Question

in part b, how is h=4 l = 10 b=8 and d=4 ??

Question Nine: For the flexible loaded area (5 mx 7 m) shown below.calculate: (a) The average consolidation settlement beneath the footing given that the load spreads evenly through the clay at a 2:1 ratio. For your calculations, divide the appropriate layer into two equal sub-layers. (b) The average immediate settlement beneath the footing given that the load spreads evenly through the day at a 2:1 ratio. 4800 KN Incompressible Drainage Layer Shim Normaly Consolidated Clay Soi Properties 16 km 1.5m 17 kN/m2 Poison's Ratio - 0.5 Void Ratio = 0.89 C = 0.07 1.5 Soi Properties Medium Dense Sand Tu - 18.7 km Poison's Ratio = 0.5 E = 30MPa Bedrock (7 + 7 = 14 marks) 8c = the log C Layer Hilm) OlkPa) 40 (kPa) 1.5 x 16+ 0.75 (17-10) = 21.25 48/5.5x7.75)= 107. TI 2 1.5 1x1b+2.25A(17-10) = 31.75 480/725x9.25) = 71.58 dc = $c Scze 1.5x0.07x log 21.25 +1077 +10g/31.75 11:58 +0.89 T 21.25 J T 31.75 Sc = 43 + 28.5 = 71.5 mm b) H-4m, L=10 m, B = 8 m, D=4m - (480x6 = 60 kPa Hg = 4/8 =0.5, lo = 1/8 = 1.25 l = 4/8=0.5 i = 0.3 do -0.852 Si = Moli qB = 0.852x0.3x60x8 = 4.09 mm 30,000

Answer 1

Please look through the attached image and rate if of help.

In Case of Saturated clays immediate settlement is insignificant Immediate Settlement here, with to medium dense sand only due here, will be As per Centre of 4800kN Trapezoidal Methad, increased stress at Elay Medium dense sand due to loos. (6+2 m2) (2+2 m2) doad spreads enenly in the deatis 2:1 or 1: 0.5 3 Comparing with lim, m=0.5 B+2m2 = 3 + 2x0.5x5 = 8m. 5+ 2x0.5x5 = com. L+2 m2 = Hm. Ho 0 = Thickness of medium dense sand, Repth ot which bense fand dayes start = 4m