ME 482 – Final Exam
Dec. 21, 2006
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Problems
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1.
Consider the technical papers and PPT presentations in class to arrive at the T/F answer for the following:
a.
DIP stands for dual integrated processor
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b.
Surface mounted components cannot be passed through a wave solder.
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c.
Bending springback lessens as the die radius is reduced.
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d.
For metals in the elastic range the stress versus strain relation is exponential.
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e.
In metal cutting the primary cutting parameters are feed, speed, and depth of cut.
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f.
Machining tolerances typically improve as the part size decreases.
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g. In FMS bottleneck analysis, fijk (operation frequency) is always equal to 1.
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h.
In high speed machining, most of the cutting energy is converted to dynamic energy.
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i.
A perfect just-in-time system with no delays between stations will have WIP.
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j.
A Kanban system is a process rather than inventory control system.
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k.
Next to diamond, cubic boron nitride is the hardest material known.
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l.
The Czochralski process is a semiconductor crystal-growing method.
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m. As friction increases in rolling the maximum pressure point moves towards the exit.
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n.
Ball grid arrays can provide more contact points than square leaded chip carriers.
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o.
In surface mounted assembly a screening process deposits the paste on the board.
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p.
In turning, surface finish is a linear function of the feed for a single point tool.
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q.
Indirect extrusion consumes more power than direct extrusion.
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r.
Stress/strain metal springback in the plastic region is at a slope equal to Young’s modulus.T_X_ F___
s.
Some surface mount assembly machines can place > 10 parts per second on the board.
t.
The electronics manufacturing paper showed an alternative process to screen templates for
attaching surface mounted components to boards.
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2. List the primary steps used to both manufacture and populate an IC board with surface mounted
components.
Manufacture:
1. Prep board (tabs, slots, holes, barcode, etc.) and drill tooling and via holes
2. Image and etch circuit patterns on board layers
Grading:
3. Sandwich/glue the layers and plate the via holes
Assemble (and Test):
1. Manufacture steps (get at least 2 steps)
2. Assembly steps (get at least 6 steps)
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Assemble screen template for board type
Use vision to align template and adjust relative to board using standoff from board
Develop proper screen paste roll and adjust mechanical parameters until paste distributed
properly on IC board pads.
4. Apply paste to each board in batch.
5. Also apply epoxy if board has through hole parts and must be passed over wave solder.
6. Assemble board using assembly machine and possible some hand assembly for odd-form
components.
7. Pass board through heat flow oven to reflow solder paste
8. Inspect board for bridging and other assembly problems.
9. Test circuit continuity using tester.
10. Thermally perform stress tests.
11. If fails 10. or 11., then pass board to rework, else process completed.
1.
2.
3.
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3. The yield of good chips in multiprobe for a certain batch of wafers is 83 %. The wafers have a nominal
diameter of 150 mm with a processable area that is 135 mm in diameter. If the defects are all assumed to
be point defects that are indistinguishable on the board, what is their density D?
Solution:
A = 14,313 mm2 = 143.13 cm2
Bose-Einstein Solution:
Eq. (35.15): Ym = 1/(1 + AD)
0.83 = 1/(1 + 143.13 D)
0.83(1 + 143.13D) = 0.83 + 118.8 D = 1
118.8 D = 1 - 0.83 = 0.17
D = 0.00143 defects/cm2
Grading:
1. Area calculation
2. Correct application of equations
3. Poisson soln
4. Bose-Einstein soln
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(20) 4.
A single pass rolling operation reduces a 20 mm thick plate to 18 mm. The starting plate is 200 mm wide,
roll radius is 250 mm, and rotational speed is 12 rpm. The work material has a strength coefficient of 600
MPa and a strain hardening exponent = 0.22. Determine (a) roll force, (b) roll torque, and (c) power
required for the operation.
Solution:
(a) Draft = d = 20 – 18 = 2 mm
Contact length = L = (250x2)0.5 = 22.36 mm =0.02236 m
True strain =  = ln(20/18) = 0.1054
Grading:
1. Contact length
2. Rolling force
3 Torque
4. Power
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Yf = 600(0.1054)0.22/1.22 = 300 MPa
Rolling force = F = 300(0.02236)(0.2) = 01.342 MN = 1,342,000 N
(b) Torque = T = 0.5(1,342,000)(0.02236) = 15,000 N-m
(c) Given N = 12 rpm, Power = P = 2(12/60)(1,342,000)(0.02236) = 37, 700 W
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5.
A titanium bar with a starting diameter of 500 mm and length of 1000 mm bar is to be turned. Cutting
conditions are f = 0.4 mm/rev and d = 3 mm. The cutting tool is cemented carbide with parameters n =
0.23 and C = 400. Using Taylor units of min for time and m/min for speed, compute the cutting speed that
will allow the tool life to just equal the cutting time for the part after one pass of length 1000 mm.
Solution:
N = v/(D)
eqn (22.1)
Tr (min) = time per revolution = 1/N =  D/v
Tp (min) = time for one pass = L /(f/T r) = LTr/f = LD/(fv) where L = 1000 mm = bar length.
Set Tp equal to the life of the tool, where vT n = C (see eqn (23.1)); thus,
Tp = T = (C/v)1/n = L D/(fv)
Solving for v,
v(1-n)= C [f/(L D)]n
v0.77 = 400[(0.4)(1000)/{(3.142)(1000)(500)}] 0.23
Solving, v = 202.19 m/min
Grading:
1. Time per rev
2. Time for pass
3. Set times equal
4. Correct answer for v
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