Question

A=4 B=7

1) For the foundation sitting on top of the 5m thick soil calculate dimension. (Factor of Safety:3 (30p) (AX100) KN (B x3) kNm SPT-N (A+2B) 5 m b XD Uniform sand JIKUM) ya 15 kN/m a. Determine the q (angle of internal friction angle) of sand by using the given formula. b. Find the dimension b based on bearing capacity c. Find settlement by using 2:1 rule (DO NOT CHANGE DIMENSION AFTER SETTLEMENT Calculation)

Answer 1

Solution - 1 Given deeta A=4; B=7 &sand= 15 kN/m3 O=Ax100 O= 400KN M= BX3=\743 M=21 KN/m N= A+2B = 4+2=) N = 18 (Spr) es => 0.0525m 21 HOO = 15.4 *NDO 4X(N)05 420 = 15.4x (1875420 Ø = 85.33° Ny=0. = 0·018 x 82.25 = 0·018 X (85.33)2-25 Ny=398.33 c=0, Df=0 for square footing, y Vu = 1:3CN +D Dg Nq +0.48 BNy oto+0.4x15xbx398-33 Vu - 2389.986 400 6x0.0525) be B + BL b 00 126 &max +

factor of tafely Vu &max Vu= a max fois 126 2389.98b- [4.00 X (3) 62 t 378 + - 2389.98b=0 b² 1200b+378-2389,9864-0 b=1.7821m Mex If Se VB EG [(1-2) q= at sm below foundation belowby 2:1 nule Es= SOO(N415) UBL Lez (Brz)(L+Z) = 500(1875) Es = 16.500 Here Vu=2389.98 x 1.7811 => 4259.18 kpa dą= (1259.18)(1-7821) ? => 294.07kpa (1-782145) La = 112@ centre) & M = 0.3&t cand) 4259.18 X 1.7821X112 (1-032) Se 17821+5) (1:12+5) Se = 1860377mm