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AISC Double Angle Weld Capacity

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jleventh

Structural
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Jun 18, 2013
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Hello, this is my first time posting on Eng-Tips, I hope you guys can help me out.

I have a question regarding the derivation of the weld capacity for double angle connections per AISC 13th edition page 10-11. I am trying to figure out where the 12.96 coefficient comes from. According to the book, the neutral axis is assumed to be (1/6)*L from the top of the angle, and the tops of the angles act in compression against each other.

If we call the vertical reaction R, then the form of the equation should be 0.928*D*L=sqrt(R^2+(R*e/a)^2) where "a" is the moment arm between the compression and tension resultants. Solving for R you get R=0.928*D*L/sqrt[1+(e^2/a^2)]. Back-solving to find "a" based on the 12.96 value, I have found that the moment arm should be (5/18)*L, but I can't come up with any distribution of loading that results in that moment arm. If the return is ignored and the load is triangularly distributed in both the tension and compression zones, I find that a=(2/3)*(L/6)+(2/3)*(5*L/6)=(2/3)*L. If the compression is a point reaction at the top of the weld and tension is a triangular distribution, a=(L/6)+(2/3)*(5*L/6)=(13/18)*L.

I hope that I'm on the right track and just missing something obvious.

Thanks in advance for any help!
 
Blodgett's Design of Welded Structures has a derivation of this equation. Salmon and Johnson do as well.
 
Thank you! I had actually looked in Blodgett when starting to try and figure this out and must have skipped right over the chapter. Do you know where the assumption of an L/6 compressive bearing length comes from? It doesn't state in Blodgett or AISC, and I don't have access to Salmon and Johnson.
 
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