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General Design Principles for DuPont Engineering Polymers - Module

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Figure 4.01 Creep<br />

Stress (S), MPa (psi)<br />

S oe o<br />

e o<br />

Strain (e), mm/mm (in/in)<br />

S oe t<br />

Creep between time T and T o = e T – e o mm/mm (in/in).<br />

The creep modulus E T Pa (psi) <strong>for</strong> design in creep<br />

applications at stress S o and time T is the slope of the<br />

secant from the origin to the point (S o e T ).<br />

The stress required to de<strong>for</strong>m a plastic material a fixed<br />

amount will decay with time due to the same creep<br />

phenomenon. This decay in stress with time is called<br />

stress relaxation.<br />

Stress relaxation is defined as the decrease, over a<br />

given time period, of the stress (Pa, psi) required to<br />

maintain constant strain. Like creep, it can occur in<br />

tension, compression, flexure or shear. On a typical<br />

stress-strain curve it is shown in Figure 4.02.<br />

Figure 4.02 Relaxation<br />

Stress (S), MPa<br />

S o<br />

S T<br />

e o<br />

Strain (e), mm/mm (in/in)<br />

S oe o<br />

S Te o<br />

e t<br />

Relaxation between<br />

time T and T o =<br />

S o – S T Pa (psi). The<br />

relaxation modulus<br />

E T Pa (psi) <strong>for</strong> design<br />

in relaxation applications<br />

(e.g., press<br />

fits) at time T is the<br />

slope of the secant<br />

from the origin to the<br />

point (S Te o).<br />

Laboratory experiments with injection molded<br />

specimens have shown that <strong>for</strong> stresses below about<br />

1 ⁄ 3 of the ultimate tensile strength of the material at<br />

any temperature, the apparent moduli in creep and<br />

relaxation at any time of loading may be considered<br />

similar <strong>for</strong> engineering purposes. Furthermore, under<br />

these conditions, the apparent moduli in creep and<br />

23<br />

relaxation in tension, compression and flexure are<br />

approximately equal.<br />

A typical problem using creep data found in the<br />

properties sections is shown below.<br />

Cylinder under Pressure<br />

Example 1: A Pressure Vessel Under Long-Term<br />

Loading<br />

As previously noted, it is essential <strong>for</strong> the designer to<br />

itemize the end-use requirements and environment of<br />

a part be<strong>for</strong>e attempting to determine its geometry.<br />

This is particularly true of a pressure vessel, where<br />

safety is such a critical factor. In this example, we will<br />

determine the side wall thickness of a gas container<br />

which must meet these requirements: a) retain pressure<br />

of 690 kPa (100 psi), b) <strong>for</strong> 10 years, c) at 65°C<br />

(150°F).<br />

The inside radius of the cylinder is 9.07 mm (0.357 in)<br />

and the length is 50.8 mm (2 in). Because the part will<br />

be under pressure <strong>for</strong> a long period of time, one<br />

cannot safely use short-term stress-strain data but<br />

should refer to creep data or, preferably, long-term<br />

burst data from actual pressure cylinder tests. Data<br />

typical of this sort <strong>for</strong> 66 nylons is shown in Fig. 4.03<br />

which plots hoop stress versus time to failure <strong>for</strong><br />

various moisture contents at 65°C (150°F). Actually,<br />

Zytel ® 101 would be a good candidate <strong>for</strong> this application<br />

as it has high impact strength in the 50% RH<br />

stabilized condition and the highest yield strength of<br />

unrein<strong>for</strong>ced nylons.<br />

Referring to the curve, we find a hoop stress level of<br />

18.63 MPa (2700 psi) at 10 years, and this can be used<br />

as the design stress. The hoop stress <strong>for</strong>mula <strong>for</strong> a<br />

pressure vessel is:<br />

t = Pr × F.S.<br />

S<br />

where:<br />

t = wall thickness, mm (in)<br />

P = internal pressure, MPa (psi)<br />

r = inside diameter, mm (in)<br />

S = design hoop stress, MPa (psi)<br />

F.S. = factor of safety = 3<br />

t = (.690) (9.07) (3) (100) (.357) (3)<br />

18.63 2700<br />

= 1.0 mm (0.040 in)<br />

The best shape to use <strong>for</strong> the ends of the cylinder is a<br />

hemisphere. Hemispherical ends present a design<br />

problem if the cylinder is to stand upright. A flat end<br />

is unsatisfactory, as it would buckle or rupture over a<br />

period of time. The best solution, there<strong>for</strong>e, is to mold<br />

a hemispherical end with an extension of the cylinder<br />

or skirt to provide stability (see Figure 4.04).

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