Grinding A

GrindingGrinding

Illegitimi non carborundumg

ver. 1ver. 1

ME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 2009

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OverviewOverview

• Processes• AnalysisAnalysis


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Horizontal Grinding


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Horizontal grinding Vertical grinding


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Centered grindingCentered grinding

Centerless grinding


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Creep Feed GrindingCreep Feed Grinding


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Grinding WheelsGrinding Wheels


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Grinding WheelsGrinding Wheels


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Grinding Wheel Information


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Correctly Mounted Wheel


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Grinding Wheel SurfaceGrinding Wheel Surface


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Grind Wheel DressingGrind Wheel Dressing


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Grinding Wheel DressingGrinding Wheel Dressing


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Grinding ChipsGrinding Chips


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Chip formation geometryChip formation geometry

w

l D

θ

t l

V

dt

θ

l

v


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Chip geometryChip geometry

• As with rolling contact length, the chip length,As with rolling contact length, the chip length, l– D = wheel diameter, d = depth of cut

dDl ⋅=• Material removal rate, MRR

– v = workpiece velocity, d = depth of cut, b = width p y pof cut

bdvMRR ⋅⋅=


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Material removal rateMaterial removal rate

Th hi h t i l• The chips have a triangular cross-section, and ratio (r) of chip thickness (t) t hi idth ( ) wto chip width (w)

2010 tw

w

• So, the average volume per chip

2010 tot

r ≈=t l

So, the average volume per chip

wtlltwVolchip 41

21

21

=⋅⋅=


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chip 422

ChipsChips

• The number of chips removed per unit• The number of chips removed per unit time (n), where c = number of cutting edges (grains) per unit area (typ 0 1 toedges (grains) per unit area (typ. 0.1 to 10 per mm2, and V = peripheral wheel velocityvelocity

cbVn ⋅⋅=


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CombiningCombining

VolnbdvMRR ⋅=⋅⋅= chipVolnbdvMRR ==

1 wtlVbcbdv41⋅=⋅⋅

trw ⋅= dDl ⋅=

dDttrbcVbdv ⋅⋅⋅⋅⋅⋅⋅=⋅⋅41


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Chip thicknessChip thickness

dv4dDrcV

dvt⋅⋅⋅

⋅=

42

or

Dd

Vvt = 4

DrcV ⋅⋅


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Specific grinding energy uSpecific grinding energy, u

• Consist of chip formation, plowing, and slidingg

uuuu ++= slidingplowingchip uuuu ++=


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Total grinding forceTotal grinding force

• Get force from power• Get force from power

MRRuPower ×= MRRuPower

bdvuVF ⋅⋅×=⋅ bdvuVFgrinding ⋅⋅×=⋅

bdvuF ⋅⋅×=

VuFgrinding ×=


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Total grinding forceTotal grinding force

F i i l lt t d• From empirical results, as t decreases, the friction component of u increases

tu 1∝

tKu 1

1 ⋅=ort

substituting

bd1V

bdvt

KFgrinding⋅⋅

⋅⋅=1

1


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Total grinding forceTotal grinding forceSubstituting for t

Vbdv

dKFgrinding

⋅⋅⋅⋅=

4

11 V

Dd

rcVv

⋅⋅⋅

4

irearranging

vrcd dDV

vrcdbKFgrinding ⋅⋅⋅⋅⋅

⋅⋅=41


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Force on a grainForce on a grain

• The force per grain can be calculatedThe force per grain can be calculated

AreauFgrain ×=g

wtuFgrain 21

×=rtw =

andKu 1

=grain 2 tKu 1 ⋅=

11 ttrt

KFgrain ⋅⋅⋅×⋅=211

1


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Force on a grainForce on a grainsubstituting for t, and rearranging

dvrKF i ⋅⋅=41

DrcVrFgrain ⋅⋅2

Dd

cVrvKFgrain ⋅⋅

⋅= 1


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Grinding temperatureGrinding temperature

• Temperature rise goes with energy delivered per unit area

inputEnergyKTΔarea

pgyKT ⋅=Δ 2

dlbu ⋅⋅× 1 dt

KKlb

dlbuKT ⋅⋅⋅=⋅⋅⋅×

⋅=Δ1

122

1

Dd

Vcrv

dKKT4

121 ⋅⋅⋅=Δ


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DVcr

Grinding temperatureGrinding temperature

• Rearranging

dDv

rcVKKT ⋅⋅⋅

⋅⋅=Δ421

• Temperatures can be up to 1600oC, but for a short time.


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Grinding – Ex. 1-1g

• You are grinding a steel, which has a specific grinding energy (u) of 35 W-s/mm3.

• The grinding wheel rotates at 3600 rpm, has a diameter (D) of 150 mm thickness (b) of 25 mm anddiameter (D) of 150 mm, thickness (b) of 25 mm, and (c) 5 grains per mm2 (c). The motor has a power of 2 kW.

• The work piece moves (v) at 1.5 m/min. The chip thickness ratio (r) is 10.

• Determine the grinding force and force per grain• Determine the grinding force and force per grain.• Determine the temperature (K2 is 0.2oK-m/N). Room

temperature is 20oC.


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Grinding Ex 1 2Grinding – Ex. 1-2

• First we need to calculate the depth of pcut. We can do this from the power.

bdvuMRRuPower ⋅⋅⋅=×=

min2mmmsWsec60

min1025min

5.1352000 26

3 ××⋅⋅⋅−

=m

mmmmdmmm

sWW

md 6104.91 −×=


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• Now for the total grinding forceg g

VbdvuFgrinding⋅⋅

×=Vgrinding

mmmmmmsWF d

25104.91min

150035

3 ⋅×⋅⋅

−=

−

mmm

revmmrevmm

Fgrinding

1000150

min3600

35 3⋅⋅

⋅=π

NFgrinding 7.70=


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Grinding Ex 1 4Grinding – Ex. 1-4• Next, the force per grain

wtuFgrain 21

×= rtw =and

ttruFgrain ⋅⋅×=21

we need twe need t

mmmm

dvt 104.91min150044 3−×⋅

mmmm

grainsmmDrcVt

150105min

1503600min

2 ⋅⋅⋅=

⋅⋅=

π

mmt 310321 −×


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mmt 1032.1 ×=


• SubstitutingSubstituting

( )231032.1102135

21 −×⋅⋅⋅=⋅⋅×= ttruFgrain ( )

22grain

NF 410053 −× NFgrain 1005.3 ×=


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• For the temperature, we need K1 and p , 1K2. K2 is given, so we need to calculate K1.1

trKttrt

KttruFgrain ⋅⋅⋅=⋅⋅⋅×⋅=⋅⋅⋅×=21

211

21

11 t 222

mKN 61

4 1032.110211005.3 −− ×⋅⋅⋅=×2

mNK 2.461 =


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Grinding Ex 1 7

• substituting

Grinding – Ex. 1-7

• substituting

dKKT =Δ1

NmK 1 6

dt

KKT ⋅⋅⋅=Δ 12

Kmmm

NN

mKT °=×⋅×

⋅⋅−

=Δ −− 640104.91

1032.112.462.0 6

6

CTTT initial °=+=Δ+= 66064020


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SummarySummary

• Overview of processes• Analysis of processAnalysis of process• Example problem


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Documents

Grinding A