Site Investigation — Worked Examples

All records are synthetic teaching cases. Keep observations and their limits separate from engineering decisions; the values do not specify a real investigation or design.

Separate TCR, SCR and RQD

A 1000 mm run yields sound full-diameter pieces of 120, 80, 250 and 100 mm, an unsound full-diameter piece of 150 mm, and fragments with an equivalent recovered length of 100 mm. Calculate the three indices using the lesson's logging convention.

Core Numerators and Common Denominator

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Threshold Inclusion and Unrounded Classification

A 150 cm run has sound full-diameter pieces of 12, 8, 25, 5, 14, 9, 32, 11, 4 and 18 cm. Validate recovery, calculate RQD and classify the unrounded value. Would a separate sound piece exactly 10 cm long qualify?

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Overburden Normalization Is a Separate Operation

A complete SPT record has already been corrected to N60=24.0N_{60}=24.0. An applicable interpretation supplies CN=1.25C_N=1.25 for this layer. Find (N1)60(N_1)_{60} and state what the number establishes.

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A Documented Mechanical Break Changes the Numerator

A 100 cm run yields sound full-diameter lengths of 5, 6, 24, 9 and 40 cm. The driller and logger document that the 5 and 6 cm pieces broke during handling, and the pieces fit as one original piece. Other breaks are natural. Determine RQD with the documented rejoining and the error if that pair is left separate.

Mechanical Rejoining and the 10 cm Cutoff

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Reject an Impossible Recovery Account

A log lists a 120 cm drilled run, 56 cm of qualifying pieces and another 90 cm of nonqualifying recovered material, without overlap. Should the logger publish an RQD of 56/120×100%=46.7%56/120\times100\%=46.7\%?

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Combine Unequal Core Runs by Drilled Length

Two valid runs have lengths of 100 and 200 cm and RQD values of 80.0% and 50.0%, respectively. Calculate their combined RQD.

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From a Complete SPT Record to N60

Three complete 150 mm increments require 6, 12 and 18 blows. Measured ER=75%ER=75\% and the adopted factors are CB=1.00C_B=1.00, CR=0.85C_R=0.85 and CS=1.00C_S=1.00. Calculate field NN and N60N_{60}.

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Preserve Partial SPT Penetration

A log records 8 blows for 150 mm seating, 12 blows for the next 150 mm, then 50 blows for only 50 mm before termination under the governing procedure. What should be retained, and can ordinary field NN or N60N_{60} be reported from this record?

Incomplete SPT Record

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Recover the Energy Ratio from a Correction

A complete test has field N=24N=24, corrected N60=27.0N_{60}=27.0, and supplied CB=CS=1.00C_B=C_S=1.00, CR=0.90C_R=0.90. What energy ratio is implied by this calculation, and does the result replace energy measurement?

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Distinguish Percentage Points from Relative Error

A 300 cm run yields 180 cm recovery, including 150 cm of qualifying sound pieces. An analyst incorrectly divides the qualifying length by recovered length. Find the true RQD, the incorrect value, their percentage-point difference and the relative overestimate.

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Match Sampling and Drilling to the Required Evidence

A preliminary profile suggests saturated soft clay below cobbly overburden. The project needs consolidation information from the clay and confirmation of the deeper material. Select an investigation response without assuming perfect samples or a particular proprietary drilling system.

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Resolve a Ground-Model Conflict Before Design

Two boreholes 20 m apart recover limestone and breccia, respectively. An intervening geophysical anomaly has no direct confirmation. Historical records also suggest nearby old workings. Does this prove an active karst cavity or establish that a proposed public park is safe?

Observed Points and an Unresolved Anomaly

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