---
title: "CU tests on overconsolidated clay"
subject: "Soil Mechanics"
chapter: 10
key: "semester_4/soil_mechanics:cu-overconsolidated-effective-parameters"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/cu-overconsolidated-effective-parameters
site: BCT Engineering
---

# CU tests on overconsolidated clay

Soil Mechanics · Chapter 10: Shear Strength of Soil

- Asked in: 2078 Chaitra, 6 marks
- Video: [YourLabs Geotechnical Triaxial CU](https://www.youtube.com/watch?v=gAfzW4xNgWk) — YourLabs
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/cu-overconsolidated-effective-parameters
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-10

## Question

The series of consolidated undrained tests on undisturbed samples of an overconsolidated clay were as below. Determine the shear parameters in terms of effective stresses.

| Cell pressure (kN/m²) | 100 | 200 | 400 | 600 |
|---|---|---|---|---|
| Deviator stress at failure (kN/m²) | 300 | 410 | 610 | 850 |
| Pore water pressure (kN/m²) | -45 | -15 | 50 | 110 |

## Short answer

Compute σ3′ = σ3 − u, σ1′ = σ3′ + σd. From p′–q plot, Kf line gives tanα=0.4421, a=18.7 kPa. Then sinφ′ = tanα → φ′ = 26.2°, c′ = a/cosφ′ = 20.8 ≈ 21 kN/m². Effective parameters: c′ ≈ 21 kN/m², φ′ ≈ 26°.

## Answer

**Step 1: Effective principal stresses at failure**

$\sigma_3' = \sigma_3 - u$ and $\sigma_1' = \sigma_3' + \sigma_d$.

| $\sigma_3$ | $\sigma_d$ | $u$ | $\sigma_3'$ | $\sigma_1'$ | $p' = \frac{\sigma_1'+\sigma_3'}{2}$ | $q = \frac{\sigma_d}{2}$ |
|---|---|---|---|---|---|---|
| 100 | 300 | -45 | 145 | 445 | 295 | 150 |
| 200 | 410 | -15 | 215 | 625 | 420 | 205 |
| 400 | 610 | 50 | 350 | 960 | 655 | 305 |
| 600 | 850 | 110 | 490 | 1340 | 915 | 425 |

**Step 2: Envelope.** Draw the four effective-stress circles and the common tangent. Alternatively, fit a straight line through the points $(p', q)$ (the $K_f$ line), $q = a + p'\tan\alpha$. A least-squares fit gives:

$$
\tan\alpha = 0.4421, \qquad a = 18.7\ \text{kPa}
$$

**Step 3: Parameters**

$$
\sin\phi' = \tan\alpha = 0.4421 \;\Rightarrow\; \phi' = 26.2^\circ
$$

$$
c' = \frac{a}{\cos\phi'} = \frac{18.7}{0.8975} = 20.8\ \text{kPa}
$$

**Answer: $c' \approx 21$ kN/m² and $\phi' \approx 26^\circ$** (in terms of effective stress).

## Check yourself

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

1. How are effective principal stresses at failure found?
   A. sigma_3' = sigma_3 + u and sigma_1' = sigma_1 + u
   B. sigma_1' = sigma_d - u
   C. sigma_3' = sigma_d / 2
   D. sigma_3' = sigma_3 - u and sigma_1' = sigma_3' + sigma_d
   Answer: D. Subtract pore pressure from total stress; for the first test sigma_3' = 100 + 45 = 145. (Key point: Effective stresses)

2. What does the negative pore pressure at low cell pressure indicate?
   A. Normally consolidated clay contraction
   B. An error in measurement
   C. Dilative behaviour of overconsolidated clay, so effective stress exceeds total
   D. Air in the sample only
   Answer: C. Overconsolidated clay tends to expand on shearing, producing suction, so sigma_3' = 100 - (-45) = 145. (Key point: Negative pore pressure)

3. What is the K_f line fit used for?
   A. Plot sigma_3 against u to get phi'
   B. Plot (p', q) points and fit q = a + p' tan alpha to get phi' and c'
   C. Plot sigma_d against time
   D. Plot e against log p
   Answer: B. The p'-q failure points lie on a line whose slope and intercept give the strength parameters. (Key point: K_f line)

4. How are phi' and c' obtained from tan alpha and a?
   A. sin phi' = tan alpha and c' = a / cos phi'
   B. phi' = alpha and c' = a
   C. tan phi' = sin alpha and c' = a cos phi'
   D. sin phi' = a and c' = tan alpha
   Answer: A. sin phi' = 0.4421 gives 26.2° and c' = 18.7/0.8975 = 20.8 kPa. (Key point: Converting parameters)

## Related in Soil Mechanics

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