Question

Soil Mechanics II Tut (5) 1) A clay has properties Ce = 0.4, Cr=0.05. Determine the final void ratio for the following situations: a) Initial stress = 50 kPa, e = 1.2, 0 = 100 kPa, Final stress = 90 kPa b) Initial stress = 50 kPa, e = 1.2.0! = 100 kPa, Final stress = 190 kPa c) Initial stress = 50 kPa, e = 1.2.0 = 50 kPa, Final stress = 90 kPa d) Initial stress = 50 kPa, e = 1.2, 0 = 100 kPa, Final stress = 30 kPa e) Initial stress = 50 kPa, e = 1.2.0 = 50 kPa, Final stress = 30 kPa

Answer for this question in British code

Answer 1

Clay brokerlêel C=0.4 Cr=0.05 To determine final voed ratio (es) in every case. (a) Initial stress (To)=50kPa - lo= Initial void ratio = 12 od= Preconsolidation stress=100 kPa final stress (of) = 90 kpa ? as to sols 50 <100 ___all_F 2 c D१० र १०० from about it is clear that soil is recompressed as initial stress (Go) on soil and final stress Cos] on soil both are less than pre consolidation Stress (old). Hence in the entire compression soil is recompressed. tence Recompression Inden (Cr) will be used. soil is preconsolidated.

Co= DE - lo-ef - log (of log (of) 0.05= 12-lf 0.012763=1.2-ef - log (18) ef = 1.187 PART(b) 1. Initial stress = 50 kPa zoo a le=12, d6 = 100kPa Of= 190kPa from (ro) gritead stress upto (oc) Preconsolidation in stress , soil is recompressed and preconsolidated tence Co will be applied. But from (oc) Preconsolidatiem stress lepto ferial stress of). Cc compression Indese will be applied as soil is normally consolidated. to coc zo 50 <100 (preconsolidated) of > rc » 190 > 100 ( Normally consolidas

Co= co-e = 1.2-e log (oc) log (lood 0.05= 1.2-e log(2) lis 1.184 d i te Now from ac to of Cc will cact. Cc = lo- ef a 1 log (6 ) Now here initial toid ratio (eo) is the void satie we have calculated in aboue step (el. Cc= e,- ef = 1.184-ef a log (6 log too 1184-es bestellen 0.4 - 1.184 - ef 1 log (1.9) ef=1.0724 h Answeer

PART (C). Initial stress = 50 kPa = 00 Co=1.2 , rc = 50 kPa final stress (85)=90 kPa Here, gnitial seress (6) and preconsolidation stress (oc) are same, thee, the entire Compression is normally consolidated Range Heree compression Index will be applied 0o=oc =50kpa corder of>oc a 90 > 50 cc = Co-ef = 1.2-ef 0.4 = 1.2- ef log C-8) ef= 1.0978 Answer PART (d) gnitead stress (60)=50kPa 60=12 , o'c= 100 kPa Of= 30 kPa of < oo 30 50

In this case of coo, that means soil is na not under compression But it is expanding so brial void ratio will be greater than Incite initial void ratio co As the soil is expanding or swelling. So swelling andex will be applõed. And oo <olc 50 <100 pro 1 of loc » 30 < 100 freda so, swelling Index is of preconsoledation Range - l2 = 6,+ es log (0) Cs= Swelling Index As In the quer question Swelling Indese is not quen we will cacheate it from previous data in the question that falls into precorded cation Range Conco) As in the guer will cacheale at falls into

0.05= be log/50) 30 De = 0.011 ef= 60+0.011 et = 1.2+0.011 ef=1211 And PART (e) oo=50 kPa , lo=1.2 rc=50ffa , of=30 grileal stress = preconsolidation stress=5 0 kPa final seless a gniteal stress » 30250 Now , final seress (of) < gneteal stress oo) soil is under Expension. And also it is in preconsolidation Range. Hence (Cs u cry will be applied. Cs= se Де o se=0.oll 0.05 = longer eta lo +0.011 ef=1.211 And