Find: Time [yr] for 90% consolidation to occur Load=300 [lb/ft2] Sand 17 26 C) 42 D) 53 γT=100 [lb/ft3] 10 [ft] Clay 25 [ft] OCR=1.0 Find the time, in years, it will take the given soil profile to settle 90% of it’s ultimate consolidation. In this problem, γT=105 [lb/ft3] cV =10 [ft2/yr] CC=0.35 CR=0.05 Impermeable Bedrock
Find: Time [yr] for 90% consolidation to occur Load=300 [lb/ft2] Sand 17 26 C) 42 D) 53 γT=100 [lb/ft3] 10 [ft] Clay 25 [ft] OCR=1.0 a 10 foot layer of sand sits above --- γT=105 [lb/ft3] cV =10 [ft2/yr] CC=0.35 CR=0.05 Impermeable Bedrock
Find: Time [yr] for 90% consolidation to occur Load=300 [lb/ft2] Sand 17 26 C) 42 D) 53 γT=100 [lb/ft3] 10 [ft] Clay 25 [ft] OCR=1.0 a 25 foot layer of clay, γT=105 [lb/ft3] cV =10 [ft2/yr] CC=0.35 CR=0.05 Impermeable Bedrock
Find: Time [yr] for 90% consolidation to occur Load=300 [lb/ft2] Sand 17 26 C) 42 D) 53 γT=100 [lb/ft3] 10 [ft] Clay 25 [ft] OCR=1.0 which is above an impermeable bedrock. γT=105 [lb/ft3] cV =10 [ft2/yr] CC=0.35 CR=0.05 Impermeable Bedrock
Find: Time [yr] for 90% consolidation to occur Load=300 [lb/ft2] Sand 17 26 C) 42 D) 53 γT=100 [lb/ft3] 10 [ft] Clay 25 [ft] OCR=1.0 It is assumed the ground water table is not present. γT=105 [lb/ft3] cV =10 [ft2/yr] CC=0.35 CR=0.05 Impermeable Bedrock
Find: Time [yr] for 90% consolidation to occur Load=300 [lb/ft2] Sand 17 26 C) 42 D) 53 γT=100 [lb/ft3] 10 [ft] Clay 25 [ft] OCR=1.0 We’ve also been given a number of the soil properties. γT=105 [lb/ft3] cV =10 [ft2/yr] CC=0.35 CR=0.05 Impermeable Bedrock
Find: Time [yr] for 90% consolidation to occur Load=300 [lb/ft2] cv * t Sand T= γT=100 [lb/ft3] 10 [ft] (HD)2 Clay 25 [ft] OCR=1.0 We begin with our time rate of consolidation equation, and solve for our time, little t. γT=105 [lb/ft3] cV =10 [ft2/yr] CC=0.35 CR=0.05 Impermeable Bedrock
Find: Time [yr] for 90% consolidation to occur Load=300 [lb/ft2] cv * t Sand T= γT=100 [lb/ft3] 10 [ft] (HD)2 T * (HD)2 Clay t= 25 [ft] OCR=1.0 We’ve been given the consolidation coefficient for the clay layer, cv γT=105 [lb/ft3] cV=10 [ft2/yr] CC=0.35 CR=0.05 Impermeable Bedrock
Find: Time [yr] for 90% consolidation to occur Load=300 [lb/ft2] cv * t Sand T= γT=100 [lb/ft3] 10 [ft] (HD)2 T * (HD)2 Clay t= 25 [ft] OCR=1.0 which is 10 feet squared per year. cv γT=105 [lb/ft3] cV=10 [ft2/yr] CC=0.35 CR=0.05 Impermeable Bedrock
Find: Time [yr] for 90% consolidation to occur Load=300 [lb/ft2] cv * t Sand T= γT=100 [lb/ft3] 10 [ft] (HD)2 T * (HD)2 Clay t= 25 [ft] OCR=1.0 We also know the drainage depth is 25 feet. This is because the impermeable bedrock would prevent water from leaving the clay layer in the downward direction. cv γT=105 [lb/ft3] cV=10 [ft2/yr] CC=0.35 CR=0.05 Impermeable Bedrock
? Find: Time [yr] for 90% consolidation to occur cv * t T= (HD)2 Load=300 [lb/ft2] cv * t Sand T= γT=100 [lb/ft3] 10 [ft] (HD)2 ? T * (HD)2 Clay t= 25 [ft] OCR=1.0 But we don’t yet know the time factor, big ‘T.’ cv γT=105 [lb/ft3] cV=10 [ft2/yr] CC=0.35 CR=0.05 Impermeable Bedrock
? Find: Time [yr] for 90% consolidation to occur cv * t T= (HD)2 Load=300 [lb/ft2] cv * t Sand T= γT=100 [lb/ft3] 10 [ft] (HD)2 ? T * (HD)2 Clay t= 25 [ft] OCR=1.0 When a soil’s average degree of consolidation, ‘U’ is less than 60% of its ultimate consolidation, cv γT=105 [lb/ft3] If U<60% cV=10 [ft2/yr] CC=0.35 CR=0.05 Impermeable Bedrock
? Find: Time [yr] for 90% consolidation to occur π * U2 cv * t T= Load=300 [lb/ft2] cv * t Sand T= γT=100 [lb/ft3] 10 [ft] (HD)2 ? T * (HD)2 Clay t= 25 [ft] OCR=1.0 the time factor, ‘T,’ is equal to PI times ‘U’ squared, all divided by 4. Where U is a decimal. cv γT=105 [lb/ft3] π * U2 If U<60% then T= cV=10 [ft2/yr] 4 CC=0.35 CR=0.05 Impermeable Bedrock
? Find: Time [yr] for 90% consolidation to occur π * U2 cv * t T= Load=300 [lb/ft2] cv * t Sand T= γT=100 [lb/ft3] 10 [ft] (HD)2 ? T * (HD)2 Clay t= 25 [ft] OCR=1.0 When a soil’s average degree of consolidation is greater than 60%, of its ultimate consolidation, cv γT=105 [lb/ft3] π * U2 If U<60% then T= cV=10 [ft2/yr] 4 CC=0.35 CR=0.05 If U>60% Impermeable Bedrock
? Find: Time [yr] for 90% consolidation to occur π * U2 cv * t T= Load=300 [lb/ft2] cv * t Sand T= γT=100 [lb/ft3] 10 [ft] (HD)2 ? T * (HD)2 Clay t= 25 [ft] OCR=1.0 the time factor, is equal to 1.781 minus 0.933 times the log of 100 percent minus ‘U.’ Where ‘U’ is a percent. cv γT=105 [lb/ft3] π * U2 If U<60% then T= cV=10 [ft2/yr] 4 CC=0.35 CR=0.05 then T=1.781-0.933*log(100%-U) If U>60% Impermeable Bedrock
? Find: Time [yr] for 90% consolidation to occur π * U2 cv * t T= Load=300 [lb/ft2] cv * t Sand T= γT=100 [lb/ft3] 10 [ft] (HD)2 ? T * (HD)2 Clay t= 25 [ft] OCR=1.0 In our problem, the average degree of consolidation is 90%, cv γT=105 [lb/ft3] π * U2 If U<60% then T= cV=10 [ft2/yr] 4 CC=0.35 CR=0.05 then T=1.781-0.933*log(100%-U) If U>60% Impermeable Bedrock
? Find: Time [yr] for 90% consolidation to occur π * U2 cv * t T= Load=300 [lb/ft2] cv * t Sand T= γT=100 [lb/ft3] 10 [ft] (HD)2 ? T * (HD)2 Clay t= 25 [ft] OCR=1.0 we we plug this into our equation, and solve for T. cv γT=105 [lb/ft3] π * U2 If U<60% then T= cV=10 [ft2/yr] 4 CC=0.35 CR=0.05 then T=1.781-0.933*log(100%-U) If U>60% Impermeable Bedrock
Find: Time [yr] for 90% consolidation to occur Load=300 [lb/ft2] cv * t Sand T= γT=100 [lb/ft3] 10 [ft] (HD)2 Clay T * (HD)2 t= 25 [ft] 25 [ft] OCR=1.0 The time factor is 0.848. We plug this value into our time rate of consolidation equation, and solve for time. cv cV=10 [ft2/yr] γT=105 [lb/ft3] π * U2 If U<60% then T= cV=10 [ft2/yr] 4 CC=0.35 CR=0.05 then T=1.781-0.933*log(100%-U) If U>60% T=0.848
Find: Time [yr] for 90% consolidation to occur Load=300 [lb/ft2] cv * t Sand T= γT=100 [lb/ft3] 10 [ft] (HD)2 Clay T * (HD)2 t= 25 [ft] 25 [ft] OCR=1.0 It takes the soil 53 years to consolidate 90% of it’s ultimate consolidation settlement. cv cV=10 [ft2/yr] γT=105 [lb/ft3] π * U2 If U<60% then T= cV=10 [ft2/yr] t=53[yr] 4 CC=0.35 CR=0.05 then T=1.781-0.933*log(100%-U) If U<60% T=0.848
Find: Time [yr] for 90% consolidation to occur Load=300 [lb/ft2] 17 26 C) 42 D) 53 cv * t Sand T= γT=100 [lb/ft3] 10 [ft] (HD)2 Clay T * (HD)2 t= 25 [ft] 25 [ft] OCR=1.0 Looking back at our possible solutions, cv cV=10 [ft2/yr] γT=105 [lb/ft3] π * U2 If U<60% then T= cV=10 [ft2/yr] t=53[yr] 4 CC=0.35 CR=0.05 then T=1.781-0.933*log(100%-U) If U<60% T=0.848
Find: Time [yr] for 90% consolidation to occur Load=300 [lb/ft2] 17 26 C) 42 D) 53 cv * t Sand T= γT=100 [lb/ft3] 10 [ft] (HD)2 Clay T * (HD)2 t= 25 [ft] 25 [ft] OCR=1.0 The answer is D. cv cV=10 [ft2/yr] γT=105 [lb/ft3] π * U2 If U<60% then T= cV=10 [ft2/yr] t=53[yr] 4 CC=0.35 CR=0.05 then T=1.781-0.933*log(100%-U) If U<60% AnswerD T=0.848
( ) ? γclay=53.1[lb/ft3] Index σ’v = Σ γ d H*C σfinal ρcn= log Find: σ’v ρc d = 30 feet (1+wc)*γw wc+(1/SG) σ’v = Σ γ d d Sand 10 ft γT=100 [lb/ft3] 100 [lb/ft3] 10 [ft] 20 ft Clay = γsand dsand +γclay dclay A W S V [ft3] W [lb] 40 ft text wc = 37% ? Δh 20 [ft] (5 [cm])2 * π/4 ( ) H*C σfinal ρcn= 1+e σinitial log ‘