INVESTIGATION OFA BEAM COLUMN
A Wl2 x 53 column with an effective length of 20 ft (6.1 m) is to carry an axial load of
160 kips (711.7 kN) and the end moments indicated in Fig. 16. The member will be se-
cured against sides way in both directions. Is the section adequate?
Calculation Procedure:
- Record the properties of the section
The simultaneous set of values of axial stress and
bending stress must satisfy the inequalities set forth
in the AISC Specification.
The properties of the section are A = 15.59 in
2
(100.586 cm
2
); Sx = 70.7 in
3
(1158.77 cm
3
); rx = 5.23
in (132.842 mm); ry = 2.48 in (62.992 mm). Also,
from the Manual, Lc = 10.8 ft (3.29 m); Ln = 21.7 ft
(6.61 m).
- Determine the stresses listed below/
The stresses that must be determined are the axial
stress fa; the bending stress fb; the axial stress F 09
which would be permitted in the absence of bending;
and the bending stress Fb, which would be permitted
in the absence of axial load. Thus,/fl = 160/15.59 =
10.26 kips/in^2 (70.742 MPa); /, = 31.5(12)770.7 = FIGURE 16. Beam column.
5.35 kips/in^2 (36.888 MPa); KLIr = 240/2.48 = 96.8;
therefore, Fa = 13.38 kips/in^2 (92.255 MPa); Lu < KL
< Lc; therefore, Fb = 22 kips/in^2 (151.7 MPa). (Al-
though this consideration is irrelevant in the present instance, note that the Specification
establishes two maximum dlt ratios for a compact section. One applies to a beam, the oth-
er to a beam column.)
- Calculate the moment coefficient C1n
Since the algebraic sign of the bending moment remains unchanged, M 1 IM 2 is positive.
Thus, Cm = 0.6 + 0.4(15.2/31.5) = 0.793.
- Apply the appropriate criteria to test the adequacy
of the section
Thus,/fl/Fa - 10.26/13.38 = 0.767 > 0.15. The following requirements therefore apply:
fJFa + [C 1 J(I -fJFi)]WFb) ^ l;/«/(0.6£) + ./i/F* < 1 where Fi = 149,000/(£Z>)^2
kips/in^2 and KL and r are evaluated with respect to the plane of bending.
Evaluating gives F'e = 149,000(5.23)^2 /240^2 - 70.76 kips/in^2 (487.890 MPa); faIF'e =
10.26/70.76 = 0145. Substituting in the first requirements equation yields 0.767 +
(0.793/0.855)(5.35/22) = 0.993. This is acceptable. Substituting in the second require-
ments equation, we find 10.26/22 + 5.35/22 = 0.709. This section is therefore satisfactory.
APPLICATION OF BEAM-COLUMN FACTORS
For the previous calculation procedure, investigate the adequacy of the Wl 2 x 53 section
by applying the values of the beam-column factors B and a given in the AISC Manual.
