Handbook of Civil Engineering Calculations

(a) Moment on riveted (b) Forces on rivets

connection in right row

FIGURE 58

/i =/4 = 96,000(5.15)7140 = 3530 Ib (15,701.4 N); and/ 2 =/ 3 = 96,000(2.92)7140 = 2000
Ib (8896.0 N). The directions are shown in Fig. 586.

4. Compute the shearing stress
   Using s = PIA 9 we find S 1 =S 4 = 3530/0.442 = 7990 lb/in^2 (55,090 kPa); also, S 2 = S 3 =
   2000/0.442 = 4520 lb/in^2 (29,300 kPa).
   5. Check the rivet forces
   Check the rivet forces by summing their moments with respect to an axis through the cen-
   troid. Thus M 1 =M 4 = 3530(5.15) = 18,180 in-lb (2054.0 N-m); M 2 =M 3 = 2000(2.92) =
   5840in-lb (659.8 N-m). Then EM = 4(18,180) + 4(5840) = 96,080 in-lb (10,855.1 N-m).

ECCENTRIC LOAD ON RIVETED

CONNECTION

Calculate the maximum force exerted on a rivet in the connection shown in Fig. 59a.

Calculation Procedure:

1. Compute the effective eccentricity
   To account implicitly for secondary effects associated with an eccentrically loaded con-
   nection, the AISC Manual recommends replacing the true eccentricity with an effective
   eccentricity.
   To compute the effective eccentricity, use ee = ea-(\ + n)/2, where ee = effective ec-
   centricity, in (mm); ea = actual eccentricity of the load, in (mm); n = number of rivets in a
   vertical row. Substituting gives ee = 8 - (1 + 3)/2 = 6 in (152.4 mm).


2. Replace the eccentric load with an equivalent system
   The equivalent system is comprised of a concentric load P Ib (N) and a clockwise mo-

3 (S) 3"

(76.2

mm)

=

9"(228.6

mm)