Skip to main content
Engineering LibreTexts

8.10: Example

  • Page ID
    34847
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    In this chapter and in Appendix W many graphs are given with a red or green rectangle giving the value of the different parameters for an α=60o blade and an internal friction angle φ=20o. The external friction angle is assumed to be δ=2/3·φ. Most graphs are dimensionless, but Figure 8-49, Figure 8-50, Figure 8-51 and Figure 8-52 are based on a compressive strength UCS=100 MPa, a blade width w=0.1 m and a layer thickness hi=0.1 m.

    8.10.1. Step 1: Brittle Shear

    The the shear angle β=43.3o, horizontal force coefficient λHF=1.912, the verticle force coefficient λVF=0.572 and the Esp/UCS ratio=0.669. This gives a horizontal force Fh=0.669 MN, a vertical force Fv=0.200 MN and a specific energy of Esp=66.9 MPa. These values are peak values, but for comparison reasons these values will be used.

    8.10.2. Step 2: The Transition Brittle Shear/Brittle Tensile

    The transitions brittle shear/brittle tensile occur for UCS/BTS=4.985 and UCS/BTS=11.75. This can also be written as BTS=0.085·UCS and BTS=0.2·UCS, so BTS=8.5 MPa and BTS=20 MPaBTS or UTS are considered positive numbers, while tensile strength is considered to be negative in this book.
    This means that below the Lower Limit BTS=8.5 MPa brittle shear failure cannot exist, so there is always brittle tensile failure. Above the Upper Limit BTS=20 MPa brittle tensile failure cannot exist, so there is always brittle shear failure. In between, both can exist , even at the same time, according to Figure 8-19 the Chip Type. Figure 8-49 and Figure 8-50 show the horizontal and vertical cutting forces as a function of the tensile strength for the case considered. Interpolation curves are shown, simulating the simultaneous occurrence of brittle shear and brittle tensile failure according to the Chip Type. For this interpolation the following method is used:

    First define a factor according to:

    \[\ \begin{array}{left}\text{BTS} < \text{LowerLimit BTS} &\Rightarrow \mathrm{f}=\mathrm{1}\\
    \text{BTS} > \text{UpperLimit BTS}&\Rightarrow \mathrm{f}=\mathrm{0}\\
    \text{BTS} > \mathrm{Lowerlimit BTS}\\
    &\Rightarrow \mathrm{f}=\left(\frac{\text { Upperlimit BTS - BTS }}{\text { Upperlimit BTS - LowerLimit BTS }}\right)^\mathrm{p}\\ \text{BTS} < \text{UpperLimit BTS}\end{array}\tag{8-140}\]

    Now the resulting cutting forces can be determined with:

    \[\ \begin{array}{left}\mathrm{F}_{\mathrm{h}}=\mathrm{F}_{\mathrm{h}, \text { TearType }} \cdot \mathrm{f}+\mathrm{F}_{\mathrm{h}, \text { ShearType }} \cdot(\mathrm{1}-\mathrm{f})\\
    \mathrm{F}_{\mathrm{v}}=\mathrm{F}_{\mathrm{v}, \text { TearType }} \cdot \mathrm{f}+\mathrm{F}_{\mathrm{v}, \text { ShearType }} \cdot(1-\mathrm{f})\end{array}\tag{8-141}\]

    The power used in Figure 8-49 and Figure 8-50 is p=1, a linear transition from tensile failure to shear failure, the Chip Type.

    8.10.3. Step 3: Applying Tensile Strengths of -5 MPa, -10 MPa and -25 MPa

    From Figure 8-49 and Figure 8-50 the horizontal and vertical peak forces can be determined. They are given in the following table. Between brackets estimated average values, based on a 60% ratio between average and peak values.

    Table 8-3: Resulting forces and specific energy.

    Tensile Strength

    Fh (N)

    Fv (MN)

    Esp (MPa)

    Esp/UCS (%)

    -5 MPa

    0.207 (0.124)

    0.062 (0.037)

    20.7 (12.4)

    20.7 (12.4)

    -10 MPa

    0.446 (0.268)

    0.134 (0.080)

    44.6 (26.8)

    44.6 (26.8)

    -25 MPa

    0.669 (0.401)

    0.200 (0.120)

    66.9 (40.1)

    66.9 (40.1)

    The UCS/BTS ratio of 10 matches the findings of Roxborough (1987) giving a specific energy of about 25% of the UCS value.

    \[\ \mathrm{E}_{\mathrm{sp}}=\mathrm{0 .2 5} \cdot \mathrm{U . C . S .}+\mathrm{0 .1 1}\tag{8-142}\]


    8.10: Example is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

    • Was this article helpful?