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worked examples

Axial Pile Capacity Example

Worked circular pile example combining average shaft resistance and toe resistance to calculate ultimate and allowable axial capacity.

Problem Statement

Estimate the axial compression capacity of one circular pile using project-specific resistance values:

  • Diameter D=0.60 mD=0.60\ \mathrm{m}.
  • Embedded length L=18 mL=18\ \mathrm{m}.
  • Average ultimate unit shaft resistance fs,avg=45 kPaf_{s,\mathrm{avg}}=45\ \mathrm{kPa}.
  • Ultimate unit toe resistance qb=2,500 kPaq_b=2{,}500\ \mathrm{kPa}.
  • Factor of safety FS=2.5FS=2.5.

The unit resistance values are assumed to have been selected by an appropriate project-specific geotechnical method. The example does not derive them from soil description.

Step 1: Shaft Surface Area

For one circular pile:

As=πDLA_s=\pi DL
As=π(0.60)(18)=33.93 m2A_s=\pi(0.60)(18)=33.93\ \mathrm{m^2}

Step 2: Toe Area

Ab=πD24A_b=\frac{\pi D^2}{4}
Ab=π(0.60)24=0.2827 m2A_b=\frac{\pi(0.60)^2}{4}=0.2827\ \mathrm{m^2}

Because 1 kPa=1 kN/m21\ \mathrm{kPa}=1\ \mathrm{kN/m^2}, multiplying unit resistance by area gives capacity in kilonewtons.

Step 3: Ultimate Shaft Resistance

Qs=fs,avgAs=45(33.93)=1,527 kNQ_s=f_{s,\mathrm{avg}}A_s =45(33.93)=1{,}527\ \mathrm{kN}

Step 4: Ultimate Toe Resistance

Qb=qbAb=2,500(0.2827)=707 kNQ_b=q_bA_b =2{,}500(0.2827)=707\ \mathrm{kN}

Step 5: Total Ultimate Capacity

Qult=Qs+Qb=1,527+707=2,234 kNQ_{\mathrm{ult}}=Q_s+Q_b =1{,}527+707=2{,}234\ \mathrm{kN}

The shaft contributes approximately 68 percent and the toe approximately 32 percent of the calculated ultimate resistance.

Step 6: Allowable Capacity

Qallow=QultFS=2,2342.5=894 kNQ_{\mathrm{allow}}=\frac{Q_{\mathrm{ult}}}{FS} =\frac{2{,}234}{2.5}=894\ \mathrm{kN}

The preliminary allowable axial geotechnical capacity is approximately 894 kN894\ \mathrm{kN} for the stated design format.

Layered-Soil Improvement

Using one average shaft value is convenient but can hide important layering. A better organization is:

Qs=fs,iπDLiQ_s=\sum f_{s,i}\pi D L_i

Each layer receives a unit shaft resistance consistent with its soil or rock condition and installation method. Layers that do not provide reliable positive resistance should not be included. Negative skin friction should be treated as load, not positive resistance.

Engineering Interpretation

The calculated capacity is not yet a pile design. The engineer must also evaluate:

  • Settlement and load-transfer response at service load.
  • Structural compression, uplift, bending, shear, and buckling.
  • Downdrag from settling soil.
  • Group efficiency, group settlement, and block behavior.
  • Lateral and cyclic response.
  • Scour or unsupported length.
  • Installation feasibility and effects.
  • Load testing, integrity testing, and acceptance criteria.
  • Compatibility with the governing allowable-stress or resistance-factor design format.

For a drilled shaft, base cleanliness and construction method can strongly influence toe resistance. For a driven pile, setup, relaxation, driving stresses, and the selected acceptance method matter.

Sensitivity To Toe Resistance

If toe resistance is ignored because base performance is uncertain, ultimate capacity becomes 1,527 kN1{,}527\ \mathrm{kN} and allowable capacity becomes about 611 kN611\ \mathrm{kN}. The 283 kN difference shows why construction and verification of toe behavior can be economically important.

References And Further Reading

  • FHWA GEC 10, Drilled Shafts.
  • FHWA GEC 12, Driven Pile Foundations.
  • FHWA GEC 15, Acceptance Procedures for Deep Foundations.
  • Project-specific geotechnical resistance and load-test criteria.

FAQ

Can average shaft resistance be used in layered soil?

It can be used for a transparent preliminary equivalent when properly weighted, but final calculations should normally sum resistance by layer and construction method.

Is factor of safety 2.5 always required?

No. Safety or resistance factors depend on the governing design framework, analysis method, testing, variability, and project requirements.

Does allowable capacity control pile quantity?

Not by itself. Group effects, cap geometry, lateral load, structural capacity, settlement, redundancy, and minimum spacing can control the layout.