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Unformatted text preview: a a w McGill University ' “in ' ' K5}; Faculty of Engineerlng Department Of Civil Engineering and Applied Mechanics Geotechnical Engineering CIVE 416 SEC 001 FINAL EXAMINATION Friday, April 20, 2007 — 2:00 am to 5:00 pm Examiner: M.A. Meguid Associat in : . Mirza Signature: @194!) J Signature: i 71, / / STUDENT NAME: I MCGILL ID. NUMBER: INSTRUCTIONS: 1. The examination is CLOSED BOOK. No notes or books are permitted. Only the attached three page crib sheets are allowed. 2. Keep your answers brief, preferably in the space provided. If you need additional space, use the reverse side of the page, clearly indicating the question number. 3. Illustrate your answers with neat sketches whenever possible. ‘ 4. If doubt exists as to the interpretation of any question, the student is urged to provide a clear statement of any assumptions made. 5. This examination consists of 10 pages, including the cover page. ALL EXAMINATION SHEETS MUST BE RETURNED Page 1 Question 1 [45 minutes, 20 marks] A retaining wall has a vertical drainage blanket as shown in Figure l. The soil is anisotropic with kh = 4 kV = 10'6 m/s. For the condition of 2D steady-state seepage i) [3 marks] Identify and clearly label ALL boundary conditions. ii) [7 marks] Draw a flow net with a minimum of five equipotential lines. Clearly label the total head for each equipotential line. iii) [5 marks] Calculate the total flow into the drainage pipe. iv) [5 marks] Determine the magnitude and location of the pore pressure at point A. ' 14333-55 2'. ‘:.‘:'.t'.4'.-.‘."l='l".w‘.i':'.1‘:'.= .".-. . ".' -.-:w‘.f:-‘.a".-:ar.\'azcr.a':-:aLoam-3.za‘acazx'2-zsaazc-x §E§£*€\'J:?:-.-' ”afllfiifi; ii? 7533:»:- 94 5m .,:, vi J‘y‘.. . .fiiflfiifi» '3 3* . :2}* “I‘d 5”. Drainage ipe Imperrneable layer ‘ Figure 1 iii) iv) CIVE 416 Page 2 Question 2 [45 minutes, 30 marks] (a) [8 marks] Using the lower bound solution, derive an expression for the ultimate bearing capacity of a strip footing of width B supported on purely cohesive material at a depth D below ground surface. (b) [22 marks] A new chimney structure for cement works will be constructed of slip-formed concrete and will weigh approximately 200 kN/m height. If it is built on a circular pad of reinforced concrete, 10.0 m in diameter and 1.0 m thick founded on a deep silty clay deposit 2 m below ground surface. (i) Determine the short-term factors of safety for a 45.0 m tall chimney using the General bearing capacity equation. ~ WWWWW/a 321602 C = . Normally consolidated Y = 20 kN/III3 silty clay deposit Cu = 50 kPa CIVE 416 Page 3 (ii) Calculate the expected consolidation settlement due to the proposed structure. Question 3 [30 minutes, 20 marks] (a) [6 marks] Derive an expression for the stability factor of safety of a finite slope in purely cohesive soil assuming a circular failure surface. (b) [14 marks] A 3:1 cut slope 6 meters deep failed immediately after construction. The material consisted of 10 meters of soft clay over shale. Calculate the undrained strength of the clay when it failed, assuming that its unit weight is 17 kN/m3. CIVE 416 Page 5 Question 4 [45 minutes, 30 marks] (a) [5 marks] Describe briefly the load carrying mechanism of two steel pipe piles, the first is driven in a deep sand deposit and the second is driven in a deep soft clay deposit. (b) [25 marks] A group of 112, 0.3 m diameter pre-cast concrete piles 15 m long consists of 7 rows of 16 spaced at 1.0 m centre to centre both ways. Soil exploration at the site revealed the following information: 0 material is variable silty clay with softer clay pockets locally and no sensitivity 0 average saturated unit weight 18.0 kN/m3 0 average undrained shear strength over the upper 30 m is 40 kPa. i) Find the maximum load that this group can carry for a factor of safety of 3. Assume on = 0.6. CIVE 416 Page 6 ii) Assuming that bedrock exists 15 m below the base of the pile tips, calculate the consolidation settlement of the pile group (average mV = 1.25 x 104 /kPa for the clay layer). CIVE 416 Page 7 Two-dimensional seepage: 62h The 2D steady state flow equation If). g Earth Pressure Theory: db = Koo’V K0 = 1 — sin (11’ a; = a; tan2(45 +¢'/2)+2c'1an (45 +¢'/2) CRIB SHEETS x k=./kxky K0 (over-consolidated) = K0 (nonnally-consolidated) ‘V OCR 0'; = O'LKa — 26K Rankine active earth pressure for sloping backfill 6'}, = 6'0 Kp + 2c' .le Ka ; cosa Coulomb active and passive earth pressure coefficient cos a — 4/0032 0: — cos2 ¢' cosa + 1/cos2 a — cos2 ¢’ 311211.11 K: Retaining walls: sm 16511105‘5{1+ Siam—6)sin(a+.3)i K = sin2 (,3 — W) sm fl 8111(15 + 6)[1 311103 + 6)sin(ar + )3) l FS (sliding) = ZFR / ZFd ZFR = (2V) tan 5 + B C'a + Pp ZFd = Pa COS 0L Braced.cuts: Soft tofirm clay: ka = l — m (4011 / yH); and pa =1.0 ka 7 H Stiff to hard fissured clay: pa = 0.2 y H to 0.4 y H Sandy soils: pa = 0.65 ka y H The factor of safety against base heave Fsb = (N, cm) / oz ‘ Stability of slopes: ' I ‘ ‘ r:" r2 6 0“ Infinite slope F = ycd seca cos eca + :2?! [1 —ru sec2 a] szte 510173 FEW N: yH Zlc'l+(W,1 cosan —ul) tan ¢' Ordinarymethod of slices F = —. 2 PK. sm an Bearing capacity and settlement of shallow foundation: Terzaghi Theory: Strip footing qu = 0’ NC + y D Nq + 0.5 B y N, Square footing qu = 1.3 c’ Nc + y D Nq + 0.4 B y N7 Circular footing qu = 1.3 c’ NO + y D Nq + 0.3 B y N, Meyerhof Theory: qu = 0’ NC (Sc ic dc) + y D Nq (Sq iq dq ) + 0.5 B 7 NY (SY iy dy) ¢ N: N. N?” Fm Value Far 0 514* 1.0 0.0 Shape: s. =3 1+ 0.210% 4:114 5 6.49 1.6 0.1 B 10 8.34 25 0.4 s., = 3., =1+0.1K,Z (12> 10 15 10.97 3.9 1.1 s, ‘= 5., = 1 4 = o 20 14.83 6.4 2.9 D 25 20.71 10.7 6.8 Dcpth: dc = 1+ 0-2 K1»; Am 26 22.25 11.8 8.0 D 28 25.79 14.7 11.2 44 = "v = ”0'1 K43 ¢> 1“ 30 30.13 18.4 15.7 d, = d, s 1 41 = o 32 35.47 23.2 22.0 . 2 34 42.14 29.4 31.1 Inclination: 1. — 1, =( - 7:5) mm 36 50.55 37.7 44.4 R v 38 61.31 48.9 64.0 9" 1. 40 75.25 64.1 93.6 3| " =( ' 5) ”0 45 133.73 134.7 262.3 . i., = 016:9 > o as = 0 50 266.50 318.5 871.7 CIVE 416 Page 8 Stress in an Elastic Mass 25-2 23 1. Point Load A0 2 ‘ 272' L5 272' (r2 +22)5/2 2. Uniformly loaded strip A02 2 i [a + sin a cos(a + 2/3)] 72' 3/2 3. Uniformly loaded circle A0- = q 1— + z 3 1+ (R / z)2 4. Uniformly loaded rectangle DEPTH RATIO, z/B DEPTH RATIO, z/B Scale of distance 00 = depth 2 at which stress is computed .- l i. m "#3" I. I"... .3. I f.’::£'."- llll O .0 O o O 1!". ll :' ‘I I:- I I n? 0.. . h. O l' t O Q ‘ t O .- .".'..'3. '9‘: ‘9'! gi’g‘g’o 9“ I .l . . ~ 0 c a. g ’ . ' ...l' .I' .é’io“ l..~“".‘.‘ I'llllllllgfl' “:9? "mun-u | 'O film: I. I I - I a 5' I: I. =- v‘ . in IE5: ‘5 3 I = i i ‘I ‘5". o“. 5.0;: :E-E. “’o ' fi‘ ‘3’}. 9".“ .fi . O. . .9: O 0: I. " n: II I I ‘ “‘ t ““\“$‘¢'f' Eva‘sw \ ‘o 03. 0:0. 0’? I: '0 3"! :ii. 0 0:: 1? II I ll ? f r I II Consolidation settlement 035 0,40 I W‘llll uni-III 1 null I I r Ihllllll-ll‘lll' . ' ‘ . 7-"! u n -=3:-'! " _— =i-III II_'K- 71 0'30 — ---III I'm-E - 1. NC soils Sc = CCH loge-fl 1+e0 0'; A C H ’ + ’ 2. 0Csoil (OCR >1 but 02> 0’0 +4109 sc =1 r “9% +60 0'0 CIVE 416 IIIIIIIIIIIIIIIIIIIIIIIIH nunmuunmmmu I'IIIII‘IIIIIIIIIII'IIII ' Illlltlllfllllia IE “K‘- I 1 Illllllllilllllfllllli ‘ flllliilllilll 3 Elllli Immediate settlement: 1- Strain influence factor: _ Z 1: S=C1C2(q_Q):E AZ 0 S I2 = strain influence factor For circular and square foundation: I2 = 0.1 at z = 0 Iz=0.5 atz=zl =0.5B Iz=0 atz=zz=2B For foundation with L/B 2 10: I2 = 0.2 at z = 0 12:05 atz=zl=B Iz=0 atz=zz=4B Values of L/B between 1 and 10 can be interpolated C1 = a correction factor for the depth of embedment = 1 — 0.5[q/( q — q )] C2 = a correction factor for creep in soil = 1 + 0.2 log (time in years/0.1) q = stress at the level of the foundation q = Y Df 2- Perfectly flexible foundation B’ = B/2 for centre of foundation = B for corner of foundation Is = shape factor If = depth factor OL = factor depends on the location of the point of interest = 4 at the centre of the foundation = l at the corner of the foundation Page 9 3. 0C soil (OCR > 1 but a; < 0’0 +41ch S C Pile Foundations _ CSH o" 1+ea Bearing capacity factor, N; 40 60 80 100 200 mm- Illllllll‘lll l C CCH I + log ' 0' 1+ e0 0' 10 0 C 0'", + A0" log 20 % Illllllllll'I Cohesionless soil: :3" 30 1’ ‘ ’1 Fl ' Bearing Capacity: Qs = 2 p AL K 0’0 tan5 and Q =At c’t Nq .3; 40 Infillllli- — E Vlfllllllll SPT: Qs=nNAs and Qt=mNAt ”50- H Settlement: S : (Qapp / Es D) 10 1{k RV Sgroup = Sindividual P D Cohesive soil: Bearing Capacity: QS = p 2 on 011 AL & Qs=pZB0’VAL Qgroup:CuI\ch2_i—4"3'ub Le Settlement: s = (D / 100) + (Qapp L / A1, Ep) 60 ”ammu- ”WWII“. [II-III...- ‘iIIIIIIII N 0-4 H — 1M------ ---is.~1I----- °'°° ----s_~3!---- -----&Q--- III-Infiagn wllllllllfia II... ”%m II... to t 00254 0-020 10 20 '79 Qt : NC Cu At For aflgén 40 BO 50 Qgroup : Ge 1’1 Qsingle \ sgroup = [C0 /(1 + 60)] [H — 2/3 L] 10g [(c'o + Ac’) / 0'0] Negative skin friction: qn = 0L cu Taylor’s Chart For [3 > 53": All circles are toe circles. For [3 < 53": Toe circle Midpoint circle — - — . - Slope circle ------ CIVE 416 Illil it at ME I ll llllll Elllll l Slope angle, [3 (deg) Page 10 ...
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