{"id":1430,"date":"2021-01-06T19:57:43","date_gmt":"2021-01-06T19:57:43","guid":{"rendered":"https:\/\/pickedshares.com\/?p=1430"},"modified":"2021-05-08T12:46:27","modified_gmt":"2021-05-08T12:46:27","slug":"engineering-mechanics-ii-exercise-3-determine-internal-forces","status":"publish","type":"post","link":"https:\/\/pickedshares.com\/en\/engineering-mechanics-ii-exercise-3-determine-internal-forces\/","title":{"rendered":"Determine internal forces"},"content":{"rendered":"\n<p>This exercise is about the determination of the internal forces of an angled beam and addresses the following questions:<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>How to determine the internal forces and moments of a beam?<\/li><li>How to calculate the shear force, the normal force and the bending moment?<\/li><li>Where do I have to divide the beam into sections?<\/li><li>How can I show the gradients of the internal forces and moments as diagrams?<\/li><\/ul>\n\n\n\n<h2 class=\"wp-block-heading\">Task<\/h2>\n\n\n\n<p>An angled beam with a <a href=\"https:\/\/pickedshares.com\/en\/reaction-forces-in-bearings-joints-and-guidances\/\">fixed bearing and a floating bearing<\/a> is loaded by the force F. The internal forces of the beam have to be determined!<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"545\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-1-1024x545.png\" alt=\"Angled beam on two supports\" class=\"wp-image-1414\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-1-1024x545.png 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-1-300x160.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-1-768x409.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-1-400x213.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-1-800x426.png 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-1.png 1513w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Angled beam on two supports<\/figcaption><\/figure>\n\n\n\n<h2 class=\"wp-block-heading\">Solution<\/h2>\n\n\n\n<p>To calculate the internal forces of the beam, the bearing loads are determined and the beam is divided into three sections. In the following, the shear forces are denoted by Q, the normal forces by N and the bending moment by M<sub>b<\/sub>. <a href=\"https:\/\/en.wikipedia.org\/wiki\/Right-hand_rule#:~:text=For%20right%2Dhanded%20coordinates%20the,the%20second%20or%20Y%20axis.&amp;text=Reversing%20two%20axes%20amounts%20to,rotation%20around%20the%20remaining%20axis.\" target=\"_blank\" rel=\"noreferrer noopener\">Left turning moments are positive<\/a>.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"691\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-2-1024x691.png\" alt=\"Bearing loads and sections of the beam\" class=\"wp-image-1418\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-2-1024x691.png 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-2-300x202.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-2-768x518.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-2-400x270.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-2-800x540.png 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-2.png 1511w\" 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\\[ \\require{cancel} \\]\n\n<p><\/p><h3>Determination of the bearing reactions<\/h3><p><\/p>\n<p>Establishing the equilibrium conditions for forces in the x and z directions as well as for the moments<\/p>\n<p>Note the downward z-axis!<\/p>\n<p>\\[ \\tag{1} \\sum F_x = 0 = -F_{Bx} \\]<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{2} \\sum F_z = 0 = - F_{Az} + F - F_{Bz} \\]<\/p>\n<p>\\[ \\tag{3} \\sum M(A) = 0 = - F \\cdot a + F_{Bz} \\cdot 2a + F_{Bx} \\cdot a \\]<\/p>\n<\/div>\n<p>From this follows<\/p>\n<p>\\[ \\tag{4} F_{Bx} = 0 \\]<\/p>\n<p>\\[ \\tag{5} F_{Bz} = \\frac{F}{2} \\]<\/p>\n<p>\\[ \\tag{6} F_{Az} = \\frac{F}{2} \\]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Determination of the internal forces in section I<\/h3>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"693\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-3-1024x693.png\" alt=\"Internal forces and moments in section I\" class=\"wp-image-1420\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-3-1024x693.png 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-3-300x203.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-3-768x520.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-3-400x271.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-3-800x542.png 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-3.png 1511w\" 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\\[ \\require{cancel} \\]\n<p>Establishing the equilibrium conditions for forces in the x and z directions as well as for the moments<\/p>\n<p>\\[ \\tag{7} \\sum F_x = 0 = -N_I -F_{Bx}\\]<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{8} \\sum F_z = 0 = -Q_I + F - F_{Bz} \\]<\/p>\n<p>\\[ \\tag{9} \\sum M(x) = 0 = -M_{bI} - F \\cdot (a-x) + \\bcancel{F_{Bx} \\cdot a} + F_{Bz} \\cdot (2a - x)\\]<\/p>\n<\/div>\n<p>With the already calculated bearing reactions, the internal forces result<\/p>\n<p>\\[ \\tag{10} N_I = 0 \\]<\/p>\n<p>\\[ \\tag{11} Q_I = \\frac{F}{2} \\]<\/p>\n<p>\\[ \\tag{12} M_{bI} = \\frac{F}{2} \\cdot x \\]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Determination of the internal forces in section II<\/h3>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"691\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-4-1024x691.png\" alt=\"Schnittgr\u00f6\u00dfen in Bereich II\" class=\"wp-image-1422\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-4-1024x691.png 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-4-300x202.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-4-768x518.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-4-400x270.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-4-800x540.png 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-4.png 1511w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Schnittgr\u00f6\u00dfen in Bereich 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\\[ \\require{cancel} \\]\n<p>Establishing the equilibrium conditions for forces in the x and z directions as well as for the moments<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{13} \\sum F_x = 0 = -N_{II} - F_{Bx}\\]<\/p>\n<p>\\[ \\tag{14} \\sum F_z = 0 = -Q_{II} - F_{Bz} \\]<\/p>\n<p>\\[ \\tag{15} \\sum M(x) = 0 = - M_{bII} + \\bcancel{F_{Bx} \\cdot a} + F_{Bz} \\cdot (2a - x) \\]<\/p>\n<\/div>\n<p>These internal forces can also be resolved to<\/p>\n\n<p>\\[ \\tag{16} N_{II} = 0 \\]<\/p>\n<p>\\[ \\tag{17} Q_{II} = - \\frac{F}{2} \\]<\/p>\n<p>\\[ \\tag{18} M_{bII} = \\frac{F}{2} \\cdot (2a - x) \\]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Determination of the internal forces for section III<\/h3>\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\/tm2-3-5-919x1024.png\" alt=\"Internal forces and moments in section III\" class=\"wp-image-1424\" width=\"460\" height=\"512\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-5-919x1024.png 919w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-5-269x300.png 269w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-5-768x856.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-5-400x446.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-5-800x891.png 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-5.png 945w\" sizes=\"auto, (max-width: 460px) 100vw, 460px\" \/><figcaption>Internal forces and moments in section III<\/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\\[ \\require{cancel} \\]\n<p>Establishing the equilibrium conditions for forces in the x and z directions as well as for the moments<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{19} \\sum F_x = 0 = -F_{Bx} + Q_{III}\\]<\/p>\n<p>\\[ \\tag{20} \\sum F_z = 0 =  N_{III} - F_{Bz} \\]<\/p>\n<p>\\[ \\tag{21} \\sum M(z) = 0 = M_{bIII} + \\bcancel{F_{Bx} \\cdot (z-a)} \\]<\/p>\n<\/div>\n<p>The internal forces and moments of the third section are<\/p>\n\n<p>\\[ \\tag{22} N_{III} = \\frac{F}{2} \\]<\/p>\n<p>\\[ \\tag{23} Q_{III} = 0 \\]<\/p>\n<p>\\[ \\tag{24} M_{bIII} = 0 \\]<\/p>\n\n\n\n<p>With this all internal forces of the beam are determined.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Graphic representation of the internal forces<\/h3>\n\n\n\n<p>The graphical representation of the normal force, shear force and bending moment is as follows:<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"690\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-6-1024x690.png\" alt=\"Graphic representation of the force and torque curve\" class=\"wp-image-1426\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-6-1024x690.png 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-6-300x202.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-6-768x518.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-6-400x270.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-6-800x539.png 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-3-6.png 1405w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Graphic representation of the force and torque curve<\/figcaption><\/figure>\n\n\n\n<p>So we can state (for this case), the internal forces of an angled beam are quite similar to an flat beam, but the directions of the normal force and the shear force switch within the angled part of the beam.<\/p>\n","protected":false},"excerpt":{"rendered":"<p> ... <a title=\"Determine internal forces\" class=\"read-more\" href=\"https:\/\/pickedshares.com\/en\/engineering-mechanics-ii-exercise-3-determine-internal-forces\/\" aria-label=\"Read more about Determine internal forces\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":1416,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[52,60],"tags":[],"class_list":["post-1430","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-engineering-mechanics-ii","category-exercises","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 - 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