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Find the capacity of a W14 X 74 column of A36 steel and a length of 20 feet.


The first section of the Manual will give the properties for the W14 X 74 column. A g , I x , I y , r y , and r x are found on pages 1-18 and 1-19 .

  • W14 X 74
  • l 20 ft
  • F y 36 ksi
  • A g 21.8
  • I x 795
  • I y 134
  • r x 6.04
  • r y 2.48


The equations and AISC guidelines for solving this and other columns and compression member problems can be found in the Manual starting on page 16.1-27.

  • First find the slenderness ratio to determine about which axis the bending will occur.
    K l r x 1 20 12 6.04 39.7
    K l r y 1 20 12 2.48 96.77
  • Since K l r y is greater than K l r y , the F cry will be less than F crx and F cry will be the governing factor.The bending will be about the y-axis. Now, we can solve for F cr and P n .
  • Section E2 of the Specifications section gives the equations to find the design compressive strength.
    • A g = gross area of member, square inches
    • F y = specified minimum yield stress, ksi
    • E = modulus of elasticity, ksi
    • k = effective length factor
    • l = laterally unbraced length of member, in.
  • Next find the design compressive strength by first finding the value for λ c .
    λ c K l r F y E 1.085
  • Since this is less than 1.5 the equation:
    F cr 0.658 λ c 2 F y
    can be used for F cr .
  • Therefore,
    F cr 21.99 ksi


Now we can use the equation

φ P n φ F cr A g 408


φ 0.85

So 408 k is the maximum load the column can sustain.

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Source:  OpenStax, Steel design (civi 306). OpenStax CNX. Jan 22, 2004 Download for free at http://cnx.org/content/col10153/1.3
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