Two level sections 85 ft apart with center heights 4.3 and 7.9 ft in fill, base width 40 ft, side slopes 2:1.
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- III. Using Atterberg's Plasticity Determination Method Determinethe following; a) Plot the Flow Curve and b) Determine the Flow Index c) Computethe Liquid Limit d) and if the Plastic Limit wa. uetermined as PL = 17.545 % what is the Plasticity Index ? Number of Blows Moisture Content,% 15 38.589 20 35.712 29 28.556 32 24.670SHOW COMPLETE SOLUTION PLS VALUE:Y = 531. Using the design chart below. Estimate the SN if W18=10x10^6, R=90%, So=0.35, Mr- 10000Psi, ΔPSI=3. Use the chart below. TL Overall standard deviation, S. TL Design serviceability loss, APSI 50 10 5.0 99.9 0.2 99 1.0 0.4 0.6 90 0.5 0.5 0.1 1.5 0.05 2.0 80 70 60 50 2.5 3.0 9 87 6|5 4 3 1 Eхample: 6.5 5.5 4.5 3.5 2.5 1.5 W18 = 5 x 106 Design structural number, SN R = 95% S, = 0.35 MR = 5000 Ib/in? APSI = 1.9 Solution: SN = 5.0 Figure 4.5 Design chart for flexible pavements based on the use of mean values for each input.
- SHOW COMPLETE SOLUTION PLS VALUES:X = 425Y = 25C = 1a. Compute the value of x b. Compute the area in the fill c. Compute thr area in cutThe head of a groundwater aquifer is described in Cartesian coordinates by 1 h(x, y) = 1+ x2 + y2 + x + xy Which of the following(s) is/are the correct strategy for determining the maximum value of the head of a groundwater aquifer. (Choose all that apply!) A) Ater putting the equation in an appropriate tonm we can apply Bisection Method B) After pasting the eigualicon in an uppoprkate fum we cian apply Ciauss Seldel Meshud C) Alliti putting tie tli inan appipriate fonm we tan apply Newien Methed for systems O D) After putting the equation in an appropriate form we can apply Lagrange Interpolation E) Alser puting thqutn in an appropriate fam we can ply Newton Dividesd Nfferne
- The figure shown is a dewatering plan to build the foundations of an office building below the water table without the sheet-piling. Assume that all b = 6 meters and all 1 = 4 meters in all grids of the middle flow channel. Dewatering wells Marshy soils Ah =Sm Excavated site OL Ah = 2m Ah = 2m Ah = 2m Lowered WT a. What is the value of n? b. How many flow channels are in the figure? c. How many potential drops are in the figure? d. The potential drop at area 1 is: e. The potential drop at area 2 is: f. The potential drop at area 3 is: g. The average hydraulic gradient in each net is:Problem: Determine oPn for the short column shown. fy 60000 psi and f'c = %3D %3D 4000 psi. 3" e = 8" • 8 #10 • 14" 20" Gy = 6" 3" 3"+Determine the oil surface elevation at section 1. Elevation = ? 60 m Oil, S = 0.9 v = 3 x 10-5 m²/s K₂ = 0.5 ww =2< * 7m + k₂ = 0.046 mm D = 0.1 m O None of the answers are correct. O Oil surface elevation at section 1 = 170.35 m 130 m Q = 0.03 m³/s Oil surface elevation at section 1 = 136 m O Oil surface elevation at section 1 = 175.35 m Elevation = 130 m
- The particle size distribution curves for three soils - A, B and C - are shown in Figure Q-2 on page 2. a. Determine the effective size, average particle size, uniformity coefficient and coefficient of curvature for soil B. b. Classify soil A according to ASTM-CS. C. Classify soil C according to the AASHTO system. The Atterberg limits for this soil are LL = 81 and PL = 35. Using sketches and notes, outline the key features of the SPT test.The figure shown is a dewatering plan to build the foundations of an office building below the water table without the sheet-piling. Assume that all b = 6 meters and all 1 = 4 meters in all grids of the middle flow channel. Dewatering wells Marshy soils Ah =m Excavated site OL Ah = 2m Ah = 2m Ah = 2m Lowered WT a. What is the value of n? b. How many flow channels are in the figure? c. How many potential drops are in the figure? d. The potential drop at area 1 is: e. The potential drop at area 2 is: f. The potential drop at area 3 is: g. The average hydraulic gradient in each net is:WA road section is 1200 m long. The cut and fill areas are to be computed between the station as shown in table below, determine the net volume of cut and fill between the stations, if you know that the material in fill will be consolidate by 12 percent after compaction. Given 100 m between each station. Station 0+00 2+00 4+00 6+00 8+00 10+00 12+00 1.0 5.0 Cut Fill 4.0 2.6 7.0 2.0 9.5 0.5 2.0 End Area (m2) 3.0 8.0 6.0