Posted  by

# Minimum Throat Thickness Of Fillet Weld

Formulae_Index
Nomenclature

The legs are the other two sides of the triangular fillet weld. The leg length is usually designated as the size of the weld. The throat of the weld is the distance from the center of the face to the root of the weld. Typically the depth of the throat should be at least as thick as the thickness of metal you are welding.

• The strength of a fillet weld is based, in the design, on the product (effective area of the weld: T x W) of the theoretical throat (design throat thickness) and effective weld length as shown in Fig. Fillet weld legs determine fillet weld sizes. Fillet weld sizes are measured by the length of the legs of the largest right triangle that may.
• Lw-minfor each length of the weld = 4.0 in. (transverse distance between welds, see J2.2b) - Given length = 5.0 in., which is Lmin.
• I am aware of the common rule that the throat thickness of the fillet weld should be roughly 0,707.minimum plate thickness. I am just in a position where the throat thickness of the fillet weld needs to be bigger than plate thickness due to fatigue reason.
Weld Strength Calculations

Introduction... Relevant Standards... Variables Associated With Welds...
Guidance Principles... Table Basic Weld Calcs... Assessment of Fillet Welds...
Examples of Fillet Welds Calcs... Properties of Fillet Welds as lines...
Example of torsion weld calc. using vectors... Capacities of Fillet Welds... Design Strength of Fillet Welds...

### Minimum Throat Thickness Of Fillet Weld Wire

Introduction

The following notes are general guidance notes showing methods ofcalculation of the strength and size of welds. Welded joints are oftencrucially important affecting the safety of the design systems. It isimportant that the notes and data below are only used for preliminary designevaluations. Final detail design should be completed in a formal wayusing appropriate codes and standards and quality reference documents

Relevant Standards
BS 5950-1:2000 .Structural use of steelwork in building. Code of practice for design. Rolled and welded sections
BS EN 10025-1:2004 - Hot rolled products of structural steels. General technical delivery conditions
Variables related to welded joints
 Strength of deposited weld material Type of joint and weld.important Size of weld .important Location of weld in relation to parts joined.important Types of stress to which the weld is subjected Conditions under which weld is carried out Type of equipment used for welding Skill of welder

Guidance Principles

A generous factor of safety should be used (3-5) and if fluctuatingloads are present then additional design margins should be included to allow for fatigue
Use the minimum amount of filler material consistent with the job requirement
Try to design joint such that load path is not not through the weld

The table below provides provides approximate stresses in, hopefully, a convenientway.
For the direct loading case the butt weld stresses are tensile/ compressive σ t for the fillet weldsthe stresses are assumed to be shear τ s applied to the weld throat.
For butt welded joints subject to bending the butt weld stresses result from a tensile/compressive stress σb and a direct shearstress τ s .
In these cases the design basis stress should be σr = Sqrt (σb2 + 4τs2)
For Fillet welded joints subject to bending the stresses in the fillet welds are all shear stresses. From bending τ b and from shear τ s
In these cases the design basis stress is generally τr =Sqrt (τ b2 + τs2)
The stresses from joints subject to torsion loading include shear stress from the applied loadand shear stresses from the torque loading. The resulting stresses should be addedvectorially taking care to choose the location of the highest stresses.

Table of bracket weld subject to direct and bending stresses
 Method of Loading Weldment Stress in Weld σbτsWeld size (h) Weldment Stress in Weld σbτs Weld size (h) Weldment Stress in WeldτbτsWeld size (h)

Assessment of Fillet Weld Groups ref notes and table Properties of Fillet Welds as lines

Important note: The methods described below is based on the simple method of calculationof weld stress as identified in BS 5950- clause 6.7.8.2 . The other method identifed in BS 5950 - 1 clause 6.7.8.3as the direction method uses the method of resolving the forces transmittedby unit thickness welds per unit length into traverse forces (FT ) andlongitudinal forces (FL ). I have, to some extent,illustrated this method in my examples below
The method of assessing fillet welds groups treating welds as lines is reasonablysafe and conservative and is very convenient to use.
a) Weld subject to bending..See table below for typical unit areas and unit Moments of Inertia
A fillet weld subject to bending is easily assessed as follows.
1) The area of the fillet weld A u.(unit thickness) is calculated assuming the weldis one unit thick.
2) The (unit) Moment of Inertia I u is calculated assuming the weldis one unit thick.
3) The maximum shear stress due to bending is determined..τb = M.y/I u
4) The maximum shear stress due to direct shear is determined. τs = P /A
5) The resultant stress τr = Sqrt (τb2 + τs2 )
6) By comparing the design strength p w with the resultant stress τr the value of the weld throat thickness is calculated and then the weld size.
i.e. if the τr /p w = 5 then the throat thickess t = 5 unitsand the weld leg size h = 1,414t
a) Weld subject to torsion..See table below for typical unit areas and unit Polar moments of Inertia
A fillet weld subject to torsion is easily assessed as follows.
1) The area of the fillet weld A u (unit thickness) is calculated assuming the weldis one unit thick
2) The (unit) Polar Moment of Inertia J u is calculated assuming the weldis one unit thick. The polar moment of inertia J = I xx + I yy
3) The maximum shear stress due to torsion is determined..τt = T.r/J u
4) The maximum shear stress due to direct shear is determined. τs = P /A u
5) The resultant stress τr is the vector sum of τt and τs. r is chosen to givethe highest value of τr
6) By comparing the design strength p w with the resultant stress τr the value of the weld throat thickness is calculated and then the weld size.
i.e. if the τr /p w = 5 then the throat thickess t = 5 unitsand the weld leg size h = 1,414.t

Examples of Fillet Weld Calculations
Example of Weld in Torsion.P = Applied load = 10 000N
P w = Design Strength = 220 N/mm 2 (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 not exceed the designstrength PwA u = Unit Throat Area = (From table below) b + d = (120 + 150) = 270mm2To obtain radius of Force from weld centre of gravityA = 250-27 =223mmMoment M = P.r = 10000.223 = 2,23.106 N.mmJ u = [(b+d)4 - 6b2d2] /12 (b+d) = 1,04.106.(From Table) It is necessary to locate the point subject to the highest shear stress.For a weld subject to only torsionthis would be simply at the point furthest from the COG. However because the weld is subject totorsion and direct 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 both positions, using the method below,and select the highest. For this example the method used is to resolve the stressesin the x and y directionsFirst considering point Z Horizontal distance from centroid r zh = 120-27= 93mm Vertical distance from centroid r zv = 42mm The vertical stress τv = τsv + τtvτsv = P /A u = 10000/270 = 37 N/mm2τtv = M.r zh /J u = 2,23.106.93/1,04.106 = 199 N/mm2τv = 236,45 N/mm2The horizontal stress τh = τsh + τthτsh = 0τth = M.r zv /J u = 2,23.106.42/1,04.106 = 90 N/mm2τh = 90 N/mm2The resultant stress on the weld at z τr = Sqrt (τh2 + τv2) = 253 N/mm2Now considering point w Horizontal distance from centroid r wh = 27mm Vertical distance from centroid r wv = 150-42= 108mm The vertical stress τv = τsv - τtvτsv = P /A u = 10000/270 = 37 N/mm2τtv = M.r wh /J u = 2,23.106.27/1,04.106 = 57,9 N/mm2τv = 20,86 N/mm2The horizontal stress τh = τsh + τthτsh = 0τth = T.r wv /J u = 2,23.106.108/1,04.106 = 231,6 N/mm2τh = 231,6 N/mm2The resultant stress on the weld at w τr = Sqrt (τh2 + τv2) = 232,5 N/mm2The maximum stress is similar but greatest at z ..The design strength p w for the weld material is 220 N/mm 2The weld throat thicknessshould be 253 /220 = 1,15mm . The weld size is therefore 1,414. 1,15 = 1,62mm use 3mm fillet weld Direction Method as BS 5950 clause 6.8.7.3L = 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 =223mmMoment M = P.r = 10000.223 = 2,23.106 N.mmJu = Polar Moment of 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.For a weld subject to only torsionthis would be simply at the point furthest from the COG. However because the weld is subject totorsion and direct 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 both positions, using the method below,and select the highest. For this example the method used is to resolve the stressesin the x and y directionsFirst considering point Z Horizontal distance from centroid r zh = 120-27= 93mm Vertical distance from centroid r zv = 42mm The vertical force /mm run F v = F sv + F tvF sv = P /L = 10000/270 = 37 N/mm runF tv = M.r zh /J u = 2,23.106.93/1,04.106 = 199 N/mm runF v = 236,45 N/mm run The horizontal force /mm run for unit(mm) weld width F h = F sh + F thF sh = 0F th = M.r zv /J u = 2,23.106.42/1,04.106 = 90 N/mm runF h = 90 N/mm run The resultant force on the weld/mm run at z F r= Sqrt (F h2 + F v2)= 253 N/mm runNow considering point w Horizontal distance from centroid r wh = 27mm Vertical distance from centroid r wv = 150-42= 108mm The vertical forces per mm run F v = F sv - F tvF sv = P /L = 10000/270 = 37 N/mm runF tv = M.r wh /J u = 2,23.106.27/1,04.106 = 57,9 N/mm runF v = 20,86 N/mm run The horizontal force /mm run = F h = F sh + F thF sh = 0F 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 v2) = 232,5 N/mm runThe maximum specific is greatest at z = 253 N/mm run..Referring to weld capacities for longitudinal stresses PL for fillet welds Capacities of Fillet Weldsthe weld capacity for a 3mm weld with and E35 Electrode S275 Steel is 462N /mm run. This weld would be more than sufficient.

Example of Weld in Bending.P= 30000 Newtons
d= 100mm
b= 75mm
y = 50mm
Design Stress p w = 220 N/mm 2 (Electrode E35 steel S275) Design Strength
Moment = M = 30000*60=18.10 5 Nmm
 Simple Method as BS 5950 clause 6.8.7.2Unit Weld Area = A u = 2(d+b) =2(100+75) =350mm 2Unit Moment of Inertia = I u = d 2(3b+d) / 6 = 100 2 (3.75 +100) / 6 =5,42.105mm4τr = Sqrt(τs2 + τb2)τs = P / A u = 30000/350 = 85,71 N/mm 2τb = M.y / I u = 18.105 . 50 / 5,42.105 = 166,05 N/mm 2τr = Sqrt(85,71 2 + 166.052) =186,86 N/mm2τr / p w = 186,86 / 220 = 0,85 = Throat Thickness...( Throat thickness for τ<>< sub=' /> = 220 N/mm2 )Leg Length = Throat thickness *1,414 = 1,2mm use 3mm weld thicknessNote : If a leg length h= 1,2mm is used in the equations in relevantpart of the ' table="" of="" bracket="" weld="" subject="" to="" direct="" and="" bending="" stresses'="" abovea="" value="" of="">τb = 198 N/mm and a valueof τs = 100 N/mm2 results with a resultantstress of Sqrt (τ b2 + τs2)= 222N/mm2.Which is in generalagreement with the above result<> Direction Method as BS 5950 clause 6.8.7.3Length of Weld of unit thickness = L = 2(d+b) =2(100+75) =350mm Moment of Inertia / mm throat thickness = I u / mm = d 2(3b+d) / 6= 100 2 (3.75 +100) / 6 =5,42.105mm 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 b2)F s = P / L = 30000/350 = 85,71 N per mm length of weldF b = M.y / I u = 18.105 . 50 / 5,42.105 = 166,05 N per mm length of weldF r = Sqrt(85,71 2 + 166.052) =186,86 N per mm length of weld. For this case for the welds under greatest loading the type of loading is traverse loading.The bending stress is in line with horizontal element and 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 throat θ = 45o + 27,5o = 72,5oReferring to weld capacities table below.Weld Capacities K is calculated at 1,36 for this resultant direction of forces.PT = a.K.pwfor a E35 Weld electrodeused with S275 steel pw = 220 N/mm2 and therefore PT = a*300N/mm2.A 3mm weld (a = 2,1mm) therefore will therefore have a design capacity of 630 N/mm run and will easily be able tosupport the load of 186,86 N per mm run

Properties of weld groups with welds treated as lines -

It is accepted that it is reasonably accurate to use properties based onunit weld thickness in calculation to determine the strength of welds as shown in the examples onthis page. The weld properties Ixx Iyy and J are assumedto be proportional to the weld thickness. The typical accuracy of this method ofcalculation is shown below..

This is illustrated in the tabled values below

 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: The error identified with this method is lower as h increasesrelative to d. This error is such that the resulting designs are conservative.

Example illustrating use of stress vectors
 Calculations based on unit valuesThis calculation uses equations from table below for Area, centroid, and Ju1) Area of weld = 0,707.5.(2b+d) Area = 0,707.5.(2.55 + 50) = 565.6mm22) There is no need to calculate the Moment of Area with this method3) The centroid v = b2 /(2b+d) v = 552/(2.55+50)= 18,9mm 4) The radii rA, rB, rC & rD are calculated .rA = rB = Sqrt ((55-18,9)^2 + 25^2 ) = 43,9rC = rD = Sqrt (18,9^2 + 25^2 ) = 31,345) The angles θA, θB, θC & θD are calculated .θA = θB = tan-1 ((55-18,9)/ 25 ) = 55,29oθC = θD = tan-1 ((18,9)/25 ) = 37o..6) The direct shear stress on the area = Force /Areaτ S = 5000/565,5 = 8,84 N/mm27) The Moment on the weld group = Force .distance to centroid M = 5000.(100+18,9) = 5.94.105Nmm 8) The Unit Polar moment of inertia of the weld group = Ju = 0.707.5.(8.b3 +6bd2+d3)/12 + b4/(2b+d)Ju = 0,707.5.(8.553+6.55.502 + 503)/12 - 553/(2.55+50) = 4,69.1059) The stress due to torsion τ TA =τ TB = M.rA/J. and .τ TC =τ TD = M.rC/Jτ TA = 5,94.105Nmm.43,9mm / 4,69.105mm4=55,6 N/mm2τ TC = τ TD = 5,945Nmm.31,34mm / 4,69.105mm4= 39,69N/mm210) The resultant stresses τ RA, = τ RB and τ RA, = τ RBare obtainedby adding the stress vectos graphically as shown belowτ RA = τ RB=48,59 N/mm2τ RC = τ RD = 45,56N/mm2

Note: The example above simply illustrates the vector method adding direct and torsional shear stresses and compares the differencein using the unit weld width method and using real weld sizes. The example calculates the stress levels in an existing weld groupit is clear that the weld is oversized for the loading scenario. The difference in the resulting values are in less than 4%. If the welds were smalleri.e 3mm then the differences would be even smaller.

Table properties of a range of fillet weld groups with welds treated as lines -
 Weld Throat AreaUnit Area Location of COG xy I xx-(unit) J-(Unit) - -

Table Of Weld Capacities

{ (FL/PL) 2 + (FT/PT) 2 } 1

The following table is in accord with data in BS 5950 part 1. Based ondesign strengths as shown in table below .. Design Strength

PL = a.pw
PT = a.K.pw
a = weld throat size.
K =1,25 (1,5 / (1 + Cos 2θ )

PT based on elements transmitting forces at 90o i.e θ = 45o and K = 1,25

 Weld Capacity E35 Electrode S275 Steel Weld Capacity E42 Electrode S355 Steel Leg Length Throat Thickness Longitudinal Capacity Transverse Capacity Leg Length Throat Thickness Longitudinal Capacity Transverse Capacity P L(kN/mm) P T (kN/mm) P L P T 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
 Electrode classification Steel Grade 35 43 50 N/mm2 N/mm2 N/mm2 S275 220 220 220 S355 220 250 250 S460 220 250 280  ### Minimum Fillet Weld Chart

 Useful Related LinksGowelding.A Real Find.this site has lots ofInformation on calculating the strength of WeldsWeld Design Notes .A set of excellent design notes
Formulae_Index

Nomenclature