--- title: "Embankment 1,00,000 m³ excavation" subject: "Soil Mechanics" chapter: 2 key: "semester_4/soil_mechanics:embankment-volume-weight-excavated" source: https://www.bctengineer.com/q/semester_4/soil_mechanics/embankment-volume-weight-excavated site: BCT Engineering --- # Embankment 1,00,000 m³ excavation Soil Mechanics · Chapter 2: Solids- Water –Air Relations and Index properties of soils - Asked in: 2073 Magh, 5 marks - Video: [Calculate Quantity of Soil and Water required for Compacted Soil| Soil Compaction Calculations](https://www.youtube.com/watch?v=JJpv_kS0BFg) — Master Civil Engineering - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/embankment-volume-weight-excavated - Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-2 ## Question An embankment of 1,00,000 m³ volume has to be constructed by compacting the soil brought from an excavation site. After compaction, the dry unit weight of the compacted soil (embankment) will be 16 kN/m³. Also, the bulk unit weight and water content of the soil at the excavation site are 12 kN/m³ and 15%, respectively. Find the volume and weight of soil to be excavated from the excavation site. Take specific gravity of soil solid as 2.70. ## Short answer Excavate 1,53,333 m³ of soil, with a moist weight of 1.84×10⁶ kN (dry weight 1.6×10⁶ kN). ## Answer #### Given Embankment: $V = 1{,}00{,}000$ m³, $\gamma_d = 16$ kN/m³. Excavation site: $\gamma = 12$ kN/m³, $w = 15\%$, $G = 2.70$. #### Dry weight needed in the embankment $$ W_s = 16 \times 100\,000 = 1.6 \times 10^{6}\ \text{kN} $$ #### Dry unit weight at the excavation site $$ \gamma_d = \frac{\gamma}{1+w} = \frac{12}{1.15} = 10.435\ \text{kN/m}^3 $$ #### Volume to be excavated $$ V_{exc} = \frac{W_s}{\gamma_d} = \frac{1.6 \times 10^6}{10.435} = 1.533 \times 10^{5}\ \text{m}^3 $$ #### Weight of soil to be excavated (moist) $$ W = \gamma \times V_{exc} = 12 \times 153\,333 = 1.84 \times 10^{6}\ \text{kN} $$ (check: $W_s(1+w) = 1.6 \times 10^6 \times 1.15 = 1.84 \times 10^6$ kN) The specific gravity is not needed for this answer; it would only be used to find void ratios. **Answer: volume $\approx 1{,}53{,}333$ m³; weight $\approx 1.84 \times 10^{6}$ kN (dry weight $1.6 \times 10^{6}$ kN).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What dry weight of soil is needed in the embankment? A. 1.6 x 10^6 kN (16 x 1,00,000) B. 1.2 x 10^6 kN C. 1.84 x 10^6 kN D. 2.7 x 10^6 kN Answer: A. Ws = dry unit weight x embankment volume = 16 x 100000 kN. (Key point: Dry weight needed) 2. How is the dry unit weight at the excavation site found? A. gamma x (1+w) = 13.8 kN/m3 B. gamma/(1-w) = 14.1 kN/m3 C. gamma/(1+w) = 12/1.15 = 10.43 kN/m3 D. gamma x w = 1.8 kN/m3 Answer: C. Dry unit weight is bulk unit weight divided by (1+w). (Key point: Dry unit weight at site) 3. How is the volume to be excavated obtained? A. Equal to the embankment volume, 1,00,000 m3 B. Embankment dry weight divided by dry unit weight at site: about 1.53 x 10^5 m3 C. Embankment dry weight divided by 12 kN/m3 D. Embankment dry weight times 10.43 Answer: B. The same dry solids must be supplied, so Vexc = Ws/gamma_d(site). (Key point: Volume to excavate) 4. How is the specific gravity 2.70 used in this answer? A. It gives the dry weight of the embankment B. It is used to find the volume to excavate C. It converts water content to moist weight D. It is not needed; it would only be used for void ratios Answer: D. The problem is solved with unit weights and w alone. (Key point: Role of G) ## Related in Soil Mechanics - [Compacted soil 17 kN/m³, OWC 16%](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Amax-dry-weight-17-void-ratio) — Soil Mechanics, chapter 5, asked 2× - [Embankment excavation quantity (18 kN/m³)](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Aembankment-excavated-quantity-2079) — Soil Mechanics, chapter 2 - [Compacted soil 18 kN/m³, porosity and ZAV](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Amax-dry-weight-18-porosity-zav) — Soil Mechanics, chapter 5, asked 2× - [Compacted soil 18 kN/m³, saturation](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Amax-dry-weight-18-saturation) — Soil Mechanics, chapter 5 - [Embankment volume from void ratios](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Aembankment-void-ratio-volume) — Soil Mechanics, chapter 2 - [Compaction test: curve, MDD, OMC, RC](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Acompaction-test-curve-mdd-omc-rc) — Soil Mechanics, chapter 5, asked 2× ## Related in other subjects - [EVA; 60 days, 25% work, Rs 12,00,000](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Aeva-numerical-60-days-25-percent) — Project Engineering, chapter 4 - [Embankment volume, 160 m, rising gradient 1 in 100](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Avolume-embankment-formation-10m-ground-levels) — Surveying I, chapter 9 - [Flexible design: four-lane, AADT 12,000](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Aflexible-design-aadt-12000-four-lane) — Transportation Engineering II, chapter 2 - [EPS and book value; ABC project Rs 6,00,000](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Aeps-bv-abc-project-600000-b79) — Project Engineering, chapter 6