--- title: "Tension crack and active pressure, layered clay" subject: "Foundation Engineering" chapter: 3 key: "semester_5/foundation_engineering:wall-cohesive-tension-crack" source: https://www.bctengineer.com/q/semester_5/foundation_engineering/wall-cohesive-tension-crack site: BCT Engineering --- # Tension crack and active pressure, layered clay Foundation Engineering · Chapter 3: Lateral Earth Pressure Theories and Retaining Walls - Asked in: 2076 Asoj, 8 marks - Video: [Tension crack zone | 3rd Civil](https://www.youtube.com/watch?v=ieygJEs7b1E) — Ahmad Maher Adam - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/foundation_engineering/wall-cohesive-tension-crack - Whole chapter: https://www.bctengineer.com/md/semester_5/foundation_engineering/chapter-3 ## Question A 6 m high vertical wall supports a saturated cohesive backfill with horizontal surface. The top 3 m of backfill weighs $18\ \text{kN/m}^3$ and has cohesion of $18\ \text{kN/m}^2$. The bulk unit weight and cohesion of the bottom 3 m of the wall are $20\ \text{kN/m}^3$ and $25\ \text{kN/m}^2$ respectively. What is the likely depth of tension crack? If the tension crack develops, what will be the active earth pressure? Draw the pressure distribution diagram and determine the point of application of the resultant pressure. ## Short answer Tension crack depth z0 = 2c1/γ1 = 2.0 m. Ignoring tension, active thrust is 111 kN/m horizontally, acting 1.24 m above the base. ## Answer **Data:** $H = 6$ m; top 3 m: $\gamma_1 = 18$, $c_1 = 18$; bottom 3 m: $\gamma_2 = 20$, $c_2 = 25$. Saturated clay is taken as $\phi_u = 0$, so $K_a = 1$ and $p_a = \gamma z - 2c$. **Depth of tension crack** (where $p_a = 0$ in the top layer): $$ z_0 = \frac{2c_1}{\gamma_1} = \frac{2\times18}{18} = 2.0\ \text{m} $$ **Pressures (kN/m$^2$)** - $z=0$: $-36$ (tension, ignored); $z = 2$ m: $0$. - $z = 3$ m (top layer): $18\times3 - 36 = 18.0$. - $z = 3$ m (bottom layer): $54 - 2\times25 = 4.0$. - $z = 6$ m: $54 + 20\times3 - 50 = 64.0$. ```text 0 | | crack zone (2 m) 2 |\ | \ 3 |--18 / 4 | \ | \ 6 |--------64 ``` **Forces (per metre) and arm above the base** | Part | Force (kN/m) | Arm (m) | Moment | |---|---|---|---| | Top layer triangle: 0.5 x 1 x 18 | 9.00 | 3.333 | 30.00 | | Bottom layer trapezoid: 3 x (4+64)/2 | 102.00 | 1.059 | 108.0 | | **Total** | **111.00** | | **138.0** | The trapezoid has parallel sides 4 (top) and 64 (at the base) and height 3 m, so its centroid is measured from the base side: $$ \bar y = \frac{3\,(2\times4+64)}{3\,(4+64)} = 1.059\ \text{m} $$ $$ \bar y = 138.0/111.0 = 1.24\ \text{m above the base} $$ **Answer:** tension crack depth $= 2.0$ m; active thrust $P_a = 111$ kN/m (horizontal), acting 1.24 m above the base of the wall. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the depth of the tension crack found for the saturated clay backfill? A. z0 = c/gamma B. z0 = 2c/gamma, where the active pressure is zero C. z0 = gamma/2c D. z0 = H/3 Answer: B. Setting pa = gamma z - 2c = 0 gives z0 = 2c/gamma = 2 m. (Key point: Tension crack depth) 2. Why is Ka = 1 used for this saturated clay? A. Because the wall is smooth B. Because cohesion is zero C. Because phi_u = 0, so tan^2(45 - 0) = 1 D. Because the surface is horizontal only Answer: C. Undrained saturated clay is taken with phi_u = 0, giving Ka = 1 and pa = gamma z - 2c. (Key point: Ka for saturated clay) 3. At 3 m depth the pressure is 18 above the interface and 4 below it. What causes the drop? A. The bottom layer is lighter B. The bottom layer has higher cohesion, so 2c is larger (50 instead of 36) C. A tension crack opens there D. Water pressure acts there Answer: B. The same overburden 54 minus 2c: 54 - 36 = 18 above, 54 - 50 = 4 below. (Key point: Pressure jump at layer interface) 4. In computing the thrust with the crack, what is done with the negative pressure above 2 m? A. It is subtracted from the thrust B. It is ignored because soil cannot take tension C. It is doubled D. It is treated as a surcharge Answer: B. The cracked zone carries no pressure, so only the positive pressure area below 2 m is used (111 kN/m). 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