Question

Question 6 It is required to construct a 4 m deep underground carpark in the heart of the city of Melbourne without disturbing the surrounding area. The required excavation for the proposed carpark and also the adjacent ground condition are shown in Figure 6.1. As can be seen from the figure, the adjacent ground consists of a 2 m deep clay deposit, resting on a 2 m thick sand deposit, where stable soil could be found below the sand deposit (see Figure 6.1). According to the conducted investigations, the water table is available at 2 m depth below the ground surface (Figure 6.1) and the capillary rise through the available clay in the clay deposit was found to be much greater than 2 m. Representative soil samples were collected from the clay and sand deposits to conduct required laboratory tests and the following results were obtained; For clay: Psa = 2000 kg/m At rest earth pressure coefficient, ko = 0.6 Drained cohesion, C' = 20 kPa Drained friction angle, -20° For sand: Pat=1800 kg/m At rest earth pressure coefficient, ko =0.5 Drained cohesion, C = 10 kPa Drained friction angle, ' = 32° In order to check the stability of the clay and soil deposits against the excavation, two soil elements, located at the center of each soil deposit, were considered for stability analysis (see Figure 6.1). Assuming the ground surface is horizontal and gravitational acceleration () is 10 m/s": (a) Draw the pore water pressure distribution (prior to the excavation) of the adjacent ground that consists of clay and sand deposits (up to 4 m depth). Show the relevant calculations. (3 marks) (b) Calculate the total horizontal and vertical stresses applying on each soil element (prior to the excavation) (5 marks) (C) Draw the total stress Mohr circles corresponding to the each soil element in Figures 6.2 and 6.3 (prior to the excavation). (2 marks) (d) The horizontal stress relief associated with the excavation will cause a 20 kPa total horizontal stress reduction in each soil element located in clay and sand deposits). The total vertical stresses applying on each soil element can be assumed to remain unchanged (as no additional vertical load will be applied) Show how the total stress acting on each soil element will change upon the excavation by drawing the new total Mohr circles corresponding to each soil element after the excavation). (4 marks) (Question 6 continued on the next page)

Answer 1

en Clay 1 Sand fsat = 2000kg/m feat= 1 acollen Ko=0.6 Ko > 0.5 Teolokta c'=20kla 0.32 p=20° am In clay pwp diagram danam am sand At A , At B, A e pore water fremue = -2X Vw 2 - 2 x 10 3 - 20kN/m2 pwp= UBSO propa les 28w a 20 kN/m2 6 Porta Vertical stresses osv SPECIALTY COPAS OVBS 2x 8 = 2X2000 8 10 - 40kN/m2 ove 28c+ arsa 2x 2000 x10d 26 18808 lo 2 76 kN/m2

**) Klo 0. 64024 SHR out to vo kN/me och kolove i od x 76. 30 kn/ue (①) since figine is not given , takley on pt element and at Rm pon al as in element elementit ovo Ixro. 2060/m2 i br> 0.68 0. 12keyou? Ov> OV-U shear stress will be 90-(0) - 30 kn/me 1 zero, because wale at horizentellene caund sustain any sheas seren

10.4. AX I Ćt stand & 208 20tan ro - 27.279 kN/m2 - tokoome 12271234) Ro 16-ing+ (27. 2897 231.14 kasus ca (12,30.919) M10,39 221) 2 61046.919 kN/me 62 = -14.919 kr/unt - 14919 T 48.999,0) (20,30.909) ONCOLOGY HYGEA ONCOLOGY NGEA SPECIALTY COPAS CRITICAL CARE

and element: Ov= ax ret lxr, 2x20t18182 58 kN/m on a 0.58582 29 kN/m2 Sula svou for Rear Ts et sſtand 88-10 248 kN/m² per a lot ud tan 32 to 40 kN/m2 relt Rx K 43-5 - 29) ²4(40)² 42.547 kim? Roryar (2940) 119.958.0) (29) candy (opo) 8,043,10) (50-40) 0.983 6N/m2 6,286.040 kN/m²,