PLASTIC DEFORMATION OF CHIPS IN METAL CUTTING

A. M. ROZENBERG and A. N. EREMIN

In the machining of ductile metals on metal-cutting machines with lathe tools, drills, milling cutters, and other metallic cutting tools, a noticeable plastic deformation of the surface machined by the tool is observed; this deformation is directly connected with the cleanliness, smoothness, and quality of the machined surface. The chip removed by the cutting tool itself proves to be very considerably plastically deformed. Although the chip itself, in the process of mechanical machining of metals in production, is of no interest to us, being waste in the manufacture of one part or another, the degree of its plastic deformation is directly connected with the work expended on the cutting process, with the cutting force arising on the cutting tool, with the amount of heat released during cutting, and therefore also with the heating, wear, and durability of the cutting tool. Therefore, in studying cutting processes it is important to be able to determine the degree of plastic deformation of the removed chip; and many recent investigations show that a number of phenomena in the cutting process can be physically explained when the degree of plastic deformation of the chip is taken into account.

By the degree of plastic deformation of the chip, investigators usually understand chip shrinkage, calling shrinkage the ratio of the distance traveled by the tool to the length of the chip removed over this distance. In this case chip shrinkage is expressed by an abstract number, which is always greater than unity. It is considered that if shrinkage were equal to unity, then in this case plastic deformation in the chip would be absent, since its length in this case would be equal to the length of the path over which it was removed. Some investigators express chip shrinkage by the inverse ratio, i.e., by the ratio of the chip length to the length of the tool path. With such notation, shrinkage takes the value of a proper fraction, which is the smaller the more the chip is deformed; again, it is assumed that in the absence of deformation the shrinkage in this case will be equal to unity.

Chip shrinkage is undoubtedly an expression of the degree of deformation of the chip, and in one of our works we showed that a change in chip shrinkage always entails a change in the cutting force, which is a function of the degree of deformation of the chip. But it is incorrect to think that chip shrinkage numerically expresses the degree of plastic deformation of the metal of the chip; it is incorrect to assume that, with shrinkage equal to unity, plastic deformation in the chip is absent; and it is also incorrect to accept that, at different cutting angles of the tool and with identical chip shrinkages, the degree of plastic deformation of the chip will be one and the same. This we shall show in the following exposition.

We shall call the chip shrinkage the ratio

\[ \zeta=\frac{l_0}{l}, \]

where \(\zeta\) is the chip shrinkage,
\(l_0\) is the path traversed by the tool in removing the given chip,
\(l\) is the length of the chip removed over the path \(l_0\).

The principal and almost the only deformation in the chip when cutting a plastic metal is compression deformation, taking place by means of shears in the plane of maximum shearing stresses \(mm\) (Fig. 1). The position of this plane and its inclination to the direction of motion of the tool determine, for a given cutting angle \(\delta\) or rake angle \(\gamma\), the magnitude of the longitudinal chip shrinkage \(\zeta\), as well as the magnitude of its transverse thickening.

In addition to the longitudinal shrinkage and transverse thickening of the chip, there is also its widening in the direction parallel to the main cutting edge of the tool. The deformation of widening may be neglected, since it usually has no appreciable magnitude. Thus, it will be correct to consider that in cutting a plastic metal, just as in ordinary compression under a press, the principal deformation is shear, and the degree of deformation must be evaluated by the magnitude of the relative shear.

It should be stipulated that in reality there cannot be such a situation in which the entire deformation of the chip takes place in one single plane (the plane of maximum shearing stresses), since in that case the rate of deformation at any value of the cutting speed would be infinitely great, which would have to lead to a colossal increase in stresses and cutting forces. We imagine that the plastic deformation of the chip proceeds in a certain finite-volume zone located ahead of the plane \(mm\), as has also been proved in a number of works. But it may be assumed that the principal part of the chip deformation is obtained in the immediate vicinity of the plane \(mm\). This assumption is confirmed by examination of microphotographs of continuous chips, which clearly show that the peculiar striated structure (texture) of the chip, which is the result of its deformation, breaks off sharply in a sufficiently narrow zone near the line “\(mm\).” Therefore, in determining the relative shear in the chip, we shall conventionally assume that the entire shear deformation occurs along the line of maximum shearing stresses \(mm\).

Fig. 1

Fig. 2

Let us find the relation between the longitudinal chip shrinkage and the magnitude of the relative shear \(\varepsilon\). In Fig. 2 a continuous chip is shown, obtained at a certain thickness of the removed layer “\(a\)” by a tool with rake angle \(\gamma\). The shear angle \(\beta_1\) determines the position of the shear plane \(ce\). If we take some layer in the chip of thickness \(x\), occupying on the lateral surface ...

of the chip surface the area of the parallelogram \(cdef\), bounded on one side by the shear plane \(ce\), and on the other side by the straight line \(df\) parallel to it, then one may imagine that before deformation this parallelogram occupied the position \(cfge\).

During chip formation, point \(f\), which lay on the outer contour of the workpiece, ended up on the outer contour of the chip at point \(d\), having moved to the segment \(fd\) in the process of shear. Similarly, point \(g\) moved to point \(f\). The relative shear in this case will be

\[ \varepsilon=\frac{B}{x}, \]

where \(B\) is the length of the line \(ce\) or \(df\) or \(fg\)

\[ B=x\cdot \operatorname{ctg}\beta_1+x\operatorname{tg}(\beta_1-\gamma). \]

Hence the relative shear is

\[ \varepsilon=\operatorname{ctg}\beta_1+\operatorname{tg}(\beta_1-\gamma). \tag{1} \]

Let us find the relation between the relative shear and the chip shrinkage. For this we shall use the equation relating the shear angle \(\beta_1\) to the chip shrinkage \(\zeta\)

\[ \operatorname{tg}\beta_1=\frac{\cos\gamma}{\zeta-\sin\gamma}. \tag{2} \]

Substitute expression (2) into equation (1)

\[ \varepsilon=\operatorname{ctg}\beta_1+ \frac{\operatorname{tg}\beta_1-\operatorname{tg}\gamma} {1+\operatorname{tg}\beta_1\cdot \operatorname{tg}\gamma}, \]

\[ \varepsilon= \frac{\zeta-\sin\gamma}{\cos\gamma} + \frac{\dfrac{\cos\gamma}{\zeta-\sin\gamma}-\operatorname{tg}\gamma} {1+\dfrac{\cos\gamma\cdot \operatorname{tg}\gamma}{\zeta-\sin\gamma}}. \]

Hence we obtain

\[ \varepsilon=\frac{\zeta^2-2\zeta\sin\gamma+1}{\zeta\cdot \cos\gamma}. \tag{3} \]

Thus, we have obtained the relation between relative shear, chip shrinkage, and rake angle. From equation (3) we have constructed a graph (Fig. 3), from which it is easy to draw the following conclusions:

a) chip shrinkage in the general case is not equal to the relative shear and therefore does not numerically express the degree of deformation of the chip metal;

b) with an increase in chip shrinkage at a constant rake angle, the relative shear, and consequently also the degree of deformation, increases;

c) at different rake angles, one and the same chip shrinkage corresponds to different and, in many cases, greatly differing relative shear. The smaller the chip shrinkage, the greater the change in relative shear with a change in rake angle;

d) when the chip shrinkage is equal to unity (\(\zeta=1\)), the relative shear is not equal to zero and may reach high values. This indicates the erroneousness of the notion that, when chip shrinkage is equal to unity, plastic deformation is absent;

e) the ratio between the values of chip shrinkage in the general case is not equal to the ratio between the magnitudes of relative shear. Therefore

one cannot carry out a quantitative comparison of the degrees of plastic deformation by comparing shrinkages.

[Figure 3: graph with vertical axis “Relative shear \(\varepsilon\)” and horizontal axis “Shrinkage \(\zeta\)”; curves are labeled by front angle values \(-15^\circ\), \(0^\circ\), \(15^\circ\), \(30^\circ\), \(45^\circ\), \(60^\circ\).]

Fig. 3

In Fig. 4 the same dependence is plotted in the coordinates \(\gamma\)—\(\varepsilon\). This gives an additional representation of the relation between chip shrinkage and the front an-

[Figure 4: graph with vertical axis \(\varepsilon\) and horizontal axis “Front angle \(\gamma^\circ\)”; curves are labeled \(\zeta = 1\), \(\zeta = 2\), \(\zeta = 3\), \(\zeta = 4\), \(\zeta = 5\), \(\zeta = 6\).]

Fig. 4

…angle and relative shear in cutting. Fig. 4 shows that, under the condition of constant shrinkage, with a change in the rake angle the relative shear changes along curves having a minimum. Therefore, with constant chip shrinkage, we do not have a monotonic change of the relative shear with a change in the rake angle.

The angle corresponding to the smallest relative shear, for each value of the shrinkage, can be found mathematically. To do this, using equation (3), one must take the first derivative \(\dfrac{d\varepsilon}{d\gamma}\), set it equal to zero, and find the value of \(\gamma\)

\[ \frac{d\varepsilon}{d\gamma} = \frac{\zeta^{2}\sin\gamma+\sin\gamma-2\zeta}{\zeta\cos^{2}\gamma} =0. \]

Hence

\[ \sin\gamma=\frac{2\zeta}{\zeta^{2}+1}. \]

Substituting here various values of \(\zeta\), we obtain \(\gamma\) corresponding to the smallest relative shear. It is not difficult to verify that these values of \(\gamma\) also correspond to the smallest relative shears in Fig. 4. For \(\zeta=1\), the smallest relative shear will correspond to \(\gamma=90^\circ\). In this case \(\varepsilon=0\).

Thus, when the shrinkage is equal to unity, plastic deformation will be absent only in the single case when the rake angle is \(\gamma=90^\circ\). This case is practically unreal.

It is of interest to show at what chip shrinkage in the cutting process the minimum possible relative shear will occur, i.e., the smallest degree of deformation of the chip.

To solve this question, let us take the first derivative \(\dfrac{d\varepsilon}{d\zeta}\),

\[ \frac{d\varepsilon}{d\zeta} = \frac{1}{\cos\gamma} - \frac{1}{\zeta^{2}\cos\gamma} =0. \]

Hence

\[ \zeta_{\varepsilon\min}=1 \]

and

\[ \varepsilon_{\min} = \frac{2-2\sin\gamma}{\cos\gamma}. \tag{4} \]

Thus, for any rake angle \(\gamma\), the smallest relative shear would occur if the chip shrinkage were equal to unity. If one imagines a case in which the chip shrinkage were less than unity, then in this case the relative shear and the degree of deformation of the chip would be greater than at \(\zeta=1\).

The relative shear \(\varepsilon\) has the smallest value at \(\zeta=1\); in this case the degree of deformation of the chip has the smallest value, but even in this case, for any value of the rake angle \(\gamma<90^\circ\), the relative shear and the degree of deformation are not equal to zero, as follows from equation (4), and only at \(\gamma=90^\circ\) do we have \(\varepsilon=0\), i.e., deformation is absent.

In Fig. 5 the ratio \(\dfrac{\varepsilon}{\zeta}\) is shown for different \(\gamma\) and \(\zeta\). It is seen from the figure that only for a narrow range of rake angles, within \(15\text{–}25^\circ\), at \(\zeta>2\), is the ratio \(\dfrac{\varepsilon}{\zeta}\cong 1\), i.e., \(\varepsilon\cong\zeta\), and only in this case can one…

...to judge from shrinkage the degree of deformation of the chip. In all other cases

\[ \frac{\varepsilon}{\zeta} \ne 1 \quad \text{and} \quad \varepsilon \ne \zeta . \]

One may cite a concrete example of how the use of chip shrinkage can lead to incorrect conclusions about the degree of its deformation. In high-speed cutting of steels in the range of very high cutting speeds, the chip shrinkage at a negative rake angle \(\gamma=-10^\circ\) exceeds only very slightly the shrinkage at a positive angle \(\gamma=+10^\circ\). The impression is created that we obtain chips with almost the same degree

Fig. 5

of deformation. From Figs. 4 and 5 it is evident how erroneous this impression is. Even if we had obtained the same shrinkage, in this case too the relative shear at \(\gamma=-10^\circ\) would be considerably greater than at \(\gamma=+10^\circ\). Thus, for \(\zeta=3\), the relative shear at \(\gamma=+10^\circ\) is \(\varepsilon=3\), while at \(\gamma=-10^\circ\), respectively, \(\varepsilon=3.7\), i.e., 23% higher.

In all the foregoing arguments and conclusions we have neglected the widening of the chip, assuming that it is insignificant and that the width of the removed chip \((b_1)\) is equal to the width of the cutting surface \((b)\). In some cases the widening of the chip is so considerable that it cannot be neglected, since this will lead to significant errors in calculating the relative shear. In these cases the calculation of the relative shear must be approached in the following way.

The equation for the relative shear

\[ \varepsilon = \operatorname{ctg}\beta_1 + \operatorname{tg}(\beta_1-\gamma) \]

remains valid also in the presence of chip widening, but the shear angle \(\beta_1\) must be determined with account taken of the chip widening.

Usually the angle \(\beta_1\) is calculated from the chip shrinkage

\[ \operatorname{tg}\beta_1=\frac{\cos\gamma}{\zeta-\sin\gamma}, \]

where \(\zeta\) is the longitudinal shrinkage of the chip.

It is considered that the longitudinal shrinkage \(\dfrac{l_0}{l}\) is equal to the transverse thickening \(\dfrac{a_1}{a}\), which follows automatically from the constancy of the volume of the metal when the chip is removed. But the equality of longitudinal shrinkage and transverse thickening proves to be correct only under the condition that there is no widening of the chip.

From the condition of constancy of the volume of the metal when the chip is removed it follows that

\[ l_0 f_0 = l f_1, \]

where \(f_0\) is the cross section of the layer of metal being removed before its deformation,
\(f_1\) is the cross section of the removed chip.

It may be assumed that, for small ratios \(\dfrac{a}{b}\), the section of the layer being removed, which before deformation had the form of a rectangle or parallelogram, and in the chip, i.e., after deformation, retained the form of a rectangle or parallelogram, but with increased thickness \((a_1>a)\) and increased width \((b_1>b)\); then \(l_0ab = la_1b_1\).

From Fig. 1 it is seen that the angle \(\beta_1\), which determines the magnitude of the relative shear, at a constant rake angle \(\gamma\) depends only on the transverse thickening of the chip. From Fig. 1 we have

\[ \frac{a}{\sin \beta_1}=\frac{a_1}{\cos(\beta_1-\gamma)}. \]

Hence

\[ \tg \beta_1=\frac{\cos \gamma}{\dfrac{a_1}{a}-\sin \gamma}. \]

It is difficult to measure the value \(a_1\) directly on the removed chip because of the irregularities on the inner side of the chip curl.

It is much more convenient to determine the longitudinal shrinkage of the chip \(\zeta=\dfrac{l_0}{l}\). In the presence of widening of the chip we must equate

\[ \frac{a_1}{a}=\frac{l_0}{l}\cdot\frac{b}{b_1}=\zeta\frac{b}{b_1}. \]

Then we shall have

\[ \tg \beta_1=\frac{\cos \gamma}{\dfrac{b}{b_1}\zeta-\sin \gamma}. \tag{5} \]

Having determined \(\beta_1\) from equation (5) and substituted it into (1), we obtain the value of the relative shear. If the relative shear is expressed through the longitudinal shrinkage, the following equation will be obtained:

\[ \varepsilon= \frac{ \left(\dfrac{b}{b_1}\zeta\right)^2 +2\dfrac{b}{b_1}\zeta\sin\gamma+1 }{ \dfrac{b}{b_1}\zeta\cos\gamma }. \tag{6} \]

In the case when the thickness of the layer being removed, \(a\), approaches in magnitude or is equal to the width of the layer being removed, \(b\), the cross section of the chip removed will be neither a rectangle nor a parallelogram. In this case it usually has the form of a trapezoid with a convex smaller base. Here the quantities \(a_1\), \(\beta_1\), and \(\varepsilon\) will vary over the width of the chip. The average values of \(\beta_1\) and \(\varepsilon\) can be calculated from some average thickness \(a_1\).