Design of continuous flushing settling basin and powerhouse

KATHMANDU UNIVERSITY
SCHOOL OF ENGINEERING
Department of Civil and Geomatics Engineering
FINAL YEAR PROJECT REPORT
On
Design of Continuous Flushing Settling Basin and Powerhouse
(Thapa khola Hydroelectric Project)
A final year project report submitted in partial fulfillment of the requirements for the Bachelor’s
Degree in Civil Engineering (Specialization in Hydropower Engineering)
Prepared By:
Tek Kathayat (016011-13)
Raj Kumar KC (016012-13)
Sailaja Poudel (016023-13)
Saroj Sapkota (016029-13)
Gokarna Sijwal (016040-13)
Project Supervisor:
Er. Sudip Jha
Submitted To:
Dhulikhel, Kavre
September, 2017

DECLARATION
We, Tek Kathayat, Raj Kumar KC, Sailaja Poudel, Saroj Sapkota and Gokarna Sijwal
declare that this project work titled “DESIGN OF CONTINUOUS FLUSHING SETTLING
BASIN AND POWERHOUSE”, submitted in partial fulfillment of Bachelor’s degree in Civil
Engineering (Specialization in Hydropower) to the Department of Civil and Geomatics
Engineering, Kathmandu University, during the academic year 2017, is a genuine work done
originally by us under the supervision of Er. Sudip Jha. Any help from other people has been
mentioned in the acknowledgement and the references, if any, have been listed in the reference
section. The report or any part of it has not been published or submitted for academic award in
any other universities or institutions. Any literature, data or works done by people other than the
group members and cited within this report have been given due acknowledgement and are listed
in the reference section.
Tek Kathayat (016011-13) ……………………………
Raj Kumar KC (016012-13) …………………………….
Sailaja Poudel (016023-13) ……………………………...
Saroj Sapkota (016029-13) ……………………………..
Gokarna Sijwal (016040-13) …………………………….
School of Engineering
Kathmandu University
Dhulikhel, Kavre, Nepal
September, 2017

CERTIFICATION
FINAL YEAR PROJECT REPORT
ON
Design of Continuous Flushing Settling Basin and Powerhouse of Thapa khola
Hydroelectric Project, Mustang
By:
Tek Kathayat (016011-13)
Raj Kumar KC (016012-13)
Sailaja Poudel (016023-13)
Saroj Sapkota (016029-13)
Gokarna Sijwal (016040-13)
Approved By:
1. Head of Department
(Signature) (Name) (Date)
2. Project Supervisor
3. External Examiner

i
ACKNOWLEDGEMENT
This study was carried out at the Department of Civil & Geomatics Engineering, Kathmandu
University, in Kavre, Nepal. The success and the final outcome of the project required a lot of
guidance and assistance from many people and we are extremely fortunate to have got this all
along the completion of the project work.
We would like to thank our acting Head of Department, Assistant Prof. Prachand Man
Pradhan, Department of Civil and Geomatics Engineering, Kathmandu University, for
providing us with the opportunity to have professional experience through this final year project.
We would like to thank Department of Civil and Geomatics engineering for providing us lab,
equipped with all necessary arrangements and facilities, to carry out our office work.
We would like to express our heartfelt gratitude to our supervisor Er. Sudip Jha, and Co-
Supervisor Er. Sanjay Dhonju Shrestha for their constant availability, encouragements, vision,
and motivation towards us regarding the project, his valuable suggestions and sharing his ideas
and pushing us to out potential and making this project what it is now.
We would also like to acknowledge Associate Dean of School of Engineering Prof. Dr. Ing.
Ramesh Kumar Maskey, Er. Shyam Sundar Khadka, for providing us with regular feedback
on our works and supporting us throughout the Project. We would like to thank Er. Devendra
Basnet and Er. Diwakar Acharya and Sudip Khadka (project manager of Thapakhola
HEP) for their constant support and encouragement throughout the duration of the project.
We are extremely thankful to Mr. Shivaji Dhakal for his constant co-operation.
We would also like to thank Mr. Chetan Susling Magar for assisting us in our lab works, and
providing us with the necessary equipment.
Last but not the least; we are grateful to our parents, family members and friends for their
encouragement and support.

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ABSTRACT
Continuous flushing settling basin is a demand for hydropower which are going to be built in
river having high sediment concentration. The main aim of continuous flushing system is to
regulate continuous production of power with continuous flushing of sand collected. While doing
this the power production would not be interrupted and becomes economical to power plant.
There are various ways of flushing the settling basin continuously. But we discussed about
Hydro suction in detail.
Building life is connected with human life in context of earthquake prone zone. So it is important
to design building resistant to those hazards. The aim of reinforced concrete design is to achieve
an acceptable probability that structures being designed will perform satisfactorily during their
intended life. With an appropriate degree of safety, the buildings should sustain the entire load
and the deformation of normal construction and have adequate durability and resistance to fire.
Particularly, in case of powerhouse, the building should also be able to withstand the loads due to
the turbine, penstock, generators and the vibration produced.
The powerhouse at Thapakhola Hydroelectric project is designed as a special Moment Resisting
Frame Structure. This project mainly focuses on seismic analysis and structural design of
different members components of a powerhouse building and economic analysis of different
sections. The project aims at obtaining a design output by limit state method. With the
involvement of all these facts, manual design of the members will have a limited approach. Thus,
the necessity of structural analysis software has been reflected. SAP2000 v.14 has been used for
analysis the members and components. Autodesk AutoCAD along with MS office like MS word,
MS excel and MS PowerPoint also been used for the successful completion of our project.
The vertical loads including dead load and live loads along with seismic loads were defined and
the resulting 14 combinations of these loads were used to analyze the structure. The lateral loads
were calculated by seismic coefficient method adopting IS 1893:2002. Structural members and
elements are designed by limit state method considering limit states of collapse and
serviceability adopting IS 456:2000. The structural members and elements are also designed
considering the ductility requirements as per IS 13920:1993. The design aids and handbooks SP-
16 and SP-34 of Indian Standard are also followed.

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SYMBOLS AND ABBREVIATIONS
SYMBOLS
D Overall depth of the section
d Effective depth of the section
B Width of the section
H Height of building
I Importance Factor
K Lateral Stiffness
L Unsupported length or clear span of element
E Structure eccentricity
P Axial load on the element
Q Design lateral force
R Response reduction factor
T Torsional moment due to lateral load
V Shear force
Wi Seismic weight of floor
Z Zone factor
Ab Area of each bar
Ac Area of concrete
Ag Gross area of section
Ah Horizontal seismic coefficient
Asc Area of compression reinforcement
Ast Area of tension reinforcement
Asv Area of vertical stirrup
bf Width of flange section
bw Width of web in T or L-section
Df Depth of flange in T or L-section
fck Characteristics compressive strength of concrete
fy Characteristic yield strength of steel
Ix, Iy Moment of inertia about X and Y axis respectively

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Lo Distance between point of inflection
Leff Effective length of the element
Ld Development length
Lx, Ly Span of slab in the shorter and longer direction respectively
Mu Factored moment, Design moment for limit state design
Mu,lim Limiting moment of resistance
Mux, Muy Factored moment about X-axis and Y-axis respectively
Mux1, Muy1 Maximum uniaxial moment capacity of the section with axial load,
about X and Y axis respectively.
Vus Strength of shear reinforcement in limit state design
Vb Total base shear
Sv Spacing of stirrup
X u,max Maximum depth of neutral axis in limit state of design
τc,max Allowable maximum shear stress in concrete with shear reinforcement
τc Shear stress
αx, αy Bending moment coefficient for slab about X-axis and Y-axis respectively
ϒm Partial safety factor
Ϭ Soil bearing pressure
Φ Diameter of bar

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ABBREVIATIONS
HSRS Hydrosuction Sediments Removal System
RCC Reinforced cement Concrete
PCC Plain Cement Concrete
IS Indian Standard
CM Center of Mass
CR Center of Rigidity
DL Dead Load
EQ Earthquake Load
LL Live Load
NBC Nepal National Building Code
IRR Internal Rate of Return
NPV Net Present Value

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List of Figures
Figure 1: Distribution of loads from slab onto the supporting beam............................................ 11
Figure 2: Hydrosuction Sediment Removal System (HSRS) ...................................................... 16
Figure 3: Flow duration curve and particle size distribution curve .............................................. 28
Figure 4 : Operating cost for Conventional and Hydrosuction flushing system........................... 45
Figure 5: Corbel Dimension........................................................................................................ 123
Figure 6: Truss analogy in corbel ............................................................................................... 124
Figure 7: Corbel Detailing .......................................................................................................... 126
Figure 8: Plan of staircase........................................................................................................... 129
Figure 9 : Reinforcement in Raft Foundation............................................................................. 144
Figure 10: Pratt truss at powerhouse........................................................................................... 156
Figure 11: Half Pratt truss at control room ................................................................................. 159
Figure 12: D.L + L.L combination in power house truss............................................................ 162
Figure 13: DL+LL combination in Control room truss .............................................................. 164
Figure 14: 3d Modeling of Powerhouse and Control Room...................................................... 178
Figure 15: Bending moment alone the grids............................................................................... 184
Figure 16: Axial force along the grids........................................................................................ 185
Figure 17: Axial force along the grids........................................................................................ 186
Figure 18: Sway along the grids ................................................................................................. 188

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List of Tables
Table 1: Installation Material Cost of the HSRS for the one unit................................................. 26
Table 2: Monthly discharge and power generation of Thapa Khola HEP.................................... 27
Table 3: Result of Suspended sediment concentration analyses (Thapa khola) ........................... 34
Table 4 : loss due to operation of conventional settling basin...................................................... 41
Table 5: Dead load calculation of control room ........................................................................... 54
Table 6: Calculation of total load above the mentioned location of Control Room..................... 62
Table 7: Calculation of Seismic weight of Control Room............................................................ 63
Table 8: Calculation of Natural time period in each direction...................................................... 64
Table 9: Calculation of Base Shear............................................................................................... 67
Table 10: Lateral Load distribution at a joint above ground level of Control Room ................... 68
Table 11: Vertical load calculation of machine hall ..................................................................... 69
Table 12: Seismic load calculation ............................................................................................... 76
Table 13: Bending Moment Coefficients (Clause 22.5.1) ............................................................ 80
Table 14: Shear Force Coefficients (Clause 22.5.1 and 22.5.2) ................................................... 81
Table 15: Base reaction of the column ....................................................................................... 179
Table 16: Storey drift calculation ............................................................................................... 187
Table 17: Estimation table .......................................................................................................... 190
Table 18: Rate analysis table ...................................................................................................... 201
Table 19: Summary of unit rates of civil works ......................................................................... 206
Table 20: Abstract of estimated cost........................................................................................... 207

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Table of Contents
ACKNOWLEDGEMENT............................................................................................................... i
ABSTRACT....................................................................................................................................ii
SYMBOLS AND ABBREVIATIONS..........................................................................................iii
SYMBOLS.....................................................................................................................................iii
ABBREVIATIONS ........................................................................................................................ v
List of Figures................................................................................................................................ vi
List of Tables ................................................................................................................................vii
Chapter 1......................................................................................................................................... 1
Introduction..................................................................................................................................... 1
1.1 Project Background............................................................................................................... 1
1.2 Features of Powerhouse........................................................................................................ 2
1.3 Features of HSRS system...................................................................................................... 2
1.3 Objectives of the Project....................................................................................................... 2
1.4 Scopes of the Project............................................................................................................. 3
1.5 Limitations of the project...................................................................................................... 3
1.6 Description of the Project ..................................................................................................... 3
1.7 Salient features...................................................................................................................... 4
Chapter 2......................................................................................................................................... 6
Methodology................................................................................................................................... 6
2.1 Methodology (part I Hydrosuction Sediment Removal system) .......................................... 6
2.1.1 Literature Review........................................................................................................... 6
2.1.2 Data Collection .............................................................................................................. 8
2.1.3 Consultation................................................................................................................... 8
2.1.4 Office Works.................................................................................................................. 8
2.2 Methodology (part II Structure Analysis of Powerhouse).................................................... 9
2.2.1 Literature Review........................................................................................................... 9
2.2.2 Data Collection ............................................................................................................ 13
2.2.3 Consultation................................................................................................................. 14
2.2.4 Office Works................................................................................................................ 14
Chapter 3....................................................................................................................................... 15
Continuous Flushing Settling Basin.............................................................................................. 15

ix
3.1 Continuous Flushing System .............................................................................................. 15
3.1.2 Serpent Sediment-Sluicing System.............................................................................. 15
3.1.3 Hydrosuction sediment removal system ...................................................................... 16
3.2 Benefits of HSRS over Conventional Method.................................................................... 17
3.3 HSRS components .............................................................................................................. 18
3.3.1 Intake............................................................................................................................ 18
3.3.2 Pipeline ........................................................................................................................ 18
3.3.3 Outlet............................................................................................................................ 18
3.3.4 Valve............................................................................................................................ 18
3.3.5 Mechanical Dredging................................................................................................... 19
3.4 Hydrosuction Dredging Case Histories .............................................................................. 19
3.5 Hydraulic Principles............................................................................................................ 20
3.5.1 Head-loss Expression................................................................................................... 20
3.5.2 Sediment-Transport Expression................................................................................... 22
3.6 Pipeline Design Steps ......................................................................................................... 22
3.8 Dimension of the Hydrosuction sediment Removal system............................................... 26
3.9 Design of Hydrosuction Sediment Removal system .......................................................... 27
3.10 Loss due to operation of conventional settling Basin ....................................................... 41
3.11 Loss due to operation of Hydrosuction sediment Removal System ................................. 44
Chapter 4....................................................................................................................................... 46
Preliminary Design ....................................................................................................................... 46
4.1 Preliminary design of slab .................................................................................................. 46
4.1.1 Control room floor slab:............................................................................................... 46
4.2 Design of Beam................................................................................................................... 47
4.2.1 Control room................................................................................................................ 47
4.2.2 Machine hall................................................................................................................. 48
4.2 Design of column................................................................................................................ 49
4.3 Preliminary Design of Penstock.......................................................................................... 50
Chapter 5....................................................................................................................................... 51
Load Calculation........................................................................................................................... 51
5.1 Vertical Load Calculations ................................................................................................. 51
5.1.1 Vertical load calculations of control room................................................................... 54

x
5.2 lateral load calculations....................................................................................................... 62
5.2.1 Seismic Weight............................................................................................................ 62
5.2.3 Calculation of Fundamental Natural Period of Vibration of the Building................... 63
5.2.4 Base Shear Calculation ................................................................................................ 65
5.2.5 Lateral Load Distribution and Storey Shear ................................................................ 67
5.3 Vertical Load Calculations of Machine Hall .................................................................... 69
5.4 Load Combination .............................................................................................................. 78
Chapter 6....................................................................................................................................... 79
Structural Design .......................................................................................................................... 79
6.1. Introduction........................................................................................................................ 79
6.2 Limit State Design: ............................................................................................................. 79
6.3 Design of structural elements.............................................................................................. 80
6.3.1 Design of Beams.......................................................................................................... 80
6.3.2 Design of Two-way Slabs.......................................................................................... 103
6.3.3 Design of Column...................................................................................................... 112
6.3.4 Corbels....................................................................................................................... 123
6.3.5 Design of Staircase .................................................................................................... 129
6.3.6 Design of Raft Foundation......................................................................................... 135
6.3.7 Shear or Flexural walls .............................................................................................. 147
6.3.8 Truss Design ............................................................................................................ 151
6.4 Centre of Mass and Centre of Rigidity ............................................................................. 171
Chapter 7..................................................................................................................................... 174
Structural Analysis...................................................................................................................... 174
7.1 Salient features of sap 2000.............................................................................................. 174
7.2 Input .................................................................................................................................. 175
7.3 Output ............................................................................................................................... 177
7.3.1 Grid for Column Location: ........................................................................................ 177
7.3.2 Base Reaction of the Column .................................................................................... 179
7.3.3 Reinforcement from SAP........................................................................................... 180
7.3.4 Bending moment........................................................................................................ 184
7.3.5 Sher force Diagram.................................................................................................... 185
7.3.6 Axial force Diagram .................................................................................................. 186

xi
7.3.7 Storey Drift Calculation............................................................................................. 187
Chapter 8..................................................................................................................................... 189
Cost Estimation........................................................................................................................... 189
8.1 Introduction....................................................................................................................... 189
8.2 Detailed estimate............................................................................................................... 189
8.3 Estimation Table............................................................................................................... 190
Chapter 9..................................................................................................................................... 201
Rate analysis ............................................................................................................................... 201
9.1 Introduction....................................................................................................................... 201
9.2 Brickwork in (1:4) Cement Sand Mortar.......................................................................... 201
9.3.2 5mm (1:4) Cement Sand Plaster .................................................................................... 202
9.4 Open Cut Excavation........................................................................................................ 202
9.5 PCC Works (1:1.5:3) ........................................................................................................ 203
9.6 M25 Concrete Works........................................................................................................ 204
9.7 Formwork for RCC........................................................................................................... 204
9.8 Reinforcement Works for Fe500 Steel ............................................................................. 205
9.9 Summary of Unit Rates of Civil Works............................................................................ 206
9.10 Abstract of Estimated Cost ............................................................................................. 207
Chapter 10................................................................................................................................... 208
Conclusion .................................................................................................................................. 208
References................................................................................................................................... 209

1
Chapter 1
Introduction
1.1 Project Background
Our project is entitled “Design of Continuous Flushing Settling Basin and Power House” which
includes design and dimensioning of HSRS system, structural analysis, design, detailing and
sizing of powerhouse. The HSRS system and powerhouse that we have designed is for Thapa
Khola Hydro Electric Project. It is run-of-river schemed project with an installed capacity of 13.6
MW.
The HSRS system has been proposed for the settling basin of Thapa Khola HEP to minimize the
loss in discharge due to conventional operation. This proposed system then add more discharge
for power generation. The design and the dimensioning of HSRS system is based upon the
journal that was published by Hotchkiss.
The design of powerhouse basically considers the dead load, live load, seismic load and wind
load. For the design of powerhouse NBC and IS 13920, IS 456:2000, IS 875-part 2 and for
seismic IS 1893:2002 has been referred. Structural analysis deals with the analyzing the internal
forces in the members of the structures. While structural analysis deals with sizing various
members of the structures to resist the internal force to which they are subjected in course of
their lifecycle.
This project has become a very good platform to transform our theoretical academic knowledge
into practical real world environment. Moreover the final year project helps to design the
structure in terms of safety, economy, stability and efficiency.

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1.2 Features of Powerhouse
 Structural System: RCC Special Moment Resisting Framed Structure System
 Building type: Industrial (heavy duty)
 Plinth area covered: 486.86 m2
 Type of foundation: Mat foundation, Machine Foundation
 No of storey: 1 storey for turbine and generator arrangement, 3 for control bay and
storage
 Type of soil: Rocky soil
 Height of the building : 20.23m
 Height of shear wall: 4.4m and 1.22m
 Tailrace canal and outlet structures
 Seismic zone: V ( from Indian standard)
1.3 Features of HSRS system
 Intake
 Pipeline, Water jet, Pumping arrangement
 Outlet
 Value
 Mechanical dredging
1.3 Objectives of the Project
The main objectives of our final year project aims at demonstrating the engineering and
managerial skills of the student for diagnosis and management of engineering projects.
Following are the points that reflect the objective of our project.
 Design of HSRS system for settling basin of Thapa Khola HEP
 Economic analysis and comparison of HSRS system against conventional operation
system
 Design and structural analysis of different components of powerhouse like beam, column,
foundation etc.
 Quantity and cost estimate of powerhouse

3
1.4 Scopes of the Project
 Lateral load considered is earthquake load and calculated by seismic coefficient method
 Design and detailing of typical structural element
1. Slab
2. Beam
3. Column
4. Staircase
5. Foundation
6. Corbel
7. Gantry girder
8. Shear wall
 Design of HSRS system for settling basin
 3-dimensional analysis is done using software SAP-2000
1.5 Limitations of the project
 Seismic and gravity loads are considered but no winds and snow loads are considered
 The approximate value of bearing capacity of soil is assumed to be 180KN/m2
due to
unavailability of data
 The design of machine foundation has not been carried out
 The deduction for wall area for opening is taken certain percentage of total area rather
calculating in detail instead
1.6 Description of the Project
Thapa Khola hydroelectric project is at Tukuche village of Mustang District in Western
Development Region of Nepal. Thapa Khola Hydroelectric Project is a run-of-river type project
with an installed capacity of 13.6 MW. The general layout of the proposed project comprises of a
diversion weir with a gated undersluice, side intake, Gravel trap, Settling basin, Headpond,
Headrace pipe, Tunnel, Surge tank, Surface penstock and a Surface powerhouse located on right
bank of Thapa Khola in Tukuche village of Mustang District. The powerhouse accommodates
two units of generating equipment. The tailrace canal finally releases water back to the Thapa
khola. The install capacity is 13.6 MW and the total project cost is 23.79 million US $. It is
estimated that the average annual energy production will be 69.767 GWh.

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1.7 Salient features
1. Project Location
Latitude : 28° 43’ 00” N to 28° 48 00” N
Longitude : 83° 34’ 00” E to 83° 38’ 00” E
Development Region : Western Development Region
Zone : Dhaulagiri
District : Mustang
Intake Site : Tukuche VDC
Powerhouse Site : Tukuche, Tukuche VDC
2. General
Name of River : Thapa Khola
Nearest Town : Tukuche Bazaar
Type of Scheme : Run-of-river
Gross Head : 388.90 m
Net Head : 375.90 m
Design Discharge : 4.20 m3
/sec
Installed Capacity : 13.60 MW
Dry Season Energy : 11.77 GWh
Wet Season Energy : 57.96 GWh
Total Energy : 69.767 GWh
3. Hydrology
Catchment Area : 46 km2
Design Discharge : 4.2 m3
/sec
Design Flood Discharge : 80.0 m3
/s (1 in 100 yr. flood)
4. Desanding Basin
Inlet transition : 20.00 m
No of chamber : 2
Dimension (L x B x H) : 55.0 m x 6.0 m x 5.0 m
Outlet transition : 9.00 m

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5. Powerhouse
Type : Surface
Dimension : 31.5 m x 13.5 m x 16.5 m
Turbine setting level : El 2620 m amsl
Number of units : Two (2)
Type of turbine : Pelton Turbine (horizontal axis)
Installed capacity : 13.60 MW
Generators : Three phase, Synchronous, 7.3 MVA
6. Tailrace Canal
Type : Concrete Conduit
Length : 38 m
Material : RCC
Size : 3.0 m x 1.2 m
7. Financial Parameters
Project Cost : 23.79 million US$
IRR : 13.37%
NPV : 40.6 million US$

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Chapter 2
Methodology
2.1 Methodology (part I Hydrosuction Sediment Removal system)
2.1.1 Literature Review
a) Topic selection
During our internship, we observe the different component of the Thapa khola hydroelectric
project carefully. The sediment concentration on the river is comparatively high. We had
designed the settling basin on the basis of design discharge and found that the constructed
settling basin can settle the particle size up to 0.15mm, on assuming 6 hour detention time, the
sediment concentration of the river was found to be 8610 ppm. From the obtained result, we
conclude that the constructed settling basin may create a problem on smooth operation of the
power plant. To solve this sedimentation problem, our external supervisor suggests us to study
on the continuous flushing system. So, we have decided to do the Final year Project on design of
the continuous flushing system (Hydrosuction Sediment Removal System).
b) Continuous flushing system
To supply the sediment free water continuously through the water conveyance system, the
sediments need to be flushed out continuously. Continuous flushing system minimizes the
manual work. On conventional system the sediment flushes at once as a result the sediments get
deposited to the downstream. In this method the sediment flushes in the continuous manner. So,
the deposition of sediment in the downstream minimizes. There are different methods for the
continuous flushing system. In this report we will discuss on serpent sediment sluicing system
and perform detail design on hydrosuction sediment sluicing system.

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i) Serpent Sediment-Sluicing System
The invention concerns a method and means for removing sediment particles from a settling
basin with a longitudinal sluice along its bottom to facilitate a controlled removal of the sediment
particles through an outlet, which sluice being sealable against said basin by at least one sealing
organ. The invention can be used for the extraction of sediments from rivers carrying sediments
in suspension, so as to control the processes on the river bed or utilize such sediments for a
variety of purposes including: sand and gravel extraction, earth filling, mineral exploration, or
the measurement of sediment-transport in rivers. A specific application of the invention is as a
water outlet in river development systems and associated constructions such as hydropower
stations, irrigation systems and normal water supply where there is a need to remove sediments
in the river water before it is lead into the pipelines, tunnels or channels that connect the river to
the end user.
The main purpose of the invention is to find a method and a means of removing sediments that
does not involve halting the fluid flow and only a restricted amount of fluid is required to remove
the sediments.
ii) Hydrosuction sediment removal system
a) HSRS Dredging: Conventional methods of hydraulic dredging use a mechanical pump to
supply the driving power to remove deposited sediment from a reservoir. Hydrosuction dredging
removes sediment from reservoirs using the hydraulic head represented by the difference
between the water levels up-stream and downstream from the dam. The potential energy thus
stored drivers water an sediment into sediment removal pipelines. No external energy is required
to transport the sediment from the intake point to the point of discharge. Two variations of
hydrosuction dredging have been used: bottom discharge and siphon dredging. In siphon
dredging, the discharge pipe is passed over the top of the dam, and in bottom dredging the pipe
passes through low-level outlets at the dam.
b) HSRS Bypassing: Hydrosuction bypassing would employ the same principle to transport
sediment, but would feature a permanent inlet station upstream from the reservoir deposition
zone to collect the sediment into a pipe or pipelines. The sediment/water mixture is transported
through the pipeline and past the dam, where it is returned to downstream receiving waters.

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2.1.2 Data Collection
For the design of the continuous flushing system, it requires the sediment concentration data and
Particle distribution curves. These data were provided by the Mount Kailash Pvt Ltd. To install
the hydrosuction sediment system it requires the different equipment like motor, pipes and
valves. To get the cost of these equipment, we have done the market survey.
2.1.3 Consultation
During our project, we have consulted with the people from respective fields. Our external
supervisor Er. Dewakar Acharya also plays important role for the topic selection, he also
suggest us suggest with different ideas. Our internal supervisor Er. Sudip Jha suggests us with
his different ideas regarding the project. We have also consulted with Prof. Ramesh Kumar
Maskey for the continuous flushing system.
2.1.4 Office Works
After the collection of the data, we determine the size of the pipe, Head loss, sediment flow rate
and discharge through the pipe. We calculate the time require to operate this system and
conventional system. On the basis of time of operation, the operation cost is determined for both
the conventional and continuous system. With the help of data collected from the market, the
installation cost for the HSRS is determined. The total cost for the installation and operation for
both the system is determined. The total cost require for the continuous flushing system is found
to be minimum.

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2.2 Methodology (part II Structure Analysis of Powerhouse)
2.2.1 Literature Review
a) Topic selection
Design of powerhouse requires both the knowledge on hydropower and structure, so we are
interested on the design of the power house. So, we decided to do the Final year Project on detail
design of power house.
b) Power house
Power house is the conspicuous and vital part of the hydroelectric plant. Two basic requirement
of the power plant is the functional efficiency and aesthetic efficiency. Depending upon the
location of the construction power house can be
i) Surface Power House
ii) Underground Power House
Our powerhouse is of surface type.
c) Dimension of the Power house
i) Machine hall or unit bay: Length of the machine hall depends on the number of units, the
distance between the units, sizes of machine and clearance. Centre to center distance between the
units is taken as (5D+2.5) m. The machine hall width is determined by the size and clearance
space from the walls gangway. The height of the machine hall is fixed by head room requirement
for the crane operation.
ii) Loading bay: Loading bay also known as erection or service bay is a space where the heavy
vehicles can be loaded and unloaded, the dismantled parts of the machines can be placed and
where small assembling of the equipment can be done. The loading bay should be of sufficient to
receive the large parts like the rotor and runner. The loading bay floor will be having a width at
least equal to the center distance between the machines.
iii) Control bay: It is the main room and controls the equipment like runner, gates, valves,
generator etc. It may be adjacent to the unit bay i.e. machine halls as it sends instructions to the
operation bay from where the operation control is achieved.

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d) Power house Structure
A hydropower station structure can be broadly visualized consisting of two main divisions:
i) Super structure
The portion of the structure that is above ground level which receives the live load, dead load
and other loads is referred to as Super structure. Following are super structures that we have
designed in our project:
a) Beam
b) Column
c) RCC gantry beam
d) Corbel
e) Slab
f) Staircase
g)Truss
ii) Sub Structure
A Sub structure is an underlying or supporting structure to superstructure. It is below ground
level. Foundation is part of substructure. Substructure is the lower portion of the building which
transmits the dead load, live loads and other loads to the underneath sub soil. Following are the
sub structures that we have designed in the project:
a) Mat foundation
b) Shear wall

11
e) Estimation of load
IS 456:2000 is taken as the references for the estimation of load. The total loads that acts on the
support beams for two way slabs may be assumed as the load within the respective area of the
slab bounded by the intersection of 45o
line from the corners with the median line of the panel
parallel to the long side. Obtained triangular and trapezoidal loading is converted into equivalent
uniformly distributed load as described in respective section as in fig.
Figure 1: Distribution of loads from slab onto the supporting beam
Load in the shaded area to be
carried by beam A
Load in the shaded area to be
carried by beam B
A
B

12
Types of Load
i) Dead load
Dead load is the load which is given by the self- weight of the structures like beam, slab, column,
wall, staircase and so on. IS 875:1987(part I) is used as the reference. Dead load acting at the
slab is transferred as trapezoidal and triangular loads on beam. Dead load from slab is transferred
in beam as uniformly distributed load. Dead load of beam is considered as UDL. Dead load of
column is considered as the point load acting at the joint
ii) Live Load
The magnitude of live load depends upon the type of occupancy of the building. IS875:1987(part
II) is used as the references for the load. In our case the occupancy is an industrial building with
heavy duty equipments. The live load distribution varies with time. Hence each member is
designed for worst combination of dead load and live loads.
iii) Earthquake Load
No building is entirely safe from earthquake. But it is in the best efforts of many building codes
to ensure a level of safety to the building by enhancing its strength and ductility. Any building
should be designed in such a way that
 It should not have structural cracks in times of minor earthquake
 It should have minor repairable cracks in moderate earthquake
 It should not be grounded in times of major earthquake
Earthquake or seismic load on a structure depends on the size of the structure, maximum
earthquake intensity or string ground motion and the local soil, the stiffness, design and
construction pattern, and its orientation in relation to the incident seismic waves. Building
experiences the horizontal distortion when subjected to earthquake motion. So building should
be designed with lateral force resisting system. For design purpose, the resultant effects are
usually represented by the horizontal and vertical seismic coefficient αh & αv. Alternatively, the
dynamic analysis of the building is required under the action of the specified ground motion or
design response spectra. Since the probable maximum earthquake occurrences are not so
frequent, buildings are designed for such earthquakes so as to ensure that they remain elastic and
the building is prone to least damage.
Seismic load is calculated by using seismic coefficient method as specified in IS 1893(part I):
2002.

13
2.2.2 Data Collection
As we are re-designing the power house of Thapa Khola HEP. For the penstock and powerhouse
sizing, design discharge and head are needed. On the basis of discharge and head, power plant
capacity is determined and plant capacity determines the size of the machine hall, loading bay
and the control bay. Discharge and head data were provided by the host organization and the
reference drawings of constructed power house of Thapa Khola HEP were provided from the
host organization.
Preliminary sizes of the flexural members of the structural system i.e. slab and beams are worked
out as per the limit state of serviceability (deflection) consideration by conforming to IS456:2000
Cl.23.2.1.Similarly, for the compression member, i.e. columns, the cross sectional area of the
column is worked out from the net vertical axial load on the column lying in the ground floor
assuming suitable percentage of steel. The net vertical axial load on each column is worked out
from the factored dead load and live load on the contributing area, which is taken as half of the
slab areas adjacent to the column under consideration. The load is increased by 20% for the
earthquake load to give the net vertical load.
The following data are used for this project work:
Concrete grade: M25
Steel grade: Fe 500
Imposed load: as per code (IS 875:1987 part-2)
Industrial building with heavy duty equipment: 10kN/m2
Staircase, passages, lobby: 4kN/m2
Roof: 0.75kN/m2
Dead load: as per code (IS 875:1987 part-1)
Density of Brickwork: 20kN/m3
Density of plaster: 20.4kN/m3
Density of concrete: 25kN/m3
Bearing capacity of soil (as per site): 150 kN/m3

14
2.2.3 Consultation
During our project, we have consulted with the people from respective fields. Our external
supervisor Er. Dewakar Acharya also plays important role for the topic selection, he also
suggest us suggest with different ideas. Our internal supervisor Er. Sudip Jha suggests us with
his different ideas regarding the project. We have also consulted with Prof.Dr.Ramesh Kumar
Maskey for the continuous flushing system.
2.2.4 Office Works
a) Preliminary Design
Preliminary sizes of the flexural members of the structural system i.e. slab and beams are
designed as per the limit state of serviceability (deflection) consideration by conforming to
IS456:2000 Cl.23.2.1. Similarly, for the compression member, i.e. columns, the cross sectional
area of the column is worked out from the net vertical axial load on the column lying in the
ground floor assuming suitable percentage of steel. The net vertical axial load on each column is
worked out from the factored dead load and live load on the contributing area, which is taken as
half of the slab areas adjacent to the column under consideration. Thus preliminary design of
column is done by considering the factored axial load on the column as stated in IS 456:2000
clause 39.3.
b) Structural Analysis
Structural analysis deals with the prediction of performance of a given structure under stipulated
loads and other external effects. The performance characteristics of the structure under the
actions of stresses and stress resultants such as axial forces, shear forces, bending moment,
deflection and support reactions has been analyzed using SAP 2000V14. The result is compared
with the preliminary design of the members.
c) Detailed Design
Limit state method is used in design of all structural members. The beam has been designed by
using IS 456:2000. Slab has been designed using coefficient given in appendix D of IS 456:2000.
The column has been designed by using SP16. Other members like corbel, staircase, mat
foundation has been designed using the respective codes. Reinforcement detailing has been
carried out by IS 13920:1993( ductile detailing).

15
Chapter 3
Continuous Flushing Settling Basin
3.1 Continuous Flushing System
To supply the sediment free water continuously through the water conveyance system, the
sediments need to be flushed out continuously. Continuous flushing system minimizes the
manual work. On conventional system the sediment flushes at once as a result the sediments get
deposited to the downstream. In this method the sediment flushes in the continuous manner. So,
the deposition of sediment in the downstream minimizes. There are different methods for the
continuous flushing system. In this report we will discuss on serpent sediment sluicing system
and perform detail design on hydrosuction sediment sluicing system.
3.1.2 Serpent Sediment-Sluicing System
The invention concerns a method and means for removing sediment particles from a settling
basin with a longitudinal sluice along its bottom to facilitate a controlled removal of the sediment
particles through an outlet, which sluice being sealable against said basin by at least one sealing
organ. The invention can be used for the extraction of sediments from rivers carrying sediments
in suspension, so as to control the processes on the river bed or utilize such sediments for a
variety of purposes including: sand and gravel extraction, earth filling, mineral exploration, or
the measurement of sediment-transport in rivers. A specific application of the invention is as a
water outlet in river development systems and associated constructions such as hydropower
stations, irrigation systems and normal water supply where there is a need to remove sediments
in the river water before it is lead into the pipelines, tunnels or channels that connect the river to
the end user.
The main purpose of the invention is to find a method and a means of removing sediments that
does not involve halting the fluid flow and only a restricted amount of fluid is required to remove
the sediments.

16
3.1.3 Hydrosuction sediment removal system
Introduction
Hydrosuction Sediment Removal System (HSRS) remove deposited or incoming sediment in
reservoirs using the energy represented by the difference between water levels upstream and
downstream from a dam.
There are two types of hydrosuction sediment removal. The first is hydrosuction dredging, in
which deposited sediment is dredged and transported to either a downstream receiving stream or
to a holding or treatment basin [Fig. 4(a)]. The second is hydrosuction bypassing, in which
incoming sediment is transported without deposition past the dam to the downstream receiving
stream [Fig. 4(b)].
Figure 2: Hydrosuction Sediment Removal System (HSRS)
a. Hydrosuction Dredging
b. Hydrosuction Bypassing

17
a) HSRS Dredging
Conventional method of hydraulic dredging use a mechanical pump to supply the driving power
to remove deposited sediment from the reservoir. Hydrosuction dredging removes sediment from
reservoir using hydraulic head represented by the difference between water level upstream and
downstream from the dam. The potential energy thus stored drives water and sediment into
sediment removal pipelines. No external energy is required to transport the sediment from the
intake point to the point of discharge.
The method uses a pipeline at or near the bottom of reservoir that extends upstream to the point
of sediment deposition to the dam. The pipe continues over or through the dam to a discharge
point downstream. The water-sediment mixture is transported through the pipeline until it is
discharged into the relatively clear water that passes the dam through outlet or hydropower
turbines. Sediments need not to be stored in spoil area. Two variations of hydrosuction dredging
have been used: bottom discharge and siphon dredging. In siphon dredging, the discharge pipe is
passed over the top of dam and in bottom dredging the pipe passes through a low-level outlets at
the dam. Both methods may employ a floating barge, which moves the pipeline inlet around the
reservoir to access a larger area.
b) HSRS Bypassing
Hydrosuction bypassing would employ the same principle to transport the sediment, but would
feature a permanent inlet station upstream from the reservoir deposition zone to collect the
sediment into the pipe or pipelines. The sediment/water mixture is transported through the
pipeline and past the dam, where it is returned to downstream receiving water.
3.2 Benefits of HSRS over Conventional Method
 It helps to return the downstream system to its more natural, pre-dam condition by
releasing the sediment in accordance with the d/s transport capacity
 It reduces the shock that associates with the flushing technique
 Less water is used, thus conserving the reservoir water storage
 In the bypassing mode, the fraction of sediment entering the bypass system never
deposits on reservoir and a better sediment balance is maintained across the dam

18
3.3 HSRS components
The principal components of HSRS are the intake, pipeline, valve, outlet works and appurtenant
facilities.
3.3.1 Intake
Intake shapes for hydrosuction dredging vary from a straight-end inlet to a dustpan shape. A tin
attachment may be useful for scarifying the bed and suspending the sediment near the inlet.
Intakes may be equipped with external power cutter heads or water jet to suspend deposits from
consolidated beds.
3.3.2 Pipeline
Sediment bypassing requires a permanent structure that excludes sediment from incoming flow
and introduces it to the bypassing pipelines.
Flexible and/or rigid pipes are used to transport sediment and water in a dredging HSRS. A
portion of the pipeline upstream from the dam is usually supported off the reservoir bottom to
make the pipeline easier to move [Fig. 4(a)]. Flexible pipe near the intake or rigid pipes
connected with rubber elbows in the pipeline can be used to facilitate movement. Rubber elbows
however, introduce additional friction losses.
Piping system for a bypassing HSRS need not be flexible. A pipe or manifold of pipes at the
sediment excluder would be attached to a permanent pipeline system that extends to a
downstream discharge point. A separate pipeline to introduce clear water into the system may be
helpful to prevent pipeline blockage and to regulate concentration to match downstream
sediment deficit conditions.
3.3.3 Outlet
The location of the HSRS outlet depends on the intended use of the sediment. For all cases, the
outlet should either be submerged or turned upward so it is always full to avoid air entering the
pipeline. The simplest discharge option is to release the sediment and water into the channel
downstream the dam. Care must be taken to avoid deposition by adding only as much sediment
to the stream as it can transport.
3.3.4 Valve
A valve location at the dam or near the outlet controls flow in the pipeline for hydrosuction
dredging. Easy excess to the valve under all operating conditions and water level is essential.
The water and soil conservation division (1989), in China, offers several practical
recommendations for locating the pipeline valve. They recommended that if the hydraulic head is
not excessive, the valve be fully open when operating, to avoid debris catching in the valve plate.
In the case of sediment bypassing, several valve to control the clear water intake (if used) and the
sediment pipe manifold would be necessary. The sediment-intake valve may need to adjust
automatically to admit only as much sediment as will make up the sediment deficit downstream.

19
Approximate valves for slurry flow are pinch and diaphragm valves, ball valve and plug valve.
Valves should be shut off during period of icing or heavy debris.
Ancillary facilities for hydrosuction dredging may include a raft or barge to move the pipeline
inlet to the reservoir, an externally powered water jet or cutter head at the inlet to breakup
consolidated sediments (if required) and instrumentation to monitor the operation. Newer
reservoir or unconsolidated sediments may not need a water jet or cutter head to loosen the
sediment.
3.3.5 Mechanical Dredging
In hydrosuction dredging it is difficult to move the pipe different position of settling basin. So
we have proposed to provide the mechanical dredging system. In this system the sediment is
transfer from inlet to the mouth of flushing opening. This is consisted of two wheel of length
40m operated with the 0.5kW (for each settling Basin). The sediment transport rate with this
system is 0.05m/s. The detail drawing is shown in annex.
3.4 Hydrosuction Dredging Case Histories
Hydrosuction dredging was first performed in the Djidiouia Reservoir and Dam, in Algeria, from
1892 to 1894. A floating pontoon barge was connected to a submerged pipeline 61cm in
diameter and 1.6 km long. A small turbine at the pipeline outlet generated power for cutting head
at the inlet that chopped up the consolidated sediment before it was mobilized. The reservoir
seriously depleted of storage in its 10-year history, yielded a total of 1,400,000 m3
of silt and
clay in two year operation period. Average sediment concentration in pipeline was 3% and
maximum concentration was 7%.
The People’s Republic of China to date has the most experience with hydrosuction dredging. The
Chinese have used either the siphon or bottom withdrawal modes in ten reservoirs beginning in
1975. In all cases, the fertile sediment-laden water was passed into irrigation canals downstream
and spread on cropland to replenish the top soil and recharge the nitrogen content. Dam height
varied from 15 to 35m, while initial reservoir storage varied from 1500000-24500000 m3
.
Incoming sediment concentration varied from 19 to 150 kg/m3
.

20
3.5 Hydraulic Principles
The problem of sediment transport in pipeline is one of two-phase flow and has been considered
by various disciplines over the last several decades. The principles and ideas has been introduced
by several engineering texts, dredging industries and most recently by chemical engineering
works by Shook and Roxo (1991).
Optimum sediment transport in pipelines occurs when the sediment particles are on the verge of
depositing in the pipe. This maximum concentration is referred to as being between a so-called
heterogeneous flow regime and a flow regime with a moving bed. Equations for head loss and
for this condition are needed to design the HSRS pipeline. The problem as applied to reservoir
sediment removal is complicated by the presence of non-uniform and cohesive sediments and by
the difficulty in obtaining detail field measurements that may confirm and verify predicted
relationships.
3.5.1 Head-loss Expression
Most expression for head-loss and transport in heterogeneous two-phase flows follow a form
first used by Durand. After 310 tests with sediment sizes ranging from 0.2 mm to 25 mm,
sediment concentrations range from 2% to 23% by volume, and pipes ranging in size from 38
mm to 710 mm in diameter (1.5 in. to 28 in.). Durand formulated a head-loss function as follows:
𝛷 = 𝐾𝜓𝑚 = (𝐽𝑚 − 𝐽)/(𝐶𝑣𝐽)………………….(1)
Where Φ= dimensionless head loss function; K= constant; ψ= dimensionless function of
hydraulic variables; m= exponent; Jm= head-loss gradient for the sediment-water mixture;
J= head-loss gradient for the clear water; and Cv= sediment concentration by volume. The
parameter ψ is defined as
Ψ= (V2
*Cd
0.2
)/ [g D(S-1)]……………………….. (2)
Where V= flow velocity; D= pipe diameter; S= specific gravity of the sediment; and Cd= particle
drag coefficient
Values for K and m have been proposed by different researchers. Durand and Condolios used K=
81 and m= -1.5 based on their experimental data. In another effort to evaluate K and m. zandi
and Govatos analyzed all available data for which the concentration by volume (Cv) was greater
than 5%, stating that concentration less than 5% produce less head loss than clear water. They
also determined from experimental plots that about one-third of the data points were for flows in
the moving-bed regime. Upon elimination of these data, they fit two lines to the data; for ψ<10,
K= 280 and m= -1.93, while for ψ>10, K= 6.3 and m= -0.354. No physical significance is
attached to the value of ψ= 10.
For non-uniform sediments it is recommended that a weighted Cd be computed from

21
√Cd = p1√Cd1 + p2√Cd2 +…………. + pn√Cdn …… (3)
Where p1, p2……., pn = decimal size fractions from the particle size distribution; and Cd1, Cd2 and
Cdn = drag coefficients of the particle diameters used to represent the sediment size fractions.
The following analysis follows Eftekharzadch(1987) but maintains the constant and exponent as
variables. Substituting from (2) into (1) yields
Jm−J
CvJ
= K [
𝑉2
√ 𝐶 𝑑
𝑔𝐷(𝑆−1)
]m
………………(4)
From the Darcy-Weisbach equation we obtain an expression for dimensionless head loss
J= fV2
/92gD)…………….. (5)
Where f= Darcy-Weisbach friction factor. By definition
Cv= Qs/Q= 4Qs/ (πD2
V)…………. (6)
Where Qs= sediment transport rate by volume. Substituting (5) and (6) in (4) and rearranging
𝐽 𝑚 =
𝑓𝑉2
2𝑔𝐷
+
2𝑓𝐾𝑉1+2𝑚 𝑄 𝑠𝐶 𝑑
0.5𝑚
𝜋𝑔1+𝑚 𝐷3+𝑚(𝑆−1) 𝑚………………………………… (7)
Equation (7) shows the variation of head-loss gradient, Jm is a function of clear-water friction
and the contribution due to the presence of sediment. The friction loss due to sediment depends
on pipe diameter, D volumetric sediment transport rate Q, sediment size and flow velocity V. the
non-flow variables may be combined into a material constant, α as
𝛼 = 𝐾𝐶 𝑑
0.5𝑚
/ [𝑔 𝐷(𝑆 − 1)] 𝑚 ………………….. (8)
Where the gravitational constant, g must be expressed in feet per second per second in
accordance with the original derivation,
Substituting α into (7) yields
𝐽𝑚 =
𝑓𝑉2
2𝑔𝐷
+
2𝑓𝛼𝑄 𝑆 𝑉1+2𝑚
𝜋𝑔𝐷3
……………….. (9)
Equation (9) can be used to estimate the head loss gradient, Jm, in a pipeline transporting a
sediment-water mixture.

22
3.5.2 Sediment-Transport Expression
In hydro suction dredging, the input energy is elevation pressure head, which is relatively fixed
and reflected by the Jm term of (9). The unknowns are flow velocity and optimum sediment
transport rate. Equation (9) suggests that flow velocity varies with sediment transport rate for a
fixed Jm. Therefore, there must exist an upper limit for a given head to transport the maximum
rate of sediment in a pipe. In other words, one can look for a minimum head-loss condition at a
fixed sediment transport rate, or locate the flow velocity at which the head loss gradient is
minimum. This is done by assuming the Darcy-Weisbach friction coefficient f is a constant,
differentiating Jm in (9) with respect to the velocity, and equating the result to zero:
𝑑𝐽𝑚/ 𝑑𝑉 = 0……………………… (10)
Which yields,
𝑉𝑚 = (−
𝐷2
2αQs(1 − 2m)
)
1
2𝑚−1
Where Vm= flow velocity when the pipe transports the maximum rate of solids with the fixed
head. Rearranging (9) and solving for Qs and simplifying yields the following expression for the
maximum sediment transport rate:
2
21
12
212
3
12
22
)
)
)21(2
(
2
2/)
)21(2
(
(
m
m
m
m
m
D
gD
f
gD
m
D
f
Jm
Qs















Where Qs = maximum sediment transport rate under the available head-loss gradient, Jm.
3.6 Pipeline Design Steps
1. Determine the approximate value of the available length, h, pipeline length, l, pipe material
and trial pipe diameter, D.
2. Collect the sediments from the area of the settling basin to HSRS operation, determine the
value of the sediments specific gravity, S, the particle size distribution, the drag coefficient for
each size fraction, and the composite drag coefficient Cd.
3. Calculate the non-flow parameters α using the equation 𝛼 =
𝐾 𝐶𝑑0.5𝑚
(𝑔𝐷(𝑆−1))
𝑚
Where g is in ft/sec2
and D is in ft.
4. Assume the pipeline velocity V, and calculate the sum of headloss gradient Jm including the
minor losses using

23
𝐽 𝑚 =
ℎ − ∑𝐾 (
𝑣2
𝑔 )
𝑙
where h is in ft ,g is in ft/sec2
and v is in ft/sec.
5. Using the trial value of the darcy-weisbach friction coefficient, f, the head-loss gradient, Jm,
the pipeline diameter D, the non-flow parameter α, and the value of the exponent m assumed to
be valid for the current solid-liquid regime, calculate the sediment flow rate Qs using the
2
21
12
212
3
12
22
)
)
)21(2
(
2
2/)
)21(2
(
(
m
m
m
m
m
D
gD
f
gD
m
D
f
Jm
Qs















Where Qs is in cfs.
6. Compute the trial value of the optimum mixture flow velocity Vm using
𝑉𝑚 = (−
𝐷2
2αQs(1 − 2m)
)
1
2𝑚−1
Where Vm is in ft/sec
7. Calculate the mixture flow Reynolds number Re= VmD/v, where v is an appropriate value of
the kinematic viscosity, and determine the roughness coefficient ϵ for the given pipe material.
8. Calculate the mixture friction coefficient fm using the explicit formula given by Swamme and
Jain
𝑓𝑚=
1.325
(ln (
ϵ
3.7D) +
5.74
𝑅 𝑒
0.5 )
2
9. Using Vm, recalculate the Jm and fm, and compare it with value of fm calculated in the step 8.
Repeat steps 3 to 8 until the difference between the fm values calculated in subsequent steps is
within acceptable tolerance
3.7 Settlement of Particle
Determination of particle size that can be settled with Settling Basin of Thapa khola
Settling basin 1 and 2 are similar taking only one to determine concentration of sediment.
Discharge=4.2/2=2.1

24
Design discharge=2.1+10%*2.1
=2.31 m3
/s
Horizontal velocity
Q= AV
V= (2.31)/ (6*3)
=0.128 m/s
We have three conditions to determine limiting velocity.
 d> 1mm
V=0.36*d^0.5
d=0.126mm<1mm Not ok.
 0.1>d>1
V=0.44*d^0.5
=0.0846 Not ok
 0.1>d
V=0.51*d^0.5
=0.063 <0.1 mm 0k
The given settling basin can settle particle size upto 0.063 mm from the horizontal velocity but
we will also check from fall velocity.
And, we know
AS = (K*Q)/Wt Let K=1.3 (range 1.2-1.5) for turbulence effect.
Wt = (1.3*2.31)/ (55*6)
=9.1 mm/s
We will determine corresponding particle from chart.
Conducting temperature to be 10 degree centigrade
Particle size settling velocity
0.1 5.12
0.2 17.11
Y 9.11
From interpolation,
Y=0.133 mm
The corresponding horizontal limiting velocity is
V=0.44*0.133^0.5

25
=0.16 m/s >0.128 m/s
Hence the settling basin is capable of settling the particle size up to 0.133 mm.
Determining the efficiency of settling basin
 From Hazen’s equation
η=1-(1+
𝑚∗𝑤
𝑣0
) ^ (-1/m)
=1-(1+
0.17∗9.1∗10^−3
7∗10^−3
) ^ (-1/0.17)
=69.13%
 From Camp’s equation
V*
= (4.2*v)/ (100*R^ (1/6))
= (4.2*0.128)/ (100*1.5^ (1/6))
=0.00502
Ratio = 0.007/0.00502
=1.394
(W*As)/ Q= (0.00698*55*6)/2.31
=1
The efficiency of settling basin is 90%. (From Camp’s chart)
 From vetter’s equation
1-η= 𝑒
−𝑤∗𝐴𝑠
𝑄
η=63.2%
The total volume that is available for sedimentation
=(2.5*2.5+3.05*1)*55
=511.5 m3
For sediment concentration determination
Q*T*C=weight of sediment
2.31*6*3600*C=511.5*(2650*0.5)
C=13.58 Kg/m3

26
3.8 Dimension of the Hydrosuction sediment Removal system
Length of pipe: 200 m
Dia. of Pipe : 0.15 m
Velocity of mixture through pipe: 2.926 m/s
Discharge of mixture through pipe: 0.0493 m3
/s
Discharge of sediment flow rate: 0.006885 m3
/s
Table 1: Installation Material Cost of the HSRS for the one unit
Materials Quantity unit
Unit
cost(Rs) Cost(Rs) Remarks
Black plastic
pipe 250 m 60 15000
Pan inlet 1 no. 2500 2500
For the inlet of
hydrosuction
Motor 1 kW 26000 26000
For the suction
initiation of
HSRS
Ball valve 2 no. 4000 8000
For opening
and closing of
valve
Mechanical
dredging 1 no. 60000 60000
Requires 1kW
to operate
Labour cost 4 no. 2500 10000
Working for 2
days
Total cost 121500

27
3.9 Design of Hydrosuction Sediment Removal system
In hydrosuction Sediment Removal system we have to first assume the trial velocity and
calculate velocity of mixture (Vm). Using this velocity of mixture calculate coefficient of
friction. Again start new trial velocity using Vm. This process is repeated till the difference of
consecutive friction coefficient is negligible. Generally the value get converge in 2 to 3 steps.
The data require for the HSRS system is shown below first and then method for the calculation
Table 2: Monthly discharge and power generation of Thapa Khola HEP
Month No. of days
River
Discharge(m3
/s)
Discharge for
power
generation(m3
/s)
Net
Head(m)
Power
Produce(kW)
Baisakh 31 2.96 2.76 382.79 9098
Jestha 31 4.15 4.01 377.15 13033
Asar 32 5.35 4.20 375.88 13594
Shrawan 31 7.27 4.20 375.88 13594
Bhadra 31 8.54 4.20 375.88 13594
Ashwin 31 6.26 4.20 375.88 13594
Kartik 30 3.4 3.20 381.04 10500
Mansir 29 1.66 1.52 386.33 5067
Poush 30 1.44 1.30 386.73 4339
Magh 29 1.37 1.23 386.73 4106
Falgun 30 1.5 1.36 386.64 4538
Chaitra 30 2 1.86 385.57 6186

28
0
2
4
6
8
10
0 0.2 0.4 0.6 0.8 1 1.2
FLOW DURATION CURVE
Source: Mount Kailash energy Pvt Ltd.
Figure 3: Flow duration curve and particle size
distribution curve
Discharge(m3
/s)
Probability percentage exceedence

29
Parameter Units Description Value Value in si
units
Step 1
h ft Available head—reservoir water surface
elevation at normal pool minus tailwater
elevaton on downstream side of dam or other
determined hydrosuction pipeb outlet location
49.2 15m
L ft Pipeline length from location of hydrosuction
operations to tailwater or other
determined outlet location
656 200m
D ft Diameter of pipe 0.492 0.15m
K Summation of coefficient of minor loss 2.6
T o
C temperature 10
rd lb/ft3
Density of reservoir in situ sediment 84 1346 kg/m3
rw lb/ft3
Density of water 62.4 1000kg/m3
units

30
Q ft/s Assume a trial flowrate through pipe 9.6 2.926m/s
g ft2
/s Acceleration due to gravity (constant) 32.2 9.81m/s2
e ft Roughness height—assuming plastic pipe
material.
8.2E-06 0.0000025
m
sg Sediment specific gravity in HSRS pipe: 2.65 for
quartz (sand and gravel); 2.8 for silt&clay
2.65 2.65
sumKi Assumed total minor energy loss coefficient for
possible minor losses in the hydrosuction piping
system. Minor losses include energy losses at
entrance, exit, bends, connections, and valves.
Losses summed.
2.6 2.6
units
Step 2
d100 mm Grain size for which 100 percent of reservoir
sediment sample is finer.
1
sediment sample is finer
0.4

31
0.18
0.133
n ft2
/s Kinematic Viscosity for assumed temperature(at
10o
C)
1.409x1
0-5
`1.31*10-6
m2
/s
W100 mm/s Particle fall velocity 106.71
Re100 Reynold’s Number for d100 using its fall
velocity
81.458
Re90 Reynold’s Number for d90 using its fall velocity 12.06
units
Cd100 Cd for d100 grain size 0.967
Cd90 Cd for d90 grain size. 3.194

32
Cd Composite Cd for sediment sample 5.22
Step 3
Ψ Calculate from Ψ Hotchkiss equation (2) 8.05
α Non-flow parameter, a, is a combination of non-
flow variables from Hotchkiss equation (8)
30847.2
Jm Calculated headloss gradient in hydrosuction
pipe using Hotchkiss Equation (7)
0.0693
Re Reynold’s number for the mixture flow in the
pipe
335047.
3
f Equation developed from Moody diamgram
yields trial friction factor value.
0.01402
Qs cfs Maximum sediment transport rate under
available headloss gradient, calculated using
Hotchkiss (1996) equation (12).
0.255 0.006885
m3
/s
units
Vm fps Calculates trial optimum mixture flow velocity
from Hotchkiss equation (11). 4.08 Rm
Calculates the mixture flow Reynold’s number.
9.6 2.926m/s
Qm cfs Discharge of mixture 1.825 0.0493 m3
/s

33
Ql cfs Discharge of clear water 1.570 0.0424 m3
/s
fm Calculates the mixture friction coefficient, fm,
using the explicit formula givenby Swamee and
Jian (Streeter and Wylie, 1985) [Hotchkiss
equation (14)].
0.01432

34
Table 3: Result of Suspended sediment concentration analyses (Thapa khola)
S.No.
Sampling
date
sampling
time Concentration(PPM)
Gauge
Height(m) Temp(o
C) Remarks
1 15-06-13 8:00 573 0.69 9 No rain
2 15-06-13 16:00 5736 0.75 10.5 Rain
3 16-06-13 8:00 5806 0.8 805 Rain
4 16-06-13 16:00 25453 0.9 8 Rain
5 17-06-13 8:00 103229 1.15 9 Rain
6 17-06-13 16:00 95717 2.2 8 Rain
7 18-06-13 8:00 44131 2.2 8 Rain
8 18-06-13 14:00 52310 2.2 7.5 Rain
9 18-06-13 16:00 8446 2.2 8 Rain
10 19-06-13 8:00 7274 1.1 7.5 Rain
11 19-06-13 16:00 12723 1.25 9 Rain
12 20-06-13 8:00 6494 1.05 7 No rain
13 20-06-13 16:00 12370 1.05 11 No rain
14 21-06-13 8:00 7020 0.99 8 No rain
15 21-06-13 13:40 7322 1 11 No rain
16 21-06-13 16:00 6458 1.05 11.5 No rain
17 22-06-13 8:00 8245 0.75 10 No rain
18 22-06-13 16:00 7235 0.86 10.5 No rain
19 23-06-13 8:00 8611 0.86 10 No rain
20 23-06-13 16:30 3989 1.1 10 No rain
21 24-06-13 8:00 17489 0.86 9 No rain
22 24-06-13 16:00 10054 1.1 9.5 No rain
23 25-06-13 8:00 19715 0.85 9 No rain
24 25-06-13 16:00 30202 1.1 9 No rain
25 25-06-13 18:30 29172 1.1 9 No rain
26 26-06-13 8:00 8199 0.86 9 No rain
27 26-06-13 16:00 6296 1 10 No rain
28 27-06-13 8:00 2091 0.85 10 No rain
29 27-06-13 14:00 41224 0.95 10.5 No rain
30 27-06-13 16:00 7436 0.97 9.5 No rain
31 28-06-13 8:00 3968 0.85 9 Rain
32 28-06-13 16:00 3483 0.95 10 No rain
33 29-06-13 8:00 1485 0.83 9.5 No rain
34 29-06-13 16:00 14270 0.95 10 Rain

35
S.No.
Sampling
date
sampling
Gauge
Height(m) Temp(o
C) Remarks
35 30-06-13 8:00 2415 0.83 9.5 No rain
36 30-06-13 16:00 1369 0.96 10.5 No rain
37 01-07-13 8:00 3291 0.83 9 No rain
38 01-07-13 14:30 6529 0.9 10.5 No rain
39 01-07-13 16:00 3543 0.92 10 No rain
40 02-07-13 8:00 1194 0.79 9 No rain
41 02-07-13 16:00 2934 0.9 10 No rain
42 03-07-13 8:00 1391 0.79 10 No rain
43 03-07-13 12:00 1767 0.79 11.5 Rain
44 03-07-13 16:00 3180 0.96 11 Rain
45 04-07-13 8:00 1,345 0.76 10 No rain
46 04-07-13 16:00 3,324 0.9 11 No rain
47 05-07-13 8:00 3607 0.9 9.5 Rain
48 05-07-13 16:00 6,734 0.98 10 Rain .,
49 06-07-13 8:00 2269 0.92 10 No rain
50 06/07/2013- 16-
00 3431 0.98 10 No rain
51 07-07-13 8:00 1686 0.85 9.5 Rain
52 07-07-13 16:00 1769 0.85 9 No rain
53 07-07-43 16:00 2,031 0.91 10 No rain
54 08-07-13 8:00 1290 0.8 9 No rain
55 08-07-13 16:00 2972 0.86 10.5 No rain
56 09-07-13 8:00 1,938 0.86 9.5 Rain
57 09-07-13 16:00 10100 1.02 10 No rain
58 10-07-13 8:00 3237 0.9 9.5 Rain
59 10-07-13 16:00 5,744 0.98 10 No rain
60 11-07-13 8:00 2,875 0.84 9.5 Rain
61 11-07-13 14:00 2,108 * 10 Rain
62 11-07-13 16:00 2,168 0.85 10 Rain
63 12-07-13 8:00 1,978 0.76 10.5 No rain
64 12-07-13 16:00 4,355 0.84 11 No rain
65 13-07-13 8:00 930 0.74 10 No rain
66 13-07-13 16:00 2,414 0.78 10.5 No rain
67 14-07-13 8:00 1,146 0.78 10.5 No rain
68 14-07-13 16:00 3,625 0.71 10 No rain
69 15-07-13 8:00 805 0.72 10 No rain
70 15-07-13 16:00 23,343 1.1 10.5 Rain
71 16-07-13 8:00 4,556 0.72 9.5 No rain
72 16-07-13 16:00 5,406 0.82 10.5 No rain

36
S.No. Sampling
date
sampling
time
Concentration(PPM) Gauge
Height(m)
Temp(o
C) Remarks
73 17-07-13 8:00 1,247 0.72 10 No rain
74 17-07-13 16:00 997 0.76 10.5 No rain
75 18-07-13 8:00 993 0.73 10.5 No rain
76 18-07-13 14:30 5,551 0.74 11 No rain
77 18-07-13 16:00 3,368 0.77 11 No rain
78 19/0712013 8:00 1,908 0.74 10.5 No rain
79 19-07-13 16:00 16,147 0.94 10 No rain
80 20-07-13 8:00 4,087 0.84 9.5 No rain
81 20-07-13 14:00 5,706 0.86 9.5 No rain
82 20-07-13 16:00 3,088 0.87 10 No rain
83 24-07-13 8:00 1,609 0.85 9.5 No rain
84 24-07-13 14:30 4,769 0.87 10 No rain
85 24-07-13 16:00 7,832 0.91 10.5 No rain
86 25-07-13 8:00 1,688 0.83 9 Rain
87 25-07-13 16:00 5,106 0.92 9 Rain
88 26-07-13 8:00 1,509 0.85 9 No rain
89 26-07-13 14:00 1,863 0.89 9.5 No rain
90 26-07-13 16:00 4,711 0.94 9.5 Rain
91 27-07-13 8:00 1,398 * * *
92 27-07-13 16:00 1,224 * * *
93 28-07-13 8:00 1,490 * * *
94 28-07-13 16:00 1,803 * * *
95 29-07-13 8:00 1,200 * * *
96 29-07-13 16:00 5,741 * * *
97 30-07-13 8:00 1,182 0.83 9 No rain

37
98 30-07-13 16:00 5,006 0.87 11 No rain
99 31-07-13 8:00 777 0.8 9.5 No rain
100 31-07-13 16:00 2,662 0.86 10 No rain
101 01-08-13 8:00 1,915 0.81 10.5 No rain
102 01-08-13 13:00 875 0.84 11 No rain
103 01-08-13 16:00 1,643 0.86 11.5 No rain
104 02-08-13 8:00 1,887 0.77 9.5 No rain
105 02-08-13 16:00 1,898 0.8 11 No rain
106 03-08-13 8:00 707 0.72 9.5 No rain
107 03-08-13 16:00 1,267 0.5 10 No rain
108 04-08-13 8:00 883 0.71 9 No rain
109 04-08-13 13:00 4,074 0.73 11 No rain
110 04-08-13 16:00 2,530 0.76 10.5 No rain
111 05-08-13 8:00 1,042 0.68 9 No rain
112 05-08-13 16:00 5,512 0.71 11 No rain
113 06-08-13 8:00 2,182 0.66 9.5 No rain
114 06-08-13 16:00 4,041 0.69 10.5 No rain
115 07-08-13 8:00 6,597 0.64 9 No rain
116 07-08-13 16:00 3,423 0.7 10 No rain
117 08-08-13 8:00 2,065 0.62 9.5 No rain
118 08-08-13 14:00 3,617 0.62 11 No rain
119 08-08-13 16:00 3,279 0.65 10 No rain
120 09-08-13 8:00 6,926 0.62 9 No rain
121 09-08-13 16:00 2,497 0.64 11 No rain
122 10-08-13 8:00 1,980 0.6 9.5 No rain
123 10-08-13 13:00 3,552 0.6 11.5 No rain
124 10-08-13 16:00 2,703 0.6 11 No rain

38
125 11-08-13 8:00 1,508 0.59 9.5 No rain
126 11-08-13 16:00 5,055 0.64 11 No rain
127 12-08-13 8:00 1,242 0.6 9.5 No rain
128 12-08-13 16:00 1,729 - * Rain
129 13-08-13 8:00 1,794 0.61 9.5 No rain
130 13-08-13 16:00 1,363 0.6 11 No rain
131 14-08-13 8:00 1,829 0.62 9.5 No rain
132 14-08-13 16:00 3,208 0.65 10 No rain
133 15-08-13 8:00 1,454 0.64 9 No rain
134 15-08-13 14:00 1,272 0.63 10 No rain
135 15-08-13 16:00 3,542 0.66 11 No rain
136 16-08-13 8:00 1,076 0.58 9 No rain
137 16-08-13 16:00 2,258 0.64 10.5 No rain
138 17-08-13 8:00 1,467 0.59 9.5 No rain
139 17-08-13 16:00 2,270 0.64 11 No rain
140 18-08-13 8:00 775 0.6 9 No rain
141 18-08-13 14:30 1,207 0.62 10.5 No rain
142 18-08-13 16:00 852 0.66 11 No rain
143 19-08-13 8:00 2,619 0.61 9.5 No rain
144 19-08-13 16:00 747 0.65 10.5 No rain
145 20-08-13 8:00 519 0.6 9 No rain
146 20-08-13 16:00 847 0.64 10 No rain
147 21-08-13 8:00 503 0.59 9.5 No rain
148 21-08-13 14:00 261 0.59 11.5 No rain
149 21-08-13 16:00 1,465 0.63 11 No rain
150 22-08-13 8:00 884 0.61 9.5 No rain

39
S.No.
Sampling
date
sampling
Gauge
Height(m) Temp(o
C) Remarks
151 22-08-13 16:00 1,723 0.64 10.5 No rain
152 24-08-13 8:00 555 * 9 No rain
153 24-08-13 16:00 1,468 0.64 9.5 No rain
154 25-08-13 8:00 417 0.6 9 No rain
155 25-08-13 13:30 252 0.62 10 No rain
156 25-08-13 16:00 1,023 0.64 10.5 No rain
157 26-08-13 8:00 446 0.6 9 No rain
158 26-08-13 16:00 1,135 0.64 11 No rain
159 27-08-13 8:00 1,085 0.59 9.5 No rain
160 27-08-13 16:00 1,679 0.66 10.5 No rain
161 28-08-13 8:00 734 0.54 9.5 No rain
162 29-08-13 14:00 269 0.54 10 No rain
163 29-08-13 16:00 1,503 0.68 11 *
164 30-08-13 8:00 915 0.55 9 c
165 30-08-13 16:00 992 0.64 10 No rain
166 31-08-13 8:00 853 0.55 9.5 No rain
167 31-08-13 16:00 3,934 0.6 10 No rain
168 01-09-13 8:00 1,354 0.57 9.5 No rain
169 01-09-13 16:00 1,465 * 10 No rain
170 02-09-13 8:00 898 0.59 9.5 rain
171 04-09-13 8:00 1,211 0.52 9.5 No rain
172 04-09-13 16:00 531 0.6 10.5 No rain
173 09-09-13 8:00 248 0.48 9.5 No rain
174 09-09-13 16:00 715 0.52 10 No rain
175 10-09-13 8:00 261 0.44 9.5 No rain
176 10-09-13 14:00 253 0.44 10 No rain
177 10-09-13 16:00 414 0.5 10.5 No rain
178 11-09-13 8:00 255 0.63 9 No rain
179 11-09-13 16:00 218 0.5 10 No rain
180 12-09-13 8:00 170 0.46 9.5 No rain
181 12-09-13 16:00 292 0.49 10.5 No rain
182 13-09-13 8:00 226 0.42 9 No rain
183 13-09-13 16:00 266 0.47 10 No rain
184 14-09-13 8:00 148 0.44 9 No rain
185 14-09-13 14:00 97 0.45 11 No rain
186 14-09-13 16:00 212 0.47 11.5 No rain

40
S.No.
Sampling
date
sampling
Gauge
Height(m) Temp(o
C) Remarks
187 15-09-13 8:00 122 0.42 9 No rain
188 16-09-13 8:00 163 0.41 9 No rain
189 16-09-13 16:00 277 0.44 10 No rain
190 17-09-13 8:00 115 0.4 8 No rain
191 17-09-13 16:00 302 0.45 10.5 No rain
192 18-09-13 8:00 108 0.4 7.5 No rain
193 18-09-13 13:30 143 0.42 11.5 No rain
194 18-09-13 16:00 163 * * No rain
195 19-09-13 8:00 114 0.41 8 No rain
196 19-09-13 16:00 172 0.43 11.5 No rain
197 20-09-13 8:00 1,000 0.48 9 No rain
198 20-09-13 16:00 356 0.44 10 No rain
199 21-09-13 8:00 336 0.42 7 No rain
200 21-09-13 16:00 6,078 0.56 10 No rain
201 22-09-13 8:00 373 0.42 8 No rain
202 22-09-13 16:00 382 0.45 11.5 No rain
203 23-09-13 8:00 207 0.41 9 No rain
204 23-09-13 12:00 375 0.42 12 No rain
205 23-09-13 16:00 260 0.43 11.5 No rain
206 24-09-13 8:00 675 0.45 9 No rain
207 24-09-13 16:00 422 0.45 11 No rain
208 25-09-13 8:00 491 0.42 9 No rain
209 25-09-13 16:00 446 0.42 11 No rain
210 26-09-13 8:00 192 0.41 8 No rain
211 26-09-13 16:00 1,400 0.45 * No rain
212 27-09-13 8:00 247 0.38 8 No rain
213 27-09-13 12:30 608 * * *
214 27-09-13 16:00 593 0.44 11.5 No rain
215 28-09-13 16:00 1,442 0.43 12 No rain
216 29-09-13 16:00 287 0.42 12 No rain
217 02-10-13 8:00 189 * * *
218 02-10-13 16:00 434 0.43 9.5 No rain
219 03-10-13 8:00 103 0.39 9 No rain
220 03-10-13 16:00 283 0.43 8 No rain
Notes:
* Data not recorded
Source: Mount Kailash Energy PVT. LTD.

41
3.10 Loss due to operation of conventional settling Basin
The process for the loss calculation is done in detail for the month of Asar. For other month loss
calculation is summarized in tabular form. Only 4 months (Asar, Shrawan, Bhadra and Ashwin)
sediment data is provided by the Mount Kailash Energy PVT LTD. For other month Kartik to
Baisakh we assume sediment data is same as for the month Ashwin while for the month Jestha
Average of Asar and shrawan. We have taken unit cost for wet season Baisakh to Mansir is
NRs 4.8. And for the dry season Poush to Chaitrais is NRs. 8.4.
Table 4 : loss due to operation of conventional settling basin
Month
Power output
of the
plant(MW)
Average monthly
sediment
concentration(ppm)©
Sediment
concentration
larger than
0.133mm(25%
of C) PPM
Average
monthly
time to
Close the
plant(hour) Cost(NRs)
Asar 13.595 10700.9 2675.225 9.44 616261.16
Shrawan 13.595 2971 742.75 2.597 169469.832
Bhadra 13.595 830.67 207.6675 0.7259 47369.3304
Ashwin 13.595 590 147.5 0.515 33606.84
Kartik 10.51 590 147.5 0.515 25980.72
Mansir 5.067 590 147.5 0.515 12525.624
Poush 4.34 590 147.5 0.515 18774.84
Magh 4.1 590 147.5 0.515 17736.6
Falgun 4.54 590 147.5 0.515 19640.04
Chaitra 6.19 590 147.5 0.515 26777.94
Baisakh 6.7 590 147.5 0.515 16562.4
Jestha 13.03 6835.95 1708.9875 6.04 377765.76

42
Calculation for monthly loss due to Closure of power plant for flushing in conventional
method ( Month Asar is taken here)
Parameters Units Description Value Remarks
Time require to
emptying the
sediment tank s
(Volume of one unit/ discharge
through flushing opening) 734
Volume of
one unit is
1813.18
m3
Time require for
flushing residual
sediment s
After emptying the sediment
basin water is allow to flow
through the settling basin for
certain time 200
Time for running
the system s
Time for operating and
informing the operator at power
house 600
Total time of
closure of power
plant hour 0.426
Energy loss due
to operation of
flushing system kWh 2896.490
The cost of the
loss energy NRs
The loss of benefit for every
33.57 hour in asar 13903.153
unit cost
in wet
season is
4.8 NRs
The loss of cost
in month NRs
total loss is
((31*24)/(33.57))*13903.15 308130.58

43
Parameters Units Description Value Remarks
Power produce
from one unit MWh
The power produce from one
unit in Asar 6.80
Sediment
Concentration PPM
Average monthly sediment
concentration is taken for design 10700.9
Sediment
Concentration
for particle
larger than
0.133mm PPM
Only 25% of sediment
concentration found to be
concentration for particle larger
than 0.133mm 2675.225
Volume
available for
storing the
sedimentation m3
sedimentation volume storage of
one unit 511.5
Average
sediment
carrying by river
in 1 second Kg
Sediment concentration x
discharge in one unit 5.61
Volume of
sediment
carrying by river
in 1 second m3
Density of sedimen in insitu is
1325 kg/m3
0.004233
Average time
require to fill up
the
sedimentation
tank hour
(Volume available for
sedimentation/ Volume of
sediment carrying in 1 second) 33.57
Discharge
through the
flushing opening m3
/s
(2/3)*Cd*b*((2g)^0.5)*(H2
3/2
-
H1
3/2
) 2.47
H2= 3.5m
H1= 3m,
Cd=0.62
b=1m

44
3.11 Loss due to operation of Hydrosuction sediment Removal System
Month
Additional
discharge
require(L/S)
Loss of power
due to
additional
discharge(kW)
Monthly Time
of
operation(hour)
Cost for
additional
dishcharge(NRs)
Power require
for the
mechanical
dredging(kW)
Cost of
mechanical
dredging
Total
cost(NRs) Remarks
Asar 0 0 744 0 1 3571.2 3571.2
Motor operated
for 24 hours for
4 months when
discharge is
sufficient and
for other month
time is
calculated by
(sediment
deposition
/hydrosuction
sediment
removal
capacity)
Capacity of
hydrosuction
9.22 kg/s
Shrawan 0 0 744 0 1 3571.2 3571.2
Bhadra 0 0 744 0 1 3571.2 3571.2
Ashwin 0 0 744 0 1 3571.2 3571.2
Kartik 82.8 265.62 24.18 30828.92 1 116.064 30829.92
Mansir 82.8 265.62 24.18 30828.92 1 116.064 30829.92
Poush 82.8 265.62 24.18 53950.61 1 203.112 53951.61
Magh 82.8 265.62 24.18 53950.61 1 203.112 53951.61
Falgun 82.8 265.62 24.18 53950.61 1 203.112 53951.61
Chaitra 82.8 265.62 24.18 53950.61 1 203.112 53951.61
Baisakh 82.8 265.62 24.18 30828.92 1 116.064 30829.92
Jestha 82.8 265.62 280.19 357235.53 1 1344.912 357236.5

45
0
100
200
300
400
500
600
700
Asar Shrawan Bhadra Ashwin Kartik Mansir Poush Magh Falgun Chaitra Baisakh Jestha
Thousands
Hydrosuction Versus Conventional method for flushing (operating
cost)
Conventional
Hydrosuction
Figure 4 : Operating cost for Conventional and Hydrosuction flushing system

46
Chapter 4
Preliminary Design
Preliminary design is the process of estimation of approximate size of the structural members
for the analysis. The preliminary design bridges the gap between the design concept and the
detailed phase. As the preliminary design is carried out with respect to the deflection criteria
of the structural element, it is not always sure that the section design during this phase will
pass the analysis. This is because other parameter like EQ load, lateral forces due to earth
pressure are not taken into account.
4.1 Preliminary design of slab
4.1.1 Control room floor slab:
Taking largest clear span = 4600 mm
Shortest span = 3950 mm
Using deflection control criteria defined in IS 456: 2000, clause 23.2.1
Design as two-way slab for continuous slab and span up to 10 m, we have
L/dmin =26* modification factor
For modification factor,
fs = 0.58*fy*
𝒂𝒓𝒆𝒂 𝒐𝒇 𝒄𝒓𝒐𝒔𝒔−𝒔𝒆𝒄𝒕𝒊𝒐𝒏 𝒐𝒇 𝒔𝒕𝒆𝒆𝒍 𝒓𝒆𝒒𝒖𝒊𝒓𝒆𝒅
𝒂𝒓𝒆𝒂 𝒐𝒇 𝒄𝒓𝒐𝒔𝒔−𝒔𝒆𝒄𝒕𝒊𝒐𝒏 𝒐𝒇 𝒔𝒕𝒆𝒆𝒍 𝒑𝒓𝒐𝒗𝒊𝒅𝒆𝒅
Assuming area of steel required tends to or equal to area of steel provided so,
fs = 0.58*500*1 = 290 kN/m2
Let, us assume % of steel for tension reinforcement be Pst =0.3% of cross section area of slab
Hence, modification factor =1.25 (CL 23.2.2(c), Fig no. (4), IS 456:2000)
So, L/dmin = 26* 1.25 = 32.5
Effective depth =
𝐿
32.5
=
4600
32.5
= 141.54 ~ 150 mm
Effective cover = 25 mm
Overall depth = 150 + 25 = 175 mm
Adopting overall depth = 175 mm

47
4.2 Design of Beam
4.2.1 Control room
Longest span of beam in X- direction is 5.1 m
Span/dmin = 26 * modification factor
Modification factor = 0.7 (for 2% of steel)
Effective depth = 5100/ (26*0.7) = 280. 22 mm~285 mm
Taking effective cover = 30 mm
Therefore, overall depth, D = 315 mm
To find the width of beam (b),
Width of beam = ½ d to 2/3 d
=310/2 to (2/3) *310
= 155 mm to 206.67 mm
Taking overall width of beam (b)= 250 mm
Section of beam = 315 mm * 350 mm
Longest span of beam in Y- direction is 4.45 m
Span/dmin =26 * modification factor
Effective depth = 4450/ (26*0.7) = 244.51 mm~250 mm
Therefore, overall depth, D=280 mm
Width of beam = ½ d to 2/3d
=280/2 to (2/3) *280
=140 mm to 186.67 mm
Taking overall width of beam (b) = 200 mm
Section of beam= 200 mm * 280 mm

48
4.2.2 Machine hall
Longest span of beam in X- direction is 5.95 m
Effective depth = 5950/ (26*0.7) =326.92mm ~ 330 mm
= 360/2 to (2/3) *360
= 180 to 240
Taking overall width of beam (b)= 240 mm
Section of beam = 240 mm * 360 mm
Longest span of beam in Y- direction is 4.65 m
Effective depth = 4650/ (26*0.7) = 255.49 ~ 260 mm
= 290/2 to (2/3) *290
=145 to 193.33
Taking overall width of beam(b) = 200 mm
Section of beam = 200 mm* 290 mm

49
4.2 Design of column
Design of column along X-axis at point(C1)
Slab load
Slab thickness = 175mm
Unit Weight of concrete = 25 kN/m3
Self-weight of slab = 25× 175/1000 =4.375 kN/m2
Load due to floor finish= 1 KN /m2
Live load = 10 KN/m2
Area of trapezoidal section =
3.95∗4.6−2∗
1
2
∗3.95∗1.975
2
∗ 2 =10.36 m2
Area of triangular section=
1
2
*3.95*1.975 =3.9 m2
Total self-weight of slab= (1+4.375+10) *(10.36+3.9) = 219.25 kN
Slab load form 3 storey = 3*219.25 = 657.75 KN
Load from beam:
= bx *Dx *Lx* density + by *Dy *Ly*density
=0.315*0.350*25*5.1 +0.2*0.280*25*4.45/2
=14.05+3.115
=17.165 *4
=68.66 KN
Load from brick
Wall thickness = 9” = 0.2286 m
Length of wall = 4.61 +
3.94
2
= 6.58 m
Load from wall = 6.58*0.2286*4.5*20.4*0.9 = 89.03 * 4 = 356.1 KN
Load from column:
Self-weight of column= B*D*5*25 = 125 BD *4 = 500 BD KN
So, total load = 657.75 + 68.66 +356.15 +500 BD = 1082.56 + 500 BD KN
Taking FOS = 3 and adding 20% for earthquake consideration
Puz = 3*1.2*(1082.56+500 BD) = (3897.22 +1.8* 10-3
Ag) KN
From IS 456:2000, CL 39.6
Puz = 0.45 fck. Ac + 0. 75fy.Asc
Assume 3% of Ag steel
fck = 25 Mpa
fy=500 Mpa
BD= Ag
(3897.22+1.8*10-3
*Ag) *103
= 0.45*25*0.97*Ag + 0.75*500*0.03*Ag
Ag = 191415.52 mm2
Take B*D = 450 mm * 450 mm

50
4.3 Preliminary Design of Penstock
A penstock is a pipe that conveys the flow from the surge tank or forebay to the turbine. The
preliminary sizing of penstock can be done using continuity equation with permissible
velocity of water in pipe.
Reference steps calculation output
Continuity equation
Pacific gas and electric
formula
Beareau of reclamation
formula
1.
2.
3.
Known data:
Head = 375.9 m
Discharge = 4.2 m3/s
Permissible velocity = 5 m/s
No of turbine units = 2
Calculation of diameter:
Cross-section area, A =
𝑑𝑖𝑠ℎ𝑎𝑟𝑔𝑒
𝑝𝑒𝑟𝑚𝑖𝑠𝑠𝑖𝑏𝑙𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
Diameter, D =√
4∗𝑄
П∗𝑣
= 1.03 m
Adopt, diameter of penstock as 1.03 m
For two turbine units,
Discharge through each division of penstock
QT = Q/2 = 2.1 m3
/s
Diameter of each penstock, D =√
4∗𝑄
П∗𝑣
=0.73 m
Adopt ,0.75 diameter of each turbine.
Calculation of thickness
Tmin =
𝐷
288
= 3.58 mm
Tmin =
𝐷+200
400
= 3.075 mm
Dia. of penstock
= 1.03 m
Dia. of each unit =
0.75 m
Adopt thickness
3.58 mm

51
Chapter 5
Load Calculation
5.1 Vertical Load Calculations
Load on beams due to slab have been calculated according to Clause 24.5 of IS 456:2000.
Loads on columns are calculated by adding reactions in the beam in both directions
(transverse and longitudinal), and self-weights of a column. Factored loads are obtained by
multiplying the loads by load factor 1.5. The thickness of wall is taken 0.228 meter and the
deduction is done for windows and doors according to its location i.e. interior 10% deduction
and exterior 10% deduction.
a) Slab Load Distribution
The load of the slab is assumed to be formed in the triangular and trapezoidal form forming at
an angle 450 as discussed earlier. Such triangular and triangular load is converted into
uniformly distributed load using the formula below:
Triangular load UDL = qLx/3
Trapezoidal load UDL= qLx/6(3-m2)
Where, q = intensity of load
Lx = short span length
Ly = Long span length
m = (Lx/Ly) = Short span/Long span
b) Self-Weight of Slab
On Second Floor, First Floor and Ground Floor at Control Room
Thickness of slab = 175mm
Unit weight of concrete = 25 kN/m3
Self-weight of slab (only) = 25*0.175 = 4.375 kN/m2
Self-weight due to floor finish= 0.816 kN/m2
(from above data)
Total self-weight = 5.191 kN/m2

52
c) Self-Weight of Wall
Height of wall= 4.4 m
Unit weight of brick masonry = 20.04 kN/m3
Thickness of wall = 0.25m
Self-weight of exterior wall= 4.4*20.04*0.228*0.9 = 18.0937 kN/m (assuming 10%
deduction for doors and windows)
Self-weight of interior wall = 4.4*20.04*0.114*0.9 = 9.047 kN/m (assuming 10% deduction
for doors and windows)
Shear Wall
Height of shear wall= 4.4 m
Height of shear wall= 1.22 m
Unit weight of concrete= 25 kN/m3
Thickness of wall= 0.175 m
Self-weight of shear wall= 4.4*0.175*25 = 19.25 kN/m
Self-weight of shear wall= 1.22*0.175*25= 5.34 kN/m
d) Self-Weight Of Beam
Size of beam along x-axis and y-axis = 0.4m*0.3m (control room)
Self-weight of beam along x-axis and y-axis = 0.4*0.3*25 = 3 kN/m
Size of beam along x-axis and y-axis = 0.5m*0.4m (machine hall except roof section, roof
section has size that of control room)
Self-weight of beam along x-axis and y-axis = 0.5*0.4*25 = 5 kN/m
e) Self-Weight of Column
Size of column= 0.5m*0.5m (control room)
Self-weight of column= 0.5*0.5*25=6.25 kN/m
Size of column = 0.6m*0.6m (machine hall)
Self-weight of column = 0.6*0.6*25 = 9 kN/m

54
5.1.1 Vertical load calculations of control room
Table 5: Dead load calculation of control room
a) Slab load converted into UDL to transform the load
Slab Dead Load on the Second, First and Control Level floor: Beam along Y-axis
Beam
ID
Beam
Length(m)
Slab
ID
Intensity
D.L (kN/m)
Lx
(m)
Ly
(m)
Lx/Ly
Slab load
kN/m
Total
UDL due
to slab
Total
Load(kN)
BEAM ALONG Y-AXIS
A(1-2) 3.95 S1 5.191 3.95 4.6 0.859 6.835 6.057 23.924
B(1-2)
3.95 S1 5.191 3.95 4.6 0.859 6.835
13.670 53.9953.95 S2 5.191 3.95 4.6 0.859 6.835
C(1-2)
3.95 S2 5.191 3.95 4.6 0.859 6.835
13.670 53.9953.95 S3 5.191 3.95 4.6 0.859 6.835
D(1-2)
3.95 S3 5.191 3.95 4.6 0.859 6.835
13.670 53.9953.95 S4 5.191 3.95 4.6 0.859 6.835
E(1-2) 3.95 S4 5.191 3.95 4.6 0.859 6.835 6.835 26.998
TOTAL 212.907

55
Slab Dead Load on the Second, First and Control Level floor: Beam along X-axis
Beam
ID
Beam
Length(m)
Slab ID
Intensity
D.L kN/m
Lx (m) Ly (m) Lx/Ly
Slab
Load
kN/m
Total
UDL due
to Slab
(kN/m)
Total
Load(kN)
BEAM ALONG X-AXIS
1(A-B) 4.6 S1 5.191 3.95 4.6 0.859 7.732 7.732 35.569
2(A-B) 4.6 S1 5.191 3.95 4.6 0.859 7.732 7.732 35.569
1(B-C) 4.6 S2 5.191 3.95 4.6 0.859 7.732 7.732 35.569
2(B-C) 4.6 S2 5.191 3.95 4.6 0.859 7.732 7.732 35.569
1(C-D) 4.6 S3 5.191 3.95 4.6 0.859 7.732 7.732 35.569
2(C-D) 4.6 S3 5.191 3.95 4.6 0.859 7.732 7.732 35.569
1(D-E) 4.6 S4 5.191 3.95 4.6 0.859 7.732 7.732 35.569
2(D-E) 4.6 S4 5.191 3.95 4.6 0.859 7.732 7.732 35.569
TOTAL 284.551
Slab Live Load on the Second, First and Control Level floor: Beam along Y-axis
Beam
ID
Beam
Length(m)
Slab ID
Intensity
D.L kN/m
Lx (m)
Ly
(m)
Lx/Ly
Slab
Load
kN/m
Total
UDL due
to Slab
(kN/m)
Total
Load(kN)
BEAM ALONG Y-AXIS
A(1-2) 3.95 S1 10 3.95 4.6 0.859 13.167 13.167 52.008
B(1-2)
3.95 S1 10 3.95 4.6 0.859 13.167
26.333 104.0173.95 S2 10 3.95 4.6 0.859 13.167
C(1-2)
3.95 S2 10 3.95 4.6 0.859 13.167
26.333 104.0173.95 S3 10 3.95 4.6 0.859 13.167
D(1-2)
3.95 S3 10 3.95 4.6 0.859 13.167
26.333 104.0173.95 S4 10 3.95 4.6 0.859 13.167
E(1-2) 3.95 S4 10 3.95 4.6 0.859 13.167 13.167 52.008
TOTAL 416.067

56
b) Wall Loads
Load from wall ground floor and first floor
Beam
ID
Beam
Length(m)
Wall Load after
deduction(kN/m)
Wall
Load
(kN)
Beam
ID
Beam
Length(m)
Wall Load after
deduction(kN/m)
Wall
Load
(kN)
External wall deduction 10% for doors and
windows
Internal wall deduction 10% for doors and
windows
1(A-B) 4.6 18.916 87.014 B(1-2) 3.95 8.122 37.359
2(A-B) 4.6 18.916 87.014 C(1-2) 3.95 8.122 37.359
1(B-C) 4.6 18.916 87.014 D(1-2) 3.95 8.122 37.359
2(B-C) 4.6 18.916 87.014
1(C-D) 4.6 18.916 87.014
2(C-D) 4.6 18.916 87.014
1(D-E) 4.6 18.916 87.014
2(D-E) 4.6 18.916 87.014
A(1-2) 3.95 16.243 74.719
E(1-2) 3.95 16.243 74.719
TOTAL 957.630

57
Load from wall second floor
Beam
ID
Beam
Length(m)
Wall Load after
deduction(kN/m)
Wall
Load
(kN)
Beam
ID
Beam
Length(m)
Wall Load after
deduction(kN/m)
Wall
Load
(kN)
External wall deduction 10% for doors and
windows
Internal wall deduction 10% for doors and
windows
1(A-B) 4.6 18.916 52.965 B(1-2) 3.95 8.122 22.741
2(A-B) 4.6 18.916 52.965 C(1-2) 3.95 8.122 22.741
1(B-C) 4.6 18.916 52.965 D(1-2) 3.95 8.122 22.741
2(B-C) 4.6 18.916 52.965
1(C-D) 4.6 18.916 52.965
2(C-D) 4.6 18.916 52.965
1(D-E) 4.6 18.916 52.965
2(D-E) 4.6 18.916 52.965
A(1-2) 3.95 16.243 45.481
E(1-2) 3.95 16.243 45.481
TOTAL 582.905
Load from Wall Basement
Beam
ID
Beam
Length(m)
Wall
Load(kN/m)
Wall
Load
(kN)
1(A-B) 4.6 21.018 96.683
2(A-B) 4.6 21.018 96.683
B(1-2) 3.95 18.048 83.021
D(1-2) 3.95 18.048 83.021
1(D-E) 4.6 21.018 96.683
2(D-E) 4.6 21.018 96.683
A(1-2) 3.95 18.048 83.021
E(1-2) 3.95 18.048 83.021
TOTAL 718.814

58
c) Column Loads
Column Load at Roof Level
Column
Name
Column
Span
(m)
Column
load
(kN/m)
Total
Column
Self
Load
(kN)
Beams
Associated
Span of
Beam (m)
Self-weight
of beam
(kN/m)
Weight of
Beam (kN)
Load from
roof truss
(kN)
Load on
column
(kN)
1A
0 0
0
1(A-B) 4.6 3 13.8
5.185 15.4180 0 A(1-2) 3.95 3 11.85
1B
0 0
0
1(A-B) 4.6 3 13.8
5.185 16.3930 0 1(B-C) 4.6 3 13.8
1C
0 0
0
1(B-C) 4.6 3 13.8
5.185 16.3930 0 1(C-D) 4.6 3 13.8
1D
0 0
0
1(C-D) 4.6 3 13.8
5.185 16.3930 0 1(D-E) 4.6 3 13.8
1E
0 0
0
1(D-E) 4.6 3 13.8
5.185 15.4180 0 E(1-2) 3.95 3 11.85
2A
0 0
0
2(A-B) 4.6 3 13.8
5.185 15.4180 0 A(1-2) 3.95 3 11.85
2B
0 0
0
2(A-B) 4.6 3 13.8
5.185 16.3930 0 2(B-C) 4.6 3 13.8
2C
0 0
0
2(C-D) 4.6 3 13.8
5.185 16.3930 0 2(B-C) 4.6 3 13.8
2D
0 0
0
2(C-D) 4.6 3 13.8
5.185 16.3930 0 2(D-E) 4.6 3 13.8
2E
0 0
0
2(D-E) 4.6 3 13.8
5.185 15.4180 0 E(1-2) 3.95 3 11.85

59
Column Load at First Floor Level
Column
Name
Column
Span (m)
Column
load
(kN/m)
Total
Column
Self
Load
(kN)
Beams
Associated
Span
of
Beam
(m)
Self-weight
of beam
(kN/m)
Weight of
Beam (kN)
Dead load
from Slab
(kN)
Live Load
from Slab
(kN)
Dead load
of Walls
(kN)
Column
Load (kN)
1A 4.43 6.25 27.688 1(A-B) 4.6 3 13.8 35.57 68.52 52.965 179.746
A(1-2) 3.95 3 11.85 23.924 52.008 45.48
1B 4.43 6.25 27.688 1(A-B) 4.6 3 13.8 35.57 68.52 52.965 267.1559
B(1-2) 3.95 3 11.85 53.995 104.017 22.74
1(B-C) 4.6 3 13.8 35.57 68.52 52.965
1C 4.43 6.25 27.688 1(B-C) 4.6 3 13.8 35.57 68.52 52.965 267.1559
C(1-2) 3.95 3 11.85 53.995 104.017 22.74
1(C-D) 4.6 3 13.8 35.57 68.52 52.965
1D 4.43 6.25 27.688 1(C-D) 4.6 3 13.8 35.57 68.52 52.965 294.1167
D(1-2) 3.95 3 11.85 53.995 104.017 22.74
1(D-E) 4.6 3 13.8 53.995 104.017 52.965
1E 4.43 6.25 27.688 1(D-E) 4.6 3 13.8 23.924 52.008 52.965 165.667
E(1-2) 3.95 3 11.85 23.924 52.008 45.48
2A 4.43 6.25 27.688 2(A-B) 4.6 3 13.8 35.57 68.52 52.965 179.746
A(1-2) 3.95 3 11.85 23.924 52.008 45.48
2B 4.43 6.25 27.688 2(A-B) 4.6 3 13.8 35.57 68.52 52.965 206.441
B(1-2) 4.6 3 13.8 53.995 104.017 22.74
2(B-C) 4.6 3 13.8 35.57 68.52 52.965
2C 4.43 6.25 27.688 2(C-D) 4.6 3 13.8 35.57 68.52 52.965 206.441
C(1-2) 4.6 3 13.8 53.995 104.017 22.74
2(B-C) 4.6 3 13.8 35.57 68.52 52.965
2D 4.43 6.25 27.688 2(C-D) 4.6 3 13.8 35.57 68.52 52.965 206.441
D(1-2) 4.6 3 13.8 53.995 104.017 22.74
2(D-E) 4.6 3 13.8 35.57 68.52 52.965
2E 4.43 6.25 27.688 2(D-E) 4.6 3 13.8 35.57 68.52 52.965 179.746
E(1-2) 3.95 3 11.85 23.924 52.008 45.48
TOTAL 2152.658

60
Column Load at Control Level Floor
Column
Name
Colu
mn
Span
(m)
Column
load
(kN/m)
Total
Column
Self
Load
(kN)
Beams
Associated
Span of
Beam
(m)
Self-weight
of beam
(kN/m)
Weight of
Beam (kN)
Dead load
from Slab
(kN)
Live
Load
from
Slab (kN)
Dead load
of Walls
(kN)
Column
Load (kN)
1A 5 6.25 31.25 1(A-B) 4.6 3 13.8 35.57 68.52 52.965 183.309
A(1-2) 3.95 3 11.85 23.924 52.008 45.48
1B 5 6.25 31.25 1(A-B) 4.6 3 13.8 35.57 68.52 52.965 178.104
B(1-2) 3.95 3 11.85 53.995 104.017 22.74
1(B-C) 4.6 3 13.8 35.57 68.52 52.965
1C 5 6.25 31.25 1(B-C) 4.6 3 13.8 35.57 68.52 52.965 178.104
C(1-2) 3.95 3 11.85 53.995 104.017 22.74
1(C-D) 4.6 3 13.8 35.57 68.52 52.965
1D 5 6.25 31.25 1(C-D) 4.6 3 13.8 35.57 68.52 52.965 196.078
D(1-2) 3.95 3 11.85 53.995 104.017 22.74
1(D-E) 4.6 3 13.8 53.995 104.017 52.965
1E 5 6.25 31.25 1(D-E) 4.6 3 13.8 23.924 52.008 52.965 169.230
E(1-2) 3.95 3 11.85 23.924 52.008 45.48
2A 5 6.25 31.25 2(A-B) 4.6 3 13.8 35.57 68.52 52.965 183.309
A(1-2) 3.95 3 11.85 23.924 52.008 45.48
2B 5 6.25 31.25 2(A-B) 4.6 3 13.8 35.57 68.52 52.965 210.004
B(1-2) 4.6 3 13.8 53.995 104.017 22.74
2(B-C) 4.6 3 13.8 35.57 68.52 52.965
2C 5 6.25 31.25 2(C-D) 4.6 3 13.8 35.57 68.52 52.965 210.004
C(1-2) 4.6 3 13.8 53.995 104.017 22.74
2(B-C) 4.6 3 13.8 35.57 68.52 52.965
2D 5 6.25 31.25 2(C-D) 4.6 3 13.8 35.57 68.52 52.965 210.004
D(1-2) 4.6 3 13.8 53.995 104.017 22.74
2(D-E) 4.6 3 13.8 35.57 68.52 52.965
2E 5 6.25 31.25 2(D-E) 4.6 3 13.8 35.57 68.52 52.965 183.309
E(1-2) 3.95 3 11.85 23.924 52.008 45.48
TOTAL 1901.452

61
Column Load at Basement
Column
Name
Column
Span
(m)
Column
load
(kN/m)
Total
Column
Self Load
(kN)
Beams
Associated
Span of Beam
(m)
Self-weight of
beam (kN/m)
Weight of
Beam (kN)
Dead load
of Walls
(kN)
Column
Load (kN)
1A 5 6.25 31.25 1(A-B) 4.6 3 13.8 52.965 93.298
A(1-2) 3.95 3 11.85 45.48
1B 5 6.25 31.25 1(A-B) 4.6 3 13.8 52.965 69.635
B(1-2) 3.95 3 11.85 22.74
1(B-C) 4.6 3 13.8 0
1C 5 6.25 31.25 1(B-C) 4.6 3 13.8 0 51.98
C(1-2) 3.95 3 11.85 22.74
1(C-D) 4.6 3 13.8 0
1D 5 6.25 31.25 1(C-D) 4.6 3 13.8 0 69.635
D(1-2) 3.95 3 11.85 22.74
1(D-E) 4.6 3 13.8 52.965
1E 5 6.25 31.25 1(D-E) 4.6 3 13.8 52.965 93.298
E(1-2) 3.95 3 11.85 45.48
2A 5 6.25 31.25 2(A-B) 4.6 3 13.8 52.965 93.298
A(1-2) 3.95 3 11.85 45.48
2B 5 6.25 31.25 2(A-B) 4.6 3 13.8 52.965 70.285
B(1-2) 4.6 3 13.8 22.74
2(B-C) 4.6 3 13.8 0
2C 5 6.25 31.25 2(C-D) 4.6 3 13.8 0 52.63
C(1-2) 4.6 3 13.8 22.74
2(B-C) 4.6 3 13.8 0
2D 5 6.25 31.25 2(C-D) 4.6 3 13.8 0 70.285
D(1-2) 4.6 3 13.8 22.74
2(D-E) 4.6 3 13.8 52.965
2E 5 6.25 31.25 2(D-E) 4.6 3 13.8 52.965 93.298
E(1-2) 3.95 3 11.85 45.48
TOTAL 757.64

62
5.2 lateral load calculations
Lateral loads or horizontal forces applied on respective floor of the building have been
computed using “Seismic Coefficient Method”. The horizontal forces due to earthquake
shock are determined and the next step was to find the design forces in the members due to
these lateral forces.
5.2.1 Seismic Weight
Seismic weight is the total dead load plus appropriate amount of specified amount of imposed
load. The weight of columns and walls in any storey shall be equally distributed to the floors
above and below the storey. The seismic weight of the building is the sum of seismic weight
of each floor.
Seismic weight = DL +0.25LL (For LL up to 3 kN/m2)
= DL +0.5LL (For LL greater than 3 kN/m2)
Table 6: Calculation of total load above the mentioned location of Control Room
Seismic load calculations Control Room
Location
DL of
Wall
(kN)
DL of
Slabs
(kN)
DL of
Beams
(kN)
Dead Load
of Column
(kN)
Load
from
Truss
(kN)
Total
DL
(KN)
Total LL
(i.e. Slab
LL) (kN)
Top of column 0 0 268.2 0 51.85 320.05 0
Second Floor
bottom
1165.8 995.565 345.15 200 0 2706.515 1947.437
First Floor
bottom
1165.8 995.565 345.15 276.875 0 2783.39 1947.437
Ground floor
bottom
1165.8 995.565 345.15 312.5 0 2819.015 1947.437

63
Table 7: Calculation of Seismic weight of Control Room
5.2.3 Calculation of Fundamental Natural Period of Vibration of the Building
According to IS 1893:2002 Clause 7.6.1
The approximate fundamental natural period of vibration (Ta), in seconds, of a moment-
resisting frame building without brick infill panels may be estimated by the empirical
expression:
Ta = 0.075 h0.75
(for RC frame building)
where,
h = height of building, in m. This excludes the basement storeys, where basement walls are
connected with the ground floor deck or fitted between the building columns. But it includes
the basement storeys, when they are not so connected.
Calculation of Seismic weight of Control Room
Joint
Dead
Load
Live
Load
Seismic Weight
(DL + 50 %of
LL)
Total Seismic
Weight (W)
Top of column 320.05 0 320.05
11550.125
Second Floor
bottom 2706.515 1947.437 3680.233
First Floor bottom 2783.39 1947.437 3757.108
Ground floor
bottom 2819.015 1947.437 3792.733

64
The approximate fundamental natural period of vibration ( Ta ), in seconds, of all other
buildings, including moment-resisting frame buildings with brick infill panels, may be
estimated by the empirical
expression:
Ta = (0.09 h)/(√d)
where,
h = height of building as defined above, and
d = Base dimension of the building at the plinth level in m, along the considered direction of
the lateral force.
Table 8: Calculation of Natural time period in each direction
Calculation of Natural time period in each direction
Direction
Height
(m)
Dimension
(m)
Ta as per
Clause
7.6.1
Ta as per
Clause
7.6.2
X-direction 12.63 20.9 0.50 0.25
Y-direction 12.63 4.95 0.50 0.51
Hence, the fundamental natural period of vibration (Ta) is taken as 0.51.

65
5.2.4 Base Shear Calculation
The total design lateral force or design seismic base shear (Vb) along any principal direction
shall be determined by the following expression:
Vb = Ah*W
Where,
Ah = Design horizontal acceleration spectrum value as per 6.4.2 (IS 1893:2002), using the
fundamental natural period T, as per 7.6 in the considered direction of vibration, and
W = Seismic weight of the building as per 7.4.2 (IS 1893:2002).
The design horizontal seismic coefficient Ah for a structure shall be determined by the
following expression:
Ah= (Z/2)*(I/R)*(Sa/g) (IS 1893 (part I):2000 Clause 6.4.2)
Provided that for any structure with T ≤ 0.1s, the value of Ah will not be taken less than Z/2
whatever be the value of I/R
Where,
Z= Zone factor for the Maximum Considered Earthquake (MCE) and service life of structure
in a zone. The factor 2 in the denominator of Z is used so as to reduce the Maximum
Considered Earthquake (MCE) zone factor to the factor for Design Basic Earthquake (DBE).
Zone Factors
Seismic Zone II III IV V
Seismic Intensity Low Moderate Severe Very Severe
Z 0.1 0.16 0.24 0.36

66
I= Importance factor, depending upon the functional use of the structures, characterized by
hazardous consequences of its failure, post-earthquake functional need, historical value, or
economic importance.
I=1.5 for important services and community buildings, such as hospitals, schools,
monumental structures, emergency building like telephone exchange, television station, radio
station, fire station buildings, large community halls and power stations.
I=1 for all other buildings.
R= Response reduction factor, depending on the perceived seismic damage performances of
the structure, characterized by ductile and brittle deformations. However the ratio (I/R) shall
not be greater than 1.
R=5 for Special RCC Moment Resisting Frame (SMRF)
Sa/g= Average response acceleration coefficient based on appropriate natural periods and
damping of the structure.
For medium Sa/g= 1+15 T {0.0<T<0.10}
=2.50 {0.10<T<0.55}
=1.36/T {0.55 <T<4.0}
Base shear= Ah*Wi
To find Ah
Z=0.36 (for seismic zone V)
R=5 (response reduction factor)
I=1.5 (importance factor)
Sa/g = 2.5 (spectral acceleration depending upon the period of vibration and
Damping 5%
g= acceleration due to gravity
Hence, Ah = (Z/2)*(I/R)*(Sa/g)
= (0.36/2)*(1.5/5)*(2.5)
=0.135

67
Table 9: Calculation of Base Shear
Calculation of Base Shear
Z I R Ta Sa/g Ah W (kN) Vb=Ah*W (kN)
0.36 1.5 5 0.511 2.5 0.135 11550.125 1559.267
5.2.5 Lateral Load Distribution and Storey Shear
Vertical distribution of base shear to different floor levels
The design base shear Vb computed can be distributed along the height of the building as per
following expressions:
Qi = (Vb*hi2
*Wi) / (ΣWi*hi2
)
Where, i is from 1 to n
n= number of storey in the building at which the mass is located
Qi= Design lateral force at each floor
hi= height of floor i measured from base
Wi= seismic weight of each floor

68
Table 10: Lateral Load distribution at a joint above ground level of Control Room
Lateral Load Distribution at a Joint above Ground Level
S.N. Storey
Weight(Wi)
(kN)
Height
(hi ) (m)
hi^2 Wi*hi2
Lateral
Force
(Qi)
(kN)
Storey
Shear(Fi)
(kN)
1 Top of column 320.05 12.63 159.517 51053.384 220.323 220.323
2
Second Floor
bottom 2706.515 9.43 88.925 240676.576 1038.649 1258.971
3
First Floor
bottom 2783.39 5 25 69584.75 300.296 1559.267
4
Ground floor
bottom 2819.015 0 0 0 0 1559.267
TOTAL 361314.71

69
5.3 Vertical Load Calculations of Machine Hall
Table 11: Vertical load calculation of machine hall
a) Wall Loads
Load from wall on Machine Hall Floor
Beam ID
Beam
length (m)
Shear wall
load(kN/m)
Total shear
wall load (kN)
Wall load after
deduction(kN/m)
Wall load (kN) Total wall load (kN)
3(A-B) 4.5 0 0 19.533 87.898 87.898
3(B-C) 4.5 0 0 19.533 87.898 87.898
3(C-D) 4.5 0 0 19.533 87.898 87.898
3(D-E) 4.5 0 0 19.533 87.898 87.898
3(E-F) 5.35 0 0 19.533 104.501 104.501
8(A-B) 4.5 19.25 86.625 19.533 0 86.625
8(B-C) 4.5 19.25 86.625 19.533 0 86.625
8(C-D) 4.5 19.25 86.625 19.533 0 86.625
8(D-E) 4.5 19.25 86.625 19.533 0 86.625
8(E-F) 5.35 19.25 102.988 19.533 0 102.988
A(3-5) 4.05 5.3375 21.617 19.533 13.803 35.420
A(5-6) 3.9 5.3375 20.816 19.533 13.803 34.620
A(6-8) 3.95 5.3375 21.083 19.533 13.803 34.886
E(3-5) 4.05 0 0 19.533 79.109 79.109
E(5-6) 3.9 0 0 19.533 76.179 76.179
E(6-8) 3.95 0 0 19.533 77.155 77.155
TOTAL 1242.952

70
Load from wall on Machine Hall first floor
Beam ID
Beam
length
(m)
Shear wall
load(kN/m)
Total shear
wall load
(kN)
Wall load after
deduction(kN/m)
Wall load (kN) Total wall load
(kN)
3(A-B) 4.5 0 0 19.099 85.945 85.945
3(B-C) 4.5 0 0 19.099 85.945 85.945
3(C-D) 4.5 0 0 19.099 85.945 85.945
3(D-E) 4.5 0 0 19.099 85.945 85.945
3(E-F) 5.35 0 0 19.099 102.179 102.179
8(A-B) 4.5 0 0 19.099 85.945 85.945
8(B-C) 4.5 0 0 19.099 85.945 85.945
8(C-D) 4.5 0 0 19.099 85.945 85.945
8(D-E) 4.5 0 0 19.099 85.945 85.945
8(E-F) 5.35 0 0 19.099 102.179 102.179
A(3-5) 4.05 0 0 19.099 77.351 77.351
A(5-6) 3.9 0 0 19.099 74.486 74.486
A(6-8) 3.95 0 0 19.099 75.441 75.441
E(3-5) 4.05 0 0 19.099 77.351 77.351
E(5-6) 3.9 0 0 19.099 74.486 74.486
E(6-8) 3.95 0 0 19.099 75.441 75.441
TOTAL 1346.474

71
Load from wall on Machine Hall second floor
Beam ID
Beam length
(m)
Shear wall
load(kN/m)
Total shear
wall load (kN)
Wall load after
deduction(kN/m)
Wall load (kN) Total wall load
(kN)
3(A-B) 4.5 0 0 20.965 94.344 94.344
3(B-C) 4.5 0 0 20.965 94.344 94.344
3(C-D) 4.5 0 0 20.965 94.344 94.344
3(D-E) 4.5 0 0 20.965 94.344 94.344
3(E-F) 5.35 0 0 20.965 112.165 112.165
8(A-B) 4.5 0 0 20.965 94.344 94.344
8(B-C) 4.5 0 0 20.965 94.344 94.344
8(C-D) 4.5 0 0 20.965 94.344 94.344
8(D-E) 4.5 0 0 20.965 94.344 94.344
8(E-F) 5.35 0 0 20.965 112.165 112.165
A(3-5) 4.05 0 0 20.965 84.910 84.910
A(5-6) 3.9 0 0 20.965 81.765 81.765
A(6-8) 3.95 0 0 20.965 82.813 82.813
E(3-5) 4.05 0 0 20.965 84.910 84.910
E(5-6) 3.9 0 0 20.965 81.765 81.765
E(6-8) 3.95 0 0 20.965 82.813 82.813
TOTAL 1478.061

72
b) Column Loads
Column Load at Machine Hall Floor
Column
Name
Column
Span (m)
Column load
(kN/m)
Total Column
Self Load (kN)
Beams
Associated
Span of
Beam (m)
Self-weight
of beam
(kN/m)
Weight
of
Beam
(kN)
Dead
load of
Walls
(kN)
Column
Load
(kN)
3A 5
9
45
3(A-B) 4.5 5 22.5 85.945
127.058A(3-5) 4.05 5 20.25 35.42
3B 5 9 45
3(A-B) 4.5 5 22.5 85.945
153.7853(B-C) 4.5 5 22.5 86.625
3C 5 9 45
3(B-C) 4.5 5 22.5 86.625
153.7853(C-D) 4.5 5 22.5 85.945
3D 5 9 45
3(C-D) 4.5 5 22.5 85.945
153.4453(D-E) 4.5 5 22.5 85.945
3E 5 9 45
3(D-E) 4.5 5 22.5 85.945
163.6873(E-F) 5.35 5 26.75 102.179
3F 5 9 45
3(E-F) 5.35 5 26.75 102.179
158.265F(3-5) 4.05 5 20.25 77.35
8A 5
9
45
8(A-B) 4.5 5 22.5 85.945
146.818A(6-8) 3.95 5 19.75 75.44
8B 5 9 45
8(A-B) 4.5 5 22.5 85.945
153.7858(B-C) 4.5 5 22.5 86.625
8C 5 9 45
8(B-C) 4.5 5 22.5 86.625
153.7858(C-D) 4.5 5 22.5 85.945
8D 5 9 45
8(C-D) 4.5 5 22.5 85.945
153.4458(D-E) 4.5 5 22.5 85.945
8E 5 9 45
8(D-E) 4.5 5 22.5 85.945
163.6878(E-F) 5.35 5 26.75 102.179
8F 5 9 45
8(E-F) 5.35 5 26.75 102.179
157.060F(6-8) 3.95 5 19.75 75.44

73
Column Load at Machine Hall first floor
Column
Name
Column
Span (m)
Column
load
(kN/m)
Total
Column
Self Load
(kN)
Beams
Associated
Span of
Beam (m)
Self-
weight of
beam
(kN/m)
Weight
of Beam
(kN)
Dead load
of Walls
(kN)
Column
Load (kN)
3A 4.9
9
44.1
3(A-B) 4.5 5 22.5 85.945
126.1575A(3-5) 4.05 5 20.25 35.42
3B 4.9 9 44.1
3(A-B) 4.5 5 22.5 85.945
152.8853(B-C) 4.5 5 22.5 86.625
3C 4.9 9 44.1
3(B-C) 4.5 5 22.5 86.625
152.8853(C-D) 4.5 5 22.5 85.945
3D 4.9 9 44.1
3(C-D) 4.5 5 22.5 85.945
152.5453(D-E) 4.5 5 22.5 85.945
3E 4.9 9 44.1
3(D-E) 4.5 5 22.5 85.945
162.7873(E-F) 5.35 5 26.75 102.179
3F 4.9 9 44.1
3(E-F) 5.35 5 26.75 102.179
157.3645F(3-5) 4.05 5 20.25 77.35
8A 4.9
9
44.1
8(A-B) 4.5 5 22.5 85.945
145.9175A(6-8) 3.95 5 19.75 75.44
8B 4.9 9 44.1
8(A-B) 4.5 5 22.5 85.945
152.8858(B-C) 4.5 5 22.5 86.625
8C 4.9 9 44.1
8(B-C) 4.5 5 22.5 86.625
152.8858(C-D) 4.5 5 22.5 85.945
8D 4.9 9 44.1
8(C-D) 4.5 5 22.5 85.945
152.5458(D-E) 4.5 5 22.5 85.945
8E 4.9 9 44.1
8(D-E) 4.5 5 22.5 85.945
162.7878(E-F) 5.35 5 26.75 102.179
8F 4.9 9 44.1
8(E-F) 5.35 5 26.75 102.179
156.1595F(6-8) 3.95 5 19.75 75.44

74
Column Load at Machine Hall second floor
Column
Name
Column
Span (m)
Column
load
(kN/m)
Total
Column
Self Load
(kN)
Beams
Associated
Span of
Beam (m)
Self-
weight of
beam
(kN/m)
Weight
of Beam
(kN)
Dead load
of Walls
(kN)
Column
Load (kN)
3A 5.33
9
47.97
3(A-B) 4.5 5 22.5 85.945
503.5275A(3-5) 4.05 5 20.25 35.42
3B 5.33 9 47.97
3(A-B) 4.5 5 22.5 85.945
530.2553(B-C) 4.5 5 22.5 86.625
3C 5.33 9 47.97
3(B-C) 4.5 5 22.5 86.625
530.2553(C-D) 4.5 5 22.5 85.945
3D 5.33 9 47.97
3(C-D) 4.5 5 22.5 85.945
529.9153(D-E) 4.5 5 22.5 85.945
3E 5.33 9 47.97
3(D-E) 4.5 5 22.5 85.945
540.1573(E-F) 5.35 5 26.75 102.179
3F 5.33 9 47.97
3(E-F) 5.35 5 26.75 102.179
534.7345F(3-5) 4.05 5 20.25 77.35
8A 5.33
9
47.97
8(A-B) 4.5 5 22.5 85.945
523.2875A(6-8) 3.95 5 19.75 75.44
8B 5.33 9 47.97
8(A-B) 4.5 5 22.5 85.945
530.2558(B-C) 4.5 5 22.5 86.625
8C 5.33 9 47.97
8(B-C) 4.5 5 22.5 86.625
530.2558(C-D) 4.5 5 22.5 85.945
8D 5.33 9 47.97
8(C-D) 4.5 5 22.5 85.945
529.9158(D-E) 4.5 5 22.5 85.945
8E 5.33 9 47.97
8(D-E) 4.5 5 22.5 85.945
540.1578(E-F) 5.35 5 26.75 102.179
8F 5.33 9 47.97 8(E-F) 5.35 5 26.75 102.179
533.5295F(6-8) 3.95 5 19.75 75.44

75
Column Load at Machine Hall at roof
Column
Name
Column
Span (m)
Column
load
(kN/m)
Total
Column Self
Load (kN)
Beams
Associated
Span of
Beam (m)
Self-weight
of beam
(kN/m)
Weight of
Beam
(kN)
Truss load
(kN)
Column
Load (kN)
3A 0
9
0
3(A-B) 4.5 3 13.5
14.258 22.954A(3-5) 4.05 3 12.15
3B 0 9 0
3(A-B) 4.5 3 13.5
14.258 23.6293(B-C) 4.5 3 13.5
3C 0 9 0
3(B-C) 4.5 3 13.5
14.258 23.6293(C-D) 4.5 3 13.5
3D 0 9 0
3(C-D) 4.5 3 13.5
14.258 23.6293(D-E) 4.5 3 13.5
3E 0 9 0
3(D-E) 4.5 3 13.5
14.258 24.9043(E-F) 5.35 3 16.05
3F 0 9 0
3(E-F) 5.35 3 16.05
14.258 24.229F(3-5) 4.05 3 12.15
8A 0
9
0
8(A-B) 4.5 3 13.5
14.258 22.804A(6-8) 3.95 3 11.85
8B 0 9 0
8(A-B) 4.5 3 13.5
14.258 23.6298(B-C) 4.5 3 13.5
8C 0 9 0
8(B-C) 4.5 3 13.5
14.258 23.6298(C-D) 4.5 3 13.5
8D 0 9 0
8(C-D) 4.5 3 13.5
14.258 23.6298(D-E) 4.5 3 13.5
8E 0 9 0
8(D-E) 4.5 3 13.5
14.258 24.9048(E-F) 5.35 3 16.05
8F 0 9 0
8(E-F) 5.35 3 16.05
14.258 24.079F(6-8) 3.95 3 11.85

76
Table 12: Seismic load calculation
Seismic load calculations Machine Hall
Location
DL of Wall (kN) DL of
Slabs (kN)
DL of
Beams
(kN)
Dead Load
of Column
(kN)
Total Load
of Truss
(kN)
Total DL (KN) Total Live
Load (kN)
Top of column 0 0 328.2 0 171.096 499.296 0
Second Floor bottom 2050.206 0 547 575.64 0 3172.846 373.5
First Floor bottom 2050.206 0 547 529.2 0 3126.406 0
Ground floor bottom 2050.206 0 547 540 0 3137.206 0
Calculation of Seismic weight of control room
Joint Dead Load Live Load
Seismic Weight
(DL + 50 %of
LL)
Total Seismic Weight
(W)
Top of column 499.296 0 499.296
10122.504
Second Floor bottom 3172.846 373.5 3359.596
First Floor bottom 3126.406 0 3126.406
Ground floor bottom 3137.206 0 3137.206

77
Calculation of Natural time period in each direction
Direction
Height (m) Dimension (m)
Ta as per
Clause 7.6.1
Ta as per Clause
7.6.1
X-direction 15.23 26.95 0.58 0.26
Y-direction 15.23 14.3 0.58 0.36
Calculation of Base Shear
Z I R Ta Sa/g Ah W (kN) Vb=Ah*W (kN)
0.36 1.5 5 0.511 2.5 0.135 10122.504 1366.538
Lateral Load distribution at a joint above ground level
S.N. Storey Weight(Wi) (kN)
Height (hi )
(m)
hi^2 Wi*hi2 Lateral Force (Qi)
(kN)
Storey Shear(Fi)
(kN)
1 Top of column 499.296 15.23 231.9529 115813.155 313.427 313.427
2 Second Floor bottom 3172.846 9.9 98.01 310970.636 841.585 1155.012
3 First Floor bottom 3126.406 5 25 78160.15 211.526 1366.538
4 Ground floor bottom 3137.206 0 0 0 0 1366.538
TOTAL 504943.9416

78
5.4 Load Combination
Different load cases and load combination cases are considered to obtain most critical element stresses in the structure in the course of
analysis.
There are all together four load cases considered for the structural analysis and are mentioned as below:
Dead Load (D.L.)
Live load (L.L.)
Earthquake Load in X-direction (E.QX)
Earthquake Load in Y-direction (E.QY)
Following load combination are adopted as per IS 1893(Part I):2002 Cl.No.6.3.1.2
1. UDCON1 (1.5 DL)
2. UDCON2 (1.5 DL + 1.5 LL)
3. UDCON3 (1.2 DL + 1.2 LL +1.2 EQx)
4. UDCON4 (1.2 DL + 1.2 LL -1.2 EQx)
5. UDCON5 (1.2 DL + 1.2 LL +1.2 EQY)
6. UDCON6 (1.2 DL + 1.2 LL -1.2 EQY)
7. UDCON7 (1.5 DL + 1.5 EQx)
8. UDCON8 (1.5 DL - 1.5 EQx)
9. UDCON9 (1.5 DL + 1.5 EQY)
10. UDCON10 (1.5 DL - 1.5 EQY)
11. UDCON11 (0.9 DL + 1.5 EQx)
12. UDCON12 (0.9 DL - 1.5 EQx)
13. UDCON13 (0.9DL + 1.5 EQY)
14. UDCON13 (0.9DL - 1.5 EQY)

79
Chapter 6
Structural Design
6.1. Introduction
It includes the manual calculations of the moment, shear force which determine the size,
reinforcement of different structural component. In case of slab, beam and mat foundation, we
have used moment coefficient method mentioned in IS 456:2000 and RCC gantry beam has been
designed as a simply supported beam. Corbel and staircase has been designed according to IS.
Design of column requires moment and for which we have to analyze the frame structure. The
moment has been taken from SAP 2000v 14 for the manual design of column calculation. As our
site is in the seismic prone zone, the detailing of component is based on IS 13920:1993(ductile
detailing). Roof truss load has been calculated based on the IS 875 part1, part2 and part 3. Truss
members and connection are designed using IS 800: 2007. These designs are based on Limit
State method.
6.2 Limit State Design:
Above design is based on Limit state method. The object of design based on limit state concept is
to achieve an acceptable probability that a structure will not become unserviceable in its lifetime
for which it is intended that is it will not intended limit state.
It consists of two parts
1) Limit state of collapse
2) Limit sate of serviceability
Limit state of collapse: The limit state of collapse of the structure or part of the structure could be
assessed from rupture of one or more critical section and from buckling due to elastic or plastic
instability (including the effects of sway where appropriate) or overturning. The resistance to
bending, shear, torsion and axial loads at every section shall not be less than the appropriate
value at that section produced by the probable most unfavourable combination of loads on the
structure using the appropriate partial safety factors.
Limit state of serviceability:
i) Deflection: The deflection of a structure or part thereof shall not adversely affect the
appropriate or efficiency of the structure or finishes or partitions.
ii) Cracking: Cracking of concrete should not adversely affect the appearance or durability of the
structure; the acceptance limits of cracking would vary with the type of structure and
environment.

80
6.3 Design of structural elements
We have designed the following structural elements in our project:
a) Design of Beam
b) Design of RCC gantry Beam
c) Design of Slab
d) Design of Column
e) Design of Corbel
f) Design of staircase
g) Design of mat foundations
h) Design of shear wall
i) Design of truss
6.3.1 Design of Beams
A reinforced concrete member should be able to resist the tensile, compressive and shear stresses
induced in it by the loads acting on the member. Concrete is fairly strong in compression but
very weak in tension. Plain concrete are thus limited in carrying capacity by the low tensile
strength. Steel is very strong in tension. Thus, the tensile strength of concrete is overcome by the
provision of the reinforcing steel in the tension zone round the concrete to make a reinforced
concrete member.
We have determined the moment and shear force in the beam through the coefficient method.
Condition for the coefficient method to use are unless more exact estimates are made, for beams
of uniform cross-section which support substantially uniformly distributed loads over three or
more spans which do not differ by more than 15 percent of the longest, the bending moments and
shear forces used in design may be obtained using coefficients given in Table 12 and Table 13
respectively of IS 456: 2000.
Table 13: Bending Moment Coefficients (Clause 22.5.1)
Type of load
Span moments Support moments
Near
Middle of
end span
At middle
of interior
span
End support (if
partially
restrained)
At support
next to the
end support
At other
interior
support
Dead load
imposed
load(fixed) 0.083 0.0625 -0.0417 -0.100 -0.083
Imposed load not
fixed 0.100 0.083 -0.0417 -0.111 -0.111
NOTE: for obtaining the bending moment the coefficients shall be multiplied by the total design load and effective span

81
Table 14: Shear Force Coefficients (Clause 22.5.1 and 22.5.2)
Type of load
At End
Support
At support next to the end support
At all other
interior
supportsOuter Side Inner Side
Dead load
imposed
load(fixed) 0.4 0.6 0.55 0.5
Imposed load not
fixed 0.45 0.6 0.6 0.6
NOTE: for obtaining the shear force, the coefficient shall be multiplied by the total design load
For single span
The bending moment coefficient at the end WL/12 and at the middle WL/24 Where W is the
total design load
After determination of bending moment and shear force the size and reinforcement is calculated.
Check for shear, stirrups spacing, anchorage length and other check are done according to IS
13920:1993
Beam design

82
Design of Beam A
References Step Calculations Output
1 Known data
Grade of steel 500N/mm2
Length of Beam = 5.1m
Width of Beam = 4.45m
Thickness of slab = 0.17m
Clear cover for beam = 30mm
2 Load
Factored Dead load of slab 6.375 kN/m2
Factored Live load of slab 15 kN/m2
Factored load(DL +LL) slab 21.37 kN/m2
Factored dead load of slab on beam 40.78 Kn
Factored Live load of slab on beam 95.95 kN
Preliminary Design
Breadth of beam(b) 0.315 m
Depth of beam(l) 0.35 m
Factored dead load of beam 4.13 kN/m
Factor dead load of beam 21.08 kN
Total factor dead load 61.86kN
Total factor live load 95.95 kN
3 Moment
IS 456:2000 clause
22.2(b) Effective span 4.6 m
IS 456:2000 clause
22.5.1 Bending moment at support
End support 30.24 kNm
Next to the end support 77.5 kNm
Bending moment at mid span 67.85 kNm
4 Size
For size of beam maximum moment is consider
assume d/b 1.5
M= 0.133fckbd2
effective depth of beam(d) 323 mm
Overall depth of beam( eff. depth + clear cover) 353 mm
Adopted overall depth of beam 400 mm
Width of beam 242 mm
Adopted overall width of beam 300 mm

83
5 Reinforcement
At support( maximum moment)
M = 0.87*fy*Ast*(d-0.42*Xumax)
Area of steel 683.68 mm2
number of rebar(16-dia) require 3.4
Adopted no. of rebar 4
At middle of span
no. of rebar(16 dia.) require 2.977
6 Check for width and reinforcement in beam
IS 13920 Clause 6.1.3 width of member > 200mm Ok
longitudinal reinforcement
IS 13920 Clause 6.2.1
Provided top as well as bottom reinforcement
of 2-16 dia. throughout the member length. Ok
Minimum reinforcement
Ast min = 0.24*
= 0.24%
= 288mm2
Provided minimum reiforcement 402.12mm2
Ok
Maximum reinforcement at any section
= 0.025bd = 3000 mm2
Provided maximum reinforcement in any
section
= 1407.43 mm2
Ok
7 Check for shear
IS 456:2000 clause
22.5.1 and 22.5.2
shear force on end support of beam A 67.93kN
shear force on interior face of first interior
support 91.6kN
Design value for shear consideration 91.6 kN
Percentage area for longitudinal steel 1.17%
IS 456:2000 Table 19 Shear strength of concrete from table 0.68N/mm2
Nominal shear stress
= (91.6*1000)/(370*300) 0.825N/mm2
IS 456:2000 Table 20 Maximum stress for the concrete(Tc) 3.1N/mm2
check 0.825
Strength of shear reinforcement
√𝑓𝑐𝑘
𝑓𝑦

84
Vus = Vu - τbd
= (91.6-0.68*300*370/1000) 16.12kN
Adopt 10 mm two legged stirrups 157.075mm2
Spacing of shear reinforcement(x)
= (0.87*fy*Asv*d)/Vus 1568.5 mm
IS 13920: 1993 Clause
6.3.5 8 Check for the spacing of stirrups
spacing of hoops over a length of 2d at either
end of a beam shall not exceed
a) d/4 92.5mm
b)8*minimum dia of longitudinal bar 128mm
but the minimum spacing of stirrups 100mm
Spacing of stirrups at 2d from the end of beam 100mm
Spacing of stirrups elsewhere
d/2 185mm
IS13920 clause 6.2.5 9 Check for Anchorage length
development length in tension(Ld)
=(0.87*fy*dia.)/4Tbd
= 48.54 ∅ 776.78 mm
Length of anchorage
=Ld +10∅
= 58.54 ∅
= 58.54*16 936.64 mm
check for max. dia of bar
Ld = 1.3(M1/V)+Lo
M1 = Moment of resistance
= 0.87*fy*Ast [d - ]
= 86.5kNm
Lo = 8 dia. ( for 90o
bend)
Shear force V = 91.6 kN
48.54 ∅=1355.62
∅ =27.92mm
use bar dia. is 16mm < 27.92 mm Ok
𝑓𝑦 𝐴 𝑠𝑡
𝑓𝑐𝑘 𝑏

85
Design Summary
Depth = 400mm
Width = 300mm
Cover = 30mm
Reinforcement at the top of both left and right of the beam = 4-16∅
Reinforcement at the bottom of the beam = 3-16∅
Reinforcement throughout at the top = 2-16∅
Reinforcement throughout at the bottom = 2-16∅

86
Design of Beam B
1 Known data
2 Load
Factored load(DL +LL) slab 21.38 kN/m2
Factored dead load of slab on beam 40.78 Kn
Preliminary Design
3 Moment
IS 456:2000 clause
IS 456:2000 clause
22.5.1 Bending moment at support
Interior spans 72.5 kNm
Next to the end support 77.5 kNm
4 Size
assume d/b 1.5
M= 0.133fckbd2

87
5 Reinforcement
At middle of span
IS 13920 Clause 6.2.1
Ast min = 0.24*
= 0.24%
= 288mm2
Ok
= 0.025bd = 3000 mm2
section
= 1407.43 mm2
Ok
7 Check for shear
IS 456:2000 clause
22.5.1 and 22.5.2
shear force on support next to the end support 94.7kN
shear force on interior support of beam 88.5kN
Design value for shear consideration 94.7 kN
= (91.6*1000)/(370*300) 0.825N/mm2
check 0.825
Vus = Vu - τbd
√𝑓𝑐𝑘
𝑓𝑦

88
= (91.6-0.68*300*370/1000) 18.2kN
Adopt 10 mm two le ged stirrups 157.075mm2
= (0.87*fy*Asv*d)/Vus 1408.53 mm
IS 13920: 1993 Clause
a) d/4 92.5mm
d/2 185mm
= 48.54 ∅ 776.78 mm
Length of anchorage
=Ld +10∅
= 58.54 ∅
= 58. 4*16 936.64 mm
Ld = 1.3(M1/V)+Lo
= 0.87*fy*Ast [d - ]
= 86.5kNm
bend)
48.54 ∅=1355.62
∅ =27.92mm
Design Summary
𝑓𝑐𝑘 𝑏

89
Depth = 400mm
Width = 300mm
Cover = 30mm

90
Design of Beam C
1 Known data
2 Load
Factored load(DL +LL) slab 21.38kN/m2
Factored dead load of slab on beam 63.12 kN
Preliminary Design
Total factor dead load 84.2 kN
3 Moment
IS 456:2000 clause
IS 456:2000 clause
22.5.1 Bending moment at support 86.99kNm
4 Size
assume d/b 1.5
M= 0.133fckbd2

91
5 Reinforcement
At middle of span
IS 13920 Clause 6.2.1
Ast min = 0.24*
= 0.24%
= 288mm2
Ok
= 0.025bd = 3000 mm2
section
= 1407.43 mm2
Ok
7 Check for shear
IS 456:2000 clause
22.5.1 and 22.5.2 shear force on support of beam 117.29kN
= (117.29*1000)/(370*300) 1.04N/mm2
check 1.04<3.1 Ok
Vus = Vu - τbd
= (117.29-0.68*300*370/1000) 45.29 kN
√𝑓𝑐𝑘
𝑓𝑦

92
= (0.87*fy*Asv*d)/Vus 565.59 mm
IS 13920: 1993 Clause
6.3.5 8 Check for the spacing f stirrups
a) d/4 92.5mm
d/2 185mm
= 48.54 ∅ 776.78 mm
Length of anchorage
=Ld +10∅
= 58.54 ∅
= 58.54*16 936.64 mm
Ld = 1.3(M1/V)+Lo
= 0.87*fy*Ast [d - ]
= 86.5kNm
bend)
48.54 ∅=1355.62
∅ =27.92mm
Design Summary
𝑓𝑐𝑘 𝑏

93
Depth = 400mm
Width = 300mm
Cover = 30mm

94
Design of Beam D
1 Known data
2 Load
Factored Dead load of slab
6.375
kN/m2
Factored load(DL +LL) slab
21.375
kN/m2
Factored dead load of slab on beam 31.56 kN
Preliminary Design
3 Moment
IS 456:2000 clause
IS 456:2000 clause
22.5.1 Bending moment at support 47.75kNm
4 Size
assume d/b 1.5
M= 0.133fckbd2
5 Reinforcement

95
At middle of span
IS 13920 Clause 6.2.1
Ast min = 0.24*
= 0.24%
= 288mm2
Ok
= 0.025bd = 3000 mm2
section
= 1407.43 mm2
Ok
7 Check for shear
IS 456:2000 clause
22.5.1 and 22.5.2 shear force on support of beam 72.93kN
= (72.93*1000)/(275*200) 1.32N/mm2
check 1.04<3.1 Ok
Vus = Vu - τbd
= (117.29-0.68*300*370/1000) 31.13 kN
sSpacing of shear reinforcement(x)
√𝑓𝑐𝑘
𝑓𝑦

96
= (0.87*fy*Asv*d)/Vus 603.64 mm
IS 13920: 1993 Clause
a) d/4 68.75mm
d/2 137.5mm
provide at a spacing 125mm
= 48.54 ∅ 776.78 mm
Length of anchorage
=Ld +10∅
= 58.54 ∅
= 58.54*16 936.64 mm
Ld = 1.3(M1/V)+Lo
= 0.87*fy*Ast [d - ]
= 43.71kNm
bend)
48.54 ∅=939.14
∅ =19.34
𝑓𝑐𝑘 𝑏

97
Design Summary
Depth = 300mm
Width = 200mm
Cover = 30mm

98
Design of RCC gantry Beam
Step Calculations Output
1 Known data
Length of gantry girder 5.95m
Width of gantry girder 2.6m
Length of crane girder 12.7m
Crane Capacity 400kN
Wt. of Crane 220kN
Grade of concrete 25N/mm2
Assume Breadth of beam(d) 600mm
Assume Breadth of beam(b) 450mm
Self wt. of beam 6.75kN/m
Clear cover 50mm
Steel grade 500mm
2 For maximum reaction on gantry girder(Ra)
Minimum gantry hook approach 1m
maximum reaction on gantry girder(Ra) 478.50kN
Load on gantry girder from each wheel 239.25kN
Factored wheel load 358.88kN
A
B

99
3 For maximum moment calculation
Case(i)
c.g of wheel load and one wheel at eual distance from
centre of girder
Reaction at A 457.37kN
Reaction at B 300.55kN
Maximum moment occurs at point c 680.53kNm
moment due to impact (25% of maximum moment) 170.13kNm
Then, total moment 850.66kNm
Case (ii)
One load is at centre
moment at C 563.70kNm
A B
A B

100
moment due to impact(25% of maximum moment) 140.93kNm
Total moment 704.63kNm
Maxumum value of moment from case(i) and case (ii) 850.66kNm
moment due to horezontal force i.e. surge(10% of crane
and load lifted) 62kN
considering the rail height 150mm
Ecentricity of transverse breaking forcei.e. hor. Force 450mm
Factored torsional moment(Tu) 41.85kNm
Equivalent Bending moment at cross section (Mf) =
Tu*(1+d/b)/1.7 57.44kNm
Total moment on beam 908.10kNm
4 Maximum shear
max shear(Sm) 581.01kN
Equivalent shear(Sf = Sm+1.6*(Mf//0.4) 810.78kN
Moment capcity of assume beam 538.65kNm
Momnt need to be bear 908.10kNm
Beam to be designed as doubly reinforce
A B

101
Additional moment to be resist 369.45kNm
ϵsc = (1-d'/Xm)*0.0035 0.00274kNm
Fsc 413 N/mm2
Area of compression steel 1656.6 mm2
No. of 20 dia to be provided 5.27mm2
Provided dia of bar 5
Provided steel 1884.9
For the determination area of rebar in tension
Ast1 2557.99 mm2
Ast2 1572.83mm2
Total area of rebar in tension 4130.82mm2
No. of 25 dia to be provided 8.42
Provided dia of bar 9
Provided area of steel 4417.73mm2
5 Check for shear
Shear force on face of beam 810.78kN
Shear strength of concrete from table 19 (IS 456:2000) 0.82N/mm2
Nominal shear stress 3.0N/mm2
Maximum stress for the concrete(Tc) for M25 3.1N/mm2
check 3.0<3.1 ok
Strength of shear reinforcement 589.38kN
Adopt 12 mm two legged stirrups 226.19 mm2
Spacing of shear reinforcement(x= (0.87*fy*Asv*d)/Vus) 100.16mm
6
Check for the spacing of stirrups( IS 13920:1993) clause
6.3.5
a) d/4 150mm

102
b)8*minimum dia of longitudinal bar (8*16mm) 192mm
Spacing of stirrups 100.16mm
Code requires maximum spacing of stirrups 300mm
Spacing of stirrups elsewhere 300mm
7 Other check according to Is13920
b/d 0.75>0.3 Ok
width of member 450>200 Ok
depth/span 0.108<1/4 Ok
tension steel ratio at any section 648 mm
provided steel bar at each section 4064.>624 Ok
maximum allowable steel 6750mm2
Ok
provided steel 5988.48m2
Ok
Design Summary
Width of beam = 450mm
Depth of beam = 600mm
Compression reinforcement = 5-20∅
Tension reinforcement = 9-25∅

103
6.3.2 Design of Two-way Slabs
When slabs are supported on four sides, two-way spanning action occurs. Such slabs may be
simply supported or continuous on any or all sides. The deflection and bending moments in a
two-way slab are considerably reduced as compared to those in a one way slab. Thus, a thinner
slab can carry the same load when supported on all the four edges. In a square slab, the two-way
action is equal in each direction. In long narrow slabs, where the length is greater than twice the
breadth, the two-way action effectively reduces to one-way action in the direction of the short
span although the end beams do carry some slab loads.
Restrained slabs
A slab may have its few or all edges restrained. The degree of restrains may vary depending
whether it is continuous over its supports or cast monolithically with its supporting beams. A
hogging or negative bending moment will develop in the top face of the slab at the supported
sides. In these slabs the corners are prevented from lifting and provision is made for torsion. The
maximum moments Mx and My at midspan on strips of unit width for spans lx and ly are given
by:
Mx = αxwlx
2
My = αywlx
2
Where, αx and αy = moment coefficients.
lx = length of shorter side
ly = length of longer side
Mx and My = maximum moments at midspan on strips of unit width and spans lx
and ly.
In table 26 of IS 456:2000 nine separate slab arrangements are given. Bending moment
coefficients given in this table were obtained by using the yield line theory.
The bending moments calculated from the coefficients are assumed to act in the middle strips of
the slab. In the edge strip the minimum quantity of main reinforcement is sufficient.

104
Reference Step Calculation Output
Slab Panel No.
1
( Two Long
Edge
Discontinuous)
IS 456:2000
Table 26
IS 456:2000
Annex D.1.1
1.
2.
3.
Thickness of slab and durability considerations
Clear spans
Lx = 4.15 m
Ly = 4.8 m
Provide d = 144 mm
Assuming clear cover = 20 mm
Providing 12mm diameter bar
Total depth of slab D = 144 + 20 + 12/2 = 170
mm
Effective Length
lx = 4150 + 300 = 4450 mm
ly = 4800 + 300 = 5100 mm
Since ly/lx =5100/4450 = 1.15 < 2
(Designed as two way slab)
Design load
Self-weight of slab = 0.17 × 25 = 4.25 KN/m2
Floor finishing load = 0.816 KN/m2
Dead load = 4.25 + 0.816 = 5.06 KN/m2
Live load = 10 KN/m2
Erection load = 1.5 KN/m2
Total load = 16.56 KN/m2
Design load = 1.5*16.56 = 24.84 KN/m2
Considering unit weight of slab = 24.84 KN/m2
Moment Calculation
Three edges discontinuous (one short edge
continuous)
Short span coefficients
αx + = 0.054
αx - = -
Long span coefficients
αy + = 0.043
αy - = 0.057
d = 144 mm
D = 170 mm
lx = 4450 mm
ly = 5100 mm

105
IS 456:2000
Clause 26.5.2.1
4.
5.
6.
For short span
Mid-span moment = αx wlx
2
= 0.054×24.84×4.452
= 26.91 KN-m
For long span
Mid-span moment = αy wlx
2
= 0.043 ×24.84×4.452
= 21.15 KN-m
Support moment = αy wlx
2
= 0.057 ×24.84×4.452
= 28.03 KN-m
Check for depth from moment consideration
d = √
𝑀𝑚𝑎𝑥
0.133×𝑓𝑐𝑘×𝑏
= √
28.03×10^6
0.133×25∗1000
= 91.81 mm
Calculation of area of steel
Min Ast = 0.12% of bD
= 0.0012×1000×170
= 204 mm2
Area of steel along short span (Mid-span)
Mut = 0.87 fy Ast (d -
𝑓𝑦∗𝐴𝑠𝑡
𝐹𝑐𝑘∗𝑏
)
26.91×106
= 0.87× 500 ×Ast (144 -
500∗𝐴𝑠𝑡
25∗1000
)
Solving the equation, we get
Ast, required = 452.44 mm2
d = 91.81mm <
140mm
Minimum area
of steel required
is 204 mm2
Provide 12mm
bars @ 200 mm
c/c
Actual Ast,provided
= 566 mm2

106
7.
8.
9.
10.
Area of steel along short-span ( at support)
Here we provide 50% of mid-span reinforcement
Ast,provided = 452.44/2 = 226.22 mm2
Area of steel along long-span (at mid-span)
𝐹𝑐𝑘∗𝑏
)
21.15×106
= 0.87× 500 ×Ast (144 -
500∗𝐴𝑠𝑡
25∗1000
)
Ast, required = 356 mm2
Area of steel along long-span ( at support)
𝐹𝑐𝑘∗𝑏
)
28.03×106
= 0.87× 500 ×Ast (144 -
500∗𝐴𝑠𝑡
25∗1000
)
Ast, required = 480 mm2
Check for shear (at short edges)
Maximum SF =
1
2
w lx
=
1
2
×24.84×4.45
= 55.27 KN/m
Provide 12mm
bars @ 300mm
c/c
Actual Ast,provided
= 377 mm2
Provide 12mm
bars @250mm
c/c
Actual Ast,provided
= 453 mm2
Provide 12mm
bars @200mm
c/c
Actual Ast,provided
= 566 mm2

107
11.
Nominal shear stress τv =
55.27×1000
1000×144
= 0.38 N/mm2
Percentage tension steel =
100∗𝐴𝑠𝑡
𝑏𝑑
=
100×377
1000×144
= 0.26 %
Shear strength of M25 concrete for 0.26 % of
reinforcement is 0.36 N/mm2
Shear strength in slabs = k × shear strength of M25
Concrete τc
Value of k = 1.26 for D = 170 mm
Hence shear strength of slab τc’ = 1.26 × 0.36
= 0.45 N/mm2
which
is greater than nominal shear stress.
Check for deflection
Along short span
Since both ends are discontinuous, the basic value α =
26
β = 1 since L<10m
Pt =
100×566
1000×144
= 0.39 %
Ƴ = 1.03, δ = 1 and λ = 1.
Allowable L/D = 26×1.03 = 26.78
Actual L/D = 4450/170 = 26.18
Nominal shear
stress = 0.38
N/mm2
Shear strength
of slab = 0.45
N/mm2
The slab is safe
in shear
Actual L/d ≈
Allowable L/D
Hence ok

108
12. Check for development length (at short edge)
Moment of resistance offered by 12 mm bars @200 mm
c/c
M1 = 0.87 fy Ast (d -
𝐹𝑐𝑘∗𝑏
)
= 0.87× 500× 566 (144 -
500∗566
25∗1000
)
= 32.66 × 106
KN-m
V = 63.34 KN
Anchorage value of bars bent at 900
including 200 mm
straight length
L0 = 8ø+400 = 496 mm
Ld ≤ 1.3
𝑀1
𝑉
+ L0
40ø ≤ 1.3
32.66×10^6
63.34×1000
+ 496
Ø ≤ 29.15 mm
Diameter of bar used is 12 mm which is less than 29.15
mm.
Hence safe.

109
Slab Moment Calculation
Area of Steel Calculation for Each Slab Panel
Calculated Area of Steel Calculated spacing (mm) for provided Ast per meter(sq mm)
Astx
+ve
Astx -
ve
Asty
+ve
Asty -
ve
Astx
+ve
Astx -
ve
Asty
+ve
Asty
-ve
Astx +ve Astx -ve Asty +ve Asty -ve
481.9
3
240.9
7
372.6
1
504.0
9
234.7
7
469.5
4
210.8
7
155.8
7
12 mm bars @
200mm c/c
12 mm bars @
300mm c/c
12mm bars at
250mm c/c
12mm bars at
200mm c/c
481.9
3
240.9
7
300.0
0
391.0
0
234.7
7
469.5
4
261.9
0
200.9
5
12 mm bars @
200mm c/c
12 mm bars @
300mm c/c
12mm bars at
250mm c/c
12mm bars at
200mm c/c
481.9
3
240.9
7
300.0
0
391.0
0
234.7
7
469.5
4
261.9
0
200.9
5
12 mm bars @
200mm c/c
12 mm bars @
300mm c/c
12mm bars at
250mm c/c
12mm bars at
200mm c/c
466.6
7
340.0
0
242.4
5
231.0
9
12mm bars
@230mm c/c
12mm bars
@230mm c/c
Slab
panel
numbe
r
Type lx ly
ly/l
x
αx
+ve
αx
-ve
αy
+ve
αy
-ve
Load (KN/m2)
Factore
d load
Moment (KNm)
Live
Loa
d
Erectio
n load
floor
finishin
g load
Dea
d
Loa
d
KN/m2 Mx
+ve
Mx
-ve
My
+ve
My
-ve
1 8.00
4.4
5
5.1
0
1.15 0.05 - 0.04 0.06
10.0
0
1.50 0.82 4.25 24.85
26.9
1
21.1
6
28.0
5
2 6.00
4.4
5
5.1
0
1.15 0.05 - 0.04 0.05
10.0
0
1.50 0.82 4.25 24.85
26.9
1
17.2
2
22.1
4
3 6.00
4.4
5
5.1
0
1.15 0.05 - 0.04 0.05
10.0
0
1.50 0.82 4.25 24.85
26.9
1
17.2
2
22.1
4
4 -
1.4
5
5.1
0
3.52 - - - -
10.0
0
1.50 0.82 4.25 24.85
26.1
2

110
Slab Shear Check for Each Panel of Conrol Room
Slab
panel No.
lx
Factored Load
(KN/m2)
Shear
Force (Vu)
τv (N/mm2) Pst τc k τc Result
1 4.45 24.85 55.29 0.38 0.26 0.36 0.45 no shear reinforcement is required
Slab Deflection Check for Each Panel of Control Room
Check for Deflection
Slab
Panel
No.
Lx
(m)
D
(mm)
Actaul Lx/D Ast Pst Modification factor Allowable Lx/D Result
1 4.45 170.00 26.18 566.00 0.39 1.03 26.78 Safe in Deflection

111
Slab Development Length Check for Each Panel of Control Room
Check for Development length
Slab Panel No. Mu Vu Ld 1.3M/V+L0 Result
1 28.05 63.34 932 1071.70 Ld < 1.3 M/V + L0
2 22.14 63.34 932 950.40 Ld < 1.3 M/V + L1
3 22.14 63.34 932 950.40 Ld < 1.3 M/V + L2
4 26.12 36.03 932 1438.44 Ld < 1.3 M/V + L3

112
6.3.3 Design of Column
A column may be defined as an element used primarily to support axil compressive loads and
with a height of at least three times its lateral dimension. A compression member subjected to
pure axial load rarely occurs in practice. All columns are subjected to some moment which may
be due to accidental eccentricity or due to end restraint imposed by which may be due to
accidental eccentricity or due to end restraint imposed by monolithically placed beams or slabs.
The strength of column depends on the strength of materials, shape and size of the cross-section,
length and the degree of positional and directional restraints at the ends. A column may be
classified based on the different criteria such as:
a) Shape of cross-section,
b) Slenderness ratio,
c) Type of loading, and
d) Pattern of lateral reinforcement
A column may be rectangular, square, circle or polygon in cross-section. The code specifies
certain minimum reinforcement bars depending on its shape. A column may be classified as
short or long depending on its effective slenderness ratio. The ratio of effective column length to
least lateral dimension is referred as effective slenderness ratio. A short column has a maximum
slenderness ratio of 12. A long column has a slenderness ratio greater than 12. However,
maximum slenderness ratio of a column should not exceed 60.
A column may be classified as follows based on the type of loading:
a) Axially loaded column,
b) a column subjected to axial load and uni-axial bending, and
c) A column subjected to axil load and bi-axial bending
A reinforcement concrete column can be classified according to the manner in which the
longitudinal bars are laterally supported, that is,
a) Tied column, and
b) Spiral column

113
COLUMN TYPE I
Material: concrete grade (fck) M25 & Steel Fe500
Length of column=4.5 m Size of column=0.5m x 0.5m Effective cover d’
=40mm
Effective length=0.65 x 4.5=2.925 m (Table 28 IS456:2000)
Slenderness ratio along x=Leffx/0.5=2.925/0.5=5.85<12 (Cl.25.1.2 IS456:2000)
Slenderness ratio along y=Leffy/0.5 =2.925/0.5=5.85<12(Cl.25.1.2 IS456:2000)
Hence, the column designed is a short column
Reference Step Calculations Output
Column 99
Cl.6.5.3.1
IS456:2000
IS 4562000
Cl 25.4

SP 16
Table 48
1.
2
3
4
Description
Pu=1537.847 kN
Mux=207 kN-m
Muy=184 kN-m
Min pt=0.8% of gross area
Max pt=4% of gross area but in extreme case it can be
extend to 6%
Minimum eccentricity
ex, min =
𝐿𝑒𝑓𝑓
500
+
𝐷
30
=
2925
500
+
500
30
= 22.52 mm
ey, min =
𝐿𝑒𝑓𝑓
500
+
𝐷
30
=
2925
500
+
500
30
= 22.52 mm
Moment calculations
Moment due to eccentricity about x
Muxe=Pux ex, min=1537.847 x 0.02252 =34.63 kN-m
Muye= Pux ey, min=1537.847 x 0.02252 =34.63 kN-m
Design
Assume the reinforcement% p=2%
p/fck=2/25 = 0.08
uniaxial moment capacity of the section about x-
axis:
d’/D = 40/500=0.08
𝑃𝑢
𝑓𝑐𝑘𝑏𝐷
=
1537.847 𝑋 1000
25 𝑋 500 𝑋 500
= 0.246
the chart for d’/d=0.1 will be used
𝑀𝑢
𝑓𝑐𝑘𝑏^2𝐷
= 0.13
Mux1 = 0.13x25x500x5002
= 406.25 kN-Uniaxial
moment capacity of the section about y-axis:
ex, min=22.52 mm
ey, min=22.52 mm
Hence, design
moment Mux=
Mux=207 kN-m
Muy=184 kN-m
Mux1=406.25 kN-
m

114
SP16 Table
48
IS 456:2000
Cl 39.6
IS 456 :2000
Table 19
5.
6.
7.
d’/B = 40/500=0.08
𝑃𝑢
=
1537.847 𝑋 1000
25 𝑋 500 𝑋 500
= 0.246
𝑀𝑢
= 0.13
Muy1 = 0.21x25x500x5002
= 406.25 kN-m
Calculation of Puz
Puz=0.45fckAc+0.75fyAs
=0.45x25x0.98Ag+0.75x500x0.02xAg
=0.45x25x0.98x500x500+0.75x415x0.02x500x500
=4631.25 kN
Pu/Puz=1537.847/4631.25 =0.33
𝛼n=2
(
𝑀𝑢𝑥
𝑀𝑢𝑥1
)2
+(
𝑀𝑢𝑦
𝑀𝑢𝑦1
)2
=0.46<1
As=2% x500x500 = 5000mm2
Now, provide 8-25mm∅ and 4- 20mm∅
Area of steel provided=5183.56 mm2
Design of lateral ties Diameter
of ties:
∅t not less than 6 mm
≥0.25x max diameter of longitudinal reinforcement
=0.25x25
=6.25mm
Hence, adopt ties of 8mm∅
Length of lateral ties(l0) is
a maximum of
-Largest lateral dimension (i.e. 500mm)
-1/6 of clear span of the member (i.e.,1/6*4500 = 750
mm)
-Minimum length of 450 mm
Spacing of hoop as special confining reinforcement
at a distance of l0 from the end of column should be
∅=8mm
L0 in both ends of
column =750mm
Provide spacing of
100mm c/c
8mm∅@250mm
c/c in central part

115
Design summary
Size of column = 0.5 mx 0.5m Cover =40mm
Main Reinforcement =8-25mm φ and 4-20mm φ
Lateral ties upto L/4 from end of column = 8 mm φ @ 100mm c/c
Lateral ties in the central part =8 mm φ @ 250 mm c/c
8.
9.
least of
¼ of minimum lateral dimension= 500/4=125mm
Should not be greater than 100mm should not be less
than 75mm
Lateral ties in central part
-provide 8mm∅ lateral ties
for the spacing of lateral ties, half of the least lateral
dimension=500/2 =250 mm
i.e., 250mm in central part
Shear check:
Vu = (data obtained from SAP) =6.44 kN
τv =
𝑉𝑢 𝑋 1000
500 𝑋 500
= 0.026 KN / mm2
Percentage of steel provide =
𝐴𝑠𝑡,𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑
𝐵𝐷
X100 % =
1.44%
τc = 0.73 N / mm2
Modification Factor = δ = 1 +
3𝑃𝑢
𝐴𝑔 𝑓𝑐𝑘
= 1+
3 𝑋 1537.847
500 𝑋 500 𝑋 25
= 1
τc’ = 0.73 x 1 = 0.73 N/mm2
τc, max = 3.1 N / mm2
So, τv< τc’< τc max (OK)

116
COLUMN TYPE II
=40mm

117
Column 99
Cl.6.5.3.1
IS456:2000
IS 4562000
Cl 25.4
SP 16
Table 48
1.
2
3
4
Description
Pu=1139.436 kN
Mux=229.38 kN-m
Muy=106.136 kN-m
extend to 6%
ex, min =
𝐿𝑒𝑓𝑓
500
+
𝐷
30
=
3087.5
500
+
600
30
= 26.18 mm
ey, min =
𝐿𝑒𝑓𝑓
500
+
𝐷
30
=
3087.5
500
+
600
30
= 26.18 mm
Moment calculations
Muxe=Pux ex, min=1139.436 x 0.02618=29.83 kN-m
Design
p/fck=1/25 = 0.04
axis:
d’/D = 40/600=0.07
𝑃𝑢
=
1139.436 𝑋 1000
25 𝑋 600 𝑋 600
= 0.126
𝑀𝑢
= 0.12
Mux1 = 0.12x25x600x6002
=1080 kN-m
Uniaxial moment capacity of the section about y-axis:
d’/B = 40/600=0.07
𝑃𝑢
=
1139.436 𝑋 1000
25 𝑋 600 𝑋 600
= 0.126
𝑀𝑢
= 0.12
Muy1 = 0.14x25x600x6002
=1080kN-m
ex, min=26.18 mm
ey, min=26.18 mm
Hence, design
moment Mux=
Mux=229.38 kN-m
Muy=106.136 kN-
m
Mux1=1080 kN-m
Muy1=1080 kN-m

118
IS 456:2000
Cl 39.6
IS 456 :2000
Table 19
5.
6.
7.
8.
Calculation of Puz
=0.45x25x0.99Ag+0.75x500x0.01xAg
=0.45x25x0.99x600x600+0.75x500x0.01x600x600
=5359.5 kN
Pu/Puz=1139.436 /5359.5 =0.21
𝛼n=2
(
𝑀𝑢𝑥
𝑀𝑢𝑥1
)2
+(
𝑀𝑢𝑦
𝑀𝑢𝑦1
)2
=0.05<1
As=1% of 600 x 600 = 3600mm2
Now, provide 8-25mm∅
of ties:
=0.25x25
=6.25mm
Length of lateral ties(l0)
is a maximum of
mm)
least of
¼ of minimum lateral dimension= 600/4=150 mm
than 75mm
L0 in both ends of
column =750mm
Provide spacing of
100mm c/c
8mm∅@150mm
c/c in central part

119
Shear check:
Vu = (data obtained from SAP) =7.28 kN
τv =
𝑉𝑢 𝑋 1000
600 𝑋 600
= 0.0202 KN / mm2
Percentage of steel provide =
𝐴𝑠𝑡,𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑
𝐵𝐷
X100 % =
1.09%
τc = 0.66 N / mm2
Modification Factor = δ = 1 +
3𝑃𝑢
𝐴𝑔 𝑓𝑐𝑘
= 1+
3 𝑋 1139.436
600 𝑋 600 𝑋 25
= 1
τc’ = 0.66 x 1 = 0.66 N/mm2
τc, max = 3.1 N / mm2
So, τv< τc’< τc max (OK)
Design summary
Main Reinforcement =8-25mm φ

120
COLUMN TYPE III
=40mm
Column 99
Cl.6.5.3.1
IS456:2000
IS 4562000
Cl 25.4
SP 16
Table 48
1.
2
3
4
Description
Pu=487.65 kN
Mux=27.88 kN-m
Muy=22.5 kN-m
extend to 6%
ex, min =
𝐿𝑒𝑓𝑓
500
+
𝐷
30
=
3087.5
500
+
600
30
= 26.18 mm
ey, min =
𝐿𝑒𝑓𝑓
500
+
𝐷
30
=
3087.5
500
+
600
30
= 26.18 mm
Moment calculations
Muxe=Pux ex, min=487.65 x 0.02618=12.76 kN-m
Design
p/fck=1/25 = 0.04
axis:
d’/D = 40/600=0.07
𝑃𝑢
=
487.65 𝑋 1000
25 𝑋 600 𝑋 600
= 0.054
𝑀𝑢
= 0.11
Mux1 = 0.11x25x600x6002
=594 kN-m
ex, min=26.18 mm
ey, min=26.18 mm
Hence, design
moment Mux=
Mux=12.76 kN-m
Muy=12.76 kN-m
Mux1=594 kN-m

121
IS 456:2000
Cl 39.6
IS 456 :2000
Table 19
5.
6.
7.
Uniaxial moment capacity of the section about y-axis:
d’/B = 40/600=0.07
𝑃𝑢
=
487.65 𝑋 1000
25 𝑋 600 𝑋 600
= 0.054
𝑀𝑢
= 0.11
Muy1 = 0.11x25x600x6002
=594 kN-m
Calculation of Puz
=0.45x25x0.99Ag+0.75x500x0.01xAg
=0.45x25x0.99x600x600+0.75x500x0.01x600x600
=5359.5 kN
Pu/Puz=1139.436 /5359.5 =0.09
𝛼n=1
(
𝑀𝑢𝑥
𝑀𝑢𝑥1
)2
+(
𝑀𝑢𝑦
𝑀𝑢𝑦1
)2
=0.08<1
As=1% of 600 x 600 = 3600mm2
Now, provide 8-25mm∅
of ties:
=0.25x25
=6.25mm
Length of lateral ties(l0)
is a maximum of
mm)
Muy1=594 kN-m
L0 in both ends of
column =750mm
Provide spacing of
100mm c/c

122
8.
least of
¼ of minimum lateral dimension= 600/4=150 mm
than 75mm
8mm∅@150mm
c/c in central part
Design summary
Main Reinforcement =8-25mm φ

123
6.3.4 Corbels
A corbel is a short cantilever projection which supports a load bearing member. The distance
between the point of application of the load and the face or root of the corbel is less than the
effective depth of the root of the corbel. Moreover, depth at the outer edge of the corbel is not
less than one-half of the depth at the support of the corbel.
The ratio of the distance of the point of application of the load from the support ‘a’ to the
effective depth of the member supporting the load is referred to as the a/d ratio or shear
span/depth ratio. When the ratio a/d is less than 2, the load is transferred to the support through
strut action rather than through flexure .this is similar to the truss action in a simple beam near
the support. Clause 28 gives design requirements for a corbel.
Figure 5: Corbel Dimension
Assumptions
1. A corbel is designed for the limit state of collapse.
2. The concrete and reinforcement in the corbel are designed on the basis of truss analogy,
that is, simple strut and tie system.
3. The resistance provided to the horizontal force should be greater than one-half of the
design vertical load on the corbel.
4. The compatibility of strains between the strut and tie should be satisfied at the root of the
corbel.

124
Truss Analogy
Figure 6: Truss analogy in corbel
Lever arm of the stress block in a corbel can be determined using the truss analogy as shown in
fig. By triangle of forces, using Lame’s theorem
P
sin(Π−θ)
=
Ft
sin(90+θ)
=
Fc
sin(90)
P
sin(θ)
=
Ft
cos(θ)
= Fc
Tension force Ft = P cot θ = P
a
z
……………………………….. (i)
Compression Force Fc = P cosec θ =
√(a2+z2)
z
P ……………………………….. (ii)
Lever Arm = d - 0.42x or x= 2.38(d-z)
The force in compression in concrete is also given by
Fc = 0.36* σck*b*(x *cos θ)
Fc = 0.36* σck*b*2.38 (d-z)*
a
√(a2+z2)
…………………………(iii)
Eqs (ii) and (iii) give,
0.86* σck*b*d (1 -
z
d
)
a
√(a2+z2)
=
√(a2+z2)
z
P
Or, 0.86* σck*b*d (1 -
z
d
)*a*z = (a2
+z2
)*P …….….……….………..(iv)
Let
P
0.86∗ σck∗b∗d
= ∝ and
a
d
= β

125
Eq. (iv) becomes, ∝( a2
+z2
) = (1-
z
d
)*a*z
Or, ∝(
a2
d2
+
z2
d2
) = (1-
z
d
)*
a
d
*
z
d
Or,
z2
d2
– (
β
∝+β
)
z
d
+(
∝
∝+β
) β2
= 0
For the given values of ∝ and β , the values of
z
d
can be calculated.

126
Detailing of Reinforcement
Figure 7: Corbel Detailing
1. The main tension reinforcement should not be less than 0.4 % of the cross sectional area
at the face of the supporting member.
2. The maximum tension reinforcement should not exceed 1.3 % of the cross sectional area
at the face of the supporting member.
3. The tension reinforcement should not be anchored at the front face of the corbel either by
welding it to a transverse bar of equal strength or, by bending back the bars to form
loops. In any case, the bearing area of the load should not project beyond the straight
portion of the bars forming the main tension reinforcement.
4. When the corbel is required to resist a horizontal force applied to the bearing plate,
because of shrinkage and temperature changes, additional tension reinforcement should
be provided to transmit this force in it’s entirely. This reinforcement should be welded to
the bearing plate and adequately anchored.
5. Theoretically compression reinforcement is not required. However, where the main
tension bars at the front face of the corbel are welded to a transverse bar, only nominal
compression steel to anchor the stirrups is provided as in fig
6. Shear reinforcement is provided in the form of horizontal stirrups distributed in the upper
two thirds of the effective depth of the corbel at the column face as shown in fig .
7. The area of shear reinforcement should be at least 50 % of the area of the main tension
reinforcement and should be adequately anchored.

127
Description Units References
Grade of concrete M25
Grade of steel fe500
Ultimate load(Pu) 564.462 kN
shear span(a) 350 mm
1.Size of Bearing plate
bearing strength of concrete(σbr) 20 Mpa 0.80*fck
Bearing area required(A) 28223.100 mm2
Pu/σbr
length of bearing plate(L) 600 mm
Width of bearing plate 47.039 mm A/L
Adopt a bearing plate of size 60*600 mm2
2.Depth of corbel at the support
for M25 concrete
Tc,max 3.1 MPa table 20,Cl 40.2.3 IS 456:2000
let, Tc 2.4 Mpa
d 391.988 mm Pu/(L*Tc)
Adopt, d 600 mm
D 650 mm
3.Depth of corbel at the face 325 mm D/2
for strut action, a/d 0.580 <0.6 ok
4.Lever arm
α 0.073
β 0.583
X2-(β/β+α)X+(α/α+β)=0 where, X=
Z/d
a 1
b -0.889
c 0.038
x1 0.845
x2 0.045
so, X 0.845
Z 506.750 mm
Z=d-0.42X
X 221.935 mm
X/d 0.370 <xu,max/d
Hence, donot Provide the compression
steel
5.Area of tension steel
Ft 389.86 kN > 0.5*Pu, ok
stress in fe500 grade tension steel(fs) 275 MPa
Area of tension steel(As) 1417.67 mm2
Ft/fs
use 20 D bars 314.16 mm2

128
provide 5 no of bars, area provided 1570.796 mm2
>As,ok
6.Check
percentage of steel 0.436 % <1.3%or >0.4%,ok
Area of shear steel
Asv(min) 708.84 mm2
0.5*As
provide 2- legged str. 12 mm-4 bars
Area provided 904.779 mm2
>Asv,oK
spacing 100 mm
7.Shear capacity of section
% of steel provided 0.44 %
Tc 0.449 MPa from table 19,IS 456:2000
increased shear strength Tc' 1.539 MPa ((2d/a)*Tc)<3.1 Mpa
total shear strength 1618.23 kN Tc'*b*d+0.87*fy*Asv*(d/X)>Pu,
oK
8.Development length
Tbd 1.4 MPa for M25,cl. 26.2.1.1 IS 456:2000
Tbd 2.24 MPa for deformed bars in tension,1.6*Tbd
Ld/D 48.55 (0.87*fy)/(4*Tbd)
Moment(M) 258.3305 kN-m 0.87*Ast*(d-0.42*X)
Shear force(V) 564.462 kN Pu
Take, Lo 400 mm
Equating 1.3(M/V)+Lo=Ld
Provide diameter less than or equal 20.491 mm
so, provide 20 diameter bar
Ld 980 mm

129
6.3.5 Design of Staircase
Staircase is designed and built for the purpose of providing access to different levels within a
building. There different types of staircases like straight stairs, quarter-turn stairs, dog-legged
stairs, spiral stairs, helicoidal stairs, open-well stairs etc. But structurally, staircases may be
classified largely into two categories depending on the predominant direction in which the slab
component of the stair undergoes flexure:
 Stair slab spanning transversely (stair widthwise).
 Stair slab spanning longitudinally.
In our project we adopt open-well stair considering structural feasibility.
Following IS 456:2000 code; design of staircase is accomplished.
Figure 8: Plan of staircase

130
Design of staircase Thapakhola HEP
Code Referred IS 456 : 2000; Plain and Reinforced Concrete Code of Practice
Space Avaibility
Constraint Dimension unit Remarks
Space Availabe 4.77m*2.83m m2
Landing provided 1.06m*1.06m m2
First Flight 2.94 m
Second Flight 3.71 m
Third Flight 2.94 m
Floor to Ceiling Height 5 m
Assumption and calculation
First Flight
Tread Dimension 250 mm
Riser Dimension 180 mm
Total No of Steps 8
Second Flight
Total No of Steps 12
Third Flight
Total No of Steps 8 No.
First Flight
No of Tread 9 No.
No of Riser 8 No.
Height of Landing from Floor level 1.52 m
Second Flight

131
No of Tread 13 No.
No of Riser 12 No.
Third Flight
No of Tread 9 No.
No of Riser 8 No.
Structural design
Assumption Value Unit Remarks
Live Load 5 kN/m2
σck 25 N/mm2
σy 500 N/mm2
Density of concrete 25 kN/m3
Width 1.06 m
Let thickness of waist slab 150 mm
Let thickness of finish 20 mm
Load calculation
Landing slab A and Going
Going
Step section 2.25 kN/m2
Inclined slab 4.783 kN/m2
Finish 0.6 kN/m2
Live load 5 kN/m2
Total load 12.633 kN/m2
Factored load 18.949 kN/m2
Taking 1.06 m width of slab load 20.086 kN/m
taking 2.01 m going, total load 40.373 kN

132
Landing slab A
self weight of slab 3.75 kN/m2
Finish load 0.6 kN/m2
live load 5 kN/m2
total load 9.35 kN/m2
Factored load 14.025 kN/m2
taking landing slab 1.19m*1.06m 17.691 kN
Landing slab B 14.025 kN/m2
taking landing slab 0.58m*0.58m 4.718 kN
Reaction at support B 23.923 kN
Reaction at support A 31.677 kN
Point where shear force is zero (x) 1.847 m
M max 31.765 kN-m
d effective 94.935 mm
d effective(adopted) 124 mm
Ast(required) 737.181 mm2
diameter of bar 12 mm
No of Bar 6.52 no.
Ast (adopted) 7-12mm diameter 791 mm2
Spacing 151.43 mm
adopted spacing 150 mm
Check for shear
Noninal shear stress(ζv) 0.241 N/mm2
Percentage Tension steel 0.602
Shear Capacity (code)(ζc) 0.523 N/mm2
shear strength of slab(ζc') 0.679 N/mm2
ζv < ζc OK

133
Check for development length
Moment of Resisitance of 12-10mm bars 34.084 kN-m
V 23.923 kN
Angle of the anchorage 45
Lo 48
1.3M/V+Lo 1900.202 mm
diameter of bar must be smaller than 32.762 mm OK
Temperature Reinforcements
1- 10 mm bars in each riser 28 no.
In waist slab(0.12%) 180 mm2
Diameter of Bar 10 mm
No of Bar 2.3 no.
spacing 277.77 mm
Provide 10mm dia bars at 350 mm spacing (Ast) 224.28 mm2
Design of Landing Slab B and C and Going
Loads on Going 18.949 kN/m2
Loads on landing slab B 14.025 kN/m2
Loads on landing slab C 14.025 kN/m2
Taking 3.13m*1.06m going, total loads 62.870 KN
Taking 0.82m*0.82m landing slab B, total loads 9.430 KN
Taking 0.82m*0.82m landing slab c, total loads 9.430 KN
Reaction at support B' 36.15 KN
Reaction at support C 36.15 KN
Point where shear force is zero (x) 2.385 m
M max 52.308 kN-m
d effective 121.825 mm
d effective(adopted) 124 mm

134
Ast(required) 1213.921 mm2
Diameter of bar 12 mm
No. of Bar 10.7 no.
Ast (adopted) 12mm diameter 1243 mm2
Spacing 100 mm
Adopted spacing 100 mm
Check for shear
Noninal shear stress(ζv) 0.275 N/mm2
Percentage Tension steel 0.946
Shear Capacity (code)(ζc) 0.625 N/mm2
Shear strength of slab(ζc') 0.812 N/mm2
ζv < ζc OK
Check for development length
Moment of Resisitance of 12-10mm bars 53.561 kN-m
V 36.15 Kn
Angle of the anchorage 45
Lo 48
1.3M/V+Lo 1974.117 mm
Diameter of bar must be smaller than 34.037 mm OK
Temperature Reinforcements
1- 10 mm bars in each riser 28 no.
In waist slab(0.12%) 180 mm2
Diameter of Bar 10 mm
No of Bar 2.3 no. adopt 3 no.
Spacing 333 mm adopt 350 mm
Provide 10mm dia bars at 350 mm spacing (Ast) 224.4 mm2

135
6.3.6 Design of Raft Foundation
If the loads transmitted by the columns in a structure are so heavy or the allowable soil pressure
so small that individual footings would either overlap or cover more than about one-half of the
area, it may be better to provide a continuous footing under all columns and walls. Such a
footing is called a raft or mat foundation. Raft foundations are also used to reduce the settlement
of structures located above highly compressible deposits. Since rafts are usually at some depth in
the ground, a large volume of excavation may be required. If weight of excavated soil is equal to
the weight of the structure and that of the raft, and the center of gravity of excavation and
structure coincide, settlement should be negligible. Where complete compensation is not
feasible, a shallower raft may be acceptable if the net increase in loads is small enough to lead to
tolerable settlement. A raft may be rectangular or circular and may be with or without an
opening.
Methods of analysis
The essential task in the analysis of raft foundation is the determination of the distribution of
contact pressure underneath the raft which is a complex function of the rigidity of the
superstructure, raft itself and the supporting soil. The following methods of analysis are
suggested which are distinguished by the assumptions involved.
Rigid foundation
Conventional method – This method is based on the assumption of linear distribution of contact
pressure. The basic assumptions of this method are as follows:
 The foundation is rigid relative to the supporting soil and the compressible soil layer is
relatively shallow
 The contact pressure variation is assumed as planar such that the centroid of contact
pressure coincides with the line of action of the resultant force of all loads acting on the
foundation
Flexible foundation
Simplified method – In this method, it is assumed that subgrade consists of an infinite array of
individual elastic springs each of which is not affected by others. The spring constant is equal to
the modulus of subgrade reaction (k). the contact pressure at any point under the raft is,
therefore, linearly proportional to the settlement at the point. This method may be used when the
following conditions are satisfied:
 The structure combined action of superstructure and the raft may be considered as
flexible with a relative stiffness factor K is less than 0.5
 Variation in adjacent column load does not exceed 20 percent of the higher value

136
Column loads of control building and their location points in Raft Foundation
Column X-cord Y-cord Load
Moment
My
Moment
Mx
(in m) (in m) (KN) (KN-m) (KN-m)
A-1 0 0 1363 0 0
A-2 5.1 0 1877 0 0
A-3 10.2 0 1875 0 0
A-4 15.3 0 1820 0 0
A-5 20.4 0 1308 0 0
B-1 0 4.45 1363 0 0
B-2 5.1 4.45 1877 0 0
B-3 10.2 4.45 1875 0 0
B-4 15.3 4.45 1820 0 0
B-5 20.4 4.45 1308 0 0

137
Design of Raft foundation of Control Building
Reference Step Calculation Output
1. Known data:
Grade of concrete = M25
Grade of Steel = Fe500
Size of column = 500 mm × 500 mm
Safe bearing capacity of soil q = 180 KN/m2
Total vertical load Q = 16486 KN
Area of footing A = 4.95 × 20.9 = 103.45 m2
Weight of footing = 10 % of Q
= 1648.6 KN
Total weight on soil = 16486 + 1648.6
= 18134.6 KN
Actual area of footing required = 18134.6/180
= 100.74 m2
<
103.45 m2
Hence ok.
total vertical load
= 18134.6 KN
Arequired = 100.74
m2
Aprovided = 103.45
m2

138
2.
3.
4.
Cantilever length from center line of column
From left column = 0.25 m
From right column = 0.25 m
From top column = 0.25 m
From bottom column = 0.25 m
Eccentricity along x-direction
Taking moment of column forces about the grid 1-1
x̄ = 10.097 m
ex = 10.097 – 10.2 = -0.103 m
Eccentricity along y-direction
Taking moment of column forces about the grid A-A
ȳ = 2.225 m
ey = 2.225 – 2.225 = 0
Ix =
20.9 ×4.95^3
12
= 211.24 m4
Iy =
4.95 ×20.9^3
12
= 3765.85 m4
A = 103.455 m2
Mxx = P.ey = 0.00 KNm
Myy = P.ex = -1703.40 KNm
P/A = 159.35 KN/m2
Soil pressure at different points is as follows:
σ =
𝑃
𝐴
±
𝑀𝑦𝑦
𝐼𝑦
. x ±
𝑀𝑥𝑥
𝐼𝑥
. y
Corner A-1
σA -1 = 163.97 KN/m2

139
5.
Corner A-2
σA -2 = 161.66 KN/m2
Corner A-3
σA -3 = 159.35 KN/m2
Corner A-4
σA -4 = 157.04 KN/m2
Corner A-5
σA -5 = 154.74 KN/m2
Columns in grid A-A and B-B are similar from
geometrical dimension and loading point of view
Corner B-1
σB -1 = 163.97 KN/m2
Corner B-2
σB -2 = 161.66 KN/m2
Corner B-3
σB -3 = 159.35 KN/m2
Corner B-4
σB -4 = 157.04 KN/m2
Corner A-5
σB -5 = 154.74 KN/m2
Maximum soil pressure = 163.97 KN/m2
<180KN/m2
Hence ok.

140
6. In the x-direction the raft is divided into two strips
(i) Strip A-A
Width = 2.475 m
Soil Pressure = 163.968 KN/m2
Span = 5.10 m
Maximum moment = 163.968 X
5.100 2
10
=
426.48
KNm/
m
(ii) Strip B-B
Width = 2.475 m
Span = 5.10 m
5.100 2
10
= 426.48 KNm/m
Cantilever Moment along X-direction
Span = 0.00 m
0.0 2
2
= 0.00 KNm/m
(iii) Strip 1-1
Maximum Soil Pressure= 163.968
KN/m
2
Span = 4.450 m
Maximum
moment
= 163.968 X
4 2
8
= 405.87 KNm/m
Cantilever Moment along X-direction
Span = 0.00 m
Maximum
moment
= 163.968 X
0.0 2
2
= 0.00 KNm/m

141
7.
Therefore,
Maximum Factored Bending Moment = 426.4 KNm/m
Limiting Moment of Resistance = 0.133 ck bd2
Therefore depth required d = 360mm
Check for Punching Shear :
Let depth required = 670mm
Shear Strength of Concrete = ks c
Where ks = 1 + c 

c  =
Short dimension of column
= 1
Long dimension of column
Therefore ks = 1

c  = 0.25 ck  N/mm2
= 1.25 N/mm2
Hence,
Shear Strength of Concrete = 1.25 N/mm2
For Corner Column
Perimeter bo = 835 + 835 + 0 + 0
= 1670 mm
Nominal Shear Stress v  =
Vu
=
1363000
bod 1118900
= 1.22 N/mm2
For Side Column
Perimeter bo = 4010
m
m
Nominal Shear Stress v  =
Vu
=
1877000
bod 2686700
= 0.70 N/mm2

142
8.
9.
Hence Effective depth is O.K.
Therefore effective depth
required = 670 mm
Adopt effective
depth = 670 mm
Overall
depth = 710 mm
Development length
Development length =
0.87 𝑓𝑦 ø
4 𝛕𝐛𝐝
Where 𝛕bd = 1.6 × 1.4 = 2.24
Therefore Ld = 49ø
For 20mm bar ( along x-direction)
Ld = 980 mm
For 20 mm bar ( along y-direction)
Ld = 980 mm
Reinforcement in long direction
M = 0.87 y Ast (d - yAst/ckb)
4.26 × 108
= 0.87×500× Ast (670-500× Ast/25×1000)
Solving
Ast = 1534 mm2/m
Minimum reinforcement required = 0.12%
= 852 mm2/m
Therefore area of steel required = 1534 mm2/m
Provide 20mm dia bar @ 200mm c/c in long direction
( At top and bottom)

143
Reinforcement in short direction
M = 0.87 y Ast (d - yAst/ckb)
4.06 × 108
= 0.87×500× Ast (670-500× Ast/25×1000)
Solving
Ast = 1456 mm2/m
Minimum reinforcement required = 0.12%
= 852 mm2/m
Therefore area of steel required = 1456 mm2/m
Provide 20mm dia bar @ 220mm c/c in long direction
Design Summary
Overall depth of footing = 710 mm
Size of footing = 4.95 m × 20.9 m
Main reinforcement in long direction = 20mm dia bar @ 200mm c/c in long direction
Main reinforcement in short direction = 20mm dia bar @ 220mm c/c in long direction

144
Figure 9 : Reinforcement in Raft Foundation
Column loads of control building and their location points in Raft Foundation

145
Column X-cord Y-cord Load Moment
My
Moment
Mx
(in m) (in m) (KN) (KN-m) (KN-m)
A-1 0 0 860 0 0
B-1 6 0 860 0 0
A-2 0 4.6 0 0 0
B-2 6 4.6 270 0 0
A-3 0 9.2 0 0 0
B-3 6 9.2 270 0 0
A-4 0 13.8 860 0 0
B-4 6 13.8 860 0 0

146
Design Summary
Overall depth of footing = 640 mm
Size of footing = 6.6 m ×14.4 m
Main reinforcement in long direction = 20mm dia bar @ 140 mm c/c in long direction
Main reinforcement in short direction = 20mm dia bar @ 310mm c/c in long direction

147
6.3.7 Shear or Flexural walls
Reinforced concrete walls in buildings are required to carry vertical loads, lateral loads as
well as bending moments in the plane of the walls. The lateral loads due to wind or
earthquake may be acting normal to the width or thickness of the wall. If the lateral loads are
acting normal to the smaller dimension of the wall in plan, it is referred to as a shear wall.
The term shear wall is a misnomer. It resists major portion of the lateral shear in buildings
through flexure deformations and not through shear deformations. Thus flexure wall is a
correct term. A wall may be added solely to resist lateral forces or concrete walls enclosing
staircases or elevator shafts may serve the same purpose.
The reinforcement and check of the shear wall is designed accordance with clause 9 of IS
13920.
Factored Axial force = 654.07kN (from SAP 2000)
Factored Bending Moment = 48.5 kNm
Length of wall = 5.1m
Thickness of wall = 180 mm ( Trial thickness)
fck = 25 N/mm2
Minimum amount of reinforcement require for each direction horizontal and vertical= 0.25%
of conc. area
Check for the neutral axis
For flexural tension failure
x
lw
=
∅+µτ
2∅+0.36
= 0.161
Where,
µ =
𝑃 𝑢
𝑓 𝑐𝑘 𝑙 𝑤 𝑡 𝑤
= 0.0285

148
∅ =
0.87𝑓𝑦 𝜌
𝑓𝑐𝑘
= 0.0435
ρ =
𝐴 𝑠𝑡
𝑙 𝑤 𝑡 𝑤
= 0.0025 = Vertical reinforcement ratio
β=
0.87𝑓𝑦
0.0035𝐸𝑠
= 0.621
x= depth of neutral axis from the extreme compression fibre
lw = horizontal length of wall in plan
tw = thickness of wall
𝑥 𝑚
𝑙 𝑤
= critical non- dimensional depth of neutral axis such that crushing of concrete and first
yielding of tension steel takes place in the extreme fibre of the section simultaneously.
=
0.0035Es
0.0035Es+0.87fy
= 0.616
Ast = area of uniformly distributed vertical reinforcement
x
lw
<
𝑥 𝑚
𝑙 𝑤
So moment capacity is determine by taking moment of all the forces about the centroid of the
section, the moment capacity is determined
𝑀 𝑈𝑣
𝑓 𝑐𝑘 𝑡 𝑤 𝑙𝑤2
= ∅[ (
∅+µ
∅
) (
0.5𝑙𝑤−0.416𝑥
𝑙 𝑤
) - (
𝑥
𝑙 𝑤
)
2
(0.168 +
β2
3
)]
= 0.03084
𝑀 𝑈𝑣 = 3610.06 kNm > 48.5 kNm Hence OK

149
The reinforcement provided
Ast = 0.0025*tw*lw = 2295mm2
Provide 10mm∅ bar at spacing of 170mm c/c in both horizontal and vertical directions.
Check for spacing according to IS 13920 clause 9.1.7
170< lw/5 = 5100/5 = 1020mm hence OK
170<3*tw = 3*180 = 360mm hence OK
Design for Shear
Shear force from lateral loading = 211.526kN
Shear force from earth pressure = 343kN
Total factored shear force for design = 554.52kN
The nominal shear stress
τv =
𝑉𝑢
𝑡 𝑤 𝑑 𝑤
Where,
Vu = factored shear force,
tw = thickness of the web, and
dw = effective depth of wall section. This may by taken as 0.8 lw for rectangular sections
τv = 0.755 N/mm2
τc = 0.36 N/mm2
from table 19 IS 456:2000
Here τc < τv
Vus = 0.87
fyAhdw
Sv
Adopt 10mm dia. Bar
So Ah = 77.5mm2
Where Vus = ( Vu – τctw dw ),

150
= 281.136 kN
Provide additional 10 dia bar at spacing of 489 mm
Hence
Provide 10mm∅ bar at spacing of 170mm c/c (spacing in horizontal) in vertical directions
and 120 mm c/c in horizontal direction ( spacing in vertical)

151
6.3.8 Truss Design
Introduction
A truss is a structure comprising one or more triangular units constructed with straight slender
members whose ends are connected at joints referred to as nodes.
 External forces and reactions to those forces are considered to act only at the nodes and
result in forces in the members which are either tensile or compressive forces.
 Moments (torsional forces) are explicitly excluded because, and only because, all the joints
in a truss are treated as revolute.
Different types of wooden and steel roof trusses
1. King Post Truss
2. Queen Post Truss
3. Howe Truss
4. Pratt Truss
5. Fink Truss
6. Fan Truss
7. North Light Roof Truss
8. Quadrangular Roof Truss
Trusses for large span construction
 Tubular steel roof truss
 Tubular monitor steel roof truss
1. King post truss
 King Post Truss is a wooden truss.
 It can also be built of combination of wood and steel.
 It can be used for spans upto 8m.
2. Queen post truss
 Queen Post Truss is also a wooden truss.
 It can be used for spans up to 10m.
3. Howe truss
 It is made of combination of wood and steel.
 The vertical members or tension members are made of steel.
 It can be used for spans from 6-30m.

152
4. Pratt truss
 Pratt Truss is made of steel.
 These are less economical than the Fink Trusses.
 Vertical members are tension and diagonal members are compression.
 Fink Trusses are very economical form of roof trusses.
5. Fan truss
 It is made of steel.
 Fan trusses are form of Fink roof truss.
 In Fan Trusses, top chords are divided into small lengths in order to provide
supports for purlins which would not come at joints in Fink trusses.
6. North light roof truss
 When the floor span exceeds 15m, it is generally more economical to change
from a simple truss arrangement to one employing wide span lattice girders which
support trusses at right angles.
 In order to light up the space satisfactorily, roof lighting has to replace or
supplement, side lighting provision must also be made for ventilation form the
roof.
 One of the oldest and economical methods of covering large areas is the North
Light and Lattice girder.
 This roof consists of a series of trusses fixed to girders. The short vertical side of
the truss is glazed so that when the roof is used in the Northern Hemisphere, the
glazed portion faces North for the best light.

153
 Used for industrial buildings, drawing rooms etc.
7. Quadrangular roof trusses
 These trusses are used for large spans such as railway sheds and Auditoriums.
8. Large span trusses
Different Terms and Terminologies in Truss
a) Bottom chord: The bottom members of a truss
b) Bottom chord splice: If the bottom chord is too long for two bottom chords, the middle
bottom chord is referred to as the splice.
c) Butt cut: A small 90 degree cut at the end of a truss – usually 1/4 .
d) Heel height: The thickness of a truss at the end of the bottom chord. Measured from the
bottom of the bottom chord to the top of the top chord at the end of truss.
e) Overall height: A vertical measurement taken at the midpoint of a truss from the bottom
of the bottom chord to the peak.
f) Overhang: The horizontal distance measured from the end of the bottom chord (butt cut)
to the end of the tail, not including the fascia. If the wall is cantilevered the measurement
is from the outside of the wall to the end of the tail.
g) Ties: Tension carrying members are called ties.
h) Struts: Compression members are called struts.
i) Span of truss: The distance between the supporting end joints of a truss is called its span.
j) Rise: The rise of a truss is the vertical distance between apex and the line joining the
supports.

154
k) Pitch: The ratio of the rise to the span is called pitch.
l) Panel: The portion of the truss lying between the two consecutives joints of the upper
chord is called a panel.
m) Bay: The portion of the roof contained between successive trusses is called bay.
n) Purlin: The member spanning from truss to truss which is meant to carry the load of the
roofing material and to transfer it to the panel joints is called purlin.
Loads on the Trusses
1) Dead loads
The dead loads of the truss include the dead load of roofing materials, purlins,
trusses and bracing systems. The unit weight of various materials are given in IS
875-part I.
G.I sheets= 85 N/m2
A.C sheets= 130 N/m2
Roof covering weight including laps, connector etc.
G.I sheeting= 100-150 N/m2
A.C sheeting= 170-200 N/m2
Weight of purlin= 100-120 N/m2 of plan area.
5 to 10 kN is added for electric fixtures, fans.
W= 20+6.6L N/m2
for a live load of 2 kN/m2
If the live load is more, the above value is to be increased by LL/2.
W= 10*{(L/3)+5}*(s/4)
Where s= spacing of trusses.
2) Imposed load
Up to 10 degree slope= 0.75 kN/m2
For more than 10 degree slope= 0.75-0.02*(θ-10)
3) Wind loads
Basic wind speed (Vb): Up to 3000 m, the value of basic wind speed is 47 m/s and
beyond that it is taken 55 m/s.
Design wind speed (Vz): The basic wind speed is modified to include the
following effects to get design wind velocity at any height (Vz) for the chosen
structure:
 Risk level;
 Terrain roughness, height and size of structure; and
 Local topography.
It can be mathematically expressed as follows:
Vz= Vb*k1*k2*k3
Where k1= design wind speed at any height z in m/s.
K2= probability factor (risk coefficient)
K3= topography factor.
Design wind pressure (pz)= 0.6*Vz
2

155
Wind load on individual members
F= (Cpe –Cpi)*A*pz
Where
Cpe= external pressure coefficient
Cpi = internal pressure coefficient
A= surface area of structural elements or cladding unit, and
Pz= design wind pressure
Load Combination
Various combination of the loads on roof trusses are considered, and the critical condition
is considered for the design. It may be noted that earthquake loads are not significant for
roof trusses because of the small self weight. The following load combinations may be
worked out:
i. Dead load + snow load
ii. Dead load + partial or full live load
iii. Dead load + wind load + internal positive air pressure
iv. Dead load + wind load + internal suction air pressure
v. Dead load + live load + wind load

156
Design of Roof Truss of Power House
Figure 10: Pratt truss at powerhouse
Description units References
panel length in plan 0.86 m
spacing of the truss 2.975 m
span of the truss 13.86 m
angle of inclination 31.2 degree
angle of inclination 0.545 radian
height of truss 4.197 m
Weight of CGI sheet
covering(thickness=1.6
mm)
131 N/m2
IS 875:part I, cl 2.1
self-weight of purlin 113 N/m ISMB 100(113 N/m),IS 808:1989
Weight of bracing 12 N/m2
CALCULATION OF
DEAD LOAD
self-weight of truss 96.2 N/m2
(span/3+5)*10
Total dead load acting
vertically at the joint
0.948 kN (weight of sheet+wt of bracing+wt of
truss)*panal length*spacing+wt of
purlin*spacing
CALCULATION OF
LIVE LOAD
live load 326 N/m2
IS 875:part II,cl 4.1(750-(angle-10)*20)
Total live load 0.834 kN

157
CALCULATION OF
WIND LOAD
basic wind speed 47 m/s
risk factor (K1) 1.07 IS 875:part III table 1(cl.5.3.1)
terrain, height and structure
size factor(k2)
0.76 IS 875:part III ,cl.5.3.2(ht of building=16.949
m, category= 4,class= B)
topography factor(k3) IS 875:part III ,cl.5.3.3
average ground level 2620.5 m
elevation of summit level 3060 m
effective height of feature
(z)
439.5 m
actual length of upwind
slope in the wind direction
1400 m
slope 0.3042 radian
slope 17.429 degree > 17
c 0.36 Is 875:part III,cl.5.3.3.1
effective horizontal length
of the hill
1465 m
height of building/effective
horizontal length
0.0116
actual length/effective
horizontal length
0.956
from figure 15 (Is 875:part
III )
s 0.35
K3 1.126 Is 875:part III,cl.5.3.3.1
Design wind speed 43.036 m/s
internal pressure
coefficient(cpi)
0.2
external pressure
coefficient(cpe)
height of structure/width of
structure
1.223
From table 5(cl.6.2.2.2) IS
875:part III
cpe on windward direction -0.8
cpe on leeward direction -0.8
Wind force on windward
direction
-2.843 kN
Wind force on leeward
direction
-2.843 kN
Calculation of load on
column due to truss
Wind load is neglected since it minimizes the
dead load and live load as wind load act in
opposite direction to the dead load and liveNo of forces on truss joints 16

158
Total dead load 15.171 kN load.
Final column load
Load on each column 14.258 kN

159
Design of Roof Truss of Control Room
Figure 11: Half Pratt truss at control room

160
Description Units References
panel length in plan 0.89 m
spacing of the truss 2.91 m
span of the truss 4.45 m
angle of inclination 24.2 degree
angle of inclination 0.422 radian
height of truss 2.000 m
Weight of CGI sheet
covering(thickness=1.6
mm)
131 N/m2
IS 875:part I, cl 2.1
self-weight of purlin 113 N/m ISMB 100(113 N/m),IS 808:1989
Weight of bracing 12 N/m2
1.CALCULATION OF
DEAD LOAD
self-weight of truss 64.8333
3
N/m2
(span/3+5)*10
Total dead load acting
vertically at the joint
0.867 kN (weight of sheet + wt of bracing +wt of
truss)*panal length* spacing+ wt of
purlin*spacing
2.CALCULATION OF
LIVE LOAD
live load 466 N/m2
IS 875:part II,cl 4.1(750-(angle-10)*20)
3.CALCULATION OF
WIND LOAD
basic wind speed 47 m/s
risk factor (k1) 1.07 IS 875:part III table 1(cl.5.3.1)
terrain, height and structure
size factor(k2)
0.76 IS 875:part III ,cl.5.3.2(ht of building=16.949
m,category= 4,class= B)
topography factor(k3) IS 875:part III ,cl.5.3.3
average ground level 2620.5 m
elevation of summit level 3060 m
effective height of feature
(z)
439.5 m
actual length of upwind
slope in the wind direction
1400 m
slope 0.3042 radian
slope 17.429 degree > 17
c 0.36 Is 875:part III,cl.5.3.3.1
effective horizontal length
of the hill
1465 m
height of building/effective
horizontal length
0.0116
actua length/effective 0.956

161
horizontal length
from figure 15 (Is 875:part
III )
s 0.35
K3 1.126 Is 875:part III,cl.5.3.3.1
Design wind speed 43.036 m/s
internal pressure
coefficient(cpi)
0.2
external pressure
coefficient(cpe)
height of structure/width of
structure
2.697
From table 5(cl.6.2.2.2) IS
875:part III
cpe on windward direction -0.884
cpe on leeward direction -0.758
Wind force on windward
direction
-3.120 kN
Wind force on leeward
direction
-2.757 kN
4.Calculation of load on
column due to truss
Wind load is neglected since it minimizes the
dead load and live load as wind load act in
opposite direction to the dead load and live
load.
No of force in truss 5
Total dead load 4.335 kN
Total live load 6.03446
7
kN
Final column load
Load on each column 5.185 kN

162
Calculation of member forces of truss
Dead load and live load combination is used as the combination for determining the forces in
members of truss.
Member forces of power house Truss
Figure 12: D.L + L.L combination in power house truss

163
Frame
number
P(kN)
Frame
number
P(kN)
Frame
number
P(kN)
6 12.474 74 -18.81 102 5.346
39 21.893 75 -20.52 104 4.455
40 21.893 76 -22.23 106 3.564
41 20.434 77 -23.94 108 2.673
42 18.974 78 -25.65 110 1.782
43 17.515 79 -13.68 112 0.891
44 16.055 80 -15.39 114 0
45 14.595 81 -17.1 115 -1.71
46 13.136 82 -18.81 117 -2.303
47 13.136 83 -20.52 119 -3.046
48 14.595 84 -22.23 121 -3.851
49 16.055 85 -23.94 123 -4.688
50 17.515 86 -25.65 125 -5.542
51 18.974 88 5.346 127 -6.406
52 20.434 90 4.455 128 -6.406
53 21.893 92 3.564 134 -1.71
54 21.893 94 2.673 144 -5.542
71 -13.68 96 1.782 145 -2.303
72 -15.39 98 0.891 147 -3.046
73 -17.1 100 0 149 -3.851
151 -4.688

164
Member forces of control room Truss
Figure 13: DL+LL combination in Control room truss
Frame
Number
P(kN)
2 -5.185
3 -5.185
4 -4.148
5 -3.111
6 -2.074
9 -1.136E-14
10 2.307
11 4.615
12 6.922
13 9.229
15 -2.53
16 -5.059
17 -7.589
18 -10.119
19 -10.119
20 5.675
21 4.747
22 3.873
23 3.102

165
Design of members of truss
Power house truss members
The maximum force in top members is 25.65 kN, compressive in nature. The maximum force in
bottom members is 21.89 kN, tensile in nature. The maximum tensile force in internal members
is 12.47 kN and compressive force is 6.41 kN.
Design of compression member for top members
Description value units References
compressive force on member 25.65 kN
factored force(Pu) 38.48 kN 1.5*force
Length of section(L) 1.4 m
Design stress 110 Mpa Assumption
sectional area(A) 349.77 mm2
Pu/design stress
select ISA 4545,6 mm
Aprovided 507 mm2
>A,ok
radius of gyration(rmin) 8.7 mm from IS 808:1989,steel table
Check for strength
Design compressive strength(Pd) 46.50 kN Aprovided*fcd>Pu,ok
Effective length(Le) 0.98 m
K*L,K=0.7,cl 7.2.2 IS
800:2007
Slenderness ratio 112.64 Le/rmin
Buckling class is c for angle section,cl 7.1.2.2
for slenderness ratio= 112.64 and fy=250
cl 7.1.2.1 IS 800:2007,table
9©
fcd 91.722 Mpa By interpolation
Design of tension member for bottom members
Description Value Units References
tensile force on member 21.89 kN
factored force(Pu) 32.835 kN 1.5*tensile force
Gross area 144.474 mm2
ϒmo*Pu/fy
Ag 180.5925 mm2
1.2 to 1.4*gross area
select ISA 2525,4mm
Agprovided 184 mm2 >Ag ,ok
check for the strength
strength against yielding
Tdg 41.81818 kN Agprovided*fy/ϒm0>Pu,ok

166
Design of compression member for internal members
sectional area(A) 87.41 mm2
Pu/design stress
Aprovided 307 mm2
>A,ok
Check for strength
K*L,K=0.7,cl 7.2.2 IS
800:2007
for slenderness ratio= 304.94 and fy=250 cl 7.1.2.1 IS 800:2007,table 9©
Design of tension member for internal members
Gross area 82.302 mm2
ϒmo*Pu/fy
Ag 102.8775 mm2
1.2 to 1.4*gross area
select ISA 2020,3mm
Tdg 25.45455 kN Agprovided*fy/ϒm0>Pu,ok,cl 6.2 IS 800:2007

167
Control Room truss members
The maximum force in top members is 10.12kN, compressive in nature. The maximum force in
bottom members is 9.23 kN, tensile in nature. The maximum tensile force in internal members is
5.68 kN and compressive force is 5.18 kN.
Design of compression member for top members
sectional area(A) 138.01 mm2 Pu/design stress
Aprovided 277 mm2 >A,ok
Check for strength
Effective length(Le) 0.735 m K*L,K=0.7,cl 7.2.4 IS 800:2007
Design of tension member for bottom members
Gross area 60.918 mm2 ϒmo*Pu/fy
Ag 76.1475 mm2 1.2 to 1.4*gross area
select ISA 2020,3mm
Tdg 25.45455 kN Agprovided*fy/ϒm0>Pu,ok

168
Design of compression member for internal members
Length of section(L) 2 m
sectional area(A) 70.64 mm2 Pu/design stress
Aprovided 277 mm2 >A,ok
Check for strength
K*L,K=0.7,cl 7.2.4 IS
800:2007
Design of tension member for internal members
Gross area 37.488 mm2 ϒmo*Pu/fy
Ag 46.86 mm2 1.2 to 1.4*gross area
select ISA 2020,3mm
Tdg 25.45455 kN Agprovided*fy/ϒm0>Pu,ok,cl 6.2 IS 800:2007

169
Design of connection
Connection design for power house truss
L1,L2 = length of longitudinal fillet weld at the top and bottom respectively on two sides of
angle section
P1,P2= factored design loads along length L1 and L2, respectively
P= Factored load acting on the centroid of the section
size of gusset plate = 8 mm thick
For Fe 410 grade of steel ,fu= 410 MPa, fy = 250MPa
for site welding: partial safety factor for the material,ϒmw= 1.5
p 38.48 kN
Section used ISA 4545,6 mm,h1=13.3 mm,h2=31.7 mm
Taking moment about the line passing through length L1
P2 27.11 kN Ph2/h
P1 11.37 kN Ph1/h
Minimum size of weld 3 mm
for 8mm thick gussetplate,cl 10.5.2.3 IS
800:2007
Maximum size of weld 4.5 mm thickness of thinner member-1.5
Provide 4 mm weld size
Effective throat
thickness(tt) 2.8 mm 0.7*weld size,cl 10.5.3.2 IS 800:2007
L1 25.74 mm 3^0.5*ϒmw*P1/(fu*tt)
L2 61.35 mm 3^0.5*ϒmw*P2/(fu*tt)
Provide 26 mm length of weld at the top
Provide 62 mm length of weld at the bottom

170
Connection design for control room truss
L1,L2 = length of longitudinal fillet weld at the top and bottom respectively on two sides of
angle section
P1,P2= factored design loads along length L1 and L2, respectively
P= Factored load acting on the centroid of the section
size of gusset plate = 6 mm thick
For Fe 410 grade of steel ,fu= 410 MPa, fy = 250MPa
for site welding: partial safety factor for the material,ϒmw= 1.5
Description
Valu
e
Unit
s References
p 15.18 kN
Section used ISA 3030,5 mm,h1=9.2 mm,h2=20.8 mm
P2 10.52 kN Ph2/h
P1 4.66 kN Ph1/h
Minimum size of weld 3 mm
for 6 mm thick gusset plate, cl 10.5.2.3 IS
800:2007
Maximum size of weld 3.5 mm thickness of thinner member-1.5
Provide 3.5 mm weld size
Effective throat
thickness(tt) 2.45 mm 0.7*weld size,cl 10.5.3.2 IS 800:2007
L1 12.04 mm 3^0.5*ϒmw*P1/(fu*tt)
L2 27.22 mm 3^0.5*ϒmw*P2/(fu*tt)
Provide 13 mm length of weld at the top
Provide 28 mm length of weld at the bottom

171
6.4 Centre of Mass and Centre of Rigidity
Control room is a four storey symmetrical building. The loading condition for 1,2 and 3 is same
while for the 4 there is truss.
Beam size = 35cm x 45 cm
Column size = 50cm x 50 cm
Floor slab = 20 cm thick including finish
Wall thickness = 23cm
Storey height= 5 m
Brick density = 20kN/m3
Live load = 10 kN/m2
Load along BB is calculated
Weight of Beam = 0.35x0.45x4.45x25+ 2x (1/2)x5.1x0.35x0.45x20=37.58 kN
Weight of slab = 2x (1/2)x4.45x5.1x0.2x25 = 113.48 kN
Weight of column =2x0.5x0.5x5x25 = 62.5 kN (since there are two numbers of column along
BB)
Weight of wall =4x (1/2)x0.23x5.1x4.5x20 = 211.14 kN
Live load = 2x (1/2)x4.45x5.1x10x0.5 = 113.48 kN ( Only 50 % kive load is consider for LL>
3kN/m2
)

172
Total weight = 398.96 kN
Weight calculation in y direction
Column
Line
Weight of
beams
Weight of
slab
Weight of
columns
weight
of walls
Live
Load
Total
Load(kN)
A-A 27.58 56.74 62.5 144.895 56.74 348.455
B-B 37.58 113.48 62.5 211.14 113.48 538.18
C-C 37.58 113.48 62.5 211.14 113.48 538.18
D-D 37.58 113.48 62.5 211.14 113.48 538.18
E-E 27.50 56.74 62.5 144.895 56.74 348.375
Total load 2311.37
Taking moment of weight about A-A
Cmx =
𝟑𝟒𝟖.𝟒𝟓𝐱𝟎+𝟓𝟑𝟖.𝟏𝟖𝐱𝟓.𝟏+𝟓𝟑𝟖.𝟏𝟖∗𝟏𝟎.𝟐+𝟓𝟑𝟖.𝟏𝟖𝐱𝟏𝟓.𝟑+𝟑𝟒𝟖.𝟑𝟕𝐱𝟐𝟎.𝟒
𝟐𝟑𝟏𝟏.𝟑𝟕
= 10.2 m
Weight calculation in x direction
Column
Line
Weight of
beams
Weight of
slab
Weight of
columns
weight
of walls
Live
Load
Total
Load(kN)
1-1 82.3 226.95 156.25 468.33 226.95 1160.78
2-2 82.3 226.95 156.25 468.33 226.95 1160.78
Total load 2321.56
Cmy =
1160.78𝑥2.225+1160.78𝑥2.225
2321.56
= 2.225m
Roof truss load is in symmetricity in the 4th
storey. So, the centre of mass is same as above
Centre of Rigidity
In x direction
Lateral stiffness of column = 12
𝐸𝐼
𝐿3
For a column having a square cross-section kx= ky = k since E,I and L are constant.
xr =
∑kyX
∑ky
=
2kx 0+2kx 5.1+2kx 10.2+2kx 15.3+2kx 20.4
10k
= 10.2 m

173
Here 2k means number of column in grid A-A, B-B, C-C,D-D,E-E respectively
In x direction
Lateral stiffness of column = 12
𝐸𝐼
𝐿3
For a column having a square cross-section kx= ky = k since E,I and L are constant.
xr =
∑kxY
∑kx
=
5kx 0+5kx 4.45
10k
= 2.225 m
Here 5k means number of column in grid 1-1, 2-2 respectively
Ecentricity
For I,II, III and IV storeys
ex = 10.2-10.2 = 0
ey = 2.225-2.225 = 0

174
Chapter 7
Structural Analysis
7.1 Salient features of sap 2000
For the structural analysis of the power house and control room SAP V14 has been use.
The SAP name has been synonymous with state-of-the-art analytical methods since its
introduction over 30 years ago. SAP2000 follows in the same tradition featuring a very
sophisticated, intuitive and versatile user interface powered by an unmatched analysis engine and
design tools for engineers working on transportation, industrial, public works, sports, and other
facilities.
From its 3D object based graphical modeling environment to the wide variety of analysis and
design options completely integrated across one powerful user interface, SAP2000 has proven to
be the most integrated, productive and practical general purpose structural program on the
market today. This intuitive interface allows you to create structural models rapidly and
intuitively without long learning curve delays. Now you can harness the power of SAP2000 for
all of your analysis and design tasks, including small day-to-day problems.
Complex Models can be generated and meshed with powerful built in templates. Integrated
design code features can automatically generate wind, wave, bridge, and seismic loads with
comprehensive automatic steel and concrete design code checks per US, Canadian and
international design standards.
Advanced analytical techniques allow for step-by-step large deformation analysis, Eigen and
Ritz analyses based on stiffness of nonlinear cases, catenary cable analysis, material nonlinear
analysis with fiber hinges, multi-layered nonlinear shell element, buckling analysis, progressive
collapse analysis, energy methods for drift control, velocity-dependent dampers, base isolators,
support plasticity and nonlinear segmental construction analysis. Nonlinear analyses can be static
and/or time history, with options for FNA nonlinear time history dynamic analysis and direct
integration.
From a simple small 2D static frame analysis to a large complex 3D nonlinear dynamic analysis,
SAP2000 is the easiest, most productive solution for your structural analysis and design needs. In
Structural analysis of the power house we use IS 1893:2002 for the seismic Considerations.
The steps for Sap analysis
1) Formation of Grid
2) Defining material
3) Defining Section
4) Assigning section to the grid
5) Defining Load pattern
6) Defining Mass source

175
7) Generating Load Combination
8) Assigning load to the Section
9) Restraint Base Joint
10) Diaphragm the Joints
11) Meshing the area section
12) Run Analysis
13) Verify the members
7.2 Input
The grid is formed according to plan of power house and control room, material we use is M25,
Fe 500 for the concrete and steel respectively.
Section assign
For power House
Sectioons
Breadth
(m) Depth(m)
Level
Z(m) X-grid Y-grid Remarks
Beam 0.4 0.5 5
Beam 0.4 0.5 9.9
Beam 0.35 0.45 15.23
Column 0.6 0.6 throughout
3-3, 8-
8
Column 0.5 0.5 throughout 5-5,6-6
Mat thickness 0.7 0 at service bay
Mat thickness 1.07 0
at machine
hall
For control room
Sectioons
Breadth
(m) Depth(m) Level Z(m) X-grid Y-grid Remarks
Beam 0.35 0.45 0,5,9.43,12.63 Size of the
beam and
column is
taken same
throughout
the control
room
Column 0.5 0.5 throughout
Slab thickness 0.17 0,5,9.43
Mat thickness 0.7 -5

176
Load pattern
Load pattern
name Type
Self
weight
multiplier Remarks
Dead Dead 1 Frame Load
Ext. Wall Dead 0
Int. wall Dead 0
Truss dead Dead 0
Stair dead Dead 0
Slab live Live 0
Stair live Live 0
Truss live Live 0
Gantry live Live 0
Eqx Quake 0
IS 1893:2003(X-
dir.)
Eqy Quake 0
IS 1893:2003(Y-
dir.)
Mass Source
Dead load is multiply with the factor 1
According to IS 456:2000
For Live Load < 3kN/m2
factor 0.25
For Live Load > 3kN/m2
Factor 0.5
Load Combinations
We use the IS 14 load combinations
1) 1.5DL
2) 1.5(DL+LL)
3) 1.2(DL+LL+EQx)
4) 1.2(DL+LL-EQx)
5) 1.2(DL+LL+EQy)
6) 1.2(DL+LL-EQy)
7) 1.5(DL+EQx)
8) 1.5(DL-EQx)
9) 1.5(DL+EQy)
10) 1.5(DL-EQy)
11) 0.9DL+1.5EQx
12) 0.9DL-1.5EQx
13) 0.9DL+1.5EQy
14) 0.9DL-1.5EQy

177
Envelope is formed using this 14 load combinations. The function of envelope is to show the
maximum value among above load combinations
Restraint the base joint
The base of the column are restrained
Diaphragm the joints
The diaphragm constraints are defined to the joints at slab level so that all these joints move
together with the slab in the same direction.
For each floor diaphragm are formed except at the position where columns are restrained.
Meshing the area
Meshing of the area is done to transfer the slab load uniformly to the beam so that slab and beam
deflect in a same pattern.
7.3 Output
7.3.1 Grid for Column Location:

178
Figure 14: 3d Model of Powerhouse and Control Room

179
7.3.2 Base Reaction of the Column
Table 15: Base reaction of the column
Joints F1(kN) F2(kN) F3(kN) M1(kNm) M2(kNm) M3(kNm)
A1 61.10 60.15 999.13 170.87 195.29 0.89
B1 66.42 64.20 1541.26 182.96 205.26 0.89
C1 65.42 65.07 1531.12 188.67 203.34 0.89
D1 69.44 69.88 1545.98 193.21 209.43 0.89
E1 45.65 67.37 1000.07 191.38 171.46 0.89
A2 61.38 47.71 998.40 191.18 196.37 0.89
B2 66.80 53.16 1541.86 199.84 206.44 0.89
C2 65.75 54.72 1529.69 204.98 204.43 0.89
D2 69.79 55.87 1424.68 213.57 210.63 0.89
E2 45.88 52.92 916.15 215.25 172.39 0.89
A3 24.33 72.41 1121.40 264.49 103.01 7.31
B3 26.13 41.21 1144.59 228.31 105.35 7.19
C3 25.92 41.34 1139.44 231.50 105.06 7.19
D3 25.85 42.35 1137.98 237.09 104.90 7.19
E3 28.34 44.27 1192.08 245.31 106.42 7.18
F3 9.34 80.69 1157.20 296.44 79.32 7.18
A5 3.07 58.55 511.94 174.85 15.90 3.52
F5 2.86 62.91 515.29 187.34 15.45 3.52
A6 8.22 59.23 512.73 174.82 3.42 3.52
F6 8.37 63.67 516.37 187.40 3.68 3.52
A8 53.03 61.26 671.18 275.19 16.64 5.13
B8 35.28 44.27 524.95 238.93 11.51 5.30
C8 44.70 44.80 520.26 243.55 13.86 5.02
D8 45.28 46.09 520.12 250.84 14.02 5.01
E8 41.91 48.21 550.78 260.35 11.58 5.00
F8 13.05 72.28 687.32 318.46 8.32 4.71

180
7.3.3 Reinforcement from SAP
Reinforcement along Grid 1-1

184
7.3.4 Bending moment
Figure 15: Bending moment alone the grids
Fig: Bending moment Along 1-1 Fig: Bending moment Along 2-2
Fig: Bending moment Along 8-8

185
7.3.5 Sher force Diagram
Fig: Shear force along 1-1
Fig: Shear force along 3-3
Fig: Axial force along 8-8 Fig: Axial force along A-A
Figure 16: Axial force along the grids

186
7.3.6 Axial force Diagram
Fig: Axial force along 1-1 Fig: Axial force along 2-2
Fig: Axial force along 3-3 Fig: Axial force along 8-8
Figure 17: Axial force along the grids

187
7.3.7 Storey Drift Calculation
Table 16: Storey drift calculation
storey drift calculation of control room
storey Along X-direction Along Y- direction Allowable
Drift(=0.004*
h)(mm)
displacement(m
m)
storey
drift(mm)
displacement(m
m)
storey
drift(mm)
fourth 30.8974 1.8724 37.365 3.065 12.8
third 29.025 7.1999 34.3 9.431 17.72
secon
d
21.8251 12.5892 24.869 14.801 20
first 9.2359 9.2359 10.068 10.068 20
groun
d
0 0 0 0 0
storey drift calculation of machine hall
storey Along X-direction Along Y- direction Allowable
Drift(=0.004*
h)(mm)
displacement(
mm)
storey
drift(mm)
displacement(m
m)
storey
drift(mm)
third 14.95 5.45 31.67 10.97 21.32
secon
d
9.5 6.51 20.7 12.66 19.6
first 2.99 2.99 8.04 8.04 20
groun
d
0 0 0 0 0

188
Sway of the power house and control room is shown below the given sway magnitude is in the
scale of 50
Fig: Sway along A-A
Fig: Sway along B-B
Fig: Sway along 3-3
Figure 18: Sway

189
Chapter 8
Cost Estimation
8.1 Introduction
Before undertaking the construction of the project, it is necessary to know tis probable cost
which is worked out by estimating. An estimate is a computation or calculation of quantities
required and expenditure likely to be incurred in the construction of the work. The primary
object of the estimate is to enable one to know beforehand, the cost of the work. The estimate is
the probable cost of a work and determined theoretically by mathematical calculations based on
the plans and drawing and current rates.
For all engineering works it is required to know beforehand the probable cost of construction
known as the estimated cost. If the estimated cost is greater than money available, then attempts
are made to reduce the cost by reducing the work or by changing the specifications. From this the
importance of estimate to engineers is known. In preparing an estimate, the quantities of different
items of work are calculated by simple measurement method and from these quantities of
different items of work are calculated. The subject of estimating is simple, nothing much to
understand, but knowledge of drawing is essential.
Accuracy in estimate is very important, if estimate is exceeded it becomes a very difficult
problem for engineers to explain, to account for and arrange for the additional money.
Inaccuracy in preparing estimate, omission of items, changes in design, improper rates, etc. are
the reasons for exceeding the estimate, through increase in rates is one of the main reason. So,
one has to take care of these things while preparing an estimate.
8.2 Detailed estimate
Preparation of detailed estimate consists of working out the quantities of different items of work
and then working out the cost i.e. the estimate is prepared in two stages:
i) Details of measurement and calculation of quantities
ii) Abstract of estimated cost
Detailed estimate method has been used as the method for estimating the quantities.

190
8.3 Estimation Table
Table 17: Estimation table
ABSTRACT OF ESTIMATE FORM
Item
No.
Particulars of item of
works
Unit Nos. Length
(L)(m)
Breadth
(B)(m)
Height
(H)(m)
Quantity Remark
1 Earthwork
1.1 Earthwork in excavation up to ground level
Earthwork in excavation cu.m. 1 - - - 14,827.94
1.2 Earthwork in excavation in foundation
a Mat foundation in Machine
Hall and service bay
cu.m. 1 35 16 5.68 3180.8
b Mat Foundation in control
room
cu.m. 1 20.9 4.45 5.5 511.53
Total cu.m. 18,520.26
2 PCC Works (1:1.5:3)
Hall and service bay
cu.m. 1 35 16 0.15 84 ( PCC 0.15m)
b Mat foundation in control
room
cu.m. 1 20.9 4.45 0.15 13.95
Total cu.m. 97.951
3 Formworks
3.1 Foundation
Hall
sq.m. 1 69.5 1 69.5

191
b Mat foundation in Service
bay
sq.m. 1 43.14 0.6 25.884
c Mat foundation in control
room
sq.m. 1 50.7 0.72 36.504
Sub Total sq.m. 131.89
3.2 For Column
1 Machine hall
i At GF
Type 1(0.6*0.6) sq.m. 12 2.4 5 144
Type 2(0.5*0.5) sq.m. 4 2 5 40
ii At 1st floor
Type 1(0.6*0.6) sq.m. 12 2.4 4.9 141.12
Type 2(0.5*0.5) sq.m. 4 2 4.9 39.2
iii At 2nd floor
Type 1(0.6*0.6) sq.m. 12 2.4 5.33 153.504
Type 2(0.5*0.5) sq.m. 4 2 5.33 42.64
2 Control room
i At GF sq.m. 10 2 5 100
ii At 1st floor sq.m. 10 2 5 100
ii At 2nd floor sq.m. 10 2 4.43 88.6
iii At 3rd floor sq.m. 10 2 3.2 64
Sub total sq.m. 913.064
3.3 For Beam
1 Machine hall
i At 1st floor
Beam along Y-axis sq.m. 2 13.7 1.8 49.32
Beam along X-axis sq.m. 2 26.35 1.8 94.86

192
ii At 2nd floor
Beam along Y-axis sq.m. 2 13.7 1.8 - 49.32
Beam along X-axis sq.m. 2 26.35 1.8 - 94.86
iii At 3rd floor -
2 Control room
3.4 For slab
i slab in Control Room sq.m. 10 5.1 4.45 226.95
3.5 For Staircase
a Waist slab sq.m. 2 6.24 1.06 13.23
b landing slab
landing slab A sq.m. 2 4.77 1.28 12.211
Landing slab B and C sq.m. 4 1.06 1.06 4.4944
3.6 Corbel sq.m. 8 Area 0.39 3.12

193
Total sq.m. 2124.9964
4 RCC Works
4.1 foundation
Mat foundation in machine
hall
cu.m. 1 20.4 14.35 1 292.74
Mat foundation in service
bay
cu.m. 1 14.35 7.22 0.6 62.1642
Mat foundation in Control
room
cu.m. 1 20.9 4.45 0.72 66.9636
Sub total cu.m. 421.868
4.2 For Column
1 Machine hall
i At GF
Type 1(0.6*0.6) cu.m. 12 0.6 0.6 5 21.6
Type 2(0.5*0.5) cu.m. 4 0.5 0.5 5 5
ii At 1st floor
Type 1(0.6*0.6) cu.m. 12 0.6 0.6 4.9 21.168
Type 2(0.5*0.5) cu.m. 4 0.5 0.5 4.9 4.9
iii At 2nd floor
Type 1(0.6*0.6) cu.m. 12 0.6 0.6 5.33 23.0256
Type 2(0.5*0.5) cu.m. 4 0.5 0.5 5.33 5.33
2 Control room
i At GF cu.m. 10 0.5 0.5 5 12.5
ii At 1st floor cu.m. 10 0.5 0.5 5 12.5
ii At 2nd floor cu.m. 10 0.5 0.5 4.43 11.075

194
iii At 3rd floor cu.m. 10 0.5 0.5 3.2 8
4.3 For Beam
1 Machine hall
i At 1st floor
Beam along Y-axis cu.m. 2 13.7 0.4 0.5 5.48
Beam along X-axis cu.m. 2 26.35 0.4 0.5 10.54
ii At 2nd floor
iii At 3rd floor
iv Gantry Beam cu.m. 2 26.35 0.75 0.6 23.715
2 Control room
4.4 For slab
i slab in Control Room cu.m. 12 5.1 4.45 0.17 46.2978
4.5 For Staircase
a Waist slab cu.m. 2 8.24 1.06 0.15 2.62
b landing slab
landing slab A cu.m. 2 5.1 1.7 0.15 2.60
Landing slab B and C cu.m. 4 1.06 1.06 0.15 0.67
4.6 Corbel cu.m. 8 0.76 Area 0.39 2.3712
Total cu.m. 723.75

195
5 Brick Works(1:5)
5.1 Machine hall
i Ground floor cu.m. 1 80.1 0.22 5 88.11
Deduction cu.m. 8.811 10% for windows and
doors
ii At 1st floor cu.m. 1 80.1 0.22 4.9 86.3478
doors
Sub total 77.713
ii At 2nd floor cu.m. 1 80.1 0.22 5.33 93.92526
doors
5.2 Control room
i Ground floor cu.m. 1 63.05 0.22 5 69.355
doors
ii At 1st floor cu.m. 1 63.05 0.22 5 15.2581
doors
ii At 2nd floor cu.m. 1 63.05 0.22 4.43 61.44853
doors

196
iii At 3rd floor cu.m. 1 63.05 0.22 3.2 153.41326
doors
Total cu.m. 511.072
6 Plastering
(12 mm thick cement
sand (1:3) plastering
6.1 For Column
1 Machine hall
i At GF
Type 1(0.6*0.6) sq.m. 12 0.6 5 36
Type 2(0.5*0.5) sq.m. 4 0.5 5 10
ii At 1st floor
Type 1(0.6*0.6) sq.m. 12 0.6 4.9 35.28
Type 2(0.5*0.5) sq.m. 4 0.5 4.9 9.8
iii At 2nd floor
Type 1(0.6*0.6) sq.m. 12 0.6 5.33 38.376
Type 2(0.5*0.5) sq.m. 4 0.5 5.33 10.66
2 Control room
i At GF sq.m. 10 0.5 5 25
ii At 1st floor sq.m. 10 0.5 5 25
ii At 2nd floor sq.m. 10 0.5 4.43 22.15
iii At 3rd floor sq.m. 10 0.5 3.2 16
6.2 For Beam
1 Machine hall
i At 1st floor

197
ii At 2nd floor
iii At 3rd floor -
2 Control room
6.3 For slab
i slab in Control Room sq.m. 20 5.1 4.45 453.9
6.4 For wall
1 Machine hall
i Ground floor sq.m. 2 80.1 5 801
Deduction sq.m. 80.1 10% for windows and
doors
ii At 1st floor sq.m. 2 80.1 4.9 784.98
doors
Sub total 706.482
ii At 2nd floor sq.m. 2 80.1 5.33 853.866

198
doors
2 Control room
i Ground floor sq.m. 2 63.05 5 630.500
Deduction sq.m. 63.05 10% for windows
and doors
ii At 1st floor sq.m. 2 63.05 5 630.5
doors
ii At 2nd floor sq.m. 2 63.05 4.43 558.623
doors
iii At 3rd floor sq.m. 2 63.05 3.2 403.52
doors
Total sq.m. 5108.491

199
Estimation of Reinforcement
Item
No.
Particulars of item of
works
Unit Nos. Length
(L)(m)
Breadth
(B)(m)
Height
(H)(m)
Volume Amount of
steel(qn)(fe
500)
Remarks
1 foundation
i Mat foundation in
machine hall
cu.m. 1 20.4 14.35 1 292.74 506.24136 100 to 120 kg
per cub.
Meterii Mat foundation in
service bay
cu.m. 1 14.35 7.22 0.6 62.1642
iii Mat foundation in
Control room
cu.m. 1 20.9 4.45 0.72 66.9636
Sub total 421.8678
2 For Column
2.1 Machine hall
i At GF 312.7465 200 to 250kg
per cub. meterType 1(0.6*0.6) cu.m. 12 0.6 0.6 5 21.6
Type 2(0.5*0.5) cu.m. 4 0.5 0.5 5 5
ii At 1st floor
Type 1(0.6*0.6) cu.m. 12 0.6 0.6 4.9 21.168
Type 2(0.5*0.5) cu.m. 4 0.5 0.5 4.9 4.9
ii At 2nd floor
Type 1(0.6*0.6) cu.m. 12 0.6 0.6 5.33 23.0256
Type 2(0.5*0.5) cu.m. 4 0.5 0.5 5.33 5.33
2.2 Control room
i At GF cu.m. 10 0.5 0.5 5 12.5
ii At 1st floor cu.m. 10 0.5 0.5 5 12.5
iii At 2nd floor cu.m. 10 0.5 0.5 4.43 11.075
iv At 3rd floor cu.m. 10 0.5 0.5 3.2 8

200
3 For Beam
3.1 Machine hall
i At 1st floor 268.873 200 to 220 kg
per cub. meterBeam along Y-axis cu.m. 2 13.7 0.4 0.5 5.48
ii At 2nd floor
ii At 3rd floor
iv Gantry Beam cu.m. 2 26.35 0.75 0.6 23.715
3.2 Control room
4 For slab
i slab in Control Room cu.m. 12 5.1 4.45 0.17 46.2978 33.925632 60 to 65 kg
per cubic
meter
4.1 For Staircase
a Waist slab cu.m. 2 8.24 1.06 0.15 2.62
b landing slab
landing slab A cu.m. 2 5.1 1.7 0.15 2.60
Landing slab B and C cu.m. 4 1.06 1.06 0.15 0.67
5 Corbel cu.m. 8 0.76 0.6 0.65 2.3712 5.21664 200 to 220 kg
per cub. meter
Total 1127.00

201
Chapter 9
Rate analysis
9.1 Introduction
The determination of rate per unit of particular item of work, from the cost of quantities of
materials, the cost of laborers and other miscellaneous petty expenses require for its completion
is known as the analysis of rate. Rates of materials are usually taken as the rates delivered at the
site of work and include the first cost (cost at origin), cost of transport, taxes etc.
For the purpose of analysis, the details about all the operations involved in carrying out the work
should be available, the quantities of materials required and their costs should be known and the
number of different categories of laborers required and the capacity of doing work per labourer
and their wages per day should be known.
9.2 Brickwork in (1:4) Cement Sand Mortar
Table 18: Rate analysis table
Brickwork in(1:4) Cement sand mortar
Analysis for 1 Cu.m.
Resources Type Qty Unit Rate Amount
Labor
Skilled 1.5 nos 1200 1,800.00
Non-Skilled 3 nos 800 2,400.00
Construction
materials
Brick 550 nos 18 9,900.00
Washing sand 0.27 m3 1,500.00 405.00
Cement 0.1 Mton 25,000.00 2,500.00
Water 120 lit 1.15 138.00
Actual rate 17,143.00
15% contractor overhead 2,571.45
Total rate 19,714.45
Rate per Cu.m. 19,714.45

202
9.3 12. 5mm (1:4) Cement Sand Plaster
12.5 mm(1:4) Cement Sand Plaster
Analysis for 100 Sq.m.
Labor
Skilled 12 Nos 1200 14,400.00
Non-Skilled 16 Nos 800 12,800.00
Construction
materials
Washing sand 1.46 m3 1,500.00 2,190.00
Cement 0.54 Mton 25,000.00 13,500.00
Rate per Sq.m. 493.235
9.4 Open Cut Excavation
Open Cut excavation
Labour
Foreman 0.04 hr. 150 6
Backhoe operater 0.15 hr. 200 30
Tractor driver 0.25 hr. 110 27.5
Helper 0.15 hr. 70 10.5
Unskilled labour 0.15 hr. 100 15
Equipment
Backhoe 0.15 hr. 2,000.00 300
Tractor 0.25 hr. 200 50
Materials
Diesel @ 25ltrs/hr 0.15 ltr 330 49.5
Lubricant 0.01 ltr 600 6
Actual rate 494.5
15% contractor overhead 74.175
Total rate 568.675
Rate per Cu.m. 568.675

203
9.5 PCC Works (1:1.5:3)
PCC Works (1:1.5:3)
Analysis for 1 Cu.m
S No. Description Nos. Unit Quantity Rate (NRs.) Amount
1 Labor
Foreman 1 hr. 8 150 1200
Skilled Labor 1 hr. 4 120 480
Unskilled labor 1 hr. 1.6 100 160
Sub-Total 1840
2 Materials
Cement kg 272.7 25 6,817.50
Sand m3
0.27 1,500.00 405.00
Aggregate 10mm m3
0.54 1,500.00 810.00
Sub-Total 8,032.50
Total A 9,872.50
Overhead and Profit 15 % of A 1,480.88
Grand Total 11,353.38

204
9.6 M25 Concrete Works
M25 Concrete works in Superstructure
Labor
Skilled 0.5 nos 1200 600
Non-Skilled 1.5 nos 800 1200
Construction
materials
Cement 0.61 Mton 25,000.00 15250
Sand 0.43 m3 1,500.00 645
20mm Aggregate 0.64 m3 1,500.00 960
10 mm Aggregate 0.21 m3 1,500.00 315
Water 300 lit 1.15 345
Diesel 3 lit 88 264
Petrol 0.1 lit 90 9
Equipment
Mixer 0.6 hr 2,300.00 1380
Vibrator 0.25 hr 115 28.75
Rate per Cubic.meter 24,146.26
9.7 Formwork for RCC
Formwork for RCC
Analysis for 1 Sq.m
S
No. Description Nos. Unit Quantity
Rate
(NRs./hr) Amount
1 Labour
Carpenter 1 hr. 2 120 240
Unskilled labour 1 hr. 1 90 90
Sub-Total 330
2 Materials
plywood m2 0.2 1035 207
Nails (25- 100 mm ) kg 0.2 180 36
Sub-Total 243
Total A 573
Overhead and Profit 15 % of A 85.95
Grand Total 658.95

205
9.8 Reinforcement Works for Fe500 Steel
Reinforcement for Fe 500 Steel
Analysis for 1 ton steel
S
No. Description Nos. Unit Quantity
Rate
(NRs./hr) Amount
1 Labour
Foreman 1 hr. 4 150 600
Electrician 1 hr. 3 120 360
Skilled Labour 3 hr. 8 360 2880
Unskilled labour 7 hr. 8 300 2400
Sub-Total 6,240.00
2 Materials
Steel bar ton 1 95,000.00 95,000.00
Binding wire kg 10 120 1,200.00
Sub-Total 96,200.00
Total A 102,440.00
Overhead and Profit 15 % of A 15,366.00
Grand Total 117,806.00

206
9.9 Summary of Unit Rates of Civil Works
Table 19: Summary of unit rates of civil works
S.No. Description Unit Rate (Rs.)
1 B/W in 1:4 Cement Sand Mortar m3
19,714.45
2 12.5 mm(1:4) Cement sand Plaster m2
493.235
3 Open Cut Excavation m3
568.675
4 Ordinary PCC Works m2
11,353.38
5 M25(1:1:2) RCC woks m3
24,146.26
6 Formwork for RCC m3
658.95
7 Reinforcement steel works for Fe500 tonn 117,806.00

207
9.10 Abstract of Estimated Cost
Table 20: Abstract of estimated cost
Item
No
Description of items Unit Quantity Rate per Rs
1 Earthwork in Excavation m3
18,520.26 568.675 m3
10,532,010.85
2 Form work plain
ordinary
m2
2,125.00 658.95 m2 1,400,266.38
3 Plane Cement
Concrete(PCC)
,M20(1:1.5:3)
m3
97.95075 11,353.38 m3 1,112,071.60
4 12mm thick
Plastering(1:3)
m2
5,108.49 493.235 m2 2,519,686.61
5 Concrete M25 for R.C.C. m3
723.75 24,146.26 m3 17,475,758.00
6 Brickworks in 1:5
cement mortar
m3
511.07215
5
19,714.45 m3 10,075,506.45
7 Reinforcement steel tonn 112.7 117,806.00 tonn 13,276,736.20
Total 56,392,036.08
Add 3% for contingencies 1,691,761.08
Add 2% for Work charged Establishment 1,127,840.72
Total 59,211,637.88

208
Chapter 10
Conclusion
The project entitled “Design of Continuous Flushing Settling Basin and Powerhouse” has been
completed. A thorough knowledge on the application of load and moment in the structural frame
of any building has been gained. The reinforcement and dimensioning obtained from software
and by manual calculation is different from one another so appropriate result has been taken
consider
ing their strength, stability and economy. The reinforcement of major components of powerhouse
has been taken from SAP 2000 as its result is found to be more accurate and reliable. The HSRS
system is found to be economical to operate for months of higher discharge and for high head
HEP. We were well trained with the use of different software like AutoCAD, SAP 2000. The
analysis, design and detailing has been done according to IS 456:2000, IS 13920:1993, IS 875
(part I, part 2, part 3):1987, IS 1893 (part 1):2002 and seismic coefficient method is used for
lateral load. Project was only limited with the practical applications of the studied course
throughout our bachelor’s degree.
The structural analysis of the building using SAP 2000 has been done. The earthquake
requirements have been fulfilled. The building was within deflection limit. Thus the powerhouse
was considered safe in all aspects of structural integrity.

209
References
 Shrestha, H. S. (2012). Application of Hydro Suction Sediment Removal System (HSRS)
on Peaking Pond
 C. Jokiel & M. Detering. An Innovative Sediment Removal Solution – Application and
Project Experiances
 Baral, Sanjeeb (2016). Fundamentals of Hydropower Engineering.Kathmandu: National
Book Center Pvt. Ltd.
 IS 456:2000 (2000). Plain & Reinforced Concrete - Code of Practice (Fourth Revision):
B I Standards.
 IS 1893(Part 1):2002 (2002). Criteria for Earthquake Resistant Design of Structures
(Fifth Revision): B I Standards.
 IS 13920:1993 (2002). Ductile Detailing of Reinforced Concrete Structures Subjected to
Seismic Forces — Code of Practice: B I Standards.
 SP 16:1980. Design Aids for Reinforced Concrete to IS 456-1978: B I Standards.
 IS 875 (Part 1, Part 2, Part 3):1987. Code of Practice for Design Loads (Other than
Earthquake) for Buildings and Structures: B I Standards.
 IS 800:2007(2007). General Construction in Steel – Code of Practice: B I Standards.
 Bhavikatti, S.S. (2015). Design of Steel structures. New Delhi: New Age International
Publishers.
 Bhavikatti, S.S. (2012). Design of R.C.C. Structural Elements. New Delhi: New Age
International Publishers.
 Jain, A.K. (2012). Reinforced Concrete, Limit State Design. Roorkee: Nem Chand &
Bros.

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