{"id":253182,"date":"2025-07-12T00:46:01","date_gmt":"2025-07-12T00:46:01","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=253182"},"modified":"2025-07-12T00:46:04","modified_gmt":"2025-07-12T00:46:04","slug":"a-cylindrical-specimen-of-steel-having-a-diameter-of-15-2-mm-0-60-in-and-length-of-250-mm-10-0-in-is-deformed-elastically-in-tension-with-a-force-of-48900-n-11000-lbf","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/12\/a-cylindrical-specimen-of-steel-having-a-diameter-of-15-2-mm-0-60-in-and-length-of-250-mm-10-0-in-is-deformed-elastically-in-tension-with-a-force-of-48900-n-11000-lbf\/","title":{"rendered":"A cylindrical specimen of steel having a diameter of 15.2 mm (0.60 in.) and length of 250 mm (10.0 in.) is deformed elastically in tension with a force of 48,900 N (11,000 lbf)."},"content":{"rendered":"\n
    \n
  1. A cylindrical specimen of steel having a diameter of 15.2 mm (0.60 in.) and length of 250 mm (10.0 in.) is deformed elastically in tension with a force of 48,900 N (11,000 lbf). The modulus of elasticity and the Poisson’s ratio for this steel is 207 GPa (
    psi) and 0.30. 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. Will the diameter increase or decrease?<\/li>\n<\/ol>\n\n\n\n

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

    To solve this, we will break it into two parts: (a) calculating the elongation in the direction of applied stress, and (b) calculating the change in diameter using Poisson’s ratio.<\/p>\n\n\n\n

    Given:<\/h3>\n\n\n\n
      \n
    • Diameter of the specimen, d=15.2\u2009mmd = 15.2 \\, \\text{mm}d=15.2mm<\/li>\n\n\n\n
    • Length of the specimen, L=250\u2009mmL = 250 \\, \\text{mm}L=250mm<\/li>\n\n\n\n
    • Force applied, F=48,900\u2009NF = 48,900 \\, \\text{N}F=48,900N<\/li>\n\n\n\n
    • Modulus of elasticity, E=207\u2009GPa=207\u00d7109\u2009N\/m2E = 207 \\, \\text{GPa} = 207 \\times 10^9 \\, \\text{N\/m}^2E=207GPa=207\u00d7109N\/m2<\/li>\n\n\n\n
    • Poisson’s ratio, \u03bd=0.30\\nu = 0.30\u03bd=0.30<\/li>\n<\/ul>\n\n\n\n

      (a) Elongation of the specimen<\/h3>\n\n\n\n

      The elongation \u0394L\\Delta L\u0394L in the direction of the applied force is given by Hooke\u2019s Law: \u0394L=F\u22c5LA\u22c5E\\Delta L = \\frac{F \\cdot L}{A \\cdot E}\u0394L=A\u22c5EF\u22c5L\u200b<\/p>\n\n\n\n

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

        \n
      • A=cross-sectional\u00a0area=\u03c0(d2)2A = \\text{cross-sectional area} = \\pi \\left( \\frac{d}{2} \\right)^2A=cross-sectional\u00a0area=\u03c0(2d\u200b)2<\/li>\n<\/ul>\n\n\n\n

        First, calculate the cross-sectional area: A=\u03c0(15.2\u2009mm2)2=\u03c0(7.6\u2009mm)2=181.46\u2009mm2A = \\pi \\left( \\frac{15.2 \\, \\text{mm}}{2} \\right)^2 = \\pi \\left( 7.6 \\, \\text{mm} \\right)^2 = 181.46 \\, \\text{mm}^2A=\u03c0(215.2mm\u200b)2=\u03c0(7.6mm)2=181.46mm2<\/p>\n\n\n\n

        Now, plug in the values into the elongation formula: \u0394L=48,900\u2009N\u00d7250\u2009mm181.46\u2009mm2\u00d7207\u00d7109\u2009N\/m2\\Delta L = \\frac{48,900 \\, \\text{N} \\times 250 \\, \\text{mm}}{181.46 \\, \\text{mm}^2 \\times 207 \\times 10^9 \\, \\text{N\/m}^2}\u0394L=181.46mm2\u00d7207\u00d7109N\/m248,900N\u00d7250mm\u200b<\/p>\n\n\n\n

        Converting mm to meters: \u0394L=48,900\u00d70.25181.46\u00d710\u22126\u00d7207\u00d7109\\Delta L = \\frac{48,900 \\times 0.25}{181.46 \\times 10^{-6} \\times 207 \\times 10^9}\u0394L=181.46\u00d710\u22126\u00d7207\u00d710948,900\u00d70.25\u200b \u0394L=0.000579\u2009m=0.579\u2009mm\\Delta L = 0.000579 \\, \\text{m} = 0.579 \\, \\text{mm}\u0394L=0.000579m=0.579mm<\/p>\n\n\n\n

        Thus, the elongation of the specimen is approximately 0.579 mm<\/strong>.<\/p>\n\n\n\n

        (b) Change in Diameter of the Specimen<\/h3>\n\n\n\n

        Using Poisson’s ratio, we know that the lateral strain (change in diameter) is related to the longitudinal strain (elongation). The formula is: \u0394dd=\u2212\u03bd\u22c5\u0394LL\\frac{\\Delta d}{d} = -\\nu \\cdot \\frac{\\Delta L}{L}d\u0394d\u200b=\u2212\u03bd\u22c5L\u0394L\u200b<\/p>\n\n\n\n

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

          \n
        • \u0394d\\Delta d\u0394d is the change in diameter<\/li>\n\n\n\n
        • ddd is the initial diameter<\/li>\n\n\n\n
        • \u03bd\\nu\u03bd is Poisson’s ratio<\/li>\n\n\n\n
        • \u0394L\\Delta L\u0394L is the elongation<\/li>\n\n\n\n
        • LLL is the original length<\/li>\n<\/ul>\n\n\n\n

          Now, substitute the known values: \u0394dd=\u22120.30\u00d70.579\u2009mm250\u2009mm=\u22120.30\u00d70.002316=\u22120.000695\\frac{\\Delta d}{d} = -0.30 \\times \\frac{0.579 \\, \\text{mm}}{250 \\, \\text{mm}} = -0.30 \\times 0.002316 = -0.000695d\u0394d\u200b=\u22120.30\u00d7250mm0.579mm\u200b=\u22120.30\u00d70.002316=\u22120.000695<\/p>\n\n\n\n

          Thus, the change in diameter is: \u0394d=\u22120.000695\u00d715.2\u2009mm=\u22120.0105\u2009mm\\Delta d = -0.000695 \\times 15.2 \\, \\text{mm} = -0.0105 \\, \\text{mm}\u0394d=\u22120.000695\u00d715.2mm=\u22120.0105mm<\/p>\n\n\n\n

          So, the change in diameter is approximately -0.0105 mm<\/strong>, which means the diameter decreases<\/strong>.<\/p>\n\n\n\n

          Conclusion:<\/h3>\n\n\n\n

          (a) The specimen will elongate by approximately 0.579 mm<\/strong> in the direction of the applied stress.
          (b) The diameter of the specimen will decrease by approximately 0.0105 mm<\/strong> due to the lateral contraction resulting from the applied tension.<\/p>\n\n\n\n

          \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

          The Correct Answer and Explanation is: To solve this, we will break it into two parts: (a) calculating the elongation in the direction of applied stress, and (b) calculating the change in diameter using Poisson’s ratio. Given: (a) Elongation of the specimen The elongation \u0394L\\Delta L\u0394L in the direction of the applied force is given […]<\/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-253182","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/253182","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=253182"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/253182\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=253182"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=253182"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=253182"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}