{"id":1515,"date":"2021-01-08T17:39:32","date_gmt":"2021-01-08T17:39:32","guid":{"rendered":"https:\/\/pickedshares.com\/?p=1515"},"modified":"2021-05-08T12:48:37","modified_gmt":"2021-05-08T12:48:37","slug":"engineering-mechanics-2-exercise-4-internal-forces-of-a-beam-under-line-load","status":"publish","type":"post","link":"https:\/\/pickedshares.com\/en\/engineering-mechanics-2-exercise-4-internal-forces-of-a-beam-under-line-load\/","title":{"rendered":"Internal forces of a beam under line load"},"content":{"rendered":"\n<p>In this exercise we investigate a beam under line load regarding the following questions:<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>How to determine the <a href=\"https:\/\/pickedshares.com\/en\/reaction-forces-in-bearings-joints-and-guidances\/\">bearing loads<\/a> of a beam under line load?<\/li><li>How to determine the internal forces of a beam under line load?<\/li><li>How to determine the maximum bending moment of an even loaded beam?<\/li><\/ul>\n\n\n\n<h2 class=\"wp-block-heading\">Task<\/h2>\n\n\n\n<p>A beam on two supports is under load of the even line load q. Determine the bearing reactions and the internal forces!<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"553\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-4-1-1024x553.png\" alt=\"Beam on two supports with an even line load\" class=\"wp-image-1501\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-4-1-1024x553.png 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-4-1-300x162.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-4-1-768x415.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-4-1-1536x830.png 1536w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-4-1-400x216.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-4-1-800x432.png 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-4-1.png 1635w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Beam on two supports with an even line load<\/figcaption><\/figure>\n\n\n\n<h2 class=\"wp-block-heading\">Solution<\/h2>\n\n\n\n<p>The beam is cut free and the bearing reactions are drawn. The horizontal component of the fixed bearing is not discussed here, as it is obviously zero. The line load q, even if it is given here as constant, is understood as a function. Note the downward z-axis!<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"550\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-4-2-1024x550.png\" alt=\"Bearing reactions of the beam\" class=\"wp-image-1505\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-4-2-1024x550.png 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-4-2-300x161.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-4-2-768x413.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-4-2-1536x826.png 1536w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-4-2-400x215.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-4-2-800x430.png 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-4-2.png 1639w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Bearing reactions of the beam<\/figcaption><\/figure>\n\n\n\n<h3 class=\"wp-block-heading\">Determination of the bearing reactions<\/h3>\n\n\n\n<p>Establishing the equilibrium conditions for forces in the z-direction as well as for the 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{1} \\sum F_x = 0  \\]<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{2} \\sum F_z = 0 = - F_{Az} + \\int_0^l{q(x)dx} - F_{Bz} \\]<\/p>\n<p>\\[ \\tag{3} \\sum M(A) = 0 = F_{Bz} \\cdot l - \\int_0^l{q(x)ldx} \\]<\/p>\n<\/div>\n\n<p>From equation (3) follows<\/p>\n\n<p>\\[ \\tag{4} F_{Bz} = \\frac{q \\cdot l}{2} \\]<\/p>\n\n<p>and so it results<\/p>\n\n<p>\\[ \\tag{5} F_{Az} = \\frac{q \\cdot l}{2} \\]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Determination of the internal forces<\/h3>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"731\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-4-3-1024x731.png\" alt=\"Internal forces of the beam\" class=\"wp-image-1507\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-4-3-1024x731.png 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-4-3-300x214.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-4-3-768x548.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-4-3-400x286.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-4-3-800x571.png 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-4-3.png 1035w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Internal forces of the beam<\/figcaption><\/figure>\n\n\n\n<p>In the above figure, all three internal forces, i.e. normal force, shear force and bending moment, are given. Since the forces in the x-direction are zero, the normal force is also zero, i.e. it is not listed further below. The function of the line load is given the substitute coordinate \u03be.<\/p>\n\n\n\n<p>Establishing the equilibrium conditions for forces in the z-direction as well as for the 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\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{6} \\sum F_z = 0 = - F_{Az} + \\int_0^x{q(\\xi)d\\xi} + Q \\]<\/p>\n<p>\\[ \\tag{7} \\sum M(x) = 0 = - F_{Az} \\cdot x + M_b + \\int_0^x{q(\\xi)\\cdot (x-\\xi)d\\xi} \\]<\/p>\n<\/div>\n<p>The already known bearing load F<sub>Az<\/sub> is used now in equation (7).<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{8} 0 = - \\frac{q \\cdot l}{2} \\cdot x + M_b + \\int_0^x{q(\\xi)\\cdot (x-\\xi)d\\xi} \\]<\/p>\n<\/div>\n<p>The bending moment results to<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{9}  M_b = \\frac{q \\cdot l}{2} \\cdot x - \\left[ q \\cdot (x\\cdot \\xi-\\frac{\\xi^2}{2}) \\right]_0^x   \\]<\/p>\n<\/div>\n<p>\\[ \\tag{10}  M_b(x) = \\frac{q \\cdot x}{2} \\cdot (l -  x) \\]<\/p>\n\n<p>In this way, all the required stress parameters are determined. In this scenario, the highest bending moment occurs in the center of the beam and is<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{11}  M_b(x=\\frac{l}{2}) = \\frac{q \\cdot \\frac{l}{2}}{2} \\cdot (l -  \\frac{l}{2}) \\]<\/p>\n<p>\\[ \\tag{12}  M_b(x=\\frac{l}{2}) = \\frac{q \\cdot l^2}{8}  \\]<\/p>\n<\/div>\n\n\n\n<p>Here you can <a href=\"https:\/\/pickedshares.com\/en\/deflection-calculator-even-line-load\/\">calculate the deflection of a beam on two supports with an even line load<\/a>.<\/p>\n","protected":false},"excerpt":{"rendered":"<p> ... <a title=\"Internal forces of a beam under line load\" class=\"read-more\" href=\"https:\/\/pickedshares.com\/en\/engineering-mechanics-2-exercise-4-internal-forces-of-a-beam-under-line-load\/\" aria-label=\"Read more about Internal forces of a beam under line load\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":1503,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[52,60],"tags":[70],"class_list":["post-1515","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 line load - Engineering mechanics<\/title>\n<meta name=\"description\" content=\"In this exercise the internal forces of a beam under line load is calculated as well as the maximum bending moment.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, 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