---
title: "Foundation stresses from moments and vertical force"
subject: "Hydropower Engineering"
chapter: 4
key: "semester_7/hydropower_engineering:gravity-dam-vertical-stress-moments"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/gravity-dam-vertical-stress-moments
site: BCT Engineering
---

# Foundation stresses from moments and vertical force

Hydropower Engineering · Chapter 4: Headworks of Storage Plants

- Asked in: 2076 Chaitra, 2+2+2 marks
- Video: [Analysis of Gravity Dams Sample Problem](https://www.youtube.com/watch?v=xyxUo_aGfNQ) — Jose Antonio Rodavia
- 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-vertical-stress-moments
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-4

## Question

Determine the maximum and minimum vertical stresses to which the foundation of the dam will be subjected from the following data: total overturning moment about toe ($\Sigma M_o$) = $1.2 \times 10^6$ kN-m; total resisting moment about toe ($\Sigma M_R$) = $2.5 \times 10^6$ kN-m; total vertical force above the base ($\Sigma V$) = $6 \times 10^4$ kN; base width of dam = 55 m; slope of d/s face = 0.8 : 1. Also calculate the maximum principal stress at the toe. Neglect tail water depth.

## Short answer

From given moments and vertical force, resultant eccentricity is 5.83 m (< B/6). Maximum vertical stress at toe is 1785 kN/m², minimum at heel is 397 kN/m². Maximum principal stress at the toe is 2928 kN/m² (no tail water).

## Answer

#### Position of the resultant
Net moment about the toe divided by the vertical load gives the distance of the resultant from the toe:

$$
\bar x=\frac{\Sigma M_R-\Sigma M_O}{\Sigma V}=\frac{2.5\times10^6-1.2\times10^6}{6\times10^4}=21.667\ \text{m}
$$

Eccentricity from the centre of the base:

$$
e=\frac{B}{2}-\bar x=27.5-21.667=5.833\ \text{m}\quad (<B/6=9.167\ \text{m},\ \text{so no tension})
$$

#### Maximum and minimum vertical stresses

$$
\sigma_{max}=\frac{\Sigma V}{B}\left(1+\frac{6e}{B}\right)=\frac{6\times10^4}{55}\left(1+\frac{6\times5.833}{55}\right)=1090.91\times1.6364=1,785.1\ \text{kN/m}^2\ (\text{at the toe})
$$

$$
\sigma_{min}=1090.91\times\left(1-0.6364\right)= 396.7\ \text{kN/m}^2\ (\text{at the heel})
$$

#### Maximum principal stress at the toe
With no tail water, $p'=0$ and $\tan\phi_d=0.8$:

$$
\sigma_1=\sigma_{max}\sec^2\phi_d=\sigma_{max}(1+\tan^2\phi_d)=1,785.1\times(1+0.8^2)=2,927.6\ \text{kN/m}^2
$$

**Answer: σ_max = 1,785 kN/m² (toe), σ_min = 397 kN/m² (heel), principal stress at the toe σ₁ = 2,928 kN/m².**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the distance of the resultant from the toe found from the given data?
   A. x = (SumM_R + SumM_O) / SumV
   B. x = (SumM_R - SumM_O) / SumV = 21.667 m
   C. x = SumM_R / SumV = 41.7 m
   D. x = SumM_O / SumV = 20 m
   Answer: B. The net moment about the toe divided by the net vertical load gives the point where the resultant cuts the base. (Key point: Locating the resultant)

2. What is the eccentricity and what does it show?
   A. e = x - B/2 = -5.83 m, so the dam overturns
   B. e = B/2 - x = 5.833 m < B/6 = 9.167 m, so there is no tension
   C. e = B/2 - x = 5.833 m > B/6, so tension exists
   D. e = B/6 = 9.167 m exactly
   Answer: B. e is measured from the centre of the base (27.5 m); since it is smaller than B/6, the resultant lies in the middle third. (Key point: Eccentricity and tension check)

3. Which expression gives the maximum vertical stress and where does it occur?
   A. (SumV / B)(1 - 6e/B), at the toe
   B. (SumV / B)(1 + 6e/B), at the toe = 1,785 kN/m^2
   C. (SumV / B)(1 + 6e/B), at the heel
   D. SumV / B only, at mid-base
   Answer: B. The resultant is nearer the toe, so the toe stress is the maximum: 1090.91 x 1.6364 = 1,785 kN/m^2. (Key point: Maximum vertical stress)

4. With no tail water and tan(phi_d) = 0.8, how is the principal stress at the toe obtained?
   A. sigma_1 = sigma_max x 0.8 = 1,428 kN/m^2
   B. sigma_1 = sigma_max x (1 + 0.8^2) = 2,928 kN/m^2
   C. sigma_1 = sigma_max x (1 - 0.8^2)
   D. sigma_1 = sigma_max / 1.64
   Answer: B. With p' = 0, sigma_1 = sigma_max sec^2(phi_d) = sigma_max (1 + tan^2 phi_d) = 1.64 x 1,785. (Key point: Principal stress at the toe)

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