Tear out of welded connection - Failure path inclined to force

[CTJ6]

thread507-445179

Hi,
For the tear out calculation of the connected member in a welded connection, would it be reasonable to use the s^2/(4g) formula (AISC 360-16 B4.3b. or CSA S16 12.3.1) for the segment of the failure path that is inclined to the force. The formula is made for bolted connection but I am wondering if one could use it in a welded connection as it would result in the same failure path (please see the attached sketch). Of course there are no holes to deduct from the area...
Thanks!
[URL unfurl="true"]https://res.cloudinary.com/engineering-com/image/upload/v1727879101/tips/Tear_out_for_welded_connection_ensf1m.pdf[/url]

 

[swearingen]

The short answer is no - for many reasons. A few of them:
- How would you define g, the transverse center-to-center between fastener gage lines? Welds do not have gage lines.
- How would you define s, the longitudinal center-to-center spacing of any two consecutive bolt holes? Welds do not have bolt holes.
- The formula is additive to account for the fact that although you may have reduced the cross section by removing bolt hole material, those bolt holes are not lined up perpendicular to the line of action and are not a wholesale reduction in total area. This does not apply to welds.


-5^2 = -25 ;-)

 

[CTJ6]

Hi swearingen,
Thanks for your response.

My assumption for the values g and s are shown on the attached sketch in the original post. My understanding was that the s^2/4g was to find the effective tension component of an inclined segment in a block shear pattern failure because there is shear and tension on the same segment. I may be wrong.

Otherwise do you have any recommandations to calculate the tension component on an inclined segment?

Thanks!
Collin