Chapter5 - Chapter 5 Slope Stability Analyses and...

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Chapter 5. Slope Stability Analyses and Stabilization Measures 1. A natural slope is 45-m high and the slope angle is 38 degrees. The surficial soil is loose silty sand with cohesion 45 kN/m 2 , internal friction angle 25 degrees, and bulk unit weight 18.0 kN/m 3 . The thickness of the topsoil is 1.0 m in the vertical direction. The slope is dry. Determine the factor of safety of the surficial soil layer against translational failure. Solution: Solution using allowable stress design approach: Since the slope is 45 m high, a think layer of loose silty sand (1.0 m in height) exists on the slope surface, the infinite slope method for dry slope is used. The factor of safety is: = 5.74 Solution using limit state design approach: It should be verified that E d ≤ R d where E d is the design effect of the actions (e.g. sliding force) and R d the design strength. Following Equation (5.20) and assuming G = 1.35 (Note that this value may change according to local design approach) E d = γ G γ k H cos β sin β = 1.35 × 18 × 1 × cos25 × sin25 = 9.3 kN/m 2 Similarly from Equation (5.19) and assuming that R = 1.10, = 1.00, = 1.00 and ´ = 1.00. R d = 1 γ R [ c k ´ γ c ´ + γ k γ γ H cos 2 β tan Φ k ´ γ Φ ] = 1 1.10 [ 45 1.0 + 18 1.0 1.0cos 2 38 tan 25 1.0 ] = ¿ 45.6 kPa Since E d ≤ R d the limit state is satisfied and the slope is safe. 2. A saturated natural slope is 45-m high and the slope angle is 38 degrees. The surficial soil is loose silty sand with effective cohesion 45 kN/m 2 , effective internal friction
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angle 25 degrees, and saturated unit weight 19.0 kN/m 3 . The thickness of the topsoil is 1.0 m in the vertical direction. The downward seepage is parallel to the slope surface. Determine the factor of safety of the surficial soil layer against translational failure. Solution: Solution using allowable stress design approach: The infinite slope method with seepage parallel to the slope face is used. The FS is based on Equation (5.23). FS f 2 c sat H sin(2 ) sat tan tan where: c = 45 kN/m 2 , = 25 , sat = 19 kN/m 3 , = 38 , = 19 9.81= 9.19 kN/m 3 , H = 1.0 m. FS 2 45 19 1.0 sin(2 38) 9.19 19 tan25 tan38 5.17 > 1.5, no slope failure. Solution using limit state design approach: Following Equation (5.31) and assuming that G = 1.35,  R = 1.10, = 1.00, = 1.00 and ´ = 1.00. Note that these partial factors of safety may change locally. E d ≤ R d = γ G γ sat ,k H cos β sin β ≤ 1 γ R [ c k ´ γ c ´ + ( γ sat ,k γ γ H cos 2 β γ w H cos 2 β ) tan Φ k ´ γ Φ ] E d = 1.35 × 19 × 1 × cos38sin 38 = 12.44 kPa R d = 1 1.10 [ 45 1.0 + ( 19 1.0 1cos 2 38 9.81 × 1cos 2 38 ) tan25 1.0 ] = 43.3 kPa Since E d ≤ R d the limit state is satisfied and the slope is safe. 3. A reservoir is 45-m deep and the side slope of the reservoir developed a loose surficial layer that is 1.0 m thick (in the vertical direction). The slope angle is 38 degrees. The surficial soil layer has cohesion 45 kN/m 2 , internal friction angle 25 degrees, and saturated unit weight 19.0 kN/m 3 . Assuming the reservoir water level is
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