chunk_19.json•1.51 kB
{
"id": "chunk_19",
"text": "= 0.84 less than 0.093 okay *** Note \nwe could reduce the base slab shear stress by about 1 to 1.5” \n \nStep 6 Compute steel reinforcement ratio, using authors derived \nformula ***see chapter three P = \n \n \n (1 ± 1 – 2.622Mu/(bd\n2\nFc’)) \nwith Mu = 107.2(12)(1.8) = 2315.52 Inch Kips and b = 12” ( 1 foot strip length) and \nd = 25.5 inches evaluating the terms separately 2.622 x 2315.52/(12(25.5)\n2\n(3) ) = \n0.2582691 and \n0.847f\nc\n’/f\ny\n = \n \n \n = 0.042235 Substituting values and taking the smaller \nnegative sign we get p = 0.042235 x (1 – 0.861237) = 0.042235(0.138763) = \n0.0058606 \nbut from code minimum is p = 200’/f\ny = \n 200/60000 = 0.0033 hence actual steel \nreinforcement ratio governs use p = 0.00586 Then area of steel As\nheel\n = p(b)(d) = \n0.00586(12)(25.5) = 1.79 in\n2\n/ft \nFor steel reinforcement in toe, let us use authors derived formula *** \nRefer to chapter 4 \n \n As = \n \n \n (b)(d) ± (0.7225Fc’\n2\nb\n2\nd\n2\n)/Fy\n2\n – 1.888Fc’(b)(Mu)/Fy\n2\n))",
"metadata": {
"book_id": 34880,
"title": "BASIC-STRUCTURAL-388457483-Retaining-Wall-Design-Analytical-and-Computer-Methods-By-Ben-David-CE",
"authors": "Ben David",
"project": "basic_structural",
"content_source": "ocr",
"content_length": 39295,
"chunk_index": 19,
"line_start": 823,
"line_end": 879,
"has_formulas": false,
"has_tables": false
}
}