   Chapter 15, Problem 15.17P Principles of Geotechnical Enginee...

9th Edition
Braja M. Das + 1 other
ISBN: 9781305970939

Solutions

Chapter
Section Principles of Geotechnical Enginee...

9th Edition
Braja M. Das + 1 other
ISBN: 9781305970939
Textbook Problem

A cut slope in saturated clay is inclined at an angle β = 26.57° (1 vertical to 2 horizontal) and has a height, H = 7 m. The undrained cohesion. cu, increases with depth, z, as shown in Figure 15.51. Assuming the critical circle to be a toe circle, determine the factor of safety, Fs using the Koppula (1984) method. Given: γsat= 18.6 kN/m3, cu(z = 0) = 9 kN/m2, and a0 = 5 kN/m3. To determine

Find the factor of safety Fs using the Koppula method.

Explanation

Given information:

The inclination β angle is 26.57°.

The height (H) of the cut is 7.0 m.

The saturated unit weight γsat of the back fill is 18.6kN/m3.

The undrained shear strength at depth z cu(z=0) is 9kN/m2.

The slope a0 of the line of plot cu(z=0) versus depth is 5kN/m2.

Calculation:

Determine the undrained shear strength cu(z) at depth z using the relation.

cu(z)=cu(z=0)+a0z

Substitute 9kN/m2 for cu(z=0) and 5kN/m2 for a0.

cu(z)=9+(5)z

Determine the value of cR using the formula.

cR=a0Hcu(z=0)

Substitute 5kN/m2 for a0, 7.0 m for H, and 9kN/m2 for cu(z=0)

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