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16.6: Chapter 6- Water Saturated Sand Cutting

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    29392
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    16.6.1. MC: Soil Mechanical Parameters

    Which of the following soil mechanical parameters play a dominant role in the cutting of saturated sand?

    The bold green answers are true.

    1. The density of water.

    2. The angle of internal friction.

    3. The permeability of the sand.

    4. The tensile strength of the sand.

    5. The porosity of the sand.

    6. The adhesion of the sand.

    7. The angle of external friction of the sand.

    8. The shear strength of the sand.

    16.6.2. MC: Dilatation 

    What is the definition of dilatation?

    The bold green answers are true.

    1. Dilatation is the increase of the pore volume of sand caused by gravitation.

    2. Dilatation is the decrease of the pore volume of sand caused by shear stress.

    3. Dilatation is the decrease of the pore volume of sand caused by dredging.

    4. Dilatation is the increase of the pore volume of sand caused by shear stress.

    5. Dilatation is the increase of the pore volume of sand caused by dredging.

    6. Dilatation is the decrease of the pore volume of sand caused by gravitation.

    16.6.3. MC: Cavitation 

    Which statements are true?

    The bold green answers are true.

    1. Cavitation is in fact the boiling of water.

    2. Cavitation at 10 degrees centigrade occurs at about 0.1 bar absolute pressure.

    3. At 100 degrees centigrade cavitation will occur at about 100 kPa.

    4. Cavitation is the vaporization of water.

    5. Cavitation at 10 degrees centigrade occurs at about 0.01 bar absolute pressure.

    6. High in the mountains cavitation will occur at a temperature lower than 100 degrees centigrade.

    7. At an atmospheric pressure of 100 kPa cavitation occurs at 10 degrees centigrade.

    8. High in the mountains cavitation will occur at a temperature higher than 100 degrees centigrade.

    16.6.4. Calc.: Porosity

    If the porosity of sand is 42% before cutting (in situ) and 50% after cutting, what is the volume increase of the sand (grains + pores) as a fraction?

    \(\ \varepsilon=\frac{\mathrm{n}_{\max }-\mathrm{n}_{\mathrm{i}}}{\mathrm{1}-\mathrm{n}_{\max }}=\frac{\mathrm{0 .5}-\mathrm{0 . 4 2}}{\mathrm{1 - 0 .5}}=\mathrm{0 .1 6}\)

    16.6.5. Calc. : Density

    Assume a water density of 1.000 ton/m3 and a quarts density of 2.650 ton/m3 . What is the density of a sand with a porosity of 40% (saturated sand)?

    \(\ \rho_{\mathrm{s}}=(1-\mathrm{n}) \cdot \rho_{\mathrm{q}}+\mathrm{n} \cdot \rho_{\mathrm{w}}=(1-\mathrm{0 .4}) \cdot \mathrm{2 .6 5}+\mathrm{0 .4} \cdot \mathrm{1 . 0 0}=\mathrm{1 .9 9}\quad (\mathrm{ton/m^3})\)

    16.6.6. Calc.: Permeability

    What is the value of the mean permeability km as used in the cutting equations for water saturated sand? Consider a sand with the following properties:

    Initial permeability: ki=0.0001 m/sec

    Maximum permeability: kmax=0.0004 m/sec

    \(\ \mathrm{k}_{\mathrm{m}}=\frac{\mathrm{k}_{\mathrm{i}}+\mathrm{k}_{\mathrm{max}}}{2}=\frac{\mathrm{0 . 0 0 0 1 + 0 . 0 0 0 4}}{2}=\mathrm{0 .0 0 0 2 5} \quad(\mathrm{m} / \mathrm{sec})\)

    16.6.7. Calc.: Dilatancy

    What is the value of the dilatancy ε?

    Consider a sand with the following properties:

    initial porosity: ni=40%

    Maximum porosity: nmax=50%

    \(\ \varepsilon=\frac{\mathrm{n}_{\mathrm{max}}-\mathrm{n}_{\mathrm{i}}}{1-\mathrm{n}_{\mathrm{max}}}=\frac{\mathrm{0 .5 - 0 .4}}{\mathrm{1 - 0 .5}}=\mathrm{0 .2 0}\quad (-)\)

    16.6.8. Calc.: Transition Velocity

    Give the equation for the transition velocity from non-cavitating cutting to cavitating cutting.

    \(\ \mathrm{v}_{\mathrm{c}}=\frac{\mathrm{d}_{1} \cdot(\mathrm{z}+\mathrm{1} \mathrm{0}) \cdot \mathrm{k}_{\mathrm{m}}}{\mathrm{c}_{\mathrm{1}} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{\varepsilon}}\)

    16.6.9. Calc.: Cutting Forces & Specific Energy

    Consider a sand with the following properties:
    Angle of internal friction: φ=36o
    Angle of external friction: δ=24o
    Initial permeability: ki=0.00005 m/sec
    Maximum permeability: kmax=0.00025 m/sec
    initial porosity: ni=42%
    Maximum porosity: nmax=50%

    The cutting blade has the following properties:
    The cutting angle: α=60o
    The shear angle: β=20o
    The blade height: hb=0.2 m
    The thickness of the layer to be cut: hi=0.1 m
    The width of the blade: w=1 m

    Coefficients for the cutting equations:
    The coefficient for non-cavitating cutting: c1=0.35
    The coefficient for cavitating cutting: d1=4.5

    General constants:
    The density of water: ρw=1.025 tons/m3 
    The gravitational constant: g=9.81 m/sec2

    A: What are the horizontal cutting forces at 0 m, 15 m and 30 m water depth for the non-cavitational cutting process at a cutting velocity of 0.5 m/s?

    \(\ \begin{array}{left} \mathrm{F}_{\mathrm{h}} &=\frac{\mathrm{c}_{1} \cdot \rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{v}_{\mathrm{c}} \cdot \mathrm{h}_{\mathrm{i}}^{2} \cdot \varepsilon \cdot \mathrm{w}}{\mathrm{k}_{\mathrm{m}}}=\frac{\mathrm{0 .3 5} \cdot \mathrm{1 .0 2 5} \cdot \mathrm{9 .8 1 \cdot 0 .} \mathrm{1}^{2} \cdot \mathrm{0 .1 6} \cdot \mathrm{1}}{\mathrm{0 .0 0 0 1 5}} \cdot \mathrm{v}_{\mathrm{c}} \\ &=\mathrm{3 7 .5 4} \cdot \mathrm{v}_{\mathrm{c}}=\mathrm{3 7 .5 4} \cdot \mathrm{0 .5}=\mathrm{1 8 .7 7} \end{array}\quad(\mathrm{kN})\)

    The non-cavitational force does not depend on the water depth.

    B: What are the horizontal cutting forces at 0 m, 15 m and 30 m water depths for the cavitational cutting process?

    \(\ \begin{array}{left} \mathrm{F}_{\mathrm{h}} &=\mathrm{d}_{1} \cdot \rho_{\mathrm{w}} \cdot \mathrm{g} \cdot(\mathrm{z}+\mathrm{1 0}) \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}=\mathrm{4 .5} \cdot \mathrm{1 .0 2 5} \cdot \mathrm{9 .8 1} \cdot \mathrm{0 .1} \cdot \mathrm{1} \cdot(\mathrm{z}+\mathrm{1 0}) \\ &=\mathrm{4 .5 2} \cdot(\mathrm{z}+\mathrm{1 0}) \end{array}\quad(\mathrm{kN})\)

    This gives for 0 m water depth a force of 45.2 kN, for 15 m water depth a force of 113 kN and for 30 m water depth a force of 180.8 kN.

    C: At which velocities are the transitions between the cavitational and the non-cavitational cutting process at 0 m, 15 m and 30 m water depths?

    \(\ \mathrm{37.54 \cdot v_{c}=4.52 \cdot(z+10) \quad\Rightarrow \quad v_{c}=0.12 \cdot(z+10)\quad(m / s e c)}\)

    This gives for 0 m water depth a transition velocity of 1.2 m/sec, for 15 m water depth a transition velocity of 3.0 m/sec and for 30 m water depth a transition velocity of 4.8 m/sec.

    D: What is the specific energy at 0 m, 15 m and 30 m water depth at a cutting velocity of 1 m/s?

    \(\ \mathrm{\mathrm{E}_{\mathrm{sp}}=\frac{\mathrm{F}_{\mathrm{h}} \cdot \mathrm{v}_{\mathrm{c}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \mathrm{v}_{\mathrm{c}}}=\frac{\mathrm{F}_{\mathrm{h}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}}=\frac{\mathrm{F}_{\mathrm{h}}}{\mathrm{0 . 1} \cdot \mathrm{1}}=\mathrm{1 0} \cdot \mathrm{F}_{\mathrm{h}}}\quad (\mathrm{kPa})\)

    At a cutting velocity of 1 m/sec the cutting process is non-cavitational at all 3 water depths, so in all 3 cases the specific energy is 375.4 kPa.

    E: What is the specific energy at a water depth of 0 m, 15 m and 30 m at a cutting velocity of 2 m/s?

    At a cutting velocity of 2 m/sec, the cutting process at 0 m water depth is cavitating, giving a specific energy of 452 kPa. At a water depth of 15 m the cutting process is non-cavitational giving a specific energy of 750.8 kPa. At a water depth of 30 m the cutting process is also non-cavitational giving a specific energy of 750.8 kPa.

    Screen Shot 2020-08-28 at 4.09.55 PM.png
    Figure 16-1: The horizontal cutting forces.
    Screen Shot 2020-08-28 at 4.10.39 PM.png
    Figure 16-2: The specific energy.

    F: Determine the pore pressure at the centre of the shear plane using the parallel resistor method for a cutting velocity of 0.5 m/sec.

    First the 4 lengths have to be determined:

    \(\ \begin{array}{left} \mathrm{s_{1}=\left(L_{\max }-L\right) \cdot\left(\frac{\pi}{2}+\frac{\pi}{2}-(\alpha+\beta)\right)+\frac{h_{b}}{\sin (\alpha)}=0.486}\\
    \mathrm{s_{2}=0.8 \cdot L \cdot(\alpha+\beta)=0.163}\\
    \mathrm{s_{3}=0.8 \cdot L \cdot(\pi-\beta)=0.327}\\
    \mathrm{s_{4}=\left(L_{\max }-L\right) \cdot(\pi+\beta)+0.9 \cdot h_{\mathrm{i}} \cdot\left(\frac{h_{\mathrm{i}}}{h_{\mathrm{b}}}\right)^{0.5} \cdot(1.85 \cdot \alpha)^{2} \cdot\left(\frac{\mathrm{k}_{\mathrm{i}}}{\mathrm{k}_{\mathrm{max}}}\right)^{0.4}=0.636}\end{array}\)

    Secondly the total resistance has to be determined:

    \(\ \mathrm{\frac{1}{R_{t}}=\left(\frac{s_{1}}{k_{\max }}\right)+\left(\frac{s_{2}}{k_{\max }}\right)+\left(\frac{s_{3}}{k_{i}}\right)+\left(\frac{s_{4}}{k_{i}}\right) \Rightarrow R_{t}=439}\)

    \(\ \mathrm{\Delta \mathrm{p}=\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \varepsilon \cdot \mathrm{v}_{\mathrm{c}} \cdot \sin (\beta) \cdot \mathrm{R}_{\mathrm{t}}=120.8(\mathrm{k} \mathrm{P} \mathrm{a})}\)

    Below which water depth will we have cavitation at this point?

    An absolute pressure of 120.8 kPa is reached at a water depth of 2.07 m, so the point on the shear plane considered will cavitated for water depths below 2.07 m.

    Screen Shot 2020-08-28 at 4.33.50 PM.png
    Figure 16-3: The pore pressures on blade and shear plane.

    16.6.10. Calc.: Cutting Forces & Specific Energy

    Consider a sand with the following properties:
    Angle of internal friction: φ=36o
    Angle of external friction: δ=24o
    Initial permeability: ki=0.000025 m/sec
    Maximum permeability: kmax=0.000125 m/sec
    initial porosity: ni=42%
    Maximum porosity: nmax=50%

    The cutting blade has the following properties:
    The cutting angle: α=60o
    The shear angle: β=20o
    The blade height: hb=0.2 m
    The thickness of the layer to be cut: hi=0.1 m
    The width of the blade: w=1 m

    Coefficients for the cutting equations:
    The coefficient for non-cavitating cutting: c1=0.45
    The coefficient for cavitating cutting: d1=5.5

    General constants:
    The density of water: ρw=1.025 tons/m3 
    The gravitational constant: g=9.81 m/sec2

    A: What are the horizontal cutting forces at 0 m, 10 m and 20 m water depth for the non-cavitational cutting process at a cutting velocity of 1.5 m/s?

    \(\ \begin{array}{left} \mathrm{F}_{\mathrm{h}} &=\frac{\mathrm{c}_{\mathrm{1}} \cdot \mathrm{\rho}_{\mathrm{w}} \cdot \mathrm{g} \cdot \mathrm{v}_{\mathrm{c}} \cdot \mathrm{h}_{\mathrm{i}}^{2} \cdot \mathrm{\varepsilon} \cdot \mathrm{w}}{\mathrm{k}_{\mathrm{m}}}=\frac{\mathrm{0} . \mathrm{45} \cdot \mathrm{1 . 0 2 5} \cdot \mathrm{9 .8 1} \cdot \mathrm{0 . 1}^{2} \cdot \mathrm{0 . 1 6} \cdot \mathrm{1}}{\mathrm{0.} \mathrm{0 0 0 0 7 5}} \cdot \mathrm{v}_{\mathrm{c}} \\ &=\mathrm{96 .5 3} \cdot \mathrm{v}_{\mathrm{c}}=\mathrm{9 6 .5 3} \cdot \mathrm{1 .5}=\mathrm{1 4 4 .8} \end{array}\quad(\mathrm{kN})\)

    The non-cavitational force does not depend on the water depth.

    B: What are the horizontal cutting forces at 0 m, 10 m and 20 m water depths for the cavitational cutting process?

    \(\ \begin{array}{left} \mathrm{F}_{\mathrm{h}} &=\mathrm{d}_{1} \cdot \rho_{\mathrm{w}} \cdot \mathrm{g} \cdot(\mathrm{z}+\mathrm{1 0}) \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}=\mathrm{5.5}  \cdot \mathrm{1 .0 2 5} \cdot \mathrm{9 .8 1} \cdot \mathrm{0 .1} \cdot \mathrm{1} \cdot(\mathrm{z}+\mathrm{1 0}) \\ &=\mathrm{5 .5 3} \cdot(\mathrm{z}+\mathrm{1 0}) \end{array}\quad (\mathrm{kN})\)

    This gives for 0 m water depth a force of 55.3 kN, for 10 m water depth a force of 110.6 kN and for 20 m water depth a force of 165.9 kN.

    C: At which velocities are the transitions between the cavitational and the non-cavitational cutting process at 0 m, 10 m and 20 m water depths?

    \(\ \mathrm{96.53\cdot v_c = 5.53 \cdot (z+10) \quad\Rightarrow \quad v_{c}=0.0573 \cdot(z+10)(m / s e c)}\)

    This gives for 0 m water depth a transition velocity of 0.573 m/sec, for 10 m water depth a transition velocity of 1.146 m/sec and for 20 m water depth a transition velocity of 1.719 m/sec.

    D: What is the specific energy at 0 m, 10 m and 20 m water depth at a cutting velocity of 1 m/s?

    \(\ \mathrm{E}_{\mathrm{sp}}=\frac{\mathrm{F}_{\mathrm{h}} \cdot \mathrm{v}_{\mathrm{c}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \mathrm{v}_{\mathrm{c}}}=\frac{\mathrm{F}_{\mathrm{h}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}}=\frac{\mathrm{F}_{\mathrm{h}}}{\mathrm{0 .1} \cdot \mathrm{1}}=\mathrm{1 0} \cdot \mathrm{F}_{\mathrm{h}} \quad (\mathrm{kPa})\)

    At a cutting velocity of 1 m/sec the cutting process is non-cavitational at all z=0 m and non-cavitational at z=10 m and z=20 m, so in the specific energy is 553 kPa at z=0 m and 965.3 kPa at z=10 m and z=20 m.

    E: What is the specific energy at a water depth of 0 m, 10 m and 20 m at a cutting velocity of 2 m/s?

    At a cutting velocity of 2 m/sec, the cutting process at 0 m water depth is cavitating, giving a specific energy of 553 kPa. At a water depth of 10 m the cutting process is cavitational giving a specific energy of 1106 kPa. At a water depth of 20 m the cutting process is also cavitational giving a specific energy of 1659 kPa.

    F: Determine the pore pressure at the centre of the shear plane using the parallel resistor method for a cutting velocity of 1.5 m/sec.

    First the 4 lengths have to be determined:

    \(\ \begin{array}{left}\mathrm{s}_{1}=\left(\mathrm{L}_{\max }-\mathrm{L}\right) \cdot\left(\frac{\pi}{2}+\frac{\pi}{2}-(\alpha+\beta)\right)+\frac{\mathrm{h}_{\mathrm{b}}}{\sin (\alpha)}=\mathrm{0 .4 8 6}\\
    \mathrm{s_{2}=0.8 \cdot L \cdot(\alpha+\beta)=0.163}\\
    \mathrm{s_{3}=0.8 \cdot L \cdot(\pi-\beta)=0.327}\\
    \mathrm{s_{4}=\left(L_{\max }-L\right) \cdot(\pi+\beta)+0.9 \cdot h_{\mathrm{i}} \cdot\left(\frac{h_{\mathrm{i}}}{h_{\mathrm{b}}}\right)^{0.5} \cdot(1.85 \cdot \alpha)^{2} \cdot\left(\frac{\mathrm{k}_{\mathrm{i}}}{k_{\max }}\right)^{0.4}=0.636}\end{array}\)

    Secondly the total resistance has to be determined:

    \(\ \mathrm{\frac{1}{\mathrm{R}_{t}}=\left(\frac{s_{1}}{k_{\max }}\right)+\left(\frac{s_{2}}{k_{\max }}\right)+\left(\frac{s_{3}}{k_{i}}\right)+\left(\frac{s_{4}}{k_{i}}\right) \Rightarrow R_{t}=878}\)

    \(\ \Delta \mathrm{p}=\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot \varepsilon \cdot \mathrm{v}_{\mathrm{c}} \cdot \sin (\beta) \cdot \mathrm{R}_{\mathrm{t}}=72 \mathrm{5}(\mathrm{k} \mathrm{P} \mathrm{a})\)

    Below which water depth will we have cavitation at this point?

    An absolute hydrostatic pressure of 725 kPa is reached at a water depth of 62.15 m, so the point on the shear plane considered will cavitated for water depths below 62.15 m.

    Screen Shot 2020-08-28 at 5.34.07 PM.png
    Figure 16-4: The horizontal cutting forces.
    Screen Shot 2020-08-28 at 5.34.45 PM.png
    Figure 16-5: The specific energy.
    Screen Shot 2020-08-28 at 5.35.41 PM.png
    Figure 16-6: The pore pressures on blade and shear plane.

    This page titled 16.6: Chapter 6- Water Saturated 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.