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Find: the soil classification
size[mm] % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 SL=25 PL=40 LL=65 4.750[mm],80% 0.850[mm], 60% 0.425[mm], 30% 0.075[mm], 10% SW SW-SM CL-ML SP-SC Find the soil classification. [pause] In this problem, --
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Find: the soil classification
size[mm] % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 SL=25 PL=40 LL=65 4.750[mm],80% 0.850[mm], 60% 0.425[mm], 30% 0.075[mm], 10% SW SW-SM CL-ML SP-SC we’ve been provided the grain-size distribution plot, ---
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Find: the soil classification
size[mm] % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 SL=25 PL=40 LL=65 4.750[mm],80% 0.850[mm], 60% 0.425[mm], 30% 0.075[mm], 10% SW SW-SM CL-ML SP-SC and the Atterberg limits.
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Find: the soil classification
size[mm] % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 SL=25 PL=40 LL=65 4.750[mm],80% 0.850[mm], 60% 0.425[mm], 30% 0.075[mm], 10% SW SW-SM CL-ML SP-SC The grain-size distribution plot shows the result from a sieve analysis test, ---
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Find: the soil classification
size[mm] % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 SL=25 PL=40 LL=65 4.750[mm],80% 0.850[mm], 60% 0.425[mm], 30% 0.075[mm], 10% SW SW-SM CL-ML SP-SC and the exact data for some of these points has been provided. The first step in classifying the soil is to ---
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Find: the soil classification
size[mm] % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 SL=25 PL=40 LL=65 4.750[mm],80% 0.850[mm], 60% 0.425[mm], 30% 0.075[mm], 10% SW SW-SM CL-ML SP-SC determine whether the majority of the soil sample by wight is fine grain material or coarse grain material.
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Find: the soil classification
size[mm] % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 soil type size range [mm] gravel sand fines >4.75 <0.075 SW SW-SM CL-ML SP-SC The percentage of fine and course grain material is determined from the plot, based on the size of each soil type.
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Find: the soil classification
4.750[mm],80% size[mm] % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 soil type size range [mm] gravel sand fines >4.75 <0.075 SW SW-SM CL-ML SP-SC Gravel is considered a course grain soil and ---
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Find: the soil classification
4.750[mm],80% size[mm] % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 soil type size range [mm] gravel sand fines >4.75 <0.075 SW SW-SM CL-ML SP-SC refers to all particles larger than 4.75 millimeters in minimum diameter. gravel
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Find: the soil classification
4.750[mm],80% size[mm] % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 soil type size range [mm] gravel sand fines >4.75 <0.075 0.075[mm], 10% SW SW-SM CL-ML SP-SC Fines is the fine grain material which consists of silt and clay, and --- gravel
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Find: the soil classification
4.750[mm],80% % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 soil type size range [mm] gravel sand fines >4.75 <0.075 0.075[mm], 10% SW SW-SM CL-ML SP-SC refers to all soil smaller than millimeters. size gravel [mm] fines
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Find: the soil classification
4.750[mm],80% % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 soil type size range [mm] gravel sand fines >4.75 <0.075 0.075[mm], 10% SW SW-SM CL-ML SP-SC Sand is a course grain soil and includes all particles smaller than gravel and larger than fines. size sand gravel [mm] fines
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Find: the soil classification
4.750[mm],80% % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 soil type size range [mm] gravel sand fines >4.75 <0.075 0.075[mm], 10% SW SW-SM CL-ML SP-SC The boundary points between soil types are traced over horizontally to the vertical axis, --- size sand gravel [mm] fines
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Find: the soil classification
4.750[mm],80% 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 soil type size range [mm] gravel sand fines >4.75 <0.075 0.075[mm], 10% SW SW-SM CL-ML SP-SC and we can compute the percentage of fine grain material to be --- size fines sand [mm] fines
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Find: the soil classification
4.750[mm],80% 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 soil type size range [mm] gravel sand fines >4.75 <0.075 0.075[mm], 10% SW SW-SM CL-ML SP-SC 10% of the entire sample, by weight. size fines sand [mm] fines =10%-0% =10%
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Find: the soil classification
4.750[mm],80% 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 soil type size range [mm] gravel sand fines >4.75 <0.075 sand 0.075[mm], 10% SW SW-SM CL-ML SP-SC The percentage of sand can be determined by --- size fines sand [mm] fines =10%-0% =10%
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Find: the soil classification
4.750[mm],80% 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 soil type size range [mm] gravel sand fines >4.75 <0.075 sand =80%-10% 0.075[mm], 10% =70% SW SW-SM CL-ML SP-SC Subtracting 10% from 80%, to get 70% sand. size fines sand [mm] fines =10%-0% =10%
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Find: the soil classification
4.750[mm],80% 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 soil type size range [mm] gravel sand fines >4.75 <0.075 gravel sand =80%-10% 0.075[mm], 10% =70% SW SW-SM CL-ML SP-SC And lastly, the the sample is composed of --- size fines sand [mm] fines =10%-0% =10%
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Find: the soil classification
4.750[mm],80% 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 soil type size range [mm] gravel sand fines >4.75 <0.075 gravel=100%-80%=20% sand =80%-10% 0.075[mm], 10% =70% SW SW-SM CL-ML SP-SC 20% gravel. size fines sand [mm] fines =10%-0% =10%
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Find: the soil classification
4.750[mm],80% 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 soil type size range [mm] gravel sand fines >4.75 <0.075 gravel=100%-80%=20% sand =80%-10% 0.075[mm], 10% =70% SW SW-SM CL-ML SP-SC Now that we have our percentages for the gravel, --- size fines sand [mm] fines =10%-0% =10%
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Find: the soil classification
4.750[mm],80% 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 soil type size range [mm] gravel sand fines >4.75 <0.075 gravel=100%-80%=20% sand =80%-10% 0.075[mm], 10% =70% SW SW-SM CL-ML SP-SC the sand, --- size fines sand [mm] fines =10%-0% =10%
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Find: the soil classification
4.750[mm],80% 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 soil type size range [mm] gravel sand fines >4.75 <0.075 gravel=100%-80%=20% sand =80%-10% 0.075[mm], 10% =70% SW SW-SM CL-ML SP-SC and the fines, --- size fines sand [mm] fines =10%-0% =10%
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Find: the soil classification
soil type % weight 4.750[mm],80% 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 gravel sand fines 20% 70% 10% gravel=100%-80%=20% sand =80%-10% 0.075[mm], 10% =70% SW SW-SM CL-ML SP-SC We know we’re dealing with a coarse grain soil. size fines sand [mm] fines =10%-0% =10%
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Find: the soil classification
soil type % weight 4.750[mm],80% 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 gravel sand fines 20% 70% 10% gravel=100%-80%=20% sand =80%-10% 0.075[mm], 10% =70% SW SW-SM CL-ML SP-SC And since there is a larger percentage of sand, than gravel, the soil is classified as a sand. size fines sand [mm] fines =10%-0% =10%
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Find: the soil classification
soil type % weight 4.750[mm],80% 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 gravel sand fines 20% 70% 10% gravel=100%-80%=20% sand =80%-10% 0.075[mm], 10% =70% SW SW-SM CL-ML SP-SC Since there is no more than 12% fines by weight in the sample, we must calculate the --- size fines sand [mm] fines =10%-0% =10%
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Find: the soil classification
soil type % weight % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 gravel sand fines 20% 70% 10% Cu coefficient of uniformity, --- size [mm]
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Find: the soil classification
soil type % weight % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 gravel sand fines 20% 70% 10% Cu Cc and the coefficient of curvature, --- size [mm]
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Find: the soil classification
soil type % weight % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 gravel sand fines 20% 70% 10% D60 Cu= D10 Cc which are based on the values of D10, --- size [mm]
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Find: the soil classification
soil type % weight % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 gravel sand fines 20% 70% 10% D60 Cu= D10 2 D30 Cc= D60*D10 D60, and D30. size [mm]
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Find: the soil classification
soil type % weight % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 gravel sand fines 20% 70% 10% D60 Cu= D10 2 D30 Cc= D60*D10 D10 refers to the diameter at which only 10% of the soil, by weight, can pass through. size [mm] 0.075[mm], 10%
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Find: the soil classification
soil type % weight % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 gravel sand fines 20% 70% 10% D60 Cu= D10 2 D30 Cc= D60*D10 Here, our D10 value is millimeters. D10=0.075[mm] size [mm] 0.075[mm], 10% D10
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Find: the soil classification
soil type % weight % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 gravel sand fines 20% 70% 10% 0.425[mm], 30% D60 Cu= D10 2 D30 Cc= D60*D10 Likewise, D30 refers to the diameter which only 30% of the soil, by weight, can pass through. D10=0.075[mm] size [mm] 0.075[mm], 10% D10
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Find: the soil classification
soil type % weight % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 gravel sand fines 20% 70% 10% 0.425[mm], 30% D60 Cu= D30 D10 2 D30 Cc= D60*D10 The D30 value is millimeters, --- D10=0.075[mm] size [mm] D30=0.425[mm] 0.075[mm], 10% D10
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Find: the soil classification
soil type % weight % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 gravel sand fines 20% 70% 10% 0.850[mm], 60% 0.425[mm], 30% D60 Cu= D30 D10 2 D30 Cc= D60*D10 and the D60 value is 0.85 millimeters. D10=0.075[mm] size [mm] D30=0.425[mm] 0.075[mm], 10% D60=0.850[mm] D10
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Find: the soil classification
soil type % weight % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 gravel sand fines 20% 70% 10% 0.850[mm], 60% 0.425[mm], 30% D60 Cu= D30 D10 2 D30 Cc= D60*D10 The values are plugged into the equations for --- D10=0.075[mm] size [mm] D30=0.425[mm] 0.075[mm], 10% D60=0.850[mm] D10
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Find: the soil classification
soil type % weight % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 gravel sand fines 20% 70% 10% 0.850[mm], 60% 0.425[mm], 30% D60 Cu= D30 D10 2 D30 Cc= D60*D10 D10, --- D10=0.075[mm] size [mm] D30=0.425[mm] 0.075[mm], 10% D60=0.850[mm] D10
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Find: the soil classification
soil type % weight % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 gravel sand fines 20% 70% 10% 0.850[mm], 60% 0.425[mm], 30% D60 Cu= D30 D10 2 D30 Cc= D60*D10 D30, --- D10=0.075[mm] size [mm] D30=0.425[mm] 0.075[mm], 10% D60=0.850[mm] D10
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Find: the soil classification
soil type % weight % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 gravel sand fines 20% 70% 10% 0.850[mm], 60% 0.425[mm], 30% D60 Cu= D30 D10 2 D30 Cc= D60*D10 and D60. D10=0.075[mm] size [mm] D30=0.425[mm] 0.075[mm], 10% D60=0.850[mm] D10
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Find: the soil classification
soil type % weight % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 gravel sand fines 20% 70% 10% 0.850[mm], 60% 0.425[mm], 30% D60 Cu= Cu=11.3 D30 D10 2 D30 Cc= D60*D10 The coefficient of uniformity computes to 11.3, --- D10=0.075[mm] size [mm] D30=0.425[mm] 0.075[mm], 10% D60=0.850[mm] D10
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Find: the soil classification
soil type % weight % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 gravel sand fines 20% 70% 10% 0.850[mm], 60% 0.425[mm], 30% D60 Cu= Cu=11.3 D30 D10 2 D30 Cc= Cc=2.8 D60*D10 And the coefficient of curvature to 2.8. D10=0.075[mm] size [mm] D30=0.425[mm] 0.075[mm], 10% D60=0.850[mm] D10
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Find: the soil classification
soil type % weight % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 gravel sand fines 20% 70% 10% 0.850[mm], 60% 0.425[mm], 30% Cu=11.3 Cc=2.8 D30 Based on the ASTM standards, --- size [mm] 0.075[mm], 10% D10
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Find: the soil classification
soil type % weight % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 gravel sand fines 20% 70% 10% 0.850[mm], 60% 0.425[mm], 30% Cu=11.3 Cc=2.8 D30 well graded if Cu 6 and 1 Cc 3 or Cu 6 1 Cc 3 poorly graded if the sand is well-graded if, --- size [mm] 0.075[mm], 10% D10
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Find: the soil classification
soil type % weight % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 gravel sand fines 20% 70% 10% 0.850[mm], 60% 0.425[mm], 30% Cu=11.3 Cc=2.8 D30 well graded if Cu 6 and 1 Cc 3 or Cu 6 1 Cc 3 poorly graded if the coefficient of uniformity is at least 6, --- size [mm] 0.075[mm], 10% D10
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Find: the soil classification
soil type % weight % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 gravel sand fines 20% 70% 10% 0.850[mm], 60% 0.425[mm], 30% Cu=11.3 Cc=2.8 D30 well graded if Cu 6 and 1 Cc 3 or Cu 6 1 Cc 3 poorly graded if and the coefficient of curvature is at least 1 and at most 3. size [mm] 0.075[mm], 10% D10
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Find: the soil classification
soil type % weight % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 gravel sand fines 20% 70% 10% 0.850[mm], 60% 0.425[mm], 30% Cu=11.3 Cc=2.8 D30 well graded if Cu 6 and 1 Cc 3 or Cu 6 1 Cc 3 poorly graded if It turns out this sample is well graded. size [mm] 0.075[mm], 10% D10
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Find: the soil classification
soil type % weight % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 gravel sand fines 20% 70% 10% 0.850[mm], 60% 0.425[mm], 30% Cu=11.3 Coarse grain soil Cc=2.8 D30 well graded if Cu 6 and 1 Cc 3 or Cu 6 1 Cc 3 poorly graded if So far we know we have a coarse grain soil, --- size [mm] 0.075[mm], 10% D10
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Find: the soil classification
soil type % weight % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 gravel sand fines 20% 70% 10% 0.850[mm], 60% 0.425[mm], 30% Cu=11.3 Coarse grain soil % sand > % gravel Cc=2.8 D30 well graded if Cu 6 and 1 Cc 3 or Cu 6 1 Cc 3 poorly graded if which is a sand, --- size [mm] 0.075[mm], 10% D10
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Find: the soil classification
soil type % weight % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 gravel sand fines 20% 70% 10% 0.850[mm], 60% 0.425[mm], 30% Cu=11.3 Coarse grain soil % sand > % gravel 5% fines 12% Cc=2.8 D30 well graded if Cu 6 and 1 Cc 3 or Cu 6 1 Cc 3 poorly graded if having 10% fine grain material, size [mm] 0.075[mm], 10% D10
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Find: the soil classification
soil type % weight % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 gravel sand fines 20% 70% 10% 0.850[mm], 60% 0.425[mm], 30% Cu=11.3 Coarse grain soil % sand > % gravel 5% fines 12% well graded Cc=2.8 D30 well graded if Cu 6 and 1 Cc 3 or Cu 6 1 Cc 3 poorly graded if and is well graded. size [mm] 0.075[mm], 10% D10
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Find: the soil classification
soil type % weight % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 gravel sand fines 20% 70% 10% 0.850[mm], 60% 0.425[mm], 30% Cu=11.3 Coarse grain soil % sand > % gravel 5% fines 12% well graded silt or clay? Cc=2.8 D30 well graded if Cu 6 and 1 Cc 3 or Cu 6 1 Cc 3 poorly graded if Lastly, we need to classify the fines as silt or clay. size [mm] 0.075[mm], 10% D10
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Find: the soil classification
SL=25 PL=40 LL=65 soil type % weight gravel sand fines 20% 70% 10% Coarse grain soil % sand > % gravel 5% fines 12% well graded silt or clay? To do so, we look at the given Atterberg limits, ---
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Find: the soil classification
SL=25 PL=40 LL=65 soil type % weight gravel sand fines 20% 70% 10% PI LL 100 60 40 20 80 10 30 50 Coarse grain soil % sand > % gravel 5% fines 12% well graded silt or clay? CL CH ML MH and plot the liquid liquid limit on the horizontal axis by the plasticity index on the vertical axis. The abbreviations on the chart stand for ---
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Find: the soil classification
SL=25 PL=40 LL=65 soil type % weight gravel sand fines 20% 70% 10% PI LL 100 60 40 20 80 10 30 50 Coarse grain soil % sand > % gravel 5% fines 12% well graded silt or clay? CL CH ML MH low plasticity clay, ---
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Find: the soil classification
SL=25 PL=40 LL=65 soil type % weight gravel sand fines 20% 70% 10% PI LL 100 60 40 20 80 10 30 50 Coarse grain soil % sand > % gravel 5% fines 12% well graded silt or clay? CL CH ML MH low plasticity silt, ---
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Find: the soil classification
SL=25 PL=40 LL=65 soil type % weight gravel sand fines 20% 70% 10% PI LL 100 60 40 20 80 10 30 50 Coarse grain soil % sand > % gravel 5% fines 12% well graded silt or clay? CL CH ML MH high plasticity clay, ---
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Find: the soil classification
SL=25 PL=40 LL=65 soil type % weight gravel sand fines 20% 70% 10% PI LL 100 60 40 20 80 10 30 50 Coarse grain soil % sand > % gravel 5% fines 12% well graded silt or clay? CL CH ML MH and high plasticity silt.
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Find: the soil classification
SL=25 PL=40 LL=65 soil type % weight gravel sand fines 20% 70% 10% PI LL 100 60 40 20 80 10 30 50 Coarse grain soil % sand > % gravel 5% fines 12% well graded silt or clay? LL=50 CL CH ML MH A liquid limit value of 50% divides the low and high plasticity soils.
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Find: the soil classification
SL=25 PL=40 LL=65 soil type % weight gravel sand fines 20% 70% 10% PI LL 100 60 40 20 80 10 30 50 Coarse grain soil % sand > % gravel 5% fines 12% well graded silt or clay? LL=50 CL CH ML MH “A” line The “A Line” separates the clay from the silt, ---
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Find: the soil classification
SL=25 PL=40 LL=65 soil type % weight gravel sand fines 20% 70% 10% PI LL 100 60 40 20 80 10 30 50 Coarse grain soil % sand > % gravel 5% fines 12% well graded silt or clay? LL=50 CL CH ML MH “A” line whose equation is shown. PI=0.73*(LL-20)
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Find: the soil classification
SL=25 PL=40 LL=65 PI=LL-PL soil type % weight gravel sand fines 20% 70% 10% PI LL 100 60 40 20 80 10 30 50 Coarse grain soil % sand > % gravel 5% fines 12% well graded silt or clay? LL=50 CL CH ML MH “A” line The plasticity index is equal to the liquid limit minus the plastic limit, --- PI=0.73*(LL-20)
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Find: the soil classification
SL=25 PL=40 LL=65 PI=LL-PL soil type % weight gravel sand fines 20% 70% 10% PI LL 100 60 40 20 80 10 30 50 Coarse grain soil % sand > % gravel 5% fines 12% well graded silt or clay? LL=50 CL CH ML MH “A” line which in this case is PI=0.73*(LL-20)
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Find: the soil classification
SL=25 PL=40 LL=65 PI=LL-PL soil type % weight gravel sand fines 20% 70% 10% PI LL 100 60 40 20 80 10 30 50 Coarse grain soil % sand > % gravel 5% fines 12% well graded silt or clay? LL=50 CL CH ML MH “A” line minus 40 PI=0.73*(LL-20)
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Find: the soil classification
SL=25 PL=40 LL=65 PI=LL-PL soil type % weight gravel sand fines 20% 70% 10% =25 PI LL 100 60 40 20 80 10 30 50 Coarse grain soil % sand > % gravel 5% fines 12% well graded silt or clay? LL=50 CL CH ML MH “A” line or, 25. PI=0.73*(LL-20)
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Find: the soil classification
SL=25 PL=40 LL=65 PI=LL-PL soil type % weight gravel sand fines 20% 70% 10% =25 PI LL 100 60 40 20 80 10 30 50 Coarse grain soil % sand > % gravel 5% fines 12% well graded silt or silt? LL=50 “A” line CH The point is plotted, and falls in the MH zone, --- CL MH PI=0.73*(LL-20) ML
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Find: the soil classification
SL=25 PL=40 LL=65 PI=LL-PL soil type % weight gravel sand fines 20% 70% 10% =25 PI LL 100 60 40 20 80 10 30 50 Coarse grain soil % sand > % gravel 5% fines 12% well graded fines=silt LL=50 “A” line CH which stands for high plasticity silt. CL MH PI=0.73*(LL-20) ML
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Find: the soil classification
SL=25 PL=40 LL=65 PI=LL-PL soil type % weight gravel sand fines 20% 70% 10% =25 PI LL 100 60 40 20 80 10 30 50 Coarse grain soil % sand > % gravel 5% fines 12% well graded fines=silt LL=50 SW SW-SM CL-ML SP-SC “A” line CH Reviewing our possible solutions, the soil is a well graded sand with silt, --- CL MH PI=0.73*(LL-20) ML
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Find: the soil classification
SL=25 PL=40 LL=65 PI=LL-PL soil type % weight gravel sand fines 20% 70% 10% =25 PI LL 100 60 40 20 80 10 30 50 Coarse grain soil % sand > % gravel 5% fines 12% well graded fines=silt LL=50 SW SW-SM CL-ML SP-SC “A” line CH the answer is B. CL MH PI=0.73*(LL-20) ML AnswerB
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( ) ? τ [lb/ft2] γclay=53.1[lb/ft3] Index σ’v = Σ φ γ Δ d ˚ H*C σfinal
Find: σ’v ρc d = 30 feet (1+wc)*γw wc+(1/SG) σ’v = Σ φ γ Δ d ˚ d Sand 10 ft γT=100 [lb/ft3] 100 [lb/ft3] φ=α1-α2 10 [ft] 20 ft Clay = γsand dsand +γclay dclay A W S V [ft3] W [lb] 40 ft text wc = 37% ? Δh 20 [ft] τ [lb/ft2] (5 [cm])2 * π/4 ( ) H*C σfinal ρcn= 1+e σinitial log ‘ φ size[mm] % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 c=0 400 1,400 σ3 Sand σ1
![( ) τ [lb/ft2] γclay=53.1[lb/ft3] Index σ’v = Σ φ γ Δ d ˚ H*C σfinal](http://slideplayer.com./slide/15046033/91/images/69/%28+%29+%CF%84+%5Blb%2Fft2%5D+%CE%B3clay%3D53.1%5Blb%2Fft3%5D+Index+%CF%83%E2%80%99v+%3D+%CE%A3+%CF%86+%CE%B3+%CE%94+d+%CB%9A+H%2AC+%CF%83final.jpg "Find: σ’v ρc d = 30 feet. (1+wc)*γw. wc+(1/SG) σ’v = Σ φ γ Δ d ˚ d. Sand. 10 ft. γT=100 [lb/ft3] 100 [lb/ft3] φ=α1-α2. 10 [ft] 20 ft. Clay. = γsand dsand. +γclay dclay. A. W. S. V [ft3] W [lb] 40 ft. text. wc = 37% Δh. 20 [ft] τ [lb/ft2] (5 [cm])2 * π/4. ( ) H*C σfinal. ρcn= 1+e σinitial. log. ‘ φ. size[mm] % passing. 0% 20% 40% 60% 80% 100% c= ,400. σ3. Sand. σ1.")
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