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Hello!
How is kF factor (EC3-5, fig. 6.1, p. 26) calculated in Detailed Design of Member Verification of Steel/Aluminium Design?
I was veryfying results from Ex. 9.2 (Arya, C. (2009) Design of structural elements, pp.389) for web buckling of an UB356x171x45, sitting on 125 mm bearings on both sides. Given ss=125mm and hw=332,6mm, the kF should be 4,25 (see the link above for the calculations).
In RSAP, however, given ss = 12,5 cm the kF becomes 2,75. I was fiddling with numbers and it seems to me as if RSAP is using hw = 100 cm (if indeed the expression given in the beginning is used), because the value of ss = 37,5 cm gives the kF = 4,25 (see the picture below) that I was trying to attain in the beginning.
This makes LamF and therefore Xf factors and ultimately FRd significantly different (handcalculations for ss=125mm give FRd=237,6kN). What am I missing here?
Thank you for indicating this situation.
The formula for Kf entered in the internal code of Robot is by mistake defined as kF = 2.0 + 6.0*(ss+c/hw) instead of kF = 2.0 + 6.0*(ss+c)/hw. This needs to be corrected.
Artur Kosakowski
Corrected in SP4 for RSA 2016.
If you find your post answered press the Accept as Solution button please. This will help other users to find solutions much faster. Thank you.
Artur Kosakowski
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