MAAE2202_Lab C

docx

School

Carleton University *

*We aren’t endorsed by this school

Course

2202

Subject

Mechanical Engineering

Date

Jan 9, 2024

Type

docx

Pages

Uploaded by LieutenantMetal13000

Carleton University Laboratory Report Course #: MAAE2202A-L3 Lab #: C Lab Section #: Torsion of a Hollow Circular Shaft

Summary The purpose of this lab is to obtain the torsional stiffness of a hollow circular shaft as well as the shear modulus of the shaft material. Another objective of this lab is to verify the simple torsion formula for shear stresses in circular shafts. These goals were accomplished using the torsion of a hollow circular shaft apparatus, with 3 separate trials to ensure accurate results. Nomenclature Table 1: Nomenclature Used in Report Symbol Parameter Unit T Applied Torque Nm J Second Moment of Area m 4 τ Shear Stress kPa D Beam Diameter mm R Beam Radius mm G Shear Modulus GPa θ Angle of Deflection rad L Beam Length mm k Torsional Stiffness Nm/rad F Force N m Mass kg g Gravitational Acceleration m/s 2 d Arm Length mm t Beam Wall Thickness mm W Weight kg ´ X Mean/Average Depends on subject %error Error Percentage % Theory and Analysis In this experiment, we aim to find torsional stiffness and shear modulus of the beam. To accomplish this, multiple steps are required. Since the torque in the beam is being applied by the weights attached to the loading bar, the applied load from the weights must be converted to an applied torque: T = mg d load Equation 1 The torque from the weights cause the beam to twist a certain amount, which is measured using the deflection arm’s gauge. The deflection arm is initially parallel with the ground. The angle of deflection is the angle of the deflection arm relative to the initial angle and can be calculated using Pythagoras ratios: θ = tan − 1 ( verticaldeflection d deflection ) Equation 2 P a g e 2 | 10

The torsional stiffness of the shaft is the rate of change of applied torque with respect the angle of deflection. It can be found by using the slope of a T vsθ graph, and is represented by the equation: k = ∆T ∆θ Equation 3 The torsional stiffness will be used to calculate the shear modulus; however, the second moment of area must be calculated first. By using the dimensions of the beam, the second moment of area can be found: J = π 2 R 4 = π 2 ( R out 4 − R ¿ 4 ) Equation 4 The shear modulus can now be found using the length, tortional stiffness, and second moment of area of the shaft. It is an intrinsic property of a material and can be calculated using the equation: G = ∆T ∆θ ∙ L J = k ∙ L J Equation 5 As the shear modulus of a material is an intrinsic property, the experimental value can be compared to published values and a percentage of error can be obtained. In this experiment, the mean of the 3 trials was used to compare to the published value. The error percentage is represented by the equation: %error = ( G ´ X − G pub G pub ) × 100% Equation 6 This experiment was performed under pure shear conditions. This allowed the experimental shear stress to be calculated using only the measured shear strain in the 45 ̊ plane. In these conditions, the shear strain in the circumferential plane is equal to 2 times the shear strain in the 45 ̊ plane. This, as well as the shear modulus is used to calculate the experimental shear stress: τ exp = G∙ ( 2 ×gauge 2 reading ) Equation 7 The shear stress can be calculated in another, more theoretical way. The theoretical shear stress uses intrinsic properties of the shaft as well as the applied torque. τ theo = T ∙ R out J Equation 8 P a g e 3 | 10

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Experimental Setup and Procedure Figure 1: Experiment apparatus with labeled components Figure 2: Set-up for vertical deflection measurements The experiment was conducted as outlined in the lab manual. Results and Discussion Requirement A One of the steps in the procedure was to adjust the shackle to ensure the vertical deflection at the tip of the beam was zero for each applied load. This was done to prevent the beam from bending and twisting as it would have affected the readings for the deflection arm, and all gauges. If this was not done, separate calculations for the curvature of the beam would have been required to achieve the same result with more inaccuracies. By zeroing the vertical deflection of the beam, it ensured that most of the deformation was in the torsional direction. The strains in the circumferential and longitudinal strain gauges were also zeroed to maximize accuracy of the strain gauge in the 45 ̊ plane (gauge 2). This is important as it is the only strain used to calculate the shear stress in pure shear conditions (this experiment). P a g e 4 | 10

Requirement B The applied torques for each trial were calculated using equation 1. The twisting angles for each trial were calculated using equation 2. The calculated results were tabulated in appendix c and plotted on a graph shown in Figure 3 . The torsional stiffness can be seen as the slope of each trial’s relation. Figure 3: Applied Torque versus Twisting Angle for Trials 1, 2, and 3 The second moment of area was calculated using equation 4 with the given dimensions of the shaft. The shear modulus was calculated using equation 5 for each trial with the torsional stiffness and the second moment of area. The results for torsional stiffness and shear modulus were tabulated in Table 2 . Table 2: Calculated Torsional Stiffness and Shear Modulus for Each Trial Torsional Stiffness, k (Nm/rad) Shear Modulus, G (GPa) Trial 1 10484 29.45 Trial 2 10575 29.71 Trial 3 11029 30.98 Requirement C The mean shear modulus across the 3 trials was calculated to be 30.0 GPa. When comparing this to the published value of 28 GPa, an error percentage of 7.3% was obtained using equation 6. There are several reasons for an error percentage of this magnitude. To start, the clearest source of error in this experiment was the jumpiness of the vertical deflection gauge dials. This provided early inaccuracies in the calculations for torsional stiffness and subsequently shear modulus. The deflection gauges also could have been improperly zeroed at the start of each trial, which would affect the results as well. P a g e 5 | 10

Requirement D The experimental stresses for trial 1 were calculated using equation 7. The shear strains on the 45 ̊ plane were used from the gauge 2 readings as well as the calculated shear modulus for trial 1. The theoretical stresses were calculated using equation 8 with the applied torques for trial 1. The calculated results for the stresses were catalogued in appendix c and plotted on the graph in Figure 4 . Figure 4: Theoretical and Experimental Stresses vs Applied Torque for Trial 1 As seen in the graph, the theoretical and experimental stresses vary from each other. The experimental stresses could have a few sources of error, one being the strain gauges attached to the shaft. As observed in the previous lab, strain gauges are extremely sensitive parts, and the slightest discrepancies can affect the results. The gauges could have also been put on imperfectly, which would result in them accounting for strains in unintentional directions. The readings for longitudinal and hoop strain were not always zero, which is an indication that the experiment could have not been under pure shear conditions, and perhaps there were deformations in other directions than the torsional. This would affect the shear strain in the 45 ̊ plane and subsequently the experimental shear stresses. Conclusion Through the use of the Torsion of a Hollow Circular Shaft apparatus, the torsional stiffness of the beam was calculated to be between 10.4 and 11.1 kN*m/rad from 3 trials. The average shear modulus across the trials was found to be 30.0 GPa, which had an error percentage of 7.3%. This is accurate enough to call the experiment successful in that regard. The experiment was also a success in gaining experience with torsion, stiffness, shear modulus, and shear stress. P a g e 6 | 10

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

References MAAE 2202 - Lab Manual pages 20-25 (MAAE 2202A/B Brightspace course page, Administration tab) Appendices Appendix A: Lab Data Beam Information Beam Length, L (mm) Deflection Arm Length d deflection , (mm) Beam Outer Diameter R out , (mm) Beam Wall Thickness t, (mm) 760 180 50.8 3.175 Trial 1: 140 mm from centre Weight (kg) Gauge 1 (microstrain) Gauge 2 (microstrain) Gauge 3 (microstrain) Vertical Deflection (mm) 4.597 -2 -11 0 0.1143 9.171 -5 -23 2 0.2286 13.766 -6 -33 2 0.3048 18.368 -6 -44 2 0.4064 22.931 -9 -57 3 0.5588 Trial 2: 118 mm from centre Weight (kg) Gauge 1 (microstrain) Gauge 2 (microstrain) Gauge 3 (microstrain) Vertical Deflection (mm) 4.597 -2 -10 1 0.0381 9.171 -5 -21 2 0.1397 13.766 -4 -28 1 0.2286 18.368 -7 -39 3 0.3048 22.931 -8 -49 3 0.4064 Trial 3: 80 mm from centre Weight (kg) Gauge 1 (microstrain) Gauge 2 (microstrain) Gauge 3 (microstrain) Vertical Deflection (mm) 4.597 -2 -8 0 0.0762 9.171 -4 -15 1 0.127 13.766 -5 -21 1 0.1778 18.368 -6 -29 2 0.254 22.931 -8 -35 2 0.3048 Appendix B: Sample Calculations for Trial 1, Weight 1 Sample calculation for applied torque from weights using equation 1 P a g e 7 | 10

T = mg d load T = ( 4.597 kg ) ( 9.81 m s 2 ) ( 140 × 10 − 3 m ) T = 6.314 N ∙m Sample calculation for angle of deflection using equation 2 θ = tan − 1 ( verticaldeflection d deflection ) θ = tan − 1 ( 0.1143 mm 180 mm ) θ = 0.000635 rad Sample calculation for torsional stiffness using equation 3 (unused value) k = ∆T ∆θ k = 25.18 N ∙m 0.002469 rad k = 10196.63 N ∙m rad Sample calculation for second moment of area using equation 4 J = π 2 R 4 = π 2 ( R out 4 − R ¿ 4 ) J = π 2 ( ( 50.8 2 × 10 − 3 m ) 4 − ( ( 50.8 2 − 3.175 ) × 10 − 3 m ) 4 ) J = 2.7056 × 10 − 7 m 4 Sample calculation for shear modulus using equation 5 G = ∆T ∆θ ∙ L J = k ∙ L J G = ( 10484 N ∙m rad ) ∙ ( 760 × 10 − 3 m 2.7056 × 10 − 7 m 4 ) G = 2.945 × 10 10 Pa = 29.45 GPa P a g e 8 | 10

Sample calculation for mean shear modulus G ´ X = 1 n ∑ i = 1 n G i G ´ X = 1 3 ( 29.45 GPa + 29.71 GPa + 30.98 GPa ) G ´ X = 30.0 GPa Sample calculation for error percentage using equation 6 %error = ( G ´ X − G pub G pub ) × 100% %error = ( 30 GPa − 28 GPa 28 GPa ) × 100% %error = 7.3% Sample calculation for experimental shear stress using equation 7 τ exp = G∙ ( 2 ×gauge 2 reading ) τ exp = ( 2.945 × 10 10 Pa ) ∙ ( 2 × 11 × 10 − 6 ) τ exp = 647888 Pa = 647.888 kPa Sample calculation for theoretical shear stress using equation 8 τ theo = T ∙ R out J τ theo = ( 6.314 N ∙m ) ∙ ( 50.8 2 × 10 − 3 m 2.7056 × 10 − 7 m 4 ) τ theo = 592709 Pa = 592.709 kPa P a g e 9 | 10

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Appendix C: Tables of Calculated Results Table #: ads Applied Torque, T (Nm) Weight Trial 1 Trial 2 Trial 3 1 6.31352 5.321395 3.607726 2 12.59545 10.61617 7.197401 3 18.90622 15.93525 10.80356 4 25.22661 21.26243 14.41521 5 31.49344 26.54447 17.99625 Table #: ads Angle of Deflection, θ (Nm) Weight Trial 1 Trial 2 Trial 3 1 0.000635 0.000212 0.000423 2 0.00127 0.000776 0.000706 3 0.001693 0.00127 0.000988 4 0.002258 0.001693 0.001411 5 0.003104 0.002258 0.001693 Table #: ads Torsional Stiffness, k (Nm/rad) Shear Modulus, G (GPa) Trial 1 10484 29.45 Trial 2 10575 29.71 Trial 3 11029 30.98 Table #: Experimental and Theoretical Shear Stresses for Trial 1 Applied Torque, T (Nm) Experimental Stress, τ exp (kPa) Theoretical Stress, τ theo (kPa) 6.31352 647.8876 592.7092065 12.59545 1354.674 1182.452933 18.90622 1943.663 1774.904272 25.22661 2591.551 2368.258148 31.49344 3357.236 2956.583601 P a g e 10 | 10