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 ?
r = √(I/A) = 10 mm ρ = L/r = 100 σc = E(π/ρ)2 = 204 MPa
From ( 2) with S = 250 MPa, the direct stress at failure is σd = 134 MPa ( note that σd must be less than both σc 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 = Sy /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.