Determine the greater of Tr /g or Tr /(db 2b): LRFD
ASD With Tr Ta Ta 5.05 kips = g 3.50 in.
With Tr Tu Tu 7.60 kips = g 3.50 in. = 2.17 kips/in. Tu 7.60 kips = d b + 2b w in. + 2 (11.0 in. – 8.00 in.
= 1.44 kips/in. Ta 5.05 kips = d b + 2b w in. + 2 (11.0 in. − 8.00 in.
)
= 1.13 kips/in. Use Tu /g in the weld size determination.
)
= 0.748 kips/in. Use Ta /(db 2b) in the weld size determination.
Using the procedure given in the AISC Manual Part 8, determine the weld size required. LRFD
ASD Rn = 0.928Dl Ω Rn Ta g l
Rn 1.392 Dl Rn Tu l g 7.60 kips 3.50 in. D ≥ 1.56 sixteenths-of-ann-inch Use --in. minimum weld size with 70-ksi filler metal. 1.392 D ≥
5.05 kips 3.50 in. D ≥ 1.55 sixteenths-of-an n-inch Use --in. minimum weld size with 70-ksi filler metal. 0.928 D ≥
Example 4.2—Through-Plate Connection Given: Design a through-plate connection as shown in Figure 4-4. Use 1-in.-diameter A325-N bolts in standard holes and 70-ksi filler metal. The applied dead and live loads are shown in Figure 4-4. The loads on the two beams can be reversed; therefore, the connection should be symmetric.
Fig. 4-4. Through-plate moment connection. AISC DESIGN GUIDE 24 / HOLLOW STRUCTURAL SECTION CONNECTIONS / 37