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o Joining process in which two (or more) parts are coalesced at
their contacting surfaces by application of heat and/or pressure
o In some welding processes a filler material is added to facilitate
coalescence
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o Fusion welding
o Joining processes that melt the base metals
o In many fusion welding operations, a filler metal is added to the
molten pool to facilitate the process and provide bulk and added
strength to the welded joint
o Heat required for the fusion of the material may be obtained by
burning of Gas (in case of Gas Welding) or by an Electric Arc (in
case of Electric Arc Welding)
o Solid State Welding
o Joining processes in which coalescence results from application of
pressure alone or a combination of heat and pressure
o If heat is used, temperature is below melting point of metals being
welded
o No filler metal is added in solid state welding
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o Provides a permanent joint: Welded components become a single entity
o Welding joints can be stronger than joining material depending upon the filler
o Usually the most economical way to join components in terms of material usage and
fabrication costs
o Not restricted to a factory environment
o Usually Lighter than riveted structures
o provide maximum Efficiency (up to 100%), not possible in case of riveted
joints
o Alterations and Additions can be easily made in the existing structures
o smooth in Appearance, therefore looks pleasing
o Difficult for riveting members (i.e. Circular Steel Pipes) can easily be welded
o It is possible to weld any part of a structure at any point. But riveting requires
enough clearance
o Takes less time than the riveting
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o Most welding operations are performed manually and are expensive
in terms of labor cost
o Most welding processes utilize high energy and are inherently
dangerous
o Welded joints do not allow for convenient disassembly
o Uneven Heating and Cooling during fabrication, therefore the
members may get distorted or additional stresses may develop
o Requires a Highly Skilled Labor and supervision
o Inspection of welding work is more difficult than riveting work
o No provision is kept for expansion and contraction in the frame →
there is a possibility of cracks developing in it
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o Butt joint
o Parts lie in same plane and are joined at their edges
o Corner joint
o Parts in a corner joint form a right angle and are joined at the corner of the angle
o Lap joint
o Consists of two overlapping parts
o Tee joint
o One part is perpendicular to the other in the approximate shape of the letter "T“
o Edge joint
o Parts in an edge joint are parallel with at least one of their edges in common,
and the joint is made at the common edge(s)
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o Each of the preceding joints can be made by welding
o Other joining processes can also be used for some of the joint
types
o There is a difference between joint type and the way it is
welded - the weld type
o
o
o
o
Fillet Weld
Groove Weld
Plug/Slot Weld
Flange Weld
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o Used to fill in the edges of plates created by corner, lap, and tee
joints
o Filler metal used to provide cross-section in approximate shape
of a right triangle
o Most common weld type in arc and oxyfuel welding
o Requires minimum edge preparation
o Can be single/double and continuous/intermittent
inside single fillet
corner joint
Outside single fillet
corner joint
double fillet lap
joint
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double fillet tee
joint
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o Usually requires part edges to be shaped into a groove to
facilitate weld penetration
o Edge preparation increases cost of parts fabrication
o The grooved shapes include square, bevel, V, U, and J, in single
or double sides
o Most closely associated with butt joints
Square groove weld
one side
Single U-groove weld
Single bevel groove weld
Single J-groove weld
Single V-groove weld
Double V-groove weld
for thicker sections
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o Attaching flat plats
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o A small fused section between surfaces of two sheets or plates
o Used for lap joints
o Most closely associated with resistance welding
Mechanical Engineering Dept. CEME NUST
o Weld on the edge of two or more parts
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o Reinforcement?
o Average normal Stress in weld
o Average shear Stress in weld
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o Transverse Fillet
o Average normal Stress at ϑ
o Average shear Stress at ϑ
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o Von Mises Stress at ϑ
o Max Von Mises
ds’/dq=0
➔
o s‘max= 2.16 F/hl
o t = 1.196 F/hl
o s = 0.623 F/hl
o Max Shear ?
dt/dq=0
➔
o t = 1.207 F/hl
o s = 0.5 F/hl
q = 62.5 O
q = 67.5 O
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o Norris, 1945
o Stress distribution on AB & BC
o AB stress was not certain
o Applies to parent metal aswell as weld metal
o Salakian and Claussen, 1937
o Stress distribution across throat
o Applies only to weld metal
o No Analytical approach exists for the prediction of
stresses
o But, weldment must be designed to be safe
o A conservative model is used which is verified
experimentally
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o The conservative model considers
o External loading is carried out as shear on the throat area (Ignoring the
normal stresses)
o Shear stresses are inflated to render the model conservative (1.17 times)
Max Shear: tmax = 1.207 F/hl
n = t / tmax = 1.17
o Use distortion energy for significant stress
o The entire force is accounted for by a shear on the min throat
area
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o Stress will be on min throat area but one should
examine
o Primary shear stresses due to external forces.
o Secondary shear stresses due to torsional and bending
moments.
o The strengths of
o the parent metal(s)
o deposited weld metal
o Permissible load(s) for
o parent metal(s)
o deposited weld metal
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o Primary shear in weld
V : Load on weld
A: Throat Area
o Secondary shear / Torsion in weld
M : Moment on centroid
J : 2nd polar moment of area of weld group
r: Distance of farthest weld from centroid
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o In fig. the rectangles represent
throat thickness of welds
o Throat area of both the welds
o 2nd moment of area of weld 1
along x and y axes?
o 2nd moment of area of weld 2
along x and y axes?
Mechanical Engineering Dept. CEME NUST
h : weld size
t1= 0.707 h1
t2= 0.707 h2
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o Centroid
o Distances of G1 and G2 from C
o Total J of the weld group
o The reverse procedure
o The allowable shear stress is given and the weld size is to be calculated.
o The procedure is to estimate a probable weld size and then to use
iteration
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o Cubes of weld thicknesses can be
neglected
o Which makes J1 and J2 linear in
the weld width
o Setting t1 = t2 = 1
o Welds can be treated as a line
o J => Ju = Unit 2nd moment of area
o As t= 0.707 h => J = 0.707 h Ju
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o Primary shear
o Same as before
o Secondary shear
o
o
o
o
o
I = Second moment of area
M = Moment causing bending
c = Max distant point from G
Iu = Unit second moment of area
I = 0.707 h Iu
o Total shear
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Minimum Weld-Metal Properties
Self Study
Stresses Permitted by the AISC Code for Weld Metal
Su / Sy of the metals
(Base + Support)
Table A:20
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For the electrode E7010, what is the
allowable load on the weldment?
Bar material is 1018CD
Support material is 1018HR
What load on the weldment is allowable
because member metal is incorporated in the
welds?
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Static
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o The permissible shear stress for the weldment illustrated is
140Mpa. Estimate the load, F, that will cause this stress in the
weldment throat.
200mm
5mm
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o A 50mm dia. steel bar is subjected to the loading indicated.
Locate and estimate the maximum shear stress in the weld
throat.
F=0
T = 2 KN-m
6mm
150 mm
50 mm dia
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o For F = 8kN & T = 2 KN-m, determine the weld size if the
maximum allowable shear stress is 120 Mpa
150 mm
50 mm dia
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o The figure shows a welded steel bracket loaded by a static force
F. Estimate the factor of safety if the allowable shear stress in
the weld throat is 120Mpa.
120 mm
7.5 kN
6mm
60
mm
120 mm
6mm
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250mm
80 kN
156mm
25mm dia
150mm
50mm dia
F=80kN
Mechanical Engineering Dept. CEME NUST
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Brackets, such as the one shown in fig-2, are used in mooring small watercraft. Failure of
such brackets is usually caused by bearing pressure of the mooring clip against the side
of the hole. Our purpose here is to get an idea of the static margins of safety involved.
We use a bracket 6 mm thick made of hot-rolled 1018 steel. We then assume wave
action on the boat will create force F no greater than 5 kN.
o Identify the moment M that produces a shear stress on the throat resisting bending
action with a “tension” at A and “compression” at C.
o Find the force component Fy that produces a shear stress at the throat resisting a
“tension” throughout the weld.
o Find the force component Fx that produces an in-line shear throughout the weld.
o Find the factor of safety guarding against shear yielding in the weldment.
o Find the factor of safety guarding against bearing failure of the hole.
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END
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