4 - ME3263 Supplements
Transcript of 4 - ME3263 Supplements
ME 3263 Manufacturing Engineering - Chp 1 Supplemental Documents
OR older edition
ME 3263 Manufacturing Engineering - Chp 1 Supplemental Documents
Derive the following equations: (1) σ = σe(1 + e) (2) ε = ln(1 + e) (3) n = ε ultimate Proof (1): Due to volume constancy: Ao.Lo = A.L → A/Ao = Lo/L By definition e = (L-Lo)/Lo = L/Lo - L/L e = L/Lo - 1 e = Ao/A - 1 → Ao/A = e + 1 By definition ε = ln(L/Lo) = ln(Ao/A) ε = ln(e + 1) [Eqn 3.8 on page 45 in your textbook] Proof (2): Due to volume constancy: Ao.Lo = A.L → A/Ao = Lo/L By definition σe = F/Ao σ = F/A σ / σe = Ao/A = e + 1 σ = σe (e + 1) [Eqn 3.9 on page 45 in your textbook] Proof (3): During a tensile test, the load increases until the ultimate point, where the necking starts. Due to volume constancy: Ao.Lo = A.L → A/Ao = Lo/L
ME 3263 Manufacturing Engineering - Chp 1 Supplemental Documents
Finding E, K, and n on the true stress-strain diagram. Remember σ = E.ε in the elastic range (Hooke’s Law) And σ = K.εn in the plastic range (Power Law) Then we can write: Log σ = Log E + Log ε in the elastic range Log σ = Log K + n. Log ε in the plastic range Using the force-length readings after a tensile test, first calculate σ and ε values. Then calculate and plot Log σ and Log ε values. It should look as follows:
Extrapolate “Log σ = Log E + Log ε” and read the Log σ value for Log ε = 0. At that point σ = E. Extrapolate “Log σ = Log K + n. Log ε” and read the Log σ value for Log ε = 0. At that point σ = K. On “Log σ = Log k + n. Log ε”, the slope of the line = n.
Log σ
Log ε = 0
Log σ = Log E + Log ε Log σ = Log K + n.Log ε
Log E
Log K
Slope=n
ME 3263 Manufacturing Engineering - Chp 1 Supplemental Documents
PROBLEMS
Problem #1 A tensile test uses a test specimen that has a gage length of 50 mm and an area = 200 mm2. During the test the specimen yields under a load of 98,000 N. The corresponding gage length = 50.23 mm. This is the 0.2 percent yield point. The maximum load = 168,000 N (ultimate) is reached at a gage length = 64.2 mm. Determine: (a) engineering and true yield strength, (b) modulus of elasticity, and (c) engineering and true tensile strength (ultimate point). Problem #2 A certain steel alloy has a yield strength of 372 MPa (54,000 psi) and modulus of elasticity of 207 GPa (30 X 106 psi). A specimen made from this steel has a rectangular cross section of 13 X 13 mm (.5 X .5 in.). A gage length of 100 mm (4 in.) is marked along the length of the specimen. Calculate the following using Engineering Stress-Strain relationship. (a) A 45 kN (10,000 lb) load is applied to the specimen and then removed from it. What is the gage length when the load is applied, and what is it after the load is removed? (b) What load would produce a stress in the specimen equal to the yield strength? (c) Under the yield load, what would be the gage length of the specimen? Assume 0.2% offset yield. (d) After removal of the yield load, what would be the gage length of the specimen? Assume 0.2% offset yield. Problem #3
ME 3263 Manufacturing Engineering - Chp 1 Supplemental Documents
Solution #1: (a) σe = 98,000/200 = 490 MPa (eng) Area at yield = 50*200/50.23 = 199.08 mm2 σ = 98,000/199.08 = 492.3 MPa (true) Note that σe and σ are very close. (b) σe = E.e or σ = E.ε Subtracting the 0.2% offset, e = (50.23 - 50.0)/50.0 - 0.002 = 0.0026 E = σ/e = 490/0.0026 = 188.5 x 103
MPa (eng)
ε = ln(1+e) = ln(1 + 0.0026) = 0.00259 σ = E.ε 492.3 MPa = E . (0.00259) E = 492.3/0.00259 = 189.6 x 103
MPa (true) Note that Eng stress-strain values are always below the True stress-strain values; therefore, Eng is “safer”. (c) σe-ultimate = F_ult / Ao = 168,000/200 = 840 MPa (eng) Volume constancy: Ao.Lo = Au.Lu Area at ultimate pt = 50*200/64.2 = 155.76 mm2 σultimate = 168,000/155.76 = 1078.58 MPa (true) Note that Eng stress-strain values are always below the True stress-strain values; therefore, Eng is “safer”.
ME 3263 Manufacturing Engineering - Chp 1 Supplemental Documents
Solution #2:
ME 3263 Manufacturing Engineering - Chp 1 Supplemental Documents
ME 3263 Manufacturing Engineering - Chp 1 Supplemental Documents
ME 3263 Manufacturing Engineering - Chp 1 Supplemental Documents
Solution #3: (a)
(b)
ME 3263 Manufacturing Engineering - Chp 1 Supplemental Documents
(c)
ME 3263 Manufacturing Engineering - Chp 1 Supplemental Documents
Self Study Questions (Try these questions either as a team or by yourself without looking at the answers first)
Q1. A test specimen in a tensile test has a gage length of 2.0 in and an area = 0.5 in2. During the
test the specimen yields under a load of 32,000 lb. The corresponding gage length = 2.0083 in.
This is the 0.2 percent yield point. The maximum load = 60,000 lb is reached at a gage length =
2.60 in. Determine: (a) yield strength Y, (b) modulus of elasticity E, (c) tensile strength TS, (d)
determine the percent elongation at ultimate point, and (e) If the specimen necked to an area =
0.25 in2, determine the percent reduction in area.
Q2. A specimen with a starting gage length = 125.0 mm and cross-sectional area = 62.5 mm2 has
been tested on a tensile test machine. Two force-length readings, when the specimen was in the
plastic zone before necking, are F=23042 N for L=131.25 mm and F=28913 N for L=147.01 mm.
Assuming uniform elongation at these two data points, determine the strength coefficient K and
strain hardening exponent n.
Q3. A copper wire of diameter 0.80 mm reaches an area reduction of 75% in a uniform manner
when the engineering stress = 248.2 MPa. Determine the true stress and true strain at this point.
ME 3263 Manufacturing Engineering - Chp 1 Supplemental Documents
Q1 Solution: (a) From volume constancy, Area at yield point = Ao.Lo / Ly = 0.5*2 / 2.0083 = 0.4979 in2 Y = 32,000/0.4979 = 64,270 lb/in2 (b) σe = E e Subtracting the 0.2% offset, e = (2.0083 - 2.0)/2.0 - 0.002 = 0.00215 E = σe /e = 64,270/0.00215 = 29.89 x 106 lb/in2 (c) From volume constancy, Area at ultimate point = Ao.Lo / Lu = 0.5*2 / 2.60 = 0.3846 in2 TS = 60,000/0.3846 = 156,006 lb/in2 (d) % elongation at ultimate pt EL= (2.60 - 2.0)/2.0 = 0.6/2.0 = 0.3 = 30% (e) % area reduction at necking AR= (0.5 - 0.25)/0.5 = 0.50 = 50%
Q2 Solution: We need two equations to solve for the two unknowns, which are K and n. Initial volume, V = A.L = 125*62.5 = 7812.5 mm2 Due to volume constancy, (A.L)F = 23042 = (A.L)F = 28913 Cross-sectional area when F = 23042 => A = 7812.5/131.25 = 59.524 mm2 Stress when F = 23042 => σ = 23042/59.524 = 387.1 MPa Strain when F = 23042 => ε = ln(Lactual / Loriginal) = ln(131.25/125) = 0.0488 mm/mm Cross-sectional area when F = 28913 => A = 7812.5/147.01 = 53.143 mm2 Stress when F = 28913 => σ = 28913/53.143 = 544.1 MPa Strain when F = 23042 => ε = ln(147.01/125) = 0.1622 mm/mm Substituting these values into the power law equation (i.e., σ = Kεn), we have: 387.1 = k(0.0488)n 544.1 = k(0.1622)n Now we have two equations and two unknowns; solution is possible ln [387.1 / 544.1] = n. ln[0.0488/0.1622] - 0.34 = n.(-1.2) n = 0.283 and K = 910 MPa and
Q3 Solution: Area reduction AR = (Ao - Af)/Ao = 0.75 Ao - Af = 0.75 Ao 1 – (Af/Ao) = 0.75 => Af/Ao = 0.25 Engineering stress, σe = F/Ao True stress, σ = F/Af => σe/σ = Af/Ao = 0.25 => σ = σe / 0.25 If engineering stress = 248.2 MPa, then true stress σ = 248.2/0.25 = 992.8 MPa True strain ε = ln(Lf/Lo) = ln(Ao/Af) = ln(1/0.25) = 1.386
ME 3263 Manufacturing Engineering - Chp 2
Supplemental Documents
Finding the dimensions of a fit (i.e., tolerances and limiting dimensions on a shaft
and a hole)
ANSI
Given: Basic Size and Class of Fit
1- Allowance Equation
(Remember d = Holemin_d and Allowance = Holemin_d - Shaftmax_d )
2- Tolerance on the Hole
3- Tolerance on the Shaft
4- Limiting Dimensions on Hole and Shaft – Use the dimensions in the table above
as “heights” of blocks… Block H… Block a… Block S
ME 3263 Manufacturing Engineering - Chp 2
Supplemental Documents
ISO
Given: Basic Size and Application
1- Determine the type of the fit such as H8/f7
EMgt 324 Fundamentals of Manufacturing - Dr. C. Saygin 15
Specifying Tolerances…ISOISO
ISO Fit diagram for SHAFTS
ISO Fit diagram for HOLES
(Basic Size)
(Basic Size)
ME 3263 Manufacturing Engineering - Chp 2
Supplemental Documents
2- Read the tolerance value from the tolerance tables (1/1000 mm)
ME 3263 Manufacturing Engineering - Chp 2
Supplemental Documents
This one is in INCHES
3- Calculate the limiting dimensions (d_min, d_max) on the shaft and hole
ME 3263 Manufacturing Engineering - Chp 2
Supplemental Documents
Self Study Questions
1) Calculate the following for the shaft and the hole given below (dimensions in inches):
Tolerance on the shaft:
Tolerance on the hole:
Allowance (min. clearance):
Max. Clearance:
Type of Fit:
2) Find limiting dimensions (S_min, S_max, H_min, H_max) of the fit for a basic size of
2.0000 inches and a Class of 5 (ANSI).
3) Find limiting dimensions (S_min, S_max, H_min, H_max) of the fit H7/h6 of for a
nominal size of 52 mm (ISO). Based on ISO classification, what type of a fit is it?
ME 3263 Manufacturing Engineering - Chp 2
Supplemental Documents
ME 3263 Manufacturing Engineering - Chp 2
Supplemental Documents
ME 3263 Manufacturing Engineering - Chp 3
Supplemental Documents
---- Ref: http://www.phy.uct.ac.za/courses/c1lab/vernier1.html ----
University of Cape Town - Department of Physics
By Written by Andy Buffler (12 September 2003)
VERNIER CALIPER
FOR (1) THROUGH (5)
(1)
ME 3263 Manufacturing Engineering - Chp 3
Supplemental Documents
(2)
(3)
(4)
ME 3263 Manufacturing Engineering - Chp 3
Supplemental Documents
(5)
MICROMETER
FOR (6) THROUGH (12)
ME 3263 Manufacturing Engineering - Chp 3
Supplemental Documents
(6)
(7)
(8)
(9)
ME 3263 Manufacturing Engineering - Chp 3
Supplemental Documents
(10)
(11)
(12)
---- Ref: http://www.phy.uct.ac.za/courses/c1lab/vernier1.html ----
ME 3263 Manufacturing Engineering - Chp 3
Supplemental Documents
ANSWERS
(1) 121.68 mm
(2) 8.10 mm
(3) 30.88 mm
(4) 34.60 mm
(5) 40.00 mm
(6) 7.72 mm
(7) 0.29
(8) 3.09 mm
(9) 3.46
(10) 3.56 mm
(11) 5.80 mm
(12) 7.38
ME 3263 Manufacturing Engineering - Chp 4 Supplemental Documents
1
Mechanical Engineering Department
ME 3263 Manufacturing Engineering
Dr. Can Saygin
Chp 4 – Manufacturing Processes
Supplement
ME 3263 Manufacturing Engineering - Chp 4 Supplemental Documents
2
Chp 4.1 Bulk Deformation
ME 3263 Manufacturing Engineering - Chp 4 Supplemental Documents
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ME 3263 Manufacturing Engineering - Chp 4 Supplemental Documents
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ME 3263 Manufacturing Engineering - Chp 4 Supplemental Documents
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ME 3263 Manufacturing Engineering - Chp 4 Supplemental Documents
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ME 3263 Manufacturing Engineering - Chp 4 Supplemental Documents
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Wire Drawing (“force F” is applied on the “product”)
Draw Stress – actual (die geometry and friction taken into account)
where with average diameter
and contact length
Draw Force:
Draw Power: where F is the draw force and v is the velocity at the exit
Process Feasibility – Maximum true strain
True strain attempted must be less than (or equal to) maximum true strain
o
fo
AAAr −
=
fo DDd −=
rAA
f
o
−==
11lnlnε
f
off
AAYY ln.. == εσ
.εYUσ fd ==
Area Reduction Draft
True Strain True Stress
Draw Stress – theoretical
f
ofd A
AY ln)tan
1( φα
μσ +=
cLD12.088.0 +=φ
2fo DD
D+
=
αsin2fo
c
DDL
−=
f
offdf A
AYAAF ln)tan
1( φα
μσ +==
vF.=Ρ
1max += nε
ME 3263 Manufacturing Engineering - Chp 4 Supplemental Documents
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Chp 4.2 Sheet Metal Working
“a” in 2nd Edition… In 3rd and 4th Editions “Ac” instead of “a” c = Ac.t See Table 20.1 for Ac
ME 3263 Manufacturing Engineering - Chp 4 Supplemental Documents
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ME 3263 Manufacturing Engineering - Chp 4 Supplemental Documents
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In 3rd and 4th Editions: α for A ; α’ for A’ ; Ab for BA ; αt’ for A’b
ME 3263 Manufacturing Engineering - Chp 4 Supplemental Documents
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ME 3263 Manufacturing Engineering - Chp 4 Supplemental Documents
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Chp 4.3 Machining
Volume of Material To be Removed Material Removal Rate = ___________________________________ Machining Time to Remove the Volume
ME 3263 Manufacturing Engineering - Chp 4 Supplemental Documents
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ME 3263 Manufacturing Engineering - Chp 4 Supplemental Documents
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PERIPHERAL MILLING
ME 3263 Manufacturing Engineering - Chp 4 Supplemental Documents
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FACE MILLING - 1
ME 3263 Manufacturing Engineering - Chp 4 Supplemental Documents
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FACE MILLING – 2
ME 3263 Chp 4 Power – Specific Cutting Energy – Material Removal Rate: Turning versus Milling
In turning, the uncut chip cross-sectional area stays constant. Therefore, we can calculate the material removal rate at “chip” level as MRR = d.f.v . Since the chip cross-section “f.d” stays constant, Specific Cutting Energy for turning can be calculated as U = Fc/f.d . Then: Power = U.MRR = (Fc/f.d).(dfv) Power = Fc. v However in milling, the chip thickness, while the cutter is in contact with the workpiece, varies. Therefore, “MRR = dfv” is not accurate for milling. We calculate the material removal rate for milling more accurately at the level of “projected area of cut” as MRR = w.d.fr . Power for milling can then be calculated as P = U.MRR or P = U.(w.d.fr)
ME 3263 Manufacturing Engineering - Chp 4 Supplemental Documents
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TUTORIAL QUESTIONS
Bulk Deformation and Sheet Metal Working
Problem 21.1: Rolling A 40-mm-thick plate is to be reduced to 30 mm in one pass in a rolling operation. Entrance speed = 16 m/min. Roll radius = 300 mm, and roll speed = 18.5 m/min. Determine (a) the minimum required coefficient of friction that would make this rolling operation possible, (b) exit velocity under the assumption that the plate widens by 2% during the operation, and (c) forward slip. Problem 21.23: Indirect Extrusion A 3.0-in. cylindrical billet with diameter = 1.5 in. is reduced by indirect extrusion to a 0.375 -in. diameter. Die angle = 90o. In the Johnson equation, a=0.8 and b=1.5. In the flow curve for the work metal, K= 75,000 lb/in2. and n = 0.25. Determine (a) extrusion ratio, (b) true strain (homogeneous deformation), (c) extrusion strain, (d) ram pressure, (e) ram force, and (f) power when v = 20 in/min
Problem 22.4: Blanking
Given: blanking part in Figure P22.4 from half-hard stainless steel, t = 5/32 in. Find dimensions of blanking punch and die.
6/15/2010
ME 3263 Manufacturing Engineering - Chp 4 Supplemental Documents
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Solution 21.1: Rolling
Given: flat rolling, t0 = 40mm, tf = 30 mm, v0 = 16 m/min, vr=18.5
m/min, R=300 mm. Find: (a) minimum μ , (b) vf, (c) s.
Maximum draft dmax = μ 2 R Given that d = t0 - tf = 40 - 30 = 10 mm.
(a) μ 2 = 10/300 = 0.0333
μ = (0.0333)0.5 = 0.1826
(b) Plate widens by 2%.
t0 w0 v0 = tf wf vf
wf = 1.02 w0
40 (w0) (16) = 30 (1.02 w0) vf
vf = 40 (w0) (16) / 30 (1.02w0) = 640/30.6 = 20.915 m/min
(c) s = (vf - vr) / vr = (20.915 - 18.5) / 18.5 = 0.13
Solution 21.23: Indirect Extrusion
Given: indirect extrusion, L0 = 3.0 in., D0 = 1.5 in., Df = 0.375 in., α = 90o,
Johnson equation: a = 0.8 and b = 1.5, flow curve: K = 75,000 lb/in2. and
n= 0.25. Find: (a) rx, (b) ε, (c) εx, (d) p, (e) F, (f) P at v = 20 in/min.
(a) rx = A0/Af = D02/Df2 = (1.5)2 / (0.375)2 = 16.0
(b) ε = ln rx = ln 16 = 2.773
(c) εx = a + b ln rx = 0.8 + 1.5 (2.773) = 4.959
(d) Yfaverage = 75,000 (2.773)0.25 / 1.25 = 77.423 lb/in2.
p= 77,423 (4.959) = 383,934 lb/in2.
(e) Ao = π Do2/4 = π (1.5)2/4 = 1.767 in2
F = (383,934)(1.767) = 678,467 lb.
(f) P= 678,467 (20) = 13,569,348 in-lb/min
HP = 13,569,348 / 369,000 = 34.3 hp. Solution 22.4: Blanking
D_die for blanking = Db = Part Dimensions
D_punch for blanking = Db – 2c
From Table 22.1 (page 504), a=0.075
Clearance c = a.t = 0.075 (5/32) = 0.0117 in.
Blanking die dimensions are same as for the part in Figure P22.4.
Blanking punch:
3.500 inch length dimension = 3.500 - 2 (0.0117) = 3.4766 in.
2.000 inch width dimension = 2.000 - 2 (0.0117) = 1.9766 in.
Top and bottom 1.00 inch extension widths = 1.0 - 2 (0.0117) = 0.9766 in.
1 inch inset dimension remains the same.
ME 3263 Manufacturing Engineering - Chp 4 Supplemental Documents
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TUTORIAL QUESTIONS Machining
Problem: Turning In a turning operation on aluminum, the cutting conditions are as
follows:
d = 0.25 in f = 0.020 in/rev v = 900 ft/min
The lathe has a mechanical efficiency of 87 %. Specific Cutting Energy
for aluminum is 100,000 in.lb/ in3.
Compute
a) Cutting Force
b) Horsepower required for the drive motor on the lathe
c) Unit horsepower
Solution
a) Specific Energy = U = 100,000 in.lb/ in3
U = Power / MRR = F_c * v / MRR where MRR = d.f.v
MRR = 0.25 in * 0.020 in/rev * 900x12 in/min = 54 in3/min
F_c = U*MRR / v = 100000 * 54 / (900*12) = 500 lb.
F_c = 500 lb.
b) hp_c = F_c * v / 33000 (ft.lb/min)/hp (Be careful with the units)
hp_c = 500 lb * 900 ft/min / 33000 (ft.lb/min)/hp
hp_c = 13.64 HP
hp_g = hp_c / E = 13.64 / 0.87
hp_g = 15.67 HP
c) hp_u = hp_c / MRR = 13.64 HP / 54 in3/min
hp_u = 0.25 HP/(in3/min)
ME 3263 Manufacturing Engineering - Chp 4 Supplemental Documents
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Turning/ A 40.0-in length is to be turned from a diameter of 5.0 in.
to 4.75 in. in one pass at a cutting speed of 400 ft/min and a feed rate
of 0.012 in/rev. Determine
a) Depth of cut
b) Machining time
c) Material Removal Rate
Solution Given:
L = 40.0 in
Do = 5.0 in
Df = 4.75 in
v = 400 ft/min
f = 0.012 in/rev.
a) d = (Do – Df) / 2 = (5 – 4.75) / 2
d = 0.125 in
b) Tm = L / fr where fr = N.f
fr = (v/π.Do).f
fr = (400*12 in/min / π*5 in) * 0.012 in/rev
fr = 3.67 in/min
Tm = 40 in / 3.67 in/min
Tm = 10.90 min
c) MRR = d.v.f
MRR = 0.125 in * 400*12 in/min * 0.012 in
MRR = 7.2 in3/min
ME 3263 Manufacturing Engineering - Chp 4 Supplemental Documents
22
Drilling/ A 0.75 in. diameter twist drill is used to drill a through hole
on a steel workpiece that has a thickness of 1.5 in. The point angle of
the drill is 118°. The cutting speed is 50 ft/min and the feed is 0.010
in./rev. Determine
a) Machining time
b) Metal removal rate
Solution Given:
D = 0.75 in
Through Hole, t = 1.5 in
θ = 118o
v = 50 ft/min
f = 0.010 in/rev
a) Tm = (t + A) / fr
where fr = N.f = (v/π.D).f
= (50*12 in/min / π*0.75 in) * 0.010 in/rev
fr = 2.55 in/min
A = 0.5*D*tan(90 - θ/2) = 0.5*0.75 in *tan(90 - 118o/2)
A = 0.225 in
Tm = (1.5 + 0.225) in / 2.55 in/min
Tm = 0.68 min = 40.6 sec
b) MRR= Area * fr
MRR = (π/4).D2 in2 * 2.55 in/min
MRR = (π/4)*(0.75)2 in2 * 2.55 in/min
MRR = 1.126 in3/min
fr
ME 3263 Manufacturing Engineering - Chp 4 Supplemental Documents
23
Face Milling/ In a face milling operation, the depth of cut is 5 mm
and the width of the workpiece is 50 mm. The length of the workpiece
is 450 mm. The chip load is 0.25 mm/tooth. The cutter has a diameter
of 100 mm and has 20 teeth. If the cutting speed is 1 m/s, calculate:
a) Material removal rate
b) Machining time
c) Power consumption if the specific cutting energy of the
workpiece is 4.1 N.m/mm3.
Solution Given:
d = 5 mm
w = 50 mm
L = 450 mm
f = 0.25 mm/tooth
D = 100 mm
nt = 20 teeth/rev
v = 1 m/s = 1000 mm/s
a) fr = N.nt.f where N = v/π.D
N = 1000 mm/s / π*100 mm N = 3.2 rev/sec = 192 rpm
fr = 192 rev/min * 20 teeth/rev * 0.25 mm/tooth
fr = 960 mm/min = 16 mm/s
MRR = w.d.fr = 50 mm * 5 mm * 16 mm/s
MRR = 4000 mm3/s
b) Tm = (L + D) / fr = (450 + 100) mm / 16 mm/s
Tm = 34.4 s
c) U = Power / MRR
U =4.1 N.m/mm3
Power = U * MRR = 4.1 N.m/mm3 * 4000 mm3/s
Power = 16,400 Nm/s = 16,400 W (Nm/s Watt)
d
w
fr
ME 3263 Manufacturing Engineering - Chp 4 Supplemental Documents
24
Peripheral Milling/ A plain milling operation is performed on the top
surface of a rectangular workpart that is 300 mm long and 100 mm
wide. The milling cutter, which is 75 mm in diameter and has four
teeth, overhangs the width of the part on both sides. Cutting
conditions are: v = 80 m/min, f = 0.2 mm/tooth, and d = 7.0 mm.
Determine:
a) Machining time for one pass
b) Material removal rate
Solution Given:
d = 7 mm
w = 100 mm
L = 300 mm
f = 0.2 mm/tooth
D = 75 mm
nt = 4 teeth/rev
v = 80 m/min
a) Tm = (L + 2A) / fr
where fr = N.nt.f = (v/π.D).nt.f
fr = (80000 mm/min / π.75mm) * 4 teeth/rev * 0.2 mm/tooth
fr = 271.6 mm/min = 4.52 mm/s
A = Sqrt{d(D-d)} = Sqrt{7 mm(75 – 7 mm)}
A = 21.8 mm
Tm = (300 + 2*21.8)mm / 4.52 mm/s
Tm = 76 sec = 1.26 min
b) MRR = w.d.fr = 100 mm * 7 mm * 4.52 mm/s
MRR = 3164 mm3/s
Top view
Cutter
100 mm
300 mm
Supplement - Chp 5 – 1 of 3 includes 5.1. Production Systems: Materials Flow, Production Volume, and Part Variety
and 5.2. Production Concepts and Mathematical Models
Mechanical Engineering Department
ME 3263 Manufacturing Engineering
Dr. Can Saygin
Chp 5 – Production Systems:
An Overview on Basics
Supplement Chp 5- 1 of 3.pdf
This chapter, including supplemental documents and slides, has been organized in 3 sections
5.1. Production Systems: Materials Flow, Production Volume, and Part Variety (Chp 5- 1 of 3.pdf)
5.2. Production Concepts and Mathematical Models (Chp 5- 1 of 3.pdf)
5.3. Analysis of Automated Flow Lines (Chp 5- 2 of 3.pdf)
Supplement - Chp 5 – 1 of 3 includes 5.1. Production Systems: Materials Flow, Production Volume, and Part Variety
and 5.2. Production Concepts and Mathematical Models
Supplement - Chp 5 – 1 of 3 includes 5.1. Production Systems: Materials Flow, Production Volume, and Part Variety
and 5.2. Production Concepts and Mathematical Models
Supplement - Chp 5 – 1 of 3 includes 5.1. Production Systems: Materials Flow, Production Volume, and Part Variety
and 5.2. Production Concepts and Mathematical Models
Supplement - Chp 5 – 1 of 3 includes 5.1. Production Systems: Materials Flow, Production Volume, and Part Variety
and 5.2. Production Concepts and Mathematical Models
Supplement - Chp 5 – 1 of 3 includes 5.1. Production Systems: Materials Flow, Production Volume, and Part Variety
and 5.2. Production Concepts and Mathematical Models
Supplement - Chp 5 – 1 of 3 includes 5.1. Production Systems: Materials Flow, Production Volume, and Part Variety
and 5.2. Production Concepts and Mathematical Models
Supplement - Chp 5 – 1 of 3 includes 5.1. Production Systems: Materials Flow, Production Volume, and Part Variety
and 5.2. Production Concepts and Mathematical Models
Supplement - Chp 5 – 1 of 3 includes 5.1. Production Systems: Materials Flow, Production Volume, and Part Variety
and 5.2. Production Concepts and Mathematical Models
Supplement - Chp 5 – 1 of 3 includes 5.1. Production Systems: Materials Flow, Production Volume, and Part Variety
and 5.2. Production Concepts and Mathematical Models
Supplement - Chp 5 – 1 of 3 includes 5.1. Production Systems: Materials Flow, Production Volume, and Part Variety
and 5.2. Production Concepts and Mathematical Models
Supplement - Chp 5 – 1 of 3 includes 5.1. Production Systems: Materials Flow, Production Volume, and Part Variety
and 5.2. Production Concepts and Mathematical Models
Supplement - Chp 5 – 1 of 3 includes 5.1. Production Systems: Materials Flow, Production Volume, and Part Variety
and 5.2. Production Concepts and Mathematical Models
Supplement - Chp 5 – 1 of 3 includes 5.1. Production Systems: Materials Flow, Production Volume, and Part Variety
and 5.2. Production Concepts and Mathematical Models
Supplement - Chp 5 – 1 of 3 includes 5.1. Production Systems: Materials Flow, Production Volume, and Part Variety
and 5.2. Production Concepts and Mathematical Models
Supplement - Chp 5 – 2 of 3 includes 5.3. Analysis of Automated Flow Lines (Chp 5- 2 of 3.pdf)
Mechanical Engineering Department
ME 3263 Manufacturing Engineering
Dr. Can Saygin
Chp 5 – Production Systems:
An Overview on Basics
Supplement Chp 5- 2 of 3.pdf
This chapter, including supplemental documents and slides, has been organized in 3 sections
5.1. Production Systems: Materials Flow, Production Volume, and Part Variety (Chp 5- 1 of 3.pdf)
5.2. Production Concepts and Mathematical Models (Chp 5- 1 of 3.pdf)
5.3. Analysis of Automated Flow Lines (Chp 5- 2 of 3.pdf)
Supplement - Chp 5 – 2 of 3 includes 5.3. Analysis of Automated Flow Lines (Chp 5- 2 of 3.pdf)
Supplement - Chp 5 – 2 of 3 includes 5.3. Analysis of Automated Flow Lines (Chp 5- 2 of 3.pdf)
Supplement - Chp 5 – 2 of 3 includes 5.3. Analysis of Automated Flow Lines (Chp 5- 2 of 3.pdf)
Supplement - Chp 5 – 2 of 3 includes 5.3. Analysis of Automated Flow Lines (Chp 5- 2 of 3.pdf)
Supplement - Chp 5 – 2 of 3 includes 5.3. Analysis of Automated Flow Lines (Chp 5- 2 of 3.pdf)
Supplement - Chp 5 – 2 of 3 includes 5.3. Analysis of Automated Flow Lines (Chp 5- 2 of 3.pdf)
Supplement - Chp 5 – 2 of 3 includes 5.3. Analysis of Automated Flow Lines (Chp 5- 2 of 3.pdf)
Supplement - Chp 5 – 2 of 3 includes 5.3. Analysis of Automated Flow Lines (Chp 5- 2 of 3.pdf)
Supplement - Chp 5 – 2 of 3 includes 5.3. Analysis of Automated Flow Lines (Chp 5- 2 of 3.pdf)
Supplement - Chp 5 – 2 of 3 includes 5.3. Analysis of Automated Flow Lines (Chp 5- 2 of 3.pdf)
Supplement - Chp 5 – 2 of 3 includes 5.3. Analysis of Automated Flow Lines (Chp 5- 2 of 3.pdf)
Supplement - Chp 5 – 2 of 3 includes 5.3. Analysis of Automated Flow Lines (Chp 5- 2 of 3.pdf)
Supplement - Chp 5 – 2 of 3 includes 5.3. Analysis of Automated Flow Lines (Chp 5- 2 of 3.pdf)
Supplement - Chp 5 – 2 of 3 includes 5.3. Analysis of Automated Flow Lines (Chp 5- 2 of 3.pdf)
Supplement - Chp 5 – 2 of 3 includes 5.3. Analysis of Automated Flow Lines (Chp 5- 2 of 3.pdf)
Supplement - Chp 5 – 2 of 3 includes 5.3. Analysis of Automated Flow Lines (Chp 5- 2 of 3.pdf)
Supplement - Chp 5 – 2 of 3 includes 5.3. Analysis of Automated Flow Lines (Chp 5- 2 of 3.pdf)
Supplement - Chp 5 – 2 of 3 includes 5.3. Analysis of Automated Flow Lines (Chp 5- 2 of 3.pdf)
Supplement - Chp 5 – 2 of 3 includes 5.3. Analysis of Automated Flow Lines (Chp 5- 2 of 3.pdf)
Supplement - Chp 5 – 2 of 3 includes 5.3. Analysis of Automated Flow Lines (Chp 5- 2 of 3.pdf)
Supplement - Chp 5 – 2 of 3 includes 5.3. Analysis of Automated Flow Lines (Chp 5- 2 of 3.pdf)
Supplement - Chp 5 – 2 of 3 includes 5.3. Analysis of Automated Flow Lines (Chp 5- 2 of 3.pdf)
Supplement - Chp 5 – 2 of 3 includes 5.3. Analysis of Automated Flow Lines (Chp 5- 2 of 3.pdf)
ME 3263 Manufacturing Engineering - Chp 5
Supplemental Documents
1
Mechanical Engineering Department
ME 3263 Manufacturing Engineering
Dr. Can Saygin
Chp 5 – Production Systems:
An Overview on Basics
Supplement Chp 5- 3 of 3.pdf
Problems
ME 3263 Manufacturing Engineering - Chp 5
Supplemental Documents
2
Section 5.2: Production Concepts: An Overview on Basics
Number of batches (lots) of various part types
[____] [____] … [____]
1 2 … nQ
Part types
…
P1 P2 … Pj
Batch (Lot) sizes of part types
[..] [..] … [..]
Q1 Q2 … Qj
Operation plans of part types
…
P1 P2 … Pj
Oper1,1 To1,1 Oper1,2 To1,2 Oper1,j To1,j
Oper2,1 To2,1 Oper2,2 To2,2 Oper2,j To2,j
… … … … … …
Opernm,1 Tonm,1 Opernm,2 Tonm,2 Opernm,j Tonm,j
Average Operation Times
…
P1 P2 … Pj
See Excel file “calculating averages.xls” to see the implementation.
1
1
1,
1
1
nm
To
To
nm
i
i
2
1
2,
2
2
nm
To
To
nm
i
i
j
nm
i
ji
j
nm
To
To
j
1
,
ME 3263 Manufacturing Engineering - Chp 5
Supplemental Documents
3
Section 5.2 of Chp 5
ME 3263 Manufacturing Engineering - Chp 5
Supplemental Documents
4
Section 5.3 of Chp 5
ME 3263 Manufacturing Engineering - Chp 5
Supplemental Documents
5
SOLUTIONS
PROBLEM 2.1
ME 3263 Manufacturing Engineering - Chp 5
Supplemental Documents
6
PROBLEM 2.2
ME 3263 Manufacturing Engineering - Chp 5
Supplemental Documents
7
PROBLEM 2.3
ME 3263 Manufacturing Engineering - Chp 5
Supplemental Documents
8
ME 3263 Manufacturing Engineering - Chp 5
Supplemental Documents
9
ME 3263 Manufacturing Engineering - Chp 5
Supplemental Documents
10
PROBLEM 2.4
ME 3263 Manufacturing Engineering - Chp 5
Supplemental Documents
11
PROBLEM 5.1
ME 3263 Manufacturing Engineering - Chp 5
Supplemental Documents
12
PROBLEM 5.5