219146825 Civil Engineering Building Project Report
Transcript of 219146825 Civil Engineering Building Project Report
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An Apartment can be defined as a structure with individual apartment units but a common
entrance and hallway. In apartment building the spaces themselves must be simple and
universal enough to adapt to a variety of life styles. It should be designed in such a way that
makes possible to move any room without crossing.
Some of the characteristics of Apartment Buildings:
a) Entering apartment: Outer clothing should be taken off the entrance like shoes,
umbrella.
b) Children coming in from play: children should be able to reach bathroom, bedroom
without crossing living room.
c) Delivery person should be paid without entering living room.
d) Passing from bedroom to bathroom
e) Passing from kitchen to bathroom
A well planned apartment is divided into living zone and sleeping zone, separated by the entry
hall. Equally important as the relation of each room to the other is the relative position it
occupies in relation to daylight and fresh air. Ideally, every room should have exterior
exposure to ensure light and air. This may however increase the perimeter of the building to
an extent that no one could afford to build it. Therefore bathrooms, invariably, kitchens, often
and dining rooms, are handled as interior spaces. Thus the apartment plan is divided into outer
and inner zones.
High Rise Apartment buildings have recently developed in massive way in context to
Kathmandu Valley. The growing population and the decrement of land for residential
buildings lead to the apartment buildings. Today, Kathmandu is a rapidly urbanizing city with
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building construction at just about every corner of the city that one can see. Kathmandu valley
is facing tremendous pressures on its population and infrastructures due to haphazard and
rapid urbanisation. The agricultural land has been converted into residential building and it is
increasing tremendously. Nevertheless, high rise building can be one of the solutions. High
Rise building is very justifiable in Kathmandu Valley as attempt to solve land use problems
by economizing precious urban territories used for service and utilization. This need for new
housing, considered against a background of continuing urbanization, clearly indicates that an
increasing proportion of an expanding housing market will be devoted to multifamily types of
housing or apartments. The inevitability of this trend contains a challenge to the architect to
do more than merely met a statistical demand.
The process of designing an apartment building may be graphically depicted in a general way
as shown in table.
Market analysis controls site characteristics utilities floor shape and site concrete steel
Distribution finding standards large scale development building height length and limitations
Building types width wind bracing systems
Building orientation
Refuse disposal spatial requirement guidelines guidelines elevators egress
Boiler room circulation core use criteria procedure plumbing
ventilating
Mail room wheeled heating andcooling
Storage commercial
Laundry and community
Chart 1: process of designing apartment
PROGRAM ZONING
AND CODES
SITE
CONSIDERATION
BUILDING
CONFIGURATIONSTRUCTUAL
SYSTEMS
VERTICAL
SERVICING
TYPICAL FLOOR
DETERMINATIOTYPICAL LIVING
UNIT DESIGN
FIRST FLOOR
ORGANIZATION
SERVICE
SPACES
Program development
Site analysis
Buildin desi n
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Nepal is an earthquake prone region. Nearly 1/3rd of the Himalayan arc marking an active
plate boundary between Eurasian and Indian plates lies in the northern part of Nepal. This
earthquake was of such immense power that it resulted in the high peaks which now
characterize Nepal – the Himalayas. Kathmandu valley, which is the capital of Nepal, has
been severely hit by earthquakes as strong as of magnitude 8.3 on Richter scale in the history
(1255, 1833 and 1934 earthquakes).
Many researchers have predicted the occurrence of strong earthquake in Kathmandu valley in
the near future. Nevertheless, most of the soil of Kathmandu valley is black cotton. Recent
years have seen an increase in the opportunities to High Rise Building in Kathmandu Valley
which lie within seismically active regions of the world. The question arises can the high rise
building resist in such seismically active zones?
. Designer deals with the design of civil engineering structures in a safe and economic way
and also the study of behavior of civil engineering structures under the effect of various kinds
of loads. Due consideration are given to the aesthetic and ecological aspects. A designer has
to deal with various structures ranging from simple ones like curtain rods and electric poles to
more complex ones like multistoried frame buildings, shell roofs bridges etc. these structure
are subjected to various load like concentrated loads uniformly distributed loads, uniformly
varying loads live loads, earthquake loads and dynamic forces. The structure transfers the
loads acting on it to the supports and ultimately to the ground. While transferring the loads
acting on the structure, the members of the structure are subjected to the internal forces likeaxial forces, shearing forces, bending and torsional moments.
Structural Analysis deals with analyzing these internal forces in the members of the
structures. Structural Design deals with sizing various members of the structures to resist the
internal forces to which they are subjected during their effective life span. Unless the proper
Structural Detailing method is adopted the structural design will be no more effective. The
Indian Standard Code of Practice should be thoroughly adopted for proper analysis, design
and detailing with respect to safety, economy, stability and strength.
:
The projected selected by our group is an apartment building located at Bafal, Kathmandu.
According to IS 1893:2002, Kathmandu lies on Vth
Zone, the severest one. Hence the effect of
earthquake is pre-dominant than the wind load. So, the building is analyzed for Earthquake as
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lateral Load. The seismic coefficient design method as stipulated in IS 1893:2002 is applied to
analyze the building for earthquake. Special reinforced concrete moment resisting frame is
considered as the main structural system of the building.
The project report has been prepared in complete conformity with various stipulations in
Indian Standards, Code of Practice for Plain and Reinforced Concrete IS 456-2000, Design
Aids for Reinforced Concrete to IS 456-2000(SP-16), Criteria Earthquake Resistant Design
Structures IS 1893-2000, Ductile Detailing of Reinforced Concrete Structures Subjected to
Seismic Forces- Code of Practice IS 13920-1993, Handbook on Concrete Reinforcement and
Detailing SP-34, Reynolds Handbook. Use of these codes have emphasized on providing
sufficient safety, economy, strength and ductility besides satisfactory serviceability
requirements of cracking and deflection in concrete structures. These codes are based on
principles of Limit State of Design.
This project work has been undertaken as a partial requirement for B.E. degree in Civil
Engineering. This project work contains structural analysis, design and detailing of a high rise
apartment building located in Kathmandu District. All the theoretical knowledge on analysis
and design acquired on the course work are utilized with practical application. The main
objective of the project is to acquaint in the practical aspects of Civil Engineering. We, being
the budding engineers of tomorrow, are interested in such analysis and design of structures
which will, we hope, help us in similar jobs that we might have in our hands in the future.
This group under the project work has undertaken the computer aided analysis and design of
high rise apartment building. The main aim of the project work under the title is to acquire
knowledge and skill with an emphasis of practical application. Besides the utilization of
analytical methods and design approaches, exposure and application of various available
codes of practices is another aim of the work.
The specific objectives of the project work are
i. Identification of structural arrangement of plan.
ii. Understanding the load assessment for the structure.
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iii. Modeling of the building for structural analysis.
iv. Detail structural analysis using structural analysis program.
v. Sectional design of structural components.
vi. Structural detailing of members and the system.
To achieve above objectives, the following scope or work is planned
i. Identification of the building and the requirement of the space.
ii. Determination of the structural system of the building to undertake the vertical and
horizontal loads.
iii. Estimation of loads including those due to earthquake
iv. Preliminary design for geometry of structural elements like slab, beam, column,
foundation, stair case
v. Determination of fundamental time period by free vibration analysis.
vi. Calculation of base shear and vertical distribution of equivalent earthquake load.
vii. Calculation of torsional moment and its additional shear
viii. Identification of load cases and load combination cases.
ix. Finite element modeling of the building and input analysis
x. The structural analysis of the building by SAP2000 for different cases of loads.
xi. Review of analysis outputs for design of individual components
xii. Design of RC frame members, walls, mat foundation, staircase, and other by limit
state method of design
xiii. Detailing of individual members and preparation of drawings as a part of working
construction document.
Building Type : Apartment Building, Located in Kathmandu
Structural System : RCC Space Frame
Plinth area covered : 12574.65 ft2
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Type of Foundation : Mat Foundation
No. of Storey : 11
Floor Height : 3.505m (Basement, semi basement and ground floor), 3.05 m
all other floors
Type of Sub-Soil : Soft Soil (Zone III)
Expansion Joints : expansion joints are provided
According to IS 456-2000, Clause 27, structures in which changes in plan dimensions take
place abruptly shall be provided with expansion joints at the section where such changes
occur. Reinforcement shall not extend across an expansion joints and the break between the
sections shall be completed. Normally structure exceeding 45m in length is designed with one
or more expansion joints.
The design is intended to serve for the following facilities in the building:-
• Basement for Parking ,
• Semi Basement for gymnasium hall, shops
• Ground floor for departmental stores
• Other floors for different apartments
• Swimming pool
• Dead loads are calculated as per IS 875 (Part 1) -1987
• Seismic load according to IS 1893 (Part 1)-2002 considering Kathmandu
located at Zone V
• Imposed loads according to IS 875(Part 2)-1987 has been taken
The building is modeled as a space frame. SAP2000 is adopted as the basic tool for the
execution of analysis. SAP2000 program is based on Finite Element Method. Due to possible
actions in the building, the stresses, displacements and fundamental time periods are obtained
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using SAP2000 which are used for the design of the members. Lift wall, mat foundation,
staircase, slabs are analyzed separately.
Following codes of practices developed by Bureau of Indian Standards were followed in the
analysis and design of building:
1. IS 456:2000 (Code of practice for plain and reinforced concrete)
2. IS 1893 (part 1):2002 (Criteria for earthquake resistant design of structures)
3. IS 13920: 1993 (Code of practice for ductile detailing of reinforced concrete structures
subjected to seismic forces)
4. IS 875 (part 1):1987 (to assess dead loads)
5. IS 875 (part 2):1987 (to assess live loads)
6. IS 875 (part 5):1987 (for load combinations)
7. SP 16, SP 24 and SP 34 (design aids and hands book)
The following materials are adopted for the design of the elements:
• Concrete Grade: M20, M25 and M30
−−−− M30 for the all columns, slabs and beams
−−−− M25 for shear walls
−−−− M20 for foundation
• Reinforcement Steel –Fe415
Limit state method is used for the design of RC elements. The design is based on IS:456-2000,
SP-16, IS:1893-2002, SP-34 and Reinforced Concrete Designer’s Handbook- Charles E.
Reynolds and James C. Stedman are extensively used in the process of design.
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The space frame is considered as a special moment resisting frame(SMRF) with a special
detailing to provide ductile behavior and comply with the requirements given in IS 13920-
1993, Hand book on Concrete Reinforcement and Detailing (SP-34) and Reinforced Concrete
Detailer’s Manual- Brian W. Boughton and Reinforced Concreter Designer’s Handbook-
Charles E. Reynolds and James C. Stedman ( for Helicoidal Staircase) are extensively used.
This project has been broadly categorized into five chapters, Summery of each chapter are
mention below:
Chapter 1 : Introduction
Chapter 2 : Preliminary load calculation and design
In this chapter, upon the preliminary load calculation is done and every
element is designed for a particular section. We generally deal with the design
of every structural element of particular floor like roof, typical floor, first floor
and basement floor. Structural arrangements is done with necessarycomputations that are performed for the vertical load calculation, preliminary
design of the structure elements, seismic load calculation and the different load
combinations that are used.
Chapter 3 Load assessment
It deals with the assessment of gravity and earthquake loads acting or likely to
be acted on the building.
Chapter 4 : Modeling and Structural Analysis
This chapter deals with the modeling techniques with SAP2000 that is
followed by the analysis of the different structural members. This includes the
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inputs given and outputs obtained in the process, the time period calculation
and storey drift of the building.
Chapter 5 : Structural Design and Comparison
It deals with the earthquake resistance design of beams, columns, slabs, shear
walls and footings considering limit state of collapse and serviceability, their
comparison with the provided ones and locating the areas of insufficient
designs. The result is compared with the results obtained from the proposed
program.
Chapter 6 : Structural Detailing and Drawings
The various structural detailing and drawings of the different members as
obtained from their respective design are listed in this chapter.
Chapter 7 Result, Conclusion and Recommendation:
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ny structure is made up of structural Elements Load carrying, such as beams and
columns and non structural elements (such as partitions, false ceilings, doors). The
structural elements put together, constitute the structural systems. Its function is to resisteffectively the action of gravitational and environmental loads, and to transmit the resulting
forces to the supporting ground without significantly disturbing the geometry, integrity and
serviceability of the structure.
The planning of the building has been done as per available land area, shape, space according
to building bylaws and requirement of commercial public building. The positioning ofcolumns, staircases, toilets, bathrooms, elevators etc are appropriately done and accordingly
Beam arrangements is carried out so that the whole building will be aesthetically, functionally
and economically feasible.
The aim of design is the achievements of an acceptable probability that structures being
design will perform satisfactorily during their intended life. With an appropriate degree of
safety, they should sustain all the loads and deformations of normal construction and use and
have adequate durability and adequate resistance to the effect of misuse and fire.
It is necessary to know the preliminary section of the structure for the detail analysis. As the
section should be given initially while doing analysis in every softwares, the need of
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preliminary design is vital. Only dead loads and live loads are considered while doing
preliminary design.
Preliminary design is carried out to estimate approximate size of the structural members
before analysis of structure. Grid diagram is the basic factor for analysis in both Approximate
and Exact method and is presented below.
Dead Load
Self Weight of the slab= 160 mm x 25 KN/m3 = 4 KN/m
2
Plaster = 25 mm x 20 KN/m3
= 0.51 KN/m2
Finishes = 25 mm x 26.70 KN/m3 = 0.67 KN/m
2
Total = 5.18 KN/m2
Imposed Load
For roof = 1.5 KN/m2
Dead load
Self weight of beam = 25×0.25×045 = 2.81 KN/m
Rectangular = 0.9 x 0.6 x 25 = 13.5 KN/m
Dead Load
Self Weight of the slab= 160 mm x 25 KN/m3 = 4 KN/m
2
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Plaster = 25 mm x 20 KN/m3
= 0.51 KN/m2
Finishes = 25 mm x 26.70 KN/m3 = 0.67 KN/m
2
Total = 5.18 KN/m2
Imposed Load
For typical floor = 3 KN/m2
b) Beam
Dead load
Self weight of beam = 25×0.3×0.5 = 3.38 KN/m
Rectangular = 0.9 x 0.6 x 25 = 13.5 KN/m
Dead Load
Self Weight of the slab= 160 mm x 25 KN/m3 = 4 KN/m
2
Plaster = 25 mm x 20 KN/m3
= 0.51 KN/m2
Finishes = 25 mm x 26.70 KN/m3 = 0.67 KN/m
2
Total = 5.18 KN/m2
Imposed Load
For roof = 5 KN/m2
Dead load
Self weight of beam = 25×0.3×0.5 5 = 3.75 KN/m
Rectangular = 0.9 x 0.6 x 25 = 13.5 KN/m
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Dead Load
Self Weight of the slab= 160 mm x 25 KN/m
3
= 4 KN/m
2
Plaster = 25 mm x 20 KN/m3
= 0.51 KN/m2
Finishes = 25 mm x 26.70 KN/m3 = 0.67 KN/m
2
Total = 5.18 KN/m2
Imposed Load
For roof = 5 KN/m2
b) Beam
Dead load
Self weight of beam = 25×0.35×0.55 = 4.38 KN/m
Rectangular = 0.9 x 0.6 x 25 = 13.5 KN/m
Dog Legged
Total thickness = 160 mm
Riser = 180 mm
Tread = 300 mm
Wt. of waist slab = 0.25 x 25 = 6.250 KN/m2
Wt. of each step = 0.50 x 0.18 x 0.3 x 25 = 0.675 KN/m
Wt. of landing = 0.25 x 25 = 6.250 KN/m2
Wt. of finishing = 0.09 x [22(0.18+0.3) + 0.18] x20 = 19.33 KN/m
Imposed load = 5 KN/m2
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Interior panel
Thickness of slab and durability consideration
Clear Spans
Lx=6 m
Ly=6 m
)(,αβγδ
SpanShorter d slabof Depth =
=26 =1
=1.65
=1.05
=1
05.1*65.126
6000
xd = = 133 mm Say D = 160 mm
Design Load
Self load of slab = 0.16 x 25 = 4KN/m
2
Live load = 1.5 KN/m2
Design load , w = 1.5(DL+LL) = 8.25 KN/m2
Considering unit width of slab , w= 8.25 KN/m
Moment Calculation
-ve Bending moment coefficient at continuous edge
x= -0.032, y= -0.032
+ve Bending moment coefficient at mid span
x= 0.024, y= 0.024
Support moment ,Ms = - xwlx2 = -0.032x 8.25 x 6
2 = -9.50 KNm
Mid span moment ,Mm = ywlx2 = 0.032 x 8.25 x 6
2 = -9.50 KNm
Check for depth from Moment Consideration
Depth of Slab,d = mm x
x
b x
M 48
1000x30138.0
105.9
fck 138.0
6max == <
133mm
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Interior panel
Thickness of slab and durability consideration
Clear Spans
Lx=6 m
Ly=6 m
αβγδ
SpanShorter d slabof Depth =)(,
=26
=1
=1.65
=1.05
=1
05.1*65.126
6000
xd = = 133 mm
Say D = 160 mm
Design Load
Self load of slab = 0.16 x 25 = 4KN/m2
Live load = 3 KN/m2
Design load , w = 1.5(DL+LL) = 10.5 KN/m2
Considering unit width of slab , w= 10.5 KN/m
Moment Calculation
-ve Bending moment coefficient at continuous edge
x= -0.032, y= -0.032
+ve Bending moment coefficient at mid span
x= 0.024, y= 0.024
Support moment ,Ms = - xwlx2
= -0.032x 10.5 x 6
2
= -12.1 KN-m
Mid span moment ,Mm = ywlx2 = 0.032 x 10.5 x 6
2 =-12.1 KN-
m
Check for depth from Moment ConsiderationDepth of Slab,d =
mm x
x
b x
M 54
1000x30138.0
101.12
fck 138.0
6max == >133mm
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Interior panel
Thickness of slab and durability consideration
Clear Spans
Lx=6 m
Ly=6 m
αβγδ
SpanShorter d slabof Depth =)(,
=26
=1
=1.65
=1.05
=1
05.1*65.126
6000
xd = = 133 mm
Say D = 160 mm
Design Load
Self load of slab = 0.16 x 25 = 4KN/m2
Live load = 5 KN/m2
Design load , w = 1.5(DL+LL) = 13.5 KN/m2
Considering unit width of slab , w= 13.5 KN/m
Moment Calculation
-ve Bending moment coefficient at continuous edge
x= -0.032, y= -0.032
+ve Bending moment coefficient at mid span
x= 0.024, y= 0.024
Support moment ,Ms = - xwlx2 = -0.032x 13.5 x 6
2 = -15.6 KN-m
Mid span moment ,Mm = ywlx2 = 0.032 x 13.5 x 6
2 =-15.6 KN-m
Check for depth from Moment Consideration
Depth of Slab,d = mm x
x
b x
M 4.61
1000x30138.0
106.15
fck 138.0
6max == <
133mm
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Interior panel
Thickness of slab and durability consideration
Clear Spans
Lx=6 m
Ly=6 m
αβγδ
SpanShorter d slabof Depth =)(,
=26
=1
=1.65
=1.05
=1
05.1*65.126
6000
xd = = 133 mm
Say D = 160 mm
Design Load
Self load of slab = 0.16 x 25 = 4KN/m2
Live load = 5 KN/m2
Design load , w = 1.5(DL+LL) = 13.5 KN/m2
Considering unit width of slab , w= 13.5 KN/m
Moment Calculation
-ve Bending moment coefficient at continuous edge
x= -0.032, y= -0.032
+ve Bending moment coefficient at mid span
x= 0.024, y= 0.024
Support moment ,Ms = - xwlx2 = -0.032x 13.5 x 6
2 = -15.6 KNm
Mid span moment ,Mm = ywlx2 = 0.032 x 13.5 x 6
2 =-15.6 KN-
m
Check for depth from Moment Consideration
Depth of Slab,d =
mm x
x
b x
M 4.61
1000x30138.0
106.15
fck 138.0
6max == < 133mm
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Deflection Criteria
Beam size-250mm*450 mm
Now,
ratiod
l
450
6000= = 13.33
15(Okay)
Depth of Beam,d =
mm x
x
b x
M 63.377
250x30138.0
1060.147
fck 138.0
6
max == < 450mm(Okay)
Deflection Criteria
Beam size-350mm*500 mm
Now,
ratiod
l
450
6000= = 13.33
15(Okay)
Depth of Beam,d =
mm x
xb x
M 51.363300x30138.0
1012.164fck 138.0
6max == <
450mm(Okay)
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Deflection Criteria
Beam size-350mm*500 mm Now,
ratiod
l
500
6000= = 12
15(Okay)
Depth of Beam,d =
mm x
x
b x
M 23.374
350x30138.0
1093.202
fck 138.0
6max == <
500mm(Okay)
Deflection Criteria
Beam size-350mm*550 mm
Now,
ratiod
l
550
6000= = 10.90 15(Okay)
Depth of Beam,d =
mm x
x
b x
M 33.396
350x30138.0
1061.227
fck 138.0
6max == <
550mm(Okay)
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Column ID: B4,basement floor
Axial column
Known data:
Axial load =5043.35KN
assume section of 600mm x 900mm
Height, L = 3.048m
38.3= D
L Hence the column can be designed as short.
Calculation:Factored Axial Load, Pu = 7565.02 KN
Assuming minimum reinforcement=0.8%
Design for section:
Pu= 0.4fck(Ag-p Ag/100)+0.67fyp Ag/100
7565.02=0.4×30×(1-0.008) Ag+0.67×415×0.008 Ag
Ag=535447.75mm2
Take B=600mm
Then,
D=892.4mm
900mm
600mm
900mm
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Column ID: O basement floor)
Known data:
Axial load =407.04KN
assume section of 400mm
Height, L = 3.048m
62.7= D
L Hence the column can be designed as
short.
Calculation:
Factored Axial Load, Pu = 610.56 KN
Assuming minimum reinforcement=0.8%
Design for section:
Pu= 1.05(0.4fckAc+0.67fyAs)
610.56 =1.05×(1-0.008) Ag+0.67×415×0.008 Ag
Ag=42923.4mm2
Then, D=234mm
D=400mm (ok)
400mm
X
Y400
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Calculations
Column ID: A11 (Basement floor)
Biaxial Column
Known data:
Axial load =237.79KN
assume section of 350mm x 350mm
Height, L = 3.048m
38.3= D
L
Hence the column can be designed as
short.
Calculation:
Factored Axial Load, Pu = 356.7 KN
Assuming minimum reinforcement=0.8%
Design for section:
Pu= 0.4fck(Ag-p Ag/100)+0.67fyp Ag/100
356.7=0.4×30×(1-0.008) Ag+0.67×415×0.008 Ag
Ag=25247mm2
Take B=350mm
Then,
D=350mm
350mm
-101.5 KNm
350mm
X
Y
101.5 KNm
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Interior panel
Thickness of slab and durability consideration
Clear Spans
Lx=6 m
Ly=1.5m
=26
=1
=1.65
=1.05
=1
05.1*65.126
6000
xd = = 133 mm
Say D = 160 mm
Design Load
Dead of flight
Calculating area
Step section =0.3*0.15/2=0.0225m2
Inclined slab = .335*.16=.0536m2
Finish =\(.15+.3)*.015=.0135m2
Total area = 0.0896m2
Dl of step section,1m width and 300mm in plan length =
2.24kN/m2
Dl per m2 on plan = 7.46kN/m
2
LL per m2 plan=4kN/m
2
Total load = 11.466kN/m2
Factored load=17.2kN/m2
Taking 1.5m width of slab, load = 25.8kN/m
2
Landing load
Self wt. of slab = .16*25 = 4kN/m2
Finish = 0.03*25 = .75kN/m2
LL = 4kN/m2
Total load = 8.75kN/m2
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Factored load = 13.125kN/m2
Taking 1.5m width, load = 19.68kN/m
Reaction at |B R b = 65.65 kN
Reaction at A, Ra = 67.18 kN
Mmax = 78.714kN-mCheck for depth from Moment Consideration
Depth of Slab, d = mm x
x
b x
M 36.106
150014.4
10254.70
14.4
6max ==
Hence adopt overall depth of slab = 160mm
Reference Steps Result
From soil report
of site
From I.S. 875_2
Table 1(1.i.e)
From I.R.C
Total plinth area of building=1257.65 sq. m
Soil bearing capacity= 90 tonnes/m2
Total load of the building
Transferred from columns=102752.62KN
From Floor of Basement
i. Live load of Garage building=2.5KN/m2
ii. Impact Factor=0.15+8/(6+L)=1KN/m2
Total load=102752.63+(2.5+1)*1257.65
= 106028.497KN
Area of foundation=Total Load/soil bearing capacity
= 106028.497/90=1178.0944m2
Since the area required for the foundation of the
building is less than the area available for foundation
construction.
Mat foundation is provided Mat foundation
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As described earlier, the building is a RCC framed structure, located in the Kathmandu valley.
Thus wind loads, snow loads, and other special types of loads described by IS 875 (part
5):1987 can be taken as negligible as compared to the dead, live and seismic loads.
According to the IS 875:1964:
The dead load in a building shall comprise the weights of all walls, partitions, floors and roofs
and shall include the weights of all other permanent features in the building.
It means the load assumed or known resulting from the occupancy or use of a building and
includes the load on balustrades and loads from movable goods, machinery and plant that are
not an integral part of the building.
These are the load resulting from the vibration of the ground underneath the superstructure
during the earthquake. The earthquake is an unpredictable natural phenomenon. Nobody
knows the exact timing and magnitude of such loads. Seismic loads are to be determined
essentially to produce an earthquake resistant design.
Seismic loads on the building may be incorporated by-
1. In this method the design earthquake forces are
determined adopting IS 1893:2002. These design forces for the buildings located along
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two perpendicular directions may be assumed to act separately along each of these two
horizontal directions.
2. In it the ground is subjected to a predetermined acceleration
and subsequent stress in the structural elements are determined by appropriate methods.
1. RCC: (IS 875 (part 1) :1987 table 1)
a) For slabs and shear walls:
RCC = 25 KN/m3
b) For columns:
RCC = 25 KN/m3
c) For Beams: RCC = 25 KN/m3
2. Plaster (12mm thickness):
plaster = 20.40 KN/m3
3. Tile (mosaic - 25mm thick):
tile = 20.40 KN/m3
4. Marble:
brick = 26.70 KN/m3
(IS 875 (part 1): 1987, table 1))
5. Cement punning:
cement = 20.40 KN/m3
(IS 875 (part 1):1987, table 17))
1. On floors: (IS 875 (part 2): 1987 table 1, (iii))
2. On Partition walls: Live Load = 1 KN/m2
(Assuming a minimum live load as per IS 875 (part 2): 1987, 3)
3. On roof slabs and slab projections: Live load = 0.75 KN/m2
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(Assuming access not provided except for the case of maintenance)
(IS 875 (part 2):1987 (table 2(i), (b))
Self Weight of the slab= 160 mm x 25 KN/m3 = 4 KN/m
2
Plaster = 25 mm x 20 KN/m3
= 0.51 KN/m
2
Finishes = 25 mm x 26.70 KN/m3 = 0.67 KN/m
2
Total = 5.18 KN/m2
Dead load
Self weight of beam = 25×0.25×045 = 2.81 KN/m
Rectangular = 0.9 x 0.6 x 25 = 13.5 KN/m
Self Weight of the slab= 160 mm x 25 KN/m3 = 4 KN/m
2
Plaster = 25 mm x 20 KN/m3
= 0.51 KN/m2
Finishes = 25 mm x 26.70 KN/m3 = 0.67 KN/m
2
Total = 5.18 KN/m2
Dead load
Self weight of beam = 25×0.3×0.5 = 3.38 KN/m
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Rectangular = 0.9 x 0.6 x 25 = 13.5 KN/m
Self Weight of the slab= 160 mm x 25 KN/m3 = 4 KN/m
2
Plaster = 25 mm x 20 KN/m3
= 0.51 KN/m2
Finishes = 25 mm x 26.70 KN/m3 = 0.67 KN/m
2
Total = 5.18 KN/m2
Dead load
Self weight of beam = 25×0.3×0.5 5 = 3.75 KN/m
Rectangular = 0.9 x 0.6 x 25 = 13.5 KN/m
Self Weight of the slab= 160 mm x 25 KN/m3 = 4 KN/m
2
Plaster = 25 mm x 20 KN/m3
= 0.51 KN/m2
Finishes = 25 mm x 26.70 KN/m3 = 0.67 KN/m
2
Total = 5.18 KN/m2
Dead load
Self weight of beam = 25×0.35×0.55 = 4.38 KN/m
Rectangular = 0.9 x 0.6 x 25 = 13.5 KN/m
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Dog Legged
Total thickness = 160 mm
Riser = 180 mm
Tread = 300 mm
Wt. of waist slab = 0.25 x 25 = 6.250 KN/m2
Wt. of each step = 0.50 x 0.18 x 0.3 x 25 = 0.675 KN/m
Wt. of landing = 0.25 x 25 = 6.250 KN/m2
Wt. of finishing = 0.09 x [22(0.18+0.3) + 0.18] x20 = 19.33 KN/m
Imposed load = 5 KN/m2
Detail load calculation of every floor is shown in table
Seismic weight is the total dead load plus appropriate amount of specified imposed
load. While computing the seismic load weight of each floor, the weight of columns and walls
in any story shall be equally distributed to the floors above and below the storey. The seismic
weight of the whole building is the sum of the seismic weights of all the floors. It has been
calculated according to IS: 1893(Part I) – 2002.
IS: 1893(Part I) – 2002 states that for the calculation of the design seismic forces of
the structure the imposed load on roof need not be considered
The seismic weights and the base shear have been computed in table
According to IS 1893 (Part I): 2002 Cl. No. 6.4.2 the design horizontal seismic
coefficient Ah for a structure shall be determined by the following expression:
gR 2
SIZA ah =
Where,
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Z = Zone factor given by IS 1893 (Part I): 2002 Table 2, Here for Zone V, Z =
0.36
I = Importance Factor, I = 1.5 for commercial building
R = Response reduction factor given by IS 1893 (Part I): 2002 Table 7, R = 5.0
Sa/g = Average response acceleration coefficient which depends on
Fundamental natural period of vibration (Ta).
For T = 0.8 and soil type IV (Soft Soil) Sa/g = 1.67/0.869797
=1.92
Now,
The design horizontal seismic coefficient, A b= Rg
ZISa
2
10368.052
05916.25.136.0==
x
x x Ah
According to IS 1893 (Part I) : 2002 Cl. No. 7.5.3 the total design lateral force or
design seismic base shear (VB) along any principle direction is given by
VB = Ah x W
Where, W = Seismic weight of the building=102752.62KN
VB = 0.1*102086.67 = 1. KN
The total base shear is firstly distributed horizontally in basement in proportion to the
stiffness. Then according to IS 1893 (Part I): 2002 Cl. No. 7.7.1 the design base shear (VB)
computed above shall be distributed along the height of the building as per the following
expression:
2
j j
n
1 j
2ii
Bi
hW
hWVQ
=Σ
=
Where,
Qi = Design lateral force at floor i
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Wi = Seismic weight of floor i
hi = Height of floor I measured from base
n = No. of storeys in the building
2
j j
n
1 j
2
iiBi
hW
hWVQ
=Σ
=
Where,
Qi = Design lateral force at floor i
Wi = Seismic weight of floor i
hi = Height of floor I measured from base
n = No. of storeys in the building
Center of Rigidity (CR) - A point through which a horizontal force is applied resulting in
translation of the floor without any rotation
W1
W2
W3
W4
W5
W1
W2
W3
W4
W5
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Center of Mass (CM) - Center of gravity of all the floor masses.
Structural eccentricity (e)
e = CMCR −
The eccentricity in building is calculated by
beeda β+α=
beedb β−δ=
Where,
eda & edb = static eccentricity at floor a & b define as the distance between
center of mass and center of rigidity.
b = maximum dimension of the building perpendicular to the direction of
earthquake under consideration
=δαand Dynamic magnification factors
=β Accidental eccentricity factor
1and05.0,5.1 =δ=β=α
The location of the center of rigidity is determined by
=y
y
r k
xk x And
=x
x
r k
yk y
33
L
EI k x = And 33 L
EI k y =
Where k x and k y are lateral stiffness of a particular element along the x and y axes.
E= Young’s Modulus of rigidity
I= Moment of Inertia
L= Length of the Member
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The total torsional stiffness of a storey I p about the center of rigidity is given by
)( 22 xk yk I y x p +=
Where,
x , y = coordinates of the centroid of a particular element in plan from
the center of rigidity.
I p = polar moment of stiffness
The additional shear on any frame on column line to a horizontal torsional moment T is given
by
xx
p
x x k
I
yT V ='
yy p
y'
y k I
xTV =
Where, ='xV Additional shear on any frame or column line in the x-direction
due to torsional moment
Vx = initial storey shear in x-direction due to lateral forces
Tx = yxeV , torsional moment due to lateral force in x-direction only
K xx = total stiffness of the column line under consideration in the x-
direction.
The subscript y represents y-direction.
The response history analysis provides structural response r(t) as a function of time, but the
structural design is usually based on the peak values of forces and deformations over the
duration of the earthquake induced response. The peak response can be determined directly
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from the response spectrum for the ground motion in case of single degree of freedom. The
peak response of multi degree freedom systems can be calculated from the response spectrum.
The exact peak calue of the nth mode response r n(t) =-r nstAn
Where An is the ordinate of the pseudo acceleration spectrum corresponding to natural period
Tn and damping ratio
The peak value r o of the total response can be estimated by combining the modal peaks r no
according to one of the modal combination rules. Because the natural frequencies of
transverse vibration of a beam are well separated, the SRSS combination rule is satisfactory.
Thus,
α
1~
2
nnor
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 together four load cases considered for the structural analysis and are mentioned as
below:
i.) Dead Load (D.L.)
ii.) Live Load (L.L)
iii.) Earthquake load in X-direction (E.Qx) Static
iv.) Earthquake load in Y-direction (E.Qy) static
v.) Earthquake load in X direction (Rx) response spectrum method
vi.) Earthquake load in Y direction (Ry) response spectrum method
Following Load Combination are adopted as per IS 1893 (Part I): 2002 Cl. No. 6.3.1.2
i.) 1.5 (D.L + L.L)
ii.) 1.5 (D.L + E.Qx)
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iii.) 1.5 (D.L - E.Qx)
iv.) 1.5 (D.L + E.Qy)
v.) 1.5 (D.L - E.Qy)
vi.) 1.2 (D.L + L.L + E.Qx)
vii.) 1.2 (D.L + L.L - E.Qx)
viii.) 1.2 (D.L + L.L + E.Qy)
ix.) 1.2 (D.L + L.L - E.Qy)
x.) 0.9 D.L + 1.5 E.Qx
xi.) 0.9 D.L -1.5 E.Qx
xii.) 0.9 D.L + 1.5 E.Qy
xiii.) 0.9 D.L -1.5 E.Qy
xiv.) 1.5 (D.L + Rx)
xv.) 1.5 (D.L - Rx)
xvi.) 1.5 (D.L + Ry)
xvii.) 1.5 (D.L - Ry)
xviii.) 1.2 (D.L + L.L + Rx)
xix.) 1.2 (D.L + L.L - Rx)
xx.) 1.2 (D.L + L.L + Ry)
xxi.) 1.2 (D.L + L.L - Ry
After checking the results, it was found that the stresses developed are most critical for the
following load combinations:
i.) 1.5 (D.L + L.L)
ii.) 1.2 (D.L + L.L + E.Qx)
iii.) 1.2 (D.L + L.L - E.Qx)
iv.) 1.2 (D.L + L.L + E.Qy)
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v.) 1.2 (D.L + L.L - E.Qy)
vi.) 1.2 (D.L + L.L + Rx)
vii.) 1.2 (D.L + L.L - Rx)
viii.) 1.2 (D.L + L.L + Ry)
ix.) 1.2 (D.L + L.L - Ry
The characteristic loads considered in the design of foundation are:
i.) Dead Load plus Live Load
To find the stress at the various points of the foundation, depth of footing and
reinforcements most critical factored loads are taken into account
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SAP2000 represents the most sophisticated and user-friendly release of SAP series of
computer programs. Creation and modification of the model, execution of the analysis, and
checking and optimization of the design are all done through this single interface. Graphical
displays of the results, including real-time display of time-history displacements are easily
produced.
The finite element library consists of different elements out of which the three dimensional
FRAME element was used in this analysis. The Frame element uses a general, three-
dimensional, beam-column formulation which includes the effects of biaxial bending, torsion,
axial deformation, and biaxial shear deformations.
Structures that can be modeled with this element include:
• Three-dimensional frames
• Three-dimensional trusses
• Planar frames
• Planar grillages
• Planar trusses
A Frame element is modeled as a straight line connecting two joints. Each element has its
own local coordinate system for defining section properties and loads, and for interpreting
output.
Each Frame element may be loaded by self-weight, multiple concentrated loads, and multiple
distributed loads. End offsets are available to account for the finite size of beam and column
intersections. End releases are also available to model different fixity conditions at the ends of
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the element. Element internal forces are produced at the ends of each element and at a user-
specified number of equally-spaced output stations along the length of the element.
Loading options allow for gravity, thermal and pre-stress conditions in addition to the usual
nodal loading with specified forces and or displacements. Dynamic loading can be in the form
of a base acceleration response spectrum, or varying loads and base accelerations.
The design of earthquake resistant structure should aim at providing appropriate dynamic
and structural characteristics so that acceptable response level results under the design
earthquake. The aim of design is the achievement of an acceptable probability that structures
being designed will perform satisfactorily during their intended life. With an appropriate
degree of safety, they should sustain all the loads and deformations of normal construction
and use and have adequate durability and adequate resistance to the effects of misuse and fire.
For the purpose of seismic analysis of our building we used the structural analysis program
SAP2000. SAP2000 has a special option for modeling horizontal rigid floor diaphragm
system.
A floor diaphragm is modeled as a rigid horizontal plane parallel to global X-Y plane, so that
all points on any floor diaphragm cannot displace relative to each other in X-Y plane.
This type of modeling is very useful in the lateral dynamic analysis of building. The base
shear and earthquake lateral force are calculated as per code IS 1893(part1)2002 and are
applied at each master joint located on every storey of the building
After the analysis of structure using SAP2000 the maximum displacement of nodes at the
expansion joint was found out. It is clear from table below that the available gap for
expansion joint is much greater relative displacement of the nodes at joint. In order to reduce
the pounding effect between the two units, the adequte spacing is provided. The separation
between the adjacent units of the same buildings in between shall be separated by a distance
equal to the amount R times the sum of the calculated storey displacements to avoid the
damaging contact when the two units deflect towards each other. Since the elevation levels of
both units are same in our case the factor R is replaced by R/2. Hence the building will not
collide at the expansion joint during earthquake condition.
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Drift
Floor Bottom Bottom
Basement 0 0 0.00077
Semi Basemen 0.00331 0.0027 0.00206
Ground 0.00749 0.0099 0.00367
First 0.0194 0.0209 0.00393
Second 0.0313 0.0327 0.00397
Third 0.0431 0.0446 0.00393
Fourth 0.055 0.0564 0.00377
Fifth 0.0668 0.0677 0.00393
Sixth 0.07801 0.0795 0.0039
Total 0.0299
spacing =0.09125/2=0.228m (in one side)
Table 13
0.39 0.09 0.0296433 0.406 0.0912
0.09 0.01199 0.0039967 0.091 0.0117
0.078 0.01121 0.0037367 0.08 0.0118
0.067 0.0118 0.0039333 0.068 0.0113
0.055 0.0119 0.0039667 0.056 0.0118
0.043 0.0118 0.0039333 0.045 0.0119
0.031 0.0119 0.0039667 0.033 0.0118
0.019 0.01191 0.00397 0.021 0.011
0.007 0.00418 0.0011943 0.01 0.0072
0.003 0.00331 0.0009457 0.003 0.0027
Top Displacement Top Displacemen
Along X Along X
Max Displaceme Relative Drift Max Displaceme Relative
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Fi : Ex ansion Joint Elevation
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`
Fig: Expansion Joint (Plan)
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In the method if design based on limit state concept, the structure shall be designed to
withstand safely all loads liable to act on it throughout its life; it shall also satisfy the
serviceability requirements, such as limitations on deflection and cracking. The acceptable
limit for the safety and serviceability requirements before failure occurs is called a ‘limit
state’. The aim of design is to achieve acceptable probabilistic that the structure will not
become unfit for the use for which it is intended, that is, that it will not reach a limit state.
Assumptions for flexural member:
i) Plane sections normal to the axis of the member remain plane after bending.
ii) The maximum strain in concrete at the outermost compression fiber is 0.0035.
iii) The relationship between the compressive stress distribution in concrete and the
strain in concrete may be assumed to be rectangle, trapezoidal, parabola or any other
shape which results in prediction of strength in substantial agreement with the result
of test. For design purposes, the compressive strength of concrete in the structure
shall be assumed to be 0.67 times the characteristic strength. The partial safety factor
m = 1.5 shall be applied in addition to this.
iv) The tensile strength of concrete is ignored.
v) The design stresses in reinforcement are derived from representative stress-strain
curve for the type of steel used. For the design purposes the partial safety factor
m =
1.15 shall be applied.
vi) The maximum strain in the tension reinforcement in the section at failure shall not
be less than: 002.0E15.1
f
s
y+
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Where, f y = characteristic strength of steel
Es = modulus of elasticity of steel
Limit state of collapse for compression:
Assumption:
In addition to the assumptions given above from i) to v), the following shall be assumed:
i.) The maximum compressive strain in concrete in axial compression is taken
as 0.002.
ii.) The maximum compressive strain at highly compressed extreme fiber in concrete
subjected to axial compressive and bending and when there is no tension on
the section shall be 0.0035 minus 0.75 times the strain at the least compressed
extreme fiber.
The limiting values of the depth of neutral axis for different grades of steel based on
the assumptions are as follows:
Fy xu,max
250 0.53
415 0.48
500 0.46
Materials adopted in our design:
M30 (1:1.5:3)
M25 (1:1:2)
Fe250-Mild Steel
Fe415
Use of SP16, IS456-2000, IS1893-2002, IS13920-1993, SP34:
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After analyzing the given structure using the software SAP2000 the structural elements are
designed by Limit state Method. Account should be taken of accepted theories, experiment,
experience as well as durability.
The code we use for the design is IS456-2000; IS1893-2002, IS13920-1993 and Design aids
are SP16 and SP34. Suitable material, quality control, adequate detailing and good
supervision are equally important during implementation of the project.
Use of different handbook for the design:
The structural elements (special staircases, lift wall, basement wall) which are not described
by the above mentioned codes and design aids were handled with the help of the handbooks
viz. Reinforced concrete Designer’s Handbook – Charles E. Reynolds
Computer aided design is the method of analyzing and designing any structure with the help
of various general use softwares and some particularly designed softwares made by using
some popular programming languages like visual basic, C++,etc.
In present time most of the building analysis and design is done by using computers. Basically
analysis and design based softwares like SAP, STAAD, etc are available in market. These
types of softwares are easy to use and can provide analysis results of complicated structures in
the matter of minutes which if calculated manually would take months.
Methodology
1. Analysis of building was done by using SAP 2000.
2. Design of slab was done by analyzing the slab of each floor on SAP 2000 in a separate
model.
3. For beam design, analysis result from SAP 2000 was arranged by using a small
program made from Visual Basics, which extracts data from SAP analysis and
arranges the required data.
4. Now beam was designed by using EXCEL and required reinforcement was calculated.
5. In case of columns, we used the design data from SAP.
6. All the other structural members were designed manually.
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A s a m p l e p r o g r a m f o r e x t r a c t i n g t h e d a t a f r o m S A P 2 0 0 0 o f t h e b e a m .
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The design includes design for durability, construction and use in service should be considered
as a whole. The realization of design objectives requires compliance with clearly defined
standards for materials, workmanship, and also maintenance and use of structure in service.
This chapter includes all the design process of sample calculation for a single element as slab,
beam, column, staircases, basement wall, lift wall, ribbed slab and mat foundation.
i.) Design of slab
ii.) Design of Beam
iii.) Design of Column
iv.) Design of Staircase
v.) Design of Basement Wall
vi.) Design of Lift Wall
vii.) Design of Mat and Foundation
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Table 14
)(,αβγδ
SpanShorter d slabof Depth =
05.1*65.126
6140
xd =
Ø
21
6140
6140
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mm x
x
b x
M 13.67
100014.4
1066.18
14.4
6
max ==
bd) bdf
M6.4
11(f
f
5.0 2ck
u
y
ck
−−
1361000)136100030
1066.186.411(
415
305.0
2
6
x x x x
x x x −−
Ø
1000xA
A
st
b 10007.399
5.78 x
2.4131000190
5.781000 == x x
S
A
v
b
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bd) bdf
M6.411(
f
f 5.0
2ck
u
y
ck −−
1361000)136100030
10146.411(
415
305.0
2
6
x x x x
x x x −−
Ø
1000xA
A
st
b 1000296
5.78 x
3021000260
5.781000 == x x
S
A
v
b
bd) bdf
M6.411(
f
f 5.0
2ck
u
y
ck −−
1381000)138100030
1087.166.411(
415
305.0
2
6
x x x x
x x x −−
Ø
1000xA
A
st
b 1000x14.358
26.50
3591000x140
26.501000x
S
A
v
b ==
bd) bdf
M6.411(
f
f 5.0
2ck
u
y
ck −−
1381000)138100030
1077.126.411(
415
305.0
2
6
x x x x
x x x −−
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Ø
1000xA
A
st
b 1000x17.267
26.50
22.2791000x180
26.501000x
S
A
v
b ==
2
14.6
136.007.307.3
5.47
−= u
V
37.0=cτ
cτ 1000
136100037.028.1 x x x
=cτ
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ovidedPr Steelof Area
quiredReSteelof Areaf 58.0 y
302
296
9.1342675.1
6140
..==
xValue Basic x F M
l x
bd
s
x4x6.1 τ
σ Φ Φ=
Φ3.40
4.146.1
41587.0
x x
x x
o1
d LV
ML +≤
o1 L
V
M+ 483.0275.0
4.45
266.18
=+
1000xA
A
sd
b 1000192
5.78 x
1000xS
A
v
b 3.1961000400
5.78= x
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Table 15
)(,αβγδ
SpanShorter d slabof Depth =
05.1*65.126
6140
xd =
Ø
216.16140
7140
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1361000)136100030
105.176.411(
415
305.0
2
6
x x x x
x x x −−
Ø
1000xA
A
st
b 10006.373
5.78 x
5.3921000200
5.781000 == x x
S
A
v
b
bd) bdf
M6.411(
f
f 5.0
2ck
u
y
ck −−
1361000)136100030
1066.186.411(
415
305.0
2
6
x x x x
x x x −−
Ø
1000xA
A
st
b
10007.399
5.78
x
2.4131000190
5.781000 == x x
S
A
v
b
bd) bdf
M6.411(
f
f 5.0
2ck
u
y
ck −−
1361000)13610003010146.411(
415305.0
2
6
x x x x x x x −−
Ø
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1000xA
A
st
b 1000296
5.78 x
3021000260
5.781000 == x x
S
A
v
b
2
14.6
136.007.307.3
5.47
−= u
V
4.0=cτ
cτ 1000
13610004.028.1 x x x
ovidedPr Steelof Area
quiredReSteelof Areaf 58.0 y
5.392
6.373
=cτ
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2.131268.1
6140
..==
xValue Basic x F M
l x
bd
s
x4x6.1 τ
σ Φ
Φ=
Φ
3.404.146.1
41587.0
x x
x x
o1
d LV
ML +≤
o1 L
V
M+ 483.0275.0
4.45
266.18
=+
1000xA
A
sd
b 1000192
5.78 x
1000xS
A
v
b 3.1961000400
5.78= x
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Table 16
)(,αβγδ
SpanShorter d slabof Depth =
05.1*65.126
4140
xd =
Ø
205.14140
4340
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mm
x
x
b x
M 3.47
100014.4
1028.9
14.4
6
max ==
bd) bdf
M6.411(
f
f 5.0
2ck
u
y
ck −−
1361000)1361000301028.96.411(
415305.0
2
6
x x x x x x x −−
Ø
1000xA
A
st
b 10007.193
5.78 x
7.2611000300
5.78
1000 == x xS
A
v
b
bd) bdf
M6.411(
f
f 5.0
2ck
u
y
ck −−
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1361000)136100030
109.66.411(
415
305.0
2
6
x x x x
x x x −−
Ø
1000xA
A
st
b 1000192
5.78 x
2621000300
5.781000 == x x
S
A
v
b
bd) bdf
M6.411(
f
f 5.0
2ck
u
y
ck −−
1361000)136100030
104.86.411(
415
305.0
2
6
x x x x
x x x −−
Ø
1000xA
A
st
b
1000192
5.78
x
2621000300
5.781000 == x x
S
A
v
b
bd) bdf
M6.4
11(f
f
5.0 2ck
u
y
ck
−−
1381000)136100030
104.66.411(
415
305.0
2
6
x x x x
x x x −−
Ø
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1000xA
A
st
b 1000192
5.78 x
2621000300
5.781000 == x x
S
A
v
b
2
14.4
136.007.207.2
32
−= uV
3.0=cτ
cτ 1000
13610003.028.1 x x x
ovidedPr Steelof Area
quiredReSteelof Areaf 58.0 y
262
192
=cτ
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6.79262
4140
..==
xValue Basic x F M
l x
1000xA
A
sd
b 1000192
5.78 x
1000xS
A
v
b 3.1961000400
5.78= x
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Table 17: Design of Beam 106ÿ(B2ÿC2)
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77/203