Based on the force analysis of the torsional penetrometer, two driving indicators can be derived from the view of interface mechanics. The resistance was derived from the energy conservation, which considered the grousers’ effect. The traction was derived by adding three main forces, rather than the only shear force. Compared with the traditional formulas, the new formulas were all derived based on the torsional penetrometer instead of two independent test devices, and more important factors were considered in the derivations rather than a simplification.

## 2.2.1. Driving resistance

According to the principle of energy conservation, the resistance based on the torsional penetrometer in the soft clay can be derived by the following processes.

The force analysis of a unit in the penetration process is illustrated in Fig. 6. In the penetration process, the dead weight *G* is balanced by the reacting pressure \({\sigma }_{1}\) at the bottom of the plate and \({\sigma }_{2}\) at the bottom of the grouser. Thus, the penetration work \({E}_{P}\) can be derived by the integral of \({\sigma }_{1}\left(x\right)\) and \({\sigma }_{2}\left(x\right)\).

$${E}_{P}=b(l-t){\int }_{0}^{x}{\sigma }_{1}\left(X\right)dX+bt{\int }_{0}^{h}{\sigma }_{2}\left(H\right)dH \left(3\right)$$

where \(b\) is the width of the unit, \(l\) is the length of the unit, \(x\) is the sinkage, \(t\) is the thickness of the grouser and plate, \(h\) is the height of the grouser.

The resistance work \({E}_{R}\) is the work done by the resistance \({R}_{c}\) per unit length of the torsional penetrometer.

$${E}_{R}={E}_{P} \left(4\right)$$

$${R}_{c}l=b(l-t){\int }_{0}^{x}{\sigma }_{1}\left(X\right)dX+bt{\int }_{0}^{h}{\sigma }_{2}\left(H\right)dH \left(5\right)$$

Finally, the resistance \({R}_{c}\) is derived as:

$${R}_{c}=\frac{b(l-t){\int }_{0}^{x}{\sigma }_{1}\left(X\right)dX+bt{\int }_{0}^{h}{\sigma }_{2}\left(H\right)dH}{l} \left(6\right)$$

## 2.2.2. Driving traction

When the actual track is activated to shear the soft clay, its driving traction coming from the soil is concentrated on the interaction interface (Wong, 2009). Several minor forces, such as the frictional force, cohesive force and pore water pressure were ignored. Three kinds of main interface forces were considered. The detailed force analysis of a unit in the shearing process is illustrated in Fig. 7. The first part \({F}_{\tau }\) is the shear force coming from all shear interfaces, the second part \({F}_{R}\) is the passive earth force coming from the action of grousers on soft clay and the third part \({F}_{B}\) is the additional force coming from the stress distribution of Boussinesq in the soft clay.

As for the shear force \({F}_{\tau }\), which could be integrated by shear stress in four shearing surfaces:

$${F}_{\tau }=2l{\int }_{0}^{t}{\tau }_{1}\left(j\right)ds+b{\int }_{0}^{l-t}{\tau }_{2}\left(j\right)ds+2h{\int }_{0}^{t}{\tau }_{3}\left(j\right)ds+b{\int }_{0}^{t}{\tau }_{4}\left(j\right)ds \left(7\right)$$

where \({\tau }_{1}\) is the shear stress at the bilateral surfaces of the plate, \({\tau }_{2}\) is the shear stress at the bottom of the plate, \({\tau }_{3}\) is the shear stress at the bilateral surfaces of the grouser and \({\tau }_{4}\) is the shear stress at the bottom of the grouser, correspondingly, \({\tau }_{max,n}\)(n = 1,2,3,4) is the maximum shear stress and \({\tau }_{res,n}\)(n = 1,2,3,4) is the residual shear stress.

Regarding passive earth force \({F}_{R}\), it can be calculated using Rankine theory (Lambe et al., 1991). In the zone of Rankine passive earth force, the earth pressure perpendicular to the grouser can be expressed as:

$$\rho \left(z\right)=\frac{1}{2}\gamma {z}^{2}{K}_{P}+2cz\sqrt{{K}_{P}} \left(8\right)$$

where \(\gamma\) is the effective stress of soft clay, \(z\) is the vertical depth along the grouser, \({K}_{P}\) is the coefficient of passive Earth pressure and \(c\) is the cohesive force of soft clay.

Thus, the passive earth force can be expressed as:

$${F}_{R}=b{\int }_{0}^{h}\rho \left(z\right)dz=b\left(\frac{1}{6}\gamma h{K}_{P}+c\sqrt{{K}_{P}}\right) \left(9\right)$$

Regarding the additional force \({F}_{B}\), it can be solved based on the Boussinesq equation (Sadd et al., 2009). According to the spatial stress distribution of Boussinesq, taking the load on the unit as a belt load acting on the semi-infinite body (see Fig. 8).

The stress distribution of the elastomer is the same as that of the viscoelastic body under a certain load, so the stress distribution of the soft clay could be considered as elastomer and the stress at any point perpendicular to the belt load could be written as:

$${\sigma }_{x}={\int }_{0}^{\frac{\pi }{2}-\theta }{\sigma }_{p}{sin}^{2}\theta d\theta =\frac{p}{2\pi }\left(\pi -2\theta -{sin}2\theta \right) \left(10\right)$$

where \({\sigma }_{x}\) is the stress at any point perpendicular to the belt load, \(\theta\) is the angle between the point and horizontal line of the belt load, and \(p\) is the belt load.

$${tan}\theta =\frac{H}{b} \left(11\right)$$

Have an integral for all stresses perpendicular to the belt load, \({F}_{B}\) is solved as:

$${F}_{B}=l{\int }_{0}^{h}{\sigma }_{x}dH=plh\left[\frac{1}{2}-\frac{arctan\left(h/b\right)}{180}\right] \left(12\right)$$

In the end, the traction \(T\) can be derived by adding the above three forces:\(T={F}_{\tau }+{F}_{R}+{F}_{B}=2l{\int }_{0}^{t}{\tau }_{1}\left(j\right)ds+b{\int }_{0}^{l-t}{\tau }_{2}\left(j\right)ds+ 2h{\int }_{0}^{t}{\tau }_{3}\left(j\right)ds+b{\int }_{0}^{t}{\tau }_{4}\left(j\right)ds+ b{h}^{2}\left(\frac{1}{6}\gamma h{K}_{P}+c\sqrt{{K}_{P}}\right)+ plh\left[\frac{1}{2}-\frac{arctan\left(h/b\right)}{180}\right] \left(13\right)\)