# 7.4: The Flow Type


# 7.4.1. The Forces

The most common failure mechanism in clay is the Flow Type as is shown in Figure 7-18, which will be considered first. The Curling Type and the Tear Type may occur under special circumstances and will be derived from the equations of the Flow Type.

Figure 7-19 illustrates the forces on the layer of soil cut. The forces shown are valid in general. The forces acting on this layer are:

1. A normal force acting on the shear surface N1 resulting from the effective grain stresses.

2. A shear force as a result of pure cohesion $$\ \tau_{\mathrm{c}}$$. This force can be calculated by multiplying the cohesion c/cohesive shear strength $$\ \tau_{\mathrm{c}}$$ with the area of the shear plane.

3. A force normal to the blade N2 resulting from the effective grain stresses.

4. A shear force as a result of pure adhesion between the soil and the blade $$\ \tau_{\mathrm{a}}$$. This force can be calculated by multiplying the adhesion a (adhesive shear strength $$\ \tau_{\mathrm{a}}$$) of the soil with the contact area between the soil and the blade.

The forces acting on a straight blade when cutting soil, can be distinguished as:

1. A force normal to the blade N2 resulting from the effective grain stresses.

2. A shear force as a result of pure adhesion between the soil and the blade $$\ \tau_{\mathrm{a}}$$. This force can be calculated by multiplying the adhesive shear strength $$\ \tau_{\mathrm{a}}$$ of the soil with the contact area between the soil and the blade.

These forces are shown in Figure 7-20.

Pure clay under undrained conditions follows the φ=0 concept, meaning that effectively there is no internal friction and thus there is also no external friction. Under drained conditions clay will have some internal friction, although smaller than sand. The reason for this is the very low permeability of the clay. If the clay is compressed with a high strain rate, the water in the pores cannot flow away resulting in the pore water carrying the extra pressure, the grain stresses do not change. If the grain stresses do not change, the shear stresses according to Coulomb friction do not change and effectively there is no relation between the extra normal stresses and the shear stresses, so apparently φ=0. At very low strain rates the pore water can flow out and the grains have to carry the extra normal stresses, resulting in extra shear stresses. During the cutting of clay, the strain rates, deformation rates, are so big that the internal and external friction angles can be considered to be zero. The adhesive and cohesive forces play a dominant role, so that gravity and inertia can be neglected.

The horizontal equilibrium of forces:

$\ \mathrm{\sum F_{h}=N_{1} \cdot \sin (\beta)+C \cdot \cos (\beta)-A \cdot \cos (\alpha)-N_{2} \cdot \sin (\alpha)=0}\tag{7-37}$

The vertical equilibrium of forces:

$\ \mathrm{\sum F_{v}=-N_{1} \cdot \cos (\beta)+C \cdot \sin (\beta)+A \cdot \sin (\alpha)-N_{2} \cdot \cos (\alpha)=0}\tag{7-38}$

The force K1 on the shear plane is now:

$\ \mathrm{N}_{1}=\frac{-\mathrm{C} \cdot \cos (\alpha+\beta)+\mathrm{A}}{\sin (\alpha+\beta)}\tag{7-39}$

The force K2 on the blade is now:

$\ \mathrm{N}_{2}=\frac{\mathrm{C}-\mathrm{A} \cdot \cos (\alpha+\beta)}{\sin (\alpha+\beta)}\tag{7-40}$

From equation (7-40) the forces on the blade can be derived. On the blade a force component in the direction of cutting velocity Fh and a force perpendicular to this direction Fv can be distinguished.

$\ \mathrm{F}_{\mathrm{h}}=\mathrm{N}_{\mathrm{2}} \cdot \sin (\alpha)+\mathrm{A} \cdot \cos (\alpha)\tag{7-41}$

$\ \mathrm{F}_{v}=\mathrm{N}_{\mathrm{2}} \cdot \cos (\alpha)-\mathrm{A} \cdot \sin (\alpha)\tag{7-42}$

Since λc and λa are almost identical, an average value λs is used in the following equations. With the relations for the cohesive force C, the adhesive force and the adhesion/cohesion ratio (the ac ratio r):

$\ \mathrm{C=\frac{\lambda_{s} \cdot c \cdot h_{i} \cdot w}{\sin (\beta)}}\tag{7-43}$

$\ \mathrm{A}=\frac{\lambda_{\mathrm{s}} \cdot \mathrm{a} \cdot \mathrm{h}_{\mathrm{b}} \cdot \mathrm{w}}{\sin (\alpha)}\tag{7-44}$

$\ \mathrm{r}=\frac{\mathrm{a} \cdot \mathrm{h}_{\mathrm{b}}}{\mathrm{c} \cdot \mathrm{h}_{\mathrm{i}}}\tag{7-45}$

The horizontal Fh and vertical Fv cutting forces can be determined according to:

$\ \begin{array}{left} \mathrm{F_{h}}&=\mathrm{\frac{C \cdot \sin (\alpha)+A \cdot \sin (\beta)}{\sin (\alpha+\beta)}=\frac{\frac{\lambda_{s} \cdot c \cdot h_{i} \cdot w}{\sin (\beta)} \cdot \sin (\alpha)+\frac{\lambda_{s} \cdot a \cdot h_{b} \cdot w}{\sin (\alpha)} \cdot \sin (\beta)}{\sin (\alpha+\beta)}}\\ &=\mathrm{\lambda_{s} \cdot c \cdot h_{i} \cdot w \cdot \frac{\frac{\sin (\alpha)}{\sin (\beta)}+r \cdot \frac{\sin (\beta)}{\sin (\alpha)}}{\sin (\alpha+\beta)}}\end{array}\tag{7-46}$

$\ \begin{array}{left}\mathrm{F}_{v}&=\mathrm{\frac{C \cdot \cos (\alpha)-A \cdot \cos (\beta)}{\sin (\alpha+\beta)}=\frac{\frac{\lambda_{s} \cdot c \cdot h_{i} \cdot w}{\sin (\beta)} \cdot \cos (\alpha)-\frac{\lambda_{s} \cdot a \cdot h_{b} \cdot w}{\sin (\alpha)} \cdot \cos (\beta)}{\sin (\alpha+\beta)}}\\ &=\lambda_{\mathrm{s}} \cdot \mathrm{c \cdot h _ { \mathrm { i } } \cdot \mathrm { w } \cdot \frac { \frac { \operatorname { cos } ( \alpha ) } { \operatorname { sin } ( \beta ) } - \mathrm { r } \cdot \frac { \operatorname { cos } ( \beta ) } { \operatorname { sin } ( \alpha ) } } { \operatorname { sin } ( \alpha + \beta ) }}\end{array}\tag{7-47}$

The normal force on the shear plane is now equal to the force K1 as is used in sand cutting, because the internal friction angle φ is zero:

$\ \mathrm{N}_{1}=\frac{-\mathrm{C} \cdot \cos (\alpha+\beta)+\mathrm{A}}{\sin (\alpha+\beta)}\tag{7-48}$

The normal force on the blade is now equal to the force K2 as is used in sand cutting, because the external friction angle δ is zero:

$\ \mathrm{N}_{2}=\frac{\mathrm{C}-\mathrm{A} \cdot \cos (\alpha+\beta)}{\sin (\alpha+\beta)}\tag{7-49}$

Equations (7-48) and (7-49) show that both the normal force on the shear plane N1 and the normal force on the blade N2 may become negative. This depends on the ac ratio between the adhesive and the cohesive forces and on the blade angle α and shear angle β. A negative normal force on the blade will result in the Curling Type of cutting mechanism, while a negative normal force on the shear plane will result in the Tear Type of cutting mechanism. If both normal forces are positive, the Flow Type of cutting mechanism will occur.

# 7.4.2. Finding the Shear Angle

There is one unknown in the equations and that is the shear angle βThis angle has to be known to determine cutting forces, specific energy and power.

$\ \mathrm{F_{h}=\lambda_{s} \cdot c \cdot h_{i} \cdot w \cdot\left(\frac{\frac{\sin (\alpha)}{\sin (\beta)}+r \cdot \frac{\sin (\beta)}{\sin (\alpha)}}{\sin (\alpha+\beta)}\right) \quad\text{ with: }r=\frac{a \cdot h_{b}}{c \cdot h_{i}}}\tag{7-50}$

Equation (7-50) for the horizontal cutting force Fh can be rewritten as:

$\ \mathrm{F}_{\mathrm{h}}=\lambda_{\mathrm{s}} \cdot \mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot\left(\frac{\sin ^{2}(\alpha)+\mathrm{r} \cdot \sin ^{2}(\beta)}{\sin (\alpha+\beta) \cdot \sin (\beta) \cdot \sin (\alpha)}\right)=\lambda_{\mathrm{s}} \cdot \mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \lambda_{\mathrm{HF}}\tag{7-51}$

Equation (7-47) for the vertical cutting force Fv can be rewritten as:

$\ \mathrm{F}_{\mathrm{v}}=\lambda_{\mathrm{s}} \cdot \mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot\left(\frac{\sin (\mathrm{\alpha}) \cdot \cos (\alpha)-\mathrm{r} \cdot \sin (\beta) \cdot \cos (\beta)}{\sin (\alpha+\beta) \cdot \sin (\beta) \cdot \sin (\alpha)}\right)=\lambda_{\mathrm{s}} \cdot \mathrm{c} \cdot \mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \lambda_{\mathrm{V F}}\tag{7-52}$

The strengthening factor λswhich is not very sensitive for β in the range of cutting velocities vc as applied in dredging, can be determined by:

$\ \begin{array}{left}\lambda_{\mathrm{s}}=\left(1+\frac{\tau_{0}}{\tau_{\mathrm{y}}} \cdot \ln \left(1+\frac{1.4 \cdot \frac{\mathrm{v}_{\mathrm{c}}}{h_{\mathrm{i}}} \cdot \frac{\sin (\alpha)}{\sin (\alpha+\beta)}}{\dot{\varepsilon}_{0}}\right) \right)\\ \text { With }: \tau_{0} / \tau_{\mathrm{y}}=0.1428 \text { and } \dot{\varepsilon}_{0}=0.03\end{array}\tag{7-53}$

The shear angle β is determined by the case where the horizontal cutting force Fh is at a minimum, based on the minimum energy principle (omitting the strengthening factor λs).

\ \begin{aligned} \frac{\partial \mathrm{F}_{\mathrm{h}}}{\partial \beta}&=\frac{\mathrm{2} \cdot \mathrm{r} \cdot \sin ^{2}(\beta) \cdot \cos (\beta) \cdot \sin (\alpha+\beta) \cdot \sin (\alpha)}{\sin ^{2}(\alpha+\beta) \cdot \sin ^{2}(\alpha) \cdot \sin ^{2}(\beta)} & \\ &+\frac{-\sin (\alpha) \cdot \sin (\alpha+2 \cdot \beta) \cdot\left(\sin ^{2}(\alpha)+\mathrm{r} \cdot \sin ^{2}(\beta)\right)}{\sin ^{2}(\alpha+\beta) \cdot \sin ^{2}(\alpha) \cdot \sin ^{2}(\beta)}=0 \end{aligned}\tag{7-54}

In the special case where there is no adhesion a=0r=0, the shear angle β is:

$\ \sin (\alpha+2 \cdot \beta)=0\text{ for }\alpha+2 \cdot \beta=\pi \operatorname{giving} \beta=\frac{\pi}{2}-\frac{\alpha}{2}\tag{7-55}$

An approximation equation for β based on curve fitting on equation (7-54) for the range 0.5<r<2 gives:

$\ \begin{array}{left}\beta=1.26 \cdot \mathrm{e}^{(-0.174 \cdot \alpha-0.3148 \cdot \mathrm{r})}\text{ in radians or}\\ \beta=72.2 \cdot \mathrm{e}^{(-0.003 \cdot \alpha-0.3148 \cdot \mathrm{r})}\text{ in degrees}\end{array}\tag{7-56}$

For a clay, with shear strength c=1 kPa, a layer thickness of hi=0.1 m and a blade width of w=1 m Figure 7-21, Figure 7-23 and Figure 7-24 give the values of the shear angle β, the horizontal cutting force Fh and the vertical cutting force Fv for different values of the adhesion/cohesion (ac) ratio and as a function of the blade angle α. The use of the ac ratio makes the graphs independent of individual values of hb and a. In these calculations the strain rate factor λs is set to 1. For different values of the strain rate factor λ, the cohesion c, the blade with and the layer thickness hi, the values found in Figure 7-23 and Figure 7-24 can be multiplied by the corresponding factors.

The horizontal cutting force Fh is at an absolute minimum when:

$\ \alpha+\beta=\frac{\pi}{2}\tag{7-57}$

This is however only useful if the blade angle α can be chosen freely. For a worst case scenario with an ac ratio r=2, meaning a high adhesion, a blade angle α of about 55o is found (see Figure 7-23), which matches blade angles as used in dredging. The fact that this does not give an optimum for weaker clays (clays with less adhesion) is not so relevant.

Figure 7-21, Figure 7-23 and Figure 7-24 show that the shear angle β is decreasing with an increasing blade angle α and an increasing ac ratio r. For practical blade angles between 45 and 60 degrees, the shear angle may vary between 35 and 60-70 degrees, depending on the ac ratio r. The horizontal force first decreases to a minimum with an increasing blade angle, after which it increases. At very large blade angles the horizontal force increases strongly to values that are not reasonable anymore. Nature will find another mechanism with smaller forces, the wedge mechanism, which will be described in Chapter 13: A Wedge in Clay Cutting. The vertical force (positive is downwards directed) is first increasing with an increasing blade angle to a maximum value, after which it is decreasing to very large negative (upwards directed) values at very large blade angles.

Figure 7-22 shows the sum of the blade angle and the shear angle. When this sum is 90 degrees, the minimum of the horizontal force is found. The graph shows clearly that this is the case for a 55 degree blade and an ac ratio r=2.

See Appendix V: Clay Cutting Charts for more and higher resolution charts.

# 7.4.3. Specific Energy

In the dredging industry, the specific cutting energy Esp is described as:

The amount of energy, that has to be added to a volume unit of soil (e.g. clay) to excavate the soil.

The dimension of the specific cutting energy is: kN/m2 or kPa for sand and clay, while for rock often MN/m2 or MPa is used. For the case as described above, cutting with a straight blade with the direction of the cutting velocity vc perpendicular to the blade (edge of the blade), the specific cutting energy Esp is:

$\ \mathrm{E}_{\mathrm{s p}}=\frac{\mathrm{F}_{\mathrm{h}} \cdot \mathrm{v}_{\mathrm{c}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \mathrm{v}_{\mathrm{c}}}=\frac{\mathrm{F}_{\mathrm{h}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w}}\tag{7-58}$

With the following equation for the horizontal cutting force Fh:

$\ \mathrm{F_{h}=\lambda_{s} \cdot c \cdot h_{i} \cdot w \cdot\left(\frac{\sin ^{2}(\alpha)+r \cdot \sin ^{2}(\beta)}{\sin (\alpha+\beta) \cdot \sin (\beta) \cdot \sin (\alpha)}\right)=\lambda_{s} \cdot c \cdot h_{i} \cdot w \cdot \lambda_{H F}}\tag{7-59}$

This gives for the specific cutting energy Esp:

$\ \mathrm{E}_{\mathrm{sp}}=\frac{\mathrm{F}_{\mathrm{h}} \cdot \mathrm{v}_{\mathrm{c}}}{\mathrm{h}_{\mathrm{i}} \cdot \mathrm{w} \cdot \mathrm{v}_{\mathrm{c}}}=\lambda_{\mathrm{s}} \cdot \mathrm{c} \cdot\left(\frac{\mathrm{s i n}^{2}(\alpha)+\mathrm{r} \cdot \sin ^{2}(\beta)}{\sin (\alpha+\beta) \cdot \sin (\beta) \cdot \sin (\alpha)}\right)=\lambda_{\mathrm{s}} \cdot \mathrm{c} \cdot \lambda_{\mathrm{HF}}\tag{7-60}$

The cohesion is half the UCS value, which can be related to the SPT value of the clay by a factor 12, so the cohesion is related by a factor 6 to the SPT value (see Table 7-1), further, the strengthening λ factor will have a value of about 2 at normal cutting velocities of meters per second, this gives:

$\ \lambda_{\mathrm{s}} \cdot \mathrm{c} \approx \mathrm{2} \cdot \mathrm{6} \cdot \mathrm{S P T}=\mathrm{1 2} \cdot \mathrm{S P T}\tag{7-61}$

Now a simplified equation for the specific energy Esp is found by:

$\ \mathrm{E}_{\mathrm{sp}}=\mathrm{1 2} \cdot \mathrm{S P T} \cdot\left(\frac{\sin ^{2}(\alpha)+\mathrm{r} \cdot \sin ^{2}(\beta)}{\sin (\alpha+\beta) \cdot \sin (\beta) \cdot \sin (\alpha)}\right)=12 \cdot \mathrm{SPT} \cdot \lambda_{\mathrm{HF}}\tag{7-62}$

Figure 7-25 shows the specific energy Esp and the production Pc per 100 kW installed cutting power as a function of the SPT value.

 SPT Penetration (blows/ foot) Estimated Consistency UCS (kPa) <2 Very Soft Clay <24 2 - 4 Soft Clay 24 - 48 4 - 8 Medium Clay 48 - 96 8 - 16 Stiff Clay 96 – 192 16 - 32 Very Stiff Clay 192 – 384 >32 Hard Clay >384

See Appendix U: Specific Energy in Clay for more graphs on the specific energy in clay.

This page titled 7.4: The Flow 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.