HW06-CE357-Consolidation_Solution - SI-Spring 22

The University of Texas at Austin Department of Civil, Architectural and Environmental Engineering Homework #6 Solution CE 357 Geotechnical Engineering 1/6 Problem 1. Given; Consolidation test results Find; Consolidation curve, C c , C r , σ´ c , and S c (a) Figure 1. Schematic diagram for problem 2 (b) C c = 𝛥? 𝛥𝑙??𝜎′ 𝑣 = | 0.691−0.889 𝑙??1280−𝑙??320 | ≈ 0.33 in virgin compression curve C r = 𝛥? 𝛥𝑙??𝜎′ 𝑣 = | 0.927−0.953 𝑙??160−𝑙??20 | ≈ 0.02879 in recompression curve According to Arthur Casagrande’s graphical method, the maximum past effective vertical stress can be determined as shown in Figure 1. 1. Choose a point of maximum curvature on the consolidation curve by eye. 2. Draw a horizontal line from the point. 3. Draw a line tangent to the curve at the point selected in step 1. 4. Draw a line bisecting the angle constructed by step 2 and 3. 5. Extend the “straight portion” of the virgin compression curve up to the bisecting line in step 4. 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1 1 10 100 1000 10000 Void Ratio, e log σ ' v (kPa) Step1. Maximum point of curvature Step 2 Step 3 Step 4 Step 5 θ θ σ ´ c or σ ´ p = 310 kPa

The University of Texas at Austin Department of Civil, Architectural and Environmental Engineering Homework #6 Solution CE 357 Geotechnical Engineering 2/6 The point where the lines in step 4 and step 5 intersect is the maximum past effective vertical stress. Figure 2. Consolidation Curve  σ´ c ≈ 300 kPa or σ´ p ≈ 300 kPa (c) First, check if the final effective vertical stress is larger or smaller than the maximum past effective vertical stress. σ´ vf = σ´ vo + Δσ v = 120 kPa+ 180 kPa= 300 kPa = σ´ max,p The stress state lies on the range of recompression curve (C r line)  S c = 𝐶 ? ? 1+? 𝑜 ×log( 𝜎′ 𝑣𝑜 +𝛥𝜎 𝑣 𝜎′ 𝑣𝑜 ) = 0.02879×6 1+0.97 ×log( 300 120 ) = 0.034m

The University of Texas at Austin Department of Civil, Architectural and Environmental Engineering Homework #6 Solution CE 357 Geotechnical Engineering 4/6 be used. Thus, σ vo,A = 𝛾 ?,?𝑎?? × d sand + 𝛾 ?,?𝑙𝑎𝑦 × d clay,center = 19 kN/m 3 x1.5m+16.9 kN/m 3 x1.5m = 53.85 kPa σ´ vo,A = σ vo,A - u = σ vo,A - 𝛾 𝑤 ×d water = 53.85 kPa – 9.81 kN/m 3 ×(1m+1.5m) = 29.325 kPa σ´ vf,A = σ´ vo,A + Δσ v = σ´ vo,A + 𝛾 ?,?𝑖𝑙𝑙 ×d fill = 29.325 kPa + 20.35 kN/m 3 x2.5m = 80.2 kPa σ vo,B = 𝛾 ?,?𝑎?? × d sand + 𝛾 ?,?𝑙𝑎𝑦 × d clay,center = 19 kN/m 3 × 1.5m + 16.9 kN/m 3 × 4.5m = 104.55 kPa σ´ vo,B = σ vo,B - u = σ vo,B - 𝛾 𝑤 ×d water = 104.55 kPa – 9.81 kN/m 3 ×(1m+4.5m) = 50.595 kPa σ´ vf,B = σ´ vo,B + Δσ v = σ´ vo,B + 𝛾 ?,?𝑖𝑙𝑙 ×d fill = 50.595 kPa + 20.35 kN/m 3 x2.5m = 101.47 kPa S c = S layer1 + S layer2 S layer1 = 𝐶 𝑐 ? 1+? 𝑜 × log( 𝜎′ 𝑣𝑜,? +𝛥𝜎 𝑣 𝜎′ 𝑣𝑜,? ) = 0.68×3 1+1.35 × log( 80.2 29.325 ) = 0.379m S layer2 = 𝐶 𝑐 ? 1+? 𝑜 × log( 𝜎′ 𝑣𝑜,? +𝛥𝜎 𝑣 𝜎′ 𝑣𝑜,? ) = 0.68×3 1+1.35 × log( 101.47 50.595 ) = 0.262m  S c = S layer1 + S layer2 = 0.379m +0.262m = 0.641m or  S c = 𝐶 𝑐 ? 1+? 𝑜 × [ log( 𝜎′ 𝑣𝑜,? +𝛥𝜎 𝑣 𝜎′ 𝑣𝑜,? ) + log( 𝜎′ 𝑣𝑜,? +𝛥𝜎 𝑣 𝜎′ 𝑣𝑜,? ) ] = 0.68×3 1+1.35 × [ log( 80.2 29.325 ) + log( 101.47 50.595 ) ] = 0.641m Problem 3. (Calculation from one layer, to compare with the calculation from two layer)

The University of Texas at Austin Department of Civil, Architectural and Environmental Engineering Homework #6 Solution CE 357 Geotechnical Engineering 5/6 Given; H = 6m, Gs = 2.7, w=50%, Cc=0.68 𝛾 ?,?𝑎?? = 19 kN/m 3 , 𝛾 ??𝑦,?𝑖𝑙𝑙 =18 kN/m 3 w fill = 10% Find; S c (Settlement of NC clay layer) The effective stress at the center of the entire layer is used to compare the result with the previous calculation. σ vo,A = 𝛾 ?,?𝑎?? × d sand + 𝛾 ?,?𝑙𝑎𝑦 × d clay,center = 19 kN/m 3 x 1.5m + 16.9 x 3 = 79.2kPa σ´ vo,A = σ vo,A - u = σ vo,A - 𝛾 𝑤 ×d water = 79.2 kPa – 9.81 kN/m 3 x4m = 39.96 kPa σ´ vf,A = σ´ vo,A + Δσ v = σ´ vo + 𝛾 ?,?𝑖𝑙𝑙 ×d fill = 39.96 + 20.35 kN/m 3 x 2.5m = 90.835 kPa  S c = 𝐶 𝑐 ? 1+? 𝑜 ×log( 𝜎′ 𝑣𝑜,? +𝛥𝜎 𝑣 𝜎′ 𝑣𝑜,? ) = 0.68×6 1+1.35 ×log( 90.835 39.96 ) = 0.619m