[amp_mcq option1=”$$\\frac{{{\\text{WL}}}}{{\\text{P}}}$$” option2=”$$\\frac{{{\\text{WL}}}}{{2{\\text{P}}}}$$” option3=”$$\\frac{{{\\text{WL}}}}{{3{\\text{P}}}}$$” option4=”$$\\frac{{{\\text{WL}}}}{{4{\\text{P}}}}$$” correct=”option4″]<\/p>\n
The correct answer is $\\frac{{{\\text{WL}}}}{{4{\\text{P}}}}$.<\/p>\n
The central dip $h$ of a bent tendon is the vertical distance between the tendon and the horizontal line passing through the supports. The tendon is required to balance a concentrated load $W$ at the center of the span $L$.<\/p>\n
The bending moment at any point on the tendon is given by $M = Wx$, where $x$ is the distance from the point to the center of the span. The maximum bending moment occurs at the center of the span, where $x = L\/2$. Therefore, the maximum bending moment is $M = WL\/2$.<\/p>\n
The stress in the tendon is given by $\\sigma = \\frac{M}{I}$, where $I$ is the moment of inertia of the tendon. The moment of inertia of a bent tendon is given by $I = \\frac{bh^3}{12}$, where $b$ is the width of the tendon and $h$ is the central dip.<\/p>\n
The maximum stress in the tendon occurs at the center of the span, where $x = L\/2$. Therefore, the maximum stress in the tendon is $\\sigma = \\frac{WL}{2} \\cdot \\frac{bh^3}{12}$.<\/p>\n
The tendon is safe if the maximum stress in the tendon is less than the yield stress of the material. The yield stress of the material is denoted by $\\sigma_y$. Therefore, the tendon is safe if $\\frac{WL}{2} \\cdot \\frac{bh^3}{12} \\leq \\sigma_y$.<\/p>\n
Solving for $h$, we get $h \\geq \\frac{2WL}{{3\\sigma_y b}}$.<\/p>\n
The load $W$ is given in the question. The yield stress $\\sigma_y$ and the width $b$ of the tendon are known from the design of the tendon. Therefore, we can calculate the minimum central dip $h$ required to balance the concentrated load $W$ at the center of the span $L$.<\/p>\n
The other options are incorrect because they do not satisfy the condition $\\frac{WL}{2} \\cdot \\frac{bh^3}{12} \\leq \\sigma_y$.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”$$\\frac{{{\\text{WL}}}}{{\\text{P}}}$$” option2=”$$\\frac{{{\\text{WL}}}}{{2{\\text{P}}}}$$” option3=”$$\\frac{{{\\text{WL}}}}{{3{\\text{P}}}}$$” option4=”$$\\frac{{{\\text{WL}}}}{{4{\\text{P}}}}$$” correct=”option4″]<\/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-5713","post","type-post","status-publish","format-standard","hentry","category-rcc-structures-design","no-featured-image-padding"],"yoast_head":"\n