the "Load Factor Design Selection Table for Beams" in Part 3 of the AISC LRFD
Manual
Calculation Procedure:
- Determine if the beam is a compact section
The W24 x 75 is a compact section. This can be verified by noting that in the Properties
Tables in Part 1 of the AISC LRFD Manual both bf/2tfand hjtw for a W24 x 76 beam are
less than the respective flange and web values of \p for Fy = 36 ksi (248 MPa). - Find the flexural design strength for minor-axis bending
For minor- (or y-) axis bending, Mny = Mpy ZyFy regardless of unbraced length (Eq. [56]).
The flexural design strength for minor-axis bending of a W24 x 75 is always equal to
<M^ =bZyFy = 0.90 x 28.6 in
3
x 36 ksi = 927 kip-in = 77kip-ft (104 kNm). - Compute the flexural design strength for major-axis bending
The flexural design strength for major-axis bending depends on Cb and Lb. For a simply
supported member, the end moments M 1 = M 2 = O; Cb = 1.0. - Plot the results
For O < Lb < (Lp = 8.0ft), QpMn = Q 1 JMp = 540 kip-ft (732 kNm).
At Lb = Lr = 23.4 ft, QpMf1x = QbM,. = 343 kip-ft (465 kNm). Linear interpolation is re-
quired for Lp<Lb< Lr, For Lb > Ln refer to the beam graphs in Part 3 of the AISC LRFD
Manual
Figure 366 shows the data plotted for this beam, after using data from the AISC table
referred to above.
Related Calculations. This procedure is the work of Abraham J. Rokach,
MSCE, Associate Director of Education, American Institute of Steel Construction. SI val-
ues were prepared by the handbook editor.
DESIGNING WEB STIFFENERS
FOR WELDED BEAMS
The welded beam in Fig. 37a (selected from the table of Built-Up Wide-Flange Sections
in Part 3 of the AISC LRFD Manual) frames into the column in Fig. 37 b. Design web
stiffeners to double the shear strength of the web at the end panel.
Calculation Procedure:
- Determine the nominal shear strength for a stiffened web
At the end panels there is no tension field action. The nominal shear strength for a stiff-
ened web is, using the AISC LRFD Manual equation, Vn = 0.644^C 11 ,. Assuming
^->234 /I C=JMOO_
tw ^F;
c
" (Mj^
Substituting, we obtain
T,K n^, r^ 44,000 A 26,400£
--
0
^'"G^-^--MF