FIGURE 59
ment M in-lb (N-m). Thus, P = 15,000 Ib (66,720.0 N), M = 15,000(6) = 90,000 in-lb
(10,168.2 N-m).
- Compute the polar moment of inertia of the rivet group
Compute the polar moment of inertia of the rivet group with respect to an axis through its
centroid. Thus, J = 2(*
2
+/) = 6(3)
2
- 4(4)
2
= 118 in
2
(761.3 cm
2
).
- Resolve the tangential thrust on each rivet into its horizontal
and vertical components
Resolve the tangential thrust/ Ib (N) on each rivet caused by the moment into its horizon-
tal and vertical components. fx and fy, respectively. These forces are as follows: fx =
MyIJ and fy = MxIJ. Computing these forces for rivets 1 and 2 (Fig. 59) yields
fx = 90,000(4)7118 = 3050 Ib (13,566.4 N);£ = 90,000(3)7118 = 2290 Ib (10,185.9 N).
- Compute the thrust on each rivet caused by the concentric load
This thrust isJ^ = 15,000/6 = 2500 Ib (11,120.0 N).
- Combine the foregoing results to obtain the total force on the
rivets being considered
The total force F Ib (N) on rivets 1 and 2 is desired. Thus, Fx =fx = 3050 Ib (13,566.4 N);
Fy =fy +fy = 2290 + 2500 = 4790 Ib (21,305.9 N). Then F= [(305O)
2
- (479O)
2
]
0 5
= 5680
Ib (25,264.6 N).
The above six steps comprise method 1. A second way of solving this problem,
method 2, is presented below.
The total force on each rivet may also be found by locating the instantaneous center of
rotation associated with this eccentric load and treating the connection as if it were sub-
jected solely to a moment (Fig. 596).
- Locate the instantaneous center of rotation
To locate this center, apply the relation h = J/(eeN), where N= total number of rivets and
the other relations are as given earlier. Then h = 118/[6(6)] = 3.28 in (83.31 m).
