4 - ME3263 Supplements

ME 3263 Manufacturing Engineering - Chp 1 Supplemental Documents

OR older edition

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

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

Slope=n

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

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”.

Solution #2:

Solution #3: (a)

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.

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)

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

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)

2- Read the tolerance value from the tolerance tables (1/1000 mm)

This one is in INCHES

3- Calculate the limiting dimensions (d_min, d_max) on the shaft and hole

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?

---- 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)

MICROMETER

FOR (6) THROUGH (12)

---- Ref: http://www.phy.uct.ac.za/courses/c1lab/vernier1.html ----

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

Mechanical Engineering Department

ME 3263 Manufacturing Engineering

Dr. Can Saygin

Chp 4 – Manufacturing Processes

Supplement

Chp 4.1 Bulk Deformation

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

AAAr −

fo DDd −=

11lnlnε

AAYY ln.. == εσ

.εYUσ fd ==

Area Reduction Draft

True Strain True Stress

Draw Stress – theoretical

AY ln)tan

1( φα

μσ +=

cLD12.088.0 +=φ

2fo DD

αsin2fo

offdf A

AYAAF ln)tan

1( φα

μσ +==

vF.=Ρ

1max += nε

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

In 3rd and 4th Editions: α for A ; α’ for A’ ; Ab for BA ; αt’ for A’b

Chp 4.3 Machining

Volume of Material To be Removed Material Removal Rate = ___________________________________ Machining Time to Remove the Volume

PERIPHERAL MILLING

FACE MILLING - 1

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)

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

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.

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)

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

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

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)

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

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 – 2 of 3 includes 5.3. Analysis of Automated Flow Lines (Chp 5- 2 of 3.pdf)

Dr. Can Saygin

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)

Dr. Can Saygin

Problems

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.

Section 5.2 of Chp 5

Section 5.3 of Chp 5

SOLUTIONS

PROBLEM 2.1

PROBLEM 2.2

PROBLEM 2.3

PROBLEM 2.4

PROBLEM 5.1

PROBLEM 5.5

4 - ME3263 Supplements

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