The document provides details on the design of the third phase of the Thannermukkom salt water barrier bridge. It includes the design of the following bridge components:
1) Deck slab using Pigeaud's curves to calculate bending moments from dead and live loads.
2) Cantilever slab, longitudinal girders, cross girders, bearings, pedestals, operating platform, pier, pier cap, pile, pile cap and apron designed based on codes and previous project details.
3) Pier design carried out using STAAD Pro software. Reinforced concrete grade M30 and steel grade Fe415 are used.
The preliminary dimensions and design loads as per IRC codes are
1. 1
CHAPTER 1
INTRODUCTION
1.1 GENERAL
The Thannermukkom salt water barrier was constructed as a part of the Kuttanad Development
Scheme to prevent tidal action and intrusion of salt water into the Kuttanad low-lands across
Vembanad Lake between Thannermukkom on South and Vechur on North. It is the largest mud
regulator in India. This barrier essentially divides the lake into two parts - one with brackish
water perennially and the other half with fresh water fed by the rivers draining in to the lake. The
TMB was planned to be of 1402m length with 93 shutters. The TMB was commissioned to
prevent saline intrusion from Cochin gut into the southern part of Vembanad Lake (where paddy
cultivation is practised) to protect the crop from damage from saltwater intrusion. The plan was
to build the Thannermukkom Barrage in three phases. The first phase at Thannermukkom end
comprising 31 shutters and 2 locks for navigation was completed in 1968. The second phase at
Vechoor end with 31 shutters and one lock was completed in 1974. When the work on the third
phase with 31 shutters was delayed, a coffer dam was erected in 1975 to stop the salt water flow.
The expected benefit from TBM was safe punja paddy and the intensification to a second
crop(virippu crop) in about 18,500 ha of kayal and lower kuttanad.
Though the partial commissioning of the barrage could prevent the saline water to the southern
side and thereby supported paddy cultivation, it created several problems in the wet land system,
the major ones are
(1) Concentration of pollutants due to inadequate tidal flushing due to closure of the bund and
due to blockage at the centre 1/3rd portion(ie coffer dam)
(2) Reduction in flood discharge during monsoon as the central 1/3rd portion is blocked
(3) Silting at the southern side of the barrage
During 2007, Union Ministry of Agriculture invited M.S.SWAMINATHAN RESEARCH
FOUNDATION (MSSRSF) for a detailed study of the economic and ecological problems of the
Alappuzha district as well as the Kuttanad wetland ecosystem as a whole and to recommend the
possible solutions. One of the recommendations given by MSSRF is as follows.
2. 2
1.2 MEASURES FOR SALINITY AND FLOOD MANAGEMENT IN KUTTANAD
1. Undertake and complete the work on phase 3 of the TMB following modern design,
compatible with the renovated phase 1 and phase 2 portions and with all shutters operatable.
2. Dismantle and remove the coffer dam without letting the soil and debris spreading on the lake
bed. Dredge the part of Lake south of the coffer dam to suitable depth based on proper
bathymetric studies and gradient analysis.
We feel that the construction of third stage of Thenneermukkom barrage is the need of the hour
and hence we choose the design of third stage as project.
Fig 1.1 : Thanneermukkom Salt Water Barrier
3. 3
CHAPTER 2
LITERATURE REVIEW
2.1 GENERAL
The main purpose of Thanneermukkom salt water barrier is to protect paddy cultivation on
Kuttanad area from saline intrusion from Cochin gut. At the same time, water transport and land
transport across the site should not be affected. To maintain water transport, lock gates are
constructed and to maintain land transport, the bridge is constructed. Since our project is limited
to central portion which is presently a coffer dam and is being replaced by a concrete bridge
structure with regulators, only the design of central portion is included. After comparison of
different types of section of girders, T beam girder is selected. The shape and height of pier
selected has also their own significance. Different papers from which information gathered for
the progress of the project work are included in this section.
2.2 ARTICLE DETAILS
Arnold W. Hendry , Leslie G. Jaeger(1955) in their article ―The Load Distribution in
Interconnected Bridge Girders with Special Reference to Continuous Beams” explained about
how a method which is usually applied to a simply supported beam, can be applied to analysis of
an interconnected continuous beam. The method outlined is for the analysis of interconnected
bridge girders having any degree of torsional rigidity and is based on two assumptions that the
transverse members can be replaced by a continuous medium and that torsion of these members
can be neglected. The solution is reached by harmonic analysis and distribution coefficients are
tabulated for single span bridges having from two to six main girders for all harmonics of the
bending moment and deflection curves for the span. The application of the method to continuous
beam systems by superposition is also explained. A method for the derivation of influence lines
for bending moments in the longitudinals of continuous bridges is also developed.
Mundzir Hasan Basri (2001), in the article “Two New Methods for Optimal Design of
Subsurface Barrier to Control Seawater Intrusion” analysed two new methods to control
seawater intrusion using a subsurface barrier through development and application of the implicit
and explicit simulation-optimization approaches and had developed implicit and explicit
4. 4
simulation-optimization models for design of a subsurface barrier that controls seawater
intrusion. No prior work has been done in which a model for optimal design of a barrier for
controlling seawater intrusion is developed. The objective of the seawater intrusion control
problem is to minimize the total construction costs while requiring that salt concentrations be
held below specified values at two control locations at the end of the design period.
Nan Hu and Gonglian Dai (2010) in the article “The Comparative Study of Portal-Frame Pier
for High-Speed Railway” explained about the performance of portal-frame pier. They proved
that it not only provides comfort of passengers but also the safety of the vehicles. The study was
made by the comparison of five different portal-frame piers and based on it, the type, selection
for structure design and the effect of different structure parameters on structure performance as
well as stability of vehicle has been analysed.
The essentiality of replacement of central bunded portion of Thanneermukkom salt water barrage
is explained in the report named “Study for modernizing the Thanneermukkom bund and
Thottappally spillway for efficient water management in Kuttanad region, Kerala” (2011)
submitted to Kerala Governmnent on August 2011, based on studies conducted by Indian
Institute of Technology, Madras, Chennai and Centre for Water resources development and
management, Kozhikode, Kerala. The study revealed that the middle bunded portion of the
barrage does not have significant effect on flooding in the study area. But due to the absence of
natural flushing action, accumulation of pollutants occurs in the lake. Improper working of
several existing gates of the barrage also creates problems. The report suggest to replace mud
regulator with a gated structure with provisions for natural flushing and at the same time
restricting the salinity to levels below critical value for normally grown paddy by proper
monitoring of salt concentration.
M.G. Kalyanshetti, C.V. Alkunte (2012) carried out a study for effectiveness of IRC live load for
various height of pier and span of bridge for different shape of pier in the article “Study on
Effectiveness of IRC Live load on R.C.R Bridge Pier”. The effectiveness of IRC live load for
various height of pier and span of bridge for different shape of pier is studied using computer
5. 5
programming. The study reveals that pier designed by considering IRC class A loading should be
checked for IRC 70R wheeled loading.
Parvin Eghbali, Amir Ahmad Dehghani, Hadi Arvanaghi, Maryam Menazadeh (2013) explained
about scouring around bridge foundations which is one of the major causes of bridge damaging
in the article “The Effect of Geometric Parameters and Foundation Depth on Scour Pattern
around Bridge Pier”. The effect of shape, level and position angle of foundation on scour pattern
were investigated experimentally and the results showed that the scour depth depends on
foundation depth, shape and foundation alignment to the approach flow. The best foundation
shape which leads to minimum scour was aerodynamic shape along the channel, square,
aerodynamic across the channel and cylindrical shape respectively, the best positioning of
foundation is that below the initial channel bed and the best angle for positioning of foundation is
zero (parallel to flow direction).
Amit Saxena and Dr.Savita Maru (2013) in the article “Comparative Study of the Analysis and
Design of T-Beam Girder and Box Girder Superstructure” compared T-beam girder and Box
girder to check which among the two is favourable. The decisions were taken based on obvious
element of Engineering - safety, serviceability and economy. They had concluded that service
load bending moments and shear force for T-beam girder are lesser than box girder which allow
the designer to have lesser heavier section for T-beam girder than Box girder for span upto 25m.
Also the cost of concrete and quantity of steel required by the T-beam girder is less.
Manjeetkumar M Nagarmunnoli and S V Itti (2014) in their article “Effect of Deck Thickness in
RCC T-beam Bridge” explained that the increased traffic demand, material ageing, cracking of
bridge components, physical damages incurred by concrete, corrosion of reinforcement and
inadequate maintenance of bridges necessitate the assessment of bridges periodically for their
performances. The non linear analysis is a tool to simulate the exact material behaviour, to
evaluate strength in inelastic range. An attempt has been made to perform non linear finite
element analysis to analyse the component of a selected road bridge. They studied about the
fatigue life evaluation of reinforced Concrete Highway Bridge and has carried out Non-linear
6. 6
analysis of the structural element using ANSYS. RCC T-beam Bridge has been chosen for
detailed 3D nonlinear analysis.
H P Santhosh, Dr. H M Rajashekhara Swamy, Dr. D L Prabhakara (2014) in their article
“Construction Of Cofferdam -A Case Study” explained about the present state of construction of
cofferdam techniques and techniques developed world wide for mitigating hydraulic forces on
the temporary structures. A cofferdam is a temporary structure designed to keep water and soil
out of the excavation in which a bridge pier or other structure is built. When construction must
take place below the water level, a cofferdam is built to give workers a dry work environment.
Sheet piling is driven around the work site, seal concrete is placed into the bottom to prevent
water from seeping in from underneath the sheet piling, and the water is pumped out.
Kavitha.N, Jaya kumari.R, Jeeva.K, Bavithra.K, Kokila.K (2015) in their article “Analysis and
Design of Flyover” conducted a traffic survey at the four road junction in Salem town and
designed all the structural parts for the grade separator. The grade separator is of 640 m length
with 21 spans, 20m per span. It consists of a deck slab, longitudinal girders, cross girders, deck
beam, pier and foundation. Slab is designed by Working stress method as per the
recommendation of IRC:21-2000, Clause 304.2.1. Cantilever slab is designed for maximum
moment due to cantilever action. Longitudinal girders are designed by Courbon‘s method. Cross
girders are designed mainly for stiffness to longitudinal girders. Elastomeric reinforced bearing
plate is used. The deck beam is designed as a cantilever on a pier. The Pier is designed for the
axial dead load and live load from the slab, girders, deck beam. Foundation designed as footing
for the safe load bearing in the soil.
Praful N K and Balaso Hanumant (2015) in their article “Comparative Analysis of T-Beam
Bridge by Rational Method and Staad Pro” explained that the bridge is a structure providing
passage over an obstacle without closing the way beneath. The required passage may be for a
road, a railway, pedestrians, a canal or a pipeline. T-beam bridge decks are one of the principal
types of cast-in place concrete decks. T-beam bridge decks consist of a concrete slab integral
with girders. The finite element method is a general method of structural analysis in which the
solution of a problem in continuum mechanics is approximated by the analysis of an assemblage
7. 7
of finite elements which are interconnected at a finite number of nodal points and represent the
solution domain of the problem. A simple span T-beam bridge was analyzed by using I.R.C.
loadings as a one dimensional structure using rational methods. The same T-beam bridge is
analysed as a three dimensional structure using finite element plate for the deck slab and beam
elements for the main beam using software STAAD ProV8i. Both FEM and 1D models where
subjected to I.R.C. Loadings to produce maximum bending moment, Shear force and similarly
deflection in structure was analysed. The results obtained from the finite element model are
lesser than the results obtained from one dimensional analysis, which means that the results
obtained from manual calculations subjected to IRC loadings are conservative.
8. 8
CHAPTER 3
DESIGN
3.1 GENERAL
A T beam bridge with suitable foundation is designed based on codal provisions. Codes used for
the design are IS 456:2000, SP 16:1980, IRC 6:2000, IRC 21: , IS 2911-1979 (Part 1) and IS
6966 (Part 1): 1989. Following components are designed in this section:
1. Deck slab
2. Cantilever slab
3. Longitudinal girder
4. Cross girder
5. Bearings
6. Pedestals
7. Operating platform
8. Pier
9. Pier cap
10. Pile
11. Pile cap
12. Apron and cut off
Design of pier is done using STAAD.Pro and the remaining components are designed manually.
Designing is done based on data obtained from previous records of similar bridges, study reports
and drawings given from the site and codal provisions.
Given Details
Clear roadway = 7.5m
Effective span of T-beam = 13.22m
Four longitudinal T-beams of thickness 625mm spaced at 2.625m intervals
Five cross beams of thickness 300mm spaced at 3.305m intervals
The preliminary dimensions may be assumed as shown in Fig 5.1. These may be checked later
and modified, if necessary. M30 grade concrete and high yield deformed bars of Fe415 grade
conforming to IS 1786 will be used. Clear cover to reinforcement is taken as 40mm.
9. 9
3.2. DECK SLAB
Step 1: Data known
The slab is supported on four sides by beams.
Thickness of slab H = 300mm
Thickness of wearing course = 56mm
Span in transverse direction = 2.625m
Effective span in transverse direction = 2.625 - 0.625 = 2m
Span in longitudinal direction = 3.305m
Effective span in longitudinal direction = 3.305 - 0.3 = 3.005m ≈ 3m
Fig 3.1: Preliminary Dimensions for T-beam Deck
Step 2: Bending moment due to dead load
Weight of deck slab = 0.3 x 24= 7.2 kN/m2
(Unit weight of concrete = 24 kN/m3
)
Weight of wearing course = 0.056 x 22= 1.232 kN/m2
(Unit weight of bitumen = 22 kN/m3
)
Total weight = 7.2 + 1.232 = 8.432 kN/m2
Since the slab is supported on all four sides and is continuous, Pigeaud‘s curves will be used to
get influence coefficients to compute moments.
Ratio K = Short span / Long span = 2/3 = 0.66
1/K = 1.5
10. 10
From Pigeaud‘s curves, m1 = 0.045 and m2 = 0.018
Total dead weight W = 8.432 x 2 x 3 = 50.6 kN
Moment along short span = W (m1 + 0.15m2) = 50.6 [0.045 + (0.15 x 0.018)] = 2.42kNm
Moment along long span = W (m2 + 0.15m1) = 50.6 [0.018 + (0.15 x 0.045)] = 1.25kNm
Fig 3.2: Class AA Track located for maximum moment on deck slab
Step 3: Live load bending moment due to IRC Class A-A tracked vehicle
Size of one panel of deck slab = 2m x 3m
One track of the tracked vehicle is placed symmetrically on the panel as shown in Fig 3.2. Track
contact dimensions are taken from Fig 3.1.
Impact factor fraction = 25% (IRC: 6-2000, Clause 211.3)
Width of load spread along short span;
u = 0.85 + 2 x 0.056 2 + 0.32 = 1.007m
Similarly,
Width of load spread along longitudinal direction;
v = 3.6 + 2 x 0.056 2 + 0.32 = 3.724m, but limited to 3m
K = 0.66
11. 11
u/B = 1.007/2 = 0.503
v/L = 3/3 = 1
Using Pigeaud‘s curve for K= 0.7,
m1 = 7.6x 10-2
m2 = 3.1 x 10-2
Total load per track including impact = 1.25 x 350 (Fig 1, Page 11, IRC:6-2000)
= 437.5kN
Effective load on the span W = 437.5 x (3/ 3.724) = 352.44 kN
Moment along shorter span = W (m1 + 0.15m2) = 352.44 [(7.6 + 0.15x3.1) x 10-2
] = 28.42kNm
Moment along longer span = W (m2 + 0.15m1) = 352.44 [(3.1 + 0.15x7.6) x 10-2
] = 14.94kNm
Fig 3.3: Disposition of Class AA Wheeled Vehicle for Maximum Bending Moment
Step 4: Live load bending moment due to IRC class AA wheeled vehicle
The class AA wheeled vehicle should be placed on the deck slab panel as shown in the
Fig 3.3 for producing the severest moments. The front axle is placed along the centre line with
the 62.5kN wheel at the centre of the panel. Only three wheels per axle, i.e., a total of six wheels,
can be accommodated within the panel. The maximum moments at the centre in the short span
and the long span directions are computed for individual wheel loads taken in the order shown.
12. 12
(a) B.M due to wheel 1
Consider the load marked 1 in fig.
Tyre contact dimensions are 300 x 150 mm
u = 0.3 + 2x0.056 2 + 0.32 = 0.509 m
v = 0.15 + 2x0.056 2 + 0.32 = 0.398m
u/B = 0.509/2 = 0.254
v/L = 0.398/3 = 0.132
K = 0.66
Using Pigeaud‘s curves,
m1 = 19 x 10-2
m2 = 15 x 10-2
Total load allowing for 25% impact, W = 1.25 x 62.5 = 78.1 kN
Moment along short span = W (m1 + 0.15m2) = 78.1 [(19+0.15x15)10-2
] = 16.596 kNm
Moment along long span = W (m2 + 0.15m1) = 78.1 [(15+0.15x19)10-2
] = 13.94 kNm
(b) B.M. due to wheel 2
Here the wheel load is placed unsymmetrically with respect to YY axis of the panel. But
Pigeaud‘s curves have been derived for loads symmetrical about the centre. Hence we use an
approximate device to overcome the difficulty. We imagine the load to occupy an area placed
symmetrically on the panel and embracing the actual area of loading, with intensity of loading
equal to that corresponding to the actual load. We determine the moments in the desired
directions for that imaginary loading. Then we deduct the moment for a symmetrical loaded area
beyond the actual loaded area. Half of the resulting value is taken as the moment due to the
actual loading.
13. 13
Fig 3.4: Details of Disposition of Wheel 2
Intensity of loading = 62.5x1.25 / 0.509x 0.398 = 385.646 kN/m2
Consider the loaded area of 2m x 0.398m
For this area,
u = 2m
v = 0.398m
u/B = 2/2 = 1
v/L = 0.398/3 = 0.132
K = 0.66
1/K = 1.5
14. 14
Therefore,
m1 = 7.8x10-2
m2 = 8x10-2
Moment along short span = W (m1+ 0.15m2) = 385.646 (7.8 + (0.15x8)10-2
) x 2 x 0.398
= 27.63 kNm
Moment along long span = W (m2 + 0.15m1) = 385.646 (8 + (0.15x7.8) x 10-2
) x 2 x 0.398
= 28.15 kNm
Next, consider the area between the real load and the dummy load
i.e., 1.491 m x 0.398 m
For this moment,
u = 1.491m
v = 0.398m
u/B = 1.491/2 = 0.745
v/L = 0.398/3 = 0.132
Therefore,
m1 = 10.3x10-2
m2 = 10.2x10-2
Moment along short span = 385.646 [10.3 + (0.15 x 10.2) x 10-2
)] x 1.491 x 0.398 = 27.07 kNm
Moment along long span = 385.646 [10.2 + (0.15 x 10.3) x 10-2
] x 1.491 x 0.398 = 26.878 kNm
Net bending moment along short span = ½ [27.63-27.07] = 0.28 kNm
Net bending moment along long span = ½ [28.15-26.878] = 0.636 kNm
By similar figure and procedure for case (b), remaining cases are done.
(c) B.M. due to wheel 3
Intensity of loading = (37.5 x 1.25)/(0.509 x 0.398) = 231.387 kN/m2
Consider the loaded area of 2m x 0.398m
For this area,
Moment along shorter span = 231.387 x [7.8 + (0.15 x 8) x 10-2
] x 2 x 0.398
= 16.576 kNm
Moment along longer span = 231.387 x [8 + (0.15 x 7.8) x 10-2
] x 2 x 0.398
= 16.89 kNm
15. 15
The area between the real and the dummy load is 0.691m x 0.398m
u = 0.691
v = 0.398
u/B = 0.691/2 = 0.345
v/L = 0.398/3 = 0.132
Therefore,
m1 = 16.9 x 10-2
m2 = 14 x 10-2
Bending moment along shorter span = 231.387 x [16.9 + (0.15x14) x 10-2
] x 0.691 x 0.398
=12.1 kNm
Bending moment along longer span = 231.387 x [14 + (0.15x16.9) x 10-2
] x 0.691 x 0.398
= 10.52 kNm
Therefore,
Net bending moment along shorter span = ½ (16.576-12.1) = 2.238 kNm
Net bending moment along longer span = ½ (16.89-10.52) = 3.185 kNm
(d) B.M due to wheel 4
Here the load is placed eccentric with respect to the XX axis.
Intensity of loading = 385.646 kN/m2
The loaded area with respect to XX axis = 2m x 0.398m
Moment along shorter span = 27.63 kNm
Moment along longer span = 28.15 kNm
Consider the area between the real load and the dummy load = 0.509m x 0.802m
u = 0.509
v = 0.802
u/B = 0.509/2 = 0.25
v/L = 0.802/3 = 0.267
Therefore,
m1 = 17.8 x 10-2
m2 = 11 x 10-2
16. 16
Moment along shorter span = 385.646 x [17.8 + (0.15 x 11) x 10-2
] x 0.509 x 0.802
= 30.62 kNm
Moment along longer span = 385.646 x [11 + (0.15 x 17.8) x 10-2
] x 0.509 x 0.802
= 21.52 kNm
Therefore,
Net bending moment along shorter span = ½ (27.63-30.62) = -1.5 kNm
Net bending moment along longer span = ½ (28.15-21.52) = 3.315 kNm
(e) B.M due to wheel 5
In this case, the loading is eccentric with respect to both XX and YY axes. A strict
simulation of the symmetric loading conditions would lead to complicated and laborious
calculations. Hence as a reasonable approximation, only the eccentricity with respect to the XX
axis is considered and the calculations are made as for case (d).
Intensity of loading = 385.646 kN/m2
Consider the loaded area of 2m x 0.389m
Moment along shorter span = 27.63 kNm
Moment along longer span = 28.15 kNm
Consider the area between the real load and the dummy load = 0.267m x 0.802m
u = 0.267
v = 0.802
u/B = 0.267/2 = 0.133
v/L = 0.802/3 = 0.267
Therefore,
m1 = 21.5 x 10-2
m2 = 13 x 10-2
Moment along shorter span = 385.646 x [21.5 + (0.15 x 13) x 10-2
] x 0.267 x 0.802
= 19.365 kNm
Moment along longer span = 385.646 x [13 + (0.15 x 21.5) x 10-2
] x 0.267 x 0.802
= 11.89 kNm
17. 17
Therefore,
Net bending moment along shorter span = ½ (27.63-19.365) = 4.132 kNm
Net bending moment along longer span = ½ (28.15-11.89) = 8.13 kNm
(f) B.M due to wheel 6
Intensity of loading = 231.387 kN/m2
Consider the loaded area of 2m x 0.398m
Moment along shorter span = 27.63 kNm
Moment along longer span = 28.15 kNm
Consider the area between the real load and the dummy load = 0.509m x 0.802m
Moment along shorter span = 30.62 kNM
Moment along longer span = 21.52 kNm
Therefore,
Net moment along shorter span = ½ (27.63-30.62) = -1.5 kNm
Net moment along longer span = ½ (28.15-21.52) = 3.315 kNm
(g) Total B.M due to all wheels on the span
The total effect is computed as the summation of individual effects.
Total B.M along shorter span = 16.596 + 0.28 + 2.238 - 1.5 + 4.132 - 1.5 = 20.246 kNm
Total B.M along longer span = 13.94 + 0.636 + 3.185 + 3.315 + 8.13 + 3.315 = 32.421 kNm
Step 5: Design B.M
The wheels causing maximum effects are adopted for design moments. The above computations
assumed as a simply supported condition along the four edges. In fact, the deck slab is
continuous. To allow for continuity, the computed moments are multiplied by a factor of 0.8.
Design B.M along shorter span = 0.8 x (2.42 + 28.42) = 24.672 kNm
Design B.M along longer span = 0.8 x (1.25 + 32.421) = 26.94 kNm
Step 6: Reinforcement Details
σcbc =
30
3
= 10 N/mm2
σst = 200 N/mm2
18. 18
k =
280
(280+3σst)
= 0.318
j = 1-
k
3
= 0.894
Q = 0.5σcbckj = 0.5 x 10 x 0.318 x 0.894 = 1.42
Effective depth required =
M
Q
=
26.94 x 106
1.42 x 103
= 137.74 mm ≈ 140 mm
Assuming 12mm diameter bars and a clear cover of 40mm,
Overall depth, D = 140 + 40 + 6 = 186mm < 300 mm
Effective depth provided, d = 300-40-6 = 254 mm
Hence safe
Area of main reinforcement = Ast =
M
σst j d
=
(26.94 x 106)
(200 x 0.894 x 254)
= 594.2 mm2
No: of bars =
Ast required
Ast provided
=
594.2
(
π
4
x122)
= 5.254 ≈ 6
Spacing =
π
4
x122x1000
594.2
= 190.33mm
Area of longitudinal reinforcement =
24.672 x 106
200 x 0.894 x 254
= 543.25 mm2
No: of bars =
543.25
(
π
4
x122)
= 4.8 ≈ 5
Spacing =
π
4
x122x1000
543.25
= 208.186 mm
Therefore, Provide 6 numbers of 12mmΦ @ 190mm c/c as main reinforcements and 5 numbers
of 12mmΦ @ 200mm c/c as transverse reinforcement.
19. 19
3.3. CANTILEVER SLAB
Step 1: Moment due to dead load
The total maximum moment due to dead load per metre width of cantilever slab is computed as
in the following table; using details from Fig 3.6.
Fig 3.5: Cantilever Slab with Class A wheel
Table 3.1: Dead load Calculation of Cantilever Slab
Sl.No Description Load (kN) Lever arm Moment (kNm)
1.
2.
3.
4.
5.
Handrails
Kerb
Footpath
Slab
rectangular
Slab
triangular
0.550 x (0.995+0.475)
x 24 = 19.4
0.225x0.225x24 = 1.215
1.475x0.275x24 = 9.735
2.25x0.2x24 = 2.7
0.5x2.25x0.05x24 = 1.35
0.550/2 = 0.275
[0.55+1.475+(0.225/2)]= 2.1375
[0.55+(1.475/2)] = 1.2875
2.25/2 = 1.125
2.25/3 = 0.75
5.34
2.6
12.53
3.04
1.0125
Total 51.88
20. 20
Step 2: Moment due to live load
Live load on footpath = 400kg/m2
= 4kN/m2
(From IRC 6: 2014)
Live load per meter width including impact = 4 x1.25 = 5 kN/m2
Moment due to live load = 5 x 1.475 x (1.475/8) = 1.36 kNm
Step 3: Reinforcement
Total moment due to dead load and live load = 51.88 + 1.36 = 53.24 kNm
Effective depth required =
M
Q
=
53.24 x 106
1.42 x 103
= 193.63 ≈ 200mm
Effective depth provided = 300 - 40 – 8 = 252 mm
Hence safe
Area of reinforcement required = Ast =
M
σst j d
=
(53.24 x 106)
(200 x 0.894 x 252)
= 1181.59 mm2
Adopt 16mmΦ bars as main reinforcements,
No: of bars =
Ast required
Ast provided
=
1181.59
(
π
4
x162)
=5.87 ≈ 6
Spacing =
π
4
x 162 x 1000
1181.59
= 170.162mm
Therefore, provide 6 numbers of 16mmΦ @ 170mm c/c as main reinforcements.
B.M for distributors = 0.2 x 51.88 + 0.3 x 1.36 = 10.784 kNm
Area of distributors =
10.784 x 106
200 x 0.89 x 252
= 240.41 mm2
Assuming 12mmΦ bars as distributors
No: of bars =
Ast required
Ast provided
=
240.41
π
4
x122
= 2.2 ≈ 4
Spacing =
π
4
x 122 x 1000
240.41
= 170.162mm = 470.96mm
Therefore, provide 4 numbers of 12mmΦ @ 450mm c/c as distributors.
21. 21
3.4. LONGITUDINAL GIRDER
Step 1: Given data
Effective span = 13.22m
Slab thickness = 0.3m
Width of rib = 0.625m
Spacing of main beams = 2.625m
Overall depth of beam = 1.13m
Step 2: BM due to DL
Dead load per m run is estimated as below
Table 3.2: Dead load Calculation of Longitudinal Girder
Sl.No Item Details Weight (kN)
1. Wearing coarse 0.625 x 0.056 x 22 0.77
2. Deck slab 2.625 x 0.3 x 24 18.9
3. Rib 0.625 x 1.13 x 24 16.95
4. Cross beams (5 x 3 x 2 x 0.3 x 24) /13.22 6.535
Total 43.155
Maximum BM =
43.155 x 13.22 x 13.22
8
= 942.766 kNm
Step 3: BM due to LL
Maximum live load bending moment will occur under class A two lane loading
Impact factor fraction =
4.5
6+13.22
= 0.234
The loading is arranged in the transverse direction as shown in the figure, allowing the minimum
clearance near the left kerb. All the four wheel loads are of equal magnitude.
Live load BM can be determined by using anyone of the following methods:
1. Courbon‘s method
2. Hendry- Jaegar method
3. Morice- little method
22. 22
Fig 3.6 : Transverse Disposition of Trains of Class A Loading for Determination of Reactions on
Longitudinal Beams
Step 4: Live load moment- Hendry-Jaegar method
MI of longitudinal girder = I = 0.2664 m4
MI of cross beams = It = 0.120 m4
Parameter A =
12
π4
(
L
h
)3nE It
EI
=
12
π4
13.22
2.625
3 5xEx 0.120
Ex0.2664
= 35.441
For the purpose of design, coefficients of F = α will be adopted. Using figures, the
distribution coefficients for A = 35 will be obtained as in Table 3.3.
Table 3.3: Distribution Factors for Unit load on Girder
Unit load on
girder
Distribution
factors
A B C D
B 0.36 0.268 0.204 0.227
C 0.268 0.283 0.193 0.218
Treating the deck slab as continuous in the transverse direction, the support moments at the
locations of the four longitudinal girders due to the loading as shown in the fig are computed
using method of moment distribution. The moments at A, B, C and D are found to be -0.246w, -
0.47w, -0.36w and 0, respectively.
Reaction RA, RB, RC and RD are determined from the support moments.
23. 23
RA = 0.471w
RB = 1.45w
RC = 1.53w
RD = 0.55w
These reactions are treated as loads on the interconnected girder system and multiplying these
by the respective distribution coefficients and adding the results under each girder and the final
reaction at each girder is obtained as shown in the table
Maximum bending moment on the intermediate beam = 952 kNm
Table 3.4: Reaction Factors on Girder
Load Girder A Girder B Girder C Girder D
1) 1.45w on
girder B
0.36x1.45w
= 0.522w
0.268x1.45w
= 0.3886w
0.204x1.45w
= 0.3w
0.227x1.45w
= 0.33w
2) 1.53w on
Girder C
0.268x1.53w
= 0.41w
0.283x1.53w
= 0.433w
0.193x1.53w
= 0.295w
0.218x1.53w
= 0.33w
Net reaction 0.932w 0.822w 0.6w 0.66w
Step 6: Design maximum bending moment
Live load bending moment obtained from Hendry Jaegar method will be adopted as
Design B.M = Moment due to D.L + Moment due to L.L
= 942.766 + 952 = 1894.766 kNm
≈ 1900 kNm
Step 7: Design of section
Effective flange width for the Tee beam section will be determined as er clause 305.15.2 of IRC
Bridge code.
24. 24
Effective flange width = thickness of web + (0.2 x 0.7 x effective span)
= 0.3+ (0.14 x 13.22)
= 2.15 m
Allowing a distance of 120mm from the bottom of T-beam to the center of gravity of rods and
assuming the centre of compression to be 120 mm below the top, and allowing a stress of
180MPa as permissible stress at the centre of gravity of steel area, area of steel reinforcement
As =
1900x106
180x 1130−120−120
= 11860.175 mm2
Provide 32 mm diameter bars,
No: of bars =
Ast required
Ast provided
=
11860.175
(
π
4
x322)
= 14.74 ≈ 16
Therefore, provide, 16 numbers of 32mm Φ bars in four rows of four bars each.
As provided = 12867.96 mm2
Fig 3.7 : Reinforcement Diagram of Deck slab, Cantilever slab and Longitudinal Girder
25. 25
3.5. CROSS GIRDER
Step 1: Given data
Spacing of cross beam = 3m
Effective span = 2.625 - 0.625 = 2 m
Impact factor for 2 m span for
Class AA tracked vehicle = 0.25
Class AA wheeled vehicle = 0.25
Class A loading = 0.55
Step 2: BM due to DL
The weight of slab and wearing course will be apportioned between the cross beams and the
longitudinal girders in accordance with the trapezoidal distribution of the loads on the panel as
shown in Fig 3.10
Fig 3.8: Deck Panel showing Trapezoidal Distribution of Dead Load
26. 26
Weight of deck slab and wearing course per m2
= (0.3x24) + (0.056x22) = 8.432 kN/ m2
Total load on cross beam due to slab by trapezoidal distribution= (2x0.5x2x1x8.432) =16.864 kN
Self weight of cross beam and weight of wearing course over the cross beam
= (2 x 0.625 x 1.13 x 24) + (2 x 0.625 x 0.056 x 22)
= 35.44 kN
Total load due to cross beam in one span = 16.864+35.44 = 52.304 kN
The cross beam is continuous over two spans. The exterior girders restrain the cross beams
at the ends, and at the middle girder, the beam approaches a fixed condition. The exact degree of
restraint at the girder locations is difficult to determine, and is somewhere intermediate between
the free and the fixed condition. Moment coefficients are listed in standard texts for free and
fixed ends and for uniform and concentrated loads, for multiple spans. Weighted coefficients are
chosen as the sum of one third of the value corresponding to the case of two span continuous
beam with free end and two thirds of the value corresponding to a single fixed ended spans.
Coefficient for maximum positive bending moment = (1/3) x 0.07 + (2/3) x 0.042
= 0.051
Coefficient for maximum negative bending moment = (1/3) x 0.125 + (2/3) x 0.083
= 0.097
Positive BM = 0.051 x 52.304 x 2 x 1 = 5.33 kNm
Negative BM = 0.097 x 52.304 x 2 x 1 = 10.17 kNm
Step 3: BM due to LL
Class AA tracked vehicles produces severe effect than the other loadings. Fig 3.10 shows the
disposition of one track on a crossbeam.
27. 27
Fig 3.9: Disposition of Class AA Track for Maximum Bending Moment on Cross Beam
Load on cross beam = 2 [ 350 x
1.65
3.6
x
2.325
3.300
] + [ 350 x
0.3
3.6
] = 255 kN
Coefficient of maximum positive BM due to concentrated load
= (1/3) x 0.203 x (2/3) x 0.125 = 0.151
Coefficient of maximum negative BM due to concentrated load
= (1/3) x 0.188 + (2/3) x 0.125 = 0.146
Positive BM due to impact = 0.151 x 255 x 2 x 1.13 = 87.02 kNm
Negative BM including impact = 0.146 x 255 x 2 x 1.13 = 84.14 kNm
Step 4: Design of section
Design positive BM = 5.33 + 87.02 = 92.35 kNm
Effective depth = 1130 – 73 = 1057 mm
28. 28
Area of steel required =
(92.35 x 106)
(200 x 0.89 x 1057)
= 490.84 mm2
Add 0.3% of area of the beam to give additional stiffness to the beam
Additional area of steel required = (0.3/100) x 200 x 1057 =634.2 mm2
Total area of steel required = 490.84 + 634.2 = 1125.04mm2
Provide 4 bars of 22mm Φ giving an area of 1520.53 mm2
Design negative BM = 10.17 + 84.14 = 94.31 kNm
Area of steel required =
(94.31 x 106)
(200 x 0.89 x 1057)
= 501.26 mm2
Provide 3 bars of 22mm Φ giving an area of 1140.4 mm2
Fig 3.10: Reinforcement Details of Cross Beam
29. 29
3.6. BEARINGS
Design an elastomeric unreinforced neoprene pad bearing to suit the following data.
Vertical DL = 877.34 + 68.8 = 946.14 ≈ 1000 kN
Vertical LL = 700 kN
Horizontal force = 0.2 x 700 = 140 kN
Modulus of rigidity of elastomer = G = 1 N/mm2
Friction coefficient = 0.3
Step 1: Design
Total vertical load = 1000+700 = 1700 kN
Horizontal force = 140 kN
From IRC: 83 1987, Part II,
Select the standard plan dimensions of elastomeric bearings,
a = 350 mm
b = 700 mm
Thickness, (t) should be less than a/5
Therefore, select a thickness of 50 mm
Area, A = 350 x 700 = 245000 mm2
tanΦ =
H
GA
=
140 x 103
1 x 245000
= 0.571
u = t tanΦ = 50 x 0.571 = 28.55
But t >1.43u
30. 30
1.43 u = 1.43 x 28.55 = 40.82 < 50 mm
Hence design is safe.
Step 2: Axial stress
Shape factor = S =
ab
2t(a+b)
=
350 x 700
2 x 50 (350+700)
= 2.33
σm =
P
A1
P = 1000 kN
A1
= (a-u)b = (350 – 28.55) x 700 = 225015 mm2
σm =
1000 x 103
225015
= 4.44
Check:
σm < 2GS
2GS = 2 x 1 x 2.33 =4.66
4.44 < 4.66
Hence design is safe
σm
1
=
PL
A1
=
700 x 103
225015
= 3.11 > 1 +
a
b
= 1+
350
700
= 1.5
3.11 > 1.5
Hence design is safe
31. 31
3.7. PEDESTALS
Step 1: Given Details
Total factored load = 3100 kN
Size of base plate = 300mm x 300mm
M30 concrete and Fe415 steel
Step 2: Design
Adopt minimum size of pedestal including 10mm clearance = (size of base plate + clearance)2
= 310mm x 310mm
Safe pressure = bearing strength x 310/300
Bearing strength = fcb = 0.45fck = 0.45x30 = 13.5 N/mm2
Therefore, safe pressure = 13.5 x (310/300) = 13.95 N/mm2
Load carried by pedestal = 13.95 x 310 x 310 = 1340.595 kN
Balance load = 3100 – 1340.595 = 1759.405 ≈ 1760 kN
Area of reinforcements required =
(1760 x 10−3)
(0.87 x 415)
= 4874.67 mm2
Assume 24 mm diameter bars,
No: of bars =
Ast required
Ast provided
=
4874.67
(
π
4
x242)
= 10.77 ≈ 12
Spacing =
π
4
x242x1000
4874.67
= 92.8 mm
Provide 6 nos of 24 mm diameter on both sides of the base plate at 90mm c/c
Step 3: Check for percentage steel
% of steel =
4874.67
(310x310)
x 100 = 5.07 > 0.4 % , Hence safe
32. 32
3.8. OPERATING PLATFORM
Step 1: Given details
Span = 13.22m
M30 grade concrete
Fe415 grade steel
Total load acting on the operating platform, W = 20 Tonne = 20 x 10 = 200 kN
Therefore, load acting on each beam =
200
2 x 13.22
= 7.56 kN/m
Assume, b/d = 0.5
Step 2: Moment of resistance
Mr = Qbd2
Q = 0.5σcbckj
σcbc =
30
3
= 10 N/mm2
k =
280
(280+3σst)
= 0.318
j = 1-
𝑘
3
= 0.894
Q = 0.5σcbckj = 0.5 x 10 x 0.318 x 0.894 = 1.42
Mr = 1.42 x 0.5d x d2
= 0.71d3
Step 3: Moment calculation
M =
W L2
8
=
7.56 x 13.222
8
= 165.156 kNm
33. 33
M = Mr
165.156 x 106
= 0.71d3
Therefore, d = 615.005 mm = 615mm
b = 0.5d = 0.5 x 615 = 307.5mm
Therefore, adopt dimensions of the beam = 0.31m x 0.62m
Step 4: Reinforcement Details
Ast =
M
σst j d
=
165.156 x 106
(200 x 0.894 x 620
= 1489.82 mm2
Adopt 16mmΦ bars,
No: of bars =
Ast required
Ast provided
=
1489.82
(
π
4
x162)
= 7.41 ≈ 8
Spacing =
π
4
x162x1000
1489.82
= 134.95mm
Therefore, provide 6 numbers of 16mmΦ @ 130mm c/c.
Ast provided = 6 x (π/4) x 162
= 1206.37 mm2
Ast remaining = 1489.82-1206.37 = 283.448 mm2
Provide 2 legged 8mm diameter vertical stirrups,
No: of stirrups = 283.448/((π/4) x 82
) = 5.64
Therefore, provide 6 nos of 8mmΦ 2 legged vertical stirrups.
34. 34
3.9. PIER
Step 1: Given details
A trestle type pier comprising of two reinforced concrete columns with a connecting cap is
adopted for the Thannermukkom third stage. For the analysis and design of pier, StaadPro V8i
software is used.
Total height of pier = 10m
Height of pier with combined H and semicircular portion = 6m (from base)
Height of pier with extending H portion = 4m
Height of circular pier = 10m
Diameter of circular pier = 1.7m
Centre to centre distance between H and circular pier = 9.65m
Centre to centre distance between pier = 13.22m
Grade of concrete = M30
Grade of steel = Fe415
Fig 3.11: H pier
Forces considered in the design are as follows,
Self weight of pier
Dead load from superstructure
Live load
35. 35
Impact allowance
Load from operating platform
Water pressure on pier
Force on pier due to water pressure from submerged shutter
Breaking force
Wind load
Effect of buoyancy
Step 2: Calculation of design loads
(a) Self weight of pier
StaadPro generates the self weight based on model created.
(b) Dead load from superstructure
Total load including self weight of deck slab, girders, wearing course, parapet =157 KN/m.
This load is equally distributed over 4 main girders.
Load per girder= 39.25 KN/m
(c) Live load
As per IRC 6-2000, clause 207.4 Table 2
2 lanes of class A loading.
(d)Impact Allowance
IRC 6-2000 clause 211.1
Impact factor fraction =
4.5
6+13.22
= 0.234
Impact Factor = 0.234
This fraction of live load has been given as impact allowance.
36. 36
(e) Load from operating platform
Load from operating platform = 70.06 KN on each pier
(f) Water pressure on pier
IRC 6-2000 clause 213.1
P= 0.5 KV2
K=0.66 For piers with semicircular ends
V= 3 m/s
Water pressure = 32.67 kN
(g) Water pressure-submerged shutter
Water pressure from submerged shutter= 235.44 KN
(h)Breaking Force
IRC 6-2000 clause 214.2, 20% of vehicular load is taken.
20% of class AA load = 0.2 x 700 = 140KN
(h) Effect of Buoyancy
The upward force is equal to the weight of water displaced by the submerged body.
Submerged volume of H pier= 6.74 x 5.5 =37.07m3
Unit weight ofwater = 10KN/m3
Buoyant force = 370.7 KN
Submerged volume of circular pier =12.48 m3
Buoyant force = 124.8 KN
37. 37
(i) Wind load
IRC 6-2000 clause 212.3
Wind load = Area of exposure x wind intensity
Total wind load = 44.1 KN
Step 3: Generation of Staad model
For Staad analysis H piers with semicircular ends have been replaced by equivalent area
rectangular section.
A Staad model has been created with given dimensions and material properties with four spans.
Fig 3.12: Staad Model of Pier
44. 44
Fig 3.16: Stress contour for H pier
Fig 3.17: Stress contour for Circular Pier
45. 45
Step 5: Staad design results
For H pier
Reqd. steel area : 54040.00 Sq.mm.
Reqd. concrete area: 6700959.50 Sq.mm.
Main reinforcement : Provide 68 - 32 dia. (0.81%, 54688.85 Sq.mm.) (Equally distributed)
Tie reinforcement : Provide 8 mm dia. rectangular ties @ 300 mm c/c
Puz : 107282.91 Muz1 : 32008.87 Muy1 : 17280.50
Interaction ratio: 0.00 (as per Cl. 39.6, IS456:2000)
Fig 3.18: Reinforcement details of H pier
For circular pier
Length: 10000.0 mm
Cross section: 1700.0 mm dia.
46. 46
Cover: 40.0 mm
Reqd. steel area : 18158.41 Sq.mm.
Reqd. concrete area: 2251642.50 Sq.mm.
Main reinforcement : Provide 37 - 25 dia. (0.80%, 18162.33 Sq.mm (Equally distributed)
Tie reinforcement : Provide 8 mm dia. circular ties @ 300 mm c/c
SECTION CAPACITY BASED ON REINFORCEMENT REQUIRED (KNS-MET)
Puz : 36048.98 Muz1 : 4483.67 Muy1 : 4483.67
INTERACTION RATIO: 0.07 (as per Cl. 39.6, IS456:2000)
Fig 3.19: Reinforcement details of Circular Pier
3.10. PIER CAP
Step 1: Data known
Grade of concrete = M30
Grade of steel = Fe415
47. 47
Fig 3.20: Pier Cap Plan
Design loads are as follows,
Superstructure dead load
Live load
Load from operating platform
Self weight
Load from deck slab
The superstructure dead load is equally shared by four girders
Dead load from superstructure = 626.44 KN per girder
Live load
As per IRC 6-2000, clause 207.4 Table 2
2 lanes of class A loading.
Analysis and design are performed in StaadPro V8i.
50. 50
Fig 3.22: Stress contour- Pier Cap
Step 4: Staad design results
M30 Fe415 (Main) Fe415 (Sec.) COVER: 75mm
Top reinforcement 30 Nos. 16mmυ
Bottom Reinfocement 30 Nos 16mmυ
8mmυ vertical stirrups 100mm c/c
4 no:s 25mmυ along each face [side face reinforcement
Fig 3.23: Reinforcement details of Pier Cap
51. 51
5.11. PILE
Cast-in-situ piles are those piles which are cast in position inside the ground. Bored cast-in-situ
piles of 1200mm diameter resting on fine sand layer is provided for foundation with a depth of
30m below bed level (as per soil report furnished by Chief Engineer, Kuttanad). Casing of 6mm
is used for installation of pile and boring operations are done by rotary type drilling rig.
All specifications for design of pile are based on IS 2911-1979 Part 1.
Step 1: Data known
Load from H pier = 3146.04 kN (STAAD.Pro result)
Load from circular column = 4013.39 kN (STAAD.Pro result)
Total resultant factored load Pu = 1.5 (3146.04 + 4013.39) = 10739.145 kN
Moment on H pier = 435.11 kNm (STAAD.Pro result)
Moment on circular column = 968.6 kNm (STAAD.Pro result)
Total factored moment Mu = 1.5 ( 435.11 + 968.6) = 2105.565 kNm
Assume 1200mm Φ pile and 32mm Φ reinforcement steel bars with a clear cover of 40mm
Effective cover = 40 + (
1
2
) x 32 = 56mm
Step 2: Longitudinal Reinforcement
From Chart 55 (Interaction Curve) of SP 16 : 1960, for
Pu
fck 𝐷2
=
10739.145 x 103
30 x 12002
= 0.248 ,
Mu
fck 𝐷3
=
2105.565 x 106
30 x 12003
= 0.041,
52. 52
d′
D
=
56
1200
= 0.046 ≈ 0.05 and
fy = 415N/mm2
,
Percentage steel required for longitudinal reinforcement in piles ‗p‘ is given by
P
fck
= 0.02
p = 0.02 x 30 = 0.6%
Minimum p = 1.25% of Ag for
Le
D
< 30 where Ag is the gross sectional area of pile
Therefore, provide p = 1.25%
Area of longitudinal steel, As =
1.25
100
x
π
4
x 12002
≈ 14137.166 mm2
Number of 32mm Φ bars =
14137.166
804.2477
= 18
Provide 22 no:s of 32mm Φ bars with total area 17694mm2
as longitudinal reinforcement.
Step 3: Lateral reinforcement
The minimum lateral reinforcement in a pile should be as follows:
a) Diameter of lateral tie should not be less than 5mm
b) Volume of lateral ties = 0.6% of gross volume of pile at ends(length 3D from endpoint)
= 0.2% of gross volume of pile at body
Use 12mm Φ Fe 415 lateral ties
1. At body of pile,
Volume of lateral reinforcement (for 1m length) =
0.2
100
x
π x 12002
4
x 1000 =2261.946 x 103
mm3
53. 53
Volume of single lateral tie = 1088π x
π x 122
4
= 386.572 x 103
mm3
No: of ties =
2261.946 x 103
386.572 x 103
= 5.85 ≈ 6
Spacing of ties =
1000
6
= 166.66 mm
Code provision for spacing of ties:
i. < least lateral dimension = 1200mm
ii. < 16 times diameter of longitudinal reinforcement = 16 x 32 = 512mm
iii. < 300mm
iv. Should not be less than 150mm
v. > actual spacing required = 166.66mm
Provide 12mm Φ ties @ 200mm c/c
2. At upper and lower ends of pile,
Volume of lateral reinforcement (for 1m length) =
0.6
100
x
π x 12002
4
x 1000 = 6785.84 x 103
mm3
Volume of single lateral tie = 1088π x
π x 122
4
= 386.572 x 103
mm3
No: of ties =
6785.84 x 103
386.572 x 103
= 17.555 ≈ 18
Spacing of ties =
1000
18
= 55.55 mm
Code provision for spacing of ties:
i. < least lateral dimension = 1200mm
ii. < 16 times diameter of longitudinal reinforcement = 16 x 32 = 512mm
iii. < 300mm
iv. Should not be less than 150mm
v. > actual spacing required = 55.55mm
54. 54
Provide 12mm Φ ties @ 200mm c/c
Fig 3.24: Reinforcement Details of Pile
5.1.11. PILE CAP
Pile cap is the thick concrete mat that rests on piles that have been driven into the soft or unstable
ground to provide a suitable stable foundation.
Step 1: Data known
Use M30 grade concrete and HYSD steel bars of grade Fe415
For M 30 Concrete, fck = 30 N/mm2
For Fe 415 Steel , fy = 415 N/mm2
Load from H pier = 3146.04 kN (STAAD.Pro result)
Load from circular column = 4013.39 kN (STAAD.Pro result)
No: of piles n = 8
55. 55
Step 2: Preliminary Dimensions
Fig 3.25: Pile cap with Piles
Fig 3.26: Pile Cap loading
56. 56
Diameter of pile, d = 1200 mm
Here, a pile cap is designed for a group of eight piles of diameter 1200 mm.
As per IS 2911, Cl 6.6.1,
Spacing between 2 piles = 2d to 2.5d = 2400 mm to 3000mm
Breadth of pile cap = 2400 + (2 × 1200) + (2 × 150)
= 5100 mm
Depth of pile cap = Development length of column bar + Cover+ Projection
As per Table 65 of SP 16: 1980,
For 25 mm dia bars, development length, Ldt = 940 mm
Assuming a clear cover of 150 mm and a 50 mm projection of pile in to the cap concrete
Depth of pile cap = 940 + 50 + 150 + 150 = 1290 mm
Provide a depth of 1800 mm for the pile cap
Length of pile cap = (4 x 1200) + (3 x 2800) + (2 x 150) = 13500mm
Hence, size of pile cap = 13500 mm x 5100 mm x 1800 mm
Step 3: Loads and bending moments
Total axial loads on H pier and circular column = 3146.04 + 4013.39 = 7159.43 kN
Considering the pile cap to be rigid,
Load taken by each pile =
7159.43
8
= 894.93 kN
Bending moment in pile cap at the face of the circular column in Y direction,
My1 = 7159.43 x 3390 = 24270.46 kNm
57. 57
Bending moment in pile cap at the face of the H pier in Y direction,
My2 = 7159.43 x 3410 = 24413.65 kNm
Bending moment in pile cap at the face of the circular column in X direction,
Mx1 = 7159.43 x (1700/2) = 6085.515 kNm
Bending moment in pile cap at the face of the H pier in X direction,
Mx2 = 7159.43 x (2000/2) = 7159.43 kNm
Step 4: Reinforcement due to truss action
If ,
Shear span
Effective depth
< 0.6 , Truss action occurs
Shear span
Effective depth
> 0.6 , Flexural action occurs
Shear span av = 7.16 m for end pile
Effective depth, d = 1800 - 50 - 150 -
25
2
= 1587.5 mm
Shear span
Effective depth
= 4.5 > 0.6, hence flexural action occurs
Step 5: Main reinforcements due to bending
Near circular column,
Mu
bd 2 =
24270.46 ×106
13500 × 1587.52 = 0.713 N/mm2
Near H pier,
Mu
bd 2 =
24413.65 ×106
13500 × 1587.52 = 0.717 N/mm2
From Table - 4 of SP 16:1980,
For Mu/bd2
= 0.717 N/mm2
, fy = 415 N/mm2
,
58. 58
Pt = 0.2047
Min Pt required = 0.2
Ast=
Ptbd
100
=
0.2047 × 5100 × 1587.5
100
= 16573.02 mm2
Ast
2
=
16573.02
2
= 8286.51 mm2
Provide 17 no: of 25mmΦ bars
Area of steel provided = 8345 mm2
Step 6: Check for shear reinforcement
Maximum shear force in pile cap = 8 x 894.93 = 7159.43 kN
τv = Vu
bd = 7159.43 × 103
5100× 1587.5 = 0.884 N/mm2
Ast = 8345 mm2
100 Ast
bd
=
100 ×1964
1700× 1587.5
= 0.0727
From Table 19 of IS 456, corresponding to M 30 concrete:
τc = 0.29 N/mm2
Since τv>τc , shear reinforcement needs to be provided
Vus = Vu - τc b d
= 7159.43 x 103
- 0.29 x (5100 x 1587.5) = 4811.517 kN
Assume 16mmΦ bars
Area of shear reinforcement effective in shear = 202 mm2
Spacing of bars, Sv =
0.87 dfyAsv
Vus =
0.87 × 1587.5 × 415 × 202
4811.517 ×103 = 24.06mm
Hence, provide 16 mm Φ bars at 150mm c/c
59. 59
Fig 3.27 : Reinforcement Details of Pile Cap
5.1.12. APRON AND CUTOFF
The third stage of regulator will have 17 spans of 13.22m centre to centre distance.The critical
differential heads under which the regulator has to be operated is as follows:
Step 1: Water head
When the shutters are down,
Water on paddy fields side = 1.22m
Water on sea side = -0.46m
Maximum differential head = 1.22+0.46 = 1.68m
In the reverse direction (shutters are open),
Water on sea side = 0.85m
Water on land side = - 0.61m
Max differential head = 0.85 - 0.61 = 0.24m
60. 60
So design the foundation apron for a critical differential head of 1.8m on either direction.
Step 2: Discharge based on maximum flood level
Maximum probable monsoon flood is calculated as follows:
Ryve‘s formula, Q = CA(2/3)
=8.45 x1124(2/3)
=913.5 cumec
Q = Discharge in cumecs
C = Flood coefficient (obtained from Table 1)
Table 3.10: Flood coefficient in Ryve‘s formula
Location of catchment C
1. Areas within 24km from coast 6.75
2. Areas within 24km to 161km from Coast 8.45
3. Limited areas near hills 10.1
Discharge through Thottapilly spillway = 400cumec (From records)
Net discharge Q‘ = Total discharge – Discharge through spillway
= 913.5- 400
= 513.5 cumec
Length of waterway L = Number of spans x span = 17 x 13.22 = 224.74m
Discharge per unit width, q =
Q′
L
=
513.5
224.74
= 2.28 cumec/m
Scour depth R = [1.35(
q2
f
)(1/3)
]x 1.9
= [1.35(
2.282
1
)(1/3)
]x 1.9
=4.44 m
61. 61
f = Silt factor [IS 6966(Part 1):1989]
Discharge over weir,
Q = CLH(3/2)
From the above equation,
H(3/2)
=
Q
LC
where, H = Total head causing the flow in m
C =1.3 (Assume 95% drowning) [ Graph- Page 5 of IS 6966(Part 1): 1989]
So,
H= [
2.28
1.3
](2/3)
= 1.45m
Depth of cutoff = R – H = 4.44 - 1.45 = 3m
Step 3: Discharge based on tidal variation
Tidal velocity measured = 0.8m/s
Maximum tidal level = 0.85m
Sill level = -4.27m
Maximum depth of water during high tide = 0.85 - (- 4.27) = 5.12m
Discharge per unit width q = 5.12 x 0.8 = 4.096cumec
Scour depth R = [1.35(
q2
f
)(1/3)
]x 1.9 = 6.05m
Depth of cutoff = 6.05 - 5.12 = 0.93m
Therefore take depth of cutoff = 3.0 m
Step 4: Exit gradient
From IS 6966( Part 1):1989 , Cl. 3.4.1
GE = (S-1)(1-n)
= (2.56 - 1)(1 - 0.4)
= 0.936 x 7
62. 62
= 6.55
where,
S = Specific gravity
n = Porosity
Safe exit gradient =
1
GE
=
1
6.55
≈
1
7
Step 5: Dimensions of apron
We have ,
GE =
H
d
x
1
π √λ
1
7
=
1.8
2.7
x
1
π √λ
i.e,
1
π √λ
= 0.21
which implies α = 4.0
From Khosla‘s curve
α =
b
d
= 4.0
b = 4.0 x 2.7 ≈ 11 m.
i.e, Length of apron = 11 m.
From Khosla‘s curve,
1
α
= 0.2
ΦD = 28
ΦE=40
ΦC1 = 100 - ΦE = 100- 40 = 60%
ΦD1 = 100 - ΦD = 100- 28 = 72%
a) Correction for mutual interference
63. 63
C = 19 x
D
b’
x
d +D
b’
b‘ =11 m
D =2.7m
d = 2.7m
C = 19 x
2.7
11
x
2.7 +2.7
11
= 4.6
b) Correction for floor thickness
Assume thickness = 1 m
Depth = 3 m
(ΦD1 − ΦC1)
d x t
=
(72−60)
2.7 x 1
= 4.44
Correction for Φc = 60 + 4.6 + 4.44 = 69.04 %
Residual Pressure head at C1
Pa = Φc1 x Hs
= 0.6904 x 1.8
= 1.24 m
Thickness of apron t,
Pa
(G−1)
=
1.24
(2.56−1)
= 0.7966 m ≈ 80 cm
Step 6: Protection works
Block protection (IS 6966 (Part 1): 1989) Cl. 20.2
Pervious block protection shall be provided just beyond both ends of impervious floor which
shall comprise of cement concrete blocks of adequate size laid over a suitably designed inverted
filter for the grade of material in the river bed. The cement concrete block shall generally be not
smaller than 1500 x 1500 x 900mm size to be laid with gaps of 75mm width, packed with gravel.
64. 64
The length of block protection shall be approximately equal to 1.5D where this length is
substantial, block protection with inverted filter may be provided in part of the length and block
protection only with loose stone spawls in remaining length.
Fig 3.28: Apron And Cutoff Wall
65. 65
CHAPTER 4
CONCLUSIONS
The Thanneermukkom Bund, which is considered as the largest mud regulator in the country,
was built as part of the Kuttanad Development Scheme. The huge construction is built across
Lake Vembanad, the largest lake of the State and divides the lake into two parts. It also connects
the two districts of Kottayam and Alappuzha through Thanneermukkom and Vechoor villages on
either side of the bund. The bund, which has a length of 1400 meters, is situated between 76° 23′
and 76° 25′E and 9° 40′B latitude. This includes a 470m-long reclaimed portion in the middle.
The construction began in 1958 but completion of the project took many years. The western and
eastern portions of the bund, with a length of 470 meters each, have 31 shutters on each side. The
construction of these parts was completed in 1967. In 1977, the government started building the
middle part. Land had to be reclaimed from the backwaters to complete the construction. The
shutters were connected on either side to control the entry of salt water. The parts of the bund
which are essential for the efficient working of the bund which have been successfully designed
are:
1. Deck slab
2. Cantilever slab
3. Longitudinal girder
4. Cross girder
5. Operating platform
6. Bearings
7. Pedestals
8. Pier
9. Pier cap
10. Pile
11. Pile cap
12. Apron and Cutoff
66. 66
CHAPTER 5
SCOPE FOR FUTURE WORK
Analysis and designs of different components are done based on general procedures and software
available. But various limitations, as listed below, are present which could be overcome in
future, if proper facilities become available.
1. Pier analysis in STAAD.Pro encountered limitations in considering sloshing effect and
earthquake effect.
2. As the time for completion of project work is limited, only a single span is considered for
analysis and design based on the single span is provided for the whole structure.
67. 67
REFERENCES
1. Amit Saxena, Dr.Savita Maru. (2013), ―Comparative Study of the Analysis and Design of T-
Beam Girder and Box Girder Superstructure‖, IJREAT, International Journal of Research in
Engineering & Advanced Technology
2. Arnold W. Hendry , Leslie G. Jaeger. (1955), “The Load Distribution in Interconnected
Bridge Girders with Special Reference to Continuous Beams”, Publisher- IABSE
publications, Memoires AIPC,IVBH Abhandlungen
3. H P Santhosh, Dr. H M Rajashekhara Swamy, Dr. D L Prabhakara. (2014), ―Construction Of
Cofferdam -A Case Study‖, IOSR Journal of Mechanical and Civil Engineering (IOSR-
JMCE)
4. IS 2911-1-1 (2010): Design and Construction of Pile Foundations — Code of Practice, Part
1: Concrete Piles
5. IS 456: 2000, Plain and Reinforced concrete
6. Kavitha.N, Jaya kumari.R, Jeeva.K, Bavithra.K, Kokila.K. (2015), ―Analysis and Design of
Flyover‖, National Conference on Research Advances in Communication, Computation,
Electrical Science and Structures (NCRACCESS-2015)
7. Manjeetkumar M Nagarmunnoli and S V Itti. (2014), ―Effect of Deck Thickness in RCC T-
beam Bridge‖, International Journal of Structural and Civil Engineering Research, ISSN
2319 – 6009, Vol. 3, No. 1
8. M.G. Kalyanshetti, C.V. Alkunte. (2012), ―Study on Effectiveness of IRC Live load on
R.C.R Bridge Pier‖, International Journal of Advanced Technology in Civil Engineering,
ISSN: 2231 –5721, Volume-1, Issue-3,4
9. Mundzir Hasan Basri. (2001), ―Two New Methods for Optimal Design of Subsurface
Barrier to Control Seawater Intrusion‖
10. Nan Hu, Gonglian Dai. (2010), ―The Comparative Study of Portal-Frame Pier for High-
Speed Railway‖
11. N. Krishna Raju. (2013), “Advanced Reinforced Concrete Design (IS:456-2000)”, Publisher-
CBS Publisher, Edition-2
68. 68
12. Parvin Eghbali, Amir Ahmad Dehghani, Hadi Arvanaghi, Maryam Menazadeh. (2013), ―The
Effect of Geometric Parameters and Foundation Depth on Scour Pattern around Bridge Pier‖,
Journal of Civil Engineering and Urbanism, Volume 3, Issue 4: 156-163
13. P.N.Modi. (2008), “Irrigation Water Resources and Water Power Engineering”, Publisher-
Standard Book House Delhi, Edition-7
14. Praful N K , Balaso Hanumant. (2015), ―Comparative Analysis of T-Beam Bridge by
Rational Method and Staad Pro‖, International Journal of Engineering Sciences & Research
Technology
15. SP 16 : 1980
16. S. Ponnuswamy (2007), “Bridge Engineering”, Publisher- Mcgraw Hill Education, Edition-2
17. ―Study for modernizing the Thanneermukkom bund and Thottappally spillway for efficient
water management in Kuttanad region, Kerala‖, Indian Institute of Technology,
Madras,Chennai and Centre for Water Resources Development and Management,
Kozhikode, Kerala (2011)
18. Victor, Johnson D. (2007), “Essentials of Bridge Engineering”, Publisher- Oxibh, Edition-6
71. 71
TYPE CONCRETE
STRENGTH FCU 30000
END DEFINE MATERIAL
MEMBER PROPERTY AMERICAN
10 20 30 40 50 PRIS YD 1.7
1 11 21 31 41 PRIS YD 3.5 ZD 1.93
2 12 22 32 42 PRIS YD 2 ZD 1.8
3 TO 9 13 TO 19 23 TO 29 33 TO 39 43 TO 49 PRIS YD 1.5 ZD 2
51 TO 66 PRIS YD 1.13 ZD 0.625
CONSTANTS
MATERIAL CONCRETE ALL
SUPPORTS
1 7 14 20 25 31 36 42 47 53 FIXED
LOAD 1 LOADTYPE None TITLE SUPER STRUCTURE DEAD LOAD
MEMBER LOAD
51 TO 66 UNI GY -39.25 0 13.22
LOAD 2 LOADTYPE Dead TITLE SELF WEIGHT OF PIERS
SELFWEIGHT Y -1
LOAD 3 LOADTYPE Live REDUCIBLE TITLE LIVE LOAD
72. 72
MEMBER LOAD
51 TO 54 UNI GY -17.4 0 6.61
51 TO 54 UNI GY -15.39 6.61 13.22
59 TO 62 UNI GY -18.169 0 6.61
59 TO 62 UNI GY -20.48 6.61 13.22
55 TO 58 UNI GY -25.75 0 6.61
55 TO 58 UNI GY -29.246 6.61 13.22
LOAD 4 LOADTYPE Live REDUCIBLE TITLE OPERATING PLATFORM
JOINT LOAD
2 15 26 37 48 FY -90.06
LOAD 5 LOADTYPE Live REDUCIBLE TITLE WATER FORCE
MEMBER LOAD
1 11 21 31 41 CON GX 32.67 3.33
LOAD 6 LOADTYPE Live REDUCIBLE TITLE WATER@SHUTTER
MEMBER LOAD
1 11 21 31 41 CON GX 235.44 4.166
LOAD 7 LOADTYPE Traffic TITLE BREAKING FORCE
MEMBER LOAD
6 16 26 36 46 CON Z 140 1.325
73. 73
LOAD 8 LOADTYPE Fluids TITLE BOUYANCY ON PIER
JOINT LOAD
2 15 26 37 48 FY 370.7
5 18 29 40 51 FY 124.8
LOAD 9 LOADTYPE Accidental TITLE IMPACT FOR LL
MEMBER LOAD
51 TO 54 UNI GY -7.67 0 13.22
55 TO 58 UNI GY -12.86 0 13.22
59 TO 62 UNI GY -9.05 0 13.22
LOAD 12 LOADTYPE Wind TITLE WIND LOAD
JOINT LOAD
3 16 27 38 49 FX 44.41
LOAD COMB 10 FLOOD CONDITION
1 1.0 2 1.0 3 1.0 4 1.0 5 1.0 6 1.0 7 1.0 8 1.0 9 1.0 12 1.0
LOAD COMB 11 DRY CONDITION
1 1.0 2 1.0 3 1.0 4 1.0 7 1.0 9 1.0 12 1.0
PERFORM ANALYSIS PRINT ALL
START CONCRETE DESIGN
CODE INDIAN
74. 74
CONCRETE TAKE
FC 30000 ALL
FYMAIN 415000 ALL
FYSEC 415000 ALL
DESIGN COLUMN 21 22 30
END CONCRETE DESIGN
PERFORM ANALYSIS PRINT ALL
FINISH
STAAD EDITOR – PIER CAP
STAAD SPACE
START JOB INFORMATION
ENGINEER DATE 22-Feb-16
END JOB INFORMATION
INPUT WIDTH 79
UNIT METER KN
JOINT COORDINATES
25 0 0 26.44; 26 0 10 26.44; 27 0 6 26.44; 28 -1 10 26.44; 29 9.65 10 26.44;
31 9.65 0 26.44; 32 12 10 26.44;
75. 75
MEMBER INCIDENCES
21 25 27; 22 27 26; 30 29 31; 31 28 26; 32 26 29; 33 29 32;
DEFINE MATERIAL START
ISOTROPIC CONCRETE
E 2.17185e+007
POISSON 0.17
DENSITY 23.5616
ALPHA 1e-005
DAMP 0.05
TYPE CONCRETE
STRENGTH FCU 30000
END DEFINE MATERIAL
MEMBER PROPERTY AMERICAN
30 PRIS YD 1.7
21 PRIS YD 3.5 ZD 1.93
22 PRIS YD 2 ZD 1.8
31 TO 33 PRIS YD 1.5 ZD 2
CONSTANTS
MATERIAL CONCRETE ALL
76. 76
SUPPORTS
25 31 FIXED
LOAD 1 LOADTYPE None TITLE SUPER STRUCTURE DEAD LOAD
MEMBER LOAD
32 CON Y -626.44 3.226
32 CON Y -626.44 5.876
32 CON Y -626.44 8.526
33 CON Y -626.44 0.825
LOAD 3 LOADTYPE Live REDUCIBLE TITLE LIVE LOAD
MEMBER LOAD
32 CON Y -216.74 3.226
32 CON Y -363.52 5.876
32 CON Y -255.64 8.526
LOAD 4 LOADTYPE Live REDUCIBLE TITLE OPERATING PLATFORM
JOINT LOAD
26 FY -90.06
LOAD 2 LOADTYPE None TITLE PIERCAP SELF WEIGHT
SELFWEIGHT Y -1 LIST 31 TO 33
LOAD COMB 5 COMBINATION LOAD CASE 5
77. 77
1 1.0 3 1.0 4 1.0 2 1.0
PERFORM ANALYSIS PRINT ALL
START CONCRETE DESIGN
CODE INDIAN
FYSEC 415000 MEMB 21 22 30
CONCRETE TAKE
FYMAIN 415000 ALL
FYSEC 415000 ALL
FC 30000 ALL
DESIGN BEAM 31 TO 33
END CONCRETE DESIGN
PERFORM ANALYSIS PRINT ALL
FINISH
