{"id":1546,"date":"2021-01-09T21:05:43","date_gmt":"2021-01-09T21:05:43","guid":{"rendered":"https:\/\/pickedshares.com\/?p=1546"},"modified":"2021-05-08T12:51:09","modified_gmt":"2021-05-08T12:51:09","slug":"engineering-mechanics-2-exercise-7-internal-forces-of-a-beam-with-exceeding-length","status":"publish","type":"post","link":"https:\/\/pickedshares.com\/en\/engineering-mechanics-2-exercise-7-internal-forces-of-a-beam-with-exceeding-length\/","title":{"rendered":"Internal forces of a beam with exceeding length"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">This exercise is about the following questions:<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>How to determine the internal forces of a beam under single loads?<\/li><li>How to calculate the function of the bending moment?<\/li><li>Where do I have to divide the beam into sections for the calculation?<\/li><\/ul>\n\n\n\n<h2 class=\"wp-block-heading\">Task<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\"><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">A beam with an exceeding end is mounted on two supports and is loaded once vertically and once at an angle by the force F. The beam itself is not interrupted in the floating bearing B. The bearing reactions and the internal forces have to be determined!<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"536\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-1-1024x536.png\" alt=\"Beam with excess length\" class=\"wp-image-1525\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-1-1024x536.png 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-1-300x157.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-1-768x402.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-1-1536x803.png 1536w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-1-400x209.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-1-800x418.png 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-1.png 1539w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Beam with excess length<\/figcaption><\/figure>\n\n\n\n<h2 class=\"wp-block-heading\">Solution<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">The beam is cut free and the bearing reactions are noted. Left turning moments are accepted positively. Note the downward z-axis!<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The beam is divided into three sections as shown.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"534\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-2-1024x534.png\" alt=\"Bearing loads and sections of the beam\" class=\"wp-image-1529\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-2-1024x534.png 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-2-300x157.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-2-768x401.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-2-400x209.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-2-800x417.png 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-2.png 1535w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Bearing loads and sections of the beam<\/figcaption><\/figure>\n\n\n\n<h3 class=\"wp-block-heading\">Determination of the bearing loads<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Establishing the equilibrium conditions for forces in the x and z directions 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{1} \\sum F_x = 0 = F_{Ax} - F \\cdot cos \\alpha\\]<\/p>\n<p>\\[ \\tag{2} \\sum F_z = 0 = - F_{Az} +F -F_{Bz} + F \\cdot sin \\alpha \\]<\/p>\n<p>\\[ \\tag{3} \\sum M(A) = 0 = -F\\cdot l +F_{Bz} \\cdot 2 l - F \\cdot sin \\alpha \\cdot 3 l \\]<\/p>\n<\/div>\n<p>F<sub>Bz<\/sub> can thus be determined directly.<\/p>\n<p>\\[ \\tag{4} F_{Bz} =F \\cdot \\frac{ 1 + 3 \\cdot sin \\alpha }{2}  \\]<\/p>\n<p>Then the first two equations are solved for F<sub>Ax<\/sub> and F<sub>Az<\/sub>.<\/p>\n<p>\\[ \\tag{5} F_{Ax} = F \\cdot cos \\alpha\\]<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{6} F_{Az} = F - F \\cdot \\frac{ 1 + 3 \\cdot sin \\alpha }{2} + F \\cdot sin \\alpha \\]<\/p>\n<\/div>\n<p>\\[ \\tag{7} F_{Az} = \\frac{F}{2} \\cdot \\left( 1 - sin \\alpha \\right) \\]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Determination of the internal forces<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">It is beneficial to choose the cutting direction in the respective sections or areas so that the equations can be kept as simple as possible. In this case this means that the first section on the right is cut off, the two following sections are cut off on the left. The sections considered are marked in green for clarity. In the following, normal forces are denoted by N, transverse forces by Q and the bending moment by 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=\"945\" height=\"797\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-3.png\" alt=\"Internal forces and moments in section I\" class=\"wp-image-1531\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-3.png 945w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-3-300x253.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-3-768x648.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-3-400x337.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-3-800x675.png 800w\" sizes=\"auto, (max-width: 945px) 100vw, 945px\" \/><figcaption>Internal forces and moments in section I<\/figcaption><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">Establishing the equilibrium conditions for forces in the x and z directions 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{8} \\sum F_x = 0 = F_{Ax} + N_I \\]<\/p>\n<p>\\[ \\tag{9} \\sum F_z = 0 = -F_{Az} + Q_I \\]<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{10} \\sum M(x) = 0 = -F_{Az} \\cdot x + M_{bI} \\]<\/p>\n<\/div>\n<p>The already known bearing loads are now used:<\/p>\n<p>\\[ \\tag{11} N_I = - F \\cdot cos \\alpha \\]<\/p>\n<p>\\[ \\tag{12} Q_I = \\frac{F}{2} \\cdot \\left( 1 - sin \\alpha \\right) \\]<\/p>\n<p>\\[ \\tag{13} M_{bI} = \\frac{F}{2} \\cdot \\left( 1 - sin \\alpha \\right) \\cdot x  \\]<\/p>\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=\"534\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-4-1024x534.png\" alt=\"Schnittgr\u00f6\u00dfen in Bereich II\" class=\"wp-image-1533\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-4-1024x534.png 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-4-300x156.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-4-768x400.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-4-1536x800.png 1536w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-4-400x208.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-4-800x417.png 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-4.png 1541w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Schnittgr\u00f6\u00dfen in Bereich II<\/figcaption><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">This section is investigated opposite to section I. This means that the internal forces are drawn negatively here.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Establishing the equilibrium conditions for forces in the x and z directions 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{14} \\sum F_x = 0 = -N_{II} - F \\cdot cos \\alpha \\]<\/p>\n<p>\\[ \\tag{15} \\sum F_z = 0 = -Q_{II} - F_{Bz} + F \\cdot sin \\alpha \\]<\/p>\n<p>\\[ \\tag{16} \\sum M(x) = 0 = -M_{bII} + F_{Bz} \\cdot (2l-x) - F \\cdot sin \\alpha \\cdot (3l-x) \\]<\/p>\n<\/div>\n\n<p>The already known support force F<sub>Bz<\/sub> is now used so that all internal forces in section II can be determined:<\/p>\n\n<p>\\[ \\tag{17} N_{II} = F \\cdot cos \\alpha \\]<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{18} Q_{II} =  - F \\cdot \\frac{ 1 + 3 \\cdot sin \\alpha }{2} + F \\cdot sin \\alpha \\]<\/p>\n<p>\\[ \\tag{19} Q_{II} =  - \\frac{F}{2} \\cdot (1 + sin \\alpha) \\]<\/p>\n<p>\\[ \\tag{20} M_{bII} = F \\cdot \\frac{ 1 + 3 \\cdot sin \\alpha }{2} \\cdot (2l-x) - F \\cdot sin \\alpha \\cdot (3l-x) \\]<\/p>\n\n<p>\\[ \\tag{21} M_{bII} = \\frac{F}{2} \\cdot \\left( 2l - x - \\sin \\alpha\\cdot x \\right)  \\]<\/p>\n<\/div>\n\n\n\n<h4 class=\"wp-block-heading\">Section III<\/h4>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"532\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-5-1024x532.png\" alt=\"Internal forces and moments in section III\" class=\"wp-image-1535\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-5-1024x532.png 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-5-300x156.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-5-768x399.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-5-400x208.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-5-800x415.png 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-7-5.png 1531w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Internal forces and moments in section III<\/figcaption><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">Establishing the equilibrium conditions for forces in the x and z directions 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{22} \\sum F_x = 0 = -N_{III} - F \\cdot cos \\alpha \\]<\/p>\n<p>\\[ \\tag{23} \\sum F_z = 0 = -Q_{III} + F \\cdot sin \\alpha \\]<\/p>\n<p>\\[ \\tag{24} \\sum M(x) = 0 = -M_{bIII} - F \\cdot sin \\alpha \\cdot (3l-x) \\]<\/p>\n<\/div>\n\n<p>The internal forces of section III are:<\/p>\n<p>\\[ \\tag{25} N_{III} = - F \\cdot cos \\alpha \\]<\/p>\n\n<p>\\[ \\tag{26} Q_{III} = F \\cdot sin \\alpha \\]<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{27} M_{bIII} = - F \\cdot sin \\alpha \\cdot (3l-x) \\]<\/p>\n<\/div>\n\n\n\n<p class=\"wp-block-paragraph\">This means that all support reactions and internal forces of the beam are determined. <\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Here we have some more interesting <a href=\"https:\/\/pickedshares.com\/en\/category\/engineering-mechanics-ii\/\">exercises and solution regarding Engineering mechanics II - theory of materials<\/a>.<\/p>\n","protected":false},"excerpt":{"rendered":"<p> ... <a title=\"Internal forces of a beam with exceeding length\" class=\"read-more\" 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