--- title: "Principal stresses at toe and heel with silt and earthquake" subject: "Hydropower Engineering" chapter: 4 key: "semester_7/hydropower_engineering:gravity-dam-principal-stress-earthquake-silt" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/gravity-dam-principal-stress-earthquake-silt site: BCT Engineering --- # Principal stresses at toe and heel with silt and earthquake Hydropower Engineering · Chapter 4: Headworks of Storage Plants - Asked in: 2076 Chaitra, 10 marks - Video: [DHS Module 4 [Part 06] - Stability Analysis of Gravity Dams - Problem](https://www.youtube.com/watch?v=NV6EbVvtpv8) — AM CIVIL ACADEMY - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/gravity-dam-principal-stress-earthquake-silt - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-4 ## Question Determine the principal stresses at the toe and heel of the dam shown in the figure for the reservoir full condition. Consider the following forces: (i) self weight ($w_c = 25$ kN/m³), (ii) water pressure ($w = 10$ kN/m³), (iii) uplift pressure, (iv) silt pressure with the depth of silt as 20 m, (v) earthquake forces, $\alpha_h = 0.1$. [Figure: dam section with dimensions 6 m, 10 m (crest width), water depth 94 m, silt depth 20 m, total height 110 m, base 100 m, downstream slope 0.8 : 1, gallery position marked with 4, 8, 8 m near the heel and 80 m.] ## Short answer At the toe, vertical stress is 1377 kN/m², major principal stress 2300 kN/m², shear stress 1127 kN/m². At the heel, vertical stress is 1259 kN/m²; the face normal stress (water, silt, earthquake) is 1076 kN/m², so major principal stress equals the vertical stress. ## Answer #### Assumptions (from the figure) - Crest width 10 m, height 110 m, base width 100 m, u/s face vertical, d/s face straight from the crest edge to the toe (run 90 m, $\tan\phi_d=90/110=0.818\approx0.8:1$). Water depth $H_w=94$ m, silt depth $h_s=20$ m. No tail water. - $\gamma_c=25$, $\gamma_w=10$ kN/m³ (given). Uplift: $\gamma_wH_w=940$ kN/m² at the heel, one-third of it at the gallery (8 m from the heel), zero at the toe. - Silt: submerged unit weight $10$ kN/m³, $\phi=30^\circ$ so $K_a=\frac13$: $P_s=\frac12K_a\gamma_sh_s^2$ acting at $h_s/3$. - Earthquake: $\alpha_h=0.1$ acting towards the downstream (worst case). Inertia force $=\alpha_hW$ at the centre of gravity of the dam. Hydrodynamic pressure by Zangar: $p_e=C_m\alpha_h\gamma_wH_w$ with $C_m=0.735$, resultant $P_e=0.726\,p_eH_w$ at $0.412H_w$ above the base. #### Calculations Dam weights: $W_1=25\times10\times110=27{,}500$ kN, $W_2=25\times\frac12\times90\times110=123{,}750$ kN, total $W=151,250$ kN/m. Centre of gravity height $\bar z=40.00$ m. $P_s=\frac12\times\frac13\times10\times20^2=666.7$ kN/m; $p_e=0.735\times0.1\times10\times94=69.1$ kN/m²; $P_e=0.726\times69.1\times94=4,715$ kN/m; $F_i=0.1\times151,250=15,125$ kN/m. | Force | Magnitude (kN/m) | Lever arm / height (m) | Moment about toe (kN·m/m) | |---|---|---|---| | Rectangle (W₁) | 27,500 | 95.00 | 2,612,500 | | Triangle (W₂) | 123,750 | 60.00 | 7,425,000 | | **ΣW** | **151,250** | | **ΣM_R = 10,037,500** | | Water thrust | 44,180 | 31.33 | 1,384,307 | | Uplift U (upward) | 19,427 | — | 1,368,640 | | Silt pressure | 667 | 6.67 | 4,444 | | Earthquake: inertia of dam (0.1W) | 15,125 | 40.00 | 605,000 | | Earthquake: hydrodynamic pressure | 4,715 | 38.73 | 182,602 | | **Total** | ΣH = 64,687 | | **ΣM_O = 3,544,993** | $$ \Sigma V = W-U = 151,250-19,427 = 131,823\ \text{kN/m} $$ $$ \bar x=\frac{\Sigma M_R-\Sigma M_O}{\Sigma V}=\frac{10,037,500-3,544,993}{131,823}=49.25\ \text{m from the toe},\quad e=50-49.25=0.75\ \text{m} $$ ($B/6=16.67$ m: the resultant is inside the middle third.) $$ \sigma_{toe}=\frac{131,823}{100}\left(1+\frac{6\times0.75}{100}\right)=1,377.4\ \text{kN/m}^2,\quad \sigma_{heel}=1,318.2\left(1-0.045\right)=1,259.0\ \text{kN/m}^2 $$ #### Principal stresses **Toe** (no tail water, $p'=0$): $$ \sigma_1=\sigma_{toe}(1+\tan^2\phi_d)=1,377.4\,(1+0.669)=2,299.5\ \text{kN/m}^2,\quad \sigma_3=0 $$ Shear stress $\tau_{toe}=\sigma_{toe}\tan\phi_d=1,127.0$ kN/m². **Heel** (u/s face vertical, so no shear on the face and the face is a principal plane). The normal stress on the face is the total water, hydrodynamic and silt pressure $p=940+69.1+66.7=1075.8$ kN/m². - Vertical principal stress: $\sigma_z=1,259.0$ kN/m² - Horizontal principal stress: $\sigma_h=p=1,075.8$ kN/m² - Major principal stress at the heel $=1,259.0$ kN/m², minor $=1,075.8$ kN/m². **Answer: toe — σ_z = 1,377 kN/m², major principal stress = 2,300 kN/m², τ = 1,127 kN/m²; heel — σ_z = 1,259 kN/m² (face normal stress 1,076 kN/m²).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In which direction is the horizontal earthquake acceleration taken and why? A. Towards the upstream, because it reduces the water thrust B. Towards the downstream, the worst case, since it adds to the water thrust C. Vertically downward only D. Either way; it makes no difference to the moments Answer: B. Downstream acceleration adds to the hydrostatic thrust and hydrodynamic pressure, so it is the critical (worst) case. (Key point: Earthquake direction) 2. How is the silt pressure calculated? A. P_s = gamma_w x h_s^2 / 2 with gamma_w = 10 B. P_s = (1/2) x K_a x gamma_s x h_s^2 with K_a = 1/3 (phi = 30 deg), acting at h_s/3 C. P_s = K_a x gamma_s x h_s, acting at h_s/2 D. P_s = 0.726 x p_e x h_s Answer: B. Silt acts as an active-pressure fluid: P_s = 1/2 K_a gamma_s h_s^2 = 666.7 kN/m, at one-third of the silt depth above the base. (Key point: Silt pressure) 3. With no tail water, how is the major principal stress at the toe related to the vertical stress? A. sigma_1 = sigma_toe x (1 + tan^2(phi_d)), with sigma_3 = 0 B. sigma_1 = sigma_toe x tan(phi_d) C. sigma_1 = sigma_toe, since the face is unloaded D. sigma_1 = sigma_toe / (1 + tan^2(phi_d)) Answer: A. The free d/s face carries no pressure (p' = 0), so it is a principal plane with zero stress and sigma_1 = sigma_toe sec^2(phi_d). (Key point: Principal stress at the toe) 4. Why is the heel treated as having no shear on the u/s face? A. Because the heel has zero stress B. Because the u/s face is vertical, so it is a principal plane; the face normal stress equals the water/silt/hydrodynamic pressure C. Because earthquake forces cancel at the heel D. Because uplift is largest at the heel Answer: B. On a vertical u/s face the water pressure acts normal to the face with no shear, so the vertical and horizontal stresses are the principal stresses. (Key point: Heel stresses with vertical face) ## Related in Hydropower Engineering - [Derive principal and shear stress equations at toe/heel](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Agravity-dam-principal-stress-derivation-tailwater-eq) — Hydropower Engineering, chapter 4 - [Gravity dam stresses with drainage gallery uplift](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Agravity-dam-stresses-2080-baishakh) — Hydropower Engineering, chapter 4 - [Gravity dam forces and stability checks (2081 Baishakh)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Agravity-dam-forces-stresses-2081-baishakh) — Hydropower Engineering, chapter 4 - [Gravity dam stresses, principal stress, FOS (2079)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Agravity-dam-stresses-2079-bhadra) — Hydropower Engineering, chapter 4 - [Gravity dam section: shear stress near toe (2082)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Agravity-dam-stability-2082-baishakh) — Hydropower Engineering, chapter 4 - [Uplift pressure diagram with drainage gallery](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Auplift-pressure-diagram-gravity-dam) — Hydropower Engineering, chapter 4 ## Related in other subjects - [Stresses in gravity dam with drain holes](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Agravity-dam-stresses-figure) — Hydro Power, chapter 4 - [Sliding and overturning check of gravity dam](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Agravity-dam-sliding-overturning-figure) — Hydro Power, chapter 4 - [Stability of gravity dam (crest 1530 m)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Agravity-dam-stability-1530) — Hydro Power, chapter 4 - [Stability checks of 50 m trapezoidal dam](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Atrapezoidal-dam-50m-checks) — Hydro Power, chapter 4