design of proposed auditorium-project report

DESIG� OF PROPOSED AUDITORIUM

ACK�OWLEDGEME�T

We are delighted to express our hearty thanks to our honorable Principal,

Dr.S.Joseph Sekhar, Ph.D for providing facilities to undertake this mini

project work.

We are grateful to Mrs.S.Judes Sujatha, B.E, M.Tech., Head of the

Department, for extending all possible help in the execution of this work.

We are extremely indebted to our internal guide,Mr.A.B.Danie Roy,

M.E., who has suggested this topic, provided all the expertise, facilities and

help in shaping this project into a successful one.

We acknowledge the help rendered by Mr.A.Jinu Antony, Engineer

associated with auditorium design

We also express our thanks to all the staff members of the Civil

Engineering department, for their support.

ABSTRACT

This project deals with the analysis and design of the Auditorium of

St.Xavier’s Catholic College of Engineering with special emphasis on Slabs,

Beams, Columns, Footing and Staircase.

Analysis is carried out using Substitute Frame Analysis and

preliminary analysis of Beams is carried out using Moment Distribution

method.

Concrete mix used for the RCC members is M20 and steel used is

high yield strength deformed bars of grade Fe415. Limit State Method is

adopted for the design of all structural members in the building.

Safe bearing capacity of soil is taken as 200kN/m2. Footing is

designed as Isolated type. Plan and detailing of reinforcement are enclosed

in this report.

TABLE OF CO�TE�TS

CHAPTER �O. TITLE PAGE �O.

ABSTRACT

LIST OF SYMBOLS

LIST OF FIGURES

1. I�TRODUCTIO� 1.1 GENERAL 1

1.2 OBJECTIVES 2

1.3 DESIGN OF RC STRUCTURES 2

1.3.1 LIMIT STATE DESIGN 2

1.4 SLABS 3

1.4.1 CLASSIFICATION OF SLABS 3

1.5 BEAMS 4

1.5.1 DESIGN OF BEAMS 4

1.6 COLUMN 5

1.6.1 SHORT COLUMN 5

1.6.2 SLENDER COLUMN 5

1.6.3 CLASSIFICATION OF COLUM 5

1.7 FOOTING 6

1.7.1 TYPES OF COLUMN FOOTING 6

1.8 STAIRCASE 7

1.8.1 CLASSIFICATION OF STAIRS 7

2. PLA��I�G OF PROPOSED AUDITORIUM

2.1 LAYOUT OF SITE

2.2 PLANS

2.3 SECTION

2.4 ELEVATION

3 METHODOLOGY

3.1 LIMIT STATE DESIGN

3.2 PARTIAL SAFETY FACTOR

4 A�ALYSIS

INTRODUCTION

METHOD OF SUBSTITTE FRAME ANALYSIS

ANALYSIS OF FRAMES

5 DESIG�

SLABS

5.1.1. DESIGN OF SLABS

BEAMS

5.2.1 TYPES OF BEAMS

5.2.2 DESIGN OF L-BEAMS

5.2.3 DESIGN OF T-BEAMS

5.3 STAIRCASE

5.3.1 TYPES OF STAIRS

5.3.2 DESIGN OF DOGLEGGED STAIRCASE

5.4 COLUMN

5.4.1 TYPES OF COLUMN

5.4.2 DESIGN OF COLUMN

5.4.2.1 DESIGN OF AXIALLY LOADED COLUMN

5.4.2.2DESIGN OF UNIAXIALLY LOADED COLUMN

5.4.2.3DESIGN OF BI AXIALLY LOADED COLUMN

5.5 FOOTING

5.5.1 TYPES OF FOOTING

5.5.2 DESIGN OF FOOTING

6 CO�CLUSIO�

REFERE�CE

LIST OF FIGURES

Serial �o. Title Page �o.

figure 1 analysis using substitute frame method - frame1 ................... 17 figure 2 analysis using substitute frame method-frame2 ..................... 31 figure 3 analysis using substitute frame method-frame3 ..................... 45 figure 4 analysis using substitute frame method-frame4 ..................... 59 figure 5 reinforcement details of two way slab-section ..................... 105 figure 6 reinforcement details of two way slab- plan ......................... 105 figure 7 reinforcement details of l-beams- longitudinal section ........ 111 figure 8 reinforcement details of l-beams- cross section.................... 111 figure 9 reinforcement details of t-beam-longitudinal design ............ 116 figure 10 cross section of t-beam ......................................................... 116 figure 11 reinforcement details of doglegged staircase ........................ 122 figure 12 reinforcement details of axially loaded column ................... 126 figure 13 reinforcement details of uniaxially loaded column .............. 129 figure 14 reinforcement details of biaxially loaded columns ............... 132 figure 15 reinforcement details of axially loaded column ................... 137

LIST OF SYMBOLS

Mx - Moment in shorter direction

My - Moment in shorter direction

d - Effective depth

D - Overall depth

Ast - Area of Steel

P - Load

Wu (or) Pu - Design load

Mu - Design moment

Asc - Area of concrete

fy - Characteristic strength of steel

fck - Characteristic strength of concrete

B.M - Bending Moment

b - Breadth of beam

D - Overall depth

Vus - Strength of shear reinforcement

L - Clear span

Le - Effective span

N.A - Neutral Axis

MF - Modification factor

Q - Angle of repose of soil

M - Modular of rupture

τc - Permissible shear stress in concrete

τv - Nominal shear stress

CHAPTER 1

1. I�TRODUCTIO�

1.1 GE�ERAL

Auditorium, Conference hall, Library and Indoor Games are

necessary for an Engineering college. In St.Xavier’s Catholic College of

Engineering, Library, Conference hall are located at different locations and

also there is no special building for Auditorium. This project reports on the

analysis and design of Auditorium, Library and Indoor Games hall in one

separate block.

All structural components for the building such as beams,

columns, slabs, staircase etc are analysed and designed. Isolated footing is

adopted for all columns. Safe bearing capacity is taken as 200kN/m2. The

structure is designed by using limit state method, adopting M20 concrete and

Fe415 HYSD bars.

Site plan, plan showing various floors, section of plan,

elevation of plan and detailing of reinforcements for Beam, Column, Slab,

Staircase and Footing are also enclosed.

OBJECTIVES

2. To analyse the frames in the building.

3. To design the structural components of the five storey building.

4. To prepare the detailed drawing for the design carried out.

1.3 DESIG� OF RC STRUCTURES

Reinforced cement concrete members can be designed by the

following methods:

1. Working stress method

2. Limit state method

1.3.1 Limit state design

• Limit state method of design is based on elastic theory.

• Partial safety factors are used in this method to determine the design

loads and design strength of materials from their characteristics

values.

• The design aids to IS:456, published by the bureau of Indian

standards. The design of limit state method is very simple and hence

widely used in practice.

• This method gives economical results when compared with the

conventional working stress method.

1.4 SLABS

• Slabs are primary members of a structure, which support the imposed

load directly on them and transfer the same safety to the supporting

elements such as beams, walls, columns etc.

• A slab is a thin flexural member used in floor and roof of a structure

to support the imposed loads.

1.4.1 Classification of slabs

1.4.1.1Solid slab

1.4.1.2Hollow slab

1.4.1.3Ribbed slab

1.5 BEAM

• A beam has to be generally designed for the actions such as bending

moments, shear forces and twisting moments developed by the lateral

loads.

• The size of the beam is designed considering the maximum moment

in it and generally kept uniform throughout its length.

• IS:456:2000 recommends that the minimum grade of concrete should

not be less than M20 in RC works.

1.5.1 Design of beams

• When there is a Reinforced concrete slab over a concrete beam, then

the beam and the slab can be constructed in such a way that they act

together.

• The combined beam and slab are called as flanged beams. It may be

‘T’ or ‘L’ beams. Here both T-beams and L-beams are designed.

1.6 COLUM�S

• Vertical members in compression are called as columns and struts.

• The term column is reserved for member which transfer load to the

ground. Classification of column, depending upon slenderness ratio is

1.6.1Short columns

1.6.2Slender columns

1.6.1 Short column

IS:456:2000 classifies rectangular column as short when the

ratio of effective length(Le) to the least dimension is less than 12.

This ratio is called slenderness ratio of the column.

1.6.2Slender columns

The ratio of Le to the least dimension is less than 12 are called

as slender column.

Classification of column

1. Axially loaded column

2. Eccentrically loaded column

3. Column subjected to axial load and moment

1.7 Footing

• Foundation is the most important component of a structure.

• It should be well planned and carefully designed to ensure the

safety and stability of the structure.

• Foundation provided for RCC columns are called as column base.

1.7.1Types of column base

1. Isolated footing

2. Combined footing

3. Strap footing

4. Solid raft foundation

5. Annular raft foundation

1.8 Staircase

A staircase is a flight of steps leading from one floor to another.

It is provided to afford the means of ascent and descent between various

floors of the building. It should be suitably located in a building. In a

domestic building the stair should be centrally located to provide easy

access to all rooms. In public buildings stairs should be located near the

entrance. In big building there can be more than one stairs. Fire

protection to stairs is important too. Stairs are constructed using timber,

bricks, stone, steel or reinforced cement concrete.

1.8.1Classification of stairs

1. Single flight stairs

2. Quarter turn stairs

3. Dog legged stairs

4. Open well type stairs

5. Biffurcated stairs

6. Circular stairs

7. Spiral stairs

CHAPTER 3

3.METHODOLOGY

Various methods are available for the design of a

structure. Limit state method is adopted in this project.

3.1 Limit state design

The acceptable limit for safety and serviceability requirement

before failure occur is called limit state. The aim of design is to achieve

acceptable probabilities that the structure will not become unfit for use.

All relevant limit state shall be considered in the design to ensure

adequate degree of safety and serviceability.

3.2Partial safety factor

The value of load which has a 95% probability of a structure of

structural member for the limit state of collapse the following values

of partial safety factor is applied for limit state of collapse.

Ym = 1.5 for concrete

Ym = 1.15 for steel

CHAPTER 4

4 A�ALYSIS

4.1 Introduction

A multistoried frame is a complicated statically

indeterminate structure. The analysis by moment distribution method is

very lengthy and difficult. Hence substitute frame analysis is adopted

for better and easier calculation.

4.2 Method of substitute frames

In this method only a part of the frame is

considered for the analysis. The part considered is called as substitute

frame. The moments for each floor are separately computed. It is

assumed that the moments transferred from one floor to another are

small. Each floor is taken as connected to columns above and below

with their far ends fixed. The frame taken this way is analysed for the

moments and shears in the beams and columns.

The column will carry the maximum bending moment when

any one series of alternate spans should be loaded. The moment

distribution for the substitute frame analysis is performed only for two

cycles and hence, the method is sometimes referred to as, the two cycle

method.

When it is required to find the maximum negative

moment at a joint, the spans meeting at the joint are loaded with dead

and live load. The other spans are loaded with dead load alone.

4.3A�ALYSIS OF FRAMES

A B C D E F G

FRAME 1

Figure 1-Analysis using substitute frame method - frame1

4.3.1DISTRIBUTIO� FACTOR

4.3.2 LOAD CALCLATIO�

Joint Member Relative stiffness Total stiffness Stiffness for each member

A

AB

A1

A2

I/3.2

I/4

I/4

0.8125I

0.38

0.31

0.31

B

BA

BC

B1

B2

I/3.2

I/3.2

I/4

I/4

1.125I

0.28

0.28

0.22

0.22

C

CB

CD

C1

C2

I/3.2

I/6.4

I/4

I/4

0.9688I

0.32

0.16

0.26

0.26

D

DC

DE

D1

D2

I/6.4

I/6.4

I/4

I/4

0.8125I

0.32

0.32

0.31

0.31

E

ED

EF

E1

E2

I/6.4

I/6.4

I/4

I/4

0.8125I

0.32

0.32

0.31

0.31

F

FE

FG

F1

F2

I/6.4

I/6.4

I/4

I/4

0.8125 I

0.32

0.32

0.31

0.31

G

GF

G1

G2

I/6.4

I/4

I/4

0.656I

0.24

0.38

0.38

35.9635.9635.9635.9626.97526.975Total load

kN/m2

17.2317.2317.2317.2314.97514.975Total DL

kN/m2

4.54.54.54.53.3753.375Dead load

due to rib

kN/m2

18.7318.7318.7318.731212Total LL

kN /m

10.310.310.310.36.66.6Total DL

kN/ m

6.46.46.46.43.23.2Length m

7.997.997.997.992.562.56Area m2

151515151515LL kN/m2

8.258.258.258.258.258.25DL kN / m2

FGEFDECDBCABBeam name

4.3.3 MOME�T CALCULATIO� AT JOI�TS

JOINT-A

-15.16Net BM due

to A

-24.45

9.29

Distribution

-1.4336Distribution

+ Carryover

+23.02-23.02FEM due to

LL

-12.78FEM due to

DL

0.320.320.160.320.280.280.38DF

DEDCCDCBBCBAABMember

DCBAjoint

JOINT-B

-20.12124.563Net BM due

to A

-17.294

-2.827

27.39

-2.827

Distribution

5.7264.37Distribution

+ Carryover

+23.02-23.02+23.02-23.02FEM due to

LL

-58.81FEM due to

DL

0.320.320.160.320.280.280.38DF


DCBAjoint

JOINT-C

-133.8136.48Net BM

due to A

-132.97

-0.8256

138.13

-1.6512

Distributio

n

-10.2315.39Distributio

n +

Carryover

122.74-122.74+122.74-122.74FEM due

to LL

-58.8112.78FEM due

to DL

0.320.320.320.320.160.320.280.28DF

EFEDDEDCCDCBBCBAMember

EDCBjoint

JOINT-D

-132.51132Net BM

due to A

-132.97

0.46

131.54

0.46

Distributi

on

-10.238.8Distributi

on +

Carryove

r

122.74-122.74+122.74-122.74FEM due

to LL

-58.8112.78FEM due

to DL

0.320.320.320.320.160.320.280.28DF

EFEDDEDCCDCBBCBAMember

EDCBjoint

JOINT-E

FE G

-132.97132.97Net BM

due to A

-132.97

0

132.97

0

Distributi

on

-10.2310.23Distributi

on +

Carryove

r

122.74-122.74122.74-122.74FEM due

to LL

-58.81+58.81FEM due

to DL

0.240.320.320.320.320.320.32DF

GFFGFEEFEDDEDCMember

Djoint

JOINT-F

FE G

-136.03134.41Net BM

due to A

-132.97

1.44

132.9

1.44

Distributi

on


on +

Carryove

r

122.74-122.74122.74-122.74FEM due

to LL

58.81FEM due

to DL

0.240.320.320.320.320.320.32DF


Djoint

JOINT-G

FE G

101.06Net BM

due to A

132.97

-31.91

Distributi

on

10.23Distributi

on +

Carryove

r

122.74-122.74FEM due

to LL

58.81FEM due

to DL

0.240.320.320.320.320.320.32DF


Djoint

MID SPAN MOMENT AB

+18.375-15.16Net moment

-6.150.545distribtion

17.594.375-1.435carry over

moment

35.18-2.878.75distribution

moment

122.74-122.7423.02-23.02FEM due to

TL

12.78-12.78FEM due to

DL

0.320.320.160.320.280.280.38DF


DCBAjoint

Free BM at the centre of the span AB

= Wl2/8

=(26.975 * 3.22) / 8

=34.53 kNm

Net BM at centre of span AB

=34.53-[ (15.16+18.375) / 2]

=17.76 kNm

MID SPAN MOMENT OF BC

30.224-30.696Net moment


10.23-1.43-5.732.43carry over

moment

20.4611.45-2.874.86distribution

moment

-122.7423.02-23.02FEM due to

TL

58.81-58.8112.78-12.78FEM due to

DL

0.320.320.160.320.280.280.38DF


DCBAjoint

Free BM at the centre of the span BC

= Wl2/8

=(26.975 * 3.22) / 8

=34.53 kNm

Net BM at centre of span BC

=34.53-[ (30.696+30.224) / 2]

=4.07 kNm

MID SPAN MOMENT CD

104.99-95.923Net

moment

-6.0881.86distributio

n

10.238.795-10.23-1.43carry

over

moment

20.46-20.4617.59-2.86distributio

n

moment

-122.74122.74-122.7423.02FEM due

to TL

58.81-58.8112.78-12.78FEM due

to DL

0.320.320.320.320.160.320.280.28DF

EFEDDE DCCD CBBC BAmember

ED C BJoint

Free BM at the centre of the span CD

= Wl2/8

=(35.96 * 6.42) / 8

=184.12 kNm

Net BM at centre of span CD

=184.12-[ (95.92+104.99) / 2]

=83.66 kNm

MID SPAN MOMENT DE

105.97-105.97Net

moment


10.2310.23-10.23-10.23carry

over

moment

20.46-20.4620.46-20.46-20.46distributio

n

moment

-122.74122.74-122.74122.74FEM due

to TL

58.81-58.8158.81-58.81FEM due

to DL

0.320.320.320.320.160.320.280.28DF

EFEDDE DCCD CBBC BAmember

ED C BJoint

Free BM at the centre of the span DE

= Wl2/8

=(35.96 * 6.42) / 8

=184.12 kNm

Net BM at centre of span DE

=184.12-[ (105.97+105.97) / 2]

=78.14 kNm

MID SPAN MOMENT EF

G

111.51-105.96Net

moment

-1.0016.55Distribtion

-7.110.23-10.23-10.23carry over

moment

-14.11-20.4620.46-20.46distributio

n moment

122.74-122.74122.74FEM due

to TL

58.81-58.8158.81-58.81FEM due

to DL

0.240.320.320.320.320.320.32DF


EDCjoint

Free BM at the centre of the span EF

= Wl2/8

=(35.96 * 6.42) / 8

=184.12 kNm

Net BM at centre of span EF

=184.12-[ (105.97+111.51) / 2]

=75.4 kNm

MID SPAN MOMENT FG

G

101 05-109.03Net

moment


10.23-14.73-10.23carry over

moment

-29.4620.46-20.46distributio

n moment

122.74-122.74122.74FEM due

to TL

58.81-58.81FEM due

to DL

0.240.320.320.320.320.320.32DF


FEDjoint

Free BM at the centre of the span FG

= Wl2/8

=(35.96 * 6.42) / 8

=184.12 kNm

Net BM at centre of span FG

=184.12-[ (109.03+101.05) / 2]

=79.08 kNm

NEGETIVE MOMENT AT CENTRE OF DE

75.59-75.12Net

moment

6.55-6.08distribtion

-10.23-10.2310.238.795carry

over

moment

-20.4620.4620.46-20.46-20.4617.59distributio

n

moment

122.74-122.74122.74-122.74FEM due

to TL

-58.8158.81-58.8112.78FEM due

to DL

0.320.320.320.320.320.320.160.32DF

FGFEEFEDDE DCCD CBmember

FE D CJoint

Free BM at the centre of the span DE

= Wl2/8

=(17.23 * 6.42) / 8

=88.21 kNm

Net BM at centre of span DE

=88.21-[ (75.12+75.59) / 2]

=12.86 kNm

4.3.4 BENDING MOMENT IN COLUMN LOADING

joint A B C D E F G

Column DF

Above floor

Below floor

0.31

0.31

0.22

0.22

0.26

0.26

0.3

0.3

0.3

0.3

0.3

0.3

0.38

0.38

Member

DF

AB

0.38

BA

0.28

BC

0.25

CB

0.32

CD

0.16

DC

0.19

DE

0.19

ED

0.19

EF

0.19

FE

0.19

FG

0.19

GF

0.24

FEM due to

DL

-12.8 12. 8 -58.8 58.8 -58.8 58.8

FEM due TL -23.02 23.02 -122.7 122.7 -122.7 122.7

Distribution+

carryover

-1.435 4.37 17.59 -1.28 -6.07 8.79 6.07 -6.07 -6.07 6.07 -7.06 6.07

Total -24.45 27.39 4.81 11.5 -128.8 131.53 -52.7 52.74 -128.8 128.81 -65.9 64.88

Distribution to column Above floor

7.58

7.084

30.5

-23.64

22.82

-18.882

24.65

Distribution to column below floor

7.58

7.084

30.5

-23.64

22.82

-18.882

24.65

Joint A B C D E F G

Column DF

Above floor

Below floor

0.31

0.31

0.22

0.22

0.26

0.26

0.3

0.3

0.3

0.3

0.3

0.3

0.38

0.38

Member

DF

AB

0.38

BA

0.28

BC

0.25

CB

0.32

CD

0.16

DC

0.19

DE

0.19

ED

0.19

EF

0.19

FE

0.19

FG

0.19

GF

0.24

FEM due to

DL

-12.8 12. 8 -58.8 58.8 -58.8 58.81

FEM due TL -

23.02

23.02 -122.7 122.7 -122.7 122.7

Distribution+

carryover

1.43 2.43 5.73 1.28 -6.07 2.86 -6.07 6.07 6.07 -6.07 -14.73 6.07

Total -11.4 15.21 -17.3 24.3 -64.9 61.7 -128.8 128.8 -52.7 52.7 -137.5 128.8


3.52

0.46

10.55

20.14

22.82

25.42

-48.9


3.52

0.46

10.55

20.14

22.82

25.42

-48.9

H

FRAME 2

I J K L M N

Figure 2-Analysis using substitute frame method-frame2

4.3.5 DISTRIBUTIO� FACTOR


H

HI

H1

H2

I/3.72

I/4

I/4

0.769I

0.35

0.33

0.33

I

IH

IJ

I1

I2

I/3.72

I/3.55

I/4

I/4

1.05I

0.26

0.27

0.24

0.24

J

JI

JK

J1

J2

I/3.55

I/6.33

I/4

I/4

0.94I

0.3

0.17

0.27

0.27

K

KJ

KL

K1

K2

I/6.33

I/6.43

I/4

I/4

0.81I

0.2

0.19

0.31

0.31

L

LK

LM

L1

L2

I/6.43

I/3.55

I/4

I/4

0.937I

0.17

0.3

0.27

0.27

M

ML

MN

M1

M2

I/3.55

I/3.72

I/4

I/4

1.05 I

0.27

0.26

0.24

0.24

N

NM

N1

N2

I/3.72

I/4

I/4

0.769I

0.35

0.33

0.33

4.3. 6 LOAD CALCLATIO�

45.7744.24579.2478.1144.24545.77Total load

kN/m2

18.4217.87531.0230.6217.87518.42Total DL

kN/m2

3.3753.3754.54.53.3753.375Dead load

due to rib

kN/m2

27.3526.3748.2247.4926.3727.35Total LL

kN /m

15.04514.526.5226.1214.515.045Total DL

kN/ m

3.723.556.436.333.553.72Length m

6.7846.2420.67220.046.246.784Area m2

151515151515LL kN/m2

8.258.258.258.258.258.25DL kN / m2

MNLMKLJKIJHIBeam name

4.3.7 MOME�T CALCULATIO� AT JOI�TS

JOINT-H

-37.18Net BM due

to H

-57.2

20.02

Distribution

-4.42Distribution

+ Carryover

+52.78-52.78FEM due to

LL

-18.77FEM due to

DL

0.190.20.170.30.270.260.35DF

KLKJJKJIIJIHHIMember

KJIHjoint

JOINT-I

-44.56455.8Net BM due

to I

-38.104

-6.46

62.02

-6.22

Distribution


+ Carryover

46.47-46.47+52.78-52.78FEM due to

LL

-102.24FEM due to

DL

0.190.20.170.30.270.260.35DF


KJIHjoint

JOINT-J

-235.87110.12Net BM

due to J

-276.2

40.33

38.94

71.18

Distributio

n

-15.39-7.53Distributio

n +

Carryover

260.81-260.8146.47-46.47FEM due

to LL

-102.24102.24FEM due

to DL

0.30.170.190.20.170.30.270.26DF

LMLKKL KJJKJIIJIHMember

LKJIjoint

JOINT-K

-292.10284.03Net BM

due to K

-294.62

2.516

281.38

2.648

Distributi

on


on +

Carryove

r

273.01-273.01260.81-260.81FEM due

to LL

-58.8118.77FEM due

to DL

0.30.170.190.20.170.30.270.26DF

LMLKKL KJJKJIIJIHMember

LKJIjoint

JOINT-L

ML N

-121.65248.56Net BM

due to L

-49.88

-71.77

298.23

-40.67

Distributi

on


on +

Carryove

r

46.47-46.47273.01-273.01FEM due

to LL

-21.24+102.2

4

FEM due

to DL

0.350.260.270.30.170.190.2DF

NMMNMLLMLKKLKJMember

Kjoint

JOINT-M

ML N

-60.545.834Net BM

due to M

-68.61

8.112

37.41

8.424

Distributi

on

-15.83-9.06Distributi

on +

Carryove

r

52.78-52.7846.47-46.47FEM due

to LL

106.88FEM due

to DL

0.350.260.270.30.170.190.2DF


Kjoint

JOINT-N

ML N

40.04Net BM due

to N

57.2

-17.16

Distribution

4.42Distribution

+ Carryover

52.78-52.78FEM due to

LL

18.77FEM due to

DL

0.350.260.270.30.170.190.2DF


Kjoint

MID SPAN MOMENT HI

41.18-37.23Net moment

-11.841.573distribution

36.39.235-4.495carry over

moment


moment

260.81-260.8152.78-52.78FEM due to

TL

18.77-18.77FEM due to

DL

0.190.20.170.30.270.260.35DF


KJIHjoint

Free BM at the centre of the span HI

= Wl2/8

=(45.77 * 3.722) / 8

=79.2 kNm

Net BM at centre of span HI

=79.2-[ (37.23+41.18) / 2]

=39.995 kNm

MID SPAN MOMENT OF IJ


-2.87-7.09distribution

17.1-7.538.3617.89carry over

moment

34.1516.73-15.0635.78distribution

moment

-273.0146.47-46.47FEM due to

TL

102.24-102.24102.24-102.24FEM due to

DL

0.190.20.170.30.270.260.35DF


KJIHjoint

Free BM at the centre of the span IJ

= Wl2/8

=(44.245 * 3.552) / 8

=69.699 kNm

Net BM at centre of span IJ

=69.7-[ (60.26+52.8) / 2]

=13.2 kNm

MID SPAN MOMENT JK

247.52-231.66Net

moment

-3.13.4distribution

-5.13520.6-15.4-4.59carry over

moment

-10.27-30.7941.15-9.18distribution

moment

-46.47260.81-260.8152.78FEM due

to TL

106.9-106.918.77-18.77FEM due

to DL

0.30.170.190.20.170.30.270.26DF

LMLKKL KJJKJIIJIHmember

LKJIJoint

Free BM at the centre of the span JK

= Wl2/8

=(78.11 * 6.332) / 8

=391.22 kNm

Net BM at centre of span JK

=391.22-[ (231.7+247.52) / 2]

=151.63 kNm

MID SPAN MOMENT KL

243.18-257.26Net

moment


4.5917.07-21.614.74carry over

moment

9.18-43.2234.159.48distribution

moment

-52.78273.01-273.0146.47FEM due

to TL

18.77-18.77102.24-102.24FEM due

to DL

0.30.170.190.20.170.30.270.26DF

LMLKKL KJJKJIIJIHmember

LKJIJoint

Free BM at the centre of the span KL

= Wl2/8

=(79.24 * 6.432) / 8

=409.52 kNm

Net BM at centre of span KL

=409.52-[ (257.26+243.18) / 2]

=159.3 kNm

MID SPAN MOMENT LM

N

34.05-62.591Net

moment


-3.717-9.06-3.41-14.63carry over

moment

-7.434-6.812-18.123-29.25distribution moment

46.47-46.47260.81FEM due

to TL

21.24-21.24106.88-106.88FEM due

to DL

0.350.260.270.30.170.190.2DF


MLKjoint

Free BM at the centre of the span LM

= Wl2/8

=(44.25* 3.552) / 8

=69.71 kNm

Net BM at centre of span LM

=69.71-[ (62.59+34.05) / 2]

=21.39 kNm

MID SPAN MOMENT MN

N

37.15-40.87Net

moment


4.42-9.25-38.15carry over

moment

-18.58.843-76.3distribution

moment

52.78-52.78273.01FEM due to

TL

18.77-18.77FEM due to

DL

0.350.260.270.30.170.190.2DF


MLKjoint

Free BM at the centre of the span MN

= Wl2/8

=(45.77 * 3.722) / 8

=79.2 kNm

Net BM at centre of span MN

=79.2-[ (40.87+37.15) / 2]

=40.19kNm

NEGETIVE MOMENT AT CENTRE OF KL

84.41-147.4Net

moment

3.2-2.19distribtion

-3.4-15.4-9.0620.6carry

over

moment

-6.81-18.12-10.27-29.25-30.7941.15distributio

n

moment

41.47-41.47260.81-260.81FEM due

to TL

-21.24106.9-106.918.77FEM due

to DL

0.260.270.30.170.190.20.170.3DF

MNMLLMLKKLKJJKJImember

MLKJJoint

Free BM at the centre of the span KL

= Wl2/8

=(31.02 * 6.432) / 8

=160.31 kNm

Net BM at centre of span KL

=160.31-[ (147.4+84.41) / 2]

=44.61 kNm


joint H I J K L M N

Column DF

Above floor

Below floor

0.33

0.33

0.24

0.24

0.27

0.27

0.31

0.31

0.27

0.27

0.24

0.24

0.33

0.33

Member

DF

HI

0.35

IH

0.26

IJ

0.27

JI

0.3

JK

0.17

KJ

0.19

KL

0.19

LK

0.17

LM

0.3

ML

0.27

MN

0.26

NM

0.35

FEM due to

DL

-18.8 18.8 -106.9 106.9 -21.24 21.24

FEM due TL -52.78 52.78 -260.8 260.8 -46.5 46.5

Distribution+

carryover

-4.421 9.24 36.31 -4.56 -14.62 20.57 -5.13 -14.6 -3.41 -9.06 -3.72 -3.28

Total -57.2 62.02 17.51 14.18 -275.4 281.4 -112.0 92.26 -49.9 37.41 -24.96 17.96


18.88

-19.09

70.54

-52.50

-11.44

-2.988

-5.93


18.88

-19.09

70.54

-52.50

-11.44

-2.988

-5.93

joint H I J K L M N

Column DF

Above floor

Below floor

0.33

0.33

0.24

0.24

0.27

0.27

0.31

0.31

0.27

0.27

0.24

0.24

0.33

0.33

Member

DF

HI

0.35

IH

0.26

IJ

0.27

JI

0.3

JK

0.17

KJ

0.19

KL

0.19

LK

0.17

LM

0.3

ML

0.27

MN

0.26

NM

0.35

FEM due to

DL

-21.24 21.24 -

102.24

102.2

4

-18.77 18.77

FEM due TL -46.5 46.5 -273 273 -52.78 52.78

Distribution+

carryover

3.28 3.72 8.37 3.41 16.22 4.74 -21.61 16.22 4.59 -38.2 -9.24 4.42

Total -17.96 24.96 -38.1 49.88 -86.02 106.9 -294.6 289.2 -14.18 -19.4 -62.02 57.2


5.93

3.15

9.76

58.17

-74.26

19.53

-18.9


5.93

3.15

9.76

58.17

-74.26

19.53

-18.9

O

FRAME 3

P Q R S T U




O

OP

O1

O2

I/3.72

I/4

I/4

0.769I

0.35

0.33

0.33

P

PO

PQ

P1

P2

I/3.72

I/3.55

I/4

I/4

1.05I

0.26

0.27

0.24

0.24

Q

QP

QR

Q1

Q2

I/3.55

I/6.33

I/4

I/4

0.94I

0.3

0.17

0.27

0.27

R

RQ

RS

R1

R2

I/6.33

I/6.43

I/4

I/4

0.81I

0.19

0.19

0.31

0.31

S

SR

ST

S1

S2

I/6.43

I/3.55

I/4

I/4

0.937I

0.17

0.3

0.27

0.27

T

TS

TU

T1

T2

I/3.55

I/3.72

I/4

I/4

1.05 I

0.27

0.26

0.24

0.24

U

UT

U1

U2

I/3.72

I/4

I/4

0.769I

0.35

0.33

0.33


54.72547.79564.0563.4647.79554.725Total load

KN/m2

48.08522.04527.192722.04548.085Total DL

KN/m2

3.3753.3754.54.53.3753.375Dead load

due to rib

KN/m2

6.6425.7536.8636.4725.756.64Total LL

KN /m

40.4114.1620.2720.0614.1640.41Total DL

KN/ m

3.723.556.436.333.553.72Length m

8.2286.09515.815.396.0958.228Area m2

3151515153LL KN/m2

18.278.258.258.258.2518.27DL KN / m2

TUSTRSQRPQOPBeam name

4

.3.11 MOME�T CALCULATIO� AT JOI�TS

JOINT-O

-44.41Net BM due

to O

-68.32

23.912

Distribution

-5.2Distribution

+ Carryover

+63.12-63.12FEM due to

LL

-23.15FEM due to

DL

0.190.190.170.30.270.260.35DF

RSRQQRQPPQPOOPMember

RQPOjoint

JOINT-P

-52.28466.376Net BM due

to P

-44.194

-8.09

74.166

-7.79

Distribution


+ Carryover

50.19-50.19+63.12-63.12FEM due to

LL

-90.16FEM due to

DL

0.190.190.170.30.270.260.35DF


RQPOjoint

JOINT-Q

-193.61101.575Net BM due

to Q

-223.13

29.521

49.48

52.095

Distribution

-11.23-0.71Distribution

+ Carryover

211.9-211.950.19-50.19FEM due to

LL

-93.6855.45FEM due to

DL

0.190.190.170.30.270.260.35DF


RQPOjoint

JOINT-R

-235.66229.754Net BM

due to R

-237.47

1.81

227.944

1.81

Distributio

n


n +

Carryover

+220.68-220.68211.9-211.9FEM due

to LL

-23.1523.15FEM due

to DL

0.260.270.30.170.190.190.170.3DF

TUTSSTSRRSRQQRQPMember

TSRQjoint

JOINT-S

-104.56201.87Net BM

due to S

-49.48

-55.08

233.08

-31.212

Distributio

n

0.7112.4Distributio

n +

Carryover

50.19-50.19+220.68-220.68FEM due

to LL

-55.4590.16FEM due

to DL

0.260.270.30.170.190.190.170.3DF

TUTSSTSRRSRQQRQPMember

TSRQjoint

JOINT-T

TS U

-64.5447.13Net BM

due to T

-74.17

9.63

37.14

9.99

Distributi

on

-11.05-13.05Distributi

on +

Carryove

r

63.12-63.1250.19-50.19FEM due

to LL

93.68FEM due

to DL

0.350.260.270.30.170.190.19DF

UTTUTSSTSRRSRQMember

Rjoint

JOINT U

TS U

44.41Net BM

due to U

68.32

-23.912

Distributi

on

5.2Distributi

on +

Carryove

r

63.12-63.12FEM due

to LL

23.15FEM due

to DL

0.350.260.270.30.170.190.19DF


Rjoint

MID SPAN MOMENT OP



28.311.05-5.195carry over

moment


moment

211.9-211.963.12-63.12FEM due to

TL

23.15-23.15FEM due to

DL

0.190.190.170.30.270.260.35DF


RQPOjoint

Free BM at the centre of the span OP

= Wl2/8

=(54.725 * 3.722) / 8

=94.66 kNm

Net BM at centre of span OP

=94.66-[ (44.4+53.55) / 2]

=45.68 kNm

MID SPAN MOMENT OF PQ



12.35-0.715.999.71carry over

moment

20.4611.99-1.4219.41distribution

moment

-220.6850.19-50.19FEM due to

TL

90.16-90.1655.45-55.45FEM due to

DL

0.190.190.170.30.270.260.35DF


RQPOjoint

Free BM at the centre of the span PQ

= Wl2/8

=(47.79 * 3.552) / 8

=75.3 kNm

Net BM at centre of span PQ

=75.3-[ (49.86+57.97) / 2]

=21.4 kNm

MID SPAN MOMENT QR

203.14-188.2Net

moment


n

-3.716.05-11.23-5.4carry

over

moment

-7.4-22.4632.1-10.79distributio

n

moment

-50.19211.9-211.963.12FEM due

to TL

93.68-93.6823.15-23.15FEM due

to DL

0.30.170.190.190.170.30.270.26DF

STSRRSRQQRQPPQPOmember

SRQPJoint

Free BM at the centre of the span QR

= Wl2/8

=(63.46 * 6.332) / 8

=317.85 KNm

Net BM at centre of span QR

=317.85-[ (188.2+203.14) / 2]

=122.18 KNm

MID SPAN MOMENT RS

196.25-210.13Net

moment


n

5.412.4-16.793.4carry over

moment

10.8-33.5824.86.8distributio

n moment

-63.12220.68-220.6850.19FEM due

to TL

23.15-23.1590.16-90.16FEM due

to DL

0.260.270.30.170.190.190.170.3DF

TUTSSTSRRSRQQRQPmember

TSRQJoint

Free BM at the centre of the span RS

= Wl2/8

=(64.05 * 6.432) / 8

=331.02 KNm

Net BM at centre of span RS

=331.02-[ (210.13+196.25) / 2]

=127.8 KNm

MID SPAN MOMENT ST

U

59-33.27Net

moment


-9.716.530.71-11.23carry over

moment

-19.411.4213.05-22.46distribution

moment

50.19-50.19211.9FEM due to

TL

55.45-55.4593.68-93.68FEM due to

DL

0.350.260.270.30.170.190.19DF


TSRjoint

Free BM at the centre of the span ST

= Wl2/8

=(47.79 * 3.552) / 8

=75.3 kNm

Net BM at centre of span ST

=75.3-[ (33.27+59) / 2]

=29.17 kNm

MID SPAN MOMENT TU

U

44.4-53.18Net

moment


5.2-11.05-29.65carry over

moment


moment

63.12-63.12220.68FEM due

to TL

23.15-23.15FEM due

to DL

0.350.260.270.30.170.190.19DF


TSRjoint

Free BM at the centre of the span TU

= Wl2/8

=(54.725 * 3.72) / 8

=94.66 kNm

Net BM at centre of span TU

=94.66-[ (53.18+44.4) / 2]

=45.87 kNm

NEGETIVE MOMENT AT CENTRE OF RS

76.83-122.29Net

moment

1.79-2.45Distributi

on

0.71-11.24-3.7-16.05carry

over

moment

1.42-13.05-7.422.46-22.4632.1distributio

n

moment

50.19-50.19211.9-211.9FEM due

to TL

-55.4593.68-93.6823.15FEM due

to DL

0.260.270.30.170.190.190.170.3DF

TUTSSTSRRSRQQRQPmember

TSRQJoint

Free BM at the centre of the span RS

= Wl2/8

=(27.19 * 6.432) / 8

=140.521 kNm

Net BM at centre of span RS

=140.521-[ (122.29+76.83) / 2]

=40.96 kNm


joint O P Q R S T U

Column DF

Above floor

Below floor

0.33

0.33

0.24

0.24

0.27

0.27

0.31

0.31

0.27

0.27

0.24

0.24

0.33

0.33

Member

DF

OP

0.35

PO

0.26

PQ

0.27

QP

0.3

QR

0.17

RQ

0.19

RS

0.19

SR

0.17

ST

0.3

TS

0.27

TU

0.26

UT

0.35

FEM due to

DL

-23.2 23.2 -93.68 93.68 -55.45 55.45

FEM due TL -63.12 63.12 -211.9 211.9 -50.19 50.19

Distribution+

carryover

-5.2 11.05 28.31 -5.4 -11.23 16.04 -3.7 -11.2 0.71 -6.52 -9.7 0.68

Total -68.32 74.17 -5.16 17.75 -223.1 227.9 -97.38 82.45 -49.48 43.67 -65.15 56.13


22.55

-16.56

55.45

-40.47

-8.9

5.15

-18.5


22.55

-16.56

55.45

-40.47

-8.9

5.15

-18.5

joint O P Q R S T U

Column DF

Above floor

Below floor

0.33

0.33

0.24

0.24

0.27

0.27

0.31

0.31

0.27

0.27

0.24

0.24

0.33

0.33

Member

DF

OP

0.35

PO

0.26

PQ

0.27

QP

0.3

QR

0.17

RQ

0.19

RS

0.19

SR

0.17

ST

0.3

TS

0.27

TU

0.26

UT

0.35

FEM due to

DL

-55.45 55.45 -90.16 90.16 -23.2 23.2

FEM due TL -50.2 50.2 -220.7 220.7 -63.12 63.12

Distribution+

carryover

-0.68 9.7 5.6 -0.71 12.4 3.4 -16.79 12.4 5.4 -29.6 -11.05 5.2

Total -56.13 65.15 -44.2 49.48 -77.76 93.56 -237.5 233.1 -17.75 -6.48 -74.17 68.32


18.52

-5.03

7.63

44.61

-58.14

19.36

-22.6


18.52

-5.03

7.63

44.61

-58.14

19.36

-22.6

FRAME 4

A1 F1 G1 J1 K1 N1H1 O1 S1R1P1M1L1I1E1C1 D1B1 Q1 T1 U1



Joint Member Relative stiffness Total stiffness Stiffness for each member A1

A1B1 A11 A12

I/3.2 I/4 I/4

0.8125I

0.38 0.31 0.31

B1

B1A1 B1C1 B11 B12

I/3.2 I/3.2 I/4 I/4

1.125I

0.28 0.28 0.22 0.22

C1

C1B1 C1D1 C11 C12

I/3.2 I/3.2 I/4 I/4

1.125I

0.28 0.28 0.22 0.22

D1

D1C1 D1E1 D11 D12

I/3.2 I/3.2 I/4 I/4

1.125I

0.28 0.28 0.22 0.22

E1

E1D1 E1F1 E11 E12

I/3.2 I/3.2 I/4 I/4

1.125I

0.28 0.28 0.22 0.22

F1

F1E1 F1G1 F11 F12

I/3.2 I/3.2 I/4 I/4

1.125I

0.28 0.28 0.22 0.22

G1

G1F1 G1H1 G11 G12

I/3.2 I/3.2 I/4 I/4

1.125I

0.28 0.28 0.22 0.22

H1

H1G1 H1I1 H11 H12

I/3.2 I/3.2 I/4 I/4

1.125I

0.28 0.28 0.22 0.22

I1

I1H1 I1J1 I11 I12

I/3.2 I/3.2 I/4 I/4

1.125I

0.28 0.28 0.22 0.22

J1

J1I1 J1K1 J11 J12

I/3.2 I/3.2 I/4 I/4

1.125I

0.28 0.28 0.22 0.22

K1J1 I/3.2 0.28

K1 K1L1 K11 K12

I/3.2 I/4 I/4

1.125I 0.28 0.22 0.22

L1

L1K1 L1M1 L11 L12

I/3.2 I/3.2 I/4 I/4

1.125I

0.28 0.28 0.22 0.22

M1

M1L1 M1N1 M11 M12

I/3.2 I/3.2 I/4 I/4

1.125I

0.28 0.28 0.22 0.22

N1

N1M1 N1O1 N11 N12

I/3.2 I/3.2 I/4 I/4

1.125I

0.28 0.28 0.22 0.22

O1

O1N1 O1P1 O11 O12

I/3.2 I/3.2 I/4 I/4

1.125I

0.28 0.28 0.22 0.22

P1

P1O1 P1Q1 P11 P12

I/3.2 I/3.2 I/4 I/4

1.125I

0.28 0.28 0.22 0.22

Q1

Q1P1 Q1R1 Q11 Q12

I/3.2 I/3.2 I/4 I/4

1.125I

0.28 0.28 0.22 0.22

R1

R1Q1 R1S1 R11 R12

I/3.2 I/3.2 I/4 I/4

1.125I

0.28 0.28 0.22 0.22

S1

S1R1 S1T1 S11 S12

I/3.2 I/6.4 I/4 I/4

0.9688I

0.32 0.16 0.26 0.26

T1

T1S1 T1U1 T11 T12

I/6.4 I/8.16 I/4 I/4

0.7788I

0.2 0.16 0.32 0.32

U1

U1T1 U11 U12

I/8.16 I/4 I/4

0.6225I

0.2 0.4 0.4


No Beam name

Dead load kN/m2

Live load kN/m2

Area m2

Length m

slab DL kN/m

Total LL kN/m

Rib DL kN/m

Total DL kN/m

Total load kN/m

1 A1B1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575

2 B1C1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575

3 C1D1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575

4 D1E1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575

5 E1F1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575

6 F1G1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575

7 G1H1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575

8 H1I1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575

9 I1J1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575

10 J1K1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575

11 K1L1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575

12 L1M1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575

13 M1N1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575

14 N1O1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575

15 O1P1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575

16 P1Q1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575

17 Q1R1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575

18 R1S1 8.25 15 5.12 3.2 13.2 24 3.375 16.575 40.575

19 S1T1 8.25 15 16.2 6.4 20.88 37.97 4.5 25.38 63.35

20 T1U1 8.25 15 13.74 8.16 13.89 25.26 4.5 18.39 43.65

JOINT-A1

-23.245Net BM

due to A1

-37.492

14.247

Distributio

n

-2.868Distributio

n +

Carryover

34.624-34.624FEM due

to LL

-12.78FEM due

to DL

0.280.280.280.280.280.280.38DF

D1E1D1C1C1D1C1B1B1C1B1A1A1B1Member

D1C1B1A1joint

JOINT-B1

-38.53340.165Net BM due

to B1

-37.494

-1.0388

41.204

-1.0388

Distribution

-2.876.58Distribution

+ Carryover

34.624-34.62434.624-34.624FEM due to

LL

-14.14FEM due to

DL

0.280.280.280.280.280.280.38DF


D1C1B1A1joint

JOINT-C1

-37.4937.49Net BM

due to C1

-37.49

0

37.49

0

Distributio

n


n +

Carryover

+34.62-34.62+34.62-34.62FEM due

to LL

-14.1414.14FEM due

to DL

0.280.280.280.280.280.280.280.28DF

E1F1E1D1D1E1D1C1C1D1C1B1B1C1B1A1Member

E1D1C1B1

joint

JOINT-D1

-37.49237.492Net BM

due to D1

-37.492

0

37.492

0

Distributi

on


on +

Carryove

r

+34.62

4

-34.624+34.62

4

-34.624FEM due

to LL

-58.8114.14FEM due

to DL

0.280.280.280.280.280.280.280.28DF

E1F1E1D1D1E1D1C1C1D1C1B1B1C1B1A1Member

E1D1C1B1

joint

JOINT-E1

-37.49237.492Net BM

due to E1

-37.492

0

37.492

0

Distributio

n


n +

Carryover

+34.624-34.624+34.62

4

-

34.624

FEM due

to LL

-14.1414.1

4

FEM due

to DL

0.280.280.280.280.280.280.280.28DF

G1H1G1F1F1G1F1E1E1F1E1D1D1E1D1C1Member

G1F1E1

D1joint

JOINT-F1

-37.49237.492Net BM

due to F1

-37.492

0

37.492

0

Distributi

on


on +

Carryove

r

+34.62

4

-34.624+34.62

4

-34.624FEM due

to LL

-14.1414.14FEM due

to DL

0.280.280.280.280.280.280.280.28DF

G1H1G1F1F1G1F1E1E1F1E1D1D1E1D1C1Member

G1F1E1

D1joint

JOINT-G1

-37.49237.492Net BM

due toG1

-37.492

0

37.492

0

Distributi

on


on +

Carryove

r

+34.624-34.624+34.624-

34.624

FEM due

to LL

-14.1414.14FEM due

to DL

0.280.280.280.280.280.280.280.28DF

I1J1I1H1H1I1H1G1G1H1G1F1F1G1F1E1Member

I1H1G1F1

joint

JOINT-H1

-37.49237.492Net BM

due to H1

-37.492

0

37.492

0

Distributi

on


on +

Carryove

r

+34.62

4

-34.624+34.62

4

-34.624FEM due

to LL

-14.1414.14FEM due

to DL

0.280.280.280.280.280.280.280.28DF

I1J1I1H1H1I1H1G1G1H1G1F1F1G1F1E1Member

I1H1G1F1

joint

JOINT-I1

-37.49237.492Net BM

due to I1

-37.492

0

37.492

0

Distributio

n


n+

Carryover

34.624-34.62434.624-34.624FEM due

to LL

-14.1414.14FEM due

to DL

0.280.280.280.280.280.280.280.28DF

K1L1K1J1J1K1J1I1I1J1I1H1H1I1H1G1Member

K1J1I1H1

joint

JOINT-J1

-37.49237.492Net BM

due to J1

-37.492

0

37.492

0

Distributi

on


on +

Carryove

r

+34.62

4

-34.624+34.62

4

-34.624FEM due

to LL

-14.1414.14FEM due

to DL

0.280.280.280.280.280.280.280.28DF

K1L1K1J1J1K1J1I1I1J1I1H1H1I1H1G1Member

K1J1I1H1

joint

JOINT-K1

-37.49237.492Net BM

due to K1

-37.492

0

37.492

0

Distributio

n


n +

Carryover

34.624-34.62434.624-34.624FEM due

to LL

-14.1414.14FEM due

to DL

0.280.280.280.280.280.280.280.28DF

M1N1

M1L1L1M1L1K1K1L1K1J1J1K1J1I1Member

M1L1K1

J1joint

JOINT-L1

-37.49237.492Net BM

due to L1

-37.492

0

37.492

0

Distributio

n


n +

Carryover

34.624-34.62434.624-34.624FEM due

to LL

-14.1414.14FEM due

to DL

0.280.280.280.280.280.280.280.28DF

M1N1M1L1L1M1L1K1K1L1K1J1J1K1J1I1Member

M1L1K1

J1joint

JOINT-M1

-37.49237.492Net BM

due to M1

-37.492

0

37.492

0

Distributi

on


on +

Carryove

r

34.624-34.62434.624-34.624FEM due

to LL

-14.1414.14FEM due

to DL

0.280.280.280.280.280.280.280.28DF

O1P1O1N1N1O1N1M1M1N1M1L1L1M1L1K1Member

O1N1M1

L1joint

JOINT-N1

-37.49237.492Net BM

due to N1

-37.492

0

37.492

0

Distributi

on


on +

Carryove

r

34.624-34.62434.624-34.624FEM due

to LL

-14.1414.14FEM due

to DL

0.280.280.280.280.280.280.280.28DF

O1P1O1N1N1O1N1M1M1N1M1L1L1M1L1K1Member

O1N1M1

L1joint

JOINT-O1

-37.49237.492Net BM

due to O1

-37.492

0

37.492

0

Distributi

on


on +

Carryove

r

34.624-34.62434.624-34.624FEM due

to LL

-14.1414.14FEM due

to DL

0.280.280.280.280.280.280.280.28DF

Q1R1Q1P1P1Q1P1O1O1P1O1N1N1O1N1M1Member

Q1P1O1

N1joint

JOINT-P1

-37.49237.492Net BM

due to P1

-37.492

0

37.492

0

Distributi

on


on +

Carryove

r

34.624-34.62434.624-34.624FEM due

to LL

-14.1414.14FEM due

to DL

0.280.280.280.280.280.280.280.28DF

Q1R1Q1P1P1Q1P1O1O1P1O1N1N1O1N1M1Member

Q1P1O1

N1joint

JOINT-Q1

-37.49237.492Net BM

due to Q1

-37.492

0

37.492

0

Distributi

on


on +

Carryove

r

34.624-34.62434.624-34.624FEM due

to LL

-14.1414.14FEM due

to DL

0.160.320.280.280.280.280.280.28DF

S1T1S1R1R1S1R1Q1Q1R1Q1P1P1Q1P1O1Member

S1R1Q1

P1joint

JOINT-R1

-29.43434.362Net BM

due to R1

-26.304

-3.13

37.492

-3.13

Distributi

on

8.322.868Distributi

on +

Carryove

r

34.624-34.62434.624-34.624FEM due

to LL

-86.6314.14FEM due

to DL

0.160.320.280.280.280.280.280.28DF

S1T1S1R1R1S1R1Q1Q1R1Q1P1P1Q1P1O1Member

S1R1Q1

P1joint

JOINT-S1

T1S1 U1

-197.2298.342Net BM

due to S1

-227.65

30.43

37.492

60.85

Distributi

on


on +

Carryove

r

216.23-216.2334.624-34.624FEM due

to LL

-102.0414.14FEM due

to DL

0.20.160.20.160.320.280.28DF

U1T1T1U1T1S1S1T1S1T1R1S1R1Q1Member

R1joint

JOINT-T1

T1S1 U1

-260.98239.207Net BM

due to T1

-266.42

5.44

232.397

6.81

Distributio

n


n +

Carryover

242.2-242.2216.23-216.23FEM due

to LL

14.14FEM due

to DL

0.20.160.20.160.320.280.28DF


R1joint

JOINT-U1

T1S1 U1

203.72Net BM

due to U1

254.65

-50.93

Distributio

n

12.45Distributio

n +

Carryover

242.2-242.2FEM due

to LL

86.63FEM due

to DL

0.20.160.20.160.320.280.28DF


R1joint

MID SPAN MOMENT A1B1

+32.814-23.24Net moment


2.876.58-2.87carry over

moment


moment

34.624-34.62434.624-34.624FEM due to

TL

14.14-14.14FEM due to

DL

0.280.280.280.280.280.280.38DF


D1C1B1A1joint

F

ree BM at the centre of the span A1B1

= Wl2/8

=(40.575 * 3.22) / 8

=51.936 kNm

Net BM at centre of span A1B1

=51.936-[ (23.24+32.814) / 2]

=23.909 kNm

MID SPAN MOMENT OF B1C1



2.87-2.872.872.685carry over

moment

5.74-5.745.745.37distribution

moment

-34.62434.624-34.624FEM due to

TL

14.14-14.1414.14-14.14FEM due to

DL

0.280.280.280.280.280.280.38DF


D1C1B1A1joint

Free BM at the centre of the span B1C1

= Wl2/8

=(40.575 * 3.22) / 8

=51.936 kNm

Net BM at centre of span B1C1

=51.936-[ (31.702+30.14) / 2]

=21.015 kNm

MID SPAN MOMENT C1D1

33.35-33.35Net

moment

1.6-1.6distributio

n

2.872.87-2.87-2.87carry

over

moment

5.74-5.745.74-5.74distributio

n

moment

-34.62434.624-34.62434.62

4

FEM due

to TL

14.14-14.1414.14-14.14FEM due

to DL

0.280.280.280.280.280.280.280.28DF

E1F1E1D1D1E1D1C1C1D1C1B1B1C1B1A1member

E1D1C1B1

Joint

Free BM at the centre of the span C1D1

= Wl2/8

=(40.575 * 3.22) / 8

=51.936 kNm

Net BM at centre of span C1D1

=51.936-[ (33.35+33.35) / 2]

=18.586 kNm

MID SPAN MOMENT D1E1

33.35-33.35Net

moment

1.6-1.6distribution

2.872.87-2.87-2.87carry over

moment

5.74-5.745.74-5.74distribution

moment

-34.62434.624-34.62434.624FEM due

to TL

14.14-14.1414.14-14.14FEM due

to DL

0.280.280.280.280.280.280.280.28DF

F1G1F1E1E1F1E1D1D1E1D1C1C1D1C1B1member

F1E1D1C1

Joint

Free BM at the centre of the span D1E1

= Wl2/8

=(40.575 * 3.22) / 8

=51.936 kNm

Net BM at centre of span D1E1

=51.936-[ (33.35+33.35) / 2]

=18.586 kNm

MID SPAN MOMENT E1F1

33.35-33.35Net

moment

1.6-1.6distribution

2.872.87-2.87-2.87carry over

moment


moment

-34.62434.624-34.62434.624FEM due

to TL

14.14-14.1414.14-14.14FEM due

to DL

0.280.280.280.280.280.280.280.28DF

G1H1G1F1F1G1F1E1E1F1E1D1D1E1D1C1member

G1F1E1

D1Joint

Free BM at the centre of the span E1F1

= Wl2/8

=(40.575 * 3.22) / 8

=51.936 kNm

Net BM at centre of span E1F1

=51.936-[ (33.35+33.35) / 2]

=18.586 kNm

MID SPAN MOMENT F1G1

33.35-33.35Net

moment

1.6-1.6distribution

2.872.87-2.87-2.87carry over

moment


moment

-34.62434.624-34.62434.624FEM due

to TL

14.14-14.1414.14-14.14FEM due

to DL

0.280.280.280.280.280.280.280.28DF

H1I1H1G1G1H1G1F1F1G1F1E1E1F1E1D1member

H1G1F1E1

Joint

F

ree BM at the centre of the span F1G1

= Wl2/8

=(40.575 * 3.22) / 8

=51.936 kNm

Net BM at centre of span F1G1

=51.936-[ (33.35+33.35) / 2]

=18.586 kNm

MID SPAN MOMENT G1H1

33.35-33.35Net

moment

1.6-1.6distribution

2.872.87-2.87-2.87carry over

moment


moment

-34.62434.624-34.62434.624FEM due

to TL

14.14-14.1414.14-14.14FEM due

to DL

0.280.280.280.280.280.280.280.28DF

I1J1I1H1H1I1H1G1G1H1G1F1F1G1F1E1member

I1H1G1F1

Joint

Free BM at the centre of the span G1H1

= Wl2/8

=(40.575 * 3.22) / 8

=51.936 kNm

Net BM at centre of span G1H1

=51.936-[ (33.35+33.35) / 2]

=18.586 kNm

MID SPAN MOMENT H1I1

33.35-33.35Net

moment

1.6-1.6distribution

2.872.87-2.87-2.87carry over

moment


moment

-34.62434.624-34.62434.624FEM due

to TL

14.14-14.1414.14-14.14FEM due

to DL

0.280.280.280.280.280.280.280.28DF

J1K1J1I1I1J1I1H1H1I1H1G1G1H1G1F1member

J1I1H1G1Joint

F

ree BM at the centre of the span H1I1

= Wl2/8

=(40.575 * 3.22) / 8

=51.936 kNm

Net BM at centre of span H1I1

=51.936-[ (33.35+33.35) / 2]

=18.586 kNm

MID SPAN MOMENT I1J1

33.35-33.35Net

moment

1.6-1.6distribution

2.872.87-2.87-2.87carry over

moment


moment

-34.62434.624-34.62434.624FEM due

to TL

14.14-14.1414.14-14.14FEM due

to DL

0.280.280.280.280.280.280.280.28DF

K1L1K1J1J1K1J1I1I1J1I1H1H1I1H1G1member

K1J1I1H1

Joint

Free BM at the centre of the span I1J1

= Wl2/8

=(40.575 * 3.22) / 8

=51.936 kNm

Net BM at centre of span I1J1

=51.936-[ (33.35+33.35) / 2]

=18.586 kNm

MID SPAN MOMENT J1K1

33.35-33.35Net

moment

1.6-1.6distribution

2.872.87-2.87-2.87carry over

moment


moment

-34.62434.624-34.62434.624FEM due

to TL

14.14-14.1414.14-14.14FEM due

to DL

0.280.280.280.280.280.280.280.28DF

L1M1L1K1K1L1K1J1J1K1J1I1I1J1I1H1member

L1K1J1I1

Joint

Free BM at the centre of the span J1K1

= Wl2/8

=(40.575 * 3.22) / 8

=51.936 kNm

Net BM at centre of span J1K1

=51.936-[ (33.35+33.35) / 2]

=18.586 kNm

MID SPAN MOMENT K1L1

33.35-33.35Net

moment

1.6-1.6distribution

2.872.87-2.87-2.87carry over

moment


moment

-34.62434.624-34.62434.624FEM due

to TL

14.14-14.1414.14-14.14FEM due

to DL

0.280.280.280.280.280.280.280.28DF

M1N1M1L1L1M1L1K1K1L1K1J1J1K1J1I1member

M1L1K1

J1Joint

F

ree BM at the centre of the span K1L1

= Wl2/8

=(40.575 * 3.22) / 8

=51.936 kNm

Net BM at centre of span K1L1

=51.936-[ (33.35+33.35) / 2]

=18.586 kNm

MID SPAN MOMENT L1M1

33.35-33.35Net

moment

1.6-1.6distribution

2.872.87-2.87-2.87carry over

moment


moment

-34.62434.624-34.62434.624FEM due

to TL

14.14-14.1414.14-14.14FEM due

to DL

0.280.280.280.280.280.280.280.28DF

N1O1N1M1M1N1M1L1L1M1L1K1K1L1K1J1member

N1M1L1K1

Joint

F

ree BM at the centre of the span L1M1

= Wl2/8

=(40.575 * 3.22) / 8

=51.936 kNm

Net BM at centre of span L1M1

=51.936-[ (33.35+33.35) / 2]

=18.586 kNm

MID SPAN MOMENT M1N1

33.35-33.35Net

moment

1.6-1.6distribution

2.872.87-2.87-2.87carry over

moment


moment

-34.62434.624-34.62434.624FEM due

to TL

14.14-14.1414.14-14.14FEM due

to DL

0.280.280.280.280.280.280.280.28DF

O1P1O1N1N1O1N1M1M1N1M1L1L1M1L1K1member

O1N1M1

L1Joint

F

ree BM at the centre of the span M1N1

= Wl2/8

=(40.575 * 3.22) / 8

=51.936 kNm

Net BM at centre of span M1N1

=51.936-[ (33.35+33.35) / 2]

=18.586 kNm

MID SPAN MOMENT N1O1

33.35-33.35Net

moment

1.6-1.6distribution

2.872.87-2.87-2.87carry over

moment


moment

-34.62434.624-34.62434.624FEM due

to TL

14.14-14.1414.14-14.14FEM due

to DL

0.280.280.280.280.280.280.280.28DF

P1Q1P1O1O1P1O1N1N1O1N1M1M1N1M1L1member

P1O1N1M1

Joint

F

ree BM at the centre of the span N1O1

= Wl2/8

=(40.575 * 3.22) / 8

=51.936 kNm

Net BM at centre of span N1O1

=51.936-[ (33.35+33.35) / 2]

=18.586 kNm

MID SPAN MOMENT O1P1

33.35-33.35Net

moment

1.6-1.6distribution

2.872.87-2.87-2.87carry over

moment


moment

-34.62434.624-34.62434.624FEM due

to TL

14.14-14.1414.14-14.14FEM due

to DL

0.280.280.280.280.280.280.280.28DF

Q1R1Q1P1P1Q1P1O1O1P1O1N1N1O1N1M1member

Q1P1O1

N1Joint

Free BM at the centre of the span O1P1

= Wl2/8

=(40.575 * 3.22) / 8

=51.936 kNm

Net BM at centre of span O1P1

=51.936-[ (33.35+33.35) / 2]

=18.586 kNm

MID SPAN MOMENT P1Q1

33.35-33.35Net

moment

1.6-1.6distribution

2.872.87-2.87-2.87carry over

moment


moment

-34.62434.624-34.62434.624FEM due

to TL

14.14-14.1414.14-14.14FEM due

to DL

0.280.280.280.280.280.280.280.28DF

R1S1R1Q1Q1R1Q1P1P1Q1P1O1O1P1O1N1member

R1Q1P1O1

Joint

F

ree BM at the centre of the span P1Q1

= Wl2/8

=(40.575 * 3.22) / 8

=51.936 kNm

Net BM at centre of span P1Q1

=51.936-[ (33.35+33.35) / 2]

=18.586 kNm

MID SPAN MOMENT Q1R1

37.084-33.35Net

moment

5.33-1.6distribution

16.162.87-2.87-2.87carry over

moment

32.33-5.745.74-5.74distribution

moment

-216.2334.624-34.62434.624FEM due

to TL

14.14-14.1414.14-14.14FEM due

to DL

0.160.320.280.280.280.280.280.28DF

S1T1S1T1R1S1R1Q1Q1R1Q1P1P1Q1P1O1member

S1R1Q1

P1Joint

F

ree BM at the centre of the span Q1R1

= Wl2/8

=(40.575 * 3.22) / 8

=51.936 kNm

Net BM at centre of span Q1R1

=51.936-[ (33.35+37.084) / 2]

=16.719 kNm

MID SPAN MOMENT R1S1

60.03-19.038Net

moment

5.8961.526distribution

15.552.878.32-2.87carry over

moment

31.1116.645.74-5.74distribution

moment

-242.234.624-34.62434.624FEM due

to TL

86.63-86.6314.14-14.14FEM due

to DL

0.160.20.160.320.280.280.280.28DF

T1U1T1S1S1T1S1T1R1S1R1Q1Q1R1Q1P1member

T1S1R1Q1

Joint

Free BM at the centre of the span R1S1

= Wl2/8

=(40.575 * 3.22) / 8

=51.936 kNm

Net BM at centre of span R1S1

=51.936-[ (19.038+60.03) / 2]

=12.402 kNm

MID SPAN MOMENT S1T1

U1

210.75-197.61Net moment


16.165-11.42-2.87carry over

moment


moment

216.23-216.2334.624FEM due to

TL

102.04-102.0414.14-14.14FEM due to

DL

0.20.160.20.160.320.280.28DF

U1T1T1U1T1S1S1T1S1R1R1S1R1Q1Member

T1S1R1joint

F

ree BM at the centre of the span S1T1

= Wl2/8

=(216.23 * 3.22) / 8

=276.77 kNm

Net BM at centre of span S1T1

=276.77-[ (197.61+210.75) / 2]

=72.594kNm

MID SPAN MOMENT T1U1

U1


-0.6880.48Distribution

-3.44-4.377.37carry over

moment


moment

43.65-43.6540.575FEM due to

TL

86.63-86.63FEM due to

DL

0.20.160.20.160.320.280.28DF

U1T1T1U1T1S1S1T1S1R1R1S1R1Q1Member

T1S1R1joint

F

ree BM at the centre of the span T1U1

= Wl2/8

=(43.65 * 8.162) / 8

=363.3kNm

Net BM at centre of span T1U1

=363.3-[ (54.42+30.79) / 2]

=320.69kNm

MID SPAN MOMENT K1L1

18.62-18.62Net

moment


-2.87-2.872.872.87carry over

moment

-5.745.745.74-5.74-5.745.74distribution

moment

34.62

4

-34.62434.624-34.624FEM due

to TL

-14.1414.14-14.1414.14FEM due

to DL

0.280.280.280.280.280.280.280.28DF

M1N1M1L1L1M1L1K1K1L1K1J1J1K1J1I1member

M1L1K1

J1Joint

Free BM at the centre of the span K1L1

= Wl2/8

=(40.575 * 3.22) / 8

=51.936 kNm

Net BM at centre of span K1L1

=51.936-[ (18.62+18.62) / 2]

=33.316 kNm


joint A1 B1 C1 D1 E1 F1 G1

Column DF

Above floor

Below floor

0.31

0.31

0.22

0.22

0.22

0.22

0.22

0.22

0.22

0.22

0.22

0.22

0.22

0.22

Member

DF

A1 B1

0.39

B1A1

0.28

B1 C1

0.28

C1 B1

0.28

C1 D1

0.28

D1 C1

0.28

D1 E1

0.28

E1D1

0.28

E1 F1

0.28

F1E1

0.28

F1 G1

0.28

G1F1

0.28

G1H1

0.28

FEM due to

DL

-14.2 14.2 -14.2 14.2 -14.2 14.2

FEM due TL -34.62 34.62 -34.62 34.62 -34.62 34.62 -34.62

Distribution+

carryover

-2.87 6.75 2.87 -2.87 -2.87 2.87 2.87 -2.87 -2.87 2.87 2.87 -2.87 -2.87

Total -37.49 41.37 -11.3 11.3 -37.49 37.49 -11.3 11.3 -37.49 37.49 -11.3 11.3 -37.49


11.62

-6.62

5.77

-5.77

5.77

-5.77

5.77


11.62

-6.62

5.77

-5.77

5.77

-5.77

5.77

joint H1 I1 J1 K1 L1 M1 N1

Column DF

Above floor

Below floor

0.22

0.22

0.22

0.22

0.22

0.22

0.22

0.22

0.22

0.22

0.22

0.22

0.22

0.22

Member

DF

H1 G1

0.28

H1 I1

0.28

I1 H1

0.28

I1 J1

0.28

J1 I1

0.28

J1 K1

0.28

K1 J1

0.28

K1 E1

0.28

L1 K1

0.28

L1 M1

0.28

M1 L1

0.28

M1 N1

0.28

N1M1

0.28

N1O1

0.28

FEM due to

DL

-14.2 14.2 -14.2 14.2 -14.2 14.2 -14.2

FEM due

TL

34.62 -

34.62

34.62 -

34.62

34.62 -

34.62

34.62

Distribution

+ carryover

2.87 2.87 -2.87 -2.87 2.87 2.87 -2.87 -2.87 2.87 2.87 -2.87 -2.87 2.87 2.87

Total 37.5 -11.3 11.3 -37.5 37.5 -11.3 11.3 -37.5 37.5 -11.3 11.3 -37.5 37.5 -11.3


-5.77

5.77

-5.77

5.77

-5.77

5.77

-5.77


-5.77

5.77

-5.77

5.77

-5.77

5.77

-5.77

joint O1 P1 Q1 R1 S1 T1 U1

Column DF

Above floor

Below floor

0.33

0.33

0.24

0.24

0.27

0.27

0.31

0.31

0.27

0.27

0.24

0.24

0.33

0.33

Member

DF

O1N1

0.28

O1P1

0.28

P1O1

0.28

P1Q1

0.28

Q1P1

0.28

Q1R1

0.28

R1Q1

0.28

R1S1

0.28

S1R1

0.32

S1T1

0.16

T1S1

0.2

T1U1

0.16

U1T1

0.2

FEM due to

DL

14.2 -14.2 14.2 -14.2 14.2 -102.1 102.1

FEM due TL -34.62 34.62 -34.62 34.62 -216.2 216.2

Distribution+

carryover

-2.87 -2.87 2.87 2.87 -2.87 -2.87 2.87 32.33 -2.87 -11.42 16.17 -10.2 -9.14

Total 11.3 -37.5 37.5 -11.3 11.3 -37.5 37.5 -11.3 11.3 227.65 232.4 -112.2 92.9


5.77

-5.77

5.77

-12.25

-62.12

-38.45

-37.2


5.77

-5.77

5.77

-12.25

-62.12

-38.45

-37.2

joint A1 B1 C1 D1 E1 F1 G1

Column DF

Above floor

Below floor

0.31

0.31

0.22

0.22

0.22

0.22

0.22

0.22

0.22

0.22

0.22

0.22

0.22

0.22

Member

DF

A1 B1

0.39

B1A1

0.28

B1 C1

0.28

C1 B1

0.28

C1 D1

0.28

D1 C1

0.28

D1 E1

0.28

E1D1

0.28

E1 F1

0.28

F1E1

0.28

F1 G1

0.28

G1F1

0.28

G1H1

0.28

FEM due to

DL

-14.2 14.2 -14.2 14.2 -14.2 14.2 -14.2

FEM due TL -34.6 34.62 -34.62 34.62 -34.62 34.62 -34.62

Distribution+

carryover

2.87 2.76 -2.87 2.87 2.87 -2.87 -2.87 2.87 2.87 -2.87 -2.87 2.87 2.87

Total 11.3 16.9 -37.5 37.49 -11.3 11.3 -37.49 37.49 -11.3 11.3 -37.49 37.49 -11.3


-3.49

4.53

-5.77

5.77

-5.77

5.77

-5.77


-3.49

4.53

-5.77

5.77

-5.77

5.77

-5.77

joint H1 I1 J1 K1 L1 M1 N1

Column DF

Above floor

Below floor

0.22

0.22

0.22

0.22

0.22

0.22

0.22

0.22

0.22

0.22

0.22

0.22

0.22

0.22

Member

DF

H1 G1

0.28

H1 I1

0.28

I1 H1

0.28

I1 J1

0.28

J1 I1

0.28

J1 K1

0.28

K1 J1

0.28

K1 E1

0.28

L1 K1

0.28

L1 M1

0.28

M1 L1

0.28

M1 N1

0.28

N1M1

0.28

N1O1

0.28

FEM due to

DL

14.2 -14.2 14.2 -14.2 14.2 -14.2 14.2

FEM due

TL

-

34.62

34.62 -

34.62

34.62 -

34.62

34.62 -34.62

Distribution

+ carryover

-2.87 -2.87 2.87 2.87 -2.87 -2.87 2.87 2.87 -2.87 -2.87 2.87 2.87 -2.87 -2.87

Total 11.3 -37.5 37.5 -11.3 11.3 -37.5 37.5 -11.3 11.3 -37.5 37.5 -11.3 11.3 -37.5


5.77

-5.77

5.77

-5.77

5.77

-5.77

5.77


5.77

-5.77

5.77

-5.77

5.77

-5.77

5.77

joint O1 P1 Q1 R1 S1 T1 U1

Column DF

Above floor

Below floor

0.33

0.33

0.24

0.24

0.27

0.27

0.31

0.31

0.27

0.27

0.24

0.24

0.33

0.33

Member

DF

O1N1

0.28

O1P1

0.28

P1O1

0.28

P1Q1

0.28

Q1P1

0.28

Q1R1

0.28

R1Q1

0.28

R1S1

0.28

S1R1

0.32

S1T1

0.16

T1S1

0.2

T1U1

0.16

U1T1

0.2

FEM due to

DL

-14.2 14.2 -14.2 14.2 -86.63 86.63

FEM due TL 34.62 -34.6 34.62 -34.62 34.62 -242.2 242.2

Distribution+

carryover

2.87 2.87 -2.87 -2.87 2.87 2.87 -2.87 8.32 2.87 15.56 4.16 -24.22 12.45

Total 37.5 -11.3 11.3 -37.5 37.5 -11.3 11.3 -26.31 37.5 -71.1 90.79 -266.4 254.7


-5.77

5.77

-5.77

3.31

8.73

56.2

-102


-5.77

5.77

-5.77

3.31

8.73

56.2

-102

CHAPTER 5

5.DESIG�

5.1SLABS

The most common type of structural element used to cover floors and

roofs of building are reinforced concrete slabs of different types. One

way slabs are those supported on the two opposite sides so that the

loads are carried along one direction only. Two way slabs are

supported on all four sides with such dimensions such that the loads

are carried to the supports along both directions.

If Ly/Lx < 2, then the slab is designed as two way slab

If Ly/Lx >2, then the slab is designed as one way slab.

Where, Ly = longer span dimension of the slab.

Lx = shorter span dimension of slab.

Restrained slabs are referred to as slabs whose corners are prevented from

lifting. They may be supported on continuous or discontinuous edges.

5.1.1 DESIG� OF SLAB

Dimensions

Lx =3.2

Ly =5.5

Span ratio =5.5 /3.2

=1.1<2

ie, Two way slab

Assume ,

Overall depth =150-20

D =130mm

Load calculation

Live load =10kN/m2

Light patrician =1kN/m2

D.L =(0.15*25)

=3.75kN/m2

Floor finish =0.75kN/m2

Total load =15.5kN/m2

Factored load =1.5*15.5

= 23.25kN/m2

1) Lx =3.2+(0.23/2)+(0.23/2)

=3.43

2) Ly =3.55+(0.23/2)+(0.23/2)

=3.78

Lx =3.33

Ly =3.78

Bending moment

Mx =αx wlx2

My =αy wlx2

Lx / Ly =1.1

Lx =0.074

Ly =0.061

Mx =0.074*23.25*3.432

=20.24kNm

My =0.061*23.25*3.432

=16.69kNm

Check for depth

Mulim =0.138fckbd2

20.24*103 =0.138*20*1000*d2

D =85.63

D < 130mm

Hence the effective depth selected is sufficient to resist the design

ultimate moment.

Find the spacing

Use 10mm dia bars

Spacing =(1000*π*102/4)/465.86

=168.59mm

Provide 150 mm c/c spacing

Find Ast provided

Ast provided =1000ast/spacing

=(1000*π*102/4)/150*4

=523.59mm2

For longer span

16069*106 =0.87*415* Ast *130[1-( Ast *415/20*1000*130)]

Ast =378mm2

Using 10mm dia bars

Spacing =(1000*π/4*102)/378

=207mm.

Adopt the spacing 200mm

Checks,

Ast min =0.12% of c/s of Fe 415

=0.12/100[1000*150]

=180mm2

Ast min< Ast for longer and shorter span,

Hence, provide 10mm dia bars @150mmc/c

Ast =465.86mm2[shorter span]

Hence, provide 10mm dia bars @ 200mm c/c

Ast =378mm2[longer span]

Check for shear

Shear stress

τv =Vu/bd

Vu =wl/2

=23.25*3.43/2=39.87Kn

τv =39.87*103/1000*100

τv =0.398N/mm2

Pt =100 Ast pro / bd

=100*523.59/1000*130

To find τc

From IS456:2000,

τc = 0.432

kτc > τv

k*τc=1.3*0.43

=0.5616

0.398 > 0.5616N/mm2

Hence it is safe.

Check for deflection

(L/d) basic =20

Pt =100 Ast pro/bd

=100*523.59/1000*130

=0.4

Fs =0.58*415*523.59/465.86

=270

kc =1

kf =1

kt =1.2

(L/d)max =(L/d)basic*kt*kc*kt

=20*1.3*1*1

=26

(L/d)act =3200/130=24.16

(L/d)act < (L/d)max

Hence safe against deflection.

Check for control

Reinforcement provided is more than, the minimum % of c/s area

Ast =(0.12/100)*1000*150

=180mm2

Spacing of main reinforcement should not be greater than 3d

ie, 3*130 =390mm

Diameter of reinforcement should be less than D/8

150/8 =18.75

Hence cracks will be with in safe permissible limits

Torsion reinforcement at corner

Area of torsion steel at each of the corners in 4 layers is computed as

=0.75* Ast along shorter span

=0.75*523.59

=393mm2

Length cover which torsion steel is provided

=1/5*shorter span

=1/5*3200

=640mm

Using 6mm dia bars

Spacing =1000ast/ Ast

=(1000*π*62/4)393

=71.9mm

Provide 6mm bars at 100mm c/c for length and 640mmat all corners in

4 layers

Reinforcement in end strips

Ast =0.12% of c/s

=180mm2

Assume 10mm dia bars

Spacing =(1000*π/4*102)/180

=436 > 300

As per code spacing should not exceed 300mm

Provide 10mm dia bars at 300mm c/c

Ast =(1000*π/4*102)/300

Ast =262mm2

10mm dia bars @ 150mm c/c


150

Figure 5-Reinforcement details of two way slab-section

REINFORCEMENT DETAILS OF TWO WAY SLAB

6mm dia @ 100mm c/c



400

3500

2400

2625

437.5

437.5

400

ly/8

ly/8

lx/8 lx/8

3ly/4

3lx/4

3200

Figure 6-Reinforcement details of two way slab- plan

5.2 BEAMS

Beams are defined as structural members subjected to transverse load

that caused bending moment and shear force along the length.

The plane of transverse loads is parallel to the plane of symmetry of the

cross section of the beam and it passes through the shear centre so that the

simple bending of beams occurs. The bending moments and shear forces

produced by the transverse loads are called as internal forces.

5.2.1Types of beams

Depending upon the supports and end condition, beams are classified as

below.

5.2.1.1simply supported beams

5.2.1.2over hanging beams

5.2.1.3cantilever beam

5.2.1.4fixed beam

The reinforced concrete beams, in which the steel reinforced is placed

only on tension side, are known as singly reinforced beams, the tension

developed due to bending moment is mainly resisted by steel reinforcement

and compression by concrete.

When a singly reinforced beam needs considerable depth to exist large

bending moment, then the beam is also reinforced in the compression zone.

The beams having reinforcement in compression and tension zone is called

as doubly reinforced beam.

5.2.2 DESIG� OF L-BEAMS

Dimensions

c/c of support = 3.2+(0.3/2)+(0.3/2) = 3.5m

Thickness of slabs = 150 mm

fy = 20 N/mm2

fck = 415 N/mm2

Width of beam = 300 mm

Overall depth = 300 mm

Effective cover = 25 mm

Effective depth = 300-25-10=265mm

Effective span

a) c/c of supports = 3.2 +(0.3/2) +(0.32/2) = 3.5 m

b) Clear span + d = 3.2 +0.265 = 3.465m

Hence, l = 3.465 m

Load calculation

Dead load of slab = (3.465/2)*0.15*25 = 6.5 kN/m

Floor finish = 0.75*(3.465/2) = 1.3 kN/m

Self weight of rib = 0.3 *0.15 *25 = 1.125 kN/m

Live load = 4*(3.465/2) = 6.93 kN/m

Total load = 16.855 kN/m

Effective flange width

a)bf = (Lo/12)+bw+3Df = 952.125 mm

b) bf = bw +0.5 times spacing b/w ribs = 1900 mm

Ultimate BM and SF

At support,

Mu = 1.5 * wl2/12 = 25.3 kNm

Vu = 1.5 * wl / 2 = 43.8 kN

At centre of span section,

Mu = 1.5 * wl2 / 24 = 12.65kNm

Vu = 1.5 * wl / 2 = 43.8 kN

Torsion moment produced due to dead load of span and live load on it

= working load parameter-rib wt = 16.855-1.125=15.73 kN/m

Ultimate load on slab = 15.73 *3.465 * 1.5 = 81.8 kN

Total ultimate load = 82/2 =41 kN

Distance of centroid of SF from the centre line of the Beam=(952.125/2)-

150

= 326.06mm

Ultimate tortional moment = 4 * 103 *326.06= 13.37 kNm

Equivalent BM and SF

According to IS456 2000 clause 41.4.2

Mel = Mu +Mt

Mt = Tu*(1+D/b)/1.7 = 15.73 kNm

Mel = 13.37 + 15.73 = 29.09 kNm

Equivalent SF

Ve = Vu + 1.6(Tu/b)

=115.1 kN

Main reinforcement

Mu (lim) = 0.138*fck*bd2 = 58.15 kNm

Mel < Mu (lim)

Hence the section is under reinforced

To find Ast

Mu = 0.87*fy* Ast *d[1-( Ast *fy/bd*fck)]

Ast = 332.83mm2

20mm dia rods are used

Ast pro = 628.32 mm2

Ast min = 0.85*bw*d/ fy = 162.83 mm2

Assume 20mm dia bars, Ast pro = 628 mm2

Provide 2 nos of 20mm dia bars @ side face

Reinforcement

Shear reinforcement

τve =Ve/ bw *d = 1.45 N/ mm2

Pt= (100* Ast)/( bw *d) = 0.79

Ref table 19 of IS456 2000

τc=0.56N/ mm2

Hence shear reinforcement are required using 10mm dia 2 legged

stirrups with side cover 25mm top+ bottom cover of 25mm

b1= 300-25-25 = 250mm

d1= 300-25-25 = 250mm

Asv= 157 mm2

σc = Asv *0.87*fy/ (τv – τc)*b = 214.6

Provide 10mm dia 2 legged stirrups @200mm spacing

Check for deflection

(L/d)max = (L/d)basic *kt * kc* kf

(L/d)basic = 20 [for simply supported]

(L/d)max = 20*1*1*1.04 = 20.8

(L/d)actual = 3200/300 = 10.66

(L/d)max > (L/d)actual

Hence the design is safe

LONGITUDINAL SECTION OF L-BEAM

2 nos of 20mm dia bars

3 nos of 20mm dia bars10mm dia 2 legged stirrups @ 200mm c/c

3500

Figure 7-Reinforcement details of L-beams- longitudinal section

CROSS SECTION OF L-BEAM

SUPPORT SECTION

2 bars of 20mm dia

3 bars of 20mm dia

10mm dia 2 legged stirrups @ 200mm c/c

150

MID SECTION

2 bars of 20mm dia

3 bars of 20mm dia


150

Figure 8- Reinforcement details of L-beams- cross section

5.2.3 DESIG� OF T- BEAM

Dimensions

Slab thickness =150mm

c/c of support =3.2+(0.3/2)+(0.3/2)

=3.5m

fy =20N/mm2

fck =415N/mm2

Cross sectional dimension

Width of beam =300mm

Overall depth =300mm

Effective cover =25mm

Effective depth =300-2-10

=265mm

Effective span

1. c/c of support =3.2+(0.3/2)+(0.3/2)

=3.5m

2. clear span+depth =3.2+0.265

=3.465m

Load calculation

Dead load of slab =(3.465/2)*0.15*25

=6.5kN/m

Floor finish =0.75*(3.465/2)

=1.3kN/m

Self weight of rib =0.3*0.15*25

=1.125kN/m

Live load =4*(3.465/2)

=6.93kN/m

Light partition =1kN/m

Total load =16.855kN/m

Ultimate moment and shear

Mu =1.5wl2/8

=(1.5*16.855*3.4652)/8

=37.935kN/m

Vu =wl/2

=(16.855*3.465)/2

=43.8kN/m

Effective width of flange

Refer page no 36 clause 23,

1. bf =L0/b+bw+6Df

=(3.465*0.7)/6+300+(6*150)

=1604.25mm

2. c/c of rib =3000-(300/2)-(300/2)

=2700mm

Ie, bf =1604.25mm

Moment capacity of flange

Assume N.A lies with in the flange

Xu(max)=Df , b=bf

Mu(limit) =(0.36*Xu(max))/d *(1-(0.42Xu(max)/d))*(bd2fck)

=0.36*(150/265) *[1-(0.42*150)/26]*(1604.25*2652*20)

=349.98kNm

Mu < Mu(limit)

Hence the section is under reinforced.

Since the section should design as a singly reinforced beam.

Find Ast

Mu =0.87fy Ast d [1-( Ast fy /bf d fck)]

37.935*106=0.87*415* Ast *265*[1-( Ast *415/1604*265*20)]

Ast =404.46mm2

Check for Ast min

Ast min/bw d =0.85/fy

Ast =(0.85*300*265) /415

=162.83mm2

Ast > Ast min

Ast =404.46mm2

N*πd2/4 =404.46

N =2 nos.

Ast provide,

Provide 2 nos of 20mm dia bars

=2*π202/4

=628mm2

And two longer bars of 12mm dia on the compression face .

Shear reinforcement

τv =Vu/bw d

=43.8*103/300*265

=0.55

Pt =100 Ast /bw d

=(100*π/4*202*2)/300*265

=0.79

τc =0.56+((0.62-0.56)/(1-0.75))*(0.79-0.75)

=0.57

τv<τc

Minimum shear reinforcement in the form of stirrups shall be provided.

Design of shear reinforcement

Asv/b Sv =0.4/0.87*fy

Sv =302.47mm

The spacing should not exceed 300mm

Sv =300mm

Provide 8mm dia bars 2 legged stirrups at 300mmc/c.

Check for deflection control

(L/d)max =(L/d)basic*kt*kf*kc

(L/d)basic =20*0.8

=16

fsc =0.58*fy* Ast req/ Ast pro

=0.58*415*404.46/628

fsc =155

kt =1.5,pt=0.78,kf=1,kc=0.8

(L/d)act =3200/265

=12.075

(L/d)max =16*1.5*1*0.8

=19.2

(L/d)act < (L/d)max .

Hence safe.

LONGITUDINAL SECTION OF T-BEAM



8mm dia 2 legged stirrups @300mm c/c

150

3500

Figure 9-Reinforcement details of T-beam-longitudinal design

CROSS SECTION OF T-BEAM

SUPPORT SECTION

2 bars of 12mm dia

3 bars of 20mm dia


150

MID SECTION

2 bars of 12mm dia

3 bars of 20mm dia


150

Figure 10- Cross section of T-beam

5.3 STAIRCASE

Stairs are needed for ascending and descending from floor to floor.

The stairs in a structure consists of a series of steps with or without landing

and give access from floor to floor. A flight between the landings is also

called as stair. There may be 3 to 12 steps in one flight. Each step has one

tread and riser. A landing is usually provided after 12 to 15 steps. The width

of landing should not be less than the width of stair.

5.3.1Types of stairs

The different types of reinforced cement concrete stairs are:

1. Straight stairs

2. Dog-legged stairs

3. Open well stairs

4. Quarter turn stairs

5. Geometrical stairs

6. Circular stairs

7. Spiral stair

Dog-legged type of stair is selected and designed for the proposed

auditorium.

5.3.2 DESIG� OF DOG-LEGGED STAIRCASE

Dimensions

Room size =6.4*3.6m

Height of room =4m

Live load =5kN/m2

fy =415N/mm2

fck =20N/mm2

Assume

Tread =300mm

Riser =125mm

It is proposed to provide 2 flights for the stairway, Hence,

The height of each flight =vertical distance/2

=4/2

=2m

No of steps required =2/0.125

=16nos

No of treads in each flight = No of riser-1

=16-1

=15

Space occupied by threads =15*0.3

=4.5m

Assume width of landing =1.25m

Hence space left for passage =6.4-4.5-1.25

=0.65m

Let as assume bearing of landing as 150mm

ie, Effective span of flight =4.5+1.25+0.15/2

=5.8m

Let the thickness of waist slab be 200mm,

This can be assumed as 40mm to 50mm/m span

Load calculation

Dead load of waist slab =w’*√(R2+T2)/T

=0.2*√(0.1252+0.32)/0.3

=5.5kN/m2

Dead load of slab =125/2

=62.5(average)

ie, dead load =0.0625*25

=1.56kN/m2

Main reinforcement

Mu=0.87fyAst*d[1-fy Ast/bdfck]

76.699*106=0.87*415*Ast*180*[1-(415*Ast/1000*20*180)]

Ast=1408.9mm2

Spacing with 10mm dia bars

Sv=(1000*¶/4*102)/1408.9

Sv=55.75mm

Provide 10mm dia bars 100mm c/c

Distributor Reinforcement

Provide Astmin =0.12% of cs area and assume 8mm dia bars

Astmin =0.12/100*1000 *200

=240mm2

Spacing =1000(π/4)*82/240

=209.44mm

Provide 8mm dia bars@200mm c/c

Assume top finish =0.1kN/m2

Live load =25kN/m2

ie, Total load =5.5+1.56+0.1+0.5

=12.16kN/m2


=18.24kN/m2

Factored moment =wl2/8

=18.24*5.82/8

Distributor Reinforcement

Provide =0.12% and assume 8mm Dia bars

Astmin =0.12/100*1000 *200

=240mm2

Spacing =1000(π/4)*82/240

=209.44mm

Provide 8mm dia bars@200c/c

Assume top finish =0.1kN/m2

Live load =25kN/m2

ie, Total load =5.5+1.56+0.1+0.5

=12.16kN/m2


=18.24kN/m2

Factored moment =wl2/8

=18.24*5.82/8

=76.699kNm

Check for depth for waist slab

D =√Mu/(0.138fckb)

=√76.699*106/(0.138*20*1000)

=166.7

But, D =200mm

D =200-20

=180>166.7

Hence ok.

Check for depth for waist slab

D =√Mu/(0.138fckb)

=√76.699*106/(0.138*20*1000)

=166.7mm

But D =200mm

D =200-20

=180>166.7

Hence ok.

DETAILING OF DOG LEGGED STAIRCASE


Tread = 300mm

Riser =125mm


Thickness of flight = 200mm12mm dia bars @ 100mm c/c


Figure 11-Reinforcement details of doglegged staircase

5.4 COLUM�S

A column is defined as a structural member subjected to compressive

force in a direction parallel to its longitudinal axis. The columns are used

primarily to support compressive load. When the compression members

are over loaded then their failure may take place in direct compression

(crushing), excessive bending combined with twisting. Failure of column

depends upon slenderness ratio.

5.4.1 Types of columns

1) Short column

2) Long column

When slenderness ratio (lex/b) is less than 12, the compression member

(lex/b) is said to be short column and if the slenderness ratio is greater than

12, it is called as long column.

5.4.2DESIG� OF COLUM�

5.4.2.1 DESIG� OF AXIALLY LOADED COLUM�

Dimensions

Factored load =1200kN

Concrete grade =M20

Characteristic strength of reinforcement =415N/mm2

Unsupported length of column =3.55m

Cross sectional area of column =400*300

Leff =k*L

k =0.65(effectively held in

position at both ends)

Leff =0.65*3550

=2307mm

Slenderness ratio

Leff /D =2307/400

=5.8<12

Leff /d =2307/300=7.7<12

ie, column is designed as short column

Minimum eccentricity

emin1 =2307/500+400/30=17.9mm

emin2 =2307/500+300/30=14.6mm

From clause 39.3 Is 456-2000

400*0.05=20>17.9

300*0.05=15>14.6

ie, codal formula for axially compressed column can be used.

Longitudinal reinforcement

Pu =[0.4fck Ac+0.67fy-0.4fck)Ast]

1200*103 =[(0.4*20*400*300)+[(.67*415)-(0.4*20)]Asc

Asc =888.7mm2

Minimum reinforcement provided

=0.008*400*300=960mm2

ie, Provide 6 nos of 20 mm dia bars of longitudinal reinforcement

Lateral ties

Tie diameter >6mm

<16mm

Provide 8mm diameter ties

Tie spacing > 16*20=320mm

ie, provide 8mm dia ties @ 300mm c/c

AXIALLY LOADED COLUMN


Lateral ties 8mm dia bars @ 300mm c/c

Figure 12-Reinforcement details of axially loaded column

5.4.2.2 DESIG� OF U�IAXIALLY LOADED COLUM�

Dimensions

Size of column = 400mm×300mm

Load, Pu = 650 kN

Factored moment= 127 kNm

Eccentricity = 127/650 = 0.19

fck = 20 N/mm2

fy = 415 N/mm2

D =400mm

b =300mm

Assuming cover, d`=50mm

d`/D = 50/400 =0.125

Pu/ fck bD = (650*103)/(20*300*400)

= 0.27

Mu/ fck bD2 = (127*106)/ (20*300*4002)

= 0.132

From graph 45 of SP16,

p/ fck = 0.07

percentage of steel = 1.4

As = pbD/100

= (1.4*400*300)/100

=1800mm2

nπd2/4 = 1800mm2

n = 6 nos

Provide 8 nos. of 20mm dia bars and they are equally arranged on all four

sides

Spacing = 400-(50+50+20)

= 280mm < 300mm

Hence safe

Design of lateral ties

Dia of lateral ties not less than 6mm and not greater than 16mm

Take 8mm dia ties

Spacing should not be greater than 300mm or 16φ=16*20= 320mm

Hence provide 8mm φ bars @ 250mm c/c

UNIAXIALLY LOADED COLUMN


Lateral ties 8mm dia bars @ 250mm c/c

Figure 13-Reinforcement details of uniaxially loaded column

5.4.2.3 DESIG� OF BI AXIALLY LOADED COLUM�

Dimension

b = 450mm

D = 600mm

fck =20 N/mm2

fy = 414 N/mm2

Pu = 590.6 kN

Mux= 150 kN

Muy= 106 kN

Reinforcements

Reinforcements are distributed equally on four sides

As a trial adopt percentage of reinforcement in the CS as p =1%

As = pbD/100 = 1*450*600/100

= 2700mm2

Provide 10 bars of 20mm dia on each face

As = 10*π*202/4

= 3141.6mm2

P = (100*3141.6)/(450*600)

= 1.16

p/ fck = 1.16/20 =0.058

d`= 40+10 = 50mm

Pu/ fck bD = (590.6*103)/(20*450*300)

= 0.12

d`/D = 50/600 =0.08

from chart 44 of SP16

Mu/ fck bD2 = 0.09

For moments about minor axis yy

b= 450mm

d`=40+10=50mm

d`/D = 50/450 =0.111

Pu/ fck bD = 0.12

From chart 44 for d`/D=0.15

Mu/ fck bD2 = 0.09

Muy1=0.09*20*600*4502

= 218.7 kNm

Puz= 0.45 fck Ac+0.75 fy As

=0.45*20[(600*450)-314.6]+[0.75*415*3141.6*10-3]

= 3379.55kN

Pu/ Puz = (590.6/3379.55)=0.175

According to IS456 clause 39.6

αn =1.04

(Mux/ Mux1) αn +( Muy/ Muy1)

αn = 0.97<1

Hence the section is safe

Design of lateral ties

According to IS456:2000,

Dia of lateral ties not less than 6mm and not greater than 16mm

Take 8mm dia ties

Spacing should not be greater than 300mm or 16φ=16*20= 320mm

Hence provide 8mm φ bars @ 300mm c/c

BIAXIALLY LOADED COLUMN


Lateral ties 8mm dia bars @300mm c/c

Figure 14-Reinforcement details of biaxially loaded columns

5.5 FOOTI�G

The portion of the structure above the plinth is called superstructure

and the portion of the structure below the plinth is called as substructure.

The footing or foundation is the part of substructure which remains in

contact with the soil or rock. The footing or foundation transmits safely the

load to soil stratum and distributes the load over large area of the stratum so

that the bearing pressure developed in the soil remains less than the safe

bearing capacity of the soil and is ensured that the any settlement which may

occur shall be nearly uniform as possible and the differential settlement of

the various parts of the structure shall be eliminated as nearly as possible

5.5.1Types of footing

1) Isolated footing

2) Combined footing

3) Spread footing

4) Raft footing

5.5.2 DESIG� OF FOOTI�G

Dimension

Factored load, Pu = 1200 kN

Size of column = 400 × 300 mm

SBC of soil = 200 kN/m2

fck = 20 N/mm2

fy = 415 N/mm2

Size footing

Load on column = 1200 kN

Weight of footing and backfill at 10% = 120 kN

Area of footing =(1200+120)/(1.5*200)

= 4.4 m2

Adopt 2.5m×2m rectangular footing

Net soil pressure at ultimate load is given by,

qu = 1320/(2.5*2)

= 264 kN/ mm2

One way shear

Critical section is at a distance ‘d’ from the column face

Factored shear force, Vu1 = (0.264*2500)(1050-d)

= 660(1050-d)

Assuming percentage of reinforced cement, Pt = 0.25%

For M20 grade concrete, from table 19 IS456:2000

τc = 0.36 N/mm2

one way shear resistance, Vc1 = 0.36*2500*d

= 900d

Equating both,

660(1050-d) = 900d

Hence d = 442.3 mm

Two way shear

Assuming effective depth = 443 mm

Two way shear resistance at a critical section (d/2) from face of

column,

Then, Vu2 = 0.264{(2500*2000)-[(400+d)(300+d)]}

= 0.264{(2500*2000)-[(400+443)(300+443)]}

= 1154.6 kN

Two way shear resistance, Vc2 is computed on,

Vc2 = Ks τc [2(400+d)+2(300+d)]d

Ks = 1

τc = 0.25√k

= 1.118 N/mm2

Hence, Vc2 = 1565.2d+4.47d2

Equating both,

1154644=1565.2d+4.47 d2

d= 362 mm

Therefore one way shear is critical

Adopt effective depth = 450mm

Overall depth = 550mm

Ultimate moment at column face = 0.264*2000*10502/2

= 291.06 kNm

291.06*106 = 0.87fyAstd[1-(Astfy/bdfck)]

291.06*106 = 162472.5 Ast – 7.49 Ast2

Ast = 1970.4mm2

Pt = 100 Ast/bd

=(100*1970.4)/(1000*450)

= 0.43>0.25

Assuming 60mm dia bars,

Spacing = 1000ast/ Ast

=1000*π*162/(4*1970.4)

=105mm

Hence, provide 16mm dia bars @100mm c/c in both directions.

FOOTING FOR AXIALLY LOADED COLUMN

Column Reinforcement

16mm dia bars @ 100mm c/c (both ways)

GL

Figure 15-Reinforcement details of axially loaded column

CHAPTER 6

6. CO�CLUSIO�

The analysis and design of the structural components of the

college auditorium envisaged planning for each floor of the building with

detailed analyses of Beams, Columns, Slabs and Stairs. Isolated footings for

Columns were considered. This work throws an insight into the structural

components of the proposed college auditorium which will be constructed

soon.

REFERE�CES

1. “Advanced Reinforced Concrete Design”, by N.Krishna Raju.

2. “Strength of Materials”, by Ramamirtham and Narayanan.

3. “Reinforced Concrete Design”, by P.P.Vargheese

4. IS:875 part , “Code of Practice for design loads for buildings and

structures – Dead Loads”.


structures – Live Loads”.


structures – Wind Loads”.

7. IS:456: 2000, “Plain and Reinforced Concrete - Code of Practice”.

8. “Design of Concrete Structures”, by Shah

design of proposed auditorium-project report

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