MEE 324 Lab 3

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Arizona State University, Tempe *

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324

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Mechanical Engineering

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Apr 3, 2024

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Lab 3: Pure Torsion Fabian Ameen Lab Number: Thursday 10:30 - 11:35 AM Date of Experiment: 2/22/2024 Due Date: 3/15/2024

Abstract: This experiment aimed to test the torsion behavior of 1018 Steel and 2024 Aluminum samples through experimental analysis. The experimental setup used a biaxial machine to apply torque to bone-shaped circular rods and hollow tubes, with torsion and twist angle measured within the linear elastic regime. Prior to testing, measurements of the sample’s geometry were taken. The measurements from the lab were then used to estimate the value of the shear modulus for each sample using a linear regression, polar moment of inertia, applied torque, and the degree of twist. Combining the estimated shear modulus with the previously obtained elastic modulus, the Poisson’s ratio was calculated for the steel samples. Additionally, the Poisson’s ratio and elastic modulus of aluminum from previous labs was used to calculate an expected shear modulus which was then compared to the estimated value from this experiment. Data Analysis: The first step of this experiment was to take measurements of the diameter and length of each of the 4 samples to use for calculations later. Solid Steel Hollow Steel Solid Aluminum Hollow Aluminum Length (in) 10.875 10.875 10.875 10.875 Diameter (in) 0.369 0.501 0.381 0.512 Figure 1: Dimensions of the samples Based on the gathered data of the applied torque and the resulting twist (in radians) 4 diagrams can be made for each sample that was tested Figure 2: Diagram of Torque vs Twist for Solid Steel sample y = 0.0006x + 0.0007 R² = 0.9999 0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0 10 20 30 40 50 60 70 Twist (radians) Torque (Lb-in) Torque vs Twist (Solid Steel)

Figure 3: Diagram of Torque vs Twist for Hollow Steel sample Figure 4: Diagram of Torque vs Twist for Solid Aluminum sample Figure 5: Diagram of Torque vs Twist for Hollow Aluminum sample ࠵? = 0.0002*T + 0.0073 R² = 0.9995 0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0 20 40 60 80 100 120 140 Twist (radians) Torque (Lb-in) Torque vs Twist (Hollow Steel) ࠵? = 0.0015*T - 0.0046 R² = 1 0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0 5 10 15 20 25 30 Twist (radians) Torque (Lb-in) Torque vs Twist (Solid Aluminum) ࠵? = 0.0007*T + 0.0007 R² = 1 0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0 10 20 30 40 50 60 Twist (radians) Torque (Lb-in) Torque vs Twist (Hollow Aluminum)

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In order to relate the given equation for the relationship between torque and twist angle (1) to the trendline’s of each diagram, first the polar moment of inertia must be calculated for each sample using the given equation (2) ࠵? = ࠵? ∗ ࠵? ∗ ∆" # (1) ∫ ∫ ࠵? $ ∗ ࠵? ∗ ࠵?࠵? ∗ ࠵?࠵? % ! % " $& ’ (2) Using the measured values of the samples’ diameters the polar moment of inertia can be found for each sample using two equations, one for a circular cross section and one for a hollow cross section (2). * * ࠵? $ ∗ ࠵? ∗ ࠵?࠵? ∗ ࠵?࠵? ( $ ( $ )’.’+, $& ’ Solid Steel Hollow Steel Solid Aluminum Hollow Aluminum J (in^4) 0.00175 0.00432 0.00206 0.00465 Figure 6: Calculated values for the polar moment of inertia Based on equation 1 the slope of the linear regression would be m = L/(G * J). Therefore, the shear modulus of each sample can be estimated using the following equation. ࠵? = ࠵? ࠵? ∗ ࠵? Solid Steel Hollow Steel Solid Aluminum Hollow Aluminum G (ksi) 10357.142 12542.101 3605.854 3336.705 Figure 7: Estimated Shear modulus based on experimental data. ࠵? = - $(/01) (3) Using the elastic modulus for steel found in Lab 1 (Figure 8) and the estimated shear modulus the following equation can be used to calculated Poisson’s ratio of the steel sample using equation 3. This gives a value of 0.25898 for the solid sample and 0.52357 for the hollow sample. Compared to the literature value of 0.29 [1] the calculated Poisson’s ratio of the solid steel sample was much closer than the hollow sample. Steel GPa ksi E 217.74 31580.517 Figure 8: Lab 1 elastic modulus and converted value Solid Steel Hollow Steel ࠵? 0.25898 0.52457

Using the elastic modulus of aluminum from lab 1, 217.74 GPa or 31580.517 ksi, and the Poisson ratio from lab 2, 0.4 and putting them into equation 3 gives an expected shear modulus of 4409.66535 ksi, which is higher than the estimated values of both aluminum samples. Conclusion In conclusion, this experiment focused on understanding how 1018 Steel and 2024 Aluminum materials behave under torsion was successful. By performing torsion tests the angle of twist for di_erent levels of applied torque was measured, and these relationships were plotted. Using the findings the shear modulus was calculated for each of the samples. Additionally, previous data was used to predict and compare values. Overall, the experiment improved my understanding of torsional behavior in materials. References AISI 1018 Steel, Cold Drawn , www.matweb.com/search/datasheet_print.aspx?matguid=3a9cc570fbb24d119f08db22a53e 2421. Accessed 15 Mar. 2024. [1]