{"id":1800,"date":"2021-01-13T14:07:54","date_gmt":"2021-01-13T14:07:54","guid":{"rendered":"https:\/\/pickedshares.com\/?p=1800"},"modified":"2021-05-08T12:56:07","modified_gmt":"2021-05-08T12:56:07","slug":"engineering-mechanics-2-exercise-9-internal-forces-of-a-beam-under-partial-line-load","status":"publish","type":"post","link":"https:\/\/pickedshares.com\/en\/engineering-mechanics-2-exercise-9-internal-forces-of-a-beam-under-partial-line-load\/","title":{"rendered":"Internal forces of a beam under partial line load"},"content":{"rendered":"\n<p>This exercise addresses the following questions:<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>How to calculate the resulting moment with a line load?<\/li><li>How to calculate the internal forces for a line load?<\/li><li>How to calculate a beam with partial line load?<\/li><\/ul>\n\n\n\n<h2 class=\"wp-block-heading\">Task<\/h2>\n\n\n\n<p>A beam with a fixed restraint in A is loaded over half of its length by the uniform line load q. The internal forces are to be determined!<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"601\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-9-5-1024x601.png\" alt=\"Beam with fixed restraint and partial line load\" class=\"wp-image-1795\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-9-5-1024x601.png 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-9-5-300x176.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-9-5-768x451.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-9-5-400x235.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-9-5-800x469.png 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-9-5.png 1239w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Beam with fixed restraint and partial line load<\/figcaption><\/figure>\n\n\n\n<h2 class=\"wp-block-heading\">Solution<\/h2>\n\n\n\n<p>The beam is divided into two sections. In order to be able to calculate the internal forces, the support reactions do not necessarily have to be determined. This depends on how the cuts are made, i.e. which area of \u200b\u200bthe carrier is being \"cut off\". In the solution shown here, the cuts are placed in such a way that the support reactions are not included in the equilibrium of forces. Nevertheless, for the sake of completeness, the bearing reactions are calculated in the first step.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"489\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-9-2-1024x489.png\" alt=\"Cut-free beam and division of the sections\" class=\"wp-image-1789\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-9-2-1024x489.png 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-9-2-300x143.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-9-2-768x366.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-9-2-400x191.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-9-2-800x382.png 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-9-2.png 1381w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Cut-free beam and division of the sections<\/figcaption><\/figure>\n\n\n\n<p>The line load is calculated along the auxiliary coordinate \u03be. The establishment of the equilibrium of forces in the x and z directions as well as the equilibrium of moments (<a href=\"https:\/\/pickedshares.com\/en\/right-hand-rules\/\">counterclockwise torques positive<\/a>) leads to:<\/p>\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>\\[ \\tag{1} \\sum F_x = 0 = F_{Ax} \\]<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{2} \\sum F_z = 0 = F_{Az} +\\int_{l}^{2l}{q\\,d\\xi}  \\]<\/p>\n<p>\\[ \\tag{3} \\sum M(A) = 0 = M_A - \\int_{l}^{2l}{q \\xi \\,d\\xi} \\]<\/p>\n<\/div>\n\n<p>The bearing reactions of the fixed restraint can thus be determined directly<\/p>\n<p>\\[ \\tag{4} F_{Ax} = 0 \\]<\/p>\n<p>\\[ \\tag{5} F_{Az} = - q \\cdot l  \\]<\/p>\n<p>\\[ \\tag{6} M_A = \\frac{3}{2} \\cdot q \\cdot l^2 \\]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Determination of the internal forces<\/h3>\n\n\n\n<p>In the following, shear forces are referred to as Q, normal forces as N and bending moments as M<sub>b<\/sub>.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Section I<\/h4>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"506\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-9-3-1024x506.png\" alt=\"Internal forces and moments in section I\" class=\"wp-image-1791\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-9-3-1024x506.png 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-9-3-300x148.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-9-3-768x379.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-9-3-400x198.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-9-3-800x395.png 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-9-3.png 1383w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Internal forces and moments in section I<\/figcaption><\/figure>\n\n\n\n<p>Establishing the equilibrium of forces in the x and z directions as well as the equilibrium of moments:<\/p>\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>\\[ \\tag{7} \\sum F_x = 0 = -N_I \\]<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{8} \\sum F_z = 0 = -Q_I +\\int_{l}^{2l}{q\\,d\\xi}  \\]<\/p>\n<p>\\[ \\tag{9} \\sum M(x) = 0 = - M_{bI} - \\int_{l}^{2l}{q \\cdot (\\xi - x) \\,d\\xi} \\]<\/p>\n<\/div>\n<p>This results in the internal forces in section I:<\/p>\n\n<p>\\[ \\tag{10} N_I = 0 \\]<\/p>\n<p>\\[ \\tag{11} Q_I = q \\cdot l  \\]<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{12} M_{bI} = - \\frac{q \\cdot l}{2} \\cdot \\left( 2 \\cdot x - 3 \\cdot l \\right)  \\]<\/p>\n<\/div>\n\n\n\n<h4 class=\"wp-block-heading\">Section II<\/h4>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"505\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-9-4-1024x505.png\" alt=\"Internal forces and moments in section II\" class=\"wp-image-1793\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-9-4-1024x505.png 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-9-4-300x148.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-9-4-768x379.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-9-4-400x197.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-9-4-800x395.png 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-9-4.png 1389w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Internal forces and moments in section II<\/figcaption><\/figure>\n\n\n\n<p>Establishing the equilibrium of forces in the x and z directions as well as the equilibrium of moments:<\/p>\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>\\[ \\tag{13} \\sum F_x = 0 = -N_{II} \\]<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{14} \\sum F_z = 0 = -Q_{II} +\\int_{x}^{2l}{q\\,d\\xi}  \\]<\/p>\n<p>\\[ \\tag{15} \\sum M(x) = 0 = - M_{bII} - \\int_{x}^{2l}{q \\cdot (\\xi - x) \\,d\\xi} \\]<\/p>\n<\/div>\n<p>This results in the internal forces in section II:<\/p>\n\n<p>\\[ \\tag{16} N_{II} = 0 \\]<\/p>\n<p>\\[ \\tag{17} Q_{II} = q \\cdot (2\\cdot l - x)  \\]<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{18} M_{bII} = - q \\cdot \\left( \\frac{x^2}{2} - 2 \\cdot l \\cdot x + 2 \\cdot l^2 \\right) \\]<\/p>\n<\/div>\n\n\n\n<p>With this all internal forces of the beam are determined. This exercise is part of the <a href=\"https:\/\/pickedshares.com\/en\/category\/engineering-mechanics-ii\/\">Engineering Mechanics II - Strength of Materials collection<\/a>.<\/p>\n","protected":false},"excerpt":{"rendered":"<p> ... <a title=\"Internal forces of a beam under partial line load\" class=\"read-more\" href=\"https:\/\/pickedshares.com\/en\/engineering-mechanics-2-exercise-9-internal-forces-of-a-beam-under-partial-line-load\/\" aria-label=\"Read more about Internal forces of a beam under partial line load\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":1787,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[52,60],"tags":[70],"class_list":["post-1800","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-engineering-mechanics-ii","category-exercises","tag-line-load","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>Internal forces of a beam under partial line load<\/title>\n<meta name=\"description\" content=\"In this exercise the internal forces of a beam with a one sided fixed restraint and a partial line load are determined.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" 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