BTH-1 Design of BTH devices - how to evaluate "strenght of welds"

[ColonelSanders83]

Hello All,

I am currently evaluating a bolt on lifting device. I am attempting to evaluate one of the welds and I have gotten myself confused.

BTH-1 para 3-3.4.1(b) states "The design strength of welds subject to shear shall be equal to the effective area of the weld multiplied by the allowable stress Fv given by eq (3-53). Stresses in the base metal shall not exceed the limits defined in para. 3-2)

where Fv=(0.6*EXX)/(1.2*Nd)=17500 psi (E70 rods and Nd = 2)

I have a 5" SCH XXS pipe with a full penetration groove at its base joining to a 1.25" plate, according to 3-3.4.2 this give an effective throat of 0.75", an effective length of 13.35" and an effective area of 10.01 in^2.

If I follow the code, I multiply the two together and get a force of 175000 pounds.

1) What am I supposed to do with this?
2) Where does the actual shear load the weld sees come into play?

3) Should I multiply the effective area by the actual shear load and then compare that value with Fv? That would be consistent with the code and would make far more sense.

Any thoughts are appreciated.


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[JStephen]

The equation referenced gives the allowable shear load for a weld subject to shear. If the weld in question isn't subject to shear, it doesn't matter what the weld strength in shear is. If it is subject to shear, you compare the actual shear force to the design shear strength.

Multiplying area times actual load and comparing to allowable load would be pretty meaningless.