Particulars of a pin-ended column made from a 250 MPa yield steel are :

L = 1 m A = 10^{3} mm^{2} I = 10^{5} mm^{4} 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.

Copyright 1999-2005 Douglas Wright,

last updated May 2005