--- title: "Annular ring, stress at five depths" subject: "Soil Mechanics" chapter: 8 key: "semester_4/soil_mechanics:annular-ring-stress-with-depth" source: https://www.bctengineer.com/q/semester_4/soil_mechanics/annular-ring-stress-with-depth site: BCT Engineering --- # Annular ring, stress at five depths Soil Mechanics · Chapter 8: Vertical Stresses Below Applied Loads - Asked in: 2079 Jestha, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/annular-ring-stress-with-depth - Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-8 ## Question The annular ring foundation of external and internal diameter 4 m and 6 m respectively [as printed] transmits a pressure of 100 kN/m². Compute the vertical stresses at the depth 0.5 m, 1 m, 2 m, 4 m and 8 m below the center. Also draw the stress distribution curve along depth. ## Short answer Stress equals outer circle (R=3 m) minus inner circle (r=2 m): Δσz = q[(1−(1+(R/z)2)−3/2) − (1−(1+(r/z)2)−3/2)]. At depths 0.5, 1, 2, 4, 8 m, vertical stresses are 0.98, 5.78, 18.29, 20.35, 9.22 kPa. Maximum ~22 kPa near z=3 m. ## Answer #### Given Ring diameters 4 m and 6 m (as printed: taken as outer $D = 6$ m, inner $d = 4$ m), so $R = 3$ m and $r = 2$ m. Pressure $q = 100$ kN/m². #### Method The stress under the centre of a ring equals that of the full circle of radius $R$ minus that of the inner circle (radius $r$) at the same pressure: $$ \Delta\sigma_z = q\left[\left(1 - \frac{1}{\left(1+(R/z)^2\right)^{3/2}}\right) - \left(1 - \frac{1}{\left(1+(r/z)^2\right)^{3/2}}\right)\right] $$ #### Calculation | $z$ (m) | Circle $R=3$: $\Delta\sigma$ (kPa) | Circle $r=2$: $\Delta\sigma$ (kPa) | Ring $\Delta\sigma_z$ (kPa) | |---|---|---|---| | 0.5 | 99.56 | 98.57 | 0.98 | | 1 | 96.84 | 91.06 | 5.78 | | 2 | 82.93 | 64.64 | 18.29 | | 4 | 48.80 | 28.45 | 20.35 | | 8 | 17.91 | 8.69 | 9.22 | #### Distribution curve ```text depth z (m) stress (kPa) 0 |* 0 0.5 |* 1 1 |* 5.8 2 |---* 18.3 3 |----* 22.2 (maximum, z ≈ 3 m) 4 |----* 20.4 8 |--* 9.2 ``` The stress along the axis is near zero at the surface (the centre is unloaded), increases to a maximum of about 22 kPa at $z \approx 3$ m, and then decreases. **Answer: $\Delta\sigma_z = 0.98,\ 5.78,\ 18.29,\ 20.35,\ 9.22$ kN/m² at 0.5, 1, 2, 4 and 8 m.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the axial stress under a ring (annular) foundation found? A. Stress of the full outer circle minus stress of the inner circle, same pressure B. Stress of the outer circle plus the inner circle C. Average of the two circles D. Stress of the inner circle only Answer: A. The ring is the outer circle minus the unloaded inner circle. (Key point: Ring method) 2. The stress under the centre of a circle of radius R at depth z is A. q (R/z) B. q / (1+(R/z)^2) C. q [1 - 1/(1+(R/z)^2)^(3/2)] D. 3q/(2 pi z^2) Answer: C. This is the Boussinesq result for the centre of a uniformly loaded circle. (Key point: Circle formula) 3. How does the stress on the axis of the ring vary with depth? A. Maximum at the surface, decreasing with depth B. Constant with depth C. Zero at all depths D. Near zero at the surface, rises to a maximum (about 20 kPa near 3 to 4 m), then decreases Answer: D. The centre is unloaded so stress is small near the surface, peaks around 3 to 4 m, then falls. (Key point: Stress profile) 4. With R = 3 m, r = 2 m, q = 100 kPa, the ring stress at z = 2 m is A. 18.29 kPa B. 64.64 kPa C. 82.93 kPa D. 5.78 kPa Answer: A. 82.93 - 64.64 = 18.29 kPa. 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