Find: cV [in2/min] Deformation Data C) 0.03 D) 0.04 Time

Find: cV [in2/min] 0.01 0.02 Deformation Data C) 0.03 D) 0.04 Time
plan view of 0.01 0.02 C) 0.03 D) 0.04 sample in ring Deformation Data Time Deformation [min] [in] 0.1 0.25 0.5 … 480 1440 0.0000 0.0035 0.0040 0.0050 … 0.0270 0.0275 Load profile view of Find the consolidation coefficient, C sub v, in inches squared per minute. In this problem, soil loaded in consolidometer Ho=1.06 [in] double drain

plan view of 0.01 0.02 C) 0.03 D) 0.04 sample in ring Deformation Data Time Deformation [min] [in] 0.1 0.25 0.5 … 480 1440 0.0000 0.0035 0.0040 0.0050 … 0.0270 0.0275 Load profile view of a sample of soil is loaded into a metal ring, soil loaded in consolidometer Ho=1.06 [in] double drain

plan view of 0.01 0.02 C) 0.03 D) 0.04 sample in ring Deformation Data Time Deformation [min] [in] 0.1 0.25 0.5 … 480 1440 0.0000 0.0035 0.0040 0.0050 … 0.0270 0.0275 Load profile view of and tested in a consolidometer. soil loaded in consolidometer Ho=1.06 [in] double drain

plan view of 0.01 0.02 C) 0.03 D) 0.04 sample in ring Deformation Data Time Deformation [min] [in] 0.1 0.25 0.5 … 480 1440 0.0000 0.0035 0.0040 0.0050 … 0.0270 0.0275 Load profile view of We also been provided soil data. soil loaded in consolidometer Ho=1.06 [in] double drain

Find: cV [in2/min] 0.01 0.02 Deformation Data C) 0.03 D) 0.04
abbreviated table plan view of 0.01 0.02 C) 0.03 D) 0.04 sample in ring Deformation Data Time Deformation [min] [in] 0.1 0.25 0.5 … 480 1440 0.0000 0.0035 0.0040 0.0050 … 0.0270 0.0275 Load profile view of As well as testing results. The deformation, or displacement, of the soil was recorded at various time intervals. soil loaded in consolidometer Ho=1.06 [in] double drain

Find: cV [in2/min] Time Deformation Time Deformation [min] [in] [min]
0.1 0.25 0.5 1 2 4 8 0.0000 0.0035 0.0040 0.0050 0.0080 0.0120 0.0165 0.0195 15 30 60 120 240 480 1440 0.0220 0.0235 0.0250 0.0260 0.0268 0.0270 0.0275 The entire set of deformation data is listed here. [pause] The general equation for ---

Find: cV [in2/min] cv * t T= (HD)2 Deformation Data Time Deformation
0.1 0.25 0.5 … 480 1440 0.0000 0.0035 0.0040 0.0050 … 0.0270 0.0275 the time rate of consolidation is solved for the consolidation coefficient, C sub v. Ho=1.06 [in] double drain

Find: cV [in2/min] cv * t T= (HD)2 T * (HD)2 cv = t Deformation Data
Time Deformation T * (HD)2 [min] [in] cv = t 0.1 0.25 0.5 … 480 1440 0.0000 0.0035 0.0040 0.0050 … 0.0270 0.0275 --- Ho=1.06 [in] double drain

Time Deformation T * (HD)2 [min] [in] cv = t 0.1 0.25 0.5 … 480 1440 0.0000 0.0035 0.0040 0.0050 … 0.0270 0.0275 We’ll use the soil data to determine the drainage height, Ho=1.06 [in] double drain

Time Deformation T * (HD)2 [min] [in] cv = t 0.1 0.25 0.5 … 480 1440 0.0000 0.0035 0.0040 0.0050 … 0.0270 0.0275 and the deformation data to determine the time factor, big T, and the duration of settlement, small t. Ho=1.06 [in] double drain

Time Deformation T * (HD)2 [min] [in] cv = t 0.1 0.25 0.5 … 480 1440 0.0000 0.0035 0.0040 0.0050 … 0.0270 0.0275 Let’s solve for the drainage height, first. Ho=1.06 [in] double drain

Time Deformation T * (HD)2 [min] [in] cv = t 0.1 0.25 0.5 … 480 1440 0.0000 0.0035 0.0040 0.0050 … 0.0270 0.0275 Load The initial height of the sample if 1.06 inches, Ho Ho=1.06 [in] double drain

Time Deformation T * (HD)2 [min] [in] cv = t 0.1 0.25 0.5 … 480 1440 0.0000 0.0035 0.0040 0.0050 … 0.0270 0.0275 Load but since the soil is doubly drained, meaning, water can exit the sample through both the top and bottom of the ring, Ho Ho=1.06 [in] double drain

Time Deformation T * (HD)2 [min] [in] cv = t 0.1 0.25 0.5 … 480 1440 0.0000 0.0035 0.0040 0.0050 … 0.0270 0.0275 Load the drainage height is half the total height, or, 0.53 inches. Ho HD=0.53 [in] Ho=1.06 [in] double drain

Time Deformation T * (HD)2 [min] [in] cv = t 0.1 0.25 0.5 … 480 1440 0.0000 0.0035 0.0040 0.0050 … 0.0270 0.0275 Next we solve for our time factor, and duration of settlement.

Find: cV [in2/min] T * (HD)2 cv = t Time Deformation [min] [in] 4 8 15
30 60 120 240 480 1440 0.0165 0.0195 0.0220 0.0235 0.0250 0.0260 0.0268 0.0270 0.0275 Time Deformation [min] [in] 0.1 0.25 0.5 1 2 0.0000 0.0035 0.0040 0.0050 0.0080 0.0120 Again, we see the entire set of data. We’ll generate a semi-log plot ---

Find: cV [in2/min] ΔH [in] Time [min]
where the vertical axis is deformation, or displacement, in inches, Time [min]

Find: cV [in2/min] ΔH [in] Time [min]
and the horizontal axis is time, in minutes, plotted on a log scale. Time [min]

Find: cV [in2/min] ΔH [in] 0.000 0.005 0.010 0.015 0.020 0.025 0.030
After the data points are plotted, 0.025 0.030 0.1 1 10 100 1000 Time [min]

Find: cV [in2/min] T * (HD)2 cv = t ΔH [in] 0.000 0.005 0.010 0.015
0.020 we return to our time rate of consolidation equation. 0.025 0.030 0.1 1 10 100 1000 Time [min]

? Find: cV [in2/min] T * (HD)2 cv = t ΔH [in] 0.000 0.005 0.010 0.015
where U=50% Find: cV [in2/min] T * (HD)2 ΔH [in] cv = t 0.000 ? 0.005 0.010 0.015 0.020 Our goal here is to find the time duration, little t, it will take the soil to consolidate 50% of it’s ultimate consolidation. 0.025 0.030 0.1 1 10 100 1000 Time [min]

? Find: cV [in2/min] T50 * (HD)2 cv = t50 ΔH [in] 0.000 0.005 0.010
where U=50% Find: cV [in2/min] T50 * (HD)2 ΔH [in] cv = t50 0.000 ? 0.005 0.010 0.015 0.020 Knowing the average degree of consolidation, U, is 50%, we can solve for the time factor, big T. 0.025 0.030 0.1 1 10 100 1000 Time [min]

? Find: cV [in2/min] π * U2 T50 * (HD)2 cv = t50 If U<60% then T= 4
where U=50% Find: cV [in2/min] T50 * (HD)2 cv = t50 ? π * U2 If U<60% then T= 4 When U is less than 60%, T is about PI times, U squared, divided by 4, where U is a decimal.

where U=50% Find: cV [in2/min] T50 * (HD)2 cv = t50 ? π * U2 If U<60% then T= 4 then T= *log(100%-U) If U>60% When U is greater than 60%, T is solved using this second equation, where U is a percent.

where U=50% Find: cV [in2/min] T50 * (HD)2 cv = t50 0.50 ? π * U2 If U<60% then T= 4 Since U is 50%, we’ll use the first equation.

where U=50% Find: cV [in2/min] T50 * (HD)2 cv = t50 0.50 ? π * U2 If U<60% then T= 4 T=0.197 The time factor is

0.015 0.020 To solve the for time it takes the soil to consolidate to half it’s ultimate consolidation, 0.025 0.030 0.1 1 10 100 1000 Time [min]

consolidation settlement 0.010 0.015 0.020 we draw a best fit line along the data points for consolidation settlement, 0.025 0.030 0.1 1 10 100 1000 Time [min]

consolidation settlement 0.010 0.015 creep settlement 0.020 and a second best fit line along the data points for creep settlement. 0.025 0.030 0.1 1 10 100 1000 Time [min]

consolidation settlement “primary consolidation” 0.010 0.015 creep settlement “secondary consolidation” 0.020 As an aside, some books refer to these two types of settlement as primary consolidation and secondary consolidation. 0.025 0.030 0.1 1 10 100 1000 Time [min]

consolidation settlement 0.010 0.015 creep settlement 0.020 Where these two lines intersect represents the displacement of the soil sample, at ultimate consolidation, 0.025 0.030 0.1 1 10 100 1000 Time [min]

consolidation settlement 0.010 0.015 creep settlement 0.020 This ultimate consolidation is inches. 0.025 0.030 0.1 1 10 100 1000 Time [min]

0.015 0.020 Next, we draw horizontal lines at the data points corresponding to the time of 0.25 minutes and 1 minute. 0.025 D100 0.030 0.1 1 10 100 1000 Time [min]

x 0.005 x 0.010 0.015 0.020 Then we add a third horizontal line above the two horizontal line, and with equal separation as between the first two. 0.025 D100 0.030 0.1 1 10 100 1000 Time [min]

0.015 0.020 This third line represents the displacement when no consolidation settlement has occurred. 0.025 D100 0.030 0.1 1 10 100 1000 Time [min]

0.015 0.020 The displacement of the soil at 50% ultimate consolidation, 0.025 D100 0.030 0.1 1 10 100 1000 Time [min]

D50=0.013 [in] 0.015 0.020 is determined by averaging the 0 percent and 100 percent displacement values. 0.025 D100 0.030 0.1 1 10 100 1000 Time [min]

D50=0.013 [in] 0.015 0.020 From the graph, we can determine the time it will take the soil to consolidate to 50% of its ultimate consolidation, 0.025 0.030 0.1 1 10 100 1000 Time [min]

Find: cV [in2/min] T50 * (HD)2 cv = t50 ΔH [in] 0.000 0.005 0.010
D50=0.013 [in] 0.015 0.020 by tracing a line horizontally, right, to the lab test data, and then vertically, down, to the time axis. Our t sub 50 is 2.7 minutes. [pause] 0.025 t50=2.7 [min] 0.030 0.1 1 10 100 1000 Time [min]

Find: cV [in2/min] T50 * (HD)2 cv = t50 0.197 0.53 [in] 2.7 [min]
Plugging in our known values,

Find: cV [in2/min] T50 * (HD)2 cv = t50 cv = 0.020 [in2/min] 0.197
We calculate the consolidation coefficient to be inches squared per minute.

Find: cV [in2/min] T50 * (HD)2 cv = t50 0.01 0.02 C) 0.03 D) 0.04
0.197 0.53 [in] T50 * (HD)2 0.01 0.02 C) 0.03 D) 0.04 cv = t50 2.7 [min] cv = [in2/min] Compared with our possible solutions,

Find: cV [in2/min] T50 * (HD)2 cv = t50 0.01 0.02 C) 0.03 D) 0.04
0.197 0.53 [in] T50 * (HD)2 0.01 0.02 C) 0.03 D) 0.04 cv = t50 2.7 [min] cv = [in2/min] Answer  B the answer is B.

( ) ? 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 ‘

Find: cV [in2/min] Deformation Data C) 0.03 D) 0.04 Time

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Find: cV [in2/min] Deformation Data C) 0.03 D) 0.04 Time

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