5.7: Usage of the Model for Dry Sand
- Page ID
- 33734
To use the model for dry sand, first the dry density ρs of the sand and the internal friction angle φ have to be known. The external friction angle δ is assumed to be 2/3 of the internal friction angle φ. Secondly the geometry of the blade, the cutting angle α, the blade height hb and the blade width w have to be chosen. Thirdly the operational parameters, the layer thickness hi and the cutting velocity vc have to be chosen. Based on the dimensionless inertial effect parameter λi the fraction of the contribution of the inertial force to the total horizontal force can be determined with:
\[\ \mathrm{f}_{\mathrm{i}}=\frac{\mathrm{1}}{1+\mathrm{e}^{-\mathrm{2} \cdot \log \left(\lambda_{\mathrm{i}} / \mathrm{5}\right)}}\tag{5-27}\]
This equation is empirically derived for a 60° blade and a 40° internal friction angle and may differ for other values of the blade angle and the internal friction angle.
\[\ \begin{array}{left} \mathrm{F_{h}=\left(1-f_{i}\right) \cdot \rho_{s} \cdot g \cdot h_{i}^{2} \cdot w \cdot \lambda_{H D}+f_{i} \cdot \rho_{s} \cdot v_{c}^{2} \cdot h_{i} \cdot w \cdot \lambda_{H I}}\\
\mathrm{=\rho_{s} \cdot g \cdot h_{i}^{2} \cdot w \cdot\left(\left(1-f_{i}\right) \cdot \lambda_{H D}+f_{i} \cdot \lambda_{i} \cdot \lambda_{H I}\right)}\end{array}\tag{5-28}\]
\[\ \begin{array}{left}\mathrm{F}_{\mathrm{v}}=\left(\mathrm{1}-\mathrm{f}_{\mathrm{i}}\right) \cdot \rho_{\mathrm{s}} \cdot \mathrm{g} \cdot \mathrm{h}_{\mathrm{i}}^{2} \cdot \mathrm{w} \cdot \lambda_{\mathrm{V D}}+\mathrm{f}_{\mathrm{i}} \cdot \rho_{\mathrm{s}} \cdot \mathrm{v}_{\mathrm{c}}^{2} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \lambda_{\mathrm{VI}}\\
=\rho_{\mathrm{s}} \cdot \mathrm{g} \cdot \mathrm{h}_{\mathrm{i}}^{2} \cdot \mathrm{w} \cdot\left(\left(1-\mathrm{f}_{\mathrm{i}}\right) \cdot \lambda_{\mathrm{V D}}+\mathrm{f}_{\mathrm{i}} \cdot \lambda_{\mathrm{i}} \cdot \lambda_{\mathrm{VI}}\right)\end{array}\tag{5-29}\]
The specific energy is now:
\[\ \mathrm{E}_{\mathrm{sp}}=\rho_{\mathrm{s}} \cdot \mathrm{g} \cdot \mathrm{h}_{\mathrm{i}} \cdot\left(\left(1-\mathrm{f}_{\mathrm{i}}\right) \cdot \lambda_{\mathrm{HD}}+\mathrm{f}_{\mathrm{i}} \cdot \lambda_{\mathrm{i}} \cdot \lambda_{\mathrm{HI}}\right)\tag{5-30}\]
In the case of saturated sand or gravel with a very high permeability (in general coarse gravel), the equations change slightly, since the weight of the soil cut is determined by the submerged weight, while the mass of the soil cut also includes the mass of the pore water. The wet density of saturated sand or gravel is usually close to ρs=2 ton/m3, while the submerged weight is close to (ρs-ρw)·g=10 kN/m3 (a porosity of 40% and a quarts density of ρq=2.65 ton/m3 are assumed). This will double the contribution of the inertial forces as determined by the following dimensionless parameter:
\[\ \lambda_{\mathrm{i}}=\frac{\mathrm{v}_{\mathrm{c}}^{2}}{\mathrm{g} \cdot \mathrm{h}_{\mathrm{i}}} \cdot \frac{\left(\rho_{\mathrm{s}}-\rho_{\mathrm{w}}\right)}{\rho_{\mathrm{s}}} \approx \frac{\mathrm{2} \cdot \mathrm{v}_{\mathrm{c}}^{\mathrm{2}}}{\mathrm{g} \cdot \mathrm{h}_{\mathrm{i}}}\tag{5-31}\]
Using this dimensionless inertial effect parameter λi, the cutting forces can be determined by:
\[\ \mathrm{F}_{\mathrm{h}}=\left(\rho_{\mathrm{s}}-\rho_{\mathrm{w}}\right) \cdot \mathrm{g} \cdot \mathrm{h}_{\mathrm{i}}^{\mathrm{2}} \cdot \mathrm{w} \cdot\left(\left(\mathrm{1}-\mathrm{f}_{\mathrm{i}}\right) \cdot \lambda_{\mathrm{H D}}+\mathrm{f}_{\mathrm{i}} \cdot \lambda_{\mathrm{i}} \cdot \lambda_{\mathrm{HI}}\right)\tag{5-32}\]
\[\ \mathrm{F}_{\mathrm{v}}=\left(\rho_{\mathrm{s}}-\rho_{\mathrm{w}}\right) \cdot \mathrm{g} \cdot \mathrm{h}_{\mathrm{i}}^{\mathrm{2}} \cdot \mathrm{w} \cdot\left(\left(1-\mathrm{f}_{\mathrm{i}}\right) \cdot \lambda_{\mathrm{V D}}+\mathrm{f}_{\mathrm{i}} \cdot \lambda_{\mathrm{i}} \cdot \lambda_{\mathrm{V I}}\right)\tag{5-33}\]
The specific energy is now:
\[\ \mathrm{E}_{\mathrm{sp}}=\left(\rho_{\mathrm{s}}-\rho_{\mathrm{w}}\right) \cdot \mathrm{g} \cdot \mathrm{h}_{\mathrm{i}} \cdot\left(\left(1-\mathrm{f}_{\mathrm{i}}\right) \cdot \lambda_{\mathrm{HD}}+\mathrm{f}_{\mathrm{i}} \cdot \lambda_{\mathrm{i}} \cdot \lambda_{\mathrm{HI}}\right)\tag{5-34}\]
Under water at high cutting velocities there may also be a drag force which has not been taken into account here.
The horizontal cutting force coefficients λHD and λHI can be found in Figure 5-11 and Figure 5-16. The vertical cutting force coefficients λVD and λVI can be found in Figure 5-12 and Figure 5-17.
The cutting forces calculated are for a plane strain 2D cutting process, so 3D side effects are not included.
