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MECHANICS - EXAMPLE

    Example


 

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, Esteel, is 29,000 ksi and the yield stress, σyld, is 36 ksi.

     
    Solution


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,

     Ixx = 110 in4
     Iyy = 37.1 in4
     A = 9.13 in2

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

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

The critical stress,

      σcr = Pcr / 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.

     

W8x31 I-Beam with Cover Plates
 

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 Iyy still has a lower value than Ixx. So, it will buckle around y axis. Critical load with the cover plates is

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

The critical stress,

      σcr = Pcr / 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,

     δPcr = (1013 - 184.4) kip
             = 828.6 kip

or nearly 450% increase.

     
   
 
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