- Compute the force on the rivets
Considering rivets 1 and 2, use the equation F = Mr'IJ, where rr = distance from the in-
stantaneous center of rotation O to the center of the given rivet, in. For rivets 1 and 2, r' =
7.45 in (189.230 mm). Then F = 90,000(7.45)7118 = 5680 Ib (25,264.6 N). The force on
rivet 1 has an action line normal to the radius OA.
DESIGN OF A WELDED LAP JOINT ____
The 5-in (127.0-mm) leg of a 5 x 3 x % in (127.0 x 76.2 x 9.53 mm) angle is to be weld-
ed to a gusset plate, as shown in Fig. 60. The member will be subjected to repeated varia-
tion in stress. Design a suitable joint.
Calculation Procedure:
- Determine the properties of the angle
In accordance with the AISC Specification, arrange the weld to have its centroidal axis
coincide with that of the member. Refer to the AISC Manual to obtain the properties of
the angle. Thus A = 2.86 in^2 (18.453 cm^2 ); yv = 1.70 in (43.2 mm); y 2 = 5.00 - 1.70 = 3.30
in (83.820 mm). - Compute the design load and required weld length
The design load P Ib (N)^=As = 2.86(22,000) = 62,920 Ib (279,868.2 N). The AISC Spec-
ification restricts the weld size to^5 /i6 in (7.94 mm). Hence, the weld capacity = 5(600)
3000 Ib/lin in (525,380.4 N/m); L = weld length, in (mm) = P/capacity, Ib/lin in =
62,920/3000 = 20.97 in (532.638 mm). - Compute the joint dimensions
In Fig. 60, set c = 5 in (127.0 mm), and compute a and b by applying the following equa-
tions: a = Ly 2 /w - c/2; b = Ly 1 Iw - c/2. Thus, a = (20.97 x 3.30)75 -^5 A =11.34 in
(288.036 mm); b = (20.97 x 1.70)75 -^5 A = 4.63 in (117.602 mm). Make a = 11.5 in
(292.10 mm) and b = 5 in (127.0 mm).
FIGURE 60angle