---
title: "Inclined specimen, total heads, gradient"
subject: "Soil Mechanics"
chapter: 6
key: "semester_4/soil_mechanics:inclined-specimen-total-head-gradient"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/inclined-specimen-total-head-gradient
site: BCT Engineering
---

# Inclined specimen, total heads, gradient

Soil Mechanics · Chapter 6: Principle of Effective Stress, Capillarity and Permeability

- Asked in: 2073 Magh, 4+1 marks
- Video: [Calculate Pressure, Elevation and Total Head in a Soil Sample For a Steady Seepage Condition|Problem](https://www.youtube.com/watch?v=QhR0lYjwJdU) — Master Civil Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/inclined-specimen-total-head-gradient
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-6

## Question

In the figure, water flows from point (1) to point (3) via the soil specimen which is inclined at an angle $\theta$. Piezometers inserted at points 1, 2 and 3 show piezometric heights $h_1$, $h_2$ and $h_3$ respectively. In the figure, $z_1$, $z_2$ and $z_3$ represent the distance of points 1, 2 and 3 from the datum level. (i) Find the total heads at points 1, 2 and 3 from the datum level. (ii) Find the hydraulic gradient for this case when water enters the specimen from point (1) and exits from point (3). [Figure: soil specimen of length L inclined at angle $\theta$ between two water tanks, with piezometers at points 1, 2 and 3.]

## Short answer

Total heads: H₁ = z₁ + h₁, H₂ = z₂ + h₂, H₃ = z₃ + h₃. Hydraulic gradient from point 1 to 3 is i = (H₁ - H₃)/L, where L is the soil specimen length measured along the incline.

## Answer

Total head at a point = elevation head + pressure head, both measured from the datum: $H = z + h_p$. The piezometer reading $h$ at each point is the pressure head.

#### (i) Total heads

$$
H_1 = z_1 + h_1, \qquad H_2 = z_2 + h_2, \qquad H_3 = z_3 + h_3
$$

The total head falls steadily along the specimen from point 1 to 3 because of friction loss in the soil; $H_2$ lies between $H_1$ and $H_3$.

#### (ii) Hydraulic gradient (entry 1 to exit 3)

Head loss between 1 and 3:

$$
\Delta H = H_1 - H_3 = (z_1 + h_1) - (z_3 + h_3)
$$

The flow path through the soil is the sample length $L$ measured along the incline, so

$$
i = \frac{\Delta H}{L} = \frac{(z_1 + h_1) - (z_3 + h_3)}{L}
$$

The inclination $\theta$ does not enter explicitly if $L$ is the length along the sample, because the elevation differences are already included in the $z$ terms. (If only the horizontal distance $L_h$ is known, $L = L_h/\cos\theta$.)

## Check yourself

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

1. Total head at point 1 is:
   A. h1
   B. z1
   C. z1 + h1
   D. z1 - h1
   Answer: C. Total head is elevation head plus pressure head: H = z + h_p. (Key point: Total head)

2. The head loss between entry 1 and exit 3 is:
   A. (z1 + h1) - (z3 + h3)
   B. h1 - h3 only
   C. z1 - z3 only
   D. (z1 + h3) - (z3 + h1)
   Answer: A. Delta H = H1 - H3 using total heads. (Key point: Head loss)

3. The hydraulic gradient is:
   A. Delta H divided by horizontal projection always
   B. Delta H divided by the sample length L along the incline
   C. Delta H times L
   D. Delta H divided by sin(theta)
   Answer: B. i = Delta H / L with L the flow path length along the specimen; theta does not enter separately. (Key point: Gradient)

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