Question

3. A soil boring was drilled in a multi-layered deposit, as shown in the boring log below. A SPT was also performed at intervals of 1 m. A donut hammer and a liner sampler were used. Care was taken to connect the rod segments firmly and to follow standard procedure. a. Find the corrected Noo at the depths of 9 and 27 feet. b. What is the (N;) at the depth of 13 feet? c. Estimate the drained friction angle obtained from soil sample 9. d. Estimate the undrained shear strength at the depth of 11 feet. BORING LOG Depth in Water table USCS Graphic Water levels z During drilling After completion Descriptions GRAVEL Samples SPT results Remarks 11.6.6 Dark brown and gray moist, staff to very stiff, SANDY CLAY with trace to little gravel 12 12.18.12 12.10.10 Dark gray and brown, slightly moist, very stiff, SILTY CLAY with trace to little sand 9.11.10 5.5.2 Dark, moist, stiff, SANDY CLAY with trace to small gravel Dark brown, very moist to wet, loose CLAYEY SAND with trace to gravel 3.3.3 2.2.3 Dark brown, very moist to wet, loose SILTY SAND with trace to gravel and clay 2.2.2 Boring caved to 21 feet upon auger removal 2.2.4 SC Dark brown, wet, medium dense, CLAYEY SAND with little to some gravel and trace silt Boring terminated at 30 feet Ground water was not encountered during drilling or upon completion

Answer 1

Idution : (a) To Calculate Noo, first we will Calcolate Navg. Naug at gft- 91111010, Naug at 29 ft- 6+6+6 6. Now Noo at g - Em. Cs Cs Ca Narg 0.6 Where Ema 0.6, efficieny = 0.681x1x di&lo of high efficieny Us style - dojnut Safety Lammer Noo at g = 8.5 Assumed CCB = 1 (Bose hole dia Correction) Noo at 27'2 0.6x1x1 X0.85x6 C-1 (Sampler Cosceticu) (p=0.85 Rod Length Corretion 0.6 sil (6) (N.) 6o at 13 feet depth (+)60 = Noo 2000 (blfth | Eleptn) Ceffective Unit Weight) at 13 depta Nauga 5+5+7= 5.67, Noo= 0.6X1X1X0.8585:67 0.6 No=4.816 = (N.) 60 = 4.816 [2000 v (13) (127-62.47 pcp) = Assumed (Not given) 71432 - 7 =) (N.) 60 at 13 ft = 7

0.6 (1,60 = 1.7 rom (C) Doained friction angle from Soil Sample I SPT results are & Narg = 2+2+2 = 2 (Nico = 0.6X1X1X0:2542 As per Peck, hanson (1974) O'- 27.11° +0.3% (N) 60 - 0100054(14) Go o= 27,6 Aug @ from Soapon blw Cu BN, Cu value Coresponding to (Narg = 5+5+7)=5167 (4= 30 (Appoox.) N60 = 0.6X/X/X0.85x5.67 =hiblar of Su 015qu or 1:6 Nfield +15,4 Sur 41.6 (5.67)+154 = 24,472 Ich or Su= 2.1 (Noo)+17.6 - 2.1(4,8195) +17.6. su= 27.72 klach