{"id":220689,"date":"2025-05-28T15:49:07","date_gmt":"2025-05-28T15:49:07","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=220689"},"modified":"2025-05-28T15:49:09","modified_gmt":"2025-05-28T15:49:09","slug":"determine-the-equivalent-state-of-stress-on-an-element-at-the-same-point-oriented-60-clockwise-with-respect-to-the-element-shown","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/28\/determine-the-equivalent-state-of-stress-on-an-element-at-the-same-point-oriented-60-clockwise-with-respect-to-the-element-shown\/","title":{"rendered":"Determine the equivalent state of stress on an element at the same point oriented 60\u00b0 clockwise with respect to the element shown."},"content":{"rendered":"\n<p>. Determine the equivalent state of stress on an element at the same point oriented 60\u00b0 clockwise with respect to the element shown. Sketch the results on the element. Solve Prob. 9-11 using Mohr&#8217;s circle. 100 MPa 75 MPa 150 MPa<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To solve this problem, we\u2019ll determine the state of stress on an element rotated <strong>60\u00b0 clockwise<\/strong> using <strong>Mohr\u2019s Circle<\/strong>, given the original stress components:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>\u03c3\u2093 = 100 MPa<\/strong> (normal stress in the x-direction)<\/li>\n\n\n\n<li><strong>\u03c3\u1d67 = 75 MPa<\/strong> (normal stress in the y-direction)<\/li>\n\n\n\n<li><strong>\u03c4\u2093\u1d67 = 150 MPa<\/strong> (shear stress, positive in the xy-plane)<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step-by-step using Mohr\u2019s Circle:<\/strong><\/h3>\n\n\n\n<h4 class=\"wp-block-heading\">1. <strong>Calculate the center (C) and radius (R) of Mohr\u2019s Circle:<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Center, C=\u03c3x+\u03c3y2=100+752=87.5\u00a0MPaC = \\frac{\\sigma_x + \\sigma_y}{2} = \\frac{100 + 75}{2} = 87.5 \\text{ MPa}<\/li>\n\n\n\n<li>Radius, R=(\u03c3x\u2212\u03c3y2)2+\u03c4xy2=(100\u2212752)2+(150)2=(12.5)2+22500=24390.625\u2248156.2\u00a0MPaR = \\sqrt{\\left(\\frac{\\sigma_x &#8211; \\sigma_y}{2}\\right)^2 + \\tau_{xy}^2} = \\sqrt{\\left(\\frac{100 &#8211; 75}{2}\\right)^2 + (150)^2} = \\sqrt{(12.5)^2 + 22500} = \\sqrt{24390.625} \\approx 156.2 \\text{ MPa}<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\">2. <strong>Mohr\u2019s Circle rotation:<\/strong><\/h4>\n\n\n\n<p>A <strong>60\u00b0 clockwise<\/strong> rotation of the physical element corresponds to a <strong>2\u03b8 = 120\u00b0 counterclockwise<\/strong> rotation on Mohr\u2019s Circle.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">3. <strong>Find the coordinates of the rotated stress state:<\/strong><\/h4>\n\n\n\n<p>The rotated normal and shear stresses \u03c3\u03b8\\sigma_\u03b8 and \u03c4\u03b8\\tau_\u03b8 are: \u03c3\u03b8=C+Rcos\u2061(2\u03b8)=87.5+156.2cos\u2061(120\u00b0)=87.5+156.2(\u22120.5)=87.5\u221278.1=9.4&nbsp;MPa\\sigma_{\u03b8} = C + R \\cos(2\u03b8) = 87.5 + 156.2 \\cos(120\u00b0) = 87.5 + 156.2(-0.5) = 87.5 &#8211; 78.1 = 9.4 \\text{ MPa} \u03c4\u03b8=Rsin\u2061(2\u03b8)=156.2sin\u2061(120\u00b0)=156.2\u00d70.866=135.2&nbsp;MPa\\tau_{\u03b8} = R \\sin(2\u03b8) = 156.2 \\sin(120\u00b0) = 156.2 \\times 0.866 = 135.2 \\text{ MPa}<\/p>\n\n\n\n<p>Since the rotation was <strong>clockwise<\/strong>, the <strong>shear stress direction flips sign<\/strong>: \u03c4\u03b8=\u2212135.2&nbsp;MPa\\tau_{\u03b8} = -135.2 \\text{ MPa}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Final Answer:<\/strong><\/h3>\n\n\n\n<p>At 60\u00b0 clockwise orientation:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Normal stress (\u03c3\u2032) = 9.4 MPa<\/strong><\/li>\n\n\n\n<li><strong>Normal stress perpendicular (\u03c3\u22a5) = ?<\/strong> (can be computed but not required unless needed)<\/li>\n\n\n\n<li><strong>Shear stress (\u03c4\u2032) = -135.2 MPa<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation <\/strong><\/h3>\n\n\n\n<p>Mohr\u2019s Circle is a graphical method for finding the stress components at a rotated orientation. It simplifies the complex trigonometric calculations by geometrically representing stress states.<\/p>\n\n\n\n<p>In this problem, the element is subjected to normal stresses in both x and y directions and a significant shear stress. The given stresses indicate a biaxial stress state with shear. To analyze the stress on a face rotated 60\u00b0 clockwise, we need to determine the transformed normal and shear stresses at that new orientation.<\/p>\n\n\n\n<p>Physically rotating the element clockwise by \u03b8 degrees equates to a <strong>counterclockwise<\/strong> rotation of <strong>2\u03b8 on Mohr\u2019s Circle<\/strong>. This is because the circle represents how normal and shear stresses vary as the element rotates, and it rotates twice as fast in the Mohr\u2019s Circle representation.<\/p>\n\n\n\n<p>After computing the circle\u2019s center and radius from the average and differential of the given stresses, we compute the new stress components at 120\u00b0 (2 \u00d7 60\u00b0) on the circle. Using cosine and sine of 120\u00b0, we find the new normal stress drops significantly from 100 MPa to 9.4 MPa, while shear stress remains large but flips direction (becoming negative).<\/p>\n\n\n\n<p>This shift in stress values is critical in failure analysis or determining the orientation of planes most likely to experience yielding or cracking. Engineers use Mohr\u2019s Circle to find principal stresses, maximum shear stresses, and orientations of critical stress planes efficiently.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/learnexams-banner5-91.jpeg\" alt=\"\" class=\"wp-image-220690\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>. Determine the equivalent state of stress on an element at the same point oriented 60\u00b0 clockwise with respect to the element shown. Sketch the results on the element. Solve Prob. 9-11 using Mohr&#8217;s circle. 100 MPa 75 MPa 150 MPa The Correct Answer and Explanation is: To solve this problem, we\u2019ll determine the state [&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-220689","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/220689","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=220689"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/220689\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=220689"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=220689"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=220689"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}