Select the INCORRECT statement in the case of a triaxial test.
Try it first: pick an option, then read the explanation.
Asked in SSC JE 28/10/2020 (Evening)
Detailed Explanation
The triaxial test is the most versatile laboratory strength test. A cylindrical specimen in a rubber membrane sits in a water-filled cell; the cell pressure gives σ₃ and the ram load gives the deviator stress, so σ₁ = σ₃ + σd.
Why “The state of stress within the specimen during any stage of test as well as at failure is completely indeterminate” Is Correct
Because both principal stresses are measured throughout, the complete state of stress is known at every stage, and a Mohr circle can be drawn at any time. So statement (a) is incorrect. Statements (b), (c) and (d) are genuine advantages of the triaxial test over the direct shear test.
| Feature | Direct shear test | Triaxial test |
|---|---|---|
| Failure plane | Forced: horizontal | Free to form on the weakest plane |
| Stress distribution | Non-uniform (higher at edges) | Uniform on the failure plane |
| Drainage control | None | Full control (UU, CU, CD) |
| Pore pressure measurement | Not possible | Possible |
| Stress state | Only on the failure plane | Known at every stage (σ₁, σ₃) |
| Best for | Sands, quick drained tests | All soils |
Why the Other Options Are Wrong
- B. The stress distribution on the failure plane is uniform: Correct: the stress on the failure plane is uniform, unlike in the direct shear box.
- C. All the three drainage conditions can be performed with complete control under the test: Correct: UU, CU and CD conditions can all be run with full drainage control.
- D. Precise measurement of pore pressure and volume change during test are possible: Correct: pore pressure and volume change can be measured accurately.
Key Point
Triaxial = stress state fully known. Direct shear = stress known only on the forced failure plane.
Key Concepts for Students
- The failure plane is free to form at 45° + φ/2.
- End friction from the platens causes some non-uniformity (barrelling).
- Area correction: A = A₀/(1 − εa) during shearing.
Practice more: this is question 31 in Shear Strength of Soils MCQs – Part 3. Try the whole set and check every answer instantly.