Reduced design strength of steel section due to shear force 1

[Sjotroll]

Hi, Eurocode 3 defines that if the design shear force is more than half the design shear resistance, then the combined effect of bending and shear is calculated by considering a reduced design strength. This reduced strength is calculated as f[sub]yr[/sub] = f[sub]y[/sub] × (1-ro). However, this reduction should only be made within the shear area. And here comes the problem which is not explained anywhere, how do I calculate one single (representative) design strength of the whole section to calculate the resistance to the combined effect of shear forces and moments? I know that eurocode 3 also defines another more complicated formula for calculating the design moment resistance considering the factor ro, but if I use this formula or apply the previously mentioned one to the whole section, the results are vastly different. The complicated formula barely reduces the strength, while fy(1-ro) greatly reduces it. Yet, I've seen calculations which take the fy(1-ro) for the whole section and go with that. This is obviously conservative, but is it justified to use it like that? What is the true way of calculating the strength of the whole section based on the reduced strength of the shear area and the non-reduced strength of the non-shear area?

 

[GeorgeTheCivilEngineer]

I can only see this being an issue at unusual connections or arrangements of beams. In practical terms, unless this was dealing with very large sections or there was a large economic reason, it is normally easier to get V_Ed to be <50% of V_Pl (cl6.2.8(2)) and eliminate this as a code check...!

 

[centondollar]

This is not an issue if you design the cross-section elastically. It is also not an issue if you design it plastically and assume that flanges resist the full bending moment - this will lead to less than, say, 10 or 15% error (on the safe side) in capacity evaluation for plate girders and a bit more error for stockier sections.

Those eurocode formulae are derived by utilizing yield criteria, so if the section you wish to design is steel only, you can make a spreadsheet to determine the equivalent stress (e.g., von Mises) due to normal stress and shear stress and skip the canned formulas entirely if you are not satisfied by them. Heck, you could even iterate the solution (assume plastic shear capacity in the web and increment strains until the combined normal and shear stress no longer satisfy the yield criterion) in such a spreadsheet.

 

[Sjotroll]

@centondollar Could you please expand a little more on the two solutions - one to design the section elastically, and the other to design it plastically with the assumptions that the flanges resist the full bending moment?
In the first case, if I use the elastic section-modulus then I can ignore the reduction because even if V = V[sub]Rd[/sub] the elastic moment will still not bring the section to plastic yielding?
In the second case, would I calculate the second-moment-of-area and the section-modulus only for the flanges? That in itself would be a reduction, but it would compensate for using the full strength instead of a reduced one over the whole section?

 

[centondollar]

Elastic:

sigma = M/W +P/A , W = elastic section modulus of entire section
tau = V*S/(I*t) , where V=shear force, S = first moment of area, I = 2nd moment of area, t = wall thickness. For an I-beam (I assume this is the shape you are designing), the shear stress in flanges is negligible, so bending-shear interaction is quite small.

Moving loads and variables loads can cause bending and shear peaks to occur simultaneously (continuous beams also have peak shear and bending at intermediate supports), which might also require checking of interaction of shear and bending.

Second case:
yes, flanges only for bending (and any axial force you may have). This way, you do not require the web to resist normal stress, and can therefore utilize the entire web (elastic or plastic) for shear.

This advice doesn't apply if your web buckles due to shear or compression, or if local buckling (flange due to web, flange in compression, point loads etc.) reduces capacity.