Full Report

ANALYSIS AND DESIGN OF COMPRESSOR SHELTER

A Project Report Submitted by:

(Team I.D : 1098) Patel Paras Madhavbhai (110050106006) Mistry Suresh Aidanram (110050106014)

In partial fulfillment for the award of the degree

of BACHELOR OF ENGINEERING

in CIVIL ENGINEERING

Guide

Dr. Paulomi B Vyas Principal

Babaria Institute of technology, Vadodara

Department of Civil Engineering BITS edu campus, Vadodara.

Gujarat Technological University, Ahemdabad.

December, 2014

CANDIDATE’S DECLARATION

I declare that final semester report entitled “Analysis And Design Of Compressor Shelter” is

our own work conducted under the supervision of the guide DR.PAULOMI VYAS (Internal,

Guide).

I further declare that to the best of my knowledge the report for B.E. final year does not

contain part of the work which has been submitted for the award of B.E. Degree either in this

or any other university without proper citation.

PATEL PARAS MADHAVBHAI 110050106006 TEAM ID 1098

I

CERTIFICATE

This is to certify that the project entitled “Analysis And Design Of Compressor Shelter” is a

bonafied report of the work carried out by Patel Paras Madhavbhai for Industry Defined

Project in Semester VII under the guidance and supervision of external guide Mr.Neerav A

Mehta and internal guide Dr.Paulomi Vyas, Principal for the partial fulfillment of award of

the Degree of Bachelor of Civil engineering at Babaria Institute of Technology, Varnama,

Vadodara, Gujarat.

To the best of my knowledge and belief, this work embodies the work of candidate

themselves, have duly completed, fulfills the requirement of the ordinance relating to the

Bachelor Degree awarded by Gujarat Technological University and is up to the standard in

respect of content, presentation and language for being referred to the Examiner.

Main Guide

Head of Department

NEERAV A MEHTA Prof. NIPA A DESAI Asst General Manager L&T Knowledge city,Vadodara

HOD Civil Department BIT-Varnama

Internal- Guide

DR. PAULOMI VYAS PRINCIPAL BIT-Varnama

II

ACKNOWLEDGEMENT

The success and final outcome of this project required a lot of guidance and assistance from

many people and I am extremely fortunate to have got this all along the completion of my

project work. Whatever I have done is only due to such guidance and assistance and I would

not forget to thank them.

I respect and thank Mr. Bhatt Anand, for giving me an opportunity to do the project and

providing me all support and guidance which made me complete the project on time . I am

extremely grateful to him for providing such a nice support and guidance though he had busy

schedule managing the company affairs.

I owe my profound gratitude to my project guide Mr. Neerav A Mehta, who took keen

interest on my project work and guided me all along, till the completion of my project work

by providing all the necessary information for developing a good system.

I would not forget to remember my co-guide Mr. Sagar Sonawane, for their unlisted

encouragement and more over for their timely support and guidance till the completion of our

project work.

I heartily thank my internal project guide, Dr Paulomi Vyas, Principal , Civil department, for

his guidance and suggestions during this project work.

Patel Paras Madhavbhai

III

ABSTRACT

Petroleum industry is often divided into the three major sectors namely upstream, midstream and downstream sector. Upstream sector includes searching for potential sources of oil and gas fields. Midstream sector includes processes, stores and transports commodities such as curd oil, natural gas for further extraction of different products whereas downstream sectors would include process plants for extraction of different petroleum products. Process plants in mid and downstream sectors mainly consist of pipe racks, equipment structures, plant buildings and compressor shelter, etc. The present study covers analysis and design of compressor shelter which consists of operating floor for compressor operation and maintenance, static equipment’s, piping for equipment and overhead traveling gantry crane. Various parameters affect the structural size and quantities of the compressor shelter which include wind & seismic loads, crane loads, maintenance & live loads and removable roof requirements etc. Other parameters also to be taken into consideration are structural quantity and size are support conditions & use of appropriate structural/built-up sections which depends on design case selected. Therefore, these parameters are considered for the parametric case study.

• Introduction and functional requirements of compressor shelter. WORKFLOW CHART

• Detail study of each component of compressor shelter. • Design load calculation as per IS codes. • Understanding various structural geometries and support conditions. • Manual analysis of each component for compressor shelter. • Staad-Pro analysis of each component of compressor shelter. • Verifying software analysis results with manual checks. • Design of foundation and foundation joints. • Connection detailing and design. • Parametric study of structural geometry with various support conditions. • Parametric study of various structural geometries. • Comparison and conclusion based on parametric study results.

IV

ACKNOWLEDGEMENT

Table of CoNTeNTS

II

ABSTRACT III

TABLE OF CONTENT IV

LIST OF FIGURE VI

LIST OF TABLE VIII

NOMENCLATURE AND ABBREVIATION IX

SR

NO

CHAPTER

NO.

PARTICULAR PAGE NO

1 1 INTRODUCTION 1

2 1.1 General 2

3 1.2 Major items of compressor shelter 2

4 1.3 Information required for the design of compressor

shelter

2

5 1.4 Objective of study 04

6 2 SCOPE OF WORK 05

7 3 CRITERIA OF DESIGN 07

8 3.1 General 8

9 3.2 Design criteria and specifications 8

10 3.3 Material of construction 8

11 3.4 Soil data 8

12 3.5 Load on compressor shelter 8

13 3.6 Connection 12

14 3.7 Load combination 12

V

15 3.8 Design parameter 18

16 4 ANALYSIS AND DESIGN OFCOMPRESSOR

SHELTER

21

17 4.1 General 22

18 4.2 Structural modelling of compressor shelter 22

19 4.3 Dead load 23

20 4.4 Live load 24

21 4.5 Wind load 25

22 4.6 Seismic load 30

23 4.7 Design of purlin 32

24 4.8 Design of gantry 34

25 4.9 Design of foundation 44

26 4.10 Design of base plate and anchor bolt 52

27 4.11 Connections 55

28 5 STAAD.PRO REPORT 57

29 5.1 Material 58

30 5.2 Basic load cases 58

31 5.3 Node displacement summary 59

32 5.4 Beam displacement detail summary 59

33 5.5 Beam end displacement summary 60

34 5.6 Beam end force summary 60

35 5.7 Beam force detail summary 61

36 5.8 Reaction summary 61

37 6 REFERENCES 63

VI

FIGURE NO.

liST of fiGUReS

TITLE PAGE NO.

3.1 Direction Of Dead Load 9

3.2 Imposed Load 9

3.3 Effect Of Wind Load 10

4.1 3D Model Of Shelter 22

4.2 Transverse Direction 23

4.3 Longitudinal Direction 23

4.4 Bending Moment Due To Dead Load 24

4.5 Bending Moment Due To Live Load 25

4.6 Bending Moment Due To Wind load In +X Direction 28

4.7 Bending Moment Due To Wind load In -X Direction 28

4.8 Bending Moment Due To Wind load In +Z Direction 29

4.9 Bending Moment Due To Wind load In -Z Direction 29

4.10 Bending Moment Due To Siesmic Load In +X Direction 31

4.11 Bending Moment Due To Siesmic Load In +Z Direction 31

4.12 Bending Moment Due To Siesmic Load In +Y Direction 32

4.13 ISMC CHANNEL 200 33

4.14 Gantry Data 35

4.15 UB 610 With MC400 at its Top 38

4.16 Geometry of Footing 45

VII

4.17 One Way Shear Check 48

4.18 Two Way Shear Check 49

4.19 Foundation Rebar

Arrangement

51

4.20 Corner of Base Plate 53

4.21 Middle of Base Plate 53

4.22 Edge of Base Plate 53

4.23 Gantry to Column Connection 55

4.24 Gantry Beam Stay 56

5.1 Reactions 62

VIII

TABLE NO.

liST of TableS

TITLE PAGE NO.

3.1 Zone factor Z 11

3.2 Support Condition 12

3.3 Loads & Load Combination 12

3.4 Design Parametrs 18

4.1 Support Condition 22

4.2 Wind Force Direction 26

4.3 Governing Load For Footing 44

4.4 Loads At Base Of Foundation 44

4.5 Bearing Capacity Check 46

4.6 Overturning Moment Check 46

4.7 Sliding Check 47

4.8 Roark’s Chart 54

4.9 Property Of Section 55

5.1 Material 58

5.2 Basic Load Cases 58

5.3 Node Displacement 59

5.4 Beam Displacement 59

5.5 Beam End Displacement 60

5.6 Beam Forces 60

5.7 Beam End Forces Details 61

5.8 Reactions 61

IX

ABBREVIATION NOTATION AND NOMENCLATURE

L = Length of member (m) I = Moment if inertia (cm4

Z = Section modulus (cm

) 3

T

)

f

T

= Thickness of flange (mm)

w

R = Radius of gyration(cm)

= Thickness of Web (mm)

Fa = Permissible axial stress (N/mm2

f

)

a = Actual axial stress (N/mm2

F

)

B = Permissible bending stress (N/mm2

f

)

B = Actual bending stress (N/mm2

B = Width of section(mm)

)

D = Depth of section(mm)

Cc

W = Uniformly distributed load (kN/mm

= Effective slenderness ratio 2

P = Axial load on member (kN)

)

Fx = Horizontal force in X-direction (kN)

Fz = Horizontal force in Z-direction (kN)

S = Spacing between two member(mm)

Cf

M = Bending moment (kNm)

= Force coefficient

V = Shear force (kN)

KZT

C

= Topographic factor

pe

C

= Wind external force coefficient

pi

G = Gust factor

= Wind internal force coefficient

QZ = Wind pressure intensity (kN/mm2

Z = Zone factor

)

hh

I = Importance factor

= Height of structure

T = Time period

Cv

= Seismic coefficient

Team ID: 1098 INTRODUCTION

CIVIL ENGINEERING DEPARTMENT, BITS EDU CAMPUS, VARNAMA. 1

CHAPTER

1

INTRODUCTION



Compressors constitute an important part of the mechanical equipment in oil and gas refineries and petrochemical plants. Compressors are used for different applications in

1.1 GENERAL

the main and auxiliary process cycles. Compressor shelter is the enclosure provided for the compressor and its associated equipment to protect them from various environment agencies such as wind, snow, heat, rain, etc. It may be with or without wall cladding. It include operating platform, hoisting devices such as hoist or crane, which are generally provided for the operation and maintenance purpose. Sometime compressor shelter is provided with two compressors. One is operating and other is provided for standby. When compressor is damaged or in case of power cut, to avoid shutdown of the plant, another compressor is required. If the main operating compressor is electric type then the other one can be of steam or diesel operating compressor. 1.1.2 DEFINITION ". Compressor shelter is the enclosure provided for the compressor and its associated equipment to protect them from various environment agencies such as wind, snow, heat, rain, etc. It may be with or without wall cladding. It include operating platform, hoisting devices such as hoist or crane, which are generally provided for the operation and maintenance purpose."

1. Vibrating Equipment such as, compressor, blowers, pump. 1.2 MAJOR ITEMS FOR WHICH THE SHELTER IS ENVISAGED.

Foundation of the vibrating equipment and foundation of the shelter are isolated from each other in order to avoid vibration in shelter.

2. Static Equipment i.e. equipment other than pump or compressor such as a. Lube oil rundown tank. b. Silencer. c. Instrument Panel. d. Electrical Panel. e. Seal oil Console. 3. Piping for the equipments

In order to avoid transfer of vibration in shelter, pulsating pipe from the compressor shall not be supported on the shelter. Separate supports shall be provided for them.

4. Hoist or overhead travelling crane. 5. Operating platform around the equipment. 6. Stair and ladder for providing access to the operating platform. 1.3 INFORMATION REQUIRED FOR DESIGN OF COMPRESSOR SHELTER



1.3.1 Job Specification Job specification contains design criteria affecting design of compressor shelter: 1. Space requirement 2. Walkway, platform, ladder, access to equipment in compressor shelter. 3. Minimum headroom clearance under overhead piping or supporting steel within

area. 4. Access roads. 5. Standards to be used for the minimum spacing of lines in compressor shelter. 6. Handling and headroom requirement for equipment positioned under

compressor shelter.

1.3.2 Information Required For Basic Design The following are the minimum information required for the basic design of the compressor shelter. 1. Configuration and dimension information

a. Dimension (i.e. width, length, eaves height, extent of roofing and wall cladding, head clearance)

b. Location of longitudinal span where vertical bracing cannot be provided.

c. location of hoist beam, travelling crane distance, lifting height. d. Location of overhead travelling crane, travelling range and lifting

height. e. Location of dropping area. f. Location of stair, walkway, and ladder. g. Required area of operating stage around the equipment for the

maintenance purpose. h. Roof drainage method. 2. Piping information a. Piping route, piping load, and amount of thrust generated due to

pulsation of fluid in pipes.

b. Diameter and number of pipes for the wind load calculation. 3. Information related to Equipment. Location and loading data of the various components such as a. Lube oil rundown tank. b. Silencer. c. instrument panel. d. Electrical panel. e. Sea oil console



4. Information related to lifting devices - hoist beam capacity, dead weight, crane capacity, maximum wheel load, maintenance stage of hoisting devices, end clearance etc.

5. Other information such as width, height and route of electrical or instrument cable tray. Or duct, method of equipment installation etc.

1.3.3 Information Required For Detail Design All information mentioned in the basic information shall be fixed in loading data and shall be the base of the detail design.

To Analyse and design a economical and stable roofed structure for the usage in industrial purpose like shelter for compressor and their equipment etc., using STAAD PRO and manual calculations. 1.4 OBEJECTIVE OF STUDY

GTU Team ID 1098 SCOPE OF WORK

CIVIL ENGINEERING DEPARTMENT, BITS EDU CAMPUS, VARNAMA.

5

CHAPTER

2

SCOPE OF WORK

GTU Team ID 1098 SCOPE OF WORK


6

The main scope of this project is to apply class room knowledge in the real world by designing a roofed compressor steel shelter structure. These steel building require large and clear areas unobstructed by the columns. The large floor area provides sufficient flexibility and facility for later change in the production layout without major building alterations. The industrial buildings are constructed with adequate headroom for the use of an overhead traveling crane.

General Steel-framed buildings are commonly in use for industrial purposes. They are classified into three broad categories:

• Warehouse and factory buildings.

• Large span storage buildings.

• Heavy industrial process plant structures. i.e: compressor steel shelter come in this category

In the design of industrial buildings, load conditions and geometrical factors will dictate the degree of complication and hence the economy. The designer should possess good knowledge about the industrial process or purpose for which the building is intended. In this way, an optimum balance between safety, function and economy can be achieved.

For the present case study, a compressor steel shelter having span of 32m and width of 18m situated at manglore karnataka is to be designed and to be modeled in STAAD PRO. Different load such as dead load, wind load, live load, earthquake load are to be calculated and to be applied to the structure and stable and economical shelter is to be design.

For optimum steel shelter various parameters affect the structural size and quantities of the compressor shelter which include wind & seismic loads, crane loads, maintenance & live loads and removable roof requirements etc. Other parameters to be taken into consideration are structural quantity and size are support conditions & use of appropriate structural/built-up sections which depends on design case selected.

thus after designing the shelter, the parametric study of structural geometry with different support conditions and parametric study of different structural geometries is to be done and studying the different parametric study ,the conclusion and results are to be made in this present case study of analysis and design of compressor steel shelter.

GTU TEAM ID 1098 DESIGN CRITERIA


7

CHAPTER

3

DESIGN CRITERIA



8

This outline the criteria for the design of the structure, loads, load combinations, and material used for the problem.

3.1 GENERAL

The design of compressor shelter is carried out in accordance with the following codes and standards.

3.2 DESIGN CRITERIA AND SPECIFICATIONS

1. DESIGN PHILOSOPHY

2. INDIAN STANDARD CODE IS-800(2007)

3. INDIAN STANDARD CODE IS-875(1987) (PART 1 TO PART 3)

4. INDIAN STANDARD CODE 456(2007)

5. INDIAN STANDARD CODE IS-1893(2002) (PART 1 & PART 4)

Entire superstructure shall be of structural steel

3.3 MATERIAL OF CONSTRUCTION

1. Structural steel: fu = 410 N/mm2 fy = 250 N/mm

2. Structural concrete: M30 grade

2

Soil bearing capacity to be considered for the design is as follows.

3.4 SOIL DATA

SBC = 250N/mm

2

3.5.1 Dead Load (DL)

3.5 LOAD ON COMPRESSOR SHELTER



9

Figure 3.1 Direction of dead load

Dead loads are permanent loads that do not change in the structure’s life. They are,

• Self-weight of the structure

• Material incorporated into the structure: walls, floors, roofs, ceilings and permanent constructions

• Permanent equipments: fixtures, fittings, electrical wiring, plumbing tubes, ducted air system.

• Partitions, fixed and movable

• Stored materials When there is significant design change, dead loads should be reassessed and followed by a fresh structural analysis.

Calculation of Dead loads is done as follows:

Dead load of component= unit weight of the component x volume of the component

3.5.2 Live Load (LL):

Figure 3.2 Imposed or live load

Live loads are the result of the occupancy of a structure. In other words, it varies with how the building is to be used.

The specified live loads are generally expressed either as uniformly distributed area loads or point loads applied over small areas.



10

3.5.3 Wind Load (WL):

The wind pressure on a structure depends on the location of the structure, height of structure above the ground level and also on the shape of the structure. The code gives the basic wind pressure for the structures in various parts of the country. Both the wind pressures viz. including wind of short duration and excluding wind of short duration, have been given. All structures should be designed for the short duration wind.

Figure 3.3 effect of wind on building

Wind load are calculated as follows as per is code 875 part-3

The wind loads are calculated using IS: 875(part3) as.

Wind pressure = 0.6 x Vz2 , where Vz

V

=design wind speed

z = k1 k2 k3Vb k1 = probability factor

k2

k

= Terrain and height factor

3

Wind force = (Cpe-Cpi) x A x Pd, where, Cpe = external pressure

= Topography factor

Cpi = internal pressure

3.5.4 Seismic Load (V):

Earthquake loading is different from wind loading in several respects and so earthquake design is also quite different from design for wind and other gravity loads. Severe earthquakes impose very high loads and so the usual practice is to ensure elastic behavior under moderate earthquake and provide ductility to cater for severe earthquakes. Steel is inherently ductile and so only the calculation of loads due to moderate earthquake is considered. This has be done as per the IS 1893 code.



11

According to code, a horizontal seismic coefficient times the weight of the structure should be applied as equivalent static earthquake load and the structure should be checked for safety under this load as specified in IS 800.

𝐴ℎ =ZISa2Rg

Where,

Ah =horizontal seismic coefficient

Z = Zone factor corresponding to the seismic zone obtained from a map

I = Importance factor,

R = Response reduction factor,

𝑆𝑎𝑔

= Spectral Acceleration Coefficient

Table 3.1 Zone factor Z

Seismic Zone II III IV V

Sesimic Intensity Low Moderate Severe Very Severe

Zone factor 0.10 0.16 0.24 0.36

3.5.5 Crane Load (C):

Weight of crane bridge = 18400kg

Weight of trolley = 6000kg

Lift capacity of the crane = 15tonne

Distance between rails = 14.9m

Hook approach = 1.3m

No. of wheel = 4 wheel

Wheel load = 16.5tonne per wheel

Vertical impact load = 0.25 x (crane weight + trolley + lifted weight)



12

Side thrust /Horizontal impact load normal to runway rails = 020 x (trolley weight + lifted weight)

Longitudinal tractive force = 0.10 x (Maximum wheel load)

The steel supports shall be taken at top of the concrete pedestal and the boundary condition are as follow:

3.6 CONNECTION

Table 3.2 Support condition

MEMBER TYPE Transverse direction Longitudinal direction

COLUMN Pinned connection Pinned connection

Table 3.3 Load combination

3.7 LOAD COMBINATION

L/C Name 51 1.5(DL+LL+CL)

52 1.2(DL+LL+CL)+0.6(WL+X)

53 1.2(DL+LL+CL)+0.6(WL+Z)

54 1.2(DL+LL+CL)+0.6(WL-X)

55 1.2(DL+LL+CL)+0.6(WL-Z)

56 1.2(DL+LL+CL)+0.6(EQ SRSS +X)+0.18(EQ SRSS +Z)+0.18(EQ SRSS +Y)

57 1.2(DL+LL+CL)+0.6(EQ SRSS +X)+0.18(EQ SRSS +Z)-0.18(EQ SRSS +Y)

58 1.2(DL+LL+CL)+0.6(EQ SRSS +X)-0.18(EQ SRSS +Z)+0.18(EQ SRSS +Y)

59 1.2(DL+LL+CL)+0.6(EQ SRSS +X)-0.18(EQ SRSS +Z)-0.18(EQ SRSS +Y)

60 1.2(DL+LL+CL)-0.6(EQ SRSS +X)+0.18(EQ SRSS +Z)+0.18(EQ SRSS +Y)

61 1.2(DL+LL+CL)-0.6(EQ SRSS +X)-0.18(EQ SRSS +Z)+0.18(EQ SRSS +Y)

62 1.2(DL+LL+CL)-0.6(EQ SRSS +X)+0.18(EQ SRSS +Z)-0.18(EQ SRSS +Y)



13

63 1.2(DL+LL+CL)-0.6(EQ SRSS +X)-0.18(EQ SRSS +Z)-0.18(EQ SRSS +Y)

















80 1.2(DL+LL+(WL+X)+CL)

81 1.2(DL+LL+(WL+Z)+CL)

82 1.2(DL+LL+(WL-X)+CL)

83 1.2(DL+LL+(WL-Z)+CL)







14





















108 1.5(DL+(WL+X)) 109 1.5(DL+(WL+Z)) 110 1.5(DL+(WL-X)) 111 1.5(DL+(WL-Z)) 112 1.5(DL)+1.5(EQ SRSS +X)+0.45(EQ SRSS +Z)+0.45(EQ SRSS +Y) 113 1.5(DL)+1.5(EQ SRSS +X)-0.45(EQ SRSS +Z)+0.45(EQ SRSS +Y) 114 1.5(DL)+1.5(EQ SRSS +X)+0.45(EQ SRSS +Z)-0.45(EQ SRSS +Y) 115 1.5(DL)+1.5(EQ SRSS +X)-0.45(EQ SRSS +Z)-0.45(EQ SRSS +Y) 116 1.5(DL)-1.5(EQ SRSS +X)+0.45(EQ SRSS +Z)+0.45(EQ SRSS +Y)



15

117 1.5(DL)-1.5(EQ SRSS +X)-0.45(EQ SRSS +Z)+0.45(EQ SRSS +Y) 118 1.5(DL)-1.5(EQ SRSS +X)+0.45(EQ SRSS +Z)-0.45(EQ SRSS +Y) 119 1.5(DL)-1.5(EQ SRSS +X)-0.45(EQ SRSS +Z)-0.45(EQ SRSS +Y) 120 1.5(DL)+0.45(EQ SRSS +X)+1.5(EQ SRSS +Z)+0.45(EQ SRSS +Y) 121 1.5(DL)-0.45(EQ SRSS +X)+1.5(EQ SRSS +Z)+0.45(EQ SRSS +Y) 122 1.5(DL)+0.45(EQ SRSS +X)+1.5(EQ SRSS +Z)-0.45(EQ SRSS +Y) 123 1.5(DL)-0.45(EQ SRSS +X)+1.5(EQ SRSS +Z)-0.45(EQ SRSS +Y) 124 1.5(DL)+0.45(EQ SRSS +X)-1.5(EQ SRSS +Z)+0.45(EQ SRSS +Y) 125 1.5(DL)-0.45(EQ SRSS +X)-1.5(EQ SRSS +Z)+0.45(EQ SRSS +Y) 126 1.5(DL)+0.45(EQ SRSS +X)-1.5(EQ SRSS +Z)-0.45(EQ SRSS +Y) 127 1.5(DL)-0.45(EQ SRSS +X)-1.5(EQ SRSS +Z)-0.45(EQ SRSS +Y) 128 1.5(DL)+0.45(EQ SRSS +X)+0.45(EQ SRSS +Z)+1.5(EQ SRSS +Y) 129 1.5(DL)-0.45(EQ SRSS +X)+0.45(EQ SRSS +Z)+1.5(EQ SRSS +Y) 130 1.5(DL)+0.45(EQ SRSS +X)-0.45(EQ SRSS +Z)+1.5(EQ SRSS +Y) 131 1.5(DL)-0.45(EQ SRSS +X)-0.45(EQ SRSS +Z)+1.5(EQ SRSS +Y) 132 1.5(DL)+0.45(EQ SRSS +X)+0.45(EQ SRSS +Z)-1.5(EQ SRSS +Y) 133 1.5(DL)-0.45(EQ SRSS +X)+0.45(EQ SRSS +Z)-1.5(EQ SRSS +Y) 134 1.5(DL)+0.45(EQ SRSS +X)-0.45(EQ SRSS +Z)-1.5(EQ SRSS +Y) 135 1.5(DL)-0.45(EQ SRSS +X)-0.45(EQ SRSS +Z)-1.5(EQ SRSS +Y) 136 0.9(DL) +1.5(WL+X) 137 0.9(DL) +1.5(WL+Z) 138 0.9(DL) +1.5(WL-X) 139 0.9(DL) +1.5(WL-Z) 140 0.9(DL)+1.5(EQ SRSS +X)+0.45(EQ SRSS +Z)+0.45(EQ SRSS +Y) 141 0.9(DL)+1.5(EQ SRSS +X)-0.45(EQ SRSS +Z)+0.45(EQ SRSS +Y) 142 0.9(DL)+1.5(EQ SRSS +X)+0.45(EQ SRSS +Z)-0.45(EQ SRSS +Y) 143 0.9(DL)+1.5(EQ SRSS +X)-0.45(EQ SRSS +Z)-0.45(EQ SRSS +Y) 144 0.9(DL)-1.5(EQ SRSS +X)+0.45(EQ SRSS +Z)+0.45(EQ SRSS +Y) 145 0.9(DL)-1.5(EQ SRSS +X)-0.45(EQ SRSS +Z)+0.45(EQ SRSS +Y) 146 0.9(DL)-1.5(EQ SRSS +X)+0.45(EQ SRSS +Z)-0.45(EQ SRSS +Y) 147 0.9(DL)-1.5(EQ SRSS +X)-0.45(EQ SRSS +Z)-0.45(EQ SRSS +Y) 148 0.9(DL)+0.45(EQ SRSS +X)+1.5(EQ SRSS +Z)+0.45(EQ SRSS +Y) 149 0.9(DL)-0.45(EQ SRSS +X)+1.5(EQ SRSS +Z)+0.45(EQ SRSS +Y) 150 0.9(DL)+0.45(EQ SRSS +X)+1.5(EQ SRSS +Z)-0.45(EQ SRSS +Y) 151 0.9(DL)-0.45(EQ SRSS +X)+1.5(EQ SRSS +Z)-0.45(EQ SRSS +Y) 152 0.9(DL)+0.45(EQ SRSS +X)-1.5(EQ SRSS +Z)+0.45(EQ SRSS +Y) 153 0.9(DL)-0.45(EQ SRSS +X)-1.5(EQ SRSS +Z)+0.45(EQ SRSS +Y) 154 0.9(DL)+0.45(EQ SRSS +X)-1.5(EQ SRSS +Z)-0.45(EQ SRSS +Y) 155 0.9(DL)-0.45(EQ SRSS +X)-1.5(EQ SRSS +Z)-0.45(EQ SRSS +Y) 156 0.9(DL)+0.45(EQ SRSS +X)+0.45(EQ SRSS +Z)+1.5(EQ SRSS +Y) 157 0.9(DL)-0.45(EQ SRSS +X)+0.45(EQ SRSS +Z)+1.5(EQ SRSS +Y) 158 0.9(DL)+0.45(EQ SRSS +X)-0.45(EQ SRSS +Z)+1.5(EQ SRSS +Y) 159 0.9(DL)-0.45(EQ SRSS +X)-0.45(EQ SRSS +Z)+1.5(EQ SRSS +Y) 160 0.9(DL)+0.45(EQ SRSS +X)+0.45(EQ SRSS +Z)-1.5(EQ SRSS +Y) 161 0.9(DL)-0.45(EQ SRSS +X)+0.45(EQ SRSS +Z)-1.5(EQ SRSS +Y) 162 0.9(DL)+0.45(EQ SRSS +X)-0.45(EQ SRSS +Z)-1.5(EQ SRSS +Y) 163 0.9(DL)-0.45(EQ SRSS +X)-0.45(EQ SRSS +Z)-1.5(EQ SRSS +Y) 164 1(DL+LL+CL) 165 1(DL)+0.8(LL+CL+(WL+X))



16

166 1(DL)+0.8(LL+CL+(WL+Z)) 167 1(DL)+0.8(LL+CL+(WL-X)) 168 1(DL)+0.8(LL+CL+(WL-Z))

169 1(DL)+0.8(LL+CL)+0.8(EQ SRSS +X)+0.24(EQ SRSS +Z)+0.24EQ SRSS +Y

170 1(DL)+0.8(LL+CL)+0.8(EQ SRSS +X)-0.24(EQ SRSS +Z)+0.24EQ SRSS +Y

171 1(DL)+0.8(LL+CL)+0.8(EQ SRSS +X)+0.24(EQ SRSS +Z)-0.24EQ SRSS +Y

172 1(DL)+0.8(LL+CL)+0.8(EQ SRSS +X)-0.24(EQ SRSS +Z)-0.24EQ SRSS +Y

173 1(DL)+0.8(LL+CL)-0.8(EQ SRSS +X)+0.24(EQ SRSS +Z)+0.24EQ SRSS +Y

174 1(DL)+0.8(LL+CL)-0.8(EQ SRSS +X)-0.24(EQ SRSS +Z)+0.24EQ SRSS +Y

175 1(DL)+0.8(LL+CL)-0.8(EQ SRSS +X)+0.24(EQ SRSS +Z)-0.24EQ SRSS +Y

176 1(DL)+0.8(LL+CL)-0.8(EQ SRSS +X)-0.24(EQ SRSS +Z)-0.24EQ SRSS +Y


















17


193 1(DL+(WL+X))

194 1(DL+(WL+Z)) 195 1(DL+(WL-X)) 196 1(DL+(WL-Z)) 197 1(DL)+1(EQ SRSS +X)+0.3(EQ SRSS +Z)+0.3(EQ SRSS +Y) 198 1(DL)+1(EQ SRSS +X)-0.3(EQ SRSS +Z)+0.3(EQ SRSS +Y) 199 1(DL)+1(EQ SRSS +X)+0.3(EQ SRSS +Z)-0.3(EQ SRSS +Y) 200 1(DL)+1(EQ SRSS +X)-0.3(EQ SRSS +Z)-0.3(EQ SRSS +Y) 201 1(DL)-1(EQ SRSS +X)+0.3(EQ SRSS +Z)+0.3(EQ SRSS +Y) 202 1(DL)-1(EQ SRSS +X)-0.3(EQ SRSS +Z)+0.3(EQ SRSS +Y)

203 1(DL)-1(EQ SRSS +X)+0.3(EQ SRSS +Z)-0.3(EQ SRSS +Y)

204 1(DL)-1(EQ SRSS +X)-0.3(EQ SRSS +Z)-0.3(EQ SRSS +Y)

205 1(DL)+0.3(EQ SRSS +X)+1(EQ SRSS +Z)+0.3(EQ SRSS +Y)

206 1(DL)-0.3(EQ SRSS +X)+1(EQ SRSS +Z)+0.3(EQ SRSS +Y)

207 1(DL)+0.3(EQ SRSS +X)+1(EQ SRSS +Z)-0.3(EQ SRSS +Y)

208 1(DL)-0.3(EQ SRSS +X)+1(EQ SRSS +Z)-0.3(EQ SRSS +Y)

209 1(DL)+0.3(EQ SRSS +X)-1(EQ SRSS +Z)+0.3(EQ SRSS +Y)

210 1(DL)-0.3(EQ SRSS +X)-1(EQ SRSS +Z)+0.3(EQ SRSS +Y)

211 1(DL)+0.3(EQ SRSS +X)-1(EQ SRSS +Z)-0.3(EQ SRSS +Y)

212 1(DL)-0.3(EQ SRSS +X)-1(EQ SRSS +Z)-0.3(EQ SRSS +Y)

213 1(DL)+0.3(EQ SRSS +X)+0.3(EQ SRSS +Z)+1(EQ SRSS +Y)

214 1(DL)-0.3(EQ SRSS +X)+0.3(EQ SRSS +Z)+1(EQ SRSS +Y)

215 1(DL)+0.3(EQ SRSS +X)-0.3(EQ SRSS +Z)+1(EQ SRSS +Y)

216 1(DL)-0.3(EQ SRSS +X)-0.3(EQ SRSS +Z)+1(EQ SRSS +Y)

217 1(DL)+0.3(EQ SRSS +X)+0.3(EQ SRSS +Z)-1(EQ SRSS +Y)

218 1(DL)-0.3(EQ SRSS +X)+0.3(EQ SRSS +Z)-1(EQ SRSS +Y)



18

219 1(DL)+0.3(EQ SRSS +X)-0.3(EQ SRSS +Z)-1(EQ SRSS +Y)

220 1(DL)-0.3(EQ SRSS +X)-0.3(EQ SRSS +Z)-1(EQ SRSS +Y)

Design of the compressor shelter is carried out in STAAD-PRO (Series 4) using IS CODE 800-2007 standards. Following are the various parameters, which are given for the design purpose. This procedure is same for all the exercise.

3.8 DESIGN PARAMETERS

Table 3.4 Design parameters

PARAMETER VALUE DESCRIPTION

CODE - Must be specified as IS800 LSD Design Code to follow. See section 5.48.1 of the Technical

Reference Manual.

ALPHA 0.8 A Factor, based on the end connection type, controlling the Rupture Strength of the NetSection, as per Section 6.3.3:

0.6 = For one or two bolts

0.7 = For three bolts

0.8 = For four or more bolts

CMX 0.9 Equivalent uniform moment factor

for Lateral Torsional Buckling(as

per Table 18, section 9.3.2.2)

CMY 0.9 Cm value in local Y axes, as per

Section 9.3.2.2.

CMZ 0.9 Cm value in local Z axes, as per

Section 9.3.2.2.

DFF None

(Mandatory

for deflection

check)

"Deflection Length" / Maximum

allowable local deflection.



19

BEAM 1.0 0.0 = design at ends and those locations specified by the SECTION command.

1.0 = design at ends and at every 1/12th point along member length (default).

0 = Minimum detail

1 = Intermediate detail level

2 = Maximum detail

DJ1 Start Joint of

Member

Joint No. denoting starting point

for calculation of "Deflection

Length”.

DJ2 End Joint of

Member

Joint No. denoting end point for

calculation of "Deflection Length".

FU 420 MPA Ultimate Tensile Strength of Steel

in current units.

FYLD 250 MPA Yield Strength of Steel in current

units.

KX 1.0 Effective Length Factor for Lateral

Torsional Buckling (as per Table‐

15, Section 8.3.1)

KY 1.0 K value in local Y‐axis. Usually, the

Minor Axis.

KZ 1.0 K value in local Z‐axis. Usually, the

Major Axis.

LX Member

Length

Effective Length for Lateral

Torsional Buckling (as per Table‐

15, Section 8.3.1)

LY Member Length to calculate Slenderness



20

Length Ratio for buckling about local Y

axis.

LZ Member Length

Same as above except in Z‐axis

(Major).

MAIN 180 Allowable Slenderness Limit for

Compression Member (as per

Section 3.8)

TMAIN 400 Allowable Slenderness Limit for

Tension Member (as per Section

3.8)

RATIO 1.0 Permissible ratio of the actual to

allowable stresses.

TRACK 2 which results are reported.

0 = Minimum detail

1 = Intermediate detail

level

2 = Maximum detail

GTU TEAM ID 1098 ANALYIS AND DESIGN OF COMPRESSOR SHELTER


21

CHAPTER 4

ANAYSIS AND DESIGN OF COMPRESSOR SHELTER



22

This chapter is focused on structural modeling ,analysis and design of compressor shelter . Analysis and design of compressor shelter is carried out using STAAD. Pro (Series 4) software.

4.1 GENERAL

4.2 STRUCTURAL MODELING OF COMPRESSOR SHELTER

Figure 4.1 3D model of shelter

1.As shown in fig. 3D model of compressor shelter is prepared in STAAD. Pro2006. 2.Support condition are as follows:

Table 4.1 Support condition

MEMBER X-DIRECTION Z-DIREXTION COLUMN BASE PINNED PINNED 3.Staility to the structure is provided by the following conditions. Transverse direction : Rigid frame Longitudinal direction :Braced frame



23

Figure 4.2 Transverse Direction

4.The longitudinal direction are usually designed with brace frame due to less access requirements. the brace frame are effective structural forms for providing stiffness.

Figure 4.3 Longitudinal Direction

4.2.1 Member And Node Nnumber For member and node number refer to staad file in attachments.

For GI sheeting Thickness = 0.80mm 4.3 DEAD LOAD

Load = 0.069kN/m2

Fixing load = 0.025kN/m

Service load = 0.100kN/m2

Total load = 0.194kN/m2

2



24

For 8 m bay, roof dead load =0.194 x 8 x 18 = 27.93kN/mweight of purlin = 0.2 x 8 = 1.60kN

2

self weight of truss = 0.130 x 8 x 18 = 18.72 kN Foe welded sheet roof truss, self weight(w) = 53.7 + 0.53(A) = 53.7 + 0.53(8 x 18) = 0.130kN/m

• Loads in staad.PRO

2

Dead load due to sheeting = Load due to weight of sheet x Distance between purlin x Distance between two columns = 0.194 x 1.4 x 8 = 2.173kN Dead load due to purlin = Weight of purlin (ISMC 400, Steel table) x distance between two columns = 0.2kN/m x 8m =1.6kN Total Dead Load = 2.173 + 1.600 = 3.773kN ……….. say 3.78kN

Figure 4.4 Bending Moment Due To Dead Load

Live load =0.75-((18.434.4 LIVE LOAD ( AS PER IS 875 PART-2)

0 - 100

=0.5814kN > 0.4 (OK) )0.02) Table 2 Pg 14

• Load in staad.PRO



25

Live Load Live Load = 0.75-((18.430 - 100)0.02) (Table 2, Clause 4.1, Pg.14)

=0.5814kN > 0.4 (OK) Total Load Applied On Node

=0.5814 x 1.4 x 8 =6.512kN

Figure 4.5 Bending Moment Due To Live Load

WIND SPEED CALCULATION

4.5WIND LOAD (AS PER IS 875 PART-3)

Place = mangalore

Basic wind speed = 39m/s

Wind force = (Cpe-Cpi) x A x P ,

Where Cpi = ±0.5

Angle of roof = 18.430

Height of building = 11.5m(h)

Short dimension of building in plan = 18m(w)

h/w = 0.6388 > 0.5

As per IS 875 Part -3



26

α = 18.43o

For 0

by interpolation method

0

Cpe = -0.85 (windward direction) Cpe = -0.5 (leeward direction)

(wind angle)

For 900

Cpe = -0.8 (windward direction) Cpe = -0.6 (leeward direction)

(wind angle)

• Design wind speed, Vz

Vz = K1 x K2 x K3 x K4

Risk co-efficient (K1) assume K1 = 1 ( i.e life is 50 year)

Terrain category = 1 and Class A1 so K2 = 1.09

K3 it is assume as 1 (as per is 875 part - 3)

• Wind pressure calculation

Total height of building = 16.14m and Vb = 39m/s

design wind speed, Vz = K1 x K2 x K3 x K4

= 1 x 1.09 x 1 x 39

= 42.51m/s

Therefore Design wind pressure Pd = 0.6Vz2 = 0.6(42.51)2 = 108.4N/m2 = 1.084kN/m

Table 4.2 Wind force in Windward and Leeward

2

WIND

ANGLE

Cpe Cpi Cpe ± Cpi A x Pd WIND FORCE

WINDWARD LEEWARD

0

0

-0.8 -0.5 -0.5 -1.3 -1 12.36 -16.07 -12.36

+0.5 -0.3 0 12.36 -3.70 0

90

-0.8

0

-0.6 -0.5 -1.3 -1.1 12.36 -16.07 -13.60

+0.5 -0.3 -0.1 12.36 -3.70 -1.24



27

• Wind Load In Staad.Pro

Wind Load On Roof

Wind load

Wind ward

Lee ward

Wind ward

Lee ward

Roof angle

Wind ward(Load

kN)

Lee ward(Load

kN) Directio

n H

sinα H

cosα H

sinα H

cosα

WL(+X) -16.07 -12.36 -16.07 -12.36

18.43°

5.08 15.245 3.9 11.7

2

WL(-X) -3.7 0 -5.08 15.245 -3.9 11.7

2

WL(+Z) -16.07 -13.6 -16.07 -13.6

5.08 15.245 4.3 12.9

WL(-Z) -3.7 -1.24 -5.08 15.245 -4.3 12.9

• Wind Force On Column (IS 875-1987, Part 3, Table 4, Clause 6.2.2.1, pg.14)

Wind Angle

Wind ward

Lee ward

Wind ward

Lee ward CPI CPE+/-CPI

A B C D A B C D

0° 0.7 -0.3 -0.7 -0.7 0.5 1.5 0.2 -0.2 -0.2

-0.5 0.2 -0.8 -1.5 -1.5

90° -0.5 -0.5 0.7 -0.1 0.5 0 0 1.5 0.4

-0.5 -1 -1 0.2 -0.6

Load(kN)

Wind Pressure(Pd) x Bay Length(l)

A B C D

1.084 x 8 = 8.872

13.008 1.734 -1.734 -1.734 1.734 -6.938 -13.008 -13.008

0 0 13.008 3.469

-8.672 -8.672 1.734 -5.203



28

Figure 4.6 Bending Moment Due To Wind Load In +X Direction

Figure 4.7 Bending Moment Due To Wind Load In -X Direction



29

Figure 4.8 Bending Moment Due To Wind Load In +Z Direction

Figure 4.9 Bending Moment Due To Wind Load In -Z Direction



30

Design Spectrum, Ah

4.6 SEISMIC LOAD IN STAAD.PRO

Ah = (Z x I x Sa) / (2 x R x g)

Z/2 = 1.0 (As per Doc. No. 6812-9-2554-0138, Page No-3)

I = 1.75 (As Structure falls under Category 2,As per IS 1893 (P-4) : 2005))

R = 4 (Considering STEEL FRAME WITH CONCENTRIC BRACES,As per IS 1893 (P-4) : 2005))

Damping = 2% (For STEEL structure)

For Seismic load in X = (Z/2) x (I/R) = 1.0 x (1.75/4) = 0.4375

For Seismic load in Z = (Z/2) x (I/R) = 1.0 x (1.75/4) = 0.4375

For Seismic load in Y = (2/3) x 0.5 = 0.291

(1) Lump mass shall be calculated in separate STAAD file by providing release of

Fx, Fz, Mx, My & Mz at the junction of beams & columns. While calculating Lump

Mass consider DL, % of LL, P(O) loads.

LL shall be reduced 50%, when LL > 3 Kn/sqm (IS 1893 (P-I) : 2002 & Cl. No. 7.3.2)

Equipment Operating Weight Shall not be considered in above calculation.

(2) Support reaction from Lump Mass STAAD File shall be provided in +ve direction in below space.



31

Figure 4.10 Bending Moment Due To Siesmic Load In +X Direction

Figure 4.11 Bending Moment Due To Siesmic Load In +Z Direction



32

Figure 4.12 Bending Moment Due To Siesmic Load In +Y Direction

4.7 DESIGN OF PURLIN

Span of purlin = 8m spacing of purlin = 1.4m Dead load = 0.194kN/mWind pressure = 1.084 x 1.3 = 1.4092 kN/m

2

2

• LOAD COMBINATION 1. DEAD LOAD + LIVE LOAD

DD+LL = 0.194 + 0.581 = 0775kN/m

2

Wz = (0.775 x cos18.430

= 1.03kN/m) x 1.4

Wy = (0.775 x sin18.432

0

= 0.34kN/m) x 1.4

Mz = 1.5 x 1.03 x 82

2

= 9.89kNm /10

My = 1.5 x 0.34 x 82

= 3.26kN/m /10

Sfz = 1.5 x 1.03 x 8/2 = 6.18kN



33

Assume ISMC-200 (purlin) Properties of ISMC 200 h=D=200mm Izz = 1819.3 x 104mmbf = 75mm Zez = 181.9 x 10

4

3mmft = 11.4mm Zey = 26.3 x 10

3 3mm

tw = 6.1mm Zpz = 211.25 x 103

3mm Zpy = 40.716 x 10

3

3mmNow,

3

Section classification bf/tf = 75/11.4 = 6.54 < 9.4 d/tw = (200-(2 x 11.4))/6.1 = 29.04 < 42 Hence section is plastic

• SHEAR CAPACITY (IS 800-2007, Clause 8.4, pg.59) Av = (200 x 6.1) = 1220mmShear capacity = 𝐴𝑣 ∗ 𝑓𝑦

√3∗ 𝛾𝑚𝑜`

= (1220 x 250)/(1.73 x 1.1 x 10

2

3

) = 160.273 > 6.3kN (OK)

Shear capacity is greater than shear force hence OK

• MOMENT CAPACITY (IS 800-2007, Clause 8.2.1.2, pg.53)

𝑀𝑑𝑧 =𝛽𝑏 ∗ 𝑍𝑝𝑧 ∗ 𝑓𝑦

γmo

Mdz = (1 x 211.25 x 103 x 250)/1.10 x 10 = 48.01kNm

6

Mz = (1.2 x 181.9 x 103 x 250)/1.10 x 10 = 49.60kNm < Mdz ….. OK

6

Hence Mdz > Mz, The assumed section is safe. Mdy = (1.0 x 40.716 x 103 x 250)/1.10 x 10 = 9.25kNm

6

My = (1.2 x 26.3 x103 x 250)/1.10 x 10 = 7.17kNm < My .....OK

6

Hence Mdy > My The assumed section is safe

Figure 4.13 ISMC CHANNEL 200



34

• CHECK FOR AXIAL BENDING (IS 800-2007, Clause 9.3.1.1, pg.70) 𝑀𝑧𝑀𝑑𝑧

+ 𝑀𝑦𝑀𝑑𝑦

≤ 1

Therefore, 0.55 < 1 ......OK

• CHECK FOR DEFLECTION 𝛿 = 5𝑊𝐿3

384𝐸𝐼

W= 1.03 x 8 = 8.24kN 𝛿 = 5∗8.24∗1000∗ 80003

384∗2∗105∗1819.3∗ 104 = 15.09mm

Deflection limit is L

150 (IS 800-2007, Table- 6, pg.31)

= 8000

150 = 53.33mm > 15.09mm ..... (OK)

2. DEAD LOAD + WINDLOAD Load normal to slope = -2.586+0.194cos18.430 =-2.39kN/mLoad parallel to slope = 0.194sin18.430 = 0.07kN/m

2

2

Wz = 2.39 x 1.4 = 3.346 kN/mWy =0.07 x 1.4 = 0.098kN/m

2

Mz = 1.5 x3.346x 8

Mdz = 48.01kNm > Mz ..... (OK) = 32.12kNm

My = 1.5 x 0.098 x 82

Mdz = 9.25kNm > My ..... (OK) /10 = 0.940kNm

Hence assumed channel section for purlin is OK.

4.8 CRANE LOAD

DATA OF GANTRY



35

Figure 4.14 Gantry Data

• DATA FOR 15T CRANE CAPACITY

Centre-to-centre distance between column =8.0 m

Crane capacity =15T

Self-Weight of the crane girder = 24400 kg

Self-Weight of Crab = 6000 kg

Minimum hook approach (L1

Distance between wheel centre (c) =3.6m

) = 1.3m

Centre-to-centre distance between gantry rail (Lc

Self-weight of rail section = 300N/m

) = 14.9m

Yield stress of steel =250Mpa

SOLUTION

1. Load and bending moment calculations

(a) Load

(i) Vertical loading

• Calculation of maximum static wheel load

Maximum static wheel load due to the weight of the crane =180.44/4 = 45.11kN

Maximum static wheel load due to crane load



36

W1 = [Wt(Lc - L1)]/(2Lc

• Total load due to weight of crane =95.25 + 45.11 =140.36kN

) =[(150+60)(14.9-1.3)/2x14.9] =95.25kN

To allow for impact this load should be multiplied with 25% (IS 800 TABLE 12.3)

Design load =165.625kN

Therefore factored design load =1.5 x 165.625 = 249 kN

(ii) Lateral (horizontal ) surge load

• Lateral load per wheel =10%( hook + crab load)/4

=0.1 x (150+60)/4

=5.225kN

• Factored lateral load = 1.5 x 5.225

= 7.8375kN

(iii)Longitudinal (horizontal ) braking load

• Horizontal force along rails = 5% of wheel load

=0.05 x 165.625

= 8.281kN

• Factored load = 1.5 x 8.281 = 12.42kN

(b)Maximum bending moment

(i)Vertical maximum bending moment

• Without considering the self weight ,

The bending moment is maximum when the two loads are in such a position that the centre of gravity of the wheel loads and one of the wheel loads are equidistant from the centre of gravity of the girder.

M1 = Wc

M

L/4 = 249 x8/4 = 498 kNm

2 = 2Wc(L/2-c/4)2/L = 2 x 249(8/2-3.6/4)2

Hence M = 598.2kNm

/8 = 598.2kNm



37

• Assume that the self weight of the gantry girder is 1.86kN/m

Total dead load = 1860 + 300 = 2.160kNm

Factored dead load =2.16 x 1.5 = 3.24 kNm

BM due to dead load = wl2

(ii)Horizontal bending moment

/8 = 25.92kNm

• Moment due to surge load = 2 x 7.83(8/2-3.6/4)2

My = 20.06kNm

/7.5

(iii)Bending moment due to drag (assuming the height of rail as 0.15m& depth of

girder as 0.6m)

• Reaction due to drag force = 12.42 x (0.3+0.15)/8 =0.698kN

M3

Therefore, Total design bending moment

= R(L/2-c/4) = 0.698(8/2-3.6/4) = 2.1638kNm

Mz

(c)Shear force

=598.2+2.1638 = 600.36

(i)Vertical shear force

• Shear force due to wheel load

WL

Shear force sue to dead load =wl/2 = 12.92kN

(2-c/L) = 249(2-3.6/8) = 385.95kN

Maximum ultimate shear force = 385.95 + 12.92 = 398.91

(ii)Lateral shear force due to surge load

Vy

Reaction due to drag force = 0.698kN

= 7.83(2-3.6/8) = 12.1365kN

And maximum ultimate reaction

Rz

= 398.91 + 0.698 = 399.608



38

For Section selection

UB BEAM 610x229x140 with MC 400 at its top

Properties of section

UB BEAM MC400

A=17800mm2 A=6380mm

t

2

f= 22.1mm tf

t

=15.3mm

w=13.1mm tw

B=230.2mm B=100mm

=8.8mm

Izz=111777 x 104mm4 Izz=15200 x 104mm

I

4

yy=4505 x 104mm4 Iyy=508 x 104mm

R=12.7mm Cy=24.2mm

4

h=617.2mm R1=15.0mm

R2=8.0mm

1. Elastic properties of the combined section

Total area = 17800 + 6380 = 24180mm

The distance of NA of the built-up section from the extreme fiber of tension flange

2

Ӯ= [17800 x 617.2/2 + 6380 x (617.2 + 24.2 - 8.8)]/24180 = 385.9621mm

h1 = Ӯ -hB

=385.96 - 617.2/2 = 77.36mm

/2

h2 = (hB + tch) - Ӯ - C

=(617.2 + 8.8) -385.96 - 24.2

y

= 215.84mm

h3 = (hB + tch) - Ӯ - tw = (617.2 + 8.8) -385.96 -8.8) = 231.24mm

Figure 4.15 UB 610 With MC400 at its Top



39

Iz = IzB + ABh12 + (Iy)ch + Achh2

= 111777 x 10

2

4 + 17800 x (77.36)2 + 508 x 104 + 6380 x (215-84)

= 1.52 x 10

2

9mm

Z

4

zb = 1.52 x 109

= 3.95 x 10

/385.96

6mm

Z

3

bt = 1.52 x 109

= 6.189 x 10

/(617.2+15.3-385.96)

6mm

I

3

yy combined = 4505 x 104 + 15200 x 10

= 19705 x 10

4

4 mm

I

4

y

I

for tension flange about y-y axis

tf = 22.1 x (230.2)3/12 = 2246.6 x 104mm

I

4

y

I

for compression flange about y-y axis

cf =2246.6x 104 + 15200 x 104 = 17446.6 x 104mm

Z

4

y(for top flange only) =17446.6 x 104

=87.233 x 10

/200

4 mm

2. Calculation of plastic modulus

3

The plastic neutral axis divides the are in to two equal area i.e 12090mm

d

2

p =6380/2x tw

Now, Plastic section modulus below the equal area axis is

=6380/2 x 13.1 = 243.511mm

Summation of AӮ = (22.31 x 230.2) x (552.11 - 22.1/2) +(552.11 - 22.1) x 13.1 x [(552.11-22.1)/2] = 4592.5 x 103mm

Plastic section modulus above the equal area axis is

3

Summation of AӮ = 6380 x (73.89 - 24.2) + 230.2 x 22.1 x(73.89 - 8.8 - 22.1/2) + (73.89 - 8.8 - 22.1) x 13.1 x(73.89 - 8.8 - 22.1)/2 = 604.05 x 103mm

3



40

Therefore, Zpz = 4592.5 x 103 + 604.05 x 103 = 5196.55 x 103mm

For top flange only

3

Zpy =22.1 x (230.2)2/4 + (400 - 2 x 15.3)2 x 8.8/4 + (2 x 100 x15.3)x (200 - 15.3/2) = 1181576.013 mm

3. Check for moment capacity

3

• Check for plastic section

b/t of the flanges of the I-beam = [(230.2-13.1)/2x22.1]

=4.9117 < 9.4

b/t of the flanges of the channel = (1000 - 8.8)/15.3

= 5.960 < 9.4

d/t of the web of the I-section = (617.2 - 2 x 22.1)/13.1

= 43.74 < 84

Hence the section is plastic

• A local moment capacity

1.2Zefy/1.1 = 1.2 x 3.95 x106

M

x (250)/1.1 = 1077.27kN

dz = fyZp/1.1 = (250/1.1) x 5196.55 x 103 x10-3

Hence take,

= 1181.03 > 1077.27kNm

Mdz

M

=1077.29kNm

dz=( fyZp/1.1) x Zp(top flange) = (250/1.1) x 1181576.013 x 106

1.2Z

= 268kNm

efy/1.1 = 1.2 x 872330 x 250/1.1 x 10-6

Hence take M

= 237.908 < 268kNm

dy

• Combined local capacity check

= 237.908kNm

600.36/1077.29 + 20.06/237.908 = 0.6415 < 1

Hence the section is right choice.



41

4. Check for buckling resistance

As per IS 800 (clause 8.2.2), the design bending strength Md= βbZpf

We have β

bd

b

H = 617.2 + 8.8 = 626

= 1.0

KL = 8000mm

E = 2 x105N/mm

T

2

f

r

= 22.1 + 8.8 = 30.9mm

y=squareroot of Iyy

I

/A

yy= 19705 x 104mm

A= 24180mm

4

r

2

y

According to clause 8.2.2.1 of IS 800 elastic lateral buckling moment

= 90.273

𝑀𝑐𝑟 = 𝐶1 𝜋2𝐼𝑦ℎ2(𝐾𝐿)2

�1 + 120

[𝐾𝐿𝑟𝑦ℎ𝑡𝑓

]2�0.5

C1 = 1.132 ( from table 42 os IS 800:2007)

Mcr = 3012.85 x 106

N/mm

Non-dimensional Slenderness ratio :

𝜆𝐿𝑇𝑍 = �𝛽𝑏𝑍𝑝𝑧𝑓𝑦𝑀𝑐𝑟

= 0.6566

Along the z- section

ø𝐿𝑇𝑍 = 0.5[1 + 𝛼𝐿𝑇(𝜆𝐿𝑇𝑍 − 0.2) + 𝜆𝐿𝑇𝑍2

= 0.5[1+021(0.7992-0.2)+0.79922

= 0.882

]



42

𝛸𝐿𝑇𝑍 =1

[ø𝐿𝑇𝑍 + (ø𝐿𝑇𝑍2 + 𝜆𝐿𝑇𝑍2 )0.5]≤ 1

=1/[0882+(08822-0.79922)

= 0.7967 ≤ 1

0.5

fbd =fy 𝛸LT/γmo γmo

f

= 1.10 (from table 5 of code)

bd =0.7635 x 250/1.1 = 173.522 N/mm

Therefore,

2

Mdz = βbZpzfbd = 1.0 x 173.522 x 5196.55 x 10

=901.71kNm

-3

Here,

Mdz

Thus the beam is satisfactory under vertical loading. now it is necessary to check it under biaxial bending

=901.71 > 600.36kNm

Mdz=( fyZp/1.1) x Zp

= (250/1.1) x 1181576.013 x 10

(top flange)

6

= 268kNm > 237.908kNm

Hence, Mdz

= 237.908kNm

(a)Check for biaxial bending

𝑀𝑍

𝑀𝑑𝑧+𝑀𝑦

𝑀𝑑𝑦< 1

So, 600.36/901.71 + 20.06/237.908 = 07501 < 1

Hence the beam is safe.

If it come more than 1 than slighty bigger size of top channel may be selected.

5.Check for shear capacity

• For vertical loading



43

Vz

Shear capacity = A

=399.608kN

vfyw

= (617.2 x 13.1) x 250(√3x 1.10) x 10

(√3x1.10)

= 1060.923kNm

-3

The maximum shear force is 472.09kN which is less than 0.6 times the shear capacity

i.e 0.6 x 1060.923 = 636.55kN

Hence it is safe in vertical shear and there is no reduction in the moment capacity

6. Check for deflection

Serviceability vertical wheel load excluding impact= 140.36kN

• Deflection in mid span

∆=Wl3[(3a/4L)-(a3/L3

Where

)]/(6EI)

(i)Vertical

Combined Izz =1.52 x 109mm

So, ∆=140.36 x10

4

3 x (8000)3[(3x 2200)/(4x 8000) - (2200)3/(8000)3]/(6 x2 x105 x 1.52 x 109

= 7.28 < L/750 =10.66 mm (Table 6 of IS 800)

)

(ii)Lateral

Only the compound top flange will be assumed to resist the applied surge load as in the bending check

I = (IZch)+ IF = 17446.6x 104mm

∆ =5.225 x 10

4

3 x(8000)3[(3 x2200)/(4x8000)-(2000)3/(8000)3]/6 x 2 x105 x 17446.6 x 10

= 2.368 < 10.66mm (Table 6 of IS 800)

4

Hence the section is capable for gantry girger beam UB BEAM 610x229x140 with MC400 channel at its top.



44

4.9 DESIGN OF FOUNDATION

1. DESIGN OF FOOTING F1

Design of footing is as per IS code :456-2000

1.1 DATA

Governing load cases (kN)

Table 4.3 Governing Loads For Footing

Load combination Fx Fy Fz

110 207.166 572.949 3.252

82 166.291 766.687 36.674

139 2.542 59.196 171.839

Table 4.4 Load at the base of foundation

Lc Fx Fy Fz Mx Mz

110 207.166 572.949 3.252 621.498 9.756

82 166.291 766.687 36.674 498.87 110.02

139 2.542 59.196 171.839 7.626 515.51

GEOMETRY OF FOOTING



45

Figure 4.16 Geometry of Footing

D1 =2.7m D2 =1.0m

H1 =0.75m H3 =0.30m

H6 =2.25m H2 =2.55m

A1 =area of footing =2.7 x 5.5 =14.85 m

A2 =area of pedestal =1 x 1.45 =1.45m

2

1.2 FOUNDATION WEIGHT

2

w1 =weight of footing =A1 x H1 x γ

= 14.85 x 0.75 x 30

c

= 334.125kN

w2 =weight of pedestal = A2 x H2 x γ

= 1.45 x 2.55 x 30

c

=110.925kN

Total foundation weight =334.125+110.925

= 445.05kN



46

1.3 BEARING CAPACITY CHECK

Section Moduls Zx=13.86m3, Zz=6.932m

Area of Footing =14.85m

3

Table 4.5 Bearing Capacity Check

2

LC CASE P/A Mx/Z Mx z/Z Pmax z Pmin A.SBC CHECK

110 38.58 44.84 1.40 84.82 7.66 250 OK

82 51.64 35.49 15.86 103.86 0.21 250 OK

139 3.93 0.58 73.57 78.08 70.22 250 OK

Footing is safe in soil bearing capacity

1.4STABILITY CHECK

1.4.1FOR OVERTURNING

FACTOR OF SAFETY=Stabilising Moment(St.Mo)/Overturning Moment(Ot.Mo)

Table 4.6 Overturning Moment Check

X –direction

CASE P D/2 St.Mo Ot.Mo FS MIN. FS CHECK

110 572.94 1.375 787.79 621.498 1.26 1.2 OK

82 766.87 1.375 1054.44 498.87 2.11 1.2 OK

139 59.196 1.375 81.39 7.626 10.67 1.2 OK

Z-direction

CASE P D/2 St.Mo Ot.Mo FS MIN. FS CHECK

110 572.949 1.375 787.7 9.756 80.75 1.2 OK

82 766.87 1.375 1054.44 110.02 9.58 1.2 OK

139 59.196 1.375 81.39 515.51 0.15 1.2 OK



47

1.4.3FOR SLIDING

FACTOR OF SAFETY = (DOWNWARD WT OF FOOTING)*FRICTION COEFFICIENT)/FORCE IN X OR ZDIECTION

Table 4.7 Sliding Check

X-direction

CASE TOTAL WEIGHT

FORCE FS MIN. FS CHECK

110 1017.9 207.166 1.9 1.2 OK

82 1211.92 166.291 2.9 1.2 OK

139 504.246 2.542 155 1.2 OK

Z-direction

CASE TOTAL WEIGHT

FORCE FS MIN. FS CHECK

110 1017.9 3.252 125 1.2 OK

82 1211.92 36.074 13 1.2 OK

139 504.246 171.839 1.2 1.2 OK

2. MEMBER DESIGN

2.1 DESIGN OF FOOTING

Assume D =750mm =0.75m

d = Effective Depth = 250mm

For higher shear criteria d =d x 2 =500mm =0.5 m

fck=30N/mm

fy=250N/mm

2

2



48

%𝑝𝑡 = 50𝑓𝑐𝑘𝑓𝑦

�1 −�1 −4.6𝑀𝑢𝑓𝑐𝑘𝑏𝑑2

�

Where,

Mu=0.148 x fck x bd2 = 0.148 x 30x 2750 x 2502

Therefore, %pt = 0.534%

= 763.125 kNm

Ast

= 7342.5mm

for X-direction =%pt/100 x bd = 0.534/100 x 2750 x 500

PROVIDE 16 BARS OF 25 DIA IN X-DIRECTION AT BASE IN X DIRECTION

2

• Ast

=0.534 x 5500 x 500

for Y-direction = %pt/100 x bd

= 14685 mm

PROVIDE 30 BARS OF 25 DIA IN Y-DIRECTION AT BASE IN Y DIRECTION

2

2.2CHECK FOR SHEAR

2.2.1 ONE WAY SHEAR CHECK IN X-DIRECTION

For one way shear check critical section is taken at distance d, from face of column

Where, d is effective depth

Vu

Where, Uplift force = Factored load/Area of footing = 766.687/2.7x 5.5 = 50.90kN/m

= uplift force x 5.5 x d

V

2

u

τv= V

= 140kN

u/bd = 140 x 103

= 0.102N/mm

/2750 x 500 2

Figure 4.17 One Way Shear Check



49

%pt= 100Ast/bd

= 100 x 7850/2750 x 500

=0.570%

From IS 456:2000, table 19

By interpolation method ,τc= 0.509N/mm

So τc > τv hence ...............OK

2

ONE WAY SHEAR CHECK IN Y-DIRECTION

Vu

Τv = V

= 50.90 x 2.75 x 0.5 = 69.98kN

u/bd = 70 x 103/5500 x 500 = 0.025N/mm

%pt= 100Ast/bd

2

= 100 x 14718.75/5500 x 500

= 0.532 %

From IS 456:2000, table 19

By interpolation method, τc= 0.50N/mm

So τc > τv hence ...............OK

2

2.2.2 TWOWAY SHEAR CHECK

For two way shear check critical section is taken at 0.5d from face of column

d= effective depth

For X-direction

Length of critical section

bo

= 4(1000+250+250)

=4b'

Figure 4.18 Two Way Shear Check



50

= 6000mm

Vu

= 50.9(10.09)

= Uplift pressure x (15.125-5.03)

= 513.835kN

Τv =Vu/bod =513.835/6000x500 = 0.171N/mm

τc = 0.25√𝑓Rck

2

= 0.25√30 = 1.369N/mm

So, τc > τv.......................OK

2

For Y-direction

Length of critical section

bo

V

=4b' = 4(1450+2500+250) = 7800mm

u

= 50.9(10.09)

= Uplift pressure x (15.125-5.03)

= 513.835kN

Τv =Vu/bo

=513.835/7800x500

d

= 0.132N/mm

τc = 0.25√𝑓Rck

2

= 0.25√30

= 1.369N/mm

So, τc > τv.......................OK

2



51

3.DESIGN OF PEDESTAL

Assume size = 1000mm x 1450mm

Axial load = 766.91kN

fck=30N/mm

fy=415N/mm

2

Assume 0.8% of steel tp provide in pedestal (IS CODE 456:2000 PG 48 cl 26.5.3.1(h))

2

Asc provided = 0.008 x Ag(100 x 1450) = 11600

PROVIDE 24 BAR OF 25 DIA

• Diameter of lateral ties is minimum of following (IS CODE 456 PG 49)

(i) 1/4 x 25 = 6.25mm

(ii) 6mm

So, use 8mm dia ties

• Pitch for lateral ties is minimum of following (IS CODE 456:2000 PG 49)

(i)Least lateral dimension= 1000mm

(ii)16 x dia of bar = 16 x 25 = 400mm

(iii)300 mm

Taking smaller of these value Therefore, pitch = 300mm

PROVIDE 8 DIA TIES @ 300MM C/C

.

Figure 4.19 Foundation Rebar Arrangement



52

DATA

4.10 DESIGN OF BASE PLATE & ANCHOR BOLTS

Anchor bolt dia. = 36mm Total nos = 8 Bolt in tension = 4 Area of one bolt = 1017mmTotal area in tension = 4068mm

2

Allowable tensile capacity = 175kN 2

Allowable shear capacity = 145.85kN Modular ratio of elasticity(m) = Es/Ec = 2x105

P / 280000 = 7.143

cC = Width of base plate = 900mm

= 766.87kN

• For value of Y, Pt

If ∑v=0, sc = s x (P& sc

t+Pc

If ∑m=0, P

) / (Y x C)

c = -Pt

Elastic behavior of concrete support , -P

(D/2 –Y/3 + f)/(D/2 – Y/3 – e)

t / (As

Solving above 3 equations, Y

x sc x m) = (D/2 – Y/3 + f)/Y

3 + 3 x (e - D/2)Y2 + (8 x m x As x (f+e) / C)Y - (8 x m x As

K

x (f+e) / C) + (D/2 + f) = 0

1

K

= 3 x (e - D/2) = -1350

2 = 8 x m x As

K

x (f+e) / C = 36677.30

3 = -k2

So Y=1322.88mm

+ (D/2 + f) = -24317049.9

Pt = 315.04kN (by 2nd

Max. pressure below base plate (sc) = 1.82

Equation)

• Tension check

= Pc x (D/2 – Y/3 - e) / (D/2 – Y/3 + f)

=31.883kN

Tension in each bolt

=31.883/4 = 7.970kN < 175kN (hence ok)

• Shear check



53

Total shear force = (fx2 + fy

2)1/2 = (207.1662 + 171.832 )1/2

Shear in each bolt = 268.49/8 =33.565kN < 143.85kN

= 268.49kN

Check for combined stress

= (ft / Ft) + (fv / Fv

• Design

) = (7.970/175) + (33.565/143.85) = 0.275 < 1.4 (hence ok)

Case:1 Corner of base plate

a = 210mm

b = 324mm

a/b = 0.648

Bending stress = 0.5

Maximum pressure below base plate (sc) = 2 x (Pt+Pc

Permissible bending stress = 0.6(f

) / (Y x C) = 1.82

y) = 150N/mm

Permissible stress = (b x sc x b

2

2)/t2

So, t = 26mm

Case:2 Middle of base plate

a = 236mm

b = 424mm

a/b = 0.55

Bending stress coefficient = 0.360

So, t = 29mm

Case: 3 Edge of base plate

a = 324mm

b = 427mm

figure 14.20 Corner of Base Plate

Figure 14.21 Middle of Base Plate

Figure 14.22 Edge of Base Plate



54

a/b = 0.75

Bending stress coefficient = 0.555

So, t = 36mm

Hence provide 36mm base plate (take max. of three cases)

• Roark’s chart

Table 4.8 Roark’s Chart

a/b Bending coefficient

0.5 0.360

0.6 0.444

0.7 0.528

0.8 0.582

0.9 0.642

1.0 0.672

1.2 0.720

1.4 0.756

1.5 0.770

2.0 0.792

>2.0 0.798



55

GANTRY BRACKET TO COLUMN FLANGE 4.11 CONNECTION

Properties of section

Table 4.9 Property Of Section

SECTION h B t tw If Izz AREA (A) yy COLUMN HEA 700

690mm 300mm 14.5mm 27mm 215301.3 mm

513.89 mm4

260.47 4

BEAM UB 533

533.1mm 209.3 10.1mm 15.6mm 55230mm 2389 mm4 117cm4 4

Let us assume throat thickness equal to unity total height of UB Beam is 533.1 Lw

= 2 x 209.3 +2 x 476.5 =total length of weld

= 1371.6mm moment of inertia of weld Izz=2 x [209.3 x 1/12 x (476.5/2)2] + 2 x (476.5)2

= 54980278.68 mm/12

Z = I

4 zz/y

=54980278.68/(533.1/2) max

= 206267mmDirect shear stress = 380 x 100/1371.6

3

=27.7048N/mm Bending stress = M/Z = 242566/206267 = 1170N/mm Resultant stress = [(27.708)2+(1170)2] =1170.32N/mm

0.5

Figure 14.23 Gantry to Column Connection



56

Now strength of weld of 1mm length = tf

= 0.7 x S x 158 x design strength of weld

therefore 0.7 x S x 158 = 1170.32 S = 10.58mm=11mm So provide 11mm weld all around the beam Design of channel use to support the gantry bracket to column.

Assume ISMC 200

A = 2850mm

r = 11mm

2

Assuming two bolts at each end and fixed connection. (from table-12 of IS-800 2007)

K1 = 0.20

K

2 = 0.35

K3

ɛ = (250/f

= 20

y)0.5

ʎ

= 1.0

vv = (L/r)/(ɛ x (pi2 x E / fy)0.5

ʎ

) = 1.98

0 = (Lw+Lf)/( ɛ x (pi2 x E / fy) / 250)0.5

ʎ

= 0.246

e = (k1 + ʎvv2

k2 + ʎ02

k3)0.5

f

= 1.66

cd = (fy/ɣmo)/(ɸ+(ɸ2-ʎe2)0.5) = 1839.92N/mm

P

2

d = fcd

(hence assumed member is ok)

x A = 5244kN > 156.97kN

Figure 14.24Gantry beam stay

GTU TEAM ID 1098 STAAD.PRO RESULTS


57

CHAPTER 5

STAAD.PRO Results



58

Input

5.1 Materials

Table 5.1 Materials

Mat Name E (kN/mm2 ν )

Density (kg/m3

α ) (/°C)

1 STEEL 205.000 0.300 7.83E 3 12E -6

5.2 Basic Load Cases

Table 5.2 Basic Load Cases

Number Name 1 DL 2 CL 3 LL 4 WL +X 5 WL –X 6 WL +Z 7 WL –Z 8 EQ SRSS +X 9 EQ SRSS +Z

10 EQ SRSS +Y



59

Output

5.3 Node Displacement Summary

Table 5.3 Node Displacement

Node L/C X (mm)

Y (mm)

Z (mm)

Resultant (mm)

rX (rad)

rY (rad)

rZ (rad)

Max X 64 136:0.9(DL) +1.5(WL+X) 97.948 0.156 0.014 97.949 -0.000 0.007 0.005

Min X 55 110:1.5(DL+(WL-X)) -97.970 0.043 -0.759 97.973 -0.000 -0.007 -0.005

Max Y 303 109:1.5(DL+(WL+Z)) -0.058 18.193 107.062 108.596 0.012 0.000 -0.000

Min Y 275 137:0.9(DL) +1.5(WL+Z) -0.189 -18.173

117.175 118.576 0.012 -0.000 -0.000

Max Z 169 137:0.9(DL) +1.5(WL+Z) 0.144 -5.761

148.961 149.072 -0.000 0.000 -0.000

Min Z 164 139:0.9(DL) +1.5(WL-Z) -0.139 -5.362 -

137.318 137.423 0.000 0.000 0.000

Max rX 3 137:0.9(DL) +1.5(WL+Z) 0.000 0.000 0.000 0.000 0.019 -0.000 0.000

Min rX 21 139:0.9(DL) +1.5(WL-Z) 0.000 0.000 0.000 0.000 -0.018 -0.000 0.000

Max rY 18 138:0.9(DL) +1.5(WL-X) 0.000 0.000 0.000 0.000 -0.000 0.007 0.024

Min rY 25 136:0.9(DL) +1.5(WL+X) 0.000 0.000 0.000 0.000 0.000 -0.007 -0.024

Max rZ 13 110:1.5(DL+(WL-X)) 0.000 0.000 0.000 0.000 -0.000 -0.007 0.024

Min rZ 24 136:0.9(DL) +1.5(WL+X) 0.000 0.000 0.000 0.000 -0.000 0.007 -0.024

Max Rst 169 109:1.5(DL+(WL+Z)) 0.130 -8.500 148.832 149.075 0.000 0.000 -0.000

5.4 Beam Displacement Detail Summary Table 5.4 Beam Displacement

Beam L/C d (m)

X (mm)

Y (mm)

Z (mm)

Resultant (mm)

Max X 32 136:0.9(DL) +1.5(WL+X) 3.120 100.143 0.139 0.020 100.143

Min X 38 110:1.5(DL+(WL-X)) 3.120 -100.163 0.038 -0.690 100.165

Max Y 533 109:1.5(DL+(WL+Z)) 1.942 -0.057 18.193 107.062 108.596

Min Y 515 137:0.9(DL) +1.5(WL+Z) 1.942 -0.189 -18.173 117.174 118.576

Max Z 144 137:0.9(DL) +1.5(WL+Z) 3.200 0.154 -5.604 149.011 149.117

Min Z 199 139:0.9(DL) +1.5(WL-Z) 0.800 -0.149 -5.204 -137.365 137.464

Max Rst 144 137:0.9(DL) +1.5(WL+Z) 3.200 0.154 -5.604 149.011 149.117



60

5.5 Beam End Displacement Summary

Table 5.5 Beam End Displacement

Beam Node L/C X (mm)

Y (mm)

Z (mm)

Resultant

(mm)

Max X 32 64 136:0.9(DL) +1.5(WL+X) 97.948 0.156 0.014 97.949

Min X 38 55 110:1.5(DL+(WL-X)) -97.969 0.043 -0.758 97.972

Max Y 533 303 109:1.5(DL+(WL+Z)) -0.057 18.193 107.062 108.596

Min Y 515 275 137:0.9(DL) +1.5(WL+Z) -0.189 -18.173 117.174 118.576

Max Z 144 169 137:0.9(DL) +1.5(WL+Z) 0.144 -5.761 148.961 149.072

Min Z 155 164 139:0.9(DL) +1.5(WL-Z) -0.140 -5.362 -137.319 137.423

Max Rst 144 169 109:1.5(DL+(WL+Z)) 0.130 -8.499 148.832 149.075

5.6 Beam End Force Summary

The signs of the forces at end B of each beam have been reversed. For example: this means that the Min Fx entry gives the largest tension value for an beam. Table 5.6 Beam End Forces

Axial Shear Torsion Bending

Beam Node L/C Fx (kN)

Fy (kN)

Fz (kN)

Mx (kNm)

My (kNm)

Mz (kNm)

Max Fx 19 20 82:1.2(DL+LL+(WL-X)+CL) 609.368 -36.674 -2.237 0.000 0.000 0.000

Min Fx 7 29 136:0.9(DL) +1.5(WL+X) -375.510 -10.087 -0.002 0.000 -0.007 37.322

Max Fy 496 302 51:1.5(DL+LL+CL) 296.064 204.181 -0.016 -0.002 -0.575 359.209

Min Fy 497 289 51:1.5(DL+LL+CL) 294.074 -204.171 -0.483 -0.000 -0.781 -358.994

Max Fz 119 79 137:0.9(DL) +1.5(WL+Z) -325.092 -0.001 33.872 0.000 -63.387 0.015

Min Fz 116 84 139:0.9(DL) +1.5(WL-Z) -310.913 0.001 -31.339 0.000 58.127 -0.015

Max Mx 427 106 108:1.5(DL+(WL+X)) 31.293 -34.860 5.163 3.738 -0.117 -26.938

Min Mx 135 107 108:1.5(DL+(WL+X)) 35.268 -29.551 -4.374 -3.740 0.301 -24.634

Max My 33 84 139:0.9(DL) +1.5(WL-Z) -308.957 0.001 9.986 -0.000 58.127 -0.015

Min My 42 79 137:0.9(DL) +1.5(WL+Z) -323.135 -0.001 -10.952 0.000 -63.387 0.016

Max Mz 15 71 137:0.9(DL) +1.5(WL+Z) -213.115 -32.677 0.372 0.000 0.016 879.939

Min Mz 458 71 137:0.9(DL) +1.5(WL+Z) -119.877 -168.001 0.344 0.015 -0.005 -879.936



61

Sign convention as diagrams:- positive above line, negative below line except Fx where positive is compression. Distance d is given from beam end A.

5.7 Beam Force Detail Summary

Table 5.7 Beam Force Details

Axial Shear Torsion Bending

Beam L/C D (m)

Fx (kN)

Fy (kN)

Fz (kN)

Mx (kNm)

My (kNm)

Mz (kNm)

Max Fx 19 82:1.2(DL+LL+(WL-X)+CL) 0.000 609.368 -36.674 -2.237 0.000 0.000 0.000

Min Fx 7 136:0.9(DL) +1.5(WL+X) 3.700 -375.510 -10.087 -0.002 0.000 -0.007 37.322

Max Fy 496 51:1.5(DL+LL+CL) 0.000 296.064 204.181 -0.016 -0.002 -0.575 359.209

Min Fy 497 51:1.5(DL+LL+CL) 0.000 294.074 -204.171 -0.483 -0.000 -0.781 -358.994

Max Fz 119 137:0.9(DL) +1.5(WL+Z) 0.000 -325.092 -0.001 33.872 0.000 -63.387 0.015

Min Fz 116 139:0.9(DL) +1.5(WL-Z) 0.000 -310.913 0.001 -31.339 0.000 58.127 -0.015

Max Mx 427 108:1.5(DL+(WL+X)) 0.000 31.293 -34.860 5.163 3.738 -0.117 -26.938

Min Mx 135 108:1.5(DL+(WL+X)) 0.000 35.268 -29.551 -4.374 -3.740 0.301 -24.634

Max My 33 139:0.9(DL) +1.5(WL-Z) 3.600 -308.957 0.001 9.986 -0.000 58.127 -0.015

Min My 42 137:0.9(DL) +1.5(WL+Z) 3.600 -323.135 -0.001 -10.952 0.000 -63.387 0.016

Max Mz 15 137:0.9(DL) +1.5(WL+Z) 3.600 -213.115 -32.677 0.372 0.000 0.016 879.939

Min Mz 458 137:0.9(DL) +1.5(WL+Z) 0.000 -119.877 -168.001 0.344 0.015 -0.005 -879.936

Table 5.8 Reactions Of Footing 5.8 Reaction Summary

Horizontal Vertical Horizontal Moment

Node L/C FX (kN)

FY (kN)

FZ (kN)

MX (kNm)

MY (kNm)

MZ (kNm)

Max FX 3 110:1.5(DL+(WL-X)) 225.300 583.681 -4.780 0.000 0.000 0.000

Min FX 4 108:1.5(DL+(WL+X)) -222.019 577.047 -4.692 0.000 0.000 0.000

Max FY 20 82:1.2(DL+LL+(WL-X)+CL) 166.291 766.687 -36.674 0.000 0.000 0.000

Min FY 4 138:0.9(DL) +1.5(WL-X) 216.968 -567.841 -7.682 0.000 0.000 0.000

Max FZ 21 139:0.9(DL) +1.5(WL-Z) -3.066 -61.918 173.944 0.000 0.000 0.000

Min FZ 3 137:0.9(DL) +1.5(WL+Z) 3.212 -84.234 -180.731 0.000 0.000 0.000

Max MX 1 1:DL 0.053 57.507 -0.547 0.000 0.000 0.000

Min MX 1 1:DL 0.053 57.507 -0.547 0.000 0.000 0.000

Max MY 1 1:DL 0.053 57.507 -0.547 0.000 0.000 0.000

Min MY 1 1:DL 0.053 57.507 -0.547 0.000 0.000 0.000

Max MZ 1 1:DL 0.053 57.507 -0.547 0.000 0.000 0.000

Min MZ 1 1:DL 0.053 57.507 -0.547 0.000 0.000 0.000



62

Figure 5.1 Reactions Of Foundation

GTU TEAM ID 1098 REFERENCES


63

REFERENCES

GTU TEAM ID 1098 REFERENCES


64

1. IS: 456- 2000

2. IS: 800- 2007

3. IS: 875 1987 (part 1 to part 3)

4. SP 16

5. SP 38

6. IS: 1893:2005 part 1 and part 4

7. NPTEL, a online material for students published by IIT-Kharagpur, Design of Steel Structures, Design of welds, Module 24.

8. NPTEL, a online material for the students developed by IIT-Kharagpur, Design of Reinforced concrete structures, Design of Footings, Module27.

9. “Design of Steel structures”, Ramachandra Vol 1 & Vol 2, Standard Book House.

10. “Design of Steel Structures”, Dayaratnam, S.Chand Publications.

11. “Design of Steel structures”, N.Subramanian Based on the limit state design as per the latest indian standard code IS 800:2007

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