# 1.2 Example finding capacity

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#### 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 1-18 and 1-19 .

• 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 16.1-27.

• 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 y-axis. 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,
${F}_{cr}=21.99\mathrm{ksi}$

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.

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