{"id":5863,"date":"2024-04-15T02:20:04","date_gmt":"2024-04-15T02:20:04","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=5863"},"modified":"2024-04-15T02:20:04","modified_gmt":"2024-04-15T02:20:04","slug":"a-simply-supported-beam-6-m-long-and-of-effective-depth-50-cm-carries-a-uniformly-distributed-load-2400-kg-m-including-its-self-weight-if-the-lever-arm-factor-is-0-85-and-permissible-tensile-stress","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-simply-supported-beam-6-m-long-and-of-effective-depth-50-cm-carries-a-uniformly-distributed-load-2400-kg-m-including-its-self-weight-if-the-lever-arm-factor-is-0-85-and-permissible-tensile-stress\/","title":{"rendered":"A simply supported beam 6 m long and of effective depth 50 cm, carries a uniformly distributed load 2400 kg\/m including its self weight. If the lever arm factor is 0.85 and permissible tensile stress of steel is 1400 kg\/cm2, the area of steel required, is A. 14 cm2 B. 15 cm2 C. 16 cm2 D. 17 cm2"},"content":{"rendered":"<p>[amp_mcq option1=&#8221;14 cm2&#8243; option2=&#8221;15 cm2&#8243; option3=&#8221;16 cm2&#8243; option4=&#8221;17 cm2&#8243; correct=&#8221;option1&#8243;]<!--more--><\/p>\n<p>The correct answer is A. 14 cm2.<\/p>\n<p>The formula for calculating the area of steel required is:<\/p>\n<p>$A = \\frac{wL}{bd}f_y$<\/p>\n<p>where:<\/p>\n<ul>\n<li>$A$ is the area of steel required (in cm2)<\/li>\n<li>$w$ is the uniformly distributed load (in kg\/m)<\/li>\n<li>$l$ is the length of the beam (in m)<\/li>\n<li>$b$ is the effective depth of the beam (in cm)<\/li>\n<li>$d$ is the lever arm factor (in dimensionless)<\/li>\n<li>$f_y$ is the permissible tensile stress of steel (in kg\/cm2)<\/li>\n<\/ul>\n<p>In this case, we have:<\/p>\n<ul>\n<li>$w = 2400 \\frac{kg}{m}$<\/li>\n<li>$l = 6 m$<\/li>\n<li>$b = 50 cm = 0.5 m$<\/li>\n<li>$d = 0.85$<\/li>\n<li>$f_y = 1400 \\frac{kg}{cm^2}$<\/li>\n<\/ul>\n<p>Substituting these values into the formula, we get:<\/p>\n<p>$A = \\frac{(2400 \\frac{kg}{m})(6 m)}{(0.5 m)(0.85)(1400 \\frac{kg}{cm^2})} = 14 cm^2$<\/p>\n<p>Therefore, the area of steel required is 14 cm2.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>[amp_mcq option1=&#8221;14 cm2&#8243; option2=&#8221;15 cm2&#8243; option3=&#8221;16 cm2&#8243; option4=&#8221;17 cm2&#8243; correct=&#8221;option1&#8243;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[640],"tags":[],"class_list":["post-5863","post","type-post","status-publish","format-standard","hentry","category-rcc-structures-design","no-featured-image-padding"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.2 (Yoast SEO v23.3) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>A simply supported beam 6 m long and of effective depth 50 cm, carries a uniformly distributed load 2400 kg\/m including its self weight. If the lever arm factor is 0.85 and permissible tensile stress of steel is 1400 kg\/cm2, the area of steel required, is A. 14 cm2 B. 15 cm2 C. 16 cm2 D. 17 cm2<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/a-simply-supported-beam-6-m-long-and-of-effective-depth-50-cm-carries-a-uniformly-distributed-load-2400-kg-m-including-its-self-weight-if-the-lever-arm-factor-is-0-85-and-permissible-tensile-stress\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A simply supported beam 6 m long and of effective depth 50 cm, carries a uniformly distributed load 2400 kg\/m including its self weight. 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If the lever arm factor is 0.85 and permissible tensile stress of steel is 1400 kg\/cm2, the area of steel required, is A. 14 cm2 B. 15 cm2 C. 16 cm2 D. 17 cm2","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/exam.pscnotes.com\/mcq\/a-simply-supported-beam-6-m-long-and-of-effective-depth-50-cm-carries-a-uniformly-distributed-load-2400-kg-m-including-its-self-weight-if-the-lever-arm-factor-is-0-85-and-permissible-tensile-stress\/","og_locale":"en_US","og_type":"article","og_title":"A simply supported beam 6 m long and of effective depth 50 cm, carries a uniformly distributed load 2400 kg\/m including its self weight. If the lever arm factor is 0.85 and permissible tensile stress of steel is 1400 kg\/cm2, the area of steel required, is A. 14 cm2 B. 15 cm2 C. 16 cm2 D. 17 cm2","og_description":"[amp_mcq option1=&#8221;14 cm2&#8243; option2=&#8221;15 cm2&#8243; option3=&#8221;16 cm2&#8243; option4=&#8221;17 cm2&#8243; correct=&#8221;option1&#8243;]","og_url":"https:\/\/exam.pscnotes.com\/mcq\/a-simply-supported-beam-6-m-long-and-of-effective-depth-50-cm-carries-a-uniformly-distributed-load-2400-kg-m-including-its-self-weight-if-the-lever-arm-factor-is-0-85-and-permissible-tensile-stress\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2024-04-15T02:20:04+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"1 minute"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-simply-supported-beam-6-m-long-and-of-effective-depth-50-cm-carries-a-uniformly-distributed-load-2400-kg-m-including-its-self-weight-if-the-lever-arm-factor-is-0-85-and-permissible-tensile-stress\/","url":"https:\/\/exam.pscnotes.com\/mcq\/a-simply-supported-beam-6-m-long-and-of-effective-depth-50-cm-carries-a-uniformly-distributed-load-2400-kg-m-including-its-self-weight-if-the-lever-arm-factor-is-0-85-and-permissible-tensile-stress\/","name":"A simply supported beam 6 m long and of effective depth 50 cm, carries a uniformly distributed load 2400 kg\/m including its self weight. 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