{"id":253179,"date":"2025-07-12T00:36:30","date_gmt":"2025-07-12T00:36:30","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=253179"},"modified":"2025-07-12T00:36:34","modified_gmt":"2025-07-12T00:36:34","slug":"a-cylindrical-bar-of-metal-having-a-diameter-of-20-8-mm-and-a-length-of-194-mm-is-deformed-elastically-in-tension-with-a-force-of-53100-n","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/12\/a-cylindrical-bar-of-metal-having-a-diameter-of-20-8-mm-and-a-length-of-194-mm-is-deformed-elastically-in-tension-with-a-force-of-53100-n\/","title":{"rendered":"A cylindrical bar of metal having a diameter of 20.8 mm and a length of 194 mm is deformed elastically in tension with a force of 53,100 N."},"content":{"rendered":"\n<p>A cylindrical bar of metal having a diameter of 20.8 mm and a length of 194 mm is deformed elastically in tension with a force of 53,100 N. Given that the elastic modulus and Poisson&#8217;s ratio of the metal are 61.2 GPa and 0.34, respectively, determine the following: (a) The amount by which this specimen will elongate in the direction of the applied stress. (b) The change in diameter of the specimen: Indicate an increase in diameter with a positive number and a decrease with a negative number.<\/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 will need to use the formulas related to elasticity, specifically for tension and the relationship between stress, strain, and the material properties.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Given:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Diameter of the cylindrical bar d=20.8\u2009mmd = 20.8 \\, \\text{mm}d=20.8mm<\/li>\n\n\n\n<li>Length of the cylindrical bar L=194\u2009mmL = 194 \\, \\text{mm}L=194mm<\/li>\n\n\n\n<li>Force applied F=53,100\u2009NF = 53,100 \\, \\text{N}F=53,100N<\/li>\n\n\n\n<li>Elastic modulus (Young&#8217;s modulus) E=61.2\u2009GPa=61.2\u00d7109\u2009PaE = 61.2 \\, \\text{GPa} = 61.2 \\times 10^9 \\, \\text{Pa}E=61.2GPa=61.2\u00d7109Pa<\/li>\n\n\n\n<li>Poisson\u2019s ratio \u03bd=0.34\\nu = 0.34\u03bd=0.34<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Calculate the cross-sectional area<\/h3>\n\n\n\n<p>The cross-sectional area of the cylindrical bar can be calculated using the formula for the area of a circle: A=\u03c0(d2)2=\u03c0(20.8\u2009mm2)2=\u03c0\u00d7(10.4\u2009mm)2A = \\pi \\left( \\frac{d}{2} \\right)^2 = \\pi \\left( \\frac{20.8 \\, \\text{mm}}{2} \\right)^2 = \\pi \\times (10.4 \\, \\text{mm})^2A=\u03c0(2d\u200b)2=\u03c0(220.8mm\u200b)2=\u03c0\u00d7(10.4mm)2 A=\u03c0\u00d7(108.16\u2009mm2)=339.45\u2009mm2=339.45\u00d710\u22126\u2009m2A = \\pi \\times (108.16 \\, \\text{mm}^2) = 339.45 \\, \\text{mm}^2 = 339.45 \\times 10^{-6} \\, \\text{m}^2A=\u03c0\u00d7(108.16mm2)=339.45mm2=339.45\u00d710\u22126m2<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Calculate the tensile stress<\/h3>\n\n\n\n<p>Tensile stress \u03c3\\sigma\u03c3 is defined as the force per unit area: \u03c3=FA=53,100\u2009N339.45\u00d710\u22126\u2009m2=156,000,000\u2009Pa=156\u2009MPa\\sigma = \\frac{F}{A} = \\frac{53,100 \\, \\text{N}}{339.45 \\times 10^{-6} \\, \\text{m}^2} = 156,000,000 \\, \\text{Pa} = 156 \\, \\text{MPa}\u03c3=AF\u200b=339.45\u00d710\u22126m253,100N\u200b=156,000,000Pa=156MPa<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Calculate the elongation (change in length)<\/h3>\n\n\n\n<p>The elongation \u0394L\\Delta L\u0394L of the specimen can be calculated using Hooke&#8217;s Law for elastic deformation in tension: \u0394L=FLAE\\Delta L = \\frac{F L}{A E}\u0394L=AEFL\u200b<\/p>\n\n\n\n<p>Substitute the known values: \u0394L=53,100\u2009N\u00d70.194\u2009m339.45\u00d710\u22126\u2009m2\u00d761.2\u00d7109\u2009Pa=0.000264\u2009m=0.264\u2009mm\\Delta L = \\frac{53,100 \\, \\text{N} \\times 0.194 \\, \\text{m}}{339.45 \\times 10^{-6} \\, \\text{m}^2 \\times 61.2 \\times 10^9 \\, \\text{Pa}} = 0.000264 \\, \\text{m} = 0.264 \\, \\text{mm}\u0394L=339.45\u00d710\u22126m2\u00d761.2\u00d7109Pa53,100N\u00d70.194m\u200b=0.000264m=0.264mm<\/p>\n\n\n\n<p>So, the specimen elongates by <strong>0.264 mm<\/strong> in the direction of the applied stress.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Calculate the change in diameter<\/h3>\n\n\n\n<p>The lateral strain (change in diameter) is related to the longitudinal strain by Poisson&#8217;s ratio. The longitudinal strain \u03f5L\\epsilon_L\u03f5L\u200b is given by: \u03f5L=\u0394LL=0.264\u2009mm194\u2009mm=0.00136\\epsilon_L = \\frac{\\Delta L}{L} = \\frac{0.264 \\, \\text{mm}}{194 \\, \\text{mm}} = 0.00136\u03f5L\u200b=L\u0394L\u200b=194mm0.264mm\u200b=0.00136<\/p>\n\n\n\n<p>The lateral strain \u03f5D\\epsilon_D\u03f5D\u200b in the direction perpendicular to the applied stress (change in diameter) is: \u03f5D=\u2212\u03bd\u00d7\u03f5L=\u22120.34\u00d70.00136=\u22120.0004624\\epsilon_D = -\\nu \\times \\epsilon_L = -0.34 \\times 0.00136 = -0.0004624\u03f5D\u200b=\u2212\u03bd\u00d7\u03f5L\u200b=\u22120.34\u00d70.00136=\u22120.0004624<\/p>\n\n\n\n<p>Now, the change in diameter \u0394d\\Delta d\u0394d is: \u0394d=\u03f5D\u00d7d=\u22120.0004624\u00d720.8\u2009mm=\u22120.0096\u2009mm\\Delta d = \\epsilon_D \\times d = -0.0004624 \\times 20.8 \\, \\text{mm} = -0.0096 \\, \\text{mm}\u0394d=\u03f5D\u200b\u00d7d=\u22120.0004624\u00d720.8mm=\u22120.0096mm<\/p>\n\n\n\n<p>Thus, the diameter decreases by <strong>0.0096 mm<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p>(a) The elongation in the direction of the applied stress is <strong>0.264 mm<\/strong>.<\/p>\n\n\n\n<p>(b) The change in diameter is a <strong>decrease of 0.0096 mm<\/strong>.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner9-93.jpeg\" alt=\"\" class=\"wp-image-253180\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>A cylindrical bar of metal having a diameter of 20.8 mm and a length of 194 mm is deformed elastically in tension with a force of 53,100 N. Given that the elastic modulus and Poisson&#8217;s ratio of the metal are 61.2 GPa and 0.34, respectively, determine the following: (a) The amount by which this specimen [&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-253179","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/253179","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=253179"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/253179\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=253179"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=253179"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=253179"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}