TGPSC HMWSSB Manager 2015 Paper 2 · Question 36

Question 36 of the TSPSC Manager (Engineering), HMWSSB, 2015: Paper 2 (Civil Engineering) (1 November 2015). Try it first, then open the answer and explanation.

1. Manager 2015 Paper 2 (Civil) · Q36 · Geotechnical Engineering · Soil Mechanics

   The shear strength of a pure clay specimen when tested in Unconfined compression Test was found to be 100 kPa. If the same specimen was tested in Tri-axial Compression Test, the deviatoric stress at which specimen will undergo failure when the confining stress was 50 kPa, will be:

   1. 50 kPa
   2. 100 kPa
   3. 200 kPa
   4. 400 kPa
   Show answer

   Answer: (3) 200 kPa

   A shear strength of 100 kPa means c_u = 100 kPa, so the unconfined compressive strength is q_u = 2c_u = 200 kPa. For a pure clay tested undrained (φ_u = 0) the deviator stress at failure does not depend on the confining pressure, so at 50 kPa confinement it is again 2c_u = 200 kPa; 100 kPa would be the answer only if 100 kPa were the compressive strength.

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