Assignment 4-solution

CVG 4108 Geotechnical Design Assignment 4 Shoring Walls Due Date: Question 1 The key differences between parts a), b), and c) will have to do with the pore water pressure distribution. Case b), is the most typical case, where seepage is allowed below the sheet pile wall, but pumps keep the water level at or below the bottom of the excavation. In this case, the pore water pressure is equal on the active and passive sides at the level of the toe of the sheet pile, but the pressure is 0 at the bottom of the excavation on the passive side. In condition a), the water is hydrostatically distributed on both the active and passive sides. In condition c), the pore water pressures are equal on both sides of the sheet pile wall below the bottom of the excavation. For the sand layer: K a = tan 2 ( 45 ° − ϕ / 2 ) = tan 2 ( 45 ° − 30 ° / 2 ) = 1 / 3 , K p = tan 2 ( 45 ° + ϕ / 2 ) = tan 2 ( 45 ° + 30 ° / 2 ) = 3 For the clay layer: Figure 1: Soil profile and embeded sheet pile 1m 5.22m D 5m Sand Clay γ sat =17 kN/m³ γ dry =15.8 kN/m³ γ sat =19.5 kN/m³

Assignment 4: Shoring Walls p. 2 of 16 K a = tan 2 ( 45 ° − ϕ / 2 ) = tan 2 ( 45 ° − 21 ° / 2 ) = 0.4723 , K p = tan 2 ( 45 ° + ϕ / 2 ) = tan 2 ( 45 ° + 21 ° / 2 ) = 2.12 a) No seepage allowed (hydrostatic distribution on both sides of the wall, though the GWT is at different levels on each side) pt 0 0.0 30 0 0.33333 3.00000 0.5 1.0 15.8 pt 1 1.0 30 0 0.33333 3.00000 3.0 4.0 19.5 pt 2 5.0 30 0 0.33333 3.00000 5.0 0.0 19.5 pt 3 5.0 21 10 0.47236 2.11705 5.6 1.2 17 pt 4 6.2 21 10 0.47236 2.11705 12.7 13.0 17 pt 5 19.2 21 10 0.47236 2.11705 Location z (m) z mid (m) Δz (m) ϕ c (kPa) Ka Kp γ (kN/m³) pt 0 0.0 0.0 0.0 0 0 0 1.0 pt 1 1.0 15.8 0.0 15.8 5.3 5.3 4.0 pt 2 5.0 93.8 40.0 53.8 17.9 57.9 0.0 pt 3 5.0 93.8 40.0 53.8 11.7 51.7 1.2 pt 4 6.2 114.5 52.2 62.3 15.7 67.9 13.0 pt 5 19.2 335.5 182.2 153.3 58.7 240.9 Location z (m) Δz (m) σ a (kPa) u a (kPa) σ a ' (kPa) p a (kPa) p a +u a (kPa) pt 0 0.0 0 0 0 0 0 1.0 pt 1 1.0 0 0 0 0 0 4.0 pt 2 5.0 0 0 0 0 0 0.0 pt 3 5.0 0 0 0 0 0 1.2 pt 4 6.2 0 0 0 29.1 29.1 13.0 pt 5 19.2 221 130 91 221.8 351.8 Location z (m) Δz (m) σ p (kPa) u p (kPa) σ p ' (kPa) p p (kPa) p p +u p (kPa) Due Date: CVG 4108

Assignment 4: Shoring Walls p. 4 of 16 c) In this case, pore pressures would be equal on both sides of the wall (below excavation) Due Date: CVG 4108 pt 0 0.0 0 0 0 1.0 pt 1 1.0 5.3 0 5.3 4.0 pt 2 5.0 57.9 0 57.9 0.0 pt 3 5.0 51.7 0 51.7 1.2 pt 4 6.2 67.9 29.1 38.8 13.0 pt 5 19.2 240.9 293.4 -52.5 Location z (m) Δz (m) p a +u a (kPa) p p +u p (kPa) p net (kPa) pt 0 0.0 0.0 0.0 0 0 0 1.0 pt 1 1.0 15.8 0.0 15.8 5.3 5.3 4.0 pt 2 5.0 93.8 40.0 53.8 17.9 57.9 0.0 pt 3 5.0 93.8 40.0 53.8 11.7 51.7 1.2 pt 4 6.2 114.5 52.2 62.3 15.7 67.9 13.0 pt 5 19.2 335.5 182.2 153.3 58.7 240.9 Location z (m) Δz (m) σ a (kPa) u a (kPa) σ a ' (kPa) p a (kPa) p a +u a (kPa) pt 0 0.0 0 0 0 0 0 1.0 pt 1 1.0 0 0 0 0 0 4.0 pt 2 5.0 0 0 0 0 0 0.0 pt 3 5.0 0 0 0 0 0 1.2 pt 4 6.2 0 52.2 -52.2 -81.4 -29.2 13.0 pt 5 19.2 221 182.2 38.8 111.2 293.4 Location z (m) Δz (m) σ p (kPa) u p (kPa) σ p ' (kPa) p p (kPa) p p +u p (kPa)

Assignment 4: Shoring Walls p. 5 of 16 The net pressure diagrams of the previous cases are illustrated as follows: Due Date: CVG 4108 pt 0 0.0 0 0 0 1.0 pt 1 1.0 5.3 0 5.3 4.0 pt 2 5.0 57.9 0 57.9 0.0 pt 3 5.0 51.7 0 51.7 1.2 pt 4 6.2 67.9 -29.2 97.1 13.0 pt 5 19.2 240.9 293.4 -52.5 Location z (m) Δz (m) p a +u a (kPa) p p +u p (kPa) p net (kPa) -200 -100 0 100 200 300 400 0 5 10 15 20 25 pa+ua (kPa) pp+up (kPa) pnet (kPa) -100 0 100 200 300 400 0 5 10 15 20 25 pa+ ua (kP a) -100 0 100 200 300 400 0 5 10 15 20 25 pa+ ua (kP a) a) b) c)

Assignment 4: Shoring Walls p. 7 of 16 The detailed calculations corresponding to the final condition illustrated in the previous table is shown below: Due Date: CVG 4108 Table 1: Consecutive Iterations for embedment depth seeking ΣΜ=0. The value of Lp/D corresponding to each iteration set is also shown. D ΣM Lp/D 13 100.58375833 0.15 15 -3842.852025 0.15 13.1 -65.06526333 0.15 13.1 813.3681096 0.16 13.35 431.9240436 0.16 13.6 36.138697867 0.16 13.6 5167.9766152 0.215 13.85 4975.2495382 0.215 20.3 -5728.025735 0.215 16.8 1675.8811848 0.215 17.6 323.40726107 0.215 17.8 6.7078966167 0.215 17.8 -6939.483993 0.178 18.05 -7677.608388 0.178 15.4 -1198.859038 0.178 14.9 -282.8229386 0.178 14.7 72.17159718 0.178 14.7 640.93288766 0.183 14.95 216.64022189 0.183 15.1 -8.419057596 0.183 15.1 101.054775 0.184 15.35 -353.3015451 0.184 15.2 -71.63599193 0.184 15.2 -194.6628252 0.183 15.45 -626.6895648 0.183 15.1 -8.419057596 0.183 Table 2: Complementary set of iterations changing Lp/D to achieve ΣF=0 Lp/D ΣF D ΣM 0.15 79.975 13.1 -65.06526333 0.16 67.6174 13.6 36.138697867 0.215 -142.44755 17.8 6.7078966167 0.178 24.5351 14.7 72.17159718 0.183 2.454815 15.1 -8.419057596 0.184 -5.69268 15.2 -71.63599193 0.183 2.454815 15.1 -8.419057596

Assignment 4: Shoring Walls p. 8 of 16 Due Date: CVG 4108 pt 0 0.0 30 0 0.3333333333 3 # 1.0 15.8 pt 1 1.0 30 0 0.3333333333 3 # 4.0 19.5 pt 2 5.0 30 0 0.3333333333 3 # 1.2 19.5 pt 3 6.2 30 0 0.3333333333 3 # 12.3 19.5 pt 4 18.6 30 0 0.3333333333 3 # 2.8 19.5 pt 5 21.3 30 0 0.3333333333 3 # Location z (m) Δz (m) ϕ c (kPa) Ka Kp γ (kN/m³) pt 0 0.0 0.0 0.0 0 0 0 1.0 pt 1 1.0 15.8 0.0 15.8 5.3 5.3 4.0 pt 2 5.0 93.8 40.0 53.8 17.9 57.9 1.2 pt 3 6.2 117.6 52.2 65.4 21.8 74 12.3 pt 4 18.6 358.2 175.6 182.6 60.9 236.5 2.8 pt 5 21.3 412.0 203.2 208.8 626.4 829.6 Location z (m) Δz (m) σ a (kPa) u a (kPa) σ a ' (kPa) p a (kPa) p a +u a (kPa) pt 0 0.0 0 0 0 0 0 1.0 pt 1 1.0 0 0 0 0 0 4.0 pt 2 5.0 0 0 0 0 0 1.2 pt 3 6.2 12.2 12.2 0 0 12.2 12.3 pt 4 18.6 252.8 168.2 84.5 253.5 421.7 2.8 pt 5 21.3 306.65 203.2 103.5 34.5 237.7 Location z (m) Δz (m) σ p (kPa) u p (kPa) σ p ' (kPa) p p (kPa) p p +u p (kPa)

Assignment 4: Shoring Walls p. 10 of 16 From the previously listed calculations the theoretical embedment depth, D = 15.1 m. Add 30 to 40% for final value. Assuming 30%, the actual embedment depth will be: 19.11 m (round up to 20 m for a 32% increase). Question 3 In order to implement 2 rows of anchors, we must first use a virtual excavation with a depth of excavation corresponding to the depth of the second row of anchors. Note that there is no pivot point to find as we are using the free earth support method, which assumes that the pivot is at the toe, and the imbalanced sum of forces will be carried by the anchor. Due Date: CVG 4108 Figure 3: Net pressure diagram for the condition depicted in Question 2 -400 -200 0 200 400 600 800 0.0 5.0 10.0 15.0 20.0 25.0 pnet (kPa) pnet (kPa) z (m)

Assignment 4: Shoring Walls p. 11 of 16 Here also, we do not achieve an exact value of ΣM=0 due to the rounding of the embedment depths. Please note that the value of D=16.8 m obtained here is strictly a computation value. Since we still have to implement the second row of anchors and the full excavation, it cannot represent an actual embedment depth. Since the imbalanced ΣF=53 kN/m, that value will become the load to be carried by the row of anchors located at a depth of 2 m. To calculate the load on the second row of anchors, we must now implement the full excavation depth, and find ΣM=0 when the forces of the first row of anchors are applied. Note that when calculating the moments, the moment arm is always relative to the currently unknown anchor forces. The embedment depth of 7.6 m is now the real theoretical embedment depth, and the ΣF=191kN/m is now the line load that must be carried by the second row of anchors. The detailed tables corresponding Due Date: CVG 4108 Table 3: Result of iterations for the first row of anchors. An hypothetical excavation depth of 3.5 m is assumed. D ΣM ΣF 13 -20844.6375 -1955.65 15 -31606.30417 -2648.975 9.1 -7422.920833 -896.325 7.3 -3923.622833 -536.835 5.3 -1541.736833 -233.605 4 -658.5875 -90.7 3 -253.9041667 -10.175 2.4 -101.5161667 26.005 2 -30.92083333 45.05 1.8 -3.656166667 53.065 Table 4: Iterations for second row of anchors D ΣM ΣF 13 -10810.62057 -667.998 15 -18912.50457 -1173.948 10.3 -3698.023233 -147.073 9.2 -1839.6339 13.352 8.1 -501.4223667 141.762 7.7 -123.1547667 181.392 7.6 -32.93096667 191.232

Assignment 4: Shoring Walls p. 13 of 16 You will have noticed that in the above table, the ΣF=244 kN/m (not 191 kN/m). this is because in the iterations table, the force due to the first row of anchors (53 kN/m) was subracted separately from the contribution of p net . Hence 244 kN/m-53 kN/m=191 kN/m. Similarly, ΣM=-112+80=-32 kN m/m as indicated in the iterations table. The following is a table showing the resulting anchor loads and their moment contributions (moments calculated with respect to the second row of anchors). Question 4 From Question 3 it was determined that the tension in the two rows of anchors is as follows: 53 kN/m for the first row (at 2 m depth) and 191 kN/m for the 2 nd row (at 3.5 m depth). These are the horizontal components of the anchor forces. The actual tension the anchors must resist are given by T = T horizontal / cos α , where α is the angle between the anchor and the horizontal. Please note that although we used 45° in the question, typical values would be much smaller. T first row = 53 / cos ( 45 ° )= 75 kN / m T second row = 191 / cos ( 45 ° )= 270 kN / m Since no spacing is specified, I will assume a spacing of 1 m. The tension in each anchor will be 75 kN for the anchors of the first row, and 270 kN for the anchors in the second row. The total length of the anchor tendon is composed of the free tendon length, and the actual anchor Due Date: CVG 4108 pt 0 0.0 0 0.0 0.0 1.0 0.0 2.7 -3.0 -2.8 0.0 -7.5 2.7 -7.5 pt 1 1.0 5.3 2.7 -7.5 2.5 13.3 41.1 -1.3 -0.8 -16.6 -34.3 54.4 -50.8 pt 2 3.5 38.2 57.0 -58.3 1.5 57.3 14.8 0.8 1.0 43.0 14.8 72.1 57.8 pt 3 5.0 57.9 129.1 -0.6 1.2 70.6 2.4 2.1 2.3 149.0 5.5 73.0 154.5 pt 4 6.2 61.8 202.1 154.0 7.6 469.7 -427.5 6.5 7.8 3062.3 -3328.8 42.2 -266.5 pt 5 13.8 -50.7 244.3 -112.5 Location z (m) Δz (m) p net (kPa) A rec (kN/m) A tri (kN/m) lrec (m) ltri (m) Mrec (kN m/m) Mtri (kN m/m) F (kN/m) M (kN/m) ΣF (kN/m) ΣM (kN m/m) 53 2 80 191 3.5 0 Force on Anchor (kN/m) Depth of Anchor (m) Moment due to Anchor (kN m/m)

Assignment 4: Shoring Walls p. 14 of 16 length L s (the bonded length). The free tendon length is the distance between the sheet pile wall insertion point and the failure plane. The proper anchor extends beyond this. To find the free tendon lenght (and the depth, and corresponding effective stress) at which the tendon begins, the simplest way is to consider the equation of the line representing the tendons and the wedge. Considering the anchor line of action z first row = 2 m +Δ x tan ( 45 ° ) z second row = 3.5 m +Δ x tan ( 45 ° ) The equation of the failure plane is given by z failure =( 6.22 + 7.6 )−Δ x tan ( 60 ° )= 13.82 −Δ x tan ( 60 ° ) The intersection with the first row of anchors is given by: 13.81 m −Δ x tan ( 60 ° )= 2 m +Δ x tan ( 45 ° ) Δ x = 13.82 – 2 tan ( 45 ° )+ tan ( 60 ° ) = 4.326 m The intersection with the second row of anchors is given by: Δ x = 13.82 – 3.5 tan ( 45 ° )+ tan ( 60 ° ) = 3.778 m The depth of intersection is given by: z first row = 2 m + 4.326 m tan ( 45 ° )= 6.326 m Similarly, z first row = 3.5 m + 3.778 m tan ( 45 ° )= 7.278 m The free length of the tendon L free , first row = √ ( 6.326 − 2 ) 2 + 4.326 2 = 6.118 m L free , second row = √ ( 7.278 − 3.5 ) 2 + 3.778 2 = 5.343 m The geotechnical factor of safety of the anchor is given by: Due Date: CVG 4108

Assignment 4: Shoring Walls p. 16 of 16 For the first row: 1 2 sin ( 45 ° ) 9.5 kN / m 3 L s 2 + 119.657 kPa L s − 2.0 × 75 kN 1.257 m 2 / m × 0.1 = 0 3.3588 kN / m 3 L s 2 + 119.657 kPa L s − 1193.317 kN / m = 0 L s = − 119.657 + √ 119.657 2 – 4 × 3.3588 ×− 1193.317 2 × 3.3588 = 8.121 m = 8.1 m For the second row: 1 2 sin ( 45 ° ) 9.5 kN / m 3 L s 2 + 138.221 kPa L s − 2.0 × 270 kN 1.257 m 2 / m × 0.1 = 0 3.3588 kN / m 3 L s 2 + 138.221 kPa L s − 4295.943 kN / m = 0 L s = − 138.221 + √ 138.221 2 − 4 × 3.3588 ×− 4295.943 2 × 3.3588 = 20.684 m = 20.7 m The anchor length will therefore be 8.1 m for the first row, and 20.7 m for the second row. The total length of the anchors will be: 6.1 m + 8.1 m = 14.2 m for the first row, and 5.3 m+20.7 m=26.0 m for the second row of anchors. The structural factor of safety requires the yield strength of the steel of the tendons: σ y =450 MPa FS = σ y A x T = σ y π 4 d 2 T d = √ FST σ y π 4 d = √ 2.0 × 75 kN 450 × 10 3 kPa π 4 = 21 mm minimum required cross-section or the first row of anchors. d = √ 2.0 × 270 kN 450 × 10 3 kPa π 4 = 39 mm minimum required cross-section or the second row of anchors Due Date: CVG 4108