Dec. 17, 2008
Name_________________________
(20) 1. Consider the technical papers, videos, and PPT presentations in class to arrive at the T/F answer for the following: a.
DIP stands for dual in-line process. (package!) b.
Surface mounted components are glued before passing through a wave solder.
T__ F_X_
T_X_ F___ c.
Surface mounted circuit boards use vision system calibration of the template and board before solder paste is applied. T_X_ F___ d.
Via holes are used to pass thru-hole component leads through a board. T___ F_X_ e.
Every via hole must pass all the way through between the top and bottom board layers. T___ F_X_ f.
Machining tolerances typically improve as the part size increases. T___ F_X_ g.
The extended bottleneck model assumes that there are always a minimum number of parts in the FMS. h.
In conventional machining most of the cutting energy is converted to thermal energy. i.
A perfect just-in-time system with no delays between stations will have WIP. j.
A Kanban system is a push-pull system of inventory control. k.
Diamond is harder than Tungsten Carbide inserts. l.
Inkjet printing techniques can be used to develop electronic circuits on small special purpose circuit boards.
T_X_ F_X_
T_X_ F___
T_X_ F__
T__ F_X_
T_X_ F___
T_X_ F_X_ m.
As friction increases in rolling the maximum pressure point moves towards the entrance. T_X_ F___ n.
A technical paper showed that electronic assembly is comprised of 4 major operations:
1) thru-hole component insertion, 2) wave soldering, 3) testing, and 4) packaging. o.
In thru-hole IC assembly a screening process deposits the paste on the board. p.
In milling the surface feed-rate is limited by the feed per tooth and the spindle rpm.
T___ F_X_
T___ F_X_
T_X_ F___ q.
Direct extrusion consumes more power than indirect extrusion. r.
All metal parts stressed in the plastic region at any non-melting temperature have strain springback at a slope equal to Young’s modulus when stress is released. s.
Thru-hole component insertion is accomplished by crimping the leads before they pass through the board. t.
The rake angle of a tool used in turning must always be greater than zero degrees.
T_X_ F___
T___ F_X_
T__ F_X_
T___ F_X_

2
(20) 2. A single crystal boule of silicon is grown by the Czochralski process to an average diameter of 110 mm with length = 1200 mm. The seed and tang ends are removed, which reduces the length to 950 mm. The diameter is ground to 100 mm. A 30 mm wide flat is ground on the surface which extends from one end to the other. The ingot is then sliced into wafers of thickness = 0.50 mm, using an abrasive saw blade whose thickness = 0.33 mm. Assuming that the seed and tang portions cut off the ends of the starting boule are conical in shape, determine: (a) the original volume of the boule, mm 3 ; (b) how many wafers are cut from it, assuming the entire 950 mm length can be sliced; and (c) what is the volumetric proportion of silicon in the starting boule that is wasted during processing?
Solution :
(a) Total volume V = V
1
V
1
= V
3
(tang) + V
2
(cylinder) + V
= 0.333

(55) 2 (125) = 395,972 mm 3
V
2
=

R 2 L =

(55) 2 (950) = 9,028,152 mm 3
3
(seed)
= (cone in which h = 0.5(1200-950) = 125, D = 110, R = 55) =

R 2 h/3
Total V = 2(395,972) + 9,028,152 = 9,820,095 mm 3
(b) Number of wafers = 950/(0.50 + 0.33) = 1144.6

1144 wafers
(c) Area of one wafer A w
= A c
- A s
, where A c
= area of the circle of radius R = 50 mm, and
A
A c s
= the area of the segment A s
created by the flat ground on the cylindrical surface.
=

R 2 =

(50) 2 = 7854.0 mm 2
The area of a segment of the circle created by the 30 mm chord As =

R 2

/360 - 0.5R
2 sin

, where
 is the angle formed by two radii of the circle and the chord.
0.5

= sin 1(15/50) = 17.46

which gives

= 34.92

.
A s
=

(50) 2 (34.92)/360 - 0.5(50) 2 sin 34.92 = 761.8 - 715.5 = 46.3 mm 2
A w
= A c
- A s
= 7854.0 - 46.3 = 7807.7 mm
Volume of one wafer V w
= A w
2 t = 7807.7(0.5) = 3903.8 mm
Volume of 1144 wafers = 1144(3903.8) = 4,465,994 mm 3
3
Grading of 2:
1. Correct volume calc
2. Number wafers
3. Remove chord area
4. Proportion wasted
Volume wasted = 9,820,095 - 4,465,994 = 5,354,101 mm 3
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Proportion wasted = 5,354,101/9,820,095 = 54.52%.
(20) 3. A 2 inch thick slab is 10 inches wide and 12 ft long. Thickness is to be reduced in three steps in a hot rolling operation. Each step will reduce the slab to 75% of its previous thickness. The slab will widen by
3% in each rolling step. If the entry speed is 40ft/min and the roll speed is the same for all three steps, determine: a) length and b) exit velocity of the plate after the final reduction.
Solution: a) After three passes, t f
= (0.75)(0.75)(0.75)(2.0) = 0.844 in. w f
= (1.03)(1.03)(1.03)(10.0) = 10.927 in.
Using the constant volume assumption: t o w o
L o
= t f
(2.0)(10.0)(12 x 12) = (0.844)(10.927)L f w f
L f
Grading of 3:
1. Const volume assumption
2. Final length
3. Volume rate eqn
4. Exit velocity
L f
= (2.0)(10.0)(12 x 12)/(0.844)(10.927) = 312.3 in. = 26.025 ft b) Given that roll speed is the same at all three stands and that t o w o v o
= t f w f v f
(volume rate eqn)
Note that because the roll speed is the same the entry speed will be 40 ft/min at each stage.
Step 3: v f
= (0.75 x 0.75 x 2.0)(1.03 x 1.03 x 10.0)(40)/(0.844)(10.927) =
51.77 ft/min.
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3
(20) 4. You worked 21.1 in your HW. Now suppose that the rake angle is changed to  = 0 deg. Assuming that the friction angle remains the same, determine (a) the shear plane angle, (b) chip thickness, and (c) the shear strain for the operation.
Solution :
From Problem 21.1,

= 15
 and

= 26.85

. Using the Merchant Equation, Eq. (21.16):
  
= 45 +

/2 -

/2; rearranging,

= 2(45) +

- 2

  
= 90 + 15 – 2(26.85) = 51.3

Now, with

= 0 and
 remaining the same at 51.3

,

= 45 + 0/2 – 51.3/2 = 19.35°
(b) Chip thickness at

= 0: t c
= t o
/tan

= 0.30/tan 19.35 = 0.854 mm
(c) Shear strain

= cot 19.35 + tan (19.35 - 0) = 2.848 + 0.351 = 3.199
Grading of 4:
1. 21.1 answers
2. Apply correct eqns
3. Chip thickness
4. Shear strain
(20) 5. In a production turning operation, the workpart is 125 mm in diameter and 300 mm long. A feed of 0.225
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5 pts mm/rev is used in the operation. If cutting speed = 3.0 m/s, the tool must be changed every 5 work parts; but if cutting speed = 2.0 m/s, the tool can be used to produce 25 pieces between tool changes. Determine the Taylor tool life equation for this job.
Solution :
(1) T m
=

(125 mm)(0.3 m)/(3.0 m/s)(0.225 mm) = 174.53 s = 2.909 min.
T = 5(2.909) = 14.54 min.
(2) T m
=

(125 mm)(0.3 m)/(2.0 m/s)(0.225 mm) = 261.80 s = 4.363 min.
T = 25(4.363) = 109.08 min.
(1) v = 3 m/s = 180 m/min.
(2) v = 2 m/s = 120 m/min.
(1) 180(14.54) n = C
(2) 120(109.08) n = C
180(14.54) n = 120(109.08) n ln 180 + n ln(14.54) = ln 120 + n ln(109.08)
5.1929 + 2.677 n = 4.7875 + 4.692 n
5.1929 - 4.7875 = (4.692 - 2.677) n
0.4054 = 2.0151 n
C = 180 (14.54) 0.2012
n = 0.2012
C = 308.43
Grading of 5:
1. Machining time calculations
2. Setup 2 Taylor’s eqns
3. Correct n
4. Correct C
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