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.
${A}_{g}$ ,
${I}_{x}$ ,
${I}_{y}$ ,
${r}_{y}$ , and
${r}_{x}$ are found on pages
118 and
119 .
 W14 X 74

$l=20\mathrm{ft}$

${F}_{y}=36\mathrm{ksi}$

${A}_{g}=21.8$

${I}_{x}=795$

${I}_{y}=134$

${r}_{x}=6.04$

${r}_{y}=2.48$
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 16.127.
 First find the slenderness ratio to determine about which axis
the bending will occur.
$\frac{Kl}{{r}_{x}}=\frac{1\times 20\times 12}{6.04}=39.7$
$\frac{Kl}{{r}_{y}}=\frac{1\times 20\times 12}{2.48}=96.77$
 Since
$\frac{Kl}{{r}_{y}}$ is greater than
$\frac{Kl}{{r}_{y}}$ , the
${F}_{cry}$ will be less than
${F}_{crx}$ and
${F}_{cry}$ will be the governing factor.The bending will be about the yaxis. Now, we can solve for
${F}_{cr}$ and
${P}_{n}$ .
 Section E2 of the Specifications section gives the
equations to find the design compressive strength.

${A}_{g}$ = gross area of member, square inches

${F}_{y}$ = specified minimum yield stress, ksi

$E$ = modulus of
elasticity, ksi

$k$ = effective length
factor

$l$ = laterally unbraced
length of member, in.
 Next find the design compressive strength by first
finding the value for
${\lambda}_{c}()$ .
$${\lambda}_{c}=\frac{Kl}{r\pi}\sqrt{\frac{{F}_{y}}{E}}=1.085$$
 Since this is less than 1.5 the equation:
${F}_{cr}=\mathrm{0.658}^{{\lambda}_{c}^{2}}{F}_{y}$
can be used for
${F}_{cr}$ .
 Therefore,
${F}_{cr}=21.99\mathrm{ksi}$
Answer
Now we can use the equation
$\phi {P}_{n}=\phi {F}_{cr}{A}_{g}=408$
where
$\phi =0.85$
So 408 k is the maximum load the column can sustain.