1.2 Example finding capacity

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,
  $F_{cr}=21.99\mathrm{ksi}$

Answer

Now we can use the equation

where

So 408 k is the maximum load the column can sustain.