Weld Strengths for Various Shapes

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Weld Strength Calculations

Introduction.....RelevantStandards .....VariablesAssociatedWithWelds .....GuidancePrinciples .....TableBasicWeldCalcs. ..... AssessmentofFilletWelds.....ExamplesofFilletWeldsCalcs.....PropertiesofFilletWeldsaslines.....Exampleoftorsionweldcalc.usingvectors .....CapacitiesofFilletWelds.....DesignStrengthofFilletWelds.....

Introduction

Thefollowingnotesaregeneralguidancenotesshowingmethodsofcalculationofthestrengthandsizeofwelds.Weldedjointsareoftencruciallyimportantaffectingthesafetyofthedesignsystems.Itisimportantthatthenotesanddatabelowareonlyusedforpreliminarydesignevaluations.Finaldetaildesignshouldbecompletedinaformalwayusingappropriatecodesandstandardsandqualityreference

documents

Relevant StandardsBS5950-1:2000..Structuraluseofsteelworkinbuilding.Codeofpracticefordesign.RolledandweldedsectionsBSEN10025-1:2004-Hotrolledproductsofstructuralsteels.Generaltechnicaldeliveryconditions

Variables related to welded joints

Strengthofdepositedweldmaterial1.Typeofjointandweld..important2.Sizeofweld..important3.

Locationofweldinrelationtopartsjoined..important4.Typesofstresstowhichtheweldissubjected5.Conditionsunderwhichweldiscarriedout6.Typeofequipmentusedforwelding7.Skillofwelder8.

Guidance Principles

Agenerousfactorofsafetyshouldbeused(3-5)andiffluctuatingloadsarepresentthenadditionaldesignmarginsshouldbeincludedtoallowforfatigueUsetheminimumamountoffillermaterialconsistentwiththejobrequirementTrytodesignjointsuchthatloadpathisnotnotthroughtheweld

Thetablebelowprovidesprovidesapproximatestressesin,hopefully,aconvenientway.

Forthedirectloadingcasethebuttweldstressesaretensile/compressive σ tforthefilletweldsthestressesareassumedtobeshear τ sappliedtotheweldthroat.

Forbuttweldedjointssubjecttobendingthebuttweldstressesresultfromatensile/compressivestress σbandadirectshearstress τ s.

Inthesecasesthedesignbasisstressshouldbe σr=Sqrt(σb2+4τs2)

ForFilletweldedjointssubjecttobendingthestressesinthefilletweldsareallshearstresses.Frombending τ bandfromshearτ s

Inthesecasesthedesignbasisstressisgenerally τr=Sqrt(τ b2+τs2)Thestressesfromjointssubjecttotorsionloadingincludeshearstressfromtheappliedloadandshearstressesfromthetorqueloading.Theresultingstressesshouldbeaddedvectoriallytakingcaretochoosethelocationofthehigheststresses.

Welding Tables Heavy duty welding tables with fume capture, extraction and filtration. www.oskarairproducts.com

Pipe Plugs, All Types Most Pressure, Chemical, Temps Standard and Custom Configurations www.petersenproducts.com

Resistance Welders Specializing in Standard and Custom Built Resistance Welding Machinery www.seedorffacme.com

Page 1 of 9Weld Stress Calculations

3/25/2012http://www.roymech.co.uk/Useful_Tables/Form/Weld_strength.html



Table of bracket weld subject to direct and bending stresses

MethodofLoading Weldment

StressinWeld

σbτs

Weldsize(h)

Weldment

StressinWeld

σbτs

Weldsize(h)

Weldment

StressinWeld

τbτs

Weldsize(h)

Assessment of Fillet Weld Groups refnotesandtablePropertiesofFilletWeldsaslines

Importantnote:ThemethodsdescribedbelowisbasedonthesimplemethodofcalculationofweldstressasidentifiedinBS5950-clause6.7.8.2.TheothermethodidentifedinBS5950-1clause6.7.8.3asthedirectionmethodusesthemethodofresolvingtheforcestransmittedbyunitthicknessweldsperunitlengthintotraverseforces(F T)andlongitudinalforces(FL).Ihave,tosomeextent,illustrated

thismethodinmyexamplesbelowThemethodofassessingfilletweldsgroupstreatingweldsaslinesisreasonablysafeandconservativeandisveryconvenienttouse.a)Weldsubjecttobending....SeetablebelowfortypicalunitareasandunitMomentsofInertia Afilletweldsubjecttobendingiseasilyassessedasfollows.1)TheareaofthefilletweldAu..(unitthickness)iscalculatedassumingtheweldisoneunitthick..

2)The(unit)MomentofInertiaI uiscalculatedassumingtheweldisoneunitthick..

3)Themaximumshearstressduetobendingisdetermined... τb=M.y/Iu





4)Themaximumshearstressduetodirectshearisdetermined.. τs=P/A

5)Theresultantstress τr=Sqrt(τb2+τs2)6)Bycomparingthedesignstrengthp wwiththeresultantstressτrthevalueoftheweldthroatthicknessiscalculatedandthentheweldsize.

i.e.iftheτr/pw=5thenthethroatthickesst=5unitsandtheweldlegsizeh=1,414ta)Weldsubjecttotorsion...SeetablebelowfortypicalunitareasandunitPolarmomentsofInertia Afilletweldsubjecttotorsioniseasilyassessedasfollows.1)TheareaofthefilletweldAu(unitthickness)iscalculatedassumingtheweldisoneunitthick

2)The(unit)PolarMomentofInertiaJ uiscalculatedassumingtheweldisoneunitthick..ThepolarmomentofinertiaJ=Ixx+Iyy

3)Themaximumshearstressduetotorsionisdetermined... τt=T.r/Ju4)Themaximumshearstressduetodirectshearisdetermined.. τs=P/Au5)Theresultantstress τristhevectorsumof τtandτs.rischosentogivethehighestvalueof τr6)Bycomparingthedesignstrengthp wwiththeresultantstressτrthevalueoftheweldthroatthicknessiscalculatedandthentheweldsize.

i.e.iftheτr/pw=5thenthethroatthickesst=5unitsandtheweldlegsizeh=1,414.t

Examples of Fillet Weld Calculations

Example of Weld in Torsion..

P = Applied load = 10 000N

P w = Design Strength = 220 N/mm2 (Electrode E35 steel S275) Design Strength

b = 120mm.d = 150 mm

x = b2 / 2(b+d) = 27mm.. (From table below)

y = d2 / 2(b+d) = 42mm..(From table below)

Simple Method as BS 5950 clause 6.8.7.2

..The vector sum of the stresses due to forces and moments should notexceed the design strength Pw

A u = Unit Throat Area

= (From table below) b + d = (120 + 150) = 270mm2 To obtain radius of Force from weld centre of gravity

A = 250-27 =223mm

Moment M = P.r = 10000.223 = 2,23.106 N.mm

J u = [(b+d)4 - 6b

2d

2] /12 (b+d) = 1,04.10

6..(From Table)

It is necessary to locate the point subject to the highest shear stress..Fora weld subject to only torsion this would be simply at the point furthestfrom the COG. However because the weld is subject to torsion anddirect shear the problem is more complicated. A normal method ofdetermining the stresses in these cases is to use vector addition.

It is generally prudent to calculate the total shear stress at bothpositions, using the method below, and select the highest.. For thisexample the method used is to resolve the stresses in the x and ydirections

Direction Method as BS 5950 clause 6.8.7.3

L = Length of weld 1 unit thick =(From table below) b + d = (120 + 150) = 270mmTo obtain radius of Force from weld Centre of Gravity (Cog) .

A = 250-27 =223mm

Moment M = P.r = 10000.223 = 2,23.106

N.mmJu = Polar Moment o f inertia for weld 1unit(mm) thick.

= [(b+d)4 - 6b2d2] /12 (b+d) = 1,04.106 mm4 /mm..(From Table)

It is necessary to locate the point subject to the highest shear stress..Fora weld subject to only torsion this would be simply at the point furthestfrom the COG. However because the weld is subject to torsion anddirect shear the problem is more complicated. A normal method ofdetermining the stresses in these cases is to use vector addition.

It is generally prudent to calculate the total shear stress at bothpositions, using the method below, and select the highest.. For thisexample the method used is to resolve the stresses in the x and ydirections





First considering point Z

Horizontal distance from centroid rzh = 120-27= 93mmVertical distance from centroid r zv = 42mm

The vertical stress τ v = τ sv + τ tv

τ sv = P /A u = 10000/270 = 37 N/mm2

τ tv = M.r zh /J u = 2,23.106.93/1,04.106 = 199 N/mm2

τ v = 236,45 N/mm2

The horizontal stressτ h = τ sh + τ th

τ sh = 0

τ th = M.r zv /J u = 2,23.106.42/1,04.10

6 = 90 N/mm

2

τ h = 90 N/mm2

The resultant stress on the weld at z

τ r = Sqrt (τ h2 + τ v2) = 253 N/mm2

Now considering point w

Horizontal distance from centroid rwh = 27mmVertical distance from centroid r wv = 150-42= 108mm

The vertical stress τ v = τ sv - τ tv

τ sv = P /A u = 10000/270 = 37 N/mm2

τ tv = M.r wh /J u = 2,23.106.27/1,04.106 = 57,9 N/mm2

τ v = 20,86 N/mm2

The horizontal stressτ h = τ sh + τ th

τ sh = 0

τ th = T.r wv /J u = 2,23.106.108/1,04.106 = 231,6 N/mm2

τ h = 231,6 N/mm2

The resultant stress on the weld at w

τ r = Sqrt (τ h2 + τ v

2) = 232,5 N/mm

2

The maximum stress is similar but greatest at z ....The design strength pw for the weld material is 220 N/mm2 The weld throat thickness should be 253 /220 = 1,15mm .The weld size is therefore 1,414. 1,15 = 1,62mm use 3mm fillet weld

First considering point Z

Horizontal distance from centroid rzh = 120-27= 93mmVertical distance from centroid r zv = 42mm

The vertical force /mm run F v = F sv + F tv

F sv = P /L = 10000/270 = 37 N/mm run

F tv = M.r zh /J u = 2,23.106.93/1,04.10

6 = 199 N/mm run

F v = 236,45 N/mm run

The horizontal force /mm run for un it(mm) weld width Fh = F

sh + F

th

F sh = 0

F th = M.r zv /J u = 2,23.106.42/1,04.10

6 = 90 N/mm run

F h = 90 N/mm run

The resultant force on the weld/mm run at z

F r = Sqrt (F h2 + F v

2) = 253 N/mm run

Now considering point w

Horizontal distance from centroid rwh = 27mm

Vertical distance from centroid r wv = 150-42= 108mm

The vertical forces per mm run F v = F sv - F tv F sv = P /L = 10000/270 = 37 N/mm run

F tv = M.r wh /J u = 2,23.106.27/1,04.106 = 57,9 N/mm run

F v = 20,86 N/mm run

The horizontal force /mm run = F h = F sh + F th F sh = 0

F th = M.r wv /J u = 2,23.106.108/1,04.106 = 231,6 N/mm runF h = 231,6 N/mm run

The specific force on the weld at w

F r = Sqrt (F h2 + F v

2) = 232,5 N/mm run

The maximum specific is greatest at z = 253 N/mm run....

Referring to weld capacities for longitudinal stresses PL for fillet welds

Capacities of Fillet Welds the weld capacity for a 3mm weld with andE35 Electrode S275 Steel is 462N /mm run. This weld would be morethan sufficient.

Example of Weld in Bending..

P= 30000 Newtonsd= 100mmb= 75mmy = 50mm

Design Stress p w = 220 N/mm2 (Electrode E35 steel S275) Design Strength

Moment = M = 30000*60=18.105 Nmm

Simple Method as BS 5950 clause 6.8.7.2 Direction Method as BS 5950 clause 6.8.7.3





Unit Weld Area = A u = 2(d+b) =2(100+75) =350mm2

Unit Moment of Inertia = I u

= d2(3b+d) / 6 = 100

2 (3.75 +100) / 6 =5,42.10

5mm

4

τ r = Sqrt(τ s2 + τ b2)

τ s = P / A u = 30000/350 = 85,71 N/mm2

τ b = M.y / I u = 18.105 . 50 / 5,42.10

5= 166,05 N/mm

2

τ r = Sqrt(85,712 + 166.05

2) =186,86 N/mm

2

τ r / p w = 186,86 / 220 = 0,85 = Throat Thickness.....( Throat thickness for τ = 220 N/mm2 )Leg Length = Throat thickness *1,414 = 1,2mm use 3mm weldthickness

Note : If a leg length h= 1,2mm is used in the equations in relevant partof the "Table of bracket weld subject to direct and bending stresses"

above a value ofτ b = 198 N/mm and a value ofτ s = 100 N/mm2

results with a resultant stress of Sqrt (τ b2 + τ s2) = 222N/mm2..Which

is in general agreement with the above result

Length of Weld of unit thickness = L = 2(d+b) =2(100+75) =350mmMoment of Inertia / mm throat thickness = Iu / mm

= d2(3b+d) / 6 = 100

2 (3.75 +100) / 6 =5,42.10

5mm

4 / mm

F r = Resultant force per unit length of weld.

F s = Shear force per unit length of weld.F b = Bending force per unit length of weld.

F r = Sqrt(F s2 + F b

2)F s = P / L = 30000/350 = 85,71 N per mm length of weld

F b = M.y / I u

= 18.105 . 50 / 5,42.105 = 166,05 N per mm length of weld

F r = Sqrt(85,712 + 166.05

2) =186,86 N per mm length of weld.

For this case for the welds under greatest loading the type of loading istraverse loading. The bending stress is in line with horizontal elementand the shear stress is in line with vertical member.

The angle of the resulting specific load to the horizontal element

= arctan(85,71/166,5)= 27,5o.

This is an angle with the weld throatθ = 45o + 27,5o = 72,5o

Referring to weld capacities table below.Weld Capacities K iscalculated at 1,36 for this resultant direction of forces.

PT = a.K.pwfor a E35 Weld electrode used with S275 steel

pw = 220 N/mm2 and therefore PT = a*300N/mm

2..

A 3mm weld (a = 2,1mm) therefore will therefore have a designcapacity of 630 N/mm run and will easily be able to support the load of186,86 N per mm run

Properties of weld groups with welds treated as lines

Itisacceptedthatitisreasonablyaccuratetousepropertiesbasedonunitweldthicknessincalculationtodeterminethestrengthofweldsasshownintheexamplesonthispage.TheweldpropertiesI xxIyyandJareassumedtobeproportionaltotheweldthickness.Thetypical

accuracyofthismethodofcalculationisshownbelow...

Thisisillustratedinthetabledvaluesbelow

d b h Ixx Iyy J= Ixx +Iyy

Accurate 3 60 50 955080 108000 1063080

Simple 3 60 50 900000 108000 1008000





Error 6% 0 5%

Note:Theerroridentifiedwiththismethodislowerashincreasesrelativetod.Thiserrorissuchthattheresultingdesignsareconservative.

Example illustrating use of stress vectors

Calculation based on real weld sizes

1)Theareaofthewelds(basedonthroatweldthicknessat0,707.5=3,5mm)

Area=(57.3,5+2.55.3,5)=584,5mm2

2)Themomentofareaaboutx-x=

MofArea=(57.3,5.3,5/2+2.55.3,5.(27.5+3,5))=12284mm3

3)Thecentroidv=MomentofArea/Area

MofArea/Area=21mm

4)Theradiir A,rB,rC&rDarecalculated..

r A=rB=Sqrt((58,5-21)^2+28,5^2)=47,1

rC=rD=Sqrt((21)^2+28,5^2)=35,40...

5)Theanglesθ A,θB,θC&θDarecalculated..

θ A=θB=tan-1((58,5-21)/28,5)=52,7 o

θC=θD=tan-1((21)/28,5)=36,4 o...

6)Thedirectshearstressonthearea=Force/Area

τ S=5000/584=8,56N/mm2

7)TheMomentontheweldgroup=Force.Distancetocentroid

M=5000.(100+21)=6,05.105Nmm

8)Thepolarmomentofinertiaoftheweldgroup=J=Ixx+Iyy

Iyy=2.[55.3,53/12+3,5.55.(50/2+3,5/2)

2]

+573.3,5/12=3,3.10

5mm

4

Ixx=2.[553.3,5/12+3,5.55.(55/2+3,5-21)

2]

+3,5357/12+3,5.57.(21-3,5/2)

2=2,097.10

5mm

4

J=Ixx+Iyy=5,4.10

5

mm

4

9)Thestressduetotorsion

τ TA=τ TB=M.r A/J..and..τ TC=τ TD=M.rC/J

τ TA=6.055Nmm.47,1mm/5,4.105mm

4=52,8N/mm

2

τ TC=τ TD=6.055Nmm.35,4mm/5,4.105mm

4=39,70N/mm

2

10)Theresultantstresses τ RA,=τ RBandτ RA,=τ RBareobtainedbyaddingthestressvectosgraphicallyasshownbelow

τ RA=τ RB=46,29N/mm2

τ RC=τ RD==45,31N/mm2

Calculations based on unit values This calculation uses equations from table belowfor Area, centroid, and Ju

1)Areaofweld=0,707.5.(2b+d)

Area=0,707.5.(2.55+50)=565.6mm2

2)ThereisnoneedtocalculatetheMomentofAreawiththismethod

3)Thecentroidv=b2/(2b+d)

v=552/(2.55+50)=18,9mm

4)Theradiir A,rB,rC&rDarecalculated..

r A=rB=Sqrt((55-18,9)^2+25^2)=43,9

rC=rD=Sqrt(18,9^2+25^2)=31,34

5)Theanglesθ A,θB,θC&θDarecalculated..

θ A=θB=tan-1((55-18,9)/25)=55,29 o

θC=θD=tan-1((18,9)/25)=37o...

6)Thedirectshearstressonthearea=Force/Area

τ S=5000/565,5=8,84N/mm2

7)TheMomentontheweldgroup=Force.distancetocentroid

M=5000.(100+18,9)=5.94.105Nmm

8)TheUnitPolarmomentofinertiaoftheweldgroup=

Ju=0.707.5.(8.b3+6bd

2+d

3)/12+b

4/(2b+d)

Ju=0,707.5.(8.553+6.55.50

2+50

3)/12-55

3/(2.55+50)=4,69.1

9)Thestressduetotorsion

τ TA=τ TB=M.r A/J..and..τ TC=τ TD=M.rC/J

τ TA=5,94.105Nmm.43,9mm/4,69.10 5mm4=55,6N/mm2

τ TC=τ TD=5,945Nmm.31,34mm/4,69.105mm

4=39,69N/mm

10)Theresultantstresses τ RA,=τ RBandτ RA,=τ RBareobtainedbyaddingthestressvectosgraphicallyasshownbelow

τ RA=τ RB=48,59N/mm2

τ RC=τ RD=45,56N/mm2





Note:Theexampleabovesimplyillustratesthevectormethodaddingdirectandtorsionalshearstressesandcomparesthedifferenceinusingtheunitweldwidthmethodandusingrealweldsizes.Theexamplecalculatesthestresslevelsinanexistingweldgroupitisclearthattheweldisoversizedfortheloadingscenario.Thedifferenceintheresultingvaluesareinlessthan4%.Iftheweldsweresmalleri.e3mmthenthedifferenceswouldbeevensmaller.

Table properties of a range of fillet weld groups with welds treated as lines

WeldThroatArea

UnitArea

LocationofCOGxy

Ixx-(unit) J-(Unit)

-





-

Table Of Weld Capacities

Thefilletweldcapacitytablesrelatedtothetypeofloadingontheweld.Twotypesofloadingareidentifiedtraverseloadingandlongitudinalloadingasshowbelow

Theweldloadingshouldbesuchthat

{(FL/PL)2+(FT/PT)

2}≤1

ThefollowingtableisinaccordwithdatainBS5950part1.Basedondesignstrengthsasshownintablebelow... DesignStrength

PL=a.pw

PT=a.K.pw

a=weldthroatsize.

K=1,25√(1,5/(1+Cos2θ)

PTbasedonelementstransmittingforcesat90 oi.eθ=45oandK=1,25

WeldCapacityE35ElectrodeS275Steel WeldCapacityE42ElectrodeS355Steel

LegLengthThroat

Thickness

LongitudinalCapacity

TransverseCapacity LegLength

ThroatThickness

LongitudinalCapacity

TransverseCapacity

PL(kN/mm) PT(kN/mm) PL PT

mm mm kN/mm kN/mm mm mm kN/mm kN/mm

3 2,1 0,462 0,577 3 2,1 0,525 0,656

4 2,8 0,616 0,720 4 2,8 0,700 0,875

5 3,5 0,770 0,963 5 3,5 0,875 1,094

6 4,2 0,924 1,155 6 4,2 1,050 1,312

8 5,6 1,232 1,540 8 5,6 1,400 1,750

10 7,0 1,540 1,925 10 7,0 1,750 2,188

12 8,4 1,848 2,310 12 8,4 2,100 2,625

15 10,5 2,310 2,888 15 10,5 2,625 3,281

18 12,6 2,772 3,465 18 12,6 3,150 3,938

20 14,0 3,08 3,850 20 14,0 3,500 4,375

22 15,4 3,388 4,235 22 15,4 3,850 4,813

25 17,5 3,850 4,813 25 17,5 4,375 5,469

Design Strength p

w

of fillet welds





Electrodeclassification

SteelGrade35 43 50

N/mm2

N/mm2

N/mm2

S275 220 220 220

S355 220 250 250

S460 220 250 280

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Weld Strengths for Various Shapes