--- title: "Excavation effect on vertical stress" subject: "Soil Mechanics" chapter: 8 key: "semester_4/soil_mechanics:excavation-effect-vertical-stress" source: https://www.bctengineer.com/q/semester_4/soil_mechanics/excavation-effect-vertical-stress site: BCT Engineering --- # Excavation effect on vertical stress Soil Mechanics · Chapter 8: Vertical Stresses Below Applied Loads - Asked in: 2079 Asoj, 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/excavation-effect-vertical-stress - Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-8 ## Question An excavation 3 m × 6 m for foundation is made. The depth of the foundation is 2.5 [m] below the ground surface. The bulk unit weight of the soil is 2 kN/m³ [as printed]. Determine the effect of this excavation on the effective vertical stress at a depth of 6 m from the ground surface (i) vertically below the center of the foundation and (ii) 6 m away from the center of the foundation. ## Short answer Excavation acts as a negative load q = γDf = 5 kPa (using 2 kN/m3). At 3.5 m below base, the reduction below centre is 2.0 kN/m2; at 6 m away, reduction is 0.2 kN/m2. If γ = 20 kN/m3, values become 20.3 and 2.0 kN/m2. ## Answer #### Principle Excavation removes soil weight, so it acts like a **negative uniform load** on the base of the excavation: $$ q = \gamma D_f = 2\times2.5 = 5\ \text{kN/m}^2 \quad (\text{unit weight as printed}) $$ The reduction in vertical stress is obtained by the rectangular area (Fadum / Newmark) corner formula $I(m,n)$ with $m = B/z$, $n = L/z$, where $z$ is the depth below the excavation base. Depth required = 6 m from the ground, so $z = 6 - 2.5 = 3.5$ m. #### (i) Below the centre Divide the 3 m × 6 m area into four rectangles of 1.5 m × 3 m (corner at the centre). $$ m = \frac{1.5}{3.5} = 0.429, \quad n = \frac{3}{3.5} = 0.857 \;\Rightarrow\; I = 0.1013 $$ $$ \Delta\sigma_z = 4\,q\,I = 4\times5\times0.1013 = 2.03\ \text{kN/m}^2 \;(\text{reduction}) $$ #### (ii) At a point 6 m from the centre (along the 6 m side direction) The point is 3 m beyond the end of the excavation. Use a rectangle from 3 m to 9 m (length) and 1.5 m (half-width), doubled for the two halves: $$ I_1 = I\!\left(\frac{9}{3.5},\frac{1.5}{3.5}\right) = I(2.571, 0.429) = 0.1212, \qquad I_2 = I\!\left(\frac{3}{3.5},\frac{1.5}{3.5}\right) = 0.1013 $$ $$ \Delta\sigma_z = 2\,q\,(I_1 - I_2) = 2\times5\times(0.1212 - 0.1013) = 0.20\ \text{kN/m}^2 \;(\text{reduction}) $$ (Along the 3 m side direction the same method gives 0.11 kN/m².) **Answer: The effective vertical stress at 6 m depth is reduced by about 2.0 kN/m² below the centre and about 0.2 kN/m² at 6 m from the centre.** If the intended unit weight is 20 kN/m³ ($q = 50$ kN/m²), all reductions are 10 times larger: 20.3 kN/m² and 2.0 kN/m². The reduction is small, so excavation has little effect on points far away. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the effect of excavation on the stress below it treated? A. As a positive load equal to the soil weight B. As a negative uniform load q = gamma Df on the excavation base C. It has no effect on stress D. As a line load along the edges Answer: B. Removing soil removes its weight, equivalent to a negative uniform load q = gamma x Df. (Key point: Principle) 2. For a point 6 m below ground and an excavation base at 2.5 m, which z is used in m = B/z and n = L/z? A. 6 m B. 2.5 m C. 8.5 m D. 3.5 m Answer: D. The excavation acts at its base, so z = 6 - 2.5 = 3.5 m. (Key point: Depth to use) 3. How is the stress reduction below the centre of the 3 m x 6 m excavation found? A. One rectangle 3 x 6 with I and multiply by q/4 B. Four rectangles of 1.5 x 3 m with corner at the centre: 4 q I C. Two rectangles and divide by 2 D. Add I for 1.5 x 3 and 3 x 6 Answer: B. The area is split into four equal rectangles sharing the centre corner, so delta_sigma = 4 q I = 4 x 5 x 0.1013 = 2.03 kN/m2. (Key point: Below the centre) 4. For a point outside the excavation, the influence is obtained by A. adding all rectangle factors B. difference of influence factors of a larger and a smaller rectangle: 2q(I1 - I2) C. ignoring the excavation D. using the centre formula Answer: B. The loaded area is represented as a large rectangle minus the part between the point and the excavation, so I1 - I2 is used, giving about 0.20 kN/m2. 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