A vertical trench 18 ft (5.50 m) deep is to be constructed in soft clay having a shear strength of 500 psf (24 kPa) and a unit weight of 112 pcf (17.6 kN/m³). What is the maximum safe depth of cut that can be made without bracing?

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Accepted Answer

From Equation (14.7),

[latex]N_{s\left(critical ight) }=\frac{\gamma H}{c} =\frac{4\cos \phi }{1- \sin \phi }=4 an \left(\frac{\pi }{4} +\frac{\phi }{2} ight)[/latex] (14.7)

[latex]N_{s\left(critical ight) } =\frac{4\cos \phi }{1- \sin \phi } =\frac{4\cos (0) }{1- \sin (0) }=\frac{4}{1} =4[/latex]

[latex]H_{c}=N_{s}\frac{c}{\gamma }=4\cdot \frac{500 lb/ft^{2} }{112 lb/ft^{3} } = 17.9 ft[/latex]

So, theoretically the maximum depth is about 18 ft (5.5 m), however, a factor of safety of at least 1.5 would ordinarily be applied, reducing this depth by 17.9 / 1.5 to about 11.9 ft (3.6 m).

Question 14.1: A vertical trench 18 ft (5.50 m) deep is to be constructed i...

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