Let's say we have a beam with pinned connections at both ends. The beam is subjected to axial load $P$. The beam has a cross sectional area $A$, radius of gyration $r$, distance from the neutral axis to the extreme fiber $z$, length $L$ and theoretical Euler buckling load $P_e$.
It is impossible to manufacture beams that are perfectly straight, so we assume the beam as an initial maximum deflection at the midpoint, $d$.
The maximal stress occurs at the midpoint of the beam, on the extreme fiber looking from the neutral axis. This stress is the sum of direct compressive stress $\sigma_{c}$ and bending stress $\sigma_{b}$:
$$\sigma_{max}=\sigma_{c}+\sigma_{b}=\frac{P}{A}+\frac{Mz}{Ar^2}=\frac{P}{A}+\frac{Pz}{Ar^2}\frac{d}{\frac{P_e}{P}-1}=\frac{P}{A}+\frac{P\mu}{\frac{P_e}{P}-1}$$
where $\mu$ is an imperfection factor that captures the effect initial imperfection.
We cannot predict $\mu$ exactly as it is related to manufacturing but we can assume its maximum values. In Eurocode it is assumed that $$\mu=\alpha(\lambda-0.2)$$
where $\lambda=\frac{L}{r}$, also called slenderness and $\alpha$ is a constant chosen from cross sectional properties. The assumption therefore is, that the imperfection factor is proportional to slenderness of the beam with some constant of proportionality $\alpha$. It is reasonable assumption; longer and thinner the beam is, more "crookedness" can be expected to be present after fabrication of the beam.
$\alpha$ is said to, among other things, to capture the effect of residual stresses due to manufacturing. Whether the beam is welded together or hot rolled affects the choice of $\alpha$. I don't see why this is a logical way of including residual stresses. The imperfection factor $\mu$ appears only in $\sigma_b$, which implies that if the initial deflection is negligeable, $\sigma_{max}$ equals just $\frac{P}{A}$ and the residual stress doesn't matter. But surely the residual stress matters in simple compression too?
Shouldn't a more logical way be something like this:
$$\sigma_{max}=\sigma_{c}+\sigma_{b}+\sigma_{residual}$$
with some appropriate expression for $\sigma_{residual}$ involving similar factor to $\alpha$? I get that the existing method does work, but I don't understand why.
