{"id":5069,"date":"2025-07-14T10:10:02","date_gmt":"2025-07-14T04:40:02","guid":{"rendered":"https:\/\/study.madeeasy.in\/?p=5069"},"modified":"2025-07-18T14:56:21","modified_gmt":"2025-07-18T09:26:21","slug":"analysis-of-strain-in-pure-bending","status":"publish","type":"post","link":"https:\/\/www.madeeasy.in\/study\/me\/strength-of-materials\/analysis-of-strain-in-pure-bending","title":{"rendered":"Analysis of strain in pure bending"},"content":{"rendered":"

Normal strain in beams<\/h2>\n

Consider a simply supported beam of length L subjected to moments at ends A and B as shown in figure.<\/p>\n

Take two section x1-x1 and x2-x2 which are dx length apart and section x1-x1 is at distance x from end A, as shown. Due to applied moments M0, the beam will sag in a manner as shown.<\/p>\n

\"Deflected<\/p>\n

As it is case of pure bending, the elastic curve will be circular in shape.
\nPoint O is the centre of curvature. Cross-section x1-x1 and x2-x2 will remain plane and normal to longitudinal fibres of beam so as to keep radius of curvature constant.
\nAs it is evident , longitudinal fibres EF are shortened while fibres AB are elongated due to bending.<\/p>\n

Thus, top fibres are in compression while bottom fibres are in tension. Somewhere between top and bottom fibres, a surface of layer exist at which there is no change in length known as neutral layer as represented by CD. The intersection of neutral surface with any cross-sectional plane is called neutral axis.
\nRadius of curvature R is taken as distance from centre of curvature O to neutral layer.<\/p>\n

As derived earlier,<\/p>\n

Radius of curvature, R = dx\/d\u03b8<\/p>\n

Curvature, C = d\u03b8\/dx<\/p>\n

To calculate longitudinal strain, take a fibre GH at distance y from neutral layer CD as shown.<\/p>\n

The length L1 of fibre GH =(R + y) d\u03b8<\/p>\n

= Rd\u03b8+ dx\/R<\/p>\n

= dx+ y dx\/R<\/p>\n

Longitudinal strain in fibre GH, \u03b5x = L1-dx\/dx=y\/R<\/p>\n

Hence, longitudinal strain in any fibre,<\/p>\n

\u03b5x = y\/R<\/p>\n

Here, y is the distance of fibre from neutral axis. Strains are positive below neutral axis and negative above neutral axis.<\/p>\n","protected":false},"excerpt":{"rendered":"

Normal strain in beams Consider a simply supported beam of length L subjected to moments at ends A and B<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1619,10],"tags":[1647,1636,1645],"class_list":["post-5069","post","type-post","status-publish","format-standard","hentry","category-strength-of-materials","category-me","tag-deflected-shape-of-beam","tag-longitudinal-strain","tag-radius-of-curvature"],"_links":{"self":[{"href":"https:\/\/www.madeeasy.in\/study\/wp-json\/wp\/v2\/posts\/5069","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.madeeasy.in\/study\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.madeeasy.in\/study\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.madeeasy.in\/study\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.madeeasy.in\/study\/wp-json\/wp\/v2\/comments?post=5069"}],"version-history":[{"count":0,"href":"https:\/\/www.madeeasy.in\/study\/wp-json\/wp\/v2\/posts\/5069\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.madeeasy.in\/study\/wp-json\/wp\/v2\/media?parent=5069"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.madeeasy.in\/study\/wp-json\/wp\/v2\/categories?post=5069"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.madeeasy.in\/study\/wp-json\/wp\/v2\/tags?post=5069"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}