A braced cut (or excavation) is made in a deposit of sand to a depth of 6m and width of 4m. The braced cut is supported by horizontal struts as shown in Figure Q4. The angle of internal friction of sand is 32° while the unit weight is 18.6kN/m². If the struts are located at 3m center to center in plan. Calculate: - a) The earth pressure envelope on the cut (7 marks) b) The struts load at level A, B and C. (3 marks) c) The strut section if the permissible axial (axial or bending) stress is 200 x 10 kPa. (8 marks) d) The required section of the wales at level A, if the axial (axial or bending) stress is 210 x 103 kPa (2 marks) 4m 1m А ЧЕ 2 m B 2m с 1m
a) To calculate the earth pressure on the cut, we can use the Rankine's theory of earth pressure. The equation for the active earth pressure is given by:
P = Ka * γ * H^2
Where P is the earth pressure, Ka is the active earth pressure coefficient, γ is the unit weight of the sand, and H is the height of the cut.
To calculate Ka, we can use the equation:
Ka = (1 - sin(φ')) / (1 + sin(φ'))
Where φ' is the effective angle of internal friction. Given that φ' = 32° = 0.559 radians, we can calculate Ka as:
Ka = (1 - sin(0.559)) / (1 + sin(0.559)) = 0.19
Substituting the given values, we have:
P = (0.19) * (18.6 kN/m²) * (6m)^2 = 115.03 kN/m
Therefore, the earth pressure on the cut is 115.03 kN/m.
b) To calculate the strut load at different levels, we need to determine the horizontal force acting on each strut.
At level A:
The horizontal force acting on each strut is equal to half of the earth pressure on the cut since the struts are located at 3m center to center in plan. Therefore, the strut load at level A is:
Load_A = (115.03 kN/m) / 2 = 57.515 kN/m
At level B:
The horizontal force acting on each strut is equal to the earth pressure on the cut since the struts are at the corners of the cut. Therefore, the strut load at level B is:
Load_B = 115.03 kN/m
At level C:
The horizontal force acting on each strut is again equal to half of the earth pressure on the cut since the struts are located at 3m center to center in plan. Therefore, the strut load at level C is:
Load_C = (115.03 kN/m) / 2 = 57.515 kN/m
c) To calculate the required section of the struts, we can use the equation for the axial stress:
σ = P / A
Where σ is the axial stress, P is the strut load, and A is the cross-sectional area of the strut. Rearranging the equation, we have:
A = P / σ
Given that the permissible axial or bending stress is 200 x 10^3 Pa, we can calculate the required section of the struts. Substituting the values, we have:
Section = (57.515 kN/m) / (200 x 10^3 Pa) = 287.58 x 10^-3 m²
Therefore, the required section of the strut is 287.58 x 10^-3 m².
d) To calculate the required section of the wales at level A, we can use the same equation as before:
Section = (Load_A) / (210 x 10^3 Pa) = (57.515 kN/m) / (210 x 10^3 Pa)
Simplifying, we have:
Section = 0.27389 x 10^-3 m²
Therefore, the required section of the wales at level A is 0.27389 x 10^-3 m².
a) The earth pressure envelope on the cut can be determined using Rankine's earth pressure theory. The earth pressure for the active condition can be calculated using the formula:
Pa = Ka * γ * H * [(1 - sinφ) / (1 + sinφ)]
Where:
Pa = earth pressure at depth H
Ka = active earth pressure coefficient
γ = unit weight of the sand
H = depth of the cut
φ = angle of internal friction
Given values:
Ka = 0.33 (from Rankine's earth pressure theory)
γ = 18.6 kN/m²
H = 6m
φ = 32°
Plugging in the values:
Pa = 0.33 * 18.6 * 6 * [(1 - sin(32°)) / (1 + sin(32°))
Pa ≈ 0.33 * 18.6 * 6 * (0.463) ≈ 17.05 kN/m²
Therefore, the earth pressure envelope on the cut is approximately 17.05 kN/m².
b) The struts load at level A, B, and C can be calculated by considering the equilibrium of forces acting on the struts.
At level A:
The total horizontal force acting on the struts can be calculated using the formula:
Fh = Pa * (H + 0.5 * d)
Where:
Fh = total horizontal force
Pa = earth pressure at depth H (17.05 kN/m²)
H = depth of the cut (6m)
d = distance between the struts (3m)
Fh = 17.05 * (6 + 0.5 * 3) ≈ 102.3 kN
At level B:
The total horizontal force is the same as at level A.
Fh = 102.3 kN
At level C:
The total horizontal force can be calculated similarly:
Fh = Pa * (H - 0.5 * d)
Fh = 17.05 * (6 - 0.5 * 3) ≈ 85.65 kN
Therefore, the struts load at level A, B, and C are approximately 102.3 kN, 102.3 kN, and 85.65 kN, respectively.
c) The strut section can be determined by considering the allowable axial stress and the applied load.
Since the allowable axial stress is given as 200 x 10 kPa (200 kN/m²), the maximum permissible load on the strut can be calculated using the formula:
Maximum Load = Area * Permissible Axial Stress
Let's assume a square strut section. The area of the strut can be calculated by dividing the load by the permissible axial stress:
Area = Maximum Load / Permissible Axial Stress
For example, if the maximum load is 102.3 kN (as calculated in part b) at level A):
Area = 102.3 kN / (200 kN/m²) = 0.5115 m²
Hence, the required strut section is approximately 0.5115m².
d) The required section of the wales at level A can be determined in a similar manner as the strut section.
Since the axial stress is given as 210 x 10³ kPa (210 MPa), the maximum permissible load on the wales can be calculated using the formula:
Maximum Load = Area * Permissible Axial Stress
Assuming a rectangular wale section with a width of 1m, we can calculate the required height of the wale (h) using the formula:
Maximum Load = h * 1m * (210 x 10³ kPa)
For example, if the maximum load is 102.3 kN (as calculated in part b) at level A):
h = 102.3 kN / (1m * 210 x 10³ kPa)
h ≈ 0.488 m
Therefore, the required section of the wales at level A is approximately 0.488m in height with a width of 1m.