Solutions to Supplementary Problems

Transcription

1 Solution to Supplmntary Problm Chaptr 5 Solution 5. Failur of th tiff clay occur, hn th ffctiv prur at th bottom of th layr bcom ro. Initially Total ovrburn prur at X : = = 7 kn/m Por atr prur at X : u = 9.8 = 9.6 kn/m rtian prur at X : = 45 kn/m Initial ffctiv prur at X : = u = = 7.4 kn/m t failur Th ffctiv prur at point X i altr only by th ight of th ovrburn rmov that i by g h = 9 h. Thrfor at failur: = g h = h = From hich, h = = =.8m 9 Thrfor, atr burt into th arthork hn th xcavation rach a pth of.8 m. Introuction to Soil Mchanic, Firt Eition. Béla Boó an Colin Jon. 0 John Wily & Son, Lt. Publih 0 by John Wily & Son, Lt INDD 6//0 4:0:5 PM

2 Introuction to Soil Mchanic Solution 5. Bfor th clayy an i xcavat to accommoat th culvrt: m 4m m Figur 5.4 GL g = 9. kn/m Rock Clayr an GWL g at = 0 kn/m Firm clay Point : = 9. = 57.kN/m u = 0 = u = 57.kN/m Point : = = 97.kN/m u = 9.8 = 9.6kN/m = u = = m Point : = = 7.kN/m u = = 9.4 kn/m = = 98.kN/m t compltion: Th prur at th thr point ar altr by th: a) xcavation for th culvrt b) ight of th culvrt an atr c) ight of th n fill a) Prur rla by th xcavation: E = 9. = 57. kn/m b) Wight of concrt culvrt pr mtr lngth: 5m m.5 m 4.5 m Volum : V = =.75 m Wight : Wc =.75 = 90kN 0 Figur 5.4 Wight of culvrt an.5 m p atr. Wc = Wc = = 56.kN c) Wight of 0 m thick fill on culvrt: W f = = 800 kn Nt prur xrt by th culvrt on th firm clay layr pr mtr lngth: W + W kn/m 5 5 c f c = E = = Final prur : t point : = = 4 kn/m c u = 0 = 4 kn/m INDD 6//0 4:0:56 PM

3 Solution to Supplmntary Problm: Chaptr 5 t point : = = 74 kn/m u = 9.8 = 9.6kN/m = u = = 54.8 kn/m t point : = = 44 kn/m u = = 9.4 kn/m = = 74.8 kn/m Rult ar tabulat for comparion. Tabl 5.8 Initially (kn/m ) Finally (kn/m ) u u Point Point Point Solution 5. Dry an (calculat g ) (.4): G + = + = G + + at at at G = = = at Voi ratio : 0.6 (6%) at (.4): G = = = 6 kn/m Silty clay (calculat g at ) (.6): (.4): mg = = = S r (48.6%) G at = = = 9.8 kn /m Figur 5.44 ho th actual unit ight of ach layr, applicabl to th calculation of prur INDD 6//0 4::05 PM

4 4 Introuction to Soil Mchanic GL Dry an m g α = 6 kn/m Silty clay 4m g at = kn/m Stiff clay Figur 5.44 GWL Prur at m blo GL : = 6 = 48kN/m u = 0 = 48 kn/m Prur at 7 m blo GL : = = kn/m u = = 9.4 kn/m = 9.4 = 9.8 kn/m Solution 5.4 Th rquir quantiti hav to b xpr in trm of g an G. From (.60): ( ) From (.): = ng n= an n= G G n n G G = = = G (5.6) (5.7) From (.6): From (.): = ( n) = = g at = g + g G G (5.8) Tabulating th timat valu of oil charactritic. Tabl 5.9 Layr No. Thickn (m) g (kn/m ) G n a) Prur in trm of aturat unit ight. Th rult ar ran in Figur g (kn/m ) g at (kn/m ) Tabl 5.0 Total: Por: u Effctiv : = -u (m) kn/m = = = = = = = = = = = = INDD 4 6//0 4:: PM

5 Solution to Supplmntary Problm: Chaptr 5 5 b) Th ffctiv prur can b trmin irctly, ithout timating th por prur, by man of th ubmrg unit ight. Tabl 5. (m) kn/m = = = 44.6 Thi vrifi th rult of Tabl 5.0. Solution 5.5 /a: Total prur at X: Por prur at X: Effctiv prur at X: Canclling h g trm: Thrfor, = h g + g at u = ( h + ) g + = u = h + ( h+ ) at = h + h at = ( g at g ) = (5.9) /b: Th moifi ubmrg nity ( g ) of th clay i givn by (5.) a: g = g i g hr i h h = = = = h (5.0) From (5.9): = = = (5.). For critical pth: = c = 0 Thrfor, c = = 0 c That i, = 0 = n 0 c c = (5.) INDD 5 6//0 4::9 PM

6 6 Introuction to Soil Mchanic. = 8kN/m = = 9.9kN/m 8 c = =.94 m < m 9.9 Thrfor, th m thick layr houl not fail. Moifi nity: 8 = = 9.9 =.9 kn/m Not: Th valu of g i an inication of th ritanc of a layr againt failur, hn ubjct to irct artian prur in th abnc of othr, ovrlying trata. > 0thlayritabl If: = 0 th layr fail ( = c) < 0thlayrhanoritanc Effctiv prur: Eithr rtian prur ha: = = = 9.87kN/m or = g =.9 = 9.87 kn/m h 8 = = = m Not: Formula (5.) i applicabl to a ingl ubmrg layr only, hich may b ovrlain by atr or ovrburn abov th groun atr tabl. For to ubmrg layr, Problm 5.6 an 5.7. Solution 5.6 Drivation. Total prur: x = g + g (at) (at) + g (at) Por prur: u x = ( + ) g W rtian prur: Effctiv prur: x = u = + [ ] + [ ] (at) (at) t th top of gravl: = x + + (5.). Uing th abov rult a ± in orr to tranform g an g into = an = Thrfor, = ( ) x ( ) ( ) = = Hnc, x = (5.a) INDD 6 6//0 4::6 PM

7 Solution to Supplmntary Problm: Chaptr 5 7 Solution 5.7 Th moifi nity for th lor clay layr i ngativ ( g 4 = 4.6). Th layr i too thin, an hnc oul iintgrat but for th oil abov it. For th ubmrg part of th mium clay g 0. Thi layr oul jut fail in th abnc of th ovrburn. It may b aum, thrfor, that th firt m thick oil, abov th gravl, contribut littl or nothing to th groun-ritanc againt. Thi can alo b vrifi by timating th rquir layr thickn at failur an comparing it ith th actual thickn. pply formula (5.)..8 For ubmrg mium clay: c = = =.9m.m 9.89 Thrfor, th xiting layr thickn i critical:.8 For th 0.8 m thick clay: c = = =.4 m > 0.8 m 0.9 Thrfor, th thickn i l than th critical valu. Effctiv prur a) t point x : x = = 64kN/m ux = = 0kN/m = u = 64 0 = 64kN/m x x x t point y : y = = 87.7kN/m uy = =.6 kn/m y = = 64 kn/m t point : Z = = 0.9kN/m uz = =.4kN/m = = 7.5 kn/m b) t point x: x = = 64 kn/m t point y: y = = 64 kn/m t point : = = 7.5 KN/m c) t point x: x = 64 kn/m t point y: pply th rlvant portion of formula (5.a): y = + + = = = 64kN/m y t point : Similarly, Z = = = ( 4.6) +.8 = 7.5kN/m INDD 7 6//0 4::4 PM

8 8 Introuction to Soil Mchanic Not: Of th thr procur (b) i th implt. long a th ubmrg niti ar knon, formula (5.) an (5.) can b appli irctly to any numbr of layr. Whilt th moifi ubmrg nity i uful a an inicator of th aquacy of layr thickn, it u for th calculation of ffctiv prur i cumbrom. lo, th + trm pn on th numbr of ubmrg layr, a hon in Figur 5.49, hr o inicat th ight of oil abov th groun atr tabl. GL 0 0 g g 0 = 0 (Four ubmrg layr) = 0 + g g = 0 + g + g g = 0 + g + g + g 4 g 4 x 4 = 0 + g + g + g + 4 g 4 + Figur 5.49 In gnral, if thr ar n numbr of ubmrg layr, thn i multipli by ( n ). Th mtho of rivation for th xprion a hon in problm 5.6, for to layr. Solution 5.8 Givn: V = 944 cm = m g = 5. kn/m n = 0.45 S r = 0.69 a) atr i a to th ry oil, it unit ight i incra by: Δ g = g g G + S r G + S r G From (.8): = Δ = G From (.4): = = ( G + S r G) + + = Sr INDD 8 6//0 4::49 PM

9 Solution to Supplmntary Problm: Chaptr 5 9 But n = Δ = nsr + = =.05 kn/m (5.4) Hnc, th unit ight g = g + Δ g = = 8.5 kn/m From (.5): V = ns r V = = m Wight Or Ma W = V g = =.87 0 kn W = V Δ g = =.88 0 kn 0 W M = = = 0.94 kg b) Taking S r = in (5.4): Δ g = n g = = 4.4 kn/m Thrfor unit ight: at = +Δ = = 9.5 kn/ m lo in (.5): V = nv = = m Wight: W = V Δ g = = kn Ma: M = = 0.44 kg 9.8 Compar th rult ith tho in Tabl INDD 9 6//0 4::5 PM