---
title: "CU test: σ3 200, σ1 550, u 80"
subject: "Soil Mechanics"
chapter: 10
key: "semester_4/soil_mechanics:cu-test-failure-plane-direction"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/cu-test-failure-plane-direction
site: BCT Engineering
---

# CU test: σ3 200, σ1 550, u 80

Soil Mechanics · Chapter 10: Shear Strength of Soil

- Asked in: 2073 Bhadra, 2+2+2+2 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/cu-test-failure-plane-direction
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-10

## Question

A consolidated undrained triaxial test was performed on a normally consolidated saturated clay and the cell pressure, $\sigma_3$ = 200 kN/m², axial stress, $\sigma_1$ = 550 kN/m² and pore water pressure, $u_w$ = 80 kN/m² were measured. Answer the following: i) Plot the Mohr circle of stresses in regard to total stress. ii) Plot the Mohr circle of stresses in regard to effective stress. iii) Assume the condition of normal consolidation and c' = 0. Then obtain the value of $\phi'$. iv) If Mohr-Coulomb's failure criterion is assumed to be valid, then determine the direction of failure plane that might occur within the specimen.

## Short answer

Effective stresses are sigma3' = 120 kPa and sigma1' = 470 kPa. With c' = 0, sin phi' = 350/590 = 0.593, so phi' = 36.4 degrees. The failure plane is inclined at 45 + phi'/2 = 63.2 degrees to the horizontal, or 26.8 degrees to the specimen axis.

## Answer

**Data:** $\sigma_3 = 200$, $\sigma_1 = 550$, $u = 80$ kN/m². Deviator stress $= 350$ kN/m². Effective stresses: $\sigma_3' = 200-80 = 120$, $\sigma_1' = 550-80 = 470$ kN/m².

#### i) Mohr circle, total stress

Centre $= \frac{200+550}{2} = 375$, radius $= \frac{550-200}{2} = 175$. The circle cuts the $\sigma$-axis at 200 and 550.

#### ii) Mohr circle, effective stress

Centre $= \frac{120+470}{2} = 295$, radius $= 175$. The circle cuts the $\sigma$-axis at 120 and 470. It is the total stress circle shifted left by $u = 80$.

```text
 tau
  |          /     /
  |        /     /   (effective circle)
  |      /     /  .-'''-.   (total circle)
  |    /     / .-'.-'''-. \
  |  /     /.-'  '       ' |
  +-+--+---+----+-------+-+-- sigma
   0  120 200  470     550
```

#### iii) Effective friction angle

For $c' = 0$:

$$
\sin\phi' = \frac{\sigma_1'-\sigma_3'}{\sigma_1'+\sigma_3'} = \frac{350}{590} = 0.593 \;\Rightarrow\; \phi' = 36.4^\circ
$$

(For information, the total stress angle is $\sin\phi = 350/750$, $\phi = 27.8^\circ$.)

#### iv) Direction of the failure plane

The failure plane is inclined to the major principal plane (horizontal) at

$$
\theta = 45^\circ + \frac{\phi'}{2} = 45^\circ + 18.2^\circ = 63.2^\circ
$$

So the plane is at about $63.2^\circ$ to the horizontal, which is $26.8^\circ$ to the axis of the specimen.

**Answer: $\phi' \approx 36.4^\circ$; failure plane at about $63^\circ$ to the horizontal.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What are the centre and radius of the total stress circle?
   A. Centre 295, radius 175
   B. Centre 375, radius 350
   C. Centre 375, radius 175 kN/m²
   D. Centre 550, radius 200
   Answer: C. (200 + 550)/2 = 375, (550 - 200)/2 = 175. (Key point: Total circle)

2. How is the effective stress circle related?
   A. Shifted right by 80
   B. It is the total circle shifted left by u = 80, spanning 120 to 470
   C. It has radius 255
   D. It is smaller than the total circle
   Answer: B. Subtracting u from both principal stresses shifts the circle without changing size. (Key point: Effective circle)

3. What is phi' for c' = 0?
   A. 36.4°, from sin phi' = 350/590
   B. 27.8°
   C. 30°
   D. 45°
   Answer: A. sigma_1' = 470 and sigma_3' = 120. (Key point: Phi')

4. What is the direction of the failure plane?
   A. 26.8° to the horizontal
   B. 45° to the horizontal
   C. 58.9° to the vertical
   D. 63.2° to the horizontal, 26.8° to the specimen axis
   Answer: D. theta = 45° + phi'/2 = 63.2° from the major principal plane. (Key point: Failure plane)

## Related in Soil Mechanics

- [CU test NC clay 200/350/80, φ and φ'](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Acu-nc-clay-200-350-80) — Soil Mechanics, chapter 10
- [CU test on NC clay, three samples](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Acu-test-nc-clay-envelopes) — Soil Mechanics, chapter 10
- [CU tests on overconsolidated clay](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Acu-overconsolidated-effective-parameters) — Soil Mechanics, chapter 10
- [Three triaxial tests, Mohr circles, φ](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Amohr-circle-three-triaxial-tests-friction) — Soil Mechanics, chapter 10
