A rectangular footing 4.0 x 3.0 m2is loaded by a vertical load of 200.0 ton. The footing is founded at a depth of 2.50 m. The bulk and saturated unit weights are 1.93 t/m3 and 2.21 t/m3. Ground water table is located at ground surface. The shear strength parameters are c = zero and ɸ = 35o. Determine the factor of safety with respect to shear failure.
Question
A rectangular footing 4.0 x 3.0 m2is loaded by a vertical load of 200.0 ton. The footing is founded at a depth of 2.50 m. The bulk and saturated unit weights are 1.93 t/m3 and 2.21 t/m3. Ground water table is located at ground surface. The shear strength parameters are c = zero and ɸ = 35o. Determine the factor of safety with respect to shear failure.
Solution
To solve this problem, we need to use the formula for the factor of safety (F) with respect to shear failure for a strip footing loaded by a vertical load. The formula is:
F = (q * B * Nq + 0.5 * γ * B^2 * Nγ) / q0
Where:
- q is the vertical load per unit area (q = P / A)
- B is the width of the footing
- γ is the bulk unit weight
- Nq and Nγ are bearing capacity factors which depend on the angle of internal friction (ɸ)
- q0 is the initial vertical stress at the foundation level
Let's calculate each term:
-
The vertical load per unit area (q) is the total load divided by the area of the footing. The area (A) is 4.0 m * 3.0 m = 12.0 m^2. So, q = 200.0 ton / 12.0 m^2 = 16.67 t/m^2.
-
The width of the footing (B) is 4.0 m.
-
The bulk unit weight (γ) is 1.93 t/m^3.
-
The bearing capacity factors (Nq and Nγ) can be calculated using the formulas Nq = e^πtanɸ * tan^2(45 + ɸ/2) and Nγ = 2 * (Nq + 1) * tanɸ. Substituting ɸ = 35° into these formulas, we get Nq = 33.17 and Nγ = 45.14.
-
The initial vertical stress at the foundation level (q0) is the bulk unit weight times the depth of the foundation. So, q0 = 1.93 t/m^3 * 2.5 m = 4.825 t/m^2.
Substituting these values into the formula for the factor of safety, we get:
F = (16.67 t/m^2 * 4.0 m * 33.17 + 0.5 * 1.93 t/m^3 * (4.0 m)^2 * 45.14) / 4.825 t/m^2 = 3.48
So, the factor of safety with respect to shear failure is 3.48.
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