design of weld
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17
Design
Static StrengthofWelded Joints
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Lecture S;cope Static stren gth of welded joints- need for strength design- types of loads carried by welds- effective weld area- stresses acti 19 on weld area- allowable str'3sses on welds- design of weld groups
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Lecture 17
Need 'for Strength Design S t r e n ~ l t h and rigidity are not critical in somewelded parts
- e.g. machine guards, furniture- In such cases only casual attention is paid to weldstrength
Many other welded structures must meetrequimments for strength and/or rigidity- B r i d g E ~ s , ships, pressure vessels- Methe ds are required for ensuring that welded jointsbetween members possess adequate strength
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Weld Joint Strength
Strength d3sign implies knowledge of -loads- load carryin area- material properties- failure criteria
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Types of Loads Carried by Welds-
Lecture 17
Full penetration groovewelds in tension
Compression nonnalto axis of weld
Tension orcompression parallelto weld axis
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Lecture 17
Weld Design Full PElnetration Groove Welds- Usuall:( treated as a continuous part of the structureandnot s p l ~ c i f i c a l l y analyzed- Welds assumed to match parent strength- Some codes reduce the allowable stress in welds notsubject to NDE by applying a "Joint Efficiency Factor"
Partial Penetration Groove, and Fillet Welds- Since ':he effective area differs from the area of the basemetal, methods are needed to ensure the welds haveadequate strength to carry the imposed loads
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Effective Weld Area Full Penetration Groove Welds .r:
- the effective area is the thickness of the thinner Lpartjoined Partial PenetlCltion Groove Welds
- the effective area is nonnally the depth of thechamfer multiplied by the weld length (may bereduced for narrowweld preparations)
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Fillet Welds- The effective area is nonnally taken as thetheoretical throat multiplied by the weld length-for equal leg fillets T =O.707*fillet size
Actual ...... ~throat ~a " ~ ~ ~ = J z : O .~ L e g
' \} Theoretical1- - - - -_ -1 Throat. T
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Lecture 17
Stresues on Weld AreaThe force t"ansmitted by a unit area of weld may bedecomposnd into shear and normal stress components (filletweld shown)
P = Force vector0.L = Stress normal toweld throat't' = shear stress acting
~ perpc:ndicular toweld axis' tI = shear stressparallel to weld axis
weld throat area
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Design Stresses on welds IIW/ISO criterion for fillet weld stresses- The IntemaHonallnstitute ofWelding has published thefollowing cri':erion (see IIW doc XV-358-74 & Welding inthe World alticle Vol 2 No 21964.)- adopted by ISO as recommendation R617a :? R ro 2 3(" 2 ? la fJy- .1 .+ . 1 .+ ,1
a = allovrable stress in the base materiala = 0.7 fJ r structural steels up to 500 MPa UTS and 0.85 forhigh-strength steels with UTS
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Lecture 17
Design Stresses on Welds---;;.---------------- PrerE!quisites for IIW approach:- Stru
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Design Stresses on Welds AWS 01.1 standard approach for fillet/partialpenetration welds in steel structures- Defines allowable values for each stress componentbased (primarily) on weld metal tensile strength- CSAW59 Ulies a similar approach but differs in detail- Always consult the applicable standard
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Design Stresses on WeldsAWS 01.1 Full Penetration Welds
Lecture 17
TenoellTepa
St
Stress in Weld Allowable Stressnsion or compression Same as base metalnnal and parallel toective areansion or compression Same as base metalrallel to axis ofweld
earon the effective area 0.3 of nominal UTS of weldmetal, but not more than 0.4"yield strength of base metal
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Design Stresses on WeldsAWS 01.1 Partial P e n E ~ t r a t i o n Groove Welds
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StressCompression ncarea Goints not dTension or compweld axis
Shearparallel to
Tension normal t= o.7*V3*(50JOO/8.5L)A2L >= 57.98 mm ( a ~ r e e s with AWS)
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Lecture 17
In the previous example, the force carried bythe welds were determined by applying theprinciples of static equilibrium
The forces carried by welds in morecompl icated connections cannot be so easilydetermined Various methods have been proposed todetermine the capacity of the weld group- Mostl', empirical rules- Either elastic or plastic analyses
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F"Uet W e ~ d Stress Distdbution Design method8 usually assume uniform stress in welds andignore stress co ncentrations at ends and residual stress OK if welds have sufficient deformation capacity
Stress in weld .....Stress concentrations_ / at ends relieved byYield stress ~ - yielding
Distance along weld::. p
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When welds W1 or W3 exist, the load carriedby each weld is not readily calculated _- depends on loading and stiffness of parts joined
~ ~ 12 S~ ~~ ~~ . t ':::;''::::::~ I::::::~ L1 W2 I ~ W 3 PI~ ,::::::-~ I ~~ ,~~ ~
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VVe'd S t n ~ s s D i s t r i b u t ~ o n Qualitative behaviour
- L2 L1, W3 not present: the load is carried by W2- L2
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23/36ecture 17
A simple method for design of planarconnections subject to shear forces andbending moments is to consider the welds as apattern of lines Known as the "polar moment method"
- Welding in World artide, P 180, Welding Handbook
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Weld GrouqosDerivationAssuming:
The welds are of equal size S The force per unit length F at any point in the weld group is proportional to thedistance from the centroid C, i.e. F = kr. F acts normal to the line joining the centroid residual stresses are negligible
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If dL is any elemental length of weld,
M = frdLoJSince F = kr. =kJwWhere Jw is the linear polar moment of inertia of the weld group aboutthe axis through C and normal to the plane.Hence:F Mr=
JwThe weld throat T required to satisfy allowable stress limits can be found using:
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F =
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Weld Groups Similar results apply for bending momentsparallel to the plane of the weld group. - Shear forces are assumed uniformly distributed
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Lecture 17
ProcHdure for polar moment method usingtabulated formulas1. Fintl the position on the welded connection where thecombined forces are maximum
usuEllly the position furthest from the centroid of the connection morE! than one load combination or position may need to be considered sheElr forces are assumed uniformly distributed
2. Final the force per unit length resulting from eachloading at this position using the tabulated fonnulas3. Add the forces vectorially4. Determine the required throat size by dividing the totalunit force by the allowable weld stress
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ExarrtuJle 2 (
I p= 80 kN r - - - l " ~ ~ - - - - ' - i . +(18000Ib)T F50- . . -(10") !Ct I
- - - -+_250__(10")
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Lecture 17
Example 2 Step 1:. Identify the most highly stressed location- The point of maximum combined unit force is at the top righthand comer
Step 2: Determine unit forces at this point- The distance from the centroid to the point of combinedstress, and the polar moment of inertia are given by thetable (handout) as:
c _ b (b +d) _ 95 mmyr - 2b+d -
J = b_3 + (b +2d) + d 2 (6b+d) = 6.3x]06 mm 3w 3 2b+d 12
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Example .2 Step 2 (cont'd)- The horizontal and vertical components of the twistingmoment are given by:
fh = Td/2 = 410 N/mmJ w
Iv = TCyr = 306 N /mmJw- The vertical shear force isjs = P/Lw = 158 N/mm
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Step 3: Determine the resultant forcer; :::: Vf/+(j:'+fsl :::: 620 N/m
Step 4: Determine the required weld sizeAllowable weld stress for 6018 electrodes = 0.3*410 = 124 MPaRequired throat T = 620/124 mm = 5mmAssuming equal leg fillets, weld size =T/O.707 = 7mmAn 8 mm fillet weld should be specified on the weld symbol
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ASME Boiler & Pressure VesselCodeCategories of weld joint in Class 3 nuclear vessels
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SEE NC-33!i12
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Categories of Butt Joints~ - ~ - ~ ~ ~0rIIt-Y.... Jel...
Lecture 17
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ASME.B&PV CodeExamples of permissible nozzle connections forClass 3 nuclear vessels
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longltudln8l Box
Support Rods forlower Skywalk
_ Support Rodsto CeiUngI Washer and Nut on Bottom
Upper Skywalk
f~ f Force PutWeld RootIn Tension
Lecture 17 p3
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~ N e l d Structural Failures Failures of w E ~ l d e d structures result in financialloss, damage to the environment, and injury.orloss of life- Hyatt Regency hotel walkway collapse- Ramsgate ferry ramp collapse Codes and standards define rules for welded jointdesign for various types of critical structure, e.g.buildings, bridges, pressure vessels based ontests, analysis and experience Use of the applicable codes is advisable and maybe legally required.
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