When calculating the factor of safety for a fillet weld that is loaded in shear what should be the maximum allowable stress?

Some books suggest using the distortion energy (Ssy = 0.577Sy for the maximum allowable stress) while others use maximum shear stress theory (Ssy = 0.3Sut for the maximum).

Are both of these ways correct? Which one is used more in the field?

  • 3
    $\begingroup$ Maximum allowable stress and/or factor of safety should be something that comes from the Code that you are using. Each code may make different trade offs. $\endgroup$
    – hazzey
    Commented Apr 19, 2016 at 14:23
  • $\begingroup$ Are these welds in steel? $\endgroup$
    – Ethan48
    Commented Aug 9, 2016 at 13:11

2 Answers 2


A lot depends on the loading situation and the quality standard of the weld. As mentioned in another answer the yield stress of the filler metal is often higher than that of the base metal.

However this needs to be qualified by the fact that the welding process can, in some circumstances, modify the material properties of the base metal in the heat affected zone (especially for cold worked or high alloy steels).

There is also the question of weld defects such as incomplete penetration, cold starts, inclusions, porosity etc etc as well as the fact that some alloys require specific pre and/or post weld treatment.

In very general terms weld defects are often related to crack propagation problems and so are more of a concern with structures where fatigue and other cyclic or impact related modes of failure are a concern.

Critical welding applications are well covered by application/industry specific standards which cove the whole process as well as testing and inspection procedures.

For good quality welds in low alloy steels with good penetration and no major defects it is generally reasonable to consider a welded joint as if it was a continuous piece of metal unless you're working to a code which states otherwise.

Having said that I recently quoted for a maintenance job for aluminium alloy generator cages where welds are a known failure point.

However this is all highly contextual which is why we have specific welding codes which define procedures and testing protocols and specific situational rather than having a single fudge factor for 'welds'.

  • $\begingroup$ good answer, your alloy cage welds failing may be affected by cyclic fatigue as well due to the vibrations - not a simple case then. Oh, good luck! $\endgroup$
    – Solar Mike
    Commented Jun 5, 2017 at 5:05

For most cases, yield strength (and also thickness) of weld metal is equal or higher than the yield stress (and also thickness) of base metal. In practical we may think they're equal. Because it's pointless to make calculations for weld metal which is 300 MPa when base metal yields in 235 MPa.

There is a document called PQR (Procedure Qualification Record). In PQR, company makes both non-destructive and destructive tests on a welded specimen and use this document as proof for it's qualified welds. You can read yield and tensile stress values on PQR. You may find values of weld metal may be lower than values of base metal. But that's an exceptional. Let's think weld metal is at least as strong as the adjoining base metal.

But the point is "welding" is a service provision which is defined in ISO 9001 7.5.2. That means you can never be sure welding process is grateful even if you have some controls and tests.

So, 0.3 Sut or 0.577 Sy they're all attitudes for precaution and they're all true. Sorry, I don't know which one is used more in field.


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