{"id":1878,"date":"2021-01-14T11:29:26","date_gmt":"2021-01-14T11:29:26","guid":{"rendered":"https:\/\/pickedshares.com\/?p=1878"},"modified":"2021-05-08T12:58:40","modified_gmt":"2021-05-08T12:58:40","slug":"engineering-mechanics-2-exercise-10-beam-with-triangular-line-load","status":"publish","type":"post","link":"https:\/\/pickedshares.com\/en\/engineering-mechanics-2-exercise-10-beam-with-triangular-line-load\/","title":{"rendered":"Beam with triangular line load"},"content":{"rendered":"\n<p>This exercise shows how to calculate the bearing loads and the internal forces for a beam with triangular line load. <\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Task<\/h2>\n\n\n\n<p>A beam with a <a href=\"https:\/\/pickedshares.com\/en\/reaction-forces-in-bearings-joints-and-guidances\/\">fixed restraint<\/a> on one side is loaded with a triangular line load. The load sizes and the internal forces have to be determined!<\/p>\n\n\n\n<p>The function for the line load is<\/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>\\[ q(x) = q_0 \\cdot \\left( 1 - \\frac{x}{l} \\right) \\]<\/p>\n\n\n\n<figure class=\"wp-block-image size-large custom-flex-image\"><img loading=\"lazy\" decoding=\"async\" width=\"863\" height=\"539\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-10-4.png\" alt=\"Beam with fixed restraint and triangular line load\" class=\"wp-image-1868\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-10-4.png 863w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-10-4-300x187.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-10-4-768x480.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-10-4-400x250.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-10-4-800x500.png 800w\" sizes=\"auto, (max-width: 863px) 100vw, 863px\" \/><figcaption>Beam with fixed restraint and triangular line load<\/figcaption><\/figure>\n\n\n\n<h2 class=\"wp-block-heading\">Solution<\/h2>\n\n\n\n<p>Since the entire beam is loaded by the line load, no division of the beam into sections is required. In order to be able to calculate the internal forces, the support reactions do not necessarily have to be determined (in this case!). This depends on how the cuts are made, i.e. which area of \u200b\u200bthe beam is being \"cut off\". In the solution shown here, the sections are placed in such a way that the support reactions are not contained in the force equilibria of the internal forces. Nevertheless, for the sake of completeness, the reactions of the fixed restraint are calculated in the first step.<\/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\/2021\/01\/tm2-10-2.png\" alt=\"Bearing loads of the fixed restraint\" class=\"wp-image-1864\" width=\"749\" height=\"414\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-10-2.png 749w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-10-2-300x166.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-10-2-400x221.png 400w\" sizes=\"auto, (max-width: 749px) 100vw, 749px\" \/><figcaption>Bearing loads of the fixed restraint<\/figcaption><\/figure>\n\n\n\n<p>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\/\" target=\"_blank\" rel=\"noreferrer noopener\">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_{0}^{l}{q_0 \\cdot \\left( 1 - \\frac{x}{l} \\right)\\,dx}  \\]<\/p>\n<p>\\[ \\tag{3} \\sum M(A) = 0 = M_A - \\int_{0}^{l}{q_0 \\cdot \\left( 1 - \\frac{x}{l} \\right) \\cdot x\\,dx} \\]<\/p>\n<\/div>\n\n<p>The support 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} = - \\frac{q_0 \\cdot l}{2}  \\]<\/p>\n<p>\\[ \\tag{6} M_A = \\frac{q_0 \\cdot l^2}{6}  \\]<\/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<figure class=\"wp-block-image size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-10-3.png\" alt=\"Internal forces of the beam\" class=\"wp-image-1866\" width=\"748\" height=\"417\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-10-3.png 748w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-10-3-300x167.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-10-3-400x223.png 400w\" sizes=\"auto, (max-width: 748px) 100vw, 748px\" \/><figcaption>Internal forces of the beam<\/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 (the line load is now calculated using the auxiliary coordinate \u03be!):<\/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 \\]<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{8} \\sum F_z = 0 = -Q +\\int_{x}^{l}{q_0 \\cdot \\left( 1 - \\frac{\\xi}{l} \\right)\\,d\\xi}  \\]<\/p>\n<p>\\[ \\tag{9} \\sum M(x) = 0 = - M_b - \\int_{x}^{l}{q_0 \\cdot \\left( 1 - \\frac{\\xi}{l} \\right) \\cdot \\left( \\xi - x \\right) \\,d\\xi} \\]<\/p>\n<\/div>\n\n<p>The normal force is quickly determined:<\/p>\n\n<p>\\[ \\tag{10} N = 0 \\]<\/p>\n\n<p>The function of the shear force results from<\/p>\n<p>\\[ \\tag{11} Q = \\left[ q_0 \\cdot \\left( \\xi - \\frac{\\xi^2}{2 \\cdot l} \\right) \\right]_{x}^{l}  \\]<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{12} Q = q_0 \\cdot \\left( l - \\frac{l^2}{2 \\cdot l} \\right) - q_0 \\cdot \\left( x - \\frac{x^2}{2 \\cdot l} \\right)  \\]<\/p>\n<p>\\[ \\tag{13} Q = q_0 \\cdot \\left( \\frac{x^2-2 \\cdot l \\cdot x}{2 \\cdot l} + \\frac{l}{2} \\right)  \\]<\/p>\n<\/div>\n\n<p>And, finally, the bending moment:<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{14} M_{b} = \\frac{{q_0} {{x}^{3}}-3 l\\, {q_0} {{x}^{2}}+3 {{l}^{2}}\\, {q_0} x-{{l}^{3}}\\, {q_0}}{6 l} \\]<\/p>\n<\/div>\n\n\n\n<p>The function can easily be checked at this point. At the point x = 0, the bending moment must correspond to the support reaction M<sub>A<\/sub>. <\/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\\[ \\require{cancel} \\]\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{15} M_{b}(x=0) = \\frac{{q_0} {{0}^{3}}-3 l\\, {q_0} {{0}^{2}}+3 {{l}^{2}}\\, {q_0} 0-{{l}^{3}}\\, {q_0}}{6 l} \\]<\/p>\n<\/div>\n<p>\\[ \\tag{16} M_{b}(x=0) = - q_0 \\cdot \\frac{l^2}{6} \\]<\/p>\n\n\n\n<p>The result is the same. The negative sign arises from the differently assumed directions of rotation for the two moments.<\/p>\n\n\n\n<p>This exercise is part of the <a href=\"https:\/\/pickedshares.com\/en\/category\/engineering-mechanics-ii\/\">collection Engineering Mechanics II<\/a>.<\/p>\n","protected":false},"excerpt":{"rendered":"<p> ... <a title=\"Beam with triangular line load\" class=\"read-more\" href=\"https:\/\/pickedshares.com\/en\/engineering-mechanics-2-exercise-10-beam-with-triangular-line-load\/\" aria-label=\"Read more about Beam with triangular line load\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":1868,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[52,60],"tags":[70],"class_list":["post-1878","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>Beam with triangular line load<\/title>\n<meta name=\"description\" content=\"In this exercise the bearing reactions and the internal forces are calculated for a beam with triangular line load.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link 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