WORLD FOR CIVIL

Topic 6.6a: Riveted Joint Selection - Example 1

Riveted Joint Selection - Example 1

A butt joint (Diagram 1) is to connect two steel plates both with a width of 7 inches and a thickness of 3/4 inch. The rivets to be used have a diameter of 5/8 inch. The maximum allowable stresses for the rivet and weld materials are as follows:

Rivets: = 15, 000 lb/in², _t = 24, 000 lb/in², _c = 26, 000 lb/in²
Plate: = 16,000 lb/in², _t = 22, 000 lb/in², _c = 24, 000 lb/in²
A. Determine the number of rivets for the most efficient joint.
B. Select the best pattern from the patterns in Diagram 2 .
C. Calculate the strength and efficiency of the joint.

Solution: Part A
Step 1: Calculate the load which would cause the joint to fail in plate tearing at row 1 - assuming one rivet in row 1. We use the plate tearing relationship:
P_row1 = (w - n d)t _all. , where
w = width of the main plate = 7 inches
n = number of rivets in row (in row 1, 1 rivet)
d = diameter of rivet = 5/8 inch
t = thickness of the main plate = 3/4 inch
_all = Maximum allowable tensile stress for the plate material = 22, 000 lb/in²,
P_row1 = (7" - 1 * 5/8")*(3/4") * 22, 000 lb/in² = 105,200 lb.

Step 2. Calculate the maximum load which one rivet could carry in rivet shear. We use the rivet shear relationship:
P_{rivet shear} = N (pi * d²/4) _all. Where:
N = Number of areas in shear. For 1 rivet in a double cover plate butt joint this will be 2 area; so N = 2
d = 5/8 inch
_all = 15,000 lb/in²
P_{rivet shear} = 2 [3.1416 * (5/8")²/4] * 15,000 lb/in² = 9200 lb./rivet

Step 3. Calculate the maximum load which one rivet (or the plate material behind one rivet) could carry in bearing (compression).
P_bearing = N (d*t) _all. , where
N = Number of rivets in compression = 1
d = Diameter of rivet = 5/8 inch
t = Thickness of the main plate = 3/4 inch
_all = Lowest allowable compressive stress of the rivet or plate material = 24, 000 lb/in²
P_bearing = 1 * [(5/8") * (3/4")] * 24, 000 lb/in² = 11,250 lb./rivet

Step 4. Divide the allowable load in plate tearing, calculated in step 1, by the smallest of the loads one rivet could carry from steps 2 and 3, and round up to determine the number of rivets which will result in the most efficient joint.
# Rivets = 105, 200 lb. / 9200 lb./rivet = 11.43 ; rounding up # rivets = 12

Solution: Part B
We use the number of rivets found in Part A above to select the best joint pattern from the given set of patterns. From Diagram 1 above we select the a lap joint pattern with correct number of rivets. (See Diagram 3.)

Solution: Part C
Finally, using the selected joint pattern from part B, we calculate the strength and efficiency of the joint.
1.) Rivet Shear: We have already determined that one rivet can carry 9,200 lb./rivet, so the total load the joint can carry in rivet shear will be the product of the number of rivets and the allowable load per rivet: P_shear = 12 * 9200 lb/rivet = 110,400 lb.

2.) Bearing Failure: Again, we have already calculated the allowable load per rivet in bearing = 9000 lb./rivet. The total load the joint can carry in bearing (compression) will be the product of the number of rivets and the allowable load per rivet: P_bearing = 12 * 11,250 lb./rivet = 135,000 lb.

3.) Plate Tearing (row 1): We have already calculated plate tearing at row 1, in step 1 above. P_row1 = (7" - 1 * 5/8")*(3/4") * 22, 000 lb/in² = 105,200 lb. However, now that we have selected the rivet pattern, we also need to calculate plate tearing at row 2 (and perhaps row 3)

4.) Plate Tearing (row 2): Row 2 carries 11/12 of the load, so we write:
(11/12)P_row2 = (7" - 2 * 5/8")*(3/4") * 22, 000 lb/in² = 94,875 lb., and then
P_row2 = (12/11) 94,875 lb. = 103,500 lb. Notice, that this loading is the lowest, up to this point, to cause failure, and so it is currently the strength of the joint. However, we will continue and check the load which causes row 3 to fail.

5.) Plate Tearing (row 3): Row 3 carries 9/12 of the load, so we write:
(9/12)P_row3 = (7" - 2 * 5/8")*(3/4") * 22, 000 lb/in² = 94,875 lb., and then
P_row3 = (12/9) 94,875 lb. = 126,500 lb. This loading is much larger than the loading which will cause row 2 to fail, so we can stop here.

Considering all the modes of failure, we see: Strength of the Joint is 103,500 lb. (plate tearing row 2),and Efficiency = Joint Strength / Plate Strength = 103,500 lb./(7"*3/4")*22,000 lb/in² = .896 = 89.6%

This is the most efficient joint for the dimensions and allowable stresses of both the plates and rivets. By changing the size of the rivets (and the material and associated allowable stresses) we could arrive at a joint of greater efficiency. It is usually possible to arrive at a joint efficiency of 90% or greater. This type of problem is easily done using current computer spreadsheet methods.