Particulars of a pin-ended column made from a 250 MPa yield steel are :
L = 1 m       A = 10³ mm²       I = 10⁵ mm⁴       η = 0.3
What load can it support if a safety factor of 2 is specified ?

Properties :
r = √(I/A) = 10 mm       ρ = L/r = 100       σ_c = E(π/ρ)² = 204 MPa
From ( 2) with   S = 250 MPa, the direct stress at failure is   σ_d = 134 MPa   ( note that   σ_d must be less than both σ_c and S ).
The failure load is therefore this mean stress times the area,   A, ie.   134 kN.
So if the safety factor is 2, the permissible load will be   134/2 = 67 kN.

Let us see what would happen if this had been carried out   incorrectly with safety factor applied to strength rather than to load.
From ( 2) with   S = S_y /n = 250/2 = 125 MPa, the direct stress is found to be   83 MPa. Multiplying by the area, the corresponding load is   83 kN. Hence the   incorrect conclusion is that the column could withstand a load of   83 kN with a safety factor of   2.
But if the   83 kN load were actually applied then the safety factor is   n = failure load/actual load = 134/83 = 1.6 - significantly less than the   2 expected.