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# Simple (dumb) question regarding weld design

## Simple (dumb) question regarding weld design

(OP)
Hey folks.
For whatever reason, weld sizing has always been kind of unclear to me. I'm using ASD in the red (14th) Steel Manual and have a couple questions:
1) EQ.8-5b and EQ.8-8b states that they calculates the "shear per linear inch" due to concentric load and moment; I take it that these values would have units of FORCE/LENGTH. Fair enough, but then EQS.8-9b, 8-10b and 8-11b calculate the "resultant force" which I take to mean a value with units of FORCE. They appear to have the same basic form so I'm wondering what units the results of EQS.9/10/11 are actually in?

2) I'm assuming that with EQS.9/10/11, I'm actually getting a result in a FORCE/LENGTH unit. It sounds stupid, but how do I then compare that to an allowable force as calculated by following Chapter J and Table J2.5? Is it as simple as multiplying my resultant FORCE/LENGTH value from EQ.8-11b by the length of weld and comparing that to the available strength? Alternatively, in calculating the available strength (0.60*Fexx*Ae), I guess I could use a unit length in the effective weld area and then I'm calculating the available strength per linear unit length which I can compare directly to the results of EQ-11b.

Thanks in advance and have a great day!

### RE: Simple (dumb) question regarding weld design

Hi

I haven’t got access to the reference you are talking about however if your weld strength is units of force/length then yes just multiply that figure by the length of the actual weld.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

### RE: Simple (dumb) question regarding weld design

(OP)
Thank you !

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