Page 196 - Handbook of Civil Engineering Calculations, Second Edition

HANGERS, CONNECTORS, AND WIND-STRESS ANALYSIS 1.179

Calculation Procedure:

1. 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 105, d 18.32 in. (465.3 mm);
b f 11.79 in. (299.5 mm); t f 0.911 in. (23.1 mm); t w 0.554 in. (14.1 mm); k 1.625
in. (41.3 mm). For the W27 84, d 26.69 in. (677.9 mm); b f 9.96 in. (253 mm); t f
0.636 in. (16.2 mm); t w 0.463 in. (11.8 mm).

2. Compute F 1
Thus, F 1 M u /(0.95d) 8100/[0.95(18.32)] 465 kips (2068.3 kN).
3. 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.
0.5
The allowable average shearing stress is taken as f y /(3) , where f y denotes the yield-point
0.5
stress. The capacity of the web 0.554(26.69)(36/3 ) 307 kips (1365.5 kN). There-
fore, use diagonal web stiffeners.
4. Design the web stiffeners
2 0.5
2
Referring to Fig. 10c, we see that ac (18.3 + 26.7 ) 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 ob-
tained by computing F 2 and considering the capacity of the web across ab.) Then, A st
2
3
192/36 5.33 sq.in. (34.39 cm ). Use two plates 4 /4 in. (101.6 19.1 mm).
5. 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
lb/lin in. (175 N/mm) times the number of sixteenths in the weld size. The welds are
generally designed to develop the full moment capacity of each member. Refer to the
AISC Specification.
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(d – 2t f )] 228/[2(18.32 – 1.82)] 6.91 kips/lin in.
7
(1210.1 N/mm). Use a /16-in. (11.1-mm) weld.
Weld at bc. Use a full-penetration butt weld.
Weld at ac. Use the minimum size of /4 in. (6.4 mm). The required total length of
1
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/sq.in. (248.2 MPa); or F 3 36(0.554)(0.636 + 2 1.625) 78 kips (346.9 kN). To
evaluate F 4 , apply the moment capacity of the 27-in. (685.8-mm) member. Or F 4 228 –
78 150 kips (667.2 kN).
5
The minimum weld size of /4 in. (6.4 mm) is inadequate. Use a /16-in. (7.9-mm)
1
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 82, and the ultimate moment at the two sec-
tions of tangency—p and q—is 6600 in.·kips (745.7 kN·m). The section of contraflexure in