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Example 12.7 · α method, layered clay

Example 12.7 of Das & Sivakugan (2017), Principles of Foundation Engineering, solved in Rischio: shaft capacity, beside the source's answers.

Source
Das, B. M., & Sivakugan, N. (2017). Principles of Foundation Engineering, SI edition, 9th ed. Cengage Learning. Example 12.7.
ID
das-pfe9-12-7
Open in the studioA copy, in your account
22 kPa30 kPa46 kPa
The model as the editor draws it, with what governs its result.

Source vs Rischio

SourceRischioDifference
Shaft capacity1 050 kN1 050 kN0.0 %

Every check

What the engine checks in this model, against Rischio's default limits.

Single pilePassesFS=3.94\mathrm{FS} = 3.94≥3.0\ge 3.0

The model

The example's data, element by element, as the model holds it.

Driven pilePile · Circular · Driven

Diameter or side dd
0.457 m
Length LL
20 m

GroupGroup · single pile

Piles along x nxn_x
1
Piles along y nyn_y
1
Spacing ss
1.5 m

LoadLoad · on the pile

Vertical load VV
300 kN

Clay 1Stratum · Undrained

Thickness tt
3 m
Unit weight γ\gamma
16 kN/m³
Undrained shear strength cuc_\mathrm{u}
25 kPa

Clay 2Stratum · Undrained

Thickness tt
7 m
Unit weight γ\gamma
17 kN/m³
Saturated unit weight γsat\gamma_\mathrm{sat}
17 kN/m³
Undrained shear strength cuc_\mathrm{u}
40 kPa

Clay 3Stratum · Undrained

Unit weight γ\gamma
18 kN/m³
Saturated unit weight γsat\gamma_\mathrm{sat}
18 kN/m³
Undrained shear strength cuc_\mathrm{u}
90 kPa

Water tableWater table · depth below ground

Depth zwz_\mathrm{w}
3 m

How it's calculated

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