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Example finding K for frames

Module by: Michael Terk. E-mail the author

Summary: (Blank Abstract)

Problem

Find the KK-value for all columns of the following frame. All columns are oriented with the web in the plane of the paper, (the plane of buckling).

Figure 1: (not drawn to scale)
Figure 1 (framepicture.bmp)

Section Properties

The section properties can be found in the Dimensions and Properties section of the Manual. The lengths are found on Figure 1.

W8 X 31

  • I=110⁢in.4 I 110 in. 4
  • L=15⁢ft L 15 ft

W14 X 30

  • I=291⁢in.4 I 291 in. 4
  • L=20⁢ft L 20 ft

W18 X 50

  • I=800⁢in.4 I 800 in. 4
  • L=30⁢ft L 30 ft and L=20⁢ft L 20 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 II, above, are with respect to the x-axis.

Figure 2: An example of a column bending in the x-direction.
Figure 2 (bentcol.bmp)

For each column:

  1. First we must determine the GG values for each corner of the frame with the equation:
    G=∑ I c L c ∑ I g L g G I c L c I g L g (1)
    where, cc is for column and gg is for girder or beam.
  2. Then we use Fig. C-C2.2a and Fig. C-C2.2b in the Manual to find the KK values for each column. If a column is sidesway inhibited, it is braced against any sideways movement; if a column is sidesway uninhibited, it is not braced in the sideways direction.

Column AB

  • G A =10 G A 10 because it is a pinned support.
  • G B =1101580030=0.274 G B 110 15 800 30 0.274
  • A⁢B A B is sidesway inhibited (the support at JJ braces any sideways motion of the column) so lining up the value of G A G A and G B G B on 16.1-191 gives: K=0.77 K 0.77

Column CD

  • G C =10 G C 10 because it is a pinned support.
  • G D =11015+1101280030+80020=0.2475 G D 110 15 110 12 800 30 800 20 0.2475
  • C⁢D C D is sidesway inhibited (the support at JJ braces any sideways motion of the column) so lining up the value of G C G C and G D G D on 16.1-191 gives: K=0.76 K 0.76

Column GF

  • G F =1 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 =11015+1101280020+80020=0.206 G G 110 15 110 12 800 20 800 20 0.206
  • G⁢F G F is sidesway inhibited (the support at JJ braces any sideways motion of the column) so lining up the value of G G G G and G F G F on 16.1-191 gives: K=0.67 K 0.67

Column ED

  • G D =0.2475 G D 0.2475 from before.
  • G E =1101229120=0.63 G E 110 12 291 20 0.63
  • E⁢D E D is sidesway uninhibited (there is no sideways bracing for the top portion of the frame) so lining up the value of G E G E and G D G D on 16.1-192 gives: K=1.15 K 1.15

Column GH

  • G G =0.206 G G 0.206 from before.
  • G H =0.63 G H 0.63 from before.
  • G⁢H G H is sidesway uninhibited (there is no sideways bracing for the top portion of the frame) so lining up the value of G G G G and G H G H on 16.1-192 gives: K=1.14 K 1.14

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