---
title: "Saturated clay shrinkage limit numerical"
subject: "Soil Mechanics"
chapter: 2
key: "semester_4/soil_mechanics:saturated-clay-shrinkage-limit-numerical"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/saturated-clay-shrinkage-limit-numerical
site: BCT Engineering
---

# Saturated clay shrinkage limit numerical

Soil Mechanics · Chapter 2: Solids- Water –Air Relations and Index properties of soils

- Asked in: 2073 Bhadra, 6 marks
- Video: [#15 numerical on shrinkage limit & shrinkage ratio || soil mechanics 1 ||](https://www.youtube.com/watch?v=GrfdYK5n-Os) — Numerical's Civil
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/saturated-clay-shrinkage-limit-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-2

## Question

A sample of saturated clay has a volume of 97 cm³ and mass of 202 gm. When completely dried, its volume is 87 cm³ and mass of 167 gm. Determine: (i) initial water content, (ii) specific gravity of soil solids, (iii) shrinkage limit.

## Short answer

Initial water content=20.96% (mass loss/dry mass). Solid volume Vs=62 cm³ gives G=2.69. Shrinkage limit=20.96%−(10/167)=15.0%.

## Answer

#### Given
Initial: $V_1 = 97$ cm³, $M_1 = 202$ g. Dry: $V_d = 87$ cm³, $M_s = 167$ g. $\rho_w = 1$ g/cm³.

#### (i) Initial water content
$$
w_1 = \frac{M_1 - M_s}{M_s} = \frac{202 - 167}{167} = 0.2096 = 20.96\%
$$

#### (ii) Specific gravity
The sample is saturated initially, so $V_1 = V_s + V_{w1}$:
$$
V_{w1} = 35\ \text{cm}^3 \Rightarrow V_s = 97 - 35 = 62\ \text{cm}^3
$$
$$
G = \frac{M_s}{V_s\rho_w} = \frac{167}{62} = 2.69
$$

#### (iii) Shrinkage limit
$$
w_s = w_1 - \frac{(V_1 - V_d)\rho_w}{M_s} = 0.2096 - \frac{(97 - 87)}{167} = 0.2096 - 0.0599 = 0.1497
$$
So $w_s = 15.0\%$.

Check using the dry volume: $V_d = V_s + w_s M_s = 62 + 0.1497 \times 167 = 87.0$ cm³ (agrees).

**Answer: (i) $w = 20.96\%$, (ii) $G = 2.69$, (iii) shrinkage limit $= 15.0\%$.**

## Check yourself

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

1. Initial water content for 202 g wet and 167 g dry?
   A. 17.3%
   B. 35%
   C. 16.5%
   D. 20.96%
   Answer: D. w = (202-167)/167 = 0.2096. (Key point: Initial water content)

2. Why does Vs = V1 - Vw1 = 97 - 35 = 62 cm3?
   A. The sample is saturated, so volume is solids plus water only
   B. The sample contains air voids that are ignored
   C. The dry volume equals the solids volume
   D. Water has density 2.69 g/cm3
   Answer: A. A saturated sample has no air, so V = Vs + Vw. (Key point: Volume of solids)

3. G is found as
   A. Ms/V1 = 1.72
   B. Ms/(Vs rho_w) = 167/62 = 2.69
   C. M1/Vs = 3.26
   D. Ms/Vd = 1.92
   Answer: B. G = Ms/(Vs rho_w). (Key point: Specific gravity)

4. How is shrinkage limit calculated?
   A. ws = w1 + (V1-Vd)/Ms = 26.9%
   B. ws = (V1-Vd)/Vd = 11.5%
   C. ws = w1 - (V1-Vd)rho_w/Ms = 15.0%
   D. ws = w1 = 20.96%
   Answer: C. Water lost while volume decreases is subtracted from w1: 20.96 - 5.99 = 15.0%. (Key point: Shrinkage limit)

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