Unconfined Compressive Strength Calculator
Unconfined compressive strength and undrained shear strength of a cohesive-soil sample from the peak load and corrected area.
Result appears here
Fill in the fields and press Calculate.
Formula, calculation, assumptions and sources
Axial strain, εa = ΔL ÷ L₀
Corrected area, Ac = A₀ ÷ (1 − εa) where A₀ = π(D₀/2)²
qu = P ÷ Ac (converted to kPa)
su = qu ÷ 2 (φ = 0 assumption for saturated cohesive soil)
Run the calculation to see each step with your own numbers.
A standard 1.5 in × 3 in specimen, 250 N peak load, 0.5 cm deformation at failure.
A₀ = π(1.905)² = 11.40 cm²; strain = 0.5 ÷ 7.62 = 0.0656
Ac = 11.40 ÷ 0.9344 = 12.20 cm²
qu = 250 N ÷ 0.00122 m² ≈ 204.9 kPa; su ≈ 102.5 kPa
- A right-cylindrical specimen loaded axially at a constant strain rate to failure, per ASTM D2166 — a specimen that fails along a distinct shear plane rather than bulging uniformly may not follow the area correction exactly.
- The su = qu ÷ 2 relationship assumes a saturated cohesive soil tested quickly enough to be undrained (φ ≈ 0) — it does not apply to unsaturated, granular or heavily fissured/cemented soils.
- This is a strength-index test, not a substitute for a triaxial (CU/CIU) test where effective-stress parameters or a confining pressure matter.
- Why correct the area at all?
- As the specimen shortens under load it also bulges slightly, so its true cross-section at failure is larger than the original A₀. Using A₀ instead of the corrected Ac would understate the stress the specimen actually carried — ASTM D2166 requires the correction.
- Why divide qu by exactly 2 for shear strength?
- In an unconfined compression test the confining (lateral) stress is zero, so the Mohr circle at failure has its centre at qu/2 with radius qu/2. For a saturated clay tested quickly (undrained, φ≈0), that radius IS the undrained shear strength — hence su = qu/2.
- Can I use this for sand or unsaturated soil?
- No — the test and the su = qu/2 relationship assume a saturated, cohesive (clay/silt) soil with negligible internal friction under undrained loading. Granular or unsaturated soils need a different test (e.g. triaxial or direct shear) to get a meaningful shear-strength parameter.
Test method, area correction & formulas
ASTM D2166/D2166M, 'Standard Test Method for Unconfined Compressive Strength of Cohesive Soil'.
About the Unconfined Compressive Strength Calculator
The Unconfined Compressive Strength Calculator applies ASTM D2166: from a cylindrical soil specimen's initial dimensions, its peak axial load, and its axial deformation at failure, it computes the area-corrected unconfined compressive strength (qu) and, using the standard φ = 0 assumption for saturated cohesive soil, the undrained shear strength (su = qu ÷ 2). The area correction accounts for the specimen bulging slightly as it's compressed.
What you enter
- Initial specimen diameter, D₀ — A standard 1.5 in (3.81 cm) Shelby-tube specimen.
- Initial specimen length, L₀ — Standard height-to-diameter ratio is 2:1 — 3 in (7.62 cm) for a 1.5 in specimen.
- Peak axial load, P
- Axial deformation at failure, ΔL
What you get back
- Unconfined compressive strength, qu — kPa
- Undrained shear strength, su = qu ÷ 2 — kPa
The formula, a worked example, the assumptions behind it and the sources are shown in full below the calculator.
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