Prepared by

Jishnu V

Engineer

BHEL-ISG

1) INTRODUCTION

2) ADVANTAGES & DISADVANTAGES OF WELDED CONNECTIONS

3) TYPES OF WELDING PROCESSES

4) TYPES OF WELDED JOINTS

5) STRESSES & STRESS CONCENTRATION FACTORS OF WELDED JOINTS

6) ANALYSIS OF UNSYMMETRICAL WELDED SECTIONS WHICH ARE AXIALLY LOADED

7) SPECIAL CASES OF FILLET JOINTS

8) SOLVED EXAMPLES

The process of permanently joining two or

more metal parts by the fusion of edges of

the metals with or without applying pressure

and a filler material is called welding

If pressure is applied Forge welding

Without pressure, with a separate weld

metal Fusion welding

In a fusion welding, heat of melting is

obtained by two ways:- a) Gas heating b)

Electric arc

Advantages:-

1. Comparatively lighter than riveted structures

2) Greater strength compared to riveted joints

3) Addition and alterations can be done easily

4) Better finish than riveted joints. Hence maintenance costs and painting costs are less

5) Lesser time consuming

6) Tension members are not weakened in welded joints compared to riveted joints

Disadvantages:-

1) Requires skilled labour

2) Possibility of additional stress

development due to uneven heating and

cooling. Or in other words, the members may

get distorted

3) Testing is difficult.

4) As there is no provision for expansion or

contraction of joints, cracks may develop

and propogate

1.1) Fusion welding is a welding in which

the parts to be jointed are held in position

while the molten metal is supplied to joint

The molten metal may come from the parts

themselves (ie parent metal) or filler metal

which normally have the composition of the

parent metal

The fusion welding can be classified into

three types based on the method of heat

generation. Viz Thermit welding, electric arc

resistance welding and gas welding

1.1.1) Thermit welding:

Thermit= Aluminium + Iron oxide

Thermit is heated, melted and then poured

into a mould made around the joint and

fuses with the parts to be welded

Advantages of thermit welding

- Simultaneous melting of thermit and parts

- Uniform cooling of molten metals and

thermit.

No residual stresses

Fundamentally, thermit welding= melting +

casting

Applications:- Fabrication of rails,

locomotive frames, stern frames, rudder

frames.

Repair applications:- Replace broken teeth,

weld necks on rolls or pinions, repair broken

shears.

Gas welding:

Applying flame of an oxy-acetylene or

hydrogen gas from a welding torch upon the

surfaces of the prepared joint

The intense heat at the white cone of the

flame heats up the local surfaces to fusion

point and using the welding rod supplies

metal for the weld

Heating rate is slow and hence is used for

thinner metals

Electric arc welding

Base metal is melted using arc stream

forming a pool of molten metal. This molten

metal is forced out of the pool by the blast

from arc

A small depression is formed in the base

metal and the molten metal is deposited

around the edge of this depression, which is

called arc crater

There are two types of arc welding

Unshielded arc welding:-Larger filler rod is

used

Here the weld metal in molten state mix

with oxygen and nitrogen in the atmosphere

forming a relatively weaker joint

It has lower ductility and corrosion resistance

In shielded metal arc welding (SMAW), the welding rods are coated with solid material

This coating helps in focusing the arc stream, which protects the globules of metal from air

Prevents the absorption of large amounts of harmful oxygen and nitrogen

The two important types of welded joints

are:-

1. Butt weld

2. Lap (Fillet joint)

3.1) Butt weld joint:-

If the edges of the two plates are touching

each other and are joined by welding, then

the joint is called butt weld joint

Let l= length of weld

t= Depth of weld

F=Tensile force

σt= Allowable tensile stress

The tensile force, F= σt x A= σt x l x t

Types of butt welds are

A) Single V butt joint

B) Double V butt joint

C) Single U butt joint

D) Double U butt joint

3.2) Fillet weld or lap joint

When the two members are overlapped and

joined by welding, then the joint is called

fillet weld joint

Let l= length of weld

t= Throat thickness

s= Size of weld

F=Tensile force

σt= Allowable tensile stress for weld metal

Throat thickness= AB sin45=s x 0.707

Throat area (minimum area of weld)= Length

of weld x throat thickness= l x s x 0.707

Tensile strength of the joint/ maximum

tensile force which the fillet joint can take

P=Allowable tensile stress of the weld x

throat area= σt x0.707 x l x s

Tensile strength of the joint for double fillet

weld= 2 x σt x0.707 x l x s= 1.414 x σt xl x s

3.3) Strength of parallel fillet joint

Parallel fillet welds are designed for shear

strength

Consider a double parallel fillet joint as

shown below

If τ is the allowable shear stress of the weld

metal, then the weld strength (shear

strength) of the joint is F= Allowable shear

stress x throat area= 2 x 0.707 x s x l x τ

Consider a combination of transverse fillet

weld and parallel fillet weld

If τ is the allowable shear stress of the weld

metal and σt is the allowable tensile stress

of the weld metal, then the weld strength

(shear strength) of the joint is F = σt x l1 x

0.707 x s + τ x l2 x 1.414 x s

For re-inforced welded joints, the throat

dimensions may be taken as 0.85 times the

plate thickness

Consider an unsymmetrical section (angle)

welded on flange edges subjected to an axial

loading

Let la = length of the weld at the top

lb = length of weld at the bottom

L= total length of weld= la+lb

P= Axial load

a= Distance of the top weld from an axis

passing through the CG of the angle (known

as the gravity axis)

b= Distance of the bottom weld from gravity

axis.

f= Resistance offered by the weld per unit

length.

Resistance offered by top weld= f x la

Moment of the resistance offered by top

weld about the gravity axis= f x la x a

Similarly the moment of resistance offered

by the bottom weld about the gravity axis= f

x lb x b

For equilibrium, sum of these moments

should be zero.

Hence f x la x a= f x lb x b

=> la x a= lb x b; l= la + lb

Hence la= l x b / (a+ b) and lb = l x a/(a + b)

6.1) Circular fillet weld subjected to

torque/ torsion

Consider a circular rod connected to a rogid

plate by a fillet joint as shown below.

Let

d= Diameter of the rod

r= Radius of the rod

T= Torque applied

s=Size of weld/ weld leg size

t= Throat thickness

J= Polar moment of inertia of the weld

section= ∏d³t/4

According to torsion equation, Maximum

shear stress generated due to torsion τ=T.r/J

τ= T x 2/∏d²t.

Throat thickness, t=s x 0.707

Hence τ= 2T/(∏d² x s x0.707)

=> τ= 2.83T/∏d²s

6.2) Circular fillet weld subjected to

bending moment

Consider a circular rod connected to a rigid

plate by a fillet weld as shown below

Let M be the bending moment to which the weld/rod is subjected to and Z be the section modulus.

Z= ∏d²t/4

Bending stress σb= M/Z= (4 x M)/ ∏d²t

t=s x 0.707

Hence b=5.66M/ ∏d²s

1) Two steel plates 10cm wide and 1.25cm

thick are to be joined by double lap weld

joint. Find the length of the weld if the

maximum tensile stress is not to exceed 75

N/mm² and maximum tensile load carried by

the plates is 100kN.

Soln) Given:-

Width of the plate= 0.1m

Thickness of the plate, s= weld leg

size=(1.25/100) m

Length of weld, l= ?

σt= Maximum tensile stress= 75 x 10^6 N/m²

Maximum tensile load, F= 10^6 N

l= F/ (1.414 x σt x t)

=> l= 0.7544m= 75.44cm

2) Two plates of width 15cm and thickness

1.25cm are welded by a single V butt joint. If

the safe stress in the weld is 135MPa, find

the permissible load carried by the plates.

Soln) Given:

s= 1.25/100 m

σt= 135 x10^6 N/m²

P= (σt x s x l)= 135 x10^6 x 1.25/100 x 0.15

(Length of weld= width of plate)= 253125N=

253.125kN

3) A plate 10cm wide and 1.15cm thick is

joined with another plate by a single fillet

weld and a double parallel fillet weld as

shown below. The maximum tensile and

shear stresses are 75N/mm² and 55N/mm²

respectively. Find the length of each parallel

fillet if the joint is subjected to a load of

80kN.

Given:-

σt= 75MPa

τ= 55MPa

Width of plate, b= 100mm

Length of single fillet weld, l1= width of plate=100mm

Thickness of plate= weld leg size= 1.15/100 m

l2= Length of each parallel fillet weld.

F1= Force carried by single fillet weld= 0.707 x σt

x l1 x s= 60978.75N

F2= Load carried by double parallel fillet weld=

1.414 x l2 x s x τ= 894355 x l2

F= F1+F2

=> 80000=60978.75 + F2

=> l2= 0.02127m= 2.13cm

4) A 200mm x 150mm x 10mm angle carrying a load of

250kN, is to be welded to a steel plate by fillet welds as

shown below. Find the length of the weld at the top and

bottom if the allowable shear stress in the weld is

102.5kPa. The distance between the neutral axis and the

edges of the angle section are 144.7mm and 55.3mm

respectively

Soln)

Given:

a= 144.7mm

b= 55.3mm

F= 250kN= F1 + F2

l=l1+ l2

F= τ x (l1 + l2) x 0.707 x s

=> (l1+ l2)= 0.345m

l1= l x b/(a+b)= 0.345 x 55.3/200

=0.0954=9.54cm

l2= l – l1= 24.96cm