A slope with β = 45° is to be constructed with a soil that has $\phi ^\prime$ = 20° and c′ = 24 kN/m². The unit weight of the compacted soil will be 18.9 kN/m³.
a. Find the critical height of the slope.
b. If the height of the slope is 10 m, determine the factor of safety with respect to strength.

Question

A slope with β = 45° is to be constructed with a soil that has [latex]\phi ^\prime [/latex] = 20° and c′ = 24 kN/m². The unit weight of the compacted soil will be 18.9 kN/m³.
a. Find the critical height of the slope.
b. If the height of the slope is 10 m, determine the factor of safety with respect to strength.

Accepted Answer

Part a
We have
[latex]m=\frac{c^\prime}{\gamma H_{cr}}[/latex]
From Figure 15.23, for β = 45° and [latex]\phi ^\prime [/latex] = 20°, m = 0.06. So
[latex]H_{cr}=\frac{c^\prime}{\gamma m}=\frac{24}{(18.9)(0.06)}=[/latex] 21.1 m
Part b
If we assume that full friction is mobilized, then, referring to Figure 15.23 (for β = 45° and [latex]\phi_d^\prime[/latex] = [latex]\phi ^\prime [/latex]= 20°), we have
[latex]m=0.06=\frac{c^\prime_d}{\gamma H}[/latex]
or
[latex]c^\prime_d=(0.06)(18.9)(10)=11.34[/latex] kN/m²
Thus,
[latex]F_{\phi^\prime}=\frac{ an \phi^\prime}{ an \phi^\prime_d}=\frac{ an 20}{ an 20}=1[/latex]
and
[latex]F_{c^\prime}=\frac{c^\prime}{c^\prime_d}=\frac{24}{11.34}=2.12[/latex]
Since [latex]FS_{c^\prime} eq FS_{\phi^\prime}[/latex], this is not the factor of safety with respect to strength.
Now we can make another trial. Let the developed angle of friction, [latex]\phi^\prime_d[/latex], be equal to 15°. For β = 45° and the friction angle equal to 15°, we find from Figure 15.23.
[latex]m=0.083=\frac{c^\prime_d}{\gamma H}[/latex]
or
[latex]c^\prime_d=(0.083)(18.9)(10)=15.69[/latex] kN/m²
For this trial,
[latex]F_{\phi^\prime}=\frac{ an \phi^\prime}{ an \phi^\prime_d}=\frac{ an 20}{ an 15}=1.36[/latex]
and
[latex]F_{c^\prime}=\frac{c^\prime}{c^\prime_d}=\frac{24}{15.69}=1.53[/latex]
Similar calculations of [latex]F_{\phi^\prime}[/latex] and [latex]F_{c^\prime}[/latex] for various assumed values of [latex]\phi^\prime_d[/latex] are given in the following table.

[latex]\phi_d^\prime[/latex]	[latex] an \phi_d^\prime[/latex]	[latex]F_{\phi^\prime}[/latex]	m	[latex]c_d^\prime[/latex] (kN/m²)	[latex]F_{c^\prime}[/latex]
20	0.364	1.0	0.06	11.34	2.12
15	0.268	1.36	0.083	15.69	1.53
10	0.176	2.07	0.105	19.85	1.21
5	0.0875	4.16	0.136	25.70	0.93

The values of [latex]F_{\phi^\prime}[/latex] are plotted against their corresponding values of [latex]F_{c^\prime}[/latex] in Figure 15.28, from which we find
[latex]FS_{c^\prime}=FS_{\phi^\prime}=FS_s=[/latex] 1.42
Note: We could have found the value of [latex]FS_s[/latex] from Figure 15.25c. Since β = 45°, it is a slope of 1V:1H. For this slope
[latex]\frac{c^\prime}{\gamma H}=\frac{24}{(18.9)(10)}=0.127[/latex]
From Figure 15.25c, for’ c′/γH = 0.127, the value of [latex]FS_s \approx[/latex] 1.4

Question 15.6: A slope with β = 45° is to be constructed with a soil that h...

Related Answered Questions

Solve Example 15.9 using Spencer’s solution (Figure 15.35). ...

Verified Answer:

Solve Example 15.9 using Michalowski’s solution (Figure 15.36). ...

Verified Answer:

A given slope under steady-state seepage has the following: H = 21.62 m, Φ′ = 25°, slope: 2H:1V, c′ = 20 kN/m², γ = 18.5 kN/m³, ru = 0.25. Determine the factor of safety, Fs. Use Table 15.3. ...

Verified Answer:

Solve Example 15.6 using Michalowski’s solution. ...

Verified Answer:

A cut slope in saturated clay (Figure 15.21) makes an angle of 56° with the horizontal. Assume kh = 0.25. a. Determine the maximum depth up to which the cut could be made. b. How deep should the cut be made if a factor of safety of 2 against sliding is required? ...

Verified Answer:

A cut slope was excavated in a saturated clay. The slope made an angle of 40° with the horizontal. Slope failure occurred when the cut reached a depth of 6.1 m. Previous soil explorations showed that a rock layer was located at a depth of 9.15 m below the ground surface. Assuming an undrained ...

Verified Answer:

Verified Answer:

A cut is to be made in a soil having γ = 105 lb/ft³, c′ = 600 lb/ft², and Φ′ = 15°. The side of the cut slope will make an angle of 45° with the horizontal. What should be the depth of the cut slope that will have a factor of safety (Fs) of 3? ...

Verified Answer:

For the slope shown in Figure 15.31, find the factor of safety against sliding for the trial slip surface AC. Use the ordinary method of slices. ...

Verified Answer:

For the infinite slope with a steady state seepage shown in Figure 15.9, determine: a. The factor of safety against sliding along the soil-rock interface. b. The height, H, that will give a factor of safety (Fs) of 2 against sliding along the soil-rock interface. ...

Verified Answer: