A

VOCATIONAL TRAINING REPORT

ON
REINFORCED CONCRETE STRUCTURES

Starting Date of Training: 10/06/2019

End Date of Training: 08/07/2019

Submitted in the partial fulfillment of

the Requirements for the award of the degree of

Bachelor of Technology

In Civil Engineering
By
YASHO DEEP SINGH
1619200104

GL BAJAJ INSTITUTE OF TECHNOLOGY & MANAGEMENT, Gr. Noida

DR. APJ ABDUL KALAM TECHNICAL UNIVERSITY
LUCKNOW
Year -2019-2020
 DECLARATION
I, Yasho Deep Singh, student of B. Tech VII semester, studying at GL
BAJAJ INSTITUTE OF TECHNOLOGY & MANAGEMENT, Gr. Noida, hereby
declare that the summer training report on “Construction of New High
Court at Ranchi, Jharkhand.” under the purview of Reinforced Concrete
Structures. Is a result of my indebtness to other work publications,
references, if any, have been duly acknowledged.

The report submitted to MECON Limited in partial fulfillment of summer

training is the original work conducted by me. The information and data
given in the report is authentic to the best of my knowledge. This summer
training report is not being submitted to any other university or company
for award of any other certificate, appreciation and fellowship.

YASHO DEEP SINGH

 ACKNOWLEDGEMENT

We express our sincere thanks to Dr. PRASHANT

MUKHERJEE, Head of Civil Engineering department for
his support and guidance for doing the project.

We express our indebtness and gratitude to our guide Ms.

Preeti Gangwar, Assistant professor, Department of Civil
Engineering, GL BAJAJ INSTITUTE OF TECHNOLOGY
AND MANAGEMENT, for her guidance and care taken by
him in helping us to complete the project work successfully.

We express our deep gratitude to Mr. Abhishek

Bhattacharya, Project Manager, MECON, Ranchi for his
valuable suggestions and guidance rendered in giving shape
and coherence to this endeavor. We are also thankful to his
team members for their support and guidance throughout the
period of project.
 ABSTRACT
The Jharkhand High Court is one of the newest High
Courts in India. It was established in 2000 under the Bihar
Reorganisation Act, 2000, after the state of Jharkhand was
carved out of the state of Bihar. The court has jurisdiction
over Jharkhand state.

The seat of the court is at Ranchi, the administrative capital of

the state. The court has a sanctioned judge strength of 20. The
new building of Jharkhand High Court has been approved.

A 165 acres land has been granted near the HEC Industrial
Complex, which will be used for development of High Court,
Residential Complex for Judges and Lawyers Chambers.

The estimated cost for the project is around Rs. 265 Crores but
its actual cost has reached upto Rs. 700 crores. The complex
will also have an auditorium with a 1000-seat capacity, four
conference rooms, eight committee meeting halls, separate
building for advocate general and government pleader.
STUDY AREA:

M/s The Building Construction Department (BCD), Government

of Jharkhand, Ranchi has undertaken the construction and
development of the proposed New High Court Building and
Residential Complex & allied facilities on a site comprising
about 165 acres of land within HEC Campus, Village Tiril,
Dhurwa, Ranchi.

Connectivity: Nearest railway station is Hatia Railway station at

an aerial distance of 4km towards East. Nearest Airport is Birsa
Munda Airport at an aerial distance of 4.6 Km towards East. The
site is well connected to the Navasarai Road, NH-75, NH-23,
Ring Road, NH-39.
 CONTENTS

1. AIM OF THE PROJECT

THEORY
2. INTRODUCTION
3. DEMAND FOR COURTS
4. CLASSIFICATION OF BUILDING BASED ON OCCUPANCY
5. SELECTION OF PLOT AND STUDY
6. SURVEY OF SITE FOR PROPOSED BUILDING
7. ADMINSTRATIVE BUILDINGS
a) BUILT UP AREA
b) VARIOUS ROOMS DIMENSIONS
8. BUILDING BYE LAWS AND REGULATIONS
9. ARRANGEMENT OF ROOMS

DESINGS
10. DESING OF SLAB
11. DESING OF BEAM
a) FRAME ANALYSIS
12. DESING OF COLUMN
13. DESING OF FOOTING

DRAWINGS
14. BEAM
15. FOOTING
16. PHOTOS

CONCLUSION
 AIM OF THE PROJECT

The aim of the project is to plan and design the framed structure an
administrative building.
INTRODUCTION
 2.INTRODUCTION
A court is any person or institution with authority to judge or
adjudicate, often as a government institution,
with the authority to adjudicate legal
disputes between parties and
carry out the administration of justice in civil, criminal,
and administrative matters in accordance with the rule of
law. In
both common law and civil law legal systems, courts are the
central
means for dispute resolution, and it is generally understood
that all
people have an ability to bring their claims before a court.
Similarly,
the rights of those accused of a crime include the right to
a defense before a court.

 Courts are important because they help protect our

constitutional rights

 Courts provide a forum to resolve disputes and to

and enforce laws in a fair and rational manner
Courts are an impartial forum, and judges are free
to apply the law without regard to the states wishes
or the weight of public opinion but in line with
human rights
.
The engineer has to keep in mind the municipal
conditions, building bye laws, environment, financial
capacity, water supply, sewage arrangement, provision of
future, aeration, ventilation etc., in suggestion a particular
type of plan to stakeholders.
 3. DEMAND FOR COURTS

The court is a tribunal which can be presided over by a judge, several

judges or a magistrate, and is tasked with hearing and giving out
judgement in civil and criminal cases. It is part of the judiciary arm of
the government whose role is to ensure that laws are adhered to or
followed strictly

The special features of the demand for courts consists of in its

unique nature and depend on the following factors.

• Availability of skilled labours.

• Availability of transport facility.

• Cost of labours & material of construction.

• Predictions of future demand.

• Rate of population growth and cases

• Supply of various resources at economical prices.

• Town planning & environmental conditions.

4.CLASSIFICATION OF BUILDINGS BASED ON OCCUPANCY

GROUP-A RESIDENTIAL BUILDINGS

GROUP-B EDUCATIONAL BUILDINGS

GROUP-C INSTITUTIONAL BULIDINGS

GROUP-D ASSEMBLY BUILDINGS

GROUP-E BUSINESS BUILDINGS

GROUP-F MERCANTILE BUILDINGS

GROUP-G INDUSTRIAL BUILDINGS

GROUP-H STORAGE BUILDINGS

GROUP-I HAZARDOUS BUILDINGS

RESIDENTIAL BUILDINGS:
These building include any building in which sleeping accommodation
provide for normal residential purposes, with or without cooking and dining
facilities. It includes single or multi-family dwellings, apartment houses,
lodgings or rooming houses, restaurants, hostels, dormitories and residential
hostels.

EDUCATIONAL BUILDINGS:
These include any building used for school, college or day-care purposes
involving assembly for instruction, education or recreation and which is not
covered by assembly buildings.

INSTITUTIONAL BUILDINGS:
These buildings are used for different purposes, such as medical or other
treatment or care of persons suffering from physical or mental illness,
diseases or infirmity, care of infants, convalescents or aged persons and for
penal detention in which the liberty of the inmates is restricted. Institutional
buildings ordinarily provide sleeping accommodation for the occupants.

ASSEMBLY BUILDINGS:
These are the buildings where groups of people meet or gather for
amusement, recreation, social, religious, assembly halls, city halls, marriage
halls, exhibition halls, museums, places of work ship, etc.

BUSINESS BUILDINGS:
These buildings are used for transaction of business, for keeping of accounts
and records and for similar purposes, offices, banks, professional
establishments, courts houses, libraries. The principal function of these
buildings is transaction of public business and keeping of books and records.
MERCANTILE BUILDINGS:
These buildings are used as shops, stores, market, for display a sale of
merchandise either wholesale or retail, office, shops, storage service
facilities incidental to the sale of merchandise and located in the same
building.

INDUSTRIAL BUILDINGS:
These are buildings where products or materials of all kinds and
properties are fabrication, assembled, manufactured or processed, as
assembly plant, laboratories, dry cleaning plants, power plants, pumping
stations, smoke houses, laundries etc.

STORAGE BUILDINGS:
These buildings are used primarily for the storage or sheltering of goods,
wares or merchandise vehicles and animals, as warehouses, cold storage,
garages, trucks.

HAZARDOUS BUILDINGS:
These buildings are used for the storage, handling, manufacture or
processing of highly combustible or explosive materials or products which
are liable to burn with extreme rapidly and/or which may produce
poisonous elements for storage handling, acids or other liquids or
chemicals producing flames, fumes and ex plosive, poisonous, irritant or
corrosive gases processing of any material producing explosive mixtures of
dust which result in the division of matter into fine particles subjected to
spontaneous ignition.
 5.SELECTION OF PLOT AND STUDY

Selection of plot is very important for construction of buildings. Site

should be in good place where there community but service is convenient but
not so closed that becomes a source of inconvenience or noisy. The
conventional transportation is important not only because of present need but
for retention of property value in future closely related to are transportation,
other facilities also necessary. One should observe the road condition
whether there is indication of future development or not in case of
undeveloped area.

The factor to be considered while selecting the building site are as

follows:-

• Access to open area.

• Agriculture polytonality of the land.

• Availability of public utility services, especially water, electricity &

sewage disposal.

• Contour of land in relation the building cost of land.

• Distance from places of work.

• Ease of drainage.

• Location with respect to school, collage & public buildings.

• Nature of use of adjacent area.

• Transport facilities.

• Wind velocity and direction.

 6.SURVEY OF THE SITE FOR PROPOSED BUILDING

Reconnaissance survey: the following has been observed during

reconnaissance survey of the site.

• Site is located nearly.

• The site is very clear planned without ably dry grass and other throne
plats over the entire area.

• No leveling is require since the land is must uniformly level.

• The ground is soft.

• Labour available near by the site.

• Houses are located near by the site.

• Detailed survey: the detailed survey has been done to determine the
boundaries of the required areas of the site with the help of
theodolite and compass.
 ADMINSTRATIVE BUILDING
Buildings which are work base of number of people should be located so as
to limit exposure of hazards.
Administrative Buildings should be situated in a safe area on the public side
of the security point. The main office block should always be near the main
entrance and other administration buildings should be near this entrance if
possible. Other buildings, such as canteen, medical centers, etc. should be in
a safe area and the latter should have ready access for food supplies.
All buildings should be upwind of plants which may give rise to
objectionable features.

a) BUILT UP AREA

Land Area: The Proposed project is being developed on the total plot area of
664994.1 m2
Built up Area: The built up area of the proposed project is 118552.98 m2
including the FAR.
Water Consumption: During the construction phase, water requirement will
be met through the treated water from private water tanker. It is estimated
that water demand during the construction phase may vary from 30 KLD.
The project site was a vacant land at the starting of construction only had
scanty vegetation mainly herbs and shrubs and possesses no existing
structure. Hence, no clearance of vegetation, no felling of trees and structure
was required. The green area has been planned to develop 10 % of open area
to provide beautiful and natural environment. The green area will be
developed as shelter belt, along with avenue plantation on both sides of road
lawns area including herbs and shrubs. The indigenous/local plants will be
planted, which will increase the aesthetic value of the area. Hence, there will
be no disturbance to the local ecology of the area.
 8.BUILDING BYE LAWS & REGULATIONS

• Line of building frontage and minimum plot sizes.

• Open spaces around the building.

• Minimum standard dimensions of building elements.

• Provisions for lighting and ventilation.

• Provisions for safety from explosion.

• Provisions for means of access.

• Provisions for drainage and sanitation.

• Provisions for safety of works against hazards.

• Requirements for off-street parking spaces.

• Requirements for landscaping.

• Special requirements for low income housing.

• Size of structural elements.

17
 9.ARRANGEMENT OF ROOMS

• COURT ROOM

• STORE ROOM

• OFFICE ROOM

• BATH & W C

• VERANDAH

• STAIR CASE
COURT ROOM:
Should be designed in a well ventilated manner for proper light and fresh air
as it will be the room with more occupancy.

OFFICE ROOM:
North aspects this makes more suitable since there will be no sun from north
side for most part of the year.

BATH & W.C:

Bath and w.c are usually combined in one room & attached to the bed room
and should be well finished. This should be filled with bath tub, shower,
wash-hand basin, w.c, shelves, towels, racks brackets, etc., all of white
glazed tiles. Floor should be mosaic or white glazed files. Instead of
providing all bed room with attached bath and W.C separated baths &
latrines may also be provided

VERANDAH:
There should verandah in the front as well as in the rear. The front verandah
serves setting place for male members & weighting place for visitors. The
back verandah serve a ladies apartment for there sitting, working controlling,
kitchen works etc., verandah project the room against direct sun, rain &
weather effect. They used as sleeping place during the summer and rainy
season & are used to keep various things verandah also give appearance to
the building. The area of a building may vary from 10% to 20% of the
building.

STAIR CASE:
This should be located easily accessible to all member people, when this is
intended for visitors it should be in the front, may be on one side of verandah
The stairs case should be well ventilated & lighted the middle to make it easy
& comfortable to climb. Rises & threads should be uniform through to keep
rhythm while climbing or descending.

Some helpful points regarding the orientation of a building are as follows:-

• Long wall of the building should face north south, short wall should
face.
 • East and west because if the long walls are provided in east facing, the
wall.

• Absorb more heat of sun which causes discomfort during night.

• A verandah or balcony can be provided to wards east & west to keep

the rooms cool.

• To prevent sun’s rays & rain from entering a room through external
doors & windows sunshades are required in all directions.

STORE ROOM
There are no such proper orientation requirements with store room but it
should be capable to resist fire, seepage, etc.
ORIENTATION
After having selected the site, the next step is proper orientation of
building. Orientation means proper placement of various rooms in relation
to sun, wind, rain, topography and out look and at the same time providing
a convenient access both to the street and back yard.

The factors that effect orientation most are as follows.

• Solar heat

• Wind direction

• Humidity

• Rain fall

• Intensity of wind site condition

• Lightings and ventilation

SOLAR HEAT:
Solar heat means sun’s heat, the building should receive maximum solar
radiation in winter and minimum in summer. For evaluation of solar
radiation, it is essential to know the duration of sunshine and hourly solar
intensity on exposed surfaces.

WIND DIRECTION:
The winds in winter are avoided and are in summer, they are accepted
in the house to the maximum extent.

HUMIDITY:
High humidity which is common phenomenon is in coastal areas, causes
perspiration, which is very uncomfortable condition from the human body
and causes more disomfort.

RAIN FALL:
Direction and intensity of rainfall effects the drainage of the site and
building and hence, it is very important from orientation point of view.
INTENSITY OF WIND:
Intensity of wind in plateau regions is high and as such window openings
of comparatively small size are recommended in such regions.

SITE CONDITIONS:
Location of site in rural areas, suburban areas or urban areas also effects
orientation, sometimes to achieve maximum benefits, the building has to be
oriented in a particular direction.

LIGHTING:
Good lighting is necessary for all buildings and three primary aims. The
first is to promote the work or other activities carried on within the
building.
The second is to promote the safety of people using the buildings. The
third is to create, in conjunction to interest and of well beings.

VENTILATION:
Ventilation may be defined as the system of supplying or removing air by
natural or mechanical mean or from any enclosed space to create and
maintain comfortable conditions. Operation of building and location to
windows helps in providing proper ventilation. A sensation of comfort,
reduction in humidity, removal of heat, supply of oxygen are the basic
requirements in ventilation apart from reduction of dust.
DESIGNS
 DESIGNS

• DESIGN OF SLABS

• LOADS ON BEAMS

• DESIGN OF BEAMS

• LOADS OF COLUMNS

• DESIGN OF COLOUMNS

• DESIGN OF FOOTINGS
 10. DESIGN OF SLAB

Slabs are to be designed under limit state method by reference of IS 456:2000.

• When the slab are supported in two way direction it acts as two way supported slab.

• A two way slab is economical compared to one way slab.

SLAB DESIGN:

fck = 15 N/mm2 fy =415 N/m2

Span
i. Shorter span:- Lx = 5.8m
longer span:-Ly =7.62m

ii. Check Lx/Ly= 7.62/5.8 =1.3<2

Hence the slab has to be designed as “two way slab”.

iii. Providing over all depth of slab as 5”, 120mm

eff. depth= D-15-Ø/2
=120-15-10/2=100mm

iv. Condition:- supported on four sides.

v. Load calculation:-
Dead load = 25x0.12x1 = 3.0KN/m
Live load =2x1 = 2.0KN/m
Floor finish =1x1 = 1x1KN/m
= 6.0 KN/m
vi. Bending moment calculation:- (as per IS code 456-2000)
Type of panel:- Two adjacent edges are discontinuous

ax(+) = 0.049 ax(-) = 0.065

ay(+) = 0.035 ay(-) = 0.047

(+ve) B.M at mid span in shorter directions.

Mx(+) = ax(+)wlx2
= 0.049x6x5.8^2= 9.9kn-m
factored B.M = 9.9x1.5 =14.85kn-m

Spacing and diameter:

As per SP-16.
Provide 8mmØ bars at 210mm spacing.

(-ve) B.M at continuous edge in shorter direction.

Mx(-) =ax (-) wlx2
=0.062x6x(5.8)^2
=13.12kn-m
factored B.M = 13.12x1.5=19.67kn-m

(+ve) B.M at mid span in longer directions.

My(+)= ay(+)wlx2
= 0.035x6x(5.8)^2
=7.06kn-m
factored B.M=7.06x1.5
=10.69kn-m

(-ve) B.M at continuous edge in longer direction.

My(-ve) = ay (-ve)wlx2
=0.047x6x(5.8)^2
=9.48kn-m
factored B.M=9.48x1.5
=14.22kn-m.
Check for depth:

Permissible depth=100mm
Mu.lim =0.36.Xumax(1-0.42Xumax)fckbd^2
d d
14.86x10^6= 0.36.Xumax (1-0.42x0.48)15x1000d^2

d= 84.71 < 100mm

Hence ok.
 11. DESIGN OF BEAMS

• Beam is a member which transfers the loads from slab to columns and then foundation
to soil.

• Beam is a tension member.

• Span of slabs, which decide the spacing of beams.

• Following are the loads which are acting on the beams.

o Dead load

o Live load

o Wind load
LOADS ON BEAMS:

B1: BEAM

SPAN=5.8m (shorter span)

Assuming beam size = 9”x16”(230x405mm)
Height of the wall-10’-3m

Load calculations
 
 Wall load - 0.23x3x19 =13.11Kn/m
 
 Self load – 0.23x0.406x25 =2.33Kn/m

 
Slab load –

W = 6KN

Lx = 5.8

WLx/3= (6x5.8)/3 = 11.6Kn/m

Total load = 13.11+2.33+11.6 = 27.04Kn/m

DESIGN OF STIRRUPS:
B1:BEAM

Calculation of shear force

Va= Vb = total load

=27.04x5.8 =78.416KN
2
 Calculation of normal shear

Tv =Vu =1.5x78.416x10^3 =1.37

Bd 230x373

Calculation of permissible shear stress

Tc = % of tension steel

Pt = Ast x 100
Bd

Ast = 2x16^2xp =402.12mm^2

4

Pt = 402.12x100 = 0.60%
230x373

Tc =0.50
Tc < Tv
0.05 < 0.76
Hence provide shear reinforcement.

Design of shear:

Vs = (Tv-Tc)bd
=(0.76-0.50)x230x373
=22.30KN

Calculation: Vus =22.30 =0.59 KN/cm

D(cm) 37.3
From sp-16 table no 62 we will get dia & spacing.

Hence provide 6mm dia @ 20 cm c/c spacing.

Check for spacing:

Spacing should be provided min of the following.

(a) 0.75d = 0.75x373 =279.75 mm

(b) Asv fy =2x(6^2xp/4)x250 =153.2mm

0.4b 0.4x230

(c) design spacing 45cm c/c

Hence provide 6mm dia stirrups @ 15 cm c/c.

LOADS ON BEAMS:

B2: BEAM

SPAN=7.62m (longer span)

Assuming beam size = 9”x16”(230x405mm)
Height of the wall-10’-3m

Load calculations

 
 Wall load - 0.23x3x19 =13.11Kn/m
 
 Self load – 0.23x0.406x25 =2.33Kn/m

 
Slab load –

W = 6KN

Ly = 7.62

WLy/3= (6x7.62)/3 = 15.24Kn/m

Total load = 13.11+2.33+15.24 = 30.68Kn/m

DESIGN OF STIRRUPS:
B2:BEAM

• Calculation of shear force

Va=Vb= total load

=30.68x7.62 =116.89KN

• Calculation of normal shear

Tv =Vu =1.5x116.89x10^3 =2.04

Bd 230x373
 • Calculation of permissible shear

stress Tc = % of tension steel

Pt = Ast x 100

Bd

Ast = 2x16^2xp =402.12mm^2

Pt = 402.12x100 = 0.60%

230x373

Tc =0.50
Tc < Tv
0.05 < 0.85
Hence provide shear reinforcement.

Design of shear:

Vs = (Tv-Tc)bd
=(0.85- 0.50)x230x373
=30.02KN

Calculation: Vus =230.02 =0.89KN/cm

D(cm) 37.3
From sp-16 table no 62 we will get dia & spacing.

Hence provide 6mm dia @ 15cm c/c spacing.

Check for spacing:

Spacing should be provided min of the following.

(a) 0.75d = 0.75x373 =279.75 mm

(b) Asv fy =2x(6^2xp/4)x250 =153.2mm

0.4b 0.4x230

(c) design spacing 45cm c/c

Hence provide 6mm dia stirrups @ 15 cm c/c

DESIGN OF BEAMS:
Mu at Left span = 11.577 KN-m

Mu at Mid span = 19.18 KN-m

Mu at Right span = 20.36KN-m

Check:-
Calculation limiting moment of resistances:

Mu = 11.577 KN-m

Mulimt =0.138 fck bd2

= 0.138x20x230x305^2

= 59.05 KN-m

Mu < Mulimit

Hence it is designed as simply reinforcement beam using sp-16

Mu =11.577x10^6 =1.39

bd^2 230x305^2

Refer table no.2 at sp-16 and read out the value of percentage of

reinforcement Corresponding to fy = 415 N/mm^2 and fck = 20N/mm^2

For Mu = 1.39 Pt = ?

bd^2

1.35 0.409

1.40 0.426

1.39 ?
Mu = 1.39 Pt = 0.422

bd^2

Pt = 0.422 %

Area of reinforcement
Pt = Astx100

Bd

=0.422x230x405

100

= 393.093 mm^2

Ast required = 393.093 mm^2

Ast provided:

Hence provide 3 bars & 12 mmdia

Ast provide =400 mm^2

Reinforcement of mid span:-

Calculate limiting moment of resistances

Mu =19.18 KN-m

Mulimt = 0.138 fck bd^2

=0.138x20x230x305^2

= 59.05 KN-m

Mu < Mulimit

Hence it is designed as singly reinforcement.

BY USING SP-16

Mu =19.18x10^6

Bd^2 230x305^2

= 0.66

Refer table no.2 at sp-16and read out the value of percentage of

reinforcement Corresponding to fy = 415N/mm^2 and fck = 20 N/mm2

Mu pt

Bd^2

0.65 0.187

0.70 0.203

0.66 ?

Pt =0.190%

Reinforcement

Pt = Astx100

Bd

=0.19x230x305

100

=133.285mm2

Ast provided

Hence provided 2mm bars & 12mm dia

Ast provided = 155.2mm2

Reinforcement of right span:-
Check:

Calculate limiting moment of resistance:-

Mu =20.36 KN-m

Mulimi = 0.138 fck bd^2

=0.138x20x230x305^2

= 59.05KN-m

Mu < Mulimit

Hence it is designed as singly reinforcement.

BY USING SP-16

Mu =20.36x10^6

Bd^2 230x305^2

=1.39

Mu Pt

Bd^2

1.35 0.409

0.426 0.426

1.39 ?

Pt = 0.422%

Reinforcement =

Pt = Ast x100

bd
Ast =0.422x230x305

100

296.033mm2

Ast provided

Hence provide 3 bars and 12mm dia

Ast provided =300mm^2.

 12.DESIGN OF COLUMNS
• Columns are compression members.

• Larger spacing columns cause stocking columns in lower stores of multi

storied buildings.

• Columns are transmitted loads which are coming from slabs to foundations. Larger
spans of beams shall also be avoided from the consideration of controlling the deflection
& cracking.

COLUMNS:

The column which takes load are:

(a) Slab loads
(b) Beam loads
(c) Wall loads
(d) Self. Wt of column

S.NO TYPE OF ROOF LOAD FLOOR LOAD

LOAD

1. Wall load (5.8+7.62)x0.115x 0.91x19 (5.8+7.62)x0.23x3x19

2 2
=12.09KN =29.32KN

2. Slab load (5.8+7.62)x6 (5.8+7.62)x6

2 2
= 40.26KN =40.26KN

3. Self wt. of beam 0.23x0.406x(5.5+7.62)x25 0.23x0.406x(5.5+7.62)x25

2 2
=25KN =25KN

Total load
77.35KN 94.58KN
Total loads on column:

Loads from roof = 77.35KN

Loads from floor = 94.58KN

Self wt. of column = 0.23x0.23x3x25

= 34.5KN

total loads = 167KN

Column Axial load:

Pu = 167 KN
Cross section--- 230x230mm
calculation: Pu = 167x10^3 = 0.15
fck*b*d 20x230x230

Calculation of Eccentricity:
e= 1 + b
500 30
= 4640 + 230 = 16.94m
50030
e≤20 mm

Mue = Pu*e
= 167*0.020
= 3.34 Kn-m

Mue = 3.34x10^6 = 0.0112

fck bd^2 20x230x230^2

d’ = 0.2
D

P = 0.02
fck
 P =0.02*fck
=0.02x20
=0.4% minimum 0.8% area of steel = 0.8 Bd =
0.8x230x230 = 423.2 mm
100 100
No. of bars for 12mm dia

= 423.2 = 4 bars
p/4x12^2

STIRRUPS SPACING:

LEAST OF THE FOLLOWING:

a) 16dia of main reinforcement=16x12 =192 mm.

b) 48dia = 48x12 = 576 mm.

Provide 6 mm dia. @ 192 mm c/c when main bars size is 12 mm

 13.DESIGN OF FOOTING

Size of column (b) 230x380(a)

Load = 400.69KN

Self wt. of footing = 10%

Bearing capacity of soil = 250 Kn/m2

Area of footing

Total load = 440.76KN

Area of footing = 440.76/250 = 1.76m2

the side of the footing be in the same ratio of

column =0.23x*0.38x =1.76

= 0.0874x^2=1.76

x=4.48m

= 1.0 m

Long side of footing = 0.38*4.48

= 1.70 m

Proved a rectangle footing 1mx1.7m

Up ward soil pressure = 440.76 = 259.27 Kn/m2 = 260 KN/m2

(1*1.7)
BENDING MOMENT CALCULATION:

Maximum bending moment along y- direction longer direction

Mxx = q x1/8 (B-b)^2
=260x1.7/8 (1-6023)^2
= 32.75 KN-m

Maximum bending moment along x- direction shorter

direction Myy = q-b/8 (B-b)2
= 260x1/8(1.7-0.38)62
= 56.62 KN-m

Depth of footing:
Depth of footing form moment consideration
d = v Myy/Qb = v 56.62x10^6/0.91x1000
d =249.43
say 250 mm
check for shear (two- way shear) V=
q[Lxb-(a+d)(b+d)]
=250[1.7x1-(0.38+250)(230+250)]
=363.37 KN

Normal shear stress:

V =363.37x1063 = 654.72 N/mm^2

[2(a+d)(b+d)d] [2(0.38+0.25)(0.23+0.25)0.25]

Tc = 0.65 N/mm2.
Allowable shear stress:

Tv = k x Tc
where k = 0.5+ 0.23
0.38
=1.10 k>1.1
Ka = 1.0 x 16 x fck
Ka = 0.78 N/ mm2
Tv < Tc safe to compute normal shear stress due to one way action area of
tensile steel required.
Ast(yy) = Myy = 56.62x10^6
0.91X bd 0.91x 250x 0.23
Ast = 1082.08 mm2
Ast x 100 = 1082.08x100 =0.43%
bd 100x250x0.23
From table 23 Tc = allowable shear stress 0.27 N/ mm2

One way shear:

The critical section along (1-1)

L – a – d =17200 - 380 - 250
2 2 2 2
=410 mm

Shear force:
Upward pressure on the hatched area
V= 260X1X0.410
=106.6
Normal shear:
Tv = V = 106.6 x10^3
Bd 1x1000x250
=0.42 N/mm2
Tv >Tc in case of one way shear
The effective depth to be increase

Let the eff. Depth be 350 mm

Tv = V =
2[(a+d)+(b+d)]d
V =260 [1.7x1-(0.38-
0.350)+(0.23+0.35)] V =101.4KN
Norminal shear Tv = 101.4x103

2[(0.38+0.35)+(0.23+0.35)0.35]
= 0.110N/mm2
Tc >Tc
0.6054 > 0.110
Hence safe
Adopt eff depth = 35 mm
Eff cover = 50 mm
-------------
Overall depth = 400 mm
---------------

Reinforcement in longitudinal direction:

Ast = 32.75 x106

0.87x230x350
=447.08 mm
Spacing of 12 mm mid steel leaving a clearance of 250mm on the either
side S = 950*p*122
447.684
=239.99 mm
Provide 12mm bars at 230 mm c/c
Reinforcement in shorter direction:

Ast = Myy = 56.62x10^6

bd 230x350x0.90
= 781.50 mm2
The reinforcement in the central band width 1.7 provide in accordance with
= Reinforcement in central band width / total reinforcement in shorter direction.
= 1.7/1 = 1.7
Reinforcement in central band =Ast x 2 = 2 =578.94 mm2
B+1 (1.7+1)
Spacing of 10 mm dia bars at 190mmc/c
The steel for the remaining width = 781.50 -578.94
=202.56 mm2
Provide 4 bars of 12mm dia on either of the central ban width

Developed length:

From IS 456-2000
Ld = dia vs
4Tbd
=0.87xfyx dia =0.87x415xdia =47 dia =47x12 =528mm
4x Tbd 4x(1.6x1.2)

= (1000 – 230) -50

2
=8035 mm>528 mm
Load transfer from column to footing:
Nominal bearing stress in column concrete.
Vbt = p = 440.76x10^3 = 5.04 N/ mm2
Ac 230x380

Bearing stress un M15 concrete

=0.25x20
=5N/ mm2

Allowable bearing stress

=5V A1
A2
=v A1 >2
A2

= 5v 1697400
230x380
= 4.40 limited 2
Allowable bearing stress = 2x5 =10 N/ mm2 >6067
The minimum steel required for dowel bars or loa transferring bar is 0.5% of
column As = 0.5 x230x380
100
=437 mm2
No.of 12mm dia = 437x12^2 =3.86
p/4
Provide 4 nos of bars of 12mmbars
development length of dowel bars
Ld =vs x dia 44 dia
4T bd
for 12 mm dia Ld =528 mm
The dowel is to be extended by 528mm into column.
Available depth in footing
Effective to the centre of 20 mm dia 350mm
Deduct ½ x 20 =10 mm
Deduct 12 mm dia

Net available distance =[350-10-12]

=328
Provide bent of bars to [528-328] =200 mm.
DRAWINGS
BEAM
FOOTING
PHOTOS
ISOLATED FOOTING

ISOLATED FOOTING
COMBINED FOOTING
 CONCLUSION

We can conclude that there is difference between the theoretical and

practical work done. As the scope of understanding will be much more when
practical work is done. As we get more knowledge in such a situation where
we have great experience doing the practical work.

Knowing the loads we have designed the slabs depending upon the ratio
of longer to shorter span of panel. In this project we have designed slabs as
two way slabs depending upon the end condition, corresponding bending
moment. The coefficients have been calculated as per I.S. code methods for
corresponding lx/ly ratio. The calculations have been done for loads on
beams and columns and designed frame analysis by moment distribution
method. Here we have a very low bearing capacity, hard soil and isolated
footing done.

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