{"id":758,"date":"2020-12-08T12:31:05","date_gmt":"2020-12-08T12:31:05","guid":{"rendered":"https:\/\/pickedshares.com\/?p=758"},"modified":"2021-05-08T12:00:30","modified_gmt":"2021-05-08T12:00:30","slug":"engineering-mechanics-1-exercise-22-calculation-of-bearing-loads-and-joint-forces","status":"publish","type":"post","link":"https:\/\/pickedshares.com\/en\/engineering-mechanics-1-exercise-22-calculation-of-bearing-loads-and-joint-forces\/","title":{"rendered":"Calculation of bearing loads and joint forces"},"content":{"rendered":"\n<p>This exercise addresses the following questions:<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>How do you calculate joint forces?<\/li><li>How do I calculate a frame with two fixed bearings?<\/li><li>What are the different approaches to calculating joint forces?<\/li><\/ul>\n\n\n\n<h2 class=\"wp-block-heading\">Task<\/h2>\n\n\n\n<p>The forces F<sub>1<\/sub> to F<sub>4<\/sub> act on a frame with two fixed bearings and a joint. Determine the bearing loads and the forces in joint C!<\/p>\n\n\n\n<p>F<sub>1<\/sub> = F<sub>4<\/sub> = 10 kN, F<sub>2<\/sub> = 30 kN, F<sub>3<\/sub> = 60 kN, a = 1 m<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"624\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201114-202919_Sketch-1024x624.jpg\" alt=\"Frame with external loads and two fixed bearings\" class=\"wp-image-455\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201114-202919_Sketch-1024x624.jpg 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201114-202919_Sketch-300x183.jpg 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201114-202919_Sketch-768x468.jpg 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201114-202919_Sketch-1536x937.jpg 1536w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201114-202919_Sketch-400x244.jpg 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201114-202919_Sketch-800x488.jpg 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201114-202919_Sketch.jpg 1553w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Frame with external loads and two fixed bearings<\/figcaption><\/figure>\n\n\n\n<h2 class=\"wp-block-heading\">Solution<\/h2>\n\n\n\n<p>The solution can be done in different ways, here two variants are shown:<\/p>\n\n\n\n<p>Solution variant 1 considers both areas to the left and right of the joint separately from the beginning and sets up the corresponding equations.<\/p>\n\n\n\n<p>Solution variant 2 sets up the equilibrium conditions for the entire system as the first step, and then releases the forces of the sub-areas.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Variant 1<\/h2>\n\n\n\n<p>The following video is in German language.<\/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 berechnen - Technische Mechanik 1, &Uuml;bung 22\" style=\"width:640px;max-width:100%;margin:5px auto;\"><div class=\"lyMe hidef\" id=\"WYL_xQk-Zbktmi0\" 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%2FxQk-Zbktmi0%2Fmaxresdefault.jpg\" \/><meta itemprop=\"embedURL\" content=\"https:\/\/www.youtube.com\/embed\/xQk-Zbktmi0\" \/><meta itemprop=\"duration\" content=\"PT12M20S\" \/><meta itemprop=\"uploadDate\" content=\"2020-12-25T07:00:07Z\" \/><\/div><meta itemprop=\"accessibilityFeature\" content=\"captions\" \/><div id=\"lyte_xQk-Zbktmi0\" data-src=\"https:\/\/pickedshares.com\/wp-content\/plugins\/wp-youtube-lyte\/lyteCache.php?origThumbUrl=https%3A%2F%2Fi.ytimg.com%2Fvi%2FxQk-Zbktmi0%2Fmaxresdefault.jpg\" class=\"pL\"><div class=\"tC\"><div class=\"tT\" itemprop=\"name\">Lagerreaktionen und Gelenkkr\u00e4fte berechnen - Technische Mechanik 1, \u00dcbung 22<\/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\/xQk-Zbktmi0\" 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%2FxQk-Zbktmi0%2F0.jpg\" alt=\"Lagerreaktionen und Gelenkkr&auml;fte berechnen - Technische Mechanik 1, &Uuml;bung 22\" width=\"640\" height=\"340\" \/><br \/>Watch this video on YouTube<\/a><\/noscript><meta itemprop=\"description\" content=\"An einem Rahmen, der auf zwei Festlagern steht, wirken 4 \u00e4u\u00dfere Kr\u00e4fte. Es sind die Lagerreaktionen und die Gelenkkr\u00e4fte zu bestimmen. Die \u00dcbung ist auch hier zu finden: https:\/\/pickedshares.com\/technische-mechanik-1-uebung-22-lagerreaktionen-und-gelenkkraefte-berechnen\/ Im Video werden folgende Schritte vorgenommen: 1. Aufteilen des Rahmens in 2 Bereiche 2. Freischneiden des Systems 3. Aufstellen der Kr\u00e4ftegleichgewichte und Momentengleichgewichte 4. Umstellen der Gleichungen und Aufl\u00f6sen der Unbekannten\"><\/div><\/div><div class=\"lL\" style=\"max-width:100%;width:640px;margin:5px auto;\"><\/div><figcaption><\/figcaption><\/figure>\n\n\n<p>The frame will be divided into two sections to determine the forces.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"594\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201115-164131_Sketch-1024x594.jpg\" alt=\"Bearing reactions and joint forces\" class=\"wp-image-456\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201115-164131_Sketch-1024x594.jpg 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201115-164131_Sketch-300x174.jpg 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201115-164131_Sketch-768x445.jpg 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201115-164131_Sketch-1536x891.jpg 1536w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201115-164131_Sketch-400x232.jpg 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201115-164131_Sketch-800x464.jpg 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201115-164131_Sketch.jpg 1688w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Bearing reactions and joint forces<\/figcaption><\/figure>\n\n\n\n<!DOCTYPE html>\n<html lang=\"\">\n  <head>\n    \n    <meta http-equiv=\"Content-Type\" content=\"text\/html; charset=utf-8\"\/>\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  <\/head>\n  <body>\n    <!-- ****************************************************** -->\n    <!-- *        Created with wxMaxima version 20.06.6       * -->\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    <div class=\"comment\">Balance of forces section I\n<\/div>\n    <!-- Text cell -->\n    <div class=\"comment\">x-direction\n<\/div>\n    <!-- Code cell -->\n    <p>\\[\\tag{1} 0={F_{\\mathit{Cx}}}+{F_{\\mathit{Ax}}}+{F_1}\\]\n<\/p>\n    <!-- Text cell -->\n    <div class=\"comment\">y-direction\n<\/div>\n    <!-- Code cell -->\n    <p>\\[\\tag{2} 0={F_{\\mathit{Cy}}}+{F_{\\mathit{Ay}}}-{F_2}\\]\n<\/p>\n    <!-- Text cell -->\n    <div class=\"comment\">Moments around A (left-turning moments are positive)\n<\/div>\n\n<div style=\"overflow:auto;\">\n    <!-- Code cell -->\n    <p>\\[\\tag{3} 0=2 {F_{\\mathit{Cy}}} a-3 {F_{\\mathit{Cx}}} a-{F_2} a-2 {F_1} a\\]\n<\/p>\n<\/div>\n    <!-- Text cell -->\n    <div class=\"comment\">Balance of forces section II\n<\/div>\n    <!-- Text cell -->\n    <div class=\"comment\">x-direction\n<\/div>\n    <!-- Code cell -->\n    <p>\\[\\tag{4} 0=-{F_{\\mathit{Cx}}}+{F_{\\mathit{Bx}}}-{F_4}\\]\n<\/p>\n    <!-- Text cell -->\n    <div class=\"comment\">y-direction\n<\/div>\n    <!-- Code cell -->\n    <p>\\[\\tag{5} 0=-{F_{\\mathit{Cy}}}+{F_{\\mathit{By}}}-{F_3}\\]\n<\/p>\n    <!-- Text cell -->\n    <div class=\"comment\">Moments around B (left-turning moments are positive)\n<\/div>\n\n<div style=\"overflow:auto;\">\n    <!-- Code cell -->\n    <p>\\[\\tag{6} 0=4 {F_{\\mathit{Cy}}} a+3 {F_{\\mathit{Cx}}} a+{F_4} a+2 {F_3} a\\]\n<\/p>\n<\/div>\n    <!-- Text cell -->\n    <div class=\"comment\">Solve equation %o1 for F<sub>Ax<\/sub>\n<\/div>\n    <!-- Code cell -->\n    <p>\\[\\tag{7} {F_{\\mathit{Ax}}}=-{F_{\\mathit{Cx}}}-{F_1}\\]\n<\/p>\n    <!-- Text cell -->\n    <div class=\"comment\">Solve equation 3 for F<sub>Cx<\/sub>\n<\/div>\n    <!-- Code cell -->\n    <p>\\[\\tag{8} {F_{\\mathit{Cx}}}=\\frac{2 {F_{\\mathit{Cy}}}-{F_2}-2 {F_1}}{3}\\]\n<\/p>\n    <!-- Text cell -->\n    <div class=\"comment\">Substitute F<sub>Cx<\/sub> in equation 7\n<\/div>\n\n<div style=\"overflow:auto;\">\n    <!-- Code cell -->\n    <p>\\[\\tag{9} {F_{\\mathit{Ax}}}=-\\frac{2 {F_{\\mathit{Cy}}}-{F_2}-2 {F_1}}{3}-{F_1}\\]\n<\/p>\n<\/div>\n    <!-- Text cell -->\n    <div class=\"comment\">Solve equation 6 for F<sub>Cy<\/sub>\n<\/div>\n    <!-- Code cell -->\n    <p>\\[\\tag{10} {F_{\\mathit{Cy}}}=-\\frac{3 {F_{\\mathit{Cx}}}+{F_4}+2 {F_3}}{4}\\]\n<\/p>\n    <!-- Text cell -->\n    <div class=\"comment\">Substitute F<sub>Cy<\/sub> in equation 8\n<\/div>\n\n<div style=\"overflow:auto;\">\n    <!-- Code cell -->\n    <p>\\[\\tag{11} {F_{\\mathit{Cx}}}=\\frac{-\\frac{3 {F_{\\mathit{Cx}}}+{F_4}+2 {F_3}}{2}-{F_2}-2 {F_1}}{3}\\]\n<\/p>\n<\/div>\n    <!-- Text cell -->\n    <div class=\"comment\">... and solve for F<sub>Cx<\/sub>\n<\/div>\n\n<div style=\"overflow:auto;\">\n    <!-- Code cell -->\n    <p>\\[\\tag{12} {F_{\\mathit{Cx}}}=-\\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{9}\\]\n<\/p>\n<\/div>\n    <!-- Text cell -->\n    <div class=\"comment\">Solve equation 4 for F<sub>Bx<\/sub>\n<\/div>\n    <!-- Code cell -->\n    <p>\\[\\tag{13} {F_{\\mathit{Bx}}}={F_{\\mathit{Cx}}}+{F_4}\\]\n<\/p>\n    <!-- Text cell -->\n    <div class=\"comment\">... and substitution of F<sub>Cx<\/sub> yields F<sub>Bx<\/sub>\n<\/div>\n\n<div style=\"overflow:auto;\">\n    <!-- Code cell -->\n    <p>\\[\\tag{14} {F_{\\mathit{Bx}}}={F_4}-\\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{9}\\]\n<\/p>\n<\/div>\n    <!-- Text cell -->\n    <div class=\"comment\">Substitution of F<sub>Cx<\/sub> in equation 10\n<\/div>\n\n<div style=\"overflow:auto;\">\n    <!-- Code cell -->\n    <p>\\[\\tag{15} {F_{\\mathit{Cy}}}=-\\frac{-\\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{3}+{F_4}+2 {F_3}}{4}\\]\n<\/p>\n<\/div>\n    <!-- Text cell -->\n    <div class=\"comment\">Solve equation 5 for F<sub>By<\/sub>\n<\/div>\n    <!-- Code cell -->\n    <p>\\[\\tag{16} {F_{\\mathit{By}}}={F_{\\mathit{Cy}}}+{F_3}\\]\n<\/p>\n    <!-- Text cell -->\n    <div class=\"comment\">... and substitution of F<sub>Cy<\/sub>\n<\/div>\n\n<div style=\"overflow:auto;\">\n    <!-- Code cell -->\n    <p>\\[\\tag{17} {F_{\\mathit{By}}}=\\frac{-\\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{3}+{F_2}+2 {F_1}}{2}+{F_3}\\]\n<\/p>\n<\/div>\n    <!-- Text cell -->\n    <div class=\"comment\">F_Cy in equation 9 gives\n<\/div>\n\n<div style=\"overflow:auto;\">\n    <!-- Code cell -->\n    <p>\\[\\tag{18} {F_{\\mathit{Ax}}}=\\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{9}-{F_1}\\]\n<\/p>\n<\/div>\n    <!-- Text cell -->\n    <div class=\"comment\">Solve equation 2 for F<sub>Ay<\/sub>\n<\/div>\n    <!-- Code cell -->\n    <p>\\[\\tag{19} {F_{\\mathit{Ay}}}={F_2}-{F_{\\mathit{Cy}}}\\]\n<\/p>\n    <!-- Text cell -->\n    <div class=\"comment\">... and substitution of F<sub>Cy<\/sub> yields the last missing parameter\n<\/div>\n\n<div style=\"overflow:auto;\">\n    <!-- Code cell -->\n    <p>\\[\\tag{20} {F_{\\mathit{Ay}}}={F_2}-\\frac{-\\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{3}+{F_2}+2 {F_1}}{2}\\]\n<\/p>\n<\/div>\n    <!-- Text cell -->\n    <div class=\"comment\">Summary of the results\n<\/div>\n\n<div style=\"overflow:auto;\">\n    <!-- Code cell -->\n    <p>\\[\\tag{21} {F_{\\mathit{Ax}}}=\\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{9}-{F_1}\\]\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{22} {F_{\\mathit{Ay}}}={F_2}-\\frac{-\\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{3}+{F_2}+2 {F_1}}{2}\\]\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{23} {F_{\\mathit{Bx}}}={F_4}-\\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{9}\\]\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{24} {F_{\\mathit{By}}}=\\frac{-\\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{3}+{F_2}+2 {F_1}}{2}+{F_3}\\]\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{25} {F_{\\mathit{Cx}}}=-\\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{9}\\]\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{26} {F_{\\mathit{Cy}}}=\\frac{-\\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{3}+{F_2}+2 {F_1}}{2}\\]\n<\/p>\n<\/div>\n    <!-- Text cell -->\n    <div class=\"comment\">Inserting the numerical values \u200b\u200bdelivers\n<\/div>\n    <!-- Code cell -->\n    <p>\\[\\tag{27} {F_{\\mathit{Ax}}}=15.56 \\mathit{kN}\\]\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{28} {F_{\\mathit{Ay}}}=43.33 \\mathit{kN}\\]\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{29} {F_{\\mathit{Bx}}}=-15.56 \\mathit{kN}\\]\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{30} {F_{\\mathit{By}}}=46.67 \\mathit{kN}\\]\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{31} {F_{\\mathit{Cx}}}=-25.56 \\mathit{kN}\\]\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{32} {F_{\\mathit{Cy}}}=-13.33 \\mathit{kN}\\]\n<\/p>\n    <hr\/>\n  <\/body>\n<\/html>\n\n\n\n<h2 class=\"wp-block-heading\">Variant 2<\/h2>\n\n\n\n<p>This solution is calculated here exclusively with the variables, since the numerical values \u200b\u200bhave already been calculated above.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"868\" height=\"594\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-22alt.png\" alt=\"\" class=\"wp-image-1805\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-22alt.png 868w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-22alt-300x205.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-22alt-768x526.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-22alt-400x274.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-22alt-800x547.png 800w\" sizes=\"auto, (max-width: 868px) 100vw, 868px\" \/><figcaption>The whole frame as a system with the bearing reactions<\/figcaption><\/figure>\n\n\n\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    <link rel=\"stylesheet\" type=\"text\/css\" href=\"tm1-22%20alternativ_htmlimg\/tm1-22%20alternativ.css\"\/>\n    <!-- ****************************************************** -->\n    <!-- *        Created with wxMaxima version 20.06.6       * -->\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>Establishing the balance of forces and moments for the entire system\n<\/p>\n\n<div style=\"overflow:auto;\">\n<!-- Code cell -->\n    <p>\\[\\tag{1} \\sum F_x = 0={F_{\\mathit{Bx}}}+{F_{\\mathit{Ax}}}-{F_4}+{F_1}\\]\n<\/p>\n    <p>\\[\\tag{2} \\sum F_y = 0={F_{\\mathit{By}}}+{F_{\\mathit{Ay}}}-{F_3}-{F_2}\\]\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{3} \\sum M(A) = 0=6 {F_{\\mathit{By}}} a+{F_4} a-4 {F_3} a-{F_2} a-2 {F_1} a\\]\n<\/p>\n<\/div>\n<!-- Text cell -->\n    <p>The vertical support reaction F<sub>By<\/sub> can thus be calculated directly:\n<\/p>\n\n<div style=\"overflow:auto;\">\n    <!-- Code cell -->\n    <p>\\[\\tag{4} {F_{\\mathit{By}}}=-\\frac{{F_4}-4 {F_3}-{F_2}-2 {F_1}}{6}\\]\n<\/p>\n<\/div>\n<!-- Text cell -->\n    <p>F<sub>By<\/sub> is now substituted into equation 1, whereby F<sub>Ay<\/sub> follows.\n<\/p>\n\n<div style=\"overflow:auto;\">\n<!-- Code cell -->\n    <p>\\[\\tag{5} 0={F_{\\mathit{Ay}}}-\\frac{{F_4}-4 {F_3}-{F_2}-2 {F_1}}{6}-{F_3}-{F_2}\\]\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{6} {F_{\\mathit{Ay}}}=\\frac{{F_4}+2 {F_3}+5 {F_2}-2 {F_1}}{6}\\]\n<\/p>\n<\/div>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"594\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201115-164131_Sketch-1024x594.jpg\" alt=\"Bearing reactions and joint forces\" class=\"wp-image-456\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201115-164131_Sketch-1024x594.jpg 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201115-164131_Sketch-300x174.jpg 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201115-164131_Sketch-768x445.jpg 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201115-164131_Sketch-1536x891.jpg 1536w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201115-164131_Sketch-400x232.jpg 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201115-164131_Sketch-800x464.jpg 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201115-164131_Sketch.jpg 1688w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Investigation of the sections I and II<\/figcaption><\/figure>\n\n\n\n<p>Next, the balance of forces in the y-direction is set up for section I in order to calculate the joint force F<sub>Cy<\/sub>:<\/p>\n\n\n\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    <!-- *        Created with wxMaxima version 20.06.6       * -->\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<div style=\"overflow:auto;\">\n<!-- Code cell -->\n    <p>\\[\\tag{7} \\sum F_y = 0={F_{\\mathit{Cy}}}+{F_{\\mathit{Ay}}}-{F_2}\\]\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{8} {F_{\\mathit{Cy}}}=-\\frac{{F_4}+2 {F_3}-{F_2}-2 {F_1}}{6}\\]\n<\/p>\n<\/div><!-- Text cell -->\n    <p>The sum of moments around A leads to F<sub>Cx<\/sub>.\n<\/p>\n\n<div style=\"overflow:auto;\">\n    <!-- Code cell -->\n    <p>\\[\\tag{9} \\sum M(A) = 0=2 {F_{\\mathit{Cy}}} a-3 {F_{\\mathit{Cx}}} a-{F_2} a-2 {F_1} a\\]\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{10} {F_{\\mathit{Cx}}}=-\\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{9}\\]\n<\/p>\n<\/div>\n<!-- Text cell -->\n    <p>The equilibrium of forces in the x-direction for section I results in the horizontal support reaction in A.:\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{11} 0={F_{\\mathit{Cx}}}+{F_{\\mathit{Ax}}}+{F_1}\\]\n<\/p>\n    <!-- Code cell -->\n\n<div style=\"overflow:auto;\">\n    <p>\\[\\tag{12} {F_{\\mathit{Ax}}}=\\frac{{F_4}+2 {F_3}+2 {F_2}-5 {F_1}}{9}\\]\n<\/p>\n<\/div>\n<!-- Text cell -->\n    <p>F<sub>Ax<\/sub> is now substituted into equation 2, whereby F<sub>Bx<\/sub> follows.\n<\/p>\n    <!-- Code cell -->\n\n<div style=\"overflow:auto;\">\n    <p>\\[\\tag{13} {F_{\\mathit{Bx}}}=\\frac{8 {F_4}-2 {F_3}-2 {F_2}-4 {F_1}}{9}\\]\n<\/p>\n<\/div>\n\n\n\n<p>The second solution leads to the desired results faster!<\/p>\n","protected":false},"excerpt":{"rendered":"<p> ... <a title=\"Calculation of bearing loads and joint forces\" class=\"read-more\" href=\"https:\/\/pickedshares.com\/en\/engineering-mechanics-1-exercise-22-calculation-of-bearing-loads-and-joint-forces\/\" aria-label=\"Read more about Calculation of bearing loads and joint forces\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":454,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[21,60],"tags":[45,46,29],"class_list":["post-758","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>Calculation of bearing loads and joint forces &#8226; pickedshares<\/title>\n<meta name=\"description\" content=\"There are several forces acting on a frame with two fixed bearings. 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