ME 303

MANUFACTURING ENGINEERING

PROBLEM SETS

Prepared by:

Kamil Özden

Ural Uluer

Salih Alan

Mehmet Bilal Atar

PROBLEM SET FOR CHAPTER 2

Q2.1

A copper bar is to be cold rolled into a section which must have a min. tensile strength of

390 MPa. If the final cross-sectional area is 20.13 mm2 and assuming the flow properties

of the workpiece material are given as K = 450 MPa, n = 0.33. Calculate the followings:

a) The tensile strength of the annealed material.

b) The initial diameter of the copper bar.

Solution

It is important to understand the question well. In this question, the copper bar is

annealed first then a cold rolling operation is done on it. Initial diameter of the copper

bar is same before and after the annealing operation.

a) Flow stress of the plastically deformed bar could be shown as . As

shown in Course Slides at UTS

Then UTS of the original bar (no annealing, no cold working) is found as below:

u = 450 (0.33)0.33 u = 312.12 MPa

Since before necking occurs ( < n) area stays constant at the whole bar, the formula

below could be used:

By putting the area ratio into the formula (i) UTS of the annealed material is found as:

b) For this question d0 can be found by using the ratio:

if < n (which means necking is not started yet) with the UTScw value given in the

question as UTScw = 390 MPa, to check the value

So > n which means that necking started, then UTScw becomes equal to flow stress of

the deformed material,

Q2.2

A steel tensile specimen with an initial diameter of 20mm and gage length of 120mm is

subjected to a load of 125 kN and a gage length of 135mm is observed. Assuming

uniform deformation at this point, calculate the followings:

a) calculate the true stress, strain and the instantaneous diameter

b) supposing the plastic behavior of the specimen material is expressed as

(MPa) estimate the yield strength of the deformed specimen.

Solution

a)

from volume constancy

b)

Q2.3

A tensile bar was machined with a stepped gage section consisting of two regions of

different diameters. The initial diameters of the two regions were 2.0 cm and 1.9 cm.

After a certain amount of stretching in tension, the diameters of the two regions were

measured as 1.893 cm and 1.698 cm, respectively. Assuming the tensile strain hardening

is described by , find n for the material.

Solution

Strains at larger and smaller diameter sections are respectively:

(

)

(

)

The load carrying capacities at two sections must be the same:

⁄

2.0 cm

ccm cm

1.9 cm

cm

1.893 cm 1.698 cm

Before:

After:

rrrrr

Q2.4

A steel test specimen (modulus of elasticiy = 205 GPa) in a compression test has a

starting height of 50mm and diameter of 38 mm. The metal yields (0.2% offset) at a load

600kN. At a load of 1100kN, the height has been reduced to 40mm. Assuming that the

cross-sectional area increases uniformly during the test, determine the followings:

a) Yield strength in MPa

b) Total strain at the offset yield point

c) Flow stress in Mpa and the plastic strain under the load of 1100kN

d) Strength coefficient K in MPa if strain hardening exponent is 0.25

Solution

a)

(

)

b)

(

)

c)

(

)

(

)

d)

Q2.5

A test specimen in a tensile test has a gage length of 50 mm and a cross-sectional

area of 316 mm2. During the test the specimen yields (0.2% offset) under a load of

140kN. The corresponding gage length is 50.2 mm. The maximum load of 200 kN is

reached at a gage length of 58 mm. Determine the followings:

a) Yield strength in MPa

b) Modulus of elasticity in GPa

c) Tensile strength in MPa

d) Percentage elongation if fracture occurs at a gage length of 66 mm

e) Percentage area reduction (ductility) if the specimen necked to an area of 156 mm2

f) Strain hardening exponent n and strength coefficient K in MPa.

g) New yield strength and tensile strength in MPa and the ductility for the specimen

with a 20% cold work.

Solution

a)

b)

(

)

c)

d)

(

) (

)

e)

(

) (

)

f) Strain hardening exponent is the same as the true strain at UTS;

(

)

(

⁄

)

g)

⁄

Since

(

)

(

) (

) (

) (

)

Q2.6

A tensile test on a certain metal is carried out at a temperature of 540oC. At a strain rate

of 10/s, the stress is measured as 160 MPa; and at a strain rate of 300/s, the stress is

measured as 310 MPa.

a) Determine the strength constant C in MPa and the strain rate sensitivity exponent m.

b) If the temperature were 480oC, what changes would you expect in the values of C and

m?

Solution

a)

b)

Since the temperature is lower, the strength constant would be higher whereas the

strain rate sensitivity exponent would become smaller.

Q2.7

A tensile test uses a test specimen that has a gage length of 50 mm and an area of 200

mm2. During the test the specimen yields under a load of 98000 N. The corresponding

gage length is 50.23 mm. This is the 0.2 percent yield point. The maximum load of

168000 N is reached at a gage length of 64.2 mm. Determine the followings:

a) yield strength,

b) modulus of elasticity, and

c) tensile strength.

d) If fracture occurs at a gage length of 67.3 mm, determine the percent elongation.

e) If the specimen necked to an area = 92 mm2, determine the percent reduction in area.

Solution

a)

b)

c)

d)

( )

e)

( )

Q2.8

In a tensile test on a metal specimen, true strain is 0.08 at a stress of 265 MPa. When

true stress becomes 325 MPa, true strain is 0.27. Determine the strength coefficient and

the strain-hardening exponent in the flow curve equation.

Solution

Solve (1) and (2) together;

(

)

Q2.9

A metal alloy has been tested in a tensile test with the following results for the flow

curve parameters: strength coefficient is 620.5 MPa and strain-hardening exponent is

0.26. The same metal is now tested in a compression test in which the starting height of

the specimen is 62.5 mm and its diameter is 25 mm. Assuming that the cross section

increases uniformly, calculate the load required to compress the specimen to a height of

50 mm.

Solution

(

)

Questions with Answers Only

Q2.10

A tensile test for a certain metal provides flow curve parameters: strain-hardening

exponent is 0.3 and strength coefficient is 600 MPa. Determine the followings:

a) the flow stress at a true strain of 1.0 and (ANS: 600 MPa)

b) true strain at a flow stress of 600 MPa. (ANS: 1)

Q2.11

During a tensile test in which the starting gage length is 125.0 mm and the cross-

sectional area is 62.5 mm2, the following force and gage length data are collected

(1) 17793 N at 125.23 mm,

(2) 23042 N at 131.25 mm,

(3) 27579 N at 140.05 mm,

(4) 28913 N at 147.01 mm,

(5) 27578 N at 153.00 mm, and

(6) 20462 N at 160.10 mm. The maximum load is 28913 N and the final data point

occurred immediately prior to failure. Plot the engineering stress strain curve and

calculate the followings:

a) yield strength, (ANS: 310.27 MPa)

b) modulus of elasticity, and (ANS: 168.6 GPa)

c) tensile strength. (ANS:462.6 MPa)

Q2.12

In a tensile test a metal begins to neck at a true strain of 0.28 with a corresponding true

stress of 345.0 MPa. Without knowing any more about the test, can you estimate the

strength coefficient and the strain-hardening exponent in the flow curve equation?

(ANS: n=0.28, K=492.7 MPa)

Q2.13

A tensile test specimen has a starting gage length of 75.0 mm. It is elongated during the

test to a length of 110.0 mm before necking occurs. Calculate the followings:

a) the engineering strain and (ANS: )

b) the true strain. (ANS: )

c) Compute and sum the engineering strains as the specimen elongates from:

(1) 75.0 to 80.0 mm,

(2) 80.0 to 85.0 mm,

(3) 85.0 to 90.0 mm,

(4) 90.0 to 95.0 mm,

(5) 95.0 to 100.0 mm,

(6) 100.0 to 105.0 mm, and

(7) 105.0 to 110.0 mm.

(ANS: )

d) Is the result closer to the answer to part (a) or part (b)?

Does this help to show what is meant by the term true strain?

Q2.14

A copper wire of diameter 0.80 mm fails at an engineering stress of 248.2 MPa. Its

ductility is measured as 75% reduction of area. Determine the true stress and true strain

at failure. (ANS: )

Q2.15

The flow curve parameters for a certain stainless steel are; a strength coefficient of 1100

MPa and a strain hardening exponent of 0.35. A cylindrical specimen of starting cross-

sectional area of 1000 mm2 and height of 75 mm is compressed to a height of 58 mm.

Calculate the force required to achieve this compression, assuming that the cross section

increases uniformly. (ANS: )

PROBLEM SET FOR CHAPTER 3

Q3.1

A 40 mm diameter billet of annealed steel with a length of 150 mm is direct cold

extruded to 25 mm diameter using a die with an included angle of 120°. The flow curve

of the material is given as 0 18

900.

eqf with initial flow stress

0550

fMPa , and the

coefficient of friction is 0.07 . Calculate the extrusion force using both ;

a) elementary theory of plasticity

b) empirical methods

Hint : 2α<90° deforms along minimum energy path (dead zone forms)

Note : In empirical case, use container friction expression in the elementary theory of

plasticity

Solution

Annealed billet :

a) Using Elementary Theory of Plasticity for Direct Impact Extrusion of Billets :

[

(

) ]

First let’s find equivalent strain and mean flow stress ;

(

)

Since

3140extrusion

F F kN

b) Using Empirical Method for forward extrusion :

where

Q3.2

Ck10 steel cans of inner diameter 70 mm, wall thickness 5 mm, base thickness 6 mm and

height of 120 mm (including the base thickness) will be manufactured using an

axisymmetrical backward extrusion process. Annealed slugs used for the process have a

diameter of 80 mm. The friction coefficient at the punch-workpiece and workpiece-die

interface is 0.05 . Determine the extrusion force by using both ;

a) Dipper’s analysis if applicable

b) Empirical methods

Note: Check for piercing as well

(For Ck 10; K=846 MPa and n=0.16)

Solution

a )

Extrusion force by Dipper’s Solution :

(

) (

)

By using volume constancy →

(

)

(

) (

)

The effective coefficient of friction :

1

2

1 0,05 0,5( 0,5) 0,275

2 2

1 70 6920,6.(1 .0,05. ) 1083.(1 0,275. ) 2540

3 6 5

.70. .2540 9773

4

zm punch

punch i punch

MPa p

F A p kN

b)

Extrusion force by Empirical Methods :

2

00 2 2

0,16

2

80. . 0,8 1,2. 0,8 1,2.ln 0,8 1,2.ln 2,54

80 70

. 846.(1,45)774

1 1,16

.802,54.774. 9882

4

e fm e eq

i

n

eq

fm

AF Q A where Q

A

KMPa

n

F kN

Q3.3

On a 250 kN drawing bench , 1015 steel bars arereduced by 25% in area. If a die with an

included angle of 16° is used and the coefficient of friction is 0.12, find the maximum initial

diameter of the bars that are to be drawn. Check whether there occurs plastic deformation after

the drawn product emerges from the die and check the existence of Chevron cracks inside the

drawn product. (For 1015 Steel; K=620 MPa and n=0.18) Note : Use elementary theory of

plasticity solution.

Solution

0 1 1 0

0 0 1

0,18

250

0,25 0,75 ln 0,29

. 620.0,29420,5

1 420,5

drawing

eq

n

eq

fm

F kN

A A A A

A A A

KMPa

n

max0

8 , 0,12

?

o

d

Drawing Force by using Elementary Theory of Plasticity :

1 1

1 1

1.

1

. . 121,9.

2.tan . . 39,4.

3

0

2. . . .106,2.

sin 2

ideal fm eq

shear fm

cont

fr

fm eqdie sh

fr

F A A

F A A

F

AF A

3 2

1 1 1

2

0 0

250.10 267,5. 935 34,5

9351246 40

0,75

drawingF N A A mm d mm

A mm d mm

Check plastic deformation and Centerburst :

0,18

1

. 620.0,29 496

. 464 250

n

yield eq

yield y

K MPa

F A kN kN

No plastic def.

0 1

0 1

2 sin 19,7

( ) / 21,89 2

d d

L mmL

h d d

L L

nonhomogenity is not larger than 2

no centerburst

Q3.4

Steel tubes with an inner diameter of 38 mm are to be drawn over a stationary mandrel

of 38 mm in diameter to reduce their outer diameters from of 50 mm to 46 mm. For this

purpose a die with included angle of 18° is used. The flow curve of the material is

expressed by 0 24

700.

eqf with initial flow stress

0240

fMPa . The coefficient of

friction between the mandrel and the workpiece is 0.04 whereas it is 0.07 at the die-

shoulder interface. The drawing velocity is 0.5 m/sec. Calculate ;

a) drawing force

b) necessary power

Solution

1 2

0,24

38

50 46

38

700.

240

i

drawing

o o

m

o

f eq

fo

d mm

d mm d mm

mm

MPa

For steel tubes

d for mandrel

9

.

0,04

0,07

0,5 /

?

?

mandrel wp

die sh

drawing

drawing

v m s

F

Power

Drawing Force :

2 2

1 2 2

0,24

2 22

1

50 38. . 111 ln 0,45

46 38

700.0,45467

1,24

.(46 38 )528

4

1.

2

ideal fm eq eq

fm

shear

F A kN

MPa

A mm

F

1

1

1.

tan . . 19,5

. . .28

tan

2. . . .50,3

sin 2

209

fm

fm eq m wpmandrel die

fr

fm eq sdie sh

fr

drawing

A kN

AF kN

AF kN

F F kN

. 209.0,5 104,4Power F v kW

Q3.5

A copper alloy (K= 450 MPa, n= 0.33) sheet with 600 mm. width and 3 mm. thick is to be cold

rolled to 2 mm. thickness in single pass using rolls of 250 mm. diameter. The coefficient of

friction between the interface surfaces is 0.08.

a) Check if the roll will pull in the sheet. If not change the roll diameters such that to pull the

sheet

b) Calculate the roll force

c) Find the power requirement when roll speed is 1.5 m/s.

Solution

K=450 MPa. n=0.33 w= 600 mm h0= 3 mm h1=2 mm

roll dimeter= 250 mm μ= 0.08 v= 1.5 m/s

a)

Rolls will pull if tan α ≤ μ

R

hh

21cos 101 => α = 5.13˚

(tan α = 0.09) > (μ = 0.08) so rolls will not pull in

therefore α = arctan μ => α = 4.57˚

R

hh

21cos 10 => 2R= 314.5 mm R= 157.25 mm

b)

In this question roll force is asked ;

(1.15)r fmP LwQ

Now, let’s find the required parameters one by one ;

mm 54.12)23(25.157 L 0 1 2.52

mean

h hh

2.50.2 1

12.54

meanh

L Full plastification so we can find Q by using the formula below;

1mean

L

hmeanhQ e

L

= 1.23

Mean flow stress is found as below,

MPa 1.25133.1

405.0450

1

33.0

n

Kn

fm

where ln(3 / 2) 0.405

MN 67.21.25123.160054.1215.1)15.1( fmr LwQP

c)

kW 400 W10400157.0

5.11054.121067.2 336

R

vLPPower r

Q3.6

A 35 mm thick plate will be rolled down to 32 mm. The width of the plate is 150 mm and

the rotational speed is 15 rev/min. The material properties are given as K = 480 MPa, n

= 0.22 and = 0.16 at roll-workpiece interface.

a) Determine the minimum roll radius to have initial grasping.

b) For a roll radius of 200 mm, calculate the rolling force and the power required for

this process.

Solution

h0=35 mm h1=32 mm w=150 mm ω=15 rev/min K=480 MPa

n=0.22 µ=0.16

a)

0 1

2(1 cos )

h hR

, To find Rmin for initial grasping we should first find the α value.

For grasping the condition tan should be satisfied.

tan 0.16 09.09

0 1

2(1 cos )

h hR

0

35 32

2(1 cos9.09 )R

119.4R mm

min 120R mm

b)

Roll force for R=200 mm is found with the formula,

(1.15)r fmP LwQ , the required parameters are found below ;

0 1( ) 200(35 32) 24.5L R h h mm

To find if the plate is subjected to full or partial plastification find hmean first,

0 1 33.52

mean

h hh mm

,

33.51

24.5

meanh

L Partial plastification

For partial plastification ;

0.3 0.7 1.110meanhQ

L

Mean flow stress is found below,

0

1

0.22

35 where ln( ) ln( ) 0.0896

1 32

480(0.0896)231.4

0.22 1

n

fm

fm

hK

n h

MPa

(1.15)(24.5)(150)(1.11)(231.4) 1086rP kN

15(1086) (24.5) 41.8

60.2r

vPower P L kW

R

Q3.7

A 42.0 mm thick plate made of low carbon steel is to be reduced to 34.0 mm in one pass in

Rolling operation. As the thickness is reduced, the plate widens 4%. The yield strength of steel

plate is 174 MPa and the tensile strength is 290 MPa. The entrance speed of the plate is 325 mm

and the rotational spee dis 49.0 rev/min. Determine

a) The minimum required coefficient of friction that would make this Rolling operation

possible,

b) Exit velocity of the plate,

c) Forward slip.

Solution

a)

Maximum draft:

Given that

b)

Plate widens by 4%

c)

Q3.8

A cylindrical part is warm upset forged in an open die. The initial diameter is 45mm and the

initial height is 40mm. the height after forging is 25mm. the coefficient of friction at the die-work

interface is 0.20. the yield strength of the work material is 285 MPa, and its flow curve is defined

by a strength coefficient of 600 MPa and a strain hardening exponent of 0.12. Determine the

force in the operation

a) Just as the yield point is reached (yield at strain=0.002)

b) At a height of 35 mm

c) At a height of 30 mm

d) At a height of 25 mm

Solution

a)

b)

c)

d)

Q3.9

A spool of wire has a starting diameter of 2.5 mm. It is drawn through a die with an opening that

is to 2.1 mm. The entrance angel of the die is 18 degrees. Coefficient of friction at the work-die

interface is 0.08. The work metal has a strength coefficient of 450 MPa and a strain hardening

coefficient of 0.26. The drawing is performed at room temperature. Determine area reduction,

draw stress, and draw force required for the operation.

Solution

(

)

(

)

(

)

Questions with Answers Only

Q3.10

A series of cold rolling operations are to be used to reduce the thickness of a plate from

50mm down to 25mm in a reversing two-high mill. Roll diameter= 700 mm and

coefficient of friction between rolls and work=0.15. The specifications are that the draft

is to be equal on each pass. Determine

a) Minimum number of passes required, (ANS: 4 passes)

b) Draft for each pass. (ANS: 6.25 mm)

Q3.11

A single-pass rolling operation reduces a 20 mm thick plate to 18 mm. The starting plate

is 200 mm wide. Roll radius=250 mm and rotational speed=12 rev/min. The work

material has a strength coefficient=600 MPa and a strength coefficient=0.22. Determine

a) Roll force, (ANS: 672000N)

b) Roll torque, (ANS: 3720 Nm)

c) Power required for this operation. (ANS: 37697 W)

Q3.12

A cold heading operation is performed to produce the head on a steel nail. The strength

coefficient for this steel is 600 MPa, and the strain hardening exponent is 0.22.

Coefficient of friction at the die-work interface is 0.14. The wire stock out of which the

nail is made is .00 mm in diameter. The head is to have a diameter 0f 9.5 mm and a

thichness of 1.6 mm. The final length of the nail is 120 mm.

a) What length of stock must project out of the die in order to provide sufficient volume

of material for this upsetting operation? (ANS: 5.78 mm)

b) Compute the maximum force that the punch must apply to form the head in this open

die operation. (ANS: 59886 N)

Q3.13

A hot upset forging operation is performed in an open die. The initial size of the work

part is: D0=25 mm, and h0=50 mm. The part is upset to a diameter=50mm. The work

metal at this elevated temperature yields at 85 MPa (n=0). Coefficient of friction at the

die-work interface=0.40. Determine

a) Final height of the part, (ANS: 12.5 mm)

b) Maximum force in the operation. (ANS: 273712 N)

Q3.14

A billet that is 75 mm long with diameter=35 mm is direct extruded to a diameter of 20

mm. the extrusion die has a die angle=75o. For the work metal, K=600 MPa and n=0.25.

In the Johnson extrusion strain equation, a=0.8 and b=1.4. Determine

a) Extrusion ratio, (ANS: 3.0625)

b) True strain, (ANS: 1.119)

c) Extrusion strain, (ANS: 2.367)

d) Ram pressure and force at L=70mm. (ANS: 3143.4 MPa, 3024321 N)

PROBLEM SET FOR CHAPTER 4

Q4.1

When a round sheet metal blank is deep drawn, it is found that it does not exhibit any earing. Its

r values in the 0° and 90° directions to rolling are 1.4 and 1.8, respectively. What is the r value in

45° direction?

Solution

For earing to occur there should be planar anisotropy. So planar anisotropy should be zero for

no earing.

90 45

90

45 45

1,4 1( 2. )

1,8 2

0 1,4 1,8 2. 1,6

o

o

rr r r

r

r r

Planar anisotropy r=

r

Q4.2

120 mm long rectangular steel strips will be cut from 6 mm thick stock having a tensile strength

of 350 MPa.

a) Calculate the necessary press capacity to cut a single strip with no blade angle.

b) If a 50 kN press is available, what minimum angle should be given to blade to cut the same

strips?

Solution

a)

b)

(

) (

)

Q4.3

A blanking operation is to be performed on 2.0 mm thick cold-rolled steel. The part is circular

with a diameter of 75.0 mm. Determine the blanking force required, if the tensile strength of the

steel is 450 MPa.

Solution

( )

Q4.4

What is the minimum bend radius to prevent fracture for a 2 mm thick sheet metal with a tensile

reduction of area of 15% ? Does the bend angle affect your answer?

Solution

For , the bending radius for no fracture is calculated as;

(

) (

)

→

Q4.5

A 5 mm thick steel sheet metal part, 500 mm long and 125 mm wide, required to be bent to 90°.

The material obeys 0 125

850.

and with modulus of elasticity 180 GPa, initial flow stress 380

MPa and tensile strength 610 MPa. Ductility of the material is 40%.

a) Calculate the minimum radius of bending die (rounded to the nearest integer value) possible

to prevent necking and fracture.

b) If the bending die radius is 17 mm, calculate the included die angle and bending angle. Note

that you should check the gentle bend assumption first.

c) Calculate the free bending force with a die opening of 100 mm.

Solution

Steel Sheet Metal Part:

5

500

125

h mm

l mm

w mm

0,125850

380

180

610

f

fo MPa

E GPa

UTS MPa

a)

For no necking;

2. 16.3

2( 1)

n

b bn

eR h R mm

e

17bR mm

For no fracture when q=0.4;

2

2

(1 ). 2.8

2

3

b b

b

qR h R mm

q q

R mm

Larger value is taken as bending radius which is 17 mm.

b)

?

?

b

b

Gentle bend assumption;

2 17 10 (gentle bend assumption is satisfied)bR h

3

1 3. . 4. .b b b

f

R R Y R Y

R h E h E

where Y=380 MPa

2 . 91.73 , 88.27

2

bo o

f b b b

f

hR

hR

where 90 , 90o o

f f

c)

Q4.6

A 0.5 mm thick sheet is to be drawn with a LDR of 1.9 to produce a container with an inside

diameter of 40 mm and a height of 60 mm. Tensile strength is 150 MPa. Assume the thickness

of the bottom and the walls of the container remain unchanged. (i.e., 0.5 mm)

a) Calculate the diameter of the initial circular blank necessary.

b) For drawing the same initial circular blank, check if single drawing is enough and if not

calculate the smallest possible punch diameter for the first step and estimate the number of

redrawing necessary.

c) Calculate the drawing force. (If single drawing is not enough calculate the drawing force of

the first step)

Solution

Thick Sheet Container

5 40 , 60 , 150drawn with an LDR=1.9i conth mm d mm h mm UTS MPa

a)

Using volume constancy;

b)

LDR=1.9

Redrawing is necessary

After first redrawing;

single redrawing is sufficient

c)

Drawing force;

(

)

Q4.7

A cup is to be drawn in a deep drawing operation. The height of the cup is 75 mm and its inside

diameter is 100 mm. If the blank diameter is 225 mm, determine

a) drawing ratio

b) reduction, and

c) Does the operation seem feasible?

Solution

a)

b)

c)

The drawing ratio is larger than 1.2 and this may lead to wrinkling. Therefore, the operation is

not feasible.

Q4.8

A circular sheet of diameter 150 mm has to be deep drawn to a cup with 123 mm inner

diameter. If the sheet thickness is 3 mm and the ultimate strength of the sheet material

300 MPa, determine the deep drawing force.

Solution

( ) (

) (

)

Questions with Answers Only

Q4.9

Prior to a sheet rolling process a 10 mm diameter circle is scribed onto a piece of sheet metal.

After the sheet is deformed the circle has changed into an ellipse. The axis of the ellipse parallel

to the sheet rolling direction is 14.2 mm, while perpendicular to the sheet rolling direction the

ellipse axis is 7.9 mm. Calculate the plastic anisotropy in the direction along the sheet rolling

direction. (ANS: 2.05)

Q4.10

A compound die will be used to blank and punch a large washer of outside diameter 80 mm and

inner diameter 40 mm, out of aluminum alloy sheet stock of 3 mm thick. Determine the

minimum tonnage of the press to perform the operation if UTS is 140 MPa. (Assume that no

shear angle is included.) (ANS: 134.6 kN)

Q4.11

5 mm thick sheet metal blank is to be V-bent (90˚ final bend angle required). The width (i.e.

length of bend) of the sheet metal is 50 mm. Assume that the material obeys = 500. 0.3 and

other material properties are given as E=210 GPa, y=200 MPa and UTS=350 MPa. Wb is given to

be 20 mm. Calculate

a) the minimum radius of bending die possible to prevent necking, (ANS: 5 mm)

b) the die angle, (ANS: 89.832)

c) bending angle, (ANS: 90.168)

d) the bending force. (ANS: 21875 N)

Q4.12

100 mm diameter and 90 mm deep cup is to be drawn from 1 mm thick steel sheet of deep

drawing quality. UTS = 410 MPa. (D =Db, d = Dp)

a) Determine the blank diameter and suitable punch diameter for

the first draw. (ANS: 114 mm)

b) Estimate the press capacity. (ANS: 172 kN)

t/D x100 D/d

0.15 1.43

0.20

0.30

0.40

0.50

Above 0.50

1.54

1.67

1.82

1.91

2.00

PROBLEM SET FOR CHAPTER 5

Q5.1

A 50 mm diameter bar is to be turned (orthogonal cutting) to a final diameter of 47.80 mm with

the following cutting parameters recorded:

n = 20 = 15

chip thickness = 0.3 mm f = 0.12 mm/rev

Ft = 100 N N = 352 rpm

E1= 1.4 W.s/mm3 (for 1 mm undeformed chip thickness)

Calculate;

a) tangential cutting force component.

b) coefficient of friction.

c) the mean shear stress on shear plane.

Solution

a)

? ? meancV mRR

2 2 (50 47,80). . . 352. . 1802,5 / 1,8 /

60 2 60 2mean

meanc mean

DV r N mm s m s

350 47,8. . 0,12 1802,5 238 /

2mRR f d V x x mm s

?cF

. .

0

0,2

. . . .

.( ) .cos 0,12.cos15 0,116

1,4.( ) 2,154 . . 284,33

where

c

c c c

V f d

c c

c c

Power F V E mRR F E f d

E E a a f mm

E a F E f d N

b)

?

.sin .cos20

.cos .sin

284,330,820

100

where f oc n t nn

n c n t n

c

t

F F F

F F F

F N

F N

c)

?ss

s

F

A

0

.cos .sin ?

.cos 0,116tan 0,387

1 .sin 0,3

22,72 223,64

. 0,12.1,10,341

sin sin sin(22,72)

654,4

s c t

c n cc

c n

o

s

cs

ss

s

F F F

r ar

r a

F N

A f dA

FMPa

A

Q5.2

Turning operation is performed on a work material with shear strength of 380 MPa. Assuming

Ernst-Merchant theory holds and according to the following cutting parameters:

v = 2.2 m/s n

d = 1.8 mm f = 0.25 mm/rev rc = 0.48

Calculate;

a) shear plane angle and the shear force.

b) tangential cutting force and the power requirement.

Solution

a)

1.cos 0,48.cos5tan 0,499 tan (0,499) 26,52

1 .sin 1 0,48.sin5c n

c n

r

r

2.. . 0,25.1,8 0,45

sin

380.0,45383

sin(26,52)

s cs s s c

s

AF A A f d mm

F N

b)

Since Ernst – Merchant Theory holds ;

2 2.(26,52) 5 41,962 2

. cos( ) cos(41,96 5)383. 685,4

sin cos( ) cos(26,52 41,96 5)

. 685.2,2 1507,9

s

n

s c nc

n

F

c

AF N

P F v W

Q5.3

In an orthogonal cutting operation a 8 mm deep groove is to be turned on a 50 mm diameter

steel bar. Spindle speed is 300 rpm and a feed rate of 0.25 mm/rev is given to the tool. Produced

chips have a width of 2 mm. Calculate the material removal rate at the beginning and at the end

of the cut.

Solution

? ? i fmRR mRR

(

)

(

)

Q5.4

A m c i i g op io is fou o fo o T y o ’s oo if qu io i cu i g sp of

400 m/min for 1 minute tool life and with the tool life exponent of 0.5. For this operation

calculate the percentage increase in tool life if the cutting speed is reduced by 50%.

Solution

1 2

1

400 m / min 200 m / min

1min

0.5

v v

T

n

0.5 0.51 1 2 2 2 2. . 400.1 200. 4minn nv T v T T T

Percent increase in tool life is ;

2 1

1

4 1.100% .100% 300%

1

T T

T

Q5.5

A 100mm diameter bar is turned in one pass on a lathe to 90mm. A minimum tool life of 8 hours

is desired when a feed of 0.35 mm/rev is given. Find the tool life exponent and calculate the

maximum cutting speed in m/min for these conditions. The appropriate tool life equation is :

18

10 8 6

3.10

. .T

v f d where T(min), v(m/min), f(mm/rev), d(mm)

Solution 100

90

8

0,35 /

i

f

d mm

d mm

t hours

f mm rev

1 2

8

1 1 10 8 61

3.10

. .. .n nn

KT

v f dv f d

where “d” is depth of cut

110 0,1 tool life exponentn

n

8

10 8 6

3.108.60 33,5m/min

.0,35 .5v

v where 5

2

i fd dd mm

Q5.6

Rods of 80 mm in diameter and 400 mm in length are to be turned to a diameter of 77.6 mm.

Triangular inserts, which cost $4.5 will be used for turning operation. Tool life exponent and

constant are 0.3 and 180m/min (for 1 min tool life), respectively. Resetting and indexing the tool

takes 1.6 min. Loading and unloading time for the part is 1.2 min. The operator wage is $6.6/hr

and the machine cost is $10.8/hr, both including overheads. Rotational speed is 400 rpm and

feed rate is set to 0.30 mm/rev.

Calculate:

a) tool life for minimum production cost and maximum production rate criteria

b) cutting speed for minimum production cost and maximum production rate criteria

c) unit production cost for the given cutting parameters.

Solution

0 80

77,6

400

400

0,3 /

i

d mm

d mm

l mm

N rpm

f mm rev

0.3

1

Triangular insert # of edge = 3

Cost of insert $ 4,5

. 180m/ min

1,6min 1,2min

$ $6,6 10,8

ch

o m

V T

t t

R Rhr hr

a)

1.

( )t

mc ch

o m

Cnt t

n R R

where 4,5

1,5$3

tC 15,8minmct

1 0,7. .1,6 3,73min

0,3mp ch

nt t

n

b)

0,3

0,3

0,3

0,3

180. 180 78,64m/ min

(15,8)

180. 180 121,24m/ min

(3,73)

mc mc mc

mp mp mp

V t V

V t V

c)

0 1 0( )( ) [ ( ) ]

1 400min3,33min

. (0,3) .400min

pr m c t m tch

c

C R R t t n t R R C

tmm revf N

rev

0,3 0,3

80 77,6 1. . . . .400 99,03

2 1000 min

. 180 (99,03). 180 7,33min

3,330,454

7,33

2,207$

m

ct

pr

mv D N

V T T T

tn

t

C

Q5.7

For a turning operation brazed carbide tip type cutting tools, which cost $6 and can be reground

5 times, are used. The cost of regrinding is $0.8. The lathe costs $36000 and is used on an 8-hour

shift per day, 5 days per week and 50 weeks per year. Its cost is to be amortized over 10 years

and the machine overheads are 100%. The operator’s wage is $6/hr and operator overheads are

100%. The governing tool life equation for the tool is given as vT0.25 = 240 (m/min). The tool

changing time is 3 min and the loading/unloading time per part is 4min. The length of the part is

400 mm and initial diameter is 42 mm. A 2 mm depth of cut is given and the feed rate is set to

0.24 mm/rev. Calculate:

a) the minimum production cost

b) the minimum production time in min.

Solution

cos t of tool = $ 6

# of regrinding = 4

cost of reground = $ 1

lathe cost = $ 36000

usage per year = 50.5.8.60=120000 min/year

Amortization time = 10 years

machine overheads = 100%

Operator's wage = $ 6 /hr

operator overheads = 100 %

0,25

0

m. 240

min

3min

4min

400

42

2

0,24 /

ch

l

V T

t

t

l mm

d mm

d mm

f mm rev

a)

?prC

0 0

0

( )( ) [ ( ) ]

6$ 6operator wage rate + overhead = 0.2$ / min

60min 60

cost of machine 36000machine overhead = .2 0,06$ / min

Amortization time 10.120000

0,26$ / min

.2

pr m c t m tl ch

o

m

o m

c

C R R t t n t R R C

R

R

R R

d dBt

V

0,25

1.

cost of tool cost of grinding 6 $ 4 $1 . 2

# of cutting edges cutting edge 4 1 regrind 5 edges edge

32,076min

(32,076) 240 100,87m/ min

4.

2

i tmc ch

o m

t

mc

c

Cl nt t

f v n R R

C

t

V V

t

32 38 400.10

. 2,08min0,24 100,847

min( )

2,08 C 1,76$

min 32,076( )

c

t pr

tpiece

n

tedge

b)

0,25

.

1 0,75. .3 9min

0,25

240138,861m/ min

9

1,51min

1,514 1,51 .3 6,01min

9

pr c tl ch

mp ch

c

pr

t t t n t

nt t

n

V

Bt

V

t

Questions with Answers Only

Q5.8

A 80 mm diameter steel bar is to be turned to 76 mm diameter at 352 rpm with the following

conditions:

normal rake angle = 15 side cutting edge angle = 10

chip thickness = 0.75 mm feed rate = 0.20 mm/rev

thrust force = 400 N α = 0.2 (for specific cutting energy equation)

specific cutting energy for 1 mm undeformed chip thickness = 1.6 W.s/mm3

Calculate;

a) cutting speed and material removal rate (ANS: 1.47 m/s, 590 mm3/s)

b) cutting force (ANS: 885.74 N)

c) coefficient of friction (ANS: 0.819)

d) the mean shear stress (ANS: 493.1 MPa)

Q5.9

In an orthogonal cutting operation a 10 mm deep groove is to be turned on a 60 mm diameter

steel bar. Spindle speed is 400 rpm and a feed rate of 0.25 mm/rev is given to the tool. Produced

chips have a width of 3 mm. Calculate the material removal rate at the beginning and at the end

of the cut. (ANS: 942.5 mm3/s, 628.3 mm3/s)

Q5.10

A machining operation is found to follow the Taylor’s tool life equation with a cutting speed of

400 m/min for 1 minute tool life and with the tool life exponent of 0.5. For this operation

calculate the percentage increase in tool life if the cutting speed is reduced by 50%. (ANS:

300%)

Q5.11

Tool life tests on a lathe have resulted in the following data:

(1) v = 100 m/min T = 10 min

(2) v = 75 m/min T = 30 min

a) Determine the tool life exponent and cutting speed for 1 min tool life in Taylor’s equation.

(ANS: n=0.262 , C=182.8 m/min)

b) based on the equation found in part (a), calculate the tool life for a speed of 90 m/min (ANS:

14.9 min)

c) based on the equation found in part (a), calculate the cutting speed corresponding to a tool

life of 20 min (ANS: 83.4 m/min)

Q5.12

A 100mm diameter bar is turned in one pass on a lathe to 90mm. The specific cutting energy is

2.48 GJ/m3. A tool life of 8 hours is desired. The appropriate tool life equations for these

conditions are:

60TV n

6810

18

dfv

103T

where T(min), v(m/min), f(mm/rev), d(mm)

Calculate:

a) the tool life exponent, n, (ANS: n=0.1)

b) the cutting speed, V, (ANS: 32.4 m/min)

c) feed rate, f, (ANS: 0.365 mm/rev)

d) main cutting force, Fc, (ANS: 4.5 kN)

e) and required power at the tool tip. (ANS: 2.4 kW)

Q5.13

A 125mm wide pulley is made of 1015 steel. The forged outer diameter is 506mm and it is to be

rough turned on an engine lathe with a depth of 3mm and a feed of 0.5mm/rev with a proper

HSS tool. If for these conditions the relationship VT0.1 = 60 is valid, what is the cutting time per

piece if the cutting speed is such that the tool life is 2h? (ANS: 10.7 min)

Q5.14

Referring to problem 5.13, determine the cutting speed that would result in maximum

production if the tool change time is 2 min. (ANS: 45 m/min)

Q5.15

Refering to problem 5.13, determine the cutting speed and unit production time for minimum

cost if the following conditions pertain to this operation: (ANS: 42.5 m/min, 11.95 min)

Machine+Operator rate: $13.00/h Overheads: $15.50/h

Tool cost: $1.25 Tool regrinds: 20 times

Tool grind time: 3 min. Grinder rate: $12.50/h

Tool change time: 2 min. Load and unload time: 2 min