# 16.7: Chapter 7- Clay Cutting


# 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.

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