{"id":714,"date":"2020-12-06T21:39:12","date_gmt":"2020-12-06T21:39:12","guid":{"rendered":"https:\/\/pickedshares.com\/?p=714"},"modified":"2021-05-08T11:43:59","modified_gmt":"2021-05-08T11:43:59","slug":"engineering-mechanics-1-exercise-16-bearing-reactions-and-joint-forces","status":"publish","type":"post","link":"https:\/\/pickedshares.com\/en\/engineering-mechanics-1-exercise-16-bearing-reactions-and-joint-forces\/","title":{"rendered":"Bearing loads and joint forces"},"content":{"rendered":"\n<h2 class=\"wp-block-heading\">Task<\/h2>\n\n\n\n<p>Determine the bearing reactions in the fixed restraint A, the forces in joint B and the bearing reactions in the floating bearing C for the beam with the load F as shown.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"529\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201105-134239_Sketch-1024x529.jpg\" alt=\"Beam with fixation, joint and floating bearing\" class=\"wp-image-419\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201105-134239_Sketch-1024x529.jpg 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201105-134239_Sketch-300x155.jpg 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201105-134239_Sketch-768x397.jpg 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201105-134239_Sketch.jpg 1314w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Beam with fixation, joint and floating bearing<\/figcaption><\/figure>\n\n\n\n<h2 class=\"wp-block-heading\">Solution<\/h2>\n\n\n\n<p>The following video is in german language, but English subtitles are available.<\/p>\n\n\n<figure class=\"wp-block-embed-youtube wp-block-embed is-type-video is-provider-youtube wp-embed-aspect-4-3 wp-has-aspect-ratio\"><div class=\"lyte-wrapper\" title=\"Lagerreaktionen und Gelenkkr&auml;fte bestimmen - Technische Mechanik 1, &Uuml;bung 16\" style=\"width:640px;max-width:100%;margin:5px auto;\"><div class=\"lyMe hidef\" id=\"WYL_oGZZet-Lf5U\" itemprop=\"video\" itemscope itemtype=\"https:\/\/schema.org\/VideoObject\"><div><meta itemprop=\"thumbnailUrl\" content=\"https:\/\/pickedshares.com\/wp-content\/plugins\/wp-youtube-lyte\/lyteCache.php?origThumbUrl=https%3A%2F%2Fi.ytimg.com%2Fvi%2FoGZZet-Lf5U%2Fmaxresdefault.jpg\" \/><meta itemprop=\"embedURL\" content=\"https:\/\/www.youtube.com\/embed\/oGZZet-Lf5U\" \/><meta itemprop=\"duration\" content=\"PT8M5S\" \/><meta itemprop=\"uploadDate\" content=\"2020-12-16T14:41:50Z\" \/><\/div><meta itemprop=\"accessibilityFeature\" content=\"captions\" \/><div id=\"lyte_oGZZet-Lf5U\" data-src=\"https:\/\/pickedshares.com\/wp-content\/plugins\/wp-youtube-lyte\/lyteCache.php?origThumbUrl=https%3A%2F%2Fi.ytimg.com%2Fvi%2FoGZZet-Lf5U%2Fmaxresdefault.jpg\" class=\"pL\"><div class=\"tC\"><div class=\"tT\" itemprop=\"name\">Lagerreaktionen und Gelenkkr\u00e4fte bestimmen - Technische Mechanik 1, \u00dcbung 16<\/div><\/div><div class=\"play\"><\/div><div class=\"ctrl\"><div class=\"Lctrl\"><\/div><div class=\"Rctrl\"><\/div><\/div><\/div><noscript><a href=\"https:\/\/youtu.be\/oGZZet-Lf5U\" rel=\"nofollow\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pickedshares.com\/wp-content\/plugins\/wp-youtube-lyte\/lyteCache.php?origThumbUrl=https%3A%2F%2Fi.ytimg.com%2Fvi%2FoGZZet-Lf5U%2F0.jpg\" alt=\"Lagerreaktionen und Gelenkkr&auml;fte bestimmen - Technische Mechanik 1, &Uuml;bung 16\" width=\"640\" height=\"340\" \/><br \/>Watch this video on YouTube<\/a><\/noscript><meta itemprop=\"description\" content=\"F\u00fcr den abgebildeten Tr\u00e4ger sollen die Lagerreaktionen in der festen Einspannung A, dem Gelenk B und im Loslager C bestimmt werden. Der Link zur \u00dcbung: https:\/\/pickedshares.com\/technische-mechanik-1-uebung-16-lagerreaktionen-und-gelenkkraefte\/ Auch wenn dieser Tr\u00e4ger und die angreifende Kraft relativ einfach zu \u00fcberschauen sind, werden die vollst\u00e4ndigen Kr\u00e4fte- und Momentenbilanzen aufgestellt. Es geht ja um die \u00dcbung ;-) Als erstes wird der Tr\u00e4ger in zwei Bereiche eingeteilt. Die Kr\u00e4fte- und Momentenbilanzen werden dann f\u00fcr beide Bereiche separat aufgestellt. Zu beachten ist, dass im Gelenk B die Kr\u00e4fte in den Bilanzen f\u00fcr Bereich 1 und 2 entgegengesetzt anzutragen sind. Die Gleichungen k\u00f6nnen dann entsprechend aufgel\u00f6st werden und man erh\u00e4lt die Lagerreaktionen.\"><\/div><\/div><div class=\"lL\" style=\"max-width:100%;width:640px;margin:5px auto;\"><\/div><figcaption>Determination of the bearing loads and joint forces (german language)<\/figcaption><\/figure>\n\n\n<p>The beam has to be divided into two sections.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"715\" height=\"443\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/tm1-16_sections.jpg\" alt=\"Beam sections\" class=\"wp-image-940\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/tm1-16_sections.jpg 715w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/tm1-16_sections-300x186.jpg 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/tm1-16_sections-400x248.jpg 400w\" sizes=\"auto, (max-width: 715px) 100vw, 715px\" \/><figcaption>Beam sections<\/figcaption><\/figure>\n\n\n\n<p>Then the sections are cut free and the bearing forces and joint forces applied. It should be noted that the joint forces are applied in opposite directions in the two different sections.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"565\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/Screenshot_20201202-121612_Sketch-1024x565.jpg\" alt=\"Cutting free areas 1 and 2\" class=\"wp-image-593\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/Screenshot_20201202-121612_Sketch-1024x565.jpg 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/Screenshot_20201202-121612_Sketch-300x165.jpg 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/Screenshot_20201202-121612_Sketch-768x424.jpg 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/Screenshot_20201202-121612_Sketch-400x221.jpg 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/Screenshot_20201202-121612_Sketch-800x441.jpg 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/Screenshot_20201202-121612_Sketch.jpg 1068w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Cutting free sections 1 and 2<\/figcaption><\/figure>\n\n\n\n<script src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-AMS_HTML\" async=\"async\">  \/\/ A comment that hinders wxWidgets from optimizing this tag too much.\n<\/script>\n\\[ \\require{cancel} \\]\n\\[ \\newcommand{\\myvec}[1]{{\\begin{pmatrix}#1\\end{pmatrix}}} \\]\n\n<p>The balances or equilibria of forces and moments are set up separately for both areas.<\/p>\n\n<h4>Section I:<\/h4>\n<p>The equilibrium of forces in the x-direction is<\/p>\n<p>\\[\\tag{1} \\sum F_x = 0 = F_{Ax} \\]<\/p>\n\n<p>The equilibrium of forces in the y-direction is<\/p>\n<p>\\[\\tag{2} \\sum F_y = 0 = F_{Ay} - F_B \\]<\/p>\n<p>The equilibrium of moments around point A is<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[\\tag{3} \\sum M(A) = 0 = M_A - F_B \\cdot a \\]<\/p>\n<\/div>\n\n<h4>Section II:<\/h4>\n\n<p>The equilibrium of forces in the x-direction is<\/p>\n<p>\\[\\tag{4} \\sum F_x = 0 \\]<\/p>\n\n<p>The equilibrium of forces in the y-direction is<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[\\tag{5} \\sum F_y = 0 = F_B - F + F_C \\]<\/p>\n<\/div>\n\n<p>The equilibrium of moments around point B delivers<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[\\tag{6} \\sum M(B) = 0 = - F \\cdot a + F_C \\cdot 2 \\cdot a \\]<\/p>\n<\/div>\n<p>\\[\\tag{7} F_C =  \\frac{F \\cdot \\bcancel{a}}{2\\bcancel{a}} \\]<\/p>\n\n<h4>Substitute and solve<\/h4>\n<p>It follows from equation 5 for F <sub> B <\/sub><\/p>\n\n<p>\\[\\tag{8} F_B = \\frac{F}{2} \\]<\/p>\n\n\n<p>Equation 2 yields<\/p>\n\n<p>\\[\\tag{9} F_{Ay} = \\frac{F}{2} \\]<\/p>\n\n<p>The moment in the fixed restraint results from equation 3:<\/p>\n<p>\\[\\tag{10} M_A = \\frac{F \\cdot a}{2} \\]<\/p>\n","protected":false},"excerpt":{"rendered":"<p> ... <a title=\"Bearing loads and joint forces\" class=\"read-more\" href=\"https:\/\/pickedshares.com\/en\/engineering-mechanics-1-exercise-16-bearing-reactions-and-joint-forces\/\" aria-label=\"Read more about Bearing loads and joint forces\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":413,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[21,60],"tags":[30,45,46,29],"class_list":["post-714","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-engineering-mechanics-i","category-exercises","tag-beams","tag-bearing-loads","tag-reaction-forces","tag-statics","infinite-scroll-item","masonry-post","generate-columns","tablet-grid-50","mobile-grid-100","grid-parent","grid-33"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Bearing loads and joint forces &#8226; pickedshares<\/title>\n<meta name=\"description\" content=\"A beam with a fixed bearing, a joint and a floating bearing is loaded with a force. 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