Metal Machining

Objectives Introduce cutting terminology and principles Review modern machining technologies and new methods (papers) Introduce cutting parameters Develop cutting models Analyze a cutting example

Machining types Turning Drilling Milling Shaping Planing Broaching

Machining tools Single point Multiple point

Machining tool materials Most modern cutting tool materials are a matrix of materials designed to be very hard. These materials will be covered in the next chapter.

Machining surface finish

Machining terminology Speed – surface cutting speed (v) Feed – advance of tool through the part (f) Depth of cut – depth of tool into part (d) Rake face – tool’s leading edge Rake angle – slant angle of tool’s leading edge (a) Flank – following edge of cutting tool Relief angle – angle of tool’s following edge above part surface

Machining terminology (cont.) Chip thickness – thickness of machined chip (tc ) Depth of cut = to Shear plane length – measured along shear plane chip (ls ) Chip width (not shown) – width of machined chip (w ) Shear angle – angle of shearing surface measured from tool direction (f) ls Orthogonal model

Cutting conditions Note: - Primary cutting due to speed Types of cuts: - Lateral motion of tool is feed - Tool penetration is depth of cut The three together form the material removal rate (MRR): MRR = v f d with units of (in/min)(in/rev)(in) = in3/min/rev (or vol/min-rev) Types of cuts: Roughing: feeds of 0.015 – 0.05 in/rev depths of 0.1 – 0.75 in Finishing: feeds of 0.005 – 0.015 in/rev depths of 0.03 – 0.075 in

Cutting geometry Obviously, the assumed failure mode is shearing of the work along the shear plane. Chip thickness ratio = r = to / tc From the shear plane geometry: r = ls sinf/[ls cos(f - a)] which can be arranged to get tan f = r cos a /[1 – r sin a]

Cutting geometry Note from the triangles in (c) that the shear strain (g) can be estimated as g = AC/BD = (DC + AD)/BD = tan(f - a) + cot f Thus, if know r and a, can determine f, and given f and a, can determine g.

Cutting forces Since R = R’ = R’’, we can get the force balance equations: F = Fc sin a + Ft cos a F = friction force; N = normal to chip force N = Fc cos a - Ft sin a Fc = cutting force; Ft = thrust force Fs = Fc cos f - Ft sin f Fs = shear force; Fn = normal to shear plane force Fn = Fc sin f + Ft cos f Forces are presented as function of Fc and Ft because these can be measured. Friction angle = b tan b = m = F/N Shear plane stress: t = Fs/As where As = to w/sin f

Cutting forces given shear strength Letting S = shear strength, we can derive the following equations for the cutting and thrust forces*: Fs = S As Fc = Fs cos ( b - a)/[cos ( f + b - a)] Ft = Fs sin ( b - a)/[cos ( f + b - a)] * The other forces can be determined from the equations on the previous slide.

Merchant equations Combining the equations from the previous slides: t = (Fc cos f - Ft sin f)/(tow/sin f ) Merchant eqn The most likely shear angle will minimize the energy. Applying dt/df = 0 gives: f = 45° + a/2 - b/2 Merchant reln What does the Merchant relation indicate? The Merchant reln is a function of a and b. Where did these variables come from? Answer - Although the Merchant eqn is not shown as a direct function of a and b, these enter from the equations for Fc and Ft from the previous slide! If we increase the shear angle, we decrease the tool force and power requirements! increase in friction angle decreases shear angle increase in rake angle increases shear angle

Cutting models The orthogonal model for turning approximates the complex shearing process: to = feed (f) w = depth of cut (d)

Cutting power Power is force times speed: P = Fc v (ft-lb/min) The cutting horsepower is hpc = Fc v/33,000 (hp) The unit horsepower is hpu = hpc/MRR units? Due to efficiency losses (E about 90%), the gross hp is hpg = hpc/E

Cutting energy Specific energy is U = Fc v/(v tow) = Fc /(tow) (in-lb/in3) The table shown contains power and specific energy ratings for several work materials at a chip thickness of 0.01 in. For other chip thicknesses, apply the figure to get a correction factor multiply U by correction factor for thickness different than 0.01”).

Machining example In orthogonal machining the tool has rake angle 10°, chip thickness before cut is to = 0.02 in, and chip thickness after cut is tc = 0.045 in. The cutting and thrust forces are measured at Fc = 350 lb and Ft = 285 lb while at a cutting speed of 200 ft/min. Determine the machining shear strain, shear stress, and cutting horsepower. Solution (shear strain): Determine r = 0.02/0.045 = 0.444 Determine shear plane angle from tan f = r cos a /[1 – r sin a] tan f = 0.444 cos 10 /[1 – 0.444 sin 10] => f = 25.4° Now calculate shear strain from g = tan(f - a) + cot f g = tan(25.4 - 10) + cot 25.4 = 2.386 in/in answer!

Machining example (cont.) In orthogonal machining the tool has rake angle 10°, chip thickness before cut is to = 0.02 in, and chip thickness after cut is tc = 0.045 in. The cutting and thrust forces are measured at Fc = 350 lb and Ft = 285 lb while at a cutting speed of 200 ft/min. Determine the machining shear strain, shear stress, and cutting horsepower. Solution (shear stress): Determine shear force from Fs = Fc cos f - Ft sin f Fs = 350 cos 25.4 - 285 sin 25.4 = 194 lb Determine shear plane area from As = to w/sin f As = (0.02) (0.125)/sin 25.4= 0.00583 in2 The shear stress is t = 194/0.00583 = 33,276 lb/in2 answer!

Machining example (cont.) In orthogonal machining the tool has rake angle 10°, chip thickness before cut is to = 0.02 in, and chip thickness after cut is tc = 0.045 in. The cutting and thrust forces are measured at Fc = 350 lb and Ft = 285 lb while at a cutting speed of 200 ft/min. Determine the machining shear strain, shear stress, and cutting horsepower. Solution (cutting horsepower): Determine cutting hp from hpc = Fc v/33,000 hpc = (350) (200)/33,000 = 2.12 hp answer!

Cutting temperatures In machining 98% of the cutting energy is converted into heat. This energy flows into the work part, chip, and tool. Cook determined an experimental equation for predicting the temperature rise at the tool-chip interface during machining: DT = 0.4 U (v to/K)0.333/(rc) where DT = mean temperature rise (°F) U = specific energy (in-lb/in3) v = cutting speed (in/s) to = chip thickness before cut (in) rc = volumetric specific heat of the work material (in-lb/(in3-°F)) K = thermal diffusivity of the work material (in2/s) Note - To get total temperature at tool-chip interface, must add in ambient temperature! Example in text calculates DT = 936° total tool temperature, given v = 200 ft/min, rc = 120 in-lb/(in3- °F) and K = 0.125 in2/s

Cutters Toroid

Cutters

Machining What did we learn?