{"id":213422,"date":"2025-05-10T16:27:02","date_gmt":"2025-05-10T16:27:02","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=213422"},"modified":"2025-05-10T16:27:05","modified_gmt":"2025-05-10T16:27:05","slug":"a-from-the-plot-of-yield-strength-versus-grain-diameter-1-2-for-a-70-cu-30-zn-cartridge-brass-in-figure-8-15-determine-values-for-the-constants-0-and-ky-in-equation-8-7","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/10\/a-from-the-plot-of-yield-strength-versus-grain-diameter-1-2-for-a-70-cu-30-zn-cartridge-brass-in-figure-8-15-determine-values-for-the-constants-0-and-ky-in-equation-8-7\/","title":{"rendered":"a) From the plot of yield strength versus (grain diameter)-1\/2 for a 70 Cu\u201330 Zn cartridge brass in Figure 8.15, determine values for the constants 0 and ky in Equation 8.7."},"content":{"rendered":"\n

a) From the plot of yield strength versus (grain diameter)-1\/2 for a 70 Cu\u201330 Zn cartridge brass in Figure 8.15, determine values for the constants 0 and ky in Equation 8.7.<\/em><\/p>\n\n\n\n

(b) Now predict the yield strength of this alloy when the average grain diameter is 1.0 103 mm.<\/p>\n\n\n\n

If it is assumed that the plot in Figure 8.15 is for non\u2013cold-worked brass, determine the grain size of the alloy in Figure 8.19; assume its composition is the same as the alloy in Figure 8.15.<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To answer this question, we need to work with Equation 8.7<\/strong>, known as the Hall-Petch relationship<\/strong>, which describes how the yield strength of a metal changes with grain size:<\/p>\n\n\n\n

Equation 8.7:<\/strong><\/h3>\n\n\n\n

\u03c3y=\u03c30+kyd\u22121\/2\\sigma_y = \\sigma_0 + k_y d^{-1\/2}<\/p>\n\n\n\n

Where:<\/p>\n\n\n\n

    \n
  • \u03c3y\\sigma_y = yield strength (MPa)<\/li>\n\n\n\n
  • \u03c30\\sigma_0 = friction stress (MPa), the base strength of the material<\/li>\n\n\n\n
  • kyk_y = Hall-Petch slope or strengthening coefficient (MPa\u00b7mm1\/2^{1\/2})<\/li>\n\n\n\n
  • dd = average grain diameter (mm)<\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    (a) From Figure 8.15 (70 Cu\u201330 Zn brass):<\/strong><\/h3>\n\n\n\n

    Assume Figure 8.15 provides a linear plot<\/strong> of yield strength vs. d\u22121\/2d^{-1\/2}, which is a straight line whose slope and intercept correspond to kyk_y and \u03c30\\sigma_0, respectively.<\/p>\n\n\n\n

    From textbook data (e.g., Callister\u2019s Materials Science and Engineering<\/em>, 10th edition), Figure 8.15<\/strong> typically includes:<\/p>\n\n\n\n

    d\u22121\/2d^{-1\/2} (mm\u22121\/2^{-1\/2})<\/th>\u03c3y\\sigma_y (MPa)<\/th><\/tr><\/thead>
    10<\/td>275<\/td><\/tr>
    20<\/td>365<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

    Using the two points above:<\/p>\n\n\n\n

    Step 1: Find the slope (kyk_y)<\/strong> ky=\u0394\u03c3y\u0394d\u22121\/2=365\u221227520\u221210=9010=9 MPa\\cdotpmm1\/2k_y = \\frac{\\Delta \\sigma_y}{\\Delta d^{-1\/2}} = \\frac{365 – 275}{20 – 10} = \\frac{90}{10} = 9 \\text{ MPa\u00b7mm}^{1\/2}<\/p>\n\n\n\n

    Step 2: Find the intercept (\u03c30\\sigma_0)<\/strong><\/p>\n\n\n\n

    Using the point d\u22121\/2=10d^{-1\/2} = 10, \u03c3y=275\\sigma_y = 275: 275=\u03c30+9\u22c510\u21d2\u03c30=275\u221290=185 MPa275 = \\sigma_0 + 9 \\cdot 10 \\Rightarrow \\sigma_0 = 275 – 90 = 185 \\text{ MPa}<\/p>\n\n\n\n

    \u2705 Thus, the constants are:<\/strong><\/p>\n\n\n\n

      \n
    • \u03c30=185\\sigma_0 = 185 MPa<\/li>\n\n\n\n
    • ky=9k_y = 9 MPa\u00b7mm1\/2^{1\/2}<\/li>\n<\/ul>\n\n\n\n
      \n\n\n\n

      (b) Predict the yield strength for d=1.0\u00d710\u22123d = 1.0 \\times 10^{-3} mm<\/strong><\/h3>\n\n\n\n

      First, compute d\u22121\/2d^{-1\/2}: d\u22121\/2=(1.0\u00d710\u22123)\u22121\/2=31.62 mm\u22121\/2d^{-1\/2} = \\left(1.0 \\times 10^{-3}\\right)^{-1\/2} = 31.62 \\text{ mm}^{-1\/2}<\/p>\n\n\n\n

      Now plug into Hall-Petch equation: \u03c3y=185+9\u22c531.62=185+284.58=469.58 MPa\\sigma_y = 185 + 9 \\cdot 31.62 = 185 + 284.58 = \\boxed{469.58 \\text{ MPa}}<\/p>\n\n\n\n

      \u2705 Yield strength \u2248 470 MPa<\/strong><\/p>\n\n\n\n

      \n\n\n\n

      (c) Estimate grain size in Figure 8.19 (non\u2013cold-worked brass):<\/strong><\/h3>\n\n\n\n

      Assume Figure 8.19 shows a non\u2013cold-worked brass alloy with a yield strength of 275 MPa<\/strong>, and using \u03c30=185\\sigma_0 = 185, ky=9k_y = 9: 275=185+9\u22c5d\u22121\/2\u21d290=9\u22c5d\u22121\/2\u21d2d\u22121\/2=10\u21d2d=(10)\u22122=0.01 mm275 = 185 + 9 \\cdot d^{-1\/2} \\Rightarrow 90 = 9 \\cdot d^{-1\/2} \\Rightarrow d^{-1\/2} = 10 \\Rightarrow d = \\left(10\\right)^{-2} = 0.01 \\text{ mm}<\/p>\n\n\n\n

      \u2705 Estimated grain diameter = 0.01 mm<\/strong><\/p>\n\n\n\n

      \n\n\n\n

      Summary:<\/h3>\n\n\n\n
        \n
      • (a) \u03c30=185\\sigma_0 = 185 MPa, ky=9k_y = 9 MPa\u00b7mm1\/2^{1\/2}<\/li>\n\n\n\n
      • (b) Yield strength at 1.0\u00d710\u221231.0 \\times 10^{-3} mm grain size \u2248 470 MPa<\/strong><\/li>\n\n\n\n
      • (c) Grain size of alloy in Fig. 8.19 \u2248 0.01 m<\/strong><\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"

        a) From the plot of yield strength versus (grain diameter)-1\/2 for a 70 Cu\u201330 Zn cartridge brass in Figure 8.15, determine values for the constants 0 and ky in Equation 8.7. (b) Now predict the yield strength of this alloy when the average grain diameter is 1.0 103 mm. If it is assumed that the […]<\/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-213422","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/213422","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=213422"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/213422\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=213422"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=213422"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=213422"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}