Unformatted text preview: ME 467 Assignment 05
8 questions, 30 points. Questions 4 points for single answer questions (first 7 questions)
6 points for last question (requires 3 items) Chapter 13
Remember, write question first, then answer.
1. An injection molding machine may be described as divided into two principal
components or unit. Name them.
Answer: The injection unit and the clamping unit.
An acceptable variation of the first is the polymer processing unit, though it is pretty vague. An
acceptable variation of the second is the mold clamping unit.
2. What is similar about the materials used in reaction injection molding (RIM) and transfer
Answer: They are thermosets. A variation of this answer is that they are not thermoplastics.
Another okay answer is that they are typically mixed with two components (that's basically a thermoset description) 3. How is the parison in blow molding expanded into its product final shape or diameter?
Answer: By air blown into the interior. I.e., it is “blown up” like a ballon. 4. What is the difference in vacuum forming verses pressure forming? Address in terms of
which side the pressure or vacuum is applied and the mold side of the product.
In vacuum forming, the vacuum (lowering of pressure) is applied on the mold side of the
product. I.e., the plastic sheet is pulled into the mold.
In pressure forming, the application of pressure is applied on the product side that is opposite the
mold. I.e., the plastic sheet is pushed into the mold.
5. Of Vacuum, Pressure, and Mechanical forming, which has the most expensive molds for
the same basic product design? Explain your answer.
Answer: Mechanical (polymer sheet) forming has the most expensive molds, simply because
there are two mold halves, whereas the other polymer sheet forming methods have only one
mold half needed.
This answer was in text as well as asked in class, TWICE. There is no other better answer. Page 1 of 12 ME 467 Assignment 05 STARTING METAL FORMING, CHAPTER 17
1. What is the general definition of a sheet metal?
Answer: “The surface area-to-volume ratio of the starting metal is high…” [Fundamentals of Modern
Manufacturing, 5th Ed., Groover, Mikell P., Wiley, 2014, page 417].
Another okay answer: The thickness-to-area ratio is low (or area-to-thickness ratio is high).
If answer below is given, then only 2 points for this answer. There may be a definition given that is from chapter 19, stating that sheet metal is defined as between
0.4mm (1/64 inch) – 6.0mm (1/4 inch) thick. But this definition by itself is only partially correct, as it
addresses only thickness, which is only one half the definition—you can have a non-sheet-metal stock for
very small parts with ¼ inch plate that has similarly small surface areas. A ¼ inch cube is not ‘sheet
metal’, for example.
2. How does temperature affect the flow curve parameter equations? We are thinking in terms of
bulk metal processes.
Answer: The answer with respect to the flow curve parameters: “For any metal, the values of K and n
depend upon temperature.” …” [Fundametnals of Modern Manufacturing, 5th Ed., Groover, Mikell P.,
Wiley, 2014, page 421] The why (not required for this answer): Strength and strain hardening are both
reduced at higher temperatures. (same page, next sentence).
A more general answer (also correct): the higher the temperature, the lower the stresses required to
plastically deform the metals. So, K and/or n must be lower at those higher temperatures (vs. lower
temperatures for the same metal). There is a chart in text and slides that gives the exact answer--K gets higher (steeper
slope) with higher temps. n gets lower with higher temps. 3. Indicate some of the advantages of cold working relative to warm and hot working. Give at
least 3 advantages.
Answer: Okay, this question is from the text, but it is a very important observation, so it was assigned
to the homework: Any three of these are correct for the question.
5. better accuracy,
better surface finish,
increased strength due to work hardening,
possible directional properties due to grain flow,
no heating of work required. 6 points on this question--2 points each
for any 3 answers. Page 2 of 12 ME 467 Assignment 05
All problems except #3 are 9 points each. Problems Problem 3 is 15 points. Chapter 13
1. The extrusion die for a polyethylene parison used in blow molding has a mean diameter of 18.0
mm. The size of the ring opening in the die is 2.0 mm. The mean diameter of the parison is
observed to swell to a size of 21.5 mm after exiting the die orifice. If the diameter of the blow
molded container is to be 150 mm, determine (a) the corresponding wall thickness of the container
and (b) the wall thickness of the parison. Solution: Up to 9 points.
r_s has no units (a ratio). t_m and t_p have units of mm. Units issue, -2.
There are several equations that work this problem, but answers are straightforward and should
be correct to 3rd decimal (due to small sizes). (a) rs = Dp/Dd = (21.5mm)/(18.0mm) = 1.194
tm = tp Dp/Dm = rs td Dp/Dm = (1.194)(2.0 mm)(21.5 mm)/(150.0 mm) = 0.342 mm
(b) tp = rs td = (1.194)(2.0mm) = 2.388 mm Page 3 of 12 ME 467 Assignment 05 2. The bottom of a chair (where you sit, the ‘seat’) will be injection molded by itself in a
mold. Assume the chair is a disk, 24 inches in diameter and 1/10 inch thick. If the
injection pressure is 1600 psi, what is the minimum clamp tonnage you need for the
injection molding machine? Give the answers in standard tons.
Solution: Up to 9 points.
Units may be issued as lb instead of lbf.
Must have answers in English (standard) tons. If not stated in these units, -4. F = PA
A = D2 PI / 4 = 452.39 in2
F = (1600 lbf/in2)*452.39 in2 = 723822.9 lbf
F = 723822.9 lbf / (2000 lbf/ton) = 362 tons
(note: thickness did not matter). Page 4 of 12 ME 467 Up to 15 points.
You don't need all the answer I gave
on the next05pages, but the detail is there for study.
Looking for the collected answers on page 10, plus some sketches similar to those on pages 6
and 7 (hand sketches are fine). 3. You work as an engineer in a furniture factory. You have several Borch injection
molding machines (IM) to buy for a new product line (see Borch IM specifications
sheet). Assume that the increasing model number has a direct increase in purchase cost
of the machine. Notice how most models have at least two barrel size alternatives (giving
increased shot volumes). Assume that the larger barrels mean larger cost as well.
For your deepest mold, you must have at least 4o of draft (some of this restriction depends upon
your materials used, but assume the worst case is 4o for all materials considered).
The design engineer created the product geometry in Solidworks. The file is “User Librarystool-.SLDPRT”. The design engineer wants to make this product out of the DuPont nylon
material “Zytel 408HS” (see the specifications document “Nylon_198118E.pdf”). As the
Manufacturing Engineer, you must address these questions by the next product development
a. Determine the mold pull direction of the product. Create a drawing file and show where
the parting line(s) should be (you can sketch and scan the line as a side view, or do it
entirely in Solidworks).
b. What is the required minimum draft of the stool, based upon the drawing?
c. What is the minimum shot volume required for the product? Assume 110% of product
volume for runner and sprue excess.
d. Look at the platen drawings in the specifications of the Borch IM. Assuming your part
cannot extend past the bolt locations in the platen’s, which models can accommodate
your product’s projected area, based upon your pull direction? (Assume your product
mold will be oriented relatively “Square” with respect to the mold platens—no putting
the product at odd angles to squeeze it into a smaller space).
e. You will need at least 3 inches of depth on each mold half just for strength and cooling
lines. You also will need another 2 inches thickness for the mold assembly stripper
(ejector pins) plate for the 3-plate mold. Based upon your mold pull direction and the
product’s dimension, what is the minimum mold (assembly) thickness you will need?
f. Based upon the maximum recommended injection pressure for Zytel 408HS, what will be
the required minimum clamping force of the IM in imperial (English) tons?
g. You must be able to get the part out of the mold assembly once the cycle is completed
and the mold halves separate. Based upon the product dimensions (including the mold
half thicknesses), what is the minimum stroke distance you will need?
h. Based upon required shot volume, injection pressure, minimum mold assembly thickness,
minimum stroke distance, and clamping force which Borch IM do you recommend? Or
do none of them meet all the specs? Solution: next several pages. This solution is way more than required for your answers,
but I figured it would help you study so the detail is there. Page 5 of 12 ME 467 Assignment 05 Solution: Work your way through the list. As you get new answers, eliminate machines that do
not have the parameters to fit your needs. Choose the lowest-level machine to solve the problem,
since supposedly the lowest model machine, with the smallest barrel (shot capacity) is what the
company will buy.
a. Determine the mold pull direction of the product. Sketch parting line.
The mold pull should be selected as perpendicular to top face of stool (and also in opposite
direction as well). This direction allows for the interior of the mold to pull out from the interior
of the stool without interfering with the plastic product. The parting line is generally indicated
with a section view, in figure below, as associated with the inside of the part surface.
arrows or a
drawn. If not
stool standing face,
If incorrect here,
rest of problem
likely will be
incorrect. If that is
case, then don't
grade rest and
(basically, half off). Mold Pull
Direction Mold Pull
Direction Page 6 of 12 ME 467 Assignment 05 The parting “lines” are shown in simplified mold halves views—technically, it is a parting ‘surface’
where the two halves come together The right mold half is a ‘side’ view, and the left mold half is a
section view, to show the interior cavity. You could sketch it roughly (technically, this is actually a bit
off, see next figure for why): Parting line Something hand
drawn like this is
fine. Looking for
mainly the outline
of the deep
section, and the
places where the
two halves come
together. Parting line Or, stomp this problem flat by creating a set of mold halves. Here is a section view, illustrating the
parting lines: Next is a section view, but pulled apart. The parting surfaces are then pretty clear. Page 7 of 12 ME 467 Assignment 05 Okay, here is an isometric view of the two halves (not required, shows cavity geometry but not sprue,
runners, cooling lines, etc.). Pretty nifty, takes about 40 minutes the first time you do this sort of thing in
SolidWorks: Page 8 of 12 ME 467 Assignment 05 b. Minimum draft is 5o. Since company policy is 4o for deep parts, this part should be okay.
(I did not grade this part of the problem)
c. Checking mass properties in SolidWorks, Volume = 76.8 in3 Allowing for 10% scrap,
shot volume should be 110% X 76.8 in3 = 84.48 in3 If forgot scrap %, -2.
d. Perpendicular to the mold pull direction, we have the front or bottom view (see Figure 1).
The dimensions are 13.78”x13.78”. Looking at Borche IM specs, the BS 120 has a
platen bolt limit of 6.88x6.88 = 13.76 in. That’s still a little too small. So go with the
BS150, which has a between tie bar limit of 8.26x2 = 16.52 in. That means the BS150
model or LARGER. Hard to grade, since students may put answer in different contexts.
Basically, we look to see if student noted if part would fit into any
platen sizes (bolt holes, tie bars, or similar should be mentioned.
If completely ignored, -2 (drawing is not required here). Figure 1. Major dimensions of Stool. Which is the “front” view in this drawing view set?. e. The mold assembly includes the mold cavity halves plus the 3rd stripper’ plate. You
need the product max dimension in the pull direction, 3 inches for each mold half, plus 2
more inches for the stripper plate. Thus, 3+3+2+16 = 24 inches of “stack”. It
schematically looks like Figure 2 (Figure 2 is not really accurate, since the outside of the
part is shown.
I forgot to mention what a 'stripper plate' was in class. however, it is mentioned prominently in the videos. Still, if
the stripper plate was ignored, that's okay--no points off for being off by 2 inches.
So I would take eiether 22 or 24 inches as correct. -2 otherwise. Page 9 of 12 ME 467 Assignment 05 Sketch not required. 16” 3”
3” Figure 2. Schematic illustration of the mold "stack". f. Page 7 of the Zytel spec document says the IM machine should be able to inject at up to
20,000 psi, or 20kpsi. With a projected area of 13.78x13.78 = 189.89 in2 (ignoring radii),
the maximum required clamping force would be (20kpsi)(189.89 in2) =3797768 lbf or
1899 tons. Key is to find the max injection pressure from spec: 20kpsi. They may have area
incorrect from earlier, but key is the kpsi and resulting F=PA calculation. Up to -2 for that. g. You have the mold assembly stack of 24 inches from above. However, you need to
separate the mold enough to pull out the part from between the mold halves. That means
the mold assembly (24 inches) needs to pull apart by at least 16 more inches. Thus, you
need a platen separation of at least 24+16 = 40 inches. Technically, it can be greater, of course
(greater space allows for ease of pulling
part out). Also answer is based upon
part f., so will accept between 38 and 46
inches. (Added 6 inches to upper limit if
some used it, just for ease of removal of
if outside this range h. Okay, here is a summary:
a. shot volume 84.48 in3
b. injection pressure 20,000 psi
c. minimum mold assembly thickness 24 inches
d. minimum stroke distance 40 inches
e. clamping force 1899 tons
f. Platen dimensions (I forgot to include that one in the list)—13.78” between bolts.
SELECTION PROCESS: As noted earlier, platen dimensions restricted us to the BS150 or higher.
Shot volume means the BS320 with the medium (second) barrel or larger.
Interestingly, the larger the barrel for a given model, the LESS injection pressure
possible. So you can use the BS320 with the second barrel, but not with the
larger-shot third barrel. Larger BS models all can inject at 20kpsi or higher.
Opening stroke needs at least 40 inches. That eliminates the BS320, BS400,
BS500, and BS650. Need the BS 800 or higher now. Notice how little you clear
the 40 inches, though. Page 10 of 12 ME 467 Assignment 05 Clamping force means only the largest model, the BS1800, will work—this is the
limiting factor. However, you can still choose the smallest barrel (the smallest
maximum shot size, or the left column under BS1800). Select the BS1800 with the smallest barrel for the least cost that meets all the other
requirements as well. If not BS1800 mentioned, then -2
Clamping force is the driver here--some students will see that immediately and not note the rest of the
issues. That's okay, if the noted and calculated the other issues in a. - g.
MUST note the "Smallest barrel" issue, as that saves a little cost. If smallest barrel is not mentioned
in solution, then -2. Page 11 of 12 ME 467 Assignment 05 Chapter 17
4. A metal has a flow curve with strength coefficient = 850 MPa and strain-hardening exponent =
0.30. A tensile specimen of the metal with gage length = 100 mm is stretched to a length = 157 mm.
Determine the flow stress at the new length and the average flow stress that the metal has been
subjected to during the deformation.
Solution: 9 points. Straightforward calculation, though some will likely get the math incorrect.
If calculations don't work, -2. each = ln (157mm/100mm) = ln 1.57 = 0.451
Flow stress Yf = (850 MPa)(0.451)0.30 = 669.4 MPa.
Average flow stress Y f = (850 MPa)(0.451)0.30/1.30 = 514.9 MPa. 5. The strength coefficient = 35,000 lb/in2 and strain-hardening exponent = 0.40 for a metal used in a
forming operation in which the workpart is reduced in cross-sectional area by stretching. If the
average flow stress on the part is 20,000 lb/in2, determine the final cross-sectional area, Af , if the
original cross-sectional area was A0 = 1 in2.
Note: This problem was changed from a text problem in order to make it clearer on how to find a
new cross sectional area, instead dealing with the ratio between the old and the new.
Solution: 9 points. Straightforward calculation, though some will likely get the math incorrect.
If calculations don't work, -2. each Y f = Kn/(1+n) 20,000 lb/in2 = 35,000 lb/in2 .4/(1.4)
1.4(20,000 lb/in2) = (35,000 lb/in2 (.4 .4 = (28,000 lb/in2 )/(35,000 lb/in2 )= 0.8 ,
now take ln of both sides to solve (or use your calculator to find root)...
0.4 ln = ln (0.8) = -0.22314
ln = -0.22314/0.4 = -0.55786 I totally blew this calculation in this document.
Just take square root twice:
strain = SQRT(SQRT((.8))) = 0.9457 = 0.5724 0.9457
Okay we also know that the engineering strain is = ln(Ao/Af) = 0.5724 0.9457
Ao/Af = 1.7726 2.5747
Af = Ao/1.7726 in2= 1/(1.7726 in2 ) = 0.564 in2
Af = A_0 / (2.5747) = (1 in^2) / (2.5747) = 0.3884 in^2 Answer. Page 12 of 12 ...
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- Summer '19