---
title: "Coarse material thickness against piping"
subject: "Soil Mechanics"
chapter: 7
key: "semester_4/soil_mechanics:piping-coarse-material-thickness"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/piping-coarse-material-thickness
site: BCT Engineering
---

# Coarse material thickness against piping

Soil Mechanics · Chapter 7: Seepage through Soils

- Asked in: 2073 Bhadra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/piping-coarse-material-thickness
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-7

## Question

A soil stratum having thickness 1.15 m, porosity = 30% and G = 2.7 is subjected to an upward seepage head of 1.95 m. Determine the thickness of coarse material required above the soil stratum to provide a factor of safety of 2 against piping, assuming that the coarse material has the same specific gravity and porosity as the soil and the head loss in the coarse material is negligible.

## Short answer

Submerged unit weight γ' = 1.19γw. With FS=2, γ'(L+z) / (γw h) = 2 → 1.19(1.15+z) = 2×1.95 → z ≈ 2.13 m. Add a 2.13 m thick coarse material layer above the soil to prevent piping.

## Answer

A layer of coarse material (a loaded filter) placed on top of the soil adds weight to resist the upward seepage force. The head loss in this material is negligible.

#### Given

Soil: $L = 1.15$ m, $n = 0.30$, $G = 2.7$. Upward seepage head at the base $h = 1.95$ m. Required factor of safety $F = 2$. Coarse layer of thickness $z$ with the same $G$ and $n$.

#### Submerged unit weight

$$
e = \frac{n}{1-n} = \frac{0.30}{0.70} = 0.4286
$$

$$
\gamma' = \frac{(G-1)\gamma_w}{1+e} = \frac{1.7}{1.4286}\gamma_w = 1.19\,\gamma_w
$$

#### Factor of safety against piping

At the base of the soil stratum, the effective stress with the load of the coarse layer:

$$
\sigma' = \gamma'(L + z) - \gamma_w h
$$

Factor of safety = submerged weight / uplift:

$$
F = \frac{\gamma'(L+z)}{\gamma_w h} = 2
$$

$$
1.19\,(1.15 + z) = 2 \times 1.95 = 3.9
$$

$$
1.15 + z = 3.277 \;\Rightarrow\; z = 2.13\ \text{m}
$$

For comparison, without the coarse layer $F = \dfrac{1.19\times1.15}{1.95} = 0.70 < 1$, so the stratum would pipe.

**Answer: Thickness of coarse material required $\approx 2.13$ m.**

## Check yourself

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

1. With n = 0.30, what is e?
   A. 0.30
   B. 0.43
   C. 0.70
   D. 2.33
   Answer: B. e = n/(1-n) = 0.3/0.7 = 0.4286. (Key point: Void ratio)

2. What is gamma' / gamma_w for G = 2.7 and e = 0.4286?
   A. 1.19
   B. 1.70
   C. 0.70
   D. 2.7
   Answer: A. (G-1)/(1+e) = 1.7/1.4286 = 1.19. (Key point: Submerged weight)

3. Which expression is used for F with the coarse layer z?
   A. gamma'(L+z)/(gamma_w h)
   B. gamma' L/(gamma_w h z)
   C. gamma_w h/(gamma'(L+z))
   D. gamma' z/(gamma_w L)
   Answer: A. Submerged weight of soil plus coarse layer divided by uplift head. (Key point: Factor of safety)

4. Solving 1.19(1.15 + z) = 3.9 gives z of about:
   A. 2.13 m
   B. 3.28 m
   C. 1.15 m
   D. 4.5 m
   Answer: A. 1.15 + z = 3.277, so z = 2.13 m. (Key point: Thickness)

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