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1. CIV3004_v02
Design of Structural Steel Elements
3-0-3
1

2. UNIT-I
INTRODUCTION TO STEEL STRUCTURES AND DESIGN OF
CONNECTIONS
2

3. What are steel structures ?
• Steel structure is an assemblage of a group of members expected to
sustain the applied forces and transfer them safely to ground.
• Depending on the orientation of the member in structure and its
structural use, the member is subjected to forces, either axial, bending
or torsion
• Ex: - I-Beam, Tee section, Channel section, Steel plate etc..
3

4. Common Steel structures
• Roof truss in factories, cinema halls, railways etc.,
• Crane girders, columns, beams
• Plate girders, bridges
• Transmission towers, water tank, chimney etc.,
4

5. 5

6. Structural Steel
6
Steel structures facilitate ease of fabrication and faster erection of structure .Bolts
and welding employed for joining .

7. Advantages of steel as structural material
• High strength
• The high ratio of strength to weight (the strength per unit weight)
• Excellent ductility and seismic resistance
• Withstand extensive deformation without failure even under high
tensile stress.
• Elasticity, uniformity of material
• Predictability of properties, close to design assumption
• Ease of fabrication and speed of erection
7

8. Disadvantages of steel
• Susceptibility to corrosion
• High Initial cost and maintenance costs
• Loss of strength at elevated temperature
• Fireproofing costs
• Susceptibility to buckling
• Fatigue and brittle fracture
8

9. Various forms of steel
9

10. Properties of Structural Steel
The properties of steel required for engineering design may be classified as :
• i) Physical properties
• ii) Mechanical Properties
i) Physical Properties : Irrespective of its grade physical properties of steel may
be taken as given below (clause 2.2.4 of IS 800)
a) Unit weight of steel : 7850 kg/m3
b) Modulus of Elasticity E = 2.0 x 105 N/mm2
c) Poisson’s ratio µ = 0.3
d) Modulus of rigidity G = 0.769x 105 N/mm2
10

11. Stress-Strain Curve for Steel
11

12. 12

13. Ultimate strength or Tensile strength
• Ultimate tensile strength is the highest stress at which a tensile
specimen fails by fracture and is given by
𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑡𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ =
𝑢𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑡𝑒𝑛𝑠𝑖𝑙𝑒 𝑙𝑜𝑎𝑑
𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛
• As per code , the characteristic ultimate tensile strength (fu)is
defined as the minimum value of stress below which not more than
a specified percentage (5%) of corresponding stresses of samples
tested are expected to occur.
13

14. Designation of Steel
• The steel is designated as Fe310, Fe 410WA, Fe540B etc….
14
Steel
Fe
Ultimate tensile stress in MPa
310
Weldable
W
Grade of Steel
A,B,C

15. GRADES OF STEEL
• Grade A - used for structures subjected to normal conditions
• Grade B - used for situations where severe fluctuations are there but
temp > 00 C
• Grade C - used upto temp – 400 C and have high impact properties
• Yield strength or yield stress is the material property defined as
the stress at which a material begins to deform plastically whereas yield
point is the point where nonlinear (elastic + plastic) deformation
begins. ... Once the yield point is passed, some fraction of the
deformation will be permanent and non-reversible.
15

16. Properties of steel………………………….
• Ductility is defined as the ability of a material to change its shape
without fracture.
• capacity of steel to undergo large inelastic deformation without
significant loss of strength or stiffness .
• The ductility of the tension test specimen is measured by
Percentage elongation =
𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑒𝑑 𝑙𝑒𝑛𝑔𝑡ℎ −𝑔𝑢𝑎𝑔𝑒 𝑙𝑒𝑛𝑔𝑡ℎ
𝑔𝑎𝑢𝑔𝑒 𝑙𝑒𝑛𝑔𝑡ℎ
x100
•
16

17. Properties of steel………………………….
• Toughness: Capacity to absorb energy, measure of fracture
resistance under impact. Area under the stress-strain curve is a
measure of toughness
• Hardness is a measure of the resistance of the material to
indentations and scratching.
• Corrosion resistance: Steel corrodes in moist air, sea water and
acid. Adopt Painting, metallic coating, plastic coating, using
corrosion resistant steel to resist corrosion.
• Fatigue Resistance: Damage of material to cyclic loading and it
may occurs due to moving loads, vibration in bridge etc.
17

18. STRUCTURAL STEEL
SECTIONS
18

19. ROLLED STEEL SECTIONS
Steel is rolled to a required shape during fabrication. Steel sections of
standard shapes, sizes and lengths are rolled in steel mills.
Various types of rolled steel sections are:
• Rolled steel I – section (Beam Section)
• Rolled steel channel section
• Rolled steel Angle section
• Rolled steel Tee section
• Rolled steel bars and flats
• Rolled steel Tubes
• Rolled steel Plates, sheets and strips
19

20. Rolled I – Sections
• I sections which are also called as steel beams or rolled steel joist
are extensively used as beams, lintels, columns etc. It consists two
flanges and a web connected as shown in figure.
Types of I- sections:
• ISJB – Indian standard junior beam
• ISLB –Light beam
• ISMB -Medium beam
• ISWB -Wide flange beam
• ISHB - Heavy beam
20

21. Rolled Channel Sections
• The channel section or C- section consists two equal flanges connected to web at
both ends. Channel sections are extensively used in steel framed structures.
Types of Channel Sections:
21

22. Rolled T- Sections
• T section consists of flange and web arranged in “T” shape. They are used in steel
roof trusses to form built up sections. Two angle sections can also be joined to get
T section.
Types of T- sections:
22

23. Rolled steel sections
• Rolled Round Bars : Round bars contain circular
cross sections and these are used as reinforcement in
concrete and steel grill work etc. Round bars are
available in various diameters varies from 5 mm to
250 mm.
• Rolled Square Bars: Square bars contain square cross
sections and these are widely used for gates, windows,
grill works etc. the sides of square cross section ranges
from 5 mm to 250 mm.
• Rolled Flat Bars: Flat bars are also used for gates,
windows, grill works etc. Flat bars are designated with
width of the bar which varies from 10 mm to 400 mm.
thickness of flat bars will be from 3 mm to 40 mm
23

24. Rolled steel sections
Corrugated Sheets
• Plain steel sheets are passed through machines which
produce bends by pressing them called corrugations. These
sheets are used for roof coverings.
Expanded Metal
• Expanded metal sheets are made from mild steel sheets.
Which are cut through machine and expanded. Generally,
Diamond shaped mesh is appeared in this type of sheets.
Rolled Steel Plates
• Steel plates are well used items in steel structures. They are
used for connecting steel beams, tensional member in roof
truss etc. They are designated with their thickness which is
varying from 5 mm to 50 mm.
•
24

25. Rolled steel sections
Ribbed Bars (HYSD)
• Ribbed HYSD bars are made of high yield strength steel. Ribs are
nothing but projections produced on bars by cold twisting of bar in
hot rolled condition. The twist is made according the standard
requirements.
Rolled Thermo-Mechanically Treated (TMT) Bars
• Thermo Mechanically Treated or TMT bars are high-strength
reinforcement bars having a hardened outer core and a soft inner
core. They are manufactured by passing hot rolled steel billets
through water, which hardens the surface and increases its tensile
strength while the inner core stays at a comparatively warmer
temperature and hence, the core becomes more ductile. This
variation in the microstructure of the cross-section of the bar
provides huge strength to the bar.
25

26. Choice of sections
• Governed by sectional properties and availability
• Popular in India – ISMB, ISMC, equal angles
• Channels are used in purlins, Tee and angles in truss, I section in
beam and column
Other forms of sections
• Built-up, stepped, wide flange, hybrid, cold
• formed (formed from light gauge steel strips)
26

27. Types of Loads
Load Combination
27

28. Types of Loads
• Dead loads (1S 875 (Part 1)
• Imposed loads (live load, crane load, snow load, dust load, wave load, earth
pressures, etc) - IS 875 (Part 2).
• Wind loads - IS 875 (Part 3)
• Earthquake loads - IS 1893 (Part 1)
• Erection loads
• Accidental loads such as those due to blast, impact of vehicles, etc;
• Secondary effects due to contraction or expansion resulting from
temperature changes, differential settlements of the structure as a whole or
of its components, eccentric connections, rigidity of joints differing from
design assumptions.
28

29. Load combinations
• Load combinations for design purposes shall be those that
produce maximum forces and effects and consequently maximum
stresses and deformations.
The following combination of loads with appropriate partial safety
factors (see Table 4) maybe considered.
• a) Dead load + imposed load
• b) Dead load + imposed load + wind or earthquake load
• c) Dead load + wind or earthquake load
• d) Dead load+ erection load.
29

30. Design philosophy
30

31. Objectives of Design
The aim of design is to decide shape, size and connection details of the
members so that the structure being designed performs satisfactorily
during its intended life.
• Sustain all loads expected on it
• Sustain deformations during and after construction
• Should have adequate durability
• Should have adequate resistance to fire and other weathering agents.
• Should not collapse under accidental loads
31

32. Design philosophy
1. Working Stress method
• Stress at which the material starts to yield is taken as permissible
stress of the section. All sections are designed not to exceed the
permissible stress
• Permissible stress = Yield stress / F.O.S
• Since steel can resist load after yield point, following this principle
results in bulky, uneconomical sections
2. Ultimate Load method (plastic design method)
• Permissible load is a load when all the fiber in the steel is yielded
• This method does not ensure serviceability
32

33. Design philosophy
3. Limit State Method (IS 800 : 2007) : Limit states are the states beyond which the
structure no longer satisfies the specified performance requirements. It includes both
limit state of strength and serviceability.
Limit state of strength:
• associated with failures (or imminent failure), under the action of probable and most
unfavorable combination of loads on the structure which may endanger the safety of
life and property.
It includes
i) Loss of equilibrium of whole or part of the structure
ii) Loss of stability of structure as whole or part of the structure
iii) Failure by excessive deformation
iv) Fracture due to fatigue and brittle fracture
33

34. Design philosophy- Limit State
Limit state of Serviceability
• Deformation and deflections, which may adversely affect the
appearance or effective use of the structure
• 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
• Repairable damage or crack due to fatigue.
• Corrosion and durability
• Fire.
34

35. CONNECTIONS
BOLTED CONNECTION
35

36. Introduction
• Connections are structural elements used for joining different
members of a structural steel frame work.
• Steel Structure is an assemblage of different member such as
“BEAMS,COLUMNS” which are connected to one other, usually at
member ends fastners, so that it shows a single composite unit.
36

37. Why connection is required ?
• Connections between different members of a steel framework not only
facilitate the flow of forces and moments from one member to another
but also allow the transfer of forces up to the foundation level
• A connection failure may lead to a catastrophic failure of the whole
structure.
• Normally, a connection failure is not as ductile as that of a steel member
• For achieving an economical design, it is important that connectors
develop full or a little extra strength of the members it is joining.
37

38. Components of connection
• BOLTS
• WELD
• Connecting Plates
• Connecting Angles
38

39. TYPES OF CONNECTIONS
Riveted Connections Bolted Connections
39
Welded Connections

40. BOLTED CONNECTIONS
• A bolt is a metal pin with a head formed at one end and shank threaded at
the other in order to receive a nut.
• Bolts can be used for making end connections in tension and compression
members and in fabrication of built-up and compound members.
40

41. Types of Bolts
Bolts are classified as
• Unfinished/ Black Bolts (IS 1363 : 2002)
• Finished/ Turned Bolts (IS 1364 : 2002)
• High Strength Friction Grip (HSFG) Bolts (IS 3757 : 1985 and IS
4000 : 1992)
41

42. Unfinished/Black Bolts
• Ordinary, unfinished, common bolts made up of mild steel and least expensive
• The bolts are available from 5 to 36mm in diameter and are designated as M16,
M20, M24 to M30
• Used for primarily light structures under static load such as small trusses,
purlins etc
• Used as temporary fasteners during erection where HSFG bolts or welding are
used as permanent fasteners.
• Not recommended for heavy or dynamic loads or impact or fatigue loads.
42

43. Finished/Turned Bolts
• These bolts are made from mild steel, hexagonal rods, which are
finished by turning to a circular shape.
• Tolerances allowed are about 0.15 mm to 0.5 mm.
• The small tolerance necessitates the use of special methods to
ensure that all the holes align correctly and hence expensive.
• Mainly used in special jobs (in some machines and where there are
dynamic loads).
43

44. High Strength Friction Grip (HSFG) Bolts
• High-strength bolts are made from bars of medium carbon steel.
• Special techniques are used for tightening the nuts to induce a
specified initial tension in the bolt, which causes sufficient friction
between the faying faces.
• The tension in the bolt ensures that no slip takes place under working
conditions and so the load transmission from plate to the bolt is
through friction and hence it is also known as non-slip or slip critical
or friction type connection.
• HSFG bolts are made from quenched and tempered alloy steels with
grades from 8.8 to 10.9.
44

45. Classification of bolts based on type of load transfer
• Bearing type - Unfinished and finished bolts belong to bearing type
since they transfer shear force from one member to the other by
bearing.
• Friction Grip type- HSFG bolts are friction type since they transfer
shear by friction.
45

46. Advantages of Bolted Connections
• Use of unskilled labour and simple tools
• Noiseless and quick fabrication
• No special equipment/process needed for installation
• Fast progress of work
• Accommodates minor discrepancies in dimensions
• The connection supports loads as soon as the bolts are tightened
• Alterations if any can be done easily
• Working area required in the field is less
46

47. Disadvantages of Bolted Connections
• Tensile strength is reduced considerably due to stress concentrations
and reduction of area at the root of the threads.
• Rigidity of joints is reduced due to loose fit, resulting into excessive
deflections
• Due to vibrations nuts are likely to loosen, endangering the safety of
the structures.
47

48. Types of Bolted Joints
• Two types of bolted joints subjected to axial forces (loads are assumed
to pass through centre of gravity of group of bolts)
• (i) Lap Joint and (ii) Butt Joint
• Lap Joint : It is the simplest joint in which plates to be connected
overlap one another.
48

49. Butt Joint : Members to be connected are placed end to end with cover
plate.
Cover plates are provided in two ways
(i) single cover plate butt joint. (ii) Double cover plate butt joint
49

50. • In case of double cover butt joint, the total shear force to be transmitted
by the members is split in two parts (as shown in fig)
• In case of lap joint, only one plane on which the force acts and hence
shear capacity of double cover joint is doubled that of lap joint.
50

51. Load transfer mechanism in bolted connection
• Load bearing type – the load transfer from one connected part to other
part by shear and bearing
• Slip-critical or slip resistance connection – the load transfer by friction
(HFSG bolts)
• In slip resistance connections, the entire forces are resisted by friction
due to large clamping forces developed in bolts.
• The bolts are snug tight and they are termed as pre tensioned joints
51

52. 52
Bolt Shear Transfer – Free Body Diagram
(a) Bearing Connection
(b) Friction Connection
T
Frictional Force T
Clamping Force, PO
Bearing stresses
Tension
in bolt
T
T
T
Clamping Force, PO
FORCE TRANSFER MECHANISM

53. Failure of Bolted Joints
The bolted joints may fail in any one of the following six ways:
1) Shear failure of bolts
2) Bearing failure of bolts
3) Bearing failure of plates
4) Tensile failure of bolts
5) Tensile or tearing failure of plates
6) Block Shear Failure
53

54. Failure of Bolted Joints
1. Shear failure of Bolts :
• Shear stresses are generated when the
plates slip due to applied forces.
• Failure occurs when shear stress in bolt is
greater than nominal shear capacity of
bolt.
• This failure take place at the bolt shear
plane (interface)
• Bolts may be in single or double shear.
54

55. Failure of Bolted Joints
2. Bearing failure of Bolts:
• The bolt is crushed half circumference
• The plate may be strong in bearing and the heaviest pressed plate may
press the bolt shank.
• It occurs when plates are made of high strength steel
55

56. Failure of Bolted Joints
3. Bearing failure of plates:
• The plate may crush in bearing, if the plate material is weaker than the
bolt material.
• The bearing problem occurs due to the presence of nearby bolt or the
proximity of an edge in the direction of the load.
56

57. Failure of Bolted Joints
4. Tension failure of bolts:
• This failure occurs when bolt in tension and tensile stress in bolt is
greater than the permissible stress of the bolt
• Tension Failure at root of thread (weak)
5.Tension or tearing failure of plates:
• tearing failures occur when the bolts are stronger than the plates.
• Plate breaks along the bolt line
57

58. Failure of Bolted Joints
6. Block shear failure :
• A combination of shear failure and Tension Failure
• A portion of plate (block) shears along the force direction
• Block of material within the bolted area breaks away from
reminder area.
58

59. Failure modes of Bolted Connection
59

60. Specifications
60

61. Specifications
61

62. Specifications
62

63. Grade Classification of Bolts
• The grade classification of bolt is indicative of the strength of the
material of the bolt. The two grades of bolts commonly used are 4.6 and
8.8.
• For a 4.6 grade 4 indicates that the ultimate tensile strength of the bolt,
fub = 4 x 100 = 400 N/mm2 and, 0.6 indicates that the yield strength of the
bolt, fyb= 0.6 x Ultimate strength = 0.6 x 400 = 240 N/ mm2.
63

64. Tensile Properties of Bolts
64

65. Strength of Bolts
65

66. Bearing type connections
In bearing type, load transferred is greater than friction resistance
Bolts in Bearing type connection are checked for
• (i) shear, and
• (ii) bearing
Strength of Bolt:
Strength of bolt = Minimum of i) Strength of bolt in bearing, and ii) Strength of bolt in
shearing
Strength of bolt connection = strength of one bolt x no. of bolts
Strength of joint = Minimum of (i) strength of bolt or bolt group, and (ii) net tensile
Strength of plate
66

67. Shearing Strength of Bolt:
Nominal Shear Capacity of the bolt is Vnsb =
𝑓𝑢𝑏
(𝑛𝑛.𝐴𝑛𝑏+𝑛𝑠.𝐴𝑠𝑏)
3
Design Shear Capacity of the bolt is Vdsb =
Vnsb
Ɣ𝒎𝒃
=
𝒇𝒖𝒃
(𝒏𝒏.𝑨𝒏𝒃+𝒏𝒔.𝑨𝒔𝒃)
𝟑 Ɣ𝒎𝒃
fub = Ultimate tensile strength of bolt
nn = No. of shear planes with threads,
ns = No. of shear planes without threads (shanks),
Asb = Nominal area of shank, Asb,
Anb = Net stress area of bolt = 0.78x Asb
Ɣm = partial safety factor for bolt material= 1.25
67
For safety of bolt
Vnsb < Vdsb

68. Bearing Strength of Bolt
• The nominal bearing strength of bolt is given by
Vpb = 2.5 x kb.d.t.
fu
Ɣmb
• The design strength of bolt is Vdpb =
Vpb
Ɣ𝑚𝑏
= 2.5 x kb.d.t.
fu
Ɣ𝑚𝑏
where
d0 = diameter of hole, d = diameter of bolt
e, p = end and pitch distances of the bolt along bearing direction
fub = Ultimate tensile strength of bolt , fu = Ultimate tensile strength of plate
t = aggregate thickness of the connected plates in bearing stress in same direction
Ɣm = partial safety factor for bolt material= 1.25
68
For
safety
of
joint
in
bearing
V
pb
<
V
dpb

69. Tensile strength of plate
• The tensile strength of the plate is given by
Tnd = 𝟎. 𝟗 𝑨𝒏
𝒇𝒖
𝜸𝒎𝟏
where fu = the ultimate stress of material in Mpa
𝛾𝑚1 = partial safety factor = 1.25
An = the effective net area of plate in mm2
An = (b-nd0).t for chain bolting
An ={ b-n.do + ∑
𝑝𝑠
2
4𝑔
}.t for staggered bolting
69
b = width of plate
t= thickness of thinner plate
d0 = diameter of hole
ps = staggered pitch
g = gauge length

70. Design of Bolt Joint – Basic specifications
70
1. Clearance for fastener holes : Table 19 (Clause 10.2.1)
• Diameter of bolt hole (d0) = dia of bolt (d) + 1mm ( dia range from 12 to 14mm)
= d + 2 ( d range from 16 to 24mm)
= d+ 3 ( d is greater than 30mm)
d+1 d+1 d+2 d+2 d+2 d+2 d+3 d+3
Nominal
diameter (d) in mm
12 14 16 20 22 24 30 36
Diameter of
hole (d0) in mm
13 15 18 22 24 26 33 39

71. Design of Bolt Joint – Basic specifications
2. Area of bolts at root (Anb) = 0.78 Asb
where Asb = area of bolt at shank = πd2/4
3. Grade of Bolts (as per IS1367)
71
For a 4.6 grade:
4 indicates that the ultimate tensile strength
of the bolt, fub = 4 x 100 = 400 N/mm2
0.6 indicates that the yield strength of the
bolt, fyb= 0.6 x Ultimate strength = 0.6 x 400
= 240 N/ mm2.

72. Efficiency of Joint
• It is defined as the ratio of strength of joint and strength of plate in
tension
Efficiency ŋ =
𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑗𝑜𝑖𝑛𝑡
𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑠𝑜𝑙𝑖𝑑 𝑝𝑙𝑎𝑡𝑒
𝑥 100
Strength of solid plate is governed by its strength in yielding
Strength of joint is the smaller of strength in shear and strength in bearing
72

73. Problems in Bolted Connections
Cover the problems on
• Strength of bolt , plate and efficiency of joint – both lap and butt
joint
• Design of Lap joint and butt joint
• Eccentric bolted connection
73

74. Q1. Calculate the strength of a 20mm diameter bolt of grade 4.6 for the following cases: the
main plates to be jointed are 12mm thick.
a) Lap Joint
b) Single cover butt joint, the cover plate being 10mm thick
c) Double cover butt joint each of cover plate being 8mm thick.
Solution: Given data
• For Fe410 grade of steel : fu = 410 Mpa
• For bolts of grade 4.6 : fub = 400 Mpa
• Partial safety factor for the material of bolt (Ɣmb)= 1.25
• Net tensile stress area of 20mm dia bolt Anb = 0.78.πd2/4 =0.78x πx202/4 = 245 mm2
74

75. Case (a): Single bolt Lap Joint ( dia of bolt d = 20mm)
• The strength of bolt in single shear Vsb = Anb
fub
√3.Ɣmb
=245x
400
√3x1.25
x 10-3= 45.26 kN
• The strength of bolt in bearing Vpb = 2.5 x kb.d.t.
fu
Ɣmb
• d0 = dia of hole = 20+2 = 22mm, e = 1.5xd0 =33mm, pitch p = 2.5xd = 50mm
i)
𝑒
3𝑑𝑜
=
33
3𝑥22
= 0.5 ii)
𝑝
3.𝑑0
- 0.25 =
50
3.𝑥22
− 0.25 =0.5, iii)
fub
fu
=
400
410
=0.975. and iv) 1.0
Hence the least of above values kb = 0.5
• Bearing strength Vpb = 2.5 x kb.d.t.
fu
Ɣmb
=2.5 x 0.5x20x12x.
410
1.25
x10-3 = 98.4 kN
• The strength of bolt will be minimum of the strength in shear and bearing and it is 45.26 kN
75

76. Case (b): Single cover butt joint with 10mm thick cover plate
• The bolt will be in single shear and bearing.
• t = least of aggregate thickness of cover plates (10mm) and min thickness
of main plates (12mm) jointed. Hence t =10mm.
• The strength of bolt in single shear (from case:a) =45.26 kN
• The strength of bolt in bearing
Vpb = 2.5 x kb.d.t.
fu
Ɣmb
=2.5 x 0.5x20x10x.
410
1.25
x10-3 = 82 kN
The strength of bolt will be minimum of the strength in shear and bearing and it
is 45.26 kN
76

77. Case (c): The bolt will be in double shear and bearing
• t = least of aggregate thickness of cover plates and min thickness
of main plates jointed.
• Sum of thickness of cover plates = 8+8 =16mm,
• thickness of main plate is 12mm
• The strength of bolt in double shear Vsb = 2xAnb
fub
√3.Ɣmb
=2x245x
400
√3x1.25
x 10-3= 90.52 kN
• The strength of bolt in bearing Vpb = 2.5 x kb.d.t.
fu
Ɣmb
= 2.5 x 0.5x20x12x.
410
1.25
x10-3 = 98.4 kN
• The strength of bolt will be minimum of the strength in shear and bearing and
it is 90.52 kN
77
Hence t =12mm

78. Q2:
• A single bolted double cover butt joint is used to connect two plates which
are 8mm thick. Assuming 16mm diameter bolts of grade 4.6 and cover
plates to be 6mm thick. Calculate the strength and efficiency of the joint, if
4 bolts are provided in the bolt line at a pitch of 45mm as shown.
• Also determine the efficiency of the joint if two lines of bolts with two
bolts in each line have been arranged to result in a double bolted double
cover butt joint
• Figure:
78

79. Solution: For Fe410 grade of steel : fu = 400 Mpa,
For bolts of grade 4.6 : fub = 400 Mpa
Dia of bolt d = 16mm , Dia of hole d0 = 16+2 = 18mm
Net area Anb = 0.78 .Asb = 0.78.π(16)2/4 = 157mm2
Ɣmb = partial safety factor for material of bolt = 1.25
Ɣm1 = partial safety factor for resistance governed by ultimate stress = 1.25
t = least of aggregate thickness of cover plates (6+6 =12mm) and minimum
thickness of main plates (8mm).
Hence take t = 8mm
Step1: Strength of Bolt :
i) The strength of bolt in double shear Vsb = 2[Anb
fub
√3.Ɣmb
]=2x157x
400
√3x1.25
x 10-3= 58 kN
79
i

80. ii) The strength of bolt in bearing Vpb = 2.5 x kb.d.t.
fu
Ɣmb
• d0 = dia of hole = 16+2 = 18mm, e = 1.5xd0 =27mm ~ 30mm, pitch p = 45mm
i)
𝑒
3𝑑𝑜
=
30
3𝑥18
= 0.55 ii)
𝑝
3.𝑑0
- 0.25 =
45
3.𝑥18
− 0.25 =0.58, iii)
fub
fu
=
400
410
=0.975. and iv) 1.0
Hence the least of above values kb = 0.55
• Bearing strength Vpb = 2.5 x kb.d.t.
fu
Ɣmb
=2.5 x 0.55x16x8x.
410
1.25
x10-3 = 57.73 kN
• The strength of bolt will be minimum of the strength in shear (i) and bearing(ii) = 57.73 kN
80
ii

81. iii) The net tensile strength of plate per pitch length
Tnd = 𝟎. 𝟗 𝑨𝒏
𝒇𝒖
𝜸𝒎𝟏
= 0.9x
𝒇𝒖
𝜸𝒎𝟏
x(p-nd0).t =
0.9x410x(45−18)x8x10−3
𝟏.𝟐𝟓
=63.76 kN
Hence the strength of joint per pitch length will be least of the strength per pitch length
in shear (i), bearing for bolts (ii) and net strength of plate (iii)
The strength of joint per pitch length = 57.73 kN
Step2: Strength of solid plate per pitch length = 𝟎. 𝟗
𝒇𝒖
𝜸𝒎𝟏
𝒑. 𝒕 = 𝟎. 𝟗 𝒙
𝟒𝟏𝟎
𝟏.𝟐𝟓
x45x8x10-3 = 106.27 kN
Efficiency of joint ŋ =
𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑗𝑜𝑖𝑛𝑡
𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑠𝑜𝑙𝑖𝑑 𝑝𝑙𝑎𝑡𝑒
𝑥 100 =
57.73
106.27
𝑥 100 = 54.32 %
81
iii

82. Case 2: when bolts are arranged in two rows
i) The strength of bolt in double shear = 2 x58 = 116 kN
ii) The strength of bolt in bearing = 2 x57.73 = 115.46 kN
iii) The net strength of plate = 63.76 kN (Only one bolt will fall in section 1-1)
Hence the strength of the joint per pitch length will be least of i), ii) and iii) is 63.76 kN
Efficiency of joint ŋ =
𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑗𝑜𝑖𝑛𝑡
𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑠𝑜𝑙𝑖𝑑 𝑝𝑙𝑎𝑡𝑒
𝑥 100
=
63.76
106.27
𝑥 100 = 59.99 %
82

83. Design a lap joint to connect two plates each of width 120 mm, if the thickness of one
plate is 16 mm and the other is 12 mm. The joint has to transfer a factored load of 160
kN. The plates are of Fe410 grade. Use bearing type of bolts and draw connection details.
Solution:
Using M16 bolts of grade 4.6
d = 16 mm, do = 16+2 =18mm, fub = 400 N/mm2
Since it is a lap joint, the bolt is in single shear, the critical section being at the roots of
the thread of the bolts.
Design strength of a bolt in shear Vsb = Anb
fub
√3.Ɣmb
=0.78x
𝜋162
4
x
400
√3x1.25
x 10-3= 28.974 kN
83
Q 3: Design Examples

84. • Minimum pitch to be provided (p) = 2.5 d = 2.5 × 16 = 40 mm
• Minimum edge distance (e) = 1.5 do = 1.5 × 18 = 27 mm
• Provide p = 40 mm and e = 30 mm
The strength of bolt in bearing Vpb = 2.5 x kb.d.t.
fu
Ɣmb
• Kb is the least of i)
𝑒
3𝑑𝑜
=
30
3𝑥18
= 0.55 ii)
𝑝
3.𝑑0
- 0.25 =
40
3.𝑥18
− 0.25 =0.4907,
• iii)
fub
fu
=
400
410
=0.975. and iv) 1.0
Hence the least of above values kb = 0.4907
• Bearing strength Vpb = 2.5 x kb.d.t.
fu
Ɣmb
=2.5 x 0.4907x16x12x.
410
1.25
x10-3 = 77.25 kN
• Design strength of bolt = least of (i) and (ii) =28.974
84

85. • Hence to transfer design force of 160 kN
• No of bolts req= Design force/ strength of bolt = 160/28.974 = 5.5
• Provide 6 no of bolts in two rows with pitch of 40mm as shown in figure.
Check for strength of plate:
• Tnd = 𝟎. 𝟗 𝑨𝒏
𝒇𝒖
𝜸𝒎𝟏
= 0.9x (b-nd0).t
𝒇𝒖
𝜸𝒎𝟏
=
0.9x(120−2x18)x12x410x10−3
𝟏.𝟐𝟓
= 297.56 kN > 160 kN ….Safe..
85

86. Q.4 Design Example
Two ISF sections 200mm x 10mm each and 1.5m long are to be jointed to make a
member length of 3m. Design a butt joint with the bolts arranged in the diamond
pattern. The flats are supposed to carry a factored tensile force of 450 kN. Steel is of
grade Fe 410. 20mm diameter bolts of grade 4.6 are used to make the connection. Also
determine the net tensile strength of the main plate and cover plates.
Solution: For Fe410 grade of steel : fu = 410 Mpa,
For bolts of grade 4.6 : fub = 400 Mpa
Dia of bolt d = 20mm , Dia of hole d0 = 20+2 = 22mm
Net area Anb = 0.78 .Asb = 0.78.π(20)2/4 = 245 mm2
Ɣmb = partial safety factor for material of bolt = 1.25
Ɣm1 = partial safety factor for resistance governed by ultimate stress = 1.25
86

87. • Design strength of a bolt in double shear Vsb = 2xAnb
fub
√3.Ɣmb
=2x245x
400
√3x1.25
x 10-3= 90.52 kN
• The strength of bolt in bearing Vpb = 2.5 x kb.d.t.
fu
Ɣmb
• d0 = dia of hole = 20+2 = 22mm, e = 1.5xd0 =33mm, pitch p = 2.5xd = 50mm
•
𝑒
3𝑑𝑜
=
33
3𝑥22
= 0.5 ii)
𝑝
3.𝑑0
- 0.25 =
50
3.𝑥22
− 0.25 =0.5, iii)
fub
fu
=
400
410
=0.975. and iv) 1.0
• Hence the least of above values kb = 0.5
• Bearing strength Vpb = 2.5 x kb.d.t.
fu
Ɣmb
=2.5 x 0.5x20x10x.
410
1.25
x10-3 = 82 kN
• Design strength of bolt = least of (i) and (ii) = 82 kN
• Number of bolts =
𝑫𝒆𝒔𝒊𝒈𝒏 𝒇𝒐𝒓𝒄𝒆
𝑺𝒕𝒓𝒆𝒏𝒈𝒕𝒉 𝒐𝒇 𝒃𝒐𝒍𝒕
=
𝟒𝟓𝟎
𝟖𝟐
= 5.48 - 6
87

88. • Arrange the bolt in the diamond pattern as shown in figure.
88

89. • As per code IS 800 :
• Thickness of cover plate : > (5/8).t = (5/8)x10 = 6.25 - 8mm
• Provide 8mm thick cover plates to make double cover butt joint.
• The tensile strength of main plate is critical at sec 1-1
• Tnd1 = 0.9x (b-nd0).t
𝒇𝒖
𝜸𝒎𝟏
=
0.9x(200−1x22)x10x410x10−3
𝟏.𝟐𝟓
= 525.45 kN
• The tensile strength of cover plate is critical at sec 3-3
• Tnd1 = 0.9x (b-nd0).t
𝒇𝒖
𝜸𝒎𝟏
=
0.9x(200−3x22)x16x410x10−3
𝟏.𝟐𝟓
= 632.90 kN
89
> 450 kN factored
tensile force
Hence safe

90. Two flats (Fe 410 Grade Steel), each 210mm x 8mm, are to be joined using 20mm
diameter, 4.6 grade bolts, to form a lap joint. The joint is designed to transfer a factored
load of 250kN. Design the joint.
90
Q 5: Design Examples

91. High Strength Friction Grip bolts (HSFG)
• (HSFG) provide extremely efficient connections and perform well under
fluctuating/fatigue load conditions.
• These bolts should be tightened to their proof loads and require
hardened washers to distribute the load under the bolt heads.
• The tension in the bolt ensures that no slip takes place under working
conditions and so the load transmission from plate to the bolt is through
friction and not by bearing.
• HSFG bolts are made from quenched and tempered alloy steels with
grades from 8.8 to 10.9
• HSFG bolts will come into bearing only after if slip takes place
91

92. Slip Resistance as per IS: 800: Cl.10.4.3
• Slip resistance per bolt is given by Vsf < Vdsf
• Vdsf = Vnsf / 𝛾𝑚𝑓
• Vnsf = nominal shear capacity of a bolt as governed by slip for friction type connection,
and is given as:
• Vnsf = µf. ne. Kh. Fo
• i) Slip resistance of bolt Vdsf =
𝜇𝑓𝑛𝑒𝑘ℎ𝐹𝑜
𝛾𝑚𝑓
where Fo = minimum bolt tension (proof load) at installation = Anb fo
fo = proof stress = 0.7 fub
µf = slip factor as specified in table 20 (0.55)
ne =number of effective interfaces offering frictional resistance to slip = 1
92

93. • Kh = 1.0 for fasteners in clearance holes,
= 0.85 for fasteners in oversized and short slotted holes loaded
perpendicular to the slot.
γmf = 1.10 (if slip resistance is designed at service load)
= 1.25 (if slip resistance is designed at ultimate load)
ii) Bearing strength of HSFG Bolts:
HSFG bolts will come into bearing only after slip takes place. If slip is not
critical, HSFG bolts will slip into bearing.
93

94. An ISA 100mm x 100mm x10mm carries a factored tensile force of 100 kN.
It is to jointed with a 12mm thick gusset plate . Design a high strength
bolted joint when a) no slip is permitted b) when slip is permitted. Steel is
of grade Fe410.
Solution: Let us provide HSFG bolts of grade 8.8 and of diameter 16mm
For 8.8 grade bolts : fub = 800 MPa, Anb = 0.78.π(16)2/4 = 157mm2
When slip is not permitted, the joint will be a slip critical connection and when it is allowed
to slip, the joint will be bearing type connection.
a) Slip-critical connection:
Proof load F0 = Anb x 0.7 fub = 157x0.7x800 x10-3
= 87.92 kN
Slip resistance of bolt Vdsf =
𝜇𝑓
𝑛𝑒
𝑘ℎ
𝐹𝑜
𝛾𝑚𝑓
94

95. • μf = 0.55
• ne = number of effective interfaces offering frictional resistance to slip = 1
• Kh = 1.0 for fasteners in clearance holes
• γmf =1.25 (if slip resistance is designed at ultimate load
• Slip resistance of bolt Vnsf =
𝜇𝑓
𝑛𝑒
𝑘ℎ
𝐹𝑜
𝛾𝑚𝑓
=
0.55𝑥1𝑥1𝑥87.9
1.25
• =38.67 kN
• No of bolts required = 100/38.67 =2.58 = 3 nos
• Provide 3-16mm diameter 8.8 grade HSFG bolts for making connection
•
95

96. ii) Slip is permitted (Bearing type connection)
Strength of a bolt in shear Vsb = Anb
fub
√3.Ɣmb
=157x
800
√3x1.25
x 10-3= 58 kN
Bearing strength Vpb = 2.5 x kb.d.t.
fu
Ɣmb
=2.5 x 0.5x16x10x.
410
1.25
x10-3 = 65 kN { assume kb =0.5)
Hence strength of bolt = 58 kN
No of bolts required = 100/58 = 1.72 = 2
Provide 2, 16mm diameter HSFG bolts
96