Find: ρc [in] from load after 2 years Load=400 [lb/ft2] 2.25 2.50 C) 2.75 D) 3.00 γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] γT=110 [lb/ft3] 20 [ft] Find the settlement from consolidation in inches from a load after 2 years. In this problem, Clay CR=0.04 CC=0.40 normally consolidated CV=5 [ft2/year] γT=118 [lb/ft3] Gravel 4[ft]

Find: ρc [in] from load after 2 years Load=400 [lb/ft2] 2.25 2.50 C) 2.75 D) 3.00 γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] γT=110 [lb/ft3] 20 [ft] A load of 400 pounds per square feet is applied to a soil profile consisting of --- Clay CR=0.04 CC=0.40 normally consolidated CV=5 [ft2/year] γT=118 [lb/ft3] Gravel 4[ft]

Find: ρc [in] from load after 2 years Load=400 [lb/ft2] 2.25 2.50 C) 2.75 D) 3.00 γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] γT=110 [lb/ft3] 20 [ft] a 6 foot thick layer of sand, Clay CR=0.04 CC=0.40 normally consolidated CV=5 [ft2/year] γT=118 [lb/ft3] Gravel 4[ft]

Find: ρc [in] from load after 2 years Load=400 [lb/ft2] 2.25 2.50 C) 2.75 D) 3.00 γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] γT=110 [lb/ft3] 20 [ft] a 20 foot thick layer of clay Clay CR=0.04 CC=0.40 normally consolidated CV=5 [ft2/year] γT=118 [lb/ft3] Gravel 4[ft]

Find: ρc [in] from load after 2 years Load=400 [lb/ft2] 2.25 2.50 C) 2.75 D) 3.00 γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] γT=110 [lb/ft3] 20 [ft] and a 4 foot thick layer of gravel. Clay CR=0.04 CC=0.40 normally consolidated CV=5 [ft2/year] γT=118 [lb/ft3] Gravel 4[ft]

Find: ρc [in] from load after 2 years Load=400 [lb/ft2] 2.25 2.50 C) 2.75 D) 3.00 γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] γT=110 [lb/ft3] 20 [ft] We’ve also been provided various soil properties. [pause] Clay CR=0.04 CC=0.40 normally consolidated CV=5 [ft2/year] γT=118 [lb/ft3] Gravel 4[ft]

Find: ρc [in] from load after 2 years Load=400 [lb/ft2] 2.25 2.50 C) 2.75 D) 3.00 γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] γT=110 [lb/ft3] 20 [ft] In this problem, we assume the clay is the only soil type which consolidates. Clay CR=0.04 CC=0.40 normally consolidated CV=5 [ft2/year] γT=118 [lb/ft3] Gravel 4[ft]

( ) Find: ρc [in] from load after 2 years H*C σfinal ρc= log Load=400 [lb/ft2] ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] γT=110 [lb/ft3] 20 [ft] Clay So we begin by solving for the ultimate consolidation in the clay layer. CR=0.04 CC=0.40 normally consolidated CV=5 [ft2/year] Grav γT=118 [lb/ft3] 4[ft]

( ) Find: ρc [in] from load after 2 years H*C σfinal ρc= log Load=400 [lb/ft2] 20 [ft] ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] γT=110 [lb/ft3] 20 [ft] Clay The height of the clay layer is 20 feet. CR=0.04 CC=0.40 normally consolidated CV=5 [ft2/year] Grav γT=118 [lb/ft3] 4[ft]

( ) Find: ρc [in] from load after 2 years H*C σfinal ρc= log Load=400 [lb/ft2] 20 [ft] ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] γT=110 [lb/ft3] 20 [ft] Clay Since the clay is normally consolidated, CR=0.04 CC=0.40 normally consolidated CV=5 [ft2/year] Grav γT=118 [lb/ft3] 4[ft]

( ) Find: ρc [in] from load after 2 years H*C σfinal ρc= log Load=400 [lb/ft2] 20 [ft] 0.40 ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] γT=110 [lb/ft3] 20 [ft] Clay we’ll use 0.40 for our consolidation index. CR=0.04 CC=0.40 normally consolidated CV=5 [ft2/year] Grav γT=118 [lb/ft3] 4[ft]

( ) ? Find: ρc [in] from load after 2 years H*C σfinal ρc= log Load=400 [lb/ft2] 20 [ft] 0.40 ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] ? 2 [ft] γT=110 [lb/ft3] 20 [ft] Clay Next, we’ll solve for the initial void ratio. CR=0.04 CC=0.40 normally consolidated CV=5 [ft2/year] Grav γT=118 [lb/ft3] 4[ft]

( ) ? Find: ρc [in] from load after 2 years γT= H*C σfinal ρc= log Load=400 [lb/ft2] 20 [ft] 0.40 ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] ? 2 [ft] (SG+e) * γW γT=110 [lb/ft3] 20 [ft] γT= Clay Since the clay layer is saturated, we know the total unit weight can be determined from the specific gravity, void ratio and unit weight of water. 1+e CR=0.04 CC=0.40 normally consolidated CV=5 [ft2/year] Grav γT=118 [lb/ft3] 4[ft]

( ) ? Find: ρc [in] from load after 2 years γT - γW γT= H*C σfinal ρc= Load=400 [lb/ft2] 20 [ft] 0.40 ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] ? 2 [ft] (SG+e) * γW γT=110 [lb/ft3] 20 [ft] γT= Clay After solving for the void ratio --- 1+e CR=0.04 CC=0.40 normally consolidated SG * γW - γT CV=5 [ft2/year] e= γT - γW Grav γT=118 [lb/ft3] 4[ft]

( ) ? Find: ρc [in] from load after 2 years γT - γW γT= H*C σfinal ρc= Load=400 [lb/ft2] 20 [ft] 0.40 ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] ? 2 [ft] (SG+e) * γW γT=110 [lb/ft3] 20 [ft] γT= Clay and plugging in our known values, assuming the specific gravity is 2.70, 1+e CR=0.04 CC=0.40 2.70 normally consolidated SG * γW - γT CV=5 [ft2/year] e= 62.4 [lb/ft3] γT - γW Grav γT=118 [lb/ft3] 4[ft]

( ) ? Find: ρc [in] from load after 2 years γT - γW γT= H*C σfinal ρc= Load=400 [lb/ft2] 20 [ft] 0.40 ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] ? 2 [ft] (SG+e) * γW γT=110 [lb/ft3] 20 [ft] γT= Clay we find the initial void ratio to be 1.23. 1+e CR=0.04 CC=0.40 2.70 normally consolidated SG * γW - γT CV=5 [ft2/year] e= 62.4 [lb/ft3] γT - γW Grav γT=118 [lb/ft3] 4[ft] e=1.23

( ) Find: ρc [in] from load after 2 years H*C σfinal ρc= log Load=400 [lb/ft2] 20 [ft] 0.40 ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] 1.23 γT=110 [lb/ft3] 20 [ft] Clay Now, all we have to solve for --- CR=0.04 CC=0.40 normally consolidated CV=5 [ft2/year] Grav γT=118 [lb/ft3] 4[ft]

( ) ? Find: ρc [in] from load after 2 years H*C σfinal ρc= log Load=400 [lb/ft2] 20 [ft] 0.40 ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] ? 1.23 γT=110 [lb/ft3] 20 [ft] Clay is the initial and final vertical effective stress values. CR=0.04 CC=0.40 normally consolidated CV=5 [ft2/year] Grav γT=118 [lb/ft3] 4[ft]

( ) ? Find: ρc [in] from load after 2 years H*C σfinal ρc= log Load=400 [lb/ft2] 20 [ft] 0.40 ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] ? 1.23 σ’i = Σ γ’ * d γT=110 [lb/ft3] 20 [ft] Clay Let’s evaluate a point in the middle of the clay layer, which would be at a total depth of 16 feet. CR=0.04 CC=0.40 normally consolidated CV=5 [ft2/year] Grav γT=118 [lb/ft3] 4[ft]

( ) ? Find: ρc [in] from load after 2 years H*C σfinal ρc= log Load=400 [lb/ft2] 20 [ft] 0.40 ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] ? 1.23 γT=110 [lb/ft3] σ’i,= 4 [ft] * 106 [lb/ft3] 20 [ft] We add the overburden stress from the unsaturated sand, CR=0.04 CC=0.40 +… normally consolidated CV=5 [ft2/year] Grav γT=118 [lb/ft3] 4[ft]

( ) ? Find: ρc [in] from load after 2 years H*C σfinal ρc= log Load=400 [lb/ft2] 20 [ft] 0.40 ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] ? 1.23 γT=110 [lb/ft3] σ’i,= 4 [ft] * 106 [lb/ft3] 20 [ft] from the saturated sand, CR=0.04 CC=0.40 + 2 [ft] * (112-62.4) [lb/ft3] normally consolidated +… CV=5 [ft2/year] Grav γT=118 [lb/ft3] 4[ft]

( ) ? Find: ρc [in] from load after 2 years H*C σfinal ρc= log Load=400 [lb/ft2] 20 [ft] 0.40 ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] ? 1.23 γT=110 [lb/ft3] σ’i,= 4 [ft] * 106 [lb/ft3] 20 [ft] Clay and from the top 10 feet of the clay later. CR=0.04 CC=0.40 + 2 [ft] * (112-62.4) [lb/ft3] normally consolidated +(10 [ft]) * (110-62.4) [lb/ft3] CV=5 [ft2/year] Grav γT=118 [lb/ft3] 4[ft]

( ) ? Find: ρc [in] from load after 2 years H*C σfinal ρc= log Load=400 [lb/ft2] 20 [ft] 0.40 ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] ? 1.23 γT=110 [lb/ft3] σ’i,= 4 [ft] * 106 [lb/ft3] 20 [ft] Clay Our initial stress is 999 pounds per square feet, CR=0.04 CC=0.40 + 2 [ft] * (112-62.4) [lb/ft3] normally consolidated +(10 [ft]) * (110-62.4) [lb/ft3] CV=5 [ft2/year] σ’i,clay= 999 [lb/ft2] Grav γT=118 [lb/ft3] 4[ft]

( ) ? Find: ρc [in] from load after 2 years H*C σfinal ρc= log Load=400 [lb/ft2] 20 [ft] 0.40 ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] ? 1.23 γT=110 [lb/ft3] σ’i,= 4 [ft] * 106 [lb/ft3] 20 [ft] Clay and after adding the load of 400 pounds per square feet, CR=0.04 CC=0.40 + 2 [ft] * (112-62.4) [lb/ft3] normally consolidated +(10 [ft]) * (110-62.4) [lb/ft3] CV=5 [ft2/year] σ’i,clay= 999 [lb/ft2] Grav γT=118 [lb/ft3] 4[ft]

( ) ? Find: ρc [in] from load after 2 years H*C σfinal ρc= log Load=400 [lb/ft2] 20 [ft] 0.40 ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] ? 1.23 γT=110 [lb/ft3] σ’i,= 4 [ft] * 106 [lb/ft3] 20 [ft] Clay our final stress is 1,399 pounds per square feet, CR=0.04 CC=0.40 + 2 [ft] * (112-62.4) [lb/ft3] normally consolidated +(10 [ft]) * (110-62.4) [lb/ft3] CV=5 [ft2/year] σ’i,clay= 999 [lb/ft2] Grav γT=118 [lb/ft3] σ’f,clay= 1,399 [lb/ft2] 4[ft]

( ) Find: ρc [in] from load after 2 years H*C σfinal ρc= log Load=400 [lb/ft2] 20 [ft] 0.40 1,399 [lb/ft2] ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] 1.23 999 [lb/ft2] γT=110 [lb/ft3] 20 [ft] Clay We update our settlement equation --- CR=0.04 CC=0.40 normally consolidated CV=5 [ft2/year] Grav γT=118 [lb/ft3] 4[ft]

( ) Find: ρc [in] from load after 2 years H*C σfinal ρc= log Load=400 [lb/ft2] 20 [ft] 0.40 1,399 [lb/ft2] ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] 1.23 999 [lb/ft2] γT=110 [lb/ft3] ρc= 6.30 [in] 20 [ft] Clay to find the consolidation settlement is 6.30 inches. This is the maximum consolidation settlement which would occur, CR=0.04 CC=0.40 normally consolidated CV=5 [ft2/year] Grav γT=118 [lb/ft3] 4[ft]

( ) Find: ρc [in] from load after 2 years H*C σfinal ρc= log Load=400 [lb/ft2] 20 [ft] 0.40 1,399 [lb/ft2] ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] 1.23 999 [lb/ft2] γT=110 [lb/ft3] ρc= 6.30 [in] 20 [ft] Clay if the soil had an infinite amount of time to consolidate. CR=0.04 CC=0.40 normally consolidated CV=5 [ft2/year] Ultimate consolidation (given infinite time) Grav γT=118 [lb/ft3] 4[ft]

( ) Find: ρc [in] from load after 2 years H*C σfinal ρc= log Load=400 [lb/ft2] 20 [ft] 0.40 1,399 [lb/ft2] ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] 1.23 999 [lb/ft2] γT=110 [lb/ft3] ρc= 6.30 [in] 20 [ft] Clay We’ll call this the ultimate consolidation. This is the consolidation which is typically referred to in consolidation settlement problems, when time is not mentioned. CR=0.04 CC=0.40 ρult= 6.30 [in] normally consolidated CV=5 [ft2/year] Ultimate consolidation (given infinite time) Grav γT=118 [lb/ft3] 4[ft]

Find: ρc [in] from load after 2 years Load=400 [lb/ft2] ρult= 6.30 [in] γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] γT=110 [lb/ft3] 20 [ft] Clay However, in this problem, were asked to find the consolidation settlement after 2 years. So, we already know the answer is between 0 and 6.30 inches. CR=0.04 CC=0.40 normally consolidated CV=5 [ft2/year] Grav γT=118 [lb/ft3] 4[ft]

Find: ρc [in] from load after 2 years Load=400 [lb/ft2] ρult= 6.30 [in] ρ2yr = U * ρult γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] γT=110 [lb/ft3] 20 [ft] Clay The consolidation settlement after a certain time, equals the ultimate consolidation times the percent of consolidation which has occurred, U. CR=0.04 CC=0.40 normally consolidated CV=5 [ft2/year] Grav γT=118 [lb/ft3] 4[ft]

<100% Find: ρc [in] from load after 2 years ρult= 6.30 [in] Load=400 [lb/ft2] ρult= 6.30 [in] ρ2yr = U * ρult γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] Average degree of consolidation 0% < <100% γT=110 [lb/ft3] 20 [ft] Clay We’ll call this ‘U’ value the ‘Average Degree of Consolidation’ which ranges from 0 and 100 percent. CR=0.04 CC=0.40 normally consolidated CV=5 [ft2/year] Grav γT=118 [lb/ft3] 4[ft]

<100% Find: ρc [in] from load after 2 years ρult= 6.30 [in] Load=400 [lb/ft2] ρult= 6.30 [in] ρ2yr = U * ρult γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] Average degree of consolidation 0% < <100% γT=110 [lb/ft3] 20 [ft] Clay U is a function of big T, the ‘Time Factor,’ CR=0.04 CC=0.40 U = f ( T ) normally consolidated CV=5 [ft2/year] Grav γT=118 [lb/ft3] 4[ft]

<100% Find: ρc [in] from load after 2 years ρult= 6.30 [in] Load=400 [lb/ft2] ρult= 6.30 [in] ρ2yr = U * ρult γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] Average degree of consolidation 0% < <100% γT=110 [lb/ft3] 20 [ft] Clay And the Time Factor, is equal to, CR=0.04 CC=0.40 U = f ( T ) normally consolidated CV=5 [ft2/year] cv * t T= Grav γT=118 [lb/ft3] 4[ft] (HD)2

<100% Find: ρc [in] from load after 2 years ρult= 6.30 [in] Load=400 [lb/ft2] ρult= 6.30 [in] ρ2yr = U * ρult γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] Average degree of consolidation 0% < <100% γT=110 [lb/ft3] 20 [ft] Clay the time rate of consolidation, which is sometimes called the consolidation coefficient, CR=0.04 CC=0.40 U = f ( T ) 5 [ft2/yr] normally consolidated CV=5 [ft2/year] cv * t T= Grav γT=118 [lb/ft3] 4[ft] (HD)2

<100% Find: ρc [in] from load after 2 years ρult= 6.30 [in] Load=400 [lb/ft2] ρult= 6.30 [in] ρ2yr = U * ρult γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] Average degree of consolidation 0% < <100% γT=110 [lb/ft3] 20 [ft] Clay times the time the soil has to consolidate, CR=0.04 CC=0.40 U = f ( T ) 5 [ft2/yr] normally consolidated 2 [yr] CV=5 [ft2/year] cv * t T= Grav γT=118 [lb/ft3] 4[ft] (HD)2

<100% ? Find: ρc [in] from load after 2 years ρult= 6.30 [in] Load=400 [lb/ft2] ρult= 6.30 [in] ρ2yr = U * ρult γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] Average degree of consolidation 0% < <100% γT=110 [lb/ft3] 20 [ft] Clay divided by the drainage distance, squared. CR=0.04 CC=0.40 U = f ( T ) 5 [ft2/yr] normally consolidated 2 [yr] CV=5 [ft2/year] cv * t T= Grav γT=118 [lb/ft3] 4[ft] (HD)2 ?

<100% ? Find: ρc [in] from load after 2 years ρult= 6.30 [in] Load=400 [lb/ft2] ρult= 6.30 [in] ρ2yr = U * ρult γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] Average degree of consolidation 0% < <100% γT=110 [lb/ft3] 20 [ft] Clay The drainage distance is the furthest distance a drop of water would have to travel to exit the consolidating layer. CR=0.04 CC=0.40 U = f ( T ) 5 [ft2/yr] normally consolidated 2 [yr] CV=5 [ft2/year] cv * t T= Grav γT=118 [lb/ft3] 4[ft] (HD)2 ?

<100% Find: ρc [in] from load after 2 years ρult= 6.30 [in] Load=400 [lb/ft2] ρult= 6.30 [in] ρ2yr = U * ρult γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] Average degree of consolidation 0% < <100% γT=110 [lb/ft3] 20 [ft] 10 [ft] Clay Since our clay layer is doubly drained, the drainage distance is half the layer layer thickness, or, 10 feet. CR=0.04 CC=0.40 U = f ( T ) 5 [ft2/yr] normally consolidated 10 [ft] 2 [yr] CV=5 [ft2/year] cv * t T= Grav γT=118 [lb/ft3] 4[ft] (HD)2 10 [ft]

<100% Find: ρc [in] from load after 2 years ρult= 6.30 [in] Load=400 [lb/ft2] ρult= 6.30 [in] ρ2yr = U * ρult γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] Average degree of consolidation 0% < <100% γT=110 [lb/ft3] 20 [ft] Clay Which makes our Time Factor equal to 0.10. [pause] CR=0.04 CC=0.40 U = f ( T ) 5 [ft2/yr] normally consolidated 2 [yr] CV=5 [ft2/year] cv * t T= =0.10 Grav γT=118 [lb/ft3] 4[ft] (HD)2 10 [ft]

Find: ρc [in] from load after 2 years Load=400 [lb/ft2] ρult= 6.30 [in] ρ2yr = U * ρult γT=106 [lb/ft3] 4 [ft] Sand T=0.10 γT=112 [lb/ft3] 2 [ft] U = f ( T ) γT=110 [lb/ft3] 20 [ft] Clay Next we solve for our average degree of consolidation. CR=0.04 CC=0.40 normally consolidated CV=5 [ft2/year] Grav γT=118 [lb/ft3] 4[ft]

Find: ρc [in] from load after 2 years Load=400 [lb/ft2] ρult= 6.30 [in] ρ2yr = U * ρult γT=106 [lb/ft3] 4 [ft] Sand T=0.10 γT=112 [lb/ft3] 2 [ft] U = f ( T ) γT=110 [lb/ft3] 20 [ft] If T<0.285± Clay When T is less than about 0.285, the average degree of consolidation is equal to the square root of 4 times T, divided by PI, all times 100%. CR=0.04 CC=0.40 Then 4T U≈ *100% π normally consolidated CV=5 [ft2/year] Grav γT=118 [lb/ft3] 4[ft]

Find: ρc [in] from load after 2 years Load=400 [lb/ft2] ρult= 6.30 [in] ρ2yr = U * ρult γT=106 [lb/ft3] 4 [ft] Sand T=0.10 γT=112 [lb/ft3] 2 [ft] U = f ( T ) γT=110 [lb/ft3] 20 [ft] If T<0.285± Clay When T is greater than about 0.285, the average degree of consolidation is this second equation. CR=0.04 CC=0.40 Then 4T U≈ *100% π normally consolidated CV=5 [ft2/year] If T>0.285± Then γT=118 [lb/ft3] U≈100-10{(1.781-T)/0.933}

Find: ρc [in] from load after 2 years Load=400 [lb/ft2] ρult= 6.30 [in] ρ2yr = U * ρult γT=106 [lb/ft3] 4 [ft] Sand T=0.10 γT=112 [lb/ft3] 2 [ft] U = f ( T ) γT=110 [lb/ft3] 20 [ft] If T<0.285± Clay For our problem, T equals 0.10, so we’ll use the first equation to solve for ‘U.’ CR=0.04 CC=0.40 Then 4T U≈ *100% π normally consolidated CV=5 [ft2/year] If T>0.285± Then γT=118 [lb/ft3] U≈100-10{(1.781-T)/0.933}

Find: ρc [in] from load after 2 years Load=400 [lb/ft2] ρult= 6.30 [in] ρ2yr = U * ρult γT=106 [lb/ft3] 4 [ft] Sand T=0.10 γT=112 [lb/ft3] 2 [ft] U = f ( T ) γT=110 [lb/ft3] 20 [ft] If T<0.285± Clay After 2 years, this 20 foot thick clay layer has consolidated 35.7% of it’s ultimate consolidation settlement. CR=0.04 CC=0.40 Then 4T U≈ *100% π normally consolidated CV=5 [ft2/year] U=35.7% Grav γT=118 [lb/ft3]

Find: ρc [in] from load after 2 years Load=400 [lb/ft2] ρult= 6.30 [in] ρ2yr = U * ρult γT=106 [lb/ft3] 4 [ft] Sand T=0.10 γT=112 [lb/ft3] 2 [ft] U = f ( T ) γT=110 [lb/ft3] 20 [ft] If T<0.285± Clay --- CR=0.04 CC=0.40 Then 4T U≈ *100% π normally consolidated CV=5 [ft2/year] U=35.7% Grav γT=118 [lb/ft3]

Find: ρc [in] from load after 2 years Load=400 [lb/ft2] ρult= 6.30 [in] ρ2yr = U * ρult γT=106 [lb/ft3] 4 [ft] Sand T=0.10 γT=112 [lb/ft3] 2 [ft] U = f ( T ) γT=110 [lb/ft3] 20 [ft] If T<0.285± Clay Or, 2.25 inches of settlement. CR=0.04 CC=0.40 Then 4T U≈ *100% π normally consolidated CV=5 [ft2/year] U=35.7% Grav γT=118 [lb/ft3] ρ2yr = 2.25 [in]

Find: ρc [in] from load after 2 years Load=400 [lb/ft2] ρult= 6.30 [in] 2.25 2.50 C) 2.75 D) 3.00 ρ2yr = U * ρult γT=106 [lb/ft3] 4 [ft] Sand T=0.10 γT=112 [lb/ft3] 2 [ft] U = f ( T ) γT=110 [lb/ft3] 20 [ft] If T<0.285± Clay Compared with our possible solutions, CR=0.04 CC=0.40 Then 4T U≈ *100% π normally consolidated CV=5 [ft2/year] U=35.7% Grav γT=118 [lb/ft3] ρ2yr = 2.25 [in]

Find: ρc [in] from load after 2 years Load=400 [lb/ft2] ρult= 6.30 [in] 2.25 2.50 C) 2.75 D) 3.00 ρ2yr = U * ρult γT=106 [lb/ft3] 4 [ft] Sand T=0.10 γT=112 [lb/ft3] 2 [ft] U = f ( T ) γT=110 [lb/ft3] 20 [ft] If T<0.285± Clay the answer is A. CR=0.04 CC=0.40 Then 4T U≈ *100% π normally consolidated CV=5 [ft2/year] U=35.7% Answer  A ρ2yr = 2.25 [in]

( ) ? 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 ‘