chunk_14.json•1.56 kB
{
"id": "chunk_14",
"text": "Kips. per Cubic Foot. Substituting values we get P\na \n= \n \n \n (0.115) (26)\n2\n (0.294) = 11.43 \nKips/Foot *** A strip of 1 foot is considered. \nP\na- horizontal\n \n= P\na\n (COS B) = 11.43(COS 10 = 11.25 Kips per Foot from code wide beam shear is \ngiven as v\nc\n= 2 f\nc\n’ = 2(0.85) (3000) = 0.09311 in K.S.I. WITH A LOAD FACTOR OF 1.8 \nP\na- horizontal\n = 1.8(11.25) The shear carried by concrete at junction is Vc = (v\nc\n) (t) \n(12) equating Vc to P\na- horizontal \nwe get P\nah\n = Vc 0.09311(12) (t) = 1.8(11.25) \nsolving for t = \n \n \n \n \n \n \n \n \n = 18.14” Allow covering of 3.5” then T at junction is 18.14” + 3.5” \n= 21.6” with a slope batter of 1/4” per foot then thickness at top of wall is t = t\njunction\n – h (0.25) \n= 21. 6 – 26(0.25) T\ntop\n = 15.1 Inches use t = 16” to maintain even dimensions let us use t = \n16 inches + 26((0.25) = 22.5” use 23” \nStep 2 Compute overturning and sliding stability of wall \nFrom figure it is evident that H’ = H + 2.42 + 9.5(Tan 10 ) = 30.1 Feet \nP\nah’\n = 1/2(y)(H’)\n2\n(K\na",
"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": 14,
"line_start": 552,
"line_end": 606,
"has_formulas": false,
"has_tables": false
}
}