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SEMINAR ON MECHANICS OF METAL CUTTING

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(1)

MECHANICS OF METAL CUTTING

(feed force)

Main Cutting force Radial

force Tool feed

direction

(2)

Topics to be covered

 Tool terminologies and geometry

 Orthogonal Vs Oblique cutting

 Turning Forces

 Velocity diagram

 Merchants Circle

 Power & Energies

(3)

Need for calculating forces, velocities and angles during machining??

• We need to determine the cutting forces in turning for Estimation of cutting power consumption, which also

enables selection of the power source(s) during design of the machine tools.

• Structural design of the machine – fixture – tool system.

• Evaluation of role of the various machining parameters (tool material and geometry) on cutting forces to make machining process more efficient and economical.

• Condition monitoring of the cutting tools and machine tools.

(4)

To o l

Wo rk p ie c e

Ch ip

Heat Generation Zones

(Dependent on sharpness of tool)

(Dependent on m) (Dependent on f)

10%

30%

60%

(5)

Tool Terminology

Side relief angle

Side cutting edge angle (SCEA) Clearance or end

relief angle

Back Rake(BR),+

Side Rake (SR), +

End Cutting edge angle (ECEA)

Nose Radius

Turning Cutting

edge Facing

Cutting edge

(6)

Cutting Geometry

(7)

Material Removal Rate

MRR vfd

Roughing(R)

f 0.4 1.25mm / rev d 2.5 20mm

Finishing(F)

f 0.125 0.4mm / rev d 0.75 2.0mm

vR  vF

(8)

METAL CUTTING

ORTHOGONAL CUTTING OBLIQUE CUTTING

Cutting Edge is normal to tool feed.

Here only two force components are considered i.e. cutting force and thrust force. Hence known as two dimensional cutting.

Shear force acts on smaller area.

Cutting Edge is inclined at an acute angle to tool feed.

Here only three force components are considered i.e. cutting force, radial force and thrust force. Hence known as three dimensional cutting.

Shear force acts on larger area.

Metal Cutting is the process of removing unwanted material from the workpiece in the form of chips

(9)

Assumptions

(Orthogonal Cutting Model)

 The cutting edge is a straight line extending perpendicular to the direction of motion, and it generates a plane surface as the work moves past it.

 The tool is perfectly sharp (no contact along the clearance face).

 The shearing surface is a plane extending upward from the cutting edge.

 The chip does not flow to either side

 The depth of cut/chip thickness is constant uniform relative velocity between work and tool

 Continuous chip, no built-up-edge (BUE)

(10)

TERMINOLOGY

(11)

TERMINOLOGY

α : Rack angle

 b : Frictional angle

ϕ : Shear angle

Ft : Thrust Force

Fc: Cutting Force

Fs: Shear Force

Fn: Normal Shear Force

F: Frictional Force

N: Normal Frictional Force

V: Feed velocity

Vc: Chip velocity

Vs: Shear velocity

(12)

FL

FC

Fr

D IR ECT IO N O F R O TAT IO N

W O R K PIEC E

CU TT IN G TO O L D IR ECT IO N O F FEED

Ve lo c i t y o f

To o l r e la t i v e t o w o r k p i e c e V

Longitudinal Force

Radial Force

‘Thrust’ Force Tangential Force 'Cutting' Force

Forces For Orthogonal Model

End view Note: For the 2D Orthogonal Mechanistic

Model we will ignore the Longitudinal component

12

(13)

Orthogonal Cutting Model

(Simple 2D mechanistic model)

Mechanism: Chips produced by the shearing process along the shear plane a

t0

f

+ Rake Angle Chip

Workpiece Clearance Angle Shear Angle

tc

depth of cut

Chip thickness

Tool

Velocity V

tool

13

(14)

Orthogonal Cutting

r to

tc lssinf ls cos(f a)

tan f r cosa 1 rsina

AC

BD AD DC

BD tan(f a) cotf

14

(15)

tool

Cutting Ratio

(or chip thicknes ratio)

As Sinf = to

AB and Cosf-a) = tc AB Chip thickness ratio (r) = t0

tc = sinf cos(f a)

f

t

c

t

o

fa)

A B

Chip

Workpiece

(16)

Experimental Determination of Cutting Ratio

Shear angle f may be obtained either from photo-micrographs or assume volume continuity (no chip density change):

Since t0w0L0 = tcwcLc and w0=wc (exp. evidence) Cu t t in g ra t io, r = t0

tc = Lc L0

i.e. Measure length of chips (easier than thickness)

w

t

L

0

0

0

wc

Lc

tc

(17)

Shear Plane Length and Angle f

Shear plane length AB = t0 sinf

Shear plane angle (f) = Tan-1 rcosa 1-rsina

or make an assumption, such as f adjusts to minimize cutting force: f = 450 + a/2 - b/2 (Merchant)

f

tc to

fa)

A B

Chip

tool

Workpiece

(18)

Velocities

(2D Orthogonal Model)

Velocity Diagram

From mass continuity: Vto = Vctc

From the Ve loci ty diagram: Vs = V cosa

cos(f a) Vc = Vr and Vc = V sinf

cos(f a)

(Chip relative to workpiece)

V = Chip Velocity

(Chip relative to tool)

Tool

Workpiece

Chip

Vs V = Cutting Velocity

(Tool relative to workpiece)

Shear Velocity

c

a

f  a

90  f f

Vc Vs

V

(19)

Cutting Forces ( 2D Orthogonal Cutting)

Free Body Diagram

Generally we know:

Tool geometry & type Workpiece material and we wish to know:

F = Cutting Force F = Thrust Force F = Friction Force N = Normal Force F = Shear Force F = Force Normal

to Shear

c t

s n

Tool

Workpiece

Chip

Dynamometer R

R

R

R

Ft Fc

f Fs

Fn N

F

19

(20)

Fs , Resistance to shear of the metal in forming the chip. It acts along the shear plane.

Fn , ‘Backing up’ force on the chip provided by the workpiece. Acts normal to the shear plane.

N, It at the tool chip interface normal to the cutting face of the tool and is provided by the tool.

F, It is the frictional resistance of the tool acting on the chip.

It acts downward against the motion of the chip as it glides upwards along the tool face.

Cutting Forces ( 2D Orthogonal Cutting)

20

(21)

Fn

Ft

CONSTRUCTION OF MERCHANT’S CIRCLE

Fs

Fc

F R

α

α φ

λ

φ λ-α

N

V

Knowing Fc , Ft , α and ϕ, all other component forces can be calculated.

Please note l is same as b in next slide = friction angle

21

(22)

Force Circle Diagram

(Merchants Circle)

R Ft

Fc

Tool

F

N a

b  a b

a

a Fs

f b  a

f

Fn

22

(23)

23

(24)

Cutting Forces

• Forces considered in orthogonal cutting include

Cutting, friction (tool face), and shear forces

Cutting force,Fc acts in the direction of the cutting speed V, and supplies the energy required for cutting

Ratio of Fc to cross-sectional area being cut (i.e. product of width and depth of cut, t0) is called: specific cutting force

Thrust force,Ft acts in a direction normal to the cutting force

• These two forces produces the resultant force, R

• On tool face, resultant force can be resolved into:

Friction force, F along the tool-chip interface

Normal force, N to to friction force 24

(25)

Cutting Forces

• It can also be shown that (b is friction angle)

• Resultant force, R is balanced by an equal and opposite force along the shear plane

• It is resolved into shear force, Fs and normal force, Fn

• Thus,

• The magnitude of coefficient of friction, m is b

b cos

sin N R

R

F

f f

f f

cos sin

sin cos

t c

n

t c

s

F F

F

F F

F

a m a

tan tan

t c

c t

F F

F F

N F

25

(26)

Cutting Forces

• The toolholder, work-holding devices, and machine tool must be stiff to support thrust force with minimal

deflections

If Ft is too high tool will be pushed away from workpiece this will reduce depth of cut and dimensional accuracy

• The effect of rake angle and friction angle on the direction of thrust force is

• Magnitude of the cutting force, Fc is always positive as the force that supplies the work is required in cutting

• However, Ft can be +ve or –ve; i.e. Ft can be upward with a) high rake angle, b) low tool-chip friction, or c) both

b a

)

Rsin Ft

26

(27)

Forces from Merchant's Circle

Friction Force F = Fcsina + Ftcosa Normal Force N = Fccosa - Ftsina

Shear Force Fs = Fccosf - Ftsinf

m = F/N and m = tanb typically 0.5 - 2.0)

Force Normal to Shear plane Fn = Fcsinf + Ftcosf

(28)

Stresses

On the Shear plane:

Normal Stress = s = Normal Force / Area = Fn

AB w = Fnsinf tow Shear Stress = s = Shear Force / Area = Fs

AB w = Fssinf tow

On the tool rake face:

 = Normal Force / Area = N

tc w (often assume tc = contact length)

 = Shear Force / Area = F tc w

(29)

Pow

•Power (or energy consumed per unit time) is the product of

er

force and velocity. Power at the cutting spindle:

•Power is dissipated mainly in the shear zone and on the rake face:

•Actual Motor Power requirements will depend on machine efficiency E (%):

Cutting Power Pc = FcV

Power for Shearing Ps = FsVs Friction Power Pf = FVc

Motor Power Required = Pc

E x 100

(30)

Material Removal Rate (MRR)

Material Removal Rate (MRR) = Volume Removed Time

Volume Removed = Lwto

Time to move a distance L = L/V Therefore, MRR = Lwto

L/V = Vwto

MRR = Cutting velocity x width of cut x depth of cut

(31)

Specific Cutting Energy

(or Unit Power)

Energy required to remove a unit volume of material (often quoted as a function of workpiece material, tool and process:

Ut = Energy

Volume Removed = Energy per unit time

Volume Removed per unit time

Specific Energy for shearing Us = FsVs Vwto Ut = Cutting Power (Pc)

Material Removal Rate (MRR) = FcV

Vwto = Fc wto

Specific Energy for friction Uf = FVc

Vwto = Fr

wto = F

wtc = 

(32)

Specific Cutting Energy

Decomposition

1. Shear Energy/unit volume (Us)

(required for deformation in shear zone) 2. Friction Energy/unit volume (Uf)

(expended as chip slides along rake face) 3. Chip curl energy/unit volume (Uc)

(expended in curling the chip)

4. Kinetic Energy/unit volume (Um) (required to accelerate chip)

Ut = Us + Uf +Uc +Um

(33)

Cutting Forces and Power measurement

Measuring Cutting Forces and Power

• Cutting forces can be measured using a force

transducer, a dynamometer or a load cell mounted on the cutting-tool holder

• It is also possible to calculate the cutting force from the power consumption during cutting (provided

mechanical efficiency of the tool can be determined)

• The specific energy (u) in cutting can be used to calculate cutting forces

33

(34)

Cutting Forces and Power

Power

34

Prediction of forces is based largely on

experimental data (right)

Wide ranges of values is due to differences in

material strengths

Sharpness of the tool tip also influences forces and power

Duller tools require

higher forces and power

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