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Unformatted text preview: WEIth 3W WW4
L&c#20 Material Cost Considerations: After identifying candidate material which satisfy requirements and preference, it is important to consider their cost relative to the parts performance. Material costs are generally provided in a cost per weight relationship ($/pound). Cost per weight does not .
always provide a good indicator as to the materials actual cost relative to your part application requirements. The effect of material density, strength, and rigidity on cost must be evaluated. Your part may require high strength, which would indicate that you need a high strength material. However, part strength can be affected by its structural design. Therefore, by establishing a required modulus too early may unnecessarily limit your ﬁnal selection and compromise your design. ' Consider a structural part where both a HDPE and a glass ﬁlled nylon satisfy the material
requirements. The cost per pound of the HDPE is considerably less than that of the nylon.
However, as the nylon has a much higher modulus, the part can be designed‘with a much
thinner wall than if the HDPE is used. The thinner wall will reduce the amount of
material used and the cooling time required. Both of these will reduce the cost of the
molded part. Which of these two materials is less expensive to use with the part is not
clear and must be evaluated. ‘ At a minimum, material costs should becontrasted based on volume rather than weight.
In a structural application, the relative required wall thickness requirements of the
candidate materials can also be contrasted. If the strength of a part is being considered, it
should be realized that a part made from a Nylon could have a thinner wall than if it were
produced from a HDPE. This will reduce the amount of material required and molding
cycle time, thereby possibly reducing total part cost despite the higher cost per pound for
the Nylon. ' Though contrasting the structure of two parts made with different materials. is best
evaluated using structural analysis programs, the following method provides a simpliﬁed
means of determining equivalent cost of plastic materials based on  structural
requirements. Candidate materialscan be evaluated based on their relative strength, as
indicated by their yield strength, or for their rigidity, as indicated by their relative moduli. Contrasting the cost of plastic materials based on strength: This process provides a means to contrast the cost of plastic materials based on their abilityto withstand stress. 1. Determine what is the primary type of stress the part will experience
(tensile, compressive, ﬂexural, shear, torsion).
2. Find the corresponding yield stress values of your candidate materials. Determine the yield strength ratio (0*) of the candidate materials utilizing one of the materials as a reference material. In this example,
there are two candidate materials — “A” and “B”. La) Yield Stress of Material " A" = 10,000 psi (Reference Material) Beaumont, l3, “Part Design,” Chapter 10 in Injection Molding Handbook, Osswald, T. A., Gramann, P., and
Turng, L. S., eds., Hanser Publisher,te—bc published in October 2001. Yield Stress of Material "B": 7,000 psi 0* (YieZdSz‘rength Ratio of "B" relativez‘o "A")= M = .7
10,000
4. Realizing that material “B” has only 70% of the strength of material “A”, an equivalent geometry can be derived which Would give a .
structure the same strength whether it were made from material “A” of
“B”. To accomplish this, a formula is derived from the stress formula
for therparticular loading case for which your part will experience in
application. Additionally the designer must consider which geometrical
dimension in the part is considered a variable. For example in a beam
which will be exposed to a ﬂexural load, is the width of the beam (wall
thickness) or it height a variable? In a ﬂexural loading case, the following standard stress formula
would be used. 'Mc
0—:—
I In a tensile or compressive loading case, the following simple
relationship can be used. F
0':—
A Example: Find the equivalent geometry for a ﬂexural load case where part
height (h) is the variable, i.e. what height is required of a part made
from material “B” so that it has the same strength of a part made from
material ”A”. As we are only looking for equivalence, one does not
need to know the actual load conditions, nor the actual part
dimensions. Regardless of the parts actual shape the materials are
contrasted assuming the structure is a simple beam with a height (h)
and a width (w). Mc M.5d 6M 6M
5:73;; =wh2 d: E
[2 M = bending moment c = distance from the neutral axis to
outer surface I = moment of Inertia As we are looking to contrast candidate materials under identical structural
loads, and part width is not a variable, then Beaumont, J., “Part Design,” Chapter 10 in Injection Molding Handbook, Osswald, T. A., Gramann, P., and
Turng, L. S., eds, Hanser Publisher, tebs published in October 2001. ‘ 9a w .
is the same for either material and can be canceled from the equation. Therefore, ﬁnding the equivalent height can be reduced to h": J.
0* Find the equivalent height (h*) of material “B” relative to material “A” h" = ,U— =‘/I=1.195h
0* .7 This indicates that a part which is produced from the material having a yield stress of
7,000 psi must be 1.195 time taller than a part being produced from the reference material which has a yield stress of 10,000 psi. 5. Find the current. Cost/Volume ($/in3) of the material from the material
suppliers. This will most likely require that cost be converted from cost
per weight (SS/pound) to cost per volume ($/in3). 6. I Find the equivalent Cost/Volume of the candidate materials based on their
relative heights and their relative costs per volume. Equivalent Cost/Volume = ($/in3)(h*) I Contrasting the cost of plastic materials based on rigidity: The following method
provides a means to contrast the cost of plastic materials based on their ability to resist deﬂection. 1. Determine what the primary type of load the part will experience in
application. (ﬂexural, tensile, compression, shear) 2. Find the corresponding modulus values for the candidate materials. Determine the modulus ratio (E *) of the candidate materials using one of the materials as a reference. In this example, there are two candidate materials — 66A!) CCB” U.) Modulus of material “A” = 350,000 psi (Reference Material)
Modulus of material “B” =. 210,000 psi _ 210,000 — = .592
355,000 Modulus Ratio of Material " B"relative to "A": E* 4. Realizing that material “B” has only 59.2% of the rigidity of material “A”, an
equivalent geometry can be derived which would give a structure with the
same rigidity whether it were made from material “A” of “B”. To accomplish Beaumont, J ., “Part Design,” Chapter 10 in Injection Molding Handbook, Osswald, T. A., Gramann, P., and
Turng, L. S., eds., Hanser Publisher,te=19c published in October 2001. this, a formula is derived from the deﬂection formula for the particular
loading case for which your part will experience in application. Again the designer must consider which geometrical dimension in the part is considered
a variable (height or width). In a ﬂexural loading case, any standard beam equation under a ﬂexural
load can be reduced to: 1 ' 1 ‘EF—7
E W
12 Again, as we are looking to ﬁnd the relative height (h *) for a part made
of material ”B”,'Which will provide the same rigidity as when produced
from material “A”, y and w will cancel out and be reduced to: h*=31_
VEae 5. Find the Cost/Volume ($/in3) 6. Find the "Equivalent Cost/Volume" based on performance = ($/in3)(h*) Beaumont, J ., “Part Design,” Chapter 10 in Injection Molding Handbook, Osswald, T. A., Gramann, P., and
Tumg, L. S., eds., Hanser Publisher, to—be published in October 2001. WW” » x Ebaw‘w ‘ x \ XV C“ » 4») $53») . ...
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 Spring '04
 TURNG
 Tensile strength, Candidate Materials

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