{"id":194612,"date":"2025-02-24T10:01:24","date_gmt":"2025-02-24T10:01:24","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=194612"},"modified":"2025-02-24T10:01:36","modified_gmt":"2025-02-24T10:01:36","slug":"determine-the-maximum-bending-moment-in-a-simply-supported-beam-of-length-4-meters-carrying-a-uniformly-distributed-load-of-500-n-m","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/24\/determine-the-maximum-bending-moment-in-a-simply-supported-beam-of-length-4-meters-carrying-a-uniformly-distributed-load-of-500-n-m\/","title":{"rendered":"Determine the maximum bending moment in a simply supported beam of length 4 meters carrying a uniformly distributed load of 500 N\/m"},"content":{"rendered":"\n

Determine the maximum bending moment in a simply supported beam of length 4 meters carrying a uniformly distributed load of 500 N\/m.<\/p>\n\n\n\n

The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

Maximum Bending Moment Calculation<\/strong><\/h3>\n\n\n\n

A simply supported beam of length ( L = 4 ) meters carries a uniformly distributed load (UDL) of ( w = 500 ) N\/m.<\/p>\n\n\n\n

The maximum bending moment<\/strong> in a simply supported beam with a UDL over the entire span occurs at the center of the beam and is given by the formula:<\/p>\n\n\n\n

[
M_{\\text{max}} = \\frac{wL^2}{8}
]<\/p>\n\n\n\n

Substituting the given values:<\/p>\n\n\n\n

[
M_{\\text{max}} = \\frac{(500)(4^2)}{8}
]<\/p>\n\n\n\n

[
M_{\\text{max}} = \\frac{500 \\times 16}{8} = \\frac{8000}{8} = 1000 \\text{ Nm}
]<\/p>\n\n\n\n

Thus, the maximum bending moment<\/strong> is 1000 Nm (or 1 kNm)<\/strong> at the midpoint of the beam.<\/p>\n\n\n\n

\n\n\n\n

Explanation<\/strong><\/h3>\n\n\n\n
    \n
  1. Reaction Forces Calculation<\/strong>
    The total uniformly distributed load over the beam is: [
    W = w \\times L = 500 \\times 4 = 2000 \\text{ N}
    ] Since the beam is simply supported, the two supports share the load equally: [
    R_A = R_B = \\frac{2000}{2} = 1000 \\text{ N}
    ]<\/li>\n\n\n\n
  2. Bending Moment Distribution<\/strong><\/li>\n<\/ol>\n\n\n\n
      \n
    • The bending moment at any point ( x ) along the beam is given by: [
      M(x) = R_A x – \\frac{w x^2}{2}
      ]<\/li>\n\n\n\n
    • At mid-span<\/strong> ( x = L\/2 = 2 ) m: [
      M(2) = (1000 \\times 2) – \\frac{500 \\times 2^2}{2}
      ] [
      M(2) = 2000 – \\frac{500 \\times 4}{2}
      ] [
      M(2) = 2000 – 1000 = 1000 \\text{ Nm}
      ]<\/li>\n\n\n\n
    • At supports (( x = 0 ) or ( x = 4 )), ( M = 0 ).<\/li>\n<\/ul>\n\n\n\n
        \n
      1. Conclusion<\/strong>
        The bending moment follows a parabolic distribution<\/strong>, with a maximum at the center. The maximum bending moment is 1 kNm<\/strong>.<\/li>\n<\/ol>\n\n\n\n
        \n\n\n\n

        Now, I’ll generate an image showing the shear force and bending moment diagram<\/strong> for the beam.<\/p>\n\n\n\n

        \"\"<\/figure>\n\n\n\n
        \"\"<\/figure>\n\n\n\n

        The images above illustrate the Shear Force Diagram (SFD)<\/strong> and Bending Moment Diagram (BMD)<\/strong> for the simply supported beam under a uniformly distributed load.<\/p>\n\n\n\n

          \n
        1. Shear Force Diagram (SFD)<\/strong><\/li>\n<\/ol>\n\n\n\n
            \n
          • The shear force starts at +1000 N<\/strong> at the left support.<\/li>\n\n\n\n
          • It linearly decreases to -1000 N<\/strong> at the right support.<\/li>\n\n\n\n
          • The change in shear force is linear because the load is uniformly distributed.<\/li>\n<\/ul>\n\n\n\n
              \n
            1. Bending Moment Diagram (BMD)<\/strong><\/li>\n<\/ol>\n\n\n\n
                \n
              • The bending moment is zero<\/strong> at the supports.<\/li>\n\n\n\n
              • It reaches a maximum of 1000 Nm<\/strong> at the mid-span (x = 2 m)<\/strong>.<\/li>\n\n\n\n
              • The shape of the bending moment diagram is parabolic<\/strong> due to the UDL.<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"

                Determine the maximum bending moment in a simply supported beam of length 4 meters carrying a uniformly distributed load of 500 N\/m. The Correct Answer and Explanation is : Maximum Bending Moment Calculation A simply supported beam of length ( L = 4 ) meters carries a uniformly distributed load (UDL) of ( w = […]<\/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-194612","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/194612","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=194612"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/194612\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=194612"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=194612"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=194612"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}