Question: Question 3 ( a ) Explain why the effective stress path of a drained triaxial compression test has a slope of 1 on the s

Question 3
(a) Explain why the effective stress path of a drained triaxial compression test has a slope of 1 on the s'-t' plane, and a slope of 3 on the q'-p' plane.
(b) The initial volume and area of a sandy soil sample (assume c'=0) are 200cm3 and 20cm2 respectively. A consolidated undrained (CU) triaxial test is conducted on the sample. The initial void ratio (e0) of the specimen is 0.661, and volume change after consolidation at effective pressure (3') of 200kPa is 2.5ml. At failure, the axial shortening is 12.5mm, the axial load is 750N, and pore water pressure increased by 100kPa compared with the beginning of compression stage. Find the volume, area and length after consolidation, and determine the friction angle of the soil.
(7 marks)
Specimen height (h**) and area (A**) during consolidation:
h**~~h0(1-v3);,A**~~A0(1-2v3)
Specimen area (A) during compression:
A=A**(1-v**1-a**)
where A0 is the initial area, v is the volumetric strain during consolidation; v** and a** are volumetric strain and axial strain during compression.
(c) For the same soil prepared under the same conditions and same initial void ratio as in Q3(b), the critical state parameters , and N are found to be 0.10,2.05 and 2.17, respectively. Another CU triaxial test is conducted with effective confining pressure (3') of 300kPa.
i. Estimate the volume change during the consolidation stage.
(4 marks)
ii. What would be the principal stress difference at failure under undrained compression?
(4 marks)
(d) The following parameters are known for a saturated normally consolidated clay: N=2.32,=0.15,=2.17 and M=0.94.
Estimate the value of void ratio (e) and principal stress difference (q) at failure if a drained triaxial compression test was conducted on the clay specimen consolidated under cell pressure (3) of 400kPa, with back pressure of 200kPa. Sketch the effective and total stress paths of the test on the q-p' plane. (6 marks)
Equation of ICL:
Equations of CSL:
v=N-lnp'
q'=Mp'
v=-lnp'
sin'=3M6+M
5
Question 3 ( a ) Explain why the effective stress

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