chunk_20.json•1.59 kB
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"text": "2\n – 1.888Fc’(b)(Mu)/Fy\n2\n)) \n \nwith Mu = 20.8 Using a load factor of 1.8 Mu\ntoe\n = 20.8(12)(1.8) = 449.28 Inch kips with d = \n25.5” and b = 12” substituting values and evaluating the mathematical expressions one by \none. .7225f\nc\n’b\n2\nd\n2\n/fy\n2\n = 0.7225(3)\n2\n(12)\n2\n(25.5)\n2\n/(60)\n2\n = 169.1300 \n\n254 \n \n0.85f\nc\n’bd/f\ny\n = \n \n \n = 13.005 \n1.88888f\nc\n’bMu/f\ny\n2 = \n \n \n \n \n \n = \n8.48636 \nSubstituting numerical values in the above equation we get \n13.05 \n \n 13.05 \n \n = 13.05 \n12.674 taking the least value we further get As = 13.005 - 12.674 = o.331 in\n2\n/ft. \nCheck actual steel reinforcement from code allowable \np = As/bd = \n \n \n = 0.0018 less than P minimum from code Therefore p \nminimum from code governs = 0.0033 \nRequired area of steel at toe = As\ntoe\n = m\ninimum\n (b)(d) = 0.0033(12)(25.5) = 1.02 in\n2\n/feet \n \n***Note author by computer use the mathematical expression below *** See \nchapter 4 \n \n*** Note given d, b and Mu depth of stress rectangular block is",
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"title": "BASIC-STRUCTURAL-388457483-Retaining-Wall-Design-Analytical-and-Computer-Methods-By-Ben-David-CE",
"authors": "Ben David",
"project": "basic_structural",
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