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Uplift pressure with Bligh's creep theory

Short answer

Bligh's theory assumes seepage follows base contact with creep length L = horizontal + 2×vertical, giving straight hydraulic gradient. Piping safety requires L ≥ C×H. Uplift pressure head at point x is hx = H(1 - Lx/L). Floor thickness for uplift safety is t = (4/3)hx/(G-1) with 33% safety factor.

Write with definition sketch how do you determine the pressure along the foundation of the structure and ensure safety against uplift pressure using Bligh's creep theory.

Bligh's creep theory

Water seeping below the structure follows the contact between the base and the subsoil (the creep path). The length of creep LL is the sum of the horizontal creep (the horizontal lengths of the contact) and the vertical creep (the lengths of the cutoffs, each counted twice because the water goes down and up):

L=Lh+LvL = L_h + L_v

The head loss is assumed to be proportional to the creep length along the path (straight hydraulic gradient line).

   H ~~~~~~~~~|  ____________   ~~~~~~~ (d/s)
   u/s    |   |  |  floor   |  |
   ===+===|===+==+==========+==+====
     pile 1                 pile 2
   point A (start) ... point B (end of creep)
   HGL: straight line from H at A to 0 at B

Safety against piping

L≥C HorHL≤1CL \ge C\,H \quad \text{or} \quad \frac{H}{L} \le \frac{1}{C}

with Bligh's coefficient CC (e.g. 18 for light sand and mud, 15 for fine micaceous sand, 12 for coarse sand, 5-9 for sand mixed with boulders and gravel).

Pressure along the foundation

The head lost up to a point is proportional to the creep length up to it:

Head lost up to point x=H LxL,hx=H−HLxL=H(1−LxL)\text{Head lost up to point } x = H\,\frac{L_x}{L},\qquad h_x = H - H\frac{L_x}{L} = H\left(1 - \frac{L_x}{L}\right)

where hxh_x is the residual (uplift) pressure head at the point and LxL_x the creep length from the upstream end to it. The uplift pressure on the floor is γwhx\gamma_w h_x (in kN/m²).

Safety against uplift

Let tt be the floor thickness and GG the specific gravity of the floor material. The floor is safe when its submerged weight balances the uplift:

γw(G−1) t≥γwhx⇒t=hxG−1\gamma_w (G-1)\,t \ge \gamma_w h_x \Rightarrow t = \frac{h_x}{G-1}

Usually 33% is added for safety, i.e. t=43hxG−1t = \dfrac{4}{3}\dfrac{h_x}{G-1}. If the thickness becomes excessive, lengthen the u/s floor or deepen the u/s pile to reduce hxh_x, or thicken the floor.

✓ Answer checked by Claude Opus 5.5 (high effort)

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