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looking for weld stregth chart 1
- Thread starter marmon
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I think if you get this very inexpensive book "The Design of Weldments" by Omer Blodgett you will have all the information you will ever need about fillet welds.
Get yourself a copy of this excellent reference book on welding;
"Design of Weldments" by Omer Blodgett
This book has charts on allowable stresses for weldments expressed in psi/inch of weld, and fillet weld size. If you deal with welding on a regular basis, you should have it for useful reference material.
"Design of Weldments" by Omer Blodgett
This book has charts on allowable stresses for weldments expressed in psi/inch of weld, and fillet weld size. If you deal with welding on a regular basis, you should have it for useful reference material.
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ok so i have a 8" long piece by 1-1/8" thick by 2-3/8" tall welded to a solid piece along the 8" length both sides. would this piece take 9000 lbf/inch? so 72000lbs?
or .4(Sty)(t) = .4(36000)(1.125)= 16,200lb?. so the plate will yield before the weld shears? The load being applied perpendicular to the weld, what about longitudinal?
Thanks
or .4(Sty)(t) = .4(36000)(1.125)= 16,200lb?. so the plate will yield before the weld shears? The load being applied perpendicular to the weld, what about longitudinal?
Thanks
For simplicity, you could assume the shear strength of the base metal adjacent to the weld leg if the load is perpendicular to your weld is still 0.40(Sty)(t), so that would be, in your case, 0.40(36000)(0.3125)(2 welds)(8) = 72000 lbf. If the load is instead parallel to your weld, the shear strength of the plate through its thickness is 0.40(36000)(1.125)(8) = 129600 lbf (which doesn't govern) and the weldment strength is 4500(2 welds)(8) = 72000 lbf.
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- #7
sorry for the questions as i am new to this
for .4(Sty)(t) you use fillet size for (t).
Now for longitudinal loading the formulae stay the same?
I thought maybe perpendicular loading would take into account bending moments. Taking into account how high the load is applied.
such as PxL=(sigma)(W)(a)(D+a)
W=length of weld
D= plate thickness
L= plate height
a= weld throat
P= load
so substituting a a failure stress in psi and solving for P will get me my failure load?
or am i looking into this too hard
this equation was what i was originally using when i was looking to find the yeild strength of a 5/16" weld
for .4(Sty)(t) you use fillet size for (t).
Now for longitudinal loading the formulae stay the same?
I thought maybe perpendicular loading would take into account bending moments. Taking into account how high the load is applied.
such as PxL=(sigma)(W)(a)(D+a)
W=length of weld
D= plate thickness
L= plate height
a= weld throat
P= load
so substituting a a failure stress in psi and solving for P will get me my failure load?
or am i looking into this too hard
this equation was what i was originally using when i was looking to find the yeild strength of a 5/16" weld
marmon: Correction: I had forgotten that you don't have to check the shear stress on the base metal adjacent to the weld leg (for steel welds), and the first couple of references I had handy at that moment didn't mention it.
Therefore, the allowable strength of the 0.3125 inch fillet weld you mentioned is 813 N/mm (4640 lbf/inch) or 0.40(Sty)(t), whichever is less, where Sty = plate tensile yield strength (250 MPa for your plate), and t = plate thickness.
For simplicity, if the load is perpendicular to your weld, you could assume the weld strength is still 4640 lbf/inch. So if your load were parallel to the plate and perpendicular to the weld, that would be, in your case, 4640(2 welds)(8) = 74240 lbf. (And obviously the plate tensile strength wouldn't govern here.) If the load is instead parallel to your weld, the shear strength of the plate through its thickness is 0.40(36000)(1.125)(8) = 129600 lbf (which doesn't govern) and the weldment strength is 4640(2 welds)(8) = 74240 lbf.
But I now see your perpendicular load is perpendicular to the plate and perpendicular to the weld. So I agree with your equation, except change the "D+a" to just "D". Hence the weld stress due to bending would be sigma_y = P*L/(b*D*a), where b = plate length = 8 inch. But you also have weld stress perpendicular to the plate, which would be sigma_z = P/(2*b*a). So I think that would be sigma = (sigma_y^2 + sigma_z^2)^0.5. Therefore, ensure sigma*b is less than the weld allowable strength.
Therefore, the allowable strength of the 0.3125 inch fillet weld you mentioned is 813 N/mm (4640 lbf/inch) or 0.40(Sty)(t), whichever is less, where Sty = plate tensile yield strength (250 MPa for your plate), and t = plate thickness.
For simplicity, if the load is perpendicular to your weld, you could assume the weld strength is still 4640 lbf/inch. So if your load were parallel to the plate and perpendicular to the weld, that would be, in your case, 4640(2 welds)(8) = 74240 lbf. (And obviously the plate tensile strength wouldn't govern here.) If the load is instead parallel to your weld, the shear strength of the plate through its thickness is 0.40(36000)(1.125)(8) = 129600 lbf (which doesn't govern) and the weldment strength is 4640(2 welds)(8) = 74240 lbf.
But I now see your perpendicular load is perpendicular to the plate and perpendicular to the weld. So I agree with your equation, except change the "D+a" to just "D". Hence the weld stress due to bending would be sigma_y = P*L/(b*D*a), where b = plate length = 8 inch. But you also have weld stress perpendicular to the plate, which would be sigma_z = P/(2*b*a). So I think that would be sigma = (sigma_y^2 + sigma_z^2)^0.5. Therefore, ensure sigma*b is less than the weld allowable strength.
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