--- title: "Trapezoidal masonry wall, sliding and overturning" subject: "Foundation Engineering" chapter: 3 key: "semester_5/foundation_engineering:trapezoidal-wall-coulomb-sliding" source: https://www.bctengineer.com/q/semester_5/foundation_engineering/trapezoidal-wall-coulomb-sliding site: BCT Engineering --- # Trapezoidal masonry wall, sliding and overturning Foundation Engineering · Chapter 3: Lateral Earth Pressure Theories and Retaining Walls - Asked in: 2080 Bhadra, 10 marks - Video: [Retaining Walls Explained | Types, Forces, Failure and Reinforcement](https://www.youtube.com/watch?v=UummY_mo85Y) — Structures Explained - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/foundation_engineering/trapezoidal-wall-coulomb-sliding - Whole chapter: https://www.bctengineer.com/md/semester_5/foundation_engineering/chapter-3 ## Question A trapezoidal masonry retaining wall 1 m wide at top and 3 m wide at its bottom is 4 m high. The vertical face is retaining soil ($\phi = 30^\circ$) at a surcharge angle of $20^\circ$ with the horizontal. Determine the maximum and minimum intensities of pressure at the base of the retaining wall using Coulomb's earth pressure theory. Unit weights of soil and masonry are $20\ \text{kN/m}^3$ and $24\ \text{kN/m}^3$ respectively. Take $\delta = 24^\circ$ on the base wall, determine the factor of safety against sliding and overturning. ## Short answer Using Coulomb Ka = 0.416, base pressures are qmax = 99.4 kN/m² and qmin = 46.7 kN/m²; FS against overturning is 5.54 and against sliding is 1.60. ## Answer **Geometry and assumptions:** top width 1 m, base width 3 m, height 4 m. The soil face (back) is vertical and the front face is inclined, so the toe is the front bottom corner. The backfill slopes at $\beta = 20^\circ$, with $\phi = 30^\circ$ and $\gamma = 20\ \text{kN/m}^3$. The only $\delta$ given, $24^\circ$, is used both as the wall friction (for Coulomb's $K_a$) and as the base friction (for sliding). Passive resistance is neglected. #### 1. Coulomb active thrust $$ K_a = \frac{\sin^2(\theta+\phi)}{\sin^2\theta\,\sin(\theta-\delta)\left[1+\sqrt{\dfrac{\sin(\phi+\delta)\sin(\phi-\beta)}{\sin(\theta-\delta)\sin(\theta+\beta)}}\right]^2} = 0.4162\quad(\theta = 90^\circ) $$ $$ P_a = \tfrac12 K_a\gamma H^2 = 0.5\times0.4162\times20\times4^2 = 66.59\ \text{kN/m} $$ It acts at $H/3 = 1.333$ m above the base, at $24^\circ$ below the horizontal: $$ P_h = P_a\cos24^\circ = 60.83,\qquad P_v = P_a\sin24^\circ = 27.08\ \text{kN/m} $$ #### 2. Weights and moments about the toe | Part | Weight (kN/m) | Arm from toe (m) | Moment | |---|---|---|---| | Rectangle 1 x 4 x 24 | 96.00 | 2.500 | 240.00 | | Triangle 0.5 x 2 x 4 x 24 | 96.00 | 1.333 | 128.00 | | $P_v$ (at back, x = 3) | 27.08 | 3.000 | 81.25 | | **Total** | **219.08** | | **449.25** | Overturning moment: $M_o = P_h\times1.333 = 60.83\times1.333 = 81.11$ kN m/m. #### 3. Base pressure $$ \bar x = \frac{M_r - M_o}{\sum V} = \frac{449.25-81.11}{219.08} = 1.680\ \text{m} $$ $$ e = \frac{B}{2}-\bar x = 1.5-1.680 = -0.180\ \text{m},\qquad |e| = 0.180\ \text{m} < B/6 = 0.5\ \text{m} $$ $$ q = \frac{\sum V}{B}\left(1 \pm \frac{6e}{B}\right) $$ The resultant falls toward the heel, so $q_{max}$ is at the heel and $q_{min}$ is at the toe: $$ q_{max} = 99.4\ \text{kN/m}^2,\qquad q_{min} = 46.7\ \text{kN/m}^2 $$ Both are compressive, so there is no tension. #### 4. Factors of safety $$ FS_{ot} = \frac{M_r}{M_o} = \frac{449.25}{81.11} = 5.54\ (>2,\ \text{safe}) $$ $$ FS_{s} = \frac{\sum V\tan24^\circ}{P_h} = \frac{219.08\times0.4452}{60.83} = 1.60\ (>1.5,\ \text{safe}) $$ **Answer:** $q_{max} = 99.4$ and $q_{min} = 46.7$ kN/m$^2$; $FS_{ot} = 5.54$; $FS_{sliding} = 1.60$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In Coulomb's method with wall friction delta = 24 degrees, how is Pa resolved? A. Horizontal only B. At delta below the horizontal, giving Ph = Pa cos delta and Pv = Pa sin delta C. Parallel to the slope D. Perpendicular to the wall face only with no vertical part Answer: B. The thrust is inclined at delta to the wall normal (horizontal here), so it has a vertical component that adds to stability. (Key point: Coulomb thrust direction) 2. How is the factor of safety against overturning computed? A. Sum V tan delta / Ph B. Resisting moment about the toe / overturning moment about the toe C. Sum V / Ph D. q_max / q_min Answer: B. FSot = Mr/Mo = 449.25/81.11 = 5.54, with Pv included in Mr. (Key point: Overturning factor of safety) 3. Which expression gives the sliding factor of safety here (passive ignored)? A. Sum V x tan(24 degrees) / Ph B. Ph / Sum V C. Sum V / (Ph tan 24) D. Mr tan 24 / Mo Answer: A. Friction resistance V tan delta divided by horizontal thrust: 219.08 x 0.4452/60.83 = 1.60. (Key point: Sliding factor of safety) 4. With e = 0.18 m < B/6 and the resultant towards the heel, where is q_max and what is the formula? A. At the toe, q = V/B (1 + 6e/B) B. At the heel, q = V/B (1 + 6e/B) C. At the heel, q = V/B (1 - 6e/B) D. At the middle only Answer: B. The resultant lies nearer the heel, so heel pressure is the maximum (99.4) and the toe the minimum (46.7). (Key point: Base pressure and eccentricity) ## Close variants set in other exams - [Trapezoidal masonry wall, overturning](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Atrapezoidal-wall-friction-overturning) — Foundation Engineering, chapter 3, asked 2× ## Related in Foundation Engineering - [Stability of cantilever wall, 6 m, q = 35](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Astability-cantilever-wall-q35) — Foundation Engineering, chapter 3 - [Active pressure on wall, cohesive soil](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Acohesive-wall-pressure-diagram) — Foundation Engineering, chapter 3 - [Stability of wall with surcharge 25 kN/m²](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Astability-retaining-wall-q25) — Foundation Engineering, chapter 3 ## Related in other subjects - [Base width of 20 m high trapezoidal gravity dam](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Agravity-dam-base-width-trapezoidal-20m) — Hydropower Engineering, chapter 4 - [3 m wall under wind: fixed top and bottom](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Awind-wall-3m-fixed-top-bottom) — Concrete Technology and Masonry Structure, chapter 9, asked 2× - [3 m wall under wind: free top, fixed bottom](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Awind-wall-3m-free-top) — Concrete Technology and Masonry Structure, chapter 9, asked 2× - [Overturning and sliding check with shear strength](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Agravity-dam-overturning-sliding) — Hydro Power, chapter 4