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Example finding capacity

Module by: Michael Terk. E-mail the author

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Problem

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

Givens

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

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

Solution

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.

  1. First find the slenderness ratio to determine about which axis the bending will occur.
    Kl r x =1×20×126.04=39.7 K l r x 1 20 12 6.04 39.7 (1)
    Kl r y =1×20×122.48=96.77 K l r y 1 20 12 2.48 96.77 (2)
  2. Since Kl r y K l r y is greater than Kl r y K l r y , the F cry F cry will be less than F crx F crx and F cry F cry will be the governing factor. The bending will be about the y-axis. Now, we can solve for F cr F cr and P n P n .
  3. Section E2 of the Specifications section gives the equations to find the design compressive strength.
    • A g A g = gross area of member, square inches
    • F y F y = specified minimum yield stress, ksi
    • EE = modulus of elasticity, ksi
    • kk = effective length factor
    • ll = laterally unbraced length of member, in.
  4. Next find the design compressive strength by first finding the value for λ c λ c .
    λ c =Klrπ F y E=1.085 λ c K l r F y E 1.085 (3)
  5. Since this is less than 1.5 the equation:
    F cr =0.658 λ c 2 F y F cr 0.658 λ c 2 F y (4)
    can be used for F cr F cr .
  6. Therefore,
    F cr =21.99ksi F cr 21.99 ksi (5)

Answer

Now we can use the equation

φ P n =φ F cr A g =408 φ P n φ F cr A g 408 (6)

where

φ=0.85 φ 0.85 (7)

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

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