# 1.5 Example with different effective lengths

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#### Problem

A W12 X 65 column, 24 feet long, is pinned at both ends in the strong direction, and pinned at the midpoint and theends in the weak direction. The column has A36 steel.

#### Number 1 - find effective length

Since the x-direction is the strong one and the y-direction is the weak one, then:

${L}_{x}=24$
${L}_{y}=12$

Notice that the effective length in the y-direction is half the total length of the member because there is alateral support at the midpoint.

Looking at the Manual on page 16.1-189 shows that the $K$ value for a column pinned at both ends is 1.0. Since the column is pinned at the ends and at the middle,

${K}_{x}=1$
${K}_{y}=1$

Now we can say that:

${K}_{x}{L}_{x}=1\times 24=24$
${K}_{y}{L}_{y}=1\times 12=12$

#### Number 2 - finding the capacity

Since, the steel is A36, you cannot use the column tables from Chapter 4 of the Third Edition Manual as the values are all given in terms of ${F}_{y}=50\mathrm{ksi}$ . However, in the Second Edition, in Chapter 3, the column tables give information forterms of ${F}_{y}=36\mathrm{ksi}$ .

From page 3-24 of the Second Edition Manual , the capacity for a W12 X 65 column with ${K}_{y}{L}_{y}=12$ is 519 kips.

Then to find ${K}_{x}{L}_{x}$ in terms of ${r}_{y}$ , ${K}_{x}{L}_{x}$ must be divided by: $\frac{{r}_{x}}{{r}_{y}}$ . This gives:

$\frac{{K}_{x}{L}_{x}}{\frac{{r}_{x}}{{r}_{y}}}=\frac{24}{1.75}=13.71$

This is close enough to 14, that we can then look in the tables for the $\mathrm{KL}$ value of 14, or interpolate for 13.71) and find the capacity forthe W12 X 65 member. The capacity is 497kips.

#### Method 2 - with buckling formulas

If you do not have the tables for A36 steel, you must use the formulas on page 16.1-27 of the Manual .

#### Number 1 - show the width-thickness ratio

In order for the equations in section E2 of the Manual to apply, the width-thickness ratio must be ${\lambda }_{r}$ .

$\frac{{b}_{f}}{2{t}_{f}}< {\lambda }_{r}$

The value for $\frac{{b}_{f}}{2{t}_{f}}$ (9.92) can be found on page 16.1-21, as well as the value for $\frac{h}{{t}_{w}}$ (24.9). The formula for ${\lambda }_{r}$ can be found on page 16.1-14/15. Then, the value for that formula can be found on page 16.1-150.

The flanges are unstiffened and in pure compression, so the formula is:

$\frac{{b}_{f}}{2{t}_{f}}=9.92< 0.56\sqrt{\frac{E}{{F}_{y}}}=15.9$

The web is stiffened and in compression, so the formula is:

$\frac{h}{{t}_{w}}=24.9\le 1.49\sqrt{\frac{E}{{F}_{y}}}=42.3$

Another way to easily find the formulas for ${\lambda }_{r}$ is to go to page 16.1-183 and look at the picture of the I-shaped member. The arrows point to either the flange orthe web and formulas correspond to the arrows giving the axial compression formulas that you need for that elementof the member.

#### Number 2 - compute slenderness ratios

The slenderness ratios can be found for both the x-axis and the y-axis. We know $K$ , and $L$ , and $r$ can be found in the properties section of the Manual on page 1-20.

$\frac{{K}_{x}{L}_{x}}{{r}_{x}}=\frac{24\times 12\times 1}{5.28}=54.54$
$\frac{{K}_{y}{L}_{y}}{{r}_{x}}=\frac{12\times 12\times 1}{3.02}=47.68$

Then, using Table 3-36 on page 16.1-143 of the Manual and interpolation, we can determine that ${\phi }_{c}{F}_{cr}=26.21$ , and that ${\phi }_{c}{P}_{n}={\phi }_{c}{F}_{cr}{A}_{g}=500k$

The capacity of the W12 X 65 column is 500 kips.

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