Topic 6.5a: Riveted Joints - Example 1
A riveted lap joint is shown in Diagram 1. The diameter of the rivets is 5/8 inch. The width of the plates is 6 inches, and the thickness of the plates is 1/2 inch. The allowable stresses are as follows:
Rivets: 
= 16, 000 lb/in2, 
t = 22, 000 lb/in2, c = 24, 000 lb/in2
Plate: = 14,000 lb/in2, t = 20, 000 lb/in2, c = 23, 000 lb/in2
We would like to determine the Strength of the Joint, and the Efficiency of the Joint.
Part 1: To determine the Strength of the Joint, we calculate the load, P, which will cause the joint to fail in each of the main modes of failure (Rivet Shear, Bearing, and Plate Tearing). The lowest load which will cause the joint to fail is known as the Strength of the Joint. Please remember that the term failure of the joint in these problems does not mean when the joint actually breaks, but rather we consider the joint to have failed when the stress in the joint exceed the allowable stress. At this point the joint may or may not actually fail depending on the safety factor used in the determining the allowable stress, and of course, depending on how much the allowable stress is exceeded.
1. Rivet Shear: The load the joint can carry before failing in rivet shear is given by:
P = N (pi * d2/4) = ( 9 rivet areas)* [3.1416 * (5/8)2/4]*16, 000 lb/in2 = 44,200 lb.
Thus at a load of 44, 200 lb., the joint will fail in shear. Please note we used the allowable shear stress for the rivet material in our equation, as it is the rivet (not the plate) which fails in shear.
2. Bearing (compression) Failure: We next determine the load the rivet or plate can carry before failing in compression.
Pbearing = N (d*t) all = ( 9 rivets) * (5/8" *1/2") * (23,000 lb/in2) = 64,700 lb.
Thus a load of 64,700 lb. will cause the joint (plate) to fail in compression. Notice we use the smaller of the allowable compressive stress between the rivet and plate.
3. Plate Tearing (Row 1): We now determine the load the joint (plate) can carry before failing in tension - at row 1.
Prow1 = (w - n d)t all = (6"- 1* 5/8") *(1/2") * (20,000 lb/in2 ) = 53,750 lb.
This is the load which will cause the plate to fail in tension at row 1.
We see that of the three main modes of failure above, the joint fails first in rivet shear at the lowest load of 44,200 lb. Up to this point, this is the Strength of the Joint, however we still need to check plate tearing failure at rivet row 2 (and perhaps row 3) to see if the joints fails at a lower load there. We do this as follows.
3a. Plate Tearing (Row 2): The main difference between Plate Tearing at row 2 and at row 1 (other than the fact that there are different numbers of rivet in the rows) is that plate material at row 2 does not carry the entire load, P. This is due to the fact that part of the load has been transferred to the bottom plate by the rivet in row 1. Since there are 9 rivets in the pattern and we assume that the rivets share the load equally, then one ninth (1/9) of the load has been transferred to the bottom plate. Thus row 2 carries of 8/9 of the load P, and we may write.
(8/9)Prow2 = (w - n d)t all = (6"- 2* 5/8") *(1/2") * (20,000 lb/in2 ) = 47,500 lb., and then:
Prow2 = (9/8) * 47,500 lb. = 53,400 lb.
This is the load which will cause shear row 2 to fail in tension. We see that it is lower than the load that will cause row 1 to fail (that is, the joint would fail at row 2 before row 1), however the joint will still fail first in rivet shear (44,200 lb.) - since that is the lowest applied load which will cause the joint to exceed the allowable stress.
3b. Plate Tearing (Row 3): Since the joint would fail at row 2 before row 1, we need now to continue and determine the load at which row 3 would fail. As there are a total of 3 rivets in row 1 and row 2, this means that 3/9 of the load has been transferred to the bottom plate, and thus the plate material at row 3 carries only 6/9 of the load, P. Then we can write:
(6/9)Prow3 = (w - n d)t all = (6"- 3* 5/8") *(1/2") * (20,000 lb/in2 ) = 41,250 lb., and then:
Prow3 = (9/6) * 41,250 lb. = 61,900 lb.
Here we see that load which causes row 3 to fail in tension is much larger than the loads which would cause row 1 and row 2 to fail. This means we can stop here.
Of all the loads which will cause the joint to fail (either in shear, compression, or tension) the lowest is still the rivet shear - 44,200 lb., and this is the final Strength of the Joint, the largest load we can safely apply.
Part II. By itself the strength of the joint does not really tell us how 'good' the joint is. The way we determine how good or efficient a joint we have is to compare the strength of the joint with the strength of the plate if it were solid (no riveted or bolted joint). The strength of the joint is 44,200 lb., found above. The strength of the plate we determine by realizing that if it is solid the joint will fail in tension (plate tearing). See Diagram 2. And we can write:
Plate Strength = area of plate cross section times allowable tensile stress for plate material, or
Pplate = (w * t) * all = (6" * 1/2") * 20,000 lb/in2 = 60,000 lb.
This is the plate strength, and we then define the joint efficiency as the ratio of the Joint Strength to the Plate Strength, or
Efficiency = Joint Strength / Plate Strength = 44,200 lb. / 60,000 lb. = .737 = 73.7 %. This tells us that the joint can carry 73.7 percent of what the solid plate could carry.
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