Calculation of pressure in a closed thin-walled pressure vessel (e.g., a drinks can) based on measured hoop strain.
", "licence": "Creative Commons Attribution 4.0 International"}, "feedback": {"showactualmark": true, "showanswerstate": true, "allowrevealanswer": true, "intro": "Calculation of can pressure based on measurement of hoop strain.
", "feedbackmessages": [], "showtotalmark": true, "advicethreshold": 0}, "duration": 0, "name": "Drinks Can Pressure Calculation", "showQuestionGroupNames": false, "percentPass": "100", "question_groups": [{"pickingStrategy": "all-ordered", "name": "Group", "pickQuestions": 1, "questions": [{"name": "Pressure in a Drinks Can", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Francis Franklin", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1887/"}], "tags": [], "metadata": {"description": "Work backwards from hoop strain measurement to calculate prior gauge pressure in a drinks can.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "It is possible to estimate the pressure in a drinks can by measuring the hoop strain before and after opening it. (The can is an example of a thin-walled pressure vessel, and the stress can be considered as plane stress, i.e., stress through the thickness of the wall is neglected.)
", "advice": "A steel drinks can ($E=209$ GPa, $\\nu=0.3$) has diameter $\\var{diameter}$ mm and wall thickness $\\var{thickness}$ mm. A strain gauge is fixed circumferentially and the difference in strain, before and after opening the can, is measured as $\\var{strain}$ μm/m (microstrain).
\nUsing $\\sigma_a = \\sigma_h/2$, Hooke's Law is:
\n$\\epsilon_h = {1 \\over E}(\\sigma_h-\\nu \\sigma_a) = {\\sigma_h \\over E} \\left(1-{\\nu \\over 2}\\right)$
\nand rearranging:
\n$\\sigma_h = {E \\epsilon_h \\over\\left(1-\\nu / 2\\right)} = {209 \\times 10^9 \\times \\var{strain} \\times 10^{-6} \\over (1-0.3 / 2)} $
\nso the hoop stress in the can wall prior to opening it was $\\var{siground(SH,3)}$MPa.
\nHoop stress is related to pressure by:
\n$\\sigma_h = {p D \\over 2 t}$
\nand rearranging:
\n$p = \\sigma_h {2 t \\over D} =\\var{siground(SH,3)}$MPa $\\times {2 \\times \\var{thickness} \\over\\var{diameter}} = \\var{siground(P/10,3)}$MPa.
\nThe pressure in the can prior to opening it was, therefore, $\\var{siground(P,3)}$ bar.
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", "templateType": "anything"}, "diameter": {"name": "diameter", "group": "Ungrouped variables", "definition": "random(50..70#1)", "description": "Diameter
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", "templateType": "anything"}, "SH": {"name": "SH", "group": "Ungrouped variables", "definition": "0.209*strain/(1-0.3/2)", "description": "Hoop stress.
", "templateType": "anything"}, "P": {"name": "P", "group": "Ungrouped variables", "definition": "10*SH*2*thickness/diameter", "description": "Pressure [bar].
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\nUsing Hooke's Law:
\n$\\epsilon_h = {1 \\over E}(\\sigma_h-\\nu \\sigma_a)$
\nand remembering that $\\sigma_a = \\sigma_h/2$:
\n