Subject : SOIL IMPROVEMENT TECHNIQUES { for MASTER Students }

i need full solution of these TWO questions.

1st question liquefication. 2nd question pile group and stone column.

please solve both of questions with full details step by step. and as soon as possible please.

EXAMPLE FOR QUESTION 1

Question 2

for file group and stone column please take some data from question 1 because these two questions are related to each other

sir please solve as soon as possible

thankyou

please solve both of questions with full details step by step. and as soon as possible please.

+ Question 1)50p. For a site in Lefke Cyprus, site survey was conducted and a tall rise building is to be designed. Check for the liquefaction potential and find the Factor of Safety for each layer of soil. Expected intensity for the earthquake is Mw7.5 and peak ground acceleration is 0.3g. Water table is on surface Layer Soil Layer Liquid Plastic Moisture Fines Unit SPT Thickness blow Limit Limit Content% % weigth kN/m 1 Sand 3 3 3 20 2 Sandy 3 5 40 29 35 52 18 Silt 3 Clay 4 11 55 30 47 92 18 4 Silty sand 4 15 30 20 27 35 20 5 Claye 5 9 23 15 20 15 21 Sand Question 2)50p. Design pile group and stone column solution as load bearing solution in order to resist 80 000 kN total building weigth on 10m x 20m foundation on the soil specified in question 1. Liquefection Tutorial for CEGED36 amax = 0.20g ER=80 M= 6 SPT Layer I Layer No Soifescription The Jeness (1) count, of Fil Send 2 3 3 20 2 5 40 25 35 52 18 Sandy Silt Clay 3 14 55 47 92 18 Silty Saud 15 30 27 35 20 N 5 6 9 23 15 20 15 21 Cayey Sound Bedrock 25 - To check liquefaction susceptibility. two conditions need to be satisfied: 1). Saturation If it's 2) If it's coarse greived soil with FC = 35%. ~ considered Busceptity soil or coarse grained with FC >35%. criteria by Bray & Saucio (2006) that is fiP1 0.85 considered susceptible grained use liquefaction susceptibility LL Layer 1 Coarse with FC-35%. & saturated o -grained susceptible Layer 2 Fiu grained (FC > 50 %). PI = 40-29 = 11 & saturated W 35-0.875 susceptible LL 40 Layer 3 Fine (FC>50%) PI-55-30-25 > 12 not susceptibl Layer 4 Coarse -grained (FC#35%) & saturated vs susceptible Layer 5 Coarse grained ( FC= 35%) & saturated~ susceptible grained! N35 + 10 4 Liguid A 6 8 x 20 - In the middle of Sand layer (1m). = 20 1 = 20 kPa. -26= 10.1 = 10 kPa 5w0 = 10 kPa = 1-0.00765x2= 0.99235 CSR=065 amay. Fre ra= 0.65.0.2. 20-8.99235= 6.258 0 Po towo 1,7 = 10 = $10 = 3.16 71.7 ~CN= 107 (N60 = . CN. CE = 3-1.700 = 6.8=7 0.6 CRRAS = 0.08 from Figure The MSI for M6.0 is 1.32 FS = (CRR ).MSF=- 0.08 0.256 CSR In the middle of sift layer f3.5m):. Ovo = 20 x2 + 16-1.5 = 67 kPa U= 103.5 - 35 kPa v=67-35= 32 k. Pa = 1-0.00765-3.5 = 0.9732 CSR=0.65 0.2.67 0-97320.265 32 100 CN= 32 = 1.77 21.7 = 1.7 (N) = 5+1.700 11 0=5 6=1.2 for FC > 35 %, ( 160, cs= a + b / N160 = 5+1-2-11-18 - 1.32 0.41 liquefaction From Figure of clean sand curve: CRRY.S=0.2 FS= 0.2 -1.32=1 liquefaction 0.265 . In the middle of Silty Sand layer (12 m) : Ovo =20*2 +18+ 3 + 18+ 5 + 20 2 = 224 kPa 2 = 1012= 120 kPa Ovd=224-120 = 104 kPa 1-1.174-00267 +12=0.8536 CSR=0.65 0.2 224 0.8536=0.239 2 104 100 CN = 1104 = 0 981 (N) 150.981-08-20 - From Fig & 35%, fines curve CRR = 0.4 FS = (0.4) 1.32 = 2.2 No liquefaction 0.239/ In the middle of clayy sand layer (17 m): Jv= 202 + 16 =3 +18 +5+20+4 +213 = 327 kPa u= 10*17=1to lupa 327-170 = 157 kPa 1.114 0.0267 17 = 0.72 CSR=0.65 0.2 327 > 0.72 = 0.195 157 100 CN = 0.6 (N) 60 = 90.000 10 From Fiqe 15% fines: CRA CRR = 0.14 FS= 0.14 Q.195 -1.32= 0.95 Liquefactio + Question 1)50p. For a site in Lefke Cyprus, site survey was conducted and a tall rise building is to be designed. Check for the liquefaction potential and find the Factor of Safety for each layer of soil. Expected intensity for the earthquake is Mw7.5 and peak ground acceleration is 0.3g. Water table is on surface Layer Soil Layer Liquid Plastic Moisture Fines Unit SPT Thickness blow Limit Limit Content% % weigth kN/m 1 Sand 3 3 3 20 2 Sandy 3 5 40 29 35 52 18 Silt 3 Clay 4 11 55 30 47 92 18 4 Silty sand 4 15 30 20 27 35 20 5 Claye 5 9 23 15 20 15 21 Sand Question 2)50p. Design pile group and stone column solution as load bearing solution in order to resist 80 000 kN total building weigth on 10m x 20m foundation on the soil specified in question 1. Liquefection Tutorial for CEGED36 amax = 0.20g ER=80 M= 6 SPT Layer I Layer No Soifescription The Jeness (1) count, of Fil Send 2 3 3 20 2 5 40 25 35 52 18 Sandy Silt Clay 3 14 55 47 92 18 Silty Saud 15 30 27 35 20 N 5 6 9 23 15 20 15 21 Cayey Sound Bedrock 25 - To check liquefaction susceptibility. two conditions need to be satisfied: 1). Saturation If it's 2) If it's coarse greived soil with FC = 35%. ~ considered Busceptity soil or coarse grained with FC >35%. criteria by Bray & Saucio (2006) that is fiP1 0.85 considered susceptible grained use liquefaction susceptibility LL Layer 1 Coarse with FC-35%. & saturated o -grained susceptible Layer 2 Fiu grained (FC > 50 %). PI = 40-29 = 11 & saturated W 35-0.875 susceptible LL 40 Layer 3 Fine (FC>50%) PI-55-30-25 > 12 not susceptibl Layer 4 Coarse -grained (FC#35%) & saturated vs susceptible Layer 5 Coarse grained ( FC= 35%) & saturated~ susceptible grained! N35 + 10 4 Liguid A 6 8 x 20 - In the middle of Sand layer (1m). = 20 1 = 20 kPa. -26= 10.1 = 10 kPa 5w0 = 10 kPa = 1-0.00765x2= 0.99235 CSR=065 amay. Fre ra= 0.65.0.2. 20-8.99235= 6.258 0 Po towo 1,7 = 10 = $10 = 3.16 71.7 ~CN= 107 (N60 = . CN. CE = 3-1.700 = 6.8=7 0.6 CRRAS = 0.08 from Figure The MSI for M6.0 is 1.32 FS = (CRR ).MSF=- 0.08 0.256 CSR In the middle of sift layer f3.5m):. Ovo = 20 x2 + 16-1.5 = 67 kPa U= 103.5 - 35 kPa v=67-35= 32 k. Pa = 1-0.00765-3.5 = 0.9732 CSR=0.65 0.2.67 0-97320.265 32 100 CN= 32 = 1.77 21.7 = 1.7 (N) = 5+1.700 11 0=5 6=1.2 for FC > 35 %, ( 160, cs= a + b / N160 = 5+1-2-11-18 - 1.32 0.41 liquefaction From Figure of clean sand curve: CRRY.S=0.2 FS= 0.2 -1.32=1 liquefaction 0.265 . In the middle of Silty Sand layer (12 m) : Ovo =20*2 +18+ 3 + 18+ 5 + 20 2 = 224 kPa 2 = 1012= 120 kPa Ovd=224-120 = 104 kPa 1-1.174-00267 +12=0.8536 CSR=0.65 0.2 224 0.8536=0.239 2 104 100 CN = 1104 = 0 981 (N) 150.981-08-20 - From Fig & 35%, fines curve CRR = 0.4 FS = (0.4) 1.32 = 2.2 No liquefaction 0.239/ In the middle of clayy sand layer (17 m): Jv= 202 + 16 =3 +18 +5+20+4 +213 = 327 kPa u= 10*17=1to lupa 327-170 = 157 kPa 1.114 0.0267 17 = 0.72 CSR=0.65 0.2 327 > 0.72 = 0.195 157 100 CN = 0.6 (N) 60 = 90.000 10 From Fiqe 15% fines: CRA CRR = 0.14 FS= 0.14 Q.195 -1.32= 0.95 Liquefactio