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
Before: 2.0 cm
1.9 cm
cm
ccm cm
After: 1.893 cm
rrrrr
1.698 cm
Strains at larger and smaller diameter sections are respectively:
(
)
(
)
The load carrying capacities at two sections must be the same:
⁄
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 crosssectional 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 crosssectional 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
f
 900
0.18
eq
with initial flow stress
f0
 550MPa , 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
Fextrusion   F  3140kN
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 :
0,05  0,5
 0,275
2
1
70
6
)  1083.(1  0,275. )  2540MPa   p punch
zm  920,6.(1  .0,05.
3
6
5
2
 .70
Fpunch  Ai . p punch 
.2540  9773kN
4
1
2
  ( 1  0,5) 
b)
Extrusion force by Empirical Methods :
F  Qe .
fm
. A0 where Qe  0,8  1,2.
eq
 A0 
 802 
 0,8  1,2.ln    0,8  1,2.ln  2
 2,54
2 
 80  70 
 Ai 
 K . eq n   846.(1,45)0,16 

  774 MPa
fm  
 n  1  
1,16


 
 .802
F  2,54.774.
 9882kN
4
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
Fdrawing  250kN
A0  A1
A
 0,25  1  0,75 
A0
A0
eq
 ln
A0
 0,29
A1
d0max  ?
n
620.0,290,18


 420,5MPa
n 1
420,5
K.
fm
  8o ,   0,12
eq
Drawing Force by using Elementary Theory of Plasticity :
Fideal  A1.
fm
.
eq
2
Fshear  .tan  .
3
cont
Ffr  0
2.
Ffrdiesh. 
Fdrawing
A0 
.
 121,9. A1
fm
. A1  39,4. A1
. . A1
 106,2. A1
sin 2
 250.103 N  267,5. A1  A1  935mm 2  d1  34,5mm
fm
eq
935
 1246mm2  d 0  40mm
0,75
Check plastic deformation and Centerburst :
yield
 K.
Fyield 
y
n
eq
 620.0,290,18  496MPa
. A1  464kN  250kN  No plastic def.
d0  d1
2  sin   L  19,7mm
L
h (d 0  d1 ) / 2

 1,89  2  nonhomogenity is not larger than 2
L
L
 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
f
 700
0.24
eq
with initial flow stress
f0
 240MPa . The coefficient of
friction between the mandrel and the workpiece is 0.04 whereas it is 0.07 at the dieshoulder interface. The drawing velocity is 0.5 m/sec. Calculate ;
a) drawing force
b) necessary power
Solution
di  38mm
drawing
d o1  50mm 
d o2  46mm
For steel tubes
die sh.  0,07
d m  38mm for mandrel
vdrawing  0,5m / s
  9o
f
 700.
fo
 240MPa
Fdrawing  ?
0,24
eq
Power  ?
Drawing Force :
Fideal  A1.
fm
.
eq
 502  382 
 111kN  eq  ln  2
 0,45
2 
 46  38 
700.0,450,24
 fm 
 467 MPa
1,24
 A1 
1
Fshear  . tan  .
2
Ffrmandrel die 
Ffrdie sh. 
Fdrawing
fm
mandrel  wp  0,04
.
fm
. A1  19,5kN
eq
.m wp . A1
tan 
2. fm . eq . s . A1
sin 2
  F  209kN
 .(462  382 )
 28kN
 50,3kN
Power  F .v  209.0,5  104,4kW
4
 528mm 2
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
roll dimeter= 250 mm
w= 600 mm
h0= 3 mm  h1=2 mm
μ= 0.08
v= 1.5 m/s
a)
Rolls will pull if


  cos 1 1 
tan α ≤ μ
h0  h1 
2 R 
=>
α = 5.13˚
(tan α = 0.09) > (μ = 0.08)
so rolls will not pull in
therefore α = arctan μ =>
α = 4.57˚
cos   1 
h0  h1
2R
=> 2R= 314.5 mm
R= 157.25 mm
b)
In this question roll force is asked ;
Pr  (1.15) LwQ
fm
Now, let’s find the required parameters one by one ;
L  157.25(3  2)  12.54 mm
hmean 
h0  h1
 2.5
2
hmean
2.5

 0.2  1  Full plastification so we can find Q by using the formula below;
L
12.54
h
Q  mean
L
 h L

e mean  1 = 1.23


Mean flow stress is found as below,
fm
K
n
n 1
 450
Pr  (1.15) LwQ
0.4050.33
 251.1 MPa where
1.33
fm
 ln(3 / 2)  0.405
 1.15 12.54  600  1.23  251.1  2.67 MN
c)
Power  Pr  L 
v
1.5
 2.67 106 12.54 10 3 
 400 103 W  400 kW
R
0.157
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
n=0.22
µ=0.16
w=150 mm
ω=15 rev/min
K=480 MPa
a)
R
h0  h1
, To find Rmin for initial grasping we should first find the α value.
2(1  cos  )
For grasping the condition tan    should be satisfied.
tan   0.16    9.090
R 
h0  h1
35  32
R 
 R  119.4mm  Rmin  120mm
2(1  cos  )
2(1  cos 9.090 )
b)
Roll force for R=200 mm is found with the formula,
Pr  (1.15) LwQ
fm
, the required parameters are found below ;
L  R(h0  h1 )  200(35  32)  24.5mm
To find if the plate is subjected to full or partial plastification find hmean first,
hmean 
h0  h1
h
33.5
 33.5mm , mean 
 1  Partial plastification
2
L
24.5
For partial plastification ;
Q  0.3
hmean
 0.7  1.110
L
Mean flow stress is found below,
fm

K n
where
n 1
fm

480(0.0896)0.22
 231.4MPa
0.22  1
 ln(
h0
35
)  ln( )  0.0896
h1
32
Pr  (1.15)(24.5)(150)(1.11)(231.4)  1086kN
Power  Pr  L 
v
15
 (1086)  (24.5) 
 41.8kW
R
60.2
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.
ro  1,4
1
Planar anisotropy  r= (ro  r90  2.r45 )
r90  1,8
2
r  0  1,4  1,8  2.r45  r45  1,6
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
 850
0.125
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:
h  5mm
l  500mm
w  125mm
 850
fo
 380MPa
0,125
E  180GPa
UTS  610MPa
a)
For no necking;
2  en
Rb 
.h  Rb  16.3mm
2(en  1)
Rb  17mm
f
For no fracture when q=0.4;
(1  q) 2
.h  Rb  2.8mm
2q  q 2
Rb  3mm
Rb 
Larger value is taken as bending radius which is 17 mm.
b)
b  ?
b  ?
Gentle bend assumption;
Rb  2h  17  10 (gentle bend assumption is satisfied)
3
Rb
R Y 
R Y 
 1  3. b .   4. b .  where Y=380 MPa
Rf
 h E
 h E

h

R

b

2  .    91.73o ,   88.27o where  f  90o ,  f  90o
f 
b
b
h b
 Rf  

2
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
drawn with an LDR=1.9
h  5mm 
 di  40mm, hcont  60mm,UTS  150MPa
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
other material properties are given as E=210 GPa,
y=200
= 500.
0.3
and
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
1.54
0.30
1.67
0.40
1.82
0.50
1.91
Above 0.50
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
3
E1= 1.4 W.s/mm (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)
Vcmean  ?
mRR  ?
Vcmean  .rmean  N .
2 Dmean
2 (50  47,80)
.
 352. .
 1802,5mm / s 1,8m / s
60 2
60
2
 50  47,8 
3
mRR  f .d .V  0,12 x 
 x1802,5  238mm / s
2


Fc  ?
Power  Fc .Vc  E. mRR  Fc  E. f .d
Vc . f .d
E  E0 .(ac )

E  1,4.(ac )
 where ac  f .cos   0,12.cos15  0,116mm
0,2
 2,154  Fc  E. f .d  284,33N
b)
 ?

Ff
Fn

Fc .sin  n  Ft .cos  n
Fc .cos  n  Ft .sin  n
Fc  284,33N 
   0,820
Ft  100 N

where
 n  20o
c)
s 
Fs
?
As
Fs  Fc .cos   Ft .sin     ?
rc .cos  n
a 0,116
 rc  c 
 0,387
1  rc .sin  n
a0
0,3
tan  
   22,72o  Fs  223,64 N
Ac
f .d
0,12.1,1


 0,341
sin  sin  sin(22,72)
F
  s  s  654,4 MPa
As
 As 
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:
n
f = 0.25 mm/rev
v = 2.2 m/s
d = 1.8 mm
rc = 0.48
Calculate;
a) shear plane angle and the shear force.
b) tangential cutting force and the power requirement.
Solution
a)
tan  
rc .cos  n
0, 48.cos5

 0, 499    tan 1 (0, 499)  26,52
1  rc .sin  n 1  0, 48.sin 5
Fs   s . As 
Fs 
 s . Ac
 Ac  f .d  0, 25.1,8  0, 45mm2
sin 
380.0, 45
 383N
sin(26,52)
b)
Since Ernst – Merchant Theory holds ;
2     n 


 2.(26,52)    5     41,96
2
2
 s . Ac cos(   n )
cos(41,96  5)
Fc 
 383.
 685, 4 N
sin  cos(     n )
cos(26,52  41,96  5)
Fs
P  Fc .v  685.2, 2  1507,9W
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
mRRi  ?
mRR f  ?
(
)
(
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
v1  400 m / min  v2  200 m / min
T1  1min
n  0.5
v1.T1n  v2 .T2n  400.10.5  200.T20.5  T2  4min
Percent increase in tool life is ;
T2  T1
4 1
.100% 
.100%  300%
T1
1
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 :
T
3.1018 where T(min), v(m/min), f(mm/rev), d(mm)
v . f 8 .d 6
10
Solution
di  100mm
d f  90mm
t  8hours
f  0,35mm / rev

T
K
1
1
1
v n . f n1 .d n2

3.108 where “d” is depth of cut
v10 . f 8 .d 6
1
 10  n  0,1  tool life exponent
n
8.60 
di  d f
3.108
 5mm
 v  33,5m/min where d 
10
8 6
2
v .0,35 .5
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
d 0  80mm
di  77,6mm
l  400mm
N  400rpm
f  0,3mm / rev
Triangular insert  # of edge = 3
Cost of insert  $ 4,5
V .T 0.3  180m/ min
tch  1,6 min
t1  1, 2 min
$
$
Ro  6,6
Rm  10,8
hr
hr
a)
tmc 

Ct
1 n 
. tch 

n 
( Ro  Rm ) 
tmp 
1 n
0,7
.tch 
.1,6  3,73min
n
0,3
where
Ct 
b)
180
 78,64m/ min
(15,8)0,3
180

 121, 24m/ min
(3,73)0,3
Vmc .tmc 0,3  180  Vmc 
Vmp .tmp 0,3  180  Vmp
c)
C pr  ( R0  Rm )(t1  tc )  nt [tch ( R0  Rm )  Ct ]
tc 
1
400 min

 3,33min
f .N (0,3) mm .400 rev
rev
min
80  77,6 1
m
.
.400  99,03
2
1000
min
V .T 0,3  180  (99,03).T 0,3  180  T  7,33min
t 3,33
nt  c 
 0, 454
t 7,33
C pr  2, 207$
v   .Dm .N   .
4,5
 1,5$
3
 tmc  15,8min
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 vT 0.25 = 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 %
V .T 0,25  240
m
min
tch  3min
tl  4 min
l  400mm
d 0  42mm
d  2mm
f  0, 24mm / rev
a)
C pr  ?
C pr  ( R0  Rm )(tl  tc )  nt [tch ( R0  Rm )  Ct ]
6$
6
  0.2$ / min
60 min 60
cost of machine
36000
Rm 
 machine overhead =
.2  0,06$ / min
Amortization time
10.120000
Ro  Rm  0, 26$ / min
Ro  operator wage rate + overhead =
tc 
B  d 0  di l
 .
.
V 2
f
v
tmc 
Ct 
1 n 
t 

n  ch Ro  Rm 
cost of tool
cost of grinding
6
$
4
$


1
.
2
# of cutting edges
cutting edge
4  1 regrind 5 edges
edge
 tmc  32,076 min
Ct 
V (32,076)0,25  240  V  100,87m/ min
tc 
 42  38 400.103
.
 2,08min
0, 24 100,847
min
tc (
)
2,08
piece
nt 

min
t(
) 32,076
edge
2
.
C pr  1,76$
b)
t pr  tl  tc  nt .tch
1 n
0,75
.t 
.3  9 min
n ch 0, 25
240
V  0,25  138,861m/ min
9
B
tc   1,51min
V
1,51
t pr  4  1,51 
.3  6,01min
9
tmp 
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:
V  T n  60
T
3  1018
v10  f 8  d 6
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