chunk_18.json•1.47 kB
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"text": "= 3.02 (1 ) = 5.3 ksf maximum at toe q = 0.7 ksf \nmax at heel. \nStep 4 Compute base slab shear and bending moments toe and heel. \nFor toe at stem face x = 3 feet. slope pressure is q = 5.3 – 0.36 – 0..32X. Neglecting \nsoil over toe and integrating the pressure diagram we get the shear at any section \nV = 4.94X – 0.32X\n2\n/2 = 13.4 Kips Integrating the shear diagram we have the moment at any \nsection M = 4.94 \n \n dx - .32/2 \n \n \n2\n dx = 4.94/2X\n2\n – 0.32/6X\n3\n = 20.8 Foot \nkips \n \nFor heel at approximately CG of tension steel X = 9.5 + \n \n \n = 9.79 Feet for moment , use 9.5 \nfeet for shear. \nUse average height of soil on heel for downward pressure includes Pav = 2.6 kips, the \npressure q is = 3.45 - .70 -.32X. The shear at any section X is V = 2.75X – \n \n \n X\n2\n + \nPavX \nM\nheel\n \n= 107.2 Foot kips \n \n\n253 \n \nStep 5. Check base slab shear using largest V, with LF = 1.8 and d = 2.417 – 0.29 = 2.13 feet \nActual shear stress V = \n \n \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",
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