ECCENTRIC LOAD ON A WELDED
CONNECTION
The bracket in Fig. 61 is connected to its support with a Vi-in (6.35-mm) fillet weld. De-
termine the maximum stress in the weld.
Calculation Procedure:
- Locate the centroid of the
weld group
Refer to the previous eccentric-load cal-
culation procedure. This situation is anal-
ogous to that. Determine the stress by lo-
cating the instantaneous center of
rotation. The maximum stress occurs at A
and B (Fig. 61).
Considering the weld as concentrated
along the edge of the supported member,
locate the centroid of the weld group by
taking moments with respect to line aa.
Thus m = 2(4)(2)/(12 + 2 x 4) = 0.8 in
(20.32 mm).
- Replace the eccentric load
with an equivalent concentric FIGURE 61
load and moment
Thus P = 13,500 Ib (60,048.0 N); M -
124,200 in-lb (14,032.1 N-m).
- Compute the polar moment of inertia of the weld group
This moment should be computed with respect to an axis through the centroid of the
weld group. Thus Ix = (1/12)(12)^3 + 2(4)(6)^2 = 432 in^3 (7080.5 cm^3 ); / = 12(0.8)^2 +
2(1/12)(4)^3 + 2(4)(2 - 0.8)^2 = 29.9 in^3 (490.06 cm^3 ). Then J = Ix + / = 461.9 in^3
(7570.54 cm^3 ).
- Locate the instantaneous center of rotation O
This center is associated with this eccentric load by applying the equation h = J/(eL),
where e = eccentricity of load, in (mm), and L = total length of weld, in (mm). Thus, e =
10 - 0.8 = 9.2 in (233.68 mm); L =12 + 2(4) = 20 in (508.0 mm); then h = 461.9/[9.2(2O)]
= 2.51 in (63.754 mm).
- Compute the force on the weld
Use the equation F = Mr'IJ, Ib/lin in (N/m), where r' = distance from the instantaneous
center of rotation to the given point, in (mm). AtA and B, r' = 8.28 in (210.312 mm); then
F= [124,200(8.28)]/461.9 = 2230 Ib/lin in (390,532.8 N/m).
- Calculate the corresponding stress on the throat
Thus, S = PIA = 2230/[0.707(0.25)] = 12,600 lb/in^2 (86,877.0 kPa), where the value 0.707
is the sine of 45°, the throat angle.
