{"id":738,"date":"2020-12-07T20:00:23","date_gmt":"2020-12-07T20:00:23","guid":{"rendered":"https:\/\/pickedshares.com\/?p=738"},"modified":"2021-02-05T11:24:46","modified_gmt":"2021-02-05T11:24:46","slug":"engineering-mechanics-1-exercise-21-bearing-loads-and-hinged-rod-force","status":"publish","type":"post","link":"https:\/\/pickedshares.com\/en\/engineering-mechanics-1-exercise-21-bearing-loads-and-hinged-rod-force\/","title":{"rendered":"Bearing loads and hinged rod force"},"content":{"rendered":"\n<h2 class=\"wp-block-heading\">Task<\/h2>\n\n\n\n<p>A beam is hold by a fixed bearing on the left and a hinged rod (pendulum support). The beam itself is stiff. The force F acts on the beam.<\/p>\n\n\n\n<p>Determine the bearing loads in A, B and C and the force in the hinged rod.<\/p>\n\n\n\nF = 15 kN<br>\n\u03b1 = 75\u00b0\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"784\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/tm1-21-1024x784eng.jpg\" alt=\"Beam with pendulum support\" class=\"wp-image-436\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/tm1-21-1024x784eng.jpg 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/tm1-21-1024x784eng-300x230.jpg 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/tm1-21-1024x784eng-768x588.jpg 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/tm1-21-1024x784eng-400x306.jpg 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/12\/tm1-21-1024x784eng-800x613.jpg 800w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Beam with pendulum support<\/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-16-9 wp-has-aspect-ratio\"><div class=\"lyte-wrapper\" title=\"Lagerreaktionen und Stabkraft - Technische Mechanik 1, &Uuml;bung 21\" style=\"width:640px;max-width:100%;margin:5px auto;\"><div class=\"lyMe hidef\" id=\"WYL_i3qe_38ya84\" 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%2Fi3qe_38ya84%2Fmaxresdefault.jpg\" \/><meta itemprop=\"embedURL\" content=\"https:\/\/www.youtube.com\/embed\/i3qe_38ya84\" \/><meta itemprop=\"duration\" content=\"PT8M46S\" \/><meta itemprop=\"uploadDate\" content=\"2020-12-21T21:14:44Z\" \/><\/div><meta itemprop=\"accessibilityFeature\" content=\"captions\" \/><div id=\"lyte_i3qe_38ya84\" data-src=\"https:\/\/pickedshares.com\/wp-content\/plugins\/wp-youtube-lyte\/lyteCache.php?origThumbUrl=https%3A%2F%2Fi.ytimg.com%2Fvi%2Fi3qe_38ya84%2Fmaxresdefault.jpg\" class=\"pL\"><div class=\"tC\"><div class=\"tT\" itemprop=\"name\">Lagerreaktionen und Stabkraft - Technische Mechanik 1, \u00dcbung 21<\/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\/i3qe_38ya84\" 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%2Fi3qe_38ya84%2F0.jpg\" alt=\"Lagerreaktionen und Stabkraft - Technische Mechanik 1, &Uuml;bung 21\" width=\"640\" height=\"340\" \/><br \/>Watch this video on YouTube<\/a><\/noscript><meta itemprop=\"description\" content=\"Ein Tr\u00e4ger wird auf einer Seite durch ein Festlager und in der Mitte durch einen Stab unter einem Winkel von 75\u00b0 abgest\u00fctzt. Am Tr\u00e4ger greift die Kraft F an. Die Lagerreaktionen in A, B und C und die Stabkraft sind zu bestimmen. Die \u00dcbung zum Nachlesen findet man hier: https:\/\/pickedshares.com\/technische-mechanik-1-uebung-21-lagerreaktionen-und-stabkraft\/\"><\/div><\/div><div class=\"lL\" style=\"max-width:100%;width:640px;margin:5px auto;\"><\/div><figcaption>Solution video (German with English subtitles)<\/figcaption><\/figure>\n\n\n<h3 class=\"wp-block-heading\">Step by step<\/h3>\n\n\n\n<p>The whole beam is divided into three sections to determine the reaction forces.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/tm1-21_solved1-1024x450.jpg\" alt=\"Free cut of the beam\" class=\"wp-image-447\" width=\"512\" height=\"225\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/tm1-21_solved1-1024x450.jpg 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/tm1-21_solved1-300x132.jpg 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/tm1-21_solved1-768x338.jpg 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/tm1-21_solved1-400x176.jpg 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/tm1-21_solved1-800x352.jpg 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/tm1-21_solved1.jpg 1282w\" sizes=\"auto, (max-width: 512px) 100vw, 512px\" \/><figcaption>Free body diagram of the beam<\/figcaption><\/figure>\n\n\n\n<figure class=\"wp-block-image size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/tm1-21_solved2.jpg\" alt=\"Reaction forces of joint B\" class=\"wp-image-448\" width=\"388\" height=\"218\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/tm1-21_solved2.jpg 776w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/tm1-21_solved2-300x169.jpg 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/tm1-21_solved2-768x432.jpg 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/tm1-21_solved2-400x225.jpg 400w\" sizes=\"auto, (max-width: 388px) 100vw, 388px\" \/><figcaption>Reaction forces of joint B<\/figcaption><\/figure>\n\n\n\n<figure class=\"wp-block-image size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/tm1-21_solved3.jpg\" alt=\"Reaction forces of bearing C\" class=\"wp-image-449\" width=\"333\" height=\"292\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/tm1-21_solved3.jpg 666w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/tm1-21_solved3-300x263.jpg 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/tm1-21_solved3-400x351.jpg 400w\" sizes=\"auto, (max-width: 333px) 100vw, 333px\" \/><figcaption>Reaction forces of bearing C<\/figcaption><\/figure>\n\n\n\n\n    <title>tm1-21<\/title>\n    <script type=\"text\/x-mathjax-config\">  MathJax.Hub.Config({\n    displayAlign: \"left\",\n    context: \"MathJax\",\n    TeX: {TagSide: \"left\"}\n  })\n<\/script>\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    \n    <noscript>\n      <div class=\"error message\">\n        <p>Please enable JavaScript in order to get a 2d display of the equations embedded in this web page.<\/p>\n      <\/div>\n    <\/noscript>\n    <p hidden=\"hidden\">\\(      \\DeclareMathOperator{\\abs}{abs}\n      \\newcommand{\\ensuremath}[1]{\\mbox{$#1$}}\n\\)<\/p>\n    \n    <!-- Text cell -->\n    <p>Section I: Balance of forces in x-direction\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{1} 0={F_{\\mathit{Bx}}}+{F_{\\mathit{Ax}}}\\]\n<\/p>\n    <!-- Text cell -->\n    <p>Section I: Balance of forces in y-direction\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{2} 0={F_{\\mathit{By}}}+{F_{\\mathit{Ay}}}-F\\]\n<\/p>\n    <!-- Text cell -->\n    <p>Section I: Balance of moments around A\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{3} 0={F_{\\mathit{By}}} l-2 F l\\]\n<\/p>\n    \n    <!-- Text cell -->\n    <p>Section II: Balance of forces in x-direction\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{4} 0=-{F_S} \\cos{\\left( \\alpha \\right) }-{F_{\\mathit{Bx}}}\\]\n<\/p>\n    <!-- Text cell -->\n    <p>Section II: Balance of forces in y-direction\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{5} 0={F_S} \\sin{\\left( \\alpha \\right) }-{F_{\\mathit{By}}}\\]\n<\/p>\n    \n    <!-- Text cell -->\n    <p>Section III: Balance of forces in x-direction\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{6} 0={F_S} \\cos{\\left( \\alpha \\right) }+{F_{\\mathit{Cx}}}\\]\n<\/p>\n    <!-- Text cell -->\n    <p>Section III: Balance of forces in y-direction\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{7} 0={F_{\\mathit{Cy}}}-{F_S} \\sin{\\left( \\alpha \\right) }\\]\n<\/p>\n    \n    <!-- Text cell -->\n    <p>Step by step solution of the found equations\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{8} {F_{\\mathit{By}}}=2 F\\]\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{9} {F_{\\mathit{By}}}=30 \\mathit{kN}\\]\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{10} {F_{\\mathit{Ay}}}=F-{F_{\\mathit{By}}}\\]\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{11} {F_{\\mathit{Ay}}}=-15 \\mathit{kN}\\]\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{12} {F_S}=\\frac{30 \\mathit{kN}}{\\sin{\\left( \\alpha \\right) }}\\]\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{13} {F_S}=31.06 \\mathit{kN}\\]\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{14} {F_{\\mathit{Bx}}}=-{F_S} \\cos{\\left( \\alpha \\right) }\\]\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{15} {F_{\\mathit{Bx}}}=-8.04 \\mathit{kN}\\]\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{16} {F_{\\mathit{Ax}}}=-{F_{\\mathit{Bx}}}\\]\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{17} {F_{\\mathit{Ax}}}=8.04 \\mathit{kN}\\]\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{18} {F_{\\mathit{Cx}}}=-{F_S} \\cos{\\left( \\alpha \\right) }\\]\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{19} {F_{\\mathit{Cx}}}=-8.04 \\mathit{kN}\\]\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{20} {F_{\\mathit{Cy}}}={F_S} \\sin{\\left( \\alpha \\right) }\\]\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{21} {F_{\\mathit{Cy}}}=30.0 \\mathit{kN}\\]\n<\/p>\n    <hr>\n    \n  \n\n\n\n\n<p><\/p>\n","protected":false},"excerpt":{"rendered":"<p> ... <a title=\"Bearing loads and hinged rod force\" class=\"read-more\" href=\"https:\/\/pickedshares.com\/en\/engineering-mechanics-1-exercise-21-bearing-loads-and-hinged-rod-force\/\" aria-label=\"Read more about Bearing loads and hinged rod force\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":436,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[21,60],"tags":[45,46,29],"class_list":["post-738","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-engineering-mechanics-i","category-exercises","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 hinged rod force &#8226; pickedshares<\/title>\n<meta name=\"description\" content=\"A beam is supported by hinged rod and has to bear the load F. 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