Structural Thesis

8/12/2019 Structural Thesis

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DEPARTMENT OF ARCHITECTURE

ANDHRA UNIVERSITY COLLEGE OF ENGINEERING (A)

VISAKHAPATNAM- 530003(AP)

CERTIFICATE

This is to certify that this is a Bona-fide project work on

"ANALYSIS AND DESIGN OF A RESIDENTIAL HOUSE"

STRUCTURAL THESIS

Submitted by:

G.HEMA (309106101008)student of 5/5 B.Arch, 1st semester batch 2009-2014

Mr. M. RAVINDRA Prof. G. VISWANADH KUMAR(Internal Examiner) (Head of department)

External Examiner:

Date: 30-12-2013


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ACKNOWLEDGEMENT

I express my deep sense of gratitude to Mr. M. Ravindra sir for his

Valuable guidance and constant encouragement for bringing out this Project.

We especially thank Prof. G.Viswanadh Kumar, Head of the Department

of Architecture for his encouragement throughout for our project work.

G.Hema 309106101008


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INDEX

1. INTRODUCTION

2. ANALYSIS OF SLABS

3. DESIGN OF SLABS

4. ANALYSIS OF BEAMS

5. DESIGN OF BEAMS

6. ANALYISIS OF COLUMNS

7. DESIGN OF COLUMNS

8. ANALYSIS OF FOOTINGS

9. DESIGN OF FOOTINGS

10. DESIGN OF STAIRCASE


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INTRODUCTION TO STRUCTURAL DESIGN

Structural design for framed R.C.C Structure may be designed by

any of the following methods:

1) WORKING STRESS METHOD

2) ULTIMATE STRENGTH METHOD

3) LIMIT STATE METHOD

Working Stress Method of Design:

It is probably the earliest codified method of design of reinforces

concrete structures, and it is based on a criterion that the actual

stresses developed in the material under the action of working loads

must be limited to a set allowable values, i.e., the concept of workingstress id use in providing for the factor of safety. The method also

constrains that the deformation of the structures or elements must be

within acceptable values an elastic linear structural analysis is

considered to be a basis in the determination of stress in materials.

However this method is simple to understand.

Ultimate Strength Design (USD):

It is primarily based on the strength concept. Multiplying with load factors to

give a hypothetical load pattern called Ultimate loads enhances the working

loads. Then the designforceson the members are obtained by an elastic

structure analysis under the action of ultimate loads. The members are

proportioned such that the strength of the


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member is not less than the design ultimate force i.e., instead of

considering the material strength only (as in W.S.D.) , the economy in

the cost of structure

Limit State Design (L.S.D):

This method is more rational in approach and is replacing

the working stress method and the ultimate load method in the design

of reinforced concert structures. The limit state method of design has

originated from the ultimate design. (in a way USD can be called as the

strength limit state of the limit state design).

The concepts of safety and service ability are rationalized better

consequently certain amount of sophistication is introduced. The

structures are design to provide adequate strength, serviceability, and

durability. Partial safety factor apply to loads and materials provide the

required safety, serviceability of the structures. The design forces and

deformation are arrived at by the limit are elastic analysis depending on

the type of the limit state considered. The object of this method is

based on the limit state concept i.e., to achieve an acceptable

probability that a structure will not become unserviceable in its lifetimefor which

it is instead (i.e., it will reach the limit state).The important states, which

must be examined in the design, are:

Limit State of Collapse: This limit state corresponds to the maximum

load carrying capacity. Violation of the limit state of collapse implies

failure in a since that a clearly define state of structural usefulness has

been exceeded. However it does not mean a complete collapse.

The limit state corresponds to:

a. Flexure b. Compression c. Shear d. Torsion

Limit state of service ability: This state corresponds to the development of

excessive deformation and is used for checking the


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members in which the magnitude of deformation may limit the use of

the structure are its components

SALIENT FEATURES OF LIMIT STATE METHOD :

Limit State Method of design is a future improvement of ultimate load of

design. The acceptable limit of safety & serviceability requirements

before failure occur I called LIMIT STATE.

Design: The designer will design based on limit state method using SP

16 and IS 456-1978 code. The structure shall be designed to withstand

safely all loads liableto act on it through out of its life.

AIM

The aim of design is to achieve acceptable port abilities that the

structure will not become unfit for the use for which it is intended that it

will not reach a LIMIT STATE. This method is appeared to be

satisfactory and acceptable and recommended by the codes of practice

of many countries.

DESIGN CONSTANTS

Type = Duplex house

Floor to floor height = 3000mm

Depth of the foundation = 1200mm

below G.L

Bearing capacity of soil = 200 KN/sqm

Assumed imposed loads on floors:

Live load - 2KN/sqm

Floor finish - 1KN/sqm

Dead load - 3KN/sqm

Total load - 6KN/sqm


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DESIGN BASIS

Limit state method is based on IS 456-1978 S.I. units.

SPECIFICATIONS OF MATERIALS

Materials used are:

a) Cement

b)Sand

c) Gravel

d)Steel

e)Bricks

FOR SLABS M15 grade concrete is used

FOR BEAMS M15 grade concrete is used

FOR COLUMNS M15 grade concrete is used

FOR FOOTINGS M15 grade concrete is used

Fe 415 steel is used as main reinforcement called as Tor steel

Fe 250 steel is used as distribution steel called as mild steel

R.C.C (REINFORCED CEMENT CONCRETE)

RCC consists of steel and concrete. Concrete is good at compressive

and tension, butitis costly .if both were used together in proportions to

bear the required loads, the structure formed would be efficient in

strength and durability. Multistoried (G +) structures can be built in

R.C.C. the advantages of R.C.C is one can go for more number of

loads and can punch number of loads and can punchnumber of

openings as required . for the structure to be effective and to carry the

heavy loads R.C.C is suggested . this section is economical and cheap.


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STEEL

Grades used Fe 250- mild steel bars , Fe 415, Fe500 cold twisted or

hot rolled High yields strength deformed bars

Steel of grade Fe 250 is commonly used for:a) One way slab upto 3m span.

b) Two way slab upto 4m span .

c) Stirrups in beams and tiles in columns.

d) Mainsteel in columns.

DEAD LOADS

It should comprise the weight of all walls, partitions, floors and roofs and

shall include the weights of all permanent constructions in the building.

Steel of grade Fee 415 / Fe500 is common for:

a) One way slab above 4m span.

b) Two way slab above 4m span

c) Main reinforcement in beams and column footings.

d) Main steel in columns with larger concrete areas.

MOMENTS DISTRIBUTION METHOD OF ANALYSIS:

In this method all the members of structure are initially

assumed fixed at the end. In all position and direction and fixed end

movements due to external loads are worked out. The joints are

assumed to be locked and external moment is applied to achieve

fixated members at the joints. The extended moment is called UN

balanced moment and the external forces called SWAY FORCES. Theexternal moment is applied to prevent rotation of the joints and external

forces are applied to prevent the displacement of the joints. The

restraints provided at a joint are released and their effects are

evaluated. This process is continued till the external movements or

forces as the joints are zero or negligible.


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STIFFNESS FACTOR:

For Prismatic members,

Stiffness K is defined as ration of moment of inertia to length of the

members.i.e., K = I/R.

CARRY OVER FACTOR:

Consider a member AB fixed AB fixed at end B and support at end A

A moment MA applied at A, without displacing A, will produce moment at B.

Let it be MB. The ratio of MB/MA is known as CARRY OVER FACTOR

SIGN CONVENTION:

For this method clock wise moments at ends are considered positive and

anticlockwise moments are considered as negative.

SLAB

Slabs are the plate elements having the depth(D) much smaller than

itsspan and width. They usually carry a uniformly distributed load and form the

floor or roof of the building. Like beams, slabs, also may be simple support,

cantilever or continuous. They are classified according to the systems of

support as under:


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1) One way reinforced slabs.

2) Two way reinforced slabs

3) Flat slabs support directly on columns with beams.

4) Circular and other shapes

5) Grid slabs or waffle slabs.If the slab is supported on all four edges

and iflx

Ly>2 ; where Ly is a longer span and Lx is a

shorter span, then the slab is said to

be One way slab

and iflx

Ly


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Beam layout first floor


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Beam layout ground floor


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ANALYSIS OF SLABS:

SLAB PANEL S1 , S7

Step 1 :

Data:

lx= 4.47 ; ly = 5.46

Breadth

Lengthratio of the slab,

=

=1.2217


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Mux(-ve)= x(-ve)wlx2

= 15.2144 KNm

Mux(+ve)= x(+ve)wlx2

=11.41712 KNm

Muy(-ve)= y(-ve)wly2

=11.8194 KNm

Muy(+ve)= y(+ve)wly2

=8.8017 KNm

Shear forces Vu=

= 27.42403 KN

Step 6:

Minimum Depth Required:

The minimum depth required to resist the bending moment.

Mu = 0.138 fckbd2

d = 74.246 mm < 115 mm (provided length)

Hence provided depth is required.

(Provided depth is not reduced to satisfy the stiffness requirements)

Step 7:

Reinforcement : Along x direction

Mux(-ve) =

fckbd

fyAdfyA stst 187.0

Ast = 273.0397 mm2

Mux(+ve) =

fckbd

fyAdfyA stst 187.0


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Ast= 202.9807 mm2

For max steel area ,Using 8mm diameter bars , spacing of bars.

S =

X 1000 = 184mm

Maximum spacing is (i) 3d = 3 X115 = 345mm

(ii) 300mm which ever is less.

No.of bars = Ast / ast

= 5.43 (approx 6)

Hence provide 6 , 8mm bars at 184 mm c/c.

Reinforcement : Along y direction

Muy(-ve) =

fckbd

fyAdfyA stst 187.0

Ast = 222.1055 mm2

Muy(-ve) =

fckbdfyAdfyA stst 187.0

Ast = 164.057 mm2

For max Ast , Using 8mm diameter bars , spacing of bars.

S =X 1000 = 226mm

Maximum spacing is (i) 3d = 3 X`125 = 375mm



= 3.79 (approx. 4)

Hence provide 4, 8mm bars at 226mm c/c.

Step 8:


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Reinforcement in edge strip :

Ast = 0.12 % of gross area.

= 190mm2

Using 8mm diameter bars, spacing of bars.

S =

= 263.3mm

Maximum spacing is (i) 5d = 5 X 125 = 625mm



= 3.79 (approx. 4)


Hence provide 8mm bars at 227.5mm c/c in edge strips in both the directions.

Step 9:

Check for deflection :

For contInous basic value of 1/d ratio = 26% of steel at mid span.

Pt=

= 0.1884

fs = 0.58 X fy

= 0.58 X 415

= 240 N/mm2

From figure 4 of IS:456 , modification factor = 1.95

Maximum permittedratio = 1.95 X 39 = 76.05

l/d provided = 39.86957

v = Vu/bd

= 0.1714


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From table 19 IS:456:2000 , for obtained % of steel and M20 concrete

grade

c (max ) = 0.28

c>v

SLAB PANEL S2,S8,S10,S15

Step 1 :

Data:

lx= 5.46 ; ly = 5.84

Breadth


=

=1.06


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Overall depth , D = (

)

= 0.184 (approx. 195)

Step 3 :

Effective span:

Effective span of simply supported slab will be the least of clear span +

effective depth and center to center distance of supports.

lx = 5.46+0.115 = 5.6

ly = 5.84+0.14 = 5.98

=1.067


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y(-ve) = 0.047

y(+ve) = 0.035


= 20.46993 KNm


=14.64449 KNm


=19.01357 KNm


= 14.15904

Shear forces Vu=

= 36.12 KN

Step 6:



Mu = 0.138 fckbd2




Step 7:


Mux(-ve) =

fckbd

fyAdfyA stst 187.0

Ast = 300.35 mm2


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Mux(+ve) =

fckbd

fyAdfyA stst 187.0

Ast= 212.82 mm2


S =X 1000 = 167.27mm

Maximum spacing is (i) 3d = 3 X140= 420mm



= 5.978 (approx 6)



Muy(-ve) =

fckbd

fyAdfyA stst 187.0

Ast = 291.044 mm2

Muy(+ve) =

fckbd

fyAdfyA stst 187.0

Ast = 214.83 mm2


S = X 1000 = 172 mm




= 3.79 (approx. 4)


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Hence provide 4, 8mm bars at 172 mm c/c.

Step 8:



= 220.8mm2


S =

= 227.53mm

Maximum spacing is (i) 5d = 5 X 140 = 700 mm



= 5.79 (approx. 6)



Step 9:


For contonous basic value of 1/d ratio = 26% of steel at mid span.

Pt=

= 0.154

fs = 0.58 X fy

= 0.58 X 415

= 240 N/mm2

From figure 4 of IS:456 , modification factor = 2

Maximum permitted

ratio = 2 X 39 = 78

l/d provided = 40


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v = Vu/bd

= 0.1629


grade

c (max ) = 0.28

c> v

SLAB PANEL S5

Step 1 :

Data:

lx= 3.76 ; ly = 4.47

Breadth


=

=1.18


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Cover = 20mm

Diameter of reinforcement bar = 8mm


)

= 0.144

Step 3 :

Effective span:



lx = 3.76+0.0964 = 3.86

ly = 4.47+0.0964= 4.57

=1.183


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x (-ve) = 0.0418

x (+ve) = 0.0312

y(-ve) = 0.037

y(+ve) = 0.028


= 7.099 KNm


=5.299 KNm


=6.2846 KNm


= 4.7559KNm

Shear forces Vu=

= 22.00 KN

Step 6:



Mu = 0.138 fckbd2




Step 7:



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Mux(-ve) =

fckbd

fyAdfyA stst 187.0

Ast = 149.08 mm2

Mux(+ve) =

fckbd

fyAdfyA stst 187.0

Ast= 110.606 mm2


S =X 1000 = 336.99 mm




= 3.43 (approx 6)



Muy(-ve) =

fckbd

fyAdfyA stst 187.0

Ast = 140.27 mm2

Muy(+ve) =

fckbd

fyAdfyA stst 187.0

Ast = 105.54 mm2


S =X 1000 = 290mm



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= 3.43 (approx. 4)


Step 8:



= 172.8mm2


S =

= 290.74mm




= 3.43 (approx. 4)



Step 9:



Pt=

= 0.1488

fs = 0.58 X fy

= 0.58 X 415

= 240 N/mm2


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Maximum permittedratio = 2X 39 = 78

l/d provided = 38.6

v = Vu/bd

= 0.1629


grade

c (max ) = 0.28

c> v

SLAB PANEL S6,S13

Step 1 :

Data:

lx= 3.76 ; ly = 5.84

Breadth


=

=1.55


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=

=96

Adopting effective depth ,d =96 mm

Cover = 20mm



)

= 0.144

Step 3 :

Effective span:



lx = 3.76+0.0964 = 3.86

ly = 5.84+0.0964= 5.94

=1.53


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Step 5 :

Design Moments and shear forces :

The slab is continous. The corners are held down. Hence moment co-

efficient are obtained from Table26 of IS: 456.

x (-ve) = 0.0576

x (+ve) = 0.0449

y(-ve) = 0.037

y(+ve) = 0.028


= 9.783 KNm


=7.62 KNm


=6.2846 KNm


= 4.7559KNm

Shear forces Vu=

= 22.00 KN

Step 6:



Mu = 0.138 fckbd2





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Step 7:


Mux(-ve) =

fckbd

fyA

dfyA st

st 187.0

Ast = 207.3319 mm2

Mux(+ve) =

fckbd

fyAdfyA stst 187.0

Ast= 160.4234 mm2


S =X 1000 = 242 mm




= 3.43 (approx 4)



Muy(-ve) =

fckbd

fyAdfyA stst 187.0

Ast = 140.27 mm2

Muy(+ve) =

fckbd

fyAdfyA stst 187.0

Ast = 105.54 mm2


S =

X 1000 = 358 mm


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= 3.43 (approx. 4)


Step 8:



= 172.8mm2


S =

= 290.74mm




= 3.43 (approx. 4)



Step 9:


For continous basic value of 1/d ratio = 26% of steel at mid span.

Pt=

= 0.186

fs = 0.58 X fy


34/82

= 0.58 X 415

= 240 N/mm2



l/d provided = 38.6

v = Vu/bd

= 0.1629


grade

c (max ) = 0.28

c> v

Hence Safe

SLAB PANEL S3

Step 1 :

Data:

lx= 2.44 ; ly = 4.27

Breadth


=

=1.75


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36/82

Total = 7.35 KN/m2

Factored load Wu = 1.5 X 7.35 =11.025 KN/m2

Step 5 :




x (-ve) = 0.0634

x (+ve) = 0.0441

y(-ve) = 0.037

y(+ve) = 0.028


= 4.4741 KNm


=3.1121 KNm


=2.611 KNm


= 1.975 KNm

Shear forces Vu=

= 13.946 KN

Step 6:



Mu = 0.138 fckbd2



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Step 7:


Mux(-ve) =

fckbd

fyAdfyA stst 187.0

Ast = 142.3616 mm2

Mux(+ve) =


Ast= 97.9878 mm2


S =X 1000 = 312 mm




= 3.2 (approx 4)



Muy(-ve) =

fckbd

fyAdfyA stst 187.0

Ast = 90.25 mm2

Muy(+ve) =

fckbd

fyAdfyA stst 187.0


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Ast = 67.908 mm2


S =

X 1000 = 312mm




= 3.2 (approx. 4)


Step 8:



= 160.8 mm2


S =

= 312.43 mm




= 3.2 (approx. 4)


Hence provide 8mm bars at 312 mm c/c in edge strips in both the directions.

Step 9:




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Pt=

= 0.223

fs

= 0.58 X fy

= 0.58 X 415

= 240 N/mm2



l/d provided = 38.6

v = Vu/bd

= 0.1549


grade

c (max ) = 0.3

c> v

Hence Safe

SLAB PANEL S9

Step 1 :

Data:

lx= 4.47 ; ly = 5.46

Breadth


=

=1.2217


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fck= 20N/mm2 (compression strength)

fy= 415N/mm2 (mild steel)

Step 2 :

Thickness of Slab:

Assuming effective depth ,d =

=

=114.6 mm

Adopting effective depth ,d =115

Cover = 20mm



)

= 0.159 (approx. 160)

Step 3 :

Effective span:



lx = 4.47+0.115 = 4.585

ly = 5.46+0.115 = 5.575

=

=1.21


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Loads per unit area of slab :

Self wt of slab = 0.16 X 25 = 3.975 KN/m2

Live load = 3 KN/m2

Floor finish = 1 KN/m2

Total = 7.975 KN/m2

Factored load Wu = 1.5 X 7.975 = 11.965 KN/m2

Step 5 :




x (-ve) = 0.0715

x (+ve) = 0.0534

y(-ve) = 0

y(-ve) = 0.043


= 17.9807 KNm


=13.4289 KNm


=0 KNm


=10.81 KNm

Shear forces Vu=

= 27.42403 KN


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Step 6:



Mu = 0.138 fckbd2




Step 7:


Mux(-ve) =

fckbd

fyAdfyA stst 187.0

Ast = 324.95 mm2

Mux(+ve) =

fckbd

fyA

dfyA

st

st 187.0

Ast= 239.92 mm2


S =X 1000 = 154mm




= 6.46 (approx 7)




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Muy(-ve) =

fckbd

fyAdfyA stst 187.0

Ast = 0 mm2

Muy(-ve) =

fckbd

fyAdfyA stst 187.0

Ast = 202.64 mm2


S =X 1000 = 247mm

Maximum spacing is (i) 3d = 3 X`125 = 375mm



= 4.03 (approx. 4)


Step 8:



= 190mm2


S =

= 263.3mm




= 3.79 (approx. 4)



44/82


Step 9:


For contInous basic value of 1/d ratio = 26% of steel at mid span.

Pt=

= 0.2198

fs = 0.58 X fy

= 0.58 X 415

= 240 N/mm2



l/d provided = 39.86957

v = Vu/bd

= 0.1714


grade

c (max ) = 0.3

c> v

SLAB PANEL S 12

Step 1 :

Data:

lx= 3.76 ; ly = 4.47

Breadth


=


45/82

=1.18


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Step 4 :


Self wt of slab = 3.6 KN/m2

Live load = 3KN/m2


Total = 8.75 KN/m2


Step 5 :




x (-ve) = 0.0586

x (+ve) = 0.044

y(-ve) = 0.047

y(+ve) = 0.035


= 9.9535 KNm


=7.4736 KNm


=7.9832 KNm


= 5.9449KNm

Shear forces Vu=

= 22.00 KN


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Step 6:



Mu = 0.138 fckbd2




Step 7:


Mux(-ve) =

fckbd

fyAdfyA stst 187.0

Ast = 211.0652 mm2

Mux(+ve) =

fckbd

fyA

dfyA

st

st 187.0

Ast= 157.126 mm2


S =X 1000 = 238 mm




= 4.2 (approx 5)




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Muy(-ve) =

fckbd

fyAdfyA stst 187.0

Ast = 179.3591 mm2

Muy(+ve) =

fckbd

fyAdfyA stst 187.0

Ast = 132.5205 mm2


S =X 1000 = 280mm




= 3.5 (approx. 4)


Step 8:



= 172.8mm2


S =

= 290.74mm




= 3.43 (approx. 4)

Hence provide 4, 8mm bars at 290.74mm c/c.


49/82


Step 9:



Pt=

= 0.186

fs = 0.58 X fy

= 0.58 X 415

= 240 N/mm2



l/d provided = 38.6

v = Vu/bd

= 0.1629


grade

c (max ) = 0.28

c> v

SLAB PANEL S4,S11

Step 1 :

Data:

lx= 2.44 ; ly = 5.84

Breadth


=


50/82


51/82

Step 4 :


Self wt of slab = 2.85 KN/m2

Live load = 2KN/m2


Total = 5.85 KN/m2


Step 5 :


Bending Moment Mu= Wl2/8

= 8.091

Shear forces Vu=

= 12.89513 KN

Step 6:



Mu = 0.138 fckbd2

d = 54.145 < 70 mm (provided length)



Step 7:

Reinforcement :

Mu =



52/82

Ast = 430 mm2

S = 116.81 mm


= 8.5 (approx. 9)


Step 8:



= 136.8 mm2


S = 116.81 mm



.

Step 9:



Pt=

= 0.6445

fs = 0.58 X fy

= 0.58 X 415

= 240 N/mm2



53/82

Maximum permittedratio = 1 X 39 = 9

l/d provided = 38.6

v = Vu/bd

= 0.184


grade

c (max ) = 0.5

c> v

Hence Safe

ANALYSIS OF BEAMS

BEAM

A beam primarily is flexural member and resist load in vertical bending.

How ever

it sometimes resists lateral loads also. It resist load by bending and

shear. The check for development length, deflection, minimum steel,

maximum steel and cracking are require in design.

GENERAL DESIGN REQUIREMENTS FOR BEAMS:

1.Effective Span : The effective span of a simply supported beam shall

be taken as clear span plus effective depth of the beam or center to center

distance between the supports which ever is less.

The effective span of a cantilever shall be taken as its length to the face ofthe support plus half the effective depth except where it forms the end of a

continuous beam where the length to the center of the support shall be

taken.

2. Limiting Stiffness: The stiffness of beams is governed by the span to

the depth ratio. As per clause 23.2 of IS :456 for spans not exceeding 10m

, the span to effective depth ratio should not exceed the limits (Basic

values) given below.

Cantilevers - 7


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Simply supported - 20

Continuous - 26

For spans above 10m , the above values may be multiplied by 10/span in

m.

Depending on the amount and type of steel , the above value be modified

by multiplying with the modification factors obtained from Fig 4 & 5 of IS :

456.

3.Minimum Reinforcement: The minimum area of tension reinforcement

should not be less than the following (Clause 26.5.1 of IS:456)

Ast/bd = 0.85/fy

This works out only 0.2% for Fe 415 steel and 0.34% for Fe 250 steel.

4.Maximum Reinforcement : The maximum area of tension

reinforcement should not exceed 4% of the gross cross sectional area

(Clause 26.5.1 of IS : 456)

Ptmax < 0.04 bD

Where D=gross depth of the beam .

5.Spacing of bars : The horizontal distance between two parallel main

reinforcing bars shall usually be not less than the greatest of the following.

(a) . Diameter of the bar if the diameters are equal .

(b). Diameter of the largest bar if the bars are unequal.

(c). 5 mm more than the nominal maximum size of the aggregate.

When there are two or more rows of bars , the bars shall be vertically in line

and the minimum vertical distance between the bars shall be 15 mm , twothirds of the nominal maximum size of aggregate or the maximum size of the

bars which ever is greater.

The maximum spacing of the bars in tension for beams is taken from Table -

15 of IS:456-2000 depending on the amount of redistribution carried out in

analysis and fy.

6.Cover of Reinforcement : Reinforcement shall have concrete cover of

thickness as follows :


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(a) . At each end of reinforcement bar not less than 25mm nor less than

twice the diameter of such bar.

(b) . For longitudinal reinforcing bar in beam , not less than 25mm nor less

than the diameter of such bar.

7. Side Face Reinforcement : Where the depth of the beam exceeds

750mm , side face reinforcement shall be provided along the two faces. The

total area of such reinforcement shall not be less than 0.1% of the beam area

and shall be distributed equally on two faces at a spacing not exceeding

300mm or width of the beam which ever is less.

ANALYSIS OF FRAMES BY MOMENT DISTRIBUTION METHOD

LOAD CALCULATIONS:

Arrangement of loads for maximum design moments:

For designing a continuous beam or slab, it is necessary to determine the maximum

positive and negative moments at critical sections i.e. at the support and at zero

shear force by considering various arrangements of loads on structural frames.

A. Combination of loads shall be as given in IS-875 (part 5)

I. Design dead load on all spans with full design live load on two adjacent

spans.

II. Design dead load an all spans with full design live load on alternative spans.

B. When design live load does not exceed 3/4 th of design dead loads, design live

loads on all spans.

LOAD INTENSITY CALCULATION:

DEAD LAODS: -

Dead load includes the weight of all permanent components of a building including

walls, partitions, roofs, finishes and fixed permanent equipment and fittings that are

an integrated part of the structure.

Dead loads are calculated in the basis of unit weight of material given in IS 875 (part

1)

The unit weight of plain cement concrete24 KN/m2


56/82

Reinforcement - 25 KN/m2

Brick masonry- 19 KN/m2

Imposed loads:

The imposed loads to be assumed in the design are the maximum loads that

probably will be produced by the intended use and occupancy but shall not be less

than the equivalent minimum loads specified in.: 875- II 1987. So for this project

following live loads are considered.

For rooms, kitchen toilet and bathrooms - 2 KN/m2

Corridors, storeroom and staircase - 3 KN/m2

LOAD DISTRIBUTION FROM SLABS ONTO BEAMS

In two way slabs, the load distribution to the long span beams is trapezoidal and the

load distribution to the short span beams is triangular. These trapezoidal and triangular loads

are converted into equivalent uniformly distributed loads by using following formulae.

Direction Equivalent U.D.L B.M criteria Equivalent U.D.L S.F criteria

SHORT SPAN w LX/3 wLX/4

LONG SPAN wLX/2(1-1/3r ) wLX/2(1-1/2r)

Where,

W = intensity of load on slab

R = Ly/Lx (aspect ratio)

Lx = short span

Ly = long span


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58/82

FRAME A

FRAME A

MOMENT OF INERTIA

I = bxd3/12

Moment of inertia of columns

IGD = IDA = IIF = IFC = IHE = IEB = 306.7 x106

Moment of inertia at beams:

IGH = IHI = IDE = IEF = 460 x 2003/12=306.7 x 106mm4


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FIXED END MOMENTS

M= -wl2/12

Fixed end moments for vertical members are equal to zero

MGD = MHE = MIF = MDA = MEF=0

And the fixed end moments are

MGH= MDE= -19.78 KN m

MHG= MED = +19.78 KN m

MHI= MEF = -47.29 KN m

MIH= MFE= +47.29 KN m

STIFFNESS CONSTANTS: EI/L

EI/L at GH = EI/L at DE = 0.2237EI

EI/L at HI= EI/L at EF = 0.1712EII/L

EI/L at GD= EI/L at DA= EI/L at HE= EI/L at EB= EI/L at IF= EI/L at FC = 0.3333EI


60/82

E D AD BE CF

EF EH EB ED DE DG DA

0.1712 0.3333 0.3333 0.2237 0.2237 0.3333 0.3333 0.3333 0.3333 0.3333

0.1613 0.3139 0.3139 0.2107 0.2512 0.3744 0.3744 - - -

-47.29 0 0 19.78 -19.78 0 0 0 0 0

4.4374 8.6354 8.6354 5.7963 4.9687 7.4056 7.4056 0 0 0

-4.8306 6.2956 0 2.4844 2.8982 5.9172 0 3.7028 4.3177 -9.4059

-0.637 -1.2397 -1.2397 -0.8321 -2.2144 -3.3004 -3.3004 0 0 0

1.3692 -0.0611 0 -1.1072 -0.4161 -2.3713 0 -1.6502 -0.6198 2.6659

-0.0324 -0.063 -0.063 -0.0423 0.7001 1.0436 1.0436 0 0 0

-0.1757 0.2664 0 0.35 -0.0211 0.5058 0 0.5218 -0.0315 -0.3421

-0.071 -0.1383 -0.1383 -0.0928 -0.1242 -0.1852 -0.1852 0 0 0

-47.2301 13.6953 7.1944 26.3363 -13.9888 9.0153 4.9636 2.5744 3.6664 -7.0821

G H I F

GD GH HG HE HI IH IF FI FC FE

EI/L 0.3333 0.2237 0.2237 0.3333 0.1712 0.1712 0.3333 0.3333 0.3333 0.1712

DF 0.5983 0.4016 0.3071 0.4677 0.2351 0.3393 0.6607 0.3978 0.3978 0.2043

FEM 0 -19.78 19.78 0 -47.29 47.29 0 0 0 47.29

BAL 11.8343 7.9436 8.4483 12.5913 6.4676

-

16.0455

-

31.2445

-

18.8119 -18.89 -9.6613

CO 3.7028 4.2242 3.9718 4.3177 -8.0227 3.2338 -9.4059

-

15.6222 0 2.2187

BAL -4.7427 -3.1834 -0.0819 -0.1221 -0.0627 2.0942 4.0779 5.3319 5.3319 2.7383CO -1.6502 -0.0409 -1.5917 -0.6198 1.0471 -0.0314 2.6659 2.0389 0 -0.3185

BAL 1.0117 0.6791 0.3576 0.5329 0.2737 -0.8938 -1.7406 -0.6843 -0.6843 -0.3514

CO 0.5218 0.1788 0.3395 -0.0315 0.4469 0.1368 -0.3421 -0.8703 0 -0.0162

BAL -0.4192 -0.2814 0.0426 0.0635 0.0326 0.0696 0.1356 0.3526 0.3526 0.1811

TOTAL 10.2585 -10.26 31.2662 16.732 -47.1075 35.8537

-

35.8537

-

28.2653

-

13.8898 42.0807


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FIXED END MOMENTS



And the fixed end moments are

MGH= MDE= -wl2/12= -22.7981 x (4.47)2/12 = -37.95 KNM

MHG= MED = 37.95 KNM

MHI= MEF = -28.95 x (5.84)2/12 = -82.27 KNM

MIH= MFE= 82.27 KNM


EI/L at GH = EI/L at DE = 0.2237EI

EI/L at HI= EI/L at EF = 0.1712EI/L

EI/L at GD= EI/L at DA= EI/L at IF= EI/L at FC = 0.3333EI

EI/L at HE= EI/L at EB= EI/L =1.76EI

G H I F

GD GH HG HE HI IH IF FI FC FE

EI/L 0.3333 0.2237 0.2237 1.76 0.1712 0.1712 0.3333 0.3333 0.3333 0.1712

DF 0.5984 0.4016 0.1038 0.8167 0.0794 0.3393 0.6606 0.3978 0.3978 0.2043

FEM 0 -37.95 37.95 0 -82.27 82.27 0 0 0 82.27

BAL 22.709 15.2407 4.6004 36.196 3.519

-

27.9142

-

54.3476 -32.727 -32.727

-

16.8077

CO 7.0123 2.3002 7.6203 9.9631 -13.7571 1.7595

-

16.3635

-

27.1738 0 0.9683

BAL -5.6264 -3.776 -0.3764 -2.9615 -0.2879 4.9551 9.6474 10.4245 10.4245 5.3537

CO -1.5856 -0.1882 -1.888 -3.2511 2.4775 0.1439 5.2122 4.8237 0 -0.316

BAL 1.0814 0.7123 0.2762 2.1737 0.2113 -1.7136 -3.3481 -1.7931 -1.7931 -0.9209

CO 0.6037 0.1381 0.3562 0.0875 -0.8598 0.1056 -0.8965 -1.674 0 0.0085

BAL -0.4438 -0.2979 0.0432 0.3399 0.033 0.2683 0.5224 0.6625 0.6625 0.3402

TOTAL 23.7506 -23.8208 48.5819 42.5476 -90.934 59.8746

-

59.5737

-

47.4572

-

23.4331 70.8961


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I=bd3/12

Moment of inertia at columns:

IGD = IDA = IIF = IFC = 460 x 2003/12 = 306.7 x 106mm4

IHE = IEB = 200 x 4603/12 = 1622.3 x 106 mm4


IGH = IHI = IDE = IEF = 460 x 2003/12=306.7 x 106mm4

FIXED END MOMENTS



And for beams:

MGH= -wl2/12= -19.49 x 4.472/12 = -32.45 KNM

MHG= 32.45KNM

MDE=-18.17KNM

MED =18.17KNM


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66/82

FRAME E:

MOMENT OF INERTIA:

Moment of inertia of columns:

IKF = IFA =ILG=IGB= IMH= IHC= INI= IID= IOJ = IJE= 1622.3 x 106mm4


IKL= ILM=IMN=INO= IFG= IGH= IHI= IIJ=306.7 x 106mm4

FIXED END MOMENTS



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MKF = MFA = MLG = MGB = MMH= MHC = MNI = MID= MOJ = MJE=0

And for beams:

MKL = MFG =-38.88KNM

MLK = MGF =38.88 KNM

MLM = MGH =-11.23 KNM

MML= MHG =11.23 KNM

MMN= MHI=-2.99 KNM

MNM= MIH =2.99 KNM

MNO= MIJ =-38.88 KNM

MON= MJI =38.88 KNM


EI/L at KL= EI/L at FG= 0.1832 EI

EI/L at LM = EI/L at GH = 0.2659 EI

EI/L at MN = EI/L at HI = 0.3788 EI

EI/L at NO = EI/L at IJ = 0.1832 EI

EI/L at KF = EI/L at FA = EI/L at LG = EI/L at GB = EI/L at MH = EI/L at HC = EI/L at NI= EI/Lat ID= EI/L at OJ = EI/L at JE = 1.76 EI

K L M N

KF KL LK LG LM ML MH MN NM NI NO

EI/L 1.76 0.1832 0.1832 1.76 0.2659 0.2659 1.76 0.3788 0.3788 1.76 0.1832

DF 0.9057 0.0942 0.0829 0.7967 0.1203 0.1105 0.7319 0.1575 0.1631 0.7579 0.0788

FEM 0 -38.88 38.88 0 -11.23 11.23 0 -2.99 2.99 0 -38.88

BAL 35.2136 3.6624 -2.2921

-

22.0281 -3.3262

-

0.9105

-

6.0308

-

1.2978 5.8536 27.2 2.828

CO 9.2375 -1.146 1.8312 -6.13 -0.4553

-

1.6631

-

1.7407 2.9268

-

0.6489 1.265 -1.831

BAL -7.3283

-

0.7622 0.394 3.7876 0.5719 0.527 0.3491 0.0751 0.1982 0.9207 0.0957

CO -4.0319 0.1971 -0.3811 2.2958 0.0263 0.2859 0.4809 0.0991 0.0375 -0.4854 0.3683

BAL 3.4732 0.3612 0.1569 1.507 0.2278

-

0.0956

-

0.6337

-

0.1364 0.0129 0.0603 0.0062

CO 0.8139 0.0785 0.1806 0.4579 -0.0478 0.1139

-

0.0878 0.0064

-

0.0682 -0.2109 0.1821

BAL 0.8082 -0.084 -0.0489 -0.4706 -0.071 0.0035 0.0237 0.0051 0.0158 0.0735 0.0076

TOTAL 38.1862

-

36.573 38.7206

-

20.5804

-

14.3043 9.4911

-

7.6393

-

1.3117 8.3909 28.8232

-

37.2231


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O J I H

ON OJ JO JE JI IJ IN ID IH HI HC HM HG

0.183

2 1.76 1.76 1.76

0.183

2

0.183

2 1.76 1.76

0.378

8

0.376

8 1.76 1.76 0.2659

0.0942

0.9507

0.4752

0.4752

0.0494

0.0448

0.4311

0.4311

0.0927

0.0909

0.4225

0.4225 0.0638

38.88 0 0 0 38.88

-

38.88 0 0 2.99 -2.99 0 0 11.23

-3.662

-

35.21

3 -18.47 -18.47

-

1.920

6 1.6 2.53 2.53 3.327

-

0.749

-

3.481

4

-

3.481

4 -0.5257

1.414

-

9.235

-

17.60

65 0 0.8

-

0.960

3 13.6 0

-

0.374

5

1.663

5 0

-

3.015

4 -0.9248

0.7367

7.0834 7.634 7.634 0.83

-

0.5494

-

0.9708

-

0.9708

-

0.2099

0.2069

0.9619

0.9619 0.1452

0.047

8 3.817

3.541

7 0

-

0.274

7 0.415

0.460

3 0

0.103

5

-

0.104

9 0

0.174

5 0.3463

-

0.364

1

-

3.500

3

-

1.552

4

-

1.552

4

-

0.161

3

-

0.043

8

-

0.421

9

-

0.421

9

-

0.090

7

-

0.037

8

-

0.175

7

-

0.175

7 -0.0265

0.003

1

-

0.776

2

-

1.750

1 0

-

0.021

9

-

0.080

6

0.030

1 0

-

0.018

9

-

0.045

3 0

-

0.316

8 0.069

0.0728

0.7001 0.842 0.842

0.0875

0.0031

0.0299

0.0299

0.0064

0.0266

0.1238

0.1238

0.01869

37.12

83

-

37.12

4

-

27.36

13

-

11.54

64

38.21

9

-

38.49

6

15.25

76

1.167

2

5.732

9 -2.03

-

2.571

4

-

5.729

1

10.332

19


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G F

GH GB GL GF FG FK FA AF BG CH DI EJ

0.2659 1.76 1.76 0.1832 0.1832 1.76 1.76 1.76 1.76 1.76 1.76 1.76

0.0669 0.4434 0.4434 0.0461 0.0474 0.4752 0.4752 0 0 0 0 0

-11.23 0 0 38.88 -38.88 0 0 0 0 0 0 0

-1.8497 -12.26 -12.26 -1.2746 1.8429 18.475 18.475 0 0 0 0 0

-0.2628 0

-

11.0141 0.9214 -0.6373 17.6068 0 9.2375 -6.13

-

1.7404 1.265 -9.235

0.6927 4.5916 4.5916 0.4773 -0.8043 -8.0639 -8.0639 0 0 0 0 0

0.0726 0 1.8938 -4.0319 0.2386 -3.6642 0 4.0319 2.2958 0.4809

-

0.4854 3.817

0.1381 0.9158 0.9158 0.0952 0.1623 1.6278 1.6278 0 0 0 0 0

-

0.01325 0 0.7543 0.08115 0.0476 1.7366 0 0.8139 0.4579

-

1.0087

-

0.2109

-

0.7762

-0.055

-

0.3645 -0.3645 -0.0379 -0.0845 -0.8478 -0.8478 0 0 0 0 0

-

12.5074

-

7.1171

-

15.4831 35.11065

-

38.1147 26.8703 11.1911 14.0833

-

3.3763

-

2.2682 0.5687

-

6.1942

CALCULATION FOR BEAM ON FRAME A

BEAM DESIGN AB:

Beam size =200 x 460

Effective depth = 460-30=430mm

Design of support moment(H):

Support moment M = 19.78 KN-m

Factored moment Mu = 1.5(M) =74.67 KN-m

For Fe-415 and M25 grade concrete

From table D of SP 16;

Mulim /bd2

Mulim =102.06 KN-m

Mu< Mulim

Hence the beam is designed as singly reinforced section.

Mu /bd2

= (74.6710001000)/(2004302)=2.019

From table-3 of SP-16

Pt=Percentage steel value corresponding to Mu /bd2

value 2.019 is

Area of tensile steel = Ast = (Ptbd)/ (100)

Ast = (0.670200430)/(100)


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Ast = 576.2mm2

Ast min = (0.85bd)/ fy = (0.85200430)/415=176.14 mm2

Ast max=0.04bD = (0.04200430) = 3440 mm

2

Hence Ast = 576.2mm2

Assuming 16 mm diameter bars;

ast = ( (16)2/4)) = 201.06mm

2

Number of bars to be provided = n = (Ast/ast) = (576./201.06) = 3

Therefore provide 3 No.s 16 mm bars.

Ast provided = 603.18 mm2

CHECK FOR SHEAR:

Design shear force Vu = WuL/2 =28.3974.47/2 = 63.467 KN-m

Factored shear force =1.563.467 =95.2 KN

Nominal shear stress v = Vu /(bd)

v = 95.21000/(200420) =1.13 N/mm2

%steel = Pt = (100 A st provided)/bd

= (100603.18)/ (200430)=0.701 %

c = 0.56N/mm2

(From table 19 of IS 456:2000, for value of P t = 0.701)

c max = 2.8 N/ mm2

(From table 20 of IS 456:20)

v > c < c max

Hence shear reinforcement is to be provided in accordance with clause 40.4 of IS 456:2000.

Provide shear reinforcement using vertical stirrups.

Design of vertical stirrups:

Vus = Vu(cbd)

Vu = (63.4671000)(0.56 200430)

= 153. 07 KN

Assuming 2- legged 8mm bars as vertical stirrups;

Asv = ( (8)2

/4) 2=100.53mm2


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Spacing S = (0.87fyd Asv )/ Vus

S = (0.87415430100.53)/153070

S =101.96 mm

Spacing from minimum reinforcement consideration as per IS 456:2000.

Asv /bSv = (0.4/(0.87fy))

Asv = ( (8)2/4) 2 = 100.53mm

2

Asv /bSv = (0.4/ (0.87fy))

Sv = (100.530.87415) /( 2000.4) =453mm

Maximum allowed spacing = 0.75d=0.75(430) = 322 or 300mm whichever is less

Spacing should be least of above.

Hence provide 2-legged 8mm stirrups @ 101.96 mm c/c.

Design of Columns

The column is a structure is usually carrying axial compressive loads. Rectangular

cross sections are selected considering the bi-axial bending for columns. All columns have

been divided in to groups. Each group of column is designed taking the factored

compressive load & Factored moments.

All columns are subjected to moment in both directions. Columns are designed for bi-

axial bending in addition to the axial compressive load. Column are designed using

interaction diagrams, which are available in SP 16: 1980 for different combination of grade of

steel and grade of concrete. Longitudinal and transverse reinforcement are provided in

accordance with same norms laid down in clause 26.5.3 of IS 456: 2000. While designing

the column it is assumed that the reinforcement is distributed equally on four sides.

Columns are so oriented that the maximum depth available perpendicular to axis ofmaximum moment is acting.

M20 grade of concrete and Fe-415 grade of steel is used for columns.

Column or strut is a compression member, the effective length of which exceeds three times

the least lateral dimension.]

A compression member is subjected to pure axial load rarely occurs in practice. All

columns are subjected to some moment, which may be due to accidental eccentricity or due

to end restraint imposed by monolithically placed beams or slabs. The strength of column


72/82

depends on the strength of the materials, shape and size of cross section, length and the

degree of positional and directional resistance and its ends. A column may be classified

based on the different criteria, such as:

Shape of cross section.

Slenderness ratio.

Type of loading.

Pattern of lateral reinforcement.

A column may be rectangular, square, circular or polygonal in cross section. The

code specifies certain minimum reinforcement bars depending on its shape. A column

classified as short or long column depending up on its effective slenderness ratio. The ratio

of effective column length to least lateral dimension is referred to as effective slenderness

ratio. A short column has maximum slenderness ratio of 2. Its design based on the strength

of the materials and the applied loads. A long column has a slenderness ratios greater than

12 .it is designed to resist the applied load plus the additional bending moment induced due

to its buckle.

A column may be classified as follows based on the types of loading.

Axially loaded columns

A column subjected to axial load and uni-axial bending

A column subjected to axial load and bi-axial bending.

GENERAL RULES:

I. Between consecutive floors there should be an equal rise for every parallel step.

Similarly there should be equal tread.

II. The sum of tread of a single step, twice the riser should be in between 550mm and

700mm.

III. The rise of the step should not be more than 200mm and thread should not be lessthan 240mm.


73/82

COLUMN DESIGN C13:

Breadth of the column, b = 200mm

Depth of the column, D = 460mm

Concrete Mix: M25 ; fck = 25 N/mm2

Characteristic strength of reinforcement fy= 415N/mm2

ECCENTRICITY:

Considering minimum eccentricity e=20mm

Eccentricity ex=( ) ( ) (along x-axis i.e shoter dimension side)

ex =

(

) (

) = 21mm>20mm

Hence eX = 21mm

Eccentricity ey=( ) ( ) (along y-axis i.e larger dimension side)

ey=( ) ( ) = 12.67mm< 20mmhence ey = 20mm

MOMENTS:

Load acting on column E: P = 1216.8 KN

Factored Load, Pu= 1.5(P) = 1825.2 KN

Mx=P ex = (1825.221) = 38.329KN-m

My= P ey = (1825.220) = 36.504KN-m

Assume Percentage reinforcement, pt = 3%

p/ fck =( ) = 0.15Effective cover provided d = 50mm

Uniaxial moment capacity of the section about X-X axis:

d/D = 50/460=0.108

Chart for d/D = 0.10 will be used.

( ) = ( ) = 0.99From chart 44 of SP-16, we get


74/82

( ) = 0.02Mux1= 0.06fckbD

2 = 0.02202004602

=169.28 KN-m > 38.329KNm

Hence safe.

Uniaxial moment capacity of the section about Y-Y axis :

d/D = 50/200= 0.25

Chart for d/D=0.25 will be used.

( )= =0.99From chart - 46 of SP-16

( ) = 0.02Muy1 = 0.0220200

2 460

= 73.6 KN-m > 36.504KNm

Uniaxial moment capacity of the section about Y-Y axis:

d/D = 45/230= 0.20

Chart for d/D=0.20 will be used.

( ) = 1.77From chart - 46 of SP-16

( ) = 0.07Muy1 = 0.0725230

2 600

= 55.54 KN-m > 47.04KN-m

Hence safe.

CALCULATION OF PUZ:

From chart-63 of SP-16, for 3% reinforcement;

( ) = 19 N/mm2

Puz= 19Ag = 19200460 =1748KN

(

) = (

) = 1.04


75/82

( ) = () = 0.226

( ) =() = 0.495

From chart 64 of SP:16

( ) = 0.545(chart) > 0.226Hence safe.

Ast= 3% Ag = (3200460)/100 =2760mm2

Assuming 25mm dia bars

Provide 6No.s of 25mm

Astprovided is 2943.75mm2

Percentage reinforcement provided, p = (1002943.75)/(200460)=3.19%

ANALYSIS OF FOOTING

An isolated footing may be square, rectangular or circular in plan. Further it may be

axially loaded to eccentrically loaded. The design of square footing involves thedetermination of size and depth of the footing and the amount of main reinforcement anddowels.

Rectangular footing may be used in locations where space is restricted and itis not possible to provide a square footing. Rectangular footings are also provided forrectangular columns of pedestals. The method of designing a rectangular footing isessentially identical to that of a square footing except that each projection has to bedesigned separately.

In designing a circular footing which supports a circular column of pedestal, the

circular column or pedestal is replaced by an equivalent square column or pedestalwhich can be inscribed within its parameter.

Then the design procedure of the footing is identical with that of the square footing.

Depth of foundationis governed by the following factors:

1) To secure safe bearing capacity.

2) To penetrate below the zone where seasonal weather changes area likely to

cause significant moment due to swelling and shrinkage of soils, and

3) To penetrate below the zone where seasonal weather changes are

likely to cause significant moment due to swelling and shrinkage of soils and4) To penetrate below the zone this may be affected by frost.


76/82

IS-1080-1962 requires that in all soils, a minimum depth of50 cms is necessary.

However if good rock is net at smaller depths, only removal of topsoil may be sufficient. As estimate of depth of footing below ground level

may be obtained by listing the rank formula i.e.

H==P (1-sin) 2Y (1+sin) 2

Where h= Minimum depth of foundationP = Cross bearing capacity = 25

Y = Density of soil = 1.8t.lcu.m

= Angle of repose of soil

Approxi. = 300/1.8(1-sin 302)

=

=

30

1.85m

BENDING MOMENT:

The critical section for computing maximum bending moment for design ofan isolated concrete footing supporting different types of structures is as follows:

1) At the face of the column, pedestal or wall for footings supportinga concrete column pedestal or wall.

2) Half way between the centerline and the edge of the wall forfootings under masonry walls and

3) Half way between the face of the column or pedestal and the edge of the

gusseted base for footings under gusseted base.

Determine the minimum depth required to resist bending moment

Calculate the depth required for bending moment and check the depth for single shear and

double shear. The depth is kept uniform, if the footing size is small and is made slopping, if

the footing is large.

The maximum bending moment is calculated at the face of the column by passing a section

extends completely across the footing as shown:

Projection of the footing = (B-b)/2

The bending moment about x-x is (as a cantilever slab, wl2)

Mu =

= qu

Where qu= upward soil pressure

B = width of footing, b = width of column


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Determine the area of reinforcement required in width B using :

Mu= 0.87 fyAst(1-

)

Using the bars of diameter not less than 10mm, find the spacing of bars.

Spacing = ( )Where ast= area of bar used

Ast= total area of steel required

B = width of the footing

d = effective depth of footing

provide same reinforcement in both the directions.

Check for one way shear:

The check for one way shear is carried out similar to that of beams or slabs. The critical

section for one way shear is at a distance d from the column extending the full width of the

footing as shown,

VU= soil pressure from the shaded area

= quB(

d )

u =


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Permissible value of punching shear stress is p = 0.25fckCheck for bond length :

since the footing Is designed as a cantilever with reinforcement subjected to deigned

strength at the column face, sufficient bond length should be available from the face of thecolumn.

Ld=

Check for bearing stress:

The compressive stress in concrete at the base of the column is transferred by bearing to

the top of the supporting footing. The bearing pressure on the loaded area shall not exceed

the permissible bearing stress.

Actual bearing pressure = < permissible value

As per clause 34.4 of IS : 4562000, the permissible bearing stress is

= 0.45fck , in which should not exceed 2Where A1 = supporting area for bearing of footing

A2 = loaded area at the column face.

FOOTING DESIGN FOR COLUM OF FRAME A

Load, P = 1216.8 KN

Safe bearing capacity = 250KN/m2

fck = 20 N/mm2

fy = 415N/mm2

SELF WEIGHT OF COLUMN:

Dimensions of column = 200460 mm

Percentage of column load = 1216.8KN

.. Total load of soil = 1216.8+250

= 1466.8 KN 1500KN

..Adopt load of soil = 1500 KN

Area of footing =


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= 1500/250

= 6m2

Area provided Ap = 2.482.85

= 7.068m2

Factored load Pu = 1.51216.8

= 1825.2 KN

Ultimate load qu =

= 1825.2/7.068

= 0.25N/mm2

DEPTH OF FOOTING FOR B.M CONSIDERATION:

Bending moment Mu = 0.252.85 (9122/2)

= 300 KN-m

Mu = 0.138bd2fck

=> = d.. d = 195.3 mm

Provide 500 mm so as to resist shear (.. d = 500 mm)

Mu = 0.87fyAstd 1(fyAst/fckBd)

..Ast = 2739.6 mm2

Let us use 16 mm dia bars

Spacing S = ast B/ Ast=209.05 210 c/c

CHECK FOR ONE WAY SHEAR:

Factored shear force = qu B

- d

= 0.252850865

= 616.3 KN

Nominal shear v = Vu/Bd

= 616.3/(2850500)

= 0.43 N/mm2


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Vu = 0.332850 525=498.25103 N

Percentage of steel Pt = (162/4) (100/(200460))

=0.167

From table-19 ( c)

c =0.303 N/mm2

.. c> v (safe)

Safe also under two way shear.

STAIRCASE DESIGN

Wall thickness= 200 mm

Height of floor = 3m

Stair width = 1.2 m

Height of one flight = 1.5m

Riser(R) = 142mm

Tread (T) = 300mm


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No. of risers =21

No.of treads = 20

width of landing = 1.3 m

Effective span = 2.64 m

Thickness of slab = span/25

= 2640/25

= 105.6

= 110

D = d + cover

Adopting 30 mm cover D = 110+30 = 140mm

Weight of waist slab = )2 X 25

= 3.87 KN/m2

Weight of Steps = Rx 25/2

= 1.77 KN/m2

Live Load = 3 KN/m2

Floor Finish = 0.6 KN/m2

Total Load = 9.24 KN/m2

Factored Load = 1.5 X 9.24 = 13.86 KN/m2

Factored B.M

Mu = Wu l2

/8

= 13.8 x (2.64)2

/8

=12.02 x 106

N-mm

Min depth required

Mu = 0.138 fck b d2

= 65 < 110

So provided depth is adequate

Tension Reinforcement :


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Mu = 0.87fyAstd[ 1(fyAst/fckbd)]

Ast = 321.29

Using 10 mm dia bars

S= 78.5/321.29 x 1000

= 244

Provide 10 mm dia bars @ 250 mm c/c

Distribution Reinforcement :

Ast = 0.12 % of gross area

= 0.12 x 1000 x 140 /100

= 170 mm2

Using 8 MM Bars , Spacing

S= 300 mm

Hence provide 8 mm dia bars @ 300 mm c/c

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