1.9 Example finding k for frames

Problem

Find the $K$ -value for all columns of the following frame. All columns are orientedwith the web in the plane of the paper, (the plane of buckling).

Section properties

The section properties can be found in the Dimensions and Properties section of the Manual . The lengths are found on .

W8 x 31

• $I=110\mathrm{in.}^4$
• $L=15\mathrm{ft}$

W14 x 30

• $I=291\mathrm{in.}^4$
• $L=20\mathrm{ft}$

W18 x 50

• $I=800\mathrm{in.}^4$
• $L=30\mathrm{ft}$ and $L=20\mathrm{ft}$

Solution

Since the columns and beams are oriented with the web as the plane of buckling, the axis of bending is the x-axis.Therefore, all values of $I$ , above, are with respect to the x-axis.

For each column:

• First we must determine the $G$ values for each corner of the frame with the equation:
  $G=\frac{\sum \frac{I_c}{L_c}}{\sum \frac{I_g}{L_g}}$
  where, $c$ is for column and $g$ is for girder or beam.
• Then we use Fig. C-C2.2a and Fig. C-C2.2b in the Manual to find the $K$ values for each column. If a column is sidesway inhibited, it is braced againstany sideways movement; if a column is sidesway uninhibited, it is not braced in the sidewaysdirection.

Column ab

• $G_A=10$ because it is a pinned support.
• $G_B=\frac{\frac{110}{15}}{\frac{800}{30}}=0.274$
• $AB$ is sidesway inhibited (the support at $J$ braces any sideways motion of the column) so lining up the value of $G_A$ and $G_B$ on 16.1-191 gives: $K=0.77$

Column cd

• $G_C=10$ because it is a pinned support.
• $G_D=\frac{\frac{110}{15}+\frac{110}{12}}{\frac{800}{30}+\frac{800}{20}}=0.2475$
• $CD$ is sidesway inhibited (the support at $J$ braces any sideways motion of the column)so lining up the value of $G_C$ and $G_D$ on 16.1-191 gives: $K=0.76$

Column gf

• $G_F=1$ because it is a pinned support that cannot move sideways, so it is like a fixed support (?check on this).
• $G_G=\frac{\frac{110}{15}+\frac{110}{12}}{\frac{800}{20}+\frac{800}{20}}=0.206$
• $GF$ is sidesway inhibited (the support at $J$ braces any sideways motion of the column) so lining up the value of $G_G$ and $G_F$ on 16.1-191 gives: $K=0.67$

Column ed

• $G_D=0.2475$ from before.
• $G_E=\frac{\frac{110}{12}}{\frac{291}{20}}=0.63$
• $ED$ is sidesway uninhibited (there is no sideways bracing for the top portion of the frame) so lining up thevalue of $G_E$ and $G_D$ on 16.1-192 gives: $K=1.15$

Column gh

• $G_G=0.206$ from before.
• $G_H=0.63$ from before.
• $GH$ is sidesway uninhibited (there is no sideways bracing for the top portion of the frame) so lining up thevalue of $G_G$ and $G_H$ on 16.1-192 gives: $K=1.14$