4.7.2 Determine the mean CBR for this particular subgrade.
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- SHOW COMPLETE SOLUTION PLS VALUES:Y = 25A = 5Show That: G, • p, (1– 4.) Pa = 1+wG, | 1) Y min Y. max 2) Dr = (x100%) Y max – Ymir3) Explain the common properties of dense, uniform, gap and open gradations. What is the main difference between them? Explain by showing sample gradations on the following chart. %pass. Sieves
- For Sample C (Below), determine the: a. Üumformity Coefficient b. Coefficient of Curvature c. Well graded, poorly graded or gap graded? Particle Size Distribution GRAVEL SAND SILT or CLAY Coarse Fime Course Medium Fine 100 100 90 90 80 80 70 70 60 60 50 50 40 40 30 30 20 20 10 10 100 10 1 0.1 0.01 0.001 Grain size in millimeters Percent finer by weightDetermine the Freundlich and Langmuir isotherm model parameters for the set of equilibrium adsorption data provided below. You may use linear regression, and plot Ce/Qad versus Ce for the Langmuir equation (qe = QaabCe)/(1+ bCe) and log qe versus log Ce for the Freundlich isotherm equation (qe = KF Ce"), or use non-linear fitting feature built in Excel. In your solution, you need to explain your procedure and circle your parameters. Q ad is the (monolayer) adsorption capacity for the Langmuir isotherm model. Adsorption isotherm data: Carbon type, F-400; chemical: tetrachloroethene, temperature: 13.8°Č Ce (umol/L): 15.7 qe (umol/g): 1246 1.27 489 0.396 298 0.225 250 0.161 213 Note: 1) Use the proper units for the isotherm parameters. 2) When you graph the isotherm data, do not connect the data points. 3) Provide the curve fitting (dashed line and isotherm parameters with 3 significant digits. 4) The x-axis and y-axis should be the bottom and the left sides of your figure. "Please see…. Given: H1 = Poisson's ratio of soil sample 1 Hz = Poisson's ratio of soil sample 2 coefficient of earth pressure at rest %3D %3D k1 for soil sample 1 k2 = coefficient of earth pressure at rest for soil sample 2 If H/42 then the value of k;/k2 will be %3D %3D = 1.5 and (1-µj)/(1-µ2) = 0.875, %3D %3D
- Problem 2 Given: Mass of moist soil Volume of soil Dry mass of soil Specific Gravity (Gs) Problem 3 Mass of moist soil Dry mass of soil Specific Gravity (Gs) Saturation 887 g 555 cm³ 798 g 2.69 186 g 140 g 2.66 100% Find: e = n = S = W = Y = Vsat = Vsub = e = n = W = Y = % % % lb/ft³ lb/ft³ lb/ft³ % % lb/ft³Problem 7 n = Gs= Saturation = 42% 2.65 70% @DY e= p= y = g/cm³ lb/ft³using tge inverse distance weighted interpolation method and the value of three known soil samples, calculate the gold concentration in the unknown point, P01. Use a power (exponentO of n=3 Samples /Au ppb / Eastings / Northings Soil 01 / 17 344937 250624 Soil 02 / 253 / 345261 / 249165 Soil 03 / 1002 / 345584 / 249137 P01 / ? 345123 / 249098 Round to the ones
- For Sample A (Below), determine the: a. Uniformity Coefficient b. Coefficient of Curvature c. Well graded, poorly graded or gap graded? Particle Size Distribution GRAVEL SAND SILT or CLAY Coarse Fine Course Medium Fine 100 100 90 90 80 80 70 70 60 60 50 50 40 40 30 30 20 20 10 110 100 10 1 0.1 0.01 0.001 Grain size in millimeters Percent finer by weightStructural Analysis Q.4, Plz, Answer.The velocity distribution in the boundary layer is given by = (*)¹" 1/7 y point velocity at distance y Ա U = free stream velocity nominal thickness What would be the displacement thickness(8*)?