MAAE2202_Lab B

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Carleton University Laboratory Report Course #: MAAE 2202 Lab #: B Lab Section #: Stresses in a Thin-Walled Pressure Vessel

2022/10/25 Summary The main objectives of this experiment are to obtain the internal pressure of a thin-walled pressure vessel by measuring the strains on the surface, and to gain practical experience in strain gauge application techniques. In this experiment, the pressure vessel used was a pop can made of aluminum 3004. The experiment was performed twice, once with an undisturbed can, and once with a disturbed, shaken can. The carbon dioxide bubbles formed in the disturbed can varied the results of the experiment. The strain in the longitudinal and circumferential directions was measured using strain gauges. From these values, the stress and pressure in both directions was calculated. For the unshaken can, the pressure in the hoop and longitudinal direction was 227kPa and 105kPa, respectively. For the shaken can, the pressures were 244kPa in the hoop direction and 98kPa in the longitudinal. Nomenclature Table 1: Nomenclature Used in the Report Symbol Parameter Unit F Internal Force kN σ Stress GPa ε Strain Unitless E Elastic/Young’s Modulus GPa P Internal Pressure GPa t wall Vessel Wall Thickness MPa d Vessel Nominal Diameter GPa R Hoop to Longitudinal Ratio Unitless SF Safety Factor Unitless %error Error Percentage % Theory and Analysis Thin-walled cylindrical pressure vessels such as the pop can used in this experiment develop stresses in both circumferential and longitudinal directions when subjected to internal pressure. These stresses are called the hoop and longitudinal stress and can be seen in Figure 1 . Figure 1: Diagram showing stresses in the circumferential and longitudinal directions of a cylinder P a g e 2 | 12

Like other stresses, the stresses in the longitudinal and circumferential direction of the vessel adhere to the Young’s Modulus relation: E = σ long ε long →σ long = E×ε long →ε long = σ long E E = σ hoop ε hoop →σ hoop = E ×ε hoop →ε hoop = σ hoop E To find the hoop stress, the forces in the circumferential direction across the can must be analyzed. This will be done using force equilibrium of Figure 2 . Figure 2: Diagram to show force equilibrium in the circumferential direction Assuming the body is in equilibrium, the force F shown in the diagram can be written in 2 ways: 1. F = PA = P hoop × L×d 2. F = 2 ×σ hoop ×t wall ×L If these equations are set equal, solving for the hoop stress we obtain: σ hoop = P hoop d 2 t wall Equation 1 By substituting hoop stress from the Young’s modulus relation into Equation 1 and solving for pressure: E×ε hoop = P hoop d 2 t wall → P hoop = 2 Eε hoop t wall d Equation 2 P a g e 3 | 12

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To find the longitudinal stress, the same method is used. This time we will use force equilibrium for Figure 3 . Figure 3: Diagram to show force equilibrium in the longitudinal direction Assuming the body is in equilibrium, similar to the hoop force, the force F can be expressed in 2 ways: 1. F = P long ×π× d 2 4 2. F = σ long ×π ×d×t wall Setting the equations equal and solving for longitudinal stress, we obtain: σ long = P long d 4 t wall Equation 3 In the same way as the hoop pressure was found, by substituting in stress from Young’s modulus relation to Equation 3, we receive: E×ε long = P long d 4 t wall →P long = 4 E ε long t wall d Equation 4 The safety factor of the can uses the ultimate stress of the can and the experimental stress in the form: SF = σ ult σ exp Equation 5 P a g e 4 | 12

The experimental ratio is the ratio of experimental hoop stress to experimental longitudinal stress: R exp = σ hoop σ long Equation 6 The experimental ratio will be used to compare to the theoretical ratio to gain an error percentage: %error = | R exp − R theo R theo | × 100% Equation 7 Experimental Setup and Procedure P a g e 5 | 12

Figure 4: Labeled diagram of experimental setup The experiment was conducted as outlined in the lab manual. Results and Discussion Requirement A To find the hoop and longitudinal stresses and pressures, equations 1-4 were used with MatWeb’s published modulus of elasticity for aluminum 3004. The results were tabulated in Table 2 . Table 2: Summary of Calculated Stresses and Pressures Experiment Hoop Stress (Pa) Longitudina l Stress (Pa) Hoop Pressure (Pa) Longitudina l Pressure (Pa) Unshaken Can 70,829,200 16,467,100 226,653 105,389 Shaken Can 76,134,500 15,295,800 243,630 97,893 The hoop stress and pressure for both the unshaken and shaken cans is significantly greater than the longitudinal stresses and pressures. One major reason for this is the geometry of the pop can. The long cylinder naturally poses for more strain in the circumferential direction as it is easier for the internal pressure to increase the can’s diameter than it is to increase the can’s length. This is because there is less support stopping the diameter from expanding than there is for the length. Analyzing a pop can, if you were to take 2 points parallel in the hoop direction, there is only empty space between them. Using the same method, if you were to take 2 points in the longitudinal direction, there would be material between the points. Another reason the hoop stresses and pressures are so much larger than the longitudinal is the nature of the equations, as the hoop stress should be twice as large as the longitudinal stress (further explained in the Requirement C section). Something else to note is that the hoop stress is greater for the shaken can, while the inverse is true for the longitudinal stress. This could mean some of the longitudinal stress was transferred to hoop stress when the can was shaken, although critically speaking, this could mean anything given only one experiment was conducted for each can. Requirement B To find the safety factor of the can, equation 5 was used with MatWeb’s published ultimate strength value for aluminum 3004 as well as the calculated hoop and longitudinal stress values. The results can be seen in Table 3 . Table 3: Safety Factors of Cans Using Different Stresses Experiment SF using Hoop Stress SF using Longitudinal Stress P a g e 6 | 12

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Unshaken Can 2.527 10.870 Shaken Can 2.351 11.703 The safety factor in the longitudinal direction is much greater than the circumferential direction for both the unshaken and shaken cans. This is because the stresses were much greater in the hoop direction (further explained in the Requirement A section). Because the safety factors in the longitudinal directions are so much greater, they are irrelevant to the can’s overall safety factor, as it does not matter in which direction it breaks. As the hoop safety factor is lower for the shaken can, we can say with certainty that the overall safety factor is equal to or less than 2.35. Requirement C When analyzing equations 1 and 3, it is possible to compare hoop and longitudinal stress only if the hoop pressure is equal to the longitudinal pressure. Theoretically, the internal pressure should be uniform throughout the can so we can assume P hoop =P long . If this is the case, if we were to solve for internal pressure in equations 1 and 3, and set them equal, we can find the theoretical ratio: 2 σ hoop t wall d = 4 σ long t wall d → 2 σ hoop = 4 σ long → σ hoop σ long = 2 = R theo To find the experimental ratio, equation 6 was used for both the unshaken and shaken cans. The error percentage relative to the theoretical value was also calculated using equation 7. The results are catalogued in Table 4 . Table 4: Experimental Stress Ratio and Error Percentage When Compared to Theoretical Experiment Ratio Error Percentage (%) Unshaken Can 4.301 115 Shaken Can 4.977 148 The ratio is really a measure of the difference in hoop stress to longitudinal stress. The ratio in the shaken can is greater than the unshaken can. Reasons for this could include more randomness in the particles, or the way in which the can was opened. If the can was opened at a different angle, or opened at a different pace, it could affect the result. Considering only 1 experiment was performed for both the unshaken and shaken cans, it would be ludicrous to come to a definite conclusion as the results could be inverted given another trial. Neither of the cans came close to the theoretical ratio, with both having a laughable error percentage. Reasons for error include the sensitivity of the strain gauge, the orientation of the strain gauge, and the way the strain gauge was put on. The slightest movement, or vibration drastically fluctuated the gauges reading, so the calculations would not be very accurate. The gauges were also lined up using the historic ‘eyeball’ technique, where no equipment was used to confirm if the gauges were perpendicular. The gauges were put on using a layer of super glue which could have led to inaccuracies as well. Another source of error could be that the theoretical value was obtained through P a g e 7 | 12

the analysis of a perfect cylinder, to which the pop can clearly is not. The material at the top of the can could be different from the body and the bottom’s concavity would affect the results. Requirement D The can’s shape naturally helps battle over-pressurization as the cylindrical walls eliminate edges which would be weak points. The curvature at the top and bottom of the can also ensure fewer sharp edges. The concave bottom allows the can to support more pressure than if the bottom was flat. It distributes some of the vertical forces into horizontal components. Another advantage of the dome shaped bottom is the ease in increasing the volume of the can without breakage. If the can is dropped, often the concave bottom will push out and become convex, increasing the total volume of the can, and lowering the pressure. I can say with confidence that this happens more often than the can spewing liquid as I work for Pepsi and see this happening all the time. Conclusion The main objectives of the experiment were met. The internal pressures of the thin-walled pressure vessels were calculated from the measured strains. Practical experience was gained from the use of strain gauges application techniques. Although definitive answers for stress and pressure were calculated, the degree of accuracy is very low for an experiment of this nature. None of the results would have any validity in the professional field as only one trial was performed for both the unshaken and shaken cans. This is further proven by the enormous percentage of error for the stress ratio when compared to the theoretical. With this being said, the experiment still provides reasonable approximations for the can’s behaviour, such as the fact that the circumferential stress and pressure is much larger than the longitudinal. References MAAE 2202 - Lab Manual pages 11-19 (MAAE 2202A/B Brightspace course page, Administration tab) P a g e 8 | 12

“Aluminum 3004-O.” MatWeb , The Aluminum Association, https://www.matweb.com/search/datasheet.aspx? matguid=b40f2ca72f504f298d6f5399ce00440f&ckck=1. Ward-Bailey, 14 April, 2015, “The Surprising Science Behind the Aluminum Soda Can” https://www.csmonitor.com/ Science/Science- Notebook/2015/0414/The- surprising-science-behind- the- aluminum-soda-can , Oct. 17, 2019, The Christian Science Monitor. Ward-Baily, Jeff. “The Surprising Science Behind the Aluminum Soda Can.” The Christian Science Monitor, 14 Apr. 2015, https://www.csmonitor.com/Science/Science- Notebook/2015/0414/The-surprising-science-behind-the-aluminum-soda-can. P a g e 9 | 12

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Ward-Bailey, 14 April, 2015, “The Surprising Science Behind the Aluminum Soda Can” https://www.csmonitor.com/ Science/Science- Notebook/2015/0414/The- surprising-science-behind- the- aluminum-soda-can , Oct. 17, 2019, The Christian Science Monitor Appendices Appendix A: Lab Data Table 5: Can Dimensions Given in Lab Data Can Dimensions (mm) Diameter, d 65.53 Length, L 121.45 Thickness, t wall 0.104 Table 6: MatWeb’s Published Aluminum 3004 Properties P a g e 10 | 12

Metal Properties Modulus of Elasticity, E 68.9 GPa Ultimate Stress, σ ultimate 179 MPa Table 7: Experimental Strain Results for Unshaken Can Unshaken BAM Reading Strain Hoop -1028 1028 x 10 -6 Longitudinal -239 239 x 10 -6 Table 8: Experimental Strain Results for Shaken Can Shaken BAM Reading Strain Hoop -1105 1105 x 10 -6 Longitudinal -222 222 x 10 -6 Appendix B: Sample Calculations Sample calculation for hoop stress using Young’s modulus relation for the unshaken can σ hoop = E×ε hoop σ hoop = ( 68.9 × 10 9 Pa ) × ( 1028 × 10 − 6 ) σ hoop = 70,829,200 Pa = 70.8 MPa Sample calculation for longitudinal stress using Young’s modulus relation for the unshaken can σ long = E×ε long σ long = ( 68.9 × 10 9 Pa ) × ( 239 × 10 − 6 ) σ long = 16,467,100 Pa = 16.5 MPa Sample calculation for internal pressure using Equation 2 for the unshaken can with hoop strain P hoop = 2 Eε hoop t wall d P hoop = 2 × ( 68.9 × 10 9 Pa ) × ( 1028 × 10 − 6 ) × ( 1.04 × 10 − 4 m ) ( 0.065 m ) P hoop = 226,653 Pa = 226.7 kPa Sample calculation for internal pressure using Equation 4 for the unshaken can with longitudinal strain P a g e 11 | 12

P long = 4 Eε long t wall d P long = 4 × ( 68.9 × 10 9 Pa ) × ( 239 × 10 − 6 ) × ( 1.04 × 10 − 4 m ) ( 0.065 m ) P long = 105,389 Pa = 105.4 kPa Sample calculation for safety factor using equation 5 for the unshaken can with hoop stress SF = σ ult σ exp SF = 179 MPa 70.8 MPa SF = 2.527 Sample calculation for experimental stress ratio using equation 6 for the unshaken can R exp = σ hoop σ long R exp = 70.8 MPa 16.5 MPa R exp = 4.301 Sample calculation for error percentage using equation 7 for unshaken can %error = | R exp − R theo R theo | × 100% %error = | 4.301 − 2 2 | × 100% %error = 115 % P a g e 12 | 12

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