- 234 Views
- Uploaded on

Download Presentation
## Machining Operations by Ed Red

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

- Introduce machining operations terminology
- Introduce machining efficiency measures
- Reconsider cutting parameters as they apply to efficiency
- Review a machining efficiency example
- Consider modern machine operations (papers)

- Chatter – interrupted cutting usually at some frequency
- Down milling – cutting speed in same direction as part feed
- Up milling – cutting speed in opposite direction as part feed
- Peripheral milling – tool parallel to work
- Face milling – tool perpendicular to work
- Ideal roughness – geometrically determined roughness
- Machinability – machining success determined by tool life, surface finish
- Optimal machining – parameter choices that increase machining throughput or reduce operational costs

(other than normal turning)

Facing

Taper

Form

Contour

Threading

Cutoff

Chamfer

Boring

Knurling

Drilling

Peripheral milling cutting positions

Face milling cutting positions

Offset face cut

Full face cut

Spindle rpm related to cutter diameter and speed:

N (rpm)= v/(p D)

Feedrate in in/min:

fr = N nt f

where f = feed per tooth

nt = number of teeth

MRR is

MRR =w d fr

Slab milling:

Approach distance, A :

A = d (D-d)

Time to mill workpiece, Tm:

Tm = (L + A)/fr

Face milling:

Allow for over-travel O where A = O:

Full face A = O = D/2

Partial face A = O = w (D – w)

Machining time:

Tm = (L + 2A)/fr

Milling time analysis - example

Problem statement:

A face milling operation is performed to finish the top surface of a steel rectangular workpiece 12 in. long by 2 in. wide. The milling cutter has 4 teeth (cemented carbide inserts) and is 3 in. in diameter. Cutting conditions are 500 fpm, f = 0.01 in./tooth, and d = 0.150 in. Determine the time to make one pass across the surface and the metal removal rate during the cut.

Milling time analysis - example

Solution? Numbers?

Full face A = O = D/2

Machining time Tm = (L + 2A)/fr

Metal removal rate MRR = w d fr

Feedrate in in/min fr = N nt f

N (rpm)= v/(p D)

Ideal roughness,

Ri = f2/(32 NR)

where NR = tool nose radius

Actual roughness,

Ra = rai Ri (about 2 x Ri )

because of edge effects, chip interactions, surface tearing, etc.

What is a free machining steel?

http://www.sandmeyersteel.com/303.html#1

- Machinability is a measure of machining success or ease of machining.
- Suitable criteria:
- tool life or tool speed
- level of forces
- surface finish
- ease of chip disposal

Problem statement:

A series of tool life tests is conducted on two work materials under identical cutting conditions, varying only speed in the test procedure. The first material, defined as the base material, yields the Taylor tool life equation

v T0.28= 1050

and the other material (test material) yields the Taylor equation

v T0.27= 1320

Determine the machinability rating of the test material using the cutting speed that provides a 60 min. tool life as the basis of comparison. This speed is denoted by v60.

Solution:

The base material has a machinability rating = 1.0. Its v60 value can be determined from the Taylor tool life equation as follows:

v60 =1050/600.28= 334 ft/min

The cutting speed at a 60 min. tool life for the test material is determined similarly:

v60 =1320/600.27= 437 ft/min

Accordingly, the machinability rating can be calculated as

MR (for the test material) = 437/374 = 1.31 (or 131%)

Cutting speed can be chosen to maximize the production rate or minimize the cost per part (or unit) produced. This is referred to as optimized machining because more than one production variable contributes to the production rate and costs.

Variables:Th - part handling time Co (Cg) – operator (grinder’s) cost rate/min

Tm – machining time Ch – cost of part handling time

Tt – tool change time Cm – cost of machining time

np – number of parts cut by Ctc – cost of tool change time tool during tool life

Tc – cycle time per part Ct – cost per cutting edge

T – tool life Ctp = Ct/np - tool cost per part

Maximum production rate - turning

Total time per part produced (cycle time):

Tc = Th + Tm + Tt/np

where Tt/np is the tool change time per part.

Consider a turning operation. The machining time is given by

Tm = p D L/(v f)

The number of parts cut per tool is given by

np = T/Tm= f C(1/n)/(p D L v(1/n -1) )

Maximum production rate - turning

Substituting, we get the total cutting time

Tc = Th + p D L/(v f) + Tt[ p D L v(1/n -1)/( f C(1/n) )]

Minimizing cycle time (dTc/dv = 0 ) gives optimum (max) cutting speed and tool life:

vmax = C/[(1 - n) Tt/n]n

Tmax = (1 - n) Tt /n

Minimum cost per unit - turning

Cost of part handling time:

Ch = CoTh

Cost of machining time:

Cm = CoTm

Cost of tool change time:

Ctc = CoTt /np

Minimum cost per unit - turning

Tool cost per part:

Ctp = Ct /np

Tooling cost per edge:

Disposable inserts Ct = Pt /nene = num of edges/insert Pt = original cost of tool

Single point grindable Ct = Pt /ng + Tg Cg “includes purchase price”ng = Num tool lives/tool Tg = time to grind tool

Minimum cost per unit - turning

Total cost per part:

Cc = Co Th + Co Tm + Co Tt /np + Ct /np

Substituting for Tm and np:

Cc = Co Th + Cop DL/fv + (CoTt + Ct )pDLv(1/n -1)/( f C(1/n) )Minimizing cost per part (dCc/dv = 0) gives cutting speed and tool life to minimize machining costs per part:

vmin = C{n Co/[(1 – n)(Ct + CoTt)]}n

Tmin = (1 – n) (Ct + CoTt)/(n Co)

Minimum cost per unit - example

Problem statement:

Suppose a turning operation is to be performed with HSS tooling on mild steel (n = 0.125, C = 200 from text table). The workpart has length = 20.0 in. and diameter = 4.0 in· Feed = 0.010 in./rev. Handling time per piece = 5.0 min and tool change time = 2.0 min. Cost of machine and operator = $30.00/hr, and tooling cost = $3.00 per cutting edge. Find (a) cutting speed for maximum production rate and (b) cutting speed for minimum cost

Minimum cost per unit - example

Solution:

Cutting speed for maximum production rate is

vmax = C/[(1 - n) Tt/n]n = 200/[(.875) 2/0.125]0.125 = 144 ft/min

Converting Co from $30/hr to $0.5/min, the cutting speed for minimum cost is given by

vmin = C{n Co/[(1 – n)(Ct + CoTt)]}n = = 200{(0.125)(0.5)/[(0.875)(3.00 + (0.5)(2))]}0.125 = 121 ft/min

What did we learn?

Download Presentation

Connecting to Server..