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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