An A-36 steel column has a length of 12 feet and is pinned at both ends. If the cross-sectional area has the dimension shown, determine the critical load.<\/p>\n\n\n\n
The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n To determine the critical load<\/strong> for an A-36 steel column pinned at both ends, use Euler\u2019s Buckling Formula<\/strong>: Pcr=\u03c02EI(KL)2P_{cr} = \\frac{\\pi^2 E I}{(KL)^2}<\/p>\n\n\n\n Since the moment of inertia depends on the cross-sectional dimensions and you mentioned \u201chas the dimension shown\u201d but did not attach an image, we will assume a typical W8 \u00d7 40 wide flange<\/strong> steel column often used with A-36 steel.<\/p>\n\n\n\n Now plug into Euler’s formula: Pcr=\u03c02\u22c529,000,000\u22c591.8(1\u22c5144)2P_{cr} = \\frac{\\pi^2 \\cdot 29,000,000 \\cdot 91.8}{(1 \\cdot 144)^2} Pcr=3.14162\u22c529,000,000\u22c591.820736P_{cr} = \\frac{3.1416^2 \\cdot 29,000,000 \\cdot 91.8}{20736} Pcr\u224826.7\u00d710920736P_{cr} \\approx \\frac{26.7 \\times 10^9}{20736} Pcr\u22481,287,031 pounds=1,287 kipsP_{cr} \\approx 1,287,031 \\text{ pounds} = 1,287 \\text{ kips}<\/p>\n\n\n\n The critical load is approximately 1,287 kips (1 kip = 1,000 pounds).<\/strong><\/p>\n\n\n\n Euler\u2019s formula determines the axial load at which a long slender column will buckle. The formula considers the length, end conditions, and stiffness (moment of inertia and modulus of elasticity) of the column. Since the ends are pinned, the column is more prone to buckling than one with fixed ends. The smaller the moment of inertia or longer the length, the lower the critical load. This analysis assumes linear elastic behavior and no initial imperfections. Always select the least moment of inertia<\/strong> (about the weaker axis) to ensure safety.<\/p>\n","protected":false},"excerpt":{"rendered":" An A-36 steel column has a length of 12 feet and is pinned at both ends. If the cross-sectional area has the dimension shown, determine the critical load. The correct answer and explanation is: To determine the critical load for an A-36 steel column pinned at both ends, use Euler\u2019s Buckling Formula: Pcr=\u03c02EI(KL)2P_{cr} = \\frac{\\pi^2 […]<\/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-28927","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/28927","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=28927"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/28927\/revisions"}],"predecessor-version":[{"id":28933,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/28927\/revisions\/28933"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=28927"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=28927"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=28927"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Given:<\/h3>\n\n\n\n
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For a W8 \u00d7 40:<\/h3>\n\n\n\n
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Final Answer:<\/h3>\n\n\n\n
Explanation:<\/h3>\n\n\n\n