Experiment #8 Shear Test on Metal Fastners

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Washington State University *

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225

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Mechanical Engineering

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Apr 3, 2024

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Met 225 Shear Test of Metal Fasteners November 20 th ,2023 Aaliyah Ashley aashl009@odu.edu

Shear Test of Metal Fasteners Objectives: The objective of this experiment is to perform a shear test on steel bolt fasteners in both single and double shear. To determine the shearing stress on all the fasteners in each test preformed. Determine the bearing stress behind each bolt on each plate in each test performed. Determine the normal stress at the critical sections for all the plates pulled in each test performed. Theory: The theory behind the experiment is to focus on the strength of fasteners in single and double shear connection. The strength of connection plates is stronger when compared to the strength of the fasteners. Equations: T = P nA T = P 2 nA Description of Apparatus: Bolted connection-Single Shear This was for single shear and the bolt, washers, nuts are holding the flat bars in place before getting placed into the machine. Bolted connection- Double shear This was double shear flat plates were added and more bolts and fasters.

Universal Testing Machine This is 60k Super “L” that was to test single and double shear. It is white has a knob to tighten the piece in the machine. Adjustable Wrench The adjustable wrench was used to tighten the fasteners and nuts into the flat heads and bars . The wrench could be adjusted to how big or small the nut is. Hex Head fasteners and nuts These were used to get the shear from when put into the fastners.

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Procedure: 1. The diameter of each of the bolts were measure, the cross-sectional area of each of the bars, and the net cross sectional areas at each drilled hole. 2. The two bars were connected using the lap joint. 3. The remaining bars as a butt joint with a cover plate on each side of the joint were connected. 4. The tensile strength of the bars were assumed and the shear strength of the bolts were 60 ksi, and maximum load required was calculated to cause failure in both test specimen setups. 5. The proper range on the testing machine for the lap joint specimen was selected. The specimen was inserted, and tension was pulled in until failure occurred. 6. Step five was repeated for the butt joint specimen. Data Table: Lap Joint: Minimum bolt diameter at shear plane: 0.250in Ultimate Load: 2637.1ibf Bolt Cross-sectional area: 0.49in Plate A(in) B(in) Hole Diameter 0.2625 0.2680 Plate Width 2.0125 2.0095 Plate Thickness 0.3720 0.3790 Net Section Area 0.651 0.660 Bearing Area 0.09765 0.1016 Butt Joint: Minimum bolt diameter at shear plane: 0.250in Ultimate Load: 5040.1 ibf Bolt Cross-sectional area: 0.49in Plate/ Position Hole Diameter Plate Width Plate Thickness Net Section Area Bearing Area C/1 0.2680 2.0095 0.3790 0.660 0.1016 C/2 0.2680 2.0095 0.3790 0.660 0.1016 C/3 0.2680 2.0095 0.3790 0.660 0.1016 D/4 0.2680 2.0095 0.3790 0.660 0.1016 E/1 0.2680 2.0095 0.3790 0.660 0.1016

E/2 0.2680 2.0095 0.3790 0.660 0.1016 E/3 0.2680 2.0095 0.3790 0.660 0.1016 E/4 0.2680 2.0095 0.3790 0.660 0.1016 F/2 0.2680 2.0095 0.3790 0.660 0.1016 F/2 0.2680 2.0095 0.3790 0.660 0.1016 F/3 0.2680 2.0095 0.3790 0.660 0.1016 F/4 0.2680 2.0095 0.3790 0.660 0.1016 Type of Stress Lap Joint Butt joint with Cover Plates-Double Shear 1 2 3 4 Bolt Shear Stress(T) 53722.6 51338.03 51338.03 51338.03 51338.03 Bearing Stress(  b) 40184.3 76076.98 76076.98 76076.98 76076.98 Cover Plate 76076.98 76076.98 76076.98 76076.98 Butt Plate Critical Section stress(  cs) 657604.7 164401.2 164401.2 164401.2 164401.2 Cover plate 164401.2 164401.2 164401.2 164401.2 Butt Plate Sample Calculations: Net Section Area : ( .3720 ∗ 2.0125 ) − 0.09765 = 0.651 Bearing Area : ( 0.2625 ∗ 0.3720 ) = 0.09765 Bolt Shear Stress : 5040.1 1 ∗ ( π 4 ) ∗ ( . 250 2 ) ∗ 2 = 51.338.03 Bearing Stress 5040.1 .250 ∗ 2650 = 76076.98 Critical SectionStr ess π 2 ∗ 2398654.4 ( .75 / .125 ) 2 = 657604.7

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Discussion: The results for this experiment are going to be talk about, how the lab could have gone better and how the fasteners looked once the failed. For starters there were not many things that went wrong in the experiment. Before starting the experiment, the fasteners were perfectly straight. When the load failed the force was around 3000 ibf for single shear and around 5000 ibf for double shear. The fastener when failed for single shear when it was being pulled it looks like if it were to break it happen towards the top but since was needed to break. The top half of the fastener has shifted to the side compared to the rest of it. The fastener for the double shear indented in the middle of it. Since it happened closer to the middle there is a section of the fastener that is sticking out compared to the rest of it. There was nothing that really went wrong with experiment expect that the fasteners got stuck in the plates which meant that the were stuck in the machine. The only with that I think should have been done differently is testing the bolts first to see which one unscrews first. That way you know which ones to unscrew first instead of trying to unscrew the one and struggling with it. Overall, This experiment went well and nothing major went wrong. The results for double and single shear failed at different loads but they were around 2000 ibf from each other. The fasteners failed in two different places.