chunk_16.json•1.55 kB
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"text": "= tan\n2\n(45 + \n \n \n ) = 3.255 The friction coefficient factor F\nr\n = R tan ’ + cB’ + P\np\n here \nthe limiting value of c’ is 0.50c to 0.75c here c = 0.40 given see datas problem with R = 43.5 \nsubstituting values we get Fr = 43.5 tan 32 + 0.67(0.4)(14.42) = 31 Kips From Bowles \nreference text book page 438 “Foundation Analysis and Design” P\np\n = 1/2YH\n2\nK\np\n + 2cHK\np\n \nsee drawing here H is the depth of soil from surface of toe up to soil surface with K\np\n = 3.255 & \n1.804 respectively substituting values in the above equation we get \nP\np = \n0.50(0.112)(3)\n2\n(3.25) + 2(0.4)(3)(1.804) = 6 kips \n \nS\nummation \nS\nFr\n \n= P\np\n + F\nr \n= 31 + 6 = 37 Kips The resulting F = \n \n \n = 37/15.1 = 2.45 \ngreater than 1.5 (i.e recommended o kay) \nNow locate resultant = Summation weight on base and the eccentricity. Taking moments at \ntoe we have \n \n = \n \n \n = \n \n \n = 5.37 ‘ Compute eccentricity e \ne = B/2 - \n \n 14.42/2 – 5.37 = 1.84 Feet less than L/6 recommended for eccentricity okay.",
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"book_id": 34880,
"title": "BASIC-STRUCTURAL-388457483-Retaining-Wall-Design-Analytical-and-Computer-Methods-By-Ben-David-CE",
"authors": "Ben David",
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