{"id":996,"date":"2020-04-27T08:00:29","date_gmt":"2020-04-27T07:00:29","guid":{"rendered":"https:\/\/blogs.ua.es\/jjrr2011\/?p=996"},"modified":"2020-04-22T12:08:09","modified_gmt":"2020-04-22T11:08:09","slug":"chapter-6-iii-problems","status":"publish","type":"post","link":"https:\/\/blogs.ua.es\/jjrr2011\/2020\/04\/27\/chapter-6-iii-problems\/","title":{"rendered":"Chapter 6: Equilibrium of rigid bodies (III). Distributed forces &#8211; Problems"},"content":{"rendered":"<p style=\"text-align: justify\"><span style=\"color: #800000\"><strong>After reading<\/strong><\/span> previous entry and <span style=\"color: #008000\"><strong>understanding<\/strong><\/span> the meaning of distributed forces, <strong>how to calculate the equivalent resultant to the distributed load<\/strong>, you should try to solve next exercises.<\/p>\n<p style=\"text-align: justify\"><span style=\"color: #800080\"><strong>EXERCISE 1:<\/strong><\/span> Calculate the <strong>supports reactions<\/strong> at A and B for the beam subjected to the asymmetric load distributions.<\/p>\n<p style=\"text-align: justify\"><a href=\"https:\/\/blogs.ua.es\/jjrr2011\/files\/2020\/04\/distributed_forces_beam_example.jpeg\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-1000\" src=\"https:\/\/blogs.ua.es\/jjrr2011\/files\/2020\/04\/distributed_forces_beam_example.jpeg\" alt=\"\" width=\"716\" height=\"302\" srcset=\"https:\/\/blogs.ua.es\/jjrr2011\/files\/2020\/04\/distributed_forces_beam_example.jpeg 716w, https:\/\/blogs.ua.es\/jjrr2011\/files\/2020\/04\/distributed_forces_beam_example-300x127.jpeg 300w\" sizes=\"auto, (max-width: 716px) 100vw, 716px\" \/><\/a><span style=\"color: #800080\"><strong>EXERCISE 2:<\/strong><\/span> The symmetric simple truss is loaded as shown. Determine the <strong>reactions<\/strong> at wall support A and the <strong>tension<\/strong> in cable CD.<\/p>\n<p><a href=\"https:\/\/blogs.ua.es\/jjrr2011\/files\/2020\/04\/distributed_forces_truss_example.jpeg\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-999\" src=\"https:\/\/blogs.ua.es\/jjrr2011\/files\/2020\/04\/distributed_forces_truss_example.jpeg\" alt=\"\" width=\"490\" height=\"423\" srcset=\"https:\/\/blogs.ua.es\/jjrr2011\/files\/2020\/04\/distributed_forces_truss_example.jpeg 490w, https:\/\/blogs.ua.es\/jjrr2011\/files\/2020\/04\/distributed_forces_truss_example-300x259.jpeg 300w\" sizes=\"auto, (max-width: 490px) 100vw, 490px\" \/><\/a><\/p>\n","protected":false},"excerpt":{"rendered":"<p>After reading previous entry and understanding the meaning of distributed forces, how to calculate the equivalent resultant to the distributed load, you should try to solve next exercises. EXERCISE 1: Calculate the supports reactions at A and B for the beam subjected to the asymmetric load distributions. EXERCISE 2: The symmetric simple truss is loaded [&hellip;]<\/p>\n","protected":false},"author":2285,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[135920,135935,9333],"tags":[135938,135928,135932],"class_list":["post-996","post","type-post","status-publish","format-standard","hentry","category-aims","category-problems","category-subject","tag-distributed-forces","tag-equilibrium","tag-rigid-body"],"_links":{"self":[{"href":"https:\/\/blogs.ua.es\/jjrr2011\/wp-json\/wp\/v2\/posts\/996","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blogs.ua.es\/jjrr2011\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blogs.ua.es\/jjrr2011\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blogs.ua.es\/jjrr2011\/wp-json\/wp\/v2\/users\/2285"}],"replies":[{"embeddable":true,"href":"https:\/\/blogs.ua.es\/jjrr2011\/wp-json\/wp\/v2\/comments?post=996"}],"version-history":[{"count":3,"href":"https:\/\/blogs.ua.es\/jjrr2011\/wp-json\/wp\/v2\/posts\/996\/revisions"}],"predecessor-version":[{"id":1001,"href":"https:\/\/blogs.ua.es\/jjrr2011\/wp-json\/wp\/v2\/posts\/996\/revisions\/1001"}],"wp:attachment":[{"href":"https:\/\/blogs.ua.es\/jjrr2011\/wp-json\/wp\/v2\/media?parent=996"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blogs.ua.es\/jjrr2011\/wp-json\/wp\/v2\/categories?post=996"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blogs.ua.es\/jjrr2011\/wp-json\/wp\/v2\/tags?post=996"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}