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Midterm, answers 2019

Principles of Manufacturing (Carleton University)

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MECH 3700 Principles of Manufacturing Engineering

CARLETON UNIVERSITY

MIDTERM
EXAMINATION
February 2019

Duration: 1.5 HOUR No. of Students: 105

Department Name & Course Number: Mechanical & Aerospace Eng. MECH 3700
Instructor(s): A. Artemev

AUTHORIZED MEMORANDA
Calculators only

Student MUST count the number of pages in this examination question paper before
beginning to write and report any discrepancy immediately to a proctor. This question
paper has 3 pages.

All questions have the same value. Answer each question in this exam paper within the

outlined boxes, which are provided after each question and before the next question (please note

that some boxes take more than one page). The answers, or parts thereof, outside of the outlined

boxes will not be graded.

1. A tensile test was performed using a specimen cut from a 1-mm thick sheet metal.
The specimen had a gauge section with the initial length of 50 mm and width of 10
mm. The lengths and widths of the gauge section obtained at two different levels of
the applied tensile load are shown in Table 1. (a) Estimate by calculation the UTS of
the metal. (b) Discuss briefly an expected accuracy of the UTS estimate.

Table 1. Deformed gauge section geometry at two load values.

Applied Load (kN) 1.4 1.9

Gauge Section Length (mm) 57 69
Gauge Section Width (mm) 9.33 8.43

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MECH 3700 Principles of Manufacturing Engineering

Solution:
a) Find engineering stresses at two levels of the applied load:
at 1.4 kN:

F1 1.4  10 3 N
S1    140 MPa
A0 10  1 mm 2

at 1.9 kN:

F2 1.9  10 3 N
S2    190 MPa
A0 10  1 mm 2

Find engineering strains at two levels of the applied load:

at 1.4 kN:
l l 57  50
e1  1 0   0.14
l0 50
at 1.9 kN:
l l 69  50
e2  2 0   0.38
l0 50
Find true strains at two levels of the applied load:
 1  lne1  1  0.131
 2  ln e2  1  0.322
Find true stress values at two levels of the applied load:
 1  S1  e1  1  159.6 MPa
 2  S 2  e2  1  262.2 MPa
Using  1 ,  2 , 1 ,  2 values find the strength hardening exponent and strength constant:

 1  K   1n (1)

 2  K   2n (2)

Take logarithm of (1) and (2) and divide one obtained equation by the other to find n as:
 
ln 2 
 1 
n  0.552
2 
ln 
 1 

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MECH 3700 Principles of Manufacturing Engineering

then find K as:


K  1  490 MPa
 1n

At the onset of necking (when the UTS point is reached)  UTS  n . Therefore, the true
stress corresponding to the UTS point on the S(e) diagram can be found as:

 UTS  K  n n  353 MPa

The UTS is the engineering stress value corresponding to  UTS and, therefore, the UTS
can be found as:
UTS   UTS  exp n   203 MPa
b) Both experimental points were obtained at strains below that corresponding to the
onset of necking (and well above the yield strain which in alloys is usually below 0.01) ,
therefore the UTS estimate should be relatively accurate.

2. (a) Compare the main advantages and disadvantages of the basic oxygen converter
process and the electric arc furnace process. Which process would you recommend
for the manufacturing of H26 tool steel (Fe - 0.5 wt. % C, 0.3 wt. % Mn, 0.3 wt. %
Si, 4.0 wt. % Cr, 1.0 wt. % V, 18.0 wt. % W) ?
(b) Identify the major products of the Bayer and Hall-Héroult processes.

Answer:
(a) The basic oxygen process is fast and has a high productivity and low cost;
however, it allows for a very limited quality control. The electric arc furnace
process is more expensive, takes longer to complete the production cycle, can use
load with 100% of scrap metal, and allows for the production of the material with
a very high quality. The electric arc furnace process should be recommended for
the manufacturing of H26 tool steel which is highly alloyed and, therefore,
expensive.
(b) Alumina and aluminum.

3. A very large 220 mm-thick iron plate is cast by pouring iron into a sand mould at the
o
temperature of 50 C above the plate’s melting temperature, so that heat is withdrawn

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MECH 3700 Principles of Manufacturing Engineering

from both faces of the solidifying plate. Estimate (by calculation) the solidification
rate df s / dt , where f s is a fraction of solid and t is time, at the moment when the
fraction of the solid phase in the casting is 0.7. The material parameters of iron are:
o
melting temperature 1540 C; density 7.9 g / cm 3 ; specific heat 0.77 J / g  C ; latent

heat of solidification 281 J / g ; thermal conductivity 73 W / m  C . The material

o
parameters of sand are: melting temperature 1640 C; density 1.52 g / cm 3 ; specific

heat 1.2 J / g  C ; latent heat of solidification 156 J / g ; thermal conductivity 0.65

o
W / m C . The environment temperature in the foundry is 20 C.

Solution:
The fraction of solid equal to 0.7 is achieved when the thickness of solid on both
sides of the plate is equal to 77 mm.
The equation for the thickness of solid, S , is:

2 TM  T0
S   k m   m  Cm  t
 
C  H f  C  CC  T 
This equation can be converted into an equation for the solidification time:

 
 S  C  H f   C  C C  T  2
t   
k m   m  Cm 2 TM  T0 

2
3.14  0.077 7900  281  1000  7900  0.77  1000  50  
t   
0.65  1520  1200  2 1540  20 

t  10.83  10 3 s

The solidification rate can be related to the growth rate, v growth , using the total

thickness of the plate as:

df s v growth
 2
dt L

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MECH 3700 Principles of Manufacturing Engineering

The growth rate can be estimated from:

dS km   m  Cm TM  T0 
v growth   
dt  t  C  H f   C  C C  T  

v growth 
0.65  1520  1200

1540  20
3.14  10.83  10 3 7900  281  1000  7900  0.77  1000  50

v growth  3.555  10 -6 m / s

Therefore, the solidification rate is:

df s 3.555  10 6 m / s
 2  3.23  10 5 s 1
dt 0.22 m

Slightly different solution:

2 TM  T0
S   k m   m  Cm  t
 
C  H f  C  CC  T 

t

S 2   C  H f  C  CC  T
2

4 TM  T0 2  km   m  Cm 

dS 1 TM  T0 k   C
   m m m
dt 
 C  H f  C  CC  T t 

By using expression for t(S):

dS 2
 
TM  T0 2 k   C
 m m m

dt  C  H f  C  CC  T 2 S 

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df s 2 dS
 
4

TM  T0 2 k   C
 m m m
dt 
L dt   L C  H f  C  CC  T 2 S 
Substituting parameters with mass in g and length in cm:

0.65
1.52 1.2
df s

4

1540  202
 100
dt   22 7.9  281  1.52  1.2  50 2 7 .7

df s
 3.23 105 s 1
dt
Alternatively we can get:

df s

8

TM  T0 2 k   C
 m m m  3.23 105 s 1

dt   L2 C  H f  C  CC  T 2 fs 

4. a) Briefly describe the main advantages of the Cosworth - Ford casting process over
the conventional sand casting methods.
b) List three material characteristics, which produce a strong effect on the castability
of alloys.

Answer.
a). The main advantages of the Cosworth - Ford casting process over conventional sand
casting methods are:
 a higher dimensional accuracy due to a lower coefficient of thermal expansion of
zircon sand;
 a finer grain structure resulting from a higher cooling rate due to a higher thermal
conductivity of zircon sand;
 lower buoyant forces on cores due to similar densities of zircon sand and Al;
 the atmosphere control reducing the danger of the hydrogen contamination;
 a tranquil flow of the metal reducing the danger of the oxide ingestion;

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MECH 3700 Principles of Manufacturing Engineering

b). The material characteristics that produce a strong effect on the castability of alloys
are:
 fluidity which depends on:
 solidification range
 viscosity
 eutectic fraction
 solid phase morphology
 solidification shrinkage
 thermal shock resistance (thermal expansion coefficient, toughness)

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MECH 3700 Principles of Manufacturing Engineering

T  2T 1 T km dQ dV
J  k m    th    c H f
x x 2  th t Cm   m dt dt


x
t
  2 
T  x   A   exp  d  B
 4   th 
erf x  
2

x
 
  exp   2 d
  0

2  x3 x5 x7 x9  T TM  T0  x 2 
erf( x)  x     ...  
 exp 
  3  1! 5  2! 7  3! 9  4!  x    th  t  4   th  t 
  

2 TM  T0 v2 p
S k m   m  Cm  t G  k  pg h   const
  c  H f 2 g   g

h 2f  y Hf h f  F l
Y*  H*  S*  a  e
u  ks    Cs C s  Tm  T0  ks A0 l0

F lf  A0  A f
t    ln    E e RA    ln(e  1)  t   a  (e  1)
A  l0  A0

f
t  K  n
 
u  f      d   tm   f Wi  V  u  
dh

h  dt l
v
0

v 1 dA Wi h1 dh
    t  C   m p av  Wi  C  V   m   C  V   m   f
l A dt t av h0 h

Q
D  G  b  b      RT
p n
  A     e
kT  d   G 

J cal
1 in  25.4 mm 1 J  0.239 cal  0.738 ft lb f 1  2.39  10 4
kg K gK

W cal m
1  2.39  103 g  9.80665 1 N  0.2248 lb f 1 MPa  145 psi
mK cm s K s2

1 kg / mm 2  1422 psi

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