{"id":635,"date":"2021-11-30T13:40:03","date_gmt":"2021-11-30T13:40:03","guid":{"rendered":"https:\/\/isostatika.com\/?p=635"},"modified":"2021-12-13T13:39:34","modified_gmt":"2021-12-13T13:39:34","slug":"calculating-section-modulus-of-a-steel-beam","status":"publish","type":"post","link":"https:\/\/isostatika.com\/en\/blog\/calculating-section-modulus-of-a-steel-beam\/","title":{"rendered":"Calculating the section modulus of a steel beam"},"content":{"rendered":"\n<p>Building rehabilitation sometimes requires to use <strong>beams that are not found in standard shapes,<\/strong> unique pieces that we usually create by means of <strong>adding steel plates<\/strong>. Before developing a complete structural model, we should err on the side of caution and check that specific shape.<\/p>\n\n\n\n<p>One of the most important features we need to know is the <strong>Section Modulus of the beam\u2019s cross section<\/strong> (and more specifically: the elastic section modulus \u2013 S, or Wel in Eurocodes-). The main purpose of getting S (or W) is that it is a very simple way of calculating the flexural resistance of that beam.<\/p>\n\n\n\n<p>And as the Section Modulus is simply defined as S = I \/ y, being I the area moment of inertia (or the second moment of the area) and y the distance from the centroid or neutral axis to the furthest point of that section, we just need to discover the value of I.<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>I_x =  \u0283 y^2 dA    \\text{ &nbsp;&nbsp; and  &nbsp;&nbsp;}      Iy =  \u0283 x^2 dA<\/pre><\/div>\n\n\n\n<p><strong>Steiner\u2019s theorem<\/strong> allows us to get it working with smaller regular pieces. The area moment of inertia of a shape related to a certain Z\u2019 axis is equal to the area moment of inertia of that shape, related to its own Z axis (the one passing through the body\u2019s center of mass) plus a product of the area and the distance between the two axis:<\/p>\n\n\n\n<figure class=\"wp-block-image size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/isostatika.com\/imagenes\/area-moment-inertia-isostatika-1.jpg\" alt=\"\" class=\"wp-image-747\" width=\"393\" height=\"412\" srcset=\"https:\/\/isostatika.com\/imagenes\/area-moment-inertia-isostatika-1.jpg 785w, https:\/\/isostatika.com\/imagenes\/area-moment-inertia-isostatika-1-286x300.jpg 286w, https:\/\/isostatika.com\/imagenes\/area-moment-inertia-isostatika-1-768x805.jpg 768w\" sizes=\"auto, (max-width: 393px) 100vw, 393px\" \/><\/figure>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>I_x = \u2211 I_x,i+ \u2211(Ai \\ x \\ di^2)<\/pre><\/div>\n\n\n\n<p>So if we get the centroid of the complete shape, we can have its <strong>area moment of inertia<\/strong> by adding every piece\u2019s I related to this new axis :<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>I_x = \u2211 I_x,i+ \u2211(Ai \\ x \\ di^2)<\/pre><\/div>\n\n\n\n<p>Once we get it, we just need to divide it between the distance to the furthest point to know S.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Example: Creation of a T section adding two plates of 150 x 10 mm&nbsp;<\/h2>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>I_x = \u2211 I_x,i+ \u2211(Ai \\ x \\ di^2)\t<\/pre><\/div>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>I_a = bh^3\/12 = 12500 mm^4 \\text{&nbsp;&nbsp;&nbsp;&nbsp;}\tAa \\ x \\ da^2 = 2400000 mm^4\n<\/pre><\/div>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>I_b = bh^3\/12 = 2812500 mm^4 \\text{&nbsp;&nbsp;&nbsp;&nbsp;}\tAb \\ x \\ db^2 = 2400000 mm^4\n<\/pre><\/div>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>\\text{So &nbsp;&nbsp;}I_x = 7625000 mm^4\n<\/pre><\/div>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\"><pre>\\text{And then:&nbsp;&nbsp;&nbsp;&nbsp;}W_{sup} = 169444 mm^3 \\text{&nbsp;&nbsp;&nbsp;&nbsp;}\tW_{sup} = 66304 mm^3\n<\/pre><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Building rehabilitation sometimes requires to use beams that are not found in standard shapes, unique pieces that we usually create by means of adding steel [&hellip;]<\/p>\n","protected":false},"author":3,"featured_media":350,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[60],"tags":[],"class_list":["post-635","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-steel"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.2 - 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