A Beginner's Guide to the Steel Construction Manual, 13^th ed. (old)

Chapter 6 - Buckling Concepts

© 2006, 2008 T. Bartlett Quimby

Section 6.4

Example Problem 6.1

Grid 3 Columns

Last Revised: 11/04/2014

The figure given here shows the support conditions for this column.  We will start from the top and work our way down for computing the effective lengths.

Third Floor Column Segment

Weak 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)_Y,33 = 1.0*12' = 12 ft.

The subscript used denotes the weak direction, "Y", on grid "3" on the third floor, "3".

Strong 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, W21x62 and W18x35 girders and the W12x65 column resist joint rotation 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 strong axis of the column.  So we use I_x for the girder term and I_x for the column term:

  G_3R = S(I_c/L_c) / S(I_g/L_g)
  G_3R = [(533 in⁴)/(12 ft)] / [(1330 in⁴)/(30 ft)+(510 in⁴)/(15 ft)]
  G_3R = 0.567

  The subscript used is "3" 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 W21x55 girder and the W12x65 column frame into the joint but niether resist joint rotation 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 column.  So we use I_x for the column term:

  G₃₃ = S(I_c/L_c) / S(I_g/L_g)
  G₃₃ = [(533 in⁴)/(12 ft)+(533 in⁴)/(10 ft)] / [0]
  G₃₃ = infinite!!!  ... use 10.0

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

Entering the sidesway uninhibited nomograph on SCM page 16.1-242 with these two values and I get something close to 1.79.  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)_X,33 = (1.79)(12 ft) = 21.48 ft

Second Floor Column Segment

Weak 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.

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

(KL)_Y,32 = 1.0*10' = 10 ft.

Strong 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.
• At the top of the column segment, only the W12x65 column resists joint rotation in plane, as noted above.  We already computed this "G" term when determining the effective length of the third floor column segment so:

  G₃₃ = 10.0
   

• At the bottom of the column segment, W24x68 and W18x35 girders and the W12x65 column resist joint rotation 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 strong axis of the column.  So we use I_x for the girder term and I_x for the column term:

  G₃₂ = S(I_c/L_c) / S(I_g/L_g)
  G₃₂ = [(533 in⁴)/(10 ft)+(533 in⁴)/(11 ft)] / [(1830 in⁴)/(30 ft)+(510 in⁴)/(15 ft)]
  G₃₂ = 1.071

Entering the sidesway inhibited nomograph on SCM page 16.1-241 with these two values and I get something close to 0.86.  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)_X,32 = (0.86)(10 ft) = 7.60 ft

First Floor Column Segment

Weak 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)_Y,31 = 1.0*5.50' = 5.50 ft.

Strong 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, W24x68 and W18x35 girders and the W12x65 column resist joint rotation 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 strong axis of the column.  So we use I_x for the girder term and I_x for the column term.  We already computed this "G" term when determining the effective length of the second floor column segment so:

  G₃₂ = 1.071
   

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

  G₃₁ = 10.0

Entering the sidesway uninhibited nomograph on SCM page 16.1-242 with these two values and I get something close to 1.88.  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)_X,31 = (1.88)(11 ft) = 20.68 ft

Column Line Summary

To summarize the results, the effective lengths for the column segments on grid 3 are:

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