{"id":1932,"date":"2021-01-15T12:46:57","date_gmt":"2021-01-15T12:46:57","guid":{"rendered":"https:\/\/pickedshares.com\/?p=1932"},"modified":"2021-05-08T13:01:06","modified_gmt":"2021-05-08T13:01:06","slug":"engineering-mechanics-1-exercise-39-method-of-sections","status":"publish","type":"post","link":"https:\/\/pickedshares.com\/en\/engineering-mechanics-1-exercise-39-method-of-sections\/","title":{"rendered":"Method of sections"},"content":{"rendered":"\n<p>This exercise addresses the following questions:<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>How do you use the method of sections (Ritter)?<\/li><li>How do you calculate individual member forces with little effort?<\/li><li>How do you recognize null members (members without force, zero members, unstrained members) in a truss?<\/li><\/ul>\n\n\n\n<p>Here is an exercise about the <a href=\"https:\/\/pickedshares.com\/en\/engineering-mechanics-1-exercise-19-truss-member-forces-and-bearing-loads\/\">calculation of a truss using the method of joints<\/a>.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Task<\/h2>\n\n\n\n<p>A truss is supported in bearings A and B and is loaded at its outer end by a horizontal force F. The truss itself should be seen as massless. The member force in the member shown in dashed lines must be calculated! The null members of the framework are to be marked!<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"732\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-1post-1024x732.png\" alt=\"Truss with horizontal load\" class=\"wp-image-1909\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-1post-1024x732.png 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-1post-300x215.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-1post-768x549.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-1post-400x286.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-1post-800x572.png 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-1post.png 1145w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Truss with horizontal load<\/figcaption><\/figure>\n\n\n\n<h2 class=\"wp-block-heading\">Solution<\/h2>\n\n\n\n<p>The method of sections is used to determine the member force. The selected cutting line is shown in the following picture. You can freely choose which side of the cut is investigated. In the solution shown here, the side with the bearing reactions is considered, which means that these must be determined 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\/tm1-39-2-1024x825.png\" alt=\"Truss with bearing reactions and the cut line\" class=\"wp-image-1913\" width=\"512\" height=\"413\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-2-1024x825.png 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-2-300x242.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-2-768x619.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-2-400x322.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-2-800x645.png 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-2.png 1207w\" sizes=\"auto, (max-width: 512px) 100vw, 512px\" \/><figcaption>Truss with bearing reactions and the cut line<\/figcaption><\/figure>\n\n\n\n<p><\/p>\n\n\n\n<p>Since the members of this framework are only arranged horizontally, vertically and at an angle of 45 \u00b0, the <a href=\"https:\/\/pickedshares.com\/en\/important-angles-and-related-results-of-the-trigonometric-functions\/\">results of the trigonometric functions can be noted in abbreviated<\/a> form.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Bearing loads<\/h3>\n\n\n\n<p>Establishing the equilibrium of forces and moments (<a href=\"https:\/\/pickedshares.com\/en\/right-hand-rules\/\">counterclockwise torques positive<\/a>):<\/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_{Bx} - F \\]<\/p>\n<p>\\[ \\tag{2} \\sum F_y = 0 = F_{Ay} + F_{By} \\]<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{3} \\sum M(A) = 0 = F_{By} \\cdot l + F \\cdot 2 \\cdot l \\]<\/p>\n<\/div>\n\n<p>So the bearing loads are<\/p>\n<p>\\[ \\tag{4} F_{Bx} = F \\]<\/p>\n<p>\\[ \\tag{5} F_{By} = - 2 \\cdot F \\]<\/p>\n<p>\\[ \\tag{6} F_{Ay} = 2 \\cdot F \\]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Zero members, first approach<\/h3>\n\n\n\n<p>Before looking at the section, it makes sense to mark the zero members that are immediately recognizable. Zero members are always created when only one member is connected to a node for a certain axis direction x or y. This means that this member cannot pass its load on to any other member.<\/p>\n\n\n\n<p>For the present truss, this applies to the members marked in blue in the following figure.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"658\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-3-1024x658.png\" alt=\"Obvious zero members of the truss\" class=\"wp-image-1915\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-3-1024x658.png 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-3-300x193.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-3-768x494.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-3-400x257.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-3-800x514.png 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-3.png 1125w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Obvious zero members of the truss<\/figcaption><\/figure>\n\n\n\n<p>We will find some more zero members later.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Member forces<\/h3>\n\n\n\n<p>Next, the bar forces (here as tensile forces) are applied to the cut bars and the static equilibrium conditions are established. Point A is chosen as the pivot point.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"657\" height=\"947\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-4.png\" alt=\"Free body diagram for the left section of the truss\" class=\"wp-image-1917\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-4.png 657w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-4-208x300.png 208w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-4-400x577.png 400w\" sizes=\"auto, (max-width: 657px) 100vw, 657px\" \/><figcaption>Free body diagram for the left section of the truss<\/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\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{7} \\sum F_x = 0 = F_{Bx} + F_1 + \\frac{F_2}{\\sqrt{2}} + F_3 \\]<\/p>\n<p>\\[ \\tag{8} \\sum F_y = 0 = F_{Ay} + F_{By} + \\frac{F_2}{\\sqrt{2}} \\]<\/p>\n<p>\\[ \\tag{9} \\sum M(A) = 0 = F_{By} \\cdot l - F_3 \\cdot l - F_1 \\cdot 2 \\cdot l \\]<\/p>\n<\/div>\n<p>The member force F<sub>2<\/sub> is determined quickly:<\/p>\n<p>\\[ \\tag{10} F_2 = 0 \\]<\/p>\n<p>So it is another zero member.<\/p>\n\n\n\n<p>The remaining member forces are now calculated. The previous equations could be rearranged and used, or we could consider the moments around point C..<\/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{11} \\sum M(C) = 0 = F_{Bx} \\cdot l - F_{Ay} \\cdot l - F_1 \\cdot l \\]<\/p>\n<\/div>\n<p>\\[ \\tag{12} F_1 = -F  \\]<\/p>\n<p>I.e. the considered member is a compression member. The last missing member force results from equation 7.<\/p>\n<p>\\[ \\tag{13} F_3 = 0 \\]<\/p>\n<p>This member is also a zero member.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Zero members, second approach<\/h3>\n\n\n\n<p>With the zero members found so far, all nodes of the framework can now be checked again. The following additional null bars result:<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"701\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-6-1024x701.png\" alt=\"All zero members of the truss\" class=\"wp-image-1921\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-6-1024x701.png 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-6-300x205.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-6-768x526.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-6-400x274.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-6-800x548.png 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-6.png 1249w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>All zero members of the truss<\/figcaption><\/figure>\n\n\n\n<p>The high number of zero members results from the direction of the applied external load. That is, if the acting load contains a vertical component, the distribution of the zero members in the truss also changes.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Alternative consideration<\/h3>\n\n\n\n<p>As mentioned above, the static equilibrium could also have been considered for the other side of the cut. In that case the cut element would have looked like this:<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"885\" height=\"577\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-5.png\" alt=\"Free body diagram for the right section of the truss\" class=\"wp-image-1919\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-5.png 885w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-5-300x196.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-5-768x501.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-5-400x261.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm1-39-5-800x522.png 800w\" sizes=\"auto, (max-width: 885px) 100vw, 885px\" \/><figcaption>Free body diagram for the right section of the truss<\/figcaption><\/figure>\n\n\n\n<p>This site has an excellent <a href=\"https:\/\/valdivia.staff.jade-hs.de\/fachwerk.html\" target=\"_blank\" rel=\"noreferrer noopener\">online calculator for 2-dimensional trusses<\/a>.<\/p>\n","protected":false},"excerpt":{"rendered":"<p> ... <a title=\"Method of sections\" class=\"read-more\" href=\"https:\/\/pickedshares.com\/en\/engineering-mechanics-1-exercise-39-method-of-sections\/\" aria-label=\"Read more about Method of sections\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":1911,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[21,60],"tags":[28],"class_list":["post-1932","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-engineering-mechanics-i","category-exercises","tag-truss","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>Method of sections &#8226; pickedshares<\/title>\n<meta name=\"description\" content=\"Here we determinate a member force using the method of sections. 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