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Category : Eurocode 3 - Steel structures - Part 1-8: Design of joints

Last answer on : 2021-06-14 20:48

Last answer on : 2021-06-14 20:48

Posted by Oliver G. on 2021-05-27 22:54

Hey,

thanks for providing your online calculator...regarding lifting lugs, how do you calculate the effective weld length L(w,eff) and F (w, Ed, includes eccentricity effects)? I use to calculate the angle projection of 2R on the weld as the effective length.

Would be great to see the formula just to clarify the weld.

Best,

Oliver

thanks for providing your online calculator...regarding lifting lugs, how do you calculate the effective weld length L(w,eff) and F (w, Ed, includes eccentricity effects)? I use to calculate the angle projection of 2R on the weld as the effective length.

Would be great to see the formula just to clarify the weld.

Best,

Oliver

Posted by Stephane P. on 2021-06-09 18:30

Hi Oliver,

Sorry for my late answer.

I'm glad you like this tool.

In the software (version 01-0209):

- the effective length of the weld is calculated as follows:

L_{w,eff} = min(H_{1}; L_{1} - a_{w}) + min(H_{1}; L - L_{1} - a_{w})

- the tensile stress of the weld is calculated as follows:

F_{v} = F * sin(angle)

R_{z,Vl} = 0.5 * F_{v} / min(L - L_{1} - a_{w}; H_{1})

R_{z,Vr} = 0.5 * F_{v} / min(L_{1} - a_{w}; H_{1})

F_{H} = F * cos(angle)

M_{FH} = F_{H} * H_{1}

R_{z,H} = M_{FH}/ (0.5 * L * (L - L_{1}))

R_{z} = max( R_{z,Vl} + R_{z,H}; R_{z,Vr})

- the shear stress of the weld is calculated as follows:

R_{x} = F_{H} / L_{w,eff}

If you have a more efficient solution, of course I am interested!

;)

See you

Stéphane

Sorry for my late answer.

I'm glad you like this tool.

In the software (version 01-0209):

- the effective length of the weld is calculated as follows:

L

- the tensile stress of the weld is calculated as follows:

F

R

R

F

M

R

R

- the shear stress of the weld is calculated as follows:

R

If you have a more efficient solution, of course I am interested!

;)

See you

Stéphane

Posted by Oliver G. on 2021-06-14 15:57

Hey Stéphane,

thanks for your detailed answer, very much appreciated.

I can follow your calculation, but have still issues with finding this in the Eurocode, especially the calculation of the effective length...and the impact of "H1" on the effective weld length.

Maybe this exceeds the scope of this forum but I just cannot find it.

best,

Oliver

thanks for your detailed answer, very much appreciated.

I can follow your calculation, but have still issues with finding this in the Eurocode, especially the calculation of the effective length...and the impact of "H1" on the effective weld length.

Maybe this exceeds the scope of this forum but I just cannot find it.

best,

Oliver

Posted by Stephane P. on 2021-06-14 16:29

Hi Oliver,

H_{1} is taken into account because a diffusion of the force at 45 ° on either side of the axis is considered in the calculation (or less if H_{1} is greater than the horizontal parts L_{1} minus the ineffective part of the weld).

Stéphane

H

Stéphane

Posted by Stephane P. on 2021-06-14 16:42

The shear stress of the weld should probably be calculated as Rx = F_{H} / (L - 2 * a_{w}).

Posted by Oliver G. on 2021-06-14 20:48

Hey,

Aaaahh..

- You have a limiting case if H1 = L1 at a 90 degrees vertical lifting lug, these two lengths form an isosceles triangle with the 45 angle

- If you increase H1 then the L's will take over as the decisive length since the 45 degrees line leaves the outer shape of the padeye

- If you decrease H1 below this limiting point then H1 is still one of the 45 degrees triangle legs and determines the "new" L1 *

- At asymmetric padeyes it becomes a little bit more complicated, but at 45 degrees force angle you have again H1= "new" L1, with H1 now being the other triangle leg.

Correct?

I do agree with the shear stress formula, with the welds around all corners we could also drop the 2 x aw maybe.

Thanks for your help! Oliver

Aaaahh..

- You have a limiting case if H1 = L1 at a 90 degrees vertical lifting lug, these two lengths form an isosceles triangle with the 45 angle

- If you increase H1 then the L's will take over as the decisive length since the 45 degrees line leaves the outer shape of the padeye

- If you decrease H1 below this limiting point then H1 is still one of the 45 degrees triangle legs and determines the "new" L1 *

- At asymmetric padeyes it becomes a little bit more complicated, but at 45 degrees force angle you have again H1= "new" L1, with H1 now being the other triangle leg.

Correct?

I do agree with the shear stress formula, with the welds around all corners we could also drop the 2 x aw maybe.

Thanks for your help! Oliver