{"id":183589,"date":"2025-01-16T17:20:30","date_gmt":"2025-01-16T17:20:30","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=183589"},"modified":"2025-01-16T17:20:32","modified_gmt":"2025-01-16T17:20:32","slug":"consider-the-brass-alloy-for-which-the-stress-strain-behavior-is-shown-in-the-figure-below","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/16\/consider-the-brass-alloy-for-which-the-stress-strain-behavior-is-shown-in-the-figure-below\/","title":{"rendered":"Consider the brass alloy for which the stress-strain behavior is shown in the figure below"},"content":{"rendered":"\n<p>Q1\/ Consider the brass alloy for which the stress-strain behavior is shown in the figure below. A cylindrical specimen of this material 10.0 mm in diameter and 101.6 mm long is pulled in tension with a force of 10,000 N. If it is known that this alloy has a value for Poisson&#8217;s ratio of 0.35, compute (a) the specimen elongation, and (b) the reduction in specimen diameter.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/01\/image-286.png\" alt=\"\" class=\"wp-image-183590\"\/><\/figure>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>To determine the elongation and reduction in diameter of a brass alloy specimen under tension, we can utilize the material&#8217;s stress-strain behavior, Poisson&#8217;s ratio, and the applied tensile force. Here&#8217;s a step-by-step approach:<\/p>\n\n\n\n<p><strong>Given Data:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Original Diameter (d\u2080):<\/strong> 10.0 mm<\/li>\n\n\n\n<li><strong>Original Length (L\u2080):<\/strong> 101.6 mm<\/li>\n\n\n\n<li><strong>Applied Force (F):<\/strong> 10,000 N<\/li>\n\n\n\n<li><strong>Poisson&#8217;s Ratio (\u03bd):<\/strong> 0.35<\/li>\n<\/ul>\n\n\n\n<p><strong>Assumptions:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The material behaves elastically under the applied load.<\/li>\n\n\n\n<li>The stress-strain curve is linear within the elastic region.<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 1: Calculate the Cross-Sectional Area (A\u2080)<\/strong><\/p>\n\n\n\n<p>The cross-sectional area of the cylindrical specimen is given by:<\/p>\n\n\n\n<p>[ A\u2080 = \\pi \\left( \\frac{d\u2080}{2} \\right)^2 ]<\/p>\n\n\n\n<p>Substituting the given diameter:<\/p>\n\n\n\n<p>[ A\u2080 = \\pi \\left( \\frac{10.0\\,\\text{mm}}{2} \\right)^2 = \\pi \\times (5.0\\,\\text{mm})^2 \\approx 78.54\\,\\text{mm}^2 ]<\/p>\n\n\n\n<p><strong>Step 2: Determine the Stress (\u03c3)<\/strong><\/p>\n\n\n\n<p>Stress is defined as the force applied divided by the cross-sectional area:<\/p>\n\n\n\n<p>[ \\sigma = \\frac{F}{A\u2080} ]<\/p>\n\n\n\n<p>Substituting the known values:<\/p>\n\n\n\n<p>[ \\sigma = \\frac{10,000\\,\\text{N}}{78.54\\,\\text{mm}^2} \\approx 127.32\\,\\text{MPa} ]<\/p>\n\n\n\n<p><strong>Step 3: Find the Strain (\u03b5) from the Stress-Strain Curve<\/strong><\/p>\n\n\n\n<p>Using the stress-strain curve for the brass alloy, locate the strain corresponding to the calculated stress. For a stress of approximately 127.32 MPa, the strain (\u03b5) is approximately 0.0015.<\/p>\n\n\n\n<p><strong>Step 4: Calculate the Elongation (\u0394L)<\/strong><\/p>\n\n\n\n<p>Elongation is the product of the strain and the original length:<\/p>\n\n\n\n<p>[ \\Delta L = \\epsilon \\times L\u2080 ]<\/p>\n\n\n\n<p>Substituting the known values:<\/p>\n\n\n\n<p>[ \\Delta L = 0.0015 \\times 101.6\\,\\text{mm} \\approx 0.1524\\,\\text{mm} ]<\/p>\n\n\n\n<p><strong>Step 5: Calculate the Lateral Strain (\u03b5\u2097) and Reduction in Diameter (\u0394d)<\/strong><\/p>\n\n\n\n<p>Lateral strain is related to the longitudinal strain by Poisson&#8217;s ratio:<\/p>\n\n\n\n<p>[ \\epsilon\u2097 = -\\nu \\times \\epsilon ]<\/p>\n\n\n\n<p>Substituting the known values:<\/p>\n\n\n\n<p>[ \\epsilon\u2097 = -0.35 \\times 0.0015 \\approx -0.000525 ]<\/p>\n\n\n\n<p>The reduction in diameter is:<\/p>\n\n\n\n<p>[ \\Delta d = \\epsilon\u2097 \\times d\u2080 ]<\/p>\n\n\n\n<p>Substituting the known values:<\/p>\n\n\n\n<p>[ \\Delta d = -0.000525 \\times 10.0\\,\\text{mm} \\approx -0.00525\\,\\text{mm} ]<\/p>\n\n\n\n<p><strong>Results:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Elongation (\u0394L):<\/strong> Approximately 0.1524 mm<\/li>\n\n\n\n<li><strong>Reduction in Diameter (\u0394d):<\/strong> Approximately -0.00525 mm<\/li>\n<\/ul>\n\n\n\n<p>These calculations indicate that under the applied tensile force, the brass alloy specimen will elongate by about 0.1524 mm and experience a reduction in diameter of approximately 0.00525 mm.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Q1\/ Consider the brass alloy for which the stress-strain behavior is shown in the figure below. A cylindrical specimen of this material 10.0 mm in diameter and 101.6 mm long is pulled in tension with a force of 10,000 N. If it is known that this alloy has a value for Poisson&#8217;s ratio of 0.35, [&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-183589","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183589","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=183589"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183589\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=183589"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=183589"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=183589"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}