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Above picture is from Eurocode 3: design of steel connections. It gives geometrical requirements for a pin ended member. Looking at situation A, we have the minimum distance $c$ from the edge of the hole to the edge of the connection. I wonder where that term $\frac{d_0}{3}$ comes from?

Neglecting the safety factor $\gamma$ for simplicity, we should have the nominal stress on each side of the hole as: $$\sigma = \frac{F_{Ed}}{A} = \frac{F_{Ed}}{2ct}$$

At the yield limit, the stress equals the yield stress $f_y$:

$$f_y = \frac{F_{Ed}}{2ct}$$

Solving for $c$:

$$c = \frac{F_{Ed}}{2t f_y}$$

Now, there is no extra term $\frac{d_0}{3}$ here. Where does it come from in the Eurocode formula? My guess is that it could have something to do with stress concentrations, which we know to be around 3 times the nominal stress around a hole, but I don't really see how we would incorporate that into the equation, besides maybe multiplying the nominal stress by 3 before solving for $c$:

$$f_y = 3\frac{F_{Ed}}{2ct}$$

$$c = \frac{3F_{Ed}}{2t f_y}$$

But that is not the same result.

  • $\begingroup$ Perhaps they just decided... $\endgroup$
    – Solar Mike
    Apr 29, 2021 at 10:44
  • $\begingroup$ I don't have the exact reference on hand, but I believe the increase in edge distance is due to concerns of stress concentration around the pinhole. You shall search the EU websites for relevant studies, from which the equations were formulated. $\endgroup$
    – r13
    Apr 29, 2021 at 11:25
  • 1
    $\begingroup$ The stress concentration factor or 3 only applies for hole in an infinte plate. For a finite width plate, the more relevant quantity is the stress ratio $\sigma_{\max}/\sigma_{\infty}$ around the hole, which increases from 3 as the hole size increases. It is about 6 for d/W = 0.6 (i.e. a hole diameter d in a plate width (5/3)d) and increases rapidly for d/W > 0.6. $\endgroup$
    – alephzero
    Apr 29, 2021 at 12:31
  • $\begingroup$ @alephzero In practice, it is not reasonable to design the bar/plate for the maximum stress but average stress yet is safe. I think the code has decided to add the term to reflect the fact that the stress is higher than deriving from simply dividing F by 2*t*(b-do) after researches. The OP needs to dig out those studies. $\endgroup$
    – r13
    Apr 29, 2021 at 16:32

1 Answer 1


I think the EURO code considered non-linear stress distribution, thus require a wider plate to flatten the stress other than the linear stress distribution assumption that would result in b_net = F/fy*t. The graphics below might help you to have a grab on where the edge parameters were coming from.

enter image description here enter image description here enter image description here



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