# 14.2: The Equilibrium of Forces

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Figure 14-2 illustrates the forces on the layer of soil cut. The forces shown are valid in general for each type of soil.

The forces acting on the layer A-B are:

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

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

3. A shear force C1 as a result of pure cohesion $$\ \tau_\mathrm{c}$$ or shear strength. This force can be calculated by multiplying the cohesive shear strength $$\ \tau_\mathrm{c}$$ with the area of the shear plane.

4. A force normal to the pseudo blade N2, resulting from the effective grain stresses.

5. A shear force S2 as a result of the soil/soil friction N2·tan(λ) between the layer cut and the wedge pseudo blade. The friction angle λ does not have to be equal to the internal friction angle φ in the shear plane, since the soil has already been deformed.

6. A shear force C2 as a result of the mobilized cohesion between the soil and the wedge $$\ \tau_\mathrm{c}$$. This force can be calculated by multiplying the cohesive shear strength $$\ \tau_\mathrm{c}$$ of the soil with the contact area between the soil and the wedge.

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

$\ \mathrm{K}_{1}=\sqrt{\mathrm{N}_{1}^{2}+\mathrm{S}_{1}^{2}}\tag{14-1}$

The forces acting on the wedge front or pseudo blade A-C 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 S2 as a result of the soil/soil friction N2·tan(λ) between the layer cut and the wedge pseudo blade. The friction angle λ does not have to be equal to the internal friction angle φ in the shear plane, since the soil has already been deformed.

3. A shear force C2 as a result of the cohesion between the layer cut and the pseudo blade $$\ \tau_\mathrm{c}$$. This force can be calculated by multiplying the cohesive shear strength $$\ \tau_\mathrm{c}$$ of the soil with the contact area between the soil and the pseudo blade.

These forces are shown in Figure 14-3. If the forces N2 and Sare 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.

$\ \mathrm{K}_{2}=\sqrt{\mathrm{N}_{2}^{2}+\mathrm{S}_{2}^{2}}\tag{14-2}$

The forces acting on the wedge bottom A-D when cutting soil, can be distinguished as:

1. A force N3, resulting from the effective grain stresses, between the wedge bottom and the undisturbed soil.

2. A shear force S3 as a result of the soil/soil friction N3·tan(φ) between the wedge bottom and the undisturbed soil.

3. A shear force C3 as a result of the cohesion between the wedge bottom and the undisturbed soil $$\ \tau_\mathrm{c}$$. This force can be calculated by multiplying the cohesive shear strength $$\ \tau_\mathrm{c}$$ of the soil with the contact area between the wedge bottom and the undisturbed soil.

The normal force N3 and the shear force S3 can be combined to a resulting grain force K3.

$\ \mathrm{K}_{3}=\sqrt{\mathrm{N}_{\mathrm{3}}^{2}+\mathrm{S}_{\mathrm{3}}^{\mathrm{2}}}\tag{14-3}$

The forces acting on a straight blade C-D when cutting soil (see Figure 14-4), can be distinguished as:

1. A force normal to the blade N4resulting from the effective grain stresses.

2. A shear force S4 as a result of the soil/steel friction N4·tan(δ).

The normal force N4 and the shear force S4 can be combined to a resulting grain force K4.

$\ \mathrm{K}_{4}=\sqrt{\mathrm{N}_{4}^{2}+\mathrm{S}_{4}^{2}}\tag{14-4}$

The horizontal equilibrium of forces on the layer cut:

$\ \sum \mathrm{F}_{\mathrm{h}}=\mathrm{K}_{\mathrm{1}} \cdot \sin (\boldsymbol{\beta}+\boldsymbol{\varphi})+\mathrm{C}_{\mathrm{1}} \cdot \cos (\boldsymbol{\beta})-\mathrm{C}_{\mathrm{2}} \cdot \cos (\boldsymbol{\theta})-\mathrm{K}_{\mathrm{2}} \cdot \sin (\boldsymbol{\theta}+\lambda)=\mathrm{0}\tag{14-5}$

The vertical equilibrium of forces on the layer cut:

$\ \sum \mathrm{F}_{\mathrm{v}}=-\mathrm{K}_{\mathrm{1}} \cdot \cos (\boldsymbol{\beta}+\varphi)+\mathrm{C}_{\mathrm{1}} \cdot \sin (\boldsymbol{\beta})+\mathrm{C}_{\mathrm{2}} \cdot \sin (\boldsymbol{\theta})-\mathrm{K}_{\mathrm{2}} \cdot \mathrm{\operatorname { cos } ( \boldsymbol { \theta } + \lambda )}=\mathrm{0}\tag{14-6}$

The force K1 on the shear plane is now:

$\ \mathrm{K}_{1}=\frac{-\mathrm{C}_{1} \cdot \cos (\theta+\beta+\lambda)+\mathrm{C}_{2} \cdot \cos (\lambda)}{\sin (\theta+\beta+\lambda+\varphi)}\tag{14-7}$

The force K2 on the pseudo blade is now:

$\ \mathrm{K}_{2}=\frac{+\mathrm{C}_{1} \cdot \cos (\varphi)-\mathrm{C}_{2} \cdot \cos (\theta+\beta+\varphi)}{\sin (\theta+\beta+\lambda+\varphi)}\tag{14-8}$

From equation (14-8) the forces on the pseudo blade can be derived. On the pseudo blade a force component in the direction of cutting velocity Fh and a force perpendicular to this direction Fv can be distinguished.

$\ \mathrm{F}_{\mathrm{h}}=\mathrm{K}_{2} \cdot \sin (\theta+\lambda)+\mathrm{C}_{2} \cdot \cos (\theta)\tag{14-9}$

$\ \mathrm{F}_{v}=\mathrm{K}_{2} \cdot \cos (\theta+\lambda)-\mathrm{C}_{2} \cdot \sin (\theta)\tag{14-10}$

The normal force on the shear plane is now:

$\ \mathrm{N}_{1}=\frac{-\mathrm{C}_{1} \cdot \cos (\alpha+\beta+\lambda)+\mathrm{C}_{2} \cdot \cos (\lambda)}{\sin (\alpha+\beta+\lambda+\varphi)} \cdot \cos (\varphi)\tag{14-11}$

The normal force on the pseudo blade is now:

$\ \mathrm{N}_{2}=\frac{+\mathrm{C}_{1} \cdot \cos (\varphi)-\mathrm{C}_{2} \cdot \cos (\alpha+\beta+\varphi)}{\sin (\alpha+\beta+\lambda+\varphi)} \cdot \cos (\lambda)\tag{14-12}$

Now knowing the forces on the pseudo blade A-C, the equilibrium of forces on the wedge A-C-D can be derived. The horizontal equilibrium of forces on the wedge is:

$\ \begin{array}{left} \sum \mathrm{F}_{\mathrm{h}}=&-\mathrm{K}_{4} \cdot \sin (\alpha+\delta)+\mathrm{K}_{3} \cdot \sin (\varphi)+\mathrm{C}_{3} \\ &+\mathrm{C}_{2} \cdot \cos (\theta)+\mathrm{K}_{2} \cdot \sin (\theta+\lambda)=\mathrm{0} \end{array}\tag{14-13}$

The vertical equilibrium of forces on the wedge is:

$\ \begin{array}{left} \sum \mathrm{F}_{\mathrm{v}}=&-\mathrm{K}_{4} \cdot \cos (\alpha+\delta)-\mathrm{K}_{3} \cdot \cos (\varphi) \\ &-\mathrm{C}_{2} \cdot \sin (\theta)+\mathrm{K}_{2} \cdot \cos (\theta+\lambda)=\mathrm{0} \end{array}\tag{14-14}$

The unknowns in this equation are K3 and K4, since K2 has already been solved. Two other unknowns are, the external friction angle δ, since the external friction does not have to be fully mobilized, and the wedge angle θ. These 2 additional unknowns require 2 additional conditions in order to solve the problem. One additional condition is the equilibrium of moments of the wedge, a second condition the principle of minimum required cutting energy. Depending on whether the soil pushes upwards or downwards against the blade, the mobilization factor is between -1 and +1.

The force K3 on the bottom of the wedge is now:

$\ \mathrm{K}_{3}=\frac{+\mathrm{K}_{2} \cdot \sin (\alpha+\delta-\theta-\lambda)+\mathrm{C}_{3} \cdot \cos (\alpha+\delta)-\mathrm{C}_{2} \cdot \cos (\alpha+\delta-\theta)}{\sin (\alpha+\delta+\varphi)}\tag{14-15}$

The force K4 on the blade is now:

$\ \mathrm{K}_{4}=\frac{+\mathrm{K}_{2} \cdot \sin (\theta+\lambda+\varphi)+\mathrm{C}_{3} \cdot \cos (\varphi)+\mathrm{C}_{2} \cdot \cos (\theta+\varphi)}{\sin (\alpha+\delta+\varphi)}\tag{14-16}$

This results in a horizontal force on the blade of:

$\ \mathrm{F}_{\mathrm{h}}=\mathrm{K}_{4} \cdot \sin (\alpha+\delta)\tag{14-17}$

And in a vertical force on the blade of:

$\ \mathrm{F}_{\mathrm{v}}=\mathrm{K}_{4} \cdot \cos (\alpha+\delta)\tag{14-18}$