Basic Twist Drill Geometry: Helix And Clearance Angles

Refer figure 2-2

Basic Twist Drill Geometry Helix and clearance angles

where ω=sin-1(W/r) web angle at a point on the lip

2π R tan δ O = L 2π r tan δ = L r tan δ = tan δ O R cos ω ( tan ClO − sin ω O . cot p ) + sin ω . cot p tan Cl = cos ω O

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Drill Point Specification Typical features identified in handbooks and standards

• D - nominal drill diameter • 2W - web thickness • δO - helix angle at outer corner • ClO - lip clearance angle at outer corner • 2p - point angle (of the lips) • ψ - chisel edge angle 04/29/09

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Refer figure 2-3

Circumferential Clearance Require hc>0 and Clc>0° for rc
As Ωc and Ωco approach zero, Clc and Clco approach Cl and Clo

hC hC tan ClC = = r.Ω C r ( Ω S + ω ) hC tan ClCO = R( Ω S + ω O ) hCO hCO tan ClCO = == R( Ω SO + ω O ) R.Ω CO 04/29/09

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Refer figure 2-4

Development of Flute Profile 1 Coordinates of a point C on the AB generator at radius r

x = r cos ω y = − r sin ω z = r cos ω cot p

Axial translation of a generator A1B1 rotated by β

β δz = L = β R cot δ O 2π

Coordinates of a point C1 on the A1B1 generator at radius r

x = r cos(ω + β ) y = − r sin(ω + β ) z = r cos ω cot p − β R cot δ O 04/29/09

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Refer figure 2-4

Rotation for C2 to cross reference normal plane (z = 0) Coordinates of point C2 on A2B2, i.e. equations for the flute profile in a normal plane to the drill axis

Development of Flute Profile 2 β=

r cos ω cot p tan δ O R

r cos ω cot p tan δ O   x = r cos ω +  R   r cos ω cot p tan δ O   y = −r sin  ω +  R   2 2  −1 W  r − W  x = r cos sin + cot p tan δ O    r R   2 2  −1 W  r − W  y = −r sin sin + cot p tan δ O    r R   z=0

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Refer figure 2-6

Conical (flank) grinding is commonly mentioned Five independent sharpening parameters

Point Sharpening It can be shown that: 2W = function (λ, Cx, Cy)

θ - semi-cone angle

2p = function(λ, χ, θ)

χ - angle between cone axis and drill axis

ψ = function(λ, χ, θ, Cx, Cy) Clo = function(λ, χ, θ, Cx, Cy, R)

Cx, Cy - drill location λ - orientation angle so that cone and flank parameters coincide 04/29/09

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Refer figure 2-6

Numerical simulation shows that low θ causes high Clco even whole flank ground and small γw and large θ causes negative Clco

Given variety of grinders actually used, range of values for features for General Purpose drills & production variability each drill behaves as an individual tool

Setting the Point Grinder Specifying Clco provides an extra equation to uniquely specify the grinder parameters Clco = function(λ, χ, θ, Cx, Cy, R)

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Refer figure 2-7

Covers a wider range of diameters than most “western” standards (0.1 ~ 100 mm)

Relates several features to diameter, rather than just giving a range (see table 2-2)

The Chinese Standard • Besides the usual point features being quantified, this standard also gives recommended values for other features such as the fluted land width, the margin width, the body clearance and projected distance between the heel corners • Extra information allows Clco to be calculated 04/29/09

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Refer figure 2-8

From the Chinese Standard • Clearer specification of point features together with Clco allows θ, χ, λ, Cx and Cy, to be found for conical grinding

.

Findings

• Up to 25 mm, large variability in setting values but beyond 25 mm angles are constant and linear distances proportional to diameter 04/29/09

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Refer figure 2-9

Suggested method of conical grinding found in a Chinese (and Russian) handbook Method is simplified

λ = 0° cannot give straight lips. Generated lips slightly curved.

Chinese Point Sharpening • For all diameters: χ = 45° and 13°<θ<15°, Cx =(.07 to .05)D and Cy, =(1.8 to 1.9)D • λ implied to be 0° • Satisfying 2W yields very small λ • Using these values will not satisfy standards for point features (table 2-3) 04/29/09

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Refer figure 2-10

Cutting Geometry and Action Apparent “instantaneous” area of cut = D.f/2 At least for some regions we will have to consider the dynamic geometry; so we need to know the cutting velocities. Since Vf << V in the lip region can let Vw = V, also for some of chisel edge

VW = V 2 + V f2 where V = 2πNr V f = f .N Vf

f .N f tan η = = = V 2π N r 2π r 04/29/09

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Refer figure 2-10

View A in Pfe, the working plane, contains V and Vf

Basic Geometry • In view A, for a point Q on a lip, can see

Pre is the working reference plane and is the radial plane through the drill axis

–δ – Cl

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Refer figure 2-10

Lip Region Geometry - 1 View B is in the normal plane Pn at Q

View C is in the working cutting edge plane, Pse

• In view B, at point Q on a lip, are shown: – normal rake and clearance angles – reference rake and clearance angles

• View C shows the cutting velocity and cutting edge in true length, hence shows inclination angle 04/29/09

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Refer figures 2-10 & 2-11

Lip Region Geometry - 2 From view A From projecting point Q onto view B

x. cos ω z.sin p − x.sin ω cos p tan δ cos ω = sin p − tan δ sin ω cos p

Vw sin ω cos p = tan ω cos p Vw cos ω

tan α n =

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z1 x

α n = α ref − ε tan ε =

Substituting for δ and ω

tan Cl =

and

tan α ref = tan α ref

From view B

x z

tan δ =

tanα n =

tan δ cos ω + sin ω tan ω cos 2 p − tan ω cos p sin p

(

)

(

)

tan δ O r 2 − W 2 sin 2 p − R.W .sin p cos p R.sin p r 2 − W 2 04/29/09

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Refer figures 2-10 & 2-11

Clref found by projecting points Q and b from view A to view B, and substituting for z1 From view B

Lip Region Geometry 3 tan Clref tan Clref

x.sin ω − z1 sin p = x. cos ω = tan ω cos p − tan Cl sin p secω

Cln = ε − Clref tan Cl sin p cos ω + tan ω cos p( sin ω cos p − tan Cl sin p ) sin p cos ω tan ClO + cos p sin ( ω − ω O ) tan Cln = cos ω cos ω O − sin ω cos p( sin p tan ClO − cos p sin ω O ) tan Cln =

tan Cln

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( R sin p tan ClO − W cos p ) =

r 2 − W 2 + W cos p R 2 − W 2

r 2 − W 2 R 2 − W 2 − W cos p( R sin p tan ClO − W cos p ) 04/29/09

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Refer figures 2-10 & 2-11

Lip Region Geometry 4 Inclination i for point Q on lips is shown in view C

Cutting action at lips is a complex oblique process with variable Vw, αn and i

VW sin ω sin p sin i = VW = sin ω sin p W = sin p r

Ambiguity of flank specification does not affect lip geometry previous

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Refer figures 2-12 & 2-13

Require detailed knowledge of flank geometry & grinding method Assume straight line normal to drill axis Cannot ignore Vf on dynamic angles Below some limit radius Clne becomes negative and edge acts as an indentor

Chisel Edge Geometry

γW γ1 = γ 2 = 2 α nst = γ 1 Clnst = 90° − γ 2

α ne = −( α n − η ) = −( γ 1 − η )

Clne = Cln − η = (90° − γ 2 ) − η

Inclination very small previous

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Typical forms of empirical equations. Given variations in design & production results can be considered only approximate.

Boston & Oxford (¼” to 1½”)

Kronenberg previous

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Forces and Power - empirical Thrust = C1 f x1 D y1 Torque = C2 f x 2 D y 2 Power = Torque.N / C3 + Thrust.Vf ≈ Torque.N / C3 Power = C2 f x 2 D y 2 N / C3

Cast Iron:

Th = C1.f 0.6.D Tq = C2.f 0.6.D2

Steel:

Th = C1’.f 0.6.D Tq = C2’.f 0.78.D1.8 Tq = C2’.f 0.803.D1.803 04/29/09

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Forces and Power - Empirical Dimensional analysis by Shaw & Oxford

Th K1 f 1− a = 2 D BHN D1+ a

 1 − 2rc / D 1− a  2 ( ) ( ) + K 2 r / D + K 2 r / D   2 c 3 c a 1 + ( 2rc / D ) 

Tq K 3 f 1− a = 3 D BHN D1+ a

 1 − 2rc / D 2−a  + K ( 2 r / D )   5 c a ( ) 1 + 2 r / D c  

where u ∝ ( f .D )

Assuming a = 0.2 for steel, and testing general purpose drills with 2rc/D =0.18

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a

Th = 0.195.(BHN).f 0.8.D0.8 + 0.0022.(BHN).D2 or Th = C4.(BHN).f 0.63.D Tq = 0.087.(BHN).f 0.8.D1.8

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Refer figure 2-14a

Mechanics of Cutting Approach - Lip Region 1

τ, rl, β, K1P & K1Q found from orthogonal tests and thin shear zone model δTh j = [( δFQ + δFQe ) cos ε sin p − ( δFR + δFRe )( cos i cos p + sin i sin p sin ε ) ] δTq j = rj ( δFP + δFPe ) M

Th  = 2∑ δTh j 1

M

Tq  = 2∑ δTq j 1

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Refer figure 2-14a

Mechanics of Cutting Approach - Lip Region 2

Required elemental geometry

δL =

R cos ω O − W cot (180° − ψ ) M  sin p

rj =

[ R cosωO − ( j − 1 2)δL sin p] 2 + W 2

f sin p cos ε t= 2 δb = δL cos i δA = t.δb previous

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Mechanics of Cutting Approach - Lip Region 3

From classical oblique cutting thin shear zone model

δFP = δFQ = δFR =

τ .δA[ cos( β n − α n ) cos i + tan ηc sin i sin β n ]

sin φn cos i cos 2 ( φn + β n − α n ) + tan 2 ηc sin 2 β n

τ .δA sin ( β n − α n )

sin φn cos i cos 2 ( φn + β n − α n ) + tan 2 ηc sin 2 β n

τ .δA[ cos( β n − α n ) sin i − tan ηc cos i cos β n ]

sin φn cos i cos 2 ( φn + β n − α n ) + tan 2 ηc sin 2 β n

tan β n = tan β cosη

rl ( cosηc / cos i ) cosα n tan φn = 1 − rl ( cosηc / cos i ) sin α n tan i cosα n tan ( φn + β n ) = tan ηc − sin α n tan i previous

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Mechanics of Cutting Approach - Lip Region 4

Elemental ‘edge force’ components

δFPe = K1P δb δFQe = K1Q δb δFRe = K1R δb ≈ K1P sin iδb or ≈ 0

Putting it all together

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Th  and Tq  = functions{ D,2W ,2 p, δ O ,ψ , f , N , M  ,τ , rl , β , K1P , K1Q }

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Refer figure 2-14b

Mechanics of Cutting Approach - Chisel Edge 1

Elemental forces

For rlimit ≤ r ≤ rc δTh ck = δFPc sin η + δFQc cosη

δTq ck = rk ( δFPc cosη − δFQc sin η ) Mc

Total thrust and torque on chisel edge

Th c = 2∑ δTh ck 1

Mc

Tq c = 2∑ δTq ck 1

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Refer figure 2-14b

Mechanics of Cutting Approach - Chisel Edge 2

Elemental cut geometry

δb =

rc − rlim it Lc − 2rlim it = Mc 2M c

rk = rc − (k − 1 2 )δb = Lc / 2 − (k − 1 2 )δb t = ( f cosη ) / 2

Force coefficients C1P and C1Q for δFPc = C1Pδb discontinuous chips found from ‘special’ δFQc = C1Qδb orthogonal database Overall, for the chisel edge

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Th c and Tq c = functions{2W ,2 p,ψ , γ w , γ 2 , f , N , M c , C1P , C1Q }

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Refer flowchart

Mechanics of Cutting Approach

Overall results:

Th t = Th  + Th c + Th I Tq t = Tq  + Tq c Th t & Tq t = fns{ D,2W ,2 p, δ O ,ψ , γ w , γ 2 , f , N , M  , M c ,τ , rl , β , K1P , K1Q , C1P , C1Q }

Analysis is complicated so computerisation is useful

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Refer figure 2-15

Predicted Characteristics From application of Mechanics of Cutting model

For Plane Flank model

Compared with empirical results

Th t = 47.5 f .545 D1.036 (2W / D).339 2 p.168δ O−.206 2γ W.311ψ .266

(N)

Tq t = 4.56 f .660 D 2.004 (2W / D).149 2 p −.241δ O−.263 2γ W−.008ψ −.188

Th t = 99.7 f .546 D1.027 (2W / D).279 2 p.518δ O−.210ψ .050 Tq t = 3.71 f .661D 2.004 (2W / D ).113 2 p −.226δ O−.263ψ −.177

(Nm)

(N) (Nm)

Th = C1’.f 0.6.D Tq = C2’.f 0.78.D1.8 Tq = C2’.f 0.803.D1.803

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Refer figure 2-17

2 p1 = 2 p2 All specified angles must be equal

Geometric Similarity

ψ1 = ψ 2 Clo1 = Clo 2 δ o1 = δ o 2

⇒

L1 D1 = L2 D2

Clco1 = Clco 2 All specified lengths must be in proportion

All angles at corresponding points r1 & r2 must be equal previous

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2W1 B1 2rc1 D1 = = = 2W2 B2 2rc 2 D2

ω1 = ω 2 , Cl1 = Cl2 , δ 1 = δ 2 , α n1 = α n 2 , Cln1 = Cln 2 , i1 = i2 04/29/09

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Refer figure 2-18

Forces for Geometric Similarity - 1 Uth and Utq are forces per unit area.

δTh1 j = 2.U th1 j δA1 j = U th1 j f .δr1;

δTq1 j = 2.δFP1 j r1 j = 2.U tq1 j δA1 j r1 j = U tq1 j f .r1 j δr1 δTq 2 j = U tq 2 j f .r2 j δr2 where D δr1 = 1 ; M

For geometrically similar annuli, fundamental angles are equal, so the specific forces are too. previous

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δTh 2 j = U th 2 j f .δr2

δr2 =

U th1 j = U th 2 j = U thj

D2 M and U tq1 j = U tq 2 j = U tqj

r1 j

δr1 D1 = = r2 j δr2 D2 04/29/09

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Refer figure 2-18

Forces for Geometric Similarity - 2 Applying equations from previous slide

Applying to whole of the drills

C1 and C2 depend on feed and work material previous

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δTh1 j = U Thj f .δr2

D1 D = δTh2 j 1 D2 D2 2

 D1   D1    δTq1 j = U Tqj f .r2 j δr2   = δTq2 j    D2   D2 

2

M

D1 M D1 Tht1 = ∑ δTh1 j = δTh2 j = Tht 2 ∑ D D2 j =1 2 j =1  D1  Tqt1 = ∑ δTq1 j =   j =1  D2  M

2

 D1  δTq2 j = Tqt 2   ∑ j =1  D2  M

2

Tht = C1D Tqt = C2 D 2 04/29/09

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