# 1.1 Example finding fcr

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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 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,
${F}_{cr}=44.7\mathrm{ksi}$

Now we can use the equation

$\phi {P}_{n}=\phi {F}_{cr}{A}_{g}=802$

where

$\phi =0.85$

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

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