Welded joints

1. WELDED JOINTS

2. 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
AXIALLYLOADED
7) SPECIAL CASES OF FILLET JOINTS
8) SOLVED EXAMPLES

3. 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

4. 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

5. 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

7. 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

8. 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.

9. 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.

10. 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

11. 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

12. 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

13. In shielded metal arc welding (SMAW),thewelding 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

14. 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

15. Let l= length ofweld
t= Depth of weld
F=Tensile force
σt= Allowable tensile stress
The tensile force, F= σt x A= σt x l x t

16. 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

17. Fillet weld or lap joint
When the two members are overlapped and joined by
welding, then the joint is called fillet weld joint

18. Let l= length ofweld t= Throat
thickness s= Size of weld
F=Tensile force
σt= Allowable tensile stress for wedmetal

21. Weld
under
Bending

25. 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

26. Strength of parallel fillet joint
Parallel fillet welds are designed for shear
strength
Consider a double parallel fillet joint as
shown below

27. 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

28. 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

30. 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

31. 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

32. 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)

33. Circular fillet weld subjected to torque/ torsion
Consider a circular rod connected to a rogid plate by a fillet joint as
shown below.

34. 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

35. According to torsion equation, Maximum
shear stress generated due to torsion τ=T.r/J
τ= T x2/∏d²t.
Throat thickness, t=s x 0.707
Hence τ= 2T/(∏d² x sx0.707)
=> 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

36. 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 x0.707
Hence b=5.66M/ ∏d²s

37. 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

38. 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

39. 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.

40. 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

41. 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
Given:
a= 144.7mm, b= 55.3mm, F= 250kN= F1 + F2
l=l1+ l2
F= τ x (l1 + l2) x 0.707 xs => (l1+ l2)= 0.345m
l1= l x b/(a+b)= 0.345 x55.3/200 =0.0954=9.54cm
l2= l – l1= 24.96cm