For the problem given in Ex. 17.1 , determine the allowable load for a settlement of 10 mm [latex]\left(=S_{a}\right)[/latex]. All the other data remain the same.

Allowable skin load Settlement ratio [latex]S_{R}=\frac{S_{a}}{d}=\frac{10}{1.5 \times 10^{3}} \times 100=0.67 \%[/latex] From Fig. 17.18 for [latex]S_{R}=0.67 \%, N_{f m}=\frac{Q_{f m}}{Q_{f}}=0.95[/latex] by using the trend line. [latex]Q_{f m}=0.95 Q_{f}=0.95 \times 8,549=8,122 kN[/latex]. Allowable base load for [latex]S_{a}=10[/latex] mm From Fig. 17.19 for [latex]S_{R}=0.67 \%, N_{b m}=\frac{Q_{b m}}{Q_{b}}=0.4[/latex] [latex]Q_{b m}=0.4 Q_{b}=0.4 \times 2,384=954 kN[/latex]. Now the allowable load [latex]Q_{a s}[/latex] based on settlement consideration is [latex]Q_{a s}=Q_{f m}+Q_{b m}=8,122+954=9,076 kN[/latex] [latex]Q_{a s}[/latex] based on settlement consideration is very much higher than [latex]Q_{a}[/latex] ( Ex. 17.1 ) and as such [latex]Q_{a}[/latex] governs the criteria for design.

Question 17.2: For the problem given in Ex. 17.1, determine the allowable l...

Geotechnical Engineering: Principles and Practices of Soil Mechanics and Foundation Engineering [EXP-45222]

For the problem given in Ex. 17.1, determine the allowable load for a settlement of 10 mm $\left(=S_{a}\right)$ . All the other data remain the same.

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Allowable skin load

Settlement ratio $S_{R}=\frac{S_{a}}{d}=\frac{10}{1.5 \times 10^{3}} \times 100=0.67 \%$

From Fig. 17.18 for $S_{R}=0.67 \%, N_{f m}=\frac{Q_{f m}}{Q_{f}}=0.95$ by using the trend line.

$Q_{f m}=0.95 Q_{f}=0.95 \times 8,549=8,122 kN$ .

Allowable base load for $S_{a}=10$ mm

From Fig. 17.19 for $S_{R}=0.67 \%, N_{b m}=\frac{Q_{b m}}{Q_{b}}=0.4$

$Q_{b m}=0.4 Q_{b}=0.4 \times 2,384=954 kN$ .

Now the allowable load $Q_{a s}$ based on settlement consideration is