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Example 6.4 · Square footing, water table

Example 6.4 of Das & Sivakugan (2017), Principles of Foundation Engineering, solved in Rischio: allowable load, FS 3, beside the source's answers.

Source
Das, B. M., & Sivakugan, N. (2017). Principles of Foundation Engineering, SI edition, 9th ed. Cengage Learning. Example 6.4.
ID
das-pfe9-6-4
Open in the studioA copy, in your account
50.7 kPa50.7 kPa
The model as the editor draws it, with what governs its result.

Source vs Rischio

SourceRischioDifference
Allowable load, FS 3667.2 kN666.9 kN0.0 %

The book solves for the width that carries 667.2 kN; Rischio checks a given width, so it takes the book's answer, 1.36 m, and gets the load back.

Every check

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

Bearing capacityPassesFS=21.31\mathrm{FS} = 21.31≥3.0\ge 3.0
Eccentricity along the widthPasseseB=0.00 me_B = 0.00\,\mathrm{m}≤0.227 m\le 0.227\,\mathrm{m}
Eccentricity along the lengthPasseseL=0.00 me_L = 0.00\,\mathrm{m}≤0.227 m\le 0.227\,\mathrm{m}

The model

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

FootingFooting · Rectangular

Width BB
1.3603 m
Embedment depth DD
1.22 m
Length LL
1.3603 m

LoadLoad · on the footing

Vertical load VV
93.9 kN
Horizontal load HH
0 kN
Moment along the width MBM_B
0 kN·m
Moment along the length MLM_L
0 kN·m

SandStratum

Unit weight γ\gamma
16.5 kN/m³
Saturated unit weight γsat\gamma_\mathrm{sat}
18.5 kN/m³
Cohesion c′c'
0 kPa
Friction angle φ′\varphi'
34 °

Water tableWater table · depth below ground

Depth zwz_\mathrm{w}
0.61 m

How it's calculated

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