# 6.4: The Equilibrium of Forces

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

Figure 6-4 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.

2. A shear force S1 as a result of internal friction N1·tan(φ).

3. A force W1 as a result of water under pressure in the shear zone.

4. A force normal to the blade N2.

5. A shear force S2 as a result of the soil/steel friction N2·tan(δ).

6. A force Was a result of water under pressure on the blade.

The normal force N1 and the shear force S1 can be combined to a resulting grain force K1.

$\ \mathrm{K}_{1}=\sqrt{\mathrm{N}_{1}^{2}+\mathrm{S}_{1}^{2}}\tag{6-8}$

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

1. A force normal to the blade N2.

2. A shear force S2 as a result of the soil/steel friction N2·tan(δ).

3. A force W2 as a result of water under pressure on the blade.

These forces are shown in Figure 6-5. If the forces N2 and S2 are combined to a resulting force K2 and the adhesive force and the water under pressures are known, then the resulting force K2 is the unknown force on the blade. By taking the horizontal and vertical equilibrium of forces an expression for the force K2 on the blade can be derived.

$\ \mathrm{K}_{2}=\sqrt{\mathrm{N}_{2}^{2}+\mathrm{S}_{2}^{2}}\tag{6-9}$

Water saturated sand is also cohesionless, although in literature the phenomenon of water under pressures is sometimes referred to as apparent cohesion. It should be stated however that the water under pressures have nothing to do with cohesion or shear strength. The shear stresses still follow the rules of Coulomb friction. Due to dilatation, a volume increase of the pore volume caused by shear stresses, under pressures develop around the shear plane as described by Miedema (1987 September), resulting in a strong increase of the grain stresses. Because the permeability of the flow of water through the pores is very low, the stresses and thus the forces are dominated by the phenomenon of dilatancy and gravitation, inertia, adhesion and cohesion can be neglected.

The horizontal equilibrium of forces is:

$\ \mathrm{\sum F_{h}=K_{1} \cdot \sin (\beta+\varphi)-W_{1} \cdot \sin (\beta)+W_{2} \cdot \sin (\alpha)-K_{2} \cdot \sin (\alpha+\delta)=0}\tag{6-10}$

The vertical equilibrium of forces is:

$\ \mathrm{\sum F_{v}=-K_{1} \cdot \cos (\beta+\varphi)+W_{1} \cdot \cos (\beta)+W_{2} \cdot \cos (\alpha)-K_{2} \cdot \cos (\alpha+\delta)=0}\tag{6-11}$

The force Kon the shear plane is now:

$\ \mathrm{K}_{1}=\frac{\mathrm{W}_{2} \cdot \sin (\delta)+\mathrm{W}_{1} \cdot \sin (\alpha+\beta+\delta)}{\sin (\alpha+\beta+\delta+\varphi)}\tag{6-12}$

The force K2 on the blade is now:

$\ \mathrm{K}_{2}=\frac{\mathrm{W}_{2} \cdot \sin (\alpha+\beta+\varphi)+\mathrm{W}_{1} \cdot \sin (\varphi)}{\sin (\alpha+\beta+\delta+\varphi)}\tag{6-13}$

From equation (6-13) 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.

$\ \mathrm{F_{h}=-W_{2} \cdot \sin (\alpha)+K_{2} \cdot \sin (\alpha+\delta)}\tag{6-14}$

$\ \mathrm{F}_{v}=-\mathrm{W}_{2} \cdot \cos (\alpha)+\mathrm{K}_{2} \cdot \cos (\alpha+\delta)\tag{6-15}$

The normal force on the shear plane is now:

$\ \mathrm{N}_{1}=\frac{\mathrm{W}_{2} \cdot \sin (\delta)+\mathrm{W}_{1} \cdot \sin (\alpha+\beta+\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\varphi)\tag{6-16}$

The normal force on the blade is now:

$\ \mathrm{N}_{2}=\frac{\mathrm{W}_{2} \cdot \sin (\alpha+\beta+\varphi)+\mathrm{W}_{1} \cdot \sin (\varphi)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\delta)\tag{6-17}$

Equations (6-16) and (6-17) show, that the normal forces on the shear plane and the blade are always positive. Positive means compressive stresses. In water saturated sand, always the shear type of cutting mechanism will occur. Figure 6-6 shows these forces on the layer cut.

This page titled 6.4: The Equilibrium of Forces 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.