The 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

A generous factor of safety should be used (35) 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 weld
The table below provides provides approximate stresses in, hopefully, a convenient
way.
For the direct loading case the butt weld stresses are tensile/ compressive σ _{t} for the fillet welds
the 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 σ _{b} and a direct shear
stress τ _{s} .
In these cases the design basis stress should be σ _{r} = Sqrt (σ _{b}^{2} + 4τ _{s}^{2})
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 τ _{r} =Sqrt (τ _{b}^{2} + τ _{s}^{2})
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.
Method of Loading  Weldment  Stress in Weld σ _{b} τ _{s} Weld size (h) 
Weldment  Stress in Weld σ _{b} τ _{s} Weld size (h) 
Weldment  Stress in Weld τ _{b} τ _{s} Weld size (h) 
Important 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 identified 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 (F_{T} ) and
longitudinal forces (F_{L} ). 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 A _{u}..(unit thickness) is calculated assuming the weld
is one unit thick..
2) The (unit) Moment of Inertia I _{u} is calculated assuming the weld
is one unit thick..
3) The maximum shear stress due to bending is determined...τ _{b} = M.y/I _{u}
4) The maximum shear stress due to direct shear is determined.. τ _{s} = P /A
5) The resultant stress τ _{r} = Sqrt (τ _{b}^{2} + τ _{s}^{2} )
6) By comparing the design strength p _{w} with the resultant stress τ _{r} the value of the weld throat thickness is calculated and then the weld size.
i.e. if the τ _{r} /p _{w} = 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 A _{u} (unit thickness) is calculated assuming the weld
is one unit thick
2) The (unit) Polar Moment of Inertia J _{u} is calculated assuming the weld
is one unit thick..
The polar moment of inertia J = I _{xx} + I _{yy}
3) The maximum shear stress due to torsion is determined...τ _{t} = T.r/J _{u}
4) The maximum shear stress due to direct shear is determined.. τ _{s} = P /A _{u}
5) The resultant stress τ _{r} is the vector sum of τ _{t} and τ _{s}. r is chosen to give
the highest value of τ _{r}
6) By comparing the design strength p _{w} with the resultant stress τ _{r} the value of the weld throat thickness is calculated and then the weld size.
i.e. if the τ _{r} /p _{w} = 5 then the throat thickness t = 5 units
and the weld leg size h = 1,414.t
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 = b^{2} / 2(b+d) = 27mm.. (From table below) y = d^{2} / 2(b+d) = 42mm..(From table below)

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

It is accepted that it is reasonably accurate to use properties based on unit weld thickness in calculation to determine the strength of welds as shown in the examples on this page. The weld properties I_{xx} I_{yy} and J are assumed to be proportional to the weld thickness. The typical accuracy of this method of calculation is shown below...
This is illustrated in the tabled values below
d  b  h  I_{xx}  I_{yy}  J= I_{xx} +I_{yy}  
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 increases relative to d. This error is such that the resulting designs are conservative.
Calculation based on real weld sizes 1) The area of the welds Area = (57.3,5 + 2.55.3,5) = 584,5mm^{2} 2) The moment of area about xx = M of Area = (57.3,5.3,5/2 + 2.55.3,5.(27.5 + 3,5)) = 12 284mm^{3} 3) The centroid v = Moment of Area/Area M of Area / Area = 21 mm 4) The radii r_{A}, r_{B}, r_{C} & r_{D} are calculated .. r_{A} = r_{B} = Sqrt ((58,521)^2 + 28,5^2 ) = 47,1 5) The angles θ_{A}, θ_{B}, θ_{C} & θ_{D} are calculated .. θ_{A} = θ_{B} = tan^{1} ((58,521)/ 28,5 ) = 52,7^{o} 6) The direct shear stress on the area = Force /Area τ _{S} = 5000/584 = 8,56 N/mm^{2} 7) The Moment on the weld group = Force.Distance to centroid M = 5000.(100+21) = 6,05.10^{5}Nmm 8) The polar moment of inertia of the weld group = J = I_{xx} + I_{yy}
I_{yy} = 2.[55.3,5^{3} /12 + 3,5.55.(50/2 + 3,5/2)^{2}] 9) The stress due to torsion τ _{TA} = 6.05^{5}Nmm.47,1mm / 5,4.10^{5}mm^{4}=52,8 N/mm^{2} 10) The resultant stresses τ _{RA}, = τ _{RB} and τ _{RA}, = τ _{RB} τ _{RA} = τ _{RB}=46,29 N/mm^{2} 
Calculations based on unit values This calculation uses equations from table below for Area, centroid, and J_{u} 1) Area of weld = 0,707.5.(2b+d) Area = 0,707.5.(2.55 + 50) = 565.6mm^{2} 2) There is no need to calculate the Moment of Area with this method 3) The centroid v = b^{2} /(2b+d) v = 55^{2}/(2.55+50)= 18,9mm 4) The radii r_{A}, r_{B}, r_{C} & r_{D} are calculated .. r_{A} = r_{B} = Sqrt ((5518,9)^2 + 25^2 ) = 43,9 5) The angles θ_{A}, θ_{B}, θ_{C} & θ_{D} are calculated .. θ_{A} = θ_{B} = tan^{1} ((5518,9)/ 25 ) = 55,29^{o} 6) The direct shear stress on the area = Force /Area τ _{S} = 5000/565,5 = 8,84 N/mm^{2} 7) The Moment on the weld group = Force .distance to centroid M = 5000.(100+18,9) = 5.94.10^{5}Nmm 8) The Unit Polar moment of inertia of the weld group = J_{u} = 0,707.5.(8.55^{3}+6.55.50^{2} + 50^{3})/12  55^{3}/(2.55+50) = 4,69.10^{5}
9) The stress due to torsion τ _{TA} = 5,94.10^{5}Nmm.43,9mm / 4,69.10^{5}mm^{4}=55,6 N/mm^{2} 10) The resultant stresses τ _{RA}, = τ _{RB} and τ _{RA}, = τ _{RB} τ _{RA} = τ _{RB}=48,59 N/mm^{2} 
Note: The example above simply illustrates the vector method adding direct and torsional shear stresses and compares the difference in using the unit weld width method and using real weld sizes. The example calculates the stress levels in an existing weld group it 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 smaller i.e 3mm then the differences would be even smaller.
Weld  Throat Area Unit Area 
Location of COG x y 
I _{xx}(unit)  J(Unit) 
  
 
The fillet weld capacity tables related to the type of loading on the weld. Two types of loading are identified traverse loading and longitudinal loading as show below
The weld loading should be such that
{ (F_{L}/P_{L}) ^{2} + (F_{T}/P_{T}) ^{2} } ≤ 1
The following table is in accord with data in BS 5950 part 1. Based on design strengths as shown in table below ... Design Strength
P_{L} = a.p_{w}
P_{T} = a.K.p_{w}
a = weld throat size.
K =1,25 √ (1,5 / (1 + Cos ^{2} θ )
P_{T} based on elements transmitting forces at 90_{o} i.e θ = 45_{o} 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 
Electrode classification  
Steel Grade  35  43  50 
N/mm^{2}  N/mm^{2}  N/mm^{2}  
S275  220  220  220 
S355  220  250  250 
S460  220  250  280 