{"id":2514,"date":"2024-08-05T14:36:43","date_gmt":"2024-08-05T09:06:43","guid":{"rendered":"https:\/\/study.madeeasy.in\/?p=2514"},"modified":"2025-07-16T15:24:15","modified_gmt":"2025-07-16T09:54:15","slug":"simple-bending-or-pure-bending","status":"publish","type":"post","link":"https:\/\/www.madeeasy.in\/study\/ce\/strength-of-material\/simple-bending-or-pure-bending","title":{"rendered":"Simple Bending or Pure Bending"},"content":{"rendered":"<p style=\"text-align: justify;\">Consider a simply supported beam AB of length L subjected to moment M0 at its ends as shown in figures.<\/p>\n<p style=\"text-align: justify;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2515 size-full\" src=\"https:\/\/study.madeeasy.in\/wp-content\/uploads\/2024\/08\/deflected-shape.jpg\" alt=\"Deflected shape\" width=\"465\" height=\"296\" srcset=\"https:\/\/www.madeeasy.in\/study\/wp-content\/uploads\/2024\/08\/deflected-shape.jpg 465w, https:\/\/www.madeeasy.in\/study\/wp-content\/uploads\/2024\/08\/deflected-shape-300x191.jpg 300w\" sizes=\"auto, (max-width: 465px) 100vw, 465px\" \/><\/p>\n<p style=\"text-align: justify;\">Consider an element of length dx at distance x from hinged support A. Due to applied moment, the beam will sag in a manner.<\/p>\n<p style=\"text-align: justify;\">It can be seen in figure the element dx will be converted into an curve. Let the normal tangents at end point of element intersect a point O . This point O is known as centre of curvature. The distance Ox<sub>1<\/sub> and Ox<sub>2<\/sub> is known as radius of curvature. Let the angle made at centre of curvature by both the normal is denoted by d\u03b8.<\/p>\n<p style=\"text-align: justify;\">Curvature, C is defined as reciprocal of radius of curvature R i.e., C = 1\/R<\/p>\n<p style=\"text-align: justify;\">As from figure,\u00a0 ds = Rd\u03b8<br \/>\nR = ds\/d\u03b8<\/p>\n<p style=\"text-align: justify;\">Here, ds is length of arch x1x2<br \/>\nAs, curvature is reciprocal of radius of curvature, so, from eq. (i)<\/p>\n<p style=\"text-align: justify;\">Curvature, C = d\u03b8\/ds<\/p>\n<p style=\"text-align: justify;\">For small deflection, as is usually case with beam, ds = dx<\/p>\n<p style=\"text-align: justify;\">Hence, curvature C = d\u03b8\/dx<\/p>\n<p style=\"text-align: justify;\">These equations are used to calculate normal or bending stress and strains in beam. Bending moment diagram for beam.<\/p>\n<p style=\"text-align: justify;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2518 size-full\" src=\"https:\/\/study.madeeasy.in\/wp-content\/uploads\/2024\/08\/bmd-2.jpg\" alt=\"BMD\" width=\"278\" height=\"70\" \/><\/p>\n<p style=\"text-align: justify;\">As clear from bending moment diagram, the beam is free from shear force. This type of bending in which there is no shear force is called as simple bending or pure bending. Bending in presence of shear force is known as non-uniform bending.<\/p>\n<p style=\"text-align: justify;\">In pure bending, radius of curvature is constant at different sections of beam. For example, as the beam in figure is the case of pure bending, radius of curvature is R at section x3 also. As circular of curvature is constant, deflection curve of beam is circular.<\/p>\n<p style=\"text-align: justify;\">Another case of pure bending can be a simply supported beam of length L subjected to concentrated loads as shown in figure<\/p>\n<p style=\"text-align: justify;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2519 size-full\" src=\"https:\/\/study.madeeasy.in\/wp-content\/uploads\/2024\/08\/sfd-1.jpg\" alt=\"SFD\" width=\"325\" height=\"263\" srcset=\"https:\/\/www.madeeasy.in\/study\/wp-content\/uploads\/2024\/08\/sfd-1.jpg 325w, https:\/\/www.madeeasy.in\/study\/wp-content\/uploads\/2024\/08\/sfd-1-300x243.jpg 300w\" sizes=\"auto, (max-width: 325px) 100vw, 325px\" \/><\/p>\n<p style=\"text-align: justify;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2520 size-full\" src=\"https:\/\/study.madeeasy.in\/wp-content\/uploads\/2024\/08\/bmd-3.jpg\" alt=\"BMD\" width=\"291\" height=\"123\" \/><\/p>\n<p style=\"text-align: justify;\">Its shear force diagram and bending moment diagram is shown in figure above.<br \/>\nAs from shear force diagram, it is cleared that there is no shear force in region CD of beam. So in region CD, pure bending will occur due to which beam will be converted into an arc of circle.<\/p>\n<h2 style=\"text-align: justify;\">Assumptions in Theory of Pure Bending<\/h2>\n<ol style=\"text-align: justify;\">\n<li>Material of the beam is homogeneous, isotropic and linear elastic in which Hooke\u2019s law is valid.<\/li>\n<li>The beam is straight before loading.<\/li>\n<li>Cross-section of beam is prismatic throughout the length.<\/li>\n<li>The plane section before bending remains plane after bending, it means, the longitudinal strain vary linearly from zero at neutral axis to maximum at the surface and longitudinal strain at any distance y is directly proportional to its distance (y) from neutral axis.<\/li>\n<li>Every layer of material is free to expand or contract longitudinally and laterally under stress and do not exert pressure upon each other. Thus, the Poisson\u2019s effect at the interface of the adjoining differently stressed fibres are ignored.<\/li>\n<li>The value of Young\u2019s modulus (E) for the material is same in tension and in compression.<\/li>\n<li>The section of the beam is symmetrical in the loading plane. If section is non symmetrical then twisting and warping may occur apart from bending.<\/li>\n<\/ol>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_79_1 ez-toc-wrap-left counter-hierarchy ez-toc-counter ez-toc-light-blue ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\" style=\"cursor:inherit\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"Toggle Table of Content\"><span class=\"ez-toc-js-icon-con\"><span class=\"\"><span class=\"eztoc-hide\" style=\"display:none;\">Toggle<\/span><span class=\"ez-toc-icon-toggle-span\"><svg style=\"fill: #999;color:#999\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #999;color:#999\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/span><\/span><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 ' ><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.madeeasy.in\/study\/ce\/strength-of-material\/simple-bending-or-pure-bending\/#Analysis-of-stress-and-strain-in-pure-bending\" >Analysis of stress and strain in pure bending<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.madeeasy.in\/study\/ce\/strength-of-material\/simple-bending-or-pure-bending\/#Limitations-of-Equation-of-Pure-Bending\" >Limitations of Equation of Pure Bending<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.madeeasy.in\/study\/ce\/strength-of-material\/simple-bending-or-pure-bending\/#SECTION-MODULUS-Z\" >SECTION MODULUS (Z)<\/a><ul class='ez-toc-list-level-4' ><li class='ez-toc-heading-level-4'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.madeeasy.in\/study\/ce\/strength-of-material\/simple-bending-or-pure-bending\/#Section-modulus-for-different-cross-section\" >Section modulus for different cross-section<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h3 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"Analysis-of-stress-and-strain-in-pure-bending\"><\/span>Analysis of stress and strain in pure bending<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p style=\"text-align: justify;\"><strong>a) Normal strain in beams<\/strong><\/p>\n<p style=\"text-align: justify;\">Consider a simply supported beam of length L subjected to moments at ends A and B as shown in figure.<\/p>\n<p style=\"text-align: justify;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2521 size-full\" src=\"https:\/\/study.madeeasy.in\/wp-content\/uploads\/2024\/08\/normal-strain.jpg\" alt=\" Normal strain\" width=\"385\" height=\"191\" srcset=\"https:\/\/www.madeeasy.in\/study\/wp-content\/uploads\/2024\/08\/normal-strain.jpg 385w, https:\/\/www.madeeasy.in\/study\/wp-content\/uploads\/2024\/08\/normal-strain-300x149.jpg 300w\" sizes=\"auto, (max-width: 385px) 100vw, 385px\" \/><\/p>\n<p style=\"text-align: justify;\">Take two section x<sub>1<\/sub>-x<sub>1<\/sub> and x<sub>2<\/sub>-x<sub>2<\/sub> which are dx length apart and section x<sub>1<\/sub>-x<sub>1<\/sub> is at distance x from end A, as shown<\/p>\n<p style=\"text-align: justify;\">Due to applied moments M<sub>0<\/sub>, the beam will sag in a manner as shown.<\/p>\n<p style=\"text-align: justify;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2522 size-full\" src=\"https:\/\/study.madeeasy.in\/wp-content\/uploads\/2024\/08\/deflected-shape-1.jpg\" alt=\"Deflected Shape\" width=\"379\" height=\"243\" srcset=\"https:\/\/www.madeeasy.in\/study\/wp-content\/uploads\/2024\/08\/deflected-shape-1.jpg 379w, https:\/\/www.madeeasy.in\/study\/wp-content\/uploads\/2024\/08\/deflected-shape-1-300x192.jpg 300w\" sizes=\"auto, (max-width: 379px) 100vw, 379px\" \/><\/p>\n<p style=\"text-align: justify;\">As it is case of pure bending, the elastic curve will be circular in shape.<\/p>\n<p style=\"text-align: justify;\">Point O is the centre of curvature. Cross-section x<sub>1<\/sub>-x<sub>1<\/sub> and x<sub>2<\/sub>-x<sub>2<\/sub> will remain plane and normal to longitudinal fibres of beam so as to keep radius of curvature constant.<\/p>\n<p style=\"text-align: justify;\">As it is evident , longitudinal fibres EF are shortened while fibres AB are elongated due to bending.<\/p>\n<p style=\"text-align: justify;\">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.<\/p>\n<p style=\"text-align: justify;\">Radius of curvature R is taken as distance from centre of curvature O to neutral layer.<\/p>\n<p style=\"text-align: justify;\">As derived earlier,<\/p>\n<p style=\"text-align: justify;\">Radius of curvature,<br \/>\nR = dx\/d\u03b8<\/p>\n<p style=\"text-align: justify;\">Curvature,<br \/>\nC = d\u03b8\/dx<\/p>\n<p style=\"text-align: justify;\">To calculate longitudinal strain, take a fibre GH at distance y from neutral layer CD as shown.<br \/>\nThe length L<sub>1<\/sub> of fibre GH = (R + y) d\u03b8<\/p>\n<p style=\"text-align: justify;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2524 size-full\" src=\"https:\/\/study.madeeasy.in\/wp-content\/uploads\/2024\/08\/longitudinal.jpg\" alt=\"Longitudinal\" width=\"646\" height=\"177\" srcset=\"https:\/\/www.madeeasy.in\/study\/wp-content\/uploads\/2024\/08\/longitudinal.jpg 646w, https:\/\/www.madeeasy.in\/study\/wp-content\/uploads\/2024\/08\/longitudinal-300x82.jpg 300w\" sizes=\"auto, (max-width: 646px) 100vw, 646px\" \/><\/p>\n<p style=\"text-align: justify;\">Here, y is the distance of fibre from neutral axis. Strains are positive below neutral axis and negative above neutral axis.<\/p>\n<p style=\"text-align: justify;\"><strong>(b) Normal Stresses in beams<\/strong><\/p>\n<p style=\"text-align: justify;\">As per Hooke\u2019s law, \u03b5x = \u03c3\/E<\/p>\n<p style=\"text-align: justify;\">Here, \u03c3 is the bending stress in longitudinal direction and E is the Young\u2019s modulus of elasticity.<br \/>\nSo, by comparing eq. (i) and (ii)<\/p>\n<p style=\"text-align: justify;\">\u03c3\/E = y\/R<\/p>\n<p style=\"text-align: justify;\">\u03c3\/y = E\/R <img loading=\"lazy\" decoding=\"async\" class=\"alignright wp-image-2525 size-full\" src=\"https:\/\/study.madeeasy.in\/wp-content\/uploads\/2024\/08\/moment-on-section.jpg\" alt=\"Moment on section\" width=\"232\" height=\"214\" \/><\/p>\n<p style=\"text-align: justify;\">In simply supported beam, when subjected to downward M loading bending stresses are compressive above neutral axis while these are tensile below the neutral axis as shown in fig. (i).<br \/>\nDue to these stress, a moment is produced which is bending moment for the section and it acts along z-axis.<\/p>\n<p style=\"text-align: justify;\">As the resultant force in x-direction is zero.<\/p>\n<p style=\"text-align: justify;\">\u222b\u03c3<sub>x<\/sub> dA = 0&#8230;(iv) [\u2235 There is no other horizontal force]<\/p>\n<p style=\"text-align: justify;\">When dA is small area element on cross-section on section.<\/p>\n<p style=\"text-align: justify;\">By eq. (ii) \u03c3\/y = E\/R<\/p>\n<p style=\"text-align: justify;\">\u03c3 = Ey\/R<\/p>\n<p style=\"text-align: justify;\">\u222bEy\/R dA<\/p>\n<p style=\"text-align: justify;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2526 size-full\" src=\"https:\/\/study.madeeasy.in\/wp-content\/uploads\/2024\/08\/resultant-moment.jpg\" alt=\"Resultant Moment\" width=\"657\" height=\"452\" srcset=\"https:\/\/www.madeeasy.in\/study\/wp-content\/uploads\/2024\/08\/resultant-moment.jpg 657w, https:\/\/www.madeeasy.in\/study\/wp-content\/uploads\/2024\/08\/resultant-moment-300x206.jpg 300w, https:\/\/www.madeeasy.in\/study\/wp-content\/uploads\/2024\/08\/resultant-moment-130x90.jpg 130w\" sizes=\"auto, (max-width: 657px) 100vw, 657px\" \/><\/p>\n<p style=\"text-align: justify;\">Hence, equation of pure bending, M\/I = \u03c3\/y = E\/R<\/p>\n<p style=\"text-align: justify;\"><strong>(c) Transverse Strain<\/strong><\/p>\n<p style=\"text-align: justify;\">In figure shown, fibres are in compression above neutral surface or layer in longitudinal direction, so in transverse direction, fibres will elongate in transverse direction. Similarly, fibres below neutral surface are in tension in longitudinal direction, therefore, fibres will shorten in transverse direction due to Poisson\u2019s effect. Consequently, a rectangular section will be transformed into a trapezoidal shape as shown:<\/p>\n<p style=\"text-align: justify;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2528 size-full\" src=\"https:\/\/study.madeeasy.in\/wp-content\/uploads\/2024\/08\/deformed-cross.jpg\" alt=\" Deformed cross\" width=\"586\" height=\"305\" srcset=\"https:\/\/www.madeeasy.in\/study\/wp-content\/uploads\/2024\/08\/deformed-cross.jpg 586w, https:\/\/www.madeeasy.in\/study\/wp-content\/uploads\/2024\/08\/deformed-cross-300x156.jpg 300w\" sizes=\"auto, (max-width: 586px) 100vw, 586px\" \/><\/p>\n<p style=\"text-align: justify;\">Axial strain \u03b5<sub>x<\/sub> below neutral axis = y\/R (As per eq. (i))<br \/>\nTherefore, transverse strain below neutral axis = \u03bdy\/R<\/p>\n<p style=\"text-align: justify;\">where \u03bd is Poisson\u2019s ratio.<br \/>\nAs the curvature is very small in transverse direction, therefore radius of curvature R<sub>1<\/sub> is higher than radius of curvature R is longitudinal direction. The relation between R<sub>1<\/sub> and R can be given as R<sub>1 = R\/\u03bd<\/sub><\/p>\n<h3 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"Limitations-of-Equation-of-Pure-Bending\"><\/span>Limitations of Equation of Pure Bending<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<ol style=\"text-align: justify;\">\n<li>Equation of bending is applicable when beam is free from shear force i.e. equation of bending is applicable to a member which is subjected to pure bending.<br \/>\ndM\/dx = 0<\/li>\n<li>In general, beams are subjected to both bending moment and shear force. So theory of bending can be applied only for those sections where bending moments are maximum because at that section of maximum bending moment shear force is always zero. Hence, the condition of pure bending is supposed to be satisfied at those sections.<\/li>\n<\/ol>\n<h3 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"SECTION-MODULUS-Z\"><\/span>SECTION MODULUS (Z)<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p style=\"text-align: justify;\">Equation of pure bending<\/p>\n<p style=\"text-align: justify;\">M\/I = \u03c3\/y<br \/>\n\u03c3 = M\/I\/y<\/p>\n<p style=\"text-align: justify;\">For a given bending moment at y = y<sub>max<\/sub><\/p>\n<p style=\"text-align: justify;\">\u03c3 = \u03c3<sub>max<\/sub><\/p>\n<p style=\"text-align: justify;\">\u03c3<sub>max<\/sub> = M\/I\/y<sub>max<\/sub><\/p>\n<p style=\"text-align: justify;\">where I is moment of inertia of cross-section about neutral axis.<\/p>\n<p style=\"text-align: justify;\">The term I\/y<sub>max\u00a0\u00a0<\/sub>is known as section modulus and is denoted by z.<\/p>\n<p style=\"text-align: justify;\">So, section modulus, z = I\/y<sub>max<\/sub><\/p>\n<p style=\"text-align: justify;\">It represents strength of section. A section having high section modulus is stronger.<\/p>\n<h4 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"Section-modulus-for-different-cross-section\"><\/span>Section modulus for different cross-section<span class=\"ez-toc-section-end\"><\/span><\/h4>\n<p style=\"text-align: justify;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2529 size-full\" src=\"https:\/\/study.madeeasy.in\/wp-content\/uploads\/2024\/08\/cross-section.jpg\" alt=\"Cross Section\" width=\"575\" height=\"358\" srcset=\"https:\/\/www.madeeasy.in\/study\/wp-content\/uploads\/2024\/08\/cross-section.jpg 575w, https:\/\/www.madeeasy.in\/study\/wp-content\/uploads\/2024\/08\/cross-section-300x187.jpg 300w\" sizes=\"auto, (max-width: 575px) 100vw, 575px\" \/><\/p>\n<p style=\"text-align: justify;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2530 size-full\" src=\"https:\/\/study.madeeasy.in\/wp-content\/uploads\/2024\/08\/diamond-section.jpg\" alt=\" Diamond section\" width=\"614\" height=\"349\" srcset=\"https:\/\/www.madeeasy.in\/study\/wp-content\/uploads\/2024\/08\/diamond-section.jpg 614w, https:\/\/www.madeeasy.in\/study\/wp-content\/uploads\/2024\/08\/diamond-section-300x171.jpg 300w\" sizes=\"auto, (max-width: 614px) 100vw, 614px\" \/><\/p>\n<p style=\"text-align: center;\"><a class=\"btn btn-danger\" role=\"button\" href=\"https:\/\/study.madeeasy.in\/ce\/strength-of-material\/shear-force-and-bending-moment-diagram\/\" target=\"_blank\" rel=\"noopener\">&lt;&lt; Previous<\/a> | <a class=\"btn btn-success\" role=\"button\" href=\"https:\/\/study.madeeasy.in\/ce\/strength-of-material\/principal-stresses\/\" target=\"_blank\" rel=\"noopener\"> Next &gt;&gt;<\/a><br \/>\n<strong> Must Read: <\/strong> <a href=\"https:\/\/study.madeeasy.in\/subjects\/what-is-strength-of-material\/\" target=\"_blank\" rel=\"noopener\"><strong>What is Strength of Material?<\/strong><\/a><\/p>\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Consider a simply supported beam AB of length L subjected to moment M0 at its ends as shown in figures.<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[685,2],"tags":[696,693,694,1173,1174,697,695,1175],"class_list":["post-2514","post","type-post","status-publish","format-standard","hentry","category-strength-of-material","category-ce","tag-flitched-beam","tag-normal-strain","tag-resultant-moment","tag-simple-bending","tag-simple-bending-theory","tag-steel-section","tag-stress-distribution","tag-what-is-pure-bending"],"_links":{"self":[{"href":"https:\/\/www.madeeasy.in\/study\/wp-json\/wp\/v2\/posts\/2514","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=2514"}],"version-history":[{"count":0,"href":"https:\/\/www.madeeasy.in\/study\/wp-json\/wp\/v2\/posts\/2514\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.madeeasy.in\/study\/wp-json\/wp\/v2\/media?parent=2514"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.madeeasy.in\/study\/wp-json\/wp\/v2\/categories?post=2514"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.madeeasy.in\/study\/wp-json\/wp\/v2\/tags?post=2514"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}