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
120 and
121 .
 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
16.127 .
 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
$\phi {P}_{n}=\phi {F}_{cr}{A}_{g}=802$
where
$\phi =0.85$
So 802 k is the maximum load the column can sustain.
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