Any form of academic dishonesty will be dealt with severely according to the limits of the code.
1.
State all assumptions clearly, wherever necessary.
2.
Show all steps while solving questions III and IV
3.
Total number of pages in exam is 6 and the total number of points is 100.
4.
Wish you all the very best of luck!
KEY
(24 POINTS)
I.
Circle T rue or F alse (provide explanations, as and if necessary)
1.
2.
T
T
F
F
Extrusion is only capable of making symmetric cross-sections.
Due to the heat generated by plastic deformation during rolling, rolls can become slightly barrel-shaped.
3. T F
4 T F
The velocity of the strip increases from its entry value as it moves through the roll gap, in flat rolling.
In edging during performing material is gathered into a localized area, while in fullering the material is distributed away from an area.
5. T F
6.
7.
8.
9.
T
T
T
T
F
F
F
F
In flat rolling, large-diameter rolls tend to deform the metal strip plastically more at its surfaces than in the bulk.
As a result of roll bending, the rolled strip tends to be thicker at its center than its edges giving it a camber.
In flat rolling plates and sheets with very high width-to-thickness ratios, the width of the strip increases as a consequence of the reduction in thickness.
Friction is necessary for rolling materials, but, energy is dissipated in overcoming friction.
It is very important that significant lubrication is applied during coining so that the larger conversions in net shape are adequately achieved.
10. T
11.
12.
T
T
F
F
F
Tolerances achieved in die casting is often better than what can be achieved with sand casting.
The centrifugal casting process uses a polystyrene pattern which evaporates upon contact with molten metal to form a cavity for the casting.
Hot rolling converts the cast structure to a wrought structure with finer grains and enhanced strength.
1

(26 POINTS)
I I. Answer in six sentences or less.
(6 Points) a. What are the problems that arise while using square dies in extrusion? How do you determine this?
How do you remedy this? (Use pictures where necessary)
There is stagnation of metal in the dead zone. Scribe technique can be used to show metal flow.
To remedy this, small angle (draft), pp. 365.
(6 points) b. What are the advantages gained by using smaller rolls in flat rolling? What are 2 rolling configurations that best utilize the advantages of smaller rolls?
Smaller rolls have smaller contact and less friction. This reduces friction. Also, they are cheaper to be replaced than larger rolls. Four- high mills and cluster mills are configurations that use smaller rolls supported by larger rolls in the back.
(6 points) c. What is Investment casting? Is it an expendable mold type process? Discuss the tolerances achievable compared to sand casting. Discuss the economics.
The investment casting also called as lost-wax process is the process of making a wax pattern, loading with refractory slurry and melted out the wax and then used to make castings.
It is an expendable mold type process as the mold is broken to remove casting. Tolerances achieved are better (tighter control) than sand casting. It is an inexpensive process compared to die casting (set up/equipment cost of dies is very expensive).
(4 points) d. What is the fundamental difference between direct and indirect extrusion? What is hydrostatic extrusion?
Direct extrusion: The driven ram and the extruded part move in the same direction.
Indirect or inverted extrusion: The die moves towards the unextruded billet.
Hydrostatic extrusion: In hydrostatic extrusion, the billet is smaller in diameter than the chamber
(which is filled with a fluid), and the pressure is transmitted to the fluid by a ram. The fluid pressure results in triaxial compressive stresses acting on the workpiece and thus improved formability.
(4 points) e.What are the two types of die casting? Explain the differences.

Hot-chamber die casting

Cold-chamber die casting
Hot-chamber die casting requires lower pressure than in cold-chamber die casting because materials at lower temperatures need higher pressure to be forced into (and fill) the mold cavity.
Hot-chamber die casting pressure ranges up to 35 MPa (5000 psi), while cold-chamber die casting pressure ranges from 70 MPa (10,000 psi) to 150 MPa (20,000 psi).
2

(20 POINTS)
III. Three cross-sections are extruded using a press. The first is a square of side 1.414 inches, the second is an equilateral triangle of side 2.5 inches and the third is a circle of radius 1.5 inch.
Calculate the CCD for each cross-section. Determine the shape factor for each cross-section. If the original billet is of 4 inches diameter, calculate the extrusion ratio for each case.
CCD
1.414
1.414
2.5
R
R
30
R
R/2
2.5
1.25 1.25 h
(i) Square
Initial diameter of billet (rod), D = 4 inches
CCD = 3
Circumscri bed circle diameter ( CCD )

( 1 .
414 )
2 
( 1 .
414 )
2 
2 in.
Shape factor (SF)

Extrusion ratio (ER) perimeter
Area

A
A f
0


 sum of all sides
D
2
/ 4 area
1 .
414 * 1 .
414

4 ( 1 .
414 )
1 .
414 * 1 .
414

(

* ( 4 )
2
) / 4
1 .
414 * 1 .
414


6 .
283
2 .
828
(ii) Equilateral triangle
Initial diameter of billet (rod), D = 4 inches
3

tan 60
 opposite adjacent
 h
1.25
 h

tan 60 * 1.25

2.165
in.
cos 30
 adjacent hypotenuse

1.25
R

R

1 .
25 cos 30

1 .
44 in.

CCD

2 * 1.44

2 .
88 in.
Shape factor (SF)

Perimeter
Area
Extrustion ratio (ER)

A
A
0 f

1 /
 sum of all sides

D 2 area
/ 4
2 * ( 2 .
5 ) * 2 .
165


1 /
3 ( 2 .
5 )
2 * ( 2 .
5 ) * 2 .
165

2 .
77
3 .
142 * ( 4 ) 2 / 4

4 .
644
2 .
706
(iii) Circle
Initial diameter of billet (rod), D = 4 inches
Circumscri bed circle diameter ( CCD )

2R

1.5
* 2

3 in.
Shape factor (SF)
 perimeter
Area

2 *


* r
* ( r )
2

2 * 3 .
142 * 1 .
5
( 22 / 7 ) * ( 1 .
5 )
2
Extrusion ratio (ER)

A
A f
0


D

2
/ 4
* ( r )
2

(


* ( 4 )
2
( 1 .
5 )
2
) / 4

1 .
33

1 .
7778
4

(30 POINTS)
IV. An annealed copper strip 8 inches wide and 1.5 inches thick, is rolled to a thickness of 0.75 inch in one pass. The roll radius is 16 inches. Calculate the roll-strip contact length. Calculate the absolute value of the true strain that the strip undergoes in this operation. Determine the average true stress of the strip in the roll gap. Calculate the roll force. Calculate the coefficient of friction.
The behavior of the material is given as the following: strength coefficient = 90,000 psi and the strain hardening exponent is 0.5. The true stress of annealed copper in the unstrained condition is 40 % of its true ultimate stress.
(a) L

R ( h
0
 h f
)

( 16 * ( 1 .
5

0 .
75 )

3 .
464 in
(b)
 max
 ln

 h f h
0


 ln 1 .
5 / 0 .
75

0 .
693 (Taken an absolute value).
 
K
 n 
90 , 000 *

0 .
5 
K

90 , 000 Psi
(c)
 max

K
 n 
90 , 000 * ( 0 .
693 ) 0 .
5 
74921 .
95 Psi
(d)
 min

40
100
* 74921 .
95

29968 .
78
(e) The average stress, Y
(average)
=
 max
Psi

2
 min 
74921 .
95

29968 .
78

52445 .
36 Psi.
2
Alternatively, Y avg

K n
 n

1
= (90000 ∗ √0.693)/(0.5 + 1) =49947.97 psi
(f) The roll force, F = LW Y
(average)
= (3.464)*(8)*(52445.36) = 1453365.8 lb or (3.464)*8* 49947.97
= 1384158.1 lb (4 points)
(g) h o
 h f
  2
R
1.5 – 0.75 =

2
*16
 
1 .
5

0 .
75

0 .
2165
16
5