Weld Strength CalculationsIntroduction.....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.....IntroductionThe following notes are general guidance notes showing methods of calculation of the strength and size of welds.Welded joints are often crucially important affecting the safety of the design systems.It is important that the notes and data below are only used for preliminary design evaluations.Final detail design should be completed in a formal way using 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 sectionsBS EN 10025-1:2004 - Hot rolled products of structural steels. General technical delivery conditions

Variables related to welded joints1. Strength of deposited weld material2. Type of joint and weld..important3. Size of weld ..important4. Location of weld in relation to parts joined..important5. Types of stress to which the weld is subjected6. Conditions under which weld is carried out7. Type of equipment used for welding8. Skill of welder

Guidance PrinciplesA generous factor of safety should be used (3-5) and if fluctuating loads 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 weldThe table below provides provides approximate stresses in, hopefully, a convenient way.

For the direct loading case the butt weld stresses are tensile/ compressivetfor the fillet welds the stresses are assumed to be shearsapplied to the weld throat.

For butt welded joints subject to bending the butt weld stresses result from a tensile/compressive stressband a direct shear stresss.In these cases the design basis stress should ber= Sqrt (b2+ 4s2)

For Fillet welded joints subject to bending the stresses in the fillet welds are all shear stresses. From bendingband from shearsIn these cases the design basis stress is generallyr=Sqrt (b2+s2)

The stresses from joints subject to torsion loading include shear stress from the applied load and shear stresses from the torque loading.The resulting stresses should be added vectorially taking care to choose the location of the highest stresses.

Table of bracket weld subject to direct and bending stressesMethod of LoadingWeldmentStress in WeldbsWeld size (h)WeldmentStress in WeldbsWeld size (h)WeldmentStress in WeldbsWeld size (h)

Assessment of Fillet Weld Groupsref notes and tableProperties of Fillet Welds as linesImportant note: The methods described below is based on the simple method of calculation of weld stress as identified in BS 5950- clause 6.7.8.2 . The other method identifed in BS 5950 - 1 clause 6.7.8.3 as the direction method uses the method of resolving the forces transmitted by unit thickness welds per unit length into traverse forces (FT) and longitudinal 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 reasonably safe 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 Au..(unit thickness) is calculated assuming the weld is one unit thick..2) The (unit) Moment of Inertia Iuis calculated assuming the weld is one unit thick..3) The maximum shear stress due to bending is determined...b= M.y/Iu4) The maximum shear stress due to direct shear is determined..s= P /A5) The resultant stressr= Sqrt (b2+s2)6) By comparing the design strength pwwith the resultant stressrthe value of the weld throat thickness is calculated and then the weld size.i.e. if ther/pw= 5 then the throat thickess t = 5 units and 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 Au(unit thickness) is calculated assuming the weld is one unit thick2) The (unit) Polar Moment of Inertia Juis calculated assuming the weld is one unit thick.. The polar moment of inertia J = Ixx+ Iyy3) The maximum shear stress due to torsion is determined...t= T.r/Ju4) The maximum shear stress due to direct shear is determined..s= P /Au5) The resultant stressris the vector sum oftands.r is chosen to give the highest value ofr6) By comparing the design strength pwwith the resultant stressrthe value of the weld throat thickness is calculated and then the weld size.i.e. if ther/pw= 5 then the throat thickess t = 5 units and the weld leg size h = 1,414.t

Examples of Fillet Weld CalculationsExample of Weld in Torsion..

P = Applied load = 10 000NPw= Design Strength = 220 N/mm2(Electrode E35 steel S275)Design Strengthb = 120mm.d = 150 mmx= 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 design strength Pw

Au= 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 = 2,23.106N.mmJu= [(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 torsion this would be simply at the point furthest from the COG. However because the weld is subject to torsion and direct shear the problem is more complicated.A normal method of determining 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 stresses in the x and y directions

First considering point Z

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

The vertical stressv=sv+tvsv= P /Au= 10000/270 = 37 N/mm2tv= M.rzh/Ju= 2,23.106.93/1,04.106= 199 N/mm2v= 236,45 N/mm2

The horizontal stressh=sh+thsh= 0th= M.rzv/Ju= 2,23.106.42/1,04.106= 90 N/mm2h= 90 N/mm2The resultant stress on the weld at zr= Sqrt (h2+v2) = 253 N/mm2

Now considering point w

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

The vertical stressv=sv-tvsv= P /Au= 10000/270 = 37 N/mm2tv= M.rwh/Ju= 2,23.106.27/1,04.106= 57,9 N/mm2v= 20,86 N/mm2

The horizontal stressh=sh+thsh= 0th= T.rwv/Ju= 2,23.106.108/1,04.106= 231,6 N/mm2h= 231,6 N/mm2The resultant stress on the weld at wr= Sqrt (h2+v2) = 232,5 N/mm2

The maximum stress is similar but greatest at z ....The design strength pwfor the weld material is 220 N/mm2The weld throat thickness should be 253 /220 = 1,15mm .The weld size is therefore 1,414. 1,15 = 1,62mm use 3mm fillet weldDirection 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 = 2,23.106N.mmJu= Polar Moment of inertia for weld 1unit(mm) thick.= [(b+d)4- 6b2d2] /12 (b+d) = 1,04.106mm4/mm..(From Table)

It is necessary to locate the point subject to the highest shear stress..For a weld subject to only torsion this would be simply at the point furthest from the COG. However because the weld is subject to torsion and direct shear the problem is more complicated.A normal method of determining 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 stresses in the x and y directions

First considering point Z

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

The vertical force /mm run Fv= Fsv+ FtvFsv= P /L = 10000/270 = 37 N/mm runFtv= M.rzh/Ju= 2,23.106.93/1,04.106= 199 N/mm runFv= 236,45 N/mm run

The horizontal force /mm run for unit(mm) weld width Fh= Fsh+ FthFsh= 0Fth= M.rzv/Ju= 2,23.106.42/1,04.106= 90 N/mm runFh= 90 N/mm runThe resultant force on the weld/mm run at zFr= Sqrt (Fh2+ Fv2) = 253 N/mm run

Now considering point w

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

The vertical forces per mm run Fv= Fsv- FtvFsv= P /L = 10000/270 = 37 N/mm runFtv= M.rwh/Ju= 2,23.106.27/1,04.106= 57,9 N/mm runFv= 20,86 N/mm run

The horizontal force /mm run = Fh= Fsh+ FthFsh= 0Fth= M.rwv/Ju= 2,23.106.108/1,04.106= 231,6 N/mm runFh= 231,6 N/mm runThe specific force on the weld at wFr= Sqrt (Fh2+ Fv2) = 232,5 N/mm run

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

Referring to weld capacities for longitudinal stresses PLfor fillet weldsCapacities 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 Newtonsd= 100mmb= 75mmy= 50mmDesign Stress pw= 220 N/mm2(Electrode E35 steel S275)Design StrengthMoment = M = 30000*60=18.105NmmSimple Method as BS 5950 clause 6.8.7.2Unit Weld Area = Au= 2(d+b) =2(100+75) =350mm2Unit Moment of Inertia = Iu= d2(3b+d) / 6 = 1002(3.75 +100) / 6 =5,42.105mm4

r= Sqrt(s2+b2)s= P / Au= 30000/350 = 85,71 N/mm2b= M.y / Iu= 18.105. 50 / 5,42.105= 166,05 N/mm2r= Sqrt(85,712+ 166.052) =186,86 N/mm2r/ pw= 186,86 / 220 = 0,85 = Throat Thickness.....( Throat thickness for= 220 N/mm2)Leg Length = Throat thickness *1,414 = 1,2mm use 3mm weld thickness

Note : If a leg length h= 1,2mm is used in the equations in relevant part of the "Table of bracket weld subject to direct and bending stresses" above a value ofb= 198 N/mmand a value ofs= 100 N/mm2results with a resultant stress of Sqrt (b2+s2) = 222N/mm2..Which is in general agreement with the above result