| 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
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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
σb
τs
Weld size (h) | Weldment | Stress in Weld
σb
τs
Weld size (h) | Weldment | Stress in Weld
τb
τs
Weld 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 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 - 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 directions
First considering point Z
Horizontal distance from centroid r zh = 120-27= 93mm
Vertical distance from centroid r zv = 42mm
The vertical stress τv = τsv + τtv
τsv = P /A u = 10000/270 = 37 N/mm2
τtv = M.r zh /J u = 2,23.106.93/1,04.106 = 199 N/mm2
τv = 236,45 N/mm2
The horizontal stress τh = τsh + τth
τsh = 0
τth = M.r zv /J u = 2,23.106.42/1,04.106 = 90 N/mm2
τh = 90 N/mm2
The resultant stress on the weld at z
τr = Sqrt (τh2 + τv2) = 253 N/mm2
Now considering point w
Horizontal distance from centroid r wh = 27mm
Vertical distance from centroid r wv = 150-42= 108mm
The vertical stress τv = τsv - τtv
τsv = P /A u = 10000/270 = 37 N/mm2
τtv = M.r wh /J u = 2,23.106.27/1,04.106 = 57,9 N/mm2
τv = 20,86 N/mm2
The horizontal stress τh = τsh + τth
τsh = 0
τth = T.r wv /J u = 2,23.106.108/1,04.106 = 231,6 N/mm2
τh = 231,6 N/mm2
The resultant stress on the weld at w
τr = Sqrt (τh2 + τv2) = 232,5 N/mm2
The maximum stress is similar but greatest at z ..
The design strength p w for the weld material is 220 N/mm 2
The 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 6.8.7.3 L = Length of weld 1 unit thick =
(From table below) b + d = (120 + 150) = 270mm
To obtain radius of Force from weld Centre of Gravity (Cog) .
A = 250-27 =223mm
Moment M = P.r = 10000.223 = 2,23.106 N.mm
Ju = 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 directions
First 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 tv
F sv = P /L = 10000/270 = 37 N/mm run
F tv = M.r zh /J u = 2,23.106.93/1,04.106 = 199 N/mm run
F v = 236,45 N/mm run
The horizontal force /mm run for unit(mm) weld width F h = F sh + F th
F sh = 0
F th = M.r zv /J u = 2,23.106.42/1,04.106 = 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 v2)= 253 N/mm run
Now 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 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 run
F 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 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 Weldsthe weld capacity for a 3mm weld with and E35 Electrode S275 Steel is 462N /mm run. This weld would be more than sufficient. |
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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 6.8.7.2 Unit Weld Area = A u = 2(d+b) =2(100+75) =350mm 2
Unit Moment of Inertia = I u
= d 2(3b+d) / 6 = 100 2 (3.75 +100) / 6 =5,42.105mm4
τr = Sqrt(τs2 + τb2)
τs = P / A u = 30000/350 = 85,71 N/mm 2
τb = M.y / I u = 18.105 . 50 / 5,42.105 = 166,05 N/mm 2
τr = Sqrt(85,71 2 + 166.052) =186,86 N/mm2
τr / 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 thickness
Note : 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 6.8.7.3 Length 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 weld
F b = M.y / I u
= 18.105 . 50 / 5,42.105 = 166,05 N per mm length of weld
F r = Sqrt(85,71 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
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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 vectorsCalculations based on unit values
This calculation uses equations from table below
for Area, centroid, and JuDtp arizona sonoraquest com. 1) Area of weld = 0,707.5.(2b+d) Area = 0,707.5.(2.55 + 50) = 565.6mm2 2) There is no need to calculate the Moment of Area with this method 3) The centroid v = b2 /(2b+d) Hoi4 player led peace conferences download. 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 ) = 43,9
rC = rD = Sqrt (18,9^2 + 25^2 ) = 31,34 5) The angles θA, θB, θC & θD are calculated . θA = θB = tan-1 ((55-18,9)/ 25 ) = 55,29o
θC = θD = tan-1 ((18,9)/25 ) = 37o.. 6) The direct shear stress on the area = Force /Area τ S = 5000/565,5 = 8,84 N/mm2 7) 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) = 4,69.105
9) The stress due to torsion
τ TA =τ TB = M.rA/J. and .τ TC =τ TD = M.rC/J τ TA = 5,94.105Nmm.43,9mm / 4,69.105mm4=55,6 N/mm2
τ TC = τ TD = 5,945Nmm.31,34mm / 4,69.105mm4= 39,69N/mm2 10) The resultant stresses τ RA, = τ RB and τ RA, = τ RB
are obtainedby adding the stress vectos graphically as shown below τ RA = τ RB=48,59 N/mm2
τ RC = τ RD = 45,56N/mm2
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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 Area
Unit Area | Location of COG
x
y | I xx-(unit) | J-(Unit) | | - | | - |
Table Of Weld Capacities
The fillet weld capacity tables related to the type of loading on the weld. Twotypes of loading are identified traverse loading and longitudinal loading as show below The weld loading should be such that { (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 |
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