WORLD FOR CIVIL

Topic 6.9b: Torsion, Rivets & Welds - Topic Exam Solutions

Rivets & Welds - Solution Problem 1

1. A riveted lap joint is shown in Diagram 1. The diameter of the rivets is 1 inch. The width of the plates is 12 inches, and the thickness of the plates is 5/8 inch. The allowable stresses are as follows:

Rivets: = 20, 000 lb/in², _t = 26, 000 lb/in², _c = 28, 000 lb/in²
Plate: = 17,000 lb/in², _t = 22, 000 lb/in², _c = 24, 000 lb/in²
Determine the Strength of the Joint, and the Efficiency of the Joint.

Solution:
Part 1
1. Rivet Shear: The load the joint can carry before failing in rivet shear is given by:
P = N (pi * d²/4) = ( 16 rivet areas)* [3.1416 * (1)²/4]*20, 000 lb/in² = 251,300 lb.
Thus at a load of 238,800 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.
P_bearing = N (d*t) _all = ( 16 rivets) * (1" * 5/8") * (24,000 lb/in²) = 240,000 lb.
Thus a load of 240,000 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 material.

3. Plate Tearing (Row 1): We now determine the load the joint (plate) can carry before failing in tension - at row 1.
P_row1 = (w - n d)t _all = (12"- 1* 1") *(5/8") * (22,000 lb/in² ) = 151,250 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 plate tearing at row 1 with the lowest load of 151,250 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): . Since there are 16 rivets in the pattern and we assume that the rivets share the load equally, then one sixteenth (1/16) of the load has been transferred to the bottom plate by the rivet in row 1. Thus row 2 carries of 15/16 of the load P, and we may write.
(15/16)P_row2 = (w - n d)t _all = (12"- 2* 1") *(5/8") * (22,000 lb/in² ) = 137,500 lb., and then: P_row2 = (16/15) * 137,500 lb. = 146,700 lb.
This is the load which will cause shear row 2 to fail in tension. We see that it is lower than the lowest load so far that will cause the joint to fail, and thus it is the strength of the joint to this point.

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 3/16 of the load has been transferred to the bottom plate, and thus the plate material at row 3 carries only 13/16 of the load, P. Then we can write:
(13/16)P_row3 = (w - n d)t _all = (12"- 3* 1") *(5/8") * (22,000 lb/in² ) = 123,750 lb., and then: P_row3 = (16/13) * 123,750 lb. = 152,300 lb.
Here we see that load which causes row 3 to fail in tension is 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 plate tearing at row 2 - 146,700 lb., and this is the final Strength of the Joint, the largest load we can safely apply.

Part 2
Plate Strength = area of plate cross section times allowable tensile stress for plate material, or
P_plate = (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 = 146,700 lb. /(12" * 5/8") * 22,000 lb/in² = 146,700 lb. / 165,000 = .89 = 89%. This says that joint can carry 89 percent of what the solid plate could carry.