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

2
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

3
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
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
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/mm2
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 Vz
2
, 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 =
ZISa
2Rg
Where,
Ah =horizontal seismic coefficient
Z = Zone factor corresponding to the seismic zone obtained from a map
I = Importance factor,
R = Response reduction factor,
𝑆𝑎
𝑔
= 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)
64
+Y)
65
+Y)
66
+Y)
67
+Y)
68
+Y)
69
+Y)
70
+Y)
71
+Y)
72
+Y)
73
+Y)
74
+Y)
75
+Y)
76
+Y)
77
+Y)
78
+Y)
79
+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)
84
+Y)
85
+Y)
86
+Y)
87
+Y)

14
88
+Y)
89
+Y)
90
+Y)
91
+Y)
92
+Y)
93
+Y)
94
+Y)
95
+Y)
96
+Y)
97
+Y)
98
+Y)
99
+Y)
100
+Y)
101
+Y)
102
+Y)
103
+Y)
104
+Y)
105
+Y)
106
+Y)
107
+Y)
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)
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)
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
177
SRSS +Y
178
SRSS +Y
179
SRSS +Y
180
SRSS +Y
181
SRSS +Y
182
SRSS +Y
183
SRSS +Y
184
+Y
185
SRSS +Y
186
SRSS +Y
187
SRSS +Y
188
SRSS +Y
189
SRSS +Y
190
SRSS +Y
191
SRSS +Y

17
192
+Y
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/m
2
Total load = 0.194kN/m
2
2

24
For 8 m bay,
roof dead load =0.194 x 8 x 18 = 27.93kN/m
weight 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-((18.43
4.4 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
00
-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
15.24
5
3.9
11.7
2
WL(-X) -3.7 0 -5.08
15.24
5
-3.9
11.7
2
WL(+Z) -16.07 -13.6
-16.07 -13.6
5.08
15.24
5
4.3 12.9
WL(-Z) -3.7 -1.24 -5.08
15.24
5
-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/m
Wind 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/m2
Wz = (0.775 x cos18.430
= 1.03kN/m
) x 1.4
Wy = (0.775 x sin18.43
2
0
= 0.34kN/m
) x 1.4
Mz = 1.5 x 1.03 x 8
2
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 104
mm
bf = 75mm Zez = 181.9 x 10
4
3
mm
ft = 11.4mm Zey = 26.3 x 10
3
3
mm
tw = 6.1mm Zpz = 211.25 x 10
3
3
mm
Zpy = 40.716 x 10
3
3
mm
Now,
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) = 1220mm
Shear 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)
𝑀𝑧
𝑀𝑑𝑧
+
𝑀𝑦
𝑀𝑑𝑦
≤ 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/m
Load parallel to slope = 0.194sin18.430 = 0.07kN/m
2
2
Wz = 2.39 x 1.4 = 3.346 kN/m
Wy =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 104
mm4
Izz=15200 x 104
mm
I
4
yy=4505 x 104
mm4
Iyy=508 x 104
mm
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 + ABh1
2
+ (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
9
mm
Z
4
zb = 1.52 x 109
= 3.95 x 10
/385.96
6
mm
Z
3
bt = 1.52 x 109
= 6.189 x 10
/(617.2+15.3-385.96)
6
mm
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 104
mm
I
4
y
I
for compression flange about y-y axis
cf =2246.6x 104
+ 15200 x 104
= 17446.6 x 104
mm
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 103
mm
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 103
mm3

40
Therefore, Zpz = 4592.5 x 103
+ 604.05 x 103
= 5196.55 x 103
mm
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 x105
N/mm
T
2
f
r
= 22.1 + 8.8 = 30.9mm
y=squareroot of Iyy
I
/A
yy= 19705 x 104
mm
A= 24180mm
4
r
2
y
According to clause 8.2.2.1 of IS 800 elastic lateral buckling moment
= 90.273
𝑀𝑐𝑟 =
𝐶1 𝜋2 𝐼 𝑦ℎ
2(𝐾𝐿)2 �1 +
1
20
[
𝐾𝐿
𝑟 𝑦
ℎ
𝑡 𝑓
]2
�
0.5
C1 = 1.132 ( from table 42 os IS 800:2007)
Mcr = 3012.85 x 106
N/mm
Non-dimensional Slenderness ratio :
𝜆 𝐿𝑇𝑍 = �
𝛽 𝑏 𝑍 𝑝𝑧 𝑓𝑦
𝑀 𝑐𝑟
= 0.6566
Along the z- section
ø 𝐿𝑇𝑍 = 0.5[1 + 𝛼 𝐿𝑇(𝜆 𝐿𝑇𝑍 − 0.2) + 𝜆 𝐿𝑇𝑍
2
= 0.5[1+021(0.7992-0.2)+0.79922
= 0.882
]

42
𝛸𝐿𝑇𝑍 =
1
[ø 𝐿𝑇𝑍 + (ø 𝐿𝑇𝑍
2
+ 𝜆 𝐿𝑇𝑍
2 )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
𝑀 𝑍
𝑀 𝑑𝑧
+
𝑀 𝑦
𝑀 𝑑𝑦
< 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 109
mm
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 104
mm
∆ =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 Pmaxz 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 = 50
𝑓𝑐𝑘
𝑓𝑦
�1 − �1 −
4.6𝑀𝑢
𝑓𝑐𝑘 𝑏𝑑2
�
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√𝑓R
ck
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√𝑓R
ck
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 = 1017mm
Total 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
c
C = 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 = (fx
2
+ 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 mm
4
260.474
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
= 206267mm
Direct 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
K2 = 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 + ʎvv
2
k2 + ʎ0
2
k3)0.5
f
= 1.66
cd = (fy/ɣmo)/(ɸ+(ɸ2
-ʎe
2
)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)
Resultan
t
(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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