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Example finding Fcr

Module by: Michael Terk. E-mail the author

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Problem

A W12 X 72 is used as a column. It is 10 feet long and the steel strength is 50 ksi. Find the maximum compressive load it can hold.

Givens

The first section of the Manual will give the properties for the W12 X 72 column. A g A g , I x I x , and I y I y are found on page 1-20 and 1-21.

  • W12 X 72
  • 1=10ft 1 10 ft
  • F y =50ksi F y 50 ksi
  • A g =21.2 A g 21.2
  • I x =597 I x 597
  • I y =195 I y 195
  • take K=1 K 1

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 r r (governing radius of gyration about the axis of buckling, in.):
    r= r y =19521.1=3.04 r r y 195 21.1 3.04 (1)
  2. Section E2 of the Specifications section gives the equations to find the design compressive strength.
    • A g =gross area of member, square inches A g gross area of member, square inches
    • F y =specified minimum yield stress, ksi F y specified minimum yield stress, ksi
    • E=modulus of elasticity, ksi E modulus of elasticity, ksi
    • K=effective length factor K effective length factor
    • l=laterally unbraced length of member, in. l laterally unbraced length of member, in.
  3. Next find the design compressive strength by first finding the value for λ c λ c .
    λ c =Klrπ F y E=0.519 λ c K l r F y E 0.519 (2)
  4. 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 (3)
    can be used for Fcr and the column is considered short and stocky.
  5. Therefore,
    F cr =44.7ksi F cr 44.7 ksi (4)

Answer

Now we can use the equation

φ P n =φ F cr A g =802 φ P n φ F cr A g 802 (5)

where

φ=0.85 φ 0.85 (6)

So 802 k is the maximum load the column can sustain.

needs figure and help on question

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