Memorandum To: Charles Riley, Ph.D., P.E. From: Nathan Miller and Sadie Brackeen Date: April 18th, 2022 Subject: Laboratory 4 – Tensile Testing of Riveted Connections Introduction: As per your request on April 18th, 2022, my colleagues and I have completed the tensile testing of each of the samples, according to the standards that were outlined in your memorandum, and using the tensile testing machinery in the Strength of Materials Laboratory at the Oregon Institute of Technology. The purpose of this memorandum is to document, analyze, and discuss the results of our experiment. This memo includes a brief description of the procedure, lab results, a discussion of said results, the potential errors that could’ve occurred, and an assessment of the designs for each type of connection and size of rivet. Methods: The testing was performed on April 18th, 2022, in Cornett Hall, Room 144, as well as the Strength of Materials Laboratory, on the Oregon Tech Campus. The goal of our testing was to determine the tensile strength, deformation, and general patterns that each of the riveted setups followed, as well as to design multiple setups for each type of connection, size of rivet, and number of rivets. The materials and tools used were A36 Steel, Vernier calipers, and our tensile testing apparatus. To test, we measured each material to get the necessary dimensions, then collected data on the type of connection, size, number, and pattern of rivets. Next, we took our specimens and put them into the tensile testing apparatus, and ran the program on the machine’s computer. As the specimens were being tested, we observed how and where they deformed, as well as the graph of data, noting the values at which certain material properties could be estimated. Once the specimens fractured, we pulled them out and analyzed how they fractured, noting certain behaviors and patterns we recognized between samples. After, we went back to our lab and analyzed the data collected on the testing machine’s computer, found certain values for maximum stress and yield stress, and finally we came up with designs for each setup that you requested. Results & Analysis: The data found in the attached document; Microsoft Excel presents raw data from the laboratory. The following data located in Figures 1 through 5, which are tables illustrating the different designs with a connection for an A36 steel tension member, are based off experimental results and calculations found in the Microsoft Excel document. The figures were calculated and formatted in Microsoft Excel. Comparing Figures 2, 3, 4, and 5, the best option for the designed connection would be the 1/4" diameter rivet and lap splice pair. Specimen Splice/Shear Type (Lap=1, Butt=2) 1 2 1 2 3 4 Rivet Diameter (1/4" or 3/8") 0.25 0.25 Number of Rivets per Row Plate Thickness (in) Plate Width (in) 12 9 0.25 0.250 4 4 Total Rivet Area (in2) 3.000 0.884 Gross Area (in2) Net Area (in2) Total Bearing Area (in2) 1.000 1.000 0.250 0.438 0.750 0.563 1 0.375 6 0.250 4 0.663 1.000 0.438 0.563 2 0.375 6 0.25 4 1.325 1.000 0.438 0.563 Figure 1. Illustrates the specimen number, the splice/shear type, rivet diameter, and number of rivets in each material specimen. The total rivet area, gross area, net area, and total bearing area can be used to find predictions of at what load and which part of the specimen will be the mechanism for failure. Specimen 1 and 2 have the same rivet diameter (1/4”), but differ according to the splice/shear type. Specimen 3 and 4 also have the same rivet diameter (3/8”), but are either lap type shear or butt. Lap Splice w= 4 t = 0.25 sy = 36 in in ksi Plate width Plate thickness Plate yield strength su = 58 ksi Plate ultimate strength tu = 34.8 ksi Rivet shear ultimate strength d = 0.25 Ar = 0.049 in in2 Rivet diameter Rivet cross-sectional area # cols = # rows = # rivets = shear type = 4 3 12 1 Number of rivets across the connection Number of rivets in a line parallel to the load Total number of rivets 1 for single shear (lap splice), 2 for double shear (butt splice) OK Required load = 1/2 yield load for the member min width = 3.75 in P = 18 kip Connection Failure Mechanisms Shear in Rivets 20.5 kip OK Bearing in Plate 27.0 kip OK Fracture of Net 43.5 kip OK Section Figure 2. Lap splice design for a 1/4” x 4” A36 steel tension member with a 1/4" rivet diameter and a total of 12 rivets. Shearing in rivets is the expected failure mechanism that occurs first. Butt Splice w= 4 t = 0.25 sy = 36 in in ksi Plate width Plate thickness Plate yield strength su = 58 ksi Plate ultimate strength tu = 34.8 ksi Rivet shear ultimate strength d = 0.25 Ar = 0.049 in in2 Rivet diameter Rivet cross-sectional area # cols = # rows = # rivets = shear type = 3 3 9 2 Number of rivets across the connection Number of rivets in a line parallel to the load Total number of rivets 1 for single shear (lap splice), 2 for double shear (butt splice) OK Required load = yield load for the member min width = 2.75 in P = 18 kip Connection Failure Mechanisms Shear in Rivets 30.7 kip OK Bearing in Plate 20.3 kip OK Fracture of Net 47.1 kip OK Section Figure 3. Butt splice design for a 1/4” x 4” A36 steel tension member with a 1/4" rivet diameter and a total of 9 rivets. The bearing in the plate is the expected failure mechanism that occurs first, before the shearing in rivets and fracturing of the net section. Lap Splice w= 4 t = 0.25 sy = 36 in in ksi Plate width Plate thickness Plate yield strength su = 58 ksi Plate ultimate strength tu = 34.8 ksi Rivet shear ultimate strength d = 0.375 Ar = 0.110 in in2 Rivet diameter Rivet cross-sectional area # cols = # rows = # rivets = shear type = 2 3 6 1 Number of rivets across the connection Number of rivets in a line parallel to the load Total number of rivets 1 for single shear (lap splice), 2 for double shear (butt splice) OK Required load = 1/2 yield load for the member min width = 2.625 in P = 18 kip Connection Failure Mechanisms Shear in Rivets 23.1 Kip OK Bearing in Plate 20.3 kip OK Fracture of Net 47.1 kip OK Section Figure 4. Lap splice design for a 1/4” x 4” A36 steel tension member with a 3/8" rivet diameter and a total of 6 rivets. The bearing in the plate is the expected failure mechanism that occurs first, then the shearing of rivets, and the fracturing of the net section would occur last. Butt Splice w= 4 t = 0.25 sy = 36 in in ksi Plate width Plate thickness Plate yield strength su = 58 ksi Plate ultimate strength tu = 34.8 ksi Rivet shear ultimate strength d = 0.375 Ar = 0.110 in in2 Rivet diameter Rivet cross-sectional area # cols = # rows = # rivets = shear type = 2 3 6 2 Number of rivets across the connection Number of rivets in a line parallel to the load Total number of rivets 1 for single shear (lap splice), 2 for double shear (butt splice) OK Required load = yield load for the member min width = 2.625 in P = 18 kip Connection Failure Mechanisms Shear in Rivets 46.1 kip OK Bearing in Plate 20.3 kip OK Fracture of Net 47.1 kip OK Section Figure 5. Lap splice design for a 1/4” x 4” A36 steel tension member with a 3/8" rivet diameter and a total of 6 rivets. The bearing in the plate is the expected failure mechanism that occurs first, then the shearing of rivets, and the fracturing of the net section would occur last. The last two mechanisms of failure are nearly the same in terms of at what point they will occur, so there is potential for them to happen at the same time. Figure 6. Failure surface photograph of the four metal specimen connections that were tested. From right to left, the first specimen failed by shearing in the rivets at 3.96 kips, the second specimen sheared in the rivets at 8.91 kips, the third specimen fractured in the net section at 13.09 kips, and for the fourth specimen, it failed through shearing in the rivets at 7.92 kips. Discussion: For the most part, our data followed our expectations, in terms of strength and ductility. Each of the designs shown above meet the requirements that you laid out in your memo, and additionally make sense physically. As mentioned above, the best option for the designed connection would be a lap splice with 0.25” diameter rivets. The reason we chose this is because through our calculations, we found that that design would have the least shear in the rivets (20.5 kips), the least load bearing in the base plate (27.9 kips), and the lowest load at fracture of the net section (43.5 kips). The second-best would be a lap splice using 0.375” diameter rivets, then a butt splice using 0.25” diameter rivets, and finally, a butt splice using 0375” diameter rivets. All the data on their failure mechanisms and at what load they occur can be seen in figures 2-5 above. Conclusions: The best design for the rivet that you requested is a lap splice using 0.25” diameter rivets because it has the least load, and by nature, the least stress out of all the designs. The primary source of error in our experiment was likely the setup of the tensile testing apparatus when testing the materials, where they could’ve been set in the jaws improperly, leading to stress points that were away from the rivets and skewed data, and/or the batch that each material being pulled from wasn’t up to standards, and so had improper material properties. Future tests should verify the purity and quality of the batch of materials prior to testing, as well as either using different types of jaws that clamp more evenly, and are easier to lineup a piece in, or using the same jaws, make sure the piece is perfectly lined up front-to-back and upto-down in the apparatus.
