{"id":3769,"date":"2021-03-07T19:20:58","date_gmt":"2021-03-07T19:20:58","guid":{"rendered":"https:\/\/pickedshares.com\/?p=3769"},"modified":"2021-05-08T14:56:08","modified_gmt":"2021-05-08T14:56:08","slug":"bearing-loads-of-a-multi-span-beam","status":"publish","type":"post","link":"https:\/\/pickedshares.com\/en\/bearing-loads-of-a-multi-span-beam\/","title":{"rendered":"Bearing loads of a multi-span 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> of a symmetrical multi-span beam supported on three columns are calculated. Here is an <a href=\"https:\/\/pickedshares.com\/en\/deflection-calculator-beam-on-3-supports-under-line-load\/\">online calculator here for calculating asymmetrical multi-span beams<\/a>.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Task<\/h2>\n\n\n\n<p>A continuous beam on a fixed bearing and two floating bearings (multi-span beam) with the symmetrical column width l is loaded by a line load q<sub>0<\/sub>. How big are the bearing loads?<\/p>\n\n\n\n<figure class=\"wp-block-image size-large custom-flex-image75\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"395\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/03\/tm2-21-1-1024x395.png\" alt=\"Multi-span beam under line load\" class=\"wp-image-3752\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/03\/tm2-21-1-1024x395.png 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/03\/tm2-21-1-300x116.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/03\/tm2-21-1-768x296.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/03\/tm2-21-1-1536x592.png 1536w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/03\/tm2-21-1-400x154.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/03\/tm2-21-1-800x308.png 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/03\/tm2-21-1.png 1601w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Multi-span beam under line load<\/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\" target=\"_blank\" rel=\"noreferrer noopener\">Left turning moments are positive<\/a>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Determination of the bearing loads<\/h3>\n\n\n\n<figure class=\"wp-block-image size-large custom-flex-image75\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"408\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/03\/tm2-21-2-1024x408.png\" alt=\"Bearing reactions and sections of the beam\" class=\"wp-image-3758\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/03\/tm2-21-2-1024x408.png 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/03\/tm2-21-2-300x120.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/03\/tm2-21-2-768x306.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/03\/tm2-21-2-1536x613.png 1536w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/03\/tm2-21-2-400x160.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/03\/tm2-21-2-800x319.png 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/03\/tm2-21-2.png 1637w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Bearing reactions 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\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{1} \\sum F_z = 0 = -F_{Az} - F_{Bz} - F_{Cz} + \\int_0^{2l}{q_0dx} \\]<\/p>\n<p>\\[ \\tag{2} 0 = -F_{Az} - F_{Bz} - F_{Cz} + 2 \\cdot q_0 \\cdot l \\]<\/p>\n<\/div>\n<p>The sum of moments at A<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{3} \\sum M(A) = 0 = F_{Bz} \\cdot l + 2 \\cdot F_{Cz} \\cdot l - \\int_0^{2l}{q_0xdx} \\]<\/p>\n<p>\\[ \\tag{4} 0 = F_{Bz} \\cdot l + 2 \\cdot F_{Cz} \\cdot l - 2 \\cdot q_0 \\cdot l^2 \\]<\/p>\n<\/div>\n\n\n\n<p>This results in a system of equations with 2 equations but 3 unknowns. The next step is to determine the internal forces in the two sections I and II. In order to be able to integrate the function of the line load from 0 to x, it is formulated as a function of the auxiliary coordinate \u03be. Normal forces are denoted by N, transverse forces by Q and bending moments by M<sub>b<\/sub>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Determination of the internal forces<\/h3>\n\n\n\n<h4 class=\"wp-block-heading\">Section I<\/h4>\n\n\n\n<figure class=\"wp-block-image size-large custom-flex-image75\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"650\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/03\/tm2-21-3-1024x650.png\" alt=\"Internal forces and moments in section I\" class=\"wp-image-3760\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/03\/tm2-21-3-1024x650.png 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/03\/tm2-21-3-300x190.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/03\/tm2-21-3-768x487.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/03\/tm2-21-3-400x254.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/03\/tm2-21-3-800x508.png 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/03\/tm2-21-3.png 1095w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><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 forces in z-direction is<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{5} \\sum F_z = 0 = -F_{Az} + Q_I + \\int_0^{x}{q_0d \\xi} \\]<\/p>\n<\/div>\n<p>\\[ \\tag{6} 0 = -F_{Az} + Q_I + q_0 \\cdot x \\]<\/p>\n<p>The sum of moments at x is<\/p>\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{7} \\sum M(x) = 0 = \\int_0^{x}{q_0 \\xi d \\xi} - F_{Az} \\cdot x + M_{bI} \\]<\/p>\n<p>\\[ \\tag{8} 0 = \\frac{1}{2} q_0 x^2 - F_{Az} \\cdot x + M_{bI} \\]<\/p><\/div>\n\n\n\n<h4 class=\"wp-block-heading\">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=\"489\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/03\/tm2-21-4-1024x489.png\" alt=\"Internal forces and moments in section II\" class=\"wp-image-3762\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/03\/tm2-21-4-1024x489.png 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/03\/tm2-21-4-300x143.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/03\/tm2-21-4-768x367.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/03\/tm2-21-4-1536x734.png 1536w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/03\/tm2-21-4-400x191.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/03\/tm2-21-4-800x382.png 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/03\/tm2-21-4.png 1551w\" 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 forces in z-direction is<\/p>\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{9} \\sum F_z = 0 = -F_{Az} -F_{Bz}+ Q_{II} + \\int_0^{x}{q_0d \\xi} \\]<\/p>\n<p>\\[ \\tag{10} 0 = -F_{Az} -F_{Bz}+ Q_{II} + q_0 x \\]<\/p><\/div>\n<p>The sum of moments at x<\/p>\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{11} \\sum M(x) = 0 = \\int_0^{x}{q_0 \\xi d \\xi} - F_{Az} \\cdot x - F_{Az} \\cdot (x-l) + M_{bII} \\]<\/p>\n<p>\\[ \\tag{12} 0 = \\frac{1}{2} q_0 x^2 - F_{Az} \\cdot x - F_{Bz} \\cdot (x-l) + M_{bII} \\]<\/p><\/div>\n\n<p>Next, the functions of the bending lines are calculated.<\/p>\n<h3>Bending lines<\/h3>\n<p>The function of the bending line is designated as w, or the first derivative as w' and the second derivative as w''. The modulus of elasticity is E (Young's modulus) and the area moment of inertia is I.<\/p>\n<h4>Section I<\/h4>\n<p>\\[ \\tag{13}  M_{bI} = -\\frac{{q_0} {{x}^{2}}-2 {F_{\\mathit{Az}}} x}{2} \\]<\/p>\n<p>\\[ \\tag{14}  w''_I = - \\frac{1}{E \\cdot I} M_{bI}   \\]<\/p>\n<p>\\[ \\tag{15}  w''_I = \\frac{{q_0} {{x}^{2}}-2 {F_{\\mathit{Az}}} x}{2 E I}  \\]<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{16}  w'_I = \\frac{\\frac{{q_0} {{x}^{3}}}{3}-{F_{\\mathit{Az}}} {{x}^{2}}}{2 E I}+\\mathit{ c_1} \\]<\/p>\n<p>\\[ \\tag{17}  w_I = \\frac{\\frac{{q_0} {{x}^{4}}}{12}-\\frac{{F_{\\mathit{Az}}} {{x}^{3}}}{3}}{2 E I}+\\mathit{ c_1} x+\\mathit{ c_2} \\]<\/p><\/div>\n<h4>Section II<\/h4>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{18} M_{bII} = -\\frac{{q_o} {{x}^{2}}+\\left( -2 {F_{\\mathit{Bz}}}-2 {F_{\\mathit{Az}}}\\right)  x+2 {F_{\\mathit{Bz}}} l}{2}\\]<\/p>\n<p>\\[ \\tag{19} w''_{II} = - \\frac{1}{E \\cdot I} M_{bII}  \\]<\/p>\n<p>\\[ \\tag{20} w''_{II} = \\frac{{q_o} {{x}^{2}}+\\left( -2 {F_{\\mathit{Bz}}}-2 {F_{\\mathit{Az}}}\\right)  x+2 {F_{\\mathit{Bz}}} l}{2 E I}  \\]<\/p>\n<p>\\[ \\tag{21} w'_{II} = \\frac{\\frac{{q_o} {{x}^{3}}}{3}+\\frac{\\left( -2 {F_{\\mathit{Bz}}}-2 {F_{\\mathit{Az}}}\\right)  {{x}^{2}}}{2}+2 {F_{\\mathit{Bz}}} l x}{2 E I}+\\mathit{ c_3}  \\]<\/p>\n<p>\\[ \\tag{22} w_{II} = \\frac{\\frac{{q_o} {{x}^{4}}}{12}+\\frac{\\left( -2 {F_{\\mathit{Bz}}}-2 {F_{\\mathit{Az}}}\\right)  {{x}^{3}}}{6}+{F_{\\mathit{Bz}}} l\\, {{x}^{2}}}{2 E I}+\\mathit{ c_3} x+\\mathit{ c_4}  \\]<\/p><\/div>\n\n<p>The constants of integration are determined via the boundary and transition conditions.<\/p>\n<h4>Boundary and transition conditions<\/h4>\n<p>The deflection at the point x = 0 is equal to zero.<\/p>\n<p>\\[ \\tag{23} 0=\\mathit{c_2} \\]<\/p>\n<p>The deflection at the point x = l is zero. This condition can be used for both bending lines.<\/p>\n<p>\\[ \\tag{24} \\frac{\\frac{{{l}^{4}}\\, {q_0}}{12}-\\frac{{F_{\\mathit{Az}}} {{l}^{3}}}{3}}{2 E I}+\\mathit{c_1} l = 0\\]<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{25}  \\frac{\\frac{{{l}^{4}}\\, {q_o}}{12}+{F_{\\mathit{Bz}}} {{l}^{3}}+\\frac{\\left( -2 {F_{\\mathit{Bz}}}-2 {F_{\\mathit{Az}}}\\right)  {{l}^{3}}}{6}}{2 E I}+\\mathit{ c_3} l+\\mathit{ c_4}=0\\]<\/p>\n<\/div>\n<p>The deflection at the point x = 2l is zero.<\/p>\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{26} \\frac{\\frac{4 {{l}^{4}}\\, {q_o}}{3}+4 {F_{\\mathit{Bz}}} {{l}^{3}}+\\frac{4 \\left( -2 {F_{\\mathit{Bz}}}-2 {F_{\\mathit{Az}}}\\right)  {{l}^{3}}}{3}}{2 E I}+2 \\mathit{ c_3} l+\\mathit{ c_4} \\]<\/p><\/div>\n<p>Both bending lines have the same angle at the point x = 1.<\/p>\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{27} \\frac{\\frac{{{l}^{3}}\\, {q_0}}{3}-{F_{\\mathit{Az}}} {{l}^{2}}}{2 E I}+\\mathit{ c_1} = \\frac{\\frac{{{l}^{3}}\\, {q_o}}{3}+2 {F_{\\mathit{Bz}}} {{l}^{2}}+\\frac{\\left( -2 {F_{\\mathit{Bz}}}-2 {F_{\\mathit{Az}}}\\right)  {{l}^{2}}}{2}}{2 E I}+\\mathit{ c_3} \\]<\/p><\/div>\n<p>There are now sufficient equations to be able to solve the unknowns.<\/p>\n<p>\\[ \\tag{28} \\mathit{ c_1}=\\frac{11 {{l}^{3}}\\, {q_o}-9 {{l}^{3}}\\, {q_0}}{96 E I} \\]<\/p>\n<p>\\[ \\tag{29} c_2 = 0 \\]<\/p>\n<p>\\[ \\tag{30} \\mathit{ c_3}=\\frac{61 {{l}^{3}}\\, {q_o}-119 {{l}^{3}}\\, {q_0}}{96 E I} \\]<\/p>\n<p>\\[ \\tag{31} \\mathit{ c_4}=-\\frac{5 {{l}^{4}}\\, {q_o}-15 {{l}^{4}}\\, {q_0}}{48 E I} \\]<\/p>\n<p>\\[ \\tag{32} {F_{\\mathit{Az}}}=\\frac{11 l\\, {q_o}-5 l\\, {q_0}}{16} \\]<\/p>\n<p>\\[ \\tag{33} {F_{\\mathit{Bz}}}=-\\frac{11 l\\, {q_o}-21 l\\, {q_0}}{8} \\]<\/p>\n<p>\\[ \\tag{34} {F_{\\mathit{Cz}}}=\\frac{11 l\\, {q_o}-5 l\\, {q_0}}{16} \\]<\/p>\n<p>And for the sake of completeness<\/p>\n<p>\\[ \\tag{35} F_{Ax} = 0 \\]<\/p>\n\n\n\n<p><\/p>\n","protected":false},"excerpt":{"rendered":"<p> ... <a title=\"Bearing loads of a multi-span beam\" class=\"read-more\" href=\"https:\/\/pickedshares.com\/en\/bearing-loads-of-a-multi-span-beam\/\" aria-label=\"Read more about Bearing loads of a multi-span beam\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":3754,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[52,60],"tags":[66,70,73],"class_list":["post-3769","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-engineering-mechanics-ii","category-exercises","tag-bending-line","tag-line-load","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 multi-span beam &#8226; 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