{"id":259539,"date":"2025-07-19T03:16:24","date_gmt":"2025-07-19T03:16:24","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=259539"},"modified":"2025-07-19T03:16:26","modified_gmt":"2025-07-19T03:16:26","slug":"the-shaft-is-supported-by-three-smooth-journal-bearings-at-a-b-and-c-assume-f-440-n","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/19\/the-shaft-is-supported-by-three-smooth-journal-bearings-at-a-b-and-c-assume-f-440-n\/","title":{"rendered":"The shaft is supported by three smooth journal bearings at A, B, and C. Assume F = 440 N."},"content":{"rendered":"\n
The shaft is supported by three smooth journal bearings at A, B, and C. Assume F = 440 N.<\/pre>\n\n\n\n
\"\"<\/figure>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
Here is the step-by-step solution to the problem.<\/p>\n\n\n\n
System Analysis and Free Body Diagram<\/h3>\n\n\n\n
First, we define a coordinate system with the shaft aligned along the x-axis, and the origin at bearing A. The forces and supports are located as follows:<\/p>\n\n\n\n
    \n
  • Supports (Smooth Journal Bearings):<\/strong>\n
      \n
    • At A (0, 0, 0): Reaction forces R_A = A_y\u00a0j<\/strong>\u00a0+ A_z\u00a0k<\/strong><\/li>\n\n\n\n
    • At B (0.9, 0, 0): Reaction forces R_B = B_y\u00a0j<\/strong>\u00a0+ B_z\u00a0k<\/strong><\/li>\n\n\n\n
    • At C (2.7, 0, 0): Reaction forces R_C = C_y\u00a0j<\/strong>\u00a0+ C_z\u00a0k<\/strong><\/li>\n\n\n\n
    • Smooth journal bearings do not exert forces along the shaft axis (x-axis) or resist moments.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
    • Applied Forces:<\/strong>\n
        \n
      • F\u2081<\/strong>: -500\u00a0k<\/strong>\u00a0N at position\u00a0r\u2081<\/strong>\u00a0= (0, 0.9, 0) m<\/li>\n\n\n\n
      • F\u2082<\/strong>: -600\u00a0k<\/strong>\u00a0N at position\u00a0r\u2082<\/strong>\u00a0= (0.9, 0, 0) m<\/li>\n\n\n\n
      • F<\/strong>: 440\u00a0k<\/strong>\u00a0N at position\u00a0r_F<\/strong>\u00a0= (1.5, 0, 0) m<\/li>\n\n\n\n
      • F\u2083<\/strong>: -900\u00a0k<\/strong>\u00a0N at position\u00a0r\u2083<\/strong>\u00a0= (1.8, -0.9, 0) m<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n
        Equilibrium Equations and Indeterminacy<\/h3>\n\n\n\n
        For the shaft to be in static equilibrium, the sum of all forces and moments must be zero.
        \u03a3F<\/strong> = 0 and \u03a3M<\/strong> = 0.<\/p>\n\n\n\n
        This problem is statically indeterminate as stated because there are six unknown reaction components (A_y, A_z, B_y, B_z, C_y, C_z) but not enough independent equilibrium equations. Furthermore, there is a net external torque about the x-axis (\u03a3M_x = (-500 N)(0.9 m) + (-900 N)(-0.9 m) = -450 + 810 = 360 N\u00b7m), which violates static equilibrium.<\/p>\n\n\n\n
        To solve this problem, we must make a common simplifying assumption for such textbook exercises: the central bearing at B acts like a hinge, meaning the bending moment at that point is zero.<\/strong> We also proceed by ignoring the torsional imbalance about the x-axis and focusing on the bending effects.<\/p>\n\n\n\n
        Solving for Reactions<\/h3>\n\n\n\n
        1. Reactions in the y-direction (and moments about the z-axis):<\/strong><\/p>\n\n\n\n
        There are no applied forces in the y-direction. The moments about the z-axis are only created by the y-component of the reaction forces.
        Setting the bending moment M_z at the “hinge” B to zero:<\/p>\n\n\n\n
          \n
        • Considering segment AB: \u03a3M_z at B = -A_y * (0.9 m) = 0 =>\u00a0A_y = 0 N<\/strong>.<\/li>\n\n\n\n
        • Considering segment BC: \u03a3M_z at B = C_y * (2.7 m – 0.9 m) = C_y * (1.8 m) = 0 =>\u00a0C_y = 0 N<\/strong>.
          From the overall force equilibrium in the y-direction:
          \u03a3F_y = A_y + B_y + C_y = 0 => 0 + B_y + 0 = 0 =>\u00a0B_y = 0 N<\/strong>.<\/li>\n<\/ul>\n\n\n\n
          2. Reactions in the z-direction (and moments about the y-axis):<\/strong><\/p>\n\n\n\n
          We apply the “hinge” assumption by setting the bending moment M_y at B to zero.<\/p>\n\n\n\n
            \n
          • Find C_z by analyzing segment BC:<\/strong>
            Sum the moments about point B for the forces acting on segment BC:
            \u03a3M_y at B (right side) = (C_z * 1.8) + (440 N * 0.6 m) + (-900 N * 0.9 m) = 0
            1.8 C_z + 264 – 810 = 0
            1.8 C_z = 546
            C_z = 303.33 N<\/strong><\/li>\n\n\n\n
          • Find A_z by analyzing segment AB:<\/strong>
            Sum the moments about point B for the forces acting on segment AB:
            \u03a3M_y at B (left side) = (A_z * -0.9 m) + (-500 N * -0.9 m) + (-600 N * 0 m) = 0
            -0.9 A_z + 450 = 0
            0.9 A_z = 450
            A_z = 500 N<\/strong><\/li>\n\n\n\n
          • Find B_z using overall force equilibrium:<\/strong>
            Sum all forces in the z-direction:
            \u03a3F_z = A_z + B_z + C_z + F_\u2082 + F + F_\u2083 = 0 (Note: F\u2081 is at x=0, its load is taken by A_z). Let’s use all forces for clarity.
            \u03a3F_z = A_z + B_z + C_z – 500 N – 600 N + 440 N – 900 N = 0
            (500) + B_z + (303.33) – 1560 = 0
            B_z – 756.67 = 0
            B_z = 756.67 N<\/strong><\/li>\n<\/ul>\n\n\n\n
            Final Answers<\/h3>\n\n\n\n
            The calculated reaction components, rounded to three significant figures, are:<\/p>\n\n\n\n
              \n
            • Part A: Components of reaction at bearing C<\/strong>
              C_y = 0 N
              C_z = 303 N
              Answer: 0, 303<\/strong><\/li>\n\n\n\n
            • Part B: Components of reaction at bearing A<\/strong>
              A_y = 0 N
              A_z = 500 N
              Answer: 0, 500<\/strong><\/li>\n\n\n\n
            • Part C: Components of reaction at bearing B<\/strong>
              B_y = 0 N
              B_z = 757 N
              Answer: 0, 757<\/strong><\/li>\n<\/ul>\n\n\n\n
              \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"
              The shaft is supported by three smooth journal bearings at A, B, and C. Assume F = 440 N. The Correct Answer and Explanation is: Here is the step-by-step solution to the problem. System Analysis and Free Body Diagram First, we define a coordinate system with the shaft aligned along the x-axis, and the origin […]<\/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-259539","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/259539","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=259539"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/259539\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=259539"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=259539"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=259539"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}