dnom 2 Substituting known values, the nominal depth of cut (in mm) is

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1 35 Given: boring number of teeth, N t = corner radius, r ε = 0 lead angle, ψ r = 15 back rake angle, γ p = 0 side rake angle, γ f = 0 feed rate, f r = 040 mm/rev spindle speed, n s =600 rpm initial workpiece diameter, D w = 96 mm tool diameter, D t = 100 mm spindle-work axis offset (mag, ε wo = 10 mm spindle-work axis offset (dir, δ wo = 10 specific cutting and thrust energy models a The axis offset causes the depth of cut to vary with tooth angle about the nominal depth of cut The nominal depth of cut is Dw Dt dnom = Substituting known values, the nominal depth of cut (in mm is d nom = = The minimum and maximum are d nom B ε wo and occur at δ wo and δ wo 180, respectively Substituting known values, the final result (in mm is d = d ε = 1 = 1 at θ = 10 and d = d + ε = + 1 = at θ = 30 min nom wo mind max nom wo maxd b The minimum and maximum tooth-local forces (cutting and thrust occur at d min and d max, respectively The cutting force acting on a tooth at angle i is FC( θi = uc( θi ftd( θi The feed per tooth (in mm/t is the feed per revolution (in mm/rev divided by the number of teeth: fr 04 ft = = = 0 Nt The specific energy requires the uncut chip thickness and normal rake angle The normal rake angle for this neutral rake case (γ p = γ f = 0 is, by inspection, zero Since the corner radius is zero, the uncut chip thickness is simply h = ft cosψ r Substituting known values, the specific cutting energy (in N/mm is C i C Therefore, the final result (in N is and F ( (0 u ( θ = u = 000 0cos(15 e = 3018 ( θ u f d( θ 3018(0( at θ θ 10 = = = = = C minc C t mind minc mind F ( θ u f d( θ 3018(0( at θ θ 30 = = = = = C maxc C t maxd maxc maxd c Based on the geometry of two offset circles, show that the exact solution for the depth of cut as a function of tooth angle θ i is The tool and workpiece are represented by circles with the tool being centered on the x-y coordinate frame and the workpiece centered at coordinates (x wc, y wc = (ε wo cosδ wo, ε wo sinδ wo Given the distance from the origin to the tip of a tooth, r t (, and the distance from the origin to the workpiece circle at that same angle r w (, the depth of cut is d( θ = r( θ r ( θ i t w

2 For the case of work-spindle offset, r t ( = R t for all (actually only relevant at each tooth The workpiece counterpart (r w ( is found using the equation of the workpiece circle w wc w wc w ( x x + ( y y = R, where (x w, y w is a point on the workpiece circle Expanding yields rearranging yields ( ( w wc w wc w wc w wc w x x x + x + y y y + y = R ; ( ( w w wc w wc w wc wc w x + y ( x x + y y + x + y R = 0 It is also known that all points on the workpiece circle satisfy xw = rwcos θ, yw = rwsin θ, and xw + yw = rw, where r w is shorthand for r w ( Substituting for x w, y w and x w + y w yields Furthermore, x wo + y wo = ε wo, so this becomes ( θ θ ( w w wc wc wc wc w r r x cos + y sin + x + y R = 0 This is a quadratic in r w, the solution of which is ( w w wc θ wc θ εwo w r r x cos + y sin + R = 0 ( xwc θ + ywc θ ± ( xwc θ + ywc θ ( εwo R w cos sin 4 cos sin 4 rw = 1/ = ( xwc cosθ + ywc sinθ ± ( xwc cosθ ywc sin θ ( εwo R w + To get this in terms of only the runout parameter ε wo and δ wo, rather than x wc and y wc, the substitution xwc = εwo cosδwo and ywc = εwo sinδwo is made This results in the presence of the term ε wo (cosδ wo cos + sinδ wo sin, which can be replaced by the trigonometric identity ε cosδ cosθ + sinδ sinθ = ε cos δ θ = ε cos θ δ After simplifying, the result is [ ] ( ( wo wo wo wo wo wo wo * Making the trigonometric substitution results in r ( cos ( ( R 1/ ( ( ( R = ε cos δ θ ± ε δ θ ε w wo wo wo wo wo w = ε cos δ θ ± ε cos δ θ 1 + wo wo wo wo w ( δ θ = ( δ θ wo cos 1 sin ( R sin ( 1/ rw = εwo cos δwo θ ± w εwo δwo θ Factoring out ε wo, replacing with i for tooth i, using the + of the ±,and subtracting from R t yields the depth of cut result being sought: ( cos( ( sin ( 1/ d θi = Rt rw = Rt εwo δwo θi + R w εwo δwo θ i wo 1/ 1/ * Continue from here in alternate solution below

3 The form of this result highlights large terms, in particular (R w /ε wo Finding the small terms to ignore is facilitated by dividing the term in braces by R w /ε wo to obtain ( ( cos( 1 ( sin ( 1/ d θi = Rt Rw ε wo Rw δwo θi + ε wo Rw δwo θ i While the term ε wo /R w is small, setting it to zero would trivialize the solution to d( i = R t R w This is not sensible since runout is a small-term phenomena However, if ε wo /R w is small, then (R w /ε wo is very small, and that is what makes sense to ignore for simplification purposes Doing so results in ( ( d( θ = R R ε cos δ θ = d ε cos δ θ i t w wo wo i nom wo wo i Another approach to the problem is to employ the Law of Cosines to the triangle shown to the right The Law of Cosines yields ( w εwo wεwocos δwo θ w r + r = R Bringing R w to the left-hand side provides a quadratic in r w, the solution of which is r w ( ± ( ( R ε cos δ θ 4ε cos δ θ 4 ε = wo wo wo wo wo w ( cos ( ( R 1/ ( ( ( R = ε cos δ θ ± ε δ θ ε wo wo wo wo wo w = εwo cos δwo θ ± εwo cos δwo θ 1 + w Continue as above from the footnoted point The Law of Cosines allows skipping the steps involved with representing the center of the workpiece circle in Cartesian coordinates d Graph the x- and y-force components F x and F y (in N as functions of spindle angle s where, by convention, the angle of tooth one is 1 = s The cutting and thrust force components are, where u C was found earlier to be 3018 N/mm and u T (in N/mm is u T 1/ ( 05 15(0 1/ = 750 0cos(15 e = 1707 The x- and y-force components require the tangential and radial force components, which are ( ( F ( θ = F ( θ = u f d( θ and F ( θ = F ( θ sin ψ( θ = u f d( θ cos ψ( θ, Tan i C i C t i Rad i T i i T t i i where the equivalent lead angle, for this case of zero corner radius, simply equals the lead angle ψ r Substituting known values, the result (in N for d( i in mm is FTan ( θi = 3018(0 d( θi = 6036 d( θi and FRad ( θi = 1707(0 d( θi sin(15 = 884 d( θi The equations for the x- and y-force components are then or F ( θ = F ( θ sin θ F ( θ cosθ and F ( θ = F ( θ cos θ F ( θ sinθ, x i Tan i i Rad i i y i Tan i i Rad i i ( ( F ( θ = 6036sinθ 884cos θ d( θ and F ( θ = 6036cosθ 884sin θ d( θ x i i i i y i i i i The total forces on the cutter are the sum of those acting on the two teeth, ie, F = F + F and F = F + F, where θ = θ and θ = θ x x1 x y y1 y 1 s 1 s Implementing these in a spreadsheet along with the equation for d( i and graphing yields the result below

4 Fx Fy Force (N Spindle Angle (deg e As the corner radius is increased (eg, from the assumed value of zero above, r ε /d increases for any depth of cut Likewise, equivalent lead angle increases and average uncut chip thickness decreases Decreasing average uncut chip thickness, through the size effect, causes all the forces to increase Increasing equivalent lead angle causes the depth-direction force (F Rad here to increase Therefore, since all effects are to increase the radial force, the radial force will increase However, the tangential force also increases from the change in average uncut chip thickness As a result, it is difficult to say the what the net effect will be on F x and F y given the angle dependence on how these changes in F Rad and F Tan, not to mention one two teeth, will affect the total forces The plots below provide a graphical assessment of the effect 1000 Fx Fy 1000 Fx Fy Force (N Force (N Corner Radius = 1 mm (Solid Corner Radius = 0 mm (Dashed Spindle Angle (deg Corner Radius = mm (Solid Corner Radius = 0 mm (Dashed Spindle Angle (deg f Even with the runout, since it is not so large that the depth of cut becomes zero at some angles, the machined diameter is simply that of the tool, or 100 mm

5 41 Given: face milling number of teeth, N t = 8 corner radius, r ε = 10mm lead angle, ψ r = 0 back rake angle, γ p = 0 side rake angle, γ f = 0 tool diameter, D t = 700 mm feed rate, f r = 16 mm/rev depth of cut, d = 10 mm spindle speed, n s =600 rpm workpiece width, W w = 400 mm workpiece offset in y-direction, ε = 75 mm specific cutting and thrust energy models a The angle associated with a point lying on the circle representing the tool is 1 y θ = sin, = en, ex, Rt The y coordinate corresponding to the entry and exit points is W w y = # # ε, where the top sign ( is for tooth entry and the bottom sign (+ is for exit Therefore, substituting known values, the results (in mm are y en = 75 = 15 and y ex = + 75 = 75 Calculating the associated angles yields the final result (in deg to be θ en = sin = / and θ ex = sin = / b The tooth spacing t is simply 360 divided by the number of teeth, or 360 /8 = 45 If the total angle of engagement, ex en, is less than the tooth spacing, there will never be more than one tooth engaged at any point in time Furthermore, if the angle of engagement falls between the tooth spacing and two times then tooth spacing, then at all times there will be either one or two teeth engaged This caries on to the general relation that the minimum and maximum number of teeth engaged at any time is θex θen θex θen Ne = ceil and N floor max e = min, θt θt where the ceil and floor functions round the argument up and down, respectively, to an integer (note that when the argument is an integer, the ciel and floor functions return the same value, the integer argument Substituting know values, 5 ( 1 = 16 N = e ceil(16 = and N = floor(16 = 1 max e min 45 Therefore, single-tooth cutting does occur in this case for some period of time, ie, the answer is not zero When tooth i is entering the cut, its angle is i = en = 1 and the angle of the tooth it is following, tooth i+1, is i+1 = en + 45 = 4 When tooth i+1 reaches the exit point at i+1 = ex = 5, i = ex 45 = 7 Tooth i will then cut by itself from 7 until the tooth before it (tooth i 1 enters the cut, at which point i = 4 Therefore, the range over which a single tooth (tooth i this illustration is cutting is from 7 to 4 or 17 c For a corner-radiused tool as is the case here, the average uncut chip thickness is the chip area divided by the equivalent width of cut, which is the length of the cutting tooth profile that is engaged with the workpiece The chip area as a function of angle is

6 a( θi = ( ft cos θi d The width of cut is simplified in this case since the corner radius equals the depth of cut and the lead angle is zero, so that by inspection the depth of cut equals the transition depth of cut Therefore, noting that the feed per tooth is f t = f r /N t, π 1 ft cosθi w( θi = ψr + sin rε rε Substituting known values, the result (in mm as a function of angle is π 1 (16 / 8 cosθi π w( 0 sin (10 sin 1 θi = + = + ((01cosθi (10 Substituting the values of en and ex from above, the final results are ((16 / 8 cos( 1 (10 h en = = 011 π 1 + sin ((01cos( 1 and h ex ( (16 / 8 cos(5 (10 = = 010 π 1 + sin ((01 cos(5 d Fu s method is equivalent to connecting the tip of the tool (not the intersection of the two profiles at the cusp as for Colwell s method to the intersection of the tooth profile with the un-machined surface By inspection, for this special case as noted above, the equivalent lead angle will be 45 As such, F Rad ( i = F Lon ( i

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Analytic Trigonometry. Copyright Cengage Learning. All rights reserved. Analytic Trigonometry Copyright Cengage Learning. All rights reserved. 7.4 Basic Trigonometric Equations Copyright Cengage Learning. All rights reserved. Objectives Basic Trigonometric Equations Solving