Skip to main content
Engineering LibreTexts

9.4: The Curling Type

  • Page ID
    29474
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    When cutting or scraping a very thin layer of rock, the Curling Type may occur. In dredging and mining usually the layer thickness is such that this will not occur, but in drilling practices usually the layer thickness is very small compared with the height of the blade. In the Zijsling (1987) experiments layer thicknesses of 0.15 mm and 0.30 mm were applied with a PDC bit with a height and width of about 10 mm. Under these conditions the Curling Type will occur, which is also named balling. Figure 9-15 shows this type of cutting mechanism.

    Now the question is, what is the effective blade height hb,m? In other words, along which distance will the rock cut be in contact with the blade? To solve this problem an additional condition has to be found. This condition is the equilibrium of moments around the blade tip as is shown in Figure 9-16. The only forces that contribute to the equilibrium of moments are the normal forces N1 and N2 and the pore pressure forces W1 and W2. The acting points of these forces are chosen as fractions of the length of the shear plane λ1 and the blade length λ2.

    The equilibrium of moments around the blade tip is:

    \[\ \left(\mathrm{N}_{\mathrm{1}}-\mathrm{W}_{\mathrm{1}}\right) \cdot \mathrm{R}_{\mathrm{1}}=\left(\mathrm{N}_{\mathrm{2}}-\mathrm{W}_{\mathrm{2}}\right) \cdot \mathrm{R}_{\mathrm{2}}\tag{9-20}\]

    For the acting points the following can be derived:

    \[\ \mathrm{R_{1}=\frac{\lambda_{1} \cdot h_{\mathrm{i}}}{\sin (\beta)}, R_{2}=\frac{\lambda_{2} \cdot h_{\mathrm{b}, \mathrm{m}}}{\sin (\alpha)}}\tag{9-21}\]

    Screen Shot 2020-08-23 at 11.56.13 PM.png
    Figure 9-15: The Curling Type or balling.
    Screen Shot 2020-08-23 at 11.56.34 PM.png
    Figure 9-16: The equilibrium of moments on the layer cut in hyperbaric rock.

    Substituting equations (9-10) and (9-11) into equation (9-20) gives:

    \[\ \begin{array}{\left}\left(\begin{array}{left}\frac{\mathrm{W}_{2} \cdot \sin (\delta)+\mathrm{W}_{1} \cdot \sin (\alpha+\beta+\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\varphi) \\ +\frac{-\mathrm{C} \cdot \cos (\alpha+\beta+\delta)+\mathrm{A} \cdot \cos (\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\varphi)\\-\mathrm{W}_1\end{array}\right) \cdot \frac{\lambda_{1} \cdot \mathrm{h}_{\mathrm{i}}}{\sin (\beta)}\\
    =\left(\begin{array}{left}\frac{\mathrm{W}_{2} \cdot \sin (\alpha+\beta+\varphi)+\mathrm{W}_{1} \cdot \sin (\varphi)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\delta) \\ +\frac{+\mathrm{C} \cdot \cos (\varphi)-\mathrm{A} \cdot \cos (\alpha+\beta+\varphi)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\delta)\\-\mathrm{W_2}\end{array}\right) \cdot \frac{\lambda_{2} \cdot \mathrm{h}_{\mathrm{b}, \mathrm{m}}}{\sin (\alpha)}\end{array}\tag{9-22}\]

    This can be written as a second degree function of the effective or mobilized blade height hb,m:

    \[\ \begin{array}{left}\mathrm{A} \cdot \mathrm{x}^{2}+\mathrm{B} \cdot \mathrm{x}+\mathrm{C}=\mathrm{0}\\
    \mathrm{h}_{\mathrm{b}, \mathrm{m}}=\mathrm{x}=\frac{-\mathrm{B}-\sqrt{\mathrm{B}^{2}-\mathrm{4} \cdot \mathrm{A} \cdot \mathrm{C}}}{\mathrm{2} \cdot \mathrm{A}}\end{array}\tag{9-23}\]

    With:

    \[\  \mathrm{A}= \frac{\lambda_{2} \cdot \mathrm{p}_{2 \mathrm{m}} \cdot \sin (\alpha+\beta+\delta+\varphi)-\lambda_{2} \cdot \mathrm{p}_{\mathrm{2 m}} \cdot \sin (\alpha+\beta+\varphi) \cdot \cos (\delta)}{\sin (\alpha) \cdot \sin (\alpha)} +\frac{+\mathrm{a} \cdot \lambda_{2} \cdot \cos (\alpha+\beta+\varphi) \cdot \cos (\delta)}{\sin (\alpha) \cdot \sin (\alpha)} \tag{9-24}\]

    And:

    \[\ \mathrm{B}= \frac{\lambda_{1} \cdot \mathrm{p}_{2 \mathrm{m}} \cdot \sin (\delta) \cdot \cos (\varphi)-\lambda_{2} \cdot \mathrm{p}_{1 \mathrm{m}} \cdot \cos (\delta) \cdot \sin (\varphi)}{\sin (\alpha) \cdot \sin (\beta)} \cdot \mathrm{h}_{\mathrm{i}} +\frac{-\mathrm{c} \cdot \lambda_{2} \cdot \cos (\delta) \cdot \cos (\varphi)+\mathrm{a} \cdot \lambda_{1} \cdot \cos (\varphi) \cdot \cos (\delta)}{\sin (\alpha) \cdot \sin (\beta)} \cdot \mathrm{h}_{\mathrm{i}} \tag{9-25}\]

    And:

    \[\  \mathrm{C}= \frac{\lambda_{1} \cdot \mathrm{p}_{1 \mathrm{m}} \cdot \sin (\alpha+\beta+\delta) \cdot \cos (\varphi)-\lambda_{1} \cdot \mathrm{p}_{1 \mathrm{m}} \cdot \sin (\alpha+\beta+\delta+\varphi)}{\sin (\beta) \cdot \sin (\beta)} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{h}_{\mathrm{i}} +\frac{-\mathrm{c} \cdot \lambda_{1} \cdot \cos (\alpha+\beta+\delta) \cdot \cos (\varphi)}{\sin (\beta) \cdot \sin (\beta)} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{h}_{\mathrm{i}} \tag{9-26}\]

    If hb,m<hb then the Curling Type will occur, but if hb,m>hthe normal Flow Type will occur.

    \[\ \begin{array}{left}\text{if } \mathrm{h}_{\mathrm{b}, \mathrm{m}}<\mathrm{h}_{\mathrm{b}}\text{ then use }\mathrm{h}_{\mathrm{b}, \mathrm{m}}\\
    \text{if }\mathrm{h}_{\mathrm{b}, \mathrm{m}} \geq \mathrm{h}_{\mathrm{b}}\text{ then use }\mathrm{h}_{\mathrm{b}}\end{array}\tag{9-27}\]

    Now in the case of full cavitation, the adhesion can be neglected and both arms are at 50% of the corresponding length. This simplifies the equations to:

    \[\ \begin{array}{left}\mathrm{A}=\frac{\mathrm{p}_{\mathrm{m}} \cdot \cos (\alpha+\beta+\varphi) \cdot \sin (\delta)}{\sin (\alpha) \cdot \sin (\alpha)}\\
    \mathrm{B}=\frac{-\mathrm{p}_{\mathrm{m}} \cdot \sin (\varphi-\delta)-\mathrm{c} \cdot \cos (\delta) \cdot \cos (\varphi)}{\sin (\alpha) \cdot \sin (\beta)} \cdot \mathrm{h}_{\mathrm{i}}\\
    \mathrm{C}=\frac{-\mathrm{p}_{\mathrm{m}} \cdot \cos (\alpha+\beta+\delta) \cdot \sin (\varphi)-\mathrm{c} \cdot \cos (\alpha+\beta+\delta) \cdot \cos (\varphi)}{\sin (\beta) \cdot \sin (\beta)} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{h}_{\mathrm{i}}\end{array}\tag{9-28}\]

    Introducing the ratio rz between the absolute hydrostatic pressure and the shear strength c:

    \[\ \mathrm{r}_{\mathrm{z}}=\frac{\rho_{\mathrm{w}} \cdot \mathrm{g} \cdot(\mathrm{z}+\mathrm{1 0})}{\mathrm{c}}\tag{9-29}\]

    Gives for the A, B and C:

    \[\ \begin{array}{left}\mathrm{A}=\frac{\mathrm{r_{z}} \cdot \cos (\alpha+\beta+\varphi) \cdot \sin (\delta)}{\sin (\alpha) \cdot \sin (\alpha)}\\
    \mathrm{B}=\frac{-\mathrm{r}_{\mathrm{z}} \cdot \sin (\varphi-\delta)-\cos (\delta) \cdot \cos (\varphi)}{\sin (\alpha) \cdot \sin (\beta)} \cdot \mathrm{h}_{\mathrm{i}}\\
    \mathrm{C}=\frac{-\mathrm{r}_{\mathrm{z}} \cdot \cos (\alpha+\beta+\delta) \cdot \sin (\varphi)-\cos (\alpha+\beta+\delta) \cdot \cos (\varphi)}{\sin (\beta) \cdot \sin (\beta)} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{h}_{\mathrm{i}}\end{array}\tag{9-30}\]

    The term is always negative. The term 4·A·is also always negative. This results in a square root that will always be bigger than |B|. Since the sum of the angles in the arguments of the cosines will always be larger than 90 degrees, the cosines will give a negative result. So will always be negative. This implies that the negative square root gives a positive answer, while the positive square root will give a negative answer. Since the mobilized blade height has to be positive, the negative square root should be used here.

    Finally the horizontal and vertical cutting forces can be written as:

    \[\ \mathrm{F}_{\mathrm{h}}=\lambda_{\mathrm{H} \mathrm{C}} \cdot \mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}\tag{9-31}\]

    \[\ \mathrm{F}_{v}=\lambda_{\mathrm{V C}} \cdot \mathrm{c \cdot h}_{\mathrm{i}} \cdot \mathrm{w}\tag{9-32}\]

    Figure 9-17 and Figure 9-18 show the ratio of the mobilized blade height to the layer thickness hb,m/hi and the shear angle β for a 60 degree blade. From Figure 9-17 it is clear that the Curling Type already occurs at normal hb,m/hi ratios. Especially at small internal friction angles this will be the case. Figure 9-19 and Figure 9-20 show the horizontal and vertical cutting force coefficients, which are not much different from the coefficients of the Flow Type and hb,m/hi=1. . Figure 9-21 shows the Esp/UCS ratio, which is very convenient for production estimation.

    Screen Shot 2020-08-24 at 12.10.55 PM.png
    Figure 9-17: The ratio hb,m/hi for a 60 degree blade.
    Screen Shot 2020-08-24 at 12.11.43 PM.png
    Figure 9-18:The shear angle β for a 60 degree blade
    Screen Shot 2020-08-24 at 12.13.34 PM.png
    Figure 9-19: The horizontal cutting force coefficient λHC for a 60 degree blade.
    Screen Shot 2020-08-24 at 12.14.17 PM.png
    Figure 9-20: The vertical cutting force coefficient λVC for a 60 degree blade. Positive downwards.
    Screen Shot 2020-08-24 at 12.15.10 PM.png
    Figure 9-21: The Esp/UCS ratio, for a 60 degree blade.

    This page titled 9.4: The Curling Type 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.