Skip to main content

Inclined specimen, total heads, gradient

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.

In the figure, water flows from point (1) to point (3) via the soil specimen which is inclined at an angle θ\theta. Piezometers inserted at points 1, 2 and 3 show piezometric heights h1h_1, h2h_2 and h3h_3 respectively. In the figure, z1z_1, z2z_2 and z3z_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 θ\theta between two water tanks, with piezometers at points 1, 2 and 3.]

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

(i) Total heads

H1=z1+h1,H2=z2+h2,H3=z3+h3H_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; H2H_2 lies between H1H_1 and H3H_3.

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

Head loss between 1 and 3:

ΔH=H1−H3=(z1+h1)−(z3+h3)\Delta H = H_1 - H_3 = (z_1 + h_1) - (z_3 + h_3)

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

i=ΔHL=(z1+h1)−(z3+h3)Li = \frac{\Delta H}{L} = \frac{(z_1 + h_1) - (z_3 + h_3)}{L}

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

✓ Answer checked by Claude Opus 5.5 (high effort)

All Soil Mechanics chapter 6 questions →