{"id":28279,"date":"2025-06-20T11:34:43","date_gmt":"2025-06-20T11:34:43","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=28279"},"modified":"2025-06-20T11:34:44","modified_gmt":"2025-06-20T11:34:44","slug":"determine-the-critical-buckling-load-for-each-of-the-columns-using-the-euler-equation","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/determine-the-critical-buckling-load-for-each-of-the-columns-using-the-euler-equation\/","title":{"rendered":"Determine the critical buckling load for each of the columns, using the Euler equation"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Determine the critical buckling load for each of the columns, using the Euler equation. E = 29,000 ksi. Proportional limit = 36,000 psi. Assume simple ends and maximum permissible L\/r = 200. A solid round bar of 2 in diameter: a. L = 4 ft 0 in b. L = 2 ft 3 in c. L = 6 ft 6 in<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-cyan-bluish-gray-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">To determine the <strong>critical buckling load<\/strong> using <strong>Euler\u2019s formula<\/strong>, we apply the equation:Pcr=\u03c02EI(KL)2P_{cr} = \\frac{\\pi^2 E I}{(KL)^2}Pcr\u200b=(KL)2\u03c02EI\u200b<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>PcrP_{cr}Pcr\u200b = critical buckling load (lb)<\/li>\n\n\n\n<li>EEE = modulus of elasticity = 29,000 ksi = 29,000,000 psi<\/li>\n\n\n\n<li>III = moment of inertia of the cross section (in\u2074)<\/li>\n\n\n\n<li>LLL = column length (in)<\/li>\n\n\n\n<li>KKK = column effective length factor (K = 1.0 for pinned-pinned ends)<\/li>\n\n\n\n<li>\u03c0\u22483.1416\\pi \\approx 3.1416\u03c0\u22483.1416<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Compute Moment of Inertia for a solid circular section<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">For a solid round bar:I=\u03c0d464I = \\frac{\\pi d^4}{64}I=64\u03c0d4\u200b<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Given diameter d=2d = 2d=2 in:I=\u03c0(2)464=\u03c0\u22c51664=\u03c04\u22480.7854&nbsp;in4I = \\frac{\\pi (2)^4}{64} = \\frac{\\pi \\cdot 16}{64} = \\frac{\\pi}{4} \\approx 0.7854 \\text{ in}^4I=64\u03c0(2)4\u200b=64\u03c0\u22c516\u200b=4\u03c0\u200b\u22480.7854&nbsp;in4<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Compute critical loads for each length<\/h3>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>a. L = 4 ft 0 in = 48 in<\/strong><\/h4>\n\n\n\n<p class=\"wp-block-paragraph\">Pcr=\u03c02\u22c529,000,000\u22c50.7854(48)2=9.8696\u22c529,000,000\u22c50.78542304\u2248224,899,102.62304\u224897,611.5&nbsp;lbP_{cr} = \\frac{\\pi^2 \\cdot 29,000,000 \\cdot 0.7854}{(48)^2} = \\frac{9.8696 \\cdot 29,000,000 \\cdot 0.7854}{2304} \\approx \\frac{224,899,102.6}{2304} \\approx 97,611.5 \\text{ lb}Pcr\u200b=(48)2\u03c02\u22c529,000,000\u22c50.7854\u200b=23049.8696\u22c529,000,000\u22c50.7854\u200b\u22482304224,899,102.6\u200b\u224897,611.5&nbsp;lb<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>b. L = 2 ft 3 in = 27 in<\/strong><\/h4>\n\n\n\n<p class=\"wp-block-paragraph\">Pcr=\u03c02\u22c529,000,000\u22c50.7854(27)2=224,899,102.6729\u2248308,434.4&nbsp;lbP_{cr} = \\frac{\\pi^2 \\cdot 29,000,000 \\cdot 0.7854}{(27)^2} = \\frac{224,899,102.6}{729} \\approx 308,434.4 \\text{ lb}Pcr\u200b=(27)2\u03c02\u22c529,000,000\u22c50.7854\u200b=729224,899,102.6\u200b\u2248308,434.4&nbsp;lb<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>c. L = 6 ft 6 in = 78 in<\/strong><\/h4>\n\n\n\n<p class=\"wp-block-paragraph\">Pcr=224,899,102.6(78)2=224,899,102.66084\u224836,957.4&nbsp;lbP_{cr} = \\frac{224,899,102.6}{(78)^2} = \\frac{224,899,102.6}{6084} \\approx 36,957.4 \\text{ lb}Pcr\u200b=(78)2224,899,102.6\u200b=6084224,899,102.6\u200b\u224836,957.4&nbsp;lb<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answers:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>(a)<\/strong> L = 4 ft 0 in \u2192 <strong>97,612 lb<\/strong><\/li>\n\n\n\n<li><strong>(b)<\/strong> L = 2 ft 3 in \u2192 <strong>308,434 lb<\/strong><\/li>\n\n\n\n<li><strong>(c)<\/strong> L = 6 ft 6 in \u2192 <strong>36,957 lb<\/strong><\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Euler&#8217;s formula estimates the axial load at which long, slender columns buckle. The moment of inertia depends on cross-sectional geometry. The slenderness ratio L\/rL\/rL\/r ensures the column behaves elastically (Euler&#8217;s theory applies). In all cases above, the slenderness ratio is below the permissible limit of 200, confirming Euler\u2019s formula is applicable. The shorter the column, the higher the critical buckling load, as buckling is inversely proportional to the square of the column length.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"852\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-201.jpeg\" alt=\"\" class=\"wp-image-28290\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-201.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-201-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-201-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Determine the critical buckling load for each of the columns, using the Euler equation. E = 29,000 ksi. Proportional limit = 36,000 psi. Assume simple ends and maximum permissible L\/r = 200. A solid round bar of 2 in diameter: a. L = 4 ft 0 in b. L = 2 ft 3 in c. [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-28279","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/28279","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=28279"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/28279\/revisions"}],"predecessor-version":[{"id":28292,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/28279\/revisions\/28292"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=28279"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=28279"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=28279"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}