International Journal of Mechanical Sciences 47 (2005) 902–921
An analysis of force distribution in shear spinning of cone
Ming-Der Chen
a, Ray-Quan Hsu
a,, Kuang-Hua Fuh
ba
Department of Mechanical Engineering, National Chiao Tung University, 1001 Ta-Hsueh Road, 300 Hsinchu, Taiwan b
Department of Mechanical and Marine Engineering, National Taiwan Ocean University, Taiwan Received 1 July 2003; received in revised form 20 December 2004; accepted 21 January 2005
Available online 7 April 2005
Abstract
An analytic model for calculation of shear spinning force incorporate factor of over-roll (press down) of the blank is derived. The effects of blank thickness, roller nose radius, mandrel revolution and roller feed on the spinning force are discussed. Results obtained from calculation were compared with the experiment and other theoretical predictions. It is found that the present findings yield optimum results.
r2005 Elsevier Ltd. All rights reserved.
Keywords: Shear spinning; Shear force; Spun cone; Surface roughness; Over-roll depth
1. Introduction
Shear spinning is a technique for manufacturing cone shape products by virtue of a roller and a rotating male form. As shown in Fig. 1, conical parts are produced by pressing the spinning roller on to the sheet blank which in turn is mounted on a rotating mandrel. Under the pressure of the roller, the material is axially displaced whereby the blank thickness is reduced. In this process, radial-axis position of the blank element was considered fix during deformation as depicted inFig. 1.
Shear spinning process is adopted by the industry for manufacturing dished ends for boilers and tanks, wheel rims, silencer parts, nozzle, etc. As indicated by Held[1], in the spinning process, the roller press a small depth on the blank, over-roll, is important to confirm the accuracy of shape
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and dimensions, on the other hand, the over-roll can eliminate the rough surface of the blank material and produce the new surface on finished part. A way to estimate forces required for the process incorporating the over-roll of blank thus is essential for the design of the tool and selection of the machinery.
Hayama et al. [2] did consider five factors namely diameter of mandrel, corner radius of mandrel, roller diameter, roller nose radius and mandrel rotation to determine the pass schedule of roller. They also investigated experimental results of three force components with respect to stroke, blank thickness, conic angle and roller feed [3], however, no equation for spinning force calculation related to spinning parameters were proposed.
Avitzur and Yang[4]first derived the tangential force equation for spinning of cones. They also showed an approximate method to calculate the power. Kalpakcioglu[5]developed the tangential force and the specific energy equations under the assumption of simple shear deformation. Kobayashi and Thomsen[6]derived the tangential force equation assuming that the infinitesimal
Nomenclature
Cs over-roll depth (mm) f roller feed (mm/rev)
w specific deformation energy (N-mm=mm3) dv strained metal volume (mm3)
Dii; d¯ increment of normal strain due to bending, infinitesimal effective strain
sm; ¯s mean effective stress; effective stress (N=mm2)
R, y; Z cylindrical co-ordinates X, Y, Z rectangular co-ordinates g fraction
rR roller nose radius (mm) rn radius of curvature (mm)
h distance of a point from the neutral surface in the thickness direction (mm) t0; tf original; final thickness of blank (mm)
N mandrel revolution, (rev/min) a half-apex angle of mandrel (deg) c roller contact angle (deg)
y0; y0 deformation zone angle and its average value (rad)
y00; y00 angle functionally related to y0 and y0 (rad)
u1; u2 ratios of strains
DR roller diameter (mm)
At; Ar; Az deformation areas of the contact surface between roller and cone in
circumfer-ential, radial, and axial directions, respectively (mm2)
Fp; Fq force components in parallel and perpendicular directions to the roller axis,
respectively (N)
Fr; Fz force components in the radial and axial directions, respectively (N)
strain ratios remain constant during the spinning process. The radial and axial force components were calculated by using the projected contact area ratios.
However, all the studies stated above concerned only the power consumption in spinning of cones, the effect of individual spinning parameters on the spinning forces were never discussed. In this study, the effects of blank thickness, roller nose radius, mandrel revolution, roller feed and over-roll depth on the spinning forces are taken into consideration and incorporated in the proposed equations. Predictions calculated from the equations are compared with the experimental results. It is observed that the proposed equations yield the better results.
2. Analysis of shear spinning process
As depicted inFig. 1, in the process of shear spinning, the spinning roller presses on a rotating sheet blank, force it to conform with the contour of the mandrel. Deformation of the sheet blank in this process is a combination of bending and shearing. In the course of forming, it is assumed that the radial distance of any point on the neutral line (point E1or E2) remains unchanged, and
the line sections AB and CD in Fig. 1 hold straight and normal to the surface throughout the deformation. It means that after bending, the neutral line section E1E2 is also perpendicular to
line sections GH and IJ. On the other hand, any point within the blank is assumed to retain its radial distance in shearing. Line sections AB and CD change to A0B0 and C0D0 and still hold
parallel to the Z-axis. Besides, AB ¼ A0B0 and CD ¼ C0D0 [4,6]. In addition, the following assumptions were adopted:
1. The material is homogeneous, isotropic rigid-plastic body, without Bauschinger effect and volumetric change and non-strain hardening.
2. Material follows von Mises yield rule.
3. The frictional force between roller and blank is negligible, strain rate effects and temperature effects are also neglected.
As shown in Fig. 2, the initial blank thickness t0 and final thickness tf can be related by the
following correlation, known as the sine law in shear spinning:
tf ¼ ðt0CsÞ sin a, (1)
here Cs is over-roll depth and a is half-cone angle. The force applied by the roller on the blank can be resolved into three mutually perpendicular components, the circumferential force, Ft; feed
force, Fp; and the radial force, Fq (perpendicular to Fp). The order of their magnitude as pointed
out by Avitzur and Kegg is Fq4Fp4Ft[4,7]. The largest of the three Fq; however, does no work
at all, because it is not associated with any displacement. The displacement in the direction of Fp
is small compared with the circumferential direction ðf sin a ðmm=rev:Þ52pR ðmm=rev:ÞÞ; thus the work done by the force Fpcan be neglected compared with the work done by Ft: Hence, most of
the power supplied by the motor driving the chuck of the spinning machine is transmitted into torque through the circumferential force Ft[4]. Thus, the external work input is approximated by
Ftdl; where dl is the roller displacement during an infinitesimal time interval dt: On the other
hand, the total deformation work in an infinitesimal time interval is expressed by
dW ¼ w dv, (2)
where w is the specific work (work of deformation per unit volume) and dv is the volume of the blank which was strained in the same time interval. Equating the external work input to the work of deformation given by Eq. (2), permits the following energy balance:
Ftdl ffi w dv ¼ dv
Z
¯s d¯, (3)
where ¯s and d¯ are effective stress and infinitesimal effective strain, respectively. Eq. (3) can also be written as Ft¼ dv dl Z ¯s d¯ ¼ ðt0CsÞ sin a f Z ¯s d¯ ¼ At Z ¯s d¯, (4)
where f is the roller feed. At is the circumferential contact area of roller and blank.
2.1. Finite strain
The strain rates of deformation in cylindrical polar coordinates ðR; y; ZÞ assume the form (see
Fig. 2): RR¼ quR qR; yy¼ 1 R quy qy þ uR R; ZZ¼ quZ qZ , Ry¼ quy qRþ 1 R quR qy uy R, yZ ¼ 1 R quZ qy þ quy qZ, RZ¼ quZ qR þ quR qZ , (5) where RR; yy; ZZ; Ry; yZ and
RZ are the components of the normal strain-rate tensor field, u
R;
uy and u
Z are the components of velocity-vector field.
Total effective plastic strain increment dp; which is proportional to the root mean
square of the differences of the principal plastic strain increments and shear strain takes the form of dp¼ 29½ðdRRdyyÞ2þ ðdyydZZÞ2þ ðdyydRRÞ2 þ13½d2Ryþd 2 yZþd 2 RZ 1=2 . (6) Because the angular velocity of the spun cone is constant, the circumferential velocity at any point on the cone becomes
uy¼2pRN, (7)
where uyis the circumferential velocity (mm/min), R the radius at the point considered (mm) and
N the rotation of the blank per minute (rev/min).
Also, since there is no radial displacement for any point on the cone during spinning, the radial velocity uR will be:
uR¼0. (8)
The geometry of contact area between roller and blank in Z-axis can be expressed in the following form:
Then dZ ¼qZ qR dR þ qZ qy dy þ qZ qn dn (10) and uZ ¼ dZ dT ¼ qZ qR dR dTþ qZ qy dy dTþ qZ qn dn dT, (11)
where n is the number of revolutions passed from time T0to the instant, T the time passed from
T0 to the instant and u
Z the velocity component in Z-direction.
Now n ¼ NðT T0Þ; dn dT ¼N, (12) dy dT ¼ uy R¼2pN, (13) dR dT ¼u R ¼0. (14)
Thus, Eq. (11) can be simplified uZ ¼ qZ qy2pN þ qZ qn N ¼N 2pqZ qy þ qZ qn . ð15Þ
Replace (7), (8), (13) and (14), into (5), it becomes RR¼0; yy¼ 1 R quy qy ; ZZ¼ quZ qZ , Ry¼0, yZ ¼ 1 R quZ qy , RZ¼ quZ qR . (16)
Fig. 3 is a radial view of roller/blank interaction. Where r is the radius of curvature of blank/ roller contact line PQ and h is the distance of a point in the thickness-direction from the neutral surface (see the right side ofFig. 3). Suppose h5rnand r ffi rnþh; where rnis the radius of the curvature PnQn; then the bending strain Daffih=r [8], since it may be assumed that DaffiDyy;
to bending is given by Dyy¼ DZZffi h rffi t0 4rn.
Substitution of Eq. (16) into Eq. (6) yields: dp¼ 29ð6D2yyÞ þ13ðD 2 yZþD 2 RZÞ 1=2 , dp¼ 1 ffiffiffi 3 p DRZ½ð1 þ u21þu22Þ1=2, where u1¼ Dyy DRZ=2 , u2¼ DyZ DRZ . (17)
Moreover, the strain DyZ of Eq. (16) in Fig. 4can be rewritten as
DyZ ¼ 1 RD quz qy
¼tan d; tan d ) tan d0 (18)
where d0 is the angle between the tangent to the blank/roller contact curve P0 nQ
0 nQ
00
n and
the horizontal line at point Q0n on Fig. 4. Assuming the curve PnQn is an arc with radius rn; its
center is located on the Z0-axis, and by neglecting the term ðgZ0 nÞ
2 as compared with ðR ny00Þ
2
(Appendix A). We obtain rnffiðRny0Þ
2
2Z0 n
, (19)
tan d0ffi 2Z 0 n Rny0 . (20) Noting that Xn0 1 Z0 n¼rR 1 sin a1 , (21) Xn0 1 DRZ¼cot a (22)
and combining Eqs. (18)–(22), Dyy¼ t0=2 ðRnynÞ2 rR 1 sin a sin a , (23) DyZ ¼ 2 Rny0 rR 1 sin a sin a . (24)
Here, y0 is the angle of the roller/blank contact line as projected on to the R2y plane given
by Fig. 5: y0¼cos1 b2 a2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ ðRb 0Þ 2 ðba221Þ q b2 a21 2 4 3 5, (25)
where a ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r2 R ðrRf cos aÞ2 q , b2¼DR ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r2
R ½ðrRf cos aÞ cos c2
q
½ðrRf cos aÞ sin c
þr2R ðrRf cos aÞ2 (26) and rR is the roller nose radius, DR is the roller diameter, a is one half the cone angle, Rn¼
ðRiþR0Þ=2 ¼ R0 ðrRcos a=2Þ; and c is the roller contact angle, respectively (Appendix B).
Combining Eqs. (22)–(24), substituting into Eq. (17), the total effective strain Rd¯ becomes Z dp¼ Xn0 1 Dn¼ cot a ffiffiffi 3 p ð1 þ u21þu22Þ1=2, (27) u1¼ t0 ðRny0Þ2 rR 1 sin a cos a , where u2¼ 2 Rny0 rR 1 sin a cos a . 2.2. Spinning force
In the experiment, the circumferential, parallel and perpendicular force components Ft; Fpand
Fq; respectively, were measured by three-channel dynamometer. The shear spinning force on the
roller can be resolved into three mutually perpendicular components as the Ft (circumferential),
Fr (radial) and Fz (axial) as indicated byFig. 6.
Force components Fr and Fz are related to Fp and Fq by
Fr ¼Fqcos a Fp sin a, (28)
Fz ¼Fqsin a þ Fp cos a. (29)
Therefore, the force components Fr and Fz can be calculated from measured force
components Fp and Fq and half-cone angle. Circumferential force Ft is measured by
dynamometer directly.
In shear spinning process, the over-roll on blank thickness is adjusted to obtain the required surface roughness. In practice, when the roller set an amount of over-roll on the blank, it will cause a small portion of blank material curl-up to envelope the outer rim of roller nose as indicated inFig. 7. To simplify the analysis, the power consumed by this curl-up deformation is neglected.
Fig. 7. Overlapof the blank material on the roller. Fig. 6. Spinning force components.
2.3. Contact deformation area
Fig. 8 depicts the circumferential deformation area of roller and blank in shear
spinning process. Details of the deformation area with reduction in blank thickness (over-roll) is shown in the lower part of the drawing. When a roller is moving with a constant feed f and an over-roll depth Cs, the roller/blank contact point is changing from A1 to A2; and A3
finally. The radial component of A123A3 is equal to ðf sin aÞ cos a: When no over-roll of
the blank is performed in the process, points B1; B2 are the first contact points of the
roller with the blank flange at one feed apart, respectively. Hence, in the case where no over-roll is taken place, the deformation area is the domain enclosed by the dots section ðA122B1Þ; ðA122B2Þ and ðB1B2Þ: When over-roll of the blank is performed, point B3 becomes
the final contact point after a feed. The amount of deformation area increased in the case where over-roll exists is indicated by the cross hatching. The distance B1B3 in the axial direction is
computed as
ðB1B3ÞZffi ðf þ Cs tan aÞ cos a. (30)
Therefore, the total contact areas indicated by dotted hatch in Fig. 8 can be approximately computed as follows:
At ffi12ðf þ Cs tan aÞ cos aðrR cos a þ f sin a cos aÞ. (31)
On the other hand,Fig. 9presents the radial contact area of roller and blank. On the left side of the drawing is a magnified details of the contact area. The area can be calculated approximately as follows:
Ar ffi12ðf þ Cs tan aÞ cos a12DRy0, (32)
where DR is the roller diameter (mm).
Similarly,Fig. 10illustrates the axial contact area of roller and blank flange. Magnification of this area is presented on left side of the drawing. Same as above, the area is approximately
Fig. 9. Radial view of the roller/blank contact area.
calculated as follows:
Az ffi32ðf þ Cs tan aÞ sin a12DRy0 (33)
Substitute Eq. (31) into Eq. (4), we obtain the circumferential force as follows: Ftffi12ðf þ Cs tan aÞ cos aðrR cos a þ f sin a cos aÞ sm
Z d¯
, (34)
where sm and ¯ are the mean effective stress and strain.
Furthermore, the radial and axial force Fr and Fz can be expressed as following:
Fr ffi14ðf þ Cs tan aÞ cos aðDRy0Þ sm
Z d¯
, (35)
Fz ffi34ðf þ Cs tan aÞ sin aðDRy0Þ sm
Z d¯
. (36)
3. Experimental set-up
Fig. 11presents the schematic diagram of shear spinning experimental set-up and spinning force measuring system. A modified CNC spinning device is driven by a 15 hp DC motor on its spindle, and longitudinal and latitude power rates are 5 and 3 hp, respectively. A special fixture holds the blank at its rim and, can move concurrently with roller. A three-channel dynamometer (Kistler 9257A) measures the shear spinning force. The force output signals were amplified through a
three-channel charge amplifier (Kistler 5807A). For the convenience of analysis, a data recorder (KYOWA RTP 670A) was connected to the charge amplifier. The recorded data were carefully analyzed through a waveform analyzer (Data Precision Model 6100). A CNC control unit adjusts the mandrel revolution and roller feed. The shear spinning processes can be executed through a numerical control program. Material selected for the experiment is Al 1100-O and its mean effective stress sm and strain ¯ relationshipis illustrated onFig. 12. The experimental conditions
employed in spinning are given inTable 1.
4. Results and discussion
Fig. 13 depicts the three force components as measured in the experiments. The experiments were repeated ten times. The axial force Fz is the largest among three components ðFz4Fr4FtÞ: Fig. 14 expresses the force components Ft; Fr and Fz as function of blank thickness. For
rR¼4:8 mm; N ¼ 60 rev= min; f ¼ 0:16 mm=rev and Cs ¼ 0:5 mm; the experimental values are indicated by solid dots, calculated results from Eqs. (34)–(36) are shown by solid lines. The results by Kobayashi and Thomsen[6] and Kegg[7], which only Ft is proposed are also shown for the
purpose of comparison. Obviously, the predictions derived from Eqs. (34)–(36), which take into consideration the factor of over-roll (Cs) is in better accordance with the experiments. The reason
Table 1
The experimental conditions employed in spinning
Blank thickness, t0(mm) 1.5; 2.59; 4.11; 6.0; 7.0 A1-1100-O
Roller nose radius, rR(mm) 2.5; 4.0; 4.8; 5.5; 7.1 SAE 4130 Mandrel revolution, N (rev/min) 40; 60
Roller feed, f (mm/rev) 0.1; 0.13; 0.16; 0.18; 0.2 Over-roll depth, Cs (mm) 0.3; 0.4; 0.5; 0.6; 0.7
Cone angle, a (deg) 50
Roller contact angle, c (deg) 60
for the experimental value Fz at t0¼7:0 mm is much larger than the calculation may be
due to the spinning force is approaching the machine limit. In addition, Fig. 15 demonstrates the effect of over-roll factor on spinning force components. When t0¼4:11 mm; rR ¼4:8 mm;
Fig. 14. Spinning force components as function of blank thickness. Fig. 13. Spinning force components vs. the serial number of repeated experiments.
N ¼ 60 rev= min and f ¼ 0:16 mm=rev; the greater over-roll depth means deeper contact between roller/blank, thus the force exerted by the roller of course increases. But from the experimental results, we noticed that the circumferential and radial force components for Cs ¼ 0:5 mm; were smaller than Cs ¼ 0:3 mm: This may be explained by the fact that, during the spinning process, a very thin oxide layer accumulated on the roller nose/blank contact area. When the roller presses on the blank, it causes this oxide layer to crack. If the over-roll is deep enough, the whole oxide layer tends to shear-off, which in turn reduces the shear force. When over-roll depth is set deeper than 0.5 mm, the reduction of force caused by the shear-off of the oxide layer is more than compensated by the greater amount of energy required by the deformation of the blank, thus the force increases again. Fig. 16 illustrates the circumferential force component with respect to roller nose radius. While t0¼2:59 mm; N ¼ 40 rev= min;
Cs ¼ 0:5 mm; f ¼ 0:13 mm=rev: (index of ’) and f ¼ 0:18 mm=rev: (index of ), it reveals that in the calculation, the dependence on the roller nose radius is much more apparent than in the experiment, however, both the prediction and the experimental results show that the circumferential force component increases with roller nose radius, since larger nose radius induces larger contact area and thus required larger force. Furthermore, the faster roller feed needs more energy to deform the material at unit revolution, thus increases the spinning force. Finally,Fig. 17 depicts the circumferential force components relative to the roller feed. At fixed t0¼4:11 mm; rR ¼4:8 mm; N ¼ 60 rev= min and Cs ¼ 0:5 mm; the larger roller feed produces
Fig. 16. Circumferential force component with respect to roller nose radius.
larger contact length which in turn induces larger contact area, thus higher spinning force. However, the increase in the circumferential force is not obviously, since the roller feed is small in the range of 0.1 and 0.2 mm/rev.
5. Conclusions
In this study of the analysis of shear spinning force, the following conclusions have been drawn. 1. An analytical model incorporating over-roll of the blank is proposed for the calculation of
shear spinning forces.
2. The equations derived contain five parameters of the shear spinning process, namely blank thickness, roller nose radius, mandrel revolution, roller feed and over-roll depth. The effect of these parameters on shear spinning forces is discussed.
3. Shear spinning force calculated from equations derived in this study yield optimum results when over-roll of the blank is taken place in the process.
Appendix A. Referring to Fig. 4
AQ0n2¼AQnAC, ðRny00Þ 2¼ ð2r ngZ 0 nÞgZ 0 n¼2rngZ 0 n ðgZ 0 nÞ 2, rn ðRny0 0 Þ2 2gZ0 n ðRny0Þ 2 2Z0 n , tan d0¼AQ 0 n AO ¼ Rny00 rn2gZ0 n ¼ Rny 0 0 ðRny00Þ 2 2g2Z02 n=2gZ 0 n Rny 0 02gZ0n ðRny00Þ 2 2Z0 n Rny0 .
Appendix B. The derivation of Eq. (25)
Referring to Fig. 5, the following relationships are obtained: Transformation: 0ðx; y; zÞ ! 00ðx0; y0; z0Þ
x ¼ x0 cos c z0 sin c DR 2 , y ¼ y0
A torus body equation: ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2þy2 p DR 2 2 r2Rþz2¼0 (B.2) or ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x0 cos c z0 sin c DR 2 2 þy02 s DR 2 2 4 3 5 2 r2Rþ ðx0 sin c þ z0 cos cÞ2¼0 (B.3)
trace of roller sectioned by the plane PP. By substitution of z0 ¼z0
0 into Eq. (B.3), ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x0 cos c z0 0 sin c DR 2 2 þy02 s DR 2 2 4 3 5 2 r2Rþ ðx0 sin c þ z00 cos cÞ2¼0. (B.4)
Lengths a and b: substituting y ¼ 0; z0¼r1¼rRf cos a and x ¼ 0; z0¼r1¼rRf cos a into
Eq. (B.4), we get a ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffi r2 0r21 q ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r2 R ðrRf cos aÞ2 q , (B.5) where r0¼rR and ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0 r1 sin c DR 2 2 þb2 s DR 2 2 4 3 5 2 r2Rþ ð0 þ r1 cos cÞ2¼0, b2¼r2Rr21þDR ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r2 R ðr1 cos cÞ2 q r1 sin c , where r1¼rRf cos a b2¼DR ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r2
R ½ðrRf cos aÞ cos c2
q
½ðrRf cos aÞ sin c
þr2R ðrRf cos aÞ2 the roller trace equation
x0 0 a 2 þ y 0 0 b 2 ¼1 (B.6)
the cross section of cone contact with roller
x002þy002 ¼2R0jx00j (B.7)
solving the equations of (B.6) and (B.7) ðb2a2Þx002¼a2b22a2R0jx00j,
substituting K ¼ a2R 0=ðb2a2Þ; we get ðx00þKÞ2 ¼ a 2b2 b2a2þK 2, jx00j ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2b2 b2a2þK 2 s K, jx00j ¼ R0 ðb2=a2Þ 1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi b R0 2 b2 a21 þ1 s 1 2 4 3 5, (B.8) y0¼cos1 R0 jx00j R0 , y0¼cos1 ðb2=a2Þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ ðb=R0Þ2ðb2=a21Þ q ðb2=a2Þ 1 2 4 3 5. (B.9) References
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[4] Avitzur B, Yang CT. Analysis of power spinning of cones. Journal of Engineering for Industry, Transactions of the ASME 1960;82:231–45.
[5] Kalpakcioglu S. On the mechanics of shear spinning. Journal of Engineering for Industry. Transactions of the ASME Series B 1961;83:125–30.
[6] Kobayashi S, Thomsen EG. A theory of shear spinning of cones. Journal of Engineering for Industry. Transactions of the ASME 1961;83:485–95.
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