1. Using the AISC allowable-load tables, select the most economic member made of A36
steel. After a trial section has been selected, it is necessary to compare the unbraced length L′ of

the compression flange with the properties Lc and Lu of that section in order to establish the allow-
able bending stress. The variables are defined thus: Lc = maximum unbraced length of the compres-
sion flange if the allowable bending stress = 0.66fy, measured in ft (m); Lu = maximum unbraced

length of the compression flange, ft (m), if the allowable bending stress is to equal 0.60fy.
The values of Lc and Lu associated with each rolled section made of the indicated grade of steel
are recorded in the allowable-uniform-load tables of the AISC Manual. The Lc value is established
by applying the definition of a laterally supported member as presented in the Specification. The
value of Lu is established by applying a formula given in the Specification.

The values of allowable uniform load given in the AISC Manual apply to beams of A36 steel sat-
isfying the first or third condition above, depending on whether the section is compact or noncompact.

Referring to the table in the Manual, we see that the most economic section made of A36 steel is
W16 × 45; Wallow = 46 kips (204.6 kN), where Wallow = allowable load on the beam, kips (kN). Also,
Lc = 7.6 > 5. Hence, the beam is acceptable.

2. Compute the equivalent load for a member of A242 steel. To apply the AISC Manual tables to
choose a member of A242 steel, assume that the shape selected will be compact. Transform the

actual load to an equivalent load by applying the conversion factor 1.38, that is, the ratio of the allow-
able stresses. The conversion factors are recorded in the Manual tables. Thus, equivalent load =

45/1.38 = 32.6 kips (145.0 N).

3. Determine the highest satisfactory section. Enter the Manual allowable-load table with the
load value computed in step 2, and select the lightest section that appears to be satisfactory. Try W16

× 36; Wallow = 36 kips (160.1 N). However, this section is noncompact in A242 steel, and the equiv-
alent load of 32.6 kips (145.0 N) is not valid for this section.

4. Revise the equivalent load. To determine whether the W16 × 36 will suffice, revise the equiv-
alent load. Check the Lu value of this section in A242 steel. Then equivalent load = 45/1.25 = 36 kips

(160.1 N), Lu = 6.3 ft (1.92 m) > 5 ft (1.5 m); use W16 × 36.

5. Verify the second part of the design. To verify the second part of the design, calculate the bend-
ing stress in the W16 × 36, using S = 56.3 in3 (922.76 cm3

) from the Manual. Thus, M = (1/8)WL =
(1/8)(45,000)(25)(12) = 1,688,000 in·lb (190,710.2 N·m); f = M/S = 1,688,000/56.3 = 30,000 lb/in2
(206,850.0 kPa). This stress is acceptable.