Deduction of Eq. (9.1 ).

A non-free cutting model with double edged cutting tool was presented in this chapter, as shown in Fig. 9.1 , and its mathematical presentation is Eq. (9.1 ). Here, we will give a brief deduction of Eq. (9.1 ).

The unit vector in the direction of resultant force on rake face is

$${\mathbf{R}}_{{\mathbf{1}}} = - \sin \beta \,\sin \psi_{\lambda } \, {\mathbf{i}} - \sin \beta \,\cos \psi_{\lambda } {\mathbf{j}} + \cos \beta \, {\mathbf{k}}\text{,}$$

(9.6)

where,

β is the friction angle on the interface between the rake and the chips, and

i ,

j ,

k are the unit vectors in the directions of axes

x ,

y ,

z , respectively (they are not shown in Fig.

9.1 ). The resultant force imposed on chip by the rake is;

$$\varvec{R} = R{\mathbf{R}}_{{\mathbf{1}}} ,$$

(9.7)

The vector of cutting velocity is,

$$\varvec{ V} = V{\mathbf{k}}\text{,}$$

(9.8)

The vector of chip-ejection velocity is,

$$\varvec{U} = r\varvec{V}(\sin \psi_{\lambda } {\mathbf{i}} + \cos \psi_{\lambda } {\mathbf{j}}),$$

(9.9)

where,

r is cutting ratio. The vector of shear velocity is,

$$\varvec{W} = \varvec{U} + \varvec{V} = V(r\,\sin \psi_{\lambda } {\mathbf{i}} + r\,\cos \psi_{\lambda } {\mathbf{j}} + {\mathbf{k}}).$$

(9.10)

Shear speed, i.e., the module of vector

W is,

$$\left| \varvec{W} \right| = V\sqrt {1 + r^{2} }$$

(9.11)

The unit vector in the shear direction is,

$${\mathbf{W}}_{{\mathbf{1}}} = \frac{{r\,\sin \psi_{\lambda } {\mathbf{i}} + r\,\cos \psi_{\lambda } {\mathbf{j}} + {\mathbf{k}}}}{{\sqrt {1 + r^{2} } }}$$

(9.12)

The shear force, i.e., the projection of

R on the shear direction

W _{ 1 } ,

$$\varvec{R} \cdot {\mathbf{W}}_{{\mathbf{1}}} = R({\mathbf{R}}_{{\mathbf{1}}} \cdot {\mathbf{W}}_{{\mathbf{1}}} ) = R\frac{(\cos \beta - r\,\sin \beta )}{{\sqrt {1 + r^{2} } }} = \tau S,$$

(9.13)

where,

τ is shear stress of the work-piece material, and

S is the area of shear plane. So,

$$R = \frac{{\tau S\sqrt {1 + r^{2} } }}{(\cos \beta - r\,\sin \beta )}.$$

(9.14)

The cutting power is,

$$A = R \cdot \,\cos \beta \cdot V = \frac{V \cdot \tau \cdot \,\cos \beta }{(\cos \beta - r\,\sin \beta )}\sqrt {1 + r^{2} } \cdot S.$$

(9.15)

Shear plane is divided into two parts: the shear plane of major edge and the shear plane of minor edge, the areas of them are denoted as S _{ I } and S _{ II } , respectively. The calculation of S _{ I } and S _{ II } are divided into three cases:

The first case: tan

ψ _{λ} <

a _{ w/ } a _{ c } , in this case, vector

W intersects with the top surface

AFCE of the work-piece, and the intersection point is

P. What is shown in Fig.

9.1 is just this case. The coordinates of point

P In the coordinate system

xyz are

\(P_{x} = a_{c} \tan \psi_{\lambda } \,\) ,

\(P_{y} = \, a_{\text{c}} \,\) and

\(P_{z} = \, {{a_{\text{c}} } \mathord{\left/ {\vphantom {{a_{\text{c}} } {(r\,{ \cos }\,\psi_{\lambda } )}}} \right. \kern-0pt} {(r\,{ \cos }\,\psi_{\lambda } )}}\) . In this case,

S _{ I } and

S _{ II } are shown as Fig.

9.14 a, b and the total area of the two shear plane

Fig. 9.14 The shear planes (tan ψ _{ λ } < a _{ w } /a _{ c } ). a Major shear plane. b Minor shear plane

$$\begin{aligned} S & = S_{\text{I}} + S_{\text{II}} = \, \square BCEO - \Delta OPE + \Delta OAP \\ & = a_{c} (a_{w} - \frac{1}{2}a_{c} \tan \psi_{\lambda } )\sqrt {1 + \frac{1}{{r^{2} \cos^{2} \psi_{\lambda } }}} + \frac{1}{2}\frac{{a_{c}^{2} }}{{r\,\cos \psi_{\lambda } }}\sqrt {1 + r^{2} \sin^{2} \psi_{\lambda } } . \\ \end{aligned}$$

(9.16)

The second case: tan

ψ _{λ} >

a _{ w/ } a _{ c } , in this case, vector

W intersects with the side surface

FCDB of the work-piece, and the intersection point is

P . The coordinates of point

P In the coordinate system

xyz are

\(P_{x} = a_{w}\) ,

\(P_{y} = \, {{a_{w} } \mathord{\left/ {\vphantom {{a_{w} } {{ \tan }\psi_{\lambda } }}} \right. \kern-0pt} {{ \tan }\psi_{\lambda } }} \,\) and

\(P_{z} \, = {{a_{w} } \mathord{\left/ {\vphantom {{a_{w} } {\text{(}r\,{ \sin }\psi_{\lambda } }}} \right. \kern-0pt} {\text{(}r\,{ \sin }\psi_{\lambda } }})\) . In this case,

S _{ I } and

S _{ II } are shown as Fig.

9.15 b and c, and the total area of the two shear plane is,

Fig. 9.15 The area of cut and shear planes (tan ψ _{ λ } > a _{ w } /a _{ c } ). a Chip sections. b Major shear plane. c Minor shear plane

$$\begin{aligned} S & = S_{\text{I}} + S_{\text{II}} = \Delta OBP +\square ACDO - \Delta OPD \\ & = \frac{{a_{w} }}{{r\,\sin \psi_{\lambda } }}\left[ {a_{c} \sqrt {1 + r^{2} \sin^{2} \psi_{\lambda } } + \frac{1}{2}a_{w} \left( {\sqrt {1 + r^{2} \cos^{2} \psi_{\lambda } } - \sqrt {\cot^{2} \psi_{\lambda } + r^{2} \cos^{2} \psi_{\lambda } } } \right)} \right]. \\ \end{aligned}$$

(9.17)

The third case, tan

ψ _{λ} =

a _{ w/ } a _{ c } , it is a critical case between the above two cases, the vector

W intersects with the side ridge

EC of the work-piece, and the intersection point is

P , which coincides with the corner point

C as shown in Fig.

9.16 a. The coordinates of point

P in the coordinate system

xyz are

\(P_{x} = a_{w}\) ,

\(P_{y} = \, a_{c} \,\) and

\(P_{z} \, = {{a_{w} } \mathord{\left/ {\vphantom {{a_{w} } {\text{(}r\,{ \sin }\psi_{\lambda } }}} \right. \kern-0pt} {\text{(}r\,{ \sin }\psi_{\lambda } }}) = {{a_{c} } \mathord{\left/ {\vphantom {{a_{c} } {\text{(}r\,{ \sin }\psi_{\lambda } }}} \right. \kern-0pt} {\text{(}r\,{ \sin }\psi_{\lambda } }})\) . In this case,

S _{ I } and

S _{ II } are shown as Fig.

9.15 b, c, and the total area of the two shear plane is,

Fig. 9.16 The area of cut and shear planes (tan ψ _{ λ } = a _{ w } /a _{ c } ). a Chip sections. b Major shear plane. c Minor shear plane

$$S = S_{\text{I}} + S_{\text{II}} = \Delta BPO + \Delta PAO = \frac{1}{2r}\left[ {a_{c} \sqrt {a_{w}^{2} (1 + r^{2} ) + a_{c}^{2} } + a_{w} \sqrt {a_{c}^{2} (1 + r^{2} ) + a_{w}^{2} } } \right],$$

(9.18)

Substituting Eqs. (9.6 )–(9.8 ) into Eq. (9.5 ), respectively, and considering the physical unit conversion result in the Eq. (9.1 ), i.e., the expression of cutting power.