A 20 foot high steel column is constructed from an I-beam W8×31. In order to strengthen the section, the engineer decided to cover each flange with a 10 in by 1 in steel plate as shown in the diagram. What is the increase in critical load capacity of the new column? Assume the steel modulus of elasticity, E_steel, is 29,000 ksi and the yield stress, σ_yld, is 36 ksi.

Basic W8x31 I-Beam

A simply supported column (pinned at each end) will buckle around the axis with the lowest moment of inertia. The moment of inertia of a standard W8×31 can be found in the Structure Shapes appendix as,

     I_xx = 110 in⁴
     I_yy = 37.1 in⁴
     A = 9.13 in²

The lowest critical load for this I-beam is around the y axis,

     P_cr = π² EI_yy / L²
           = π² (29,000 ksi)(37.1 in⁴) / {(20 ft) (12)}²
           = 184.4 kip

The critical stress,

      σ_cr = P_cr / A
           = 184.4 / 9.13
           = 20.20 ksi

This is less than the yield stress of 36 ksi so application of Euler's equation is appropriate.

To increase the critical load of the section, the I-Beam is covered with a plate 10 in × 1 in. For this new section, the moment of inertia and area are,

The improved section I_yy still has a lower value than I_xx. So, it will buckle around y axis. Critical load with the cover plates is

     P_cr = π² EI_yy / L²
           = π² (29,000 ksi)(203.8 in⁴) / {(20 ft)(12)}²
           = 1,013 kip

The critical stress,

      σ_cr = P_cr / A
           = 1013 / 29.13
           = 34.78 ksi

This stress is still below the yield stress so Euler's equation is still appropriate.

The increase in critical load capacity,

     δP_cr = (1013 - 184.4) kip
             = 828.6 kip

or nearly 450% increase.