---
title: "Derive principal and shear stress equations at toe/heel"
subject: "Hydropower Engineering"
chapter: 4
key: "semester_7/hydropower_engineering:gravity-dam-principal-stress-derivation-tailwater-eq"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/gravity-dam-principal-stress-derivation-tailwater-eq
site: BCT Engineering
---

# Derive principal and shear stress equations at toe/heel

Hydropower Engineering · Chapter 4: Headworks of Storage Plants

- Asked in: 2075 Chaitra, 8 marks
- Video: [002_DHS- Principal and Shear Stress in a Dam](https://www.youtube.com/watch?v=ueimyVYj_mo) — Athul Mangalath Madhu
- 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-derivation-tailwater-eq
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-4

## Question

Derive the equations for principal stress and shear stress at the toe and heel of a gravity dam with tail water present and also considering hydrodynamic pressure produced by an earthquake.

## Short answer

Toe: shear τ = (σ_z − p′′) tan φ_d, principal stress σ₁ = σ_z sec² φ_d − p′′ tan² φ_d, with p′′= tail water pressure minus earthquake pressure. Heel: τ = (p′′ − σ_z) tan φ_u, σ₁ = σ_z sec² φ_u − p′′ tan² φ_u. For vertical upstream face τ=0.

## Answer

Consider a gravity dam section with a u/s face making an angle $\phi_u$ and a d/s face making an angle $\phi_d$ with the vertical. Let $\sigma_z$ be the vertical normal stress on a horizontal plane and $\tau$ the shear stress on that plane. Compression is taken as positive. Per unit length of dam.

#### 1. Toe (with tail water and earthquake)
At the toe the d/s face is loaded by the tail water pressure $p'=\gamma_wh_t$ and, in an earthquake, by the hydrodynamic pressure $p_e'$ of the tail water. In the critical case (the dam accelerating towards the upstream, so that the reservoir pressure increases), the dam moves away from the tail water and the pressure on the d/s face falls. The face then carries only a normal stress $p''=p'-p_e'$ and no shear stress (water cannot carry shear). So the face is a principal plane with $\sigma_3=p''$.

The unit normal to the d/s face makes an angle $\phi_d$ with the horizontal: $n=(\cos\phi_d,\ \sin\phi_d)$. The traction on the face must be purely normal, $\sigma\cdot n=p''n$. With the stress components $\sigma_z$ (vertical), $\sigma_x$ (horizontal) and $\tau$:

$$
\tau\cos\phi_d+\sigma_z\sin\phi_d=p''\sin\phi_d\ \Rightarrow\ \tau=-(\sigma_z-p'')\tan\phi_d
$$

$$
\sigma_x\cos\phi_d+\tau\sin\phi_d=p''\cos\phi_d\ \Rightarrow\ \sigma_x=p''+(\sigma_z-p'')\tan^2\phi_d
$$

The minus sign only shows that the shear on the horizontal plane acts in the opposite sense to the one assumed. Taking the shear at the toe as positive (the usual textbook convention):

**Shear stress at the toe:**

$$
\boxed{\tau_{toe}=(\sigma_z-p'')\tan\phi_d}
$$

The sum of normal stresses is invariant: $\sigma_1+\sigma_3=\sigma_x+\sigma_z$, with $\sigma_3=p''$:

$$
\sigma_1=\sigma_x+\sigma_z-p''=\sigma_z(1+\tan^2\phi_d)-p''\tan^2\phi_d
$$

**Major principal stress at the toe:**

$$
\boxed{\sigma_1=\sigma_z\sec^2\phi_d-p''\tan^2\phi_d},\qquad p''=p'-p_e'
$$

(When the tail water is absent, $p''=0$ and $\sigma_1=\sigma_z\sec^2\phi_d$, $\tau=\sigma_z\tan\phi_d$.)

#### 2. Heel (reservoir pressure and earthquake)
On the u/s face the normal pressure is $p''=p+p_e$, where $p=\gamma_wh$ is the hydrostatic pressure at the heel and $p_e$ the hydrodynamic pressure intensity (Westergaard/Zangar, $p_e=C\alpha_h\gamma_wh$). In the same way, with the face normal pointing upstream at angle $\phi_u$ to the horizontal, the shear has the opposite sense to that at the toe:

$$
\tau_{heel}=-(\sigma_z-p'')\tan\phi_u=(p''-\sigma_{z})\tan\phi_u,\qquad \sigma_1=\sigma_z\sec^2\phi_u-p''\tan^2\phi_u
$$

For a vertical u/s face $\phi_u=0$: $\tau=0$ and the principal stresses are $\sigma_z$ and $p''$.

#### 3. Vertical stress $\sigma_z$
$\sigma_z$ comes from the combined vertical load and moment, including the tail water weight and the earthquake effect:

$$
\sigma_z=\frac{\Sigma V}{B}\left(1\pm\frac{6e}{B}\right),\qquad e=\frac{B}{2}-\frac{\Sigma M_R-\Sigma M_O}{\Sigma V}
$$

where $\Sigma M_O$ includes the moment of the hydrodynamic thrust $P_e$ (acting at about $0.4h$ above the base) and the inertia force $\alpha_hW$ at the centre of gravity, and $\Sigma V$ includes the weight of the tail water on the d/s slope. The plus sign gives the toe stress and the minus sign the heel stress for the reservoir-full case.

#### 4. Conditions
- The shear stress and principal stress are compared with the permissible values: $\sigma_1\le\sigma_{perm}$ and $\tau\le\tau_{perm}$.
- At the toe, tail water reduces both $\tau$ and $\sigma_1$ (because $p''>0$), and the earthquake reduces this relief ($p''=p'-p_e'$). At the heel the hydrodynamic pressure increases $p''$, which reduces $\sigma_1$ and, when $p''>\sigma_z$, increases $\tau$.

## Check yourself

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

1. Why is the d/s face (at the toe) treated as a principal plane?
   A. Because the dam is symmetrical
   B. Because the face carries only the water pressure p'' normal to it and no shear stress, as water cannot carry shear
   C. Because the vertical stress is zero there
   D. Because earthquake forces act only horizontally
   Answer: B. The face is loaded only by normal water pressure, so shear is zero and sigma3 = p''. (Key point: Principal plane at the face)

2. The shear stress at the toe with tail water and earthquake is
   A. tau = sigma_z tan(phi_d)
   B. tau = (sigma_z + p'') tan(phi_d)
   C. tau = (sigma_z - p'') tan(phi_d)
   D. tau = (sigma_z - p'') sec(phi_d)
   Answer: C. Resolving the traction along the face gives tau = (sigma_z - p'') tan(phi_d); it reduces to sigma_z tan(phi_d) when p'' = 0. (Key point: Shear stress at toe)

3. Using the invariance of the sum of normal stresses, the major principal stress at the toe is
   A. sigma1 = sigma_z sec^2(phi_d) - p'' tan^2(phi_d)
   B. sigma1 = sigma_z sec^2(phi_d) + p'' tan^2(phi_d)
   C. sigma1 = sigma_z tan^2(phi_d) - p''
   D. sigma1 = sigma_z cos^2(phi_d) - p''
   Answer: A. sigma1 = sigma_x + sigma_z - p'' gives sigma_z(1+tan^2 phi) - p'' tan^2 phi = sigma_z sec^2 - p'' tan^2. (Key point: Major principal stress)

4. What is p'' at the toe, and what does it include?
   A. Only reservoir pressure at the heel
   B. p'' = tail water pressure + hydrodynamic pressure of tail water due to the earthquake
   C. Only the uplift pressure
   D. The vertical stress sigma_z
   Answer: B. p'' = gamma_w h_t + p_e' is the normal pressure on the d/s face. (Key point: Face pressure p'')

## Related in Hydropower Engineering

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