{"id":1612,"date":"2021-01-11T21:41:05","date_gmt":"2021-01-11T21:41:05","guid":{"rendered":"https:\/\/pickedshares.com\/?p=1612"},"modified":"2021-05-08T12:53:44","modified_gmt":"2021-05-08T12:53:44","slug":"engineering-mechanics-2-exercise-8-stress-in-a-tension-rod-with-dead-load","status":"publish","type":"post","link":"https:\/\/pickedshares.com\/en\/engineering-mechanics-2-exercise-8-stress-in-a-tension-rod-with-dead-load\/","title":{"rendered":"Stress in a tension rod with dead load"},"content":{"rendered":"\n<p>This exercise is about the following questions:<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>How to calculate the stress in a rod und load of dead weight?<\/li><li>How to draw the first circle of Mohr for the uniaxial stress state?<\/li><\/ul>\n\n\n\n<h2 class=\"wp-block-heading\">Task<\/h2>\n\n\n\n<p>A lean rod with the cross-sectional area q is loaded by the force F and its own weight. The MOHR stress circle is to be drawn for the position y = l \/ 2. The stresses \u03c3<sub>\u03c6<\/sub> and \u03c4<sub>\u03c6<\/sub> are to be determined from the stress circle!<\/p>\n\n\n\n<p>The following parameters are given:<\/p>\n\n\n\n<ul>\n<li>F = 1 N<\/li>\n<li>l = 0,4 m<\/li>\n<li>\u03c1 = 10000 kg\/m\u00b3<\/li>\n<li>q = 1 cm\u00b2<\/li>\n<li>\u03c6 = 30\u00b0<\/li>\n<li>g = 9,81 m\/s\u00b2<\/li>\n<\/ul>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"508\" height=\"1024\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-8-2-508x1024.png\" alt=\"Rod with tensile force and dead load\" class=\"wp-image-1603\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-8-2-508x1024.png 508w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-8-2-149x300.png 149w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-8-2-400x806.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-8-2.png 603w\" sizes=\"auto, (max-width: 508px) 100vw, 508px\" \/><figcaption>Rod with tensile force and dead load<\/figcaption><\/figure>\n\n\n\n<h2 class=\"wp-block-heading\">Solution<\/h2>\n\n\n\n<p>In the first step, the internal forces at y = l \/ 2 must be determined. Since there are no shear forces or moments, there is a uniaxial stress state in the rod.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"778\" height=\"1024\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-8-3-778x1024.png\" alt=\"Free cut rod\" class=\"wp-image-1605\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-8-3-778x1024.png 778w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-8-3-228x300.png 228w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-8-3-768x1011.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-8-3-400x526.png 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-8-3-800x1053.png 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-8-3.png 911w\" sizes=\"auto, (max-width: 778px) 100vw, 778px\" \/><figcaption>Free cut rod<\/figcaption><\/figure>\n\n\n\n<p>In addition to the normal force at the point y = l \/ 2, the weight G acts, which is expressed here by integrating the term q\u03c1g along the coordinate \u03be running in the y direction. Of course, the different dimensions must be taken into account. The stress at the location y = l \/ 2 is referred to below as \u03c3<sub>0<\/sub>.<\/p>\n\n\n\n<p>The balance of forces is made up of this<\/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_y = 0 = -\\int_{l\/2}^l{q\\cdot \\rho \\cdot g \\,d\\xi} - N + F \\]<\/p>\n<\/div>\n<p>Since the stress is sought for the cross-section under consideration, this equation can be converted into the stress form by dividing it by the cross-sectional area q.<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{2} 0 = -\\int_{l\/2}^l{\\rho \\cdot g \\,d\\xi} - \\frac{N}{q} + \\frac{F}{q} \\]<\/p>\n<\/div>\n<p>The principal stress \u03c3<sub>0<\/sub> is<\/p>\n<p>\\[ \\tag{3} \\sigma_0 = \\frac{N}{q} \\]<\/p>\n\n<p>so equation (2) becomes<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{4} 0 = -\\int_{l\/2}^l{\\rho \\cdot g \\,d\\xi} - \\sigma_0 + \\frac{F}{q} \\]<\/p>\n<\/div>\n\n<p>So the sought stress can now be calculated.<\/p>\n<p>\\[ \\tag{5} \\sigma_0 = \\frac{F}{q}-\\rho \\cdot g \\cdot \\frac{l}{2}   \\]<\/p>\n\n<p>For further consideration it makes sense to use Pascal as the unit for the stress. Using generous rounding results in the normal stress to<\/p>\n\n<p>\\[ \\tag{6} \\sigma_0 = - 10000Pa \\]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">The first circle of Mohr<\/h3>\n\n\n\n<p>The calculated normal stress forms the diameter of the Mohr stress circle. This is to be plotted on the left side of the vertical axis due to the negative stress. The finished circle looks like this:<\/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\/tm2-8-4-1.png\" alt=\"Mohr's first circle for -10000 Pa and 30\u00b0\" class=\"wp-image-1610\" width=\"400\" height=\"471\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-8-4-1.png 533w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-8-4-1-255x300.png 255w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/01\/tm2-8-4-1-400x471.png 400w\" sizes=\"auto, (max-width: 400px) 100vw, 400px\" \/><figcaption>Mohr's first circle for -10000 Pa and 30\u00b0<\/figcaption><\/figure>\n\n\n\n<p>The stresses we are looking for are already included in the above figure.<\/p>\n\n\n\n<p>With this <a href=\"https:\/\/pickedshares.com\/en\/mohrs-first-circle-uniaxial-stress\/\">online generator you can draw easily Mohr's first circle<\/a>.<\/p>\n","protected":false},"excerpt":{"rendered":"<p> ... <a title=\"Stress in a tension rod with dead load\" class=\"read-more\" href=\"https:\/\/pickedshares.com\/en\/engineering-mechanics-2-exercise-8-stress-in-a-tension-rod-with-dead-load\/\" aria-label=\"Read more about Stress in a tension rod with dead load\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":1601,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[52,60],"tags":[],"class_list":["post-1612","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 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Stress in a tension 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