Calculation Procedure:
- Record the relevant properties of the two sections
Refer to the AISC Specification and Manual. It is assumed that the moment in each mem-
ber is resisted entirely by the flanges and that the distance between the resultant flange
forces is 0.95 times the depth of the member.
Record the properties of the members: for the W18 x 105, d= 18.32 in (465.3 mm); bf
= 11.79 in (299.5 mm); tf= 0.911 in (23.1 mm); tw = 0.554 in (14.1 mm); k = 1.625 in
(41.3 mm). For the W27 x 84, d = 26.69 in (677.9 mm); Zy= 9.96 in (253 mm); tf= 0.636
in (16.2 mm); tw = 0.463 in (11.8 mm).
- Compute F 1
Thus, F 1 MJ(0.95d) = 8100/[0.95(18.32)] = 465 kips (2068.3 kN).
- Determine whether web stiffeners are needed to transmit F 1
The shearing stress is assumed to vary linearly from zero at a to its maximum value at d.
The allowable average shearing stress is taken as^/(3)
0
- 5
, where^J, denotes the yield-point
stress. The capacity of the web = 0.554(26.69)(36/3
05
) = 307 kips (1365.5 kN). There-
fore, use diagonal web stiffeners.
- Design the web stiffeners
Referring to Fig. 1Oc, we see that ac = (18.3^2 + 26.7^2 )^0 -^5 = 32.4 in (823 mm). The force in
the stiffeners = (465 - 307)32.4/26.7 = 192 kips (854.0 kN). (The same result is obtained
by computing F 2 and considering the capacity of the web across ab.) Then, Ast = 192/36 =
5.33 in^2 (34.39 cm^2 ). Use two plates 4 x % in (101.6 x 19.1 mm).
- Design the welds, using E60 electrodes
The AISC Specification stipulates that the weld capacity at ultimate load is 1.67 times the
capacity at the working load. Consequently, the ultimate-load capacity is 1000 Ib/lin in
(175 N/mm) times the number of sixteenths in the weld size. The welds are generally de-
signed to develop the full moment capacity of each member. Refer to the AISC Specifica-
tion.
Weld at ab. This weld transmits the force in the flange of the 27-in (685.8-mm) mem-
ber to the web of the 18-in (457.2-mm) member. Then F = 9.96(0.636)(36) = 228 kips
(1014.1 kN), weld force = 228/[2(J - 2^)] = 228/[2(18.32 - 1.82)] = 6.91 kips/lin in
(1210.1 N/mm). Use a 7i6-in (11.1 -mm) weld.
Weld at be. Use a full-penetration butt weld.
Weld at ac. Use the minimum size of^1 A in (6.4 mm). The required total length of weld
is L = 192/4 = 48 in (1219.2 mm).
Weld at dc. Let F 3 denote that part of F 2 that is transmitted to the web of the 18-in
(457.2-mm) member through bearing, and let F 4 denote the remainder of F 2. Force F 3 dis-
tributes itself through the 18-in (457.2-mm) member at 45° angles, and the maximum
compressive stress occurs at the toe of the fillet. Find F 3 by equating this stress to 36
kips/in^2 (248.2 MPa); orF 3 = 36(0.554)(0.636 + 2 x 1.625) = 78 kips (346.9 kN). To eval-
uate F 4 , apply the moment capacity of the 27-in (685.8-mm) member. Or F 4 = 228 - 78 =
150 kips (667.2 kN).
The minimum weld size of^1 A in (6.4 mm) is inadequate. Use a^5 /i6-in (7.9-mm) weld.
The required total length is L = 150/5 = 30 in (762.0 mm).
CURVED KNEE OF RIGID BENT
In Fig. 11 the rafter and column are both W21 x 82, and the ultimate moment at the two
sections of tangency—p and q—is 6600 in-kips (745.7 kN-m). The section of contraflex-
