3.1 Thrust
The thrust of a disc cutter is the force on the cutter applied by the TBM hydraulic cylinder through the cutter head.
Fig. 3 shows a schematic of a traditional disc cutter and the profile obtained when its penetration depth is h and the rock surface is the cutting plane, with its area denoted as Az. Fig. 4 presents the analogous schematic for the newly designed disc cutter and its resultant profile for a penetration depth of h and rock surface of area Ax.
According to the above analysis, prior to leapfrog fragmentation, rock deformation occurs in accordance with the plasticity law. This lays a foundation for the following calculation. In accordance with Ref. [20], when the penetration of the disc cutter is h, the contact area is in a state of plasticity, which is tantamount to a force acting on plane B–B. For the ith disc (location number) of the newly designed disc cutter, where the force is expressed as q
_{
xi
}, then:
$$q_{xi} = 2k\left( {1 + \frac{{\uppi }}{2}  \alpha_{i} } \right).$$
(1)
where k is the yield condition. In terms of Mises theory, \(k = {{\sigma_{s} } \mathord{\left/ {\vphantom {{\sigma_{s} } {\sqrt 3 }}} \right. \kern0pt} {\sqrt 3 }}\); in terms of Tresca theory, k = σ
_{
s
}/2, where σ
_{
s
} is the uniaxial compressive strength of the rock; α
_{
i
} = arctan(r/R
_{
i
}), where r is the radius of a disc cutter and R
_{
i
} is the radius of the ith disccutter locus.
Likewise, when penetration of a traditional disc cutter is h, the force q
_{
zi
} acting on the plane B–B is:
$$q_{zi} = 2k\left( {1 + \frac{{\uppi }}{2}} \right).$$
(2)
According to the theory presented in Ref. [22], the thrust F
_{
zi
} on each traditional disc cutter can be expressed as:
$$F_{zi} = q_{zi} A_{z} = 2k\left( {1 + \frac{\uppi}{2}} \right)A_{z} .$$
(3)
Likewise, the thrust F
_{
xi
} on each newly designed disc cutter is given by:
$$F_{xi} = q_{xi} A_{x} = 2k\left( {1 + \frac{\uppi}{2}  \alpha_{i} } \right)A_{x} .$$
(4)
3.2 Energy Consumption
Indentation tests, fullscale linear cutting tests, and analysis of actual cutting by disc cutters showed that the cutting process can be categorized as comprising penetration cutting, fullscale linear cutting, and sideslip cutting. The corresponding energy consumption can be calculated.

(1)
Energy consumption of penetration cutting
Energy consumption of penetration cutting is the work accomplished by the thrust for a certain length of excavation. If the length is L, then W
_{
zqi
}, the energy consumption of traditional disc cutters, is given by:
$$W_{zqi}^{{}} = F_{zi} L = 2k\left( {1 + \frac{\uppi}{2}} \right)A_{z} L,$$
(5)
while W
_{
xqi
}, the corresponding energy consumption of the newly designed cutters, is:
$$W_{xqi} = F_{xi} L = 2k\left( {1 + \frac{\uppi}{2}  \alpha_{i} } \right)A_{x} L.$$
(6)

(2)
Energy consumption of fullscale linear cutting
Energy consumption of fullscale linear cutting is the energy consumed during the work of the disc cutters. Linear cutting tests revealed that the resistance to disc cutters usually appeared at twothirds of the penetration depth (shown in Fig. 5); hence, the friction coefficient, η, of the disc cutters is given by:
$$\eta = \frac{{\sqrt {6rh  h^{2} } }}{3}.$$
(7)
M
_{
zi
}, the frictional torque of traditional disc cutters, is expressed as:
$$M^{\prime}_{zi} = F_{zi} \eta = 2k\left( {1 + \frac{\uppi}{2}} \right)A_{z} \frac{{\sqrt {6rh  h^{2} } }}{3}.$$
(8)
whereas M
_{
xi
}, the corresponding friction torque of the newly designed disc cutters, is given by:
$$M_{xi}^{'} = F_{xi} \eta = 2k\left( {1 + \frac{\uppi}{2}  \alpha_{i} } \right)A_{x} \frac{{\sqrt {6rh  h^{2} } }}{3}.$$
(9)
When the excavation length is L, the revolution number of the cutterhead is L/h; the length of the helix covered is L/h × 2πR
_{
i
}; and the angle passed by a disc cutter is given by:
$$\psi = \frac{L}{h} \times 2{\uppi}R_{i} \div \left( {2{\uppi}r} \right) \times 2{\uppi} = \frac{{2LR_{i} {\uppi}}}{hr}.$$
(10)
The energy consumed by linear cutting of traditional disc cutters, W
_{
zgi
}, is:
$$W_{zgi} = M^{\prime}_{zi} \psi = 4{\uppi}R_{i} k\left( {1 + \frac{{\uppi}}{2}} \right)A_{z} \frac{{L\sqrt {6rh  h^{2} } }}{3hr},$$
(11)
while the corresponding value for the newly designed cutter, W
_{
xgi
}, is given by:
$$W_{xgi} = M^{\prime}_{xi} \psi = 4{\uppi}R_{i} k\left( {1 + \frac{{\uppi}}{2}  \alpha_{i} } \right)A_{x} \frac{{L\sqrt {6rh  h^{2} } }}{3hr}.$$
(12)

(3)
Energy consumption of sideslip cutting
Side slip is the sideways motion of disc cutters in the direction perpendicular to the plane of the cutter edge. Without revolution of the cutterhead, i.e., without transport motion or penetration of the disc cutters, there would be no side slip; therefore, side slip is the result of both cutterhead revolution and disccutter penetration. The energy consumption is that consumed by disc cutters undergoing side slip, as shown in Fig. 6.
In Fig. 6, O
_{1}
O
_{2} is the axis of a disc cutter, O
_{2}
O
_{3} is the axis of the cutterhead, his the penetration, D is the instantaneous disccutter maximal penetration point, and C is the instantaneous disccutter cutting point. Because it is transport motion (revolution of the cutterhead) that leads to side slip, an analysis of transport motion is presented.
When a disc cutter passes through angle dγ (not shown in Fig. 6 for clarity), the corresponding point on the cutterhead also passes through angle dγ. In accordance with the geometric relationship shown in Fig. 6, the magnitude of the transport displacement of point C, denoted by dΛ
_{
i
}, is expressed as:
$${\text{d}}\varLambda_{i} = \overline{CF} = \overline{AC} {\text{d}}\gamma = \frac{{\overline{{O_{3} D}} }}{\cos \gamma }{\text{d}}\gamma = \frac{{R_{i} }}{\cos \gamma }{\text{d}}\gamma$$
(13)
and that of sideslip displacement of point C, represented by dΔ, is given by:
$${\text{d}}\Delta_{i} = \overline{GF} = \overline{CF} \sin \gamma = R_{i} \tan \gamma {\text{d}}\gamma .$$
(14)
The length of arc of the locus corresponding to one revolution of a disc cutter is 2πr and the angle corresponding to the arc between focal radii is 2πr/R
_{
i
}, so the corresponding sideslip displacement, Δ_{
i
}, can be expressed as:
$$\Delta_{i} = \int\limits_{0}^{{\frac{{2{\uppi}r}}{{R_{i} }}}} {{\text{d}}\Delta_{i} } = \int\limits_{0}^{{\frac{{2{\uppi}r}}{{R_{i} }}}} {R_{i} { \tan }\gamma {\text{d}}\gamma =  R_{i} {\text{lncos}}\frac{{2{\uppi}r}}{{R_{i} }}} .$$
(15)
When excavation length is L and the revolution number of cutterhead is L/h, then for each revolution of the cutterhead, the revolution number of the disc cutter is 2πR
_{
i
}/2πr = R
_{
i
}/r. The corresponding disccutter side slip is:
$$\Delta_{iL} = \frac{L}{h} \times \frac{{R_{i} }}{r}\Delta_{i} =  \frac{{R_{i}^{2} L}}{rh}\ln \cos \frac{{2{\uppi}r}}{{R_{i} }}.$$
(16)
The sideslip energy consumption of a traditional disc cutter, W
_{
zhi
}, is therefore given by:
$$W_{zhi} = F_{zi} f\Delta_{iL} =  \frac{{2kR_{i}^{2} L}}{rh}\left( {1 + \frac{{\uppi}}{2}} \right)A_{z} f\ln \cos \frac{{2{\uppi}r}}{{R_{i} }},$$
(17)
while that of a newly designed disc cutter, W
_{
xhi
}, is given by:
$$W_{xhi} = F_{xi} f\Delta_{iL} =  \frac{{2kR_{i}^{2} L}}{rh}\left( {1 + \frac{{\uppi}}{2}  \alpha_{i} } \right)A_{x} f\ln \cos \frac{{2{\uppi}r}}{{R_{i} }}.$$
(18)