{"id":2583,"date":"2024-08-10T21:42:47","date_gmt":"2024-08-10T19:42:47","guid":{"rendered":"https:\/\/jureknawrocki.com\/?p=2583"},"modified":"2026-03-14T08:05:49","modified_gmt":"2026-03-14T07:05:49","slug":"mechanika-pekania-w-praktyce-cz-i","status":"publish","type":"post","link":"https:\/\/jureknawrocki.com\/en\/mechanika-pekania-w-praktyce-cz-i\/","title":{"rendered":"Practical Fracture Mechanics Part I"},"content":{"rendered":"<p>Let me start by pointing out that this is a purely theoretical example because it is difficult to imagine the existence of pressure in a thin-walled object that has a gap running through the entire thickness of the wall. <\/p>\n\n\n\n<p>There is nothing more practical than a good theory!<\/p>\n\n\n\n<p>Suppose we have a pipe or cylindrical pressure vessel that has a slot that is inclined to make it more difficult at a certain angle. To determine if the component can work, the stress intensity factor (SIF) must be determined and compared to the critical value for the material. The pressure is constant and there are no vibrations.    <\/p>\n\n\n\n<p>For stresses, the typical formulas apply: <\/p>\n\n\n\n<figure class=\"wp-block-image aligncenter size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"246\" height=\"113\" src=\"https:\/\/jureknawrocki.com\/wp-content\/uploads\/Mech-pekania-I-2-2.webp\" alt=\"\" class=\"wp-image-2612\" srcset=\"https:\/\/jureknawrocki.com\/wp-content\/uploads\/Mech-pekania-I-2-2.webp 246w, https:\/\/jureknawrocki.com\/wp-content\/uploads\/Mech-pekania-I-2-2-18x8.webp 18w\" sizes=\"(max-width: 246px) 100vw, 246px\" \/><\/figure>\n\n\n\n<figure class=\"wp-block-image aligncenter size-full is-resized\"><img loading=\"lazy\" decoding=\"async\" width=\"770\" height=\"663\" src=\"https:\/\/jureknawrocki.com\/wp-content\/uploads\/Mech-pekania-I-1.webp\" alt=\"\" class=\"wp-image-2620\" style=\"width:517px;height:auto\" srcset=\"https:\/\/jureknawrocki.com\/wp-content\/uploads\/Mech-pekania-I-1.webp 770w, https:\/\/jureknawrocki.com\/wp-content\/uploads\/Mech-pekania-I-1-480x413.webp 480w\" sizes=\"(min-width: 0px) and (max-width: 480px) 480px, (min-width: 481px) 770px, 100vw\" \/><\/figure>\n\n\n\n<p>The formula for a crack in a shell undergoing biaxial tension can be obtained from fracture mechanics tables. The coordinate system is attached at the center of the crack. <\/p>\n\n\n\n<figure class=\"wp-block-image aligncenter size-full is-resized\"><img loading=\"lazy\" decoding=\"async\" width=\"496\" height=\"182\" src=\"https:\/\/jureknawrocki.com\/wp-content\/uploads\/Mech-pekania-I-3.webp\" alt=\"\" class=\"wp-image-2590\" style=\"width:362px;height:auto\" srcset=\"https:\/\/jureknawrocki.com\/wp-content\/uploads\/Mech-pekania-I-3.webp 496w, https:\/\/jureknawrocki.com\/wp-content\/uploads\/Mech-pekania-I-3-480x176.webp 480w\" sizes=\"(min-width: 0px) and (max-width: 480px) 480px, (min-width: 481px) 496px, 100vw\" \/><\/figure>\n\n\n\n<figure class=\"wp-block-image aligncenter size-full is-resized\"><img loading=\"lazy\" decoding=\"async\" width=\"317\" height=\"220\" src=\"https:\/\/jureknawrocki.com\/wp-content\/uploads\/Mech-pekania-I-4.webp\" alt=\"\" class=\"wp-image-2592\" style=\"width:317px;height:auto\" srcset=\"https:\/\/jureknawrocki.com\/wp-content\/uploads\/Mech-pekania-I-4.webp 317w, https:\/\/jureknawrocki.com\/wp-content\/uploads\/Mech-pekania-I-4-300x208.webp 300w, https:\/\/jureknawrocki.com\/wp-content\/uploads\/Mech-pekania-I-4-18x12.webp 18w\" sizes=\"(max-width: 317px) 100vw, 317px\" \/><\/figure>\n\n\n\n<p>After substituting the shell stress formulas into the stress intensity factor formulas and using simple trigonometric transformations, we can finally write that:<\/p>\n\n\n\n<figure class=\"wp-block-image aligncenter size-full is-resized\"><img loading=\"lazy\" decoding=\"async\" width=\"602\" height=\"342\" src=\"https:\/\/jureknawrocki.com\/wp-content\/uploads\/Mech-pekania-I-5.webp\" alt=\"\" class=\"wp-image-2615\" style=\"width:314px;height:auto\" srcset=\"https:\/\/jureknawrocki.com\/wp-content\/uploads\/Mech-pekania-I-5.webp 602w, https:\/\/jureknawrocki.com\/wp-content\/uploads\/Mech-pekania-I-5-480x273.webp 480w\" sizes=\"(min-width: 0px) and (max-width: 480px) 480px, (min-width: 481px) 602px, 100vw\" \/><\/figure>\n\n\n\n<p>Below is the calculation for a DN 250 pipe (273 x 10 mm) under pressure of 2 MPa. <\/p>\n\n\n\n<figure class=\"wp-block-image aligncenter size-full is-resized\"><img loading=\"lazy\" decoding=\"async\" width=\"461\" height=\"518\" src=\"https:\/\/jureknawrocki.com\/wp-content\/uploads\/Mech-pekania-I-6.webp\" alt=\"\" class=\"wp-image-2601\" style=\"width:255px;height:auto\" srcset=\"https:\/\/jureknawrocki.com\/wp-content\/uploads\/Mech-pekania-I-6.webp 461w, https:\/\/jureknawrocki.com\/wp-content\/uploads\/Mech-pekania-I-6-267x300.webp 267w, https:\/\/jureknawrocki.com\/wp-content\/uploads\/Mech-pekania-I-6-11x12.webp 11w\" sizes=\"(max-width: 461px) 100vw, 461px\" \/><\/figure>\n\n\n\n<figure class=\"wp-block-image aligncenter size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"681\" height=\"215\" src=\"https:\/\/jureknawrocki.com\/wp-content\/uploads\/Mech-pekania-I-7-1.webp\" alt=\"\" class=\"wp-image-2624\" srcset=\"https:\/\/jureknawrocki.com\/wp-content\/uploads\/Mech-pekania-I-7-1.webp 681w, https:\/\/jureknawrocki.com\/wp-content\/uploads\/Mech-pekania-I-7-1-480x152.webp 480w\" sizes=\"(min-width: 0px) and (max-width: 480px) 480px, (min-width: 481px) 681px, 100vw\" \/><\/figure>\n\n\n\n<p>Regarding the temptation to obtain K-factors by simulation. In Ansys at the current stage of its development there is a certain limitation, namely the crack must be perpendicular to the surface. In addition, a local coordinate system for the crack must be created, in which the X-axis is directed towards the crack axis. This can be cleverly achieved by selecting \"Hit Point Normal\" and clicking on the surface of the pipe.   <\/p>","protected":false},"excerpt":{"rendered":"<p>Na wst\u0119pie zaznacz\u0119, \u017ce jest to przyk\u0142ad czysto teoretyczny bo trudno sobie wyrazi\u0107 istnienie ci\u015bnienia w obiekcie cienko\u015bciennym, kt\u00f3ry ma szczelin\u0119 przez ca\u0142\u0105 grubo\u015b\u0107 \u015bcianki. Nie ma nic bardziej praktycznego ni\u017c porz\u0105dna teoria ! Przypu\u015b\u0107my, \u017ce mamy ruroci\u0105g lub cylindryczny zbiornik ci\u015bnieniowy, kt\u00f3ry posiada szczelin\u0119 nachylon\u0105, aby by\u0142o trudniej, pod pewnym k\u0105tem. Aby okre\u015bli\u0107 czy [&hellip;]<\/p>\n","protected":false},"author":3,"featured_media":2627,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_et_pb_use_builder":"off","_et_pb_old_content":"","_et_gb_content_width":"","footnotes":""},"categories":[124,15],"tags":[109,32,16],"class_list":["post-2583","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mechanika-pekania","category-projekty","tag-mechanika-pekania","tag-mes","tag-naprezenia"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.6 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Mechanika p\u0119kania w praktyce cz. 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Przypu\u015b\u0107my, \u017ce mamy ruroci\u0105g lub cylindryczny zbiornik ci\u015bnieniowy, kt\u00f3ry posiada szczelin\u0119 nachylon\u0105, aby by\u0142o trudniej, pod pewnym k\u0105tem. 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