219146825 civil engineering building project report

8/18/2019 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


2


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= 0.51 KN/m2


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


Dead Load


2


= 0.51 KN/m2


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



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Dead Load

Self Weight of the slab= 160 mm x 25 KN/m

3

= 4 KN/m

2


= 0.51 KN/m2


2

Total = 5.18 KN/m2

Imposed Load

For roof = 5 KN/m2

b) Beam

Dead load



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


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



Moment Calculation


x= -0.032, y= -0.032


x= 0.024, y= 0.024

Support moment ,Ms = - xwlx2

= -0.032x 10.5 x 6

2

= -12.1 KN-m


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


Clear Spans

Lx=6 m

Ly=6 m

αβγδ


=26

=1

=1.65

=1.05

=1

05.1*65.126

6000

xd = = 133 mm

Say D = 160 mm

Design Load


Live load = 5 KN/m2



Moment Calculation


x= -0.032, y= -0.032


x= 0.024, y= 0.024


2 = -15.6 KN-m


2 =-15.6 KN-m


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


Clear Spans

Lx=6 m

Ly=6 m

αβγδ


=26

=1

=1.65

=1.05

=1

05.1*65.126

6000

xd = = 133 mm

Say D = 160 mm

Design Load


Live load = 5 KN/m2



Moment Calculation


x= -0.032, y= -0.032


x= 0.024, y= 0.024


2 = -15.6 KNm


2 =-15.6 KN-

m


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


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


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:


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


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:


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


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


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))


2


= 0.51 KN/m

2


2

Total = 5.18 KN/m2

Dead load

Self weight of beam = 25×0.25×045 = 2.81 KN/m



2


= 0.51 KN/m2


2

Total = 5.18 KN/m2

Dead load



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2


= 0.51 KN/m2


2

Total = 5.18 KN/m2

Dead load

Self weight of beam = 25×0.3×0.5 5 = 3.75 KN/m



2


= 0.51 KN/m2


2

Total = 5.18 KN/m2

Dead load




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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


41/203

Fi : Ex ansion Joint Elevation


42/203

`

Fig: Expansion Joint (Plan)


43/203

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


57/203

Table 14

)(,αβγδ


05.1*65.126

6140

xd =

Ø

21

6140

6140


58/203

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


59/203

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


61/203

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


62/203

Table 15

)(,αβγδ


05.1*65.126

6140

xd =

Ø

216.16140

7140


63/203


64/203

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



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


67/203

Table 16

)(,αβγδ


05.1*65.126

4140

xd =

Ø

205.14140

4340


68/203

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 −−


69/203

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



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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219146825 civil engineering building project report

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