{"id":265275,"date":"2025-07-21T20:01:51","date_gmt":"2025-07-21T20:01:51","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=265275"},"modified":"2025-07-21T20:01:53","modified_gmt":"2025-07-21T20:01:53","slug":"for-the-following-w-21x62-beam-determine-the-normal-stress-at-point-a-and-on-the-neutural-axis-on-a-a-cross-section-which-is-in-the-midlle-of-length-cd","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/21\/for-the-following-w-21x62-beam-determine-the-normal-stress-at-point-a-and-on-the-neutural-axis-on-a-a-cross-section-which-is-in-the-midlle-of-length-cd\/","title":{"rendered":"For the following W 21&#215;62 beam, determine the normal stress at point A and on the neutural axis on a-a cross section which is in the midlle of length CD"},"content":{"rendered":"\n<p>For the following W 21&#215;62 beam, determine the normal stress at point A and on the neutural axis on a-a cross section which is in the midlle of length CD. Length AC = 9 ft, CD = 11 ft, and DB = 5 ft. The dimensions for the W 21&#215;62 beam are in inches: tf = .615, tw = .4, bf = 8.24, d = 21. Supports at C and D are pin support and rollar support, respectively. (30 points)<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/image-631.png\" alt=\"\" class=\"wp-image-265279\"\/><\/figure>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Here are the calculated normal stresses for the specified points on the W 21&#215;62 beam.<\/p>\n\n\n\n<p><strong>Final Answer:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The normal stress at Point A (top fiber) is\u00a0<strong>-30.2 ksi (Compression)<\/strong>.<\/li>\n\n\n\n<li>The normal stress on the neutral axis is\u00a0<strong>0.554 ksi (Tension)<\/strong>.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation:<\/strong><\/h3>\n\n\n\n<p>The total normal stress (\u03c3) at any point on the cross-section is the superposition of the stress due to the axial force (P) and the stress due to the bending moment (M). The governing equation is:<\/p>\n\n\n\n<p><strong>\u03c3 = P\/A &#8211; My\/I<\/strong><\/p>\n\n\n\n<p>Here is the step-by-step determination of the stresses:<\/p>\n\n\n\n<p><strong>1. Calculate Support Reactions<\/strong><\/p>\n\n\n\n<p>First, we find the reactions at the pin support C and roller support D using static equilibrium equations on the free-body diagram of the entire beam.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Sum of moments about C = 0:\n<ul class=\"wp-block-list\">\n<li>(20 kips * 9 ft) + D_y * (11 ft) &#8211; (20 kips * 16 ft) = 0<\/li>\n\n\n\n<li>11 * D_y = 180 kip-ft + 320 kip-ft = 500 kip-ft<\/li>\n\n\n\n<li><strong>D_y = 45.45 kips<\/strong>\u00a0(upward)<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Sum of horizontal forces = 0:\n<ul class=\"wp-block-list\">\n<li>C_x + 10 kips = 0 =>\u00a0<strong>C_x = -10 kips<\/strong>\u00a0(acting to the left)<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Sum of vertical forces = 0:\n<ul class=\"wp-block-list\">\n<li>C_y + D_y &#8211; 20 kips &#8211; 20 kips = 0<\/li>\n\n\n\n<li>C_y + 45.45 kips &#8211; 40 kips = 0 =>\u00a0<strong>C_y = -5.45 kips<\/strong>\u00a0(acting downward)<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<p><strong>2. Determine Internal Forces at Cross-Section a-a<\/strong><\/p>\n\n\n\n<p>The cross-section a-a is in the middle of CD, which is 5.5 ft from C (or 14.5 ft from the left end A). We make a cut at this section and analyze the left portion.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Axial Force (P):<\/strong>\u00a0The internal axial force P must balance the external horizontal reaction C_x.\n<ul class=\"wp-block-list\">\n<li>P + C_x = 0 => P &#8211; 10 kips = 0 =>\u00a0<strong>P = 10 kips<\/strong>\u00a0(Tension)<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Bending Moment (M):<\/strong>\u00a0Summing moments about the cut (at x = 14.5 ft):\n<ul class=\"wp-block-list\">\n<li>M &#8211; (20 kips * 14.5 ft) &#8211; (5.45 kips * 5.5 ft) = 0<\/li>\n\n\n\n<li>M = 290 kip-ft + 29.975 kip-ft = 319.975 kip-ft<\/li>\n\n\n\n<li><strong>M \u2248 320 kip-ft<\/strong>\u00a0(This positive moment causes compression at the top and tension at the bottom).<\/li>\n\n\n\n<li>Convert to kip-inches: M = 320 kip-ft * 12 in\/ft =\u00a0<strong>3840 kip-in<\/strong>.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<p><strong>3. Calculate Cross-Sectional Properties<\/strong><\/p>\n\n\n\n<p>Using the given dimensions for the W 21&#215;62 beam (d=21, b_f=8.24, t_f=0.615, t_w=0.4, all in inches):<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Area (A):<\/strong><br>A = 2*(b_f * t_f) + (d &#8211; 2<em>t_f)t_w = 2(8.24<\/em>0.615) + (21 &#8211; 2*0.615)*0.4 =\u00a0<strong>18.04 in\u00b2<\/strong><\/li>\n\n\n\n<li><strong>Moment of Inertia (I):<\/strong><br>I = I_web + 2*(I_flange_centroid + A_flange * d_y\u00b2)<br>I = [ (1\/12)<em>0.4<\/em>(19.77)\u00b3 ] + 2*[ (1\/12)<em>8.24<\/em>(0.615)\u00b3 + (8.24<em>0.615)<\/em>(10.1925)\u00b2 ]<br>I = 257.6 + 2*(0.16 + 526.5) =\u00a0<strong>1310.8 in\u2074<\/strong><\/li>\n<\/ul>\n\n\n\n<p><strong>4. Calculate Normal Stresses<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Stress at Point A (Top Fiber):<\/strong><br>Point A is at the top of the beam, so y = d\/2 = +10.5 in.<br>\u03c3_A = P\/A &#8211; My\/I = (10 kips \/ 18.04 in\u00b2) &#8211; (3840 kip-in * 10.5 in \/ 1310.8 in\u2074)<br>\u03c3_A = 0.554 ksi &#8211; 30.76 ksi =\u00a0<strong>-30.21 ksi (Compression)<\/strong><\/li>\n\n\n\n<li><strong>Stress at the Neutral Axis:<\/strong><br>At the neutral axis, the distance y = 0. Therefore, the bending stress term is zero.<br>\u03c3_NA = P\/A &#8211; M(0)\/I = P\/A = 10 kips \/ 18.04 in\u00b2<br>\u03c3_NA =\u00a0<strong>+0.554 ksi (Tension)<\/strong><\/li>\n<\/ul>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner6-1443.jpeg\" alt=\"\" class=\"wp-image-265284\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>For the following W 21&#215;62 beam, determine the normal stress at point A and on the neutural axis on a-a cross section which is in the midlle of length CD. Length AC = 9 ft, CD = 11 ft, and DB = 5 ft. The dimensions for the W 21&#215;62 beam are in inches: tf [&hellip;]<\/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-265275","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/265275","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=265275"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/265275\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=265275"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=265275"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=265275"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}