--- title: "Trench bottom heave safety factor" subject: "Foundation Engineering" chapter: 4 key: "semester_5/foundation_engineering:trench-heave-rock" source: https://www.bctengineer.com/q/semester_5/foundation_engineering/trench-heave-rock site: BCT Engineering --- # Trench bottom heave safety factor Foundation Engineering · Chapter 4: Arching in Soils and Braced Cuts - Asked in: 2074 Chaitra, 1+3 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/foundation_engineering/trench-heave-rock - Whole chapter: https://www.bctengineer.com/md/semester_5/foundation_engineering/chapter-4 ## Question What is arching effect in soils? A long 5 m wide and 10 m high vertical trench has to be constructed in a deep deposit of cohesive soil with $c = 35\ \text{kN/m}^2$ and $\gamma = 18\ \text{kN/m}^3$. The safety of the bottom of trench against heave is to be checked before protecting the trench walls using sheet piles. If the excavation is to be completed rapidly, determine the factor of safety against bottom heave. What will be the factor of safety if a hard rock is present at 2.5 m from the bottom of the trench? ## Short answer Arching is pressure transfer from yielding to stationary soil via shear. For deep clay, factor of safety against heave FS = 5.7c/(γH − √2 cH/B) = 2.46. With hard rock at 2.5 m below base, FS = 5.7c/(γH − cH/T) = 4.99. ## Answer #### Arching effect Arching is the transfer of pressure from a yielding part of a soil mass to the adjacent stationary parts through the shear resistance of the soil. The pressure on the yielding part decreases and that on the stationary part increases (Terzaghi's trap door experiment). #### Factor of safety against bottom heave (Terzaghi) Rapid excavation means undrained conditions, so $\phi_u = 0$ and $c = c_u = 35$ kN/m$^2$. The heave is checked as a bearing failure of the base under the weight of soil beside the cut, resisted by shear on vertical planes. The bearing capacity of the base is taken as $5.7c$. **(a) Deep clay** (no hard layer; $T \ge 0.7B$): the width of the failing strip is $B' = B/\sqrt2 = 0.7B$: $$ FS = \frac{5.7c}{\gamma H - \sqrt{2}\,cH/B} = \frac{5.7\times35}{18\times10 - \sqrt2\times35\times10/5} = \frac{199.5}{81.01} = 2.46 $$ **(b) Hard rock 2.5 m below the base:** here $T = 2.5\ \text{m} < 0.7B = 3.5$ m, so the failure zone is limited by the rock and $B' = T = 2.5$ m: $$ FS = \frac{5.7c}{\gamma H - cH/T} = \frac{199.5}{180 - 35\times10/2.5} = \frac{199.5}{40.0} = 4.99 $$ **Answer:** $FS = 2.46$ for deep clay. This is above the usual minimum of about 1.5, so the base is safe against heave, although the margin is modest. With the rock 2.5 m below the trench bottom, $FS = 4.99$, which is more than adequate. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is arching in soil? A. Increase of pressure on the yielding part and decrease on the stationary part B. Curved failure surface under a footing C. Formation of arch-shaped cracks in clay D. Transfer of pressure from a yielding part to adjacent stationary parts through shear resistance Answer: D. Shear resistance transfers pressure away from the yielding mass; the yielding part's pressure falls and neighbours' rises. (Key point: Arching effect) 2. Why is phi_u = 0 and c = c_u used for the heave check? A. Because the soil is sandy B. Because rapid excavation gives undrained conditions C. Because the trench is narrow D. Because sheet piles are not yet driven Answer: B. Rapid excavation does not allow drainage, so undrained strength with phi_u = 0 applies. (Key point: Undrained condition) 3. For deep clay, what is the FS against bottom heave? A. 4.99 B. 1.50 C. 2.46 D. 5.7 Answer: C. FS = 5.7c/(gamma*H - sqrt2*c*H/B) = 199.5/81.01 = 2.46. (Key point: FS in deep clay) 4. With rock 2.5 m below the base (T < 0.7B), what happens to the width of failing strip and FS? A. B' = T = 2.5 m, and FS rises to about 4.99 B. B' = 0.7B, and FS stays 2.46 C. B' = T = 2.5 m, and FS falls D. B' = B, and FS rises to 5.7 Answer: A. Rock limits the failure zone so B' = T; the resisting shear term cH/T becomes larger, giving FS = 199.5/40 = 4.99. 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