Final Exam Material - MMAE 485 PDF

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Prof. John Cesarone...

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Chapter 21: Fundamentals of Machining Broad term for wide variety of subtractive manufacturing processes Turning, milling, boring Material is removed by a cutting tool Removed material forms chips of various shapes + sizes Relative motion between work and tool is different for each type of machining But chip-making is all the same Material being removed will have some volume, so must have 3 dimensions Dimensions come from a speed, feed, depth of cut Speed has a “per time” in it, so together we get a Material Removal Rate, MRR, in3/min Speed describes how fast the tool is moving relative to the surface of the workpiece. V Depth of cut (DoC) describes how deeply we are cutting into workpiece Feed describes how much we increment tool on each pass Many independent variables to select o Speed, feed, DoC o Tool material, shape, coatings o Use of cutting fluid Dependent variables (results) o Types of chip produced o Forces in tool + workpiece o Temperature rise o Tool wear + failure o Surface finish The Orthogonal (2D) Cutting Model o Rake angle Alpha is angle of rake face from vertical, shown positive in diagram o Relief angle behind + below tool o Tool slices int part at a depth of t0 which is also the DoC o Metal is removed by shearing + slides up rake force o Shearing takes place in a shear plane at some shear angle, φ o Or sometimes a shear zone, usually considered inferior for quality o Chip does not bend up rake face, but shears like a deck of cards o Chip thickness, tc will be greater than t0 o Chip thickness is dependent on Alpha, φ, and t0 o We can calculate φ if we measure chip thickness, tc o Define a cutting ratio (chip thickness ratio): r = t0/tc and then tan φ = [r*cos(Alpha)] / [1 – r*sin(Alpha)] o If we already know φ, we can predict chip thickness:  r = t0/tc = (sin φ) / (cos[φ-Alpha]) o Since tc > t0, r < 1.00 o Reciprocal or r is chip compression ratio o DoC = t0 = “undeformed chip thickness” o Shear strain: ϒ = cot φ + tan(φ - Alpha) o Also, since tc > t0, chip velocity, Vc must be less than cutting speed V  Vc = V*r = [V*sinφ] / [cos(φ - Alpha)] Types of Chips Produced o Chips are scrap, but their condition can give us clues to other conditions o They can cause problems o Note rake-face side of chip will be smooth + shiny o Other side will be rough + jagged o (1) Continuous Chips  Cutting ductile material at high speeds + high rake angles, we can get long continuous chips  May indicate a shear zone, leading to poor finish + residual stresses  Can cause tangles in machinery  Chip breakers are features that force chips to bend + break

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(2) Built Up Edge (BUE)  Sometimes material gradually builds up on tool tip, dulling edge  As it gets larger, becomes unstable, breaks off, starts building up again  Cycle reduces surface quality  Sometimes we get a thin, stable BUE  Protects rake face, reduces wear (3) Serrated (Segmented) Chips  Semi-continuous, with alternating zones of high + low shear strain  Metals with high thermal softening (4) Discontinuous Chips  May or may not be connected  Occur with brittle materials  Variable cutting forces, and vibrations + chatter  poor quality  greater tool wear

Lecture 18, Monday 10/21/19   

Exam 2 moved from 11/6/19 to 11/11/19 due to a seminar, which Professor recommends to attend Professor solved Homework Set 6 on the board Homework Set 7 is assigned

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Oblique Cutting o Generally used instead of orthogonal o Tool enters work at an inclination angle, i o Chip becomes helical, gets pushed to the side o “rake angle” gets more complicated Cutting Forces + Power o Fc = Cutting Force, in direction of tool speed, V o Ft = thrust force, normal to travel  Positive if toward workpiece  Negative if away from o R = Resultant (force) of Fc + Ft Sometimes we resolve R into other components o F = friction force along tool/chip interface o N = normal force perpendicular to tool/chip interface o F = R*sin(Beta) o N = R*cos(Beta) o Beta = “friction angle” between R + N Resolving R for Shear Plane o Fs = shear force in shear plane o Fn = normal force, perpendicular to shear plane o Fs + Fn can be found from Fc + Ft by:  Fs = Fccos φ - Ftsinφ  Fn = Fcsin φ + Ftcosφ o Coefficient of friction, “mu” found as:  “mu” = F/N = [Ft + Fctan(Alpha)] / [Fc – Fttan(Alpha)] o Note that Fc (cutting force) is always positive, toward tool o But Ft (thrust force) can be positive (toward work) or negative (up + away) o High rake angles + low friction can make it go negative Power needed for cutting is P=FcV o This is dissipated by shearing in the shear plane/zone, and friction on rake face o P = Pshearing + Pfriction = FsVs + FVc o Specific energy (w-s/mm 3 or hp-mm/in3) is:

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Ut = (FsVs + FVc)/(wt0V) w = width of the cut

Example 21.1 o Orthogonal cutting, t0 = 0.005 in, V = 400 ft/min, Alpha = 10°, w = 0.25 in o Force transducers on tool tell us that:  Fc = 125 lb, Ft = +50 lb o Chip thickness = 0.009 inch o Find Percent of total energy that goes into overcoming friction o (Friction Energy/Total Energy) = (FVc/FcV) = ([F*r]/[FcV]), r = Vc/V o r also = t0/tc = 5/9 = 0.555 o R = (Ft2 + Fc2)0.5 = (502 + 1252)0.5 = 135 lb o Fc = R*cos(Beta - Alpha)  125 = 135cos(Beta – 10)  Beta = 32° o F = R*sinBeta = 135sin(32) = 71.5 lb o (F*r)/Fc = (71.5*0.555)/125 = 32% Temperatures in Cutting o Dissipated energy turns to heat o Temperature rise lowers strength + wear resist o May deform tool, causing inaccuracies o Could cause expansion of workpiece o Could cause metallurgical changes in work o Estimate of mean temperature:  T = ( [1.2Yf] / [ρc] ) * (Vt0/k)1/3 where:  Yf = psi  ρc = (inch-lb)/(in3-°F)  k = thermal diffusivity in in2/second o Could require iterative solution

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Tool Wear + Failure o All tools wear, gradually but unavoidably o Many types of wear:  (1) Flank Wear  Occurs on relief surface (flank) of tool, behind cutting point  Caused by rubbing of tool on surface  (2) Crater Wear – on rake force of tool  Affects tool-chip contact geometry  (3) Nose Wear – rounding + dulling of the sharp cutting edge  (4) Plastic Deformation, from high temperature  (5) Notches, groves, chips, etc.  Most are accelerated by faster speeds + high temperatures o Speed, feed, DoC  Gradual deterioration of workpiece quality, and increased cutting force Lecture 19, Wednesday 10/23/19 

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Taylor Tool Life Equation (1890s) o VTn = c, o V = cutting speed o T = tool life, minutes o n,c = experimentally determined  for various materials Tool Life o T = (c/v)1/n

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o Speed (V) most important o Feed, DoC also matter For turning: o (VTn)(dx)(fy) = C o T = (C1/n)(V-1/n)(d-x/n)(f-y/n) o If typical values were inserted, then it would be: o T = (C7)(V-7)(d-1)(f-4) o Curves compiled for various materials, used for process design Example 21.2 o Assume n = 0.5, c = 400 o How much will tool life increase if we cut speed in half? o VT0.5 = 400 o V2 = 0.5V1 o V1(T1)0.5 = 400; V2(T2)0.5 = 400 o V1(T1)0.5 = V2(T2)0.5  V1(T1)0.5 = 0.5V1(T2)0.5 o (T2/T1) = 1/0.25 = 4.00 o Tool life quadruples at half speed o Didn’t really need c o Remember this is an approximation Machinability o Qualitative assessment of how “easy” a material is to machine o Based on four factors:  (1) Surface finish + integrity of machined part  (2) Tool Life  (3) Force + Power Required  (4) Chip Control Chapter 22: Tool Materials

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Tools need high hardness, strength, + wear resistance, at high temperatures High toughness + impact strength High thermal shock resistance Chemical stability + inertness with respect to work piece material Lots of research! (for over a hundred years!) Several classes (1) High Speed Steels o One of the first tool steels, named for ability to cut at high speeds o Ironically, it is now one of the slowest tool materials o Early 1900s o Highly alloyed o Molybdenum-based, and tungsten-based (2) Cast Cobalt Alloys (1915) o Good for deep, continuous roughing cuts at high speeds o High MRR, but poor surface finish (3) Carbides – 1930s o Higher speeds + temperatures than category 1 and 2 o Tungsten carbides + titanium carbides o New ones all the time (4) Coated Tools – 1960s o Coatings reduce friction + wear, reduce chemical reactivity, extend tool life o Popular coatings:  Titanium nitride (TiN)  Titanium carbide (TiC)

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 Titanium carbonitride (TiCN)  Aluminum oxide (Al2O3)  2 to 15 micro-meters thick coatings  Applied by chemical vapor deposition (CVD) or physical vapor deposition (PVD) (5) Alumina-Based Ceramics – 50s + 60s o Ceramic + cermet (ceramic-metal) based o Strong, inert, but brittle (6) Cubic Boron Nitride (CBN) – 1962 o A 0.5 to 1.0 mm layer of CBN bonded to a carbide substrate o Very wear resistant o Substrate provides toughness o CBN provide wear resistance, chemical inertness (7) Silicon-nitride based ceramics (1970s) o Good for super-alloys (8) Diamond o Synthetic mostly, more predictable, fewer flaws Cutting Fluids o Sprayed into cut zone, several purposes:  (1) Reduce Friction + Wear (lubrication)  Improves tool life + surface finish  (2) Cool the tool + workpiece (coolant)  Improves tool life, reduces thermal distortion  (3) Flush away chips (safety)  (4) Protect against corrosion Types of Fluids o (1) Oils – mineral, animal oils, vegetables oils, synthetic  Good lubes, not good coolants o (2) Emulsions – oil + water + additives  Good for lube + cooling o (3) Semisynthetic – synthetic + mineral + water o (4) Synthetic oil + water Methods of Application o (1) Flooding – squirt it at the cutting zone under pressure – very common o (2) Mist – fluid is misted toward cutting zone, gets into inaccessible areas  Does not impede visibility  Limited cooling capacity o (3) High Pressure Systems  Like flooding, but higher pressure  Very fast heat removal o (4) Through the Cutting Tool  Hole drilled through tool-holder and tool, fluid pumped through it  Can weaken tool + holder Dry Machining o Trend toward minimizing or eliminating use of cutting fluids o Economic + environmental reasons o Use very fine mist, or nothing o For some situations, this works fine o Sometimes, compressed air is used  Blow away chips  Little bit of cooling Final Note: o Must choose speed, feed, DoC, tool material, fluid strategy

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Published guidelines exist, but just a starting point Really should optimize your process by trial + error

Lecture 20, Monday 10/28/19    

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Midterm 2; November 11 in class Manufacturing Research Lecture on November 6, 3:30 PM, RE-104 Professor solved Homework Set 7 on board Homework Set 8 is assigned Chapter 23: Turning and Hole Making Turning, etc. o Machining of cylindrical shapes o Uses cylindrical reference frame o Axial distance, radial distance, angle o “turning” is biggest category o Uses rotating workpiece (V) o Tool approaches radially, feeds axially o Usually performed on a lathe (or “turning center”) o Can produce straight, conical, curved, or grooved workpieces o The speed is determined by the surface speed of the rotating part o Will decrease as diameter decreases o Depth of cut determined by how much the tool is moved towards axis on each pass o Feed is determined by axial motion of tool per revolution o Lathes can be large or small, CNC or “conventional” (not CNC) o Workpiece is held in a chuck Material Removal Rate o Volume is removed per time o For each revolution, a ring-shaped volume is removed o Cross-sectional area is distanced traveled per revolution (f) times depth of cut, d o Volume of ring is this area times circumference of ring o Ideally, use (pi)*(Davg) where Davg = 0.5*(D0 + Df) o For light cuts on large diameters, D0 approximately = Davg approximately = Df o If N is in RPM, then  MRR = (pi)*(Davg)*(d)*(f)*(N) = d*f*V  f = dist/rev  t = l/(f*N) Forces o o o o

Fc = cutting force, downward on tool tip Fc*radius = torque on spindle Torque*spindle speed = power Ft = thrust force in axial direction  Also called feed force  Pushes tool away from chuck o Fr = radial force pushes tool away from workpiece Roughing + Finishing o Usually start with roughing cuts o Large feeds + DoC  high MRR, poor tolerances o When close to final dimensions, use finishing cuts o Lower feed + DoC  good tolerance + surface finish Example 23.1

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6 in long, 0.5 in diameter, 304 SS rod Turn down to 0.48 in. diameter N = 400 rpm, feed = 8 in/min, one pass Find: cutting speed, MRR, time, power, Fc Cutting speed V = (pi)*(D0)*(N) = (pi)*(0.5)*(400) = 628 in/min at bottom of cut, V = (pi)*(.48)*(400) = 603 in/min Vavg = (pi)*(.49)*(400) = 51.3 ft/min DoC = (0.50 – 0.48)(1/2) = 0.010 in feed = (8 in/min)/(400 rev/min) = 0.02 in/rev **calculate both feeds and write them down side by side. This way you won’t pick the wrong one** MRR = V*f*DoC = (615.7 in/min)*(0.02 in)*(0.01 in) = 0.123 in3/min Cutting time, t = l/(f*n) = 6 in /(0.02 in/rev * 400 rev/min) = 0.75 min Power, see table 21.2 SS, between 0.8 + 1.9 (hp-min)/(in3) Assume 1.47 Power = (1.47 hp-min/in3)*(0.123 in3/min) = 0.181 hp 1 hp is 396,000 in-lb/min, so 0.181 hp = 71,676 lb-in/min, or 71, 700 (rounded) This power is torque times N in radians/time, T = (71,700)/(2*pi*400) = 29 lb-in Torque is also T = Fc*Davg*(0.5), so Fc = (29 lb-in) / (0.490 in * 0.5) = 118 lb

Lecture 21, Wednesday 10/30/19 

More About Lathes o Very old process (BC) o First powered lathes were called “engine lathes,” now used for very basic models o The bed, cast iron or welded steel plates o Ways are very flat, smooth surface on top of bed o Carriage slides longitudinally on the ways o Cross-slide slides along carriage perpendicular to axis, moves tool to workpiece o Headstock contains motor + gears to turn workpiece at various rotational speeds o Older lathes used a one or two speed DC motor and lots of gears to change speed o Newer ones use variable-speed AC motors o Headstock includes the chuck which holds the workpiece o Tailstock at opposite end, can support workpiece, or can hold a tool o Leadscrews and split-nut move carriage along the ways Specifications of lathes:  Swing is maximum diameter work that can be accommodated  “gap-bed” lathes have a gap in the bed to fit larger part  Distance between centers is maximum distance between head and tail stock Types of lathes o Bench lathe – small lathe on a work bench o Toolroom lathe – high precision o Turret Lathe – contains a (usually hexagonal) turret with multiple tools, rotated as needed o VTL – vertical turret lathe  Vertical axis of rotation  Work rest on turntable  For very large work Hole Making o Drilling is a common was way to start a hole o Long, thin, prone to breakage, not very precise o Difficult chip removal, poor surface finish o

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Usually, hole improved with reaming, honing, boring Twist drill has double helix shape  Angled point MRR (material removal rate) for drilling  MRR = [ (pi*D2) / (4) ]*(f)*(N)  D = diameter of drill  f = feed in inch/rev  N = rpm Example 23.4  10 mm drill, f = 0.2 mm/rev, 800 rpm  Magnesium alloy  MRR = [ (pi*D2) / (4) ]*(f)*(N) = [ (pi*102) / (4) ]*(0.2)*(800) = 12,566 mm2/min = 209 mm3/sec  Table 21.1  0.3  0.6 w-s/mm 3  Let’s use 0.5  Power = (209 mm3/sec)*(0.5 w-s/mm3) = 104.7 w  Rot speed = (800 rpm)*([2pi]/[60]) = 83.77 rad/sec  Torque = Power/Rot speed = 104.7 w / 83.77 rad/sec = 1.25 N-m

Boring + Boring Machines o Enlarges/improves of pre-existing hole o Or produces internal profiles o Cutting tool mounted on a boring bar, which has to be very rigid to avoid deflections o Larger parts done on a boring mill Reaming + Reamers o Improves accuracy + surface finish of existing hole o Tool is similar to a drill, but without pointed tip o Helical or straight flutes o Hand reamers operated manually o Machine reamers mounted on machine tool o “Dreamer” has drill features at front, reamer at back o Need “allowance” during drilling, 0.1 – 0.2 mm Tapping o Cutting internal thread in a hole o 2,3, or 4 flutes, non-continuous o Non-helical, with gaps for chips o Must be rotated in reverse to remove from hole  Unless it’s collapsible, but that weakens it o “dropping” combines drilling + tapping o If a top is used to clean up a warn thread, it’s called “chasing the thread”

Lecture 22, Monday 11/4/19  

Midterm will be on the 11th in classroom, from 3:15 to 4:30 PM o Ch. 13-16 Professor showed videos during the rest of lecture

Lecture 23, Wednesday 11/6/19  No lecture held  Seminar Lecture 24, Monday

11/11/19  Midterm (Exam 2) Lecture 25, Wednesday 11/13/19  Professor went over Homework set 8  Homework Set 9 is assigned

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Chapter 24: Milling, etc. Previous chapter was about round, rotational parts More complicated parts also need prismatic machining Milling includes several types of processes that use a rotating, cylindrical, multi-toothed cutter (1) Peripheral Milling (aka plain milling) o Cuts with curved OD of cutter o Axis of rotation is parallel to work surface o Cutter has multiple teeth, either straight or helical o Operation (gradually engage + disengage) o When cutter length is greater than part width “slab milling” Up vs. Down Milling o Cutter rotates + translates at the same time o Same direction? Or in opposite direction? o Conventional (More Common) (or “up”) milling, the cutter scrapes up from the bottom; the chip thickness is small at first, maximum as cutter leaves at top o Advantage: smooth engagement tool life not affected by oxides or scale o In climb (or “down”) milling, cutter champs downward onto surfaces of part, so chip is thickest at beginning o Advantage: cutting force helps hold part in place o Disadvantage: greater impacts, worse tool life Cutting Parameters o Speed, V, is surface speed of tool relative to part, where o V = (pi)*D*N  D = diameter of tool  N = rpm of spindle o Chip thickness max = 2*f*(d/D)0.5  d = depth of cut (DoC)  f = feed per tooth = v/(N*n)  v = linear feed  n = # of teeth  cutting time = t = (l + lc)/v  l = workpiece length  lc = engagement length  MRR = material removal rate = [l*w*d]/t = w*d*v  Example 24.1  Slab milling, 12x4 inch block steel  f = 0.01 inch/tooth, DoC = 1/8 inch  Cutter diameter = 2 inch, 20 straight teeth  N = 100 rpm, cutter wider than block  Find MRR, Power + Torque, cutting time  Find linear speed  v = f*N*n = (0.01 in/tooth)*(100 rev/min)*(20 teeth/rev) = 20 in/min  MRR = w*d*v = (4 in)*(1/8 in)*(20 in/min) = 10 in3/min  Table 21.2  pick 1.1 [hp-min]/in3

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Power = (1.1)*(10) = 11 hp Power = (Torque)*(Rotation speed) o Torque = [(11 hp)*(33,000 lb-ft/min-hp)] / [100 rpm*2*pi] = 578 lb-ft Since D is much greater than d, then lc = (D*d)0.5 = (2 in*1/8 in)0.5 = 0.5 inch Time = [l + lc]/v = (12+0.5)/(20 in/min) = 0.625 min = 37.5 seconds

Lecture 26, Monday 11/18/19 

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(2) Face Milling o Cutting points on flat, circular face of cutter o Axis of rotation perpendicular to work surface o Cutting points are usually inserts o Tends to make rough surfaces o Still have “up” vs. “down” o Example 24.2  D = 150 mm, N = 100 RPM, w = 60 mm, l = 500 mm, d = 3 mm, v = 0.6 m/min  Ten teeth, high strength aluminum alloy  Solution: cross-section of cut = w*d = 60*3 = 180 mm2  Speed = 600, MRR = (180)*(600) = 108,000 mm3/min  Time = [l + 2*lc] / v = [500 + 150 (two halves of diameter)] / 600 = 1.08 min  Feed per tooth ...