{"id":3605,"date":"2021-03-01T09:10:00","date_gmt":"2021-03-01T09:10:00","guid":{"rendered":"https:\/\/pickedshares.com\/?p=3605"},"modified":"2021-05-08T13:22:50","modified_gmt":"2021-05-08T13:22:50","slug":"bearing-loads-of-a-statically-indetermined-beam","status":"publish","type":"post","link":"https:\/\/pickedshares.com\/en\/bearing-loads-of-a-statically-indetermined-beam\/","title":{"rendered":"Bearing loads of a statically indetermined beam"},"content":{"rendered":"\n<p>In this exercise, the <a href=\"https:\/\/pickedshares.com\/en\/reaction-forces-in-bearings-joints-and-guidances\/\">support reactions<\/a> for a statically indetermined beam are calculated using the bending line.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Exercise<\/h2>\n\n\n\n<p>A beam with a fixed restraint A and a sliding guide B is loaded centrally by the force F. Which bearing loads occur?<\/p>\n\n\n\n<figure class=\"wp-block-image size-large custom-flex-image75\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"594\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/02\/tm2-20-1-1024x594.png\" alt=\"Beam with fixation and guidance\" class=\"wp-image-3576\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/02\/tm2-20-1-1024x594.png 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/02\/tm2-20-1-300x174.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/02\/tm2-20-1-768x445.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/02\/tm2-20-1-400x232.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/02\/tm2-20-1-800x464.png 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/02\/tm2-20-1.png 1157w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Beam with fixation and guidance<\/figcaption><\/figure>\n\n\n\n<h2 class=\"wp-block-heading\">Solution<\/h2>\n\n\n\n<p>In the first step, the equations for the bearing reactions are determined. For the subsequent determination of the internal forces, the beam is divided into two areas. The horizontal forces are not considered further in the following equations, since they are obviously zero. <a href=\"https:\/\/en.wikipedia.org\/wiki\/Right-hand_rule#Direction_associated_with_a_rotation\" target=\"_blank\" rel=\"noreferrer noopener\">Left turning moments are positive<\/a>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Bearing reactions<\/h3>\n\n\n\n<figure class=\"wp-block-image size-large custom-flex-image75\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"529\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/02\/tm2-20-2-1024x529.png\" alt=\"Bearing loads and sections of the beam\" class=\"wp-image-3582\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/02\/tm2-20-2-1024x529.png 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/02\/tm2-20-2-300x155.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/02\/tm2-20-2-768x397.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/02\/tm2-20-2-400x207.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/02\/tm2-20-2-800x413.png 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/02\/tm2-20-2.png 1303w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Bearing loads and sections of the beam<\/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<p>The sum of forces in z-direction is<\/p>\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{1} \\sum F_z = 0 = -F_{Az} + F - F_{Bz} \\]<\/p><\/div>\n<p>The sum of moments regarding point A<\/p>\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{2} \\sum M(A) = 0 = M_A-F \\cdot l+2 \\cdot F_{Bz} \\cdot l+M_B \\]<\/p><\/div>\n\n\n\n<p>In the case of a statically indetermined beam, the bearing reactions cannot be determined from these equations alone, since there are too many unknowns. Therefore, in the next step, the functions of the bending moments and the bending lines are determined.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Bending moments<\/h3>\n\n\n\n<h4 class=\"wp-block-heading\">Bending moment in section I<\/h4>\n\n\n\n<figure class=\"wp-block-image size-large custom-flex-image75\"><img loading=\"lazy\" decoding=\"async\" width=\"1021\" height=\"657\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/02\/tm2-20-3.png\" alt=\"Internal forces and moments in section I\" class=\"wp-image-3584\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/02\/tm2-20-3.png 1021w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/02\/tm2-20-3-300x193.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/02\/tm2-20-3-768x494.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/02\/tm2-20-3-400x257.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/02\/tm2-20-3-800x515.png 800w\" sizes=\"auto, (max-width: 1021px) 100vw, 1021px\" \/><figcaption>Internal forces and moments in section I<\/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<p>The sum of moments regarding x is<\/p>\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{3} \\sum M(x) = 0 = M_A + M_{bI}-F_{Az} \\cdot x \\]<\/p><\/div>\n<p>\\[ \\tag{4} M_{bI} = F_{Az} \\cdot x - M_A \\]<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Bending moments in section II<\/h4>\n\n\n\n<figure class=\"wp-block-image size-large custom-flex-image75\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"566\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/02\/tm2-20-4-1024x566.png\" alt=\"Internal forces and moments in section II\" class=\"wp-image-3586\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/02\/tm2-20-4-1024x566.png 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/02\/tm2-20-4-300x166.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/02\/tm2-20-4-768x425.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/02\/tm2-20-4-400x221.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/02\/tm2-20-4-800x442.png 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/02\/tm2-20-4.png 1163w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Internal forces and moments in section II<\/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<p>The sum of moments regarding x is<\/p>\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{5} \\sum M(x) = 0 = M_A - F_{Az} \\cdot x+F \\cdot (x-l)+M_{bII} \\]<\/p>\n<p>\\[ \\tag{6} M_{bII} = (F_{Az}-F) \\cdot x+F \\cdot l-M_A \\]<\/p><\/div>\n\n\n\n<h3 class=\"wp-block-heading\">Deflection lines<\/h3>\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<p>The bending line is determined on the basis of the Bernoulli beam bending. The basic relation is<\/p>\n<p>\\[ w'' = \\frac{-M_b}{E \\cdot I} \\]<\/p>\n<p>Here w'' is the second derivative of the bending line, E is the (Young's) modulus of elasticity and I is the area moment of inertia. The bending line w is obtained by integrating twice over x. It follows from this for the present case<\/p>\n<h4>Bending line for section I<\/h4>\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{7} w''_1 = \\frac{1}{E \\cdot I} \\cdot \\left( M_A - F_{Az} \\cdot x \\right) \\]<\/p>\n<p>\\[ \\tag{8} w'_1 = \\frac{1}{E \\cdot I} \\cdot \\left(  M_A \\cdot x - \\frac{1}{2} F_{Az} \\cdot x^2 \\right) + c_1 \\]<\/p>\n<p>\\[ \\tag{9} w_1 = \\frac{1}{E \\cdot I} \\cdot \\left(  \\frac{1}{2} M_A \\cdot x^2 - \\frac{1}{6} F_{Az} \\cdot x^3 \\right) + c_1 \\cdot x + c_2 \\]<\/p><\/div>\n<h4>Bending line for section II<\/h4>\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{10} w''_2 = \\frac{1}{E \\cdot I} \\cdot \\left( (F-F_{Az}) \\cdot x - F \\cdot l + M_A  \\right) \\]<\/p>\n<p>\\[ \\tag{11} w'_2 = \\frac{1}{E \\cdot I} \\cdot \\left( \\frac{1}{2}(F-F_{Az}) \\cdot x^2 - F \\cdot l \\cdot x + M_A \\cdot x  \\right) + c_3 \\]<\/p>\n<p>\\[ \\tag{12} w_2 = \\frac{\\frac{1}{6}(F-F_{Az}) \\cdot x^3 - \\frac{1}{2} F \\cdot l \\cdot x^2 + \\frac{1}{2} M_A \\cdot x^2 }{E \\cdot I}  + c_3 \\cdot x + c_4 \\]<\/p><\/div>\n\n\n\n<p>Next, the boundary and transition conditions are established.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Boundary and transition conditions<\/h4>\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\n<h5>The deflection at the point x = 0 is equal to zero<\/h5>\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{13} w_1(x=0) = 0 = \\frac{1}{E \\cdot I} \\cdot \\left(  \\frac{1}{2} M_A \\cdot 0^2 - \\frac{1}{6} F_{Az} \\cdot 0^3 \\right) + c_1 \\cdot 0 + c_2 \\]<\/p><\/div>\n<p>\\[ \\tag{14} c_2 = 0 \\]<\/p>\n<h5>The deflection at the point x = 2l is zero<\/h5>\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{15} w_2(x=2l) = 0 = \\frac{ \\frac{4}{3}(F-F_{Az}) \\cdot l^3 - 2 F \\cdot l^3 + 2 M_A \\cdot l^2}{E \\cdot I} + c_3 \\cdot 2l + c_4 \\]<\/p><\/div>\n<h5>The angle at the point x = 0 is equal to zero<\/h5>\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{16} w'_1(x=0) = 0 = \\frac{1}{E \\cdot I} \\cdot \\left(  M_A \\cdot 0 - \\frac{1}{2} F_{Az} \\cdot 0^2 \\right) + c_1 \\]<\/p><\/div>\n<p>\\[ \\tag{17} c_1 = 0 \\]<\/p>\n<h5>The angle at the point x = 2l is equal to zero<\/h5>\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{18} w'_2(x=2l) = 0 = \\frac{1}{E \\cdot I} \\cdot \\left( 2(F-F_{Az}) \\cdot l^2 - F \\cdot l^3 + 2M_A l  \\right) + c_3 \\]<\/p><\/div>\n<h5>The deflection of both bending lines at the point x = l is the same<\/h5>\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{19} \\frac{\\frac{1}{2} M_A \\cdot l^2 - \\frac{1}{6} F_{Az} \\cdot l^3 }{E \\cdot I}  = \\frac{ \\frac{1}{6}(F-F_{Az}) \\cdot l^3 - \\frac{1}{2} F \\cdot l^3 + \\frac{1}{2} M_A \\cdot l^2 }{E \\cdot I}  + c_3 \\cdot l + c_4 \\]<\/p><\/div>\n<h5>The angle of both bending lines at the point x = l is the same<\/h5>\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{20} \\frac{ M_A \\cdot l - \\frac{1}{2} F_{Az} \\cdot l^2}{E \\cdot I}  = \\frac{\\frac{1}{2}(F-F_{Az}) \\cdot l^2 - F \\cdot l^2 + M_A \\cdot l }{E \\cdot I}  + c_3\\]<\/p><\/div>\n\n<p>The equations can now be solved. The constants of integration are given by<\/p>\n\n<p>\\[ \\tag{21} c_3 = \\frac{F \\cdot l^2}{2 \\cdot E \\cdot I} \\]<\/p>\n<p>\\[ \\tag{22} c_4 = - \\frac{F \\cdot l^3}{6 \\cdot E \\cdot I} \\]<\/p>\n\n\n<p>The bearing loads are<\/p>\n<p>\\[ \\tag{23} M_A = \\frac{F \\cdot l}{4} \\]<\/p>\n<p>\\[ \\tag{24} M_B = - \\frac{F \\cdot l}{4} \\]<\/p>\n<p>\\[ \\tag{25} F_{Az} = \\frac{F}{2} \\]<\/p>\n<p>\\[ \\tag{26} F_{Bz} = \\frac{F}{2} \\]<\/p>\n<p>And for the sake of completeness<\/p>\n<p>\\[ \\tag{27} F_{Ax} = 0 \\]<\/p>\n\n\n\n<p>The bearing reactions for the statically indetermined beam are thus determined.<\/p>\n","protected":false},"excerpt":{"rendered":"<p> ... <a title=\"Bearing loads of a statically indetermined beam\" class=\"read-more\" href=\"https:\/\/pickedshares.com\/en\/bearing-loads-of-a-statically-indetermined-beam\/\" aria-label=\"Read more about Bearing loads of a statically indetermined beam\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":3578,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[52,60],"tags":[66,73],"class_list":["post-3605","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-engineering-mechanics-ii","category-exercises","tag-bending-line","tag-statically-overdetermined","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 of a statically indetermined beam &#8226; pickedshares<\/title>\n<meta name=\"description\" content=\"How to calculate the bearing loads for a statically indetermined beam using the Bernoulli beam theory. 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