all attached

Page 1 of 13 Question 1 Figure 1 shows the plan of a raft foundation of a building cast at ground level. A sewer pipe is located at a depth of 4 m below the ground level and at a distance of 3 m from the corner of the raft foundation as shown in Figure 1 below. The raft foundation carrie s a uniform pressure on the soil below. The soil profile at the site is given in Table 1 . Your assigned values of the uniform pressure and the Young’s modulus of soil are given in Table 3 . Calculate: a) The vertical stress increase at point A on top of the p ipe using Fadum’s (1948) method. (1 2 marks) b) The vertical stress increase at a depth of 5 m below the centre of the raft foundation (point B in Figure 1 ) using Newmark’s (1942) chart. Assume that the influence factor for Newmark’s chart (I N) is equal to 0.005. (1 2 marks) c) Calculate the total settlement under the centre of the raft foundation (point B in Figure 1 ). (1 4 marks) Table 1: Soil profile Soil type Depth (m) Sand 0 - 10 Alluvial Clay 10 - 14 Gravel >10 Figure 1 Pipe A 3 m 5 m m 3 m B 3 Page 2 of 13 Question 2 A sheet pile wall is embedded into 5m of sand and has a retained height of 3 m as shown in Figure 2 . The water table is 4.5m below the ground surface. Your assigned values of the sand’s angle of frict ion, bulk unit weight and saturated unit weight are give n in Table 4 . A surcharge of 30 kN/m 2 is applied on the ground surface behind the retaining wall. Assuming that the interface friction between the soil and the wall can be ignored (δ=0º) and that the unit weight of water is 10 kN/m 3: a) Calculate the lateral earth and pore water pressures behind the sheet pile wall according to Coulomb’s method and plot their distribution along the retaining wall . (1 2 marks) b) Calculate the lateral earth and pore water pressures in front of the sheet pile wall according to Coulomb’s method and plot their distribution along the retaining wall . (1 2 marks) Figure 2 30 kN/ m2 sand sand 3 m 5 m 4.5 m 4.5 m Page 3 of 13 Question 3 a) The results obtained from an oedometer test on a sample of clay are given in Table 2 below: Table 2 i) Plot a graph of the void ratio against the logarithm of the vertical effective stress. Indicate the one -dimensional normal compression line on the graph. (12 marks) ii) Calculate the pre -consolidation pres sure . (5 marks) iii) Calculate the slope of the one -dimensional normal compression line. (5 marks) b) At a site location there is a layer of sand overlying a 4 m thick layer of clay ; the water table is at the ground surface . A thick layer of fill material is placed on the ground surface which increases the vertical effective stress at the mid -depth of the clay layer. Your assigned values of the thickness of the layer of sand as well as the values of the increase in th e vertical effective stress at the mid -depth of the clay layer are given in Table 5 . A soil sample with an initial height of 20mm was taken from the mid -depth of the clay layer and was tested in the oedometer apparatus; the results obtained are given in Ta ble 2 above . Assuming the saturated unit weight of both soil layers is 20 kN/m 3 and the unit weight of water is equal to 10kN/m 3, determine the final settlement due to the consolidation of the clay. (1 2 marks) c) Explain what would happen if the fill material is removed once the consolidation of the clay layer has been completed. (4 marks) v (kPa) 24 48 96 192 383 766 Void ratio e 1.112 1.105 1.080 0.985 0.8 50 0.731 Page 4 of 13 Table 3: Individualised values for Question 1 Note : in the unlikely event that your student number does not appear on this table, take the values of one of the last rows labelled as ‘extra’ . Student number Uniform pressure q (kPa) Young’s modulus of sand E' 1 (MPa) Young’s modulus of alluvial clay E' 2 (MPa) 14816934 110 11 1 16850261 120 12 2 17839276 130 13 3 18827956 140 14 4 18830027 150 15 5 19815622 160 16 6 19817712 170 17 7 19837277 180 18 8 Extra 1 190 19 9 Extra 2 100 10 1.5 Table 4: Individualised values for Question 2 Note : in the unlikely event that your student number does not appear on this table, take the values of one of the last rows labelled as ‘extra’ . Student number Angle of friction φ (degrees) Bulk unit weight γb (kN/m 3) Saturated unit weight γsat (kN/m 3) 14816934 31 17.1 20.1 16850261 32 17.2 20.2 17839276 33 17.3 20.3 18827956 34 17.4 20.4 18830027 35 17.5 20.5 19815622 36 17.6 20.6 19817712 37 17.7 20.7 19837277 38 17.8 20.8 Extra 1 39 17.9 20.9 Extra 2 30 17 20 Table 5: Individualised values for Question 3 Note : in the unlikely event that your student number does not appear on this table, take the values of one of the last rows labelled as ‘extra’ . Student number Thickness of sand H (mm) Increase in vertical effective stress Δσ’ v (kPa) 14816934 21 51 16850261 22 52 17839276 23 53 18827956 24 54 18830027 25 55 19815622 26 56 19817712 27 57 19837277 28 58 Extra 1 29 59 Extra 2 20 50 Page 5 of 13 FORMULAE SHEET AND DESIGN CHARTS Soil Physical Relations TERM SYMBOL UNITS FORMULAE Moisture content w % Void ratio (Partially saturated) e ratio Void ratio (Fully saturated) e ratio Porosity n ratio Specific volume v ratio Degree of saturation Sr % Specific Gravity Gs ratio Bulk density ρb kg/m 3 Dry density ρd kg/m 3 Saturated density ρsat kg/m 3 Submerged density ρ kg/m 3 Bulk unit weight b kN/m 3 Dry unit weight d kN/m 3 Saturated unit weight sat kN/m 3 Submerged unit weight  kN/m 3 s w m m w = r s s v S wG V V e = = s s v wG V V e = = e e V V n v + = = 1 e v + =1 v w r V V S = w s s w s s s V m V w G   = = ) 1( ) ( e eS G r s w b + + =   ) 1( ) 1( w e G b s w d + = + =    ) 1( ) ( e e Gs w sat + + =   ) 1( )1 ( e Gs w + − =   ) 1( ) ( e eS G r s w b + + =   ) 1( ) 1( w e G b s w d + = + =    ) 1( ) ( e e Gs w sat + + =   ) 1( )1 ( e Gs w + − =   Page 6 of 13 Soil strength Mohr -Coulomb equation Deviator stress q = σ 1 - σ3 = σ' 1 - σ'3 Mean effective stress p'= ( σ'1+σ'2+σ'3) / 3 Mean effective stress in a triaxial test p'= σc + q/3 - u Undrained shear strength su = q / 2 Critical state line (q - p') q = M p' Critical state line (v - lnp') v = Γ – λ (ln p') Hydraulic permeability Apparent velocity Flow rate Properties of flownets – Isotropic material Total flow q = k H N f / N e Total head loss H = Δh N e Flow per channel per unit length Δq = k Δh = k H / N e Total seepage flow q = Δq N f φ tan σ c' φ u)tan (σ c' τ n n   + = − + = t A Q v  = A v t Q q  = = Page 7 of 13 Consolidation Consolidation settlement Consolidation settlement Consolidation settlement Consolidation settlement for normally consolidated soils Consolidation settlement for over -consolidated clays for (σ'o + Δσ ') ≤ σ' c Consolidation settlement for over -consolidated clays for σ'o < σ' c < σ'o + Δσ ' Compression index 1-D normal compression line v = v o – λ ln σ'v Coefficient of consolidation Time factor ) e- (e e 1 H ρ 1 0 0 c + = ) σ/ σ log( )e- (e C 0 1 1 0 c   = H σΔ m ρ v v c  = E H σΔ ρ v c   =         + + = o c o c σ log C e 1 H ρ  o         + + = o o r o c σ σΔ σ log C e 1 H ρ         + + = c o c o c σ σΔ σ log C e 1 H ρ         + + o c r o σ σ log C e 1 H w ) E (k ) t(4 ) d (3 c o x 2 v  = = 2v d t) (c T = Page 8 of 13 Lateral earth pressures Coulomb’s active earth pressure coefficient Coulomb’s passive earth pressure coefficient Coulomb’s active earth pressure at depth z pa = K a q + K a γ z - Kac c Coulomb’s passive earth pressure at depth z pp = K p q + Kp γ z + Kpc c ) 1( 2 Kac c c K w a + = ) 1( 2 Kpc c c K w p + = Page 9 of 13 Active earth pressure coefficients according to Kerisel and Absi (1990); redrawn by Craig ( 2004). Note: For φ=0º: K a= 1 Page 10 of 13 Pass ive earth pressure coefficients according to Kerisel and Absi (1990); redrawn by Craig ( 2004). Note: For φ=0º: Kp=1 Page 11 of 13 STRESS DISTRIBUTION Stress due to a point load on the surface of a semi -elastic half -space : Influence Factors I p p z I z P 2 =  ) 2 ( ) ( ) 2 1 ( 2 3 2 5 3 3 2         + − − = z R R z v R r z z P r   ) 3 ( ) ( 2 ) 1 2 ( 2 3 3 2         + − − = z R R z R z z P     2 / 5 2 5 5 ) / ( 1 1 2 3 2 3       + = = z r R z I p   Page 12 of 13 Fadum’s (1948) Chart σv = q∙ Ir Page 13 of 13 Newmark’s (1942) chart For a uniform loading intensity of q the stress at the required depth below the point in question is computed as σv = N ∙ IN ∙ q DETACH AND SUBMITT CHART WITH ANSWER BOOK