rengan a than

1

A PROJECT REPORT

on

DESIGN OF INDUSTRIAL BUILDING BY LIMIT STATE AND

WORKING STRESS METHODS

Submitted in partial fulfillment of the requirements

for the award of the degree of

BACHELOR OF TECHNOLOGY

in

CIVIL ENGINEERING

by

RENGANATHAN. K

MAY-2011

2

ABSTRACT

In General, steel structures are being used for Industrial buildings since it has lot of

advantages over reinforced concrete structures. The working stress method of design had

been followed worldwide for designing of steel structures. It was the traditional method

and it’s conceptual basis is simple. The method basically assumes that structural material

behaves in a linear elastic manner. The first attainment of yield stress of steel was

generally taken to be the onset of failure as it represents the point from which the actual

behaviour will deviate from the analysis results.

An improved design philosophy to make allowances for the shortcomings in the

Working Stress Method was developed in the late 1970’s and has been extensively

incorporated in design standards and codes. The probability of operating conditions not

reaching failure conditions forms the basis of Limit State Method (LSM). The Limit

State is the condition in which a structure would be considered to have failed to fulfill

the purpose for which it was built.

An attempt has been made in this project to design an industrial building which is

served as a steel fabrication unit for construction works and it’s designed by using two

design methods (limit state & working stress method) and finding out the saving of steel

material by comparing the results. The major components of the industrial building such

as Roof truss, column, gantry girder, purlin, beam, bracing & base plate are considered

for analysis & design.

3

CONTENTS

Page No.

ABSTRACT iii

ACKNOWLEDGEMENT iv

LIST OF TABLES x

LIST OF FIGURES xi

ABBRIVATIONS & SYMBOLS x

CHAPTER 1 INTRODUCTION 01

1.1 GENERAL 01

1.2 SCOPE OF WORK 01

1.3 DETAILS OF THE BUILDING 01

1.4 LOADS 02

1.5 METHODOLOGY 02

CHAPTER

2

LITERATURE REVIEW

03

2.1 GENERAL 03

2.2 LITERATURES 03

2.2.1 ANALYSIS OF TRUSSES 03

2.2.2 BUCKLING OF MEMBER 04

4

2.2.3 PLASTIC ANALYSIS 04

2.2.4 DESIGN REQUIREMENTS 05

CHAPTER 3 ANALYSIS AND DESIGN OF ROOF 06

3.1 GENERAL 06

3.2 DESIGN OF PURLIN 06

3.2.1 LOAD CALCULATION 06

3.2.2 LOAD COMBINATION & ANALYSIS 08

3.2.3 DESIGN 10

3.3 ANALYSIS OF ROOF TRUSS 12

3.3.1 LOAD CALCULATION 13

3.3.2 TRUSS ANALYSIS 13

3.4 DESIGN OF ROOF

TRUSS 24

3.5 DESIGN OF ROOF BRACINGS 27

3.5.1 DESIGN OF RAFTER BRACING 27

3.5.2 DESIGN OF EAVE BRACING 28

5

3.6 DESIGN OF BEARER BEAM 30

CHAPTER 4 DESIGN OF GANTRY GIRDER 33

4.1 GENERAL 33

4.2 LOAD CALCULATION AND ANALYSIS 33

4.3 DESIGN 36

CHAPTER 5 ANALYSIS AND DESIGN OF COLUMN 42

5.1 LOAD CALCULATION 42

5.2 STAAD ANALYSIS 45

5.3 DESIGN OF COLUMN 60

5.4 DESIGN OF BASE PLATE 63

CHAPTER 6 DRAWINGS 66

CHAPTER 7 RESULTS AND CONCLUSION 72

7.1 RESULTS 72

7.2 COMPARISON OF RESULTS 74

7.3 CONCLUSION 75

6

LIST OF TABLES

TABLE DESCRIPTION Page No.

3.1 EXTERNAL & INTERNAL PRESSURE CO-EFFICIENT

FOR ROOF

9

3.2 WIND LOAD DETAILS FOR ROOF TRUSS (DL+LL) 16

3.3 WIND LOAD DETAILS FOR ROOF TRUSS (DL+LL) 17

3.4 MEMBER FORCES FOR ROOF TRUSS 28

5.1 EXTERNAL PRESSURE CO-EFFICIENT (Cpe) 54

7.1 LIMIT STATE METHOD RESULTS 72

7.2 WORKING STRESS METHOD RESULTS 73

7.1 OVERALL RESULTS 74

7

LIST OF FIGURES

FIGURE NO. DESCRIPTION Page No.

5.1 COLUMN WIND LOAD DIAGRAM (FOR 0 ̊) 43

5.2 COLUMN WIND LOAD DIAGRAM (FOR 90 ̊) 44

5.3 BENDING MOMENT DIAGRAM OF FRAME

(LSM)

50

5.4 BENDING MOMENT DIAGRAM OF COLUMN

(LSM)

50

5.5 AXIAL FORCE DIAGRAM OF COLUMN (LSM) 51

5.6 BENDING MOMENT DIAGRAM OF FRAME

(WSM)

58

5.7 BENDING MOMENT DIAGRAM OF COLUMN

(WSM)

58

5.8 AXIAL FORCE DIAGRAM OF COLUMN (WSM) 59

6.1 LAYOUT OF COLUMN 66

6.2 ROOF TRUSS DETAILS 67

6.3 LAYOUT OF ROOF TRUSS 68

6.4 LAYOUT OF RAFTER BRACING 69

6.5 LAYOUT OF EAVES BRACING DETAILS 70

6.6 TYPICAL ELEVATION 71

8

ABBRIVATIONS & SYMBOLS

A Area of cross section; surface area.

Ag Gross cross sectional area.

An Net area of cross section

B Breadth of section, width of flange.

Cpe External pressure co-efficient

Cpi Internal pressure co-efficient

D Overall depth

d Effective depth; Depth of web

E Young’s modulus

e Eccentricity

F Net wind force

Fbr Strength of lateral bracing

fb Actual bending stress

fbc Actual bending stress in compression

fbd Actual bending stress corresponding to lateral buckling

fcr Elastic buckling stress of column

fcr,b A Extreme fibre comp. Elastic buckling stress

fcd Design compressive stress of column

fu Characteristic ultimate tensile stress

fy Characteristic Yield stress

k1 Risk co-efficient

k2 Terrain factor

k3 Topography factor

L Unsupported length; span

LSM Limit state method

M Bending Moment

9

CHAPTER 1

INTRODUCTION

1.1 GENERAL

Steel structures are normally used in industrial and manufacturing shelters which

is economic for large spans length when compared to reinforced cement concrete

(R.C.C) buildings. It’s main advantages are high strength resulting in the reduction of

dead weight, speed of erection, ductile in it’s property and effective dismantling.

Industrial buildings are mostly of one story high, and are used for manufacturing of

heavy machinery, for storage of raw materials and finished products. Modern industries

requires large column free areas to have the potential for easy change of layout several

times during the building life. The industrial buildings are constructed with adequate

headroom for the use of an overhead traveling crane.

1.2 SCOPE OF WORK

Nowadays the conventional working stress method of design has been replaced

by limit state method. So the building is designed by using above mentioned two

methods and a comparative study has to be carried out by comparing the results. The

primary items of the industrial building such as Roof truss, column, Gantry girder,

beams, roof bracings, purlin & base plate are considered for design and comparative

studies.

1.3 DETAILS OF THE BUILDING

The proposed building is located in Chennai, the size of the building is 48m x

18m in plan and the overall height is 15m. Fink truss of height 3.5m is provided for roof

at 4m spacing. Columns are provided at the spacing of 8m.Intermediate roof trusses

between columns are supported by using bearer beam. Gantry girder is provided to

support an overhead travelling crane of capacity 10 MT. Bracings are provided at rafter

level and eaves level at the end the end bays.

10

1.4 LOADS

Dead load, live load, wind load and it’s combinations are considered for the

analysis. Dead load is considered as per guidelines provided in the code IS 875-1987

(Part-1), live load is taken by considering the roof is inaccessible and it’s further reduced

based on the roof slope as per IS 875-1987(Part-2). Wind load is calculated as per IS

800-1987(Part-3). Wind load is calculated based on the for 0 & 90 degree winds.

1.5 METHODOLOGY

The structure is Analyzed manually and using Staad Pro. For roof truss analysis,

method of joints has been used and column analysis is carried out with the help staad

pro. Structure is designed by limit state & working stress method of design. IS 800-2007

& IS 800-1984 is followed for limit state method & working stress method respectively.

Structural Components to be designed:

• Purlin

• Roof truss

• Roof bracings

• Bearer beam

• Gantry girders

• Column

• Base plate

11

CHAPTER 2

LITERATURE REVIEW

2.1 GENERAL

Studies are being carried out in the field of steel structures to find out the

economical & safe method of design since the late of 19th

century. The conventional

methods are slowly replaced by modern day concepts of design. An attempt has been

made in this project to find out the advantages of new design method which is followed

worldwide. The following journals is taken as the refernce and the guidelines given has

been followed in this project.

2.2 LITERATURES

2.2.1 ANALYSIS OF TRUSSES

“Design of steel structures” (chapter 2.4.1 ‘Analysis of trusses’) by Prof.

S. R. Satish Kumar and Prof. A. R. Santha Kumar (Indian Institute of

Technology, Madras)

“Generally truss members are assumed to be joined together so as to

transfer only the axial forces and not moments and shears from one member to

the adjacent members (they are regarded as being pinned joints). The loads are

assumed to be acting only at the nodes of the trusses. From the analysis based on

pinned joint assumption, one obtains only the axial forces in the different

members of the trusses. However, in actual design, the members of the trusses

are joined together by more than one bolt or by welding, either directly or

through larger size end gussets. Further, some of the members, particularly chord

members, may be continuous over many nodes. Generally such joints enforce not

only compatibility of translation but also compatibility of rotation of members

meeting at the joint.

12

2.2.2 BUCKLING OF MEMBER

“The behavior and design of steel structures” (chapter 3.3.4 (4th

Edition)

‘Buckling of members with residual stresses’) by N.S. Trahair and

M.A. Bradford.

“The presence of residual stresses in an intermediate length steel

compression member may cause a significant reduction in its buckling resistance.

Residual stresses are established during the cooling of a hot-rolled or welded

steel member (and during plastic deformation such as cold-rolling). The

shrinking of the late-cooling regions of the member induces residual compressive

stresses in the early-cooling regions, and these are balanced by equilibrating

tensile stresses in the late-cooling regions. In hot-rolled I-section members, the

flange – web junctions are the least exposed to cooling influences, and so these

are the regions of residual tensile stress, while the more exposed flange tips are

regions of residual compressive stress. In a straight intermediate length

compression member, the residual compressive stresses cause premature yielding

under reduced axial loads, and the member buckles inelastically at a load which

is less than the elastic buckling load.”

2.2.3 PLASTIC ANALYSIS

“Plastic analysis design of steel structures” (chapter 5.2.2 [4th

Edition]

‘Static Theorem (Lower Bound Theorem’) by M. Bill Wong

“This theorem states that the collapse load obtained for a structure that

satisfies all the conditions of static equilibrium and yield is either less than or

equal to the true collapse load. In other words, the collapse load, calculated from

a collapse mode other than the true one, can be described as conservative when

the structure satisfies these conditions. The true collapse load can be found by

choosing the largest value of the collapse loads obtained from all cases of

possible yield conditions in the structure. The yield conditions assumed in the

structure do not necessarily lead to a collapse mechanism for the structure. The

13

use of this theorem for calculating the collapse load of an indeterminate structure

usually considers static equilibrium

through a flexibility approach to produce free and reactant bending moment

diagrams. It is usually referred to as the statical method.”

2.2.4 DESIGN REQUIREMENTS

“Analysis and design of steel structures” (chapter 7.1 (3rd Edition) ‘Design

requirements’) by D.A. Nethercot, and L. Gardner

The principal design requirement of a structure is that it should be effective; that

is, it should fulfil the objectives and satisfy the needs for which it was created.

The structure may provide shelter and protection against the environment by

enclosing space, as in buildings; or it may provide access for people and

materials, as in bridges; or it may store materials, as in tanks and silos; or it may

form part of a machine for transporting people or materials, as in vehicles, or for

operating on materials. The design requirement of effectiveness is paramount, as

there is little point in considering a structure which will not fulfil its purpose.

14

CHAPTER 3

ANALYSIS AND DESIGN OF ROOF

3.1 GENERAL

Building details for design:

Over all size of building = 48.5m X 18.5m x 15m

C/C length = 48m

C/C Span = 18m

Overall Height = 15m

Height between Ground to U/S of Roof = 11.5m

Height of Roof Truss = 1/5 of Span

= 1/5 x18 = 3.6m (Say 3.5m)

Truss spacing = 1/4th

to 1/5th

of span

= 18/4 to 18/5

= 3.6 to 4.6 m (say 4m)

Number of bays = 48/4 = 12

Roof slope = Tan-1

(3.5/9)

= 21.25°

3.2 DESIGN OF PURLIN

3.2.1 LOAD CALCULATION

Dead load:

GI Sheeting (Including fixing & services) = 0.25 KN/m2

Self Wt of Purlin = 0.15 KN/m2

Total load = 0.40 KN/m2

15

Live load:

Since the roof is inaccessible & slope is more than 10 degree,

As per IS 875-1987 (Part-2)

Reduce 0.02 KN/m for every increasing degree over 10.

Live load = 0.75 – (21.25 - 10) x 0.02

= 0.525 KN/m2

Wind load:

The building is going to be proposed in Chennai,

Basic wind speed Vb = 50 m/s

The building is considered under important category

Risk co-efficient K1 = 1.05

Building comes under Terrain 2 category & class B

Terrain Facto K2 = 1.07

The land where the building is going to be proposed is flat,

Topography Factor K3 = 1

Design Wind speed VZ = Vb x K1 x K2 x K3

= 50 x 1.05 x 1.07 x 1

VZ = 56.175 m/s

Design wind pressure Pd = 0.6 Vz2

= 0.6 x 56.1752

= 1.893 KN/m2

External & Internal pressure Co-efficient (CPe & CPi):

h/w = 11.5/18.5 = 0.622

(0.5 < h/w < 1.5)

From Table 5 of IS 875-1987 (Part III)

Wind Angle for 0 degree:

For 21.25 degree, Windward side Cpe = -0.675 & Leeward side

Cpe = -0.5

16

Wind Angle for 90 degree:

For 21.25 degree, Windward side Cpe = -0.8 & Leeward side

Cpe = -0.625

Spacing between Roof Truss = 4m

Spacing between Purlin = 1.8m

A x Pd = 4 x 1.8 x 1.893

= 13.630 KN

Wind Load F = (Cpe +/- Cpi) x A x Pd

TABLE-3.1 EXTERNAL&INTERNAL PRESSURE CO-EFFICIENT FOR ROOF

Wind

Angle

Cpe

Cpi

Cpe +/- Cpi

A x

Pd

Wind load, F

(KN)

WW LW WW LW WW LW

0 -0.675 -0.50 -0.5

+0.5

-1.175

-0.175

-1.0

0

13.63

13.63

-16.02

-2.39

-13.630

0

90 -0.80 -0.625 -0.5

+0.5

-1.30

-0.30

-1.125

-0.125

13.63

13.63

-17.72

-4.90

-15.34

-1.717

3.2.2 LOAD COMBINATIONS & ANALYSIS

1.5(DL + LL)

Load Normal to the Roof Wz = 1.5 (0.40+0.525) cos 21.25°

= 1.293 KN/m²

Load parallel to the Roof Wy = 1.5 (0.40+0.525) sin 21.25°

= 0.503 KN/m²

Load per metre length Wz = 1.293 X 2

= 2.586 KN/m

Load per metre length Wy = 0.503 X 2

= 1.006 KN/m

17

Bending Moment:

Moment (Z direction):

Mz = Wz X L² /10 (Purlin is continuous)

= 2.586 X 4² /10

Mz = 4.14 KN.m

Moment (Y direction):

My = Wy X L² /10 (Purlin is continuous)

= 1.006 X 4² /10

My = 1.61 KN.m

Shear Force:

Force (Z direction): SFz = Wz X L/2

= 2.586 X 4/2

= 5.172 KN

Force (Y direction): SFy = Wy X L/2

= 1.006 X 4/2

= 2.012 KN

1.5(DL + WL)

Load Normal to the Roof Wz = 1.5 (0.3 cos 21.25° - 2.018)

= 2.607 KN/m²

Load parallel to the Roof Wy = 1.5 (0.3 sin 21.25°+0)

= 0.164 KN/m²

Load per metre length Wz = 2.607 X 2

= 5.214 KN/m

Load per metre length Wy = 0.164 X 2

= 0.328 KN/m

Factored Moment:

Moment (Z direction):

Mz = Wz X L² /10 (Purlin is continuous)

18

Mz = 5.214 X 4² /10 = 8.342 KN.m

Moment (Y direction):

My = Wy X L² /10 (Purlin is continuous)

= 0.328 X 4² /10

My = 0.523 KN.m

Shear Force:

Force (Z direction): SFz = Wz X L/2

= 5.214 X 4/2

= 10.428 KN

Force (Y direction): SFy = Wy X L/2

= 0.328 X 4/2

= 0.624 KN

3.2.3 DESIGN

LIMIT STATE METHOD

Assume ISMC 125 as a purlin section & check for the maximum load case

(DL+WL).

Properties of ISMC 150:

D = 125mm;

B= 65mm;

Tw = 5.0mm ;

Tf = 8.1mm ;

Izz = 416.4 x 104 mm4

Zez = 66.6 X 103 mm

3 ;

Zey = 13.1 X 103

mm3

Zpz = 77.8 X 103 mm3 ;

Zpy = 40.4 X 103

mm3;

Check for Shear capacity:

As per IS 800:2007, Clause 8.4,

19

Av = 125 X 5 = 625mm²

Av X Fwy/(√3 X rmo) = 625 X 250/(√3 X 1.1 X 103)

= 82.01 KN > 10.428 KN

IIIy, In Y-direction also shear capacity is too large,

Hence safe in shear.

Moment capacity of section:

Z-Direction : Mdz = B X Zpz X Fy/ rmo

= 1 X 77.8 X 250 X 103/(1.1 X 106)

= 17.68 KN.m > 8.342 KN.m

Y-Direction : Mdy = B X Zpy X Fy/ rmo

= 1 X 40.4 X 250 X 103/(1.1 X 10

6)

= 9.18 KN.m > 0.523 KN.m

Hence Safe

Check for Bi-Axial Bending:

(Mz/Mdz)+(My/Mdy) <=1.0

= (8.342/17.68)+(0.523/9.18)

= 0.52 < 1.0

Hence safe.

WORKING STRESS METHOD

Load combinations

Step 1: (DL+LL)

Bending moment Mz = 2.76 kN.m

Bending moment My = 1.07 kN.m

Shear force Fz = 3.5 kN

Shear force Fy = 1.34 kN

Step 2:.(DL+WL)

Bending moment Mz = 5.56 kN.m

Bending moment My = 0.35 kN.m

Shear force Fz = 6.95 kN

20

Shear force Fy = 0.42 kN

Design of Purlin :

Assume Zs/Zy = 7

Zx required = (5560 + 7 x 350) x 103/165

= 48.545 x 103 mm

3

Use MC 175 @ 191N/m – Zx = 139.5 x 103 mm3

Zy = 22.8 x 103 mm

3

fbt = 5560 x 103 / (139x103) + 350x103/(22.8x103)

= 55.35 N/mm2

Hence safe.

Check for Deflection:

Wz = 3.476 X 4 = 13.904 KN

Actual Deflection = 5WL3/(384EI)

= 5x13.904x1000x40003/(384x2.1x10

5x416.4x10

4)

= 13.25 mm

Deflection limit = L/180

= 4000/180 = 22.22mm > 13.25mm

Hence safe in deflection.

3.3 ANALYSIS OF ROOF TRUSS

3.3.1 LOAD CALCULATION

Dead loads

G.I. Sheet = 0.25 kN

Self weight of Roof truss = ( Span/3 + 5) x W

= ( 18/3 + 5) x 10 = 110 N/m

Purlin self weight = 0.13 kN/m

[ (0.25 x 0.11 ) x 4 x 1.8 ) + ( 0.13 x 4 )) = 3.112 kN

Load on Panel = 1.56 kN

Live load = 0.525 x 4 x 1.8 = 3.78 kN

21

Live load on end panel = 3.78 / 2 = 1.89 kN

TABLE 3.2 WIND LOAD DETAILS FOR ROOF TRUSS (DL+LL)

Wind system Wind word side (KN) Leaword side (KN)

0o -16.02 -8.01 -13.61 -6.815

90o -17.72 -8.86 -15.34 -7.67

LOAD COMBINATIONS

Load case 1 : 1.5 (DL + LL)

1.5 (3.52 + 4.2) = 11.58 say 11.6 kN

1.5 (1.56 + 1.89) = 5.79 say 5.8 kN

Load case 2 : 1.5 (DL + WL) – for 0o & 90

o

TABLE 3.3 WIND LOAD DETAILS FOR ROOF TRUSS (DL+WL)

Wind system Wind word side (kN) Leaword side (kN)

Mid point End point Mid point End point

0o -12.44 4.85 -10.58 -4.13

90o 22.3 11.15 18.57 9.285

3.3.2 TRUSS ANALYSIS

Roof truss is analyzed by using method of joints. By considering the each and every

joint of the truss and solving the equilibrium equations. For load combination (DL+LL),

analysis is given here. The remaining load cases solved similarly and the member forces

are tabulated at the end.

22

For load case (DL+LL)

11.6 11.6

11.6 11.6

11.6 11.6

11.6 11.6

5.6 5.6

58 58

F

E G

H

D O R

C J

B K

L

A Q P M N

11.6

JOINT A:

5.8

21.3 o

58

∑v = 0

58-5.8-ABSin21.3 =0

AB = 143.7 kN

AB =58-5.8

Sin 21.3

`

23

AB =143.7 kN

∑H = 0

AQ = AB Cos21.3

AQ= 133.9 kN

JOINT B:

11.6 C

68.7

143.7

A Q .

∑v = 0

-11.6+ABSin21.3-BCSin21.3+BQCos= 0

BCSin21.3-BQCos = 143.7 x Sin21.3-11.6

0 x BC - 1 x BQ = -11.6 kN

∑H = 0

ABCos21.3 =BC Cos21.3+BQ Cos

143.7 kN = 1 x BC - 1 x BQ

Solving above equation

BC = 133.4 kN

BQ = 12.1 kN

JOINT Q:

52.1 32.7

84.8

57.3

∑v = 0

BQ Cos= CQ Cos32.7

111.3

59.1

120.9

37.9

CQ =12.1x Cos52.1

Cos 32.7

24

CQ = 8.8 kN

∑H = 0

AQ + BQ Sin= PQ + CQ Sin32.7

PQ = 133.9+ 12.1 x Sin52.1 + 8.8 x Sin32.7

PQ = #### kN

JOINT C:

11.6

68.7

21.3

84.9 .

∑v = 0

BCSin21.3 - 11.6 - CDSin21.3 - CQ Cos+CP Cos=0

0

53.36 - 11.6 - 0.363 CD - 8.8+1CP = 0

0.363 CD - 1CP = 28.06

∑H = 0

BC Cos21.3 =CD Cos21.3 + CQ Sin + CP Sin

124.29 = CD Cos21.3 + 0 + CP Sin

124.29 = 0.932 CD + 0 CP


CD = 113 kN

CP = 18.6 kN

JOINT D:

11.6

68.7 21.3

32.3

11

21.3

.

111.3 59.1

36

133.4 x Sin21.3 - 11.6 - CDSin21.3 -

8.8 x Cos+CP Cos =

111.3

68.7 79

25

∑v = 0

CD Sin21.3 - 11.6 - DESin21.3 - DO Sin11+DP = 0

29.266 - DESin21.3 - DO Sin11+DP = 0

-29.27

∑H = 0

CD Cos21.3 + DO Cos11 - DESin21.3 = 0

104.815 + 0.982 DO - 0.932 DE = 0

0.932 DE - 0.982 DO = 104.8

JOINT P: 11.6

52.1 45.8

44.2

∑v = 0

CP Sin37.9 + DP = OP Sin44.2

0.697 OP -DP =

∑H = 0

PQ = PM + CP Cos37.9 + OP Cos44.2

138.7 = PM + 14.677 + OP Cos44.2

PM + 0.717 OP = 124.0

JOINT O:

79 45.8

11

55.2 44.2

.

∑v = 0

OP Sin44.2 + EO - DO Sin11 - FO Sin44.2 = 0

0.697 OP - 0.697 FO + 0.191 DO = 11.600

∑H = 0

OP Cos44.2 + DO Cos11 - FO Cos44.2 = 0

0.717 OP - 0.717 FO + 0.982 DO = 0

-0.363 DE - 0.191 DO+DP =

37.9

11.426

26


DO = 10.12 kN

Substitue "DO" in equation

0.697 OP - 0.697 FO = 9.669

0.717 OP - 0.717 FO = -9.934

FO - OP = 13.85


FO = 55.20 kN

OP = 41.30 kN

Substitue "OP" in equation

PM + 0.717 OP = 124.02

PM + 29.608 = 124.02

PM =

0.697 OP -DP = 11.426

28.793-DP = 11.426

DP =

Substitue "DP & DO" in equation

-0.363 DE - 0.191 DO+DP = -29.27 kN

-0.363 DE - 1.931+17.367 = -29.27 kN

0.363 DE = 44.70

DE =

JOINT E: 11.6

21.3

21.3

.

∑v = 0

EF Sin21.3 + 11.6 - DESin21.3 - EO = 0

EO =

EF Sin21.3 = DESin21.3

EF = DE

123.146 kN

94.415 kN

17.367 kN

11.600 kN

27

JOINT L:

5.8

21.3

58

∑v = 0

58-5.8- Sin21.3 = 0

= 143.7 kN

∑H = 0

Q = Cos21.3

Q = 133.9 kN

JOINT K:

11.6 J

68.7

59.1

L .

∑v = 0 `

-11.6+KLSin21.3-JKSin21.3+KNCos= 0

JKSin21.3-KNCos = 143.7 x Sin21.3-11.6

0 x JK - 1 x KN = -11.6 kN

∑H = 0

KLCos21.3 =KJ Cos21.3+KN Cos52.8

143.7 kN = 1 x KJ - 1 x KN

Solving the above equation

JK = 133.4 kN

KN = 12.1 kN

=58-5.8

Sin 21.3

111.3

120.9

`

28

JOINT N:

32.7 52.1

84.8

37.9

∑v = 0

KN Cos= JN Cos52.1

JN = 8.8 kN

∑H = 0

LN + KN Sin= MN + JN Sin52.1

MN =133.9+ 12.1 x Sin32.7 + 8.8 x Sin52.1

MN = 138.7 kN

JOINT J:

11.6

68.7 111.3

59.1

21.3

84.9 36 .

∑v = 0

JKSin21.3 - 11.6 - HJSin21.3 - JN Cos+JM Cos=0

0

53.36 - 11.6 - 0.363 HJ - 8.8+1JM = 0

0.363 HJ - 1JM = 28.06

∑H = 0

JK Cos21.3 =HJ Cos21.3 + JN Sin + JM Sin

124.29 = HJ Cos21.3 + 0 + JM Sin

124.29 = 0.932 HJ + 0 JM

133.4 x Sin21.3 - 11.6 - HJSin21.3 - 8.8

x Cos+JM Cos =

57.3

JN =12.1x Cos52.1

Cos 32.7

29

Solving the above equation

HJ = 112.5 kN

JM = 18.6 kN

JOINT H:

68.7 11.6

21.3

32.3 21.3 111.3

11 79 68.7

.

∑v = 0

HJ Sin21.3 - 11.6 - GHSin21.3 - HR Sin11+HM = 0

29.266 - GHSin21.3 - HR Sin11+HM = 0

-29.27

∑H = 0

JH Cos21.3 + HR Cos11 - GHSin21.3 = 0

104.815 + 0.982 HR - 0.932 GH = 0

0.932 GH - 0.982 HR = 104.8

JOINT M: 11.6

45.8 52.1

37.9

∑v = 0

JM Sin37.9 + HM = RM Sin44.2

0.697 RM -HM =

∑H = 0

MN = MP + JM Cos37.9 + RM Cos44.2

138.7 = MP + 14.677 + RM Cos44.2

MP + 0.717 RM = 124.0

-0.363 GH - 0.191 HR+HM =

44.2

11.426

30

JOINT R:

45.8 79

55.2

11

44.2

.

∑v = 0

RM Sin44.2 + GR - HR Sin11 - FR Sin44.2 = 0

0.697 RM - 0.697 FR + 0.191 HR = 11.600

∑H = 0

RM Cos44.2 + HR Cos11 - FR Cos44.2 = 0

0.717 RM - 0.717 FR + 0.982 HR = 0

Solving the above equatin

HR = 10.12 kN

Substitute "DO" in equation

0.697 RM - 0.697 FR = 9.669

0.717 RM - 0.717 FR = -9.934

FR - RM = 13.85

Solving the equation

FR = 55.20 kN

RM = 41.30 kN

Substitue "OP" in the above equation

MP + 0.717 RM = 124.02

MP + 29.608 = 124.02

MP =

0.697 RM -HM = 11.426

28.793-HM = 11.426

HM =

Sub "DP & DO" in above equation

-0.363 GH - 0.191 HR+HM = -29.27 kN

-0.363 GH - 1.931+17.367 = -29.27 kN

0.363 GH = 44.70

GH = 123.146 kN

94.415 kN

17.367 kN

31

JOINT G: 11.6

21.3

21.3

.

∑v = 0

FG Sin21.3 + 11.6 - GHSin21.3 - GR = 0

GR =

FG Sin21.3 = HGSin21.3

FG = GH

11.600 kN

Joint forces are calculated for load combinations (DL+WL) in the similar manner

followed for (DL+LL).

TABLE 3.4 MEMBER FORCES FOR ROOF TRUSS

Member Forces (kN)

DL+LL DL+WL (0) DL+WL (90)

AB -143.70 -230.10 -262.46

BC -133.40 -212.08 -241.96

CD -112.50 -176.03 -192.48

DE -123.15 -194.12 -213.00

EF -123.15 -194.12 -213.00

AQ +133.90 +218.025 240.90

PQ +138.70 +192.05 234.00

PM +94.92 +117.46 141.00

BQ -12.10 -117.46 -24.10

CQ 8.80 21.22 17.60

CP -18.60 -15.50 -41.20

DP -17.367 -31.81 -33.50

DO 10.12 29.422 19.50

EO 11.60 17.12 22.30

OP 41.30 11.60 84.30

FO 55.20 70.130 111.10

KL -143.70 -209.60 -253.40

JK -133.40 -198.90 -225.60

32

HJ -112.50 -195.04 -213.44

GH -123.15 -151.95 -192.43

FG -123.15 -151.95 -192.43

LN +133.90 198.90 212.56

MN +138.70 219.60 241.47

MP +94.92 155.75 216.23

KN -12.10 -52.31 -121.34

JN 8.80 38.20 87.34

JM -18.60 -43.39 -112.57

HM -17.36 2.13 5.45

HR 10.12 30.12 56.40

GR 11.60 15.90 34.67

MR 41.30 81.53 123.40

FR 55.20 85.52 145.43

3.4 DESIGN OF ROOF TRUSS

Limit State Method:

Design of Top Chord Member

Factored compressive force = 143.7kN

Factored tensile force = 262.46 kN

Trying ISA 90 x 90 x 8mm @ 0.108 kN/m

Sectional properties :

Area of cross section A = 1379 mm2

Radius of gyration ru = 27.5 mm

Maximum unrestrained length L = 2000 mm

KL = 0.85 x L = 0.85 x 2000 = 1700 mm

λy = 1700/27.5 = 61.82 < 180

Hence, λy is within the allowable limits.

From Table 9c of the code for KL/r = 61.8 and fy = 250 MPa,

Fcd = 166.37 N/mm2

Axial capacity = 166.37 x 1379/1000 = 229.42 kN > 143.70kN

Hence, section is safe against axial compression.

33

Axial tension capacity of the section = 1379 x 250/1.10

= 387.05 kN > 262.46 kN

Hence, section is safe in tension.

Design of Bottom Chord Member

Factored compressive force = 240.90 kN


Trying ISA 90 x 90 x 10mm @ 0.134 kN/m



Radius of gyration ru = 27.3 mm

Maximum unrestrained length = L = 2000 mm

KL = 0.85 x L = 0.85 x 2000 = 1700 mm

Note The effective length of top chord member may be taken as 0.7-1.0 times the

distance between centres of connections as per clause 7.2.4 of IS 800. We have

assumed the effective length factor as 0.85.

λy = 1700/27.3 = 62.27 < 180

Hence, λy is within the allowable limits. From Table 9c of the code for KL/r =

62.27 and fy = 250MPa,

Fcd = 164.97 N/mm2

Axial capacity = 164.97 x 1703/1000 = 280.94 kN > 240.90kN



= 387.05 kN > 138.70 kN


Design of Web Member

Factored compressive force = 84.30 kN


Trying ISA 65 x 65 x 8 @ 0.077 kN/m

34



Radius of gyration ru = 24.7mm

Maximum unrestrained length = L = 2510 mm

KL = 0.85 x L = 0.85 x 2510 = 2133.5 mm

λy = 2133.5/24.7 = 86.37 < 180

Hence, λy is within the allowable limits. From Table 9c of the code for KL/r =

86.37 and fy = 250MPa,, Fcd = 126 N/mm2

Axial capacity = 126x 976/1000 = 122.98 kN > 84.30kN



= 221.82 kN > 55.20 kN


Working Stress Method:

Top chord member

Max. axial force P = 175 kN

Length of member = 2 m

Assume slenderness ratio of λ = 60 & corresponding axial stress бac = 122

N/mm2

Area required = P/бac

= 175 x 103/122 = 1434.42 mm

2

Choose section of ISA 100 X 100 X 10 Whose properties are,

A = 1903 mm2 & rv = 30.5mm

Actual slenderness ratio of λ = L/r

= 2000/30.2 = 65.57

corresponding allowable stress бac = 117 N/mm2

Actual stress бc = P/A = 175 x 103/1903

= 91.96 N/mm2 < 117 N/mm2

Hence it’s safe. Provide ISA 100 X 100 X 10 as Top chord member

35

Bottom chord member

Max. axial force P = 160.6 kN

Length of member = 2.7 m


N/mm2


= 160.6 x 103/127= 1264.57 mm

2

Choose section of ISA 100 X 100 X 10 Whose properties are,

A = 1903 mm2

& rv = 30.5mm


= 2700/30.2= 92


Actual stress бc = P/A = 160.6 x 103/1903

= 84.39 N/mm

2 < 88 N/mm

2

Hence it’s safe. Provide ISA 100 X 100 X 10 as Bottom chord member

Intermediate (strut) member

Max. axial force P = 56.2 kN

Length of member = 2.51 m


N/mm2


= 56.2 x 103/107

= 525.23 mm2

Choose section of ISA 75x75x8 Whose properties are,

A = 1138 mm2

& rv = 22.8mm


= 2510/22.8

= 110.1


36

Actual stress бc = P/A = 56.2 x 103/1138

= 49.38 N/mm

2 < 72 N/mm

2

Hence it’s safe. Provide ISA 75 X 75 X 8 as strut Intermediate member

3.5.1 DESIGN OF RAFTER BRACING

Limit state method

Design wind pressure = 1.893 kN/m2

Maximum force coefficient = -1.3

Factored wind load on rafter bracing = 1.5x1.893x1.3x1.931x3x5/2xsec21.250

= 57.36 kN

Length of bracing = √ (1.9312 x3+5

2) =7.652 m

Try 100 x 100 x 8,

A = 1539 mm2,

rmin = 19.5 mm, and

L/r = 7652 / 19.5

= 392.4 < 400 (Table 3 of IS 800 : 2007)

Tension bracing will be effective

Axial tensile capacity :

Design strength of member due to yielding of gross section

Tdg = Agfy/γmo

= 1531 x (250 / 1.1) / 1000

= 349.77 kN > 57.6 kN

Design strength due to rupture of critical section

Tdn = αAnfy /γmt

α = 0.6 (Assuming two bolts of 16mm diameter at the ends)

An = 1539 – (18 x 8) = 1395 mm2

Tdn = 0.6 x 1395 x (250 / 1.25) x 1000 = 167.4 kN > 57.36 kN

Hence L 100 x 100 x 8 is safe. The member has been found to be safe for block

shear failure.

37

Working stress method


Maximum force coefficient = -1.3

Factored wind load on rafter bracing = 1.893x1.3x1.931x3x5/2xsec 21.250

= 38.24 kN


2) =7.652 m

Net area required = T /σat = 38240 / 150 = 254.933 mm2

Provide ISA 100 x 100 x 10 is safe

3.5.2 DESIGN EAVE BRACING

Limit state method


Maximum force coefficient = 1.2

Spacing of column at gable end = 4.5m

Factored wind load on tie manner = 1.5 x 1.893 x 1.2 x 4.5 (15/2)

= 115 kN (T)

Reaction = 115 x 4 / 2 = 230 kN

Maximum force in bracing Fbr = (230 – 57.5)/cos 29.36 = + 198 kN

Length of bracing = √(42+2.25

2) = 4.6m

Require γ min = 4600 / 300 = 13.14

Try section 2L 90 x 90 x 8

Area = 2758 mm2

γ min = 27.5 mm

Section classification

B/T = 150/10 = 15 < 15.7£

It is semi compact

Maximum unsupported length = 4600 mm

λ = 4600 x 0.85 / 27.5 = 142.8

From table 9c of IS 800 with λ = 143

Fy = 250 mpa

Fcd = 72.5MPa

38

Axial capacity in compression = 72.5 x 2758/ 1000

= 200 kN > 198 kN

Hence safe.



Maximum force coefficient = 1.2

Factored wind load on tie manner = 1.893 x 1.2 x 4.5 (15/2)

= 76.67 kN (T)

Reaction = 76.67 x 4 / 2 = 153.33 kN


2) =7.652 m

Net area required = T /σat = 76670 / 3806

= 20.67 mm2

Hence the Provided section ISA 2L 100 x 100 x 8 is safe

3.6 DESIGN OF BEARER BEAM

Limit state method

C/C Length of beam = 8 m

Load from roof truss = 93.4 KN

Self weight of beam = 0.993 KN/m

Total factored load = 93.4+(0.993 x 8 x 1.5)

= 105.316 KN

Factored bending moment, M = WL/4

= 105.316 x 8/4

= 210.63 KN.m

Factored shear force F = 105.316/2

= 52.66 KN

Trial section:

Choose a section of ISMB400 with 2Nos - 200x16thk plates.

39

Section properties:

Overall depth D = 400 + (2 x 16) = 432mm

Width of flange B = 200mm

Thickness of flange tf = 27.2mm

Thickness of web tw = 27.2mm

Depth of web = 432 – (2 x 27.2) = 377.6mm

Moment of inertia about major axis Iz = 48161 x 104 mm

4

Moment of inertia about minor axis Iy = 2755.4 x 104 mm4

Elastic section modulus Zez = 2229.7 x 103 mm

3

Plastic section modulus Zpz = 2651 x 103 mm

3

Least radius of gyration ry = 44 mm

Section classification:

Outstand of compression flange = 100/27.2

= 3.676 < 9.4

Web with N.A at mid depth = 377.6 / 8.9

= 42.43 <83.9

Therefore the section is plastic.

Calculation of lateral torsional buckling moment:

Mcr = 2E Iy h/2 (KL)

2 )((1+(1/20)((KL/ry)/(h/tf))

2)0.5

KL/r = 8000/44 = 181.82

h/tf = 432/27.2 = 15.88

Mcr =( 2

x 2 x 105x 2775.4 x 10

4 x 432/2/8000

2)

((1+(1/20)((8000/44)/(432/27.2))2)0.5

= 290.265 KN.m

λLT = √(Zp x fy/Mcr)

= √(2651 x 104 x 250/290.265 x 106)

λLT = 1.51

ФLT = 0.5 (1+0.21 (1.51-.2) +1.512) = 1.78

ΧLT = 1/( Ф LT + (Ф LT 2 - λLT

2)0.5

40

ΧLT = 1/( 1.78 + (1.78 2 – 1.51

2)0.5

= 0.36 < 1.0

Fbd = Χ LT x fy/rmo

= 0.36 x 250/1.1= 83.59 N/mm2

Md = Zp x fbd

= 2651 x 103 x 83.59 = 221.61 KN.m

Md > Max moment (210.63 KN.m)

Hence safe

Shear capacity of section :

Vd = fy/(rmo x √3) x D x tw

= 250/(1.1 x √3) x 432 x 8.9 = 504.5 KN

0.6 Vd = 504.5 x 0.6

= 302.7 KN > Max shear force (52.658 KN)


Bending moment, M = 140.42 kN.m

Shear force F = 35.11 kN

Allowable бbc = 165 N/ mm2

Z reqd, = 140.42 x 106 / 165

= 851.03 x 103

mm3

Provide ISMB 500 whose properties are,

Zxx = 1808.7 x 103

mm3

Check for deflection :

Actual deflection = (WL3/48EI)

= 105.316/1.5 x 80003/(48 x 2 x 10

5 x 48161 x 10

4)

= 7.77 mm

Allowable Deflection = L/300

= 8000/300

= 26.67mm> 7.77mm

Safe in deflection.

41

CHAPTER 4

DESIGN OF GANTRY GIRDER

4.1 GENERAL

Crane capacity = 100 kN

Self – weight of the crane girder excluding trolley = 170 kN

Self – weight of the trolley, electric motor, hook, etc. = 35 kN

Span of gantry beam = 8m

Span of crane beam = 17m

Approximate min. approach of the crane hook = 0.8m

Wheel base c/c distance = 4m

Self-weight of rail section = 300N/m

Diameter of crane wheels = 100mm

For hand –operated crane

Lateral loads = 5% of maximum static wheel load

Longitudinal loads = 5% of weight of crab and weight lifted

4.2 LOAD CALCULATION AND ANALYSIS

Maximum wheel load:

202.5 KN

15kN/m

A

B

1.2m

42

Max. Point load on crane = 100+35 = 135 kN

Max. factored load on crane = 1.5 x 135 = 202.5 kN

crane self weight UDL load =170/17 = 10 kN/m

Factored uniform load = 1.5 x 10 = 15 kNm

For maximum reaction on the gantry girder the loads are placed on the crane girder as

shown in Fig.

Taking moment about B,

RA x 17 = (202.5x16)+(15x17x8.5)

RA = 318.10 kN

Similarly RB = (15 x 17) + 202.5 - 318

RB = 139.40 kN

The reaction from the crane girder is distributed equally on the wheels at the end

of the crane girder.

Maximum wheel load on each wheel of crane = 318.1/2 = ~ 160 kN

Maximum bending moment:

It consists of maximum moments Caused by the moving wheel loads On the

gantry girder and self-weight Of the gantry girder. For maximum a Bending moment,

the wheel loads shall Be placed as shown in Fig. The calculation of maximum bending

moments due to wheel loads and self weight of gantry try girder has been done

Assume self weight of gantry girder as 1.5 kN/m.

Total dead load, w = 1.5+0.3=1.8 kN/m

Factored dead load = 1.5 x 1.8 = 2.7 kN/m

The position of one wheel load from the mid-point of span

= wheel base/4 = 4/4 = 1 m

Bending moment due to live load :

Taking moments about D,

Rc x 8 = 160 x (8-1)+160 x 3

43

Rc = 200 kN

RD = 320 – 200 = 120 kN

Max. bending moments due to live load = 120 x 3 = 360 kN m

Bending moment due to impact = 0.10 x 360 = 36 k Nm

Total bending moment due to live and impact loads = 360+36 = 396 kN.m

Bending moment due to dead load = 2.7 x 8² = 21.6 kN.m

Maximum bending moment = 396 + 21.6 = 417.6kN.m

Maximum shear force

Taking moment about D,

Rc x 8 = 160 x 8+160 x 4

Rc = 240 kN

RD = 80 kN

Hence maximum shear force due to wheel loads = 240 kN

Lateral forces

Lateral force transverse to the rails = 5% of weight of crab and weight lilfted

= 0.05 x 135 = 6.75 kN

Factored lateral force = 6.75 x 1.5 = 10.13 kN

Lateral force on each wheel = 10.13/2 = 5.065 kN

Maximum horizontal reaction due to the lateral force by proportion at C

lateral force x reaction at C / max. wheel load = 5.065 x200/160 = 6.33 kN

Horizontal reaction due to lateral force at D = 10.13-6.33 = 3.8 kN

Maximum bending moment due to lateral load by proportion

= 5.065 / 360 x 160 = 11.4 kN.m

Maximum shear force due to lateral load by proportion

= 240/160 x 5.065 = 7.60 kN

44

4.3 DESIGN OF GANTRY GIRDER

Limit state method

Preliminary trial section

Let us try ISMB 500 @ 869 N/m with ISMC 250@ 304 N/m on its top

Property I –Section ISMB 500 Channel section ISMC 250

Area, A 11074 mm² 3867 mm²

Thickness of flange, tf 17.2 mm 14.1 mm

Thickness of web, tw 10.2 mm 7.1 mm

Width of flange, bf 180 mm 80 mm

Moment of inertia Iz 45218.3x104 mm

4 3816.8x 10

4 mm

4

Iy 1369.8 x 104 mm4 219.1x 104 mm4

Depth of section, h 500 mm 250mm

Radius at root R1 17 mm

cyy 23 mm

Moment of inertia of gantry girder

The distance of NA of built –up section from the extreme fibre of compression flange

Ў = ΣAY/ ΣA

= 11074 x (250 +7.1)+3867 x23/(3867 + 11074)

= 196.51 mm

Gross moment of inertia of the built –up section

ΙΖ gross = Ιzbeam + Ιz channel

Iz,gross = 61144.71x 104 mm

4

Iy,gross = 5186.60x 104 mm

4

Zez = Iz = 61144.71 x 104/(500 + 7.1 -196.51)= 1968.67x 10³ mm³

Plastic modulus of section ( Ignoring the fillets)

Equal area axis

3867 + 180 x 17.2 + y¯ (11.2) = 180 x 17.2 + (500-2 x 17.2 – y1 ) x 10.2

45

Z = 43.24 mm

Zpz1 = 332.94 x 10³ mm³

Plastic section modulus of the section below equal area axis,

Zpz2 = 2240.02 x 10³ mm³

Zpz = Zpz1+Zpz2 = 2576.96 x 10³ mm³

Plastic section modulus of compression flange about yy-axis,

Zpfy = 492.74 x 10³ mm³

Classification of section;

The entire section is plastic (βь = 1.0)

Check for moment capacity

Local moment capacity :

Mdz = βь Zpz f y/ m0 1.2 Ze f y/ m0

Mdz = 1.0 x 492.74 x 10³ x 250/1.10 x 10-6

= 585.67kNm

1.2 Ze f y/ m0 = 536.91 <585.91 kNm

Hence, moment capacity of the section,

Mdz = 585.671> 417.60 kNm

Which is safe.

Moment capacity of compression flange about y-axis,

Mdy, f = βь Zpyf f y/ m0 1.2 Zey,f f y/ m0

= 1.0 x 492.74x 10³ x 250/1.10 x 10-6 = 111.99 kN.m

1.2 x 360.14 x10³ x 250/1.10 x 10-6

= 98.22 kN.m

Hence, moment capacity of flange Mdy, f =98.22 kNm

Combined check for local moment capacity,

(Mz/Mdz)+ My,f/Mdy, f 1.0

417.6/536.91+11.4/98.22 = 0.89 < 1.0

Which is safe.

46

Bukling resistance in bending check

The elastic lateral buckling moment,

Mcr = c1 ²EI yhf/ (2L²LT [ 1+ 1/20 (LLT lry/hfltf)²]0.5)

Overall depth of the section, hf ~ h = 500+7.1 = 507.1 mm

Effective length, LLT = 8 x 10³ mm

Thickness of flange tf = 21.3 + 7.6 = 28.9 mm

Radius of gyration, ry = Iy/A = 5186.6 x 104/ (11074+3867) = 58.92mm

The coefficient, C1 = 1.132

Mcr = 1.132 x ² x 2 x 105 x 5186.6x 10

4 x 507.6/2 x (8 x 10³)²

x [ 1+1/20 (8x10³/58.92)²]0.5

/(507.6/28.9)

= 810.58 x 106 N mm

Non-dimensional slenderness ratio,

λLTz = √(βь Zpz f y/Mcr)= √(1 x 492.74 x 10³ x 250/810.58x106)

= 0.89

ФLTz = 0.5 [1+ LT ( LT z – 0.2 ) λ² λLTz]

αLT = 0.21

ФLTz = 0.5 x [ 1 + 0.21 x (0.809-0.2) + 0.809²] = 0.89

XLTz = 1/ ФLTz +( Ф²LTz - ² LT z)0.5 = 1/0.891 + (0.891²-0.809²)0.5 = 0.97

Design bending compressive stress

fbd = XLtz fy/ mo = 0.791 x 250 / 1.1 =168.07 N/mm²

The design bending strength,

Mdz = b Zpz fbd = 1.0 x 257.70x 10³ x 168.07 x 10-6

= 433.11kNm

> 417.6 kNm.

Which is all right.

The beam is safe in bending under vertical loads.

47

Since lateral forces are also acting, the beam must be checked for bi-axial bending.

The bending strength about y-axis will be provided by the top flange only as the lateral

loads are applied there only.

Mdy = Zyt fy/γmo

Zyt = Section modulus of top flange about yy-axis

= 360.136 x 10³ mm³

(Assuming the moment of inertia of top flange to be half of the moment of inertia of I-

section).

Μdy = 360.136 x 10³ x 250 x 10⁶ = 81.85 KN.m

Hence, the section is safe.

Check for shear force due to wheel load = 240 kN

Impact load = 0.1 x 240 = 24 kN (10% of wheel load)

Design shear force = 240 + 24 = 264 kN

Shear capacity = AV yw/ 3 mo = (500 x 10.2) x 250 x10-³ / 3 1.10

= 669.20 kN

Maximum shear, V = 264kN < 401.52 kN (0.6 Vd = 0.6 x 669.20= 401.52 kN)

Since V 0.6 Vd, the case is of low shear. No reduction will be therefore in the moment

capacity.

Deflection check

= WL³x(3a/4L-a³/L³) /6EI

W = maximum static wheel load = 160/1.5 = 106.67kN.

A = L-c = 8 x 10³-4 x 10³ = 1 x 10³ mm

Vertical deflection = 12.79mm

Permissible maximum deflection = L = 8000/500 = 16 mm > 12.79 mm

Hence it’s safe.

48


Use MB 600 @ 1226 N/m for primary section.

Bf = 210 mm Ix = 91813 x 104

mm4

Tf = 20.8 mm Iy = 2651 x 104

mm4

tw = 12 mm Z = 1810 cm³

A = 15621 mm² k = 34.2 mm

For compound section,

B = 1.50 Bf = 1.5 x 210 = 315, say 350 mm

L/B = 8000/ 350 = 22.86, Fbc = 165 N/mm²

Assume ƒbc = 2/3 x Fbc = 110 N/mm²

Zc = 417.6x106/110 = 3796.36x10

3 mm³

Ap =1.2(3035-1810)x10³/600 = 1471.92 mm2

Use MC 300, area 4360 mm².

Approximate check for tensile stress=

Zc = 3060.4 + 5/3 x 43.60 x 30 =5240.4 x10³ mm³

m = 43.60/156.21 = 0.280

Zt = 5240.4x10³ /(1 + 2 x 0.28) =3359.23x10³ cm³

ƒbt = 417.6 x 10³ /(3359.23 x 10³) = 124.3 N/mm² <165 N/mm²

Hence ok.

Exact check for ƒbc, ƒbt, ƒs and Fbc

Compound Section

MC 300

A = 4630 mm² Ix = 6420 x 104 mm

4

B = 90.0 mm Iy = 313 x104 mm

4

T= 13.6 mm Cy = 23.5 mm

t = 7.8 mm

e = 4630(307.8-23.5)/(4630+156.21) = 65 mm

49

c1 =307.8– 65 = 242.8 mm

c2 = 300 + 65 = 365 mm

I = 91813 + 156.21 x 6.5² + 313 + 46.3 (24.28– 2.35)²

= 120992 x 104 mm

4

Zc = 120992 x 104 /242.8 = 4983.2 x 10

3 mm³

Zt = 120992 x 104 /365 = 3314.85 x 10

3 mm³

ƒbc = 417 x 106 /4983 x 10³ = 83.8 N/mm²

ƒbt = 83.8 x 4983.2/3314.85 = 125.98 N/mm²

ƒs = 240 x 10³/ 600 x 12 = 33.33N/mm² < 0.4 X 250 =100 N/mm²

Hence ok.

Deflection

S = 161 x 10³ x 8000³/48 x 200 x 10³ x 120992 x 104

= 7.1 mm

Slim = 8000/750 = 10.67 mm > 7.1mm

Hence safe.

50

CHAPTER 5

ANALYSIS AND DESIGN OF COLUMN

5.1 LOAD CALCULATION

Dead load 1:

Self weight of Roof truss = 7.6 kN

Reaction on Each support = 7.6/2 = 3.8 kN

Dead load 2:

Self weight of Roof purlin = 3 kN

Self weight of bracing = 5/9.66 = 0.52 kN

Self weight of sheetings = 0.25 x 4 = 1 kN

Dead load 2 (Total load) = 1 + 0.52 = 1.52 Kn

Live load = 0.75 x 4 = 3 kN/m

Wind load calculation:

h/w = 11.5 /8.5 = 0.622

l/w = 48.5 / 18.5 = 2.622; < 4 & > 1.5

0.5 < h/w < 1.5

1.5 < l/w < 2 & 1.5 < l/w < 2

4th

case of table

TABLE 5.1 EXTERNAL PRESSURE CO-EFFICIENT (Cpe):

For 0o For 90o

ROOF -0.675 -0.80

-0.50 -0.625

CLADDING 0.70 -0.50

-0.30 -0.50

Internal pressure coefficient = + 0.5

Design Wind pressure = 1.893 kNm2

Wind word direction for 0

W11 = (0.7

W12 = (0.7 + 0.5) x 8 x 1.893


W21 = (-0.675

W22 = (-0.675+0.5) x 4 x 1.893

Leeword direction for 0

W31 = (-0.3

W32 = (0.3 + 0.5) x 8 x 1.893


W41 = (-0.5

W42 = (-0.5+0.5) x 4 x 1.893

FIGURE-5.1 COLUMN WIND LOAD DIAGRAM (FOR 0

Wind word direction for 0o

= (0.7 – 0.5) x 8 x 1.893 = 3.03 kN/m

= (0.7 + 0.5) x 8 x 1.893 = 18.17 kN/m


0.675 – 0.5) x 4 x 1.893 = -8.80 kN/m

0.675+0.5) x 4 x 1.893 = -1.33 kN/m

Leeword direction for 0o

0.3 – 0.5) x 8 x 1.893 = -12.12 kN/m

= (0.3 + 0.5) x 8 x 1.893 = 3.03 kN/m


0.5 – 0.5) x 4 x 1.893 = -7.57 kN/m

0.5+0.5) x 4 x 1.893 = 0 kN/m

5.1 COLUMN WIND LOAD DIAGRAM (FOR 0 )

51

= 18.17 kN/m

= 3.03 kN/m



W51 = (-0.5

W52 = (-0.5 + 0.5) x 8 x 1.893


W61 = (-0.8

W62 = (-0.8+0.5) x 4 x 1.893


W71 = (-0.5

W72 = (-0.5 + 0.5) x 8 x 1.893


W81 = (-0.625

W82 = (-0.625+0.5) x 4 x 1.893

FIGURE-5.2 COLUMN WIND LOAD DIAGRAM (FOR 90



0.5 – 0.5) x 8 x 1.893 = -15.14 kN/m

0.5 + 0.5) x 8 x 1.893 = 0 kN/m


0.8 – 0.5) x 4 x 1.893 = -9.84 kN/m

0.8+0.5) x 4 x 1.893 = -2.27 kN/m


0.5 – 0.5) x 8 x 1.893 = -15.14 kN/m

0.5 + 0.5) x 8 x 1.893 = 0 kN/m


0.625 – 0.5) x 4 x 1.893 = -8.55 kN/m

0.625+0.5) x 4 x 1.893 = -0.95 kN/m

5.2 COLUMN WIND LOAD DIAGRAM (FOR 90 )

52

53

5.2 STAAD ANALYSIS:

Limit state method

STAAD SPACE

START JOB INFORMATION

ENGINEER DATE 18-Apr-11

END JOB INFORMATION

INPUT WIDTH 79

UNIT METER KN

JOINT COORDINATES

1 0 0 0; 2 18 0 0; 3 0 9 0; 4 18 9 0; 5 0 11.5 0; 6 18 11.5 0; 7 9 15 0;

MEMBER INCIDENCES

1 1 3; 2 3 5; 3 2 4; 4 4 6; 5 5 7; 6 7 6;

DEFINE MATERIAL START

ISOTROPIC STEEL

E 2.05e+008

POISSON 0.3

DENSITY 76.8195

ALPHA 1.2e-005

DAMP 0.03

END DEFINE MATERIAL

MEMBER PROPERTY INDIAN

1 TO 4 TABLE ST ISMB600

5 6 TABLE ST ISA75X75X8

54

CONSTANTS

MATERIAL STEEL ALL

SUPPORTS

1 2 PINNED

LOAD 1 LOADTYPE Dead TITLE DL

SELFWEIGHT Y -1 LIST 1 TO 4

JOINT LOAD

5 6 FY -3.8

MEMBER LOAD

5 6 UNI GY -1.52

LOAD 2 LOADTYPE Live TITLE LL

MEMBER LOAD

5 6 UNI GY -3

LOAD 7 LOADTYPE Live TITLE CL1

JOINT LOAD

3 MZ -85.3


JOINT LOAD

4 MZ 85.3

LOAD 3 LOADTYPE Wind TITLE WL 0 -I

MEMBER LOAD

1 2 UNI GX 3.03

5 UNI Y 8.9

3 4 UNI GX 12.12

6 UNI Y 7.57

LOAD 4 LOADTYPE Wind TITLE WL 0 +I

MEMBER LOAD

1 2 UNI GX 18.17

5 UNI Y 1.33

55

3 4 UNI GX -3.03


MEMBER LOAD

1 2 UNI GX -15.14

5 UNI Y 9.84

3 4 UNI GX 15.14

6 UNI Y 8.55


MEMBER LOAD

5 UNI Y 2.27

6 UNI Y 0.95

LOAD COMB 101 DL+LL

1 1.5 2 1.5

LOAD COMB 102 DL+CL1+WL 0-I

1 1.5 3 1.5 7 1.5

LOAD COMB 103 DL+CL1+WL 0+I

1 1.5 4 1.5 7 1.5


1 1.5 5 1.5 7 1.5


1 1.5 6 1.5 7 1.5


1 1.5 3 1.5 8 1.5


1 1.5 4 1.5 8 1.5


1 1.5 5 1.5 8 1.5


1 1.5 6 1.5 8 1.5

56

LOAD COMB 110 DL+LL+CL1

1 1.5 2 1.5 7 1.5


1 1.5 2 1.5 8 1.5

LOAD COMB 112 DL+LL+CL1+CL2

1 1.5 2 1.5 7 1.5 8 1.5

PERFORM ANALYSIS

LOAD LIST 101 TO 112

PRINT MAXFORCE ENVELOPE LIST 1 TO 4

FINISH

STAAD OUTPUT

MEMBER FORCE ENVELOPE

---------------------

ALL UNITS ARE KN METER

MAX AND MIN FORCE VALUES AMONGST ALL SECTION LOCATIONS

MEMB FY/ DIST LD MZ/ DIST LD

FZ DIST LD MY DIST LD FX DIST LD

1 MAX 223.05 0.00 103 402.27 6.00 104

0.0 0.00 101 0.00 0.00 101 98.98 C 0.00 111

MIN -135.18 0.00 108 -912.66 8.25 107

0.00 9.00 112 0.00 9.00 112 168.90 T 9.00 102

2 MAX 125.99 2.50 104 296.83 0.00 108

0.0 0.00 101 0.00 0.00 101 82.78 C 0.00 111

MIN -90.38 2.50 107 -1031.59 0.00 103

0.00 2.50 112 0.00 2.50 112 173.40 T 2.50 102

3 MAX 157.84 0.00 102 0.00 0.00 101

57

0.00 0.00 101 0.00 0.00 101 129.32 C 0.00 103

MIN -75.99 9.00 108 -684.25 9.00 106

0.00 9.00 112 0.00 9.00 112 99.66 T 9.00 108

4 MAX 83.40 2.50 103 153.01 2.50 108

0.00 0.00 101 0.00 0.00 101 113.12 C 0.00 103

MIN -132.76 2.50 108 -684.25 0.00 102

0.00 2.50 112 0.00 2.50 112 104.16 T 2.50 108

********** END OF FORCE ENVELOPE FROM INTERNAL STORAGE **********

58

FIGURE-5.3 BENDING MOMENT DIAGRAM OF FRAME (LSM)

\

FIGURE-5.4 BENDING MOMENT DIAGRAM OF COLUMN (LSM)

59

FIGURE-5.5 AXIAL FORCE DIAGRAM OF COLUMN (LSM)

60


STAAD SPACE

START JOB INFORMATION

ENGINEER DATE 20-Apr-11

END JOB INFORMATION

INPUT WIDTH 79

UNIT METER KN

JOINT COORDINATES

1 0 0 0 ; 2 18 0 0; 3 0 9 0; 4 18 9 0; 5 0 11.5 0; 6 18 11.5 0;

7 9 15 0;

MEMBER INCIDENCES

1 1 3; 2 3 5; 3 2 4; 4 4 6; 5 5 7; 6 7 6;

DEFINE MATERIAL START

ISOTROPIC STEEL

E 2.05e+008

POISSON 0.3

DENSITY 76.8195

ALPHA 1.2e-005

DAMP 0.03

END DEFINE MATERIAL

MEMBER PROPERTY INDIAN

1 TO 4 TABLE ST ISMB600

5 6 TABLE ST ISA 75X75X8

61

CONSTANTS

MATERIAL STEEL ALL

SUPPORTS

1 2 PINNED

LOAD 1 LOADTYPE Dead TITLE DL

SELFWEIGHT Y -1 LIST 1 TO 4

JOINT LOAD

5 6 FY -3.8

MEMBER LOAD

5 6 UNI GY -1.52

LOAD 2 LOADTYPE Live TITLE LL

MEMBER LOAD

5 6 UNI GY -3


JOINT LOAD

3 MZ -85.3


JOINT LOAD

4 MZ 85.3


MEMBER LOAD

1 2 UNI GX 3.03

5 UNI Y 8.9

62

3 4 UNI GX 12.12

6 UNI Y 7.57


MEMBER LOAD

1 2 UNI GX 18.17

5 UNI Y 1.33

3 4 UNI GX -3.03


MEMBER LOAD

1 2 UNI GX -15.14

5 UNI Y 9.84

3 4 UNI GX 15.14

6 UNI Y 8.55


MEMBER LOAD

5 UNI Y 2.27

6 UNI Y 0.95

LOAD COMB 101 DL+LL

1 1.0 2 1.0


1 1.0 3 1.0 7 1.0


1 1.0 4 1.0 7 1.0

63


1 1.0 5 1.0 7 1.0


1 1.0 6 1.0 7 1.0


1 1.0 3 1.0 8 1.0


1 1.0 4 1.0 8 1.0


1 1.0 5 1.0 8 1.0


1 1.0 6 1.0 8 1.0


1 1.0 2 1.0 7 1.0


1 1.0 2 1.0 8 1.0

LOAD COMB 112 DL+LL+CL1+CL2

1 1.0 2 1.0 7 1.0 8 1.0

PERFORM ANALYSIS

LOAD LIST 101 TO 112

PRINT MAXFORCE ENVELOPE LIST 1 TO 4

FINISH

64

STAAD OUTPUT

MEMBER FORCE ENVELOPE

---------------------

ALL UNITS ARE KN METER

MAX AND MIN FORCE VALUES AMONGST ALL SECTION

LOCATIONS

MEMB FY/ DIST LD MZ/ DIST LD

FZ DIST LD MY DIST LD FX DIST LD

1 MAX 148.70 0.00 103 268.18 6.00 104

0.00 0.00 101 0.00 0.00 101 65.99 C 0.00 111

MIN -90.12 0.00 108 -608.4 8.25 107

0.00 9.00 112 0.00 9.00 112 112.60 T 9.00 102

2 MAX 83.99 2.50 104 197.88 0.00 108

0.00 0.00 101 0.00 0.00 101 55.19 C 0.00 111

MIN -60.25 2.50 107 -687.73 0.00 103

0.00 2.50 112 0.00 2.50 112 115.60 T 2.50 102

3 MAX 105.23 0.00 102 0.00 0.00 101

0.00 0.00 101 0.00 0.00 101 86.22 C 0.00 103

MIN -50.66 9.00 108 -456.17 9.00 106

0.00 9.00 112 0.00 9.00 112 66.44 T 9.00 108

65

4 MAX 55.60 2.50 103 102.01 2.50 108

0.00 0.00 101 0.00 0.00 101 75.42 C 0.00 103

MIN -88.51 2.50 108 -456.17 0.00 102

0.00 2.50 112 0.00 2.50 112 69.44 T 2.50 108

****** END OF FORCE ENVELOPE FROM INTERNAL STORAGE *****

66

FIGURE-5.6 BENDING MOMENT DIAGRAM OF FRAME (WSM)

FIGURE-5.7 BENDING MOMENT DIAGRAM OF COLUMN (WSM)

67

FIGURE-5.8 AXIAL FORCE DIAGRAM OF COLUMN (WSM)

68

5.3 DESIGN OF COLUMN

Limit state method

From the analysis results

Fx = 173.40 kN

Mz = 1031.6 kN.m

Consider Built up Section of 300 x 20 + 700 x 16

Width of flange = 300 mm

Over all depth = 700 mm

Thickness of Flange = 20 mm

Thickness of Web = 16 mm

Izz = 177092x104 mm

4

Iyy = 9022.53 x 103

Zzz = 5059.79 x103 mm3

Zyy = 257.7 x 103 mm

3

Ryy = 62.36 mm

Rxx = 276.28 mm 2

A = 23200 mm2

Buckling Class Classification

D/bf = 700 / 300 = 2.33

Tf = 20 mm � 40 mm

As per table 10 IS 800 – 2007

Buckling curve ‘a’ z – z axis

Design Strength

Pd = A x Fcd

λz = 11500 / 276.28

= 41

From table 9 (c) Fcd = 198 N/mm2

Pd = 23200 x 198

= 4594 > 173.40 kN

69

Moment Capacity

For major axis bending as per class 8.22 IS800 – 2007

ʎLT = ��/��

ʎ = 41

h / Lf = 700 / 20 = 35

Elastic critical shear fcrb = 6448

ʎLT = �250/6448 = 0.19 < 0.4

Hence no need for checking lateral torsional buckling

Moment Capacity About Major Axis

= ��. ��/��

= (5059.79 x 103 x 250)/1.1

= 1149.9 kN.M > 1031.6 kN.m

Hence the built up section is safe


Length of column L = 11.5m

Maximum moment M = 687.73 kNm

Maximum axial force P = 115.60 kN

Maximum shear force F = 148.70 kN

Try built up section of 400 x 25 + 800 x 20,

Properties:

Width of flange b = 400 mm

Overall depth D = 800 mm

Thickness of flange Tf = 25 mm

Thickness of web Tw = 20 mm

A =36000mm2

Rx = 320.91mm

Ry = 86.15mm

Zx = 9268.2 x 103 mm

3

Zy = 667.9 x 103 mm

3

70

Provide cladding rails at 2m spacing.

λx = L/Rx

= 11500/320.91 = 35.84

λy = L/Ry

= 2000//86.15= 23.22

D/T = 800/25 = 32

Actual stress are,

Comp stress = P/A

= 115.6 x 1000/36000

= 3.2 N/mm2

Bending stress = M/Z

= 687.73 X 106/(9268.23 x 10

3)

= 74.20 N/mm2

The allowable stresses are comp & bending are 85 N/mm2 & 120 N/mm2

Elastic bucking stress is

Fcr = 2E/λ2

= 2

x 2 x 105

/35.842

= 370 N/mm2

Ratio between the actual comp & bending stresses are,

=3.2/85 = 0.038

The interaction criterion gives,

= 0.038+ 74.2/120 = 0.66 < 1

Since the value is less than 1 the chosen section section is safe.

Note : column size may be reduced above gantry girder beam since the moment above

that portion is comparatively small and also the gantry girder may be supported on the

column.

Column bracing are provided at the end bays and it’s design is similar roof bracings.

71

5.4 DESIGN OF BASE PLATE

Limit state method

For M20 grade of concrete

Bearing strength concrete = 0.45*fck

= 0.45 x 20

= 9 N/mm2

Partial Safety factors:

rmo = 1.10

Properties of Built up Section of 300 x 20 + 700 x 16





Design axial comp. load =152.7 KN

Required area of slab base A = 152.7 X 103/9.0

= 0.017 m2

Side of base plate required L = B = √0.017

= 130mm

But the built up column size is 300 x 700.

So provide minimum size of base plate 500mm x 800mm

The bearing pressure of concrete w = 152.7 x 103/(500x800)

= 0.382 N/mm2

< 9 N/mm2

The maximum projection a = 100 mm

The minimum projection b = 50 mm

Thickness of base plate = √(2.5w(a2 – 0.3 b

2) x γmo/fy)

= 6.35mm

Provide minimum thickness of 16mm

Hence provide the base plate of size 500 x 800 x 16mm

72

Working Stress Method

Axial force = 101.8 kN





Pedastal concrete = M20

Minimum size of base plate is controlled by size of column and edge distance

Assuming 24mm bolt, the bolt hole = 25.5 mm

Edge distance = 44 mm to outer face

= 38 mm to rolled edge

Therefore the minimum size of plate is

B (min) = 400 + 2 x (44 + 38) = 564 mm

D (min) = 800 + 2 x (44 + 38) = 964 mm

Say B = 580 mm and D = 980 mm

Size of pedestal is controlled by cover reinforcement (40 mm) plus diameters of

reinforcement (20 mm) and Half of the ferrle (75mm), Bolt spacing 500 x 900.

The size of the pedestal is

Bp = 500 + 2 x (40 + 20 + 75/2) = 695 mm, Take 750 mm

Dp = 900 + 2 x (40 + 20 + 75/2) = 1095 mm, Take 1150 mm

Allowable direct compression on concrete

σcc = 4 N/mm2

73

Plate Area = Ab = 580 x 980 mm

Concrete Area = Ab = 750 x 1150 mm

Allowable bearing stress = σcc √(Ac / Ab) = 4 x √(750 x 1150 / 580 x 980)

= 4.927 < 2 σcc = 8 N/mm2

Bearing pressure w = P / Ab

=101800/ 580 x 980 = 0.179 N/mm2 < 4.927 mm

2

Thickness of plate

a = Plate cantilever = ( 580 – 400) / 2 = 90 mm

b = ( 900 – 800 ) / 2 = 90 mm

a = b = 90 mm

Bending moment on plate is

m = w x a2 / 2 = 4 x 902 / 2 = 16200

Thickness required t > √( 6m / σbs ) = a √( 3w / σbs ) = 24.3 mm

Provide base plate of size 580 x 980 x 25 mm

74

CHAPTER 6

DRAWINGS

FIG. 6.1 LAYOUT OF COLUMN

75

FIG. 6.2 ROOF TRUSS DETAILS

76

FIG. 6.3 LAYOUT OF ROOF TRUSS

77

FIG. 6.4 LAYOUT OF RAFTER BRACING

78

FIG. 6.5 LAYOUT OF EAVES BRACING DETAILS

79

FIG. 6.6 TYPICAL ELEVATION

80

CHAPTER 7

RESULTS AND CONCLUSION

7.1 RESULTS

Limit State Method

The design results obtained using limit state method is tabulated below.

TABLE 7.1 LIMIT STATE METHOD RESULTS

SL.

NO MEMBER

LIMIT STATE METHOD

1 PURLIN ISMC 125

2 ROOF TRUSS – TOP CHORD ISA 90 X 90 X 8

3 ROOF TRUSS – BOTTOM

CHORD ISA 90 X 90 X 10

4 ROOF TRUSS – INTERMEDIATE

STRUT ISA 65 X 65 X 8

5 ROOF RAFTER BRACING ISA 100 X 100 X 8

6 ROOF EAVES BRACING ISA 90 X 90 X 8

7 BEARER BEAM ISMB 400

8 GANTRY GIRDER ISMB 500 + ISMC 250

9 COLUMN 300 X 20 + 700 X 16

10 BASE PLATE 500 X800 x 16

81

Working Stress Method

The design results obtained Working stress method is tabulated below.

TABLE 7.2 – WORKING STRESS METHOD RESULTS

SL.

NO MEMBER

WORKING STRESS

METHOD

1 PURLIN ISMC 175

2 ROOF TRUSS – TOP CHORD ISA 100 X 100 X 10


CHORD ISA 100 X 100 X 10

4 ROOF TRUSS – INTERMEDIATE

STRUT ISA 75 X 75 X 8

5 ROOF RAFTER BRACING ISA 100 X 100 X 10

6 ROOF EAVES BRACING ISA 100 X 100 X 8

7 BEARER BEAM ISMB 500

8 GANTRY GIRDER ISMB 600 + ISMC 300

9 COLUMN 400 X 25 + 800 X 20

10 BASE PLATE 580 X 980 X 25

82

7.2 COMPARISON OF RESULTS

The design results obtained using limit state method and working stress method are

tabulated below.

TABLE 7.3 – COMPARISON OF RESULTS

SL.

NO MEMBER

METHOD OF DESIGN

LIMIT STATE

WORKING STRESS

1 PURLIN ISMC 125 ISMC 175

2 ROOF TRUSS – TOP

CHORD ISA 90 X 90 X 8 ISA 100 X 100 X 10


CHORD ISA 90 X 90 X 10 ISA 100 X 100 X 10

4 ROOF TRUSS –

INTERMEDIATE STRUT ISA 65 X 65 X 8 ISA 75 X 75 X 8

5 ROOF RAFTER BRACING ISA 100 X 100 X 8 ISA 100 X 100 X 10

6 ROOF EAVES BRACING ISA 90 X 90 X 8 ISA 100 X 100 X 8

7 BEARER BEAM ISMB 400 ISMB 500

8 GANTRY GIRDER ISMB 500 + ISMC 250 ISMB 600 + ISMC 300

9 COLUMN 300 X 20 + 700 X 16 400 X 25 + 800 X 20

10 BASE PLATE 500 X 900 X16 580 X 980 X25

83

7.3 CONCLUSION

By comparing the results, the limit state method of design is considered to be

economical over the conventional working stress method. In working stress method it

was ensured that the stresses used by the working loads are less than an allowable stress

obtained by dividing the yield stress by a factor of safety. The factor of safety

represented a margin for uncertainties in strength and load. The value of factor of safety

in most cases is taken to be around 1.67.

The major innovation in the Limit State Method is the introduction of the partial safety

factor format which essentially splits the factor of safety into two factors – one for the

material and one for the load. In accordance with these concepts, the safety format used

in Limit State Codes is based on probable maximum load and probable minimum

strengths, so that a consistent level of safety is achieved. But the uncertainties affecting

the safety of a structure are due to

• Uncertainty about loading

• Uncertainty about material strength and

• Uncertainty about structural dimensions and behaviour.

These uncertainties together make it impossible to guarantee that a structure will be

absolutely safe. All that we can ensure is that the risk of failure is extremely small,

despite the uncertainties.

So the limit state method of design is more effective where the uncertainty about the

loading is less. Limit state design philosophy takes into account the statistical nature of

loads and material strengths, there by providing consistent levels of safety. Its also

considers the other requirements such as serviceability and durability.

84

REFERENCES

1. Design of Steel Structures – Prof. Dr. A. Varma (chapter 2)

2. Plastic analysis design of steel structures by M. Bill Wong (Chapter 5)

3. The behavior and design of steel structures by N.S Trahair, M.A Bradford.

4. Steel structures design & behavior (LRFD Method, 4th

Edition) by Charles

G.Salmon & John E. johnson

5. Analysis and design of steel structures by N.S Trahair, M.A Bradford, D.A.

Nethercot and L.Gardner.

6. Stuctural Analysis by L.S Negi & R.S Jangid, Tata Mcgraw-Hill publishing

company limited.

7. Design of Steel structures By N.Subramaniyan, Published by Oxford University

Press

8. Limit state design of steel structures by S K Duggal Published by Tata Mcgraw-

Hill publishing company limited.

9. Design of steel structures by A.S. Arya & J.L. Ajmani, Published by Nem Chand

& Bros.

10. Design of steel structures by M. Raghupathi Published by Tata Mcgraw-Hill

publishing company limited.

11. IS 800-2007 “Indian standard code of practice for general construction in steel”

(LSM)

12. IS 800-1984 “Indian standard code of practice for general construction in steel”

(WSM)

13. IS 875-1987 part I, part II & part III, “Indian code of practice for loads”

14. SP 38(S&T)-1987 “Handbook of typified design for structures with steel roof

trusses”

rengan a than

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