A dry sand sample in a triaxial test gave Ø = 32° under cell pressure of 100 kPa. The deviator stress at failure is:

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OSSC JE Civil Mains (Re-Exam) Official Paper: (Held On: 3rd Sept 2023)
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  1. 100 kPa
  2. 225 kPa
  3. 325 kPa
  4. 53 kPa

Answer (Detailed Solution Below)

Option 2 : 225 kPa
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Concept:

Relation between major and minor principal stress at failure on the basis of Mohr-Coulomb criteria of failure:

\({\sigma _{1f}} = {\sigma _{3f}}{\tan ^2}\left( {45 + \frac{\phi }{2}} \right) + 2C\tan \left( {45 + \frac{\phi }{2}} \right)\)

\({\sigma _{3f}} = {\sigma _{1f}}{\tan ^2}\left( {45 - \frac{\phi }{2}} \right) - 2C\tan \left( {45 - \frac{\phi }{2}} \right)\)

Calculation:

σ3f = Cell pressure = 100 kPa

σ1f = Cell pressure + Deviator stress = (100 + x) kPa

For dry sand, C = 0

The deviator stress at failure is:

\({\sigma _{1f}} = {\sigma _{3f}}{\tan ^2}\left( {45 + \frac{\phi }{2}} \right) \)

\({100+x} = {100\times }{\tan ^2}\left( {45 + \frac{32 }{2}} \right) \)

\(100 + x = 100 \times 3.2545\)

\(x = 225.45\) kPa

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