Question

(a) A shallow foundation 20 m x 16 m carries a uniform pressure of 200 kN/m²at one quarter of the foundation and 100 kN/m² at the rest of foundation. Determine the vertical stress at a point 10 m below the mid-point of one of the longer sides (i) using influence factors, (5 marks) (ii) by means of Newmark's chart (please submit the Newmark chart). (5 marks) q = 200 kN/m2 L = 20 m b = 10 m q = 100 kN/m z = 10 ml 0.28 4-mz 0.26 nz a 0.24 2.0 -T Z 11.4 0.221.1 1.0 0.20 0,- al 10.8 0.18 0.16 0.6 10.14 0.5 0.12 0.4 0.10 0.3 0.08 0.2 0.06 0.04 0.02 0 0.1 10 m

Answer 1

ten Tom a, below the Mid point of longer sides į Divide the plot into three parts lom, 7. K 8m lomat om 8m ช |(397687412 Z 12.2576 m za 10m Y 1:22576 m hoad convert UDL to point &Q=(200) (87087 = 180ook N 3 WikiTu, Iga सा ( 1413,7% NOW 3 In = 2 &T [ 1+ (1.22.57) J5/2 in 0.048196 Hoz = (16)(Q) = (0.048196) (16000) = 7-71136 kPa 근르 100 I Q = (100))(0) Klom 8m IE = 8000 KN & IB a lom 5/2 at [1+ 100 50.2022 r=[5²+42 r ry (0-2022) (8000) 100 = 16.176 kPa CS Scanned with CamScanner

ו סן 2 11 8 + 5 k 89 Pent 16m rula 89 10 lom 3 » IB गा [1+/0-8977587 0097226 » Oz = 10.897226) (100 xID X16) = 15.556 kPa Cu 100 f +620 wy rz☺ oz - =39.4436 kPa 7.711367 16.176 + 15:556 ü Newmark's chart re-drawn as :- Given plot is lom lom Е — Th 11 ft 10 m kt a look 16m ܕ ܛܝ 16M lookra + at t 200 8m Pey > 1 'lom rom lom 22 I CS Scanned with CamScanner

be used when Note :- Here ,Newmark's chart can the loading loading determing point is at end kom edok Ru 10m Z 10 1.6 n = 16 10 lom From Newmaiks Chast (giveu), Ig ~ 0195 » Oz (27) q 101195)(200) = 39 kPa rom K - + Tookla 8 8m. у 22 dvd ma = 0.8 10 in- ) 10 DU 10 m From Newmark's chart, I, 20:16 12 = (5,719) = 10:16) (100) = 16 kPa CS Scanned with CamScanner

nom Tool old 10 T = 16m ܐ ܐ ✓ nz 16 70 16 - From Newmark's chart, Iy ~0.195 z = (I819 = 10195)(100) = 19.5 kPa irz: 67 20+ 39 -16 +19.5 42.5 kPa Note: - Minute Intardation difference depicts value at two methods. in og differena CS Scanned with CamScanner