DETERMININGIFAGIVENBEAM

IS COMPACT OR NON-COMPACT

A designer plans to use a W6 x 15 and a W12 x 65 beam in (a) A6 steel (Fy = 36 ksi [248

MPa)], and (b) with Fy = 50 ksi [344.5 MPa]) and wishes to determine if the beams are

compact or non-compact.

Calculation Procedure:

For the W6 x 15 beam

1. Analyze the W6 x 15 beam
   Refer to the AISC Manual table, namely "Limiting Width-Thickness Ratios for Beams"
   and its illustration "Definition of widths (b and h) and thickness", the flanges of a W
   shape are given by

A^65

Xp=VF

y

where A^ = limiting width-thickness ratio for compact section.
Substituting for each of the two beams, we have

{

—P= = 10.8 if Fy = 36 ksi (248 MPa)

65

—7= =9.2 if Fy = 50 ksi (344.5 MPa)

2. Compute the data for the web of a W shape
   Using the same equation as in Step 1, for the web of a W shape

r 640

—/= = 106.7 if Fv = 36 ksi (248 MPa)

__64p__J V36 y

Xp~VF

y-\^640

[ V50

=90

'

5 lf Fy = 5

°

ksl (344

'

5 M?a)

3. Determine if the beam is compact
   From the Properties Tables for W Shapes, in Part 1 of the AISC LRFD Manual (Compact
   Section Criteria): for a W6 x 15
   b bf
   flange 7 = ^f = 11.5
   f

hc

web —=21.6

*W

Since flange (bit = 11.5) > (A 7 , 10.8), the W x 15 beam is noncompact in A36 steel. Like-

wise, it is noncompact ifFy = 50 ksi (344.4 Mpa).

For the WU x 65 beam

4. Compute the properties of the beam shape
   From the AISC Manual "Properties Tables for W Shapes", for a W12 x 65