Question

The figure below shows an infinite-slope model. If \(\tau\) and SS denote the driving shear stress and the resisting shear strength, respectively, then \(\tau\) along the failure plane is \(\times 10^3\) kg/ms\(^2\) (round off to two decimal places).
[Use: \(d = 2\) m; \(\rho_{soil} = 2.0\) g/cm\(^3\); \(g = 9.8\) m/s\(^2\)]

Hint:

Use the infinite-slope formula tau = rho times g times d times sin(beta) times cos(beta), with d as the vertical soil thickness from the figure, and convert density to kg/m^3 before multiplying.

Answer 1

Correct Answer: 16.97

Step 1: Set up the infinite-slope model.
In an infinite-slope model, a soil layer of thickness \(d\) rests on a slope inclined at angle \(\beta\) to the horizontal, above a failure plane running parallel to the ground surface. The weight of a soil column above the failure plane can be resolved into two parts, one acting along the slope, which drives sliding, and one acting perpendicular to the slope, which presses the soil onto the failure plane.

Step 2: Write the weight per unit plan area.
For a soil column of vertical thickness \(d\) and density \(\rho_{soil}\), the weight per unit horizontal (plan) area is
\[ W = \rho_{soil}\,g\,d \]
where \(g\) is the acceleration due to gravity.

Step 3: Resolve this weight along and normal to the slope.
Standard infinite-slope theory resolves this weight into a component parallel to the slope, which is the driving shear stress \(\tau\), and a component normal to the slope, which sets up the resisting friction. The along-slope (driving) component per unit area of the failure plane works out to
\[ \tau = \rho_{soil}\,g\,d\,\sin\beta\cos\beta \]
This is the standard infinite-slope formula for driving shear stress, and it uses \(d\) as the vertical thickness of the soil slab shown in the figure.

Step 4: Substitute the given values.
Here
\[ \rho_{soil} = 2.0\ \text{g/cm}^3 = 2000\ \text{kg/m}^3, \quad d = 2\ \text{m}, \quad g = 9.8\ \text{m/s}^2, \quad \beta = 30^{\circ} \]
\[ \sin 30^{\circ} = 0.5, \quad \cos 30^{\circ} = 0.8660 \]

Step 5: Compute the driving shear stress.
\[ \tau = 2000 \times 9.8 \times 2 \times 0.5 \times 0.8660 \]
\[ \tau = 39200 \times 0.5 \times 0.8660 = 19600 \times 0.8660 \]
\[ \tau \approx 16974.1\ \text{kg/ms}^2 \]

Step 6: Convert and round off.
Expressed as a multiple of \(10^3\),
\[ \tau \approx 16.97 \times 10^3\ \text{kg/ms}^2 \]

Step 7: Final answer.
The driving shear stress along the failure plane is about 16.97 times \(10^3\) kg/ms\(^2\), which falls in the accepted range of 16.80 to 17.10.
\[ \boxed{16.97} \]