GEO-11

Skempton undrained bearing

Net undrained capacity qu = cu Nc with Skempton's Nc(D/B, B/L).

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Bearing capacitySkempton 1951

Governing equation

N_c=5\bigl(1+0.2\tfrac{B}{L}\bigr)\bigl(1+0.2\tfrac{D}{B}\bigr)\le 7.5,\quad q_u=c_u N_c

$D \in \left[0, 10\right]$

where

c_u: Undrained strength (kPa)
B: Width (m)
L: Length (m)
D: Embedment (m)
N_c: Skempton Nc (—)
q_u: Net undrained bearing (kPa)

The figure, part by part

Each numbered marker wears the color of its card. Hover a card to light the marker. Tap a marker to jump to that card. The slider walks a red point along the curve; Animate drawing traces it like a pen.

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①

Governing law — red

N_c=5\bigl(1+0.2\tfrac{B}{L}\bigr)\bigl(1+0.2\tfrac{D}{B}\bigr)\le 7.5,\quad q_u=c_u N_c

$D \in \left[0, 10\right]$

$D = 0$ $N_c = 220$

Along the curve

②

Domain of the variable — blue

D_{\min} = 0,\quad D_{\max} = 10

$D \in \left[0, 10\right]$

$D = 0$ $D = 0$

Along the curve

③

Response curve — green

N_c=5\bigl(1+0.2\tfrac{B}{L}\bigr)\bigl(1+0.2\tfrac{D}{B}\bigr)\le 7.5,\quad q_u=c_u N_c

$D \in \left[0, 10\right]$

$D = 0$ $N_c = 220$

Along the curve

④

Rate of change — amber

\frac{\mathrm{d}N_c}{\mathrm{d}D}

$D \in \left[0, 10\right]$

$D = 0.063$ $N_c^{\prime} = 22$

Along the curve

Lecture brief

Historical brief

Soil mechanics became a quantitative laboratory after Karl von Terzaghi’s 1925–1943 work on effective stress, consolidation and bearing. These sheets still size shallow foundations, retaining walls, piles and drainage in SI. This sheet (GEO-11 — Skempton undrained bearing) is the form associated with Skempton 1951. Working symbols: $c_u$ , $B$ , $L$ , $D$ $\rightarrow$ $N_c$ , $q_u$ . For φu = 0, Prandtl's Nc = 5.14 (strip). Skempton fitted Nc = 5(1+0.2B/L)(1+0.2D/B) ≤ 7.5 for rectangles.

Purpose

Purpose: compute

N_c

q_u

from

c_u

B

L

D

in Geotechnical engineering via

N_c=5\bigl(1+0.2\tfrac{B}{L}\bigr)\bigl(1+0.2\tfrac{D}{B}\bigr)\le 7.5,\quad q_u=c_u N_c

Net undrained capacity qu = cu Nc with Skempton's Nc(D/B, B/L). Use it when a real geotechnical engineering question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given

c_u = 40.000\,\mathrm{kPa}

B = 2.000\,\mathrm{m}

L = 4.000\,\mathrm{m}

D = 1.500\,\mathrm{m}

, the governing relation

N_c=5\bigl(1+0.2\tfrac{B}{L}\bigr)\bigl(1+0.2\tfrac{D}{B}\bigr)\le 7.5,\quad q_u=c_u N_c

yields

N_c = 6.325\,\mathrm{—}

q_u = 253.000\,\mathrm{kPa}

. Saturated clay, short-term undrained loading. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

Undrained strength c_u40 kPaWidth B2 mLength L4 mEmbedment D1.5 m

Outputs

Skempton Nc N_c6.325 —
Net undrained bearing q_u253.000 kPa

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Narration of this film

Saturated clay, short-term undrained loading.

For φu = 0, Prandtl's Nc = 5.14 (strip). Skempton fitted Nc = 5(1+0.2B/L)(1+0.2D/B) ≤ 7.5 for rectangles.

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