Design of steel structure as per is 800(2007)

Types of connections
Based on the rigidity of the joint
1. Simple or flexible
• It does not offer
resistance against
rotation and also termed
as a hinged or pinned
connections.
• It transfers only axial or
shear forces and it is not
designed for moment
• It is generally connected
by single bolt/rivet and
therefore full rotation is
allowed
3. Semi-rigid
• Designed in such a
way that it offers
resistance against
rotation but less than
rigid connections
• Along with axial or
shear forces it partially
transfers moment.
• The rotations are
allowed but the
magnitude is less than
the rigid connections
2. Rigid or moment
resisting
• Designed in such a way that
it offers resistance against
rotation and also termed as
a moment resisting
connections.
• Along with axial or shear
forces it transfers moment.
• It is generally connected by
multiple bolts/rivets and
therefore rotation is not
allowed.
Note: In actual practice, connections are neither simple nor perfectly rigid,
but they can be idealized as simple, rigid and semi-rigid.

Types of connections
Based on the internal forces transferred
2. Axially loaded
connections
•These are assumed to
be hinged
•Ideally, members
should be connected by
one bolt.
•If, connected by two or
more bolts or welds,
results in to the
development of small
partial fixity which can
be ignored.
3. Eccentrically loaded
connections:
I. Connections in which
moment acting in the plane
of joint (i.e. joint subjected to
shear force and torsional
moment)
II. Connections in which
moment acting perpendicular
to the plane of joint (i.e. joint
subjected to shear force and
bending moment)
1. Beam end connections:
I. Framed connections:
Web of the beam is
connected directly to
the column or another
beam
II. Seated connections:
a) Stiffened
b) Unstiffened
flange of the beam is
connected to column
with the help of seating
arrangement
Answer
Question: In which category
it will fall in previous
classification?
Answer :
Truss member joints
Question: Give category
and an example of this
classification?
Question: In which category
it will fall in previous
classification?
Answer :
2. Rigid or moment
resisting

Eccentric Connections
Applied load does not pass through C.G of connections
Fig 2

Bolted Bracket Connections Type 2

Welded Bracket Connections Type 1

Welded Bracket Connections Type 2

• Designed more conservatively than members because they are
more complex to analyse and discrepancy between analysis and
design is large
• In case of overloading, failure in member is preferred to
failure in connection
• Connections account for more than half the cost of structural
steel work
• Connection design has influence over member design
• Similar to members, connections are also classified as idealised
types effected through rivets, bolts or weld
• Codal Provisions
Design considerations for connections

Types of Fasteners
Rivets:
• Use of rivets is becoming obsolete as it requires preheating, skilled
supervision, more labour and riveting equipment.
• Therefore, emphasis is given on design of bolts and design of rivets
is similar to design of bolts.

Bolts
Black Bolts: (IS:1363 -Part 1, 2 and 3, – 2002)
• made up of mild steel bars with square of hexagonal
head
• commonly used and less expensive.
• commonly used for light structures
• not recommended for connections subjected to
impact, fatigue and dynamic loads.
Types of Fasteners
Strength of black bolt
Bolt Grade: Grade 4.6 :- fu = 400 N/mm2 and fy = 0.6*400 = 240 N/mm2

Bolts
High strength bolts:
They are made up of bars of
medium carbon steel bars
The bolts of property class 8.8 and
10.9 are commonly used in steel
construction and identified by
‘8.8S’ and ‘10.9S’ marking on the
bolt head. Suffix ‘S’ denotes high
strength bolt.
grade 4.6 to 8.8
Types of Fasteners

Bolts
High strength friction grip (HSFG) bolts:
They conform to IS:3757-1985
These are the bolts with induced initial tension.
They do not allow slip between the connected members
Due to the high strength, no. of bolts required are less and size of gusset plate required
is also less.
grades from 8.8 to 10.9
Types of Fasteners

• Grip is the distance from behind the bolt head to the back of the nut or washer
 It is the sum of the thicknesses of all the parts being joined exclusive of washers
• Thread length is the threaded portion of the bolt
• Bolt length is the distance from behind the bolt head to the end of the bolt
Parts of the Bolt Assembly
Head
Shank
Washer
NutWasher
Face
Grip
Thread
Length
Parts of the Bolt Assembly

• The bolting operation is
very silent.
• Bolting is cold process
hence there is no risk of
fire.
• Bolting operation is more
quicker than riveting.
• Less man power is required
in making the connections.
• If subjected to vibratory
loads, results in reduction in
strength get loosened.
• Unfinished bolts have lesser
strength because of non
uniform diameter.
Advantages Disadvantages
Types of Fasteners

Essential background
• Table 1 Tensile Properties of Structural Steel Products, IS 800:2007, (Clauses
1.3.113, 1.3.119 and 2.2.4.2)
240
320

Essential background
• Table 5 Partial Safety Factor for Materials, m
• (Clause 5.4. 1)

Shear Connections
a) Lap Connection b) Butt Connection
(a)
(b)
Tension Connection Tension plus Shear Connection
Some typical arrangement
Single
shear
Double shear
Classification based on type of force in the bolts

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

Failure Of Connections
(a) Shearing of Bolts
(b) Bearing on Bolts
(c) Bearing on Plates
Zone of
plastification
Shear Connections with Bearing Bolts
Bearing
Yield
Bearing
Fracture
Bearing
Fracture
Bearing
Yield 19

• In a bearing joint the connected elements are assumed to slip into
bearing against the body of the bolt
• If the joint is designed as a bearing joint the load is transferred through
bearing whether the bolt is installed snug-tight or pretensioned
Bearing Joints

• The shear plane is the plane
between two or more pieces
under load where the pieces
tend to move parallel from
each other, but in opposite
directions
• The threads of a bolt may
either be included in the shear
plane or excluded from the
shear plane
• The capacity of a bolt is greater
with the threads excluded
from the shear plane
Threads in the Shear Plane
Threads Included In The Shear Plane
Threads Excluded From The Shear Plane

Codal Requirements of Bolted Connections
 Clearance of fastener holes (10.2.1):
Sr. No. Nominal
size of
Fastener in
mm (d)
Standard
clearance in
diameter and
width of slot in
mm
Oversize
clearance in
diameter in
mm
Clearance in
length of the slot
in mm
Short
slot
Long
slot
1 12-14 1.0 3.0 4.0 2.5d
2 16-22 2.0 4.0 6.0 2.5d
3 24 2.0 6.0 8.0 2.5d
4 >24 3.0 8.0 10.0 2.5d
 Minimum Spacing (10.2.2):
 Minimum pitch = 2.5d,
 Where d = nominal diameter of the fastener

 Maximum Spacing (10.2.3):
• In general, max. pitch = 32t or 300 mm whichever is
less
• For tension member, maximum pitch = 16t or 200 mm
whichever is less
• For compression member, maximum pitch = 12t or 200
mm whichever is less
• For consecutive fasteners in a line adjacent and parallel
to an edge of an outside plate
• maximum gauge = 100+4t or 200 mm whichever is less
• where t = thickness of thinner plate in connection

 Edge and end distances (10.2.4):
Minimum edge and end distances = 1.5 times the hole
diameter in case of rolled, machine-flame cut, sawn and
planed edges measured from centre of any hole to the nearest
edge of a plate.
Minimum edge and end distances = 1.7 times the hole
diameter in case of sheared or hand-flame cut edges

 Effective areas of bolts (10.3.1)
 For shear plane in plain part of the bolt, effective
areas of bolt is given by diameter of bolt at
shank. (i.e𝐴 𝑒 = 𝐴 𝑠𝑏 =
𝜋𝑑2
4
)
 For shear plane in threaded part of the bolt,
effective areas of bolt is given by diameter of
bolt at root of the thread (i.e𝐴 𝑒 = 𝐴 𝑛𝑏 =
𝜋𝑑 𝑛
2
4
)
 In absence of data, 𝐴 𝑛𝑏 = 0.78 𝐴 𝑠𝑏

 Shear capacity of the bolt (10.3.3):
The nominal shear capacity of the bolt is given
by:
𝑉𝑛𝑠𝑏 =
𝑓 𝑢𝑏
3
(𝑛 𝑛. 𝐴 𝑛𝑏 + 𝑛 𝑠. 𝐴 𝑠𝑏) where
Asb = Nominal plain shank area of the bolt
An
b
= Net shear area of bolt at threads corresponding to the
root diameter of the thread
fub = Ultimate tensile strength of the bolt
nn = No. of shear planes with threads intercepting the shear plane
ns = No. of shear planes without threads intercepting the shear plane
Design Strength of Bolt

 Shear capacity of the bolt (10.3.3):
The design shear capacity of the bolt is given
by:
𝑉𝑑𝑠𝑏 =
𝑉𝑛𝑠𝑏
𝛾 𝑚𝑏
=
𝑓𝑢
1.25 3
𝑛 𝑛. 𝐴 𝑛𝑏 + 𝑛 𝑠. 𝐴 𝑠𝑏
= 0.462𝑓𝑢 𝑛 𝑛. 𝐴 𝑛𝑏 + 𝑛 𝑠. 𝐴 𝑠𝑏

 Bearing capacity of the bolt (10.3.4)
The nominal bearing capacity of the bolt on any
plate is given by:
𝑉𝑛𝑝𝑏 = 2.5 𝑘 𝑏 𝑑 𝑡 𝑓𝑢where
kb =
e, p = End and pitch distances of fasteners along bearing
direction
do = Diameter of bolt hole
fub,fu = Ultimate tensile strength of the bolt & plate respectively
t = Thickness of the connected plates experiencing bearing stress in the
same direction

 Design strength of the bolt (10.3.2):
The design strength of the bolt, Vdb :
Shall be taken as the smaller of the value as
governed by,
1. Shear, Vdsb and
2. Bearing, Vdpb.

 Reduction in design strength of the bolt:
Long joints (10.3.3.1):
When the length of the joint, lj containing more than two
bolts (i.e. the distance between the first and last rows of
bolts in the joint measured in the direction of the load)
exceeds 15d in the direction of load, the nominal shear
capacity Vdb shall be reduced by the factor βlj given by
𝛽𝑙𝑗 = 1.075 −
𝑙𝑗
200𝑑
𝑏𝑢𝑡 0.75 ≤ 𝛽𝑙𝑗 ≤ 1.0

Large grip lengths (10.3.3.2):
When the grip length, lg (equal to the total thickness of
connected plates) exceeds the connected plates) exceeds
5 times the diameter, d of the bolts, the design shear
capacity shall be reduced by a factor βlg given by
𝛽𝑙𝑔 =
8𝑑
3𝑑 + 𝑙 𝑔
𝑏𝑢𝑡 𝛽𝑙𝑔 ≤ 𝛽𝑙𝑗 𝑎𝑛𝑑 𝑙 𝑔 ≤ 8𝑑

Packing plates (10.3.3.3):
The design shear capacity of bolts carrying shear
through a packing plate in excess of 6 mm shall be
decreased by a factor βpk given by
𝛽 𝑝𝑘 = 1 − 0.0125 𝑡 𝑝𝑘
Where tpk = thickness of the thicker packing, in mm.

 Tension capacity of the bolt (10.3.5):
The nominal tension capacity of the bolt is given by:
𝑇𝑛𝑏 = 0.9 𝑓𝑢𝑏 𝐴 𝑛 ≤
𝑓𝑦𝑏
𝐴 𝑠𝑏
𝛾 𝑚𝑏
𝛾 𝑚𝑜
The design tension capacity of the bolt is given by:
𝑇𝑑𝑏 =
𝑇𝑛𝑏
𝛾 𝑚𝑏
Where
fub = Ultimate tensile strength of the bolt
An = Net tensile area of the bolt
Asb = Shank area of the bolt

Example 1
Q. Determine the bolt value of 16 mm diameter bolt
4.6 grade connecting 8 mm thick plate of 410
grade in
(i) Single shear and (ii) Double shear
For 4.6 grade bolt,
fub = 400 Mpa, Asb = 201 mm2
fyb = 60% of fub Anb = 0.78 x Asb
= 0.6x400 = 156 mm2
= 240 Mpa

Example 1
𝑉𝑑𝑠𝑏 = 0.462𝑓𝑢 𝑛 𝑛. 𝐴 𝑛𝑏 + 𝑛 𝑠. 𝐴 𝑠𝑏 𝑉𝑑𝑝𝑏 = 2.0 𝑘 𝑏 𝑑 𝑡 𝑓𝑢
= 0.462x400x(1x156)x10-3 = 2x1x16x8x410x10-3
= 28.8 kN in single shear = 104.96 kN in bearing
= 57.6 kN in double shear
Bolt value = 28.8 kN in single shear

Example 2
Q. Design a lap joint between the plates of size 80x10 mm and
80x8 mm thick so as to transfer a factored load of 80 kN
using single row of bolts of grade 4.6 and grade 410 plate
For 4.6 grade bolt of 16 mm diameter,
fub = 400 Mpa, d = 16 mm Asb = 201 mm2
fyb = 60% of fub d0 = 18 mm Anb = 0.78 x Asb
= 0.6x400 e = 1.7x18 = 30.6 mm = 156 mm2
= 240 Mpa Say 35 mm
p = 2.5x16 = 40 mm

Example 2
𝑉𝑑𝑠𝑏 = 0.462𝑓𝑢 𝑛 𝑛. 𝐴 𝑛𝑏 + 𝑛 𝑠. 𝐴 𝑠𝑏
= 0.462x400x(1x156)x10-3
= 28.8 kN in single shear
𝑒
3𝑑 𝑜
=
35
3𝑥18
= 0.65,
𝑝
3𝑑 𝑜
− 0.25 =
40
3𝑥18
− 0.25 = 0.49,
𝑓𝑢𝑏
𝑓𝑢
=
400
410
= 0.98
Kb = Smaller of
𝑒
3𝑑 𝑜
,
𝑝
3𝑑 𝑜
− 0.25,
𝑓 𝑢𝑏
𝑓𝑢
𝑎𝑛𝑑 1.0
Kb = 0.49

Example 2
𝑉𝑑𝑝𝑏 = 2.0 𝑘 𝑏 𝑑 𝑡 𝑓𝑢
= 2x0.49x16x8x410x10-3
= 51.43 kN in bearing
Bolt value = 28.8 kN
No. of Bolts =
80
28.8
= 2.8 𝑠𝑎𝑦 3 𝑛𝑜𝑠.
Check: Lj = 2x40 = 80 mm < (15d = 15x16 = 240 mm)
Therefore no reduction in strength as per long joint
Lg = 10+8 = 18 mm < (5d = 5x16 = 80 mm)
Therefore no reduction in strength as per large grip
No reduction for packing plate.

Example 3
Q. A bridge truss diagonal carries an axial pull of 250 kN. Two
plates of 200x10 mm and 200x16 mm are required to be
joined together by butt joint. Design suitable double cover
butt joint with bolts of grade 4.6 and grade 410 plate
For 4.6 grade bolt of 20 mm diameter,
fub = 400 Mpa, d = 20 mm Asb = 314 mm2
= 0.6x400 e = 1.7x22 = 37.4 mm = 245 mm2
= 240 Mpa Say 40 mm
p = 2.5x20 = 50 mm

Example 3
200x16 plate
200x10 plate
250 kN
250 kN
35 40 4035 35
Packing plate 6 thick
Cover plates 8 thick

Example 3
= 0.462x400x(2x245)x10-3
𝑒
3𝑑 𝑜
=
40
3𝑥22
= 0.61,
𝑝
3𝑑 𝑜
− 0.25 =
50
3𝑥20
− 0.25 = 0.58,
𝑓𝑢𝑏
𝑓𝑢
=
400
410
= 0.98
Kb = Smaller of
𝑒
3𝑑 𝑜
,
𝑝
3𝑑 𝑜
− 0.25,
𝑓 𝑢𝑏
𝑓𝑢
𝑎𝑛𝑑 1.0
Kb = 0.58

Example 3
For connection on 16 mm plate𝑉𝑑𝑝𝑏 = 2.0 𝑘 𝑏 𝑑 𝑡 𝑓𝑢
= 2x0.58x16x20x410x10-3
= 152.19 kN in bearing on 16mm
Bolt value = 90.55 kN
No. of Bolts =
250
90.55
= 2.76 𝑠𝑎𝑦 3 𝑛𝑜𝑠.
Check: Lj = 0 < (15d = 15x20 = 300 mm)
Lg = 20+8+8 = 36 mm < (5d = 5x20 = 100 mm)

Example 3
= 2x0.58x20x10x410x10-3
For packing plate 𝛽 𝑝𝑘 = 1 − 0.0125 𝑡 𝑝𝑘 = 1-0.0125x10 = 0.875
Bolt value = 0.875x90.55 = 79.23 kN
No. of Bolts =
250
79.23
= 3.15 𝑠𝑎𝑦 4 𝑛𝑜𝑠.
Check: Lj = 50 < (15d = 15x20 = 300 mm)
Lg = 20+8+8 = 36 mm < (5d = 5x20 = 100 mm)

Example 4
Q. Figure shows the joint in the bottom chord member of a
truss, design the connection using M16 black bolt of
property class 4.6 and grade 410 steel sections.
135 kN (ISA75X75X8)
75 kN (ISA50X50X8)
370 kN (2-ISA100X100X8) 250 kN (2-ISA100X100X8)
O
B C
DA

Example 4
For 4.6 grade M16 bolt,
fub = 400 Mpa, d = 16 mm Asb = 201 mm2
= 0.6x400 e = 1.7x18 = 30.6 mm = 156 mm2
= 240 Mpa Say 35 mm
p = 2.5x16 = 40 mm
= 0.462x400x(1x156)x10-3

Example 4
Let gusset plate be 10 mm thick
𝑒
3𝑑 𝑜
=
35
3𝑥18
= 0.65,
𝑝
3𝑑 𝑜
− 0.25 =
40
3𝑥18
− 0.25 = 0.49,
𝑓𝑢𝑏
𝑓𝑢
=
400
410
= 0.98
Kb = Smaller of
𝑒
3𝑑 𝑜
,
𝑝
3𝑑 𝑜
− 0.25,
𝑓 𝑢𝑏
𝑓𝑢
𝑎𝑛𝑑 1.0
Kb = 0.49

Example 4
= 2x0.49x16x8x410x10-3
= 2x0.49x16x10x410x10-3
For single angle members, bolt value = 28.8 kN
For single angle members, bolt value = 57.6 kN

Example 4
For member OB
No. of Bolts =
75
28.8
= 2.60 𝑠𝑎𝑦 3 𝑛𝑜𝑠.
Check: Lj = 80 < (15d = 15x16 = 240 mm)
Lg = 10+8 = 18 mm < (5d = 5x16 = 80 mm)

Example 4
For member OC
No. of Bolts =
135
28.8
= 4.69 𝑠𝑎𝑦 5 𝑛𝑜𝑠.
Check: Lj = 160 < (15d = 15x16 = 240 mm)
Lg = 10+8 = 18 mm < (5d = 5x16 = 80 mm)

Example 4
For member OA if OA and OD are discontinuous at joint
No. of Bolts =
370
57.6
= 6.42 𝑠𝑎𝑦 7 𝑛𝑜𝑠.
Check: Lj = 240 < (15d = 15x16 = 240 mm)
Lg = 10+20 = 30 mm < (5d = 5x16 = 80 mm)

Example 4
For member OD if OA and OD are discontinuous at joint
No. of Bolts =
250
57.6
= 4.34 𝑠𝑎𝑦 5 𝑛𝑜𝑠.
Check: Lj = 160 < (15d = 15x16 = 240 mm)
Lg = 10+20 = 30 mm < (5d = 5x16 = 80 mm)

Example 4
For member OA and OD if they are continuous at joint
No. of Bolts =
370−250
57.6
= 2.08 𝑠𝑎𝑦 3 𝑛𝑜𝑠.
Check: Lj = 80 < (15d = 15x16 = 240 mm)
Lg = 10+20 = 30 mm < (5d = 5x16 = 80 mm)

Efficient and direct way of connecting is by welding
Metallurgical bond by heat or pressure or both
• Types of weld
Gas welding - Oxyacetelene welding , simple , slow,
repair and maintenance work
Arc welding - All structural welding
Electric arc by use of electric energy
Types of Fasteners
Welds: They conform to IS:816-1969, IS:9595-1996.

Classification of welds
Types of Fasteners
A
Ends shall be semi
circular
A
Section A-A
(c) Slot weld
(a) Groove welds
(b) Fillet welds
AA
Section A-A
(d) Plug weld

• Fillet welds: Advantages
Ease of fabrication and adaptability
Less precision
No special edge preparation
Throat of a weld
Concave and convex surfaces
Types of Fasteners
Theoretical throat (t=0.707s)
t
Te
s
Root of weld
Face of weld
Weld and leg size

Requirements of Welded Connections
 Size of the weld – S (10.5.2):
Minimum size of weld =3 mm
Depending on the thickness of the thicker plate to
be connected, minimum size of weld should be
Thickness of
thicker plate
Minimum size of weld
in mm
Upto 10 mm 3
11 to 20 mm 5
21 to 32 mm 6
33 to 50 mm 10 - 8 (first run)

 General (10.5.1):
• Minimum end return ≥2S
• (important for sides carrying bending tension)
• For lap joint Minimum lap ≥4t_ or 40 mm whichever is greater
• Single fillet weld is not subjected to moment about its
longitudinal axis

 Size of the weld – S (10.5.2):
 If minimum size of weld is greater than the
thickness of thinner member, minimum size of
weld should be thickness of thinner member
 Size of butt weld is specified by effective throat
thickness

 Effective throat thickness - t (10.5.3):
 For fillet weld t ≥ 3 mm
t ≥ 0.7x tmin of member
 For stress calculation in fillet weld, t = K.
Angle between
fusion faces in
degrees
60-
90
91-
100
101-
106
107-
113
114-
120
K 0.70 0.65 0.60 0.55 0.50

 Effective throat thickness - t (10.5.3):
 For full penetration butt weld, t = tmin of connected
members
 For incomplete penetration butt weld t = minimum
thickness of weld metal common to the parts joined.

 Effective length of weld - lw (10.5.4):
 For fillet weld, actual length of weld = lw+2S ≥4S
 For butt weld, lw ≥4S
 Intermittent welds (10.5.5):
 For fillet weld,
lw ≥ 4S or minimum of 40 mm
Clear spacing between weld lengths
≤12t or maximum of 200 mm for compression member
≤16t or maximum of 200 mm for tension member
Where t is the thickness of the thinner part.

 Intermittent welds (10.5.5):
 For butt weld,
lw ≥4S or minimum of 40 mm
Clear spacing between weld lengths
≤16 times the thickness of the thinner member
Note: Intermittent welds shall not be used for
members subjected to dynamic loads, and reversal of
stresses.

Design Strength of Weld
 Design stresses in welds (10.5.7):
 Fillet welds
Design strength of fillet weld
𝑓𝑤𝑑 =
𝑓 𝑤𝑛
𝛾 𝑚𝑤
Where 𝑓𝑤𝑛 =
𝑓𝑢
√3
γmw = 1.25 and 1.5 for shop weld and field weld respectively
 Butt welds
Stresses in welds are same as that of parent metal

Design Strength of Weld
 Reduction Factor
 When the length of the welded joint, lj is greater than 150 tt,
the design capacity of weld fwd shall be reduced by the factor
𝛽𝑙𝑗 = 1.2 −
0.2𝑙𝑗
150𝑡𝑡
≤ 1.0
Where
lj = length of the joint in the direction of the force transfer
tt = throat size of the weld.

Example 5
Q. An ISA 125x95x10 carries a factored tensile force of 420 kN
and it is connected by its longer leg to 8 mm thick gusset
plate. Design a suitable welded joint using
(i) 6 mm fillet weld on toe and heel
(ii) 6 mm fillet weld on toe heel and at end also
(iii) 6 mm fillet weld on toe and 8 mm fillet weld at heel and
end.

Example 5
Let Pu1 and Pu2 be forces acting on toe and heel
P is acting at centroid of angle section i. e. at 38.9
mm from heel.
Taking moment of forces about heel
Pu1 =
420𝑥38.9
125
= 130.7 𝑘𝑁
Pu2 = 420 − 130.7 = 289.3 𝑘𝑁

Example 5
Smin = 3 mm < 6 mm
Smax =
3
4
𝑥10 = 7.5 𝑚𝑚> 6 mm
Therefore we can use 6 mm weld i.e. S = 6 mm
Throat thickness = 0.7x6 = 4.2 mm
Strength of weld, 𝑓𝑤𝑑 =
𝑓 𝑤𝑛
𝛾 𝑚𝑤
𝑓𝑢
√3
=
410
√3
= 236 Mpa
𝑓𝑤𝑑 =
𝑓 𝑤𝑛
𝛾 𝑚𝑤
=
236
1.25
= 189 MPa

Example 5
Strength of weld per mm length
= 𝑓𝑤𝑑 𝑥 𝑡ℎ𝑟𝑜𝑎𝑡 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 = 189 x 4.2 = 795 N/mm
Length of weld required on heel
=
𝑃 𝑢2
795
=
289300
795
= 363 𝑚𝑚 𝑠𝑎𝑦 370 𝑚𝑚
Length of weld required on toe
=
𝑃 𝑢1
795
=
130700
795
= 165 𝑚𝑚 𝑠𝑎𝑦 170 𝑚𝑚

Example 5
(ii) 6 mm fillet weld on toe, heel and end
Let Pu1 Pu2 and Pu3 be forces acting on toe, heel and end
P is acting at centroid of angle section i. e. at 38.9 mm from heel.
Smin = 3 mm < 6 mm
Smax =
3
4
𝑥10 = 7.5 𝑚𝑚> 6 mm
Therefore we can use 6 mm weld i.e. S = 6 mm

Example 5
(ii) 6 mm fillet weld on toe , heel and end
Strength of weld per mm length
= 𝑓𝑤𝑑 𝑥 𝑡ℎ𝑟𝑜𝑎𝑡 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠
= 189 x 4.2 = 795 N/mm
Length of weld required on heel
=
𝑃 𝑢2
795
=
239700
795
= 302 𝑚𝑚 𝑠𝑎𝑦 310 𝑚𝑚
Length of weld required on toe
=
𝑃 𝑢1
795
=
81100
795
= 102 𝑚𝑚 𝑠𝑎𝑦 110 𝑚𝑚

Example 5
(iii) 6 mm fillet weld on toe and 8 mm on heel and end
Let Pu1 Pu2 and Pu3 be forces acting on toe, heel and end
P is acting at centroid of angle section i. e. at 38.9 mm from heel.
On heel and toe, S = 8 mm
Throat thickness = 0.7x8 = 5.6 mm
Strength of weld, 𝑓𝑤𝑑 =
𝑓 𝑤𝑛
𝛾 𝑚𝑤
𝑓𝑢
√3
=
410
√3
= 236 MPa 𝑓𝑤𝑑 =
𝑓 𝑤𝑛
𝛾 𝑚𝑤
=
236
1.25
= 189 MPa
Pu3 = fwd x 0.7S x lw= 189 x 5.6 x 125 = 132300 N = 132.3 kN

Example 5(iii) 6 mm fillet weld on toe and 8 mm on heel and end
Taking moment of forces about heel
Pu1 =
420𝑥38.9−132.3𝑥
125
2
125
= 64.6 𝑘𝑁Pu2 = 420 − 132.3 − 64.6 = 223.1 𝑘𝑁
Strength of weld per mm length on heel = 𝑓𝑤𝑑 𝑥 𝑡ℎ𝑟𝑜𝑎𝑡 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠
= 189 x 5.6 = 1058 N/mm
Length of weld required on heel =
𝑃 𝑢2
795
=
223100
1058
= 211 𝑚𝑚 say 220 mm

Example 5
(iii) 6 mm fillet weld on toe and 8 mm on heel and end
Strength of weld per mm length on toe = 𝑓𝑤𝑑 𝑥 𝑡ℎ𝑟𝑜𝑎𝑡 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 = 189 x 4.2
= 795 N/mm
Length of weld required on toe =
𝑃 𝑢1
795
=
64600
795
= 82 𝑚𝑚 say 90 mm

When torque or twisting moment is in the
plane of onnection the connections may be
termed as a bracket connection type-I This
situation occurs when line of action of load is in
the plane of bolted connection and centre of
gravity of connection is the centre of rotation
Forces on bolts bracket connection type-I
Bracket Connection - Type -1:

The forces due to direct shear F1 and due to torque F2 are given by,

Examples: 6
Determine safe load P that can be carried by a bolted bracket connection
as shown in Fig. The bolts are 20 mm diameter of grade 4.6. The
thickness of the flange of l-section is 10.6 mm and that of bracket plate is 12 mm.

Design a bracket connection using 4.6 grade black bolt of suitable size to
transmit a factored load of 135 kN (applied on a 12 mm thick bracket plate)
to the flange of a column ISHB 225. The load eccentricity is 200 mm
measured from the column axis as shown in figure. (MU 2014, 10 Mark)
Solution
(ii) Design of bolts :
Using M20 black bolts of grade 4.6
Strength of bolt: (cl. 10.3.2, IS 800-2007)
Fub = 400N/mm2
d = 20 mm do = 20 + 2 = 22 mm
emin = 1.5 do = 1.5x22 = 33 mm,
Pmin = 2.5 xd = 2.5x20 = 50mm = 60 mm
(i) Load P for which bracket designed :
The design load P is always considered 50% of
the reaction transferred by the beam. But
in this case P is directly given.
P = 135 kN
Examples: 7

When the bending moment of eccentric force
P is in a plane perpendicular to the plane
of connection then the bracket connection is
termed as bracket connection type-II
Forces on bolts bracket connection type-II
Bracket Connection Type-ll :

Determine P. Use 2ISA 100 x 100 x 12 mm angle. Use 20 mm dia bolts for
connection. The thickness of bracket plate is 16 mm as shown in
Fig. (MU 2013, 10 Mark)
Examples: 7

Design of Welded Bracket Connection :
Plate is conA10 mm thick bracket
nected with the flange of an ISHB
250 @ 51 kg/m column using 8
mm size of shop weld as shown in
Figure Find the maximum value of
load P that can safely be applied
on the bracket.
Examples: 8

A bracket plate is welded to the flange of a
column section ISHB 300@ 618 N/m as
shown in Fig. Calculate the size of weld
required to support a factored load 220 kN.
Assume shop welding.
Examples: 9

Design of steel structure as per is 800(2007)

Design of steel structure as per is 800(2007)

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