1.1 Example finding fcr

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 loadit can hold.

Givens

The first section of the Manual will give the properties for the W12 X 72 column. $A_g$ , $I_x$ , and $I_y$ are found on page 1-20 and 1-21 .

• W12 X 72
• $1=10\mathrm{ft}$
• $F_y=50\mathrm{ksi}$
• $A_g=21.2$
• $I_x=597$
• $I_y=195$
• take $K=1$

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 .

• First, find $r$ (governing radius of gyration about the axis of buckling, in.):
  $r=r_y=\sqrt{\frac{195}{21.1}}=3.04$
• Section E2 of the Specifications section gives the equations to find the design compressive strength.
  • $A_g=\text{gross area of member, square inches}$
  • $F_y=\text{specified minimum yield stress, ksi}$
  • $E=\text{modulus of elasticity, ksi}$
  • $K=\text{effective length factor}$
  • $l=\text{laterally unbraced length of member, in.}$
• Next find the design compressive strength by first finding the value for ${\lambda }_c()$ .
  $${\lambda }_c=\frac{Kl}{r\pi }\sqrt{\frac{F_y}{E}}=0.519$$
• Since this is less than 1.5 the equation:
  $F_{cr}=0.658^{{\lambda }_c^2}{F}_y$
  can be used for Fcr and the column is considered short and stocky.
• Therefore,
  $F_{cr}=44.7\mathrm{ksi}$

Answer

Now we can use the equation

where

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

needs figure and help on question