A Beginner's Guide to the Steel Construction Manual, 15^th ed.

Chapter 6 - Buckling Concepts

© 2006, 2008, 2011, 2017 T. Bartlett Quimby

Section 6.5

Example Problem 6.3

Grid 4 Columns

Last Revised: 04/19/2021

Figure 6.5.3.5 shows the support conditions for this column. We will start from the top and work our way down for computing the effective lengths.

Figure 6.5.3.5 Grid 4 Column Diagrams

Third Floor Column Segment

Strong Direction: At this level and direction, there are no beams fixed to the column to create joint rotational restraint and the column is part of a frame that inhibits sidesway (i.e., it is braced) as stated on the drawings.

For these conditions, K = 1.00. This gives an effective length,

(KL)_X,43 = 1.0*12' = 12 ft.

The subscript used denotes the strong direction, "X", on grid "4" on the third floor, "3".

Weak Direction: At this level, the support conditions are:

• Sidesway is uninhibited. In the frame profile there are no braces on the third floor level so all the columns in this level are unbraced.
• At the top of the column segment, a W18x35 girder and the W14x82 column resist joint translation in plane. Note that joint rotation in the plane of the frame, in this case, causes bending about the strong axis of the girder and bending about the weak axis of the column. So we use I_x for the girder term and I_y for the column term:

  G_4R = S(I_c/L_c) / S(I_g/L_g)
  G_4R = [(148 in⁴)/(12 ft)] / [(510 in⁴)/(15 ft)]
  G_4R = 0.363

  The subscript used is "4" for the grid line and "R" for the roof level. Note that "G" term need only be computed once for a joint. Joints are often shared by two column segments.

  Also note that we did not change the units so that they are all feet or inches. All the I terms are in⁴ and the L units are feet. Even without the unit conversion, all the units cancel, so save yourself the extra computational step! Check it out.
   

• At the bottom of the column segment, a W18x35 girder frames into the column however, due to the pinned connection, it does not provide in plane rotational restrain. The W14x82 column is the only rotational restraint. In this case the column is both above and below the joint, though it is of different lengths in each level. Note that joint rotation in the plane of the frame, in this case, causes bending about the strong axis of the girder and bending about the weak axis of the column. So we use I_x for the girder term and I_y for the column term:

  G₄₃ = S(I_c/L_c) / S(I_g/L_g)
  G₄₃ = [(148 in⁴)/(12 ft)+(148 in⁴)/(10 ft)] / [0]
  G₄₃ = infinite!!! ... use 10.0

We use a G of 10 here because this is essentially a pinned lateral support for the column. The girder provides lateral support but no rotational support.

Entering the sidesway uninhibited nomograph on SCM page 16.1-513 with these two values and I get something close to 2.38. It is difficult to get a precise value from the nomographs so your value may be slightly different.

The effective length, then, for the 3rd floor column segment is:

(KL)_Y,43 = (2.38)(12 ft) = 28.56 ft

Second Floor Column Segment

Strong Direction: At this level and direction, there are no beams fixed to the column to create rotational restraint and the column is part of a frame that inhibits sidesway (i.e., it is braced) as stated on the drawings. There are actually two column lengths on this level, but they are both the same, so we will only compute the effective length once.

For these conditions, K = 1.00. This gives an effective length,

(KL)_X,42 = 1.0*10' = 10 ft.

Weak Direction: At this level, the support conditions are:

• Sidesway is inhibited by a brace that prevents the second floor from moving laterally with respect the second floor. The brace is located between grids 2 and 3.
• We already computed this "G" term when determining the effective length of the third floor column segment so:

  G₄₃ = 10.0
   

• At the bottom of the column segment, a W18x35 girder and the W14x82 column resist joint translation in plane. In this case the column is both above and below the joint, though it is of different lengths in each case. Note that joint rotation in the plane of the frame, in this case, causes bending about the strong axis of the girder and bending about the weak axis of the column. So we use I_x for the girder term and I_y for the column term:

  G₄₂ = S(I_c/L_c) / S(I_g/L_g)
  G₄₂ = [(148 in⁴)/(10 ft)+(148 in⁴)/(11 ft)] / [(510 in⁴)/(15 ft)]
  G₄₂ = 0.831

Entering the sidesway inhibited nomograph on SCM page 16.1-512 with these two values and I get something close to 0.84. It is difficult to get a precise value from the nomographs so your value may be slightly different.

The effective length, then, for the 2nd floor column segment is:

(KL)_Y,42 = (0.84)(10 ft) = 7.40 ft

First Floor Column Segment

Strong Direction: At this level and direction, there are no beams fixed to the column to create rotational restraint and the column is part of a frame that inhibits sidesway (i.e., it is braced) as stated on the drawings. Also, the column at this level, and in this direction, is braced at mid-height, as stated on the drawing.

For these conditions, K = 1.00. This gives an effective length,

(KL)_X,41 = 1.0*5.50' = 5.50 ft.

Weak Direction: At this level, the support conditions are:

• Sidesway is uninhibited. In the frame profile there are no braces on the third floor level so all the columns in this level are unbraced.
• At the top of the column segment, a W1x35 girder and the W14x82 column resist joint translation in plane. Note that joint rotation in the plane of the frame, as discussed above, causes bending about the strong axis of the girder and bending about the weak axis of the column. So we use I_x for the girder term and I_y for the column term. We already computed this "G" term when determining the effective length of the second floor column segment so:

  G₄₂ = 0.831
   

• At the bottom of the column segment, the base is fixed to the foundation. We will assume that the fixed connection is properly designed so that we can use the G specified on SCM pg 16.1-241, second paragraph:

  G₄₁ = 1.00

Entering the sidesway uninhibited nomograph on SCM page 16.1-513 with these two values and I get something close to 1.29. It is difficult to get a precise value from the nomographs so your value may be slightly different.

The effective length, then, for the 1st floor column segment is:

(KL)_Y,41 = (1.29)(11 ft) = 14.19 ft

Column Line Summary

Table 6.5.3.5 summarizes the resulting effective lengths for the column segments on grid 4.

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