{"id":188384,"date":"2025-02-06T19:49:46","date_gmt":"2025-02-06T19:49:46","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=188384"},"modified":"2025-02-06T19:49:48","modified_gmt":"2025-02-06T19:49:48","slug":"draw-the-shear-diagram-for-the-beam","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/06\/draw-the-shear-diagram-for-the-beam\/","title":{"rendered":"Draw the shear diagram for the beam"},"content":{"rendered":"\n<p>Draw the shear diagram for the beam. Draw the moment diagram for the beam. Determine the shear throughout the beam as a functions of x, where 0 lessthanorequalto x &lt; 6 ft. Determine the moment throughout the beam as a functions of x, where 0 lessthanorequalto x &lt; 6 ft. Determine the shear throughout the beam as a functions of x, where 6 ft &lt; x lessthanorequalto 10 ft. Determine the moment throughout the beam as a functions of x, where 6 ft &lt; x lessthanorequalto 10 ft.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-108.png\" alt=\"\" class=\"wp-image-188385\"\/><\/figure>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>I&#8217;m unable to view the specific image you&#8217;ve provided, but I can guide you through the general process of drawing shear and moment diagrams for a simply supported beam subjected to various loads. Let&#8217;s consider two common scenarios:<\/p>\n\n\n\n<p><strong>1. Simply Supported Beam with a Point Load at Midspan:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Shear Force Diagram (SFD):<\/strong><\/li>\n\n\n\n<li>he beam experiences a vertical point load ( P ) at its midpoint. &#8211; he reactions at the supports are equal and can be calculated as ( R_A = R_B = \\frac{P}{2} ). &#8211; he shear force is constant between the left support and the point load, equal to ( R_A ), and then drops to ( -R_B ) after the load.<\/li>\n\n\n\n<li><strong>Bending Moment Diagram (BMD):<\/strong><\/li>\n\n\n\n<li>he bending moment increases linearly from the left support to the point load, reaching a maximum at the load&#8217;s location. &#8211; he maximum bending moment is ( M_{\\text{max}} = \\frac{P \\times L}{4} ), where ( L ) is the length of the beam. &#8211; eyond the point load, the bending moment decreases linearly back to zero at the right support.<br><strong>2. Simply Supported Beam with a Uniformly Distributed Load (UDL):<\/strong><\/li>\n\n\n\n<li><strong>Shear Force Diagram (SFD):<\/strong><\/li>\n\n\n\n<li>he beam carries a uniform load ( w ) per unit length over its entire span. &#8211; he total load is ( W = w \\times L ). &#8211; he reactions at the supports are equal and can be calculated as ( R_A = R_B = \\frac{W}{2} = \\frac{w \\times L}{2} ). &#8211; he shear force decreases linearly from ( R_A ) at the left support to ( -R_B ) at the right support.<\/li>\n\n\n\n<li><strong>Bending Moment Diagram (BMD):<\/strong><\/li>\n\n\n\n<li>he bending moment increases quadratically from the left support, reaching a maximum at the midpoint of the beam. &#8211; he maximum bending moment is ( M_{\\text{max}} = \\frac{w \\times L^2}{8} ). &#8211; eyond the midpoint, the bending moment decreases quadratically back to zero at the right support.<br><strong>General Steps to Draw Shear and Moment Diagrams:<\/strong><\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Determine Support Reactions:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>se equilibrium equations to find the reactions at the supports.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Construct the Shear Force Diagram:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>tart from one end of the beam and move across, adding or subtracting loads as you encounter them. &#8211; lot the shear force at key points, such as the locations of point loads or changes in the distributed load.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Construct the Bending Moment Diagram:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>ntegrate the shear force diagram to find the bending moment at various points along the beam. &#8211; lot the bending moment at key points, ensuring it is zero at the supports for a simply supported beam.<br>For a visual demonstration of these concepts, you might find the following video helpful:<\/li>\n<\/ul>\n\n\n\n<p>\ue200video\ue202Shear and Moment Diagram- Simply Supported Beam (Uniform Load)\ue202turn0search0\ue201<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Draw the shear diagram for the beam. Draw the moment diagram for the beam. Determine the shear throughout the beam as a functions of x, where 0 lessthanorequalto x &lt; 6 ft. Determine the moment throughout the beam as a functions of x, where 0 lessthanorequalto x &lt; 6 ft. Determine the shear throughout the [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-188384","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188384","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=188384"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188384\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=188384"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=188384"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=188384"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}