Weld between Web and flange shear flow Built up box

[msboobathi]

Dear All
I am designing the weld for Built up box RHS 300x120x16x12 ( subjected to Axial force +small flexure)

The shear capacity = 2*0.6*0.9*fy*t*D = 2*0.6*0.9*355*12*300 = 1380 kN
Equivalent horizontal shear =Vmax*1000 *AF/I = 1380*1000N*[120X16]*150/116009344 = 3426 kN
Required fillet weld = 3426 kN /(2*0.707X217 N/mm2) = 11.15 mm almost equal to wall thickness

Is this correct ? I doubt there shall be less weld thickness required

please suggest me good reference regarding this

Regards
M.S.Boobathi

 

[Agent666]

I'd design the weld for the actual maximum shear force from your analysis, not the shear capacity? I'm not aware of any standards requiring you to design for the full shear capacity of the member.

 

[phamENG]

The only reason to design for shear capacity rather than shear demand would be for future flexibility. If this is going in a building and there's a chance that an addition, renovation, or other modification in the future may place greater shear demand on some section of the beam. It makes it easier for the structural engineer 40 years from now to estimate the capacity by just looking at the size of the plates and ensures your beam will "hold together" and provide a more ductile failure mode if overloaded.

If neither of these are a concern, then design for shear demand (with appropriate resistance factors or factors of safety as your jurisdiction requires) rather than full capacity as Agent666 suggested.

 

[msboobathi]

Thanks for All. It is roof skylight member already surrounded by Permanent building -No possibility for expansion
The vertical shear max is only 400 mm i can mange with 5 mm(Wall thickness 12mm) Partial penetration Butt weld . Where as capacity demands almost equal to FPBW

Regards
M.S.Boobathi

 

[msboobathi]

Dear All
Thanks. I have few more questions on this
1) it is a built up box of RHS 300x120x16x12
2) Do we need to consider Axial force in the member . If not why. Any references for the same
3) If there is torsion how to consider as a part of built up weld

Regards
M.S.Boobathi

 

[KootK]

2) Do we need to consider Axial force in the member .

Sure, unless you're confident that the load is being delivered uniformly to all parts of the cross section. Many engineers will just include some continuous welds at the ends of the member to deal with any axial. You know, so long as the axial loads actually come in at the end of the member.

3) If there is torsion how to consider as a part of built up weld

Computed torsional stress multiplied by wall thickness will create another shear flow demand.

 

[KootK]

Even the need to restrain euler bucking creates a demand on the welds. No much though.

 

[msboobathi]

Thanks Mr.Kootk

1.The section is supporting beam. Refer the sketch attached
2.The weld used is 7 mm Partial Penetration Butt weld continuously
Design of weld between web and Flange RHS 300x120x16x12

3).This is beam supporting secondary beams in diagonal pattern of concave pattern. The member is continuous. I have explained the approach in sketch. I request experts to share their views

4)Torsion
Shear Stress = Ƭ=[Torsion/2bht] Where b= breadth;=Depth; t=thickness.
Shear force= Ƭ*t
This will be added with Minor shear -week axis shear weld demand shall be calculated