19.13
-1.525
339.63
0 265.8906
971.77
62..1
1524.75
428.55
Responses of abutment at its bottom in Seismic Combination of loads
Load
(KN)
ϒf
Distance
from
bottom(m)
Eccentricity
x
(m)
y (m)
Pu
(KN)
Mux
(KN-m)
Muy
(KNm)
Hx
(KN)
Hy
(KN)
DLss
87.38
1.35
-0.210
117.963
-24.77
0
DLwc
12.43
1.75
-0.210
21.75
-4.568
0
DLAp.s
15.14
1.35
-0.5
20.439
-10.22
0
LL
143.375
0.2
-0.210
28.675
-6.022
0
FbrH
16.635
0.2
0
0
31.61
3.327
FbrV
6.314
0.2
-0.210
1.263
-0.265
0
FshL
14.94
1.5
0
0
212.895
22.41
9.5
9.5
122
Design of Bridge Over Kerunga Khola, Chitwan
Load
(KN)
Distance
from
bottom(m)
ϒf
vL
2.241
1.5
Fcst
3.398
1
DLAb
300.51
1.35
45.08
1.5
PEPV(D)
168.42
1
PEPH(D)
366.68
1
PsurV
23.77
0.2
PsurH
51.75
0.2
Fs
2069 – AB Bridge
Eccentricity
x
(m)
9.5
Pu
(KN)
y (m)
Muy
(KNm)
Mux
(KN-m)
Hx
(KN)
Hy
(KN)
-0.210
3.362
-0.706
0
0
0
32.281
3.398
0
405.69
0
0
0
0
287.385
67.62
-0.625
168.42
-105.263
0
0 1848.067
366.68
FS
Abt
hL
4.25
5.04
0
4.25
-0.625
4.754
-2.971
0
0
0
43.988
10.35
772.316 2301.441
473.785
Total
Responses of abutment at 4 m from its bottom in basic combination of loads
Load
(KN)
ϒf
Distance
from
bottom(m)
Eccentricity
x (m)
Pu
(KN)
y (m)
Mux
(KN-m)
DLss
87.38
1.35
-0.210
117.963
-24.77
DLwc
12.43
1.75
-0.210
21.7525
-4.57
DLAp.s
13.25
1.35
-0.5
17.8875
-5.93
143.375
1.5
-0.210 215.0625
-45.16
LL
12.5-1-24=
1.15 5.5
FbrH
16.615
FbrV
6.314
1.15
FwL
1.728
1.5
Fcst
3.398
1.5
DLAb
175.51
PEPV(S)
39.85
0
0
105.1
-0.210
7.2611
-1.525
5.5
0
0
5.5
0
0
28
1.35
0
349.245
0
1.5
-0.625
56
-37.36
0
0
355.32
-0.625
17.65
-11.03
PEPH(S)
86.77
1.5
PsurV
14.7
1.2
0.42*(10.54)=2.73
123
14.26
Muy
(KN-m)
Hx
(KN)
Hy
(KN)
19.10725
2.592
5.097
130.15
Design of Bridge Over Kerunga Khola, Chitwan
Load
(KN)
PsurH
Distance
from
bottom(m)
ϒf
32.04
1.2
2069 – AB Bridge
Eccentricity
x (m)
Pu
(KN)
y (m)
3.25
Mux
(KN-m)
0
Total
0
802.82
Muy
(KN-m)
Hx
(KN)
Hy
(KN)
125
38.45
497.335
195.4
Responses of abutment at 4 m from its bottom in Seismic Combination of loads
Load
(KN)
Distance
from
bottom(m)
ϒf
Eccentricity
x (m)
y (m)
Pu
(KN)
Muy
(KNm)
Mux
(KN-m)
Hx
(KN)
Hy
(KN)
DLss
87.38
1.35
-0.210
117.963
-24.77
0
DLwc
12.43
1.75
-0.210
21.75
-4.568
0
DLAp.s
15.14
1.35
-0.5
20.439
-10.22
0
LL
143.375
0.2
-0.210
28.675
-6.022
0
FbrH
16.635
0.2
0
0
31.61
3.327
FbrV
6.314
0.2
-0.210
1.263
-0.265
0
FshL
14.94
1.5
0
0
212.895
22.41
vL
2.241
1.5
-0.210
3.362
-0.706
0
Fcst
3.398
1
0
0
32.281
3.398
175.512
1.35
0
236.94
0
0
0
0
107.58
39.495
-0.625
64.54
-40.338
0
0
0
548.028
140.52
-0.625
6.408
-4.005
0
0
0
9.568
2.944
501.34
851.09
212.094
Fs
DLAb
FS AbthL
26.33
PEPV(D)
64.54
PEP
H(D)
140.52
5.5
5.5
5.5
(12.5-0.30.752-21.5 4)/2= 2.724
1
(12.5-21 4)*0.6= 3.9
PsurV
32.04
0.2
PsurH
14.72
0.2
(12.5-24)/2 = 3.25
Total
Design of Abutment Stem
Results of analysis shows that maximum design axial load (Pu = 971.77 KN) is less than 0.1 f ck
Ac (0.1 × 20 × 1000 × 1100 × 10-3 = 2200 KN ).
124
0
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
For the case, where Pu ≤ 0.1 fck Ac, compression member is treated as a flexure
member. So, abutment stem has been designed as a cantilever slab.
Since design bending moment is higher in seismic combination of loads, design of abutment
stem has been carried out for seismic combination of loads only.
Design of bottom section
Check depth of slab
d = D – CC – ø/2 = 1250 – 50 – 25/2 = 1187.5 mm
ππ’
2301.441∗106
ππππ = √ ππ = √
2.766∗1000
= 912.2 mm
where Q = 0.36 fck× 0.48 × (1 – 0.416 × 0.48) = 2.766
∴ dprov > dbal OK
Find reinforcing bars
Since dprov > dbal, section is designed as Singly Reinforced Under-Reinforced Section
(SRURS). So, section design has been carried out by using SP 16.
a. Main vertical bars (vertical bars in the side of backfill)
Find
ππ’
ππ2
2301.441∗106
= 1000∗(1187.5)2 = 1.63, pt = 0.503%
π΄sreq. = 0.00503 ×1000×1187.5 = 5973.125mm2
Provide 32 mm ∅ bar @ 130 mm c/c, π΄sprov. = 6434 mm2, ptprov. = 0.54 %
By increasing dia., effective depth d decreases.
Initially if we use 32mm dia., d=1250-50-32/2 = 1184mm
ππ’
ππ2
= 1.64
As per SP 16, (pt)req. = 0.508% < ptprovided(=0.54%)
b. Outer vertical reinforcement (vertical bars in the side of river)
[Refer detailing criteria of IRC 112-2011 and IS 4]
Take 0.12 % of gross sectional area of abutment as outer vertical reinforcement
As = 0.12/100 ×1000×1184 = 1421 mm2
Provide 16 mm ∅ bar @ 140 mm c/c, π΄stprov = 1436.15 mm2, ptprov.= 0.13 %
c. Horizontal Reinforcement
Take, As = 0.1% of stem area of abutment or 25% of main vertical bars
= 0.001 × 1250 × 9136 = 11420 mm2
As on each face of abutment =
11420
2
= 5710 mm2
Provide 29 - 16 mm ∅ @300mm c/c on each face of abutment
Check bottom section for shear
Check πuv≤ K πuc
Where,
125
Design of Bridge Over Kerunga Khola, Chitwan
πuv = π»y/ππ =
473.785∗103
1000∗1184
2069 – AB Bridge
= 0.4 N/mm2
πuc = 0.493 N/mm2 for M20 and pt = 0.54 %
πuc.max = 2.8 N/mm2
K =1
[Refer table 19 & 20, IS 456]
πππππ πuv< K πuc, shear reinforcement is not required.
Design of abutment section at 4 m from its bottom
Check depth of slab
d = D – CC – ø/2 = 1250 – 50 – 25/2 = 1187.5 mm
851.09∗106
Mu
dbal = √Q∗b = √2.766∗1000 = 554 mm
Where, Q = 0.36 fck × 0.48 × (1 – 0.416 × 0.48) = 2.76
∴ dprov > dbal OK
Find Reinforcing bars
Since dprov > dbal, section is designed as Singly Reinforced Under-Reinforced Section (SRURS).
Section design has been carried out by using SP 16.
a. Main vertical bars (vertical bars in the side of backfill)
ππ’
851.09∗106
Find ππ2 =1000∗(1187.5)2 = 0.6, pt = 0.172 % > pt min = 0.12%
Asreq = 0.172/100×1000×1187.5 = 2043mm2
Provide 25 mm ∅ bar @ 240 mm c/c, Asprov = 2045 mm2, ptprov = 0.172%
Curtail two-third of main vertical bars (bars designed for bottom section of abutment) at 5.2
m (4m + d) from the bottom of abutment
IV.
Design of Dirt Wall
Design the dirt wall as a cantilever slab of span 0.764 m. Consider basic combination and
seismic combination of loads to determine the responses of dirt wall.
Here seismic combination is considered for design. Dirt wall have been designed as a
cantilever slab of unit width. Detailing of wall is carried out prescribed by IRC 112 - 2011 Cl.
16.3.
Surcharge load = 1.2 × 20 ×KADYN × 0.764 × 1 = 6.71 KN/m
Load due to earth pressure = ½ × KADYN × 20 × 0.7642× 1=2.14 KN/m
Seismic load due to weight of dirt wall (Ah × W) = 0.15 × 0.764 × 0.25 × 25 × 1 = 0.72 KN/m
Mu at bottom = 6.71 × .764/2 + 2.14 × 0.6 × .764 + 0.72 × 0.764/2 = 3.82 KN-m
Use M20 grade concrete,
ππ’
ππ2
3.82∗106
=1000∗2042 = 0.09, pt = 0.085 % < pt min = 0.12 %
Take pt = 0.12% Ast = 0.12/100×1000×250 = 300 mm2
Provide 12 mm ∅ bar @ 200 mm c/c at both sides of dirt wall
Provide 10 mm ∅ bar @ 250 mm c/c as horizontal reinforcement
Check bottom of slab for shear
126
Design of Bridge Over Kerunga Khola, Chitwan
Total shear at bottom of dirt wall = 6.71 + 2.14 + 0.72 = 9.57 KN
ππ’
9.57×103
Nominal Shear Stress ππ’π£ = ππ = 1000×204 = 0.05N/mm2
πππππ ππ’π£< K πuc, no shear reinforcement requires.
127
2069 – AB Bridge
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Stability Check
Stability of abutment is checked for overturning and sliding. In the following table
overturning moment and restoring moment about the toe of footing of abutment and
shear at the base of footing have been calculated.
Stability check for basic combination of loads
ϒf
Over
Turning
Load
KN/m
Restoring
Lever Arm
m
Overturning Restoring
Moment
Moment
kNm
kNm
Shear
Force kN
Vertical
Load kN
DLss
87.38
0.95
2.835
0
235.34
0
83.01
DLwc
12.43
1
2.835
0
35.24
0
12.43
DLAp.s
13.25
0.95
3.125
0
39.34
0
12.59
LL
143.375
0
2.835
0
0
0
0
FbrH
16.615
0 19.10725
0
FbrV
6.314
FwL
1.728
Fcst
3.398
DLAb
258.7
116.22
PEP
V(S)
PEP
H(S)
253.03
12.5-1=11.5
1.15
0
219.73
2.835
0
0
0
0
1.5
11.5
29.8
0
2.592
0
1.5
11.5
58.62
0
5.097
0
0.95
2.625
0
645.13
0
245.77
0
3.25
0
0
0
0
1992.6
0
379.545
0
0
0
0
0
12.5/2=6.25
410.325
0
65.652
0
0.42*12.5
=5.25
1.5
PsurV
25.13
PsurH
54.71
WBF
1275
0.95
6.375
0
7721.72
0 1211.25
475
0.95
4.75
0
2143.44
0
451.25
2711.075
10820.7
471.99
2016.3
Wfooting
0
1.2
3.25
Total
Total Overturning moment = 2711.075 KN-m
Total Restoring moment= 10820.7 KN-m
Total Shear at base of footing= 471.99 KN
Total Vertical Load at base of footing = 2016.3 KN
Check
π»ππππ πΉππππππππ ππππππ 10820.7 KN−m
a. Total Overturning moment =
= 4 > 2 (Safe in overturning)
2711.075 KN−m
b.
π»ππππ πΉππππππππ πππππ π∗ππππ 2016.3∗tan 37
=
=
= 3.22> 1.5 (Safe in sliding)
Total Shear at base
π»
471.99
Stability check for seismic combination of loads
128
Design of Bridge Over Kerunga Khola, Chitwan
ϒf
Over
turning
Load
KN/m
restoring
Lever Arm
(m)
2069 – AB Bridge
Overturning
Moment
kNm
Restoring
Moment
kNm
Shear
Force
kN
Vertical
Load kN
DLss
87.38
0.95
2.835
0
235.34
0
83.01
DLwc
12.43
1
2.835
0
35.24
0
12.43
DLAp.s
15.14
0.95
3.125
0
44.95
0
14.38
LL
143.375
0
2.835
0
0
0
0
FbrH
16.635
38.26
0
3.327
0
FbrV
6.314
2.835
0
0
0
0
FshL
14.94
11.5
257.72
0
22.41
0
FsvL
2.241
2.835
0
0
0
0
Fcst
3.398
11.5
37.12
0
3.23
0
DLAb
300.51
2.625
0
749.4
0
285.48
12.5-1=
11.5
0.2
0
1.5
0
0.95
0.95
36.05
FS AbthL
45.08
PEPV(D)
168.42
PEPH(D)
366.68
(9.148/2+2)
= 6.574
1.5
0
1
444.53
0
67.62
0
3.25
0
0
0
0
8.3
3043.44
0
366.68
0
PsurV
23.77
3.25
0
0
0
0
PsurH
51.75
6.25
0
0
0
0
WBF
1275
0.95
6.375
0
7721.72
0
1211.25
475
0.95
4.75
0
2143.44
0
451.25
3821.07
10930.09
499.32
2057.8
Wfooting
Total
Total Overturning moment = 3821.07KN-m
Total Restoring moment= 10930.08 KN-m
Total Shear at base of footing= 499.32 KN
Total Vertical Load at base of footing = 2057.8 KN
Check
π»ππππ πΉππππππππ ππππππ 10930.08 KN−m
a. Total Overturning moment =
= 2.86 > 2 (Safe in overturning)
3821.07KN−m
b.
π»ππππ πΉππππππππ πππππ π∗ππππ 2057.8∗tan 37
=
=
= 4.12> 1.5 (Safe in sliding)
Total Shear at base
π»
499.32
Thus, Abutment is safe in stability for basic and seismic combination of loads.
129
Design of Bridge Over Kerunga Khola, Chitwan
V.
2069 – AB Bridge
Analysis and Design of Spread Footing
Responses of footing at its base in basic combination of loads
Load
(KN)
Distance
from
bottom(m)
ϒf
Eccentricity
x'
(m)
y'
(m)
Pu
(KN)
Muy'
(KN- Hx'
m) (KN)
Mux'
(KN-m)
DLss
87.38
1.35
1.92
117.963
226.49
DLwc
12.43
1.75
1.92
21.7525
41.76
DLAp.s
15.14
1.35
1.625
17.8875
33.21
143.375
1.5
1.92 215.0625
412.92
LL
FbrH
16.635
1.15
FbrV
6.314
1.15
FwL
1.728
1.5
Fcst
3.398
1.5
DLAb
300.51
104
PEPV(S)
PEP
H(S)
12.5-2=
10.5
Hy'
(KN)
0
0
200.87
1.92
7.2611
13.94
10.5
0
0
27.22
2.592
10.5
0
0
53.52
5.097
1.35
2.125
405.69
862.1
1.5
1.5
156
234
0
0
1640.42
1.5
30.156
42.786
0
0
326.025
1.625
1556.82
-2529.83
226.42
1.5
PsurV
23.77
1.2
PsurH
51.75
1.2
WBF
1153.2
1.35
Wfooting
318.75
1.35
0.42*11.5=
4.83
12.5/21=5.25
19.11
339.63
62.1
0 430.3125
Total
2958.9
1585.43
428.529
Responses of footing at its base in seismic combination of loads
Load
(KN)
Distanc
e from
bottom
(m)
ϒf
Eccentricity
x
(m)
Pu
(KN)
y (m)
Mux
(KN-m)
DLss
87.38
1
1.92
87.38
167.77
DLwc
12.43
1
1.92
12.43
23.87
DLAp.s
15.14
1
1.625
15.14
24.602
LL
143.375
0.2
1.92
28.675
55.056
FbrH
16.635
0.2
FbrV
6.314
0.2
FshL
14.94
1.5
FsvL
2.241
1.5
10.5
34.934
1.92
1.263
10.5
130
3.362
Hx
(KN)
Hy
(KN)
3.327
2.424
235.31
1.92
Muy
(KNm)
6.654
22.41
Design of Bridge Over Kerunga Khola, Chitwan
Load
(KN)
Distanc
e from
bottom
(m)
ϒf
Fcst
3.398
1
DLAb
300.51
1
FS AbthL
45.08
1.5
PEPV(D)
168.42
1
H(D)
366.68
1
PsurV
23.77
0.2
PsurH
51.75
0.2
PEP
WBF
Wfooting
2069 – AB Bridge
Eccentricity
x
(m)
Pu
(KN)
y (m)
10.5
Muy
(KNm)
Mux
(KN-m)
Hx
(KN)
35.68
2.125
300.51
5.574
168.42
7.3
4.754
5.25
1275
1
475
1
0
Total
366.68
7.131
54.337
1.625
67.62
252.63
2676.764
1.5
3.398
638.584
376.913
1.5
Hy
(KN)
1275
2071.875
475
0
2371.754
2520.784
10.35
473.785
Upward pressure of soil
Basic Combination
pu =
Pu
±
A
Mux
2958.9
Ixx
9.5
*y =
=416.8
±
1585.43∗4.75
71.45
KN/m2
or 206 KN/m2 < ultimate bearing capacity of soil
(=1.5x300 KN/m2)
1
Ixx = 12 * 9.53 = 71.45 m4
Seismic Combination
pu =
Pu
±
A
Mux
2371.754
Ixx
9.5
*y =
±
2520.784∗4.75
71.45
= 417.24 KN/m2 or 82.08 KN/m2 < ultimate bearing capacity of soil
(=1.5x300 KN/m2)
Critical section for BM
Critical section for SF
d=2m
II
II
V
206 KN/m
M
I
M
416.8 KN/m
2
465.65
2
KN/m
Soil Upward Pressure
131
2
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Analysis of Footing
372.42+416.8
BMmax at IM
=
BMmax at IIM
SMmax at IIv
=
2
206+344.7
2
206+344.7
=
2
2
× 2× 2 = 789.22 KNm
× 6.25×
6.25
2
= 5377.9 KNm
× 4.25 = 1170.24 KN
Design of Footing
Check depth of footing
Mu
5377.9∗10^6
dbal = √Q∗b = √
2.76∗1000
= 1396 mm
Since d > dbal , design as SRURS.
Adopt D = 2000mm
Find Reinforcing Bar
o
Find bottom reinforcing bars @ critical section IIM
Provide reinforcement in 2 layers, d = 2000-60-20 = 1920 mm
ππ’
5377.9 ∗10^6
=
ππ^2 1000∗(2000−60−20)^2
o
=1.46 ο pt = 0.444 % (from SP 16)
Astreqd = 0.444 % of (1000* 1920) = 8525 mm2
Provide 2 layers of 20 mm dia @ 70 mm c/c, Astprovided = 8976 mm2
Bottom Reinforcing bars @ IM
ππ’
789.22 ∗10^6
=
ππ^2 1000∗(2000−60−10)^2
= 0.211, Thus pt = 0.085%
Astreqd = 0.085 % of (1000* 1930) =1621.2 mm2
Astmin = 0.12 % of bD = 2400 mm2, adopt Ast = 2400 mm2
Provide a layer of 20 mm dia bars @ 130 mm c/c , As provided = 2416 mm2
o
Top distribution Bars
Take Astmin = 0.12 % of bD = 2400 mm2, adopt Ast = 2400 mm2
Provide a layer of 20 mm dia bars @ 130 mm c/c, Astprovided = 2416 mm2
Check for One Way Shear
Nominal Shear ππ’π£ =
Vu
=
bd
1170.24∗103
1000∗(2000−60−10)
= 0.6 MPa
ππ’c = 0.48 MPa for M20 and pt = 0.45 % and ππ’c,max = 2.8 MPa
Since ππ’v > ππ’c, shear reinforcement is needed.
Design of hat reinforcement in slab
0.87 ππ¦×π΄π π£×π
Sv = (τuv−τuc)bw×d=
0.87×415×15×π×8^2/4×1930
(1.074−0.46)×1000×1930
= 440 > 0.75d and >300 mm
Adopt Sv = 300mm.
Provide 8 mm ∅ 15-legged vertical stirrups @300 mm c/c.
Check Development Length beyond the face of abutment
[IRC 21 cl. 305.6.3]
Design development length ld = α1α2l0
Where α1 = 1 for straight ends
132
Design of Bridge Over Kerunga Khola, Chitwan
α2 =
2069 – AB Bridge
ππππ ππππ’ππππ
= 0.99
πππ ππππ£ππππ
l0 = 46 Ο = 920 mm
Thus, ld = 910.8 mm
Shorter length of footing provided beyond the face of abutment = 2000 –75 = 1925 mm
Since provided length > ld, additional anchorage for bars are not required.
133
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
D. Design of RCC Solid Pier
I.
Planning and preliminary design
A. Selection of type of pier
For the bridge, we have selected solid pier. Hammer head pier, though have been popularly
used in bridges in Nepal, is technically unsuitable for the zone of high seismicity due to its
higher center of gravity. Hence, that type of pier has been avoided.
B. Material Selection
Following materials are used in the pier:
i.
M20 grade of concrete for pier stem
ii.
M30 grade of concrete for pier cap
iii.
Fe 415 HYSD bars for all RC work
C. Geometry of Pier
o Size of Pier cap =
Length of Cap (L)
L = c/c distance of external main girders + bearing widths + 2 × 0.05
= 7.5 + 0.8 + 2 × 0.4 + 2 × 0.05
= 9.20 m
Width of Cap (B)
B = 4 × 150 + bear. widths
= 4 × 150 + 250 + 400= 1250 mm
Thickness of Cap (T)
T = 0.5 m thick
Check the thickness of cap for punching Shear (Below bearing of truss)
ππ’π£≤ππ ππ’π
Vu
1190.09 × 1.5 × 1000
Where, ππ’π£ = b0× d = 2 ×(400 + 800+450 + 450 )× 450 = 0.944 N/mm2
ππ’ = Maximum Vertical load from bearings × 1.5 [From bearing design]
d = 500 – 40 – 20/2 = 450 mm, effective depth of cap below bearing
ππ’π = 0.25√fck = 0.25 ×√30 = 1.369 N/mm2
ππ = 1
πππππ ππ’π£≤ thickness of cap is safe for punching shear.
Check the thickness of cap for punching Shear (Below bearing of T-girder)
ππ’π£≤ππ ππ’π
Vu
434.22 × 1.5 × 1000
Where, ππ’π£ = b0× d = 2 ×(400 + 250+450 + 450 )× 450 = 0.467 N/mm2
ππ’ = Maximum Vertical load from bearings × 1.5 [From bearing design]
134
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
d = 500 – 40 – 20/2 = 450 mm, effective depth of cap below bearing
ππ’π = 0.25√fck = 0.25 ×√30 = 1.369 N/mm2
ππ = 1
πππππ ππ’π£≤ thickness of cap is safe for punching shear.
o Stem
Length of stem (L) = Length of pier cap – 2 x 0.05 m = 9.10 m
Width of stem at top = 1150 mm
Width of stem at bottom = 1700mm
II.
Load Calculation
1. DL from superstructure (without WC)
From RC-Tee Bridge, Total DL = (1038.08 – 129.36)
= 908.72 KN
From Truss Bridge, Total DL = (2776.2 – 383.33)
= 2392.87 KN
DL on pier from both span (DLSS) = (908.72 + 2392.87)/2 = 1650.79 KN
2. Weight of Wearing Coat
DL due to WC = (129.36 + 383.33)/2 = 256.34 KN
3. Live load from superstructure (LL)
ο·
LL due to class AA tracked load = (700/3.6) × 1.25 × Area in the ILD
= 700/3.6 × 1.25 × 3.42
= 831.25 KN
ο·
LL due to class AA wheeled load = (200 + 200 × 0.967 + 200 + 200 × 0.9) ×
1.25
= 966.75 KN
ο·
LL due to class A Loading = 2 × (27 × (0.542 + .633) + 114 × (0.9 + 1) + 68 ×
(0.88 + 0.797 + 0.714)) × 1.25
= 1027.28 KN
Take LL = 1027.28 KN
4) Load from Braking Load
From Conjugate Beam Method
135
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Stiffness of each Pier = 119.93 m-1
Stiffness of each Abutment = 74.15 m-1
ο·
Total Class A loads possible on 60 m span = 4 x 554
= 2216 KN
ο·
Total class AA tracked load possible = 700 KN
ο·
Total class AA tracked load possible = 400 KN
Taking Class A loads on overall bridge, Braking load = 0.2 × 2216
= 443.2 KN
119.93
Horizontal braking load acting on pier ( FbrH) = 2 ×119.93+2 ×74.15 × 443.2
= 136.936 KN
From SAP2000 analysis,
Vertical braking load acting on pier ( FbrV) = 18.62 KN
5. Wind load (Superstructure) [From Bearing Design]
From RC-Tee Bridge,
FWT = 71.90 KN
FWL = 17.975 KN
FWV = 124.44 KN
From Truss Bridge,
FWT = 365.04 KN
FWL = 91.26 KN
FWV = 396.18 KN
For Pier,
FWT = ½ × (71.90 + 365.04) = 218.47 KN
136
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
FWL = ½ × (17.975 + 91.26) = 54.612 KN
FWV = ½ × (124.44 + 396.18) = 260.31 KN
6. Wind load (Sub Structure - dry season)
Pd = 940.60 N/mm2
For t/b =
h/b =
9100
1150+1700
2
11500
1150+1700
2
= 6.38 > 4 &
= 7.07
CD = 0.9
A = ½ × (1.15 + 1.7) × 11 = 15.67 m2
Wind load in transverse direction of bridge FWT (sub) = Pd × A × G × CD
= 940.60 × 15.67 × 2 × 0.9 = 26.53 KN
Wind load in longitudinal direction FWL(sub)= 0.25 × FWT (sub) = 6.63 KN
7. Seismic load (Super Structure)
π
πΌ
Seismic load = 2 × π
×
Sa
g
×W
Where,
Z = 0.36, I = 1, R = 4, ππ/π = 2.5
π
2
πΌ
Sa
π
g
× ×
= 0.1125
WLTruss = 2776.2 KN
WLRC-Tee = 1035.83 KN
WLOverall = WLTruss + 2 × WLRC-Tee
= 2776.2 + 2 × 1035.83
= 4847.86 KN
Maximum LL on Overall Bridge = 2216 KN (from calculation of Braking Load)
WTOverall = 4847.86 + 0.2 × 2216 = 5291.06 KN
Longitudinal Seismic weight concentrating on a pier,
119.93
WLPier = 2 ×119.93+2 ×74.15 × 4847.86
= 1497.69 KN
π
πΌ
Seismic load in longitudinal direction of bridge (FSL) = 2 × π
×
137
Sa
g
× WLPier
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
= 0.1125 × 1497.69
= 168.49 KN
π
πΌ
Seismic load in transverse direction of bridge (FST) = 2 × π
×
Sa
g
× WTPier
= 0.1125 × (WTTruss + WTRC-Tee)/2
= 0.1125 × (2892.1 + 1208.63)/2
= 230.66 KN
V.Reaction due to seismic load in longitudinal direction (FSvL) = FSvL Truss+ FSvL RC-Tee
= 4.52 × 2 + 8.74 × 3
= 35.26 KN
vT
V.Reaction due to seismic load in transverse direction (FS ) = FSvT Truss+ FSvT RC-Tee
= 20.03 + 86.485
=107.32 KN
8. Seismic load (Sub Structure)
Seismic loads due to self-weight of pier in longitudinal and transverse direction of bridge are
equal
π
πΌ
FST(Sub) = FSL(Sub) = 2 × π
×
ππ
π
× Wpier = 0.1125 × 3709.81 = 417.35 KN
Where,
Z = 0.36, I = 1, R = 4,
Sa
π
= 2.5
Wpier =1/2 × (1.15 + 1.7) × 11 × 9.1 × 25+.5 x 1.25 x 9.2 x 25 = 3709.81 KN
9. Load due to temperature variation, creep and shrinkage effect
FCST = FCSTTruss bearing × 2 - FCSTRC-Tee bearing × 3
= 26.23 × 2- 5.44 × 3
= 36.14 KN
10. Self-Weight of Pier
Wpier =1/2 × (1.15 + 1.7) × 11 × 9.1 × 25+.5 x 1.25 x 9.2 x 25 = 3709.81 KN
11. Load due to Water Current
FWCT = 52 × K × (V × cos20β°)2 × A
= 52 × 1.5 × (1.5 × 0.203 × √2 × cos20β°)2 × 14.5
= 1.82KN
FWCL = 52 × K × (V × sin20β°)2 × A
= 52 × 1.5 × (1.5 × 0.203 × √2 × sin20β°)2 × 91
= 1.51 KN
138
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
12. Load due to hydrodynamic pressure
FhydL = C × ah× W = 0.74 × 0.1125 × 6503.88 = 541.45 KN
Where, ah=
π
2
πΌ
×π
×
ππ
π
= 0.1125
C = 0.74 for π»/π
= 7.47, [Refer Table 9.10, Swami Saran, Design of Sub Structure]
W = π × (9.1/2)2 × 10 × 10 = 6503.88 KN
FhydT = C × ah× W = 0.74 × 0.1125 ×165.13 =13.75 KN
W = π × (1.45/2)2 × 10 × 10 = 165.13 KN
13. Load due to buoyancy
Fbuoy = Submerged vol. of pier × gw = ½ (1.2 + 1.7) × 10 × 9.1 × 10 = 1319.5 KN
III.
Analysis and Design of Pier Cap
Maximum Shear Force (S.F) at face of pier stem (Basic combination)
SF at face due to DL from superstructure, LL from superstructure, VL and Self wt. of Cap
= (1650.79 × 1.35 + 256.34 × 1.75) + 1027.28 × 1.5 + (18.62 × 1.15 + 260.31 × 1.5) + (9.2×
1.25 × 0.5 × 25 × 1.35)
= 4824.02 KN
Design
Check thickness of pier cap for punching shear under a (truss) bearing
ππ’π£≤ ππ ππ
4824.02 × 1000
4824.02× 1000
Where, ππ’π£ = (2 × 400+2 × d+2 × 800+2 × d) × d = (2 × 400+2 × 452+2 × 800+2 × 452) × 452
= 2.536 N/mm2
ππ ππ =1.369 N/mm2, ππ =1+π½π=1 , ππ =0.25 √πππ = 0.25√30 = 1.369 N/mm2
d= 500 – 40 – 16/2 = 452 mm
Hence not safe
Take thickness of pier cap =800 mm
d=752 mm
4824.02 × 1000
4824.02× 1000
ππ’π£= (2 × 400+2 × d+2 × 800+2 × d) × d = (2 × 400+2 × 752+2 × 800+2 × 752) × 752
=1.186 N/mm2
Take area of steel AS = 1 % of area of cap and distribute these bars equally at top and
bottom of cap.
οΆο As in transverse direction of pier cap
= 1% of 800 × 1250 = 10000 mm2,
139
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
As on one side = 10000/2 = 5000 mm2
Take 30 mm ∅ bar, 8 no’s
Spacing = 156 mm Adopt 140 mm.
∴ 30 dia @ 140 mm c/c.
ο
οΆο As in longitudinal direction of pier cap
= 1% of 800 × 9100 = 72800 mm2
As on one side = 72800/2 = 36400 mm2
Take 30 mm ∅ bar n = 52 no’s
Spacing = 175 mm, Adopt 150 mm
∴ 30 dia @ 150 mm c/c.
In transverse direction bars are provided in the forms of stirrups.
In addition, two layers of mesh reinforcement, each consisting 6 mm ∅ @ 75 mm c/c in both
directions one at 20 mm and other at 100 mm from the top of cap are provided directly
under the bearing.
IV.
Analysis and Design of Pier Stem
In the example, responses of pier at bottom for basic combination and seismic combination
of loads have been calculated. Loads taken are vertical and longitudinal loads in first case
and vertical and transverse loads in second case.
140
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Case I
Basic combination of loads
CASE I --Basic combination of loads
Load
(KN)
DLss
DLw
c
1650.
79
256.34
Distance
from
bottom(m)
ϒf
1.3
5
1.7
5
1.1
5
1.1
5
Eccentricity
x
y (m)
(m)
Muy
(KN-m)
Hx
(KN
)
Pu
(KN)
Mux
(KN-m)
Hy
(KN)
2228.56
7
0
0
448.595
0
0
0
1858.22
2
157.476
4
21.413
0
0
0
966.632
4
81.918
5008.24
4
0
0
FbrH
136.93
6
FbrV
18.62
FwL
54.612
1.5
Wpie
r
3709.8
1
1.3
5
FwcT
1.82
1
6.67
0
0
FwcL
1.51
1
2/3*10=6.6
7
0
10.0717
1.51
Fbuo
1319.5
0.1
5
0
0
y
11.8
11.8
197.925
7508.89
3
Total without LL
LL'
690.42
1.5
1.25/20.125.15=0.3
5
Total with LL'
LL''
1027.2
8
1.5
0
Total WITH LL''
2834.92
6
1035.63
362.470
5
8544.52
3
3197.39
6
1540.92
0
9049.81
3
2834.92
6
12.139
4
1.8
2
12.139
4
1.8
2
240.904
4
12.139
4
1.8
2
240.904
4
12.139
4
1.8
2
240.904
4
i. When loaded on only one span
ii. When loaded on both span
Total Axial Load (Pu) = 8544.523 KN
Total Mux = 3197.396 KN-m
Total Muy = 12.1394 KN-m
Total Hx = 1.82 KN
Total Hy = 240.9044 KN
Total Axial Load (Pu) = 9049.813 KN
Total Mux = 2834.926 KN-m
Total Muy = 12.1394 KN-m
Total Hx = 1.82 KN
Total Hy = 240.9044 KN
141
0
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Case I
Seismic combination of loads
CASE I -- Seismic Combination of Loads
Load
(KN)
Distance
from
bottom(m)
ϒf
Eccentricity
x
(m)
y (m)
Pu
(KN)
Mux
(KN-m)
Muy
(KN-m)
Hx
(KN)
Hy
(KN)
DLss
1650.79
1
1650.79
0
0
DLwc
256.34
1
256.34
0
0
FbrH
136.936
0.2
0
323.169
27.3872
FbrV
18.62
0.2
3.724
0
0
FsL
168.49
1
0
1988.182
168.49
35.26
1
35.26
0
0
417.35
1
5.9
0
2462.365
417.35
FhydL
541.45
1
5
0
2707.25
541.45
Wpier
3709.81
1
3709.81
0
0
FWCT
1.82
1
6.67
0
0
L
1.51
1
6.67
0
10.0717
1.51
-197.925
5457.999
0
7491.038
0
1.82 1156.187
138.084
48.3294
5596.083
205.456
5663.455
7539.367
12.1394
7491.038
12.1394
FSVL
Fs(sub)
FWC
Fbuoy
LL
'
LL''
L
11.8
11.8
-1319.5 0.15
Total without LL
690.42
0.2
Total with LL'
1027.28 0.2
Total WITH LL''
0.35
12.1394
12.1394
0
0
i. When loaded on only one span
ii. When loaded on both span
Total Axial Load (Pu) = 5596.083 KN
Total Mux = 7539.367 KN-m
Total Muy = 12.1394 KN-m
Total Hx = 1.82 KN
Total Hy = 1156.187 KN
Total Axial Load (Pu) = 5663.455 KN
Total Mux = 7491.038 KN-m
Total Muy = 12.1394 KN-m
Total Hx = 1.82 KN
Total Hy = 1156.18 KN
142
1.82
1.82 1156.187
0
1.82 1156.187
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Case II
Basic combination of loads
Load
(KN)
Distance Eccentricity
from
x
bottom(m) (m) y (m)
ϒf
Pu
(KN)
Mux
(KN-m)
Muy
(KN-m)
Hx
(KN)
Hy
(KN)
DLss
1650.79
1.35
2228.567
0
0
DLwc
256.34
1.75
448.595
0
0
FbrH
136.936
1.15
0
1858.222
157.4764
FbrV
18.62
1.15
21.413
0
0
FwT
218.47
1.5
3709.81
1.35
FwcT
1.82
1
FwcL
1.51
1
Wpier
Fbuoy
LL'
11.8
11.8
0
5008.244
0
6.67
0
0
6.67
0
8.6372
-1319.5 0.15
Total without LL
690.42
1.5
-197.925
7508.893
0.35
1035.63
0
8544.523
1540.92
9049.813
'
LL''
0 3866.919
Total with LL
1027.28
1.5
Total WITH LL''
0
12.1394
0
1866.859 3879.058
1.82
0
1.51
0
329.525 158.9864
284.7983
2151.657 3879.058
0
1866.859 3879.058
i. When loaded on only one span
ii. When loaded on both span
Total Axial Load (Pu) = 8544.523KN
Total Mux = 2151.657 KN-m
Total Muy = 3879.058 KN-m
Total Hx = 329.525 KN
Total Hy = 158.986 KN
Total Axial Load (Pu) = 9049.813 KN
Total Mux = 1866.859 KN-m
Total Muy = 3879.058 KN-m
Total Hx = 329.525 KN
Total Hy = 158.986 KN
143
327.705
329.525 158.9864
329.525 158.9864
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Case II
Seismic combination of loads
CASE II -- Seismic Combination of Loads
Load
(KN)
DLss
ϒf
FbrV
FsT
1650.7
9
256.34
136.93
6
18.62
230.66
FSVT
Fs(sub)
107.32
417.35
1
1
13.75
3709.8
1
1.82
1.51
1319.5
1
1
DLwc
FbrH
Distance
from
bottom(
m)
Eccentricit
y
x y (m)
(m
)
1
1
0.2
0.2
1
Pu
(KN)
FWCT
FWCL
Fbuoy
LL'
LL''
690.42
1027.2
8
Muy
(KN-m)
Hx
(KN)
Hy
(KN)
1650.79
0
0
11.8
256.34
0
0
323.169
0
27.3872
0
0
11.8
3.724
0
5.9
107.32
0
0
2462.36
5
68.75
0
T
FhydT
Wpier
Mux
(KN-m)
5
0
3709.81
1
6.67
1
6.67
0.1
5
Total without LL
0.2
Total with LL'
0.2
Total WITH LL''
0.35
2721.78
8
230.6
6
0
417.35
13.75
0
0
0
197.925
5530.05
9
138.084
5668.14
3
205.456
0
10.0717
0
12.1394
1.82
0
1.51
0
2864.35
6
48.3294
2912.68
5
2733.92
7
232.4
8
2733.92
7
232.4
8
459.997
2
0
459.997
2
0
5735.51
5
2864.35
6
2733.92
7
232.4
8
i. When loaded on only one span
ii. When loaded on both span
Total Axial Load (Pu) = 5668.143 KN
Total Mux = 2912.685 KN-m
Total Muy = 2733.927 KN-m
Total Hx = 232.48 KN
Total Hy = 459. 997 KN
Total Axial Load (Pu) = 5735.515 KN
Total Mux = 2864.356 KN-m
Total Muy = 2733.927 KN-m
Total Hx = 232.48 KN
Total Hy = 459. 997 KN
144
459.997
2
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Design and Detailing of Pier Stem at bottom
Check slenderness ratio of column
Effective Length
Slenderness Ratio of Column = Radius of gyration =
1.2 × 11.8 ×1000
331
= 42.78 > 30
Design Pier stem
Design pier stem as a biaxially loaded long column.
a. Longitudinal Reinforcement
As critical case is difficult to know, we simply calculate longitudinal reinforcement for all the
cases and accept the percent of steel that satisfies every cases.
For Case I, Basic condition (i),
Pu = 8544.523 KN
Mux = 3197.396 KN-m
Muy = 12.1394 KN-m
Take full width of Pier, D=1.15 m (along y- direction) and unit length B=1 m (along xdirection)
Hence,
Pu = 8544.523/9.1 = 938.96 KN
Mux = 3197.396/9.1 = 351.36 KN-m
Muy = 12.139/9.1 KN-m =1.334 KN-m
Calculating accidental eccentricities,
πΏ
π·
ey,min = 500 +30
> 5% of D= 57.5 mm
= 14.16/500 + 1.15/30
= 0.0667m
= 66.7 mm
πΏ
π΅
ex,min = 500 +30
> 5% of B= 50 mm
= 14.16/500 +1/30
= 61.65 mm
Moment due to accidental eccentricities,
Mux,e = Pu × ey,min
= 938.96 × 61.65/1000
= 57.88 KN-m
Muy,e = Pu × ex,min
=938.96 × 66.67/1000
=62.6 KN-m
145
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
As the column is long, additional Bending Moment acts, given as
M’ax = k × Max
M’ay = k × May
Max =
=
ππ’ × D
2000
938.96 ×1.15
2000
πππ₯
× ( π· )2
14.16
× ( 1.15 )2
=81.85 KN-m
May =
=
ππ’ × B
πππ¦
× ( π΅ )2
2000
938.96 ×1
14.16 2
×(
)
2000
1
= 94.13 KN-m
ππ’π§−ππ’
k= ππ’π§−ππ ≤ 1
Puz = 0.446fckBD + (0.75fy – 0.446 fck)pBD
(Take p= 1.8%)
= 0.446 × 20 × 1000 × 1150 + (0.75 × 415 – 0.446 × 20) × 1.8/100 × 1000 × 1150
=16516.23 KN
Pb = k1 x fck × B × D + k2 × p × B × D
d’ = Clear cover +diameter of transverse reinforcement + diameter of longitudinal
reinforcement/2
= 40 + 10 + 32/2
= 66 mm
d’/D = 66/1150 = 0.06 , Take 0.05
From Table 60, SP16, for d’/D = 0.05 & Rectangular section
k1 = 0.219
For d’/D = 0.05, fy = 415 N/mm2 and rectangular section with reinforcement distributed
equally on to opposite sides,
k2 = 0.096
Pb = k1 x fck × B × D + k2 × p × B × D
= 0.219 × 20 × 1000 × 1150 + 0.096 × 1.8 × 1000 × 1150
= 5235.72 KN
146
Design of Bridge Over Kerunga Khola, Chitwan
k=
ππ’π§−ππ’
ππ’π§−ππ
2069 – AB Bridge
≤1
16516.23−938.96
= 16516.23−5235.72
=1.381
∴ Take k = 1
M’ax = k × Max = 1 × 81.85 = 81.85 KN-m
M’ay = k × May = 1 × 94.13 = 94.13 KN-m
Now, Mux’= Mux + M’ax
= 351.36+81.85
= 433.21 KN-m
Muy’ = Muy,e + M’ay
= 1.334 + 94.13
= 95.46 KN-m
ππ’
938.96 ×1000
Now, πππ ×B ×D = 20 ×1000 ×1150 = 0.041
p/fck = 1.8/20 = 0.09
d’/D = 0.06, Take 0.05 & d’/B = 0.066, Take 0.05
fy = 415 N/mm2
From SP-16 chart 31,
ππ’π₯,π
πππ ×B ×D2
= 0.170
Mux,l = 4496.5 KN-m
ππ’π¦,π
πππ ×B2 ×D
= 0.17
Muy,l = 3910.00 KN-m
an = 0.667 + 1.667 × Pu/Puz
≥1 & ≤ 2
= 0.762
Take an=1
Now, for p to satisfy,
ππ’π₯′
ππ’π¦′
(ππ’π₯,π)an + (ππ’π¦,π)an ≤ 1
433.21
95.46
0r, (4496.5)1 + ( 3910 )1 ≤ 1
∴ 0.121 ≤ 1
OK
147
Design of Bridge Over Kerunga Khola, Chitwan
Case
I
I
I
I
II
II
II
II
Condition
Basic I
Bacic II
Seismic I
Seismic II
Basic I
Bacic II
Seismic I
Seismic II
2069 – AB Bridge
Pu(KN)
Mux(KNm) Muy(KNm Mux,e
Muy,e
Max
May
p(%)
938.9586
351.3622
1.334 57.8868 62.62854 81.87719 94.17754
1.8
938.9586
236.4458 426.2701 57.8868 62.62854 81.87719 94.17754
1.8
614.9542
828.5019
1.334 37.91192 41.01744
53.624 61.6799
1.8
622.8729
320.0753 300.4315 38.40011 41.54562 54.31451 62.47415
1.8
994.4849
311.5303
1.334
61.31 66.33215 86.71909 99.74684
1.8
994.4849
205.1493 426.2701
61.31 66.33215 86.71909 99.74684
1.8
622.6874
823.1932
1.334 38.38868 41.53325 54.29834 62.45554
1.8
630.2764
314.7644 300.4315 38.85654 42.03943 54.9601 63.21672
1.8
Case
Condition
Max'
May'
Mux'
Muy'
I
I
I
I
II
II
II
II
Basic I
Basic II
Seismic I
Seismic II
Basic I
Basic II
Seismic I
Seismic II
81.87719
81.87719
53.624
54.31451
86.71909
86.71909
54.29834
54.9601
94.17754
94.17754
61.6799
62.47415
99.74684
99.74684
62.45554
63.21672
433.2394
318.323
882.1259
374.3898
398.2494
291.8684
877.4915
369.7245
95.51154
520.4477
63.0139
362.9057
101.0808
526.0169
63.78954
363.6483
Pu/fckBD Mux,l/fckBD2 Muy,l/fckB2D
0.0408
0.0408
0.0267
0.0271
0.0432
0.0432
0.0271
0.0274
0.17
0.17
0.16
0.16
0.17
0.17
0.16
0.16
148
0.17
0.17
0.16
0.16
0.17
0.17
0.16
0.16
∝n
1
1
1
1
1
1
1
1
Mux,l
4496.5
4496.5
4232
4232
4496.5
4496.5
4232
4232
Muy,l
3910
3910
3680
3680
3910
3910
3680
3680
(Mux'/Mux,l)∝n + (Muy'/
Muy,l)∝n
0.1208
0.2039
0.2256
0.1871
0.1144
0.1994
0.2247
0.1862
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
For 1m Strip of pier, Ast = 1.8 % × 1150 × 1000 = 20700 mm2
Taking 32 mm ∅ on two faces, n =
20700
32 2
2
= 12.86
2× π ×( )
1000
Spacing = 12.86 = 77.71 mm
∴ 32 mm ∅ @ 75 mm c/c
b. Transverse Reinforcement
The transverse reinforcement of pier stem has not been governed by shear force. So,
transverse reinforcement is provided by detailing rules. [From IRC 112, Cl. 16.2]
Take lateral tie of ∅ = 10 mm (∅≥∅ππππ/4)
Spacing of tie (Sv) ≤ Width of pier = 1150 mm
≤ 12 × 32 = 384 mm
≤ 200 mm
Provide spacing of tie (Sv) = 200 mm.
For potential plastic hinge region i.e. 1200 mm from the bottom of pier,
Spacing of tie (Sv) ≤ 5 × 40 = 200 mm
≤ (1200 – 2 × 100)/5 = 200 mm
Adopt spacing of tie (Sv) = 200 mm for potential plastic hinge region.
149
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
DRAWINGS
General Arrangements
Drawing No. 1
General Elevation
General Plan
Sheet No. 1
Sheet No. 2
T-Girder (12m span)
Drawing No. 2
Plan
Transverse Section
Longitudinal Section
Reinforcement Detailing in Transverse Section
Reinforcement Detailing along Main Girder
Reinforcement Detailing in Cross Girders (X-Section)
Reinforcement Detailing along Cross Girders
Top Reinforcement in Deck Slab
Bottom Reinforcement in Deck Slab
Transverse Section Overall Detailing
Sheet No. 1
Sheet No. 2
Sheet No. 3
Sheet No. 4
Sheet No. 5
Sheet No. 6
Sheet No. 7
Sheet No. 8
Sheet No. 9
Sheet No. 10
Steel Truss (36m span)
Drawing No. 3
Elevation and X-Section
Deck Slab Reinforcement Detailing
Details at Truss joints I
Details at Truss joints II
Details at Truss joints III
Top and Bottom Bracing
Portal and Sway Bracing
Steel Sections Used
Sheet No. 1
Sheet No. 2
Sheet No. 3
Sheet No. 4
Sheet No. 5
Sheet No. 6
Sheet No. 7
Sheet No. 8
Elastomeric Bearings for T-Girder and Truss
Drawing No. 4
Abutment
Drawing No. 5
Reinforcement Detailing in X-Sections
Reinforcement Detailing in L-Sections
Reinforcement Detailing in Spread Footing and Dirt Wall
Sheet No. 1
Sheet No. 2
Sheet No. 3
RCC Solid Pier
Drawing No. 6
Plan and Elevations
Reinforcement Detailing in Pier Cap
Reinforcement Detailing along X-Sections
Sheet No. 1
Sheet No. 2
Sheet No. 3
150
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
APPROXIMATE ESTIMATES
APPROXIMATE ESTIMATE OF QUANTITY IN T-BEAM BRIDGE
Measurement
Item
no
Description of work
1
M25 grade concrete for superstructure
1.1
Deck slab (restrained)
1.2
Deck slab (cantilever)
1.3
T-beam
main
(rectangular)
1.4
No
Quantity
Length
Breadth Height
2
12.25
4.3
0.2
4
12.25
1.45
0.17
6
12.25
0.3
0.8
17.64
Cross girder (interior)
12
1.7
0.25
0.55
2.805
1.5
Cross girder (exterior)
8
1.7
0.25
0.4
1.36
1.6
Railing post
18
0.225
0.225
1.1
1.002375
1.7
Kerb
4
12.25
0.6
0.3
8.82
1.8
Wearing coat
2
12.25
6
0.0875
12.8625
1.9
Fillet (Interior) of main girder
8
12.25
0.3
0.15
2.205
1.10
Fillet (Interior) of cross girder 32
1.7
0.3
0.15
1.224
1.11
Fillet (exterior) T-beam
4
12.25
0.85
0.15
3.12375
1.12
Approach slab
2
3.5
6
0.3
12.6
1.13
Railing post in Approach slab
8
0.225
0.225
1.1
0.4455
girders
21.07
12.0785
Total= 97.237
2
Formwork
2.1
T-Girder bottom
6
12.25
2.2
T-Girder side
12
12.25
2.3
Interior fillet of T-girder
8
12.25
0.335
32.83
2.4
End Cross girder bottom
8
1.7
0.25
3.4
2.5
End Cross girder side
8
1.7
2.6
interior Cross girder bottom
12
1.7
2.7
Interior Cross girder side
12
1.7
2.8
Exterior fillet of T-girder
4
12.25
0.863
42.287
2.9
interior fillet of cross beam
32
1.7
0.34
18.496
2.10
Deck slab (restrained)
Bottom
2
12.25
2.2
53.9
151
0.3
22.05
0.8
0.4
0.25
117.6
5.44
5.1
0.55
Unit
11.22
m3
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
APPROXIMATE ESTIMATE OF QUANTITY IN T-BEAM BRIDGE
Item
no
Description of work
Measurement
No
Length
Quantity
Breadth Height
2.11
Deck slab (cantilever)
Bottom
4
12.25
0.6
29.4
2.12
Deck slab (cantilever) sides
8
12.25
0.17
16.66
2.13
Railing post
26
2.14
approach slab bottom
2
3.5
2.15
approach slab side
4
3.5
0.225
1.1
6.435
6
42
0.3
4.2
Total=
3
Steel work in superstructure
4
48.3 mm GI pipe in railing
5
Concrete work in abutment
5.1
1% of concrete work
411.018
MT
171
rm.
14.25
Abutment foundation
2
5.2
8.5
2
176.8
5.2
Abutment portion (stem)
2
5.2
1.1
8.25
94.38
5.3
Abutment cap
2
5.2
1.175
0.3
3.666
5.4
Abutment Dirt wall
2
5.2
0.25
0.752
1.9552
Total= 276.801
Steel work in abutment
1% of concreting (Mt)
7
Pcc in pier
7.1
Well foundation
7.2
Pcc in stem
2
7.3
Pcc in Pier cap
2
21.729
Steel works in pier
m3
MT
Not Calculated
9.15
9.25
Area of Trap. Sec.=
242.475
0.5x(1.1+1.065)x9.8=13.25
1.2
1.3
Total
8
sq.
m
7.633
12
6
Unit
2% of concreting quantity
Approximate estimate for T-Beam Bridge:
Total Concrete work in Superstructure=97.237m3
Rebar for Superstructure = 7.633MT
GI pipe=171 m
Cost of concrete work = 97.237x12,000 = Rs. 11,66,844 /Cost for GI pipe = 171x760 = Rs 1,29,960 /Cost of rebar in slab = 7.633x95000 = Rs.7,25,135 /152
28.86
271.335
m3
42.599
MT
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Total cost of bridge Superstructure = Rs. 20,21,939 /Cost per meter length (Superstructure) = Rs. 84,248 /-
APPROXIMATE ESTIMATE OF QUANTITY IN TRUSS
No
.
Item no. Description of work
L
B
H
Qty.
uni
t
1 M20 grade concrete for superstructure
1.1 For slab
1
36
6
0.
2
43.2
1.2 For kerb
2
36
0.6
0.
3
12.96
18
0.225
0.225
1.
1
1.00238
1.3 For railing post
Total =
57.1624
m3
216
rm
4.487
MT
4.1 Main truss
31.42
MT
4.2 Bottom bracing
4.606
MT
4.3 Top bracing
5.88
MT
4.4
15.7
MT
18.46
MT
2 48.3 mm GI pipe in railing
6
36
1% of concrete
work
3 Steel rod for RCC and slab
4 Steel for truss
Stringer
4.5 Cross girder
Total=
Approximate estimate for Truss bridge:
Total Concrete work = 57.162m3
Rebar for slab = 4.487MT
Cost for GI pipe = 216x760=Rs 1,64,160 /Steel work in steel truss = 76.066 MT
Cost of concrete work = 57.162x12,000 = Rs. 6,85,944 /Cost of steel work in truss = 76.066x1,30,000=Rs.98,88,580 /Cost of rebar work in slab = 4.487x95,000=Rs. 4,26,265 /Total cost of bridge superstructure = Rs.1,10,00,789 /Cost per meter length = Rs. 3,05,578/-
153
76.066
MT
Design of Bridge Over Kerunga Khola, Chitwan
Cost of Substructure:
Total concrete work in abutment & pier = 548.136m3
Total rebar work in abutment & pier = 64.328MT
Cost of concrete work = 548.136x12,000 = Rs.65,77,632/Cost of rebar work in slab = 64.328x95,000 = Rs. 61,11,160 /Total cost of bridge Substructure = Rs.1,26,88,792 /-
154
2069 – AB Bridge
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
CONCLUSION AND RECOMMENDATIONS
With our relentless effort for about a year, we finally have concluded our
final year project work entitled ‘Design of Bridge over Kerunga Khola’. This
would not have been possible without the valuable guidance from our
supervisor to understand some difficult concepts of Bridge Analysis and Design.
It was our teamwork that helped us overcome the lengthy design procedures
and surpass our own expectations towards the final year project. The field visit
helped us collect a real time experience to be a Bridge Engineer while the
incorporation of two different superstructure types helped us explore the
possibilities in bridge design in terms of safety, economy and aesthetics. The
report encompasses all our work and has been prepared with the best of our
knowledge and skills.
The project work is apparently complete but we have not been able to
include analysis results from SAP2000 in this report. As the use of computers
helps make structural analysis easier and quicker, use of computer packages
along with manual calculations cannot be averted and hence, their use highly
recommended. But it should be forethought that the results from such software
are only as reliable as the knowledge and acquaintance that the user has of the
software thus needing a very meticulous application.
This hereby concludes our final year project work.
155
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
BIBLIOGRAPHY
1. Victor, D.J. 2012. Essentials of Bridge Engineering, Oxford and IBH Publishing
Company Pvt. Ltd., New Delhi
2. Design Examples Provided by Asso. Prof. N.C. Sharma, IOE, Pulchowk
3. Krishnaraju, N. Design of Bridges, Oxford and IBH Publishing Company Pvt. Ltd., New
Delhi
4. Chandra, R. 1981. Design of Steel Structures Vol. II, Standard Book House, New Delhi
5. Jain, A.K. 2002. Reinforced Concrete Limit State Design, Nem Chand and Bros,
Roorkee, India (Reprint 2009)
Codes/Standards
1. “IRC: 6-2014”
2. “IRC 21-2000 Concrete Bridge”
3. “Nepal bridge standard-2067”, Government of Nepal, Ministry of Physical Planning
and works, Department of Roads
4. “Standard specification of roads and bridges”, Government of Nepal, Ministry of
Physical Planning and works, Department of Roads
5. “IRC 83-1987(part II)”
6. “IRC 78-2000 Foundations and substructures”
7. “IRC 5-1998 Loads and Stresses”
156
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
ANNEXES
Detail Design of Connections at Truss Joints
We have, for M20 HSFG bolts,
Slip Resistance = 78478.4 kN
Joint L3
Member L2-L3
Axial force =1712.85 KN
No. of bolts= 1712.85x10^3/(2x78478.4)
= 11
Provide 12 bolts.
Try double channel section of DC300.
Check:
1. Design strength due to yielding of gross section:
Tdg=Agxfy/ ϒm0
Ag=gross area of the cross section
= 2x45.64+2x30x0.8
=139.28 cm^2
fy= yield strength of the material
=250 Mpa
ϒm0=partial safety factor for the yielding =1.1
Tdg=139.28x100x250/(1.1x1000)
=3165.45 KN >1712.85 KN ok
2. Rupture strength of the channel section
Where
Tdn = 0.9xAncxfu/ ϒm1+βAgofy/ ϒm0
β=1.4-0.076x(w/t)x(fy/fu)x(bs/Lc)
but 0.7≤ β≤(fu x ϒm0)/(fy x ϒm1)
in which
w = width of outstanding leg =90mm
bs = shear lag width = w+wt-t
= 90+75-13.6 = 151.4mm
Lc = length of end connection, i.e., distance between the outermost bolts or length of the weld along the
direction of load =150mm
Anc = net sectional area of the connected leg
Ago=gross area of the outstanding leg
t=thickness of the angle
β=1.4 - 0.076x(90/13.6)x(250/410)x(151.4/150)
=1.09 > 0.7
< 1.44
157
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Tdn = 0.9x{7.8x(300-3x22-13.6)} x410/1.25+1.09x {2x(90-7.8/2)x13.6}250/1.1
= 1087.6KN > 856.42KN (1712.85/2=856.42) OK
Member L3-L4
Axial force =2139.97 KN (T)
No. of bolts= 2139.97x10^3/(2x78478.4)
= 14
Provide 15 bolts
Try double channel section of DC300.
Check:
1. Design strength due to yielding of gross section:
Tdg=Agxfy/ ϒm0
Ag=gross area of the cross section
= 2x45.64+2x30x0.8
=139.28 cm^2
fy= yield strength of the material
=250 Mpa
ϒm0=partial safety factor for the yielding =1.1
Tdg=139.28x100x250/(1.1x1000)
=3165.45 KN >2139.97 KN ok
2. Rupture strength of the channel section
Tdn=0.9xAncxfu/ ϒm1+βxAgoxfy/ ϒm0
Where
β=1.4-0.076x(w/t)x(fy/fu)x(bs/Lc)
but 0.7≤ β≤(fux ϒm0)/(fyx ϒm1)
in which
w= width of outstanding leg =90mm
bs=shear lag width =w+wt-t
=90+75-13.6 = 151.4mm
Lc=length of end connection, i.e., distance between the outermost bolts or length of the weld along the
direction of load =150mm
Anc=net sectional area of the connected leg
Ago=gross area of the outstanding leg
t=thickness of the angle
β=1.4-0.076x(90/13.6)x(250/410)x(151.4/200)
=1.168 >0.7
<1.44
Tdn=0.9x{7.8x(300-3x22-13.6)} x410/1.25+1.168x{2x(90-7.8/2)x13.6}250/1.1
= 1129.158KN >1069.985KN (2139.97/2=1069.985) OK
Member U2-L3
158
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Axial force = 891.194 KN(T)
No. of bolts= 891.194 x10^3/(2x78478.4)
=6
Provide 6 bolts
Try with angle section 4-ISA 55x55x8.
Check
1. Rupture strength of a single angle connected through one leg and affected by
shear lag is given by
Tdn=0.9xAncxfu/ ϒm1+βxAgoxfy/ ϒm0
Where
β=1.4-0.076x(w/t)x(fy/fu)x(bs/Lc)
but 0.7≤ β≤(fux ϒm0)/(fyx ϒm1)
in which
w= width of outstanding leg =55
bs=shear lag width =w+wt-t
=55+30-8 = 77mm
Lc=length of end connection, i.e., distance between the outermost bolts or length of the weld along the
direction of load =100mm
Anc=net sectional area of the connected leg
Ago=gross area of the outstanding leg
t=thickness of the angle
β=1.4-0.076x(55/8)x(250/410)x(77/100)
=1.155 >0.7
<1.44
Tdn=0.9x{8x(55-22-8/2)} x410/1.25+1.155x{8x(55-4)}250/1.1
= 175.58KN <222.8KN (891.194 /4=222.8) Not safe
Try with 4-ISA 60x60x10
1. Rupture strength
β=1.4-0.076x(60/10)x(250/410)x(60+35-10)/100
=1.1637 >0.7
<1.44
Tdn=0.9x{10x(60-22-10/2)} x410/1.25+1.1637x{10x(60-5)}250/1.1
= 242.88KN >222.8KN (891.194 /4=222.8) safe
2. Design strength due to yielding of gross section:
Tdg=Agxfy/ ϒm0
Ag=gross area of the cross section
= 4x1100+300x8
=6800 mm^2
fy= 250 Mpa
ϒm0=partial safety factor for the yielding =1.1
Tdg=6800x250/(1.1x1000)
=1545.45 KN >891.194 KN ok
3. Block failure
The block shear strength of a bolted connection is the least of
159
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Tdb=Avgxfy/(ϒm0x√3)+0.9xAtnxfu/ϒm1
Or
=0.9xAvnxfu/(ϒm1x√3)+ Atgxfy/ϒm0
Where Avg and Avn are the minimum grosss and net areas in shear along the bolt line parallel to line of
action of force respectively.
Atg, Atn are the minimum grosss and net areas in tension from the bolt hole to the edge of a plate,
perpendicular to the line of action of force
Avg=50x3x10 =1500 mm^2
Avn=(50x3-2.5x22)x10 =950 mm^2
Atg=25x10 =250 mm^2
Atn=(25-22/2)x10 =140 mm^2
Tdb={1500x250/(1.1x√3)+0.9x140x410/1.25}x4/1000 =952.6KN
Or ={0.9x950x410/(1.25x√3)+250x250/1.1}x4/1000 = 875 KN
Least Tdb=875 KN<891.194KN Not safe
Try with 4-ISA 65x65x10
1. Rupture strength
β=1.4-0.076x(65/10)x(250/410)x(65+35-10)/100
=1.129 >0.7
<1.44
Tdn=0.9x{10x(65-22-10/2)} x410/1.25+1.129x{10x(65-5)}250/1.1
= 266.13KN > 222.8KN (891.194 /4=222.8) safe
2. Design strength due to yielding of gross section:
Tdg=Agxfy/ ϒm0
Ag=gross area of the cross section
= 4x1200+300x8
=7200 mm^2
Tdg=7200x250/(1.1x1000)
=1636.4 KN >891.194 KN ok
3. Block failure
The block shear strength of a bolted connection is the least of
Tdb=Avgxfy/(ϒm0x√3)+0.9xAtnxfu/ϒm1
Or
=0.9xAvnxfu/(ϒm1x√3)+ Atgxfy/ϒm0
Avg=50x3x10 =1500 mm^2
Avn=(50x3-2.5x22)x10 =950 mm^2
Atg=30x10 =300 mm^2
Atn=(30-22/2)x10 =190 mm^2
Tdb={1500x250/(1.1x√3)+0.9x190x410/1.25}x4/1000 =1011.6KN
Or ={0.9x950x410/(1.25x√3)+300x250/1.1}x4/1000 = 920.4 KN
Least Tdb=920.4 KN>891.194KN Safe
Member L3-U3
Axial force = 402.945 KN(C)
160
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
No. of bolts=402.945 x10^3/(2x78478.4)
=4
Provide 4 bolts
Try with angle section 4-ISA80x80x10
with A = 9600 mm2 and ry = 30.1 mm
Check for Compression capacity
For
πΎπΏ
π
=
0.8∗600
3.01
= 159.47 and buckling class ‘c’,
fcd = 53.3 MPa
So, Pd = fcdxAg = 510.72 kN >402.945 kN
1. Rupture strength of angle section
OK
β=1.4-0.076x(80/10)x(250/410)x(80+45-10)/50
=0.54 >0.7
<1.44
Tdn= 0.9x{10x(80-22-10/2)} x410/1.25+0.7x{10x(80-5)}250/1.1
= 168.39KN >100.73KN (402.945 /4=100.73) safe
2. Block failure
The block shear strength of a bolted connection is the least of
Tdb=Avgxfy/(ϒm0x√3)+0.9xAtnxfu/ϒm1
Or
=0.9xAvnxfu/(ϒm1x√3)+ Atgxfy/ϒm0
Avg=50x2x10 =1000 mm^2
Avn=(50x2-1.5x22)x10 =670 mm^2
Atg=40x10 = 400 mm^2
Atn=(40-22/2)x10 = 290 mm^2
Tdb={1000x250/(1.1x√3)+0.9x290x410/1.25}x4/1000 = 867.3KN
Or ={0.9x670x410/(1.25x√3)+400x250/1.1}x4/1000 = 820.4 KN
Least Tdb = 820.4 KN > 402.945 KN safe
JOINT U2
Member U2-L2
Axial force = 711.925 KN
No. of bolts= 711.925 x10^3/(2x78478.4)
=5
Provide 6 bolts
Try with angle section 4-ISA100x100x10
with A = 112 cm2 and ry = 3.8 cm
Check for Compression capacity
For
πΎπΏ
π
=
0.8∗600
3.8
= 126.32 and buckling class ‘c’,
fcd = 77.76 MPa
So, Pd = fcdxAg = 870.9 kN > 711.925 kN
OK
1. Rupture strength of single angle section
161
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
β=1.4-0.076x(100/10)x(250/410)x(100+60-10)/100
= 0.71 >0.7
<1.44
Tdn=0.9x{10x(100-22-10/2)} x410/1.25+0.71x{10x(100-5)}250/1.1
= 368.8 KN >178KN (711.925/4 = 178) safe
2. Block failure
The block shear strength of a bolted connection is the least of
Tdb=Avgxfy/(ϒm0x√3)+0.9xAtnxfu/ϒm1
Or
=0.9xAvnxfu/(ϒm1x√3)+ Atgxfy/ϒm0
Avg=50x3x10 =1500 mm^2
Avn=(50x3-2.5x22)x10 =950 mm^2
Atg=40x10 =400 mm^2
Atn=(40-22/2)x10 =290 mm^2
Tdb={1500x250/(1.1x√3)+0.9x290x410/1.25}x4/1000 = 1129.7KN
Or ={0.9x950x410/(1.25x√3)+400x250/1.1}x4/1000 = 1011.28 KN
Least Tdb= 1011.28 KN > 711.925 KN safe
Member U1-U2
Axial force =1713.01 KN (C)
No. of bolts= 1713.01x10^3/(2x78478.4)
= 11
Provide 12 bolts
Try double channel section of DC250
Check:
1. Design strength due to yielding of gross section:
A = 112 cm2 and ry = 11.18 cm
Check for Compression capacity
For
πΎπΏ
π
=
0.8∗600
11.18
= 53.67 and buckling class ‘c’,
fcd = 177.49 MPa
So, Pd = fcdxAg = 2236.37 kN > 1713.01 kN
OK
Tdg=Agxfy/ ϒm0
Ag=gross area of the cross section
= 2x39+2x30x0.8
=126cm2
fy= yield strength of the material
=250 Mpa
ϒm0=partial safety factor for the yielding =1.1
Tdg=126x100x250/(1.1x1000)
=2863.64KN >1713.01 KN ok
2. Rupture strength of the channel section
Tdn=0.9xAncxfu/ ϒm1+βxAgoxfy/ ϒm0
162
Design of Bridge Over Kerunga Khola, Chitwan
Where
2069 – AB Bridge
β=1.4-0.076x(w/t)x(fy/fu)x(bs/Lc)
but 0.7≤ β≤(fux ϒm0)/(fyx ϒm1)
in which
w= width of outstanding leg =80mm
bs=shear lag width =w+wt-t
=80+50-14.1 = 115.9mm
Lc=length of end connection, i.e., distance between the outermost bolts or length of the weld along the
direction of load =150mm
Anc=net sectional area of the connected leg
Ago=gross area of the outstanding leg
t=thickness of the angle
β=1.4-0.076x(80/14.1)x(250/410)x(115.9/150)
=1.197 >0.7
<1.44
Tdn=0.9x{7.2x(250-3x22-14.1)} x410/1.25+1.197x{2x(80-7.2/2)x14.1}250/1.1
= 947.23KN >856.505KN (1713.01/2=856.505) safe
Member U2-U3
Axial force =2139.97 KN (C)
No. of bolts= 2139.97x10^3/(2x78478.4)
= 14
Provide 15 bolts
Try double channel section of DC300
Check:
3. Design strength due to yielding of gross section:
A = 140.6 cm2 and ry = 11.64 cm
Check for Compression capacity
For
πΎπΏ
π
=
0.8∗600
11.64
= 30.93 and buckling class ‘c’,
fcd = 210 MPa
So, Pd = fcdxAg = 2952.6 kN > 2139.97 kN
OK
4. Rupture strength of the channel section
Tdn=0.9xAncxfu/ ϒm1+βxAgoxfy/ ϒm0
Where
β=1.4-0.076x(w/t)x(fy/fu)x(bs/Lc)
but 0.7≤ β≤(fux ϒm0)/(fyx ϒm1)
in which
w= width of outstanding leg =90mm
bs=shear lag width =w+wt-t
=90+75-13.6 = 151.4mm
Lc=length of end connection, i.e., distance between the outermost bolts or length of the weld along the
direction of load =150mm
Anc=net sectional area of the connected leg
Ago=gross area of the outstanding leg
163
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
t=thickness of the angle
β=1.4-0.076x(90/13.6)x(250/410)x(151.4/200)
=1.168 >0.7
<1.44
Tdn=0.9x{7.8x(300-3x22-13.6)} x410/1.25+1.168x{2x(90-7.8/2)x13.6}250/1.1
= 1129.158KN >1069.985KN (2139.97/2=1069.985) OK
JOINT U1
Member L0-U1
Axial force =1665.33 KN (C)
No. of bolts= 1665.33x10^3/(2x78478.4)
= 11
Provide 12 bolts
Try double channel section of DC250
Check:
1. Design strength due to yielding of gross section:
A = 112 cm2 and ry = 11.18 cm
Check for Compression capacity
For
πΎπΏ
π
=
0.8∗600
11.18
= 53.67 and buckling class ‘c’,
fcd = 177.49 MPa
So, Pd = fcdxAg = 2236.37 kN > 1713.01 kN
OK
2. Rupture strength of the channel section
Tdn=0.9xAncxfu/ ϒm1+βxAgoxfy/ ϒm0
Where
β=1.4-0.076x(w/t)x(fy/fu)x(bs/Lc)
but 0.7≤ β≤(fux ϒm0)/(fyx ϒm1)
in which
w= width of outstanding leg =80mm
bs=shear lag width =w+wt-t
=80+50-14.1 = 115.9mm
Lc=length of end connection, i.e., distance between the outermost bolts or length of the weld along the
direction of load =150mm
Anc=net sectional area of the connected leg
Ago=gross area of the outstanding leg
t=thickness of the angle
β=1.4-0.076x(80/14.1)x(250/410)x(115.9/150)
=1.197 >0.7
<1.44
Tdn=0.9x{7.2x(250-3x22-14.1)} x410/1.25+1.197x{2x(80-7.2/2)x14.1}250/1.1
= 947.23KN >832.665KN (1665.33/2=832.665) safe
164
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Member U1-L1
Axial force = 736.93 KN (C)
No. of bolts= 736.93 x10^3/(2x78478.4)
=5
Provide 6 bolts
Try with angle section 4-ISA 100x100x10
Strength on yielding = 870.9 kN > 736.93 kN
OK
Check:
1. Rupture strength of single angle section
β=1.4-0.076x(100/10)x(250/410)x(100+60-10)/100
= 0.71 >0.7
<1.44
Tdn=0.9x{10x(100-22-10/2)} x410/1.25+0.71x{10x(100-5)}250/1.1
= 368.8 KN >178KN (711.925/4 = 178) safe
2. Block failure
The block shear strength of a bolted connection is the least of
Tdb=Avgxfy/(ϒm0x√3)+0.9xAtnxfu/ϒm1
Or
=0.9xAvnxfu/(ϒm1x√3)+ Atgxfy/ϒm0
Avg=50x3x10 =1500 mm^2
Avn=(50x3-2.5x22)x10 =950 mm^2
Atg=40x10 =400 mm^2
Atn=(40-22/2)x10 =290 mm^2
Tdb={1500x250/(1.1x√3)+0.9x290x410/1.25}x4/1000 = 1129.7KN
Or ={0.9x950x410/(1.25x√3)+400x250/1.1}x4/1000 = 1011.28 KN
Least Tdb= 1011.28 KN > 711.925 KN safe
JOINT L2
Member L1-L2
Axial force =998.38 KN (T)
No. of bolts= 998.38x10^3/(2x78478.4)
=7
Provide 9 bolts
Try double channel section of DC250
Check:
1. Design strength due to yielding of gross section:
Tdg=Agxfy/ ϒm0
Ag=gross area of the cross section
= 2x39+2x30x0.8
=126cm2
fy= yield strength of the material
=250 Mpa
165
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
ϒm0=partial safety factor for the yielding =1.1
Tdg=126x100x250/(1.1x1000)
=2863.64KN >998.38 KN ok
2. Rupture strength of the channel section
Tdn=0.9xAncxfu/ ϒm1+βxAgoxfy/ ϒm0
Where
β=1.4-0.076x(w/t)x(fy/fu)x(bs/Lc)
but 0.7≤ β≤(fux ϒm0)/(fyx ϒm1)
in which
w= width of outstanding leg =80mm
bs=shear lag width =w+wt-t
=80+50-14.1 = 115.9mm
Lc=length of end connection, i.e., distance between the outermost bolts or length of the weld along the
direction of load =150mm
Anc=net sectional area of the connected leg
Ago=gross area of the outstanding leg
t=thickness of the angle
β=1.4-0.076x(80/14.1)x(250/410)x(115.9/150)
=1.197 >0.7
<1.44
Tdn=0.9x{7.2x(250-3x22-14.1)} x410/1.25+1.197x{2x(80-7.2/2)x14.1}250/1.1
= 947.23KN > 499.19KN (998.38/2=499.19KN) safe
Member U1-L2
Axial force = 1063.369 KN (T)
No. of bolts= 1063.369 x10^3/(2x78478.4)
=7
Provide 8 bolts
Try with angle section 4-ISA 70x70x10
Check:
1. Rupture strength(for single angle)
β=1.4-0.076x(70/10)x(250/410)x(70+40-10)/150
=1.1837 >0.7
<1.44
Tdn=0.9x{10x(70-22-10/2)} x410/1.25+1.1837x{10x(70-5)}250/1.1
= 301.801KN >265.84KN (1063.369KN /4=265.84KN) safe
2. Design strength due to yielding of gross section:
Tdg=Agxfy/ ϒm0
Ag=gross area of the cross section
= 4x1300+300x12
=8800 mm^2
fy= 250 Mpa
166
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
ϒm0=partial safety factor for the yielding =1.1
Tdg=8800x250/(1.1x1000)
=2000KN >1063.369KN ok
3. Block failure
The block shear strength of a bolted connection is the least of
Tdb=Avgxfy/(ϒm0x√3)+0.9xAtnxfu/ϒm1
Or
=0.9xAvnxfu/(ϒm1x√3)+ Atgxfy/ϒm0
Where Avg and Avn are the minimum gross and net areas in shear along the bolt line parallel to line of
action of force respectively.
Atg, Atn are the minimum gross and net areas in tension from the bolt hole to the edge of a plate,
perpendicular to the line of action of force
Avg=50x4x10 =2000 mm^2
Avn=(50x4-3.5x22)x10 =1230 mm^2
Atg=30x10 =3000 mm^2
Atn=(30-22/2)x10 =190 mm^2
So,
Tdb={2000x250/(1.1x√3)+0.9x190x410/1.25}x4/1000 =1274.078KN
Or ={0.9x1230x410/(1.25x√3)+300x250/1.1}x4/1000 = 1111.262 KN
Least Tdb=1111.262 KN > 1063.369 KN safe
Joint L0
Member L0-L1
Axial force =998.38 KN (T) ( same as member L1-L2)
No. of bolts= 998.38x10^3/(2x78478.4)
=7
Provide 9 bolts
Provide double channel section of DC250
JOINT U3
Member U3-U4
Axial force =2283.18 KN (C)
No. of bolts= 2283.18x10^3/(2x78478.4)
= 15
Provide 15 bolts
Try double channel section of DC300
Check:
1. Design strength due to yielding of gross section:
A = 140.6 cm2 and ry = 11.64 cm
Check for Compression capacity
For
πΎπΏ
π
=
0.8∗600
11.64
= 30.93 and buckling class ‘c’,
fcd = 210 MPa
So, Pd = fcdxAg = 2952.6 kN > 2139.97 kN
OK
2. Rupture strength of the channel section
167
Design of Bridge Over Kerunga Khola, Chitwan
2069 – AB Bridge
Tdn=0.9xAncxfu/ ϒm1+βxAgoxfy/ ϒm0
Where
β=1.4-0.076x(w/t)x(fy/fu)x(bs/Lc)
but 0.7≤ β≤(fux ϒm0)/(fyx ϒm1)
in which
w= width of outstanding leg =90mm
bs=shear lag width =w+wt-t
=90+75-13.6 = 151.4mm
Lc=length of end connection, i.e., distance between the outermost bolts or length of the weld along the
direction of load =150mm
Anc=net sectional area of the connected leg
Ago=gross area of the outstanding leg
t=thickness of the angle
β=1.4-0.076x(90/13.6)x(250/410)x(151.4/200)
=1.168 >0.7
<1.44
Tdn=0.9x{7.8x(300-3x22-13.6)} x410/1.25+1.168x{2x(90-7.8/2)x13.6}250/1.1
= 1129.158KN >1141.59KN (2283.18/2=1141.59KN) OK
Member U3-L4
Axial force = 505.25 KN (T)
No. of bolts= 505.25 x10^3/(2x78478.4)
=4
Provide 4 bolts
Try with angle section 4-ISA 55x55x8
Check:
5. Rupture strength of a single angle connected through one leg and affected by
shear lag is given by
Tdn=0.9xAncxfu/ ϒm1+βxAgoxfy/ ϒm0
Where
β=1.4-0.076x(w/t)x(fy/fu)x(bs/Lc)
but 0.7≤ β≤(fux ϒm0)/(fyx ϒm1)
in which
w= width of outstanding leg =55
bs=shear lag width =w+wt-t
=55+30-8 = 77mm
Lc=length of end connection, i.e., distance between the outermost bolts or length of the weld along the
direction of load =100mm
Anc=net sectional area of the connected leg
Ago=gross area of the outstanding leg
t=thickness of the angle
β=1.4-0.076x(55/8)x(250/410)x(77/100)
=1.155 >0.7
<1.44
168
Design of Bridge Over Kerunga Khola, Chitwan
Tdn= 0.9x {8x(55-22-8/2)} x410/1.25+1.155x{8x(55-4)}250/1.1
= 175.58KN < 126.313KN (505.25 /4=126.313KN) safe
2. Design strength due to yielding of gross section:
Tdg=Agxfy/ ϒm0
Ag=gross area of the cross section
= 4x818+300x12 (size of plate may be reduced)
=6872 mm2
fy= 250 Mpa
ϒm0=partial safety factor for the yielding =1.1
Tdg=6872x250/(1.1x1000)
=1561.81 KN >505.25 KN ok
3. Block failure
The block shear strength of a bolted connection is the least of
Tdb=Avgxfy/(ϒm0x√3)+0.9xAtnxfu/ϒm1
Or
=0.9xAvnxfu/(ϒm1x√3)+ Atgxfy/ϒm0
Avg=50x2x8 =800 mm^2
Avn=(50x2-1.5x22)x8 =536 mm^2
Atg=25x8 =200 mm^2
Atn=(25-22/2)x8 =112 mm^2
Tdb={800x250/(1.1x√3)+0.9x112x410/1.25}x4/1000 =552.14KN
Or ={0.9x536x410/(1.25x√3)+250x200/1.1}x4/1000 = 547.23 KN
Least Tdb=547.23 KN> 505.25 KN safe
169
View publication stats
2069 – AB Bridge
