Solutions to Supplementary Problems

Transcription

1 Solution to Supplementary Problem Chapter 9 Solution 9. a) Bearing capacity factor from Chart 9. for f = 8 : N = 7.5 c N = q N = 0.5 Effective overburden preure at foundation level: 0 = 0 = D g = 7 = 7 kn/m From (9.): q = [ cnc + 0 ( Nq ) BN ] + 0 F = ( ) = ( ) + 7 = 5.kN/m < 00kN/m A q < = 00 kn/m, the oft clay i too eak to carry the propoed loading. The tiff clay layer i only m belo the foundation level, hence it i poible to replace the oft clay entirely ith compacted material. b) Soil tructure: In order to calculate the bearing capacity, the unit eight, in the to tate pecified, have to be determined from the given: d = 6.kN/m m = 8% Introduction to Soil Mechanic, Firt Edition. Béla Bodó and Colin Jone. 0 John Wiley & Son, Ltd. Publihed 0 by John Wiley & Son, Ltd INDD Chapter No.: Title Name: <TITLENAME> 6//0 4:4: PM Comp. by: <USER> Date: Jun 0 Time: 04:4: PM Stage: <STAGE> Page Number:

2 Introduction to Soil Mechanic m Soft clay Compacted oil g d = 6. kn/m c = 5 kn/m f =5 m = 8% G =.75 Stiff boulder clay Soft clay m Figure 9.56 Step : It ha to be acertained that the replacement oil i partially aturated. From (.4): G d = + e Epreing, G e = = = 0.68 d 6. From (.6): mg e = S r Epreing, mg Sr = = = 0. 7 ( 7% ) e 0.68 Therefore, the material i partially aturated. From (.40): d = + m Epreing, g = ( + m ) g d =.8 6. = 9 kn/m G + e From (.4): at = = 9.8 = 0kN/m + e.68 Step : Bearing capacity at partial aturation. Fill.5 m g = 9 kn/m f =5 c = 5 kn/m Stiff boulder caly Figure 9.57 GL m m Bearing capacity factor: For f = 5 Nc = 46 Nq = N = 40 Alo, 0 = 0 = 9 = 9 kn/m (9.): q = [ ( ) ] = ( ) + 9= + 9= 488 kn/m > 00 kn/m Therefore, the compacted fill i atifactory INDD Chapter No.: Title Name: <TITLENAME> 6//0 4:4:6 PM Comp. by: <USER> Date: Jun 0 Time: 04:4: PM Stage: <STAGE> Page Number:

3 Solution to Supplementary Problem: Chapter 9 Step : Bearing capacity after flooding WL Water 0.5 m GL Fill.5 m m g at = 0 kn/m f = m c = 5 kn/m Effective overburden preure at foundation level. 0 = = 4.9 kn/m u = (0.5 + ) 9.8 = 4.7 kn/m = = 0. kn/m 0 Submerged denity = = 0 9.8= 0. kn/m at Stiff boulder clay Figure 9.58 Formula (9.) i modified to take g into account. q = [ cnc + 0 ( Nq ) βn ] + o = [ ( ) ] = ( ) = = kn/m > 00 kn/m Therefore, the effect of long-term flooding i to aturate the fill, thu decreaing the bearing capacity by 6%, in thi eample. Solution 9. The afe bearing capacity for hallo rectangular footing i given by formula (9.): B q = c + 0. Nc + 0 ( Nq ) BN + 0 F L Where F = From Chart 9. N c = 0 B = for f = 0 N q =.5 L = 4 N = A there i no ground ater table indicated, the effective overburden preure equal the total one at the bae. Therefore, 0 = 0 = = 6. kn/m. Hence, q = (.5 ) = ( ) + 6. = 4 kn/m INDD Chapter No.: Title Name: <TITLENAME> 6//0 4:4:6 PM Comp. by: <USER> Date: Jun 0 Time: 04:4: PM Stage: <STAGE> Page Number:

4 4 Introduction to Soil Mechanic The loading i eccentric, therefore the reultant force and it eccentricity are calculated a decribed in Appendi D. (a) m M=500kNm L =4m W = 000 kn In thi cae, the only vertical force i W = 00 kn R = W= 000 kn Replacing W and M by R at eccentricity e: M = Re 500 = 000e 500 e = = (b) Lc 0.5 R = 000 kn The maimum and minimum preure can no be calculated from (8.57), taking b = B and d = L. (c) f ma min R 6e = ± BL L (d) 0.5 R B =m f f ma min = ± 4 4 = 5 ± 94 = = 9kN/m = 5 94 = kn/m q ma f, therefore the oil i marginally overtreed. kn m Preure diagram 9 kn/m Figure INDD Chapter 4No.: Title Name: <TITLENAME> 6//0 4:4: PM Comp. by: <USER> Date: Jun 0 Time: 04:4: PM Stage: <STAGE> Page Number: 4

5 Solution to Supplementary Problem: Chapter 9 5 Solution 9. a) Straight pile End bearing: From (9.0): π A e = = m 4 Q e = 9 c u A e = = 89 kn d Shaft reitance: A = π. = π. = 4.78 m From (9.0): Q = αcua = = 7kN Alloable carrying capacity i the leer value of: (9.): (9.): b) Under-reamed pile End bearing: Shaft reitance: Ultimate capacity: Q Q e Q a = + = + = Q Q + Q e a = = = Accept Q a = 044 kn.5 π Ae = = 4.9m 4 e u e 794 kn 044 kn Q = 9c A = = 489 kn A = π 7= m Q = α cua = = 7kN Q u = Q e + Q = = kn Alloable carrying capacity i the leer value of: Reult: Qa = + = 4978kN.5 Qu Qa = = = 585kN.5.5 Accept Q a = 4978 kn Q a (kn) a) traight 044 b) under-reamed INDD Chapter 5 No.: Title Name: <TITLENAME> 6//0 4:4:6 PM Comp. by: <USER> Date: Jun 0 Time: 04:4: PM Stage: <STAGE> Page Number: 5

6 6 Introduction to Soil Mechanic Solution 9.4 From (.): n 0.9 e = 0.64 n = 0.9 = From Chart 9.: For f = 9 N q = 0 and N = 7 Cae : The and i dry, hence the dry bulk denity i applied. G From (.40): d = = = 5.9kN/m + e.64 From (9.): = 0 ( Nq ) dbn + o F Where, 0 = 0 = z g d = = 7.95 kn/m Subtituting: ( ) = = ( ) = = 48.4kN/m Cae : z =.5 m Figure 9.6 B =m z = 0.5 m Sand e = 0.64 G =.65 g at = 9.7 kn/m Gravel h = h A =.5m Seepage GWL = GL A = h A =.5 9. =4.7 kn/m The modified ubmerged denity i found from (5.): The arteian preure caue upard eepage, thu the ubmerged denity, modified to take the hydraulic gradient into account, i applied. The hydraulic gradient i given by (.): h i = z.5 = = The ubmerged denity from (.4): G.65 = = 9.8 = 9.87kN/m + e.64 = = 9.7kN/m = i = =.98 kn/m Therefore, the afe load that can no be carried by the and i: = = ( N ) BN + F 0 q 0 at INDD Chapter 6No.: Title Name: <TITLENAME> 6//0 4:4:48 PM Comp. by: <USER> Date: Jun 0 Time: 04:4: PM Stage: <STAGE> Page Number: 6

7 Solution to Supplementary Problem: Chapter 9 7 Where, = z = = kn/m and = = 9.85 kn/m 0 0 = ( ) = 45 kn/m Therefore, eepage ha reduced = q by 70%. Solution 9.5 From (9.8): From (9.4): From (9.5): q u = z = z q n = q u 0 = z 8 z = z q q 4 + 6z 7. z n n = = = + F From (9.6): From (9.): From (9.): q = q n + o = 7. + z + 8 z = z q = n = o = q 8 z = z 8 z = 7. + z From (9.) and (9.6): ( ) For = = q q q = q n o n n n = qn = 7. + z n = = + = = 55kN/m z z.48 m And = q = = 8.7 kn/m Note: a) The afe bearing preure i to be compared ith the afe bearing capacity of oil q o, that q b) The net bearing preure n i to be applied hen etimating the:. Preure induced by the net bearing preure at ome depth belo a footing.. Effective preure at a depth.. Conolidation of oil belo a footing and it conequent ettlement. c) For urface load n = INDD Chapter 7 No.: Title Name: <TITLENAME> 6//0 4:4:58 PM Comp. by: <USER> Date: Jun 0 Time: 04:4: PM Stage: <STAGE> Page Number: 7

8 8 Introduction to Soil Mechanic Solution 9.6. From Chart 9.6, the alloable bearing preure for and above the GWL for 4 m ide footing at N = 50 i a = 470 kn/m. For the and belo the GWL the value of N = 50 ha to be corrected in accordance ith formula (9.): ( N ) ( ) N = = =.5 blo The correponding alloable bearing preure i = 00kN/m The ultimate bearing capacity of the oft clay i determined from (9.8): q =.c N = = 48.kN/m u u c qu 0 And the afe B.C. from (9.7): q = + 0 F Where from Chart (9.): N c = 5.7 for f u = 0 The effective overburden preure on the top of clay i: 0 = 9+ = 57+ a Where, from (.4): G.66 + e = = = kN/m 0 = = 78.5kN/m Alo, o = = 97.8 kn/m Hence, q = = 88kN/m. The applied effective preure at depth m, m and 5 m ha to be determined. Thee hould be le or equal to the bearing trength of the oil belo. At m depth (That i at bae level) the net foundation preure hould be le or equal to the alloable 470 kn/m. Therefore, n = 88 9 = 50 kn/m i atifactory. At m depth ( m belo bae), the effective preure i given by: v ( ) = v + = + 9 here v i the maimum vertical preure induced by n at the level of the ater table. Thi occur under the centre of the quare footing and may be etimated by mean of:. Chart 4.7 (Steinbrenner). Chart 4.8 (Fadum). Chart 4.9 (Nemark) In thi problem, Chart 4.7 i ued to etimate the vertical preure m belo point c INDD Chapter 8No.: Title Name: <TITLENAME> 6//0 4:4:04 PM Comp. by: <USER> Date: Jun 0 Time: 04:4: PM Stage: <STAGE> Page Number: 8

9 Solution to Supplementary Problem: Chapter 9 9 4m c a =m b=m a b = = z 0.5 b = = v 7 n 7 I = 0./quare ( I ) = 4 = = kn/m = + 9 = 4kN/m > 00kN/m 4m Figure 9.66 Therefore, the induced preure eceed the alloable one. At 5 m depth ( m belo the footing) the effective preure i: From Chart 4.7: ( ) = = = v v v a z = = =.5 7 I = 0. b b v = = 70.8 kn/m = = kn/m > 88 kn/m 5 Therefore, the effective preure eceed the bearing capacity of the oft clay. It i adviable to reduce the induced maimum vertical preure at m and 5 m depth, in order to prevent poible failure of the eaker layer. Thi may be done by changing the: a) Size or hape of the footing b) Width of the footing c) Loading. Of coure, any uitable combination of thee three alteration may be choen. In thi cae to change are made a hon: (a) GL n = 48 kn/m (b) 0.6 m.4 m GWL m C 4m Sand g =9kN/m a = 470 kn/m g = 0.57 kn/m a = 00 kn/m Shape of bae area b =m a =m 6m c Soft clay q = 56 kn/m 4m Figure INDD Chapter 9 No.: Title Name: <TITLENAME> 6//0 4:4:5 PM Comp. by: <USER> Date: Jun 0 Time: 04:4: PM Stage: <STAGE> Page Number: 9

10 0 Introduction to Soil Mechanic Change : The footing i no baed only 0.6 m belo the ground urface. Thi ha the effect of raiing the bulb of preure diagram, thu reducing the induced load ( v ) on the oft clay. Change : The quare bae-area i changed into a rectangle in order to reduce the applied net bearing preure from 50 kn/m. The ne value of n i determined firt. Total load on quare bae: W = 88 4 = 608 kn. Total load on rectangle: W = 4 6 = 4 kn 608 Equating, W = 4 = 608 = = 59kN/m 4 Therefore, n = 0.6 g = = 47 kn/m Preure on the oft clay at 5 m depth (4.4 m belo the centre c) i: = + ( ) = v v a z 4.4 For v : = =.5 = =. 7 I = b b = = 94.85kN/m v Hence, 5 = kn/m < 88 kn/m. Thi i atifactory. The trength of the and belo the GWL i obviouly ufficient to carry the revied preure. Hoever, it i prudent to check it. a z.4 = =.5 = =. 7 I = 0.5 b b = = 50kN/m v The effective preure, m belo ground urface, i: = v = kn/m < 00kN/m Therefore, the rectangular arrangement i atifactory. Solution 9.7 In vie of the arteian preure, it i neceary to check hether the ecavation for the footing approache or i ithin the critical thickne (z c ) of the oil above the gravel layer. Thi i given by the effective eight of oil per m (That i the effective preure at ), balancing A = 6 kn/m. 6.6 = y 0.95 m + A = 0 y = = 7.5 or 7.5y = 0 z = + y =.95m c INDD Chapter 0 No.: Title Name: <TITLENAME> 6//0 4:4:4 PM Comp. by: <USER> Date: Jun 0 Time: 04:4: PM Stage: <STAGE> Page Number: 0

11 Solution to Supplementary Problem: Chapter 9 Thi mean that the bae of the ecavation ould be jut above the critical thickne. A the foundation preure i not applied at thi tage, the ground could fail and the ork flooded. The propoed contruction i not feaible therefore. The uggeted olution i hon belo, that i raiing the level of the bae a ell a the ground urface by 0.5 m, thu maintaining the m foundation depth. The topoil i firt removed of coure. Ne GL 0.7 m 0. m Ecavation Compacted fill clay 0.55 m.5 m GWL g = 7.5 kn/m z c =.95 m m g = 9.9 kn/m 6 kn/m Gravel Figure 9.69 Final effective preure at the top of the gravel belo the ecavation. = = 9.64 kn/m Solution 9.8 Step : Calculate the unit eight of each layer from the given oil characteritic. Stiff clay S r = 0.76 From (.8): G + Se e.8 r = = = kN/ m S r = From (.4): G.7 + e.8 = = = kN/m And g at = g + g = = 9. kn/m Compact and.67 S r = From (.4): = 9.8 =.4 kn/m. = =.kn/m at INDD Chapter No.: Title Name: <TITLENAME> 6//0 4:4: PM Comp. by: <USER> Date: Jun 0 Time: 04:4: PM Stage: <STAGE> Page Number:

12 Introduction to Soil Mechanic Medium clay.8 S r = From (.4): = 9.8 = 0kN/m.76 Step : Find the bearing capacity factor for the to clay layer, from Chart 9., and calculate the afe bearing capacitie at m and m depth. Stiff clay From UU triaial tet: c u = 79 kn/m z =m Figure 9.7 Effective overburden preure: 0 = 0 = 8. = 8. kn/m From (9.): ( ) From (9.): B =m φ u= 5 Nc = Nq = N =.5 = q = c unc + 0 Nq BN + 0 F = [ ] + 8. = 4 kn/m n = 0 = 4 8. = 06 kn/m > 45 kn/m atifactory Note, that although the ground ater table i jut belo the rupture zone ( D > B ), the clay i found to be aturated belo the bae, due to capillary ater movement. For thi reaon g = 9.9 kn/m intead of g = 8. kn/m i applied, reulting in loer value of etimated oil trength. The choice i on the afe ide therefore. In thi cae the tiff clay i trong enough to carry the net bearing preure. Medium clay From vane tet: c u = 68 kn/m Nc = 5.7 φu = 0 Nq = N = 0 GL n =45 kn/m g = 8. kn/m Note: The bearing capacity formulae are valid at foundation level. In order to determine the trength of the oil at a depth belo, the bae i imagined to be eated at that level. In thi problem, formula (9.9) i implified by the bearing capacity factor. Stiff clay Sand Medium clay Figure 9.7 B =m m g = 8. kn/m.4 m g at = 9.kN/m.6 m g at =. kn/m g at = 0 kn/m qn = cunc 0( ) 0.5 B 0 F + + = ( cn u c) F = ( ) q = 9kN/m n INDD Chapter No.: Title Name: <TITLENAME> 6//0 4:4:7 PM Comp. by: <USER> Date: Jun 0 Time: 04:4: PM Stage: <STAGE> Page Number:

13 Solution to Supplementary Problem: Chapter 9 Overburden preure at z = m: 0 = = 58.kN/m And the afe bearing capacity: q = = 87 kn/m Note: In thi cae, the idth of footing i irrelevant, becaue N g = 0. Step : Determine the approimate value of preure on the top of the medium clay, due to the net foundation load, plu the overburden. Thi i uually done in one of four ay: a) By the Bouineq-baed formula, or bulb of preure diagram, appropriate to the hape of the bae. b) By auming that the load diperion i linear ith depth at 0 to the vertical. c) By auming 45 linear diperion of the foundation preure. d) By a combination of (b) and (c). a) Bouineq-Michell olution m.4 m n = 45 kn/qm Stiff clay b = m g = 9. kn/m z =m z A the oil i loaded belo ground level, the net foundation preure i applied. 0.6 m Sand g =.4 kn/m Medium clay r r Figure 9.7 Effective overburden preure at z = m belo bae. 0 = = 0.4 kn/m The vertical preure on the top of the medium clay are calculated by formulae: r b r (4.): α = tan = tan z (4.): r b r + + β = tan α = tan α Z β+ inβco( α + β) β+ inβco( α + β) (4.8): v = n = 45 π π = 78 [ β+ inβco ( α + β) ] Tabulating the reult, hich can be checked by Chart 4.. Maimum value of induced preure: v = 4 kn/m Final maimum effective preure at ( z = m): Table 9. = 0 + v = = 54 kn/m In thi cae therefore, < q = 68 kn/m. r (m) α β v INDD Chapter No.: Title Name: <TITLENAME> 6//0 4:4:44 PM Comp. by: <USER> Date: Jun 0 Time: 04:4: PM Stage: <STAGE> Page Number:

14 4 Introduction to Soil Mechanic b) 0 diperion m n = 45 kn/m B = m.4 m z = m 0 Stiff clay From (4.): B = 0 + n B+.5z 45 = = 4 kn/m Medium clay = 4 kn/m B z = 4. m Sand Figure 9.74 c) 45 diperion n From (4.): 45 B = m B = 0 + n B+ z 45 = kn/m B z = 6m = 0 kn/m Figure 9.75 d) 0 /45 diperion n From (4.4): z = m m 45 0 B = m B =.5 m 6 kn/m B n = B+ z 45 = = 6 kn/m B z = 5.5 m Figure 9.76 Concluion: A the induced preure are le than the bearing capacity at m and m depth, the oil i not overtreed. Alo, the diperion method yield lo value in comparion ith Michell INDD Chapter 4 No.: Title Name: <TITLENAME> 6//0 4:4:5 PM Comp. by: <USER> Date: Jun 0 Time: 04:4: PM Stage: <STAGE> Page Number: 4