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16.5: Chapter 5- Dry Sand Cutting

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    29391
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    16.5.1. MC: Soil Mechanical Parameters

    Which of the following statements is true? The cutting of dry sand is:

    The bold green answers are true.

    1. Dominated by pore pressures.

    2. Dominated by adhesion and cohesion.

    3. Dominated by the permeability of the sand.

    4. Dominated by the weight of the layer cut.

    5. Influenced by the porosity of the sand.

    6. Dominated by the inertial forces.

    7. Influenced by the angle of external friction of the sand.

    8. Dominated by the shear strength of the sand.

    16.5.2. MC: The Shear Angle 

    Which of the following statements are true in the case only the weight of the soil is considered if dry sand is excavated with blade angles above 30o?

    The bold green answers are true.

    1. If the blade angle α increases, the shear angle β also increases.

    2. If the angle of internal friction φ increases, the shear β angle also increases.

    3. If the external friction angle δ increases, the shear β angle also increases.

    4. If the blade angle α increases, the shear angle β decreases.

    5. If the angle of internal friction φ increases, the shear angle β decreases.

    6. If the external friction angle δ increases, the shear angle β decreases.

    16.5.3. Calc.: The Shear Angle 

    Suppose the cutting of dry sand is completely dominated by the inertial forces. Derive the analytical solution for the shear angle based on the minimum energy principle.

    \(\ \mathrm{F_{h}=\rho_{s} \cdot v_{c}^{2} \cdot h_{i} \cdot w \cdot \frac{\sin (\alpha)}{\sin (\alpha+\beta)} \cdot \frac{\cos (\varphi)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \sin (\alpha+\delta)}\)

    This is at a minimum when the denominator is at a maximum, so:

    \(\ \begin{array}{left}\frac{\partial}{\partial \beta}(\sin (\alpha+\beta) \cdot \sin (\alpha+\beta+\delta+\varphi))=0\\
    \cos (\alpha+\beta) \cdot \sin (\alpha+\beta+\delta+\varphi)+\sin (\alpha+\beta) \cdot \cos (\alpha+\beta+\delta+\varphi)=\sin (2 \cdot \alpha+2 \cdot \beta+\delta+\varphi)=0\\
    2 \cdot \alpha+2 \cdot \beta+\delta+\varphi=\pi\\
    \beta=\frac{\pi}{2}-\frac{2 \cdot \alpha+\delta+\varphi}{2}\end{array}\)


    This page titled 16.5: Chapter 5- Dry Sand Cutting is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Sape A. Miedema (TU Delft Open Textbooks) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.