--- title: "Relative consistency, shrinkage numerical" subject: "Soil Mechanics" chapter: 2 key: "semester_4/soil_mechanics:relative-consistency-shrinkage-numerical" source: https://www.bctengineer.com/q/semester_4/soil_mechanics/relative-consistency-shrinkage-numerical site: BCT Engineering --- # Relative consistency, shrinkage numerical Soil Mechanics · Chapter 2: Solids- Water –Air Relations and Index properties of soils - Asked in: 2079 Jestha, 1+5 marks - Video: [26.Numerical On Liquidity,Plasticity, Shrinkage,Consistency And Toughness Index Of Soil | All Index](https://www.youtube.com/watch?v=cOlwb5P8CRM) — Mk Plus Point - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/relative-consistency-shrinkage-numerical - Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-2 ## Question Define relative consistency. The values of liquid limit, plastic limit and shrinkage limit of a soil were reported as follows: $\omega_L = 60\%$, $\omega_P = 30\%$, $\omega_S = 20\%$. If a sample of this soil at liquid limit has a volume of 40 cc and its volume measured at shrinkage limit was 23.5 cc, determine the specific gravity of the solids. What is its shrinkage ratio? Also draw the phase diagram of the soil at liquid limit and at shrinkage limit as per the given values. ## Short answer Relative consistency is (wL-w)/Ip. Using the volume difference between liquid and shrinkage limits, solids weight is 41.25 g and solids volume is 15.25 cc, giving specific gravity 2.70 and shrinkage ratio 1.76. ## Answer #### Relative consistency The relative consistency (consistency index) $I_c$ is the ratio of the difference between the liquid limit and the natural water content to the plasticity index: $$ I_c = \frac{w_L - w}{I_p} $$ $I_c = 1$ means the soil is at the plastic limit (stiff), and $I_c = 0$ means it is at the liquid limit (very soft). #### Given $w_L = 60\%$, $w_P = 30\%$, $w_S = 20\%$, $V_{LL} = 40$ cc, $V_{SL} = 23.5$ cc, $I_p = 60 - 30 = 30\%$. #### Mass of solids The soil is saturated at the liquid limit and at the shrinkage limit, and the volume of solids is the same. The volume difference is the volume of water lost between $w_L$ and $w_S$ (density of water 1 g/cc): $$ \begin{aligned} V_{LL} - V_{SL} &= \frac{(w_L - w_S) W_s}{\rho_w} \\ 40 - 23.5 &= (0.60 - 0.20) W_s \\ W_s &= \frac{16.5}{0.40} = 41.25\ \text{g} \end{aligned} $$ #### Specific gravity At the shrinkage limit the soil is just saturated: $V_{SL} = V_s + w_S W_s$. $$ \begin{aligned} 23.5 &= V_s + 0.20 \times 41.25 = V_s + 8.25 \\ V_s &= 15.25\ \text{cc} \\ G &= \frac{W_s}{V_s \rho_w} = \frac{41.25}{15.25} = 2.70 \end{aligned} $$ #### Shrinkage ratio $$ R = \frac{W_s}{V_{SL}\,\rho_w} = \frac{41.25}{23.5} = 1.76 $$ **Answer: $G = 2.70$, $R = 1.76$.** #### Phase diagrams (per the sample) ```text At liquid limit At shrinkage limit V = 40 cc V = 23.5 cc +--------+ +--------+ | Water | 24.75 cc | Water | 8.25 cc | 24.75 g| | 8.25 g | |--------| |--------| | Solids | 15.25 cc | Solids | 15.25 cc | 41.25 g| | 41.25 g| +--------+ +--------+ W = 66.0 g W = 49.5 g S = 100%, Va = 0 S = 100%, Va = 0 ``` Weights of water: $0.60 \times 41.25 = 24.75$ g at the liquid limit and $0.20 \times 41.25 = 8.25$ g at the shrinkage limit. At both limits there is no air. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Relative consistency Ic is: A. Ip/wL B. (w - wP)/Ip C. (wL - w)/Ip D. (wL - wP)/w Answer: C. Ic = (wL - w)/Ip; 1 at plastic limit, 0 at liquid limit. (Key point: Relative consistency) 2. How is the mass of solids found in the numerical? A. From VLL - VSL = (wL - wS) Ws / rho_w B. From Ws = G x V C. From Ip = Ws D. From wP only Answer: A. The volume loss between LL and SL equals the volume of water lost: 16.5 = 0.40 Ws, Ws = 41.25 g. (Key point: Method for Ws) 3. The specific gravity of solids obtained is: A. 1.76 B. 2.40 C. 2.70 D. 3.00 Answer: C. Vs = 23.5 - 8.25 = 15.25 cc; G = 41.25/15.25 = 2.70. (Key point: Specific gravity) 4. Shrinkage ratio R = Ws/(V_SL rho_w) equals: A. 1.76 B. 2.70 C. 0.57 D. 1.40 Answer: A. 41.25/23.5 = 1.76. 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