Unit 1 Class Notes-2019 pat..pptx

Design of Steel Structures
TE Civil
1
UNIT-1
a) Introduction to design steel structures
b) Design of Connections
c) Design of tension Member
Prof. R.M. Raut
ME (Structural Engineering)

UNIT-1
a) Introduction to design of steel structures
b) Design of tension member
2

A) Introduction to design of steel structures
3
•Industrial building, some commercial buildings like ware house, some public building like
stadiums, transport terminals, bridges of railway lines, various towers are constructed with steel
•Now days pre-engineered steel building are popular in all countries. In pre-engineered building
various components are manufactured in a factory and are erected at a site of construction by
which building completed very quickly. These buildings are combination of built up section, hot
rolled sections & cold formed elements. Pre-engineered building can be used for various
purpose such as factories, ware houses, supermarkets & offices.
•To design steel structures, civil (structural) engineer play important role. The design should suit
the purpose, safety, economy, & durability. It is difficult to access safety and economy due to
uncertainty about various factors like uncertainty about Loading, material strength, structural
dimensions & behaviors hence standardization of all design based on experience are required.
For that purpose IS 800:2007 code for general construction in steel design & introduced method
of limit state design which overcome on working stress method (IS 800-1984)
•In IS 800:2007 switching allowable stress (working stress) design approach to limit state design

Steel Pre-Engineered Building
4

Steel Pre-Engineered Building
5
Industrial Steel Building

Types of Steel Structures & Structural Members:
6
Question : State & Explain different type of steel structures with sketches.
SPPU-Dec-2013, May-2014, Aug-2014
1. Towers :-
Towers may be self supported or cable-stayed. Tower are made of steel angles or
tubes, bolted at site. Tower are different type such as telephone tower, wind mill tower,
observation tower,lighting tower, power transmission tower etc
The function of tower is to provide supports

2. Roof truss :-
8
A truss is a framed structure consisting of different member like tie, principal rafter, strut and
slings
The steel structure are used when
i) Span is very large and the beam construction is not economical
ii) The building is in area of heavy rainfall
The various types of trusses are
i Howe truss ii Pratt truss iii Simple fink truss
v Compound French truss vi Simple fan truss vii North light truss etc
iv Compound fink truss

3.Water tank :-
• They may be circular, rectangular or spherical.
• They may rest on ground or elevated
Fig. Water Tank
10

4. Bridges :-
• The truss & plate girder bridges are commonly used for small to moderate span, cable
stayed & suspension bridges for long spans. Foot bridges is very common in a railway
station.The function of bridges is to provide runway
Fig. Steel Bridge
11

5. Gantry Girder :-
• The travelling overhead crane are commonly used in work shop & Factories. The function of
gantry girder is to lift and move the heavy materials and machinery form one place to other
Fig. Gantry girder
12

6. Columns :-
• These are the members that resist compressive components of loads in a truss & bridge
piers. In building structure load and moments are transmitted to column through beam. The
function of column is to give supports
Fig. Steel Column
13

7. Chimney :-
• The basic use of chimney is for venting hot gases. It enables a smooth supply of these gases
Fig. Steel Chimney
14

8. Building frames :-
• The may include rigid, semi rigid or simple connected frames. Building frames may be
simple multistoried with single of many span. The function of building frames is to give
enclosure.
Fig. Building Frame
15

Advantages & Disadvantages of steel structures:
16
Question : State the advantages and disadvantages of steel structures.
SPPU-May-2014, Dec-2011, May-2012 May-2013
Advantages :-
 It has high strength per unit mass
 it has assured quality & high durability
 Speed of construction is high
 ductility :- steel being ductile material does not fail suddenly
 By using bolted connection steel structures can be easily dismantled & transported in other
sites quickly
Prefabrication :-as steel is in light weight it can be manufactured at the factory and transported at
site
Durability:- The properties of steel mostly do not change with time. This makes the steel most
durable material
Material is reusable
Retrofitting & strengthening :- Repairs, retrofit & strengthening of steel members is much simple.

Advantages & Disadvantages of steel structures:
17
Question : State the advantages and disadvantages of steel structures.
SPPU-May-2014, Dec-2011, May-2012 May-2013
Disadvantages :-
 Cost :- The cost of steel structures is more than RCC structures. Maintenance cost is high
 Corrosion :- Steel structure are susceptible to corrosion
Fire roof treatment :- structures steel members are non-combustible but they lose their strength
rapidly during fire therefore it required fire proof treatment which increases cost.
Aesthetic View :- Steel does not give pleasing appearance unless it is painted

Designation Yield Strength (Mpa) Ultimate Strength (Mpa)
Fe 410 250 410
Fe 440 300 440
Fe 490 350 490
Fe 540 410 540
Fe 570 450 570
18
Grade of Structural Steel :
IS Codes :
1. IS 800:2007- Code for practice for general construction in steel
2. IS 875:1987- Code for practice for design loads (for building & structures)
Part I : Dead load
Part II : Live load ( Imposed Load )
Part III :Wind load
Part IV : Special loads & load Combinations
3. IS 808:1989 - Dimension of hot rolled steel section

Pre-Engineered Steel Building
19
Industrial Steel Building

Types of steel sections :
20
• Structural steel section are hot rolled into different shapes to form the different sections.
• The use of single section or combined (built up ) sections with angles, channels, I section with or
with out cover plates & or lacing or battening are depends upon the loads to be resisted both in
magnitude and nature by the section
• IS 808-1989 – gives steel table in witch the general properties of the standard hot rolled sections are
given
Various types of rolled steel sections manufactured are bellow :
1. Rolled Steel I- Section
2. Rolled steel Channel Section
3. Rolled Steel Angle Section
4. Rolled Steel Tee- Section
5. Rolled Steel Plates

1. Rolled Steel I – Sections :
• I – Sections are designed with the depth (mm) and weight (Kg/m)
a) Indian Standard Junior Beam – ISJB
b) Indian Standard Light Beam – ISLB
c) Indian Standard Medium Beam – ISMB
d) Indian Standard Wide-Flange Beam – ISWB
e)Indian Standard Heavy Beam – ISHB
Example :- ISMB 300 @ 44.2 Kg/m
Fig. Rolled Steel I – Sections
21

Rolled Steel I – Sections
22

Rolled Steel I – Sections
23

2. Rolled Steel Channel Sections :
• Classified as
a) Indian Standard Junior Channel – ISJC
b) Indian Standard Light Channel – ISLC
c) Indian Standard Medium Channel – ISMC
d)Indian Standard Special Channel – ISHC
Channel Sections are designed with the depth
(mm) and weight (Kg/m)
Example :- ISMC 300 @ 35.1 Kg/m
Fig. Rolled Steel Channel Sections
24

Rolled Steel Channel Sections
25

3. Rolled Steel Angle Sections :
• Classified as
a) Indian Standard Equal Angle – ISA
b) Indian Standard Unequal Angle – ISA
Angle Sections are designed by series of names ISA followed by length thickness of leg.
Example : ISA 90x90x6 @ 8.2 Kg/m
: ISA 90x60x6 @ 6.8 Kg/m
Fig. Rolled Steel Equal Angle Sections Fig. Rolled Steel Unequal Angle Sections
PROF A V WAKCHAURE 26

Rolled Steel Angle Sections
27

4. Rolled Steel Tee - Sections :
• Classified as
a) Indian Standard Normal Tee Bars – ISNT
b) Indian Standard Heavy Flanged Tee Bars – ISHT
c) Indian Standard Special Legged Tee Bars – ISLT
d) Indian Standard Light Legged Tee Bars – ISLT
e) Indian Standard Junior Tee Bars – ISJT
Tee - Sections are designed depth (in mm) and
weight per meter length (in Kg/m)
Example : ISNT 60 @ 5.3 kg/m.
Fig. Rolled Steel Tee - Sections
28

Rolled Steel Tee - Sections
29

5. Rolled Steel Plates : -
30
• Classified as
a) Indian Standard Plate (ISPL) –
These plate are designated by ISPL, followed by dimension in mm ie length, width and thickness
eg- ISPL 2000x1100x8. Thickness of plates are more than 5 mm.
b) Rolled steel Strips –
These strips are designated as Indian Standard strips ISST, followed by width (mm) and thickness (mm)
eg. ISST 160x1.4. Thickness of strips are less than 5 mm.
C) Rolled Steel Flats –
They are designated by width followed by Indian standard Flats ISF and thickness e.g. 50 ISF 5 Flats have
thickness more than 5 mm
d) Rolled Steel Bars – Classified as a) Indian Standard Round Bar (ISRO)
b) Indian Standard Square Bar (ISSQ)
These are designed by ISRO, Followed by dimeter in case of round bar and width in case of square
e)g. ISRO – 12, ISRO – 16.

Various Design load C ombination: -
31
Various loads are expected to act on a structure
may be classified as given bellow :
a) Dead Loads (DL)
b) Imposed loads (IL)
c) Wind Loads (WL)
d) Earthquake Loads (EL)
e) Erection Loads (ER)
f) Accidental loads (AL)
g) Secondary Effects.
Various load combination are given as bellow :
1. DL
3. DL + WL
5. DL + TL
7. DL + IL + EL
9. DL + NL + TL
11. DL + IL + EL + TL
2. DL + IL
4. DL + EL
6. DL + IL + WL
8. DL + NL + TL
10. DL + EL + TL
12. DL + IL + EL + TL
Where TL – Temperature Loads

Design Philosophies : -
32
To decide size, shape and connection details of the member the following design philosophies are used
1. Working Stress Method (WSM)
2. Ultimate Stress Method (ULD)
3. Limit State Method (LSM)
1. Working Stress Method :-
• It is purely based on linear elastic theory it does not consider material non linearly.
The stress-strain behavior of material is considered as linear till the yield stress
• In this method yield stress at a maximum stressed fibre is considered to be a benchmark for estimating
capacity of steel section
• In this method uncertainties such as correct estimation of load, Possibility of occurrence of higher design
load, Material imperfection were taken in to account in one single factors known as factor of safety (FOS)
Allowable or Permissible stress = Yield Stress
Factor of Safety

Stress strain curve for Mild Steel
33

Advantages of WSM :-
• This method is simple.
• This method is reasonably reliable.
• As the working stress are low, the serviceability requirements are satisfied automatically.
Limitations of WSM :-
• Assumption of linear elastic behavior and the stresses under working loads can kept within
the permissible stresses are not found to be realistic.
• Many factors like the effect of stress concentrations, creep, shrinkage, residual stresses and
other secondary effect are not considered.
• This method does not consider material non-linearity and non non-linearity of structural
members.
• It gives uneconomical sections.
34

2. Ultimate Load Method (Load Factor Method) (ULM) :-
• This method is based on analysis of formation of plastic hinge when all fibres yield
• But structures continuous to resist load until sufficient no of plastic hinges are formed to transform the
structure in to collapse mechanism
• The load corresponding to this state is known as ultimate load.
Advantages of ULM :-
• Due to plastic hinge concept, redistribution of internal forces are accounted,
• It allows varied selection of load factors.
Limitations of ULM :-
• This method is more comprehensive method to take care of strength.
• It does not guarantee of serviceability performance.
35

2. Limit State Method of design (LSM):- Ref. Cl. No. 5.1/PN28/IS 800:2007
Question :- Explain advantages of limit state method over working stress method (Aug. 2015 May 2016)
• The limit state method of design developed to take account of all uncertainties (limitation of WSM and
ULM) that can be make the structure unfit for use by considering actual behavior of material and
structure.
Advantages of Limit state method over working stress method :-
• In this method, possibility of material non-linearity, structural non-linearity, calculation error are
considered.
• Factor of safety are called partial safety factors determined by statistical analysis.
• This method provides a realistic measures of the actual factor of safety.
• This method design any structural element for the strength as well as serviceability criterial.
• This method considered reserve strength of material beyond elastic limit of stress-strain curve.
36

Limit State of Strength and Serviceability:-
This method adopted two types :-
1. Limit states of strength
2. Limit state of serviceability
37
Ref. Cl. No. 5.2.2/ PN 28/ IS 800:2007
‘Limit states’ are the acceptable limit for the safety and serviceability requirements of a structure before
failure occurs
1. Limit States of Strength :- Ref. Cl. No. 5.2.2.1/ PN 28/ IS 800:2007
Question :- Explain in brief design philosophy of limit state of design for strength.
Dec. 10, May 13, Aug. 16, Dec. 16
The limit states of strength are those associated with failures (or imminent failure), under the action of
probable and most unfavorable combination of loads on the structure using the appropriate partial safety
factors, which may endanger the safety of life and property.
The limit state of strength includes:
i. Loss of equilibrium of the structure as a whole or any of its parts or components.

ii. Loss of stability of the structure (including the effect of sway where appropriate and overturning) or
any of its parts including supports and foundations.
iii. Failure by excessive deformation, rupture of the structure or any of its parts or components,
iv. Fracture due to fatigue,
v. Brittle fracture.
2. Limit States of Serviceability :- Ref. Cl. No. 5.2.2.2/ PN 28/ IS 800:2007
38
Question :- Explain in brief limit state Serviceability.
Dec. 10, May 11, Dec. 12, May 13, Dec. 13, Aug. 16, Dec. 16
These limit states are associated with the functioning (Performance) of structure or its components under
service or working loads.
The limit state of Serviceability includes:
i. Deformation and deflections, which may adversely affect the appearance or effective use of the
structure or may cause improper functioning of equipment or services or may cause damages to
finishes and non-structural members.

ii. Vibrations in the structure or any of its components causing discomfort to people, damages to the
structure, its contents or which may limit its functional effectiveness. Special consideration shall be
given to systems susceptible to vibration, such as large open floor areas free of partitions to ensure that
such vibrations are acceptable for the intended use and occupancy (see Annex C).
iii. Repairable damage or crack due to fatigue.
iv. Corrosion, durability.
v. Fire.
39

Partial Safety Factor for Loads and Resistance: -
40
1. Partial Safety Factor for Loads : -
It account for
Cl. No. 3.5.1 & 5.3.3 / PN 29 / IS 800:2007
i. Possibility of unfavorable deviation of the load from the characteristic value.
ii. Possibility of inaccurate assessment of the load.
iii. Uncertainty in the assessment of effect of the load.
iv. Uncertainty in the assessment of the limit being considered.
The load or load effect shall be multiplied by the relevant factor (FOS) γf
Table no. 4. / Cl. No. 3.5.1 & 5.3.3 / PN 29 / IS 800:2007
given in

2. Partial Safety Factor for Material γm : -
It account for
i. Possibility of unfavorable deviation of material strength from the characteristic value.
ii. Possibility of unfavorable Variation of member sizes.
iii. Possibility of unfavorable reduction in member strength due to fabrication and tolerance.
iv.Uncertainty to the calculation of strength of members.
Ref. Table no. 5. / Cl. No. 5.4.1 / PN 30 / IS 800:2007
42

Classification of Sections : - Ref. Cl No. 3.7 / PN 17./ IS 800:2007
43
Question : Explain the classification of cross-section.
Explain classification of cross section and draw the stress distribution.
SPPU-Dec. 2010, May-2011, Dec-2011, May-2012, Dec-2012,Dec-2013, May-2014, Dec-2015
Plate elements of a cross-section may buckle locally due to compressive stresses. The local buckling can be
avoided before the limit state is achieved by limiting the width to thickness ratio of each element of a cross-
section subjected to compression due to axial force, moment or shear.
On basis of the above, four classes of sections are defined as follows:
a. Class 1 (Plastic) — Cross-sections, which can develop plastic hinges and have the rotation capacity
required for failure of the structure by formation of plastic mechanism. The width to thickness ratio of
plate elements shall be less than that specified under Class 1 (Plastic), in Table 2.
b. Class 2 (Compact) — Cross-sections, which can develop plastic moment of resistance, but have
inadequate plastic hinge rotation capacity for formation of plastic mechanism, due to local buckling. The
width to thickness ratio of plate elements shall be less than that specified under Class 2 (Compact), but
greater than that specified under Class 1 (Plastic), in Table 2.

c. Class 3 (Semi-compact) — Cross-sections, in which the extreme fiber in compression can reach yield
stress, but cannot develop the plastic moment of resistance, due to local buckling. The width to
thickness ratio of plate elements shall be less than that specified under Class 3 (Semi-compact), but
greater than that specified under Class 2 (Compact), in Table 2.
d. Class 4 (Slender) — Cross-sections in which the elements buckle locally even before reaching yield
stress. The width to thickness ratio of plate elements shall be greater than that specified under Class 3
(Semi-compact), in Table 2. In such cases, the effective sections for design shall be calculated either
by following the provisions of IS 801 to account for the post-local-buckling strength or by deducting
width of the compression plate element in excess of the semi-compact section limit.
When different elements of a cross-section fall under different classes, the section shall be classified as
governed by the most critical element. The maximum value of limiting width to thickness ratios of
elements for different classifications of sections are given in Table 2.
44

Examples on Classification of cross-sections :-
Examples 1:- Classify the following sections where Fy = 250 MPa.
a) ISLB 300 @ 37.7 kg/m
Solution :
ISLB 300@37.7 kg/m
Using Steel Table,
bf = 150 mm, tf = 9.4 mm, h1 = 245.1 mm and tw = 6.7 mm
SPPU Dec. 2010,2011
Classification of section Ref. 3.7.2, 3.7.4 and Fig 2 / Table 2 / PN 18/ IS 800:2007
i. ∴ ε = √250/fy = 1
ii. Outstanding element of compression flange =
b
tf
= bf /2
tf
=
150/2
49
9.4
= 7.978<9.4 ε
∴ Flange is Plastic.

iii. Web of an I Section = d = h1
50
tw tw
=
245.1
6.7
= 36.58 < 84 ε.
.
. . Web is Plastic.
Therefore the section is plastic. ….................Ans

b) ISLB 450 @ 65.3 kg/m
Solution :
ISLB 450@65.3 kg/m
Using Steel Table,
bf = 170 mm, tf = 13.4 mm, h1 = 384 mm and tw = 8.6 mm
.
. . ε = √250/Fy = 1
i.
ii. Outstanding element of compression flange = b/ tf
bf /2 85
= = = 6.34 < 9.4 ε.
tf 13.4
.
. . Flange is Plastic.
d h1 384
iii. Web of an I Section = = = = 44.65 < 84 ε.
tw tw 8.6
.
. . Web is Plastic.
Therefore, the section is plastic. ….................Ans
51

c) ISMB 300 @ 44.2 kg/m
Solution :
ISMB 300@44.2 kg/m
Using Steel Table,
.
. . ε = √250/Fy = 1
i.
bf /2 70
= = = 5.64 < 9.4 ε.
tf 12.4
.
. . Flange is Plastic.
d h1 241.5
iii. Web of an I Section = = = = 32.5 < 84 ε.
tw tw 7.5
.
. . Web is Plastic.
Therefore, the section is plastic. ….................Ans
52

d) ISHB 200 @ 37.3 kg/m
Solution :
ISHB 200@ 37.3 kg/m
Using Steel Table,
.
. . ε = √250/Fy = 1
i.
bf /2 100
= = = 11.11
tf
10.5 ε < 11.11 < 15.7 ε .
9
. . Flange is Semi Compact.
d h1 158.4
= =
tw tw 6.1
iii. Web of an I Section = = 25.96 < 84 ε.
.
. . Web is Plastic.
Therefore most critical element is flange
Therefore, the section is Semi Compact ….................Ans
53

e) ISHB 400 @ 77.4 kg/m
Solution :
ISHB 450@77.4 kg/m
Using Steel Table,
.
. . ε = √250/Fy = 1
i.
bf /2 125
= = = 9.84
tf
9.4 ε < 9.84 < 10.5 ε .
12.7
. . Flange is Compact.
d h1 340.1
= =
tw tw 9.1
iii. Web of an I Section = = 37.37 < 84 ε.
.
. . Web is Plastic.
Therefore most critical element is flange
Therefore,the section is Compact. ….................Ans
54

Classify the following angle sections. SPPU Dec. 2010,2011
a. ISA 90 x 60 x 8 @ 8.9 kg/m
Solution :
ISA 90 x 60 x 8 @ 8.9 kg/m
Using Steel Table, Properties of section
A = 90 mm, B = 60 mm, t = 8 mm.
.
. . ε = √250/Fy = 1
i.
ii. Single angle section =
b B 60
= = = 7.5 < 15.7 ε
t
d
t
A
8
90
= = = 11.25 < 15.7 ε
t
(B+A) 150
t 8
(b+d)
= = = 18.75 < 25 ε
t T 8
Therefore, the section is semi-compact. ….................Ans
55

Classify the following angle sections.
b. ISA 150 x 150 x 12 @ 27.2 kg/m
SPPU Dec. 2010,2011
Solution :
ISA 150 x 150 x 12 @ 27.2 kg/m
A = 150 mm, B = 150 mm, t = 12 mm.
.
. . ε = √250/Fy = 1
i.
b B 150
= = = 12.5 < 15.7 ε
t
d
t
A
12
150
= = = 12.5 < 15.7 ε
t
300
t 12
(b+d) (B+A)
= = = 25 = 25 ε
t T 12
56

c. ISA 100 x 100 x 6 @ 9.2 kg/m
Solution :
ISA 100 x 100 x 6 @ 9.2 kg/m
A = 100 mm, B = 100 mm, t = 6 mm.
.
. . ε = √250/Fy = 1
i.
b B 100
= = = 16.66 > 15.7 ε
t
d
t
A
6
100
= = = 16.66 > 15.7 ε
t
200
t 6
(b+d) (B+A)
= = = 33.33 > 25 ε
t T 6
Therefore, the section is slender. ….................Ans
57

d. ISA 75 x 75 x 10 @ 11.0 kg/m
Solution :
ISA 75 x 75 x 10 @ 11.0 kg/m
A = 75 mm, B = 75 mm, t = 10 mm.
.
. . ε = √250/Fy = 1
i.
b B 75
= = = 7.5 < 15.7 ε
t
d
t
A
10
75
= = = 7.5 < 15.7 ε
t
150
t 10
(b+d) (B+A)
= = = 15 < 25 ε
t T 10
58

B) Design of Connections :-
Connections :-
The following three types of connections may be made in steel
a) Riveted connections
b) Bolted connections
c) Welded connection
a) Riveted Connections :-
A rivet is a ductile steel pin with a manufactured head at one end and a straight portion
known as shank.
Riveting is the process of the driving of rivets in to holes made
in the metal parts so that metal part are inseparable
The size of rivet hole is kept slightly more (1.5 to 2 mm )that the
Size of rivet.
Fig. Riveting
59

b) Bolted Connections:-
A bolt may defined as a metal pin with head at one end and a shank threaded at the other end to
receive a nut.
Steel washers are usually provided under the bolt as well as under the nut to serve two purposes
1. To distribute the clamping pressure of the bolted member.
2. To prevent the threaded portion of the bolt from bearing on the connecting pieces.
Fig. Bolted joint
61

Types of bolt :-
a) Unfinished or black bolts
b) Finished or turned bolts
c) High strength bolts
a) Unfinished or Black Bolts :-
sb 4
• These are bearing type of bolts. Unfinished bolts are also called ordinary, common, rough or black bolts.
• These bolts are made from mild steel rods with square or hexagonal head.
• The shank is left unfinished i.e. rough as rolled shown in fig b
• The black bolts of nominal diameters are 16, 20, 24, 30 and 36 are commonly used and they are
designated as M16, M20, M24 etc
• The ratio of net tensile area at thread to nominal plain shank area of bolt is 0.78 given in IS 1367 ( Part-1)
• Nominal Shank area A =
𝜋
𝑑2
• And Net tensile area at threads A nb 4
65
= 0.78 𝜋
𝑑2

b) Finished Bolt or Turned Bolts :-
• These are similar to unfinished bolts, with the difference that the shank of these bolts is formed from a
hexagonal rod.
• The surface of the bolts are prepared carefully and are machined to fit in the hole.
• The specification for turned bolts are given in IS-2591-1969
c) High Strength Friction Grip Bolts (HSFG):-
• The HSFG bolts are made from high strength steel rods.
• The surface of shank kept unfinished as in the case of black bolts.
• These provide a rigid joint. There is no slip between the elements connected.
• The shank of the bolt is not subjected to any shearing these result in to no-slippage in the joint
• Alterations can be done easily
66

Bolted Joint :-
There are two types of bolted joint
a) Lap joint
b) Butt joint
a) Lap joint :-
• When one member is placed above the other and both are connected by means of bolts, the joint is
known as lab joint
• The minimum lab shall not be less than four times thickness of the thinner part jointed or 40 mm
whichever is less
67

b) Butt joint :-
The two members to be connected are placed end to end and additional plate provided on either one or both
sides, called butt joint
If cover plate is provided on one side called single cover butt joint fig – e, f, g.
If the cover plates are provided on both the sides of main plates it is called a double cover butt joint fig-h, i, j.
69

Terminology :-
Q. Explain in detail gauge line, gauge distance, pitch, edge distance, and end distance with sketch
SPPU Dec. 12, May 13, May 14
i. Pitch :- It is the center to center distance between two
consecutive bolts in the direction of applied force/stress.
ii. Gauge Distance :- It is the centre to centre distance
between two consecutive bolts perpendicular in the
direction of applied force/stress
iii. Gauge line :- An imaginary line passing from bolts in the
direction of applied force/stress. Bolt line are called as
gauge line
iv. End Distance (e) :- It is the distance from end of the
member up to centre of first bolt in the direction of
applied force/stress
v. Edge distance :- (e’) :- It is the distance from edge of the
section up to centre of bolt perpendicular to the direction
of applied force/stress.
71

Failure of Bolted joint :-
The bolted joint may fail in any one of the following six ways
a) Shear Failure of Bolt :-
A bolt may fail due to shearing. The shearing may take place at one section (known as single shear) or two
section (known as double shear) depending on the type of connection.
73

b) Bearing Failure of Bolt / Plate
The force in the connecting parts is transferred through the bolt by bearing action. Due to this action, the
portion of the plate in contact with the bolt may get crushed and/or the portion of the bolt in contact with
the hole edge may get deformed.
74

c) Rupture/Tearing failure of plate :-
This type of failure takes place along the weakest section of the plate due to the presence of holes.
Fig. Rupture/Tearing failure of plate
d) Tension Failure of Bolt :-
In Some Connections, the bolts may be subjected to tension. Fracture may takes place at the root of the
thread since it is weakest section.
Fig. Tension Failure of Bolt
75

e) Splitting of Plates/Shear Failure of Plates End :-
Bolt may have been placed at a lesser edge distance than required causing the plates of split or shear dot.
Fig. Splitting of Plates
f) Block (share) Failure of Plate :-
A portion of the plane may fail by shearing and rupture known as block failure.
Fig. Block (share) Failure
76

 Design Specification for Bolted Connection as per IS 800-2007
77
1. Shear Capacity of a Bolt :- Ref. Cl No. 10.3.3/PN 75/IS 800:2007
2. Bearing Capacity of a Bolt :- Ref. Cl No. 10.3.4/PN 75/IS 800:2007
3. Tension Capacity of a Bolt :- Ref. Cl No. 10.3.5/PN 76/IS 800:2007
4. Tension Capacity of Plate :- Ref. Cl No. 6.3.1/PN 32/IS 800:2007
5. Design Strength due to Block Shear :- Ref. Cl No. 6.4/PN 33/IS 800:2007
6. Bolt Subjected to Combined Shear and Tension :- Ref. Cl No. 10.4.6/PN 77/IS 800:2007

Design Specification for Bolted Connection as per IS 800-2007
1. Shear Capacity of a Bolt :- Ref. Cl No. 10.3.3/PN 75/IS 800:2007
The design strength of the bolt, Vdsb as governed shear strength is given by:
Where :-
Vdsb = Vnsb / γmb
Vnsb - nominal shear capacity of a bolt, calculated as follows:
=
fu
√ 3 ( nn Anb + ns Asb )
Vdsb √ 3
= [ fu ( n A + n A
n nb s sb ) ] / 1.25
Where :- fu = ultimate tensile strength of a bolt.
nn = number of shear planes with threads intercepting the shear plane;
ns = number of shear planes without threads intercepting the shear plane;
sb sb 4
A = nominal plain shank area of the bolt; A =𝜋
𝑑2
Anb = net shear area of the bolt at threads, may be taken as the area
nb 4
corresponding to root diameter at the thread. A = 0.78 𝜋
𝑑2
γmb = Partial safety factor for bolt =1.25
78

Fig. Bolt
79

80

2. Bearing Capacity of a Bolt :- Ref. Cl No. 10.3.4/PN 75/IS 800:2007
The design strength of the bolt in bearing Vdpd as governed shear strength is given by:
Vdpd = Vnpb / γmb
Where :-
Vnpb = nominal bearing strength of a bolt, calculated as follows:
= 2.5 kb d .t.fu
b
𝑒
3𝑑0
𝑝
3𝑑0 fu
− 0.25),
fu𝑏
, 1
K = is the smallest of , (
e = End Distance
P = Pitch
d0 = Diameter of the hole
fub = Ultimate tensile strength of bolt.
Fu = Ultimate tensile strength of plates.
d = Nominal diameter of the bolt.
t = Least thickness of connection part or plates.

3. Tension Capacity of Bolt :- Ref. Cl No. 10.3.5/PN 76/IS 800:2007
i. Design Strength Due to Yielding of Gross Section :-
The design of strength of the bolt due to yielding of the gross section (i.e. the shank)
Tdbg = fyb Asb / γm0
ii. The design strength of bolt due to rupture at the net section (i.e. at the root of the threads)
Tdbn = 0.9 fub . Anb / γmb
Where :-
fub = ultimate tensile stress of the bolt;
fyb = yield stress of the bolt,
nb 4
A = Net tensile Area = 0.78 𝜋
𝑑2
Asb = shank area of the bolt.
82

4. Tension Capacity of Plate :- Ref. Cl No. 6.3.1/PN 32/IS 800:2007
The design strength of a plate in tension due to rupture at the net section is given by:
Tdn
= 0.9 An.fu
γm1
Where, An = Net sectional Area of plate.
Fu = Ultimate tensile strength of plate.
γm1 = Partial Safety Factor. = 1.25
a) If bolt holes are not staggered
An = (b – n d0 ) × t
b) If the holes are staggered
𝑖
=1
An = [ 𝑏 − 𝑛𝑑ℎ + σ𝑚 𝑝2𝑠
𝑖
4𝑔𝑖
] × 𝑡
Where b, t = width and thickness of the plate respectively
dh = Diameter of the bolt hole (Ref. Table 19 IS 800:2007)
g = Gauge length between the bolt holes as shown in fig
Ps = staggered pitch between line of bolt hole as shown in fig
n = Number of bolt holes in critical section i = Subscript for summation of all the inclined edges.
83

5. Design Strength due to Block Shear :-
The block shear strength Tdb of the bolted connection is the least of :
Ref. Cl No. 6.4/PN 33/IS 800:2007
db
T =
Avg.f𝑦
√3 γm0
+ 0.9 Atn fu
γm1
or
Tdb =
0.9 Atn fu
√3 γm1
+
Atg fy
γm0
Where,
Avg, Atn = are the minimum gross and net area in the shear along
the bolt line parallel to the line of action of force, resp.
Atg, Atn = are the minimum gross and net area in tension from the
bolt hole to the edge of a plate or between bolt hole
perpendicular to line of action of force, respectively
Fu, Fy = are the ultimate yield strength of the material of the
plates respectively.
84

6. Bolt Subjected to Combined Shear and Tension :- Ref. Cl No. 10.4.6/PN 77/IS 800:2007
A bolt subjected to shear and tension simultaneously should satisfy the condition
Vdf Tdf
( Vsf )2 + Tf
≤ 1.0
Where,
Vsf = Factored shear force acting on the bolt
Vdf = design shear strength,
Tf = factored tensile force acting on the bolt
Tdf = design tension strength.
85

Specifications :-
1. The diameter of hole should be the nominal diameter of the bolt plus the clearance given in
Ref. Cl No. 10.2.1/Table no-19/PN 73/IS 800:2007
86

2) The minimum pitch :- Ref. Cl No. 10.2.3/PN 74/IS 800:2007
87

3. The maximum pitch :- Ref. Cl No. 10.2.3/PN 74/IS 800:2007
88

3. The edge and end distance:- Ref. Cl No. 10.2.4.2, 10.2.4.3/PN 74/IS 800:2007
89

Examples 1:- Determine the bolt value of 20 mm diameter bolt connecting 10 mm plate in
i) Single shear, ii) Double shear. Bolt used are 4.6 grade, plate of 410 grade. Take area of bolt = 245 mm2
Solution :-
For 4.6 grade fu= 400 MPa
Single Shear Capacity of a Bolt :
Vdsb = Vnsb / γmb
Ref 1.3.113, 1.3.119 and 2.2.4.2/tab no.1/PN. 13/IS800:2007
Ref. Cl No. 10.3.3/PN 75/IS 800:2007
Vnsb =
fu
√ 3
( nn Anb + ns Asb )
dsb √ 3
90
V = [ fu ( n A
n nb s sb
+ n A ) ] / 1.25
dsb √ 3
V = [ 400 (1 × 245 + 0) ] / 1.25
Vdsb = 45.26 kN
Single Shear Capacity = 45.26 kN
Double Shear Capacity = 2 x 45.26 = 90.52 kN
∴ nn= 1, ns =0

Bearing Capacity of a Bolt :- Ref. Cl No. 10.3.4/PN 75/IS 800:2007
Where :-
Vdpd = Vnpb / γmb
= 2.5 kb d .t.fu
b
K = is the smallest of
e
3d0
p
3d0 fu
, ( − 0.25),
fub
, 1.0
Vdpd = 2.5 × 1 × 20 × 10 × 410 / 1.25
= 164000 N
= 164 kN
Bolt Value for single shear = 45.26 kN
Bolt Value for Double Shear = 90.52 kN
Bearing Capacity of Bolt = 164 kN … … … … … … … … … … … . . A n s
91

Examples 2:- Determine the ultimate load carrying in tension of lap joint shown in fig 2. If bolt thread are
outside of the shear plane. Use M 16 bolts of the product ‘c’ and property class 4.6 the yield and ultimate
strength of the plate strengths of the flats 250 MPa and 410MPa respectively.
fig 2.
92

Solution :-
Diameter of hole (d0)=16+2=18 mm
i. Shear Capacity of a Bolt :
Vdsb = Vnsb / γmb
Ref. Cl No. 10.3.3/PN 75/IS 800:2007
Vnsb =
fu
√ 3
( nn Anb + ns Asb )
dsb √ 3
V = [ fu ( n A
n nb s sb
+ n A ) ] / 1.25
dsb √ 3 4
V = [ 400 (0 +1 ×
π
162 ) ] / 1.25 ∴ nn= 0, ns =1
Vdsb = 37.146 kN
Shear Capacity of bolt = 37.146 k N … … … … … … … … … … … … … … … . . ( 1 )
93

ii. Bearing C apacity of a Bolt :- Ref. Cl No. 10.3.4/PN 75/IS 800:2007
Vdpd = Vnpb / γmb
Where :-
=108.24/1.25=86.592 k N … … … … … … … … … … ( 2 )
= 2.5 kb d .t.fu = 2.5 × 0.55 ×16 × 12 × 410 = 108.24 kN
b
𝑒
3𝑑0
𝑝
3𝑑0 fu 0
, ( − 0.25),
fu𝑏
, 1 » 𝑒
=
30
3𝑑 3X18
=0.55, » (
𝑝
3𝑑0
− 0.25)= 80
3X18
− 0.25=1.23,
fu𝑏
= 400
= 0.97, 1
fu 410
e = End Distance
P = Pitch
d0 = Diameter of the hole
b
∴ k = 0.55
fub = Ultimate tensile strength of bolt.
Fu = Ultimate tensile strength of plates.
d = Nominal diameter of the bolt.
t = Least thickness of connection part or plates.
PROF A V WAKCHAURE 94

iii. Tension Capacity of Plate :- Ref. Cl No. 6.3.1/PN 32/IS 800:2007
The design strength of a plate in tension due to rupture at the net section is given by:
Tdn
= 0.9 An.fu
=
γm1 1.25
0.9 ×1008 ×410
= 297.6 k N… … … … … … … … ( 3 )
Where, An = Net sectional Area of plate.
Fu = Ultimate tensile strength of plate.
γm1 = Partial Safety Factor. = 1.25
a) If bolt holes are not staggered
An = (b – n d0 ) x t = 120 − 2 × 18 × 12 =1008 mm2
Where b, t = width and thickness of the plate respectively
dh = Diameter of the bolt hole (Ref. Table 19 IS 800:2007)
g = Gauge length between the bolt holes as shown in fig
Ps = staggered pitch between line of bolt hole as shown in fig
n = Number of bolt holes in critical section i = Subscript for summation of all the inclined edges.
95

iv. Design Strength due to Block Shear :-
Ref. Cl No. 6.4/PN 33/IS 800:2007
db
T =
Avg.f𝑦
√3 γm0
+ 0.9 Atn f u
γm1
or
Tdb =
0.9 Avn fu
√3 γm1
+ Atg f y
γm0
Where,
Avg, Avn = are the minimum gross and net area in the shear along

Avg = 2 (80 + 30) x 12 = 2640 mm2
Avn = 2(80 + 30 – 1.5 x 18) x 12 =1992 mm2
Atg = 60 x 12 = 720 mm2
Atn = (60-18) x 12 = 504 mm2
(along 1-2-3 and 4-5-6)
(along 1-2-3 and 4-5-6)
(along 3-4)
(along 3-4)
db
T =
Avg.fy
√3 γm0 γm1
√3 x 1.1
+ 0.9 Atn fu
= 2640 × 250 +
0.9 × 504 × 410
= 495.19 kN
1.25
… … … … … … … ( a )
Or
db
T =
0.9 Avn fu
1
√3 γm γm0
+ Atg fy =
0.9 × 1992 × 410
√3 ×1.25 1.1
97
+ 730 × 250 = 503.140 kN……………..(b)
∴ Design strength due to block shear = 495.19 kN least of (a) and (b)

∴ Ultimate load Carrying capacity of the joint is the
least of
98
Shear Capacity of a Bolt Vdsb
Vdpb
Tdn
= 4 x 37.146 = 148.58 kN
Bearing Capacity of a Bolt = 4 x 86.59 = 346.36 kN
= 297.61 kN
Tension Capacity of Plate
Design Strength due to Block Shear Tdb = 495.19 kN
∴ Ultimate load carrying capacity = 148.58 k N … … … … … … … … … … … … … … … . . A n s

3. Design a double cover plate butt joint using M 24 bolts of product grade ‘c’ and property class 5.6 to
connect two flats of size 350 mm x 16 mm for maximum efficiency. Assume that one shear plane intercepts
the thread of bolts. The yield and ultimate tensile strengths of the flats are 250 Mpa & 410 Mpa respectively.
Solution :- For 5.6 grade fu= 500 MPa Ref 1.3.113, 1.3.119 and 2.2.4.2/tab no.1/PN. 13/IS800:2007
i. Design shear strength of bolt / Shear Capacity of a Bolt :
Vdsb = Vnsb / γmb
Ref. Cl No. 10.3.3/PN 75/IS 800:2007
Vnsb =
fu
√ 3
( nn Anb + ns Asb )
dsb √ 3
V = [ fu ( n A
n nb s sb
+ n A ) ] / 1.25
Since the bolt are in double shear, there are two shear planes.
It is given that one shear plane intersects the threads of bolts.
n s sb nb
4 4
99
So N = 1 and N = 1 , A =π
d2 = 453, A = 0.78 π
d2 = 353 mm2
dsb
V = [ 500 (1 × 353 + 1 × 453) ] / 1.25
Vdsb
√ 3
= 186.13 kN………………………………(1)

2. Design bearing strength of the bolt / Bearing Capacity of a Bolt :- Ref. Cl No. 10.3.4/PN 75/IS 800:2007
Where :-
Vdpd = Vnpb / γmb
= 2.5 kb d .t.fu
b
𝑒
3𝑑0
𝑝
3𝑑0 fu
, ( − 0.25),
fu𝑏
, 1
a) Minimum end distance e = 1.5 x 26 = 39 ≈ 40 mm
b)Minimum pitch distance = P = 2.5 x 24 =60 mm ≈ 80 mm
Provide pitch p = 80 mm and end distance e = 40 mm
b
K = is least of
40 80
3× 26 3 × 26 410
= 0.513, ( − 0.25)= 0.78 , 500
= 1.22, 1
∴ Kb = 0.513
Vnpb = 2.5 kb d .t.fu = 2.5 × 0.513 × 24 × 16 × 410 = 201.916 kN
Vdpd = Vnpb / γmb =
201.916
1.25
= 161.53 kN … … … … … … … … … … . ( 2 )
100

From Equation (1) and (2) least value will be the strength of bolt.
Design Strength of bolt = 161.53 kN
3. Design Strength of Plate :- Ref. Cl No. 6.2,6.3/PN 32/IS 800:2007
i. The design strength of plate due to yielding of the gross section :-
Tdg = Ag fy / γm0
=
350 ×16 × 250
1.1
= 1272.72 kN………………..(3)
ii. The design strength of plate due to rupture at the net section :-
Tdn = 0.9 An fu / γm1
=
0.9 × 4768 × 410
1.25
= 1407.51 kN………………..(4)
Considering two row of the bolt in connections
An = (b – n d0 ) × t =(350 – 2 × 26 ) × 16 = 4768 mm 2
Design strength of plate is the least value between (3) and (4)
∴ Design strength of plate = 1272.72 kN

4. Design of Connection :- Ref. Cl No. 10.3.5/PN 76/IS 800:2007
Number of bolts required = Design strength of plate
Design Strength of bolt 161.533
=
1272.72
= 7.88 ≈ 8 bolts
Provide 8 bolts. They are two row with pitch of 80 mm and end distance 40 mm
Fig.
102

5. Design of cover plates :-
Design Strength of Plate :- Ref. Cl No. 6.2,6.3/PN 32/IS 800:2007
It t1 is the thickness of each cover plate, its design strength due to yielding
i. The design strength of plate due to yielding of the gross section :-
Tdg = Ag fy / γm0
1.1
350× 𝑡1× 250
= = 79.545 t1 kN ………………..(5)
ii. The design strength of plate due to rupture at the net section :-
Tdn = 0.9 An fu / γm1 An = (b – n d0 ) × t
1
0.9 × 350 −2 ×26 × 𝑡 410
=
1.25 1
= 87.969 t kN ………..(6)
Least of (5) and (6) will be the critical value
Since there are two plates.
∴ 2 × 79.545 t1 = 1272.72
∴ t1 = 8 mm
Provide two cover plates of each 10 mm thickness.
103

• Welded connection are widely used now days.
• Welded connection is efficient and economical.
• Welded connection does not require holes to
be made in member (like riveted and bolted
connection)
• For welded connection extra materials like
plates and angles are not required
Welded Connection :-
Ref. Cl. No. 10.5/PN-
78/IS 800:2007
ButtWeld
Types of Welds
Fillet Weld
106
Welded Connection :-

Butt weld is also known as groove weld. The various shape of the groove made are :-
i. Square butt weld ii. Single V - butt weld iii. Single J – butt weld
iv. Double V – Butt weld v. Single U – butt weld
108

Size of butt weld :- The size of butt weld is the effective throat thickness. In case of double
butt weld the thickness of butt weld is taken as thickness of thinner plate.
• For single butt weld the thickness of weld is taken as 5/8 times thickness of thinner plate.
• Effective length of butt weld is equal to length of full size weld
• Minimum length of weld = 4 × size of weld.
109

Minimum size of fillet weld:-
111
Ref. Cl.No. 10.5.2/Table no. 21/PN 78/IS 800:2007

Design strength of fillet weld (Fwd):- Ref. Cl.No. 10.5.7/PN 79/IS 800:2007
112

Design of Welding Connections
Examples 1:-
A 20 mm thick plate is connected to 18 mm thick plate by 240 mm long butt weld. Determine the strength of
joint if (i) Single U – butt weld is used. (ii) Double U – butt weld is used. Consider fu = 410 N/mm2
Solution: Given Le = 240 mm
(i) For single U – butt weld
t = thickness of weld
=
5
8
× thickness of thinner plate
=
5
8
× 18 = 11.25 𝑚𝑚
Design strength of weld ( fwd ) : Ref. Cl.No. 10.5.7.1.1/PN 79/IS 800:2000
= 𝑓𝑢 /√3
=
410 /√3
𝛾𝑚𝑤
1.25
= 189.37 N/mm2
113

Total strength of weld = Design strength × 𝑡 × 𝐿𝑒
114
= 189.37 × 11.25 × 240
= 511301.3 N
= 511.30 kN … … … … … … … … … … … … … … … A n s
(i) For double U – butt weld
t = thickness of weld
thickness of weld = 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑡ℎ𝑖𝑛𝑛𝑒𝑟 𝑝𝑙𝑎𝑡𝑒
= 18 𝑚𝑚
Total strength of weld = Design strength × 𝑡 × 𝐿𝑒
= 189.37 × 18 × 240
= 818078.4 N
= 818.07 kN … … … … … … … … … … … … … … … … A n s

Examples 2:-
A tie member in a roof truss consists of 2 ISA 125 × 75 × 10 mm connected on either side of gussets plate
12 mm thick. It is subjected to a factored load of 480 kN. Assume connection are made on the field and
fu = 410 Mpa, Cxx = 42.4 mm. find length of weld
Solution :
Consider size of weld (S) 𝑘 thickness of member – 1.5
𝑘 10 – 1.5
𝑘 8.5 mm
Say S = 8 mm
2
Load carried by each angle =
480
= 240 kN
Design strength of weld ( fwd ) : Ref. Cl.No. 10.5.7.1.1/PN 79/IS 800:2000
= 𝑓𝑢 /√3
=
410 /√3
𝛾𝑚𝑤
1.5
= 157.8 N/mm2
....................... 𝑚𝑤
𝛾 = 1.5 for field weld
115

Strength of weld = Design strength of weld fwd × tickness of weld
= 157.8 × 0.7 S
= 157.8 × 0.7 × 8 = 883.73 N/mm
load carried by single angle
Strength of weld
Weld length required for one angle = =
240 ×103
883.73
= 271.57 say = 280 mm
r
∴ L1 + L2 = 280 m m … … … … … … … … … ( 1 )
Tacking moment of weld length at c. g. and equating fo
equilibrium
L1 × 42.4 = L2 × 82.6
L1 = 1.948 L2 … … … … … … … … . ( 2 )
From Equation (1) & (2)
1.948 L2 + L2 = 280
L2 = 94.98 mm
L1 = 185.09 mm
say 100 mm
say 200 mm
116

Steps for design of welded joints
Given : load to be carried by the member Pu (factored load) member details and Cxx
Soln :
Step 1 : Assume maximum size of weld = S
= thickness of member – 1.5 mm.
Step 2 : Calculate design shear strength of weld
fwd
=
𝑓𝑢 /√3
𝛾𝑚𝑤
Ref. Cl.No. 10.5.7.1.1/PN 79/IS 800:2007
Consider fu = 410 MPa if not given
𝛾𝑚𝑤 = 1.25 for shop weld
= 1.5 for field weld
Step 3 : Calculate strength of weld
= Design strength of weld fwd
= fwd × 0.7 S
Step 4 : Calculate length of weld required Lw
× tickness of weld
Lw
= Ultimate load carried by the member
strength of weld
117
Step 5 : Arrange weld length ; taking moment of weld lengths @ c.g. and equating them for equilibrium

Examples 3:-
A tie member in a roof truss consists of 2 ISA 100 × 75 × 8 mm. The angles are connected to either side of a 10
mm gusset plates and the member is subjected to a working pull of 350 kN. Design the welded connection.
Assume shop weld used.
Solution :
Factored load = 1.5 × 350 = 525 kN
Assume maximum size of weld = S
𝑘 8 – 1.5
𝑘 6.5 mm
Hence provide / use 6 mm size weld
Calculate design shear strength of weld
fwd
=
𝑓𝑢 /√3
𝛾𝑚𝑤
Ref. Cl.No. 10.5.7.1.1/PN 79/IS 800:2007
= 𝑓𝑢 /√3
=
410 /√3
𝛾𝑚𝑤
1.25
= 189.37 N/mm2
....................... 𝑚𝑤
𝛾 = 1.25 for shop weld
118

Strength of weld = Design strength of weld fwd × tickness of weld
= 189.37 × 0.7 S
= 189.37 × 0.7 × 6 = 795.354 N/mm
2
Load carried by single angle = = 525
= 262.5 × 103 N
Weld length required for one angle =
load carried by single angle
Strength of weld
=
262.5 ×103
795.354
= 330 mm
∴ L1 + L2 = 330 m m … … … … … … … … … ( 1 )
Tacking moment of weld length at c. g. and equating for
equilibrium
L1 × 31 = L2 × 69
L1 = 2.23 L2 … … … … … … … … . ( 2 )
2.23 L2 + L2 = 330
L2 = 102.167 mm
L1 = 227.84 mm ∴ provide 6 mm weld of L1 = 228 mm and L2 = 103 mm as shown in fig……………..Ans
119

C) Design of Tension Member :-
As per IS 800:2007, the design of tension member are considered on mode of
failure of tension member.The different modes of failure in tension members are :
1. Gross section yielding
2. Net section rupture
3. Block shear failure
The design strength of a member under axial tension Td is the lowest of the above
three values.
120

1. Design Strength Due to Gross SectionYielding :- Ref. Cl. No. 6.2/PN 32/IS 800:2007
May 2017
Q. Explain in brief limit strength due to yielding with suitable sketch.
121

2. Design Strength Due to Net Section Rupture :-
122
Ref. Cl. No. 6.3/PN 32/IS 800:2007
May 2017
Ref. Cl. No. 6.3.1/PN 32/IS 800:2007
Q. Explain in brief limit strength due to rupture with suitable sketch.
2.1 Design Strength Due to Net Section Rupture in Plates

2.2 Net Section Rupture in Threaded Rod :- Ref. Cl. No. 6.3.2/PN 33/IS 800:2007
124

2.3 NetSection Rupture in Single Angles :- Ref. Cl. No. 6.3.3/PN 33/IS 800:2007
Q. Explain in brief limit net section rupture in plate with suitable sketch. Aug. 2016
125

2.4 Net Section Rupture in Other Sections:- Ref. Cl. No. 6.3.4/PN 33/IS 800:2007
126

3. Design Strength Due to Block Shear Failure:- Ref. Cl. No. 6.4/PN 33/IS 800:2007
May 2017
Q. Explain in brief limit strength due Block Shear Failure with suitable sketch.
127

Design ofTension Member :-
128
As per IS 800:2007, the design of tension member are considered on mode of failure of tension
member.The different modes of failure in tension members are :
1 . G ross section yielding Ref. Cl. No. 6.2/PN 32/IS 800:2007
2. Net section rupture Ref. Cl. No. 6.3/PN 32/IS 800:2007
3. Block shear failure Ref. Cl. No. 6.4/PN 33/IS 800:2007
The design strength of a member under axial tension Td is the lowest of the above three values.

Examples on Load Carrying Capacity of a given section :-
Examples 1:- A single ISA 90 × 60 × 6 @ 6.8 kg/m is connected to 8 mm thick gusset plate at the ends with 4
numbers of 20 mm bolts to transfer tension. Determine the design tensile strength of angle section if the gusset
plate is connected to the longer leg. SPPU Dec. 2010, 10 Marks
Solution :
Fig.
129

The design tensile strength Td of the angle is calculated based on the following criteria
i. Gross section yielding Ref. Cl. No. 6.2/PN 32/IS 800:2007
The design strength Tdg of the angle limited to the yielding of gross cross section Ag is given by,
dg
Ag× fy
T =
γm0
Here fy = 250 Mpa,
Ag = (90+60-6) × 6 = 864 mm2
Or
From Steel table, Ag = 864 mm2
𝛾𝑚0 = 1.1
Tdg =
Hence,
864 × 250
1.1
= 196.36 kN
Tdg = 196.36 kN … … … … … … … . . ( 1)
130

ii. Net Section Rupture in Single Angles :- Ref. Cl. No. 6.3.3/PN 33/IS 800:2007
131

ii. Net section rupture Ref. Cl. No. 6.3.3/PN 33/IS 800:2007
The design strength Tdn of angle governed by rupture of net cross sectional area is given by
dn
T =
0.9 ×fu ×Anc
γm1
β ×Ago ×fy
+
γm0
𝑡 𝑓𝑢
𝐿𝑐 𝑓𝑦 . 𝛾𝑚1
𝛽 = 1.4 - 0.076 (𝑤
) (𝑓𝑦
) (𝑏𝑠
) ≤ (𝑓𝑢 . 𝛾𝑚0
) ≥
0.7
Here, fu = 410 Mpa, fy = 250 Mpa,
𝛾𝑚1 = 1.25 𝛾𝑚0 = 1.1,
w = 60 mm , t = 6 mm
w1 = 0.6 × 90 = 54 ≈ 50 mm
132

∴ bs = w + w1 – t = 60 + 50 – 6 = 104 mm
Assume Pitch ‘P’ = 50 mm , e = 30 mm
∴ Lc = 3 × 50 = 150 mm
𝛽 = 1.4 - 0.076 (𝑤
) (𝑓𝑦
) (𝑏𝑠
) ≤ (𝑓𝑢 . 𝛾𝑚0
) ≥ 0.7
𝑡 𝑓𝑢 𝐿𝑐 𝑓𝑦 . 𝛾𝑚1
6 410 150 250×1.25
𝛽 = 1.4 - 0.076 (60
) (250
) (104
) ≤ ( 410×1.1
) ≥ 0.7
𝛽 = 1.07 ≤ 1.44 ≥ 0.7
∴ 0.7 ≤ 1.07 ≤ 1.44
𝛽 = 1.07
133

nc 2
A = ( 90 – 22 – 6
) × 6 = 390 mm2
go 2
A = ( 60 – 6
) × 6 = 342 mm2
Tdn
= 0.9 ×fu ×Anc
+ β ×Ago× fy
γm1 γm0
Tdn
= 0.9 ×410 ×390
+ 1.07 ×342 ×250
1.25 1.1
Tdn = 198.296 kN ………………..(2)
134

iii. Design Strength due to Block Shear :-
Ref. Cl No. 6.4/PN 33/IS 800:2007
db
T =
Avg.fy
√3 γm0
+ 0.9 Atn fu
γm1
or
Tdb =
0.9 Avn fu
√3 γm1
+ Atgfy
γm0
Where,
Fu, Fy
= are the ultimate yield strength of the material of the

Here, A vg = 180 × 6 = 1080 mm2
A vn = ( 180 – 3.5 × 22 ) × 6 = 618 mm2
A tg = 40 × 6 = 240 mm2
A tn = ( 40 – 0.5 × 22 ) × 6 = 174 mm2
db
T =
or
Avg.f𝑦
√3 γm0 γm1 √3×1.1
1080×250
+ 0.9 Atn fu
= + 0.9 × 174×410
1.25
=193.07 kN
db
T =
0.9 Avn fu
√3 γm1
+ Atg fy
γm0
=
0.9 ×618 ×410
√3× 1.25
+ 240 ×250
1.1
=159.87 kN
Hence,
The block shear strength Tdb of the bolted connection,T db = 159.87 kN … … … … … … … . ( 3 )
136

Form Equation (1), (2) and (3)
The design tensile strength Td of the angle is the lowest of the equation (1), (2) & (3)
Hence ,
Td = Tdb = 159.87 kN … … … … … … … … … … … … … … . A n s
137

Examples 2:- A tension member consist of single angle ISA 100 × 100 × 10 mm @ 10.8 kg/m is connected to a
12 mm thick gusset plate using M20 bolt of class 4.6 in a single line. Determine the design strength using
bolted connection. SPPU - Dec. 2014, 13 Marks
Solution :
138
1. Gross section yielding Ref. Cl. No. 6.2/PN 32/IS 800:2007
2. Net section rupture Ref. Cl. No. 6.3/PN 32/IS 800:2007
3. Block shear failure Ref. Cl. No. 6.4/PN 33/IS 800:2007

Tdg =
Ag×fy
γm0
Here fy = 250 Mpa,
Ag = (100+100 – 10) × 10 = 1900 mm2
Or
𝛾𝑚0 = 1.1
Tdg =
1903 ×250
1.1
= 432.5 kN
Hence
Tdg = 432.5 kN … … … … … … … . . ( 1)
139

ii. Design of bolted connection
Providing M 20 bolts of class 4.6
∴ do = 20 + 2 = 22
Assuming e = 1.5 × 22 = 33 mm ≈ 35 mm
P = 2.5 × 20 = 50 mm ≈ 55 mm
140

i. Design shear strength of bolt / Shear Capacity of a Bolt : Ref. Cl No. 10.3.3/PN 75/IS 800:2007
Vdsb = Vnsb / γmb
Vnsb =
fu
√ 3
( nn Anb + ns Asb )
dsb √ 3
V = [ fu ( n A
n nb s sb
+ n A ) ] / 1.25
So Nn = 0 and Ns = 1 ,
sb 4
A =𝜋
𝑑2 = 314.15, nb 4
A = 0.78 𝜋
𝑑2 = 245 mm2
dsb
V = [ 400 ( 0 + 1 × 314.15) ] / 1.25
√ 3
= 58.04 kN………………………………(1)
Vdsb
141

2. Design bearing strength of the bolt / Bearing Capacity of a Bolt :- Ref. Cl No. 10.3.4/PN 75/IS 800:2007
Where :-
Vdpd = Vnpb / γmb
= 2.5 kb d .t.fu
b
𝑒
3𝑑0
𝑝
3𝑑0 fu
, ( − 0.25),
fu𝑏
, 1
b)Minimum pitch distance = P = 2.5 x 20 = 50 ≈ 55 mm
Provide pitch p = 55 mm and end distance e = 35 mm
b
K = is least of
35 55
3×22 3 × 22
= 0.53, ( − 0.25)= 0.58 ,
400
250
= 1.6, 1
∴ Kb = 0.53
Vnpb = 2.5 kb d .t.fu = 2.5 × 0.53 × 20 × 10 × 410 = 108.650 kN
108.650
1.25
= 86.92 kN … … … … … … … … … … . ( 2 )
142

Number of bolts required =
Design strength of plate
Design Strength of bolt 58.04
143
=
432.5
= 7.45 ≈ 8 bolts
∴ Provide 8 bolts of M 20 of class 4.6

144

dn
T =
0.9 ×fu×Anc
γm1
β ×Ago ×fy
+
γm0
𝑡 𝑓𝑢
𝛽 = 1.4 - 0.076 (𝑤
) (𝑓𝑦
) (𝑏𝑠
) ≤ (𝑓𝑢 . 𝛾𝑚0
) ≥
0.7
𝛾𝑚1 = 1.25 𝛾𝑚0 = 1.1,
w = 100 mm , t = 10 mm
w1 = 0.6 × 100 = 60 mm
145

∴ bs = w + w1 – t
Pitch P = 55 mm ,
= 100 + 60 – 10 = 150 mm
e = 35 mm
∴ Lc = 7 × 55 = 385 mm
𝛽 = 1.4 - 0.076 (𝑤
) (𝑓𝑦
) (𝑏𝑠
) ≤ (𝑓𝑢 . 𝛾𝑚0
) ≥ 0.7
10 410 385 250×1.25
146
𝛽 = 1.4 - 0.076 (100
) (250
) (150
) ≤ ( 410×1.1
) ≥ 0.7
𝛽 = 1.21 ≤ 1.44 ≥ 0.7
∴ 0.7 ≤ 1.21 ≤ 1.44
𝛽 = 1.21

nc 2
A = ( 100 – 22 – 10
) × 10 = 730 mm2
go 2
A = ( 100 – 10
) × 10 = 950 mm2
Tdn
= 0.9 ×fu × Anc
+ β ×Ago× fy
γ γ
m1 m0
Tdn =
0.9 ×410 ×730
1.25
+
1.21 × 950×250
147
1.1
Tdn = 476.75 kN … … … ………..(2)

Ref. Cl No. 6.4/PN 33/IS 800:2007
Tdb =
Avg.fy
√3 γm0
+ 0.9 Atn fu
γm1
or
Tdb =
0.9 Avn fu
√3 γm1
+
Atg fy
γm0
Where,

Here, A vg = ( 35 + 55 × 7 ) × 10 = 4200 mm2
A vn = [( 35 + 55 × 7 ) – (7.5 × 22 )] × 10 = 2550 mm2
A tg = 40 × 10 = 400 mm2
A tn = ( 40 – 0.5 × 22 ) × 10 = 290 mm2
db
T =
or
Avg.fy
√3 γm0 γm1
+ 0.9 Atn fu
=
4200×250
√3×1.1
+ 0.9 ×290×410
= 636.715 kN
1.25
db
T =
0.9 Avn fu
√3 γm1
+
Atg fy
=
γm0 √3× 1.25
0.9 ×2550 ×410
1.1
149
+ 400 ×250
= 525.515 kN
Hence,
The block shear strength Tdb of the bolted connection, T db = 525.515 kN … … … … … … … . ( 3 )

∴ Design tensile strength of ISA 100 × 100 × 10 is Td = 432.5 k N … … … … … … … … … … … … … A n s
150

Examples 3 :- Determine design tensile strength due to yielding and rupture of an ISA 125 × 95 × 10 @ 16.5
kg/m in which longer leg is connected to the 10 mm thick gusset plate by 3 number of M 20 black bolts of 4.6
grades. SPPU Dec. 2015, 6 Marks
Solution :
dg
Ag× fy
T =
γm0
Here fy = 250 MPa,
Ag = (125+95-10) × 10 = 2100 mm2
Or
𝛾𝑚0 = 1.1
Tdg =
2102 × 250
151
1.1
= 477.72 kN
Hence
Tdg = 477.72 kN … … … … … … … . . ( 1) Ans

152

Tdn = ×
0.9 ×fu Anc
γm1
+
β ×Ago ×fy
γm0
t fu
Lc fy . γm1
β = 1.4 - 0.076 (w
) (fy
) (bs
) ≤ (fu . γm0
) ≥ 0.7
𝛾𝑚1 = 1.25 𝛾𝑚0 = 1.1,
w = 95 mm , t = 10 mm
w1 = 0.6 × 125 = 75 mm
a) Minimum end distance e min = 1.5 d0 = 1.5 x 22 = 33 ≈ 40 mm
b) Minimum pitch distance = Pmin = 2.5 d = 2.5 x 20 = 50 ≈ 60 mm
153

∴ bs = w + w1 – t = 95 + 75 – 10 = 160 mm
Provide pitch P = 60 mm and end distance e = 40 mm
∴ Lc = 2 × 60 = 120 mm
𝑡 𝑓𝑢 𝐿𝑐
𝑓𝑦 . 𝛾𝑚1
𝛽 = 1.4 - 0.076 (𝑤
) (𝑓𝑦
) (𝑏𝑠
) ≤ (𝑓𝑢 . 𝛾𝑚0
) ≥
0.7
250×1.25
154
𝛽 = 1.4 - 0.076 (95
) (250
) (160
) ≤ ( 410×1.1
) ≥ 0.7
10 410 120
𝛽 = 0.813 ≤ 1.44 ≥ 0.7
∴ 0.7 ≤ 0.813 ≤ 1.44
𝛽 = 0.813

nc 2
A = ( 125 – 22 – 10
) × 10 = 980 mm2
go 2
A = ( 95 – 10
) × 10 = 900 mm2
Tdn = ×
0.9 ×fu Anc
γm1
+
β ×Ago× fy
γm0
Tdn
= 0.9 ×410 ×980
+ 0.813 ×900 ×250
1.25 1.1
155
Tdn = 455.58 kN … … … ………..(2) Ans

Examples 4 :- Determine design tensile Td of the angle 2 ISA 90 × 60 × 6 × @ 6.8 kg/m, connected to the 10 mm
thick gusset plate of 10 mm thickness back to back by 3 bolt of 16 mm diameter. SPPU May. 2015, 6 Marks
Solution :
i. G ross section yielding Ref. Cl. No. 6.2/PN 32/IS 800:2007
dg
Ag× fy
T =
γm0
Here fy = 250 Mpa,
Ag = 1730 mm2
𝛾𝑚0 = 1.1
Tdg =
1730 × 250
156
1.1
= 393.18 kN
Hence,
Tdg = 393.18 kN … … … … … … … . . ( 1) Ans

157

dn
T =
0.9 ×fu ×Anc
γm1
β ×Ago ×fy
+
γm0
𝑡 𝑓𝑢
𝛽 = 1.4 - 0.076 (𝑤
) (𝑓𝑦
) (𝑏𝑠
) ≤ (𝑓𝑢 . 𝛾𝑚0
) ≥
0.7
𝛾𝑚1 = 1.25 𝛾𝑚0 = 1.1,
w = 60 mm , t = 6 mm
w1 = 0.6 × 90 = 54 ≈ 50 mm
a) Minimum end distance e min = 1.5 d0 = 1.5 x 18 = 27 ≈ 35 mm
b) Minimum pitch distance = Pmin = 2.5 d = 2.5 x 16 = 50 ≈ 50 mm
158

∴ bs = w + w1 – t = 60 + 50 – 6 = 104 mm
Assume pitch ‘p’ = 50 mm , e = 30 mm
∴ Lc = 2 × 50 = 100 mm
𝛽 = 1.4 - 0.076 (𝑤
) (𝑓𝑦
) (𝑏𝑠
) ≤ (𝑓𝑢 . 𝛾𝑚0
) ≥ 0.7
6 410 100 250×1.25
𝛽 = 1.4 - 0.076 (60
) (250
) (104
) ≤ ( 410×1.1
) ≥ 0.7
𝛽 = 0.918 ≤ 1.44 ≥ 0.7
∴ 0.7 ≤ 0.918 ≤ 1.44
𝛽 = 0.918
159

nc 2
A = 2 ( 90 – 18 – 6
) × 6 = 828 mm2
go 2
A = 2 ( 60 – 6
) × 6 = 684 mm2
Tdn =
0.9 ×fu × Anc
γm1
+
β ×Ago × fy
γm0
Tdn =
0.9 ×410 ×828
1.25
+
0.918 ×684 ×250
160
1.1
Tdn = 387.133 kN … … … ………..(2)

Ref. Cl No. 6.4/PN 33/IS 800:2007
db
T =
Avg. fy
√3 γm0
+ 0.9 Atn fu
γm1
or
Tdb =
0.9 Avn fu
√3 γm1
+ Atg fy
γm0
Where,

Here, A vg = 2 ( 35 + 2 × 50 ) × 6 = 1620 mm2
A vn = 2 ( 135 – 2.5 × 18 ) × 6 = 1080 mm2
A tg = 2 ( 40 × 6 ) = 480 mm2
A tn = 2 ( 40 – 0.5 × 18 ) × 6 = 372 mm2
db
T =
Avg.fy
√3 γm0 γm1 √3×1.1
1620 × 250
+ 0.9 Atn fu
= + 0.9 × 372 ×410
=322.38 kN
1.25
or
db
T =
0.9 Avn fu
√3 γm1
+ Atg fy
γm0
=
0.9 ×1080 ×410
√3× 1.25 1.1
+ 480 ×250
=293.15 kN
Hence,
The block shear strength Tdb of the bolted connection,T db = 293.15 kN … … … … … … … . ( 3 )
162

Hence ,
Td = Tdb = 293.15 kN … … … … … … … … … … … … … … . A n s
163

Examples 5 :-Determine the design strength of member of roof truss consist of 2 ISA 90 × 90 × 12 mm connected
to both side of 12 mm thick plate by filet weld. Assume fy = 250 Mpa. SPPU May. 2011, May 2017 13 Marks
Solution :
dg
Ag× fy
T =
𝛾𝑚0
Here fy = 250 Mpa,
Ag = 2 × 2019 = 4038 mm2 ,
𝛾𝑚0 = 1.1
Tdg =
4038 × 250
164
1.1
= 916.36 kN
Hence,
Tdg = 916.36 kN … … … … … … … . . ( 1) Ans

ii. Length of Weld ( Lw )
𝑘 12 – 1.5
𝑘 10.5 mm
Throat Thickness = te = 0.7 S = 0.7 × 6 = 4.2 mm
Calculate design shear strength of weld Ref. Cl.No. 10.5.7.1.1/PN 79/IS 800:2007
fwd
=
𝑓𝑢 /√3
𝛾𝑚𝑤
= 410 /√3
1.25
= 189.37 N/mm2
................... 𝑚𝑤
165

Tota Design Strength of weld = Design strength of weld fwd × thickness of weld × Lw
= 2 × 189.37 × 0.7 S × Lw
= 2 × 189.37 × 0.7 × 6 × Lw = 1590.708 Lw N
Equating shear strength of weld to strength at yielding od gross section .
1590.708 Lw = 916.36 × 103 N
∴ Lw = 576.07 mm
∴ L1 + L2 = 576.07 mm … … … … … … … … . ( 1 )
Tacking moment of weld length at c. g. and equating for
equilibrium
C .G. of angle section is at C xx = 26.6 mm form top.
L1 × 26.6 = L2 × ( 90 – 26.66 )
L1 = 2.383 L2 … … … . . … … … … … … … … . ( 2 )
L2 = 170.53 mm
L1 = 406.38 mm
≈ 175 mm
≈ 410 mm ∴ Provide 6 mm weld of L1 = 410 mm and L2 = 175 mm as shown in fig.
166

167

dn
T =
0.9 ×fu×Anc
γm1
β ×Ago ×fy
+
γm0
𝑡 𝑓𝑢
𝛽 = 1.4 - 0.076 (𝑤
) (𝑓𝑦
) (𝑏𝑠
) ≤ (𝑓𝑢 . 𝛾𝑚0
) ≥
0.7
𝛾𝑚1 = 1.25 𝛾𝑚0 = 1.1,
w = 90 mm , t = 12 mm
∴ Lc = 410 mm
∴ bs = w = 90 mm
168

𝛽 = 1.4 - 0.076 (𝑤
) (𝑓𝑦
) (𝑏𝑠
) ≤ (𝑓𝑢 . 𝛾𝑚0
) ≥ 0.7
250×1.25
𝛽 = 1.4 - 0.076 (90
) (250
) ( 90
) ≤ ( 410×1.1
) ≥ 0.7
12 410 410
𝛽 = 1.32 ≤ 1.44 ≥ 0.7
∴ 0.7 ≤ 1.32 ≤ 1.44 𝛽 = 1.32
nc go 2
A = A = 2 ( 90 – 12
) × 12 = 2016 mm2
Tdn
= 0.9 ×fu ×Anc
+ β ×Ago× fy
Tdn
γm1 γm0
= 0.9 ×410 ×2016
+ 1.32 ×2016 ×250
1.25 1.1
Tdn = 1200 kN ………………..(2 )
The Design strength T d is least of eq (1) & (2)
Hence, Td = T dg = = 916.36 kN ………...……Ans.
169

Examples 6 :- Determine the tensile strength of roof truss consist of 2 ISA 80 × 80 × 12 mm connected 12 mm
thick gusset plate by filet weld. Assume fy = 250 MPa. SPPU Dec 2010, May. 2012, 15 Marks
Solution :
dg
Ag× fy
T =
𝛾𝑚0
Here fy = 250 MPa,
Ag = 2 × 1781 = 3562 mm2 ,
𝛾𝑚0 = 1.1
Tdg =
3562 × 250
170
1.1
= 809.54 kN
Hence,
Tdg = 809.54 kN … … … … … … … . . ( 1) Ans

ii. Length of Weld ( Lw )
𝑘 12 – 1.5
𝑘 10.5 mm
Throat Thickness = te = 0.7 S = 0.7 × 6 = 4.2 mm
fwd
=
fu /√3
γmw
= 410 /√3
1.25
= 189.37 N/mm2
................... 𝑚𝑤
171

Total Design Strength of weld = Design strength of weld fwd × thickness of weld × Lw
= 2 × 189.37 × 0.7 S × Lw
= 2 × 189.37 × 0.7 × 6 × Lw = 1590.708 Lw N
Equating shear strength of weld to strength at yielding od gross section .
1590.708 Lw = 809.54 × 103 N
∴ Lw = 508.92 mm
∴ L1 + L2 = 508.92 mm … … … … … … … … . ( 1 )
Tacking moment of weld length at c. g.
and equating for equilibrium
L1 × 24.2 = L2 × ( 80 – 24.2 )
L1 = 2.305 L2 … … … . . … … … … … … … … . ( 2 )
L2 = 153.98 mm
L1 = 354.93 mm
≈ 155 mm
≈ 355 mm ∴ Provide 6 mm weld of L1 = 355 mm and L2 = 155 mm as shown in fig.
172

173

dn
T =
0.9 ×fu×Anc
γm1
β ×Ago ×fy
+
γm0
𝑡 𝑓𝑢
𝛽 = 1.4 - 0.076 (𝑤
) (𝑓𝑦
) (𝑏𝑠
) ≤ (𝑓𝑢 . 𝛾𝑚0
) ≥
0.7
𝛾𝑚1 = 1.25 𝛾𝑚0 = 1.1,
w = 80 mm , t = 12 mm
∴ Lc = 355 mm
∴ bs = w = 80 mm
174

𝛽 = 1.4 - 0.076 (𝑤
) (𝑓𝑦
) (𝑏𝑠
) ≤ (𝑓𝑢 . 𝛾𝑚0
) ≥ 0.7
250×1.25
𝛽 = 1.4 - 0.076 (80
) (250
) ( 80
) ≤ ( 410×1.1
) ≥ 0.7
12 410 355
𝛽 = 1.33 ≤ 1.44 ≥ 0.7
∴ 0.7 ≤ 1.32 ≤ 1.44 𝛽 = 1.33
nc go 2
A = A = 2 ( 80 – 12
) × 12 = 1776 mm2
Tdn
= 0.9 ×fu×Anc
γm1
+ β ×Ago ×fy
γm0
Tdn
= 0.9 ×410 ×1776
+ 1.33 ×1776 ×250
1.25 1.1
Tdn = 1061 kN ………………..(2 )
The Design strength T d is least of eq (1) & (2)
Hence, Td = T dg = = 809.54 kN ………...……Ans.
175

Examples on Design of Tension Member :-
Examples 1:- A tension member 3.4 m long between centre to centre of intersection subjected to a factored
pull of 200 kN. Design economical section using double equal angle section on either side of gusset plate.
SPPU Dec. 2015, 6 Marks
Solution :
Given : L = 3.4 m = 3400 mm, T = 200 kN
Step I : Selection of section : Using yielding of gross section : Ref. Cl. No. 6.2/PN 32/IS 800:2007
u
T =
Ag×fy
γm0
T × γ 200 ×10 3
×1.1
176
Area required = m0
=
f y 250
= 880 mm2
Area required = 880 mm 2
From Steel table ,
Select 2 ISA 50 × 50 × 6 @ 9 kg/m
∴ Area provided = 1136 mm2

Step II : Design of Connection :
Assume bolted connection used in design.
d = 20 mm ∴ d0 = 22 mm
e min = 1.5 d0 = 1.5 × 22 = 33 ≈ 35 mm
P min = 2.5 d = 2.5 × 20 = 50 ≈ 55 mm
a) Design shear strength of bolt Ref. Cl No. 10.3.3/PN 75/IS 800:2007
Vdsb = Vnsb / γmb
Vnsb =
fu
√ 3
( nn Anb + ns Asb )
dsb √ 3
V = [ fu ( n A
n nb s sb
+ n A ) ] / 1.25
As double angle connected to plate therefore bolt is in double shear.
n s sb nb
4 4
∴ n = 1, n = 1 A = π
d2 = 314.15, A = 0.78 π
d2 = 245 mm2
dsb √ 3
V = [ 400 ( 1 × 245 + 1 × 314.15 ) ] / 1.25 = 103.31 kN…………...(1)
177

b) Design bearing strength of the bolt / Bearing Capacity of a Bolt :- Ref. Cl No. 10.3.4/PN 75/IS 800:2007
Where :-
Vdpd = Vnpb / γmb
= 2.5 kb d .t.fu
b
𝑒
3𝑑0
𝑝
3𝑑0 fu
, ( − 0.25),
fu𝑏
, 1
b
K = is least of
35 55
3×22 3 × 22
= 0.53, ( − 0.25)= 0.58 ,
400
250
= 1.6, 1
∴ Kb = 0.53
Vnpb = 2.5 kb d .t.fu = 2.5 × 0.53 × 20 × 10 × 410 = 108.650 kN
108.650
1.25
= 86.92 kN … … … … … … … … … … . ( 2 )
178

∴ Number of bolts =
T 200
Vd 86.92
= = 2.30 ≈ 3 bolts
Step III : Check for Yielding - Ref. Cl. No. 6.2/PN 32/IS 800:2007
dg
Ag× fy
T =
γm0
Tdg =
1138 × 250
1.1
= 258.63 kN
Tdg = 258.63 kN > T = 200 kN
∴ Ok & Safe
179

Step III : Check for rupture : Ref. Cl. No. 6.3.3/PN 33/IS 800:2007
dn
T =
0.9 ×fu ×Anc
γm1
β ×Ago ×fy
+
γm0
𝑡 𝑓𝑢
𝛽 = 1.4 - 0.076 (𝑤
) (𝑓𝑦
) (𝑏𝑠
) ≤ (𝑓𝑢 . 𝛾𝑚0
) ≥
0.7
𝛾𝑚1 = 1.25 𝛾𝑚0 = 1.1,
w = 50 mm , t = 6 mm
w1 = 0.6 × 50 = 30 mm
180

∴ bs = w + w1 – t = 50 + 30 – 6 = 74 mm
Pitch ‘P’ = 55 mm , e = 35 mm
∴ Lc = 2 × 55 = 110 mm
𝛽 = 1.4 - 0.076 (𝑤
) (𝑓𝑦
) (𝑏𝑠
) ≤ (𝑓𝑢 . 𝛾𝑚0
) ≥ 0.7
6 410 110 250×1.25
181
𝛽 = 1.4 - 0.076 (50
) (250
) ( 74
) ≤ ( 410×1.1
) ≥ 0.7
𝛽 = 1.14 ≤ 1.44 ≥ 0.7
∴ 0.7 ≤ 1.07 ≤ 1.44
𝛽 = 1.14

nc 2
A = 2 ( 50 – 22 – 6
) × 6 = 300 mm2
go 2
A = 2 ( 50 – 6
) × 6 = 564 mm2
Tdn
= 0.9 ×fu×Anc
+ β ×Ago×fy
γm1 γm0
Tdn
= 0.9 ×410 ×300
+ 1.14 ×564 ×250
1.25 1.1
182
Tdn = 234.68 kN > T = 200 kN
∴ Ok & Safe

Step V : Check for Block Shear :-
Ref. Cl No. 6.4/PN 33/IS 800:2007
db
T =
Avg.fy
√3 γm0
+ 0.9 Atn fu
γm1
or
db
T =
0.9 Avn fu
√3 γm1
+
Atg fy
γm0
Here, A vg = 2 ( 35 + 55 × 2 ) × 6 = 1740 mm2
A vn = 2 [( 35 + 55 × 2 ) – ( 2.5 × 22)] × 6 = 1080 mm2
A tg = 2 (20 × 6) = 240 mm2
A tn = 2 ( 20 – 0.5 × 22 ) × 6 = 108 mm2

db
T =
or
Avg.f𝑦
√3 γm0
=
1740×250
γm1 √3×1.1
+ 0.9 Atn fu
+ 0.9 × 108 × 410
1.25
= 260.19 kN
db
T =
0.9 Avn fu
√3 γm1
+ Atg fy
γm0
=
0.9 ×1080 ×410
√3× 1.25
+ 240 ×250
1.1
184
= 238.6 kN
Hence,
The block shear strength Tdb of the bolted connection,
T db = 238.6 kN > T = 200 kN
∴ Ok & Safe
Hence, Provide 2 ISA 50 × 50 × 6 mm as a tension member

Step VI : Detailing
Provide 2 ISA 50 × 50 × 6 @ 9 kg/m on both side of gusset plate by 3 # M 20 bolts with pitch = 50 mm and
distance 35 mm.
Fig.
185

Examples 2:- Design a double angle tension member connected on each side of 10 mm thick gusset plate to
carry an axial force of 340 kN. Also design connection using M 20 black bolts of 4.6 grade. Take fy = 250 MPa.
SPPU May 2011, Dec. 2011, May 2015, 15 Marks
Solution :
Given : Tu = 340 × 1.5 = 510 kN
u
T =
g
A × fy
γm0
Tu×γ 510 ×10 3
×1.1
Area required = m0
=
f y 250
186
= 2244 mm2
From Steel table ,
Select 2 ISA 75 × 75 × 8 @ 8.9 kg/m ∴ Area provided = 2280 mm2

Step II : Design of Connection :
Assume bolted connection used in design.
d = 20 mm ∴ d0 = 22 mm
e min = 1.5 d0 = 1.5 × 22 = 33 ≈ 35 mm
P min = 2.5 d = 2.5 × 20 = 50 ≈ 55 mm
a) Design shear strength of bolt Ref. Cl No. 10.3.3/PN 75/IS 800:2007
Vdsb = Vnsb / γmb
Vnsb =
fu
√ 3
( nn Anb + ns Asb )
dsb √ 3
V = [ fu ( n A
n nb s sb
+ n A ) ] / 1.25
As double angle connected to plate therefore bolt is in double shear.
n s sb nb
4 4
∴ n = 1, n = 1 A = π
d2 = 314.15, A = 0.78 π
d2 = 245 mm2
dsb √ 3
V = [ 400 ( 1 × 245 + 1 × 314.15 ) ] / 1.25 = 103.31 kN…………...(1)
187

b) Design bearing strength of the bolt / Bearing Capacity of a Bolt :- Ref. Cl No. 10.3.4/PN 75/IS 800:2007
Where :-
Vdpd = Vnpb / γmb
= 2.5 kb d .t.fu
b
𝑒
3𝑑0
𝑝
3𝑑0 fu
, ( − 0.25),
fu𝑏
, 1
b
K = is least of
35 55
3×22 3 × 22
= 0.53, ( − 0.25)= 0.58 ,
400
250
= 1.6, 1
∴ Kb = 0.53
Vnpb = 2.5 kb d .t.fu = 2.5 × 0.53 × 20 × 10 × 410 = 108.650 kN
108.650
1.25
= 86.92 kN … … … … … … … … … … . ( 2 )
188

∴ Number of bolts =
T 510
Vd 86.92
= = 5.86 ≈ 6 bolts
dg
Ag× fy
T =
γm0
Tdg =
2280 × 250
1.1
= 518.18 kN
Tdg = 518.18 kN > T = 510 kN
∴ Ok & Safe
189

Tdn = ×
0.9 ×fu Anc
γm1
+
β ×Ago ×fy
γm0
𝑡 𝑓𝑢
𝛽 = 1.4 - 0.076 (𝑤
) (𝑓𝑦
) (𝑏𝑠
) ≤ (𝑓𝑢 . 𝛾𝑚0
) ≥
0.7
Here, fu = 410 MPa, fy = 250 Mpa,
𝛾𝑚1 = 1.25 𝛾𝑚0 = 1.1,
w = 75 mm , t = 8 mm
w1 = 0.6 × 75 = 45 mm ≈ 40 mm
190

∴ bs = w + w1 – t = 75 + 40 – 8 = 107 mm
Pitch ‘P’ = 55 mm , e = 35 mm
∴ Lc = 5 × 55 = 275 mm
𝛽 = 1.4 - 0.076 (𝑤
) (𝑓𝑦
) (𝑏𝑠
) ≤ (𝑓𝑢 . 𝛾𝑚0
) ≥ 0.7
8 410 275 250×1.25
191
𝛽 = 1.4 - 0.076 (75
) (250
) (107
) ≤ ( 410×1.1
) ≥ 0.7
𝛽 = 1.23 ≤ 1.44 ≥ 0.7
∴ 0.7 ≤ 1.23 ≤ 1.44
𝛽 = 1.23

nc 2
A = 2 ( 75 – 22 – 8
) × 8 = 784 mm2
go 2
A = 2 ( 75 – 8
) × 8 = 1136 mm2
Tdn
= 0.9 ×fu ×Anc
+ β ×Ago×fy
γm1 γm0
Tdn
= 0.9 ×410 ×784
+ 1.23 ×1136 ×250
1.25 1.1
192
Tdn = 549 kN > T = 510 kN
∴ Ok & Safe

Step V : Check for Block Shear :- Ref. Cl No. 6.4/PN 33/IS 800:2007
db
T =
Avg.fy
√3 γm0
+ 0.9 Atn fu
γm1
or
Tdb =
0.9 Avn fu
√3 γm1
+ Atg fy
γm0
Here, A vg = 2 ( 35 + 55 × 5 ) × 8 = 4960 mm2
A vn = 2 [( 35 + 55 × 5 ) – ( 5.5 × 22)] × 8 = 3024 mm2
A tg = 2 (35 × 8) = 560 mm2
A tn = 2 ( 35 – 0.5 × 22 ) × 8 = 384 mm2

db
T =
or
Avg.f𝑦
√3 γm0
+ 0.9 Atn fu
γm1
=
4960 ×250
√3×1.1
+ 0.9 ×384 × 410
= 764.18 kN
1.25
db
T =
0.9 Avn fu
√3 γm1
+
Atg fy
γm0
=
0.9 ×1080 ×410
√3× 1.25
+ 240 ×250
1.1
194
= 642.66 kN
Hence,
The block shear strength Tdb of the bolted connection,
T db = 642.66 kN > T = 510 kN
∴ Ok & Safe
Hence, Provide 2 ISA 75 × 75 × 8 @ 8.9 kg/m as a tension member

Step VI : Detailing
Provide 2 ISA 75 × 75 × 8 @ 8.9 kg/m on both side of gusset plate by 6 # M 20 bolts with pitch = 50 mm and
distance 35 mm.
Fig.
195

Examples 3 :- Design the tie of a roof truss subjected to factored design tension, T = 230 kN using unequal
angle section. Centre to centre length of intersection is 2.8 m. Also design the welded connection and draw the
design details. SPPU Dec. 2012, May 2013, 15 Marks
Solution :
Given : Tu = 230 kN
Step I : Selection of section : using yielding of gross section : Ref. Cl. No. 6.2/PN 32/IS 800:2007
u
T =
g
A × fy
γm0
Tu×γ 230 ×10 3
×1.1
Area required = m0
=
f y 250
196
= 1012 mm2
From Steel table,
Select 2 ISA 75 × 50 × 6 @ 5.6 kg/m ∴ Area provided = 1432 mm2, Cxx = 24.4 mm

II. Design of welded connection
From table no 21/ PN 78 / IS 800 : 2007
Provide 5 mm size weld
fwd
=
fu /√3
γmw
= 410 /√3
1.25 ................... 𝑚𝑤
= 189.37 N/mm2 𝛾 = 1.25 for shop weld
Total Strength of weld = Design strength of weld fwd × thickness of weld × Lw
= 2 × 189.37 × 0.7 S × Lw
= 2 × 189.37 × 0.7 × 5 × Lw
By equating
= 1325.59 Lw N
Fwd = T
1325.59 Lw = 230 × 103 N
Lw = 173.5 mm
197

∴ L1 + L2 = Lw
∴ L1 + L2 = 173.5 mm … … … … … … … … . ( 1 )
Tacking moment of weld length at c. g. and equating for equilibrium
L1 × 24.4 = L2 × ( 75 – 24.4 )
L1 = 2.07 L2 … … … . . … … … … … … … … . ( 2 )
L2 = 56.51 mm
L1 = 117 mm ≈
≈ 60 mm
120 mm
∴ Provide 5 mm weld of L1 = 120 mm and L2 = 60 mm as shown in fig.
198

dg
Ag× fy
T =
γm0
Tdg =
1432 × 250
1.1
= 325.45 kN
Tdg = 325.45 kN > T = 230 kN
∴ Ok & Safe
199

dn
T =
0.9 ×fu ×Anc
γm1
β ×Ago ×fy
+
γm0
𝑡 𝑓𝑢
𝛽 = 1.4 - 0.076 (𝑤
) (𝑓𝑦
) (𝑏𝑠
) ≤ (𝑓𝑢 . 𝛾𝑚0
) ≥
0.7
𝛾𝑚1 = 1.25 𝛾𝑚0 = 1.1,
w = 50 mm , t = 6 mm
200

∴ bs = w
∴ bs = 50
Pitch ‘P’ = 55 mm , e = 35 mm
∴ Lc = 120
𝛽 = 1.4 - 0.076 (𝑤
) (𝑓𝑦
) (𝑏𝑠
) ≤ (𝑓𝑢 . 𝛾𝑚0
) ≥ 0.7
6 410 120 250×1.25
201
𝛽 = 1.4 - 0.076 (50
) (250
) ( 50
) ≤ ( 410×1.1
) ≥ 0.7
𝛽 = 1.24 ≤ 1.44 ≥ 0.7
∴ 0.7 ≤ 1.24 ≤ 1.44
𝛽 = 1.24

nc 2
A = 2 ( 75 – 6
) × 6 = 864 mm2
go 2
A = 2 ( 50 – 6
) × 6 = 564 mm2
Tdn
= 0.9 ×fu ×Anc
+ β ×Ago×fy
γm1 γm0
Tdn
= 0.9 ×410 ×864
+ 1.24 ×564 ×250
1.25 1.1
202
Tdn = 413.20 kN > T = 230 kN
∴ Ok & Safe

Examples 4 :- Design a suitable single equal angle section to carry a factored tensile force of 250 kN. Use 5 mm
size of fillet weld. SPPU Dec. 2016, 6 Marks
Solution :
Given : Tu = 250 kN
u
T =
g
A × fy
γm0
Tu×γ 250 ×10 3
×1.1
Area required = m0
=
f y 250
203
= 1100 mm2
From Steel table,
Select ISA 60 × 60 × 10 @ 8.6 kg/m
∴ Area provided = 1100 mm2, Cxx = 18.5 mm

fwd
=
fu /√3
γmw
= 410 /√3
1.25 ................... 𝑚𝑤
Total Strength of weld = Design strength of weld fwd
= 189.37 × 0.7 S × Lw
= 189.37 × 0.7 × 5 × Lw
× thickness of weld × Lw
By equating
= 662.795 Lw N
Fwd = T
662.795 Lw = 250 × 103 N
Lw = 377.19 mm
204

∴ L1 + L2 = Lw
205
∴ L1 + L2 = 377.19 mm … … … … … … … … . ( 1 )
L1 × 18.5 = L2 × ( 60 – 18.5 )
L1 = 2.24 L2 … … … . . … … … … … … … … . ( 2 )
L2 = 116.41 mm
L1 = 257.19 mm
≈ 120 mm
≈ 260 mm

dg
Ag× fy
T =
γm0
Tdg =
1100 × 250
1.1
= 250 kN
Tdg = 250 kN = T = 250 kN
∴ Ok & Safe
206

Step IV : Check for rupture : Ref. Cl. No. 6.3.3/PN 33/IS 800:2007
dn
0.9 ×fu ×Anc
T =
γm1
β ×Ago ×fy
+
γm0
𝑡 𝑓𝑢
𝛽 = 1.4 - 0.076 (𝑤
) (𝑓𝑦
) (𝑏𝑠
) ≤ (𝑓𝑢 . 𝛾𝑚0
) ≥
0.7
𝛾𝑚1 = 1.25 𝛾𝑚0 = 1.1,
w = 60 mm , t = 10 mm
207

∴ bs = w
∴ bs = 60
Pitch ‘P’ = 55 mm , e = 35 mm
∴ Lc = 260
𝛽 = 1.4 - 0.076 (𝑤
) (𝑓𝑦
) (𝑏𝑠
) ≤ (𝑓𝑢 . 𝛾𝑚0
) ≥ 0.7
250×1.25
208
𝛽 = 1.4 - 0.076 (60
) (250
) ( 60
) ≤ ( 410×1.1
) ≥ 0.7
10 410 260
𝛽 = 1.33 ≤ 1.44 ≥ 0.7
∴ 0.7 ≤ 1.33 ≤ 1.44
𝛽 = 1.33

Anc 2
= ( 60 – 10
) × 10 = 550 mm2
go 2
A = ( 60 – 10
) × 10 = 550 mm2
Tdn
= 0.9 ×fu ×Anc
+ β ×Ago×fy
γm1 γm0
Tdn
= 0.9 ×410 ×550
+ 1.24 ×550 ×250
1.25 1.1
209
Tdn = 328.61 kN > T = 250 kN
∴ Ok & Safe

Examples 5 :- A tie member of a truss, 2 ISA 65 × 65 × 6 @ 5.8 kg/m back to back on either side is welded to
gusset plate. Design a weld to transmit a load equal to the full strength of the member.
SPPU Aug. 2014, 6 Marks
Solution :
Design tensile strength due to yielding of gross cross section :
u
T =
g
A × fy
γm0
=
1.1
210
2 ×744×250
= 338.18 kN

From table no 21/ PN 78 / IS 800 : 2007
fwd
=
fu /√3
γmw
= 410 /√3
1.25 ................... 𝑚𝑤
Total Strength of weld = Design strength of weld fwd × thickness of weld × Lw
= 2 × 189.37 × 0.7 S × Lw
= 2 × 189.37 × 0.7 × 6 × Lw
By equating
= 1590.17 Lw N
Fwd = T
1590.71 Lw = 338.18× 103 N
Lw = 212.59 mm
211

∴ L1 + L2 = Lw
∴ L1 + L2 = 212.59 mm … … … … … … … … . ( 1 )
L1 × 18.1 = L2 × ( 65 – 18.1)
L1 = 2.59 L2 … … … . . … … … … … … … … . ( 2 )
L2 = 59.20 mm
L1 = 153.39 mm
≈ 60 mm
≈ 155 mm
∆∆∆
212

UNIT-1
a) Introduction to design of steel structure
b) Design of tension member
213

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