--- title: "Trapezoidal masonry wall, overturning" subject: "Foundation Engineering" chapter: 3 key: "semester_5/foundation_engineering:trapezoidal-wall-friction-overturning" source: https://www.bctengineer.com/q/semester_5/foundation_engineering/trapezoidal-wall-friction-overturning site: BCT Engineering --- # Trapezoidal masonry wall, overturning Foundation Engineering · Chapter 3: Lateral Earth Pressure Theories and Retaining Walls - Asked in: 2075 Asoj, 8 marks - Video: [EX 02 | RETAINING WALL | TRAPEZOIDAL SECTION | BENDING STRESS | PRESSURE AT BASE | ECCENTRICITY](https://www.youtube.com/watch?v=gZP1zXhWZnA) — Structural Funda by:- KheNiSiR - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/foundation_engineering/trapezoidal-wall-friction-overturning - 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. Unit weights of soil and masonry are $20\ \text{kN/m}^3$ and $24\ \text{kN/m}^3$. Assuming the coefficient of friction at the base of the wall as 0.45, determine the factor of safety against overturning. ## Short answer Rankine Ka = 0.414, base pressures are 92.2 kN/m² (heel) and 50.9 kN/m² (toe); factor of safety against overturning is 5.25. ## Answer **Geometry and assumptions:** top width 1 m, base 3 m, height 4 m, vertical back face against the soil and an inclined front face (the toe is the front bottom corner). The back is vertical and no wall friction is given, so Rankine's theory for a surcharge angle $\beta = 20^\circ$ is used. Passive resistance is ignored. #### 1. Active thrust (Rankine, sloping backfill) $$ K_a = \cos\beta\,\frac{\cos\beta-\sqrt{\cos^2\beta-\cos^2\phi}}{\cos\beta+\sqrt{\cos^2\beta-\cos^2\phi}} = 0.4142 $$ $$ P_a = \tfrac12 K_a\gamma H^2 = 0.5\times0.4142\times20\times4^2 = 66.27\ \text{kN/m} $$ It acts at $H/3 = 1.333$ m above the base, parallel to the backfill slope ($20^\circ$): $$ P_h = 62.28\ \text{kN/m},\qquad P_v = 22.67\ \text{kN/m} $$ #### 2. Moments about the toe | Part | Force (kN/m) | Arm (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 x = 3) | 22.67 | 3.000 | 68.00 | | **Total** | **214.67** | | **436.00** | $$ M_o = 62.28\times1.333 = 83.03\ \text{kN m/m} $$ #### 3. Base pressures $$ \bar x = \frac{436.00-83.03}{214.67} = 1.644\ \text{m},\qquad e = \frac{B}{2}-\bar x = 1.5-1.644 = -0.144\ \text{m},\qquad |e| = 0.144\ \text{m} < B/6 = 0.5\ \text{m} $$ $$ q_{max} = 92.2\ \text{kN/m}^2\ (\text{heel}),\qquad q_{min} = 50.9\ \text{kN/m}^2\ (\text{toe}) $$ #### 4. Factors of safety $$ FS_{ot} = \frac{M_r}{M_o} = \frac{436.00}{83.03} = 5.25\ (>2,\ \text{safe}) $$ Sliding (for information) with $\mu = 0.45$: $FS_s = 214.67\times0.45/62.28 = 1.55$. **Answer:** $q_{max} = 92.2$ kN/m$^2$, $q_{min} = 50.9$ kN/m$^2$, $FS_{overturning} = 5.25$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is Rankine's theory with a sloping backfill used here? A. The back is vertical and no wall friction is given B. Coulomb cannot handle slopes C. Wall is trapezoidal D. Soil is cohesive Answer: A. Without wall friction and with a vertical back, Rankine's sloping-backfill Ka = 0.4142 applies, with thrust parallel to the slope. (Key point: Why Rankine is used) 2. Pa = 66.27 kN/m acts parallel to the 20 degree slope. What are Ph and Pv? A. Ph = Pa sin 20, Pv = Pa cos 20 B. Ph = Pa cos 20 = 62.28, Pv = Pa sin 20 = 22.67 C. Both equal Pa D. Ph = Pa, Pv = 0 Answer: B. Resolving along the 20 degree inclination gives 62.28 and 22.67 kN/m. (Key point: Components of thrust) 3. Which moments are used for FS against overturning? A. Weights and Pv about the toe over Ph x H/3 B. Weights only about the heel C. Friction over Ph D. Pv only Answer: A. FSot = 436.00/83.03 = 5.25; Pv at the heel end adds to resisting moment. (Key point: Overturning check) 4. Since e = 0.144 m < B/6, what does this tell us about the base pressure? A. Tension develops at the heel B. Entire base in compression, with q_min > 0 C. The wall overturns D. q_max equals q_min Answer: B. The resultant is within the middle third, so q_min (50.9) is positive. (Key point: Base pressure with small eccentricity) ## Close variants set in other exams - [Trapezoidal masonry wall, sliding and overturning](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Atrapezoidal-wall-coulomb-sliding) — Foundation Engineering, chapter 3, asked 2× ## Related in Foundation Engineering - [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 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 ## 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×