Problem 4.11. Stresses of soil under square foundation

A square foundation of size 3 × 3 m is loaded by a normal force N = 1000 kN and a moment M = 1000 kNm. Assuming plastic deformation of the soil determine the stresses of the soil, if
a) the vector of the moment is parallel to the side of the foundation;
b) the vector of the moment is parallel to the diagonal of the foundation.

Solve Problem

Solve

Problem a)

Stress in the soil, σ [kN/m²]=

Problem b)

Stress in the soil, σ [kN/m²]=

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Steps

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Problem a)

Step 1. Draw normal stress diagram. Calculate the length of the compressed part of the foundation.

Check compressed lengthHide compressed length

See Figure 4.16.

$e = \frac{M}{N} = \frac{1000}{1000} = 1.0 m x = 2 (\frac{a}{2} - e) = 2 (\frac{3.0}{2} - 1.0) = 1.0 m$

Step 2. Determine stress in the soil.

Check stressHide stress

Assuming uniform stress distribution, the stress can be determined from the force equilibrium:

$σ = \frac{N}{a x} = \frac{N}{a 2 (\frac{a}{2} - e)} = \frac{1000}{3.0 \times 1.0} = 333.3 \frac{kN}{m^{2}}$

Problem b)

Step 1. Draw normal stress diagram. Determine the geometry of the compressed part of the foundation.

Check compressed partHide compressed part

The stress resultant (which equilibrates the normal force) acts at the centre of gravity of the compressed part (see shaded area in the Figure above).

$e = \frac{M}{N} = \frac{1000}{1000} = 1.0 m e = a \frac{\sqrt{2}}{2} - \frac{2}{3} x \to x = \frac{3}{2} (a \frac{\sqrt{2}}{2} - e) = \frac{3}{2} (3.0 \frac{\sqrt{2}}{2} - 1.0) = 1.682 m$

Step 2. Determine stress in the soil.

Check stressHide stress

Assuming uniform stress distribution, the stress can be determined from the force equilibrium:

$σ = \frac{N}{x^{2}} = \frac{N}{\frac{3^{2}}{2^{2}} {(a \frac{\sqrt{2}}{2} - e)}^{2}} = \frac{1000}{1 . 68^{2}} = 353.5 \frac{kN}{m^{2}}$

Results

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Problem a)

The normal stress diagram and the length of the compressed part of the foundation is shown in the Figure below.

See Figure 4.16.

$e = \frac{M}{N} = \frac{1000}{1000} = 1.0 m x = 2 (\frac{a}{2} - e) = 2 (\frac{3.0}{2} - 1.0) = 1.0 m$

Assuming uniform stress distribution, the stress can be determined from the force equilibrium:

$σ = \frac{N}{a x} = \frac{N}{a 2 (\frac{a}{2} - e)} = \frac{1000}{3.0 \times 1.0} = 333.3 \frac{kN}{m^{2}}$

Problem b)

The normal stress diagram and the geometry of the compressed part of the foundation is given in the Figure below.

The stress resultant (which equilibrates the normal force) acts at the centre of gravity of the compressed part (see shaded area in the Figure above).

$e = \frac{M}{N} = \frac{1000}{1000} = 1.0 m e = a \frac{\sqrt{2}}{2} - \frac{2}{3} x \to x = \frac{3}{2} (a \frac{\sqrt{2}}{2} - e) = \frac{3}{2} (3.0 \frac{\sqrt{2}}{2} - 1.0) = 1.682 m$

Assuming uniform stress distribution, the stress can be determined from the force equilibrium:

$σ = \frac{N}{x^{2}} = \frac{N}{\frac{3^{2}}{2^{2}} {(a \frac{\sqrt{2}}{2} - e)}^{2}} = \frac{1000}{1 . 68^{2}} = 353.5 \frac{kN}{m^{2}}$