INGENIA

GEO-35

CPT undrained strength from qc

su = (qc − σv) / Nkt. Nkt ≈ 10–18 in clay.

Reading speed
In situRobertsonNkt

Governing equation

su=(qcσv)/Nkts_u=(q_c-\sigma_v)/N_{kt}

where

q_c
Cone resistance (MPa)
\sigma_v
Total overburden (kPa)
N_{kt}
Cone factor Nkt ()
s_u
Undrained strength (kPa)

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-35 — CPT undrained strength from qc) is the form associated with Robertson · Nkt. Working symbols: qcq_c, σv\sigma_v, NktN_{kt} \rightarrow sus_u. A cone is a deep undrained bearing test. Nkt plays the role of Nc, calibrated on nearby su.

Purpose

Purpose: compute sus_u from qcq_c, σv\sigma_v, NktN_{kt} in Geotechnical engineering via su=(qcσv)/Nkts_u=(q_c-\sigma_v)/N_{kt} su = (qc − σv) / Nkt. Nkt ≈ 10–18 in clay. 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 qc=1.200MPaq_c = 1.200\,\mathrm{MPa}, σv=80.000kPa\sigma_v = 80.000\,\mathrm{kPa}, Nkt=14.000N_{kt} = 14.000\,\mathrm{—}, the governing relation su=(qcσv)/Nkts_u=(q_c-\sigma_v)/N_{kt} yields su=80.0kPas_u = 80.0\,\mathrm{kPa}. A cone, a tip resistance, an su estimate. Move a slider: the numbers are this situation, not a canned story.

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Inputs

Outputs

  • Undrained strength s_u80.0 kPa
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GEO-35 · gauge
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Narration of this film

A cone, a tip resistance, an su estimate.

A cone is a deep undrained bearing test. Nkt plays the role of Nc, calibrated on nearby su.

Reading speed

Watch on YouTube