When the roller comes into contact with the workpiece, the workpiece surfaces will undergo the elastoplastic deformation. However, compared with the plastic deformation, the elastic deformation is smaller and will rebound after application of USRP. Therefore, the displacement of elastic recovery is ignored and only the plastic deformation is analyzed. The specific force analysis is shown in Fig. 3.

Where, *F* is the static force; *N* is the pressure exerted by the workpiece on the roller; *f* is the friction force exerted by the workpiece on the roller; *R*T is the radius of the roller; *θ* and *ø* are the contact angle in the x-z plane and in the x-y plane, respectively. According to the Hertz contact theory, the contact shape is an approximately spherical cap [18]. In order to obtain the *N* acting on the roller in the contact zone, the contact area is divided into *j* discrete equally spaced units. The area of each interval is very small, so the *N*i is assumed to be approximately constant: *dN* in the *i*th interval. The *N* acting on the roller can be calculated by integrating over the contact zone:

$$\vec {N}=\iint {\overrightarrow {dN} }$$

3

Where, d*N* = *P*md*A*=*P*m(*R*Tdθ)(*R*Tsin*θ*d*φ*) = *P*m*R*T2sin*θ*d*θ*d*φ*, *P*m is the average surface stress in the contact zone. Then, the *N* is distributed along the axes:

$$\vec {N}={\vec {N}_x}+{\vec {N}_y}+{\vec {N}_z}$$

4

$${\vec {N}_x}= - \iint {dN\sin \theta \sin \phi = - {P_m}}{R_T}^{2}\int_{0}^{\theta } {{{\sin }^2}} \alpha d\alpha \int_{0}^{\pi } {\sin \phi d\phi }$$

5

$${\vec {N}_y}= - \iint {dN\cos \theta = - {P_m}}{R_T}^{2}\int_{0}^{\theta } {\sin \alpha \cos \alpha d\alpha \int_{0}^{\pi } {d\phi } }$$

6

Where, the minus sign means that \({\vec {N}_x}\)and \({\vec {N}_y}\)are opposite to the *x-*axis and *y*-axis, respectively.

According to Coulomb’s law, the forces of friction along the axis are calculated as follows.

$$\vec {f}=\iint {\overrightarrow {df} }$$

8

$$\vec {f}={\vec {f}_x}+{\vec {f}_y}$$

9

\({\vec {f}_x}=\mu {\vec {N}_y}={\text{-}}\mu {P_m}R_{T}^{2}\int_{0}^{\theta } {\sin \alpha \cos \alpha d\alpha \int_{0}^{\pi } {d\phi } }\) (10)

$${\vec {f}_y}=\mu {\vec {N}_x}= - \mu {P_m}R_{T}^{2}\int_{0}^{\theta } {{{\sin }^2}\alpha d\alpha \int_{0}^{\pi } {\sin \phi d\phi } }$$

11

Where,\(d\vec {f}\)and\(d\vec {N}\)are perpendicular, and *µ* is the friction coefficient between the roller and the workpiece.

According to the relationship of the Newtonian law, the forces acting on the roller are balanced in USRP, so the net force along the *y-*axis is zero:

$$\vec {F}+\vec {N}{}_{y}+{\vec {f}_y}=0$$

12

Substituting Eq. (6) and Eq. (11) into Eq. (12):

$$F=\frac{1}{4}\pi R_{{\text{T}}}^{{\text{2}}}{P_m}\left[ {\left( {1 - \cos 2\alpha } \right) - \frac{2}{\pi }\mu \left( {2\alpha - \sin 2\alpha } \right)} \right] \approx \pi {R_{\text{T}}}{P_m}{h_a}\left[ {1 - \mu \sqrt {\frac{{{h_0}}}{{2{R_T}}}} } \right]$$

13

Because *h**a* is much less than 2*R**T*, is approximated to 0. *µ* is very small, and *P*m is approximately set as the workpiece surface micro-hardness Hv, then the Eq. (13) is simplified as follows:

$$F=\pi {R_T}{P_m}{h_a}=\pi {R_T}Hv{h_a}$$

14

It can be seen from the Eq. (14) that there is a linear relationship between the rolling depth and the rolling force.