MFGE 307 – Theory of Manufacturing
Technology II
Fall 2015-2016
FINAL EXAM
January 7, 2016
Part 2: 4 Comprehensive Questions
(60 Points)
NAME & SURNAME
SIGNATURE:
STUDENT NO:
Remarks
1. This is a Closed Book, Closed Notes examination.
2. The time allowed is only 90 minutes. Please budget your time…
3. Use ball point pen or dark pencil. Write clearly and neatly
4. Write the unit of every result
5. Read all questions very carefully and try to answer all the questions. If you do not understand
the meaning of a question, please feel free to ask.
6. Thrust on your knowledge / original thinking and do not attempt cheating. Academic
dishonesty will be dealt with pertinent University regulations.
GOOD LUCK…
S. E. KILIÇ
QUESTIONS
(15 points) 1. In an orthogonal cutting test on mild steel the following data were obtained:
Uncut chip thickness : 0.25 mm
Chip thickness : 0.75mm
Width of cut : 2.5mm
Chip-tool contact length: 0.5 mm
Tangential (main) cutting force : 900N
Feed (thrust) force : 450N
Normal rake angle : 0°
Side cutting edge angle : 0°
Calculate
a)
b)
c)
d)
The mean angle of friction on the tool face in degrees.
The shear angle in degrees.
The mean shear strength of the work material in MPa on the shear plane.
The mean frictional strength of the work material in MPa on the tool face.
SOLUTION 1
a) Since n=0 Ff= Fc=900N and Fn=Ft= 450N
Then, tan= Ff/Fn=450/900= 0.5  =26.6o
b)
tan ∅ =
rc cos 𝛾𝑛
1−𝑟𝑐 sin 𝛾𝑛
and rc=
𝑎𝑐
𝑎0
rc= 0.25/0.75 =1/3
tan ∅ = 𝑟𝑐 =1/3  ∅=18.4o
c) 𝐹𝑠 =𝐹𝑐 cos ∅ − Ft sin ∅
𝐹𝑠 = 900𝑥 cos 18.4 − 450𝑥 sin 18.4 = 712 𝑁
𝜏𝑠 =
𝐹𝑠
𝐴𝑠
𝐴𝑠 =
0.25𝑥2.5
sin 18.4
= 1,98𝑚𝑚2
s=712/1.98= 360 MPa
d)
Ff= 450N Af= 𝑎𝑤 𝑙𝑓 = 2.5𝑥0.5 = 1.25𝑚𝑚2
𝜏𝑓 = 𝐹𝑓 /𝐴𝑓 =450/1.25= 360 Mpa
(15 points) 2. A 1.5 m diameter disk with a 600 mm hole in the center is to be faced, starting at the
outside diameter, on a vertical boring machine. The rotational frequency of the table is
0.5s-1, the feed is 0.25 mm and the back engagement (depth of cut) is 6 mm. The
specific cutting energy for the work material and the particular cutting conditions is
3.5 GJ/m3.
Calculate
a) The machining time in kiloseconds.
b) The power consumption at the beginning of the operation in kW.
c) The power consumption just before the end of the operation in kW.
SOLUTION 2
𝑙𝑤
a) 𝑡𝑐 =
𝑛𝑤 𝑓
𝑙𝑤 =
1500−600
2
= 450𝑚𝑚 𝑡𝑐 =
450
0.5𝑥0.25
= 3600𝑠 = 3.6𝑘𝑠
b) 𝑃𝑚𝑎𝑥 = 𝑝𝑠 𝑍𝑚𝑎𝑥
𝑍𝑚𝑎𝑥 = 𝜋𝑎𝑝 𝑓𝑛𝑤 𝑑𝑜 = 6𝑥0.25𝑥0.5𝑥1500 = 3534 𝑚𝑚3
𝑃𝑚𝑎𝑥 = 3.5 ∗ 3534 = 12.4𝑘𝑊
c) 𝑃𝑚𝑖𝑛 = 𝑝𝑠 𝑍𝑚𝑖𝑛
𝑍𝑚𝑖𝑛 = 𝜋𝑎𝑝 𝑓𝑛𝑤 𝑑𝑖 = 6𝑥0.25𝑥0.5𝑥600 = 1414 𝑚𝑚3
𝑃𝑚𝑖𝑛 = 3.5 ∗ 1414 = 5𝑘𝑊
(15 points) 3. It can be shown that for orthogonal cutting metal with a tool of zero rake angle, the rate of
heat generation Ps in the shear (primary deformation) zone is given by
Ps=Fcv(1-rc)
where
Fc= (tangential) cutting force
v= cutting speed
= mean coefficient of friction on the tool face
rc= cutting ratio
a) For these conditions, calculate the mean shear zone temperature rise θs in degrees
Celcius (Centrigrade), when the metal has a specific cutting energy of 2.8 GN/m2,
=1.0, rc=0.2, and 10% of the shear-zone heat is conducted into the workpiece. Assume
for the work material that density is 7200kg/m3, and specific heat capacity is 500J/kgK.
b) For the same conditions in (a) find the mean temperature rise of the chip in degrees
Celsius (Centigrade) from the secondary deformation (friction) zone neglecting the heat
conducted into the cutting tool.
c) What would be the value of θs if the cutting speed were doubled and the proportion of
the shear-zone heat conducted into the workpiece remained the same?
SOLUTION 3
a)
𝜃𝑠 =
(1−Γ)𝑃𝑠
𝜌𝑐𝑣𝑎𝑐 𝑎𝑤
Substituting for Ps=Fcv(1-rc) and 𝐹𝑐 = 𝑝𝑠 𝑎𝑐 𝑎𝑤
𝜃𝑠 =
b) 𝜃𝑓
(1 − Γ)𝑝𝑠 (1 − 𝜇𝑟𝑐 ) (1 − 0.1)𝑥2.8𝑥(1 − 1.0𝑥0.2)𝑥109
=
= 5600 𝐶
𝜌𝑐
7200𝑥500
𝑃𝑓
=
𝜌𝑐𝑣𝑎𝑐 𝑎𝑤
Substituting for Pf=Fcvrc and 𝐹𝑐 = 𝑝𝑠 𝑎𝑐 𝑎𝑤
𝜃𝑓 =
𝜇𝑟𝑐 𝑝𝑠
𝜌𝑐
=
1.0𝑥0.2𝑥2.8𝑥109
7200𝑥500
= 1560 𝐶
𝑐) It would remain the same; since  remains unchanged and 𝜃𝑠 does not depend on the
cutting speed.
(15 points) 4. A high-speed steel (HSS) tool is used to turn a steel workpiece that is 300mm long and
80mm in diameter. Taylor tool life parameters are estimated as C= 75 m/min and
n=0.13. The operator and the machine tool rate including overheads is 90 TL/hr and the
tooling cost per sharp cutting edge is 12 TL. It takes 2 minutes to load and unload the
workpart and 3.50 minutes to change tools. Feed rate is adjusted to 0.25 mm/rev.
Determine the following to maximize the production rate (i.e. to minimize the unit
production time).
a) the cutting speed in m/min
b) the tool life in min
c) unit production cost in TL
d) unit production time in min
SOLUTION 4
a) Cutting speed to maximize production rate:
n
v mp
 n tr 
1 
 0.13
 C
   75
x

 1  0.13 3.5 
1  n t ch 
0.13
 50m / min
b) Tool life to maximize production rate:
t mp 
1 n
1  0.13
t ch 
x3.5  23.4 min
n
0.13
c) Unit production cost:
C pr  R 0  R m t l  t m   n t t ch R 0  R m   C t 
tm 
nt 
d w l w
vf

x80 x300
50 x1000 x0.25
 6 min
tm
6

 0.26
t
23.4
90
 90 


C pr   2  6  0.263.5 x  12  16.50TL
60
 60 


d) Unit production time:
t pr  t l  t m   t ch  nt  (2  6)  3.5 x0.26  8.9 min
FORMULA SHEET
Ft
Fc
Fs  Fc cos  - Ft sin 
tan(  -  n ) =
Fns  Fc sin   Ft cos
Ff  Fc sin  n + Ft cos  n
Fn  Fc cos  n - Ft sin  n
𝐹𝑠
𝐴𝑠
𝐹𝑓
𝜏𝑓 =
𝐴𝑓
𝐴𝑠 = 𝑎𝑤 𝑙𝑠
𝜏𝑠 =
ac  f  cos  s
d wl w
tm 
vf
v  d w nw

2

4
(1 − Γ)𝑃𝑠
𝜃𝑠 =
𝜌𝑐𝑣𝑎𝑐 𝑎𝑤
𝑃𝑓
𝜃𝑓 =
𝜌𝑐𝑣𝑎𝑐 𝑎𝑤
𝑎𝑐
𝑟𝑐 =
𝑎0
𝜃𝑚 = 𝜃𝑠 + 𝜃𝑓 + 𝜃𝑎
𝑃𝑚 = 𝑃𝑠 + 𝑃𝑓
tm
t
t
tm
K
t
2     n 
dw
nw f
vt  C
Nt 
r cos  n
tan   c
1  rc sin  n

n
nt 
𝐴𝑓 = 𝑎𝑤 𝑙𝑓
𝑎𝑐
𝑙𝑠 =
sin 𝜑
    n 
Zw  v  f  ap
v1 / n  f 1 / n1  d 1 / n2
t pr  t l  t m   t ch  nt
C pr  R 0  R m t l  t m   n t t ch R 0  R m   C t 
1 n
t ch
n

Ct
1 n 

t ch 
R 0  R m 
n 
t mp 
t mc
n
vmp
 n tr 
 C
 
1  n tch 
vmc
 n

( Ro  Rm )t r
 C

1  n [tch ( Ro  Rm )  Ct ] 
n