16.7: Chapter 7- Clay Cutting
- Page ID
- 35239
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)16.7.1. Calc.: Cutting Forces
Consider clay with cohesion of 200 kPa and an adhesion of 50 kPa. The strengthening factor is 2. A blade angle of 55 degrees is used and a blade height of 0.1 m and blade width w=1 m. The layer thickness is 0.1 m. Assume the Flow Type.
What is the ac ratio r?
\(\ \mathrm{r=\frac{a \cdot h_{b}}{c \cdot h_{i}}=\frac{50 \cdot 0.1}{200 \cdot 0.1}=\frac{1}{4}\quad(-)}\)
What is the shear angle?
See Figure 7-21 (Figure 7.20, 1st edition), blade angle α=55 degrees and r=0.25 gives a shear angle β of about 57 degrees.
What are the horizontal and the vertical cutting forces?
Figure 7-23 (Figure 7.22, 1st edition) gives a horizontal cutting force coefficient λHF of about 1.3 and Figure 7-24 (Figure 7.23, first edition) gives a vertical cutting force coefficient λVF of 0.6. This gives for the Flow Type:
\(\ \mathrm{F}_{\mathrm{h}}=\lambda_{\mathrm{s}} \cdot \mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \lambda_{\mathrm{HF}}=\mathrm{2} \cdot \mathrm{2} \mathrm{0} \mathrm{0} \cdot \mathrm{0.1} \cdot \mathrm{1} \cdot \mathrm{1 .3}=\mathrm{5} \mathrm{2} \mathrm{k} \mathrm{N}\)
\(\ \mathrm{F}_{\mathrm{v}}=\lambda_{\mathrm{s}} \cdot \mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \lambda_{\mathrm{V F}}=\mathrm{2} \cdot \mathrm{2} \mathrm{0} \mathrm{0} \cdot \mathrm{0.1} \cdot \mathrm{1} \cdot \mathrm{0.6}=\mathrm{2} \mathrm{4} \mathrm{k} \mathrm{N}\)
16.7.2. Calc.: Cutting Forces & Mechanisms
Consider clay with cohesion of 200 kPa and an adhesion of 10 kPa. The strengthening factor is 2. A blade angle of 55 degrees is used and a blade height of 0.1 m and blade width w=1 m. The layer thickness is 0.1 m. Assume the Flow Type.
What is the ac ratio r?
\(\ \mathrm{r}=\frac{\mathrm{a} \cdot \mathrm{h}_{\mathrm{b}}}{\mathrm{c} \cdot \mathrm{h}_{\mathrm{i}}}=\frac{\mathrm{1 0} \cdot \mathrm{0.1}}{\mathrm{2 0 0} \cdot \mathrm{0 . 1}}=\frac{\mathrm{1}}{\mathrm{2 0}} \quad\mathrm{( - )}\)
What is the shear angle?
See Figure 7-21 (Figure 7.20, 1st edition), blade angle α=55 degrees and r=0.05 gives a shear angle β of about 62 degrees.
What are the horizontal and the vertical cutting forces?
Figure 7-23 (Figure 7.22, 1st edition) gives a horizontal cutting force coefficient λHF of about 1.1 and Figure 7-24 (Figure 7.23, first edition) gives a vertical cutting force coefficient λVF of 0.7. This gives for the Flow Type:
\(\ \mathrm{F_{h}=\lambda_{s} \cdot c \cdot h_{i} \cdot w \cdot \lambda_{H F}=2 \cdot 200 \cdot 0.1 \cdot 1 \cdot 1.1=44 \mathrm{kN}}\)
\(\ \mathrm{F}_{\mathrm{v}}=\lambda_{\mathrm{s}} \cdot \mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \lambda_{\mathrm{V F}}=\mathrm{2} \cdot \mathrm{2 0 0} \cdot \mathrm{0 .1} \cdot \mathrm{1} \cdot \mathrm{0 .7}=\mathrm{2 8} \mathrm{k N}\)
If the tensile strength is -20 kPa, will we have the Tear Type or the Flow Type?
A tensile strength of -20 kPa gives a σT/c ratio of -0.1. With an ac ratio of r=0.05 this ratio should be below -0.5 according to Figure 7-27 (Figure 7-26, 1st edition) for the Flow Type, it is not, so we have the Tear Type.
