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Unformatted text preview: stress at the sand surface in the dense, saturated state. Take the unit weight of water as 9.81kN/m 3 . Q1.5 Solution (a) In the loose dry state, the vertical total stress is given by the unit weight of the sand the depth h. The depth of the sand is given by the volume, 1200cm 3 , divided by the cross-sectional area of the measuring cylinder, 28.27cm 2 , giving h = 42.448cm. Hence v = 16.35kN/m 3 0.4245m = 6.94kPa. As the sand is dry, the pore water pressure u = 0 and ' v = v = 6.94kPa (Alternatively, the total weight of sand is 2kg 9.81 10-3 kN/kg = 0.01962kN. This is spread over an area of ( π × 0.06 2 m 2 ) ÷ 4 = 0.002827m 2 . Hence the total stress v = 0.01962kN 0.002827m 2 = 6.94 kPa.) (b) In the loose, saturated state, the pore water pressure u = 0.4245m 9.81kN/m 3 u = 4.164 kPa...
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This note was uploaded on 12/28/2011 for the course CHM 4302 taught by Professor Stuartchalk during the Fall '11 term at UNF.
- Fall '11
- Analytical Chemistry